Trigonometry: A Unit Circle Approach

ISBN·13" 9-8'{)· 3·232903-3 O- 3·232903-.! ISBN-l0 i m » z Upper Saddle River, New Jersey 07458 www.prenhall.com 9 ...

Author: Michael Sullivan

204 downloads 2355 Views 143MB Size Report

This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
Report copyright / DMCA form

DOWNLOAD PDF

ISBN·13" 9-8'{)· 3·232903-3 O- 3·232903-.! ISBN-l0 i

m » z

Upper Saddle River, New Jersey 07458 www.prenhall.com

9

801 3 2u3 2 9::: 3:; -------'

Eighth Edition

Algebra

&

Trigonometry

Michael Sullivan Chicago State University

Upper Saddle River, New Jersey 07458

Library of Congress Cataloging-in-Publication Data

Sullivan, Michael Algebra

&

trigonometry / Michael Sullivan . ....,-

Now Work

• PROBLEM 23

'" The set of real numbers is a subset of the set of complex numbers. We discuss complex numbers in

Chapter 1, Section 1 .3 .

6

CHAPTER R

Review

Approxi m ations

Every decimal may be represented by a real number (either rational or irrational), and every real number may be represented by a decimal. In practice, the decimal representation of an irrational number is given as an approximation. For example, using the symbol ;::; (read as " approximately equal to"), we can write

V2 ;::;

1 .4142

7T

;::;

3.1416

In approximating decimals, we either round off or truncate to a given number of decimal places.* The number of places establishes the location of the final digit in the decimal approximation. Truncation:

Drop all the digits that follow the specified final digit in the

decimal. Rounding: Identify the specified final digit in the decimal. If the next digit is 5 or more, add 1 to the final digit; if the next digit is 4 or less, leave the final digit as it is. Then truncate following the final digit.

EXAM P L E 5

Approximating a Decimal to Two P l aces

Approximate 20.98752 to two decimal places by (a) Truncating (b) Rounding Solution

For 20.98752, the final digit is 8, since it is two decimal places from the decimal point. (a) To truncate, we remove all digits following the final digit 8. The truncation of 20.98752 to two decimal places is 20.98. (b) The digit following the final digit 8 is the digit 7. Since 7 is 5 or more, we add 1 to the final digit 8 and truncate. The rounded form of 20.98752 to two deci mal places is 20.99. n

E XA M P L E 6

Approximating a Decimal to Two and Four Places

Number

Rounded to Two Decimal Places

Rounded to Four Decimal Places

Truncated to Two Decimal Places

Truncated to Four Decimal Places

(a) 3.14159 (b) 0.056128 (c) 893.46125

3.14 0.06 893.46

3.1416 0.0561 893.4613

3.14 0.05 893.46

3.1415 0.0561 893.4612

=-""'

Now Work

.. PROBLEM 27

Calcu lators

Calculators are finite machines. As a result, they are incapable of displaying deci mals that contain a large number of digits. For example, some calculators are capa ble of displaying only eight digits. When a number requires more than eight digits, '" Sometimes we say "correct to a given number of decimal places" instead o f "truncate."

SECTION R.l

7

Real N u m bers

the calculator either truncates or rounds. To see how your calculator handles deci mals, divide 2 by 3. How many digits do you see? Is the last digit a 6 or a 7? If it is a 6, your calculator truncates; if it is a 7, your calculator rounds. There are different kinds of calculators. An arithmetic calculator can only add, subtract, multiply, and divide numbers; therefore, this type is not adequate for this course. Scientific calculators have all the capabilities of arithmetic calculators and also contain function keys labeled In, log, sin, cos, tan, xY , inv, and so on. As you pro ceed through this text, you will discover how to use many of the function keys. Graphing calculators have all the capabilities of scientific calculators and contain a screen on which graphs can be displayed. For those who have access to a graphing calculator, we have included comments, examples, and exercises marked with a ii� , indicating that a graphing calculator is required. We have also includ,E. d an appendix that explains some of the capabilities of a graphing calculator. The ·t comments, examples, and exercises may be omitted without loss of continuity, if so desired. O pe rations

In algebra, we use letters such as x, y, a, b, and c to represent numbers. The symbols used in algebra for the operations of addition, subtraction, multiplication, and divi sion are + , - , . , and /. The words used to describe the results of these operations are sum, difference, product, and quotient. Table 1 summarizes these ideas. Table 1

Operation

Symbol

Words

Addition

a

Sum:

Subtraction M u ltiplication

Division

+

b

a - b

a plus b

a ' b, (a) . b, a . (b), (a) . (b),

Difference:

a minus b

Product:

times

a a/b or t;

Quotient: a

ab, (a)b, a(b), (a)(b)

a

b

d i vid ed by b

In algebra, we generally avoid using the multiplication sign X and the division sign -;- so familiar in arithmetic. Notice also that when two expressions are placed next to each other without an operation symbol, as in ab, or in parentheses, as in ( a ) (b), it is understood that the expressions, called factors, are to be multiplied. We also prefer not to use mixed numbers in algebra. When mixed numbers are 3

3

means 2 + 4 ' In algebra, use of a 4 mixed number may be confusing because the absence of an operation symbol used, addition is understood; for example, 2

3

between two terms is generally taken to mean multiplication. The expression 2 is 4 11 therefore written instead as 2.75 or as 4 ' The symbol called an equal sign and read as "equals" or "is," is used to express the idea that the number or expression on the left of the equal sign is equiv alent to the number or expression on the right. =

E XA M P L E 7

,

Writing Statements Using Sym bols

(a) The sum of 2 and 7 equals 9. In symbols, this statement is written as 2 + 7 9. (b) The product of 3 and 5 is 15. In symbols, this statement is written as 3 · 5 15 . =

=

•

C'J

- Now Work

PROBLEM 39

8

CHAPTER R

Review

3

('

r

r

Eva l u ate N u merica l Express io n s

Consider the expression 2 + 3 · 6. It is not clear whether we should add 2 and 3 to get 5, and then multiply by 6 to get 30; or first multiply 3 and 6 to get 18, and then add 2 to get 20. To avoid this ambiguity, we have the following agreement. We agree that whenever the two operations of addition and multiplication separate three numbers, the multiplication operation always will be performed first, followed by the addition operation.

In Words

M u ltiply first, then add.

For 2 + 3 · 6, we have 2 + 3 · 6 = 2 + 18 = 20 EXA M P LE 8

F i n d i ng the Value of an Expressi o n

Evaluate each expression. ( a) 3 + 4 · 5 Sol ution

(c) 2 + 2 · 2

(b) 8 · 2 + 1

(b) 8 · 2 + 1

(a) 3 + 4 · 5 = 3 + 20 = 23 i

t

16 + 1 = 17

M u ltiply fi rst

M u ltiply fi rst

(c) 2 + 2 · 2 = 2 + 4 = 6 .1 "'iii

.,, - Now Work

PROBLEM S 1

To first add 3 and 4 and then multiply the result by 5 , we use parentheses and write (3 + 4) · 5. Whenever parentheses appear in an expression, it means "perform the operations within the parentheses first! " E XAM P L E 9

F i n d i ng the Val ue o f an Expression

(a) ( 5 + 3 ) . 4 = 8 · 4 = 32 (b) (4 + 5 ) · (8 - 2) = 9 · 6 = 54 II

When we divide two expressions, as in 2 + 3 4 + 8 it is understood that the division bar acts like parentheses; that is, 2 + 3 4 + 8

(2 + 3 ) (4 + 8)

The following list gives the rules for the order of operations. Rules for the Order of Operations 1. Begin with the innermost parentheses and work outward. Remember that

in dividing two expressions the numerator and denominator are treated as if they were enclosed in parentheses. 2. Perform multiplications and divisions, working from left to right. 3. Perform additions and subtractions, working from left to right.

SECTION R.1

EXAM P L E 1 0

Real N u mbers

9

F in d i n g the Val ue of an Expression

Evaluate each expression. (a) 8 ' 2 + 3 2 + 5 (c) 2 + 4,7 Solution

(b) 5 ' (3 + 4) + 2 (d) 2 + [4 + 2 · (10 + 6 ) ]

(a) 8 · 2 + 3 = 16 + 3 i

(b) 5 · ( 3

=

19

M u lti ply first

+

4) + 2 = 5 · 7 + 2 = 35 + 2 = 37 i i Pa rentheses first

M u lti ply before adding

7 2 + 5 2 + 5 ( c) 2 + 4·7 2 + 28 30 (d) 2 + [4 + 2 ' ( 10 + 6 ) ] = 2 + [ 4 + 2 ' ( 1 6 ) ] = 2 + [ 4 + 32] = 2 + [36] = 38 • Figure 7

B e careful if you use a calculator. For Example 10(c), you need to use paren theses. See Figure 7.* If you don't, the calculator will compute the expression

( 2+5 ) / ( 2+4*7 ) . 2333333333 Ans � Fr.ac.

2

+

5 "2 + 4 · 7 = 2 + 2.5 + 28 = 32.5

glvmg a wrong answer. am:: = = >-

4

Now Work

PROBLEMS 57 AND 65

Work with Properties of Rea l N u m be rs

We have used the equal sign to mean that one expression is equivalent to another. Four important properties of equality are listed next. In this list, a, b, and c repre sent real numbers. 1.

The reflexive property states that a number always equals itself; that is, = a. The symmetric property states that if a = b then b = a. The transitive property states that if a = b and b = c then a = c. The principle of substitution states that if a = b then we may substitute b for a in any expression containing a . a

2. 3. 4.

-1

Now, let's consider some other properties of real numbers. We begin with an example. E XA M P L E 1 1

C o m mutative P roperties

(a) 3 + 5 5 + 3 3 + 5

= =

=

8 8 5 +3

(b) 2 · 3 = 6 3·2 = 6 2,3 = 3'2

•

This example illustrates the commutative property of real numbers, which states that the order in which addition or multiplication takes place will not affect the final result. '" Notice that we converted the decimal to i ts fraction form. Consult your manual to see how your calculator does this.

10

CHAPTER R

Review

Commutative Properties

a+b=b+a a-b = boa

(la) (lb)

Here, and in the properties listed next and on pages 1 1-13, a, b, and c represent real numbers. EXAM P L E 1 2

Associative P roperties

(a) 2 + (3 + 4) (2 + 3 ) + 4 2 + (3 + 4)

=

=

=

(b) 2 - (3 - 4) 2 - 12 = 24 (2 - 3 ) - 4 = 6 - 4 = 24 2 - (3 - 4) = (2 - 3) - 4

2 +7 9 5 +4 9 (2 + 3) + 4

=

=

=

•

The way we add or multiply three real numbers will not affect the final result. Expressions such as 2 + 3 + 4 and 3 - 4 - 5 present no ambiguity, even though addi tion and multiplication are performed on one pair of numbers at a time. This prop erty is called the associative property_ Associative Properties

a + (b + c) = (a + b) + c a + b + c a - (b - c) = (a - b) - c = a - b - c =

(2a) (2b)

The next property is perhaps the most important. Distributive Property

a - (b + c) (a + b) - c

=

=

a-b + a-c a-c + b-c

(3a) (3b)

The distributive property may be used in two different ways. E XA M P L E 1 3

D i stributive P roperty

(a) 2 - (x + 3 ) = 2 - x + 2 - 3 = 2x + 6 Use to remove parentheses. (b) 3x + 5x = (3 + 5)x = 8x Use to combine two expressions. (c) (x + 2 ) ( x + 3 ) = x(x + 3 ) + 2(x + 3 ) (x2 + 3x) + (2x + 6) 2 x + (3x + 2x) + 6 = x2 + 5x + 6 =

=

Cl!l!1C = �-

Now Work

q

PROBLEM 87

The real numbers 0 and 1 have unique properties. EXAM P LE 1 4

I dentity P roperties

(a) 4 + 0 = 0 + 4 = 4

(b) 3 - 1 = 1 - 3 = 3

The properties of 0 and 1 illustrated in Example 14 are called the properties_

.. identity

SECTION R .1

Real N umbers

1 1

Identity Properties

O+a=a+O=a a'1 = 1'a = a

(4a) (4b)

We call 0 the additive identity and 1 the multiplicative identity. For each real number a, there is a real number -a, called the additive inverse of a, having the following property: Additive I nverse Property

a + ( -a) = -a + a = 0

E XA M P L E 1 S

( Sa)

F i n d i ng an Additive I nverse

(a) (b )

The additive inverse of 6 is -6, because 6 + ( -6 ) = O. The additive inverse of -8 is - ( -8) = 8, because -8 + 8 = O.

•

The additive inverse of a, that is, -a, is often called the negative of a or the opposite of a. The use of such terms can be dangerous, because they suggest that the additive inverse is a negative number, which may not be the case. For example, the additive inverse of - 3, or ( -3), equals 3, a positive number. -

For each nonzero real number a, there is a real number l, called the multiplicata of a, having the following property:

ive inverse

M ultiplicative Inverse Property

1 1 a'- = -'a = 1 a a

if a

'*

0

(5b)

1

The multiplicative inverse - of a nonzero real number a is also referred to as the a reciprocal of a. E XA M P L E 1 6

F i n d i n g a Reci procal

(a)

The reciprocal of 6 is

�, because 6 � = 1 . .

1 1 - 3 ' because 3 · - 3 = 1 . 2 3 2 3 ( c ) The reciprocal of '3 IS. '2' because '3 '2 = 1 .

(b )

.

The reciprocal of -3

.

IS

.

•

With these properties for adding and multiplying real numbers, we can now define the operations of subtraction and division as follows: DEFINITION

The

difference

a - b, also read "a less b" or "a minus b," is defined as a - b = a + ( -b)

(6)

I�

�--------------------------------�

12

CHAPTER R

Review

To subtract b from a, add the opposite of b to a. a If b is a nonzero real number, the quotient - , also read as "a divided by b" b or "the ratio of a to b," is defined as

DEFINITION

I

a 1 - = a'if b oF 0 (7) b b � � ---------------------------------- .� EXAM P L E 1 7

Working with Diffe rences and Quotients

(a) 8 - 5 = 8 + ( -5 ) = 3 (b) 4 - 9 = 4 + ( -9) = -5 5 1 (c) - = 5 ' 8 8 • r

r

r ,....

For any number a, the product of a times 0 is always 0; that is, In Words

M ultiplication by Zero

The result of m u ltiplying by zero is zero.

a'O = 0

(8)

For a nonzero number a, Division Properties

a - = 1 a

(9)

if a oF O

2 NOTE Division by 0 is not defined. One rea son is to avoid the following difficu lty: find

2 o

x

such that

- = �

o·

x

=

2.

But

o·x

equ a l s 0 for all

x,

o

= x

so there is no u n ique n u m ber

x

means to such that

•

Rules of Signs

a( -b) = - (ab) - ( -a) = a

E XA M P L E 1 8

( -a)b = - ( ab) a -a a -b b b

( -a)( -b) = ab -a a -b b

(10)

Applying the Rules of Signs

(a) 2 ( -3 ) = - ( 2 ' 3 ) = -6 -3 3 3 (c) = -2 = ' 2 -2 I I x (e) = = - 2x x ' _2 -2

(b) ( -3 ) ( -5 ) -4 4 (d) =9 -9

=

3 · 5 = 15

,.

SECTION R.1

Real N umbers

13

Cancellation Properties

ae = be implies a = b if e "* 0 ae a if b "* 0, e "* 0 be b

E XA M P L E 1 9

U s i ng the Cancel l ation P roperties

(a) If 2x = 6, then

NOTE We follow the common prac tice of using slash ma rks to i n d icate

r r r

r

(b)

•

cancellations.

In Words

3 ·S i 18 = = 12 2'S 2 i Cancel

2x = 6 2x = 2 · 3 x =3

Factor 6. Cancel the 2's.

the 6's.

•

Zero-Product Property

If a product equ a l s 0, then one o r

If ab

both o f the factors is O.

EXAM P L E 2 0

= 0, then a = 0, or b = 0, or both.

(12)

U s i ng the Zero- Product Property

If 2x = 0, then either 2 = 0 or x = O. Since 2 "* 0, it follows that x = O. Arithmetic of Quotients

---

a e ad be ad + be -+-=-+-= b d bd bd bd a e ae b d bd a b a d ad - = -.- = e b e be d

E XA M P L E 2 1

(11)

if b "* 0, d "* 0

(13 )

if b "* 0, d "* 0

(14)

if b "* 0, e "* 0, d "* 0

(IS)

Addi ng, Su btracting, M u lti plyi ng, and Dividing Quotients

(a)

(b)

2 5 + 3 2

-

-

= i

3·5 2·2 + 3·2 3·2

-

i

_

�)

By equation (6)

i

=

By equation (13)

� �= �+( _

-

= i

2·2 + 3·5 3·2

4 + 15 6

19 6

� + �2

By equation (10)

3 · 3 + 5 ' ( -2) 5·3 By equation (13)

9 + ( - 10) 15

-1 15

-

1 15

-

•

14

CHAPTER R

Review

NOTE S l a nting the cancellation ma rks i n d ifferent d i rectio n s for d ifferent fac

( c)

tors, as shown here, is a good p ractice

� . 15 3 4

to follow, since it wi l l help in checking for

•

errors.

i

8 · 15 3·4

2. = 2 . 2. "2 5 7 9

=

By equation

3 (d)

=

r

2 · .4" '3- ' 5 '3- ' .4' · 1

=

i

(14)

2·5 = 10 1

By equation

(11)

3·9 27 = i s ' 7 35

=

By equatio n

By equation

(15)

(14) ..

NOTE In writing quotients, we s h a l l follow the usual convention and write the quotient in lowest terms. That is, we write it so that a ny common factors of the n u merator and the denominator have been removed using the cancellation properties, equation

15 15 · ..6 4 ' ..6 4 4 · ..6 · x · )(. 3 · ..6 · )(.

90

24 24>f 18x - Now Work

4x 3

(11).

x

As exa m ples,

•

# 0

PROBLEMS 67, 7 1 , AND 81

Sometimes it is easier to add two fractions using least common multiples (LCM). The LCM of two numbers is the smallest number that each has as a com mon multiple. E XA M P L E 2 2

F i nd i n g the Least Common M u lti ple of Two N u mbers

Find the least common multiple of 15 and 12. Sol ution

To find the LCM of 15 and 12, we look at multiples of 1 5 and 12. 15, 30, 45, 60, 75, 90, 105, 120, . . . 12, 24, 36, 48, 60, 72, 84, 96, 108,

120, . . .

The common multiples are in blue. The least common multiple is 60. •

E XA M P L E 2 3

Using the least Common M u ltiple to Add Two F ractio n s

8 5 F·In d : - + 12 15 Solution

We use the LCM of the denominators of the fractions and rewrite each fraction using the LCM as a common denominator. The LCM of the denominators (12 and 15) is 60. Rewrite each fraction using 60 as the denominator. 8 15

- +

5 12

8 .4 +5 .5 15 4 12 5 32 25 = + 60 60 32 + 25 60 57 60 19 20

=

-

= = = -

Now Work

PROB

l

EM 75

-

•

Real Num bers

S ECTION R . l

15

�is;torical Feature

T

he real n u m be r system h a s a history that stretches back at least

turned away from the n u m be r concept, expressing facts about whole

to the a n cient Babyl o n i a n s { 1 800 Bc). l t is rem a rka ble how much

n u m bers in terms of line segments.

the a n cient Babylonian attitudes resemble o u r own. As we stated

I n astronomy, however, Babylo n i a n methods, including the Babylon

in the text, the fu nda mental difficulty with irrational n u m bers is that

ian n u m ber system, conti nued to be used. Simon Stevin ( 1 548-1 620),

they can not be written a s quotients of integers or, equivalently, as re peating or term i n ating deci mals. The Babyl o n i a n s wrote their n u m bers in a system based o n 60 i n the same way that we write ours based on

1 0. They wou l d carry as m a ny p l a ces for 7T as the accuracy of the prob

lem demanded, just as we now use 7T �

1 37

7T �

or

or

7T �

3.141 6

or

7T �

3 . 1 4 1 592 6 53 58979

probably using the Babyl o n i a n system as a model, invented the decimal

system, complete with rules of calcu lation, in 1 585. [Others, for example,

a l-Kashi of Samarkand (d. 1 429), had made some progress in the same

d i rection.] The decimal system so effectively conceals the difficulties that the need for more logical precision began to be felt only i n the early 1 800s. Around 1 880, Georg Cantor ( 1 845- 1 9 1 8) and Rich ard Dedekind

( 1 83 1 - 1 9 1 6) gave precise definitions of real n u mbers. Cantor's defini

3 . 1 41 59

tion, a lthough more a bstract and precise, has its roots in the decimal (and hence Babylo n ia n ) n u merica l system.

depending on how accu rate we need to be.

Sets and set theory were a spi n-off of the research that went into

Things were very different for the Greeks, whose n u m ber system a l lowed only rational n u m bers. W h e n it was d iscovered that

V2 was

not

cla rifyi ng the fou n d ations of the real n u mber system. Set theory has de veloped into a large disci p l i n e of its own, and many mathematicians re

a rational n u m ber, this was regarded as a fu n d a mental flaw i n the num

gard it as the fou ndation u po n which modern mathematics is built.

ber concept. So serious was the matter that the Pythagorean Brother

Cantor's discoveries that infinite sets can also be counted and that there

hood (an early mathematical society) is s a id to have drowned one of its

are diffe rent sizes of infin ite sets a re among the most astounding results

members for revea l i n g this terrible secret. Greek mathematicians then

of modern mathematics.

H i storical Problem s

The Babylonian n u m ber system was based on 60. Thus 2,30 means 2

+

30

60

2. What are the fo llowing Babylonian n u mbers when written as frac tions a n d as deci m a l s ?

2.5, a n d 4,2 5, 1 4 means

=

4

+

25 60

+

14 2 60

=

4 +

(b) 4,52,30

(a) 2,20

1 51 4 =

3 600

(e) 3,8,29,44

4.42 0 5 5 5 5 5 . .

1 . What a re the fol lowing n u m bers i n Babylo n i a n notation? 1

5

(a) 1 -

(b) 2 6

3

R.l Assess You r Understandi ng Concepts a nd Voca b u lary 1.

The numbers in the set {x I x and b

*'

O } , are called

a

b ' where a, b are integers

=

____

2.

The value of the expression 4

3.

The fact that 2x + 3x Property.

=

+

5. True or False

Rational numbers have decimals that either terminate or are nonterminating with a repeating block of digits.

numbers.

5·6

-

3 is

6. True or False

The Zero-Product Property states that the product of any number and zero equals zero.

_ _ _ _ _

(2 + 3)x is a consequence of the

____

4.

"The product of 5 and x

+

3 equals 6" may be written as

7. True or False

The least common multiple of 12 and 18 is 6.

8. True or False

No real number is both rational and irrational.

Skill B uilding In Problems 9-20, use U

=

universal set

each set.

=

{O, 1, 2, 3, 4, 5, 6, 7, 8, 9 } , A

=

{ I , 3, 4, 5, 9 } , B

=

{2,

4,

6, 7, 8 } , and e

=

AUB

10.

Aue

11.

AnB

12.

Ane

13.

(A U B) n e

14.

(A n B) U e

15.

A

16.

e

17.

AnB

18.

Bue

19.

AUB

20.

Bne

9.

{ I , 3, 4,6 }

to find

16

CHAPTER R

Review

In Problems 21-26, list the numbers in each set that are (a) Natural numbers, (b) Integers, (c) Rational numbers, (d) Irrational numbers,

{ { 2I ' 3I ' 4I } { \12, \12

(e) Real numbers.

21.

A

23.

C

25.

E

_

-

=

=

1 , -6 ' 2 , - 1 .333 . . . (the 3 s repeat ) , ?T, 2, 5

}

0, 1,

?T,

+ 1, ?T +

�}

{ �,

22.

B

-

24.

D

=

26.

F

=

=

2.060606 . . . (the block 06 repeats ), 1 .25, 0, 1, Vs

{ - I , - 1.1 , - 1 .2, - 1 .3 }

{

- \I2 ' ?T +

\12,�

+ 10.3

}

In Problems 2 7-38, approximate each number (a) rounded and (b) truncated to three decimal places.

27. 1 8.9526

28. 25 .86134

33. 9.9985

34. 1 .0006

29. 28.65319 3 35. "7

30. 99.05249 5 36. "9

32. 0.05388 81 38. 5

31. 0.06291 521 37. 15

In Problems 39-48, write each statement using symbols.

39. The sum of 3 and 2 equals 5.

40. The product of 5 and 2 equals 10.

41. The sum of x and 2 is the product of 3 and 4.

42. The sum of 3 and y is the sum of 2 and 2.

43. The product of 3 and y is the sum of 1 and 2.

44. The product of 2 and x is the product of 4 and 6.

45. The difference x less 2 equals 6.

46. The difference 2 less y equals 6.

47. The quotient x divided by 2 is 6.

48. The quotient 2 divided by x is 6.

In Problems 49-86, evaluate each expression.

49. 9 - 4 + 2

50. 6 - 4 + 3

53. 4 + 5 - 8 ' 57. 6 - [3 ' 5 + 2 ' (3 - 2 ) ]

51 . -6 + 4 · 3

52. 8 - 4 · 2

54. 8 - 3 - 4

1 55. 4 + 3

1 56. 2 - 2

58. 2 · [8 - 3 (4 + 2 ) ] - 3

59. 2 · (3 - 5 ) + 8 · 2 - 1

60. 1 - (4 ' 3 - 2 + 2 )

"

61. 10 - [6 - 2 · 2 + (8 - 3 ) J · 2 1

62. 2 - 5 · 4 - [6 · (3 - 4) J

63. (5 - 3 ) 2

1 64. (5 + 4) 3

65.

4 + 8 5 - 3

66.

2 - 4 5 - 3

3 10 67 . 5 ' 21

5 3 68. - . 9 10

69.

� 10 . 25 27

70.

21 . 100 25 3

3 2 71. 4 + 5

4 1 72. - + 3 2

5 9 73. (5 + 5

15 8 74. - + 9 2

5 1 75. - + 12 18

2 8 76. - + 15 9

1 7 77. - - 18 30

3 2 78. - - 21 14

3 2 79. - - 15 20

6 3 80. - - 35 14

81. 11

4 1 2 84. - + - ' 3 5 6

3 3 85. 2 · - + 4 8

1 3 7 83. - . - + 2 5 10

5 18

5 82. � 2 35

27

5 1 86. 3 · - - 6 2

1n Problems 87-98, use the Distributive Property to remove the parentheses.

", 87. 6(x + 4)

(

91. 2 � r - 1:. 4� 2

)

95. (x - 2)(x + 1 )

88. 4(2x - 1 )

(

92. 3 �X + 1:. 3 6 96.

(x

)

- 4 ) (x + 1 )

89. x(x - 4) 93.

(x

+ 2)( x + 4 )

97. ( x - 8 ) (x - 2)

90. 4x(x + 3) 94.

(x

+ 5 ) (x + 1 )

98. ( x - 4 ) ( x - 2)

}

SECTION R.2

Algebra Essentials

17

Discussion and Writing 99. 100.

Explain to a friend how the D istributive Property is used to justify the fact that 2x + 3 x = 5x. Explain to a friend (2 + 3 ) ' 4 = 20.

why

2 + 3·4

=

14, whereas

101.

Explain why 2(3 . 4) is not equal to (2 · 3) . (2 ' 4).

102.

Explam why 2

103.

Is subtraction commutative? Support your conclusion with an example.

104. 105. 106. 107. 108.

4+3. 4 3 5 I S not equal to 2 + "5 ' +

.

109.

110. 111.

Is subtraction associative? Support your conclusion with an example. Is division commutative? Support your conclusion with an example. Is division associative? Support your conclusion with an example. If 2

=

x, why does x

If x

=

5, why does

=

x2

+

2? x =

3 0?

112. 113.

Are there any real numbers that are both rational and irra tional? Are there any real numbers that are neither? Explain your reasoning. Explain why the sum of a rational number and an irrational number must be irrational. A rational number is defined as the quotient of two integers. When written a s a decimal, the decimal will either repeat or terminate. By looking at the denominator of the rational number, there is a way to tell in advance whether its decimal representation will repeat or terminate. Make a list of ratio nal numbers and their decimals. See if you can discover the pattern. Confirm your conclusion by consulting books on number theory at the library. Write a brief essay on your findings. The current time is 12 noon CST. What time (CST) will it be 1 2 ,997 hours from now? a . 0 . Both 0 ( a "* 0 ) and 0 are undefmed, but for dIfferent reasons. Write a paragraph or two explaining the different reasons.

R.2 Algebra Essentials OBJECTIVES

G ra p h I n e q u a l ities (p.

1 8)

2 Find Distance on the Real N u m ber Line (p. 1 9) 3 Eva l u ate Algebra ic Expressions (p. 20)

4 Determine the Domain of a Va riable (p. 2 1 ) 5 Use the Laws of Exponents (p. 2 1 )

6 Eva l u ate S q u are Roots (p. 23)

7 Use a Ca l c u l ator to Eva l u ate Exponents (p. 24)

8 Use Scie ntific Notation (p. 24)

The Rea l N u m ber Line

Figure 8

Rea l n u m ber l i n e

I(

2 units

Scale

�

-

3

-2

1

-1

I

-!

1 Unit o I

0

I

!

I I

•

I

1� 2

II

3TI

J

The real numbers can be represented by points on a line called the real number line. There is a one-to-one correspondence between real numbers and points on a line. That is, every real number corresponds to a point on the line, and each point on the line has a unique real number associated with it. Pick a point on the line somewhere in the center, and label it O. This point, called the origin, corresponds to the real number O. See Figure 8. The point 1 unit to the right of 0 corresponds to the number 1. The distance between 0 and 1 deter mines the scale of the number line. For example, the point associated with the number 2 is twice as far from 0 as 1 . Notice that an arrowhead on the right end of the line indicates the direction in which the numbers increase. Points to the left of the origin correspond to the real numbers - 1 , -2, and so on. Figure 8 also shows the points associated with the rational numbers numbers V2 and 7T.

DEFINITION

-

� and � and with the irrational

TIle real number associated with a point P is called the coordinate of P, and the line whose points have been assigned coordinates is called the real number line'--1 1l'I!lilI 1mII =-

Now Work

PROBLEM 1 1

18

CHAPTER R

The real number line consists of three classes of real numbers, as shown in Figure 9.

Figure 9

-3

[

-

Review

!

!

2 -� -1

I -

�

o !

O

I

�

[

1

I

J

�

2

I

I

3

1. The negative real numbers are the coordinates of points to the left of the origin O.

I

-y-------' • '------v--

Negative rea l n u mbers

I

The real number zero is the coordinate of the origin O. 3. The positive real numbers are the coordinates of points to the right of the origin O.

Positive

Zero

2.

rea. n u mbers

Negative and positive numbers have the following multiplication properties: M u ltiplication Properties of Positive and Negative Numbers

The product of two positive numbers is a positive number. The product of two negative numbers is a positive number. 3. The product of a positive number and a negative number is a negative number. 1.

2.

1

Figure 1 0

a mean the same thing. It does not matter whether we write 2 < 3 or 3 > 2. Furthermore, if a a, then the difference b - a is positive. Do you see why? U s i ng I nequality Sym bols

(a) 3 < 7 (d) -8 < -4

(b) -8 > - 16 (e) 4 > - 1

(c) - 6 < 0 (f) 8 > 0 •

In Example l (a), we conclude that 3 < 7 either because 3 is to the left of 7 on the real number line or because the difference, 7 - 3 = 4, is a positive real number. Similarly, we conclude in Example l (b) that -8 > -16 either because -8 lies to the right of -16 on the real number line or because the difference, -8 - ( -16 ) = -8 + 16 = 8, is a positive real number. Look again at Example 1. Note that the inequality symbol always points in the direction of the smaller number. Statements of the form a a are called strict inequalities, whereas statements of the form a :::; b or b 2': a are called nonstrict inequalities. An inequali1iY is a statement in which two expressions are related by an inequality symbol. The expressions are referred to as the sides of the inequality. B ased on the discussion so far, we conclude that a > 0 is equivalent to a is positive a < 0 is equivalent to a is negative

SECTION R.2

We sometimes read a > 0 by saying that "a is positive." If a or a = 0, and we may read this as "a is nonnegative." '-'''"

;z;...-

Now Work

2:

19

Algebra Essentials

0, then either a > 0

PROBLEMS 1 5 AND 25

We shall find it useful in later work to graph inequalities on the real number line. G rap h i n g I nequalities

EXA M P L E 2

(a) On the real number line, graph all numbers x for which x > 4. (b) On the real number line, graph all numbers x for which x :::; 5. (a) See Figure 11. Notice that we use a left parenthesis to indicate that the number 4 is not part of the graph. (b) See Figure 12. Notice that we use a right bracket to indicate that the number 5 is part of the graph.

Solution Figure 1 1

-2 - 1

0

2

3

� 4

Figure 1 2 If I

-2 - 1

0

I

2

3

I

3 5

6

�. 7

• 6

I

7

I

t;I!l!!: = =--

2

Figure 1 3

_

4

5

4 units

•

• -5 - 4 -3 -2 - 1

1_ 3

• 0

1

u n its 2

.1

E!; 3

I.

4

DEFINITION

Now Work

PROBLEM 3 1

Find D ista n ce on t h e Rea l N u m ber Line

The absolute value of a number a is the distance from 0 to a on the number line. For example, -4 is 4 units from 0, and 3 is 3 units from O. See Figure 13. Thus, the absolute value of -4 is 4, and the absolute value of 3 is 3. A more formal definition of absolute value is given next.

The absolute value of a real number a, denoted by the symbol l a l , is defined by the rules

lal = a

if a

2:

0

and

l a l = -a

if a

6

Now Work

PROBLEM 87

Eva l u ate Squ a re Roots

A real number is squared when it is raised to the power 2. The inverse of squaring is finding a square root. For example, since 62 = 36 and (-6)2 = 36, the numbers 6 and 6 are square roots of 36. The symbol .y, called a radical sign, is used to denote the principal, or non negative, square root. For example, V36 = 6. -

DEFINITION

If a is a nonnegative real number, the nonnegative number b, such that b2 is the principal square root of a, is denoted by b = Va.

=

a

-.J

The following comments are noteworthy: 1. Negative numbers do not have square roots (in the real number system), be

cause the square of any real number is nonnegative. For example, v'=4 is not a real number, because there is no real number whose square is -4. 2. The principal square root of 0 is 0, since 02 = O. That is, YO = O. 3. The principal square root of a positive number is positive.

4. If c EXAMPLE 11

2::

0, then (vc?

= c.

For example, (V2? =

2 and

(V3?

=

3.

Evaluating Square Roots

(a)

V64 =

8

(b)

0= 1 'f16 "4

( c)

(V1.4y = 1.4

•

24

CHAPTER R

Review

()

Examples l1(a) and (b) are examples of square roots of perfect squares, since 1 1 2 64 82 and 16 "4 . Consider the expression W. Since a2 2: 0, the principal square root of a2 is defined whether a > 0 or a < O. However, since the principal square root is non negative, we need an absolute value to ensure the nonnegative result. That is,

=

=

W= lal

a any real number

(2)

Using Equation ( 2)

EXAMPLE 1 2

V(2.3)2 12.31 2.3 (b) V( -2.3)2 = 1-2.31 2.3 (c) # = Ixl (a)

=

=

=

� Now Work

7

a

P R O B L E M 83

Use a Calcu lator to Eva l u ate Exponents

G or an [2] key, which is used for computa

Your calculator has either a caret key, tions involving exponents.

II

EXAM PLE 1 3

I

Exponents on a Graphing Calculator

(2.3)5

Evaluate:

Figure 15 shows the result using a T I-84 graphing calculator.

Solution Figure 15

�=>

64.36343

8

-

Now Work

a

PROBLEM 1 13

Use Scientific Notation

Measurements of physical quantities can range from very small to very large. For exa mple, the mass of a proton is approximately 0.00000000000000000000000000167 kilo gram and the mass of Earth is about 5,980,000,000,000,000,000,000,000 kilograms. These numbers obviously are tedious to write down and difficult to read, so we use exponents to rewrite each. DEFINITION

W hen a number has been written as the product of a number x, where 1 ::; x < 10, times a power of 10, it is said to be written in scientific notation,...]

In scientific notation, Mass of a proton Mass of Earth

= 1.67 X 10-27 kilogram

=

5.98 X 1024 kilograms

SECTION R.2 Algebra Essentials

25

Conve rting a Deci mal to Scientific Notation

To change a positive number into scientific notation: 1. Count the number N of places that the decimal point must be moved to arrive at a number x, where 1 � x < 10. 2. If the original number is greater than or equal to 1, the scientific notation

is x X ION. If the original number is between 0 and 1, the scientific no tation is x X lO-N.

EXAMPLE 14

Using Scientific Notation

Write each number in scientific notation. (a) 9582 Solution

(b) 1.245

(c) 0.285

(d) 0.000561

(a) The decimal point in 9582 follows the 2. We count left from the decimal point 9 5 8 2 t t t 3

2

1

=

9.582 X 103

stopping after three moves, because 9.582 is a number between 1 and 10. Since 9582 is greater than 1, we write 9582

(b) The decimal point in 1.245 is between the 1 and 2. Since the number is already between 1 and 10, the scientific notation for it is 1.245 X 10° = 1.245. (c) The decimal point in 0.285 is between the 0 and the 2. We count o

2 8 5

4-J'

.

stopping after one move, because 2.85 is a number between 1 and 10. Since 0.285 is between 0 and 1, we write 0.285 = 2.85 X 10-1 (d) The decimal point in 0.000561 is moved as follows: 0 . 0 0 0 5 6 1 I) t t t t 1

As a result,

t;m::;== -

EXAM PLE 1 5

2

4

0.000561 = 5.61 X 10-4

Now Work

•

PROBLEM 1 1 9

Changing from Scientific Notation to Decimals

W rite each number as a decimal. (b) 3.26 X 10-5 (a) 2.1 X 104 Solution

3

(a) 2.1 X 104

=

(b) 3.26 X 10-5

2

=

(c) 1 X 10-2 = 0

� == >'--

1

I)

0

t

t

0

Now Work

2

0

1:

5

1

t t

0 2

0 4

(I

t 1:

0 3

0 3

t 1:

0 4

0 2

t 1:

(c) 1 X 10-2

X 104 = 21,000 3

(I

2

6 X 10-5

=

0.0000326

o X 10-2 = 0.01

PROBLEM 1 2 7

•

26

CHAPTER R

Review Using Scientific Notation

EXAMPLE 16

(a) The diameter of the smallest living cell is only about 0.00001 centimeter (cm).* Express this number in scientific notation. (b) The surface area of Earth is about 1.97 X 108 square miles.t Express the sur face area as a whole number. (a) 0.00001 cm = 1 X 10-5 cm because the decimal point is moved five places and the number is less than 1. (b) 1.97 X 108 square miles = 197,000,000 square miles.

Solution

•

Now Work

== "L.'l!l:

COMMENT

P R O B L E M 1 53

On a calculator, a n u m ber such as

3.615 X 1012 is usually displayed as 13.615E12·1

•

'" Powers of Ten, Philip and Phylis Morrison.

t1998 Informalion Please Almanac.

f.-li�toriC81 Feature al-jabr. This

number i s added t o one side o f a n equation, then i t must also b e added

word is a part of the title of a ninth century work, "Hisab al-jabr

to the other side in order to "restore" the equality. The title of the work,

T

he word

algebra

I

is derived from the Arabic word

w'al-muqabalah," written by Mohammed ibn Musa al-Khowarizmi.

The word

freely translated, is "The Science of Reduction and Cancellation." Of

al-jabr means "a restoration," a reference to the fact that, if a

course, today, algebra has come to mean a great deal more.

R.2 Assess Your Understanding Concepts a n d Vocabulary

1.

A(n)

7.

is a letter used in algebra to represent any

number from a given set of numbers.

2.

On the real number line, the real number zero is the coordinate of the

8.

3. An inequality of the form

a

>

b

is called a(n)

9.

____

inequality. and 4 is called the

5. 6.

2\

True or False

True or False

The absolute value of a real number is always

the number

2

is called the

1234.5678

When a number is expressed in scientific no

and a power of 10.

____

10. =

Trite or False

tation, it is expressed as the product of a number x, 0 :s x < 1,

____

In scientific notation,

The distance between two distinct points on

greater than zero.

___ _

4. In the expression

Trite or False

the real n umber line is always greater than zero.

___ _

True or False

To multiply two expressions having the same

base, retain the base and multiply the exponents.

The product of two negative real numbers is

always greater than zero.

Skill B u i lding

11.

On the real number line, label the points with coordinates 0, 1, -1,

12.

Repeat Problem

In Problems 13.

18.

�

13-22,

?0

v2? 1.41

11

for the coordinates

0,

-2,2, -1.5,

replace the question mark by 14.

5

?6

19.

�?

0.5

,

or

�,�, � and

=,

whichever is correct.

.15. - 1 ? -2 20.

�, -2 . 5 , �, and 0.25. 2 4

�

? 0.33

16. -3?. - 2 5

2 21. "3?.0. 67

17.

7T? 3.14

22.

�

? 0.25


In Problems 23-28, write each statement as an inequality. 24. z is negative 23. x is positive 26.

Y is greater than-5

27

. 25. x i s less than 2

27. x is less than or equal to 1

In Problems 29-32, graph the numbers x on the real number line. 29. x � -2 30. x < 4

28. x is greater than or equal to 2

-1

31. x>

32. x:=:;7

In Problems 33-38, use the given real number line to compute each distance. A

33.

d(C, D )

B

C

-4 -3 -2 -1

0

34. d(C, A)

D

35. d(D,E)

In Problems 39-46, evaluate each expression if x=-2 and y= 3. 39. x+2y 40. 3x+y 43.

2x x-y

-

44.

x+y x-y

-

E 2

3

4

hl y

.37. d(A, E)

41. 5xy +2 45.

53. 1 4x- 5yl

6

36. d(C, E)

3x + 2y 2+Y

54. 1 3x+ 2yl

38. d(D, B)

42. -2x+xy 46.

---

In Problems 47-56, find the value of each expression if x= 3 and y= -2. 47. Ix+yl 48. Ix - yl 49. Ixl+ Iyl 52.

5

2x- 3

--

Y

50. Ixl- Iyl

Ixl 51. x

55. 11 4xl- 15yll

56. 31xl+2 1yl

In Problems 57-64, determine which of the valuers)(a)through (d),if any, must be excluded from the domain of the variable in each expression: (a)x= 3 (c)x=O (b)x=1 (d)x= -1 x x x2- 1 x2+1 59. -2- 60. -1 57. - 58. -x 9 x-+ 9 x x 62.

x3 x -I

2-

63.

x2+ 5x- 10 x3-x

-----

64.

In Problems 65-68, determine the domain of the variable x in each expression. x 4 67. - 65. - 66. � x+ 4 x-5 x+4

68.

-9x2-X+ 1 ----:;, -- x�+x

X -? x-6

--

%

In Problems 69-72, use the formulaC = (F - 3 2) for converting degreesFahrenheit into degreesCelsius to find theCelsius measure of eachFahrenheit temperature. 72. F = -4° 70. F =212° 69. F = 3 2° 71. F = 77° In Problems 73-84, simplify each expression. 75. 4-2 74. -42 73. (-4? 81. v2s

76. -4-2 82. V36

83.

�

84.

�

In Problems 85-94, simplify each expression. Express the answer so that all exponents are positive. Whenever an exponent is 0 or negative, we assume that the base is not O. x2l 89. --4 86. (-4x2rl 85. (8x3)2 xy 93.

(3X-1 )-2 4y-1

94.

(5[2)-3 6y-2

28

CHAPTER R Review

In Problems 95-106, find the value of each expression if x 96. -3[1 Y ,. 95. 2xy-l

99. (xy) 2

100. (x

+

=

2 and y

=

-l.

97. 101.

y?

x2 + l

0

103.

yx2 + l

107.

Find the value of the expression

2x3 - 3x2 + 5 x - 4 if x 2. What is the value if x

lOS.

Find the value of the expression

4x3 + 3x2 - + 2 if x

109.

What is the value o f

104.

# + Vl

106. yt

105. xl' =

X

=

1. What is the value if

--?

(666)4 (222)4

110.

x

=

=

I?

2?

What is the value of

(0.1 )3(20)3?

In Problems 111-118, use a calculator to evaluate each expression. Round your answer to three decimal places. 114. (2.2f5 111. (8.2) 6 113. (6.1) -3 112. (3.7)5

115. "

(-2.8) 6

116.

- (2.8) 6

In Problems 119-126, write each number in scientific notation. 119. 454.2 120. 32.14

123.

32,155

124.

21,210

In Problems 127-134, write each number as a decimal. 127. 6.15 x 104 12S. 9.7 x 103

131.

1.1 X

108

132.

4.112 x 102

117.

(-8.11f4

11S.

-(8.1 1f4

121.

0.013

122.

0.00421

125.

0.000423

126.

0.0514

129.

1.214 x 10-3

130.

9.88 X

133.

8.1 x 10-2

134.

6.453 x 10-1

10-4

Appl ications a n d Extensions

1n Problems 135-144, express each statement as an equation involving the indicated variables. 139. Area of an Equilateral Triangle The area A of an equilat135. Area of a Rectangle The area A of a rectangle is the product of its length I and its width w. V3 eral triangle is - times the square of the length x of one

� �w

136.

Perimeter of a Rectangle

137.

Circllmference of a Circle

13S.

7T

Area of a Triangle

product of its base

P of a rectangle is w.

The perimeter

twice the sum o f its length I and its width

is the product o f

4

side.

T h e circumference

and its diameter

d.

g

x

C of a circle

140.

141.

Perimeter of an Equilateral Triangle

The perimeter

eq uilateral triangle is

3 times the length

Volume of a Sphere

The volume V o f a sphere is

times the cube of the radius

x

P of an

of one side.

r.

� times 7T 3

The area A of a triangle is one-half the

b and its height h.

b

142.

Surface Area of a Sphere

4 times

7T

The surface area

times the square of the radius

r.

S of a sphere is


143.

The volume V of a

Volume of a Cube

0-0'

cube is the cube of the length x of a side.

of this stated radius are acceptable. If x is the radius of a ball bearing, a formula describing this situation is Ix - 31

x

144.

The surface area

Surface Area of a Cube

S of a

cube is

6 times the square of the length x of a side.

145.

Manufacturing Cost

where the variable

146.

The weekly production cost

ufacturing x watches is given by the formula

C

=

152.

+

=

2.999 acceptable?

(b) Is a ball bearing of radius x

=

2.89 acceptable?

Body Temperature

�

1.5

(a) What is the cost of producing 1000 watches?

(a) Show that a temperature of 97°F is unhealthy.

(b) What is the cost of producing 2000 watches?

(b) Show that a temperature of 100°F is not unhealthy.

At the beginning of the month,

Balancing a Checkbook

and the other for $32. He was also assessed a monthly ser vice charge of $5. What was his balance at the end of the month?

In Problems 147 and 148, write an inequality using an absolute value to describe each statement. 147. x is at least 6 units from 4.

153.

x is more than 5 units from 2. U.S. Voltage

154. 155.

Ix - 1101

s

156.

158.

Diameter of an Atom

159.

Diameter of Copper Wire

The smallest commercial copper

Smallest Motor

The smallest motor ever made is less than

Astronomy

One light-year is defined by astronomers to be

the distance that a beam of light will travel in 1 year (365 days). If the speed of light is 186,000 miles per second, how many miles are in a light-year? Express your answer in

x

to differ from normal by at most 8 volts. A formula that describes this is

scientific notation.

160.

Astronomy

8

Earth? Express your answer in seconds, using scientific notation.

(b) Show that a voltage of 209 volts is not acceptable. The F ireBall Company

161.

Does

162.

Does

manufactures ball bearings for precision equipment. One of its products is a ball bearing with a stated radius of 3 cen timeters (cm). Only ball bearings with a radius within 0.01 cm

How long does it take a beam of light to reach

Earth from the Sun when the Sun is 93,000,000 miles from

(a) Show that a voltage of 214 volts is acceptable.

Making PI'ecision Ball Bearings

The diameter of an atom is about

1 X 10-10 meter.* Express this diameter as a decimal.

notation.

5

voltage is 220 volts. It is acceptable for the actual voltage

s

The wavelength of visible

using scientific notation.

In other countries, normal household

- 2201

Wavelength of V isible Light

light is about 5 X 10-7 meter.* Express this wavelength as a

wire is about 0.0005 inch in diameter.T Express this diameter

(b) Show that a voltage of 104 volts is not acceptable.

Ix

The height of Mt. Everest is 8872 me

0.05 centimeter wide.t Express this width using scientific

(a) Show that a voltage of 108 volts is acceptable.

Foreign Voltage

Height of Mt. Everest

decimal.

In the United States, normal household volt

this is

The distance from Earth

ters: Express this height in scientific notation.

157.

fer from normal by at most 5 volts. A formula that describes

Distance from Earth to Its Moon

to the Moon is about 4 X 108 meters.* Express this distance

as a whole number.

age is 110 volts. It is acceptable for the actual voltage x to dif

151.

Normal human body temperature is

Ix - 98.61

the next month, he deposited $80, wrote a check for $120,

150.

(a) Is a ball bearing of radius x

C is in dollars.

made another deposit of $25, wrote two checks: one for $60

149.

0.01

1.5°F is considered unhealthy.A formula that describes this is

2x,

Mike had a balance of $210 in his checking account. During

148.

s

98.6°F. A temperature x that differs from normal by at least

C of man

4000

29

� .J

�

equal 0.333? If not, which is larger? By how much?

equal 0.666? If not, which is larger? By how much?

Discussion and Writing

163. 164.

Is there a positive real number "closest" to O? Number game

165.

I'm thinking of a number! It lies between 1

to"

and 10; its square is rational and lies between 1 and 10. The number is larger than

7T.

Correct to two decimal places (that

is, truncated to two decimal places) name the number. Now think of your own number, describe it, and challenge a fellow student to name it.

Morrison. 1998 Information Please Almanac.

':' Powers of Ten, Philip and Phylis t

Write a brief paragraph that illustrates the similarities and differences between "less than"

166.

(s).

«) and "less than or equal

Give a reason why the statement 5 < 8 is true.

30

CHAPTER R

Review

R.3 Geometry Essentia ls OBJECTIVES

1 Use the Pythagorean T h eorem and Its Converse (p. 30) 2

Know Geometry Formulas (p. 31)

3

Understand Congruent Triangles and Similar Triangles (p. 32)

In this section we review some topics studied in geometry that we shall need for our study of algebra.

1

Figure 16

b Leg a

Leg

PYTHAGOREAN THEOREM

Use the Pythagorean Theorem a n d Its Converse

The Pythagorean Theorem is a statement about right triangles. A right triangle is one that contains a right angle, that is, an angle of 90°. The side of the triangle opposite the 90° angle is called the hypotenuse; the remaining two sides are called legs. In Figure 16 we have used c to represent the length of the hypotenuse and a and b to represent the lengths of the legs. Notice the use of the symboli to show the 90° angle. We now state the Pythagorean Theorem. In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. That is, in the right triangle shown in Figure 16, (1)

I�

�--------------------------------�.

A proof of the Pythagorean Theorem is given at the end of this section. EXAMPLE 1

Finding the Hypotenuse of a Right Triangle

In a right triangle, one leg has length 4 and the other has length 3. What is the length of the hypotenuse? Solution

Since the triangle is a right triangle, we use the Pythagorean Theorem with a = 4 and b = 3 to find the length c of the hypotenuse. From equation (1), we have

c2 = a2 + b2 c2 = 42 + 32 = 16 c = \I25 = 5 � ==> .-

Now Work

+ 9

= 25 •

P R O BLE M 1 3

The converse of the Pythagorean Theorem is also true. CONVERSE OF THE PYTHAGOREAN THEOREM

In a triangle, if the square of the length of one side equals the sum of the squares of the lengths of the other two sides, the triangle is a right triangle. The 90° angle is opposite the longest side.

�

A proof is given at the end of this section. EXAMPLE 2

Verifying That a Triangle Is a Right Tri angle

Show that a triangle whose sides are of lengths 5,12, and 13 is a right triangle. Iden tify the hypotenuse. Solution

We square the lengths of the sides.

52

=

25,

122 = 144,

SECTION R.3 Geometry Essentials Figure 17

31

Notice that the sum of the first two squares (25 and 144) equals the third square (169). Hence, the triangle is a right triangle. The longest side, 13, is the hypotenuse. See Figure 17.

5

"i!=,-.

Now Work

•

PROBLEM 2 1

12

EXAMPLE 3

Applying the Pythagorean Theorem

Excluding antenna, the tallest inhabited building in the world is Taipei 101 in Taipei, Taiwan. If the indoor observation deck is 1437 feet above ground level, how far can a person standing on the observation deck see ( with the aid of a telescope)? Use 3960 miles for the radius of Earth. See Figure 18.

Figure 18

Source: Council o n Tall Buildings and Urban Habitat, 2006.

Solution From the center of Earth, draw two radii: one through Taipei 101 and the other to the farthest point a person can see from the observation deck. See Figure 19. Apply the Pythagorean Theorem to the right triangle.

Since 1 mile

=

5280 feet, then 1437 feet =

d2 + (3960)2 d2

=

=

( 3960 ( 3960

+ +

) )

1437 2 5280 1437 2 5280

��!� miles. So we have (3960? :::::: 2155.57

d :::::: 46.43 A person can see about 46 miles from the observation tower. Figure 19

d

1437 It

• 1l'I!>:=="'"

2

Now Work

P R O B L E M 53

Know Geometry Form u las

Certain formulas from geometry are useful in solving algebra problems. We list some of these next. For a rectangle of length l and width w, Area = lw For a triangle with base

b

b

Perimeter

and altitude Area =

h,

1

2

-bh

=

2l + 2w

32

CHAPTER R Review

For a circle of radius r (diameter d Area

=

71-,2

=

2r),

Circumference

2'1Tr

=

=

'1Td

For a closed rectangular box of length l, width w, and height h, Volume

G

=

lwh

Surface area

=

2lh

+

2wh

+

2lw

For a sphere of radius r, Surface area

=

4'1Tr2

For a right circular cylinder of height h and radius r, Surface area

h C.'!Jn:==-

Now Work

=

2'1Tr2 + 2'1Trh

P R O BLE M 2 9

Using Geometry Formulas

EXAMPLE 4

A Christmas tree ornament is in the shape of a semicircle on top of a triangle. How many square centimeters (cm) of copper is required to make the ornament if the height of the triangle is 6 cm and the base is 4 cm? See Figure 20. The amount of copper required equals the shaded area. This area is the sum of the area of the triangle and the semicircle. The triangle has height h = 6 and base b = 4. The semicircle has diameter d = 4, so its radius is r = 2.

Solution Figure 20

Area

=

Area of triangle + Area of semicircle

= -bh =

1 1 ? 1 1 2 + -'1Tr = -(4)(6) + -'1T · 2 2 2 2 2 2 12 + 2'1T 18.28 cm

b

=

4; h

=

6;

r =

2

�

About 18.28 cm2 of copper is required. ...'!I!l:===-- -

3 ('

r r

In Word s

Two triangles are congruent if r they have the same size and r shape.

DEFINITION

Now Work

•

PROBLEM 47

U n derstand Con g ruent Triang les a n d S i m ila r Tria ng les

Throughout the text we will make reference to triangles. We begin with a discussion of congruent triangles. According to dictionary. com, the word congruent means coinciding exactly when superimposed. For example, two angles are congruent if they have the same measure and two line segments are congruent if they have the same length. Two triangles are congruent if each of the corresponding angles is the same measure and each of the corresponding sides is the same length.

.J

In Figure 21, corresponding angles are equal and the lengths of the corre sponding sides are equal: a = d, b = e, and c = f. We conclude that these triangles are congruent.

SECTION R.3 Geometry Essentials

33

Figure 21

Congruent triangles

c

It is not necessary to verify that all three angles and all three sides are the same measure to determine whether two triangles are congruent. Determining Con g ruent Tria ngles

Two triangles are congruent if two of the an gles are equal and the lengths of the corresponding sides between the two angles are equal. For example, in Figure 22 ( a ) , the two triangles are congruent because two angles and the included side are equal. 2. Side-Side-Side Case Two triangles are congruent if the lengths of the corresponding sides of the triangles are equal. For example, in Figure 22 (b) , the two triangles are congruent because the three corresponding sides are all equal. 3. Side-Angle-Side Case Two triangles are congruent if the lengths of two corresponding sides are equal and the angles between the two sides are the same. For example, in Figure 22 ( c) , the two triangles are congruent because two sides and the included angle are equal. 1. Angle-Side-Angle Case

Figure 22

(b)

(e)

We contrast congruent triangles with similar triangles. DEFINITION

Two triangles are similar if the corresponding angles are equal and the lengths of the corresponding sides are proportional.

..J

r

r

r r

r

In

Word s

Two triangles a re similar i f they have the same shape. but (possi bly) different sizes.

For example, the triangles in Figure 23 are similar because the corresponding angles are equal. In addition, the lengths of the corresponding sides are proportional because each side in the triangle on the right is twice as long as each corresponding side in the triangle on the left. That is, the ratio of the corresponding sides is a constant:

d

-

a

=

e

-

b

=

f

-

c

=

2.

34

CHAPTER R

Review Figure 23

It is not necessary to verify that all three angles are equal and all three sides are proportional to determine whether two triangles are congruent. Determ ining Similar Triang les 1. Angle-Angle Case

Two triangles are similar if two of the correspond ing angles are equal. For example, in Figure 24(a), the two triangles are similar because two angles are equal. 2. Side-Side-Side Case Two triangles are similar if the lengths of all three sides of each triangle are proportional. For example, in Figure 24(b), the two triangles are similar because 10 30

6

5 15

18

1 3

3. Side-Angle-Side Case Two triangles are similar if two corresponding sides are proportional and the angles between the two sides are equal. For example, in Figure 24(c), the two triangles are similar because 4 12 2 . = "3 and the angles between the sIdes are equal. = "6 18 Figure 24

//

80

K. EXAMPLE 5

(a)

(b)

(c)

Using Similar Triangles

Given that the triangles in Figure 25 are similar, find the missing length angles A, B, and C. Figure 25

60� 90°

x

and the

3S

SECTION R.3 Geometry Essentials Solution

Because the triangles are similar, corresponding angles are equal. So A

and C 30°. Also, the corresponding sides are proportional. That is, � 5 . th·IS equatlOn for x. =

:)X _

3 5 3 • -

5

=

90°, B

=

60°,

i. We solve x

6 x 5x'

=

3x = 30 x

=

6 x

-

Simplify

10

=

Multiply both sides by 5x.

Divide both sides by 3.

The missing length is 10 units. L'l!l: ==_ -

Now Work

•

PROBLEM 41

We begin with a square, each side of length

Proof of the Pythagorean Theorem

a

b. In this square, we can form four right triangles, each having legs equal in length to a and b. See Figure 26. All these triangles are congruent (two sides and +

their included angle are equal). As a result, the hypotenuse of each is the same, say

c, and the pink shading in Figure 26 indicates a square with an area equal to c2. Figure 26

Area

�ab

=

b

a

a

rr--------�----"

b

b Area

=

�ab

a

b

The area of the original square with sides a + b equals the sum of the areas of the 1 four triangles (each of area 'lab) plus the area of the square with side c. That is,

(a + b)?

a2 + 2ab a2

+ +

b2 b2

= = =

1 1 1 1 ab + ab + ab + ab + c2 2 2 2 2 -

-

-

2ab + c2

c2

The proof is complete.

•

Figure 27

b a

(a)

We begin with two trian gles: one a right triangle with legs a and b and the other a triangle with sides a, b, and c for which c2 = a2 + b2. See Figure 27. By the Pythagorean Theorem, the length x of the third side of the first triangle is Proof of the Converse of the Pythagorean Theorem

x2

b

-

=

a2

+

x

=

c

b2

The two triangles have the same sides and are therefore congruent. This means corre sponding angles are equal, so the angle opposite side c of the second triangle equals 90°. • The proof is complete.

36

CHAPTER R Review

R.3 Assess Your Understanding Concepts and Voca bulary 1. A(n)

2.For a triangle with base area A is

9. True or False

The triangles shown are similar.

10. True or False

The triangles shown are similar.

triangle is one that contains an angle of

90 degrees. The longest side is called the

____

b and altitude h, a formula for the

____

3.The formula for the circumference

4.Two triangles are

C of a circle of radius r is

if corresponding angles

are equal and the lengths of the corresponding sides are proportional. 5. True or False

In a right triangle, the square of the length of

the longest side equals the sum of the squares of the lengths of the other two sides. 6. True or False

The triangle with sides of length 6, 8, and 10

is a right triangle. 4

r is 37Tr2.

7. True or False

The volume of a sphere of radius

8. True or False

The triangles shown are congruent.

Ski l l Building

In Problems 11-16, the lengths of the legs of a right triangle are given. Find the hypotenuse. 12. a 6, b 8 11. a 5, b 12 3 14. a 4, b 15. a 7, b 24 =

=

=

=

=

=

=

=

13. a 16. a

=

=

10,

b

14,

b

=

24

=

48

In Problems 17-24, the lengths of the sides of a triangle are given. Determine which are right triangles. For those that are, identify the hypotenuse. 17.3, 4, 5

18.6, 8, 10

19.4, 5, 6

20.2, 2, 3

21. 7, 24, 25

22.10, 24,26

23.6, 4, 3

24.5, 4, 7

25.Find the area A of a rectangle with length 4 inches and width 2 inches. 26.Find the area A of a rectangle with length 9 centimeters and width 4 centimeters. 27.Find the area A of a triangle with height 4 inches and base 2 inches. 28.Find the area A of a triangle with height 9 centimeters and base 4 centimeters. 29. Find the area A and circumference

C of a circle of radius 5 meters.

30.Find the area A and circumference

C of a circle of radius 2 feet.

31.Find the volume V and surface area

S of a rectangular box with length 8 feet, width 4 feet, and height 7 feet.


S of a rectangular box with length 9 inches, width 4 inches, and height 8 inches.


S of a sphere of radius 4 centimeters.


S of a sphere of radius 3 feet.


S of a right circular cylinder with radius 9 inches and height 8 inches.


S of a right circular cylinder with radius 8 inches and height 9 inches.

SECTION R.3 Geometry Essentials

37

In Problems 37-40, find the area of the shaded region. 37. 38. 2 2

'0

'0

60 t? 90° 30°

10

16

In Problems 41-44, each pair of triangles is simila!: Find the missing length x and the missing angles A, B, and C. 41. 42. 44. 43. 4

v7 6y A

Application s a n d Extensions

45.

How many feet does a wheel with a diameter o f 1 6 inches travel after four revolutions?

46.

49.

In the figure shown,

ABCD

needed to enclose the window?

is a square, with each side of

length 6 feet. The width of the border (shaded portion) be

tween the outer square

6'

EFGH and ABCD is 2 feet. Find the

area of the border.

E

F A

B

0

H

48.

��

0

Refer to the figure. Square

C

-+-

G

ABCD has an area of 100 square

CGF?

o

c

11--+---11

4'

50.

feet; square BEFG has an area of 1 6 square feet. What is the

area of the triangle

A Norman window consists of a rectangle sur

dow shown in the illustration. How much wood frame is

How many revolutions will a circular disk with a diameter of 4 feet have completed after it has rolled 20 feet?

47.

Architecture

mounted by a semicircle. Find the area of the Norman win

Construction

A circular swimming pool, 20 feet in diame

ter, is enclosed by a wooden deck that is 3 feet wide. What is the area of the deck? How much fence is required to enclose the deck?

38

51.

CHAPTER R

Review

HowT.'111 Is the Great Pyramid'?

The ancient Greek philoso

52.

The Bermuda Triangle

Karen is doing research on the

Bermuda Triangle which she defines roughly by Hamilton,

pher Thales of Miletus is reported on one occasion to have visited Egypt and calculated the height of the Great Pyramid

Bermuda; San Juan, Puerto Rico; and Fort Lauderdale,

of Cheops by means of shadow reckoning. Thales knew that

Florida. On her atlas Karen measures the straight-line dis tances from Hamilton to Fort Lauderdale, Fort Lauderdale

each side of the base of the pyramid was 252 paces and that

to San Juan, and San Juan to Hamilton to be approximately

his own height was 2 paces. He measured the length of the pyramid's shadow to be 114 paces and determined the length

57 millimeters (mm), 58 mm, and 53.5 mm respectively. If

of his shadow to be 3 paces. See the illustration. Using simi

the actual distance from Fort Lauderdale to San Juan is

lar triangles, determine the height of the Great Pyramid in

1046 miles, approximate the actual distances from San Juan

terms of the number of paces.

to Hamilton and from Hamilton to Fort Lauderdale. Source:

Source: www.anselm.edulhomepageidbanachl thales. htm.This site

Source: www.en. wikipedia. orglwikilBermuda_Triangle.

references another source: Selections, from Julia E. Diggins, String,

Straightedge, and Shadow, Viking Press, New York,

Illustrations by Corydon Bell.

www. worldatlas.com

1965,

In Problems 53-55, use the facts that the radius of Earth is 3960 miles and 1 mile 5280 feet. How far can a person see from the bridge, which is 150 feet 53. How Far Can You See? The conning tower of the U.S.S. above sea level? Silvers ides, a World War II submarine now permanently sta tioned in Muskegon, Michigan, is approximately 20 feet above 56. Suppose that 117 and n are positive integers with 117 > n. If sea level. How far can you see from the conning tower? a 1172 - n2, b 2mn, and c 1172 + /12, show that a, b, 54. How Far Can You See'? A person who is 6 feet tall is stand and c are the lengths of the sides of a right triangle. (This for =

=

ing on the beach in Fort Lauderdale, Florida, and looks out

integers, such as 3,4, 5; 5,12, 13; and so on. Such triplets of

zon. How far is the ship from shore? How Far Can You See?

=

mula can be used to find the sides of a right triangle that are

onto the Atlantic Ocean. Suddenly, a ship appears on the hori

55.

=

integers are called P ythagorean triples.)

The deck of a destroyer is 100 feet

above sea level. How far can a person see from the deck?


57.

You have 1000 feet o f flexible pool siding and wish to con struct a swimming pool. Experiment with rectangular-shaped

58.

The Gibb's Hill Lighthouse, Southampton, Bermuda, in op

eration since 1846, stands 117 feet high on a hill 245 feet high,

pools with perimeters of 1000 feet. How do their areas vary?

so its beam of light is 362 feet above sea level. A brochure

What is the shape of the rectangle with the largest area? Now

states that the light itself can be seen on the horizon about

compute the area enclosed by a circular pool with a perime

26 miles distant. Verify the correctness of this information.

ter (circumference) of 1000 feet. What would be your choice

The brochure further states that ships 40 miles away can see

of shape for the pool? If rectangular, what is your preference

the light and planes flying at 10,000 feet can see it 120 miles

for dimensions? Justify your choice. If your only considera

away. Verify the accuracy of these statements. What assump

tion is to have a pool that encloses the most area, what shape

tion did the brochure make about the height of the ship?

should you use?

SECTION R04 Polynomials

39

R.4 Polynomia ls OBJECTIVES

1 Recognize Monomials (po 39) 2

Recognize Polynomials (po 40)

3

Add and Subtract Polynomials (p.4l)

4

Multiply Polynomials (p. 42)

5

Know Formulas for Special Products (p.43)

6

Divide Polynomials Using Long Division (po 44)

7

Work with Polynomials in Two Variables (po 47)

We have described algebra as a generalization of arithmetic in which letters are used to represent real numbers. From now on, we shall use the letters at the end of the alphabet, such as x, y, and z, to represent variables and the letters at the begin ning of the alphabet, such as a, b, and c, to represent constants. In the expressions 3x + 5 and ax + b, it is understood that x is a variable and that a and b are con stants, even though the constants a and b are unspecified. As you will find out, the context usually makes the intended meaning clear.

1

Recog n ize Monom ials

DEFINITION

A monomial in one variable is the product of a constant and a variable raised to a nonnegative integer power. A monomial is of the form

NOTE

The nonnegative i ntegers a re the integers 0, 1, 2, 3, ... . •

k ax

where a is a constant, x is a variable, and k ;:::: 0 is an integer. The constant a is called the coefficient of the monomial. If a *' 0, then k is called the degree of the monomial.

EXAMPLE 1

Examples of Monomials

(a) 6x2

Coefficient

Monomial

(b) - \I2x3

(c)

'" .)

(d) - 5 x (e)

x4

Degree

6

2

'" .)

o

Since 3 = 301

-5

1

Since -5x

-\12

1

3

4

=

=

3;;, x

"* 0

-5x1 •

Now let's look at some expressions that are not monomials.

EXAMPLE 2

Examples of Nonmonomial Expressions

(a) 3X1/2 is not a monomial, since the exponent of the variable x is 2" and 2" is not a nonnegative integer. (b) 4x-3 is not a monomial, since the exponent of the variable x is -3 and -3 is not a nonnegative integer. �

-

Now Work

1

1

•

PROBLEM 7

40

CHAPTER R

Review

2

Recognize Polynom ials

Two monomials with the same variable raised to the same power are called like terms. For example, 2X4 and -5x4 are like terms. In contrast, the monomials 2x3 and 2x5 are not like terms. We can add or subtract like terms using the Distributive Property. For example,

The sum or difference of two monomials having different degrees is called a binomial. The sum or difference of three monomials with three different degrees is called a trinomial. For example,

x2 - 2 is a binomial. x3 - 3x + 5 is a trinomial. 2x2 + 5x2 + 2 = 7x2 + 2 is a binomial. DEFINITION

r r

r

A polynomial in one variable is an algebraic expression of the form ( 1)

In Words

where an , al1 - 1 , . . . , al , ao are constants,* called the coefficients of the poly nomial, n ;::: ° is an integer, and x is a variable. If an =f. 0, it is called the leading coefficient, and n is called the degree of the polynomial.

A

polynomial is a sum of monomials.

.J

The monomials that make up a polynomial are called its terms. If all the coeffi cients are 0, the polynomial is called the zero polynomial, which has no degree. Polynomials are usually written in standard form, beginning with the nonzero term of highest degree and continuing with terms in descending order according to degree. If a power of x is missing, it is because its coefficient is zero. EXAMPLE 3

Examples of Polynomi als

Polynomial

Coefficients

Degree

8x3 + 4x2 - 6x + 2 3x2 - 5 = 3x2 + 0 · x + ( -5 ) 8 - 2x + x2 = 1 · x2 + ( -2)x + 8 5x + v2 = 5x1 + v2 3 = 3 · 1 = 3 · xo

-8, 4, -6, 2 3, 0, -5 1, -2, 8 5, v2 3

3 2 2 1

-

°

°

°

No degree

•

Although we have been using x to represent the variable, letters such as y or z are also commonly used.

3x4 - x2 + 2 is a polynomial (in x) of degree 4. 9y3 - 21 + y - 3 is a polynomial (in y) of degree 3.

Z5 +

7T

is a polynomial (in z) of degree 5.

Algebraic expressions such as 1

x ':' The notation

a"

is read as

"

a

sub

"

n.

x +5

and

The number

n

is called a subscript and should not be confused

with an exponent. We use subscripts to distinguish one constant from another when a large or undeter mined number of constants is required.

SECTION R.4 Polynomials

are not polynomials. The first i s not a polynomial because 1:.

x

=

41

X-I has an exponent

that is not a nonnegative integer. Although the second expression is the quotient of two polynomials, the polynomial in the denominator has degree greater than 0, so the expression cannot be a polynomial.

- Now Work

'1'l

3

PROBLEM 1 7

Add a n d S u btract Polynomials

Polynomials are added and subtracted by combining like terms.

EXAMPLE 4

Adding Polynomials

Find the sum of the polynomials:

8x3 - 2x2 + 6x - 2 and 3x4 - 2x3 + x2 + X

Solution

We shall find the sum in two ways. Horizontal Addition: The idea here is to group the like terms and then combine

them.

(8x3 - 2x2 + 6 x - 2) + (3X4 - 2x3 + x2 + x ) = 3x4 + (8x� - 2 x3 ) + ( - 2x- + x-) + (6x + x ) - 2 3x4 + 6x3 - x2 + 7 x - 2 ?

0

?

=

Vertical Addition: The idea here is to vertically line up the like terms in each poly nomial and then add the coefficients.

8x3 - 2X2 + 6x - 2 + 3x4 - 2x3 + x2 + X 3x4 + 6x3 - x2 + 7 x - 2

•

We can subtract two polynomials horizontally or vertically as well.

EXAMPLE 5

Subtracting Polynomials

Find the difference: Solution

(3X4 - 4x3 + 6x2 - 1 ) - (2X4 - 8x2 - 6x + 5 )

Horizontal Subtraction:

(3x4 - 4x3 + 6x2 - 1 ) - ( 2X4 - 8x2 - 6x + 5 ) =

3x4 - 4x3 + 6x2 - 1 + ,( _ 2X4 + 8x + 6x - 5)

�

,

Be s u re to change the sig n of each

term in the second polynomial.

=

i

(3X4 - 2X4 ) + ( -4x3 ) + (6x2 + 8x2 ) + 6x + ( - 1 - 5 )

Group like terms. =

X4 - 4x3 + 14x2 + 6x - 6

42

CHAPTER R

Review

COMMENT Vertical subtraction will be _ used when we divide polynomials,

Vertical Subtraction: We line up like terms, change the sign of each coefficient of the second polynomial, and add.

3x4 - 4x3 + 6X2 - 1 - 8X2 - 6x + 5 ] [2X4

=

+

=

3x4 - 4x3 + 6x2 - 1 + 8x2 + 6x - 5 -2x4 x4 - 4x3 + 14x2 + 6x - 6

•

The choice of which of these methods to use for adding and subtracting poly nomials is left to you. To save space, we shall most often use the horizontal format. 1.1'111::

4

\1"'-

Now Work

PROBLEM 29

M ulti ply Polynom ials

Two monomials may be multiplied using the Laws of Exponents and the Commutative and Associative Properties. For example,

Products of polynomials are found by repeated use of the Distributive Property and the Laws of Exponents. Again, you have a choice of horizontal or vertical format.

EXAMPLE 6

Multiplying Polynom i als

Find the product: Solution

(2x + 5 ) (x2 - X + 2 )

Horizontal Multiplication:

( 2x

+

5 ) (x2 - X + 2 ) = 2x (x2 - X

i

+

2)

+

5 ( x2 - X

2)

+

Distributive Property

i

=

(2x · x2 - 2x · x + 2x · 2 )

+

(5 · x2 - 5 . x + 5 . 2)

Distributive Property =

i

(2x3 - 2x2

+

4x) + (5x2 - 5x + 10)

Law of Exponents =

i

2x3

+

3x2 - X + 10

Combine like terms.

Vertical Multiplication: The idea here is very much like multiplying a two-digit number by a three-digit number.

x2 - x + 2 2x + 5 This lin e is 2x(J2 x + 2) , 2x3 - 2X2 + 4x 5x2 - 5x + 10 This line i s 5(J2 x + 2) . (+) 2x3 + 3x2 - X + 10 S u m o f t h e above two lin es. -

-

'I!�

Now Work

PROBLEM 4S

•

SECTION R.4 Polynomials

5

43

Know Form u l a s for Special Prod u cts

Certain products, which we call special products, occur frequently in algebra. We can calculate them easily using the FOIL (First, Outer, Inner, Last) method of multiplying two binomials.

c l i

� l I

Outer

F i rst

(ax + b)(ex + d) = ax(ex + d) + b(ex + d)

Il L

l n ne r

EXAMPLE 7

l

Last

J

Fi rst

Outer

�

�

Inner

,------"---;

Last

,------"---;

= ax . ex + ax . d + b . ex + b . d = aex2 + adx + bex + bd = aex2 + (ad + be)x + bd

Using FO i l

(a) ( x - 3 ) (x + 3 ) = x2 + 3x - 3x - 9 = x2 - 9

F

(b) (c) (d) (e)

( x + 2)2 = (x + 2 ) (x + ( x - 3 )2 = (x - 3 ) (x (x + 3 ) (x + 1 ) = x2 + (2x + 1 ) (3x + 4) = 6x2

0

I

L

2) x2 + 2x + 2x + 4 = x2 + 4x + 4 3) x2 - 3x - 3x + 9 = x2 - 6x + 9 + 3x + 3 = x2 + 4x + 3 + 8x + 3x + 4 6x2 + l lx + 4 =

=

X

==c- 't;'!I!l:

Now Work

=

•

PROBLEMS 47 AND 55

Some products have been given special names because of their form. The fol lowing special products are based on Examples 7(a), (b), and (c). Difference of Two Squares

(2)

(x - a) (x + a) = x2 - a2

Squares of Binomials, or Perfect Squares

(x + a f (x - a) 2

EXAMPLE 8

=

=

x2 + 2ax + a2 x2 - 2ax + a2

(3a) (3b)

Using Special Product Form ulas

(a) (x - 5 ) (x + 5 ) = x2 - 52 = x2 - 25 (b) (x + 7f = x2 + 2 ' 7 ' x + 72 = x2 + 14x + 49 (c) ( 2x + 1 ) 2 = (2x) 2 + 2 · 1 · 2x + 12 = 4x2 + 4x + 1 (d) (3x - 4) 2 = (3x i - 2 · 4 · 3x + 42 = 9x2 - 24x + 16 � _ &::i"" -

Now Work

Difference of two squares Square of a binomial Notice that we used 2x in place of x in formula (3a). Replace x by 3 x in formula (3b),

PROBLEMS 65, 67, AND 69

Let's look at some more examples that lead to general formulas.

•

44

CHAPTER R

Review Cubing a Binomial

EXAMPLE 9

(a ) ( x + 2 ? = (x

+

2)(x

+

2) 2

(b) (x - 1 )3 = (x - l ) (x - 1 ) 2

(x + 2)(x2 + 4x + 4) Formula (3a) = (x3 + 4x2 + 4x) + (2x2 + 8x + 8) = x3 + 6x2 + 1 2x + 8 =

= =

=

(x - 1 ) (x2 - 2x + 1 ) Formula (3b) (x3 - 2x2 + x) - (x2 - 2x + 1 ) x3 - 3x2 + 3x - 1

111

Cu bes of Binomials, or Perfect Cubes

f.'!f = fii: =- -

(x + a)3

=

(x - a)3

=

Now Work

x3 + 3ax2 + 3a2 x + a3 x3 - 3ax2 + 3a2 x - a3

(4a ) (4b)

PROBLEM 85

Forming the Difference of Two C ubes

EXAMPLE 1 0

(x - 1 ) (x2 +

X

+

1)

=

=

=

x (x2 + + 1 ) - 1 (x2 + + 1 ) x3 + x2 + - x2 - x - I x3 - 1 X

X

X

Forming the Sum of Two C ubes

EXAMPLE 1 1

(x + 2) (x2 - 2x

+

4) = x (x2 - 2x + 4) + 2( x2 - 2x + 4) = x3 - 2x2 + 4x + 2x2 - 4x + 8 = x3 + 8

•

Examples 10 and 1 1 lead to two more special products.

Difference of Two Cubes

(5)

S u m of Two Cubes

(6)

6

Divide Polynomials Using long Division

The procedure for dividing two polynomials is similar to the procedure for dividing two integers.

SECTION R.4 Polynomials EXAMPLE 1 2

4S

Dividing Two I ntegers

Divide 842 by 15. Solu tion

Divisor

->

56 15)842 75 92 90

f + 1)

=

3;(5 + 3x

Notice that we a lig n the 3x term under the x to make the next step easier.

STEP 3: Subtract and bring down the remaining terms.

3x

x- + 1 ) 3x� + 4x- + x + 7 ?

0

?

+ 3x 4x2 - 2x + 7

Subtract

-->

x2 - 2x - 3 x2 - X + 1 )x4 - 3x3 X4 - x3 + x2 -2x3 - x2 -2x3 + 2x2 -3x2 -3x2

+ 2x - 5 + + +

2x 2x 4x 3x x

.-

2

Now Work

=

(x - 2 ) (x + 2 ) (x2 + 4)

•

P R O B L E M S 1 5 A N D 33

Factor Perfect Squa res

When the first term and third term of a trinomial are both positive and are perfect squares, such as x2, 9x2, 1, and 4, check to see whether the trinomial is a perfect square. EXAMPLE 6

Factoring Perfect Sq uares

Factor completely: Solution

x2 + 6x + 9

The first term, x2, and the third term, 9 = 32, are perfect squares. Because the middle term 6x is twice the product of x and 3, we have a perfect square.

x2 + 6x + 9 EXAMPLE 7

(x + 3)2

9x2 - 6x + 1

The first term, 9x2 (3x f, and the third term, 1 = 12, are perfect squares. Because the middle term, -6x, is -2 times the product of 3x and 1, we have a perfect square. =

9x2 - 6x + 1 = (3x - 1 )2

EXAMPLE 8

•

Factoring Perfect Squ ares


•

Factoring Perfect Squ ares


=

25x2 + 30x + 9

The first term, 25x2 (5x f, and the third term, 9 = 32, are perfect squares. Because the middle term, 30x, is twice the product of 5x and 3, we have a perfect square. =

25x2 + 30x + 9 = (5x + 3 f

'q'=> -

Now Work

P R O B L E M S 2 5 A N D 93

•

52

CHAPTER R

Review

If a trinomial is not a perfect square, it may be possible to factor it using the technique discussed next. 3

Factor a Second-deg ree Polynom i a l : x2 + Bx + C

The idea behind factoring a second-degree polynomial like x2 + Bx + C is to see whether it can be made equal to the product of two, possibly equal, first-degree polynomials. For example, we know that

(x + 3 ) ( x

+

4)

=

x2 + 7x + 12

The factors of x2 + 7 x + 12 are x + 3 and x + 4. Notice the following:

x2 + 7x + 12

L

ab

In general, if x2 + Bx + C C and a + b = B.

=

(x + 3)(x + 4)

L- 12 is t h e product o f 3 and 4 7 is the s u m of 3 a n d 4

=

(x + a)(x + b )

x2 + (a + b)x + ab, then

=

=

To factor a second-degree polynomial x2 + Bx + C, find integers whose prod uct is C and whose sum is B. That is, if there are numbers a, b, where ab = C and a + b B, then =

x2 + Bx + C = (x + a ) (x + b)

EXAMPLE 9

Factoring Trinomials


x2 + 7x + 1 0

First, determine all integers whose product is 10 and then compute their sums. Integers whose product is 1 0

1, 10

-1, -10

2, 5

-

Sum

11

-11

7

-7

2, - 5

The integers 2 and 5 have a product of 10 and add up to 7, the coefficient of the middle term. As a result, we have

x2 + 7x + 1 0 = (x + 2 ) (x + 5 )

EXAMPLE 1 0



•

x2

-

6x + 8

First, determine all integers whose product is 8 and then compute each sum. Integers whose product is 8

1,8

- 1 , -8

2, 4

- 2, -4

Sum

9

-9

6

-6

Since -6 is the coefficient o f the middle term,

x2 - 6x + 8

=

(x

-

2) ( x

-

4)

•

SECTION R.S

Factoring Polynomials

S3


EXAMPLE 1 1

Factor completely:

x2 - x - 12

First, determine all integers whose product is - 1 2 and then compute each sum.

Solution

Integers whose product is

-

12

Sum

1. - 1 2

-1. 12

2. - 6

- 2. 6

3. - 4

- 3.

-11

11

-4

4

-1

1

4

Since - 1 is the coefficient of the middle term,

x2 - x - 1 2

=

(x + 3 ) (x - 4)

•


EXAMPLE 1 2

Factor completely:

x2 + 4x - 12

The integers -2 and 6 have a product of -12 and have the sum 4. Thus,

Solution

x2 + 4x - 12

=

(x - 2 ) (x + 6)

•

To avoid errors in factoring, always check your answer by mUltiplying it out to see if the result equals the original expression. When none of the possibilities works, the polynomial is prime. I dentifying Prime Polynomials

EXAMPLE 1 3

Show that x2 + 9 is prime. First, list the integers whose product is 9 and then compute their sums.

Solution

Integers whose product is 9

1,9

-1. -9

Sum

10

-10

3, 3

6

- 3. - 3 -6

Since the coefficient o f the middle term in x2 + 9 = x2 + Ox + 9 i s 0 and none of the sums equals 0, we conclude that x2 + 9 is prime. •

Example 13 demonstrates a more general result: THEOREM

Any polynomial of the form x2 + a2, a real, is prime. W!l: = = >-

4

Now Work

.J

PROBLEMS 39 AND 77

Factor by Grouping

Sometimes a common factor does not occur in every term of the polynomial, but in each of several groups of terms that together make up the polynomial. When this happens, the common factor can be factored out of each group by means of the Distributive Property. This technique is called factoring by grouping. EXAMPLE 1 4

Factoring by Grouping

Factor completely by grouping: Solution

(x2 + 2)x + (x2 + 2) 3 .

Notice the common factor x2 + 2 . By applying the Distributive Property, we have

(x2 + 2) x + (x2 + 2) . 3

=

(x2 + 2) (x + 3 )

Since x2 + 2 and x + 3 are prime, the factorization is complete.

•

54

CHAPTER R

Review (f. The next example shows a factoring problem that occurs in calculus. Factoring by Grouping

EXAMPLE 1 5

Factor completely by grouping:

3(x - 1 )2(x + 2)4 + 4(x - 1 )3(x + 2)3

Here, (x - 1 f(x + 2)3 is a common factor of 3(x - 1 )2(x + 2)4 and of 4(x - 1?(x + 2)3. As a result,

Solution

3(x - 1 f(x + 2)4 + 4(x - 1 )3(x + 2?

(x - 1 ?(x + 2)3 [ 3(x + 2) + 4(x - 1 ) J (x - 1 )2(X + 2?[3x + 6 + 4x - 4 J = (x - 1 f(x + 2?(7x + 2) •

=

=

Factoring by Grouping

EXAMPLE 16

Factor completely by grouping:

x3 - 4x2 + 2x - 8

To see if factoring by grouping will work, group the first two terms and the last two terms. Then look for a common factor in each group. In this example, we can factor x2 from x3 - 4x2 and 2 from 2x - 8. The remaining factor in each case is the same, x - 4. This means that factoring by grouping will work, as follows:

Solution

x3 - 4x2 + 2x - 8 = (x3 - 4x2) + (2x - 8 ) x2( X - 4) + 2( x - 4) = (x - 4 ) (x2 + 2) =

Since x2 + 2 and x - 4 are prime, the factorization i s complete. "I! �-

5

Now Work

PROBLEMS 5 1

•

AND 1 21

Factor a Second-deg ree Polynom i a l : Ax2 + Bx + C, A =1= 1

To factor a second-degree polynomial Ax2 + Bx + C, when A C have no common factors, follow these steps:

1=

1 and A, B, and

Steps for Factori ng Ax2 + Bx + C, when A -=1= 1 and A, B, and C Have No Com mon Factors STEP 1: Find the value of AC. STEP 2: Find integers whose product is AC that add up to

and b so that ab = AC and a + b = B. STEP 3: Write Ax2 + Bx + C = Ax2 + ax + bx + C . STEP 4 : Factor this last expression b y grouping. EXAMPLE 1 7

Factoring Trinom ials


B. That is, find a

2x2 + 5x + 3

Comparing 2x2 + 5x + 3 to Ax2 + Bx + C, we find that A = 2, B = 5, and C

STEP 1: The value of AC is 2 · 3 = 6. STEP 2: Determine the integers whose product is AC

=

=

3.

6 and compute their sums.

Integers whose product is 6

1,6

- 1 , -6

2, 3

- 2, - 3

Sum

7

-7

5

-5

SECTION R.S STEP 3: The integers whose product is 6 that add up to B

J

2X2 + 5x + 3

STEP 4: Factor by grouping.

=

J

=

SS

Factoring Polynomials

5 are 2 and 3.

l'

t

w

2X2 + 2x + 3x + 3

r

2x2 + 2x + 3x + 3 = (2x2 + 2x) + (3x + 3 ) = 2x(x + 1 ) + 3 ( x + 1 )

= ( x + 1 ) (2x + 3 )

As a result,

EXAMPLE 1 8

•


Factor completely: Sol u tion

2X2 + 5x + 3 = (x + 1 ) ( 2x + 3 )

2x2 - x - 6

Comparing 2x2 - x - 6 to Ax2 + Bx + C, we find that A = 2, B C = -6.

STEP 1: The value of AC is 2 . ( -6) = - 12. STEP 2: Determine the integers whose product is AC

=

=

- 1 , and

-12 and compute their sums.

Integers whose

product is - 1 2

1, -12

-1, 12

2, - 6

Sum

-1 1

11

-4

- 2, 6

3, - 4

- 3, 4

4

-1

1

STEP 3: The integers whose product is - 12 that add up to B

J

2X2 - X - 6

STEP 4: Factor by grouping.

J

=

=

- 1 are -4 and 3. w

l'

t

2X2 - 4x + 3x - 6

r

2x2 - 4x + 3x - 6 = (2X2 - 4x) + (3x - 6 ) = 2x( x - 2 ) + 3 ( x - 2) =

(x - 2 ) (2x + 3 )

As a result, 2X2 - X - 6 " "� -

Now Work

=

(x - 2) (2x + 3 )

•

PROBLEM 57

SUM MARY Type of Polynomial

Method

Example

Any polynomial

Look for common monomial factors. ( Always do this first!)

6x2 + 9 x

Binomials of degree 2 or higher

Check for a special product: Difference of two squares, x2 - a2 Difference of two cubes, x3 - a3 Sum of two cubes, x3 + a3

Trinomials of degree 2

Check for a perfect square, (x

Three or more terms

±

Factoring x2 + Bx + C (p. 52) Factoring Ax2 + Bx + C (p. 54) Grouping

af

=

3x(2x + 3 )

x2 - 1 6 = ( x - 4 ) ( x + 4 ) x 3 - 6 4 = ( x - 4 ) (x2 + 4 x + 16) x3 + 27 = (x + 3 ) (x2 - 3x + 9) x2 + 8x + 16 = ( x + 4)2 x2 - lOx + 25 = ( x - 5 ) 2 x 2 - X - 2 = ( x - 2) (x + 1 ) 6x2 + x - I = (2x + 1 ) (3x - 1 ) 2x3 - 3x2 + 4x - 6

=

(2x - 3 ) (x2 + 2)

S6

CHAPTER R

Review

R.S Assess You r Understa nding Concepts and Voca b u l a ry

)

\>�

If factored completely,

3x3 - 12x =

� True or False

__.

)1', True or False

If a polynomial cannot be written as the product of two other polynomials (excluding

1 and - 1 ) , then the polynomial is said

x2 + 4 is prime. 3x3 - 2x2 - 6x + 4 = (3x - 2)(x2 + 2).

The polynomial

to be

Skill B u i lding

In Problems 5-14, factor each polynomial by removing the common monomial factor. 8. 3x + 6 6. 7x - 14 ax2 + a

�

�

12.

ax - a

3x2 - 3x

In Problems 15-22, factor the difference of two squares. t'( x2 - 1 16. x2 - 4

K x2 - 16

20.

x2 - 25

In Problems 23-32, factor the perfect squares.

�x2 + 2x + 1 28.

24.

� x2 + 4x + 4

x2 - 4x + 4

� 4x2 + 4x + 1

x2 + lOx + 25

30.

18.

9x2 - 1

22.

36x2 - 9

�X2 - lax + 25

M. 16x2 + 8x + 1

9x2 + 6x + 1

'I'

32.

\

25x2 + lax + 1

In Problems 33-38, factor the sum or difference of two cubes.

. � x3 - 27

34.

� x3 + 27

x3 + 125

In Problems 39-50, factor each polynomial.

� x2 + 5x + 6 '2m. x2 + 7x + 10 V '-

40.

x2 + 6x + 8

44.

x2 + llx + 10

48.

x2 - 2x - 8

In Problems 51-56, factor by grouping.

� 2x2 + 4x + 3x + 6 54.

52.

58. 62. 66.

2x2 + 3x +

1

3x2 + lax + 8 3x2 - 14x + 8

38.

�. x2 + 7x + 6

,5\ x2 - lax + 16 )� x2 + 7x - 8

64 - 27x3

42.

x2 + 9x + 8

46.

x2 - 17 x + 16

50.

x2 + 2x - 8

� 2X2 - 4x + x - 2

3x2 - 3x + 2x - 2

In Problems 57-68, factor each polynomial.

��Ux2 + 14x + 8

27 - 8x3

)'t 6x2 + 9x + 4x + 6

3x2 + 6x - x - 2

�3X2 + 4x + 1 �. 3x2 + 2x - 8

36.

56.

)� 2z2 + 5z + 3 � 3x2 - 2x - 8 /?Z..,3x2 + lax - 8

9x2 - 6x + 3x - 2 60.

pz,2 + 5z + 1

64.

3x2 - lax + 8

68.

3x2 - lax - 8

I

In Problems 69-11 6, factor completely each polynomial. If the polynomial cannot be factored, say it is prime.

�ti.. x2 - 36 x2 + llx + 10 � 4x2 - 8x + 32 •. 15 + 2x - x2 �/ l + ll i + 30i .� 6x2 + 8x + 2 . � x6 - 2x3 + 1 16x2 + 24x + 9

� •

�

70.

x2 - 9

74.

x2 + 5x + 4

78.

3x2 - 12x + 15

82.

14 + 6x - x2

86.

3i - 18i - 48y

94.

8x2 + 6x - 2 x6 + 2x3 + 1

98.

9x2 - 24x + 16

90.

IJ(. 2 - 8x2 '7J x2 - lax + 21 ....

?{ x2 + 4x + 16 �x2 - 12x - 36 );t. 4x2 + 12x + 9 X X4 - 81

)� x7 - x5 5 + 16x - 16x2

�

72.

3 - 27x2

76.

x2 - 6x + 8

80.

x2 + 12x + 36

84.

x3 + 8x2 - 20x

88.

9x2 - 12x + 4

92.

X4 - 1

96.

x8 - xS

100.

5 + llx - 16x2

SECTION R.6 Synthetic Division

�. 41 - 16y 15 � x(x 3 ) - 6(x + 3 ) �'9. (3x - 2)3 - 27 +

+

112.

7(x2 - 6x

+

+

9)

+

102.

91

106.

5 (3x - 7) 110.

5(x -

+

x(3x - 7 )

(5x +

�. x3

3)

�

9y - 4

+

W.

1 - 8x2 - 9x4 (x

2f - 5(x + 2)

+

�. 3 (x2

1 )3 - 1

2x2 -

X

-2

114.

116.

x3

X4 +

+

X

104.

4 - 14x2 - 8x4

108.

( x - 1 )2 - 2 ( x - 1 )

l Ox

+

25)

x3 - 3x2 -

X

+

+

- 4(x

+

57

5)

3

+1

Applications a n d Extensions

(fi In Problems 1 1 7-126, expressions that occur in calculus are given. Faclor completely each expression. 117. 2(3x + 4)2 + (2x + 3 ) ' 2(3x + 4) ' 3 118 . 5 (2x + 1 )2 + (5x - 6 ) · 2 (2x 119. . 1 21.

2x(2x 2(x

+

+

5) +

(4x - 3 )2

125.

2 (3x - 5) ' 3 (2x x2

'

+

+

(x

2 ( 4x -

123.

127. Show that

x

1) ' 2

x2 . 2

3 ) (x - 2)3 +

+

+

1 )3

+ 3 )2 . 3 (x - 2)2 3) · 4

+

(3x

- 5)2 . 3 ( 2x

+

1 )2 · 2

4 i s prime.

+

122 .

4(x

124.

3x2(3x

126.

3(4x

5)3(x _

+

+

4)2

+

1 )2

+

x3 . 2(3x

5f · 4 ( 5x +

128. Show that

x2

+

x

+

1 )2

5 t · 2 (x -

+

(x

+

+

1)

4) · 3

(4x

+

5)3 · 2(5x +

1) ' 5

1 i s prime.

Discussion and Writing 129. M a k e u p a polynomial t h a t factors i n t o a perfect square.

1 30. Explain to a fel l o w student what you look for first when presented with a factoring problem. What do you do next?

OBJECTIVE

1

1

Divide Polynomials Using Synthetic Division (p. 57)

Divide Pol ynomials Using Synth etic Division

To find the quotient as well as the remainder when a polynomial of degree 1 or higher is divided by x - c, a shortened version of long division, called synthetic division, makes the task simpler. To see how synthetic division works, we will use long division to divide the polynomial 2x3 - x2 + 3 by x - 3. 2X2 + 5x + 15 + 3 x - 3 hx3 - x2 2 2x3 - 6x 5x2 5x2 - 15x 15x + 3 15x - 45 48

v"CHECK:

-

Now Work

PRO B L E M 7 1

The next two examples illustrate some algebra that you will need to know for certain calculus problems.

EXAMPLE 9

Writing an Expression as a Single Quotient Write the following expression as a single quotient in which only positive exponents appear.

(x-+ 1)1/2 +X' -1 (x-+ 1r l/2 . 2x 2 x2 1 2 + 1r1/7-'Zx=(x7 +1) 1/-+ 7 (x-+ 1) 1/7-+x'-(x Z (x2 + 1)1/-7 (x2 + 1) l/2 (x2 + 1) 1/2 +x2 (x2 + 1)l/2 (x2 + 1) +x2 (x2 + 1//2 2x2 + 1 (x2 + 1)1/2 7

Solution

7

&(IJ !_ we·

EXAMPLE 10

- Now Work

•

PRO B L E M 77

Factoring an Expression Containing Rational Exponents

4 x1/3 (2x + 1) +2x4/3 .) We begin by writing 2x4/3 as a fraction with 3 as denominator. /3 4 x1/.. ' (2x 6x4/3 = 4xl/\2x + 1) +6x4/3 /3 = 4xl (2x + 1) +-+ 1) + 2x4 3 3 3 3 Factor:

Solution

7

::;-

0

----'-----'-----

i

Add the two fractions

2 and

1l'1!I::=� - Now Work

)13

i

2x1/3[2(2x + 1) +3x] 3

are common factors

PRO B L E M 89

i

2x1/\7x +2) 3

Simplify

•

SECTION R.S

nth Roots; Rational Exponents

77

f-H�torica l Feature

T

he radical sign,

V, was first used in print by Christoff Rudolff in

1525. It is thought to be the manuscript form of the letter =

the Latin word radix

r

(for

root),although this is not quite conclusively

confirmed. It took a long time for

V to become the standard symbol

for a square root and much longer to standardize

-0', {/, �,

3

and

-0'8. The notation

V V16 was popular for

The bar on top of the present radical symbol, as follows,

Va2 + 2ab + b2

is the last survivor of the vinculum, a bar placed atop an expression to

indicate what we would now indicate with parentheses. For example,

V8 3

ab + c = a(b + c)

R.S Assess Your Understanding

Answers are given at the end of these exercises. If you get a wrong answel; read the pages in red. ; -32 ; v(=4)2 (pp. 2 1-24) Vi6 = (pp. 21-24)

'Are You Prepared?'

( )2 )( -3

=

__

=

\

__

=

__

__

Concepts and Vocabulary

�

In the symbol -Va, the integer n is called the

\We call .;ya the

__ __

.

__

of a.

4. True or False

�

6 . True o r False

�

=

-2 =

-3

Skill Building

In Problems 7-42, simplify each expression. Assume thai all variables are positive when they appear.

'K Vii :R: Vs

ts� � f

K�

NVs�)2

�3v2 + 4v2 P\ (\1'3 + 3)(\1'3 '[t (Vx ) - 1 2

1

)

8.

fu

jf, �

10.

\V=l

12.

\154

}� �

14.

�

16.

�

20.

y'9x5

24.

(\13 ViOt

32.

(Vs - 2) (Vs

36.

(Vx + Vs)2

)( v£3 - 3v5(h 'r\�-3X�+5�

18.

( V3.?VUx X. (3V6)(2v2) Jf: -V18 + 2Vs +

3)

22.

26.

2vTI - 3V27 34. V24 - V'sl 9

X·�-� 40. 3xv9Y + 4 v25.Y 8xy - V25x2/

(5Vs)( -3\1'3)

30.

�. 5\12 - 2\154

42.

\& y'2Ox3

+

38.

\Y32x +

�

V8x3i

In Problems 43-54, rationalize the denominator of each expression. Assume that all variables are positive when they appear. -\1'3 -\1'3 2 1 46.

� v2

44.

48.

\1'3

v2

r.:.

v7 + 2

,

ffl.

By the 1700s, the index had settled where we now put it.

and so on.

The indexes of the root were placed in every conceivable position, with

V8, V@8,

all being variants for

�

Vs

Vs

50.

\1'3t::. 2 v3 + 3 ,

- 1

CHAPTER R

78

Review

52.

"

-2

\Y9

In ProbLems 55-66, simpLify each expression. 2--6. 43/2 � 82/3 62.

Vx+h. - \IX Vx+h.+ \IX

�. � -27) 1/3

58.

+� 5 4. Vx+h � - ,vx ,vx+h � -h

163/4

60.

253/2

( )

66.

(-8 ) -2/3

27 2/3 . 8

64

16-3/2

27

In Problems 67-74, simplify each expression. Express your answer so that only positive exponents occur. Assume that the variables are positive. � x3/4XI/3x-I/2

Applications and Extensions

In Problems 75-88, expressions thaI occur in calculus are given. Write each expression as a single quotient in which only positive exponents and/or radicals appear. \. X 1 +x 7 + 2(1+x) /2 x> -1 7J£. 6 +x1- /2 x>o f' (1+x)l/2x 1/2 I

I

?

•

. (x+ 1)1/3+x. '!3'(x+ 1t2/3

78

�.�.

1 +�. 1 2� 5V4x+3

� - x ' -= -1= 2� "T' (, 1+x

¥-

(x+4)1/2 - 2x(x +4tl/2 x+4

x"* 2,x"* - 8'1

x>5 ,V� x"+ 1

x>-1

82.

2x - x ·----=== 2� r+1

-� � -

---

-3 < x

x>-4

x < -1

x"* -1

or

1

x"*-1,x"*1

x>O

In Problems 89-98, expressions that occur in calculus are given. Factor each expression. Express your answer so that only positive exponents occur. � (x + 1)3/2+x' (x+1) /2 X -1 90. (x2+4)4/3+x' (x2 + 4//3 '2x

�

1

2:

�

SECTION R.8

92.6x1/2 (2x + 3)

x?:O

+

nth Roots; Rational Exponents

79

x3/2. 8

94.2x(3x + 4)413 + x2. 4(3x + 4) 1/3 .. •

?A

4(3x + 5)1/3(2x + 3)3/2 + 3(3x + 5)4/3(2x + 3)1/2 "

3 2

r> -

-

96. 6(6x + 1)1/3(4x - 3)3/2 + 6(6x + 1)4/3 (4x - 3)1/2 " - 3 4

r> -

9S. 8xl/3 - 4x-2/3

x>O

x '1-0

In Problems 99-106, use a calculator to approximate each radical. Round your answer to two decimal places. 99. V2 100. V7 V4 102. v=s l.j

1. 2 + \13 3-

104.

Vs

Vs

lrtz � 3Vs - V2

2 V2 + 4 -

5.

-=---

--

\13

106.

2\13V2 - V4


'f/:( Calculating the Amount of Gasoline in a Tank

1

r

A Shell station stores its gasoline in underground tanks that are right circular cylinders lying on their sides. See the illustration. The volume V of gasoline in the tank (in gallons) is given by the formula V

=

40h2

)�: - 0.608

where h is the height of the gasoline (in inches) as measured on a depth stick. (a) If h = 12 inches, how many gallons of gasoline are in the tank? (b) If h = 1 inch, how many gallons of gasoline are in the tank? lOS. Inclined Planes

height

h feet is

The final velocity v =

V64h +

v

of an object in feet per second (ft/sec) after it slides down a frictionless inclined plane of

V

B

where Va is the initial velocity (in ft/sec) of the object. (a) What is the final velocity v of an object that slides down a frictionless inclined plane of height 4 feet? Assume that the initial velocity is O. (b) What is the final velocity v of an object that slides down a frictionless inclined plane of height 16 feet? Assume that the initial velocity is O. (c) What is the finaL-velocity v of an object that slides down a frictionless inclined plane of height 2 feet with an initial velocity of 4 ftlsec?

Problems

109-112

require the following information. The period T, in seconds, of a pendulum of length l, in feet, may be approximated using the formula

Period of a Pendulum

T

=

27r

-ff

In Problems 109-112, express your answer both as a square roo/ and as a decimal. T of a pendulum whose length is 64 feet. 110. Find the period T of a pendulum whose length is 16 feet. W.. Find the period T of a pendulum whose length is 8 inches. 112.Find the period T of a pendulum whose length is 4 inches.

�. Find the period


�

GiVe an example to show that

W is not equal to a. Use it to explain why W = lal ·

'Are you Prepared?' Answers 1. 9 ; - 9

�'

0 and y

>

1164

0, when they appem: �( 5Vs - 2 V32

4V12+ 5V27

In Problems 8 7-90, simplify each expression. x2 �t1 0, Y > 0 871 . - - x .." 25/

HI;

2:

88.

90.

YhvSx; ,v;:;-;;lOy

x

2:

0, Y > 0

� x > 0,Y > 0 , 4 ;::;-Q V 3x/

In Problems 91-96, rationalize the denominator of each expression. -2 92. �� V3 v· Vs -4 94. ��1+Vs � 1+ ,v3 1 - Vs

. -/ 2 J� 1 - v'2

96.

4V3+ 2 2V3+ 1

In Problems 9 7-102, write each expression as a single quotient in which only positive exponents and/or radicals appear. � (2+x2)112+x . (2+x2)-1I2·2x 98. (x2 + 4)2/ + x. (x2+4 r1/3·2x

�

3

(x2

+

�

�

4) 1 /2·2 x - x2. (x2+4r /2 . 2x

100.

1

x2+ 4 4+x2 - 2xVX 2VX (4+ ? ) ?

-102.

In Problems 103 and 104, factor each expression. t'I 6. 3(x2+4)4/3+ x·4(x2+4) 1/3·2x "

�. U.S. Population

According to the US. Census Bureau, the "- US. population in 2000 was 281,421,906. Write the US. population in scientific notation. 106. Find the hypotenuse of a right triangle whose legs are of lengths 5 and 8. 1 The lengths of the sides of a triangle are 12,1 6, and 20. Is this f a right triangle?

�

104.

r -

2x(3x+5)5/3+x2. 4(3x+5)2/3 The weekly production cost C of man

108. Manufacturing Cost

x calculators is given by the formula x2 C 3000+6x - 1000 What is the cost of producing 1000 calculators? What is the cost of producing 3000 calculators?

ufacturing

=

(a) (b)

CHAPTER R

84

Review

� Quarterly Corporate Earnings

In the first quarter of its fis cal year, a company posted earnings of $1.20 per share. Dur ing the second and third quarters, it posted losses of $0.75 per share and $0.30 per share, respectively. In the fourth quarter, it earned a modest $0.20 per share. What were the annual earnings per share of this company? 110. Design A window consists of a rectangle surmounted by a trian gle. Find the area of the window shown in the illustration. How much wood frame is needed to enclose the window?

On a recent flight to San Francisco, the pilot announced that we were 139 miles from the city, flying at an altitude of 35,000 feet. The pilot claimed that he could see the Golden Gate Bridge and beyond. Was he telling the truth? How far could he see?

112. How far can a pilot see'!

6 f! I}---I-t----I

�.

5f!

113. Use the material in this chapter to create a problem that uses

each of the following words: (a) Simplify (b) Factor

A statue with a circular base of radius 3 feet is enclosed by a circular pond as shown in the illustration. What is the surface area of the pond? How much fence is required to enclose the pond? Construction

(c) Reduce

CHAPTER TEST

� List the numbers in the set that are (a) Natural numbers,

(b) Integers, (c) Rational numbers, (d) Irrational numbers, (e) Real numbers.

{

0,1.2,

� Evaluate each expression if x (a) 3x-1i

� }

\12, 7, , 1T =

y

-3 and

y

=

4.

�

� A

erform the indicated operation. (a) (-2x3 + 4 + 10) (b) (2x - 3)(-5x + 2)

x2 - 6x

� Factor each polynomial. (a) x2 - 6x + 8 (b) 4x2 - 25 (c) 6x2 - 19x - 7

'\

Simplify each expression.

- (6x3 - 7x2 + 8x

V27

(b)

-0'=8

(c)

(x;2)�3 (x-y) 32

.. Rationalize the denominator of

(c) Vx2+7 (d) 5x3 - 3x2 + 2x - 6 \:i."The triangles below are similar. Find the missing length x and t the missing angles A, E, and C.

4 0°

divided by x - 2.

(a)

(b) 12x - 3 l

2

� Find the quotient and remainder if x3 - 3x2

C

-

1)

(d)

\13.

j\

b

=

x

>

-3.

The Zero-Product Property states that if 0 then either or , or both.

Fill in the blanks

a·

x

8x - 10 is

(16x4/3y-2/3)3/2

5 + 3 � Graph the numbers on the real number line:

�

+

__

__

Construction A rectangular swimming pool, 20 feet long and 10 feet wide, is enclosed by a wooden deck that is 3 feet wide. What is the area of the deck? How much fence is re quired to enclose the deck?

Equations and Inequalities

Interest Rates Fall on 15- and 30-Year Loans From the Associated Press May 12, 2006 Rates on 3D-year fixed-rate mortgages averaged 6.58% this week, down from 6.59% last week, mortgage company Freddie Mac said. Rates on 15-year, fixed-rate mortgages, a popular choice for refi nancing a home mortgage, fell to 6.17%, down from 6.22% last week. One-year adjustable-rate mortgages (ARM) fell to 5.62%, down from 5.67%. Rates on 5-year hybrid adjustable-rate mortgages averaged 6.22%, up slightly from 6.21% last week. The mortgage rates do not include add-on fees known as points. The 1-year ARM carried a nationwide average fee of 0.7 point, and the three other mortgage categories had an average fee of 0.5 point. Source: Adapted from Martin Cn.llsingel; "Rates on 30-year Mongages Edge Down Slightly." Associated Press Financial Wire, May 11, 2005. ©2006 The

Associated Press.

-See the Chapter Project-

A Look Ahead

Outline

Algebra. If your instructor decides to exclude complex numbers from the course,

1 .2

Chapter 1, Equations and Inequalities, reviews many topics covered in Intermediate

don't be alarmed. The book has been designed so that the topic of complex numbers can be included or excluded without any confusion later on.

1.1 linear Equations

Quadratic Equations

1.3 Complex Numbers; Quadratic Equations in the Complex Number System 1.4 Radical Equations; Equations Quadratic

1 .5

in Form; Factorable Equations Solving Inequalities

1.6 Equations and Inequalities Involving Absolute Value 1.7 Problem Solving: Interest, Mixture, Uniform Motion, and Constant Rate Job Applications Chapter Review Chapter Test Chapter Projects

85

86

CHAPTER 1


1.1 Linear Equations Before getting started, review the following:

PREPARING FOR THIS SECTION •

•

Properties of Real Numbers (Section R.l, pp. 9-14) Now Work

Domain of a Variable (Section R.2, p. 21)

the 'Are You Prepared?, problems on page 94. 1 Solve a Linear Equation (p.

OBJECTIVES

88)

2 Solve Equations That Lead to Linear Equations (p.90)

3 Solve Applied Problems Involving Linear Equations (p. 92)

An

equation in one variable is a statement in which two expressions, at least one sides of the equa

containing the variable, are equal. The expressions are called the

tion. Since an equation is a statement, it may be true or false, depending on the value of the variable. Unless otherwise restricted, the admissible values of the variable are those in the domain of the variable. These admissible values of the variable, if any, that result in a true statement are called solutions, or roots, of the equation. To solve an equation means to find all the solutions of the equation. For example, the following are all equations in one variable, x:

x +5 = 9

x2 + 5x = 2x - 2

x2-4 =0 x + 1

,V� x2 + 9=5

---

x + 5= 9, is true when x =4 and false for any other x + 5 = 9. We also say that 4 x + 5= 9, because, when we substitute4 for x, a true state

The first of these statements, choice of

x. That

is,4 is a solution of the equation

satisfies the equation ment results.

Sometimes an equation will have more than one solution. For example, the equation

x2-4 =0 x + 1

--

has

x = -2

and

x= 2

as solutions.

Usually, we will write the solution of an equation in set notation. This set is called the

solution set of the equation. For example, the solution set of the equation

x2

Ois

-

9=

{-3,3}.

Some equations have no real solution. For example,

x2 + 9=5

solution, because there is no real number whose square when added to

has no real

9equals5.

An equation that is satisfied for every value of the variable for which both sides are defined is called an

identity. For example, the equation

3x + 5= x + 3 + 2x + 2 is an identity, because this statement is true for any real number

x.

One method for solving an equation is to replace the original equation by a succession of equivalent equations until an equation with an obvious solution is obtained. For example, all the following equations are equivalent.

2x + 3 = 13 2x = 10 x=5

We conclude that the solution set of the original equation is

{5}.

How do we obtain equivalent equations? In general, there are five ways.

SECTION 1.1

Linear Equations

87

Procedures That Result in Equivalent Equations

1. Interchange the two sides of the equation: Replace

3

by

x

=

x

=

3

2. Simplify the sides of the equation by combining like terms, eliminating parentheses, and so on:

(x + 2) + 6 x + 8

Replace by

= =

2x + (x + 1) 3x + 1

3. Add or subtract the same expression on both sides of the equation:

5 (3x - 5) + 5

Replace

3x

by

-

=

=

4 4 +5

4. Multiply or divide both sides of the equation by the same nonzero expression:

6 x -I

3x x -I

Replace

x*- 1

3x 6 ·(x - 1) '(x -1) x -I x -I equation is 0 and the other side can be

by

S. If one side of the

--

=

--

factored, then

we may use the Zero-Product Property" and set each factor equal to WARNING Squaring both sides of an

Replace

equation does not necessarily lead to

by

_

an equivalent equation.

x(x - 3) x

=

0

or

x

-

3

=

0

=

0

0:

Whenever it is possible to solve an equation in your head, do so. For example,

8 is x

The solution of

2x

The solution of

3x - 15

� == ....

Now Work

=

=

=

4.

0 is x

=

5.

PRO B L EM 9

Often, though, some rearrangement is necessary.

EXAMPLE 1

Solving an Equation Solve the equation:

Solution

3x - 5

=

4

We replace the original equation by a succession of equivalent equations.

(3x

-

3x - 5 5) + 5 3x 3x 3 x

The last equation, lent, so

3

x

=

3,

=

4

=

4 + 5

=

9

=

9 3 3

Add 5 to both sides. Simplify. Divide both sides by 3. Simplify.

has the single solution

3. All these equations 3x - 5 4.

is the only solution of the original equation,

'" The Zero-Product Property says that if ab

=

0 then

a =

0 or b

=

=

0 or both equal O.

are equiva

88

CHAPTER 1


"'CheCk:

It is a good practice to check the solution by substituting

3 for x

in the

original equation.

3x -5 = 4 3(3) 51, 4 -

9-51, 4 4= 4

The solution checks. The solution set is OJD'

§ >-

Now Work

PRO B L E M

{3}.

•

23

Steps for Solving Equations STEP 1: List any restrictions on the domain of the variable. STEP 2: Simplify the equation by replacing the original equation by a succes sion of equivalent equations following the procedures listed earlier.

STEP 3: If the result of Step 2 is a product of factors equal to 0, use the Zero Product Property and set each factor equal to

STEP 4: Check your solution ( s) .

1

° (procedure5).

Solve a Linear Equation Linear equations are equations such as 3x + 12

=

°

-2x + 5= °

1

-x 2

A general definition is given next.

DEFINITION

A

linear equation in one variable is equivalent to an equation of the form

ax + b = ° where

a

and

b

are real numbers and

a

i= 0.

Sometimes, a linear equation is called a side is a polynomial in

x of degree l.

first-degree equation, because the left

It is relatively easy to solve a linear equation. The idea is to

ax + b

=

°

ax = -b x= The linear equation formula

EXAMPLE 2

b x = --. a

-b a

-

ax + b = 0, a

a of- 0

Subtract b from both sides. Divide both sides by a, a of- O.

i= 0, has the single solution given by the

Solving a Linear Equation Solve the equation:

1

-(x + 5) - 4 2

isolate the variable:

=

1

-(2x - 1) 3

SECTION 1.1

Solution

Linear Equations

89

To clear the equation of fractions, we multiply both sides by 6, the least common multiple of the denominators of the fractions

1 1 - (x + 5 ) - 4 = - (2 x - 1 ) 2 3 -

6

[� (x

] [� (2x - 1 ) ]

+ 5) - 4 =

3(x + 5 ) - 6·4

=

3x + 15 - 24 = 3x - 9 = 3x - 9 + 9 = 3x = 3x - 4x = -x = x= Check:

� and �.

6

M ultiply both sides by 6, the LCM o f 2 a nd 3.

2(2x - 1)

Use the Distributive Property on the left a nd the Associative Property on the right.

4x 4x 4x 4x + 4x + 7 -7

2 2 2+9 7 7 - 4x

Use the Distributive Property. Combine like terms. Add 9 to each side. Sim p lify. Subtract 4x from each side. Simplify M u ltiply both sides by -1.

1 1 1 - (x + 5 ) - 4 = - ( -7 + 5 ) - 4 = - ( -2) - 4 = -1 - 4 = -5 2 2 2 1 1 1 1 - (2x - 1 ) -[2( -7) - 1] = - ( - 14 - 1 ) - ( - 1 5 ) = -5 3 3 3 3 =

=

Since the two expressions are equal, the solution x = -7 checks and the solution set is { -7 } . � ==....

EXAM P L E 3

Now Work

•

PR O B L E M 3 3

Solving a Linear Equation Using a Calculator

.

2.78x +

Solve the equatIOn:

2 = 54.06 17.931

Round the answer to two decimal places. Solution

To avoid rounding errors, we solve for x before using the calculator.

2.78x +

2 = 54 . 06 17.931

2 17 . 931 2 54.06 17.93 1 x= 2.78

2.78x

=

54.06 -

Subtract

2

--

17.931

from each side.

--

------

Divide each side by 2.78.

Now use your calculator. The solution, rounded to two decimal places, is 19.41. Check: We store the unrounded solution

2 to evaluate 2.78x + 17.931 ' (2.78)(19.405921 34)

1J!l!l:-== =:�

Now Work

PR O B L E M

65

1 9.40592134 in memory and proceed

�

+ 17. 31 = 54.06

•

90

CHAPTER 1

Equations a n d Inequalities

2

Solve Equations That Lead to Linear Eq uations

The next three examples illustrate equations that lead to linear equations upon simplification. Solving Equations

EXAM P L E 4

( 2y + l ) (y - 1 ) = (y + 5) (2y - 5) (2y + l)(y - 1 ) = (y + 5 ) ( 2 y - 5 ) 21 - y - 1 = 21 + 5 y - 25 Multiply and combine like terms. Subtract 21 from each side. -y - 1 = 5y - 25 Add 1 to each side. -y= 5y - 24 Su btract 5y from each side. -6y= -24 Divide both sides by -6 y= 4 Check: (2y + l ) (y - 1 ) = [2 (4) + 1 ] (4 - 1) = (8 + 1) ( 3) = (9) (3) = 27 (y + 5) (2y - 5 ) = (4 + 5 )[2(4) - 5] = ( 9 ) ( 8 - 5 ) = (9)(3) = 27

Solve the equation: Solution

Since the two expressions are equal, the solution The solution set is {4}. EXAM P L E 5

y= 4 checks. •

Solving Equations

.

Solve the equatIOn:

3 1 + 7 -= -x - 2 x - 1 (x - l ) (x - 2 )

First, we note that the domain of the variable is {xix *" 1, x *" 2}. We clear the equation of fractions by multiplying both sides by the least common multiple of the denominators of the three fractions, (x - 1 ) (x - 2) .

Solution

3 1 7 -- = -- + ----x - 2 x - I (x - l) ( x- 2 ) ( x-

�= ( x - 1 ) ( x - 2) [ x � 1 + ( x - ;( x - 2) ]

Multiply both sides by (x - 1)(x - 2). Ca ncel on the left.

1 + �7 � ---3x - 3 = �(x - 2 ) .x---r

Use the Distributive Property on each side; cancel on the right.

1 )�

1

3x - 3 = ( x - 2 ) + 7 3x - 3 = x + 5 2x = 8

Combine l ike terms. Add 3 to each side. S ubtract x from each side.

x= 4

Divide by 2.

3 = 3 = -3 -x - 2 -4-2 2 1 + 7 1 7 1 +-= 7 2 +-=-= 7 9 -3 -= -- + = x - I (x - 1 ) (x - 2 ) 4 - 1 (4 - 1) (4 - 2) 3 3·2 6 6 6 2 Since the two expressions are equal, the solution x= 4 checks. The solution set is {4}. Check:

• �== -

Now Work

PROB L E M

59

The next example is of an equation that has no solution.

SECTION 1.1

EXA M P L E 6

91

A n Equation with No Solution . S oI ve the equatIOn:

Solution

Linear Equations

3x + 2 - 3 -x- 1 x -1 =

First, we note that the domain of the variable is {x I x "* 1}. Since the two quotients in the equation have the same denominator, x - 1, we can simplify by multiplying both sides by x - 1. The resulting equation is equivalent to the original equation, since we are multiplying by x - 1, which is not 0. (Remember, x -:/= 1).

3 3x -x - 1 + 2= - x-1 3 3X + 2 '(x - 1 (--'� ) ) x =

- 1

Multiply both sides by 1; cancel on the right.

.x----r

x

Use the Distributive Property on the left side; cancel on the left..

3x '� + 2'(x - 1 ) = 3 -.x----r 3x + 2x - 2 = 3 5x 2 = 3 5x = 5 x=1

Simplify.

-

Combine like terms. Add

2 to

each side.


The solution appears to be 1. But recall that able. The equation has no solution.

x

=

1 is not in the domain of the vari •

'."

EXA M P L E 7

";..,..- Now Work

PROB

LEM 49

Converting to Fahrenheit from Celsius In the United States we measure temperature in both degrees Fahrenheit (OF) and degrees Celsius (0C), which are related by the formula C

=

t (F - 32 ) . What are

the Fahrenheit temperatures corresponding to Celsius temperatures of 0°, and 30°C? Solution

We could solve four equations for F by replacing C each time by

10°, 20°,

0, 10, 20, and 30. Instead, it is much easier and faster first to solve the equation C = (F - 32) for F and then substitute in the values of C. 5 -(F - 32) 9

C

=

9C

=

9C

=

5(F - 32 ) 5F - 160

=

9C

=

9C

=

9 -C

5F - 160 5F F

Multiply both sides by 9. Use the Distributive Property. Intercha nge sides.

+ 160

5 + 32

Add 160 to each side. Divide both sides by 5.

t

92

CHAPTER 1

Equations and Ineq ualities

We can now do the required arithmetic. O°C: lOoC: 20°C: 30°C: 3

9 + 32 = 32°F -(0) 5 F = -59 (10) + 32 = 50°F

F

=

-9 (20) + 5 9 F = -(3 5 0) +

F

=

=

68°F

32 =

86°F

32

•

Solve Applied Problems I nvolving Linear Equations

Although each situation has its unique features, we can provide an outline of the steps to follow to solve applied problems. Steps for Solving Applied Problems STEP

1: Read the problem carefully, perhaps two or three times. Pay particu

STEP

2:

STEP

3:

STEP

4:

STEP

5:

lar attention to the question being asked in order to identify what you are looking for. If you can, determine realistic possibilities for the answer. Assign a letter (variable) to represent what you are looking for, and, if necessary, express any remaining unknown quantities in terms of this variable. Make a list of all the known facts, and translate them into mathe matical expressions. These may take the form of an equation (or, later, an inequality) involving the variable. If possible, draw an appropri ately labeled diagram to assist you. Sometimes a table or chart helps. Solve the equation for the variable, and then answer the question, usually using a complete sentence. Check the answer with the facts in the problem. If it agrees, congrat ulations! If it does not agree, try again.

Let's look at two examples. EXAM P LE 8

Investments

A total of $18,000 is invested, some in stocks and some in bonds. If the amount in vested in bonds is half that invested in stocks, how much is invested in each category? Solution

STEP

1: We are being asked to find the amount of two investments. These amounts

must total STEP

STEP

$18,000. (Do you see why?)

2: If we let x equal the amounts invested in stocks, then the rest of the money, 18,000 - x, is the amount invested in bonds. Do you see why? Look at Step 3. 3 : We set up a table: Amount in Stocks

Amount in Bonds

x

18,000

Reason

-x

Total invested is $18,000

We also know that Total amount invested in bonds

18,000

- x

is

one-half that in stocks

1(

-

2

x

)

SECTION STEP

X = -x21 18,000 = X + "21 x 3 18,000 = "2x (�) 18,000 = (�)(%X) 12,000 = X

Solution

93

Add

x

to both sides.

Simplify.

2

Multiply both sides by 3' Simplify.

So, $12,000 is invested in stocks and $18,000 - $12,000 = $6000 is invested in bonds. The total invested is $12,000 + $6000 $18,000, and the amount in bonds, $6000, is half that in stocks, $12,000.

=

� ==-

EXA M P L E 9

Linear Equations

18,000 -

4:

STEP 5:

1.1

•

Now Work

PROB

LEM 8 3

Determining an Hourly Wage

Shannon grossed $435 one week by working 52 hours. Her employer pays time-and a-half for all hours worked in excess of 40 hours. With this information, can you de termine Shannon's regular hourly wage? STEP 1: We are looking for an hourly wage. Our answer will be in dollars per hour. STEP 2: Let x represent the regular hourly wage; x is measured in dollars per hour. STEP 3: We set up a table: Hours Worked

STEP

4:

STEP 5:

Salary

Hourly Wage

Regular

40

x

40x

Overtime

12

l.5x

12(1.5x)

=

=

18x

The sum of regular salary plus overtime salary will equal $435. From the table, 40x + 18x 435. 40x + 18x 435 58x = 435 x 7.50 Shannon's regular hourly wage is $7.50 per hour. Forty hours yields a salary of 40(7.50 ) = $300, and 12 hours of overtime yields a salary of 12(1.5) (7.50) = $135, for a total of $435.

�== O>-

Now Work

P ROB

= =

LEM 87

•

Steps for Solving a Linear Equation To solve a linear equation, follow these steps: STEP 1: List any restrictions on the variable. STEP 2: If necessary, clear the equation of fractions by multiplying both sides by the least common multiple CLCM) of the denominators of all the fractions. STEP 3: Remove all parentheses and simplify. STEP 4: Collect all terms containing the variable on one side and all remaining terms on the other side. STEP 5: Simplify and solve. STEP 6: Check your solution(s) . SUMMARY

94

C H A PTER 1

�

Equations and Inequal ities

OIVing equations is among the oldest of mathematical activities,

The solution of this problem u s i ng only words is the earliest form of

and efforts to systematize this activity determined much of the

algebra. Such problems were solved exactly this way i n Babylonia i n BC

W e know almost nothing o f mathematical work before this

shape of modern mathematics.

1800

Consider the followi n g problem and its solution u s i ng only words:

date, although most authorities bel ieve the sophistication of the earli

Solve the problem of how many apples Jim has, given that

est known texts i n d i cates that a long period of previo u s d evelopment must have occurred. The method of writing out equations in words

"Bob's five apples and Jim's apples together make twelve apples" by

persisted for thousands of years, and although it now seems extremely

thinking,

cumbersome, it was used very effectively by many generations of math

"Jim's apples are all twelve apples less Bob's five apples" and then

ematicians. The Arabs d eveloped a good deal of the theory of cubic equation s while writing out all the equation s in words. About

concluding,

AD

1500,

the tendency to abbreviate words in the written equations began to

"Jim has seven apples."

lead in the direction of modern notation; for example, the Latin word et

The mental steps translated into algebra are

5

+x

=

(mean i n g and) developed i nto the plus sign, +. Although the occasional use of letters to represent variables dates back to

12

x = 12 = 7

-

d i d not become common u ntil about

5

AD

AD

1200, the practice

1600. Development thereafter

was rapid, and by 1635 algebraic notation d id not differ essentially from what we use now.

1.1 Assess Your Understanding 'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.

� The

.�The

fact that 2(x + 3 ) = 2x + 6 is because of the Property. (p. 10) � ��... The fact that 3x = 0 implies that x = 0 is a result of the V Property. (p. 13) ____

x domain of the variable in the expression _ _ is x - 4

. (p.21)

____


4.

\/..

True or False Multiplying both sides of an equation by any number results in an equivalent equation.

True or False

. 3

IS

An equation that is satisfied for every value of the variable for "\ which both sides are defined is called a(n) . 6. An equation of the form ax + b equation or a(n) _____

=

8.

0 is called a(n) equation.

The solution of the equation 3x - 8

.

S

True or False

Some equations have no solution.

Skill Building In

}( �. � :

....�7 ... x

Problems 9-16, mentally solve each equation. = 21 10. 6x = - 24

�2X -

In

3

Problems

\t 3x + 4 20. 5Y + 6

=

14. 3x + 4

0

17-64,

=

x

=

-18

solve each equation.

- Y

" ¥- 3 + 2n = 4n + 7 26. 3(2 - x)

=

2x

-

=

3x + 15

0

x

18. 2x + 9 121. 6 - x \1"-

=

24. 6 - 2m

1

X

=

1 30. "3x

=

2 2 - "3x

2 16. - x 3

2

2t - 6

=

22. 3 - 2x

=

'\

3m + 1

8x - (3x + 2)

12. 6x + 18

0

\1'9.

5x

2x + 9 =

=

=

=

3x - 10

9 2

=-

3 - t

� 2(3 + 2x) 28. 7 - (2x

=

2 - x

-

=

1)

3(x - 4) =

10

0

=

0

SECTION 1.1

32.

1

-2:1 x= 6 36.

1 1=-4 2 --p 3 3 �. x ; 1+x ; 2= 2

0 .9t 1+t =

2x- -+ + 1 16=3x 3

)i �+= � 3

(x + 2)(x- 3) (x+3)2

-x3 13 61 3) (2x + l)(x-4) �\ x(2x- = 42•

44.

95

34.-

lit' 0.9t =0.4+O.lt ��

38.

Linear Equations

=

� Z ( Z2+1)=3+

Z3

�•

2x -6 x+3 x+3

50. -=---2

-4 -3 -x+4-=x+6 4 7 -3 �. -x-2=-x+5+---(x+5)(x-2) -5 +-4 =--3 62. 5z-11 2z - 3 5-z

=

-

2x 4 3 x+2 x2-4--x2-4--3x=2 54. -x-I S � 6t+7 =2t3t+ -4 �\ 4t-1 1 1-= 60. � +2x+3 x-I (2x+3)(x-1) ,

.

56.

- -

40.

--

�,,-' x2-1- x x

x+3 -_ -3 x2 + 2

- X

4 -5 5 2y y

-

=-

� (x+7)(x ,M.

-

1)

=

(x+1)2

=(2x-l)(x-2) x(l+2x) x +3= -2 . �. -x-2 x-2 x 4 3 52. --+-x2-9 x+3=r-9 46.

-? -

5

�. 2x _ 3= x ! 5

Sw+5 4w- 3 -7 5w+7

58. --lOw

64.

X

x+1 x+4 +2x x2+

-3 x2+3x+2

-- - --2 X X

In Problems 65-68, use a calculator to solve each equation. Round the solution to two decimal places.

JJ. 'V\.

66. 6.2x - -S19.1 3-= .72 0.195 21.2=-x 14 68. lS.6 3x-2.6 2. 32 -20

21.3 =19.2 3 3.2x+65.S71

IS

2.11 +2.4 �. 14.72-21.5Sx= -x Applications and Extensions

In Problems 69-74, solve each equation. The letters a, b, and c are constants.

� x = a - b

72.

�

0

a '*

c,

a -+ -xb c, x =

70.

1 - ax =b,

a '*

Find �he number equatI On

a '*

0 , b '* 0

,

a '* -b

74. -=-- ,

0

c '*

14 x-+= ' .'\ a xb- c'

0

a

for which

x= 4

is a solution of the

76. Find

b + c b- c x+a x-a c '* 0, a '* 0 the number b for which = x 2 is a solution of the

equation

x+2b= x-4+2bx

x+2a =16+ax-6a

Problems 77-82 list some formulas that occur in applications. Solve each formula for the indicated variable.

\--l], .

EI ect' 'tY rICI

1= 1 +1

-

R

RJ

R2

= A P(l+rt) for r mv2 lro � Mechanics = F for R R 78.

80.

Finance

Chemistry

PV

!:t{ Mathematics

=

nRT a __

for T

= 1-r for r 82. Mechanics = -gt+ for t � Finance A total of $20,000 is to be invested, some in bonds K

S

v

$ 3000, how

invested in bonds is to exceed that in CDs by much will be invested in each type of investment?

for R

Vo

and some in certificates of deposit ( CDs ) . If the amount

84.

$10,000

Finance A total of is to be divided between Sean and George, with George to receive less than Sean. How much will each receive?

��. Internet Searches /

$ 3000

2005,

In November Google and Yahoo! search engines were used to conduct a total of billion online searches. Google was used to conduct billion more searches than Yahoo!. How many searches were conducted on each search engine?

0.5 3

Source: comScore Networks

3.57

96

CHAPTER 1

Equations and I nequalities

86. Sharing the Cost of a Pizza

certain four-door model has a discounted price of $8000, what was its list price? How much can be saved by purchasing last year's model? 93. Business: Marking up the Price of Books A college book store marks up the price that it pays the publisher for a book by 35 %. If the selling price of a book is $92.00, how much did the bookstore pay for this book?

Judy and Tom agree to share the cost of an $18 pizza based on how much each ate. If Tom 2 ate:3 the amount that Judy ate, how much should each pay? [Hint: Some pizza may be left.] Tom's portion

Judy's portion

87.

88.

89.

90.

91.

92.

Computing Hourly Wages Sandra, who is paid time-and-a half for hours worked in excess of 40 hours, had gross weekly wages of $442 for 48 hours worked. What is her regular hourly rate? Computing Hourly Wages Leigh is paid time-and-a-half for hours worked in excess of 40 hours and double-time for hours worked on Sunday. If Leigh had gross weekly wages of $342 for working 50 hours, 4 of which were on Sunday, what is her regular hourly rate? Computing Grades Going into the final exam, which will count as two tests, Brooke has test scores of 80, 83, 71, 61, and 95. What score does Brooke need on the final in order to have an average score of 80? Computing Grades Going into the final exam, which will count as two-thirds of the final grade, Mike has test scores of 86, 80, 84, and 90. What score does Mike need on the final in order to earn a B, which requires an average score of 80? What does he need to earn an A, which requires an average of 90? Business: Discount Pricing A builder of tract homes re duced the price of a model by 1 5 % . If the new price is $425,000, what was its original price? How much can be saved by purchasing the model? Business: D iscOlmt Pricing A car dealer, at a year-end clear ance, reduces the list price of last year's models by 15% . If a

94. Personal Finance: Cost of a Car

95.

96.

97. 98. 99.

Source: comScore Networks


100. One step in the following list contains an error. Identify it

and explain what is wrong. x= 2 3x - 2x= 2 3x = 2x+2 x2 + 3x= x2 + 2x + 2 x2 + 3x - 1 0 x2 + 2x - 8 (x - 2 ) (x + 5 ) = (x - 2) ( x+4 ) x + 5= x + 4 1= 0 =

'Are You Prepared?' Answers

1. Distributive

2. Zero-Product

101. The equation

5 8+x + 3= x+3 x+3

(1) (2) (3) (4)

--

-

has no solution, yet when we go through the process of solv ing it we obtain x -3. Write a brief paragraph to explain what causes this to happen. =

(5)

(6)

102. Make up an equation that has no solution and give it to a fel

(7)

low student to solve. Ask the fellow student to write a cri tique of your equation.

(8)

3. {xix

The suggested list price of a new car is $ 18,000. The dealer's cost is 85% of list. How much will you pay if the dealer is willing to accept $100 over cost for the car? Business: Theater Attendance The manager of the Coral Theater wants to know whether the majority of its patrons are adults or children. One day in July, 5200 tickets were sold and the receipts totaled $29,961 . The adult admission is $7.50, and the children's admission is $4.50. How many adult patrons were there? Business: Discount Pricing A wool suit, discounted by 30% for a clearance sale, has a price tag of $399. What was the suit's original price? Geometry The perimeter of a rectangle is 60 feet. Find its length and width if the length is 8 feet longer than the width. Geometry The perimeter of a rectangle is 42 meters. Find its length and width if the length is twice the width. Internet Users In March 2006, 152 million people in the United States were Internet users, which accounted for 21.9% of the world's online audience. How many people worldwide were Internet users in March 2006?

-t=

4}

SECTION 1.2

Quadratic Equations

97

1.2 Quadratic Equations

Before getting started, review the following:


•

•

Factoring Polynomials (Section R.5, pp. 49-55) Zero-Product Property (Section R.I, p. 13) Now Work

Square Roots (Section R.2, pp. 23-24)

the 'Are You Prepared?' problems on page 106.

OBJECTIVES

1

Solve a Quadratic Equation by Factoring (p. 97)

2

Know How to Complete the Square (p. 99)

3

Solve a Quadratic Equation by Completing the Square (p. 1 00)

4

Solve a Quadratic Equation Using the Quadratic Formu la (p. 1 02)

5

Solve Applied Problems Involving Quadratic Equations (p. 1 05)

Quadratic equations are equations such as 2x2 + x + 8 = 0 3x2 - 5x + 6 = 0 x2 - 9 = 0 A DEFINITION

general definition is given next. A quadratic equation

is an equation equivalent to one of the form

+ bx + c = 0 where a, b, and c are real numbers and a "* O. ax2

A

quadratic equation written in the form ax2

standard form.

(1)

+

bx

+

c=0

is said to be in

Sometimes, a quadratic equation is called a second-degree equation, because the left side is a polynomial of degree 2. We shall discuss three ways of solving qua dratic equations: by factoring, by completing the square, and by using the quadratic formula.

1

EXAM P L E 1

Solution

S olve a Quadratic Equation by Factoring

When a quadratic equation is written in standard form ax2 + bx + c = 0, it may be possible to factor the expression on the left side into the product of two first-degree polynomials. Then, by using the Zero-Product Property and setting each factor equal to 0, we can solve the resulting linear equations and obtain the solutions of the quadratic equation. Let's look at an example. Solving a Quadratic Equation by Factoring

Solve the equation: (b) 2x2 = x + 3 (a) x2 + 6x = 0 (a) The equation is in the standard form specified in equation ( 1 ) . The left side may be factored as x2 + 6x = 0 x(x + 6) = 0

Factor.

98

CHAPTER 1


Using the Zero-Product Property, we set each factor equal to 0 and then solve the resulting first-degree equations. x= 0 or x + 6= 0 Zero-Product Property X= 0 or x= -6 Solve. The solution set is {O, -6}. (b) We put the equation 2x2 = x + 3 in standard form by adding -x - 3 to both sides. 2X2= X + 3 Add 3 to both sides. 2X2 - X - 3= 0 The left side may now be factored as (2x - 3)(x + 1 ) = 0 Factor. so that 2x - 3= 0 or x + 1= 0 Zero-Product Property x = -1 Solve. x= -32 -

.

x

-

. { 3}

The solutIOn set IS -1'"2 .

•

When the left side factors into two linear equations with the same solution, the quadratic equation is said to have a repeated solution. We also call this solu tion a root of multiplicity 2, or a double root. EXAM P LE 2 Solution

Solving a Quadratic Equation by Factoring

Solve the equation: 9x2 - 6x + 1 = 0 This equation is already in standard form, and the left side can be factored. 9x2 - 6x + 1 = 0 (3x - 1 ) (3x - 1 ) = 0 so 1 x= -3 or x= -31 This equation has only the repeated solution

�. The solution set is {�}.

• I! =:;> -

Now Work P R O B L E M S 1 1

AND

21

The Square Root Method

Suppose that we wish to solve the quadratic equation x2= p (2) where p 2: 0 is a nonnegative number. We proceed as in the earlier examples. Put in sta nda rd form. x2 - p= 0 (x - vp ) (x + vp ) = 0 Factor (over the rea l numbers) . x= vp or x = -vp Solve. We have the following result: If x2= p and p

2:

0, then x= vp or x

= - vp.

(3 )

SECTION 1.2

Quadratic Equations

99

When statement (3) is used, it is called the Square Root Method. In statement (3), note that if p > 0 the equation x2 = p has two solutions, x = vp and x = -vp. We usually abbreviate these solutions as x = ± vp, read as "x equals plus or minus the square root ofp." For example, the two solutions of the equation x2 = 4

are

x = ± V4

and, since V4 = 2, we have

x = ±2

The solution set is { -2, 2}. EXA M P L E 3

Use the Square Root Method.

Solving a Quadratic Equation Using the Square Root M ethod

Solve each equation. (a) x2 = 5 (b) (x - 2) 2 = 1 6 (a) We use the Square Root Method to get

Solution

x2 = 5 x = ± Vs x = Vs or x =

- Vs


The solution set is { - v'5 , v'5}. (b) We use the Square Root Method to get (x - 2) 2 x-2 X - 2 x-2 x

The solution set

= 16 = ± Vi6 = ±4 = 4 or x - 2 = -4 x = -2 = 6 or is { -2, 6}.


•

31 2

Know How to Complete the Square

We now introduce the method of completing the square. The idea behind this method is to adjust the left side of a quadratic equation, ax2 + bx + c = 0, so that it becomes a perfect square, that is, the square of a first-degree polynomial. For example, x2 + 6x + 9 and x2 - 4x + 4 are perfect squares because x2 + 6x + 9 = (x + 3 ) 2 and x2 - 4x + 4 = (x - 2 ) 2 How do we adjust the left side? We do it by adding the appropriate number to the left side to create a perfect square. For example, to make x2 + 6x a perfect square, we add 9. Let's look at several examples of completing the square when the coefficient of x2 is 1 : Start

x2 + 8x x2 + 2x x2 - 6x x2 + x

Add

16 1 9 1 4

Result

x2 + 8x + 1 6 = (x + 4) 2 x2 + 2x + 1 = (x + 1 ) 2 x2 - 6x + 9 = (x - 3 ) 2 1 1 x2 + x + 4 = x + 2

(

Y

100

CHAPTER 1


Do you see the pattern? Provided that the coefficient of square by adding the square of of the coefficient of

�

x.

x2 is 1, we complete the

Procedure for Completing a Square

m)2 x2 mx (m)2 2 = (x 2

Result

Add

Start

+

+

+

Completing the Square

EXAM P L E 4

Determine the number that must be added to each expression to complete the square. Then factor the expression. Add

Start

? x2

+

G·8r G Y G·(-20)Y C2·(-5)Y

By

=

+12x

·12

02 - 200 p2 - 5p

16

=

36

Factored Form

y2+8y+16

(y+ 4)2

x2

=

=

Result

+

12x + 36

02 - 200+100

100

25 p2 -5p+4

25

4"

(x+ 6)2 (0 - 10)2

( %y p-

•

Notice that the factored form of a perfect square is either 4

Y

y 4

Area

Area

x2 + bx (%y (x + %y =

+

Figure 1

=

=

y2

'!I't

Area =

4y

3

Solution

x2

-

bx +

(%y = ( x %y -

Now Work P R O B l E M 3 5

Are you wondering why we call making an expression a perfect square "com pleting the square"? Look at the square in Figure 1. Its area is (y + 4 f The yellow area is l and each orange area is 4y (for a total area of 8y) . The sum of these areas is l + 8y. To complete the square we need to add the area of the green region: 4·4 = 16.As a result, l + 8y + 16 = (y + 4l

4y

EXAM P L E 5

, >-

or

Solve a Quadratic Eq uation by Completing the Square

Solving a Quadratic Equation by Completing the Square

5x

Solve by completing the square: x2 + + 4 = 0 We always begin this procedure by rearranging the equation so that the constant is on the right side.

x2 5x x2 + 5x +

x2

+ 4 =

0

= -4

Since the coefficient of is 1, we can complete the square on the left side by adding 1 = 4. Of course, in an equation, whatever we add to the left side also must be added to the right side. So we add to both sides.

( 5)2 25 "2.

2:

SECTION 1.2

25 -4 x2 + 5x + -= 4

25 4

Add

+ -

9

25 -

4

Quadratic Equations

10 1

to both sides.

Factor.

4

f2. x + �2 =± \/"4 5 3 x + -= 2 ±-2 5 3 x= - -±2 2

Use the Squa re Root Method.

5 3 -4 x = - - + -3 = -l or x = - -5 - -= 2 2 2 2 The solution set is { -4, I } -

•

.

I' !'l!'i:==""- THE SOLUTION OF THE EQUATION

CAN BE OBTAINED BY

IN EXAMPLE

FACTORING. REWORK

5

ALSO

EXAMPLE

5

USING THIS TECHNIQUE

The next example illustrates an equation that cannot be solved by factoring. EXAM P L E 6 Solution

Solving a Quadratic Equation by Completing the Square

Solve by completing the square: 2x2 - 8x - 5 = 0 First, we rewrite the equation. 2X2 - 8x - 5 = 0 2x2 - 8x = 5 Next, we divide both sides by 2 so that the coefficient of x2 is to complete the square at the next step.) x2 - 4x = -52 Finally, we complete the square by adding 4 to both sides. ?

1.

(This enables us

5 4 = -2 + 4 13 (x - 2? = 2

x- - 4x

+

x - 2= ±

.J¥


v26 x - 2 = ±-2

x= 2± v26 2

(13= 'V 2

v13 Vz

v13 Vz Vz Vz .

=

=

Vz6 2

--

NOTE If w e wanted a n approximation,

say rounded to two decimal places, of these solutions, we would use a calcula • tor to get { -0.55,4.55}.

{

v26 The solution set is 2 - 2- , 2 I.l'l. = _-

Now Work

PROBL EM

+

v26 2-

41

} •

102

CHAPTER 1


4

Solve a Quadratic Eq uation Using the Quadratic Form u l a

We can use the method of completing the square to obtain a general formula for solving any quadratic equation ax2 + bx + c NOTE There i s no loss i n generality to assume that a > 0, since if a < ° we can multiply by -1 to obtain an equiva lent equation with a positive leading coefficient. •

=0

a*" O

As in Examples 5 and 6, we rearrange the terms as ax2 + bx = -c

Since a

>

0, we can

a > 0

divide both sides by a to get b x2 + - x a

=

c a

--

Now the coefficient of x2 is 1. To complete the square on the left side, add the square of of the coefficient of x; that is, add

�

to both sides. Then if

b2 - 4ac 4a2

Provided that b2 - 4ac

)

2::

0, we now

if

c

4ac

( 4)

can use the Square Root Method to get

b2 - 4ac b x + - =± 2a 4a2 b ± Yb2 - 4ac x + - = ----2a 2a yb2 - 4ac b x = --± 2a 2a -b ± yb2 - 4ac 2a

-----

The square root of a quotient equals the quotient of th e square roots. Also,

�

=

b

2a since a > O.

Add - - to both sides. 2a Combine the quotients on the right.

What if b2 - 4ac is negative? Then equation (4) states that the left expression (a real number squared) equals the right expression (a negative number). Since this occurrence is impossible for real numbers, we conclude that if b2 - 4ac < 0 the quadratic equation has no real solution. (We discuss quadratic equations for which the quantity b2 - 4ac < 0 in detail in the next section.) We now state the quadratic formula. THEOREM

Consider the quadratic equation ax2 + bx + c = 0

a*" O

If b2 - 4ac < 0, this equation has no real solution. If b2 - 4ac 2:: 0, the real solution(s) of this equation is (are) given by the

quadratic formula. Quadratic Formula

x=

-b±

yb2 - 4ac 2a

( 5)

I

�--------------------------------��

SECTION 1.2

Quadratic Equations

103

The quantity b2 - 4ac is called the discriminant of the quadratic equation, because its value tells us whether the equation has real solutions. In fact, it also tells us how many solutions to expect. Discriminant of a Quadratic Equation

For a quadratic equation ax2 + bx + c

= 0:

b2 - 4ac > 0, there are two unequal real solutions. 2. If b2 - 4ac = 0, there is a repeated solution, a root of multiplicity 2. 3. If b2 - 4ac < 0, there is no real solution. 1. If

When asked to find the real solutions, if any, of a quadratic equation, always evaluate the discriminant first to see how many real solutions there are. EXA M P L E 7

Solving a Quadratic Equation Using the Q u adratic Formula Use the quadratic formula to find the real solutions, if any, of the equation

+ 1 =

3x2 - 5x

Solution

0

The equation is in standard form, so we compare it to ax2 + bx + c = b, and c.

+ 1 =0 + bx + c = 0

0

to find a,

3x2 - 5x ax

With a =

3,

b=

-5,

2

and c =

1,

> 0,

3, b

4(3)(1)

=

-( -5) ±

{

.

The solutIOn set is

=

-5, C

1

=

25 - 12

4a

c.

= 13

there are two real solutions, which can be found using the qua

x=

E XA M P L E 8

=

we evaluate the discriminant b2 -

b2 - 4ac = ( -5? Since b2 - 4ac dratic formula.

a

5

-

Vi3

6

5

+

'

Vi3}

6

2(3)

Vi3

5

± Vi3 6

.

•

Solving a Quadratic Equation Using the Quadratic Formula Use the quadratic formula to find the real solutions, if any, of the equation 25 -x2 - 30x 2

Solution

+ 18 =

0

The equation is given in standard form. However, to simplify the arithmetic, we clear the fractions. 25

x2 -

2 -

30x + 18 = 0

+ 36 = 0 + bx + c = 0

2 25x - 60x ax2

With a =

25,

b=

-60,

and c =

36,

b2 - 4ac = (-60)2

Clear fractions; multiply by

2.

Compare to standard form.

we evaluate the discriminant.

- 4(25)(36)

= 3600

- 3600

=

0

104

C HAPTER 1


The equation has a repeated solution, which we find by using the quadratic formula.

x= The solution set is

EXAM P LE 9

-b ±

yb2 - 4ac

60± Yo 50

2a

{�}.

60 50

6 5 •

Solving a Quadratic Equation Using the Quadratic Form u la Use the quadratic formula to find the real solutions, if any, of the equation

3x2 + 2 = 4x Solution

The equation, as given, is not in standard form.

3x2 + 2 = 4x 2 3x - 4x + 2 = 0 ax2 + bx + c = 0

Put in standard form. Compa re to standard form.

With a = 3, b = -4, and c = 2, we find

b2 - 4ac = ( -4f - 4(3) (2) = 16 - 24 = -8 Since b2 - 4ac < 0, the equation has no real solution. 1JI\OiZ . _ >-

Now Work

PROBLEMS

51

AND

•

61

Sometimes a given equation can be transformed into a quadratic equation so that it can be solved using the quadratic formula. EXAM P L E 1 0

Solving a Quadratic Equation Using the Quadratic Formula Find the real solutions, if any, of the equation:

Solution

2 3 9 + -- - = Ox7"oO ' x x2

In its present form, the equation

3 2 9 + - -- = 0 x x2 is not a quadratic equation. However, it can be transformed into one by multiply ing each side by x2 . The result is

9x2 + 3x - 2 = 0 Although we multiplied each side by x2 , we know that x2

* 0 ( do you see why? ) , so this quadratic equation is equivalent to the original equation. Using a = 9, b = 3, and c = -2, the discriminant is

b2 - 4ac = 32 - 4(9 ) ( -2) = 9 + 72 = 81 Since b2 - 4ac

0, the new equation has two real solutions.

>

x= x=

-b±

Yb2 - 4ac 2a

-3 + 9 18

The solution set is

6 18

{ -�, �}.

-3 ± \I8i -3 ± 9 2(9) 18 1 -3 - 9 -12 or x = 3 18 18

2 3

•

SECTION 1.2

SUMMARY

Quadratic Equations

105

Procedure for Solving a Quadratic Equation

To solve a quadratic equation, first put it in standard form:

ax2

+

bx +

c =

0

Then:

STEP 1: Identify a, b, and c. STEP 2: Evaluate the discriminant, b2 - 4ac. STEP 3: (a) If the discriminant is negative, the equation has no real solution. (b) If the discriminant is zero, the equation has one real solution, a repeated root. (c) If the discriminant is positive, the equation has two distinct real solutions. If you can easily spot factors, use the factoring method to solve the equation. Otherwise, use the quadratic formula or the method of completing the square.

5

So lve Applied Problems I nvolving Q u a d ratic Eq uations

Many applied problems require the solution of a quadratic equation. Let's look at one that you will probably see again in a slightly different form if you study calculus.

Constructing a Box

EXAM P L E 1 1

From each corner of a square piece of sheet metal, remove a square of side 9 cen timeters. Turn up the edges to form an open box. If the box is to hold 144 cubic centimeters (cm 3), what should be the dimensions of the piece of sheet metal? Solution

We use Figure 2 as a guide. We have labeled by x the length of a side of the square piece of sheet metal. The box will be of height 9 centimeters, and its square base will measure x - 18 on each side. The volume V ( Length X Width X Height) of the box is therefore v =

Figure

2

(x - 18)(x - 18) · 9

=

9(x - 18?

I��---- xem ---+-. I i I

1

I I

1 I I

gem:

gem

_______

1

I I I

gem

9 em

x- 18 Volume 9(x - 18)(x- 18) =

Since the volume of the box is to be 144 cm3 , we have

9(x - 18) 2 (x - 18) 2 x - 18 x

= = =

=

144 16 ±4 18 ± 4

x = 22

or

V = 144

Divide each side by 9. Use the Square Root Method. x =

14

We discard the solution x = 14 (do you see why?) and conclude that the sheet metal should be 22 centimeters by 22 centimeters.

106

CHAPTER 1


Check: If we begin with a piece of sheet metal

22 centimeters by 22 centime ters, cut out a 9 centimeter square from each corner, and fold up the edges, we get a box whose dimensions are 9 by 4 by 4, with volume 9 X 4 X 4 = 144 cm3, as required.

•

C/q'=>-

Now Work

PRO

BLEM 1 05

{-ti�torical Feature

P

roblems using quadratic equation s are found in the oldest

Thomas Harriot (1560-1621) introduced the method of factoring to ob

known mathematical literature. Babylonia n s and Egyptia n s were

tain solutions, a n d Franc;:ois Viete (1540-1603) introduced a method that

solving such problems before 1800 Be. Euclid solved quadratic

is essentia lly completing the square.

equations geometrica l ly in his Data (300 BC). a n d the Hindus a n d Arabs

Until modern times it was usual to neglect the negative roots (if there

gave rules for solving a ny quadratic equation with real roots. Because

were a ny). a n d equation s involving square roots of negative quantities

negative n umbers were not freely used before

were regarded as unsolvable until the 1500s.

AD

1500, there were sev

eral differen t types of quadratic equations, each with its own rule.

Historical Problems 1. One of al-Khwdiizmi solutions

2

+

=

85 by draw

is a difference of two squares. If you factor this difference of two

ing the square shown .The area of the four white rectangles and the 2 yellow square is x + 12x. We then set this expression equal to 85

moreover, the quadratic expression is factored, which is sometimes

to get the equation

useful.

x2

We solve x

12x

+ 12x= 85. If we ad d the four blue squares,

squares, you will easily be able to get the quadratic formula, and,

we wil l have a larger square of known area. Complete the solution. We solve x2

2. Viete's method Then

(u + u2

+

(2z

d

+

Now select z so that 2z

+

12x - 85= 0 by letting x

+ 12( u + z) - 85

12) u +

(Z2 +

+ 12= 0 a n d

u

+

,

z.

3:J�

___

0

finish the solution.

( Vb2 - 4ac )2

Look a t equation

2a

I

x

_________

:�� : 3

I I I I I I I IX Area = I I I I I I I I I I X ___ �___________ J___ _

12z - 85)= 0

3. Another method to get the quadratic formula on page 102. Rewrite the right side as

=

=

(4)

3

a n d then

subtract it from each side. The right side is now 0 a n d the left side

i3 I

Area

x2 x

=

3x 31 i

3

1.2 Assess Your Understanding 'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red.

�.

Factor: x2-5x - 6 (pp. 49-55)

The solution set of the equation (x -3)(3x . (p. 13)

� True or False

)'\. Factor:2x2-x -3 (pp. 49-55)

______

�

=

+

5)

=

0 is

Ixl. (pp. 23-24)


'?'\ To complete the square of the expression x2 the number _____

would

+

5x, you

6. The quantity b2 - 4ac is called the of a quadratic equation. If it is , the equation has no real solution.

V. True or False Quadratic equations always have two real solutions.

8. True or False If the discriminant of a quadratic equation is positive, then the equation has two solutions that are nega tives of one another.

Skill Building

In Problems

9-28, solve

'i1lx2-9x �.\ Z2 Z -

=

+

6

0 =

0

each equation by factoring.

10. x2 + 4x = 0 14. v2 + 7v

+ 6

=

0

¥(x2-25= O �. 2x2 - 5x -3

12. x2- 9 =

0

=

16. 3x2 + 5x

0 +2

=0

SECTION 1.2

'¥.z

� 4x2 + 9 'i�. f'\

It x(x - 8) + 12 = 0

18. 2/ - 50= 0

3t 2 - 48 = 0

22. 2 5x2 + 1 6 = 40x

= 12 x

. 6 ( p 2 - 1)

12 26. x + - = 7

6x - 5 = .§. x

27.

x

4(x -2) x -3

=

+

�

=

)( (2y

32. ( x + 2)2 = 1

24. 2(2u2 - 4u ) + 3 = 0

-3 3 - = --x x(x - 3)

):(.

.

5 3 = 4 + -x + 4 x -2

--

28.

(x - 1)2 = 4

34. (3 z - 2)2

+ 3)2 = 9

107

20. x(x + 4) = 12

5p

In Proble ms 29-34, so lve each eq uatio n by the Sq uare Root Method. x2 25 30. x2 = 3 6

Quadratic Equations

=

4

I n Prob le ms 35-40, what number sho uld be added to co mp lete the sq uare ofeach e xpre ssio n? x2 - 8x 36. x2 - 4x

)i. x2 + � X

�

1 3

38. x2 - -x I n Proble ms

40

41-46,

1\. x2 + 4x = 2 1

2

2

- -7

5�

42. x2 - 6x= 13 46. 2 X2 -3 x - 1 = 0

3

3

7 �

solve each eq uatio n by co mpleti ng the sq uare.

2 1 44. v-? + -"v -- = O .>

.

I n Prob le ms 47-70, fi nd the rea l so lutio ns, ifa ny , ofeach eq uatio n. Use the q uadratic formula .

'

"

W. x2 - 4x - 1 = 0

48. x2 + 4x + 2 = 0

x2 - 4x + 2 = 0

51. 2 x2 - 5x + 3 = 0

52. 2X2 + 5x + 3

55. 4x2 = 1 - 2x

56. 2X2

59. 9t 2 - 6t +

1

60. 4u2 - 6u + 9

5 63. :;-x2 - x .)

1 3

= 0

3

=

1

64. 5x2 - x

= -

68. 4 +

=

0

-2 x

=

=

54. 4t 2 + t + 1= 0

57. 4x2 = 9 x

58. 5x = 4x2

3 1 , 61. -x2 --x 4

1

5

4

1 -

2

-

=

2

62. 3" x-? - x -3= 0

0

65. 2 x( x + 2) = 3

�

-x1 - ?1 = 0 .r

53. 41 -y + 2 = 0 ""

0

50. x2 + 6x + 1 = 0

.

-

66. 3 x( x + 2) =

3x 1 +-=4 x -2 x

--

1

1 2x +-=4 x -3 x

--

70.

I n Prob le ms 71-78, fi nd the rea l so lutio ns, i fa ny , of each eq uatio n. Use the q uadratic formula a nd a ca lc ulato l: Expre ss a ny so lutio ns ro unded to two deci ma l place s.

71. x2 - 4. 1 x + 2 .2 = 0

72. x2 + 3.9 x + 1.8 = 0

73. x2 + V3 x -3= 0

74. x2 + V2 x -2 = 0

75. 7TX2 - x - 7T = 0

76. 7TX2 +

78. 7Tx2 - 1 5 V2x + 2 0

77. 3 x2 + 87TX + v29 = 0 I n P rob le ms 79. x2 - 5

79-92, fi nd the

=

TTX

-2 = 0

0

rea l so lutio ns, i fa ny , o feach eq uatio n. Use a ny method.

80. x2 - 6 = 0

81. 16x2 - 8x + 1 = 0

82. 9 x2 - 12 x + 4 = 0

83. 10x2 - 19 x - 15 = 0

84. 6x2 + 7x - 2 0 = 0

85. 2 + z = 6z2

86. 2 = y

88. 1:. x2 = V2 x + 1 2 x 2 91. -- +

89. x2 + x = 4

x -2

=

0

--

x + l

=

+

1

;-;:. x = 2 87. r? + .V2

61

90. x2 +

7x + 1 x2 - x -2

-:-

92.

X =

1

3x 1 4 - 7x + -- = ----=--- x +2 x-I x2 + x - 2

--

I n Prob le ms 93-98, use the di scri mi na nt to determi ne whether each q uadratic eq ua cio n ha s two uneq ua l rea l so lutio ns, a re peated rea l sol utio n, or 11. 0 rea l so lutio n, wi tho ut solvi ng the eq uatio n.

93. 2 x2 - 6x + 7 = 0

94. x2

96. 2 5x2 - 2 0x

97. 3 x2 + 5x - 8 = 0

+

4=0

+

4x

+

7 =0

95. 9 x2 -3 0x + 2 5 = 0 98. 2 x2 - 3 x - 7 = 0

108

CHAPTER 1



99.

College Tuition and Fees The average annual published undergraduate tuition-and-fee charges C, in dollars, for pub lic four-year institutions from academic years 2000- 2001 through 2005- 2006 can be estimated by the equation C 20.2x 2 +3 14.5x +3 467.6, where x is the number of years afer the 2000- 2001 academic year. Assuming the model will remain valid beyond 2005- 2006, in what academic year will average annual tuition-and-fee charges be $8000?

109.

=

Source: College Board, Trendsin College P ri cing 100.

2005

The median weekly earnings E, in dollars, for full-time women wage and salary workers ages 16 years and older from 2000 through 2004 can be estimated by the equation E 0. 14x 2 + 7.8x + 5 40, where x is the nurnber of years after 2000. Assuming the model will remain valid beyond 2004, in what year will the median weekly earn ings be $63 2?

Women's Weekly Earnings

(b) When will it strike the ground? (c) Will the object reach a height of 100 meters? Reducing the Size of a Candy Bar A jumbo chocolate bar with a rectangular shape measures 12 centimeters in length, 7 centimeters in width, and 3 centimeters in thickness. Due to escalating costs of cocoa, management decides to reduce the volume of the bar by 10%. To accomplish this reduction, management decides that the new bar should have the same 3 centimeter thickness, but the length and width of each should be reduced an equal number of centimeters. What should be the dimensions of the new candy bar?

=

Source: U.S. Department of Labor, Highlight s of Women 's Ea rning s in

1. 102.

2004,

September 2005

The area of the opening of a rec tangular window is to be 143 square feet. If the length is to be 2 feet more than the width, what are the dimensions? Dimensions of a Window

110.

1 04.

Watering a Field

A circular pool mea-

V\ sures 10 feet across. One cubic yard of concrete is to be used to create a circular border of uniform width around the pool. If the border is to have a depth of3 inches, how wide will the border be? (1 cubic yard 27 cubic feet) See the illustration.

The area of a rectangular window is to be 3 06 square centimeters. If the length exceeds the width by 1centimeter, what are the dimensions? Geometry Find the dimensions of a rectangle whose perimeter is 26 meters and whose area is 40 square meters.

Rework Problem 109 if

ill. Constructing a Border around a Pool

Dimensions of a Window

103.

Reducing the Size of a Candy Bar

the reduction is to be 20% .

=

An adjustable water sprinkler that sprays water in a circular pattern is placed at the center of a square field whose area is 125 0 square feet (see the figure). What is the shortest radius setting that can be used if the field is to be completely enclosed within the circle?

112.

Constructing a Border arollnd a Pool

113.

Constructing a Border around a Garden A landscaper, who just completed a rectangular flower garden measuring 6 feet by 10 feet, orders 1 cubic yard of premixed cement, all of which is to be used to create a border of uniform width around the garden. If the border is to have a depth of3 inches, how wide will the border be? ( 1 cubic yard 27 cubic feet)

if the depth of the border is 4 inches.

Rework Problem 111

=

l�. Constructing a Box An open box is to be constructed from 1J\... a square piece of sheet metal by removmg a square of Side

1 foot from each corner and turning up the edges. If the box is to hold 4 cubic feet, what should be the dimensions of the sheet metal?

106.

Rework Problem 105 if the piece of sheet metal is a rectangle whose length is twice its width.

Constructing a Box

�. Physics A ball is thrown vertically upward from the top of � a building 9 6 feet tall with an initial velocity of 80 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s 9 6 + 80t - 16r2. (a) After how many seconds does the ball strike the ground? (b) After how many seconds will the ball pass the top of the building on its way down? =

108.

Physics An object is propelled vertically upward with an initial velocity of 20 meters per second. The distance s (ill me ters) of the object from the ground after t seconds is s -4.9t 2 + 20t . (a) When will the object be 15 meters above the ground? =

114.

�.

Dimensions of a Patio A contractor orders 8 cubic yards of premixed cement, all of which is to be used to pour a patio that will be 4 inches thick. If the length of the patio is speci fied to be twice the width, what will be the patio dimensions? (1 cubic yard 27 cubic feet) Comparing TVs The screen size of a television is deter mined by the length of the diagonal of the rectangular screen. =

trad itional

4 :3

LCD 1 6:9

SECTION 1.3

Complex Numbers; Quadratic Equations in the Complex Number System

Traditional TVs come in a 4 : 3 format, meaning the ratio of the length to the width of the rectangular is 4 to 3. What is the area of a 37-inch traditional TV screen? What is the area of a 37-inch LCD TV whose screen is in a 16 : 9 format? Which screen is larger? 2 [Hint: If x is the length of a 4 : 3 format screen, then x is 4 the width.]

116.

�.

c a

121. Find k such that the equation kx2 + x + k

� -

=

0 has a

=

0 has a

repeated real solution.

122. Find k such that the equation x 2 - kx + 4 repeated real solution.

123. Show

that the real solutions of the equation ax 2 + bx + c 0 are the negatives of the real solutions of the equationax 2 -bx + c O. Assume thatb 2 - 4ac ;:0: O. =

=

124. Show that the real

solutions of the equation ax 2 + bx + c 0 are the reciprocals of the real solutions of the equation cx 2 + bx + a O. Assume thatb2- 4ac ;:0: O.

starting with 1, must be added to get a sum of 666? If a polygon ofn sides has n (n

-�.

. lS - .

Comparing TVs Refer to Problem 1 15 . Find the screen area of a traditional 50-inch TV and compare it with a 50-inch Plasma TV whose screen is in a 16 : 9 format. Which screen is larger?

Geometry

Show that the sum o f the roots of a quadratic equation is

120. Show that the product of the roots of a quadratic equation

. The sum of the consecutive integers 1, 2 , 3, . . . , n is given by 1 the formula "2n (n + 1). How many consecutive integers,

118.

109

=

3) diagonals,

=

how many sides will a polygon with 65 diagonals have? Is there a polygon with SO diagonals? Discussion and Writing

128. Create three quadratic equations: one having two distinct so

. Which of the following pairs of equations are equivalent? Explain. (a) x 2 = 9; x 3 (b) x v9 ; x 3 (c) (x - 1)(x -2) (x - 1 f; x - 2 x - 1 126. Describe three ways that you might solve a quadratic equa tion. State your preferred method; explain why you chose it.

lutions, one having no real solution, and one having exactly one real solution.

=

=

129. The word quadra tic seems to imply four (quad), yet a qua

=

=

dratic equation is an equation that involves a polynomial of degree 2 . I nvestigate the origin of the term quadra tic as it is used in the expressionquadra tic equa tion . Write a brief essay on your findings.

=

127. Explain the benefits of evaluating the discriminant of a qua dratic equation before attempting to solve it. 'Are You Prepared?' Answers

1. (x - 6)(x + 1)

2. (2x - 3)(x + 1 )

3.

{-� 3} 3'

4. True

1 .3 Com plex Num bers; Quadratic Equations in the Com plex Number System* PREPARING FOR T H I S SECTION •

Before getting started, review the following: •

Classification of Numbers (Section R.1, pp. 4-5 ) Now Work the 'Are You

Prepared?' problems

OBJECTIVES

on page

Rationalizing Denominators (Section R.S, p. 74)

1 1 6.

1

Add, Su btract, M u ltiply, and Divide Complex N u mbers (p. 1 1 0)

2

Solve Quadratic Equations in the Complex N u mber System (p. 1 1 4)

Complex N u m bers

One property of a real number is that its square is nonnegative. For example, there is no real number x for which To remedy this situation, we introduce a new number called the imaginary unit. '" This section may be omitted without any loss of continuity.

1 10

CHAPTER 1

Equations and I nequa lities

The imaginary unit, which we denote by i, is the number whose square is - 1 . That is,

DEFINITION

�I�

2 = -1 i_ ___

L-____________________

__ __ __ __ __ __ __ __ __ __

This should not surprise you. If our universe were to consist only of integers, there would be no number x for which 2x = 1 . This unfortunate circumstance was

1

2

remedied by introducing numbers such as "2 and "3 ' the rational numbers. If our universe were to consist only of rational numbers, there would be no x whose square equals 2. That is, there would be no number x for which x2 = 2. To remedy this, we introduced numbers such as V2 and .,ys , the irrational numbers. The real numbers, you will recall, consist of the rational numbers and the irrational numbers. Now, if our universe were to consist only of real numbers, then there would be no number x whose square is - 1 . To remedy this, we introduce a number i, whose square is -1. I n the progression outlined, each time w e encountered a situation that was unsuitable, we introduced a new number system to remedy this situation. And each new number system contained the earlier number system as a subset. The number sys tem that results from introducing the number i is called the complex number system. Complex numbers are numbers of the form a + hi, where a and b are real numbers. The real number a is called the real part of the number a + bi; the real number b is called the imaginary part of a + bi; and i is the imaginary

DEFINITION

unit, so P

=

-1.

�

For example, the complex number -S + 6i has the real part -S and the imagi nary part 6. When a complex number is written in the form a + bi, where a and b are real numbers, we say it is in standard form. However, if the imaginary part of a complex number is negative, such as in the complex number 3 + ( -2)i, we agree to write it instead in the form 3 2i. Also, the complex number a + Oi is usually written merely as a. This serves to remind us that the real numbers are a subset of the complex numbers. The complex number 0 + bi is usually written as bi. Sometimes the complex number bi is called a pure imaginary number. -

1

Add, Su btra ct, Multiply, a n d Divide Complex N u m bers

Equality, addition, subtraction, and multiplication of complex numbers are defined so as to preserve the familiar rules of algebra for real numbers. Two complex num bers are equal if and only if their real parts are equal and their imaginary parts are equal. That is, Equality of Complex Numbers

a + bi

= c +

di

if and only if a

=

c and b

=

d

(1)

Two complex numbers are added by forming the complex number whose real part is the sum of the real parts and whose imaginary part is the sum of the imagi nary parts. That is, Sum of Complex Numbers

(a + bi) + (c + di)

(a + c) + (b + d)i

(2)

SECTION 1.3

111

Complex Numbers; Quadratic Equations i n the Complex Number System

To subtract two complex numbers, we use this rule: Difference o f Complex Numbers

(a

EXAM P L E 1

+

bi) - (e + di)

=

(a - e)

+

(b - d)i

(3)

Adding and Subtracting Complex N u mbers

(a) (3 + 5i) + ( -2 + 3i) = [3 + ( -2 ) ] + (5 + 3)i = 1 + 8i (b) (6 + 4i) - (3 + 6i) = (6 - 3 ) + (4 - 6) i = 3 + (- 2 ) i = 3 - 2i = � = -

Now Work

•

1 3

PROBLE M

Products of complex numbers are calculated as illustrated in Example 2. EXAM P L E 2

M ultiplying Complex N u m bers (5

+

3i) ' (2 + 7i)

5 · (2 + 7i)

=

i

3i(2 + 7i) = 10 + 35i + 6i + 21P

+

i



= 10 + 41i + 21( - 1 )

i i2

=

-1

= -11 + 41i

•

Based on the procedure of Example 2, we define the product of two complex numbers as follows: Product of Complex Numbers

(a

+

bi) · (e + di) = (ae - bd) + (ad + be)i

(4)

Do not bother to memorize formula ( 4). Instead, whenever it is necessary to multiply two complex numbers, follow the usual rules for multiplying two binomi als, as in Example 2, remembering that i2 - 1 . For example, =

(2i) (2i)

=

4z2

=

-4

(2 + i)(1 - i) = 2 - 2i + i - P Q!I!l: = = ""

Now Work

PROBLE M

=

3-i

19

Algebraic properties for addition and multiplication, such as the commutative, associative, and distributive properties, hold for complex numbers. The property that every nonzero complex number has a multiplicative inverse, or reciprocal, requires a closer look. DEFINITION

If z = a + bi is a complex number, then its defined as

conjugate,

= a + bi = a - bi _ __ _ _

z__ �___________________

For example, 2 + 3i

=

__ __

__

__

denoted by

Z,

�I�

__ __ __ __ __ __ __ __ __

2 - 3i and -6 - 2i = -6 + 2i.

is

1 12

CHAPTER 1

Equations and Inequa lities

EXA M P L E 3

M u ltiplying a Complex Number by Its Conjugate Find the product of the complex number z

Solution

Since z

zz

=

=

3 + 4i and its conjugate z.

=

3 - 4i, we have

(3 + 4i) (3 - 4i)

=

9 - 12i + 12i - 16P

=

9 + 16

= 25

•

The result obtained in Example 3 has an important generalization. THEOREM

The product of a complex number and its conjugate is a nonnegative real number. That is, if z = a + bi, then

(5)

I

�J�

�__________________________________

If z

Proof

=

zz

a + bi, then

=

(a + bi) (a - bi)

=

a2 - (bi?

=

a2 - b2i2

=

a2

+

b2

•

To express the reciprocal of a nonzero complex number z in standard form,

1

multiply the numerator and denominator of - by z. That is, if z z nonzero complex number, then

1 a + bi

l

z

I z

z

z

z zz

i

EXA M P L E 4

(5).

a a2 + b2

-

b . l a2 + b2

Writing the Reciprocal of a Complex N u mber i n Standard Form Write

Solution

a + bi is a

a - bi a2 + b2

Use

=

=

3

� 4i in standard form a + bi; that is, find the reciprocal of 3 + 4i.

The idea is to multiply the numerator and denominator by the conjugate of 3 + 4i, that is, by the complex number 3 - 4i. The result is

1 3 + 4i

1 3 - 4i 3 + 4i 3 - 4i

-- = -- . -- =

3 - 4i 9 + 16

3

4 .

= - - -l 25 25

•

To express the quotient of two complex numbers in standard form, we multi ply the numerator and denominator of the quotient by the conjugate of the denominator.

EXA M P L E 5

Writing the Quotient of a Complex N u m ber in Standard Form Write each of the following in standard form. (a)

1 + 4i 5 - 12 i

(b)

3i 4 - 3i

2 -

SECTION 1 . 3

Solution

Complex Num bers; Quadratic Equations in the Complex Number System

1 13

1 + 4i 5 + 12i 5 + 12i + 20i + 48i2 5 - 12i 5 + 12i 25 + 144 43 32 -43 + 32i + i 169 169 1 69 2 - 3i 2 - 3i 4 + 3i 8 + 6i - 12i - 9P = . = 4 - 3i 4 - 3i 4 + 3i 16 + 9 17 - 6i = 17 - 6 . l 25 25 25

4i (a) 51 -+ 12i

=

(b)

-

�==- Now Work

E XA M P L E 6

27

Writing Other Expressions i n Standard Form

If z = 2 - 3i and form. (a) Wz

-

Solution

PROBLEM

•

W = 5 + 2i,

write each of the following expressions in standard

(b) z +

W

(c) z + Z

(2 - 3i) (5 - 2i) 10 - 4i - lSi + 6P z·w w · w (5 + 2i) (5 - 2i) 25 + 4 4 19 . 4 - 19i = - - -1 29 29 29 z + w = (2 - 3i) + (5 + 2i) = 7 - i = 7 + i z + Z = (2 - 3i) + (2 + 3i) = 4

(a) wz

-

(b) (c)

•

The conjugate of a complex number has certain general properties that we shall find useful later. For a real number a = a + Oi, the conjugate is a = a + Oi = a - Oi = a. That is, THEOREM

The conjugate of a real number is the real number itself. Other properties of the conjugate that are direct consequences of the definition are given next. In each statement, z and w represent complex numbers.

THEOREM

The conjugate of the conjugate of a complex number is the complex number itself. (z) = z

(6)

I

(7)

I

The conjugate of the sum of two complex numbers equals the sum of their conjugates. z+w=z+w

The conjugate of the product of two complex numbers equals the product of their conjugates. z·w = z·w

(8)

I�

�----------------------------------�

We leave the proofs of equations (6), (7), and (8) as exercises.

1 14

CHAPTER 1


Powers of ;

The powers of i follow a pattern that is useful to know. is i6

i1 = i P = -1 P = P · i = - I · i = -i

= =

i4 • i = 1 · i = i i4 • P = - 1

i7 = i4 • i3 = - i i8 = i4 • i4 = 1

i4 = p . p = ( - 1 ) ( - 1 ) = 1

And so on. The powers of i repeat with every fourth power. EXA M P L E 7

Evaluating Powers of ; (a) p7 = p4 . p = (i4) 6 . i3 = 16 · i3 = -i (b) i 1 0 1 = i100 • i 1 = ( i4) 25 . i = 125 . i = i

•

Writing the Power of a Complex N u m ber in Standard Form

EXA M P L E 8

Write ( 2 + i)3 in standard form. We use the special product formula for (x + a )3.

Solution

(x + a ) 3 = x3 + 3ax2 + 3a2 x + a3 Using this special product formula, (2 + i)3 = 23 + 3 · i . 22

+

3·P.2

= 8 + 12i + 6( - 1 )

+

+

P

(-i)

= 2 + l 1i.

•

c;rn:==- Now Work P R O B L E M 4 1

2

Solve Quadratic Equations in the Com p lex N um be r System

Quadratic equations with a negative discriminant have no real number solution. However, if we extend our number system to allow complex numbers, quadratic equations will always have a solution. Since the solution to a quadratic equation involves the square root of the discriminant, we begin with a · discussion of square roots of negative numbers. DEFINITION

WARNING In writing

v=N

=

If N is a positive real number, we define the principal square root of denoted by V- N , as

Villi be

sure to place i outside the V symbol.

EXA M P L E 9

_

where i is the imaginary unit and F = - 1 .

- N,

.J

Evaluating the Square Root of a Negative N umber (a) v=I = VIi = i (c) v=s = Y8 i = 2 Y2 i

(b) v=4 = v4 i = 2i •

SECTION 1.3

EXA M P L E 1 0

Solution

1 15

Complex Numbers; Quadratic Equations i n the Complex Number System

Solving Equations

Solve each equation in the complex number system. (a) x2= 4 (b) x2= -9 (a) x2 = 4 x = ± V4 = ± 2 The equation has two solutions, -2 and 2. The solution set is ( - 2, 2). (b) x2 = -9 x = ± \.1-9 = ± V9i = ±3i The equation has two solutions, -3i and 3i. The solution set is {-3i, 3i}.

.... .

-

:z:: """ .- Now Work ''!l!l;:cs:=

PROBLEM S

49

A ND

53

WARNING When working with square roots of negative numbers, do not set the square root of

a product equal to the product of the square roots (which can be done with positive n umbers). To see why, look at this calculation: We know that V1c5O = 10. However, it is also true that 100 so =

10

=

( -25) ( -4),

ViOO

=

V( -25)( -4)

'*

i

V=25 v=4

=

(V25i)(V4i)

=

(5i)(2i)

=

10?

=

-10 •

Here is the error.

Because we have defined the square root of a negative number, we can now restate the quadratic formula without restriction. THEOREM

Quadratic Formula

In the complex number system, the solutions of the quadratic equation ax2 + bx + c = 0, where a, b, and c are real numbers and a =f:. 0, are given by the formula x=

-b ± Vb2 - 4ac 2a

(9)

I

�

� . � -------------------

EXA M P L E 1 1 Solution

Solving Quadratic Equations i n the Complex N u m ber System

Solve the equation x2 - 4x + 8= 0 in the complex number system. Here a = 1, b = -4, c = 8, and b2 - 4ac = 16 - 4( 1 ) (8 ) = -16. Using equa tion (9), we find that - ( -4) ± v=i6 4 ± V16 i 4 ± 4i = 2 ± 2i = -= x= 2 2( 1 ) 2 The equation has two solutions 2 - 2i and 2 + 2i. The solution set is {2 - 2i, 2 + 2i} . Check :

2 + 2i:

(2 + 2i ) 2

2 - 2i:

(2

-

4( 2 + 2i ) + 8

- 2i ) 2 - 4 ( 2

1l'l!l'O=="".- Now Work

PROBLEM

- 2i)

+

4 + '8i. + 4F - .8' - '8i. + .8' =4- 4 = 0 8 = 4 - '8i. + 4F - .8' + '8i. + .8' = 4 - 4= 0 =

•

59

The discriminant b2 - 4ac of a quadratic equation still serves as a way to deter mine the character of the solutions.

1 16

CHAPTER 1


Character of the Solutions of a Quadratic Equation

In the complex number system, consider a quadratic equation ax2 + bx + c = ° with real coefficients. 1. If b2 - 4ac > 0, the equation has two unequal real solutions. b2 - 4ac = 0, the equation has a repeated real solution, a double root. 3. If b2 - 4ac < 0, the equation has two complex solutions that are not real. The solutions are conjugates of each other.

2. If

The third conclusion in the display is a consequence of the fact that if

b2 - 4ac = - N < ° then, by the quadratic formula, the solutions are x=

-b +

-b

Yb2 - 4ac

-b

2a

and

x=

Yb2 - 4ac

-------

-b -

2a

+

v=N

-

v=N

2a

2a

-b + VN i -b VN = - + -- i 2a 2a 2a

-----

-b

-

VN i = -b VN . l

2a

2a

--

2a

which are conjugates of each other. EXA M P L E 1 2

Determi ning the Character of the Solution of a Quadratic Equation Without solving, determine the character of the solution of each equation. (b) 2x2 + 4x + 1

(a) 3x2 + 4x + 5 = ° Solution

=

°

(c) 9x2 - 6x + 1 = °

(a) Here a = 3, b = 4, and c = 5, so b2 - 4ac = 1 6 - 4 ( 3 ) ( 5 ) = -44. The solutions are two complex numbers that are not real and are conjugates of each other. (b) Here a = 2, b = 4, and c = 1, so b2 - 4ac = 16 - 8 = 8. The solutions are two unequal real numbers. (c) Here a = 9, b = -6, and c = 1 , so b2 - 4ac = 3 6 - 4(9 ) ( 1 ) = 0. The solution is a repeated real number, that is, a double root. •

w = = = "",·-

Now Work

P ROBLEM

73

1.3 Assess Your Understanding 'Are You Prepared?' An swe rs a re g iven at t he end of t he se exe rcise s. If you get a w rong an swe r, read t he page s l isted in re d.

� Name the integers and the rational numbers in

}

Y True or False

the set

6 -3 , 0, v2 ,-, 1T . (pp. 4-5) 5

{

/..

'"

Rational numbers and irrational numbers are in the set of real numbers. (pp. 4 -5 ) 3 ( p 74 ) Rationalize the denominator of 2 + V3 3' .


6. True or False The conjugate of2 + 5 i is -2 - Si. True or False All real numbers are complex numbers. 8. True or False If2 -3 i is a solution of a quadratic equation with real coefficients, then -2 + 3 i is also a solution.

4. In the complex number 5

+ 2 i, the number 5 is called the part; the number 2 is called the ____ part; the number i is called the ____ ____

X The equation x 2

=-4

X

has the solution set ____

Skill Building

In P roblem s 9-46, w rite ea ch ex p re ssion in t he standa rd fo rm a + b i . .t. (2 -3 i )

}\

+ (6 + 8i)

(2 - 5 i) - (8 + 6i)

10. (4 + 5 i ) + (-8 + 2 i) 14. ( - 8 + 4i) - (2 -2 i)

� (-3 + 2 i ) - (4 -4 i ) � 3 (2 - 6i)

12. (3 -4 i) - (-3 -4 i) 16.

-4(2

+

8i)

SECTION 1.3

� 2i(2 - 3i) �. ( -6 + i ) ( -6 - i) Y. 2 7 i 1\ '

_ 2 -_i 26. _

, i23 i6 - 5

34. 38. 42. 46.

"1f- ( 1 + i)3 A i6 + i4 + P + 1 � V-4 50. v=64

47-52,

-

13 5 - 12i 2 + 3i 28. 1 - !

24. ___

' ''' l + i

(� � y

" i)

i

1 17

20. ( 5 + 3i) ( 2 - i )

O

\.� �

-2!

30.

(3 - 4i ) ( 2 + i )

}( 3 � 4i

22. ( -3 + i ) ( 3 + i)

i

In Problems

J(

18. 3i( -3 + 4i)

�G � y " 'j( +

Complex Numbers; Quadratic Equations in the Complex Number System

32. ( 1 - i f

�. ( 1 + i) 2

il4

36. i-23 40. 4i3 - 2P + 1 44. 2i4( 1 + i2)

4 + i3 (3i)4 + 1

P + i5 + P + i

perform the indicated operations and express your answer in the form a + bi.

� V-2s -

48. vC9

�, Y(3 + 4i) (4i - 3)

52. yr(4-) (3i-+ 3i4) --

In Problems 53-72, solve each equation in the complex number system. . ,2 + 4 = 0 54. x2 - 4 = 0 x2 - 16 0 x2 - 6x + 13 = 0 58. x2 + 4x + 8 = 0 x2 - 6x + 10 = 0 8x2 - 4x + 1 = 0 62. l Ox2 + 6x + 1 = 0 . 5x2 + 1 = 2x x2 + X + 1 = 0 66. x2 - x + 1 = 0 . x3 - 8 = 0 I . X4 = 1 6 70. X4 1 X4 + 1 3x2 + 36 = 0

�

�. �

�

•

=

X

·

=

56. 60. 64. 68. 72.

x2 + 25 = 0 x2 - 2x + 5 = 0 1 3x2 + 1 = 6x

x3 + 27

=

0

X4 + 3x2 - 4

=

0

In Problems 73-78, without solving, determine the character of the solutions of each equation in the complex number system. 3x2 - 3x + 4 0 74. 2x2 - 4x + 1 = 0 2x2 + 3x = 4 76. x2 + 6 = 2x 9x2 - 12x + 4 = 0 78. 4x2 + 12x + 9 = 0 2 + 3i is a solution of a quadratic equation with real coefficients. Find the other solution. 80. 4 - i is a solution of a quadratic equation with real coefficients. Find the other solution.

.�

�

In Problems 81-84, z z +Z

'Yk

�

�

=

=

3 - 4i and w 8 + 3i. Write each expression in the standard form a + bi. 82. w - IV zz 84. z - w

�

=


5.

Electrical Circuits

The impedance Z, in ohms, of a circuit element is defined as the ratio of the phasor voltage V, in volts, across

the element to the phasor current I, in amperes, through the elements. That is, Z

=

f.

If the voltage across a circuit element is 18 + i

volts and the current through the element is 3 - 4i amperes, determine the impedance.

86

·

)t

Parallel C ircuits

1 . . . ' . . . I n an ac CirCUit WitI1 two para I I el pathways, the total I mpedance Z, 111 ohms, satls fles the formula Z

=

1 1 + -, ZI Z2

where ZI is the impedance of the first pathway and Z2 is the impedance of the second pathway. Determine the total impedance if the impedances of the two pathways are ZI = 2 + i ohms and Z2 = 4 - 3i ohms. Use z

=

a + bi to show that z +

z

88. Use z

=

=

z.

( Use z

90. Use z

=

=

a + bi to show that z a + bi and w a + bi and w

=

=

=

2a and

z

-z

=

c + di to show that z + w c + di to show that z · w

=

=

2bi.

z+

W.

z · w.


91. Explain to a friend how you would add two complex numbers and how you would multiply two complex numbers. Explain any dif ferences in the two explanations.

92. Write a brief paragraph that compares the method used to rationalize the denominator of a radical expression and the method used to write the quotient of two complex numbers in standard form. 'Are You Prepared?' Answers

{

1. Integers: { -3, O } ; rational numbers: -3, 0,

�}

2. True

3. 3 ( 2 -

\13)

118

CHAPTER 1


1 .4 Radical Equations; Equations Quadratic in Form; Factorable Equations PREPARING FOR THIS SECTION •

•


Square Roots (Section R.2, pp. 23-24) Factoring Polynomials (Section R.5, pp. 49-55)

,NOW Work

nth Roots; Rational Exponents (Section R.8, pp. 72-76)


OBJECTIVES

1

1

Solve Radical Equations (p. 1 1 8)

2

Solve Equations Quadratic in Form (p. 1 1 9)

3

Solve Equations by Factoring (p. 1 2 1 )

Solve Rad ical Equations

When the variable in an equation occurs in a square root, cube root, and so on, that is, when it occurs in a radical, the equation is called a radical equation. Sometimes a suitable operation will change a radical equation to one that is linear or quadratic. A commonly used procedure is to isolate the most complicated radical on one side of the equation and then eliminate it by raising each side to a power equal to the index of the radical. Care must be taken, however, because apparent solutions that are not, in fact, solutions of the original equation may result. These are called extraneous solutions. Therefore, we need to check all answers when working with radical equations. EXA M P L E 1

Solving a Radical Equation Find the real solutions of the equation:

Solution

V2x - 4 - 2

=

0

The equation contains a radical whose index is 3. We isolate it on the left side.

V2x - 4 - 2 V2x - 4

= =

0 2

Now raise each side to the third power (the index of the radical is 3) and solve.

( V2x - 4) 3 = 23 2x - 4 = 8 2x = 12 X = 6

Raise each side to the power 3. Sim plify. Add 4 to both sides. Divide both sides by 2.

Check:

\12 ( 6) - 4 - 2 = V12 - 4 - 2 The solution set is {6}. Li!l!

EXA M P L E 2

>-

Now Work

PRO

VB - 2

=

B LE M 7

Solving a Radical Equation Find the real solutions of the equation:

Solution

=

Vx"="l = x - 7

We square both sides since the index of a square root is 2.

Vx"="l = x - 7 (\/�-=-ly = ( x - 7? X - 1 = x2 - 14x x2 - 15x + 50 = 0

Square both sides. +

49

Remove pa rentheses. Put in standa rd form.

2-2

=

o. •

SECTION 1 . 4

Radical Equations; Equations Quadratic i n Form; Factorable Equations

(x - 10) (x - 5 ) = 0 x = 10 or x = 5

1 19

Factor. Apply the Zero-Product Property and solve.

Check:

VlO-=---i = V9 = 3 and x - 7 = 10 - 7 = 3 = 5: = v'S=J:" = v4 = 2 and x - 7 = 5 - 7 = -2 The solution x = 5 is extraneous; the only solution of the equation is x = 10. The solution set is /10j. •

x x

= 1 0:

� �

&:."l!l:==- Now Work

=

PRO B LEM

19

Sometimes we need to raise each side to a power more than once in order to solve a radical equation. Solving a Radical Equation

EXA M P L E 3

Find the real solutions of the equation: Solution

V2x

+

3

-

Vx+2 2 =

First, we choose to isolate the more complicated radical expression (in this case, v'2x + 3) on the left side.

Now square both sides (the index of the radical on the left is

( V2x + 3) 2 2x + 3 2x + 3 2x + 3

= = = =

(Vx+2 + 2)2 ( Vx+2) 2 + 4 Vx+2 + 4 x + 2 + 4Vx+2 + 4 x + 6 + 4 Vx+2

2).

Square both sides. Remove parentheses. Simplify. Combine like terms.

Because the equation still contains a radical, we isolate the remaining radical on the right side and again square both sides.

x - 3 = 4 Vx+2 ( x - 3 ? = (4 Vx+2)2 x2 - 6x + 9 = 16x + 32 x2 - 22x - 23 = 0 ( x - 23 ) (x + 1 ) = 0 X = 23 or x = -1

Isolate the radica l on the right side. Square both sides. Remove pa rentheses. Put i n standard form. Factor.

The original equation appears to have the solution set { -1, 23}. However, we have not yet checked. Check:

x x

= =

23:

- 1:

V2x + 3 V2x + 3 -

Vx+2 V2(23) + 3 - V23 + 2 Vx+2 V2( - 1 ) + 3 - V + 2

v49 - v'2s = 7 - 5 = 2 -1 = Vi - Vi = 1 - 1 = 0 = The equation has only one solution, 23; the solution -1 is extraneous. The solution set is /23j. • =

===- Now Work 2

=

P RO B L EM

29

Solve Eq uations Q u a d ratic in Form

The equation X4 + x2 - 12 = 0 is not quadratic in x, but it is quadratic in x2. That is, if we let = x , we get + - 12 = 0, a quadratic equation. This equation can be solved for and, in turn, by using = x , we can find the solutions x of the orig inal equation.

u

u

2

u2 u

u

2

120

CHAPTER 1


In general, if an appropriate substitution u transforms an equation into one of the form

au2 + bu +

0

c =

a *- O

then the original equation is called an equation of the quadratic type or an equation quadratic in form.

The difficulty of solving such an equation lies in the determination that the equation is, in fact, quadratic in form. After you are told an equation is quadratic in form, it is easy enough to see it, but some practice is needed to enable you to rec ognize such equations on your own. E XA M P LE 4

Solution

Solving Equations That Are Quadratic in Form Find the real solutions of the equation:

(x + 2)2 + 1 1 (x + 2) - 12 = 0

For this equation, let u = x + 2. Then u2

=

(x + 2? + 1 1 ( x + 2) - 12

=

(x + 2 ) 2 , and the original equation,

0

becomes

u2 + 1 1u - 12 = 0 (u + 12)(u - 1 ) = 0 u = - 12 or u = 1

Let

u

= x

+ 2.

Then

u2

= (x

+ 2) 2 .

Factor. Solve.

But we want to solve for x. Because u = x + 2, we have

x + 2 = -12 o r x + 2 = 1 x = -1 x = - 14

Check:

x

-14: ( - 14 + 2)2 + 1 1 ( - 14 + 2) - 12 = ( - 12? + 1 1 ( -12) - 12 = 144 - 132 - 12 = 0 x = - 1 : ( - 1 + 2 ) 2 + 1 1 ( - 1 + 2) - 12 = 1 + 1 1 - 12 = 0 =

The original equation has the solution set { -14, - I } .

EXA M P L E 5

Solving Equations That Are Quadratic in Form Find the real solutions of the equation:

Solution

•

(x2 - 1 ) 2 + (x2 - 1 ) - 12

=

0

For the equation (x2 - 1 ) 2 + (x2 - 1 ) - 12 = 0, we let u = x2 - 1 so that u2 = (x2 - 1 ) 2 . Then the original equation,

(x2 - 1 ) 2 + (x2 - 1 ) - 12 = 0 becomes

u2 + u - 12 = 0 (u + 4)(u - 3 ) = 0 u = -4 or u = 3

Let

u

=

�

-

1. Then u2

=

(�

-

1{

Factor. Solve.

But remember that we want to solve for x. Because u = x2 - 1, we have

x2 - 1 = -4 or x2 - 1 = 3 x2 = -3 x2 = 4 The first of these has no real solution; the second has the solution set { -2 2} ,

Check:

x = -2: x = 2:

.

(4 1 )2 + (4 - 1) - 12 = 9 + 3 - 12 = 0 (4 - I ? + (4 - 1 ) - 12 = 9 + 3 - 12 = 0 -

The original equation has the solution set { -2, 2}.

•

SECTION 1.4

EXAM P L E 6

Radical Equations; Equations Quadratic i n Form; Factora ble Equations

12 1

Solving Equations That Are Quadratic i n Form x + 2 Vx - 3 = 0 For the equation x + 2 Vx - 3 = 0, let u = Vx. Then u2 equation,

Find the real solutions of the equation: Solution

x + 2Vx - 3

=

=

x, and the original

0

becomes u2 + 2u - 3 = 0 ( u + 3) (u

1) = 0 u = -3 or u = 1 -

Let u = \IX. Then u2 = x. Factor. Solve.

Since u = Vx , we have Vx = -3 or Vx = 1. The first of these, Vx = - 3, has no real solution, since the square root of a real number is never negative. The second, Vx = 1, has the solution x = 1 . Check: 1

+ 2 v1

-

3 = 1 + 2 - 3 = 0

The original equation has the solution set ( I J . ,,-===,...

ANO T H ER M ET H OD

F OR SOLV I NG EX A M PLE

•

6

WOULD

B E TO T REAT IT AS A R AD ICAL EQUATION. SOLV E WAY

IT T H IS

FOR PRACT I CE .

The idea should now be clear. If an equation contains an expression and that same expression squared, make a substitution for the expression. You may get a quadratic equation. 'e'

3

.�

Now Work

PRO B LE M

S

1

So lve Equations by Factoring

We have already solved certain quadratic equations using factoring. Let's look at examples of other kinds of equations that can be solved by factoring. EXAM P L E 7 Solution

Solvi n g Equations by Factoring Solve the equation: X4 = 4x2 We begin by collecting all terms on one side. This results in 0 on one side and an expression to be factored on the other. X4 x2 ( x2 - 4 ) = 0 or x2 - 4 = 0 x2 = 4 -

x2 = 0 x = 0

or

X4 = 4x2 4x2 = 0

x = -2

or

Factor. Apply the Zero-Prod uct Property.

x = 2

The solution set is { -2, 0, 2}. Check: x = -2:

x = 0: x = 2:

( _2 )4 = 1 6 and 4 ( -2 ? = 16 S o -2 is a solution. 04 0 and 4 · 02 = 0 So 0 is a solution . 2 24 = 16 and 4 · 2 = 1 6 S o 2 i s a solution. =

•

122

CHAPTER 1


EXA M P L E 8

Solving Equations by Factoring Solve the equation:

Solution

x3 - x2 - 4x + 4

=

0

Do you recall the method of factoring by grouping? (If not, review pp. 53-54.) We group the terms of x3 - x2 - 4x + 4 = 0 as follows: (x3 - x2) - (4x - 4)

0

=

Factor out x2 from the first grouping and 4 from the second. x2(x - 1 ) - 4( x - 1 )

=

This reveals the common factor (x - 1 ) , so we have (x2 - 4) ( x - 1 ) = 0 (x - 2)(x + 2)(x - 1) = 0 x - 2 = 0 or x + 2 = 0 or x

=

x = -2

2

0

Factor again.

x - I x

= =

0

Set each factor equa l to

1

0.

Solve.

The solution set is { -2, 1, 2 } . Check:

x

=

x

=

x

=

1'"

•

-2:

1:

2: >-

( -2)3 - ( -2f 4( -2) + 4 = -8 - 4 + 8 + 4 1 3 - 12 - 4( 1 ) + 4 = 1 - 1 - 4 + 4 = 0 -

23 - 22 - 4(2) + 4 Now Work

=

PRO B L E M

8 - 4 - 8 + 4

=

=

0

0 -2 is a sol ution. 1 is a sol ution . 2 is a solution .

•

79

1 .4 Assess Your Understanding 'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.

'%

=

True or False The principal square root of any nonnegative real number is always nonnegative. (pp. 23-24)

v=s �� Factor

6x3 -

(pp. 72-76)

2X2 ( pp. 49-55)


6. True or False Radical equations sometimes have extraneous

4. When an apparent solution does not satisfy the original equa-

�

solutions.

tion, it is called a(n ) solution. If u is an expression involving x, the equation au2+bu+ = c 0, a oF 0, is called a(n) equation ____

Skill Building

In Problems

'/(

7-40,

find the real solutions of each equation.

=1 V2t=1

10. v5t+3 =-2

=2 *- � =Vs 16. \Yx2+16 ( VIS - 2x =x 22. = x 2V- x - 1 ¥- 3+� =X

8. v3t+4 =2

)&� VI

- 2x -

= 3 ° 3 -1 14. �2x -= it(= x 8vX 20. � =x

)I: y

x2

- X

x

+2 =x 26. 2+V12 - 2x -4

=

� v3t+4=-6

12. � - 1

�. �

x2 +

=

°

2x=-1 18. = x 3vX 1( X = 2� 24. Y3 x + = x2 X - 2 ,*. � - Vx+l = 1 -

SECTION 1.4

)(.

28. � + Vx+2 = 1

¥ V3 - 2 VX = VX

Radical Equations; Equations Quadratic in Form; Factora ble Equations

�- �=2

30. V3x - 5

�

32. VlO + 3 VX = VX

34. (3x - 5 ) 1/2 = 2

35. (5x - 2 ) 1/3 = 2

37. ( x2 + 9) 1/2 = 5

38. ( x2 - 1 6 ) 1 /2

=

4

2 X - 5x2 - 12

=

0

47. (x + 2 ) 2 + 7(x + 2 ) + 12 = 0

0

=

)(

53. x - 4xVX = 0

60. XI/2 - 3xl/4 + 2

65.

= x

63. x2 + 3x +

1 I ? = __ + 2 x + 1 ( x + 1 )-

66.

68. 2x-2 - 3x-1 - 4 = 0 71.

0

(-V )2 + -2v v+1 v+ 1

=

1

= 0

49. ( 3x + 4f - 6(3x + 4) + 9 = 0 52. 3 ( 1 - y)2 + 5 ( 1 - y) + 2 = 0

3

=

55. x + VX

59. 4X I /2 - 9X I/4 + 4 = 0

\y4 - 5x2

=

54. x + 8VX = 0 57. t l/2 - 2tl/4 + 1

-

46. x6 - 7x3 - 8 = 0

2(s + I f - 5 ( s + 1 )

56. x + VX = 6

62.

40. X3/4 - 9xl/4 = 0 43. 3x4 - 2x2

48. (2x + 5 ) 2 - (2x + 5 ) - 6

50. (2 - x)2 + (2 - x ) - 20 = 0

2

(3x + 1 ) 1/2 = 4

39. x3/2 - 3xl/2 = 0

9

45. x6 + 7x3 - 8

=

36. (2x + 1 ) 1/3 = - 1

In Problems 41-72, find the real solutions of each equation. 41. x4 - 5x2 + 4 = 0 42. X4 - 1 0x2 + 25 = 0 44.

- Vx+7

123

58. Z I /2 - 4 Z I/4

0 0

=

61.

Vx2 + 3x = 6

--

1 1 + x - I ( x - 1 )2

=

72.

\Y5x2

+

4 = 0

- 6 = x

64. x2 - 3x -

12

69. 2x2/3 - 5xl/3 - 3 = 0

= 8

20

=

Vx2 - 3x = 2

67. 3x-2 - 7x-1 - 6 = 0 70. 3x4/3 + 5x2/3 - 2 = 0

(-) (-) + y 2 Y = 6 y - 1 y - 1

In Problems 73-88, find the real solutions of each equation by factoring. 73. x3 - 9x = 0 74. X4 - x2 = 0 75. 4x3

3x2

=

�.

7

77. x3 + x2 - 20x = 0

78. x3 + 6x2 - 7x = 0

80. x3 + 4x2 - X - 4 = 0

81. x3 - 3x2 - 4x + 12 = 0

83. 2x3 + 4 = x2 + 8x

84. 3x3 + 4x2 = 27x + 36

85. 5x3 + 45x = 2x2 + 18

87. x (x2 - 3x) I /3 + 2(x2 - 3x)4/3 = 0

88. 3x(x2 + 2X) I /2 - 2(x2 + 2x)3/2 = 0

86. 3x3 + 1 2x = 5x2

+

20

In Problems 89-94, find the real solutions of each equation. Use a calculator

89. x - 4xl/2 + 2 = 0

95. If k =

-- and x + 3 x - 3

90. x2/3

+

4x l/3 + 2

=

0

82. x3 - 3x2 - X + 3 = 0

LO

express any solutions rounded to two decimal places.

93. 7T ( 1 + t)2 = 7T + 1 + t . k2 - k = 12, fmd x.

x3 + x2 - x - 1 = 0

96. If k =

91. X4 + v3 x2 - 3 = 0

-- and x+ 3 x - 4

94. 7T ( 1 + r)2

=

2 + 7T ( 1 + r)

k2 - 3k = 2 8, find x.

Ap plicatiens

97.

Physics: Using Sound to Measure Distance The distance to the sur face of the water in a well can sometimes be found by dropping an ob ject into the well and measuring the time elapsed until a sound is heard. If I I is the time (measured in seconds) that it takes for the object to strike the water, then t1 will obey the equation s = 16ft , where s is the

distance (measured in feet). It follows that

II

Vs

= 4 . Suppose that t

2

is the time that it takes for the sound of the impact to reach your ears. Because sound waves are known to travel at a speed of approximately 1 1 00 feet per second, the time t to travel the distance s will be

--

s t = See the I. 11 ustratlOn. . 2 1 100 .

2

Sound waves:

Falling object:

t1 = {S "4

t2 =

II�O

124

CHAPTER 1


Now tl + tz is the total time that elapses from the moment that the object is dropped to the moment that a sound is heard. We have the equation . Total time elapse d

=

Vs

4+

s 1 00 1

Find the distance to the water's surface if the total time elapsed from dropping a rock to hearing it hit water is 4 seconds.

98.

Crushing Load

load L, in tons, using the model

99.

H, in feet, of a square wooden pillar to its crushing H2 1L If a square wooden pillar is 4 inches thick and feet high, what is its crushing load?

A civil engineer relates the thickness T, in inches, and height T

=

1

25

0

The period of a pendulum is the time it takes the pendulum to make one full swing back and forth. The

Foucault's Pendulum

32: where I is the length, in feet, of the pendulum. In 1 85 1 , lean-Bernard-Leon period T, in seconds, is given by the formula T 271"'YIT =

Foucault demonstrated the axial rotation of Earth using a large pendulum that he hung in the Pantheon in Paris. The period of Foucault's pendulum was approximately 1 6.5 seconds. What was its length?


100. Make up a radical equation that has no solution. 101 . Make up a radical equation that has an extraneous solution. 1 02. Discuss the step in the solving process for radical equations that leads to the possibility of extraneous solutions. Why is there no such

possibility for linear and quadratic equations?


2. -2

1. True



Algebra Essentials (Section R.2, pp. 1 7 1 9)

'\.Now work

-

the 'Are You Prepared?' problems on page 1 32.

OBJECTIVES

1

Use Interval Notation (p. 1 25)

2

Use Properties of Inequalities (p. 1 26)

3

Solve Inequalities (p. 1 28)

4 Solve Combined Inequalities (p. 1 29)

Suppose that a and b are two real numbers and a < b. We shall use the notation a < x < b to mean that x is a number between a and b . The expression a < x < b is equivalent to the two inequalities a < x and x < b. Similarly, the expression a :::; x :::; b is equivalent to the two inequalities a :::; x and x :::; b. The remaining two possibilities, a :::; x < b and a < x :::; b, are defined similarly. Although it is acceptable to write 3 2': X 2': 2, it is preferable to reverse the inequality symbols and write instead 2 :::; x :::; 3 so that, as you read from left to right, the values go from smaller to larger. A statement such as 2 :::; x :::; 1 is false because there is no number x for which 2 :::; x and x :::; 1 . Finally, we never mix inequality symbols, as in 2 :::; x 2': 3.

SECTION 1.5

1

Solving I n equalities

125

Use I nterva l Notation

Let a and b represent two real numbers with a < b. DEFINITION

A closed interval, denoted by [a, b], consists of all real numbers x for which a ::;

x

::;

b.

An open interval, denoted by ( a, b ) , consists of all real numbers x for which

a

< x < b.

The half-open, or half-closed, intervals are (a, b], consisting of all real num bers x for which a < x ::; b, and [a, b) , consisting of all real numbers x for which a ::; x < b.

..J

In each of these definitions, a is called the left endpoint and b the right endpoint of the interval. The symbol 00 (read as "infinity") is not a real number, but a notational device used to indicate unboundedness in the positive direction. The symbol - 00 (read as "negative infinity") also is not a real number, but a notational device used to indi cate unboundedness in the negative direction. Using the symbols 00 and - 00 , we can define five other kinds of intervals:

[ a, 00 ) ( a, 00 ) ( - 00 , a 1 ( - 00 , a) (- 00 , 00 )

Consists of all real numbers x for which x 2: Consists of all real numbers x for which x > Consists of aU real numbers x for which x ::; Consists of all real numbers x for which x < Consists of all real numbers x

a a

a

a

Note that 00 and -00 are never included as endpoints, since neither is a real number. Table 1 summarizes interval notation, corresponding inequality notation, and their graphs. Ta ble 1

I nterval

The open interval ( a, The closed interval

b)

[a, bl

The h a lf-open interval The h a lf-open interval

a < x a

The interval

( - co, a]

x :!S a

The interval

( -co, co)

The interval

The interval ( -co,

a)

Graph

a :!S x :!S b

[a, co)

The interval

EXAM P L E 1

Inequality

x< a

( a

) b

E a

) b

E a

� b � b

( a

[ a

( a

•• :.

.

� a

) a

All rea l numbers

Writing I nequalities Using I nterval N otation Write each inequality using interval notation. (a) 1

Solution

::;

x

::;

3

(b) -4 < x < 0

(c) x > 5

(d) x

::;

1

(a) 1 ::; x ::; 3 describes all numbers x between 1 and 3, inclusive. In interval notation, we write [1, 3 ] ' ( b ) I n interval notation, -4 < x < 0 i s written ( -4, 0).

126

CHAPTER 1


5 consists of all numbers x greater than 5. In interval notation, we write (5, 00 ) . (d) In interval notation, x � 1 is written ( - 00, 1 )' X >

(c)

•

EXAM P LE 2

Writing I ntervals Using I nequality N otation Write each interval as an inequality involving x. (a) [ 1 , 4)

Solution

(a) (b) (c) (d)

(0

)

(d) ( - 00 , -3J

(c) [2, 3 ]

[ 1 , 4) consists o f all numbers x for which 1 :::; x < 4. (2, 00 ) consists of all numbers x for which x > 2. [2, 3 J consists of all numbers x for which 2 � x � 3. ( - 00 , -3J consists of all numbers x for which x � -3.

1.£==> -

2

(b) (2,

Now Work

P ROBLEM S

11 , 23,

•

31

A ND

Use Properties of I n eq u a l ities

The product of two positive real numbers is positive, the product of two negative real numbers is positive, and the product of 0 and 0 is O. For any real number a, the value 2 2 of a is 0 or positive; that is, a is nonnegative. This is called the nonnegative property. f

f f

In

Word s

The s quare of a real number is

Nonnegative Property

For any real numbers

a,

never negative.

(1) If we add the same number to both sides of an inequality, we obtain an equiva lent inequality. For example, since 3 < 5, then 3 + 4 < 5 + 4 or 7 < 9. This is called the addition property of inequalities. Addition Property of Inequalities

For real numbers

a,

b, and

c,

If a < b, then a + If a

>

b, then a +

c

The addition property states that the sense, or direction, of an inequality remains unchanged if the same number is added to each side. Figure 3 illustrates the addition property (2a). In Figure 3 ( a), we see that a lies to the left of b. If c is posi tive, then a + c and b + c each lie c units to the right of a and b, respectively. Consequently, a + c must lie to the left of b + c; that is, a + c 0,

then a + c < b + c.

1I!l� - D R A W

�

- c units

� • • •

a+c

b+c

(b)

•

b

If a < b an d c < O, then a + c < b + c.

A N ILLU S T R A T I ON S IMIL A R T O

T H A T ILLU S T R A T E S

a

F IGURE

T HE A D D I T ION P RO P E R T Y

3 (2b)

SECTION 1.5

EXA M P L E 3

Solving Inequalities

127

Addition P roperty of I nequalities

(a) If x < -5, then x + 5 < -5 + 5 or x + 5 < 0. (b) If x > 2, then x + ( -2 ) > 2 + ( -2) or x - 2 > 0. II!IT::=� .-

Now Work

PROB L E M

•

39

We will use two examples to arrive at our next property. E XA M P L E 4

Solution

EXA M P L E 5

Solution

r

r

In Word s

r

reverses the inequality.

r

Multiplying by a negative number

M u ltiplying an I nequality by a Positive N u m ber

Express as an inequality the result of multiplying each side of the inequality 3 < 7 by 2. We begin with 3 2 by -4. We begin with 9>2 MUltiplying each side by -4 yields the numbers -36 and -8, so we have - 36 < -8 • Note that the effect of multiplying both sides of 9 > 2 by the negative number -4 is that the direction of the inequality symbol is reversed. Examples 4 and 5 illustrate the following general multiplication properties for

inequalities:

Multiplication Properties for Inequalities

For real numbers a, b, and e, If a If a If a If a

 0, then ae < be. be.

(3a)

> b and if e > 0, then ae > be. > b and if e < 0, then ae < be.

(3b)

The multiplication properties state that the sense, or direction, of an inequality remains the same if each side is multiplied by a positive real number, whereas the direction is reversed if each side is multiplied by a negative real number. EXA M P L E 6

M ultiplication Property of I nequ alities

1 (a) If 2x < 6, then "21 (2x) < "2 (6) or x < 3. (b) If x > 12, then -3 x < -3 ( 12 ) or x < -36. -3 -3

-

(-)

128

CHAPTER 1


-4x -8 > or x > 2. -4 -4 ( d ) If -x > 8, then ( - l ) ( -x) < ( - 1 ) (8) or x < -8.

( c ) If -4x < -8, then

c;1l'!:= : ==-- Now Work

PROB L E M

•

45

The reciprocal property states that the reciprocal of a positive real number is positive and that the reciprocal of a negative real number is negative. Reciprocal Property for Inequalities

1 a

If a > 0, then - > 0

1 a

If a < 0, then - < 0

3

1 a 1 If - < 0, then a < 0 a If - > 0, then a > 0

(4a) (4b)

Solve I nequalities

An inequality in one variable is a statement involving two expressions, at least one containing the variable, separated by one of the inequality symbols < , :5 , > , or ;::: . To solve an inequality means to find all values of the variable for which the state ment is true. These values are called solutions of the inequality. For example, the following are all inequalities involving one variable x:

x+50 x-2

x2 - 1 :5 3

--

As with equations, one method for solving an inequality is to replace it by a series of equivalent inequalities until an inequality with an obvious solution, such as x < 3, is obtained. We obtain equivalent inequalities by applying some of the same properties as those used to find equivalent equations. The addition property and the multiplication properties form the bases for the following procedures. Procedures That Leave the I nequality Symbol Unchanged

1. Simplify both sides of the inequality by combining like terms and eliminating parentheses: Replace by

(x + 2) + 6 > 2x + 5(x + 1 ) x + 8 > 7x + 5

2. Add or subtract the same expression on both sides of the inequality:

Replace by

3x - 5 < 4 (3x - 5 ) + 5 < 4

+

5

3. Multiply or divide both sides of the inequality by the same positive

expression: Replace

4x > 1 6 by

4x 16 >4 4

-

Procedures That Reverse the Sense or Direction of the Inequality Symbol

1. Interchange the two sides of the inequality: Replace

3 < x by x > 3

2. Multiply or divide both sides of the inequality by the same

expression: Replace

6 -2x -2x > 6 by -- < -2 -2

negative

SECTION 1.S

S o l v i n g Inequalities

129

As the examples that follow illustrate, we solve inequalities using many of the same steps that we would use to solve equations. In writing the solution of an inequality, we may use either set notation or interval notation, whichever is more convenient. Solving an I nequality

E XA M P L E 7

Solve the inequality: 3 - 2x < 5 Graph the solution set. Solution

3 - 2x < 5 3 - 2x - 3 < 5 - 3 -2x < 2 2 -2x -- > -2 -2

Simplify. Divide both sides by - 2. (The sense of the inequal ity symbol is reversed.)

x > -1 Figure 4

-3

-2

(

-1

Simplify

The solution set is { x i x > - I } or, using interval notation, all numbers in the interval ( - 1 , 00 ) . See Figure 4 for the graph.

[ )I ,

2

a

Subtract 3 from both sides.

•

Solving an I neq uality

EXAM P L E 8

Solve the inequality: 4x + 7 2: 2x - 3 Graph the solution set. 4x + 7 2: 2x - 3

Solution

4x + 7 - 7 2: 2x - 3 - 7

Subtract 7 from both sides.

4x 2: 2x - 10

Sim pl ify.

4x - 2x 2: 2x - 10 - 2x

Su btract 2x from both sides.

2x 2: -10

Si m plify.

-10 2x 2 2

- > -X

Figure 5

-6

[

-5

-4

-3

-2

!I

-5

The solution set is { x i x 2: - 5 } or, using interval notation, all numbers in the in terval [ - 5 , 00 ) . See Figure 5 for the graph.

-1

•

�= ::::EC-- '-

4

EXA M P L E 9

2:

Divide both sides by 2. (The direction of the inequality sym bol is unchanged.) Simplify.

Now Work

PROBLEM

53

So lve Com b i ned I n eq u a l ities

Solving Combined I nequalities Solve the inequality: -5 < 3x - 2 < 1 Graph the solution set.

Solution

Recall that the inequality -5 < 3x - 2 < 1 is equivalent to the two inequalities -5 < 3x - 2

and

3x - 2 < 1

130

CHAPTER 1

Equations and I nequa lities

We will solve each of these inequalities separately. -5 < 3x - 2 -5 + 2 < 3x - 2 + 2 -3 < 3x 3x -3 - - 3 (pp. 1 7-1 9)


3. If each side of an inequality is multiplied by a(n)

In Problems 6-9, assume that a < b and c < o. 6. Tl'ue 01' False a + c < b + c

number, then the sense of the inequality sym bol is reversed.

____

, denoted [ a, b], consists of all

4. A(n)

real numbers x for which a ::; x ::; b.

5. The

state that the sense, or direction, of an inequality remains the same if each side is multiplied by a positive number, while the direction is reversed if each side is multiplied by a negative number.

7. Tl'Ue or False 8. Tl'Ue 01' False

a - c bc C

C

10. True or False The square of any real number is always nonnegative.

Skill Building In Problems II-16, express the graph shown in blue using interval notation. A lso express each as an inequality involving x.

'- 11.

14.

]

[

-1

0

2

12.

3

� � � I____�__� ] _�__L-� 2 -1 -2 0

(

-2

-1

-1

o

)

0

13.

2

[--�---.� I --� )�� 1 5. --�---r 3

2

16.

-1

0

-1

o

€

[

I

2

3

2

3

••

[n Problems 1 7-22, an inequality is given. Write the inequality obtained by: (a) A dding 3 to each side of the given inequality. (b) Subtracting 5 from each side of the given inequality. (c) Multiplying each side of the given inequality by 3. (d) Multiplying each side of the given inequality by -2.

17. "

18.

3 < 5

2 >

19.

1

4 > -3

20.

- 3 > -5

21.

2x + 1 < 2

22. 1 - 2x > 5

In Problems 23-30, write each inequality using interval notation, and illustrate each inequality using the real number line.

23. 0 27.

::; x ::; 4

24. - 1 28.

x :2:: 4

< x < 5

x ::; 5

25. 4 29.

::; x < 6

x < -4

26.

-2 < x
1

0

In Problems 31-38, write each interval as an inequality involving x, and illustrate each inequality using the real number line.

' 31.

35.

[2, 5 ]

32.

( 1 , 2)

33. (-3 , - 2 )

34. [0, 1 )

[ 4, 00 )

36.

( - 00 , 2 ]

37.

38.

( - 00 , -3 )

In Problems 39-52, fill in the blank with the correct inequality symbol.

' 39.

40. If x

< -4, then x + 4 ___

O.

42. If x

> 6, then x - 6

-4, then 3 x

- 12.

44. If x

::; 3 , then 2x

If x > 6, then -2x

- 12.

46. If x

> -2, then -4x

-20.

48. If x

::;

41. If x

> -4, then x + 4

43. If x

:2::

' 45.

O.

If x < 5, then x - 5

47. If x

:2:: 5, then -4x

49.

I f 2x >

51.

It

.

1

- "2 x

6, then x ::; 3 , then x

3.

_

50. If 3 x -6.

( -8, 00 )

1

52. If - "4 x

o.

6. 8.

-4, then - 3 x

::; 1 2 , then x > 1 , then x

O.

12.

4. -4.

SECTION 1.5

Solvi ng Inequalities

133

In Problems 53-88, solve each inequality. Express your answer using set notation or interval notation. Graph the solution set.

54. x - 6 < 1

55. 1 - 2x :::; 3

56. 2 - 3x :::; 5

57. 3 x - 7 > 2

58. 2x + 5 > 1

59. 3 x - 1 2: 3 + x

60. 2x - 2

62. -3( 1

63. 4 - 3 ( 1 - x) :::; 3

53. x

+

1< 5

-

x) < 12

1

2:

3 +x

61. - 2 ( x + 3 ) < 8 64. 8 - 4(2 - x) :::; -2x

1

65. "2 (x - 4) > x + 8

66. 3 x

68. ::

69. 0 :::; 2x - 6 :::; 4

70. 4 :::; 2x + 2 :::; 10

72. -3 :::; 3 - 2x :::; 9

73. -3
3 (x - 2)

2

1

77. (x + 2 ) (x - 3 ) > ( x - l ) ( x + 1)

78. (x - l ) ( x + 1) > (x - 3 ) ( x + 4)

-

5 ) :::; (3x - 1) 2

2 3

2

3

>

2x - 1 < 0 4

--

79. x(4x + 3 ) :::; ( 2x + 1)2 +

1

3 4

82. 3 < -- :::; 3

0

85. 0 < - < x 5

1

< -

x

2

2

2

1

87. 0 < (2x - 4r l < "2

86. 0 < - < x

1

x

--

84. (2x - 1)- 1

83. (4x + 2rl < 0 4

+

1

81. - :::;

4

76. 0 < 1 - 3 x < 1

75. 1 < 1

80. x(9x

1 - ::

1

"2 x < 4

74. 0 < -- < 4

-

2:

3

1 3

88. 0 < (3x + 6r 1 < -


expect to live at least 49 .66 more years and an average 30-year-old female in 2005 could expect to live at least 53.58 more years. (a) To what age can an average 30-year-old male expect to live? Express your answer as an inequality. (b) To what age can an average 30-year-old female expect to live? Express your answer as an inequality. (c) Who can expect to live longer, a male or a female? By how many years?

In Problems 89-98, find a and b.

89. If -1 < x < 1, then a < x + 4 < b. 90. If -3 < x < 2, then a < x - 6 < b. 91. If 2 < x < 3, then a < -4x < b. 1

92. If -4 < x < 0, then a < "2 x < b. 93. If 0 < x < 4, then a < 2x + 3 < b. 94. If -3 < x < 3, then a < 1 - 2x < b.

Source: Actuarial Study No. 120, August 2005

1 95. If -3 < x < 0, then a < -- < b.

x+4

96. If 2 < x < 4, then a
t

a

If a is a positive number and if u is an algebraic expression, then

lui < a is equivalent to - a < u < a

(2)

lui :5 a is equivalent to -a :5 u :5 a

(3)

In other words, lui < a is equivalent to -a < u and u < a.

a

] .

..J

See Figure 10 for an illustration of statement (3).

a

Solving an I nequality I nvolving Absolute Value

EXAM P L E 3

Solve the inequality: 12x + 41 :5 3 Graph the solution set. This follows the form of statement (3); the expression u 2x + 4 is inside the a bsolute value bars.

12x + 4 1 :5 3

Solution

=

Apply statement (3).

-3 :5 2x + 4 :5 3 -3 - 4 :5 2x + 4 - 4 :5 3 - 4 :5 - 1 2x :5

-7 -7 7

-

2

2x

:5

Figure 1 1

-5

I

EI 7 -2 2

I3 I -1 a

I

2

2

4

•

-1

2

TlIe soIutlOn · set

IS .

Divide each part by 2.

1

X

2

Sim plify.

:5 2

-

- - :5

Subtract 4 from each part.

7

I

Sim plify.

:5 - 2

{ X - 2 :5

X

1

:5

-

2

}

.

.

.

[ 7 1]

, that IS, all x m the mterval - 2' - 2 . See

Figure 1 1 for the graph of the solution set.

•

Solving an I nequality I nvolving Absolute Value

EXA M P L E 4

Solve the inequality: 1 1 - 4xl < 5 Graph the solution set.

11 - 4xl < 5

Solution

This expression follows the form of statement (2); the expression u = 1 4x is inside the a bsolute value bars. Apply statement (2). -

-5 -5 - 1 -6 -6 -4

-

< < < >

3

Figure 1 2

-5 -4

(

-3 -2 -1

a

I

)

I

H 2 2

3

I

4

.

1

1 - 4x - 4x -4x -4x -4

1

-1

-1
a is equivalent to

Figure 1 4

IuI

-a

< -a or > a lui 2:: a is equivalent to u :::; -a or u 2:: a os a, a > !

0

a

EXA M P L E 6

(4)

(5 )

�

See Figure 14 for an illustration of statement (5). Solve the inequality: 12x - 51 > 3 Graph the solution set. 12x - 5 1 > 3 This fol lows the form of statement (4); the expression u

2x

==

2x

-

5 is inside the a bsol ute val ue bars.

2x - 5 < -3 or 2x - 5 > 3 - 5 + 5 < -3 + 5 or 2x - 5 + 5 > 3 + 5 2x < 2 or 2x > 8 2x 2 < -2 2

or

2x

8

> -2 2

Apply statement (4). Add 5 to each part. Simplify. Divide each part by 2.

Simplify. x>4 4 } , that is, all x in ( - 00 , 1 ) U (4, Figure 15 for the graph of the solution set.

x

Figure 1 5 •!

u

Solving an I nequality Involving Absolute Value

Solution

-2 - 1

u

�------�

[ ..

o

137

Equations and I nequalities Involving Absolute Value

)

0

1

2

3

4

5

6

7

(0

).

See •

WA RNING A common error to be avoided is to attempt to write the solution

the com bined inequal ity 1 and x > 4. C!ll!:= : ::::f.i> -

Now Work

" Recall that the symbol

>x>

x < 1 or x > 4 as 4, which is incorrect, since there are no n um bers x for which 1 > x •

PROBLEM 43

U stands for the union o f two sets. Refer t o p. 2 i f necessary.

1.6 Assess Your Understanding 'Are You Prepared?' A nswers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red.

1.

1 -2 1

==

2. True or False Ixl � 0 for any real number x. (p. 19)

(p. 19)


3. The solution set of the equation Ixl 5 is { 4. The solution set of the inequality Ixl < 5 is { xl }. ==

____

}.

5. True or False The equation Ixl == -2 has no solution. has the set of real 6. True or False The inequality Ixl � numbers as solution set.

-2

138

CHAPTER 1


Skill Building

,,

In Problems 7-34, solve each equation.

11. 1 1 - 4t l

+

S=

13

15. 1 -21x = 4 19.

12. 1 1 - 2z1 16. 1 3 1 x

1 � �I = 2 +

23. 4 - 12xl = 3 27. Ix2 - 2xl = 3 31.

" 9. 12x

8. 13xl = 1 2

7. 12xl = 6

1 32xX -- 23 1 = 2

=

6

+

9

2

I�2 - �I3 1 24. 5 - I� X I = 3 20•

32.

= 12

+ xl

1 23x 41 1 = x

+

--

=

=

2

=

11 1 9

21. l u - 21 = - -

22. 12 - vi = - 1

25. I x2 - 91 = 0

26. I x2 - 1 6 1

29. I x2

+

x - II = 1

30. Ix2

2

+

3x l

33. I x

1

+

3 18. 4 1 x l

9

=

1 2

=

28. Ix2

14. I - x l

13. 1 -2xl = l s i 17. 3" l x l

9

=

10. 13 x - 1 1

31 = 5

+

=

I x2 - 2x l

+

=

0

3x - 2 1 = 2

34. Ix2 - 2x l = I x

2

+

6x l

In Problems 35-62, solve each inequality. Express your answer using set notation or interval notation. Graph the solution set.

35. 12xl < S ' 39. I x - 21

37. 1 3 x l

36. 13xl < 15

+

2 O

x > O, y > O

III x< 0, y< 0

Quadrant IV

Quadrant

x> O , y < O

We locate a point on the real number line by assigning it a single real number, called the coordinate of the point. For work in a two-dimensional plane, we locate points by using two numbers. We begin with two real number lines located in the same plane: one horizontal and the other vertical. We call the horizontal line the x-axis, the vertical line the y-axis, and the point of intersection the origin O. See Figure 1. We assign coordi nates to every point on these number lines using a convenient scale. We usually use the same scale on each axis. In applications, however, different scales appropriate to the application may be used. The origin 0 has a value of 0 on both the x-axis and y-axis. Points on the x-axis to the right of 0 are associated with positive real numbers, and those to the left of o are associated with negative real numbers. Points on the y-axis above 0 are asso ciated with positive real numbers, and those below 0 are associated with negative real numbers. In Figure 1, the x-axis and y-axis are labeled as x and y, respectively, and we have used an arrow at the end of each axis to denote the positive direction. The coordinate system described here is called a rectangular or Cartesian* coordinate system. The plane formed by the x-axis and y-axis is sometimes called the xy-plane, and the x-axis and y-axis are referred to as the coordinate axes. Any point P in the xy-plane can be located by using an ordered pair (x, y) of real numbers. Let x denote the signed distance of P from the y-axis (signed means that, if P is to the right of the y-axis, then x > 0, and if P is to the left of the y-axis, then x < 0); and let y denote the signed distance of P from the x-axis. The ordered pair (x, y), also called the coordinates of P, then gives us enough information to locate the point P in the plane. For example, to locate the point whose coordinates are ( - 3, 1 ) , go 3 units along the x-axis to the left of 0 and then go straight up 1 unit. We plot this point by plac ing a dot at this location. See Figure 2, in which the points with coordinates ( -3, 1 ) , ( -2, -3 ) , (3, - 2), and (3, 2) are plotted. The origin has coordinates (0, 0). Any point on the x-axis has coordinates of the form (x, 0), and any point on the y-axis has coordinates of the form (0, y). If (x, y) are the coordinates of a point P, then x is called the x-coordinate, or abscissa, of P and y is the y-coordinate, or ordinate, of P. We identify the point P by its coordinates (x, y) by writing P ( x, y). Usually, we will simply say "the point (x, y ) " rather than "the point whose coordinates are (x, y ) . " The coordinate axes divide the xy-plane into four sections called quadrants, as shown in Figure 3. In quadrant I, both the x-coordinate and the y-coordinate of all points are positive; in quadrant II, x is negative and y is positive; in quadrant III, both x and y are negative; and in quadrant IV, x is positive and y is negativel Points on the coordinate axes belong to no quadrant. =

x

1;21!:1 = = ....

Now Work P R O B L E M 1 1

':' Named after Rene Descartes (1596-1650), a French mathematician, philosopher, and theologian.

SECTION 2.1

I

The Distance and M i d point Formulas

157

CO M M ENT On a graph ing calculator, you can set the scale on each axis. Once th is has been done,

you obtain the viewing recta ngle. See Figure 4 for a typical v i ewing rectangle. You should now read

_

Section 1, The Viewing Rectangle, in the Appendix.

Figure 4

1

Use the Distance Formula

If the same units of measurement, such as inches, centimeters, and so on, are used for both the x-axis and y-axis, then all distances in the xy-plane can be measured using this unit of measurement. E XA M P L E 1

F i n d i n g the Distance betwee n Two Poi nts

Find the distance d between the points ( 1 , 3 ) and (5, 6 ) . S o l ution

First w e plot the points ( 1 , 3 ) and (5, 6 ) and connect them with a straight line. See Figure Sea). We are looking for the length d. We begin by drawing a horizontal line from ( 1 , 3 ) to (5, 3 ) and a vertical line from (5, 3 ) to (5, 6), forming a right triangle, as shown in Figure S(b). One leg of the triangle is of length 4 (since Is 1 1 = 4), and the other is of length 3 (since 16 - 31 = 3). By the Pythagorean Theorem, the square of the distance d that we seek is d2 = 42 + 32 = 16 + 9 = 25 -

d = V2S = S

� .LJ

Figure 5

( 1 , 3) 4 ( 5 , 3)

6 x

3

3 (b)

(a)

6

x •

The distance formula provides a straightforward method for computing the distance between two points. THEOREM

r

To

r

ference

r

r

( '

r

The distance between two points PI = (Xl , Y I ) and P2 = (X2 ' Yz ), denoted by d ( PI , P2 ), is

I n Word s

r

compute

the

distance

between two points, find the dif of the

Distance Formula

x-coordinates,

square it, and add t h i s to the square of the difference of the y-coordinates. The square r oot of this sum is the distance.

d(PI , P2 ) = V(X2

-

xd

+ ( Y2

-

yd

(1)

I�

�--------------------------------�

Proof of the Distance Formula Let (Xl , Y l ) denote the coordinates o f point PI and let ( X2 ' Y2 ) denote the coordinates of point P2 ' Assume that the line joining PI and P2 is neither horizontal nor vertical. Refer to Figure 6(a). The coordinates of P3 are (X2 ' Y l ) . The horizontal distance from P I to P3 is the absolute value of

1 58

CHAPTER 2

Graphs

the difference of the x-coordinates, I X2 - xI I . The vertical distance from P3 to Pz is the absolute value of the difference of the y-coordinates, IY2 - Y l i . See Fig ure 6(b). The distance d( P l , P2 ) that we seek is the length of the hypotenuse of the right triangle, so, by the Pythagorean Theorem, it follows that 2 2 [ d( P l , P2 ) f = I X2 - xl l + IY2 - Y l l 2 = (X2 - x d + (Yz - Yl ) d( P l , P2 ) =

Figure 6

V(X2 - Xl? + (Y2 - yd

y

L--L

-L

__ __ __ __

�x

__ __

(b)

(a)

Now, if the line joining PI and P2 is horizontal, then the y-coordinate of PI equals the y-coordinate of P2 ; that is, Y l = Y2 . Refer to Figure 7(a). In this case, the distance formula (1) still works, because, for Yl = Yz , it reduces to d( P 1 , P2 ) = Figure 7

Y Y1

P1 =

(X1 ' Y1) •

V(X2 - X l ? + 02 = V( X2 - x d = I X2 - XI I Y2

y

d(P1, P2)

•

Y1 x

I Y2 � Y1 1 d(P� , P2) T P1 - (X1 , Y1)

r�

I,\ , Y,I

x1

(a)

x

(b)

A similar argument holds if the line joining PI and P2 is vertical. See Figure 7(b). The distance formula is valid in all cases. • E XA M P L E 2

U s i n g the Distance Formula

Find the distance d between the points ( - 4 5 ) and (3, 2). ,

Solution

Using the distance formula, equation (1), the solution is obtained as follows : 2 2 d = \1[3 - ( 4 ) f + (2 - 5 ? = \17 + ( - 3 ) -

=

, ;!I,I!

>-

V49+9 = v5s

�

7.62 •

Now Work P R O B L E M S 1 5 A N D 1 9

The distance between two points PI = (X l , Y l ) and P2 = ( X2 ' Y2 ) is never a neg ative number. Furthermore, the distance between two points is 0 only when the points are identical, that is, when Xl = X2 and Y l = Y2 . Also, because (X2 - Xl ) 2 = (Xl - X2 ) 2 and ( Yz - Y l ) 2 = ( Yl - Y2?, it makes no difference whether the dis tance is computed from PI to P2 or from P 2 to PI; that is, d(Pl , P z ) d( P2 , PI). The introduction to this chapter mentioned that rectangular coordinates enable us to translate geometry problems into algebra problems, and vice versa. The next example shows how algebra (the distance formula) can be used to solve geometry problems. =

SECTION 2.1

E XA M P L E 3

(a) (b) (c) (d)

(-2, 1 ) eo::- ____ C (3, 1) -+-

,

,

Plot each point and form the triangle ABC. Find the length of each side of the triangle. Verify that the triangle is a right triangle. Find the area of the triangle.

d(A, C)

=

3

( - 2 1 ) B = (2, 3 ) , and C = (3, 1 ) .

d(A, B) = V[ 2 - ( -2) f + (3 2 2 d ( B, C) = V(3 - 2) + ( 1 - 3 )

y

-3

=

(a) Points A, B, C and triangle ABC are plotted in Figure 8. (b) We use the distance formula, equation ( 1).

S o l ution

A=

1 S9

U s i ng Algebra to Solve G eometry P ro blems

Consider the three points A

Figure 8

The Distance and M i d point Formulas

x

=

V[3 - ( -2) f +

1 ) 2 v16+4 = V20 = 2Vs Vl+4 = Vs =

(1 - 1? = V25+O = 5 =

( c) To show that the triangle is a right triangle, we need to show that the sum of the squares of the lengths of two of the sides equals the square of the length of the third side. (Why is this sufficient?) Looking at Figure 8, it seems rea sonable to conjecture that the right angle is at vertex B. We shall check to see whether [ d(A, B) f + [ d(B, C) f [ d(A, C) f We find that 2 2 [ d(A, B) f + [ d(B, C) f ( 2 Vs ) + ( Vs ) = 20 + 5 = 25 = [ d ( A, C) f =

=

so it follows from the converse of the Pythagorean Theorem that triangle ABC is a right triangle. (d) Because the right angle is at vertex B, the sides AB and BC form the base and height of the triangle. Its area is Area m:::=:;;;;"..2

=

�

(Base ) ( Height) =

� ( 2 Vs) ( Vs )

=

5 square units •


Use the Midpoint Formula

We now derive a formula for the coordinates of the midpoint of a line segment. Let P I = ( Xl , Y 1 ) and P2 (X2 ' Y2) be the endpoints of a line segment, and let M = ( x, y) be the point on the line segment that is the same distance from P l as it is from P2 . See Figure 9. The triangles Pl AM and MBP2 are congruent. Do you see why? d(PJ , M) = d(M, P2) is given; L AP1 M L BMP2 ,* and L P1 MA L MP2B. So, we have angle-side-angle. Because triangles P l AM and MBP2 are congruent, corresponding sides are equal in length. That is,

Figure 9 Y

=

=

y

and x

x

2x = X l + X2 X l + X2 X= 2

Y - Yl

=

=

Y2

-Y

+ Y2 Yl + Y2 y= 2

2y =

Yl

':' A postulate from geometry states that the transversal P I P2 forms congruent corresponding angles with the parallel line segments P I A and M B.

160

CHAPTER 2

Graphs

THEOREM r

r r r

r

In

Word s

=

( x, y) of the line segment from

M

ment, average the x-coordinates the endpoints.

L-

+ l ( X, y ) _- X +2 Xz ' Y I 2 Yz

(

( X l , YI) to

)

Find the midpoint of a line segment from PI = ( -5, 5 ) to points P I and Pz and their midpoint. Check your answer.

y (-5 , 5 )

-

F i n d i ng the M idpoint of a Line Segment

Sol ution

Figure 1 0

=

_

=

(2)

I

��

__ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __

E XA M P L E 4

P1

The midpoint M Pz = (Xz , Yz) is

fo find the midpoint o f a line seg and average the y-coordinates of

PI

M i d point Formula

We apply the midpoint formula (2) using X l = -5, Y l = 5, Xz the coordinates ( x, y) of the midpoint M are x

5

=

Xl

+

2

Xz

-

5 + 3 2

_ _ _ _

=

- 1 and

Y

=

Yl

Pz =

= (3, 1 ) . Plot the

3, and Yz = 1 . Then

+ Y2 = 5 + 1 2

2

--

=3

That is, M = ( - 1, 3). See Figure 10. 5

-5

x

Check:

Beca u se M is the d(P1, M) = d( M, P2 ):

mid poi nt, we check the a n swer by verifyi ng that

d(Pl, M ) V[ -1 - (-5)J2 + (3 - 5)2 \11 6 + 4 V20 d( M, P2 ) V[3 (-1 ) J 2 + (1 - 3) 2 \116 + 4 = V20 =

=

=

"l' =--

-

=

=

•


2 . 1 Assess Your Understanding

'Are You Prepared?' A nswers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red. 1 . On the real number line the origin is assigned the number (p. 17-26) 2.If -3 and 5 are the coordinates of two points on the real num ber line, the distance between these points is . (p. 17-26) 3. If 3 and 4 are the legs of a right triangle, the hypotenuse is . (pp. 30-35) __.

4. Use the converse of the Pythagorean Theorem to show that a triangle whose sides are of lengths 11, 60, and 61 is a right triangle. (p. 30-35)

__

__

Concep�s and Vocabulary 5. If (x, y) are the coordinates of a point P in the xy-plane, then x is called the of P and y is the of P 6. The coordinate axes divide the xy-plane into four sections called 7. If three distinct points P, Q, and R all lie on a line and if d(P, Q) = d(Q, R ) , then Q is called the of the line segment from P t a R. __

__

__

The distance between two points is some times a negative number. 9. True or False The point ( - 1 , 4) lies in quadrant IV of the Cartesian plane. 10. True or False The midpoint of a line segment is found by av eraging the x-coordinates and averaging the y-coordinates of the endpoints. 8. True or False

SECTION 2.1

The Distance a nd Midpoint Formulas

161

Skill Building In Problems 1 1 and 12, plot each point in the xy-plane. Tell in which quadrant or on what coordinate axis each point lies. 11.

(a) A = ( -3, 2) (b) B = (6, 0) (c) C = ( -2, -2)

12. (a) A ( 1 , 4) (b) B ( -3, -4) (c) C = ( -3, 4)

(d) D = (6, 5 ) (e) E = (0, -3) (f) F = (6, -3)

(d) D (4, 1 ) (e) E (0, 1 ) (f) F = ( -3, 0)

=

=

=

=

13. Plot the points (2, 0), (2, -3), (2, 4 ) , (2, 1 ) , and (2, - 1 ) . Describe the set of all points of the form (2, y ) , where y is a real number. 14.

Plot the points (0, 3), ( 1 , 3), ( -2, 3), (5, 3 ) , and ( -4, 3). Describe the set of all points of the form ( x, 3 ) , where x is a real number.

In Problems 15-28, find the distance d ( P ] , P 2 ) between the points PI and P2 . 17. 15. Y 16. Y.

-2

2

P1 = (0, 0 )

1 9. P I

=

-1

P2 = (2, 1 )

2

�

P2 = (-

-2

x

-1

P = (0 ' 0 )

-/ 1 I I

2

I

-2 I

x

(3, -4); P 2 = (5, 4)

20.

PI

/1

22. PI = (2, -3); P2

23. PI = (4, -3); P2

24. P I

25.

PI

27. P I

(6, 4)

Y

-2 I

�I -1

I

=

(2, 2) 2

I

x

= ( - 1 , 0); P2 = (2, 4)

21. PI = (-3, 2); P2 = (6, 0) =

(1 , 1)

I . 2 x =

"' -:1 P2

18.

=

= ( -4, -3); P 2

(4, 2) =

(6, 2)

= ( -0.2, 0.3 ); P 2 = (2.3, 1 . 1 )

26.

PI = ( 1. 2, 2.3 ); P2 = ( -0.3, 1 . 1 )

= (a, b); P 2 = (0, 0)

28.

PI

= ( a , a ) ; P2

=

(0, 0)

In Problems 29-34, plot each point and form the triangle ABC. Verify that the triangle is a right triangle. Find its area. 29. A = ( -2, 5 ) ; B = ( 1 , 3 ) ; C = ( - 1 , 0) 30. A = ( -2, 5); B = ( 12, 3 ) ; C = ( 10, - 1 1 ) 31. A = ( -5, 3); B = (6, 0); C = (5, 5 )

32. A = ( -6, 3); B = (3, -5); C = ( - 1 , 5)

33. A = (4, -3); B = (0, -3); C = (4, 2)

34.

A = (4, -3); B = (4, 1 ) ; C = (2, 1 )

In Problems 35-44, find the midpoint of the line segment joining the points PI and P2 · 36. PI = ( -2, 0); P 2 = (2, 4) 35. P I = (3, - 4); P2 = (5, 4) 37.

PI

= ( -3, 2); P2

=

(6, 0)

39. PI = (4, -3); P2 = (6, 1 )

38.

PI

40. P I

= (2, -3); P2 =

=

( -4, -3); P 2

(4, 2) =

(2, 2)

41.

P I = ( -0.2, 0.3) ; P2 = (2.3, 1 . 1 )

42.

PI = ( 1 .2, 2.3) ; P 2 = ( -0.3, 1 . 1 )

43.

PI = ( a , b ) ; P2 = (0, 0)

44.

P I = (a, a ) ; P2 = (0, 0)

The midpoint of the line segment from PI to P2 is (5, -4). If P2 = (7, -2), what is PI?

Applications and Extensions 45.

Find all points having an x-coordinate of 2 whose distance from the point ( -2, - 1 ) is 5.

50.

46.

Find all points having a y-coordinate of -3 whose distance from the point ( 1 , 2 ) is 13.

47.

Find all points on the x-axis that are 5 units from the point (4, -3).

51 . Geometry The medians of a triangle are the line segments from each vertex to the midpoint of the opposite side (see the figure). Find the lengths of the medians of the triangle with vertices at A (0, 0), B = (6, 0), and C = (4, 4).

48.

Find all points on the y-axis that are 5 units from the point (4, 4).

49.

The midpoint of the line segment from Pi ta P2 is (-1, 4). If p] ( -3, 6), what is P2? =

=

C

A

B

162

CHAPTER 2

Graphs

52.

Geometry An equilateral triangle is one in which all three sides are of equal length. If two vertices of an equilateral tri angle are (0, 4) and (0, 0), find the third vertex. How many of these triangles are possible?

54.

Geometry Verify that the points (0, 0), (a, 0), and

53.

Geometry Find the midpoint of each diagonal of a square with side of length s. Draw the conclusion that the diagonals of a square intersect at their midpoints. [Hin t: Use (0, 0), (0, s), (s, 0), and (s, s) as the vertices of the square.]

(%' V;a ) are the vertices of an equilateral triangle. Then show that the mid

points of the three sides are the vertices of a second equilateral triangle (refer to Problem 52).

In Problems 55-58, find the length of each side of the triangle determined by the three points PI , P2 , and P3 . State whether the triangle is an isosceles triangle, a right triangle, neither of these, or both. (An isosceles triangle is one in which at least two of the sides are of equal length.) 55. 56. 57.

Pl = (2, 1 ) ; P2 = ( -4, 1 ) ; P3 = ( -4, -3) p] = ( - 1, 4); P = (6, 2 ) ; P3 = (4, -5) 2 Pl = ( -2, - 1 ) ; P2 = (0, 7 ) ; P3 = (3, 2)

58.

PI = (7, 2 ) ; P2 = ( -4, 0); P3 = (4, 6)

59.

Baseball A major league baseball "diamond" is actually a square, 90 feet on a side (see the figure). What is the distance directly from home plate to second base (the diagonal of the square )?

lies in the direction from home plate to first base, and the positive y-axis lies in the direction from home plate to third base. (a) What are the coordinates of first base, second base, and third base? Use feet as the unit of measurement. (b) If the right fielder is located at ( 1 80, 20) , how far is it from the right fielder to second base? (c) If the center fielder is located at ( 220, 220), how far is it from the center fielder to third base? 63.

Distance between Moving Objects A Dodge Neon and a Mack truck leave an intersection at the same time. The Neon heads east at an average speed of 30 miles per hour, while the truck heads south at an average speed of 40 miles per hour. Find an expression for their distance apart d (in miles) at the end of t hours.

64.

Distance of a Moving Object from a Fixed Point A hot-air balloon, headed due east at an average speed of 15 miles per hour and at a constant altitude of 100 feet, passes over an in tersection (see the figure). Find an expression for the distance d (measured in feet) from the balloon to the intersection t seconds later.

65.

Draft ing Error When a draftsman draws three lines that are to intersect at one point, the lines may not intersect as in tended and subsequently will form an error triangle. If this error triangle is long and thin, one estimate for the location of the desired point is the midpoint of the shortest side. The figure shows one such error triangle. Source: www. uwgb.eduldutchsISTRUCTGElsIOO.htm

2 n d base

60.

Li ttle League Baseball The layout of a Little League playing field is a square, 60 feet on a side: How far is it di rectly from home plate to second base (the diagonal of the square )? Source: Uttle League Baseball, Official Regulations and Play ing Rules, 2006.

61.

Baseball Refer to Problem 59. Overlay a rectangular coor dinate system on a major league baseball diamond so that the origin is at home plate, the positive x-axis lies in the di rection from home plate to first base, and the positive y-axis lies in the direction from home plate to third base. (a) What are the coordinates of first base, second base, and third base? Use feet as the unit of measurement. (b) If the right fielder is located at (310, 1 5 ) , how far is it from the right fielder to second base? (c) If the center fielder is located at (300, 300), how far is it from the center fielder to third base?

62.

Little League Baseball Refer to Problem 60. Overlay a rec tangular coordinate system on a Little League baseball dia mond so that the origin is at home plate, the positive x-axis

SECTION 2.2

Graphs of Equations in Two Variables; Intercepts; Symmetry

1 63

Inc., in 2004. How does your result compare to the re ported value of $256 billion? Source: Wal-Mart Stores, Inc., 2006 Annual Report

y

1.7 1. 5 1. 3

Wal-Mart Stores, Inc.

1 .4

2.6 2.7

Net sales (in $ billions)

� , :� i 350 1 · ··· · ·· ····· ·········· ·· ·········· ···· ·· .. 2. i · . . . . . . . . . ... . ... . . .. . . . · . . . . . . . . . · . . . .. . . . .. . . ... . . .. . . . ... . � 300 1 � f.-"' � 2 50 r ;;;:;,; :::::;,..;;;;- --� 200 I L _ � �................................................................................................, � 1 50 1 i � 1 00 1 ······ · ········· ······························ .. . . . . . . . J � 5 0 1· · ····· ········ ············ ··················· .. .. . ! . . .. . . . .. . . . . . . . . .. . . . . . . . .. . . . . . . . . ........ ) 1 ····· · · · ·· · · ·· · · ·· · · · ··· · · · · ·· · · · ·· · ··· · · ·· · ·· 0

x

........ ...................................

en

(a) Find an estimate for the desired intersection point. (b) Find the length of the median for the midpoint found in part (a). See Problem 51.

.. .

....

........................................................................................................

66.

Ne t Sa les The figure illustrates how net sales of Wal-Mart Stores, Inc., have grown from 2002 through 2006. Use the mid point formula to estimate the net sales of Wal-Mart Stores,

'"

..... ... . ...... .. . ........ ............ ...

..... ..

2002

'Are You Prepared?' A nswers 2. 8

1. 0

3. 5

2003

2004 Year

...... .........

2005

.

...

....

2006

4. 1 12 + 602 = 121 + 3600 = 3721 = 61 2

2.2 Graphs of Equations i n Two Variables; Intercepts; Symmetry PREPARING FOR THIS SECTION •


Solving Equations (Section 1.1, pp. 86-93) Now Work the 'Are You Prepared' problems OBJECTIVES

on

page 1 7 1 .

1 Gra p h Eq uations by Plotting Points (p. 163)

2 Find I ntercepts from a G raph (p. 165) 3

1

Find I ntercepts from an Eq uation (p. 166)

4

Test an Eq uation for Symmetry with Respect to the x-Axis, the y-Axis, a n d the Origin (p. 167)

5

Know How to Gra ph Key Equations (p. 169)

Graph Equations by Plotting Points

An equation in two variables, say x and y, is a statement in which two expressions involving x and y are equal. The expressions are called the sides of the equation. Since an equation is a statement, it may be true or false, depending on the value of the variables. Any values of x and y that result in a true statement are said to satisfy the equation. For example, the following are all equations in two variables x and y: 2x - y = 6

The first of these, x2 + l = 5, is satisfied for x = 1, y = 2, since 1 2 + 22 = 1 + 4 = 5 . Other choices of x and y, such as x = - 1 , y = -2, also satisfy this equa tion. It is not satisfied for x = 2 and y = 3, since 22 + 32 = 4 + 9 = 13 "* 5. The graph o f a n equation i n two variables x and y consists of the set of points in the xy-plane whose coordinates ( x, y) satisfy the equation. Graphs play an important role in helping us visualize the relationships that exist between two variable quantities. Table 1 shows the average price of regular y

=

2x + 5

1 64

CHAPTER 2

Graphs

Table 1 Adjusted average price (in dollars) of gasoline in Cal ifornia , 1 978-2005.

Price

Year

1 979

1 .9829

1 993

1 981

2.4977

Year 1 978

1 980 1 982 1 983

1 984

1 .5651

2.4929

1 994

2 . 1 795

1 996

1 .831 0

1 998

1 .8782

1 985

1 .7540

1 987

1 .3274

1 986 1 988 1 989 1 990 1 991

1 992

1 995 1 997

1 999

1 .3459

2000

1 .31 1 1

2002

1 .4656

2004

1 .3589

1 .4973

2001

2003 2005

Figure 1 1

Price

1 .3969

1 .5 1 89

�

3

.

Cal ifornia gasoline prices based on 2005 dollars, adj usted for inflation, 1 978-2005

0

1 .5 ... .

� 2 .5

�m 1 . 0

1 .4709

,

- ··"·"············································.......................................

....

� 2.0

; -;;;- 5 .. .. .. .. . . . .. ! . .g O. LJ. -'..L..J L...L -'-'--' L...L -'-:':--' L...L � � OO O N � � OO O N � � OO O N v ID �

1 .4669 1 .5397

0

1 .5329

I

· ·

f

··································· ················...................................................... · · ····· ···················· ···················· ······

. ..

.... . ...... .. .... .......... ... ...........

0

r- cc CO CO C:O OO (J") Q") (J") (J") Q) o o o o C) O') O') O') O') Q) m m m O") m o o o o 'T'"""" T""" ,.- .,.- r- r- r- .,-- 'T'"""" T""" T"'"" N N N N

1 .3246

Year Source: Statistical Abstract of the

1 .5270

1 .8249

United States.

1 .7536

1 .5955 1 .8950

2.1 521

2.4730

gasoline in California based on 2005 dollars (adjusted for inflation) for the years 1978-2005. In Figure 1 1 , we graph the data from the table, using the year along the x-axis and the price along the y-axis. From the graph, we can see the price was high est in 1980, 1981, and 2005 at about $2.50 per gallon. EXAM P L E 1

Determi n i n g Whether a Point Is on the G raph of an E quation

Determine if the following points are on the graph of the equation 2x - y (b) (2, -2) (a) (2, 3 ) Solution

=

6.

(a) For the point (2, 3), we check to see if x = 2, y = 3 satisfies the equation 2x - y = 6. 2x - y 2(2) - 3 = 4 - 3 = 1 *- 6 The equation is not satisfied, so the point (2, 3) is not on the graph of 2x - y 6. (b) For the point ( 2 , -2), we have 2x - y = 2(2) - ( -2) = 4 + 2 = 6 The equation is satisfied, so the point (2, -2) is on the graph of 2x - y = 6 . =

=

�,

E XA M P L E 2

'.>-

•


G raphi n g an E quation by P l otti ng Poi nts

Graph the equation: y = 2x + 5 Solution

Figure 1 2

We want to find all points (x, y) that satisfy the equation. To locate some of these points (and get an idea of the pattern of the graph), we assign some numbers to x and find corresponding values for y. If

x = o x = 1

x = -5

25 x

x = 10

Then

y = 2 ( 0) + 5

Point on Graph = 5

Y = 2( 1 ) + 5 = 7

Y = 2( -5 ) +

5 = -5

Y = 2 ( 1 0) + 5 = 25

( 0, 5 ) ( 1 , 7)

( - 5, - 5 ) ( 1 0, 25 )

By plotting these points and then connecting them, we obtain the graph o f the equa tion (a line), as shown in Figure 12.

•

SECTION 2.2

E XA M P L E 3

Graphs of Equations in Two Variables; I ntercepts; Symmetry

1 65

G rap h i n g an E q u ation by P l otti n g P o i nts

Graph the equation: y

x2

=

Table 2 provides several points on the graph. In Figure 13 we plot these points and connect them with a smooth curve to obtain the graph (a parabola).

Solution

Table 2

x

Y = J?

(x, y)

-4

16

( - 4, 1 6)

-3

9

(-3, 9)

-2

4

( - 2, 4)

a

a

(0, 0)

2

4

(2, 4)

3

9

(3, 9)

4

16

(4, 1 6 )

-1

Figure 1 3

(- 1 , 1 )

(1 , 1 )

•

The graphs of the equations shown in Figures 12 and 13 do not show all points. For example, in Figure 12, the point (20, 45 ) is a part of the graph of y 2x + 5, but it is not shown. Since the graph of y = 2x + 5 could be extended out as far as we please, we use arrows to indicate that the pattern shown continues. It is impor tant when illustrating a graph to present enough of the graph so that any viewer of the illustration will "see" the rest of it as an obvious continuation of what is actual ly there. This is referred to as a complete graph. One way to obtain a complete graph of an equation is to plot a sufficient num ber of points on the graph until a pattern becomes evident. Then these points are connected with a smooth curve following the suggested pattern. But how many points are sufficient? Sometimes knowledge about the equation tells us. For exam ple, we will learn in the next section that, if an equation is of the form y = mx + b, then its graph is a line. In this case, only two points are needed to obtain the graph. One purpose of this book is to investigate the properties of equations in order to decide whether a graph is complete. Sometimes we shall graph equations by plot ting points. Shortly, we shall investigate various techniques that will enable us to graph an equation without plotting so many points. =

CO M M ENT Another way to obtain the graph of an equation

is to

use a graphing util ity. Read Sec _

tion 2, Using a Graphing Utility to Graph Equations, i n the Appendix.

Figure 1 4 G raph crosses y-axis

Two techniques that sometimes reduce the number of points required to graph an equation involve finding intercepts and checking for symmetry.

y

�

G raph

2 x

G raph touches x-axis

E XA M P L E 4

Find Intercepts from a Graph

The points, if any, at which a graph crosses or touches the coordinate axes are called the intercepts. See Figure 14. The x-coordinate of a point at which the graph cross es or touches the x-axis is an x-intercept, and the y-coordinate of a point at which the graph crosses or touches the y-axis is a y-intercept. For a graph to be complete, all its intercepts must be displayed. F i n d i n g I ntercepts from a G raph

Find the intercepts of the graph in Figure 15 shown on p. x-intercepts? What are its y-intercepts?

166.

What are its

1 66

CHAPTER 2

Figure 1 5

y 4

Graphs

The intercepts of the graph are the points

Sol ution

( -3, 0), (0, 3), (0, 3)

(0, -3.5 ) , (4.5, 0)

3 4 . . The x-mtercepts are -3, 2 ' and 4.5; the y-mtercepts are -3.5, - "3 ' and 3.

•

In Example 4, you should notice the following usage: If we do not specify the type of intercept (x- versus y-), then we report the intercept as an ordered pair. However, if we specify the type of intercept, then we only report the coordinate of the intercept. For x-intercepts, we report the x-coordinate of the intercept; for y-intercepts, we report the y-coordinate of the intercept.

( 0 , - 3.5)

� ... ....". -

3

� �

(%, O} (o, -�}

COM M ENT For many equations, finding

Find Intercepts from an Equation

The intercepts of a graph can be found from its equation by using the fact that points on the x-axis have y-coordinates equal to 0 and points on the y-axis have x-coordinates equal to

O.

intercepts m a y not be so easy. In such

Procedure for Finding Intercepts 1.

cases, a graphing utility can be used. Read the first part of Section 3, Using

2.

a Graphing Utility to Locate Intercepts and Check for Symmetry, in the Appen dix, to find out how a graph ing utility _

locates intercepts.

EXAM P L E 5

Now Work P R O B L E M 3 9 ( a )

To find the x-intercept(s), if any, of the graph of an equation, let y = 0 in the equation and solve for x, where x is a real number. To find the y-intercept(s), if any, of the graph of an equation, let x = 0 in the equation and solve for y, where y is a real number.

F i n d i n g I ntercepts from an E qu ation

Find the x-intercept(s) and the y-intercept(s) of the graph of y y = x2 - 4 by plotting points. Solution

To find the x-intercept(s), we let y

=

=

x2 - 4. Then graph

0 and obtain the equation

x2 - 4 = 0 y = x2 - 4 with Y = 0 (x + 2)(x - 2) 0 Factor. x+2=0 x - 2 = 0 Zero-Product Property or x -2 or Solve. X = 2 The equation has two solutions, -2 and 2. The x-intercepts are -2 and 2. To find the y-intercept(s), we let x = 0 in the equation. y = x2 - 4 = 02 - 4 = -4 The y-intercept is -4. Since x2 2:: 0 for all x, we deduce from the equation y x2 - 4 that y 2:: -4 for all x. This information, the intercepts, and the points from Table 3 enable us to graph y = x2 - 4. See Figure 16. =

=

=

Ta ble 3

x

y = x2 - 4

(x, y)

-3

5

( - 3, 5)

-1

-3

(- 1 , - 3)

-3

( 1 , - 3)

3

5

(3, 5)

Figure 1 6

5 x

• &:m!J:= =--

Now Work

PROBLEM 2 1

SECTION 2.2

4


167

Test an Equation for Symmetry with Respect to the x-Axis, the y-Axis, and the Origin

We just saw the role that intercepts play in obtaining key points on the graph of an equation. Another helpful tool for graphing equations involves symmetry, particu larly symmetry with respect to the x-axis, the y-axis, and the origin. DEFINITION

A graph is said to be symmetric with respect to the x-axis if, for every point ( x, y) on the graph, the point (x, - y) is also on the graph.

-.J

Figure 17 illustrates the definition. When a graph is symmetric with respect to the x-axis, notice that the part of the graph above the x-axis is a reflection or mirror image of the part below it, and vice versa. Figure 1 7 Symmetry with respect to the x-axis

P o i nts Symm etric with Respect to the x-Axis

EXAM P L E 6

If a graph is symmetric with respect to the x-axis and the point (3, 2) is on the graph, then the point (3, 2 ) is also on the graph. -

DEFINITION

Figure 1 8

A graph is said to be symmetric with respect to the y-axis if, for every point ( x, y) on the graph, the point ( -x, y) is also on the graph.

Symmetry with respect to the y-axis y

(-x, y)

( x, y)

x

E XA M P L E 7

•

-.J

Figure 18 illustrates the definition. When a graph is symmetric with respect to the y-axis, notice that the part of the graph to the right of the y-axis is a reflection of the part to the left of it, and vice versa.

P o i nts Sym m etric with Respect to the y-Axi s

If a graph is symmetric with respect to the y-axis and the point (5, 8) is on the graph, then the point ( -5, 8) is also on the graph. DEFINITION

•

A graph is said to be symmetric with respect to the origin if, for every point (x, y) on the graph, the point ( -x, - y) is also on the graph.

Figure 1 9 Sym metry with respect to the origin y ( x, y)

Figure 19 illustrates the definition. Notice that symmetry with respect to the ori gin may be viewed in three ways : 1.

x ( - x, -y)

-.J

2.

3.

As a reflection about the y-axis, followed by a reflection about the x-axis As a projection along a line through the origin so that the distances from the origin are equal As half of a complete revolution about the origin

1 68

CHAPTER 2

Graphs

EXAM P L E 8

Poi nts Sym m etric with Respect to the Origin

If a graph is symmetric with respect to the origin and the point (4, 2) is on the graph, then the point ( - 4, -2) is also on the graph. .... ,Ii .

>-

•

Now Work P R O B L E M S 2 9 A N D 3 9 ( b )

When the graph of an equation is symmetric with respect to a coordinate axis or the origin, the number of points that you need to plot in order to see the pattern is reduced. For example, if the graph of an equation is symmetric with respect to the y-axis, then, once points to the right of the y-axis are plotted, an equal number of points on the graph can be obtained by reflecting them about the y-axis. Because of this, before we graph an equation, we first want to determine whether it has any symmetry. The following tests are used for this purpose. Tests for Symmetry To test the graph of an equation for symmetry with respect to the

x-Axis y-Axis Origin

E XA M P L E 9

Testin g an Equation for Symm etry

Test y Solution

Replace y by -y in the equation and simplify. If an equivalent equation results, the graph of the equation is symmetric with respect to the x-axis. Replace x by -x in the equation and simplify. If an equivalent equation results, the graph of the equation is symmetric with respect to the y-axis. Replace x by - x and y by -y in the equation and simplify. If an equivalent equation results, the graph of the equation is symmetric with respect to the origin.

=

4x2 --- for symmetry. x2 + 1

x-Axis: To test for symmetry with respect to the x-axis, replace y by -y. Since 4x2 4x2 -Y = - - is not equivalent to y = ---, the graph of the equation x2 + 1 x2 + 1 is not symmetric with respect to the x-axis. y-Axis: To test for symmetry with respect to the y-axis, replace x by -x. Since 4( -x f 4x2 4x2 = --- is equivalent to y = --- , the graph of the Y = 2 ( -x ) + 1 x 2 + 1 x2 + 1 equation is symmetric with respect to the y-axis. Origin: To test for symmetry with respect to the origin, replace x by -x and y by -y o 4( - X ) 2 -y = Replace x by -x and y by -yo ( -x)2 + 1 -

-Y =

4x2 x2 + 1

Sim pl ify.

4x2 M u ltiply both sides by 1 x2 + 1 Since the result is not equivalent to the original equation, the graph of the 4x2 equation y = --- is not symmetric with respect to the origin. x2 + 1 • y

=

-

-

.

SECTION 2.2


169

Seeing the Concept

4x2 Figure 20 shows the g raph of y = -2-- using a graphing utility. Do you see the symmetry with rex + 1 spect to the y-axis?

Figure 20

5

5

-5 Iiml::==:- '- Now Work P R O B l E M S 9 5

Know How to Graph Key Equations

The next three examples use intercepts, symmetry, and point plotting to obtain the graphs of key equations. It is important to know the graphs of these key equations because we use them later. The first of these is y = x3. EXAM P L E 1 0

G raphi n g the E q u ation y = x3 by F i n d i ng I ntercepts, C hecking for Symmetry, and P l otting Points

Graph the equation y = x3 by plotting points. Find any intercepts and check for symmetry first. Solution

First, we seek the intercepts. When x = 0, then y 0; and when y = 0, then x = O. The origin (0, 0) is the only intercept. Now we test for symmetry. =

Replace y by -yo Since -y x3 is not equivalent to y x3, the graph is not symmetric with respect to the x-axis. Replace x by -x. Since y = ( - x)3 = - x3 is not equivalent to y x3, the graph is not symmetric with respect to the y-axis. Replace x by -x and y by -y. Since -y = ( - x ) 3 - x3 is equivalent to y = x3 (multiply both sides by - 1), the graph is symmetric with respect to the origin.

x-Axis:

=

y-Axis:

=

=

Origin:

=

To graph y = x3, we use the equation to obtain several points on the graph. Be cause of the symmetry, we only need to locate points on the graph for which x ;:::: O. See Table 4. Since (1, 1 ) is on the graph, and the graph is symmetric with respect to the origin, the point ( - 1 , -1) is also on the graph. Plot the points from Table 4. Figure 21 shows the graph.

Table 4

x

y=r

(x, y)

a

a

(0, 0)

1

(1, 1 )

2

8

(2, 8)

3

27

(3, 27)

Y

Figure 2 1

8

(2 , 8)

6

-6

(-2, -8)

-8

x

•

1 70

CHAPTER 2

Gra phs G raphing the E q u ation x = y

E XA M P L E 1 1

Graph the equation x =

y 6

Figure 22 x = y2

(9, 3)

-2 ( 1 , -1 )

Figure 23 Y = \IX

Y 6

�

r.�. il

(9, -3)

If we restrict y so that y � 0, the equation x = l, y � 0, may be written equiv alently as y = \IX. The portion of the graph of x = l in quadrant I is therefore the graph of y = \IX . See Figure 23. Y, = Vx a nd 6

-6

1 10

-3 -2 2 3 1

1

10

y = x 1

10

3 2 1

-2 -3 1

10

(x, y)

1 , 10 10

( ) G, 3 ) G' 2 ) ( 2,D ( 3,�) ( �)

(1 , 1 )

1 0,

1

=

.

Vx

.

x

=

y2 on a g ra ph i n g calculator, you wi l l need to graph

We discuss why in Chapter 3. See Fig u re 24.

�

-�

10

\

G raph ing the E q u ation y

Graph the equation y

Solution

Y2

-----

---

-2

Table 5

-

•

CO M M ENr To see the graph of the equation

two equations:

Figure 24

x

l. Find any intercepts and check for symmetry first.

The lone intercept is (0, 0). The graph is symmetric with respect to the x-axis. (Do you see why? Replace y by - y.) Figure 22 shows the graph.

Solution

E XA M P L E 1 2

2

=

•

x 1

1.-. Find any intercepts and check for symmetry first. x

We check for intercepts first. If we let x = 0, we obtain ° in the denominator, which makes y undefined. We conclude that there is no y-intercept. If we let y = 0, we get the equation

1.-x = 0, which has no solution. We conclude that there is no

x-intercept. The graph of y =

1.-x does not cross or touch the coordinate axes.

Next we check for symmetry:

- -x1 ,

' · y b y - y YIelds y x-Axis: Rep I acmg

=

W

' h IS h IC " not eqUivalent to y =

-x1 .

1 1 y-Axis: Replacing x by -x yields y = - = - - , which is not equivalent to x -x 1 y = �. Origin: Replacing x by -x and y by -y yields - y _ 1.-, which is equivalent to x 1 y = . The graph is symmetric with respect to the origin. x

-

=

Now, we set up Table 5, listing several points on the graph. Because of the sym metry with respect to the origin, we use only positive values of x. From Table 5 we

SECTION 2.2

infer that if x is a large and positive number, then y

Figure 25 y

3

(-}, 2 )

(-2 -.1.) 2 (-1 . - 1 ) •

(-t. -2)

=

1 71

1.x is a positive number close

to O. We also infer that if x is a positive number close to 0, then

y =

1. is a large x

and positive number. Armed with this information, we can graph the equation. Figure 25 illustrates some of these points and the graph of y

x

-3 3

f f f f

Graphs of Equations in Two Variables; I ntercepts; Symmetry

=

1.. Observe how x

the absence of intercepts and the existence of symmetry with respect to the origin were utilized.

•

COM M ENT Refer to Example 3 in the Appendix, Section 3, for the graph of y =

-3

graphing utility.

1

x

using a •

2.2 Assess Your Understanding

'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red. 1. Solve the equation 2(x + 3) 2. Solve the equation x2 - 9 0. (p. 86-93) 1 = -7. (pp. 86-93)

-

=

Concepts and Vocabulary 3. The points, if any, at which a graph crosses or touches the coordinate axes are called

7. If the graph of an equation is symmetric with respect to the origin and (3, -4) is a point on the graph, then is also a point on the graph.

_ _ _ _

_ _ _ _

4. The x-intercepts of the graph of an equation are those x-values for which

8.

_ _ _ _ _

5. If for every point (x, y ) on the graph of an equation the point ( - x, y) is also on the graph, then the graph is symmetric with respect to the

9.

_ _ _ _ _

6. If the graph of an equation is symmetric with respect to the yaxis and -4 is an x-intercept of this graph, then is also an x-intercept.

True or False

equation, let x

To find the y-intercepts of the graph of an and solve for y.

= °

True or False The y-coordinate of a point at which the graph crosses or touches the x-axis is an x-intercept.

If a graph is symmetric with respect to the x-axis, then it cannot be symmetric with respect to the y-axis.

10. True or False

_ _ _ _

Skill Building In Problems ll-I6, determine which of the given points are on the graph of the equation. 12. Equation: y x3 - 2 Vx 11. Equation: y = X4 Vx Points: (0, 0) ; ( 1 , 1 ); ( - 1 , 0) Points: (0, 0); ( 1 , 1 ); ( 1, - 1 )

-

14.

=

=

15. Equation: x2 + / 4 Points: (0, 2 ) ; ( -2, 2); (V2, V2)

Equation: i = x + 1 Points: ( 1 , 2 ); (0, 1 ); ( - 1 , 0)

13. Equation: / = x2 + 9 Points: (0, 3); ( 3 , 0 ) ; ( - 3 , 0 ) 16. Equation: x2 + 4/

=

4

(D

Points: (0, 1 ) ; (2, 0); 2,

In Problems 17-28, find the intercepts and graph each equation by plot/ing points. Be sure to label the intercepts. 20. Y 3x - 9 19. y = 2 x + 8 17. y = x + 2 18. y = x - 6 21. Y =

22. y

x2 - 1

25. 2x + 3y

=

6

=

x2 - 9

26. 5 x + 2y

=

23. y

=

-x2 + 4

27. 9x2 + 4y = 36

10

=

24.

Y =

-x2 + 1

28. 4x2 + y

=

4

In Problems 29-38, plot each point. Then plot the point that is symmetric to it with respect to (a) the x-axis; (b) the y-axis; (c) the origin. 33. (5, -2) 32. (4, -2 ) 29. (3, 4) 30. (5, 3 ) 31. ( -2, 1 ) 34. (-1, - 1 )

35. ( -3, -4)

36. (4, 0)

37. (0, -3)

38. ( -3, 0)

1 72

CHAPTER 2

Graphs

In Problems 39-50, the graph of an equation is given. (a) Find the intercepts. (b) Indicate whether the graph is symmetric with respect to the x-axis, the y-axis, or the origin.

40.

39.

41.

-3

43.

Y+

44.

y

3

�

-4 45.

3

3x

-3 1

'

-6

3 x

-2

48.

8

-2 3

/

8

49.

1 /,

50.

\I�

-2

y-axis

52.

-8

x-axis

-8

53.

y

y 4

y

(0, 0\

-5

59. x

2

x +

4

+ y - 9

=0

63. Y

= x3 - 27

67. Y

=

3x 2

x

+

9

56.

/=

x + 9

60.

x2 -

Y

64. Y =

68. y

X4

-

4

- 1

57. y

=

0

2 x - 4 = 2:;:-

In Problems 71-74, dra w a quick sketch of each equation. 71. y = x

3

72. x = /

75.

If (3, b ) is a point on the graph of y =

4x +

77.

If ( a, 4) is a point on the graph of y =

x

2

+

54. y -a xis

('IT,

0)

( 2 , 2)

x

3X

-3

-2

= -0X 2

+ 4/

65. Y

=

x2

69. y

=

73. Y

= Vx

61. 9x

-

58. y

= 36

x

_

62.

= -0'X

4x2

66. y

9

70. Y =

74. Y

/=4

+

3x - 4

-x3 2

14 '�-

/

-4

In Problems 55-70, list the intercepts and test for symmetry.

/=

+ -4

Origin

-5

55.

x

4

-4

2

In Problems 51-54, draw a complete graph so that it has the type of symmetry indicated. 51.

6

-3

W \\

�

46.

-3

3'

6

,] 47.

3 x

-3

3x

4-

-3

42.

y

= x2

+ 4

X4

+ 1

5 2x

=�

1 , what is b?

76.

If ( -2, b) is a point on the graph of 2x

3 x , what is a?

78.

If (a, -5) is a point on the graph of y

1

+

=

3 y = 2, what is b ? x2

+ 6x,

what is a?

SECTION 2.3

Lines

1 73


80.

81.

82. 83.

Given that the point ( 1 , 2) is on the graph of an equation that is symmetric with respect to the origin, what other point is on the graph? If the graph of an equation is symmetric with respect to the y-axis and 6 is an x-intercept of this graph, name another x-intercept. If the graph of an equation is symmetric with respect to the origin and -4 is an x-intercept of this graph, name another x-intercept. If the graph of an equation is symmetric with respect to the x-axis and 2 is a y-intercept, name another y-intercept. Micro phones In studios and on stages, cardioid micro phones are often preferred for the richness they add to voices and for their ability to reduce the level of sound from the sides and rear of the microphone. Suppose one such cardiod pattern is given by the equation (x2 + l - xf = x2 + i. (a) Find the intercepts of the graph of the equation. (b) Test for symmetry with respect to the x-axis, y-axis, and origin.

84.

Solar Energy The solar electric generating systems at Kramer Junction, California, use parabolic troughs to heat a heat-transfer fluid to a high temperature. This fluid is used to generate steam that drives a power conversion system to produce electricity. For troughs 7.5 feet wide, an equation for the cross-section is 161 120x - 225. =

(a) Find the intercepts of the graph of the equation. (b) Test for symmetry with respect to the x-axis, y-axis, and origin. Source: u.s. D epartment of Energy

Source: www.notaviva.com

Discussion and Writing lxi, and y = ( Vx )2, not y = 0 , y = x, y ing which graphs are the same. (b) Explain why the graphs of y 0 and y = Ixl are the same. (c) Explain why the graphs of y x and y = ( Vx)2 are not the same. (d) Explain why the graphs of y 0 and y x are not the same. Explain what is meant by a complete graph. Draw a graph of an equation that contains two x-intercepts; at one the graph crosses the x-axis, and at the other the graph touches the x-axis.

In Problem 85, y o u m ay u s e a graphing utility, but i t is n o t required.

85. (a) G raph

=

88.

=

89.

=

=

86. 87.

=

90.

Make up an equation with the intercepts (2, 0), (4, 0), and (0, 1 ). Compare your equation with a friend's equation. Com ment on any similarities. Draw a graph that contains the points ( -2, - 1 ) , (0, 1 ) , ( 1 , 3 ) , and (3, 5 ) . Compare your graph with those of other students. Are most of the graphs almost straight lines? How many are "curved"? D iscuss the various ways that these points might be connected. An equation is being tested for symmetry with respect to the x-axis, the y-axis, and the origin. Explain why, if two of these symmetries are present, the remaining one must also be present.

'Are You Prepared ?' Answers 1.

{ -6 }

2.3

2.

{-3,3}

Lines OBJECTIVES

1

Calcu late a n d I nterpret the S lope of a Li n e (p. 1 74)

2

G raph Lines Given a Poi nt a n d the Slope (p. 1 76)

4

Use the Point-Slope Form of a Line; Identify Horizontal Li n es (p. 1 78)

6

Write the Equation of a Line in Slope-I ntercept Form (p. 1 79)

3

5

7

8 9

Find the Equation of a Vertical Line (p. 1 7 7) Find the Equation of a Line Given Two Poi nts (p. 179)

I dentify the Slope a n d y-Intercept of a Line from Its Eq uation (p. 180) Graph Li n es Written in General Form Using I ntercepts (p. 1 8 1) Find Equations of Para l lel Lines (p. 1 82)

1 0 Find Equations of Perpendicular Li n es (p. 1 83)

1 74

CHAPTER 2

Graphs

In this section we study a certain type of equation that contains two variables, called a linear equation, and its graph, a line. 1

Calculate and Interpret the Slope of a Line

Consider the staircase illustrated in Figure 26. Each step contains exactly the same horizontal run and the same vertical rise. The ratio of the rise to the run, called the slope, is a numerical measure of the steepness of the staircase. For example, if the run is increased and the rise remains the same, the staircase becomes less steep. If the run is kept the same, but the rise is increased, the staircase becomes more steep. This important characteristic of a line is best defined using rectangular coordinates.

Figure 26

DEFINITION

Let

P

slope

=

m

formula

(Xl , Y l ) and Q = (X2 ' Y2 ) be two distinct points. If Xl "* X2 , the of the nonvertical line L containing P and Q is defined by the ( 1)

Y2 - Y l X2 - X l If X l = X2 , L is a vertical results in division by 0).

line

and the slope

In

of L is

undefined

..J

(since this

Figure 27(a) provides an illustration of the slope of a nonvertical line; Figure 27(b) illustrates a vertical line.

L

Figure 27

(a) Slope of L is m

=

y _) -y 2

x 2

(b)

1

Slope is u ndefined;

L is vertical

As Figure 27(a) illustrates, the slope m of a nonvertical line may be viewed as n1 =

Y2 - Yl

Rise Run

Two comments about computing the slope of a nonvertical line may prove helpful: 1.

Any two distinct points on the line can be used to compute the slope of the line. (See Figure 28 for justification.)

Triangles ABC and PQR are similar (equ a l ang les), so ratios of correspond ing sides are proportional. Then

Figure 28

.

y

Y2 - Yl

Slope uSing P and Q = -- =

d(B, C) d(A, C)

-

=

X2 - Xl

Slope using A and B

x

SECTION 2.3

2.

Lines

1 75

The slope of a line may be computed from P = (X l , Y1 ) to Q = (X2 , Yz) or from Q to P because Y2 - YI x2 - X I Finally, we can also express the slope 111 of a nonvertical line as 111 =

Yz - Y1

Change in Y Change in X

�Y �X

That is, the slope 111 of a non vertical line L measures the amount that Y changes as �Y X changes from Xl to X2 ' The expression is called the average rate of change �x of Y with respect to x. Since any two distinct points can be used to compute the slope of a line, the average rate of change of a line is always the same number. EXAM P L E 1

F i n d i n g and I nterpreting the Slope of a L i n e G iven Two Points

The slope 111 of the line containing the points ( 1 , 2) and (5, - 3) may be computed as In

=

2 - ( -3 ) -3 - 2 -5 5 = - = - - or as n1. = 5 - 1 4 4 1 -5

5 -4

5 4

For every 4-unit change in x, Y will change by -5 UIilts. That is, if X increases by 4 units, then y will decrease by 5 units. The average rate of change of y with respect 5 . to X IS - - . 4 • Wi

*--

Now Work

P R O B L E M S

1 1 AND 1 7

To get a better idea of the meaning of the slope following example. E XA M P L E 2

111

of a line L, consider the

F i n d ing the S lopes of Various Lines Contai n ing the Same Point (2, 3)

Compute the slopes of the lines L1 , L2 , L3 , and L4 containing the following pairs of points. Graph all four lines on the same set of coordinate axes. L1 : L2 : L3: L4: Solution

Figure 29

P = (2, 3)

P = (2, 3)

P = (2, 3) P = (2, 3)

Ql Q2 Q3 Q4

= ( - 1, -2) = (3, - 1 )

= (5 3 ) = (2, 5 ) ,

Let 111 1 , 1112 , 1113 , and 1114 denote the slopes of the lines L I , L� , L3 , and L.. , respectively. Then -2 - 3 -5 5 A rise of 5 divided by a - 1 - 2 -3 3 - 1 3 -4 In" = = - = -4 3-2 1 3 -3 0 In, = =-=0 J 5 -2 3 111-+ is undefined because XI = X2 = 2 n1' 1 =

run of 3

-

---

--

m

1

=

� 3

m4

u n defined

m2 = -4

The graphs of these lines are given in Figure 29.

•

1 76

CHAPTER 2

Graphs

Figure 29 illustrates the following facts: 1.

2.

3. 4.

Figure 30

2

j

'�II Y6

= Ys

6x =

.f "

-3

. -. t-

=

x

..... Y3

........

�::::-:: �--:;.-:::O;�

.-- ....-:'--;j ... ./ /j ....

.:.

2x

Y4

./" '"

�

-2

"

./

3 Y1

� �

Seeing the Concept

=

Y1 = Y2

x

0

Y3

\2

4

1 = -x

1 = -x 2

Y4 = x See Figure 30.

�'. �

0

On the same screen, graph the following equations:

= x

-6x -2x -x \

= = =

Y6 Y4

=

I

Figure 31 Ys

Y2

When the slope of a line is positive, the line slants upward from left to right (Ll) ' When the slope of a line is negative, the line slants downward from left to right ( L2 ) ' When the slope is 0 , the line is horizontal ( L3)' When the slope is undefined, the line is vertical (L4) '

Ys

= 2x

Y6

= 6x

Slope of l ine is

O.

4 1

S lope of l ine is - .

Slope of l ine is

1

"2 '

Slope of l i n e is 1 .

Slope of line is 2. Slope of line is 6.

Seeing the Concept 0

On the same screen, graph the following equations: Y1

=

4 1 1

Slope of line is

O.

4 1

Y2

= - -x

S lope of line is - - .

Y3

= - -x 2

Slope of line is

- "2 '

Y4

= -x

Slope of l i ne is

-1 .

Ys

=

- 2x

Slope of l i ne is - 2 ,

Y6

= - 6x

Slope of l i ne is - 6,

See Figure 3 1 .

1

Figures 30 and 31 illustrate that the closer the line is to the vertical position, the greater the magnitude of the slope. 2

Graph Lines Given a Point and the Slope

The next example illustrates how the slope of a line can be used to graph the line. EXAM P L E 3

G raphing a Line Give n a Point and a Slope

Draw a graph of the line that contains the point (3, 2) and has a slope of: (a) Solution

3 4

4 (b) - 5

-

Rise . 3 . . The fact that the slope 1S - means that for every honzontal Run 4 movement (run) of 4 units to the right there will be a vertical movement (rise) of 3 units. Look at Figure 32. If we start at the given point (3, 2 ) and move 4 units to the right and 3 units up, we reach the point (7, 5 ) . By drawing the line through this point and the point (3, 2), we have the graph.

(a) Slope

= -

SECTION 2.3

y 6

Figure 32

Lines

1 77

(b) The fact that the slope is

(3 , 2��= Run

5

=

4

4 5

10

-4 5

Rise Run

means that for every horizontal movement of 5 units to the right there will be a corresponding vertical movement of -4 units (a downward movement). If we start at the given point (3, 2) and move 5 units to the right and then 4 units down, we arrive at the point ( 8, -2). By drawing the line through these points, we have the graph. See Figure 33. Alternatively, we can set

x

4 5

Figure 33

4 -5

Rise Run

so that for every horizontal movement of -5 units (a movement to the left) there will be a corresponding vertical movement of 4 units (upward). This ap proach brings us to the point ( -2, 6 ) , which is also on the graph shown in Figure 33. • ".".

3

.;>-


Find the Equation of a Vertical Line

Now that we have discussed the slope of a line, we are ready to derive equations of lines. As we shall see, there are several forms of the equation of a line. Let's start with an example. EXAM P L E 4

G raphing a Line

Graph the equation : x = 3 Solution

To graph x = 3, recall that we are looking for all points (x, y) in the plane for which x = 3. No matter what y-coordinate is used, the corresponding x-coordinate always equals 3. Consequently, the graph of the equation x = 3 is a vertical line with x-intercept 3 and undefined slope. See Figure 34. y 4

Figure 34

-1

-1

(3, 3) (3, 2 ) (3, 1 )

(3, 0) (3, -1 )

5

x

As suggested by Example 4, we have the following result: THEOREM

Equation of a Vertical Line

A vertical line is given by an equation of the form x=a where a is the x-intercept.

•

1 78

CHAPTER 2

Graphs

{ expression in x } . But x = 3 cannot be put in this form. To overcome this, most graph

COMM ENT To graph an equation using a g ra ph i n g util ity, we need to express the equation in the

form y =

ing utilities have special commands for drawing vertical l i nes. DRAW, LINE, PLOT, and VERT a re

a mong the more common ones. Consult your m a n u a l to determ ine the correct methodology for your •

g ra p h i n g util ity.

4

Figure 3S Y

L

( Xl , Yl ) See

Use the Point-Slope Form of a Line; I dentify Horizontal Lines

Now let L be a nonvertical line with slope m and containing the point Figure 35. For any other point ( X , y) on L, we have m=

Y - Yl

X - Xl

---

or Y - Y l = m(x -

Xl

'

)

x

THEOREM

Point-Slope Form of an Equation of a Line

An equation of a nonvertical line with slope m that contains the point ( X l , Y l ) is Y - Yl = m(x - Xl )

(2)

I�

� -----------------�

Using the Poi nt-Slope Form of a Line

EXAM P L E 5

An equation of the line with slope 4 and containing the point ( 1 , 2) can be found by using the point-slope form with m = 4, x l = 1, and Yl = 2.

Figure 36

( 1 , 2)

10

-2

Y - YI = m(x - X l ) Y - 2 = 4(x - 1 ) Y = 4x - 2 x

=

1, Y1 = 2

Solve for y.

See Figure 36 for the graph. I II ,'

EXAM P L E 6

m = 4, X1

•

;",... Now Work P R O B L E M 4 5

F i n d i n g the E q u ation of a Horizontal Line

Find an equation of the horizontal line containing the point (3, 2 ) . Solution

Figure 37 y 4 ( 3 , 2)

-1

3

5 x

Because all the y-values are equal on a horizontal line, the slope of a horizontal line is O. To get an equation, we use the point-slope form with m = 0, Xl = 3, and Yl = 2. Y - Yl = m(x - X l ) Y - 2 = O ' (X - 3 ) y-2=0 Y=2

See Figure 37 for the graph.

m = 0, X1 = 3, a n d Y1 = 2

•

SECTION 2.3

Lines

1 79

As suggested by Example 6, we have the following result: THEOREM

Equation of a Horizontal Line

A horizontal line is given by an equation of the form y=b where b is the y-intercept. 5

Find the Equation of a Line Given Two Poi n ts

We use the slope formula and the point-slope form of a line to find the equation of a line given two points. EXAM P L E 7

F i n d ing an Equation of a Line G iven Two Points

Find an equation of the line containing the points (2, 3 ) and ( -4, 5 ) . Graph the line. We first compute the slope of the line.

Solution

n1

Figure 38

=

5 -3 -4 - 2

-6

2

---

We use the point (2, 3) and the slope m = equation of the line.

1

1

3

_ l. to get the point-slope form of the 3

y - 3 = - - (x - 2) 3 See Figure 38 for the graph.

•

In the solution to Example 7, we could have used the other point, ( -4, 5), instead of the point (2, 3 ) . The equation that results, although it looks different, is equivalent to the equation that we obtained in the example. (Try it for yourself.) '1'= ;

6

'''''''


Write the Equation of a Line in Slope-Intercept Form

Another useful equation of a line is obtained when the slope m and y-intercept b are known. In this event, we know both the slope m of the line and a point (0, b) on the line; then we may use the point-slope form, equation (2), to obtain the follow ing equation: y - b = m(x - 0 ) o r y = mx + b THEOREM

Slope-Intercept Form of an Equation of a Line

An equation of a line with slope m and y-intercept b is y = mx + b

(3)

I�

�----------------------------------� = =t a,x 2:: a

The graph in Figure 54 illustrates the relationship between y and x if y varies directly with x and Ie > 0, X :2: O. Note that the constant of proportionality is, in fact, the slope of the line. If we know that two quantities vary directly, then knowing the value of each quantity in one instance enables us to write a formula that is true in all cases.

x

E XA M P LE

1

Solution

P

E iii' 600 a. >-

�

::2:

o

800

The monthly payment p on a mortgage varies directly with the amount borrowed B. If the monthly payment on a 30-year mortgage is $6.65 for every $1000 borrowed, find a formula that relates the monthly payment p to the amount borrowed B for a mortgage with these terms. Then find the monthly payment p when the amount borrowed B is $120,000. Because p varies directly with B, we know that

for some constant k. Because

Figure SS

c '"

Mortgage Payments

p =

6.65 (120 000,798)

Since p

=

kB, we have

In particular, when B

400

=

Ie

=

=

p

=

kB

6.65 when B

=

k( 1000) 0.00665 Solve for k. p

=

0.00665B

$120,000, we find that p = 0.00665( $120,000)

200

1000, it follows that

=

$798

Figure 55 illustrates the relationship between the monthly payment

-;;of--'-----;'4o;;-'--:8::'::o:--;-;12�0--;1-=-60::--..,B a mou n t borrowed B. Amount borrowed (OOO's) == 1l'I!liI: -

Now Work P R O B L E M S

3 AND 2 1

p

and the •

SECTION 2.5

Figure 56

y =

;k >

k

x

-

y

2 0, x

> °

DEFINITION

Variation

1 97

Construct a Model Using Inverse Variation

Let x and y denote two quantities. Then y varies inversely with x, or y is inversely proportional to x, if there is a nonzero constant k such that =

I

�

�-------- ---------y - - (- ----------------�

�

The graph in Figure 56 illustrates the relationship between y and x if y varies inversely with x and k > 0, x > O.

x

M aximum Weight That Can Be Supported by a Piece of Pine

EXA M P L E 2

See Figure 57. The maximum weight W that can be safely supported by a 2-inch by 4-inch piece of pine varies inversely with its length t. Experiments indicate that the maximum weight that a 10-foot-long 2-by-4 piece of pine can support is 500 pounds. Write a general formula relating the maximum weight W (in pounds) to length l (in feet). Find the maximum weight Wthat can be safely supported by a length of 25 feet.

Figure 57

Solution Because Wvaries inversely with I, we know that =

W

k 1

-

for some constant k. Because W = 500 when l 500

k S'mce W

/

=

= =

k 10 5000

k

5000 l

In particular, the maximum weight Wthat can be safely supported by a piece of pine 25 feet in length is

W

5000 W = 2s

600 500

=

200 pounds

Figure 58 illustrates the relationship between the weight Wand the length l.

400 300 200

li'l!IOl=� ....

Now Work P R O B L E M

100 o

1 0, we have

/' we have

W=

Figure 58

=

5

10

15

20

25

3

3 1

•

Construct a Model Using Joint Variation or Combined Variation

When a variable quantity Q is proportional to the product of two or more other variables, we say that Q varies jointly with these quantities. Finally, combinations of direct and/or inverse variation may occur. This is usually referred to as combined variation.

Let's look at an example.

1 98

CHAPTER 2

Graphs

EXA M P L E

3

Loss of Heat Through a Wall The loss of heat through a wall varies jointly with the area of the wall and the dif ference between the inside and outside temperatures and varies inversely with the thickness of the wall. Write an equation that relates these quantities.

Solution

We begin by assigning symbols to represent the quantities: L = Heat loss A = Area of wall

T = Temperature difference d = Thickness of wall

Then AT L = k d

where k is the constant of proportionality.

•

In direct or inverse variation, the quantities that vary may be raised to powers. For example, in the early seventeenth century, Johannes Kepler (1571-1630) discov ered that the square of the period of revolution T around the Sun varies directly with the cube of its mean distance a from the Sun. That is, T2 = ka3 , where k is the constant of proportionality.

EXAM P L E 4

Force of the Wind on a Window The force F of the wind on a flat surface positioned at a right angle to the direc tion of the wind varies jointly with the area A of the surface and the square of the speed v of the wind. A wind of 30 miles per hour blowing on a window measuring 4 feet by 5 feet has a force of 150 pounds. See Figure 59. What is the force on a window measuring 3 feet by 4 feet caused by a wind of 50 miles per hour?

Figure 59

Solution

Since F varies jointly with A and v2, we have

�\. �

where k is the constant of proportionality. We are told that F = 150 when A 4· 5 20 and v = 30. Then we have

� Wind �

=

=

150

=

F = kAv2, F

k(20 ) ( 900)

=

150, A

=

20, v = 30

1 k =120 Since F = kAv2; we have

F

=

1 - Av-? 1 20

For a wind of 50 miles per hour blowing on a window whose area is A = 3· 4 = 12 square feet, the force F is

F 1,1

,,_-

=

1 ( 12 ) ( 2500) = 250 pounds 120


•

SECTION 2.5

Variation

1 99


Concepts and Vocabulary 1. If x and y are two quantities, then y is directly proportional

to x if there is a nonzero number k such that

2.

___ _

True or False

k is a constant.

If y varies directly with x, then y

=

'5:.-, where x

Skill Building In Problems 3-14, write a general formula t o describe each variation. 3.

4. 5.

6. 7. 8. 9. 10.

1 1. 12. 13.

14.

y varies directly with x; y = 2 when x = 1 0 v varies directly with t; v = 16 when t = 2 A varies directly with x2 ; A = 47i when x = 2 V varies directly with x3; V 367i when x = 3 F varies inversely with d2 ; F = 1 0 when d = 5 y varies inversely with \IX; y = 4 when x = 9 z varies directly with the sum of the squares of x and y; z = 5 when x = 3 and y = 4 T varies jointly with the cube root of x and the square of d; T = 1 8 when x 8 and d 3 9 and d 4 M varies directly with the square of d and inversely with the square root of x; M = 24 when x z varies directly with the sum of the cube of x and the square of y; z = 1 when x = 2 and y 3 The square of T varies directly with the cube of a and inversely with the square of d; T 2 when a = 2 and d The cube of z varies directly with the sum of the squares of x and y; z = 2 when x = 9 and y = 4 =

=

=

=

=

=

=

=

4

Appl ications and Extensions In Problems

15-20,

write an equation that relates the quantities.

The volume V of a sphere varies directly with 47i . the cube of its radius r. The constant of proportionality is 3 Geometry The square of the length of the hypotenuse c of a right triangle varies jointly with the sum of the squares of the lengths of its legs a and b. The constant of proportional ity is 1 .

$ 10 00 borrowed, find a linear equation that relates the monthly payment p to the amount borrowed B for a mort gage with the same terms. Then find the monthly payment p when the amount borrowed B is $ 145,000.

15. Geometry

16.

22. Mortgage Payments The monthly payment p on a mortgage

varies directly with the amount borrowed B. lf the monthly payment on a IS-year mortgage is $8.99 for every $1000 bor rowed, find a linear equation that relates the monthly pay ment p to the amount borrowed B for a mortgage with the same terms. Then find the monthly payment p when the amount borrowed B is $1 75,000.

The area A of a triangle varies jointly with the lengths of the base b and the height h. The constant of 1 proportionality is "2 '

17. Geometry

The perimeter p of a rectangle varies jointly with the sum of the lengths of its sides I and w . The constant of proportionality is 2.

18.

Geometry

19.

Physics: Newton's Law The force F (in newtons) of attrac tion between two bodies varies jointly with their masses m and M (in kilograms) and inversely with the square of the distance d (in meters) between them. The constant of pro portionality is G = 6.67 X 10- 1 1 .

TI1e period of a pendulum is the time required for one oscillation; the pendulum is usually re ferred to as simple when the angle made to the vertical is less than 5°. The period T of a simple pendulum (in seconds) varies directly with the square root of its length l (in feet). . 27i . Th e constant 0f proportJOna I'Ity IS ;;::;: . v 32 Mortgage Payments The monthly payment p on a mort gage varies directly with the amount borrowed B. If the monthly payment on a 30-year mortgage is $6.49 for every

20. Physics: Simple Pendulum

_

21.

23.

Physics: Falling Objects The distance s that an object falls is directly proportional to the square of the time t of the fall. If an object faBs 16 feet in 1 second, how far will it faB in 3 sec onds? How long will it take an object to fall 64 feet?

The velocity v of a falling object is directly proportional to the time t of the fall. If, after 2 sec onds, the velocity of the object is 64 feet per second, what will its veloci ty be after 3 seconds?

24. Physics: Falling Objects

25.

Physics: Stretching a Spring The elongation E of a spring balance varies directly with the applied weight W (see the figure). If E = 3 when W = 20, find E when W = 15.

T E

1

200

CHAPTER 2

Graphs

26. Physics: Vibrating String The rate of vibration of a string

under constant tension varies inversely with the length of the string. If a string is 48 inches long and vibrates 256 times per second, what is the length of a string that vibrates 576 times per second?

constant of proportionality is 1T. See the figure. Write an equa tion for V.

T 1 h

At the corner Shell station, the rev enue R varies directly with the number g of gallons of gaso line sold. If the revenue is $47 AO when the number of gallons sold is 12, find a linear equation that relates revenue R to the number g of gallons of gasoline. Then find the revenue R when the number of gallons of gasoline sold is 10.5.

27. Revenue Equation

28. Cost Equation The cost C of chocolate-covered almonds

varies directly with the number A of pounds of almonds pur chased. If the cost is $23.75 when the number of pounds of chocolate-covered almonds purchased is 5, find a linear equa tion that relates the cost C to the number A of pounds of al monds purchased. Then find the cost C when the number of pounds of almonds purchased is 3.5. Suppose that the demand D for candy at the movie theater is inversely related to the price p . (a) When the price of candy is $2.75 per bag, the theater sells 156 bags of candy. Express the demand for candy in terms of its price. (b) Determine the number of bags of candy that will be sold if the price is raised to $3 a bag.

The volume V of a right circular cone varies jointly with the square of its radius r and its height h. The 1T constant of proportionality is -. See the figure. Write an 3 equation for V.

36. Geometry

I 1 h

29. Demand

The time t that it takes to get to school varies inversely with your average speed s. (a) Suppose that it takes you 40 minutes to get to school when your average speed is 30 miles per hour. Express the driving time to school in terms of average speed. (b) Suppose that your average speed to school is 40 miles per hour. How long will it take you to get to school?

30. Driving to School

The volume of a gas V held at a constant tem perature in a closed container varies inversely with its pres sure P. If the volume of a gas is 600 cubic centimeters (cm3) when the pressure is 1 50 millimeters of mercury (mm Hg), find the volume when the pressure is 200 mm Hg.

31. P ressure

The current i in a circuit is inversely propor tional to its resistance Z measured in ohms. Suppose that when the current in a circuit is 30 amperes the resistance is 8 ohms. Find the current in the same circuit when the resis tance is 10 ohms.

32. Resistance

33. Weight The weight of an object above the surface of Earth

varies inversely with the square of the distance from the cen ter of Earth. If Maria weighs 125 pounds when she is on the surface of Earth (3960 miles from the center) , determine Maria's weight if she is at the top of Mount McKinley (3.8 miles from the surface of Earth). 34. Intensity of Light The intensity 1 of light (measured in foot

candles) varies inversely with the square of the distance from the bulb. Suppose that the intensity of a 100-watt light bulb at a distance of 2 meters is 0.075 foot-candle. Determine the intensity of the bulb at a distance of 5 meters.

The volume V of a right circular cylinder varies jointly with the square of its radius r and its height h. The

35. Geometry

The weight of a body above the surface of Earth varies inversely with the square of the distance from the center of Earth. If a certain body weighs 55 pounds when it is 3960 miles from the center of Earth, how much will it weigh when it is 3965 miles from the center?

37. Weight of a Body

38. Force of the Wind on a Window The force exerted by the

wind on a plane surface varies jointly with the area of the surface and the square of the velocity of the wind. If the force on an area of 20 square feet is 1 1 pounds when the wind ve locity is 22 miles per hour, find the force on a surface area of 47. 125 square feet when the wind velocity is 36.5 miles per hour. The horsepower (hp) that a shaft can safely transmit varies jointly with its speed (in revolutions per minute, rpm) and the cube of its diameter. If a shaft of a certain material 2 inches in diameter can transmit 36 hp at 75 rpm, what diameter must the shaft have in order to trans mit 45 hp at 125 rpm?

39. Horsepower

40. Chemistry: Gas Laws The volume V of an ideal gas varies directly with the temperature T and inversely with the pres sure P . Write an equation relating V, T,and P using k as the

constant of proportionality. If a cylinder contains oxygen at a temperature of 300 K and a pressure of 1 5 atmospheres in a volume of 100 liters, what is the constant of proportion ality k? If a piston is lowered into the cylinder, decreasing the volume occupied by the gas to 80 liters and raising the temperature to 310 K, what is the gas pressure?

41. Physics: Kinetic Energy The kinetic energy K of a moving

object varies jointly with its mass m and the square of its ve locity v . If an object weighing 25 kilograms and moving with a velocity of 10 meters per second has a kinetic energy of 1250 joules, find its kinetic energy when the velocity is 15 me ters per second.

Chapter Review

The electrical resistance of a wire varies directly with the length of the wire and inversely with the square of the diameter of the wire. If a wire 432 feet long and 4 millimeters in diameter has a resistance of 1 .24 ohms, find the length of a wire of the same material whose resistance is 1 .44 ohms and whose diameter is 3 millimeters. 43. Measuring the Stress of Materials The stress in the mater ial of a pipe subject to internal pressure varies jointly with the internal pressure and the internal diameter of the pipe and inversely with the thickness of the pipe. The stress is 100 pounds per square inch when the diameter is 5 inches, 42. Electrical Resistance of a Wire

201

the thickness is 0.75 inch, and the internal pressure is 25 pounds per square inch. Find the stress when the internal pressure is 40 pounds per square inch if the diameter is 8 inch es and the thickness is 0.50 inch. 44. Safe Load for a Beam The maximum safe load for a hori zontal rectangular beam varies jointly with the width of the beam and the square of the thickness of the beam and in versely with its length. If an 8-foot beam will support up to 750 pounds when the beam is 4 inches wide and 2 inches thick, what is the maximum safe load in a similar beam 10 feet long, 6 inches wide, and 2 inches thick?

Discussion and Writing 45. In the early 17th century, Johannes Kepler discovered that

the square of the period T varies directly with the cube of its mean distance a from the Sun. Go to the library and research this law and Kepler's other two laws. Write a brief paper about these laws and Kepler's place in history.

46. Using a situation that has not been discussed in the text, write


a real-world problem that you think involves two variables that vary inversely. Exchange your problem with another stu dent's to solve and critique.


a real-world problem that you think involves three variables that vary jointly. Exchange your problem with another stu dent's to solve and critique.

a real-world problem that you think involves two variables that vary directly. Exchange your problem with another stu dent's to solve and critique.

CHAPTER REVIEW

Things to Know Formulas

Distance formula (p. 157) Midpoint formula (p. 160) Slope (p. 174) Parallel lines (p. 182) Perpendicular lines (p. 1 83) Direct variation (p. 196) Inverse variation (p. 197) Equations of Lines and Circles

Vertical line (p. 177) Horizontal line (p. 179)

Equal slopes (117. 1 = 117.2 ) and different y-intercepts (b 1 Product of slopes is - 1 ( m1 117.2 = -1 ) y = kx k y= x

=F

b2 )

•

x = a; a is the x-intercept y = b; b is the y-intercept

y - Y I = m(x - X l ) ; m is the slope of the line, (x] , Y I ) is a point on the line Slope-intercept form of the equation of a line (p. 179) Y = mx + b; m is the slope of the line, b is the y-intercept Ax + By = C; A, B not both 0 General form of the equation of a line (p. 181) ( x - hf + (y - k f = ,2; r is the radius of the circle, (h, k ) is the Standard form of the equation of a circle (p. 1 90) center of the circle Equation of the unit circle (p. 190) X2 + l = 1 Point-slope form of the equation of a line (p. 178)

General form of the equation of a circle (p. 192)

2 x

+

l+

ax

+ by +

c

= 0, with restrictions on a, b, and c

Objectives --------. Section

2.1

2

You should be able to . . .

Review Exercises

Use the distance formula (p. 157) Use the midpoint formula (p. 1 59)

1(a)-6(a), 48, 49(a), 50 1(b )-6(b), 50 (continued)

202

CHAPTER 2

Graphs

Section


Review Exercises

2.2

Graph equations by plotting points (p. 163) Find intercepts from a graph (p. 165) Find intercepts from an equation (p. 166) Test an equation for symmetry with respect to the x-axis, the y-axis, and the origin (p. 167) Know how to graph key equations (p. 169)

7 8 9-16 9-16 45-46

Calculate and interpret the slope of a line (p. 174) Graph lines given a point and the slope (p. 176) Find the equation of a vertical line (p. 177) Use the point-slope form of a line; identify horizontal lines (p. 178) Find the equation of a line given two points (p. 179) Write the equation of a line in slope-intercept form (p. 1 79) Identify the slope and y-intercept of a line from its equation (p. 180) Graph lines written in general form using intercepts (p. 181) Find equations of parallel lines (p. 182) Find equations of perpendicular lines (p. 183)

l(c)-6(c), I (d)-6(d), 49(b), 51 47 29 27, 28 30-32 27, 28, 30-36 37-40 41-44 33, 34 35, 36

2

Write the standard form of the equation of a circle (p. 189) Graph a circle (p. 191) Work with the general form of the equation of a circle (p. 192)

1 7-20 21-26 23-26

2

Construct a model using direct variation (p. 196) Construct a model using inverse variation (p. 197) Construct a model using joint or combined variation (p. 197)

52, 53, 55 54 55

2

3

4 5

2.3 2 3

4 5

6

7 8

9

10

2.4

3

2.5

3

Review Exercises In Problems 1-6, find the following for each pair of points: (a) The distance between the points (b) The midpoint of the line segment connecting the points (c) The slope of the line containing the points (d) Interpret the slope found in part (c) 1. (0, 0); (4, 2)

2. (0, 0) ; ( -4, 6)

3. ( 1 , - 1 ) ; ( -2, 3 )

4. ( -2, 2 ) ; ( 1 , 4)

5 . (4, -4); (4, 8)

6. (-3, 4); (2, 4)

In Problems 9. 2x

=

9-16,

3/

7. Graph y

=

x2 + 4 by plotting points.

8. List the intercepts of the given graph.

x

list the intercepts and test for symmetry with respect to the x-axis, the y-axis, and the origin. 10. y = 5x 11. x2 + 41 16 12. 9x2 - l 14. y

=

=

x3 -

15. x2 + x +

X

l + 2y

=

=

16. x2 + 4x +

0

9

l - 2y

=

0

In Problems 1 7-20, find the standard form of the equation of the circle whose center and radius are given. 1 7. ( h,

k)

=

In Problems 21. x2

+

24. x2 +

( -2, 3 ) ; r

21-26,

(y - If

=

18. ( h, k)

4

=

(3, 4) ; r

=

19. (h, k)

4

=

( -1 , -2 ) ; r

=

1

20. ( h, k )

=

(2, -4) ; r

find the center and radius of each circle. Graph each circle. Find the intercepts, if any, of each circle. 23. x2 + l - 2x + 4y - 4 22. (x + 2)2 + l 9 = 4

l + 4x - 4y - 1

=

=

0

25. 3x2 + 31 - 6x

+

12y

=

0

26. 2x2 + 21 - 4x

=

0

=

0

=

3

C h a pter Review

203

In Problems 27-36, find an equation of the line having the given characteristics. Express your answer using either the general form or the slope-intercept form of the equation of a line, whichever you prefa 27. Slope

=

-2; containing the point (3, - 1 )

28. Slope

containing the point ( -3, 4)

29. Vertical;

31. y-intercept

=

34. Parallel to the line x + y

=

=

0; containing the point ( -5, 4)

30. x-intercept

-2; containing the point (5, -3)

33. Parallel to the line 2x - 3y

=

=

2; containing the point (4, -5)

32. Containing the points (3, -4) and (2, 1 )

-4; containing the point ( -5, 3)

2; containing the point (1, -3)

35. Perpendicular to the line x + y

=

36. Perpendicular to the line 3x - y

2; containing the point (4, - 3)

=

-4; containing the point ( -2, 4)

In Problems 3 7-40, find the slope and y-intercept of each line. Graph the line, labeling any intercepts. 37. 4x

-

5y

=

38. 3x + 4y

-20

=

1 2

In Problems 41-44, find the intercepts and graph each line. 41. 2x - 3y = 12 42. x - 2y 8

43

=

45. Sketch a graph of y 46. Sketch a graph of y

=

=

•

=

1 6

3

40. - '4 x

- -

+

1

2' y

=

0

2

52. Mortgage Payments

Yx. 2

= (3, 4 ) , B = ( 1 , 1 ) , and C ( -2, 3) are the vertices of an isosceles triangle.

48. Show that the points A

= ( -2, 0 ) , B = ( -4, 4), and C (8, 5 ) are the vertices of a right triangle in two ways: (a) By using the converse of the Pythagorean Theorem (b) By using the slopes of the lines joining the vertices

49. Show that the points A

=

R

At the corner Esso station, the revenue varies directly with the number g of gallons of gasoline sold. If the revenue is $46.67 when the number of gaUons sold is 13, find an equation that relates revenue R to the number g of gallons of gasoline. Then find the revenue R when the number of gallons of gasoline sold is 1 1 .2.

53. Revenue Function

=

50. The endpoints of the diameter of a circle are ( -3, 2) and

54. Weight of a Body The weight of a body varies inversely with

(5, -6) . Find the center and radius of the circle. Write the standard equation of this circle. that the points A = (2, 5 ) , B (8, - 1 ) lie on a line by using slopes.

51. Show =

1 1 + -y 2" 3

-,

=

The monthly payment p on a mortgage varies directly with the amount borrowed B. If the monthly payment on a 30-year mortgage is $854.00 when $ 130,000 is borrowed, find an equation that relates the monthly payment p to the amount borrowed B for a mortgage with the same terms. Then find the monthly payment p when the amount borrowed B is $165,000.

x3.

47. Graph the line with slope '3 containing the point ( 1 , 2 ) .

C

1 3

39. -x - - y

12

=

(6, 1 ) ,

the square of its distance from the center of Earth. Assuming that the radius of Earth is 3960 miles, how much would a man weigh at an altitude of 1 mile above Earth's surface if he weighs 200 pounds on Earth's surface?

and

55. Kepler's Third Law of Planetary Motion Kepler's Third Law of Planetary Motion states that the square of the period of revolu tion T of a planet varies directly with the cube of its mean distance a from the Sun. If the mean distance of Earth from the Sun is

93 million miles, what is the mean distance of the planet Mercury from the Sun, given that Mercury has a "year" of 88 days?

.

/

_

Mercury T 88 da�s

_ - - /

=

,

"

---

_ _ - -

/

/

/

Earth days

T = 365

204

C H A PTER 2

Graphs

56. Create four problems that you might be asked to do given the two points ( -3, 4) and (6, 1 ) . Each problem should involve a dif

ferent concept. Be sure that your directions are clearly stated.

57. Describe each of the following graphs in the xy-plane. Give justification. (a) x = 0 (b) y = 0 (c) x + Y = 0 (d) xy = 0 (e) x2 +

l=0

58. Suppose that you have a rectangular field that requires watering. Your watering system consists of an arm of variable length that

rotates so that the watering pattern is a circle. Decide where to position the arm and what length it should be so that the entire field is watered most efficiently. When does it become desirable to use more than one arm? [Hint: Use a rectangular coordinate system positioned as shown in the figures. Write equations for the circle(s) swept out by the watering arm(s).] y

y

Rectangular field, one arm

Square field

Rectangular field, two arms

CHAPTER TEST In Problems

1 -3,

use PI = ( - 1 , 3 ) and P = (5, - 1 ).

1. Find the distance from PI to P .

2

2. Find the midpoint of the line segment joining p] and P .

2

3. (a) Find the slope of the line containing PI and P . 4.

(b) Interpret this slope. Graph y = x2 - 9 by plotting points.

2

2

5. Sketch the graph of l = x.

6. List the intercepts and test for symmetry: x2 + y = 9.

7. Write the slope-intercept form of the line with slope -2 con

taining the point (3, -4). Graph the line. 8. Write the general form of the circle with center (4, -3) and radius 5.

9. Find the center and radius of the circle x2 + l + 4x - 2y - 4 = O. Graph this circle.

10. For the line 2x + 3 y = 6, find a line parallel to it containing

the point ( 1 , - 1 ) . Also find a line perpendicular to it con taining the point (0, 3). 1 1 . Resistance due t o a Conductor The resistance (in ohms) of a circular conductor varies directly with the length of the conductor and inversely with the square of the radius of the conductor. If 50 feet of wire with a radius of 6 X 10-3 inch has a resistance of 10 ohms, what would be the resistance of 1 00 feet of the same wire if the radius is increased to 7 X 10-3 inch?

C U M U LATIVE REVI EW In Problems

1 -8,

find the real solution (s) of each equation. X - 12 = 0 x2 2x - 2 = 0 6. � = 3

1. 3x - 5 = 0

2. x2 -

3. 2x2 - 5x - 3 = 0 5. x2 + 2x + 5 = 0

4.

7. Ix - 21

=

1

8.

-

Yx2 + 4x = 2

In Problems 9 and 1 0, solve each equation in the complex number system. 9. x2 = - 9

In Problems

1 1-14,

10. x2 - 2x + 5 = 0

solve each inequality. Graph the solution set.

7 12. - 1 < x + 4 < 5 14. 12 + xl > 3 1 15. Find the distance between the points P = ( - 1 , 3 ) and Q = (4, -2). Find the midpoint of the line segment from P to Q.

11. 2x - 3

:s;

13. Ix - 21

:s;

16. Which of the following points are on the graph of

y = x3 - 3x + 1 ? (a) ( -2, - 1 )

(b) (2, 3 )

(c) (3, 1 )

17. Sketch the graph of y = x3. 18. Find the equation of the line containing the points ( - 1 , 4)

and (2, -2). Express your answer in slope-intercept form. 19. Find the equation of the line perpendicular to the

line

y = 2x + 1 and containing the point (3, 5 ) . Express your an swer in slope-intercept form and graph the line. 20. Graph the equation x 2 + l 4x + 8y - 5 O. -

=

Chapter Project

205

CHAPTER PROJ ECT 1. Treating year as the independent variable and the winning

2.

3.

4. Predicting Olympic Performance Measurements of human performance over time sometimes follow a strong linear re lationship for reasonably short periods. In 2004 the Summer Olympic Games returned to Greece, the home of both the ancient Olympics and the first modern Olympics. The fol lowing data represent the winning times (in hours) for men and women in the Olympic marathon. Year

Men

Women

1 984

2 .1 6

2.41

1 988

2. 1 8

2.43

1 992

2.22

2.54

1 996

2.2 1

2.43

2000

2.1 7

2.39

SOl/ree: www.hiekoksports. col11lhistorylo!mtan.dfshtml

5.

6.

value as the dependent variable, find linear equations re lating these variables (separately for men and women) us ing the data for the years 1992 and 1996. Compare the equations and comment on any similarities or differences. Interpret the slopes in your equations from part 1. Do the y-intercepts have a reasonable interpretation? Why or why not? Use your lines to predict the winning time in the 2004 Olympics. Compare your predictions to the actual results (2. 18 hours for men and 2.44 hours for women). How well did your equations do in predicting the winning times? Repeat parts 1 to 3 using the data for the years 1996 and 2000. How do your results compare? Would your equations be useful in predicting the winning marathon times in the 2104 Summer Olympics? Why or why not? Pick your favorite Winter Olympics event and find the winning value (that is distance, time, or the like) in two Winter Olympics prior to 2006. Repeat parts 1 to 3 using your selected event and years and compare to the actual results of the 2006 Winter Olympics in Torino, Italy.

Functions a nd Their Graphs

u.s. Consumers Make Wireless Top Choice

July 1 1 , 200S-WASHINGTON,

D.C.,

eTIA-The Wireless Association

President and CEO Steve Largent highlights the Federal Communi cations Commission (FCC) report on local telephone competition, which indicates for the first time there are more wireless subscribers in the United States than wireline service lines. According to the FCC report, as of December 3 1 , 2004 there were 181.1 million wireless subscribers in the U.S., compared to a combined 178 million incum bent local exchange carrier switched access lines and competitive local exchange carrier switched access lines. Also according to the report, there was a 1 5 % increase in wireless subscribers for the 12-month period ending December 31, 2004. "This significant milestone reflects consumers' approval of a vi brant, ultracompetitive industry that lets us communicate how we want, where we want and with whom we want," said Largent. "Wireless is so popular because it lets people communicate on their terms. And it's not j ust about talking anymore. You can browse the web, take pictures and video, download music, play games or conduct business. Wireless can satisfy a lot of different communication needs and desires and that's extremely popular to millions of consumers." Source: www. CTIA. org

- See the Chapter Project -

A Look Back

So far, we have developed tech n i q u es for g ra p h i ng equations containing two vari ables.

A Look Ahead

In this chapter, we look at a specia l type of eq uation i nvolving two va riables ca l led a

function. This chapter deals with what a fu nction is, how to g raph functions, proper

ties of functions, and how fu nctions a re used in appl ications. The word function a p pa rently was introduced by Rene Descartes in 1637. For h i m, a function s i m ply meant a ny positive integral power of a variable x . Gottfried Wilhelm Leibniz ( 1646- 1 7 1 6), who always emphasized the g eo metric side of mathematics, used the word function to denote a ny q u a ntity a ssociated with a cu rve, such as the coordi nates of a point on the cu rve. Leon h a rd Euler ( 1707-1 783) employed the word to mean any eq uation or form u l a i nvolving va riables and constants. His idea of a func tion is s i m i l a r to the one m ost often seen in cou rses that precede ca lculus. Later, the use of functions i n investigating heat flow equations led to a very broad defi n ition, due to Lejeune Dirichlet ( 1805-1859), which describes a function as a correspon dence between two sets. It is h i s defi n ition that we use here.

Outline 3.1

Functions

3.2 The Graph of a Function

3.3 Properties of Functions

3.4 Library of Functions; Piecewise-defined Functions

3 .5 Graphing Techniques: Transformations 3.6 Mathematical Models: Building Functions

Chapter Review Chapter Test Cumulative Review Chapter Projects

207

CHAPTER 3

208

Functions a n d Their Graphs

3 . 1 Functions PREPARING FOR THIS SECTION •

•


Intervals (Section 1 .5, pp. 125-126) Evaluating Algebraic Expressions, Domain of a Variable (Chapter R, Section R.2, pp. 20-21 )

"NOW Work

•

Solving Inequalities (Section 1 .5, pp. 128-13 1 )

the 'Are You Prepared?' problems on page 2 1 9.

OBJECTIVES

1 Determine Whether a Relation Represents a Function (p. 208)

2

3

4

1

Find the Va lue of a Fu nction (p. 2 12) Find the Domain of a Function (p. 2 15) Form the Sum, Difference, Product, and Quotient ofTwo Functions (p. 2 17)

Determine Whether a Relation Represents a Function

We often see situations where one variable is somehow linked to the value of another variable. For example, an individual's level of education is linked to annual income. Engine size is linked to gas mileage. When the value of one variable is related to the value of a second variable, we have a relation. A relation is a correspondence between two sets. If x and y are two elements in these sets and if a relation exists between x and y, then we say that x corresponds to y or that y depends on x, and we write x � y. We have a number of ways to express relations between two sets. For example, the equation y = 3 x - 1 shows a relation between x and y. It says that if we take some number x, multiply it by 3, and then subtract 1 we obtain the corresponding value of y. In this sense, x serves as the input to the relation and y is the output of the relation. We can also express this relation as a graph as shown in Figure 1. Not only can a relation be expressed through an equation or graph, but we can also express a relation through a technique called mapping. A map illustrates a rela tion by using a set of inputs and drawing arrows to the corresponding element in the set of outputs. Ordered pairs can be used to represent x � y as ( x, y ) . We illustrate these two concepts in Example 1 .

Figure 1

-4

EXA M P L E

1

Figure 2

M aps and Ordered Pairs as Relations Figure 2 shows a relation between states and the number of representatives each state has in the House of Representatives. The relation might be named "number of representatives." State

N umber of Representatives

Alaska Arizona California Colorado Florida --+-?""-=----==:::.....d-_+_ North Dakota

_::::�-.o:::::�/;:Z:>--==F:

7 8 25 53

In this relation, Alaska corresponds to 7, Arizona corresponds to 8, and so on. Us ing ordered pairs, this relation would be expressed as { (Alaska, 7 ) , (Arizona, 8), (California, 53 ) , (Colorado, 7), (Florida, 25) , (North D akota, I ) }

•

We now present one of the most important concepts in algebra - the function. A function is a special type of relation. To understand the idea behind a function,

SECTION 3.1

Figure 3

Functions

209

let's revisit the relation presented in Example 1. If we were to ask, "How many rep resentatives does Alaska have?," you would respond 7. In other words, each input state corresponds to a single output number of representatives. Let's consider a second relation where we have a correspondence between four people and their phone numbers. See Figure 3. Notice that Colleen has two tele phone numbers. If asked, "What is Colleen's phone number?", you cannot assign a single number to her. Person

Phone number

Dan

555 - 2345

Gizmo

549 - 9402

Colleen

... 930 - 3956 555 - 8294

Phoebe

839 - 901 3

Let's look at one more relation. Figure 4 is a relation that shows a correspon dence between animals and life expectancy. If asked to determine the life expectancy of a dog, we would all respond "11 years." If asked to determine the life expectancy of a rabbit, we would all respond "7 years." Figure 4

Animal

Life Expectancy

Dog

11

Duck

10

Rabbit

7

Notice that the relations presented in Figures 2 and 4 have something in com mon. What is it? The common link between these two relations is that each input corresponds to exactly one output. This leads to the definition of a function. DEFINITION

Let X and Y be two nonempty sets.'" A function from X into Y is a relation that associates with each element of X exactly one element of Y.

.J

The set X is called the domain of the function. For each element x in X, the cor responding element y in Y is called the value of the function at x, or the image of x. The set of all images of the elements in the domain is called the range of the func tion. See Figure 5. Figure 5

':' The sets X and Y will usually b e sets o f real numbers, i n which case a (real) function results. The two sets can also be sets of complex numbers, and then we have defined a complex function. In the broad definition (due to Lejeune Dirichlet), X and Y can be any two sets.

21 0

CHAPTER 3


Since there may be some elements in Y that are not the image of some x in X, it follows that the range of a function may be a subset of Y, as shown in Figure 5. Not all relations between two sets are functions. The next example shows how to determine whether a relation is a function or not.

EXAM P L E 2

Determining Whether a Relation Represents a Function Determine which of the following relations represent a function. If the relation is a function, then state its domain and range. (a) See Figure 6. For this relation, the domain represents the level of education and the range represents the unemployment rate.

Figure 6 United State:,; 2006

Level of Education

Unemployment Rate

No High School Diploma

8.5%

High School Diploma

5.0%

Some College

4.2%

College Graduate

2.7%

SOURCE: Statistical A bstract

of the

(b) See Figure 7. For this relation, the domain represents the number of calories in a sandwich from a fast-food restaurant and the range represents the fat con tent (in grams). Figure 7 SOURCE: Each company's Web site

Calories

Fat

(Wendy's Single) 4 1 0

19

(Wendy's B i g Bacon Classic) 580

29

(Burger King Whopper) 540

24

(Burger King Chicken Sandwich) 750 (McDonald's Big Mac) 600 (McDonald's McChicken) 430

�

�

-.....

33

� 23

(c) See Figure S. For this relation, the domain represents the weight (in carats) of pear-cut diamonds and the range represents the price (in dollars).

Solution

(a) The relation in Figure 6 is a function because each element in the domain corresponds to exactly one element in the range. The domain of the relation is { No High School Diploma, High School Diploma, Some College, College Graduate } and the range of the relation is { S. 5%, 5.0 % , 4.2 % , 2.7%}. (b) The relation in Figure 7 is a function because each element in the domain cor responds to exactly one element in the range. The domain of the relation is {410, 5S0, 540, 750, 600, 430 } . The range of the relation is { 19, 29, 24, 33, 23} .

SECTION 3.1

Functions

21 1

Notice that it is okay for more than one element in the domain to correspond to the same element in the range (McDonald's and Burger King's chicken sand wich both have 23 grams of fat). (c) The relation in Figure 8 is not a function because each element in the domain does not correspond to exactly one element in the range. If a 0.86-carat dia mond is chosen from the domain, a single price cannot be assigned to it. �==�·- Now Work P R O B L E M 1 5 r r

r r

r

r

r

r

In

Word s

In

Word s

For a function, no input has more than one output.

For a function, the domain is the set of inputs, and the range is . the set of outputs.

EXA M P L E 3

•

The idea behind a function is its predictability. If the input is known, we can use the function to determine the output. With "nonfunctions," we don't have this pre dictability. Look back at Figure 7. The inputs are {410, 580, 540, 750, 600, 430 } . The correspondence is num ber offat grams, and the outputs are { 19, 29, 24, 33, 23 } . If asked, "How many grams of fat are in a 410-calorie sandwich?", we can use the cor respondence to answer " 19." Now consider Figure 8. If asked, "What is the price of a 0.86-carat diamond?", we could not give a single response because two outputs result from the single input "0.86". For this reason, the relation in Figure 8 is not a function. We may also think of a function as a set of ordered pairs ( x , y) in which no ordered pairs have the same first element and different second elements. The set of all first elements x is the domain of the function, and the set of all second elements y is its range. Each element x in the domain corresponds to exactly one element y in the range.

Determining Whether a Relation Represents a Function Determine whether each relation represents a function. If it is a function, state the domain and range. (a) { ( I , 4) , (2, 5 ) , (3, 6 ) , (4, 7 ) } (b) { (1 , 4) , (2, 4) , (3, 5 ) , (6, 10) } (c) { ( -3, 9), (-2, 4) , (0, 0), ( 1,1 ), (-3, 8) }

Solution

(a) This relation is a function because there are no ordered pairs with the same first element and different second elements. The domain of this function is { I, 2, 3, 4} , and its range is { 4, 5, 6, 7 } . ( b ) This relation i s a function because there are n o ordered pairs with the same first element and different second elements. The domain of this function is { I , 2, 3, 6 } , and its range is { 4, 5, 10} . (c) This relation is not a function because there are two ordered pairs, (-3, 9) and (-3, 8), that have the same first element and different second elements.

•

In Example 3(b), notice that 1 and 2 in the domain each have the same image in the range. This does not violate the definition of a function; two different first ele ments can have the same second element. A violation of the definition occurs when two ordered pairs have the same first element and different second elements, as in Example 3(c). li!I!i:: = =

... -.


Up to now we have shown how to identify when a relation is a function for rela tions defined by mappings (Example 2) and ordered pairs (Example 3). We know that relations can also be expressed as equations. We discuss next the circumstances under which equations are functions.

21 2

CHAPTER 3


To determine whether an equation, where y depends on x, is a function, it is often easiest to solve the equation for y. If any value of x in the domain corresponds to more than one y, the equation does not define a function; otherwise, it does define a function.

Determining Whether an Equation Is a Function

EXAM P L E 4

Determine if the equation y

=

2x

-5

defines y as a function of x.

The equation tells us to take an input x, multiply it by 2, and then subtract s . For any input x, these operations yield only one output y. For example, if x = 1 , then y = 2(1) 5 = -3. If x = 3, then y = 2 ( 3 ) 5 = 1 . For this reason, the equation is a function.

Solution

-

-

•

Determining Whether an Equation Is a Function

EXA M P LE 5

Determine if the equation x2 + l = 1 defines y as a function of x.

To determine whether the equation x2 + l = 1 , which defines the unit circle, is a function, we need to solve the equation for y .

Solution

x2 + l

=

1

l= 1 Y =

-

x2

± -v1=7

For values of x between - 1 and 1 , two values of y result. For example, if x 0, then y = ± 1 , so two different outputs result from the same input. This means that the equation x2 + l = 1 does not define a function. =

-

m=

Figure 9

x Domain

2

_ f(x) = x2

Range

(a)

f(x) = x2

•


Find the Value of a Function

Functions are often denoted by letters such as f, P, g, G, and others. If f is a func tion, then for each number x in its domain the corresponding image in the range is designated by the symbol f(x), read as "f of x" or as "f at x." We refer to f(x) as the value of f at the number X; f(x) is the number that results when x is given and the function f is applied; f(x) is the output corresponding to x or the image of x; f(x) does not mean "f times x." For example, the function given in Example 4 may be written as y

=

f(x)

=

2x

- S.

Then f

(%)

=

-2.

Figure 9 illustrates some other functions. Notice that, in every function, for each x in the domain there is one value in the range.

�-"r -3 = G(O) = G( - 2) = G(3) X Domain

(b)

_ F(x) = xl F(x) = �

Range

x Domain

- g(x) = -Yx

Range

(c)

g(x) = -Yx

x - G(x) = 3 Domain

Range (d)

G(x) = 3

SECTION 3 . 1

Figure 1 0 Input x

Functions

21 3

Sometimes it is helpful to think of a function f as a machine that receives as input a number from the domain, manipulates it, and outputs the value. See Figure 10. The restrictions on this input/output machine are as follows:

�

1.

2.

Output Y = fIx)

EXA M P L E 6

It only accepts numbers from the domain of the function. For each input, there is exactly one output (which may be repeated for differ ent inputs).

For a function y = f(x), the variable x is called the independent variable, because it can be assigned any of the permissible numbers from the domain. The variable y is called the dependent variable, because its value depends on x. Any symbol can be used to represent the independent and dependent variables. For example, if f is the cube function, then f can be given by f (x) = x3 or f (t) = t3 or f ez) = Z3. All three functions are the same. Each tells us to cube the indepen dent variable to get the output. In practice, the symbols used for the independent and dependent variables are based on common usage, such as using C for cost in business. The independent variable is also called the argument of the function. Thinking of the independent variable as an argument can sometimes make it easier to find the value of a function. For example, if f is the function defined by f(x) = x3, then f tells us to cube the argument. Thus, f(2) means to cube 2, f(a) means to cube the number a, and f(x + h) means to cube the quantity x + h.

F inding Values of a Function For the function f defined by f(x)

Solution

=

(a) f(3) (d) f e -x)

(b) f(x) + f ( 3 ) (e) -f(x)

(g) f(x + 3 )

(h)

f(x

+

2X2 - 3x, evaluate

l

h - f(x)

h

*

(c) 3f(x) (f) f(3x) 0

(a) We substitute 3 for x in the equation for f, f(x) = 2X2 - 3x, to get f(3 )

=

2(3? - 3 (3 )

=

18 - 9

(b) f(x) + f(3) = (2x2 - 3x) + (9) = 2X2 - 3x + 9 (c) We multiply the equation for f by 3 . 3f(x)

=

3(2x2 - 3 x )

=

=

9

6x2 - 9x

(d) We substitute -x for x in the equation for f and simplify. f e -x ) =

=

2 ( -x )2 - 3 ( -x ) = 2X2

+

3x

_ (2X2 - 3x) = -2x2 + 3x (f) We substitute 3x for x in the equation for f and simplify. (e) -f(x)

f(3x )

=

2(3x )2 - 3(3x )

=

2(9x2) - 9x = 1 8x2 - 9x

(g) We substitute x + 3 for x in the equation for f and simplify. f(x + 3 )

=

=

= =

2(x + 3? - 3 (x + 3 ) 2(x2 + 6x + 9) - 3x - 9 2x2 2X2

+ +

12x + 18 - 3x - 9 9x + 9

Notice the use of parentheses here.

214

CHAPTER 3


(h)

+

f(x

[ 2(x + h ) 2 - 3(x + h ) ] - [ 2X2 - 3x] h

h) - f(x) h =

f(x + h)

l' 2(x + h) 2 - 3(x + h)

2 ( x2

+

2xh + h2) - 3x - 3h - 2x2 + 3x Simplify. h

2X2 + 4xh + 2h2 - 3h - 2x2 Distribute and combine like terms. h

=

4xh + 2h2 - 3h Combine like terms. h h ( 4x + 2h - 3) Factor out h. h 4x + 2h - 3 Divide out the h's.

•

Notice in this example that f(x + 3 ) =F f(x) + f ( 3 ) , f e -x) =F -f(x), and 3f(x) =F f(3x). The expression in part (h) is called the difference quotient of f, an important expression in calculus. � :::: _ ::l -

Now Work P R O B L E M S 3 9 AND 7 3

Most calculators have special keys that enable you to find the value of certain commonly used functions. For example, you should be able to find the square func tion f(x) = x2, the square root function f(x) = Vx, the reciprocal function

f(x)

=

! x

=

x-I, and many others that will be discussed later in this book (such as

In x and log x). Verify the results of Example 7, which follows, on your calculator.

EXA M P LE 7

Finding Values of a Function on a Calculator (a) f(x) x2 f ( 1.234) 1.2342 1.522756 =

(b) F (x) (c) g(x)

=

=

=

�

F ( 1.234)

=

=

�

�

1 . 34 Vx g(1.234) = v'1 .234

0.8103727715 �

1 . 1 10855526

•

Graphing calculators can be used to evaluate any function that you wish. Figure I) shows the result obtained in Example 6(a) on a TI-84 Plus graphing calculator with the function COM M E NT

to be evaluated, f(x) Figure 1 1

Plotl

P lot2

Plot3

'·.Y 1 ElZ"; 2 -3�< .... Ii ;:: = I...I � = '·S ... = '·.Y � = ,..v Ii = .... Y 7 =

=

11

2x2 - 3x, in Yl ' Y1 (3)

9

•..•

•

SECTION 3.1

Functions

21 5

I m plicit Form of a Function

In general, when a function f is defined by an equation in x and y, we say that the function f is given implicitly . If it is possible to solve the equation for y in terms of x, then we write y = f(x) and say that the function is given explicitly. For example, Implicit Form

3x

1:1 tion is the form The explicit form of a func re uired by a graphing CO M M ENT

q

x2

_

calculator.

+

y=

- Y

5

Explicit Form

y f(x) = -3x + y = f(x) = x2 - 6 4 y = f(x) =. x =

=6

xy = 4

5

We list next a summary of some important facts to remember about a func tion f. S U M M A RY

I m porta nt Facts about Functions

(a) For each x in a domain of a function f, there is exactly one image f(x) in the range; however, an element in the range can result from more than one x in the domain. (b) f is the symbol that we use to denote the function. It is symbolic of the equation that we use to get from an x in the domain to f (x) in the range. (c) If y = f(x), then x is called the independent variable or argument of f, and y is called the dependent variable or the value of f at x. 3

Find the Dom ain of a Function

Often the domain of a function f is not specified; instead, only the equation defin ing the function is given. In such cases, we agree that the domain of f is the largest set of real numbers for which the value f( x) is a real number. The domain of a func tion f is the same as the domain of the variable x in the expression f(x).

EXA M P L E

8

Finding the Domai n of a Function Find the domain of each of the following functions: (a) f(x)

Solution

=

-

3x (b) g(x) = X2 - 4

x2 + Sx

(c) h ( t)

=

v'4=3t

(a) The function tells us to square a number and then add five times the number. Since these operations can be performed on any real number, we conclude that the domain of f is the set of all real numbers. (b) The function g tells us to divide 3x by x2 - 4. Since division by 0 is not de fined, the denominator x2 - 4 can never be 0, so x can never equal -2 or 2. The domain of the function g is { x i x =1= -2, x =1= 2 } . (c) The function h tells u s to take the square root of 4 - 3t. But only nonnegative numbers have real square roots, so the expression under the square root must be nonnegative (greater than or equal to zero). This requires that

{ �}

The domain of h is t I t

:s

4 - 3t 2: 0 -3t 2: -4 4 t :S 3 or the interval

( � ]. - 00,

•

21 6

CHAPTER 3


For the functions that we will encounter in this book, the following steps may prove helpful for finding the domain of a function whose domain is a subset of real numbers and is defined by an equation. Start with the set of real numbers l. If the equation has a denominator, exclude any numbers that glve a zero denominator. 2. If the equation has a radical of even index, exclude any numbers that cause the expression inside the radical to be negative. ",.' m==> _


If x is in the domain of a function f, we shall say that f is defined at x, orf(x) If x is not in the domain of f, we say that f is not defined at x, or f(x) does

exists.

not exist .

For example, if f(x)

=

� - 1 r

,

then f(O) exists, but f ( l ) and f ( - l ) do

not exist. (Do you see why?) We have not said much about finding the range of a function. The reason is that when a function is defined by an equation it is often difficult to find the range.* Therefore, we shall usually be content to find just the domain of a function when the function is defined by an equation. We shall express the domain of a function using inequalities, interval notation, set notation, or words, whichever is most convenient. When we use functions in applications, the domain may be restricted by physi cal or geometric considerations. For example, the domain of the function f defined by f(x) x2 is the set of all real numbers. However, if f is used to obtain the area of a square when the length x of a side is known, then we must restrict the domain of f to the positive real numbers, since the length of a side can never be 0 or negative. =

EXA M P L E

9

Finding the Domain in an Application Express the area of a circle as a function of its radius. Find the domain.

Figure 1 2

Solution

G

See Figure 12. We know that the formula for the area A of a circle of radius r is 7Tr2. If we use r to represent the independent variable and A to represent the A dependent variable, the function expressing this relationship is =

A(r)

=

7TT2

In this setting, the domain is { r l r > O } . (Do you see why?)

•

Observe in the solution to Example 9 that we used the symbol A in two ways: It is used to name the function, and it is used to symbolize the dependent variable. This double use is common in applications and should not cause any difficulty. �-


;, In Section 6.2 we discuss a way to find the range for a special class of functions.

SECTION 3.1

4

Functions

217

Form the Sum, Differen ce, Product, and Quotien t of Two Functions

Next we introduce some operations on functions. We shall see that functions, like numbers, can be added, subtracted, multiplied, and divided. For example, if f(x) = x2 + 9 and g(x) = 3x + 5, then f(x)

+

g(x) = (x2

The new function y = x2

+

3x

+

+

f(x) · g (x) = ( x2

+

9)

+

+

(3x + 5) = x2

3x

+

14 is called the sum function f

9) (3x

+

5 ) = 3x3

+

5x2

+

27x

14 +

+

g. Similarly, 45

The new function y = 3x3 + 5x2 + 27x + 45 is called the product function f · g. The general definitions are given next. DEFINITION

If f and g are functions: The

sum f + g

is the function defined by (.f

+

�------

g) (x) = f(x)

+

I

g(x)

------------��

The domain of f + g consists of the numbers x that are in the domains of both f and g. That is, domain of f + g = domain of f n domain of g. DEFINITION

The difference f -

g

is the function defined by ( .f - g) (x) = f(x) - g(x)

I

----��

�--------------

The domain of f - g consists of the numbers x that are in the domains of both f and g. That is, domain of f - g = domain of f n domain of g. DEFINITION

The product f· g is the function defined by (.f . g ) ( x )

�------

I

f(x) . g(x)

------------�� =

The domain of f . g consists of the numbers x that are in the domains of both f and g. That is, domain of f . g = domain of f n domain of g. DEFINITION

The

quotient

!..g is the function defined by

(f}X) �� =

g(x) * 0

L---____

I

_

_

�

The domain of L consists of the numbers x for which g( x) * 0 that are in the g domains of both f and g. That is, domain of L = { x J g(x) * O} g

n

domain of f n domain of g.

21 8

CHAPTER 3


EXAM PLE

10

Operations on Functions Let I and g be two functions defined as I(x)

=

--

--

1 x and g(x) = x- 1 x+2

Find the following, and determine the domain in each case.

Solution

(a) (f + g ) ( x )

(b) (f - g ) ( x )

(c) (f · g) (x)

(d)

(?)c X )

The domain of I is { x i x '" -2} and the domain of g is { x i x '" I } . (a) (f + g ) ( x )

=

I(x) + g(x) =

x 1 + x+2 x- I

x(x + 2) x- I + = (x + 2 ) ( x - 1 ) (x + 2) (x - 1 )

x

=

2

+ 3x - 1 (x + 2 ) ( x - 1 )

------

The domain of I + g consists o f those numbers x that are in the domains of both I and g. Therefore, the domain of I + g is { x i x '" -2, x '" I } . (b) (f - g ) ( x )

=

I(x) - g(x)

=

-- -1 x x+2 x- I

x- I (x + 2 ) ( x - 1)

X

- (x2 + + 1 ) ( x + 2) (x - 1 )

x(x + 2) (x + 2) (x - 1)

The domain of I - g consists of those numbers x that are in the domains of both I and g. Therefore, the domain of I - g is { x i x '" -2, x '" I } . (c) (f . g ) (x)

=

I(x ) · g(x )

1 =

x+2

.x

x - I

=

x (x + 2)(x - 1 )

The domain of I , g consists of those numbers x that are i n the domains of both I and g. Therefore, the domain of I ' g is { x i x '" -2, x '" I } . (d)

(g) I

(x)

=

I(x) g(x)

=

1 x- I 1 x+2 = ' xx+2 x x- I

=

x - I x(x + 2)

The domain of L consists of the numbers x for which g(x) '" ° that are in the g domains of both I and g. Since g(x) ° when x 0, we exclude ° as well as =

=

-2 and 1 from the domain. The domain of L is { x i x '" -2, x '" 0, x '" I } . g �


•

In calculus , it is sometimes helpful to view a complicated function as the sum, difference, product, or quotient of simpler functions. For example, F(x) H(x)

=

=

+ \IX is the sum of I(x) x2 and g(x) \IX . x - 1 - is the quotient of I(x) x2 1 and g(x) 2 x + 1

2 x

-

2

=

=

-

=

=

x2 +

l.

SECTION 3.1

Functions

21 9

SUMMARY

We list here some of the important vocabulary introduced in this section, with a brief description of each term. A relation between two sets of real numbers so that each number x in the first set, the do main, has corresponding to it exactly one number y in the second set. A set of ordered pairs (x, y) or (x, f( x ) ) in which no first element is paired with two dif ferent second elements. The range is the set of y values of the function that are the images of the x values in the domain. A function f may be defined implicitly by an equation involving x and y or explicitly by writing y = f(x) . Numbers for which f(x) 0 are the zeros of f.

Function

=

Unspecified domain

If a function f is defined by an equation and no domain is specified, then the domain will be taken to be the largest set of real numbers for which the equation defines a real number.

Function notation

y = f(x)

f is a symbol for the function. x is the independent variable or argument. y is the dependent variable. f(x) is the value of the function at x, or the image of x.

3 . 1 Assess Your Understanding

'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red. 1 . The inequality - 1 < x < 3 can be written in interval nota tion as . (pp. 125-] 26)

2. If x

=

__

__

1 -2, the value of the expression 3x2 - 5x + - is x . (pp. 20-21)

3. The domain of the variable in the expression

x \s x + 4

__

.

(pp. 20-2 1 ) 4. Solve the inequality: 3 - 2x > 5. G raph the solution set. (pp. ] 28-131)

Concepts and Vocabu lary S.

If f is a function defined by the equation y = f(x), then x is called the variable and y is the variable. 6. The set of all images of the elements in the domain of a func tion is called the 7. If the domain of f is all real numbers in the interval [0, 7] and the domain of g is all real numbers in the interval [ -2, 5], the domain of f + g is all real numbers in the interval . __

__

__

8. The domain of L consists of numbers x for which g(x) g o

9.

that are in the domains of both If f(x) = x + 1 and g(x) x3 ,

__

and

__

.

Every relation is a function.

1 0.

True or False

1 1.

True or False The domain of (f . g) (x) consists of the nUlll bel's x that are in the domains of both f and g.

1 2.

True or False The independent variable is sometimes referred to as the argument of the function.

1 3.

True o r False If no domain is specified for a function f, then the domain of f is taken to be the set of real numbers.

1 4.

True or False

__

=

{xix

#-

±2}.

The domain of the function f(x) =

= x3 - (x + l ) .

then

Skill Building In Problems 15.

15-26,

Person Elvis

determine whether each relation represents a function. For each function, state the domain and range. Birthday

__ Jan. 8

Colieen ---

16.

Father

Bob _

Kaleigh

Mar. 1 5

John

Marissa

Sept. 1 7

Chuck

Daughter Beth

I--t->-- Diane Linda

Marcia

--

x2

4

x

is

220

17.

CHAPTER 3

Functions and T h e i r Gra p h s

Average I ncome

$200

Less than 9th grade

$1 8,1 20

$300

9th· 1 2th grade

$23,251

$350

High School Graduate

$36,055

Some College

$45,8 1 0

College Graduate

$67, 1 65

Salary

20 Hours

-r.--

30 Hours

18.

Level of Education

Hours Worked

$425

40 Hours

1 9. { (2, 6), ( -3, 6 ) , (4, 9 ) , (2, 10) }

20. { ( -2, 5 ) , ( -1 , 3), (3, 7 ) , (4, 12) }

21. { ( I, 3 ) , (2, 3 ) , (3, 3 ) , (4, 3 ) }

22. { (O, -2) , ( 1 , 3 ) , (2, 3 ) , (3, 7 ) }

23. { ( -2, 4), ( -2, 6) , (0, 3 ) , (3, 7) }

24. { ( -4, 4), ( - 3, 3 ) , ( -2, 2 ) , ( - 1 , 1 ) , ( - 4, 0 ) }

25. { ( -2, 4), ( -1,

26. { ( -2, 16), ( - 1 , 4), (0, 3 ) , ( 1 , 4) }

1 ) , (0, 0), ( 1 , I ) }

In Problems 27-38, determine whether the equation defines y as a function of x.

27.

Y

=

- x2 35. y = 2 x2 - 3x + 4 31. l = 4

In Problems

39-46,

(a) f(O)

+

+

43. f(x) = I x l In Problems
t(x2;)

t(X1)< t(X2);

tis decreasing on 1

tis increasing on 1 . , I'l'>l::m:l.....,..

4

(c) For all x in I, the values of tare equal; tis constant on 1

(b) For x1< X2 in I,

(a) For X1< x2in I,


1 1

I

1 3, 1 5

I

AND 2 1

Use a Graph to Locate Local Maxima and Local Minima

When the graph of a function is increasing to the left of X = c and decreasing to the right of x c,then at c the value of / is largest. This value is called a local maxi mum of f. See Figure 20(a). When the graph of a function is decreasing to the left of x = c and is increasing to the right of x = c,then at c the value of / is the smallest. This value is called a local minimum of f. See Figure 20(b). =

Figure 20

(c)

Y

t(e)

y

(e, t(e))

-71\ e

t(e)

x

--

V I (e, I I I

t(e))

e

increasing decreasing

decreasing increasing

The local maximum is f(e) and occurs at X= e.

The local minimum is t ( e) and occurs at X= e.

(a)

(b)

x

SECTION 3.3

235

Properties of Functions

A function f has a local maximum at c if there is an open interval I containing c so that, for all x not equal to c in I, f(x) < f(c). We call f(c) a local

DEFINITIONS

maximum off

A function f has a local minimum at c if there is an open interval I containing c so that, for all x not equal to c in I,f(x) > f(c). We call f(c) a local minimum off

�

If f has a local maximum at c, then the value of f at c is greater than the values of f near c. If f has a local minimum at c,then the value of f at c is less than the val ues of f near c. The word local is used to suggest that it is only near c that the value f (c) is largest or smallest. E XA M P L E 4

F i n d i n g Local M axima and Local M i n i m a from the G raph of a F un ction and Determi n i ng Where the Function Is Increasing, Decreasing, o r Constant

Figure 2 1

Figure 21 shows the graph of a function f. (a) At what number(s), if any, does f have a local maximum? (b) What are the local maxima? (c) At what number(s), if any, does f have a local minimum? (d) What are the local minima? (e) List the intervals on which f is increasing. List the intervals on which f is decreasing.

x

Solution

The domain of f is the set of real numbers. (a) f has a local maximum at 1, since for all x close to 1, x =f. 1, we have f(x) < f(l ) . (b) The local maximum is f(l ) 2. (c) f has a local minimum at -1 and at 3. (d) The local minima are f( -1) = 1 and f(3) = O. (e) The function whose graph is given in Figure 21 is increasing for all values of x between -1 and 1 and for all values of x greater than 3. That is, the function is increasing on the intervals (-1, 1) and (3, 00 ) or for - 1 < x < 1 and > 3. The function is decreasing for all values of x less than -1 and for all values of x between 1 and 3. That is, the function is decreasing on the intervals ( - , -1) and (1,3) or for x < -1 and 1 < x < 3. =

WARNING The y-value i s the local max

x

imum or local minim um and it occurs at some x-value. For Figure local maximum is

2

21,

we say the

00

and that the local

maximum occurs at x =

1.

•

•

I;!l!>: == ::> '-

I

5

Now Work

PRO B L EMS 1 7 AND 1 9

Use a Graphing Utility to Approximate Local Maxima and Local Minima and to Determine Where a Function Is Increasing or Decreasing

To locate the exact value at which a function f has a local maximum or a local min imum usually requires calculus. However, a graphing utility may be used to approx imate these values by using the MAXIMUM and MINIMUM features. E XA M P L E 5

U s i n g a G raph ing Util ity to Approxim ate Local M axima and M in i m a and to Determine Where a Functio n Is Increasing or Decreasing

(a) Use a graphing utility to graph f(x) = 6x3 - 12x + 5 for -2 < x < 2. Ap proximate where f has a local maximum and where f has a local minimum. (b) Determine where f is increasing and where it is decreasing.

236

CHAPTER 3

Functions and Their Graphs

(a) Graphing utilities have a feature that finds the maximum or minimum point of a graph within a given interval. Graph the function f for -2 < x < 2. See Figure 22(a). Using MAXIMUM, we find that the local maximum is 11 .53 and it occurs at x -0.82, rounded to two decimal places. See Figure 22(b). Using MINIMUM, we find that the local minimum is - 1 .53 and it occurs at x = 0.82, rounded to two decimal places.

Solution

=

Figure 22

30

30

-2

,r"�o;.-......_ I

�

"(J iriIUf"i1 1-:= -.B16'19'11

l /

.........

--

I

/

.......

,l.----

2

"

-2

�

"i if"iluf"il 1-:=.B16'19'1S:

'�=11_S:"19n

-1 0 (a)

-

.......

;11:".

)

2

','= -1.S:�19n

10

(b)

(b) Looking at Figures 22(a) and (b), we see that the graph of f is increasing from x -2 to x -0.82 and from x = 0.82 to x 2, so f is increasing on the in tervals ( -2, -0.82 ) and (0.82, 2) or for -2 < x < -0.82 and 0.82 < x < 2. The graph is decreasing from x = -0.82 to x 0.82, so f is decreasing on the interval ( -0.82, 0.82) for -0.82 < x < 0.82. =

=

=

=

or

....... Now Work

' I!.

6

•

PRO B LEM 4 5

Find the Average Rate of Change of a Function

In Section 2.3, we said that the slope of a line could be interpreted as the average rate of change of the line. To find the average rate of change of a function between any two points on its graph, we calculate the slope of the line containing the two points. DEFINITION

If a and b, a #- b, are in the domain of a function y of change of f from a to b is defined as

=

f(x), the average rate

ily f(b) - f(a) Average rate of change = - = --'---'----'--'ilx b-a

a#- b

(1 )

�

�------�

The symbol ily in (1) is the "change in y" and ilx is the "change in x." The aver age rate of change of f is the change in y divided by the change in x. E XA M P L E 6

F i n d i n g the Average Rate of Change

Find the average rate of change of f(x) = 3x2: (a) From 1 to 3 Solution

(b) From 1 to 5

(c) From 1 to 7

(a) The average rate of change of f(x) = 3x2 from 1 to 3 is ily ilx

f(3 ) f (I ) 27 - 3 = 3-1 3-1

_ -_ _ _ _

_ _

(b) The average rate of change of f(x) ily ilx

f(5) - f(l ) 5 - 1

=

=

=

24 = 12 2

3x2 from 1 to 5 is 75 - 3 5 - 1

=

72 4

=

18

SECTION 3.3

Figure 23

(c) The average rate of change of f(x)

Y

�y �x

1 60

120

=

Properties o f Functions

237

3x2 from 1 to 7 is

f(7) - f(l) 7 1 147 - 3 144 = 7 1 6 -

=

=

24

-

•

80

See Figure 23 for a graph of f(x) = 3x2 . The function f is increasing for x > O. The fact that the average rates of change are getting larger indicates that the graph is getting steeper; that is, it is increasing at an increasing rate.

40

:>I!

-

Now Work

PRO B L E M S 3

x

(0, 0)

The Secant Line

The average rate of change of a function has an important geometric interpretation. Look at the graph of y = f(x) in Figure 24. We have labeled two points on the graph: (a,f(a)) and (b,f (b) ). The line containing these two points is called the secant line; its slope is feb) - f(a) 111sec = b -

Figure 24

y

y=

f(x)

b

a

THEOREM

a

x

Slope of the Secant Line

The average rate of change of a function from a to b equals the slope of the secant line containing the two points (a,f(a)) and (b, f(b)) on its graph. EXAM P L E 7

.-J

F i n d i n g the Equation of a Secant Line

Suppose that g(x)

=

3x2 - 2x + 3.

(a) Find the average rate of change of g from -2 to l. (b) Find an equation of the secant line containing ( -2, g( 2 ) ) and ( 1, g(l)). -

S o l ution

(a) The average rate of change of g(x) Average rate of change

=

3x2

g(l) - g( - 2 ) 1 (-2) 4 19 3 15 -- = -5 3 _

-

=

=

-

2x + 3 from -2 to 1 is

g(l) 3(12) - 2(1) + 3 = 4 g(-2) 3(-2)2 - 2(-2) + 3 =

=

= 19

238

CHAPTER 3


(b) The slope of the secant line containing (-2, g(-2)) (-2, 19) and (l, g(l) ) (1, 4) is msec -5. We use the point-slope form to find an equation of the secant line. =

=

=

Y Y Y

-

Yl

19 19 Y

k �IIImI_ "" '-

=

=

Insec ( x - Xl) -5 ( x - ( -2 ) )

- 5x - 10 = -5x + 9

X,

=

-2'YI

=

g(-2)

=

19,

msec

=-5

Simplify.

=


Point-slope form of the secant line

Slope-intercept form of the secant line

•

59

3.3 Assess Your Understanding 'Are You Prepa red?' Answers are given at the end of these exercises. Ifyou get a wrong answel; read the pages listed in 1.

2.

The interval (2, 5) can be written as the inequality . (pp.125-L26) The slope of the line containing the points ( -2, 3 ) and (3, 8) is . (pp. 174-176) Test the equation y = 5x2 1 for symmetry with respect to the x-axis, the y-axis, and the origin. (pp. L67- 168) __

__

3.

red.

Write the point-slope form of the line with slope 5 contain ing the point (3, -2). (pp. L76-179) 2 . 5. The intercepts of the equation y = x - 9 are (pp. 165-167)

4.

__

-

Concepts a n d Vocabu lary 6.

7.

A function f is on an open interval I if, for any choice of x, and X2 in I, with X, < X2, we have f( x,) < f( X2)' A(n) function f is one for which f( -x) = f( x) for every x in the domain of f; (an) function f is one for which f( -x) = -f(x) for every x in the domain of f. True or False A function f is decreasing on an open inter val! if, for any choice of Xl and X2 in I, with x, < X2 , we have f( x,) > f( X2)' __

__

__

8.

A function f has a local maximum at c if there is an open interval! containing c so that, for all X not equal to c in I,f( x) < f( c). True or False Even functions have graphs that are sym metric with respect to the origin.

9. True or False

10.

Skil l Bu ilding !n Problems 11-20, use the graph o f the f�lI1ction f given.

(-8 , -4)

. 11. 13.

Is f increasing on the interval ( -8, -2)?

12.

Is f decreasing on the interval (-8, -4)?

Is f increasing on the interval (2, IO) ?

14.

Is f decreasing on the interval (2, 5)?

16.

List the interval(s) on which f is decreasing.

18.

Is there a local maximum at 5? If yes, what is it?

15. List the interval(s) o n which f is increasing. . 17.

-6

Is there a local maximum at 2?

If yes,

what is it?

19.

List the numbers at which f has a local maximum. What are these local maxima?

20.

List the numbers at which f has a local minimum. What are these local minima?

SECTION 3.3


239

I n Problems 21-28, the graph o f a function i s given. Use the graph t o find: (a) The intercepts, if any (b) The domain and range (c) The intervals on which it is increasing, decreasing, or constanl (d) Whether it is even, odd, or neither '21.

y

(-2,0) (2,0)

-4

,,_:l.

4

2

(33 , )

23.

y

26.

2

(t,1)

3x

-3

(- t,

-

-2

1)

3

(2,2)

3x

1 / (2 (3,0) (1,-1) ,

-1)

y

28.

3

(0,1) ('IT,

-3

y

27.

-3 (-'IT, - 1 )

y

24.

3

-3(-1,0)(1,0) 3x

x

y

y

(-3,3)

22.

4

-

1)

(-3,-2)

-3

-2

In Problems 29-32, the graph of a function f is given. Use the graph to find: (a) The n umbers, if any, at which f has a local maxim um. What are these local maxima? (b) The n umbers, if any, at which f has a local minim um. What are these local minima?

y

29.

30.

y

31.

-4

(-2,0) (2,0)

4

-3(-1,0)(1,0)

x

y

32.

2

4

2

(t,1)

3x

(-re, -1)

(- t,-1 ) -2

(re,

-2

-1)

In Problems 33-44, determine algebraically whether each function is even, odd, or neither. 33. f(x) = 4x3 37. F(x) =

41. g(x)

llirl

-\YX 1

=?

34. f(x) =

2X4 -

38. G(x) =

\IX

42. h(x)

[

=

x

x2

-

2

x

-I

3

36. h(x) = 3x3 + 5

35. g(x) = - x 2 - 5 39. f(x) = x +

43. h(x)

=

40. f(x) =

Ixl

-x3

-? .:LC - 9

-�

44. F(x) =

\12 x2

+ 1

2x

N

In Problems 45-52, use a graphing utility to graph each function over the indicated interval and approximate any local maxima and local minima. Determine where the function is increasing and where it is decreasing. Ro und answers 10 two decimal places.

,.45.

'

f(x) = x' - 3x +

47. f(x) = x5 - x3 49. f(x)

=

'

2 (-2,2)

(-2,2)

-0.2x3 - 0.6x2 + 4 x -

?

46. f(x) = x' - 3[ + 5 48. f(x)

6

51. f(x) = 0.25x4 + 0.3x3 - 0.9x2 +

(-6,4)

3 ( -3,2)

=

x4 - x2

(-1,3)

( -2,2)

50. f(x) = -0.4x3 + 0.6x2 + 3x

-2

52. f(x) = -0.4x4 - 0.5x3 + 0.8x2 -

(-4,5)

2 ( -3,2)

240

CHAPTER 3


, 53. Find the average rate of change of f(x) = _2X2 + 4 (a) From 0 to 2 (b) From 1 to 3 (c) From 1 to 4 54. Find the average rate of change of f(x) = -x3 + 1 (a) From 0 to 2 (b) From 1 to 3 (c) From -1 to 1 55. Find the average rate of change of g(x) = x3 - 2x + 1 (a) From -3 to -2 (b) From -1 to 1 (c) From 1 to 3 56. Find the average rate of change of h(x) = x2 - 2x + 3 (a) From -1 to 1 (b) From 0 to 2 (c) From 2 to 5 57. f(x) = 5x - 2 (a) Find the average rate of change from 1 to 3. (b) Find an equation of the secant line containing (1,f(l) ) and (3,/(3)).

58.

.59.

f(x) = -4x

+1 (a) Find the average rate of change from 2 to 5. (b) Find an equation of the secant line containing and (5,/ (5)).

(2,f(2))

g(x) = x2 - 2

(a) Find the average rate of change from -2 to l. (b) Find an equation of the secant line containing (-2,g( -2)) and (l,g(l)). 60. g(x) = x2 + 1 (a) Find the average rate of change from -1 to 2. (b) Find an equation of the secant line containing (-1, g( - 1)) and (2,g(2) ).

61.

h(x) = x2 - 2x

(a) Find the average rate of change from 2 to 4. (b) Find an equation of the secant line containing (2,h(2)) and (4, h( 4) ). 62. h(x) = -2x2 + x (a) Find the average rate of change from 0 to 3. (b) Find an equation of the secant line containing (0, h(O)) and (3, h(3) ).

Appl ications a n d Extensions 63.

An open box with a square base is to be made from a square piece of cardboard 24 inches on a side by cutting out a square from each corner and turning up the sides. See the figure. Constructing an Open Box

x x

'I ,1

24in.

x x

x

1 � 24in. - 1

r•.:

.,?

(a) Express the volume V of the box as a function of the length x of the side of the square cut from each corner. (b) What is the volume if a 3-inch square is cut out? (c) What is the volume if a lO-inch square is cut out? (d) Graph V = V(x). For what value of x is V largest?

(a) Use a graphing utility t o graph s = s(t). (b) Determine the time at which height is maximum. (c) What is the maximum height? On July 1,2004, the Cassini ":;: 66. Maximum Height of a Ball probe became the first spacecraft to orbit the planet Saturn. Although Saturn is about 764 times the size of Earth, it has a very similar gravitational force. The height s of an object thrown upward from Saturn's surface with an initial velocity of 100 feet per second is given as a function of time t (in sec onds) by s(t) = -17.28t2 + lOOt. (a) Use a graphing utility to graph s = s(t). (b) Determine the time at which height is a maximum. (c) What is the maximum height? (d) The same object thrown from the surface of Earth would have a height given by s(t) = -16t2 + lOOt. De termine the maximum height of the object on Earth and compare this to your result from part (c). ,,1 67. Minimum Average Cost The average cost per hour in dol lars of producing x riding lawn mowers is given by

-C(x) = 0.3x2 + 21x - 251 + 2500 x Use a graphing utility to graph C = C(x)

64. Constructing an Open Box An open box with a square base is required to have a volume of 10 cubic feet. (a) Express the amount A of material used to make such a box as a function of the length x of a side of the square base. (b) How much material is required for a base 1 foot by 1 foot? (c) How much material is required for a base 2 feet by 2 feet? ;it (d) Use a graphing utility to graph A = A (x). For what .. value of x is A smallest?

(a) (b) Determine the number of riding lawn mowers to pro duce in order to minimize average cost. (c) What is the minimum average cost? ii.r 68. Medicine Concentration The concentration C of a medica tion in the bloodstream t hours after being administered is given by C(t) = -0.002X4 + 0.039t3 - 0.285t2 + 0.766t + 0.085

65. Maximum Height of a Ball The height s of a ball (in feet) thrown with an initial velocity of 80 feet per second from an initial height of 6 feet is given as a function of the time t (in seconds) by s(t) = -16t2 + 80t + 6

(a) After how many hours will the concentration be highest? (b) A woman nursing a child must wait until the concen tration is below 0.5 before she can feed him. After tak ing the medication, how long must she wait before feed ing her child?

--

SECTION 3.3

69. E-coli Growth A strain of E-coli Beu 397-recA441 is placed into a nutrient broth at 30° Celsius and allowed to grow. The data shown below are collected. The population is measured in grams and the time in hours. Since population P depends on time t and each input corresponds to exactly one output, we can say that population is a function of time; so pet ) rep resents the population at time t. ( a ) Find the average rate o f change o f the population from o to 2.5 hours. (b) Find the average rate of change of the population from 4.5 to 6 hours. (c) What is happening to the average rate of change as time passes?

70.

Time (hoursl.

t

Population (gramsl. P

o

0.09

2.5

0.18

3.5

0.26

4.5

0.35

6

0.50

241

(d) What is happening to the average rate of change as time passes?

�------------� o�


Year

Percentage of returns e-filed

1998

20.7

1999

23.5

2000

27.6

2001

30.7

2002

35.6

2003

40.2

2004

46.5

2005

51.1

2006

57.1

SOURCE: Internal Revenue Service

For the function f(x) = x2, compute each average rate of change: (a) From 0 to 1 (d) From 0 to 0.01 (b) From 0 to 0.5 (e) From 0 to 0.001 (c) From 0 to 0.1 ij> (f) Use a graphing utility to graph each of the secant lines along with f. (g) What do you think is happening to the secant lines? (h) What is happening to the slopes of the secant lines? Is there some number that they are getting closer to? What is that number? 72. For the function f(x) = x2, compute each average rate of change: (d) From 1 to 1.01 (a) From 1 to 2 (e) From 1 to 1.001 (b) From 1 to 1.5 (c) From 1 to 1.1 hi? (0 Use a graphing utility to graph each of the secant lines along with f. (g) What do you think is happening to the secant lines? (h) What is happening to the slopes of the secant lines? Is there some number that they are getting closer to? What is that number? 71.

e-Filing Tax Returns The Internal Revenue Service Restructuring and Reform Act (RRA) was signed into law by President Bill Clinton in 1 998. A major objective of the RRA was to promote electronic filing of tax returns. The data in the table show the percentage of individual income tax re turns filed electronically for filing years 1 998-2006. Since the percentage P of returns filed electronically depends on the fil ing year y and each input corresponds to exactly one output, the percentage of returns filed electronically is a function of the filing year; so P(y) represents the percentage of returns filed electronically for filing year y. (a) Find the average rate of change of the percentage of e-filed returns from 1998 to 2000. (b) Find the average rate of change of the percentage of e-filed returns from 2001 to 2003. (c) Find the average rate of change of the percentage of e-filed returns from 2004 to 2006.

(f. Problems 73-80 require the following discussion of a secant line. The slope of the secant line containing the two points (x,f(x ) ) and (x + h,f(x + h)) on the graph of a function y = f(x) may be given as f(x + h) - f(x) (x + h) - x

f(x

+

h ) - f(x) h

h*O

In calculus, this expression is called the difference quotient of f. (a) Express the slope ofthe secant line of each function in terms ofx and h. Be sure to simplify your answer. (b) Find msecfor h = 0.5,0.1, and 0.01 at x = 1. What value does mscc approach as h approaches O? (c) Find the equation for the secant line at x = 1 with h = 0.01. ro.: (d) Use a graphing utility to graph f and the secant line found in part (c) on the same viewing window. 73. f(x) = 2x + 5 74. f(x) = -3x + 2 76. f(x) 75. f(x) x2 + 2 x =

77.

f(x)

=

2x2 - 3x

+1

78.

f(x)

=

-x2

+

3x

- 2

79.

f(x)

=

� x

80.

f(x)

=

2x 2

1 =-

x2

+

X

242

C H A PTER 3


Discussion a n d Writing 81.

82.

83.

Draw the graph of a function that has the following proper ties: domain: all real numbers; range: all real numbers; inter cepts: (0,-3) and (3,0); a local maximum of -2 is at -1; a local minimum of -6 is at 2. Compare your graph with oth ers. Comment on any differences. Redo Problem 81 with the following additional information: increasing on (-00,-1),(2,00); decreasing on (-1,2). Again compare your graph with others and comment on any differences. How many x-intercepts can a function defined on an interval have if it is increasing on that interval? Explain.

84.

85.

L;TI 86.

Suppose that a friend of yours does not understand the idea of increasing and decreasing functions. Provide an explana tion, complete with graphs, that clarifies the idea. Can a function be both even and odd? Explain. Using a graphing utility, graph y = 5 on the interval (-3,3). Use MAXIMUM to find the local maxima on ( 3,3). Com ment on the result provided by the calculator. -

'Are You Prepa red?' Answers 1.2<x 1

(a) Find f(O), f ( 1 ) , and f(2). (c) Graph f. S o l ution

=

(b) Determine the domain of f. (d) Use the graph to find the range of f.

(a) To find f(O), we observe that when f(x) -x + 1 . So we have

x

0 t he equation for f is given by

=

=

f(O) When x

=

=

-0 + 1

1, the equation for f is f(x) f(1)

When x

Figure 38

5 );

(-1 , 2)

,\1

(0 , 1 ) �' ,

-3

(2 4)

•

/

3

22

2 =

=

x2. So 4

(b) To find the domain of f, we look at its definition. We conclude that the domain of f is { x i x 2: - 1 } , or the interval [ - 1 , ) (c) To graph f, we graph "each piece." First we graph the line y -x + 1 and keep only the part for which -1 :S x < 1 . Then we plot the point ( 1 , 2) be cause, when x = 1 , f(x) = 2. Finally, we graph the parabola y = x2 and keep only the part for which x > 1. See Figure 38. .

=

/

-x+ 1

=

1

2. Thus,

(0

(1 , 2

, Y=

2, the equation for f is f(x) f(2)

Y = x2

Y

=

=

=

=

x

248

CHAPTER 3


(d) From the graph, we conclude that the range of f is { y l y > O } , or the interval (0,00 ) .

•


1l'l

EXAM PLE 4

29

Cost of E lectricity

In May 2006, Commonwealth Edison Company supplied electricity to residences for a monthly customer charge of $7.58 plus 8.275¢ per kilowatt-hour (kWhr) for the first 400 kWhr supplied in the month and 6.208¢ per kWhr for all usage over 400 kWhr in the month. (a) What is the charge for using 300 kWhr in a month? (b) What is the charge for using 700 kWhr in a month? (c) If C is the monthly charge for x kWhr,express C as a function of x. Source: Commonwealth Edison Co., Chicago, Illinois, 2006.

Sol ution

(a) For 300 kWhr, the charge is $7.58 plus 8.275¢ Charge

=

$7.58

+

=

$0.08275(300)

$0.08275 per kWhr. That is, =

$32.41

(b) For 700 kWhr, the charge is $7.58 plus 8.275¢ per kWhr for the first 400 kWhr plus 6.208¢ per kWhr for the 300 kWhr in excess of 400. That is, Charge (c)

=

$7.58 + $0.08275(400) + $0.06208(300)

=

$59.30

If 0 ::; x ::; 400, the monthly charge C (in dollars) can be found by multiply ing x times $0.08275 and adding the monthly customer charge of $7.58. So, if o ::; x ::; 400, then C(x) 0.08275x + 7.58. For x > 400, the charge is 0.08275 (400) + 7.58 + 0.06208(x - 400) , since - 400 equals the usage in excess of 400 kWhr, which costs $0.06208 per kWhr. That is, if x > 400, then =

x

Figure 39 �

C(x)

80 1++H1++H+H++++l-I++HI++H+H+H++I++H-H

� 60 1++t11++H+H++++l-1-H o en

"0

-;; 40

�

u

..c

7

=

= =

.58 +++l-1+l+1+l+1+I±iw-H-i'ffi-!-H-H+H

1n+1++H��I++H�TItt��-H

The rule for computing C follows two equations: _ C(x) -

20 l+t1tft1.!1'H+ttt l

1 00 200 300 400 500 600 700

Usage (kWhr)

0.08275(400) + 7.58 + 0.06208(x - 400) 40.68 + 0.06208(x - 400) 0.06208x + 15.848 0.08275X { 0.06208x

+ 7.58 if 0 ::; x ::; 400 + 1 5.848 if x > 400

See Figure 39 for the graph. •


Answers are given at the end ofthese exercises. [{you get a wrong answel; read the pages listed in red. 3. List the intercepts of the equation y = - 8. (pp. 1 65-1 67) Sketch the graph of y \IX . (p. l70) 1 Sketch the graph of y - . (p. 170) x

'Are You Prepa red?' 1. 2.

x3

=

=


4. The function f(x) = x2 is decreasing on the interval __. S. When functions are defined by more than one equation, they functions. are called 6. True or False The cube function is odd and is increasing on the interval ( - (X), (0 ) .

The cube root function is odd and is de creasing on the interval ( - (X) , (0 ) . True or False The domain and the range o f the reciprocal function are the set of all real numbers.

7. True o r False

8.

SECTION 3.4

Library of Functions; Piecewise-defined Functions

249

Ski l l Bu ilding In Problems

9-16,

match each graph to its function.

A. ConSLanl function D. Cube function G. Absolute value function

"· 13.

V /

C.

B. Identity function E. Square root function H. Cube root fitnction

F

. 10.

Square function Reciprocal funClion

11.

12.

/

i�

15.

V

16.

�

1

_ _ _ - - - _ _ - _ _ _ _ _

In Problems

1 7-24,

\1

sketch the graph of each function. Be sure to label three points on the graph.

17. f(x) = x 1 21. f(x) = x

{

x2 2 2x + 1 find : (a) f ( -2)

25. If f(x) =

18. f(x) = x2

19. f(x) = x3

22. f(x) = Ix l

23. f(x) = -0X

if x < 0 if x = O if x > 0 (b) f ( O )

26. If f(x) = (c) f(2)

In Problems

28. If f(x) =

-

38. f(x) =

+

1

{ X3

3x + 2

if x < - 1 if x = - 1 if x > - 1

(c ) f( O )

(b) f( -1)

if -2 � x < 1 if l � x � 4 (c) f( l ) (b) f( O )

{ 2X 1

X

{I

+

2

if x of. a if x = 0

30. f(x) =

3- 3 if

x < -2 if x 2: -2

x

{2 - x �

{3X if

33. f(x) =

x of. 0 if x = 0

4

r J {� 5

-x

+

if x < 0 if x 2: 0

3 1

39. f(x) = 2 int(x)

-3

�

VX

31. f(x) =

if -2 � x < 1 if x = 1 if x > 1

2

if x < 0 if x

34. f(x) =

37. f(x) =

2: a

{ 2X � .)x

r-3 x

+ ?

-

+

5 - x

(0, 0)

�

(2, 1 )

( 2, 1 )

(-1 , 1 )

y

2

2

-2

( d) f (3 )

(0, 0)

-2 ( - 1 , - 1)

2

x

-2

(0, 0)

_

5

x< 1 x 2: 1

-3

if �x< if x = a if x > a

if -2 � x < 0 if x > a

{ Ix l x3

40. f(x) = int(2x)

U

y

�

y

3

(b) Locate any intercepts. (d) Based on the graph,find the range.

In Problems 41-44, the graph of a piecewise-defined function is given. Write a definition for each function. U

3

29-40:

{x + 32. f(x) = -2x 35. f(x) =

�

2x2

find: (a) f( - 1 )

( d) f( 3 )

(a) Find the domain of each function. (c) Graph each function.

29. f(x) =

0

24. f(x) =

find: (a) f( -2)

if - 1 � x � 2 { 2X - 4 if 2 < x � 3 x3 2 find: (a) f( O ) (b) f( l ) (c) f(2)

27. If f(x) =

{-3l:

20. f(x) = \IX

-2

a

250 45.


CHAPTER 3

=

If f(x)

int(2x), find (a) f( 1.2),

46. If f(x) = in

{1} find

(a) f ( 1 .2),

(b) f( 1 .6),

(c) f( - 1 .8).

(b) f ( 1 .6),

(c) f( - 1 .8).

Appl ications a n d Extensions 47.

Sprint PCS offers a monthly cellular phone plan for $35. It includes 300 anytime minutes and charges $0.40 per minute for additional minutes. The follow ing function is used to compute the monthly cost for a sub scriber: Cell Phone Service

C(x)

=

{

35 0.40x - 85

49.

if 0 < x :$ 300 if x > 300

where x is the number of anytime minutes used. Compute the monthly cost of the cellular phone for use of the follow ing anytime minutes: (c) 301 (a) 200 (b) 365 Source: 48.

Sprint PCS

The short-term (no more than 24 hours) parking fee F (in dollars) for park ing x hours at O'Hare International Airport's main parking garage can be modeled by the function

Source:

Parking at O'Hare International Airport

F(x)

=

{�

int(X + 1 ) + 1 50

if 0 if 3 if 9

< < :$

50.

x :$ 3 x 20

where v represents the wind speed (in meters per second) and t represents the air temperature (0C). Compute the wind chill for the following: (a) An air temperature of 1 0°C and a wind speed of 1 me ter per second ( m/sec) (b) An air temperature of 10°C and a wind speed of 5 m/sec (c) An air temperature of lOoC and a wind speed of 1 5 m/sec (d) An air temperature of lOoC and a wind speed of 25 m/sec (e) Explain the physical meaning of the equation corre sponding to 0 :S V < 1.79. (f) Explain the physical meaning of the equation corre sponding to v > 20. Wind Chill Redo Problem 57(a)-(d) for an air temperature of -10°C. First-class Mail In 2006 the U.S. Postal Service charged $0.39 postage for a first-class letter weighing up to 1 ounce, plus $0.24 for each additional ounce up to 13 ounces. First-class rates do not apply to letters weighing more than 13 ounces. Write a piecewise-defined function C for the first-class postage charged for a letter weighing x ounces. Graph the function. Source:

United States Postal Service

Discussion a n d Writing

� In Problems 60-67, use a graphing utility. 60. Exploration Graph y = x2. Then on the same screen graph y = x2 + 2, followed by y = x2 + 4, followed by y = x2 - 2. What pattern do you observe? Can you predict the graph of y = x2 - 4? Of y = x2 + 5? 61. Exploration Graph y = x2 Then on the same screen graph y = (x - 2)2, followed by y = (x - 4)2, followed by y = (x + 2f What pattern do you observe? Can you predict the graph of y = (x + 4)2? Of y = (x - Sf? 62. Exploration Graph y = I xl. Then on the same screen graph 1 y = 21xl, followed by y = 41xl, followed by y = 2: lxl. What

63.

64.

pattern do you observe? Can you predict the graph of y = � Ixl ? Of y = 51xl? 4 Exploration Graph y x2. Then on the same screen graph y = -x2. What pattern do you observe? Now try y = Ixl and y = - Ixl. What do you conclude?

65.

y = 2x

+ 1 and y

=

2( -x)

+

1 . What do you conclude?

Graph y = x2, Y = X4, and y = x6 on the same screen. What do you notice is the same about each graph? What do you notice that is different?

66.

Exploration

67.

E xploration

6S.

Consider the equation

Graph y = x3, Y = xS, and y = x7 on the same screen. What do you notice is the same about each graph? What do you notice that is different?

y=

{�

i f x is rational if x is irrational

Is this a function? What is its domain? What is its range? What is its y-intercept, if any? What are its x-intercepts, if any? Is it even, odd, or neither? How would you describe its graph?

=

E xploration Graph y = \IX. Then on the same screen graph y = V=X. What pattern do you observe? Now try

Graph y = x3 Then on the same screen graph (x - 1 )3 + 2. Could you have predicted the result?

Exploration

y=

69.

Define some functions that pass through (0, 0) and ( 1 , 1 ) and are increasing for x 2: O. Begin your list with y = \IX, y = x, and y = x2. Can you propose a general result about such functions?

252

CHAPTER 3


'Are You Prepa red' Answers 2.

1. y

2

~

3.

(0,-8), (2, 0)

(1 , 1 ) 2

x

(-1 , -1 )

x

OBJECTIVES

1 Graph Functions Using Vertical and Horizontal Shifts (p. 252) 2 Graph Functions Using Compressions and Stretches (p.255) 3 Graph Functions Using Reflections about the x-Axis and the y-Axis

(p.258)

At this stage, if you were asked to graph any of the functions defined by y = x, y = x2 , Y = x3 , Y = Vx, Y = \Y.X, y = I x!, or y 1.. , your response should be, x "Yes,I recognize these functions and know the general shapes of their graphs." (If this is not your answer, review the previous section, Figures 29 through 35.) Sometimes we are asked to graph a function that is "almost" like one that we already know how to graph. In this section,we look at some of these functions and develop techniques for graphing them. Collectively, these techniques are referred to as transformations. =

1 EXAM P L E 1

Graph Functions Using Vertical and Horizontal Shifts Vertical Shift U p

Use the graph of f(x) = x2 to obtain the graph of g(x) Solution

=

x2 + 3.

We begin by obtaining some points on the graphs of f and g. For example, when x = 0, then y f(O) = 0 and y = g(O) = 3. When x = 1, then y = f(l) = 1 and y = g ( l ) 4. Table 7 lists these and a few other points on each graph. Notice that each y-coordinate of a point on the graph of g is 3 units larger than the y-coordinate of the corresponding point on the graph of f. We conclude that the graph of g is identical to that of f, except that it is shifted vertically up 3 units. See Figure 40. =

=

Table 7 x

Figure 40 y = fIx) = J?

-2 -1 0

4

2

4

0

y = g(x) = J? + 3

7 4 3 4 7

-3 •

SECTION 3.5

I

Graphing Techniques: Transformations

253

Exploration

Figure 41

On the same screen, graph each of the fol lowing functions: Y1 = x2

Y1

=

Y = x2 + 2 2 Y3 = x2 - 2

X2 _ - 6 1----�� 6

\

.

.

\

Result Fig u re 41 illustrates the g raphs. You should have observed a general pattern. With Y1 = x2 on the screen, the g raph of Y2 = x2 + 2 is identical to that of Y1 = x2, except that it is shifted vertically u p 2 units. The g raph of Y3 = x2 - 2 is identical to that of Y1 = x2, except that it is shifted vertically down 2 u nits.

We are led to the following conclusions: If a positive real number k is added to the right side of a function y = f(x), the graph of the new function y = f (x) + k is the graph of f shifted verti cally up k units. If a positive real number k is subtracted from the right side of a function y = f (x), the graph of the new function y = f (x) - k is the graph of f shifted vertically down k units. Let's look at another example. EXAM P L E 2

Solutio n

Vertical S hift Down

Use the graph of f(x) = x2 to obtain the graph of g(x) = x2 - 4.

Table 8 lists some points on the graphs of f and g. Notice that each y-coordinate of g is 4 units less than the corresponding y-coordinate of f. The graph of g is identical to that of f, except that it is shifted down 4 units. See Figure 42. Figure 42

Table 8 X

y= =

-2 -1 0

4

2

4

0

f(x) x2

y = =

g(x) x2 -

0 -3 -4 -3 0

4

Down 4 units

Y = X2 (2, 4)

1 -5

(0, - 4) •

_ �_ C!lll

E XA M P L E 3


H orizontal Shift to the Right

Use the graph of f(x) = x2 to obtain the graph of g(x) = (x - 2f Solution

The function g ( x) = (x - 2 f is basically a square function. Table 9 lists some points on the graphs of f and g. Note that when f(x) = 0 then x = 0, and when g(x) = 0, then x = 2. Also, when f(x) = 4, then x = -2 or 2, and when g(x) = 4, then x = 0 or 4. We conclude that the graph of g is identical to that of f, except that it is shifted 2 units to the right. See Figure 43.

254

CHAPTER 3


Table 9

Figure 43

x

g(x)

= (x

4 0 4 16

-2 0 2 4

;J

Y =

y = f(x) =�

-

Y

2)2

16 4 0 4

Y = x2

Y = (x- 2)2

( - 2, 4)

( 4, 4)

R ight 2 units

(0, 0)

(2, 0)

4

x •

Exploration

Figure 44

On the same screen, graph each of the following functions:

x2

y1 =\

Yl Y2 Y3

\ 6

x2 = (x - 3 / = (x + 2) 2

=

Result Figure 44 ill ustrates the graphs. You should observe the fol lowing pattern. With the graph of Y1 x2 on the screen, the graph of Y2 = (x - 3)2 is identical to that of Y1 x2, except that it is shifted horizontally to the right 3 u nits. The graph of Y3 = (x + 2) 2 is identical to that of Y1 x2, except that it is shifted horizonta l ly to the left 2 u n its.

=

=

=

We are led to the following conclusion. If the argument x of a function f is replaced by x h, h > 0, the graph of the new function y f(x h ) is the graph of f shifted horizontally right h units. If the argument x of a function f is replaced by x + h, h > 0, the graph of the new function y = f(x + h ) is the graph of f shifted horizontally left h units. -

=

E XA M P L E 4

Horizontal S h ift to the Left

Use the graph of f(x) Solution

-

=

x2 to obtain the graph of g(x)

=

(x + 4f

Again, the function g(x) = (x + 4) 2 is basically a square function. Its graph is the same as that of f , except that it is shifted horizontally 4 units to the left. See Figure 45. Figure 45

Y

Y = (X t 4) 2

5

( - 6 , 4)

(2, 4)

x Left 4 units • = =� -


Vertical and horizontal shifts are sometimes combined.

SECTION 3.5

E XA M P L E 5


255

C o m b i n i n g Vertical and H orizontal Sh ifts

Graph the function: f(x)

=

(x

+

3? - 5

We graph f in steps. First,we note that the rule for f is basically a square function, so we begin with the graph of y x2 as shown in Figure 46(a). Next, to get the graph of y (x + 3?, we shift the graph of y = x2 horizontally 3 units to the left. See Figure 46(b). Finally, to get t he graph of y (x + 3? - 5,we shift the graph of y (x + 3)2 vertically down 5 units. See Figure 46(c). Note the points plotted on each graph. Using key points can be helpful in keeping track of the transformation that has taken place.

Solution

=

=

=

=

Figure 46

Y 5

( -2, 1 ) -5

5 x

5 x -5

5

( -2, -4) -5 2 y = ( x + 3)

Replace x b y x + 3 ; Horizontal shift left 3 units

(b)

(a)

I

Check: Graph

=

2

3)2 -

•

(e)

•

5 and compare the g raph to Figu re 46(c).


Graph Functions Using Compressions and Stretches Vertical Stretch

Use the graph of f(x) Solution

(x +

2 Y = (x + 3) - 5

Subtract 5; Vertical shift down 5 units

In Example 5, if the vertical shift had been done first, followed by the horizon tal shift, the final graph would have been the same. Try it for yourself. ,,�-

E XA M P L E 6

f(x)

Y1 =

Vertex ( - 3, - 5)

=

Ixl to obtain the graph of g(x)

=

2 1 xl·

To see t he relationship between the graphs of f and g, we form Table 10, listing points on each graph. For each the y-coordinate of a point on t he graph of g is 2 times as large as the corresponding y-coordinate on t he graph of .f. The graph of f (x) = Ixl is vertically stretched by a factor of 2 to obtain the graph of g( ) 2 1 xl [for example, ( 1, 1 ) is on the graph of f, but ( 1 , 2) is on the graph of g]. See Figure 47. x,

x

Table 1 0

x -2

Figure 47 Y =

fIx) I xl

= 2

y =

=

g(x) 21xl

4 2

-1

0

=

0

0 2

2

2

4 •

256

C H A PTER 3


E XA M P L E 7

Vertical Compression

Use the graph of I(x) Solution

=

.

Ixl to obtam the graph of g(x)

1

=

"2 lxl.

1 For each x, the y-coordinate of a point on the graph of g is "2 as large as the corresponding y-coordinate on the graph of f. The graph of I(x)

�

compressed by a factor of to obtain the graph of g(x)

=

=

Ixl is vertically

� Ixl [for example, (2, 2)

is on the graph of I,but (2, 1 ) is on the graph of g.]. See Table 1 1 and Figure 48. Table 1 1 y = ((x) = I xl

x

2

-2 -1 o

o

2

2 •

When the right side of a function y I(x) is multiplied by a positive number a, the graph of the new function y = al(x) is obtained by multiplying each y-coordinate on the graph of y = I(x) by a. The new graph is a vertically com pressed (if 0 < a < 1 ) or a vertically stretched (if a > 1 ) version of the graph of y I(x). =

=

""

..... Now Work P R O B l EM

43

What happens if the argument x of a function y = I(x) is multiplied by a posi tive number a, creating a new function y = I(ax)? To find the answer, we look at the following Exploration. Exploration On the same screen, g raph each of the fol lowing functions: Y1

Y 2

Result

of Y1

Figure 49

=

=

((x)

=

= ((2x)

x2

= (2X) 2

You should have obtained the g raphs shown in Figure 49. The graph of Y = (2x)2 is the graph 2 x2 compressed horizontally. Look also at Table 1 2(a).

Table 1 2 X '1 1 (0 (0 .� .,� 1 1 , '1 '1 iii B Ii '1 iii 2�6 Y , a ( 2X ) 2 (a)

M

(0 1 '1 i ii Ii '1 2�1i 1(02'1

(b)

SECTION 3.5

257


Notice i n Table 1 2(a) that (1, 1 ), (2, 4), (4, 1 6), and ( 1 6, 256) are points on the graph of Y1 = x2• Also, (0.5, 1 ), ( 1 , 4). (2, 1 6), and (8, 256) are points on the graph of Y2 (2X)2. For equal y-coordinates, the =

1

x-coordinate on the graph of Y2 is - the x-coordi nate of Y1 . The graph of Y2 2 1 2 tiplying the x-coordi nate of each point on the graph of Y1 = x by "2 ' The graph of Y3

=

2

(�x) i s the graph of Y1

(2X)2 is obtained by mul-

x2 stretched horizontal ly. Look a t Table 1 2(b). Notice

and (4, 1 6) are points on the graph of Y1 x2. Also, (1 , 0.25), (2, 1 ) , (4, 4), 2 and (8, 1 6) are points on the graph of Y3 = (]"x) . For equal y-coordinates, the x-coordinate on the 2 graph of Y3 is 2 times the x-coordinate on Y1 . The graph of Y3 = '2 x is obtained by mUltiplying the that (0.5, 0.25), (1 , 1 ),

(2, 4),

=

=

=

x-coordinate of each point on the graph of Y1

(1 )2

2

x by a factor of 2.

=

If the argument x of a function y = f(x) is multiplied by a positive number the graph of the new function y = f(ax ) is obtained by multiplying each x-coordinate of y = f(x) by � . A horizontal compression results if > 1, and a,

a

a

a horizontal stretch occurs if 0

2

(0,0)

x

S o l ution

Express the area A of the rectangle as a function of x. What is the domain of A? Graph A = A(x). For what value of x is the area largest?

(a) The area A of the rectangle is A xy, where y = 25 - x2. Substituting this expression for y, we obtain A(x) = x(25 - x2) = 25x - x3 . (b) Since (x, y) is in quadrant I, we have x > O. Also, y = 25 - x2 > 0, which im plies that x2 < 25, so -5 < x < 5. Combining these restrictions, we have the domain of A as { x i O < x < 5 } , or (0, 5 ) using interval notation. (c) See Figure 59 for the graph of A = A (x). (d) Using MAXIMUM, we find that the maximum area is 48.11 at x = 2.89, each rounded to two decimal places. See Figure 60. =

Figure 60

Figure 59

50

50

.

....-. -.....

,/

/

.'

o

1//' o

,= -

. ...,;-

.......

....\

\

\

Now Work P R O B L E M 7

5

o

f, /\ \ f /1

\ HoJxif'ilu�'1 l-:=2:.8867�2:1 _ V= '18.112:52:2: � 5

o

•

266

CHAPTER 3


E XA M P L E 3

M aking a P l aypen

'"

A manufacturer of children's playpens makes a square model that can be opened at one corner and attached at right angles to a wall or, perhaps, the side of a house. If each side is 3 feet in length, the open configuration doubles the available area in which the child can play from 9 square feet to 18 square feet. See Figure 6l. Now suppose that we place hinges at the outer corners to allow for a configu ration like the one shown in Figure 62. Figure 6 1

Figure 6 2

(a) Express the area A of the configuration shown in Figure 62 as a function of the distance x between the two parallel sides. (b) Find the domain of A. (c) Find A if x = 5. (d) Graph A = A( x). For what value of x is the area largest? What is the maximum area? Sol ution

(a) Refer to Figure 62. The area A that we seek consists of the area of a rectangle (with width 3 and length x) and the area of an isosceles triangle (with base x and two equal sides of length 3). The height h of the triangle may be found using the Pythagorean Theorem. h2 +

(�y h2 h

=

=

=

32 32

-

( X2 )2 -

_ 9 -

x2 4

-

A(x)

=

36 - x2 4

---

2

area of rectangle + area of triangle 3x +

-

l Y36 - x2

The area A enclosed by the playpen is A =

_

xV36 - x2 4

=

� (� V36 - X2 )

3x + x

Now the area A is expressed as a function of x. (b) To find the domain of A, we note first that x > 0, since x is a length. Also, the expression under the square root must be positive, so 36 - x2 > ° x2 < 36 -6 < x < 6 Combining these restrictions, we find that the domain of A is ° < x < 6, or (0, 6) using interval notation. * Adapted from Proceedings, Summer Conference for College Teachers on Applied Mathematics (University of Missouri, Rolla), 197 1 .

SECTION 3.6

Figure 63

(c) If x

=

Mathematical Models: Building Functions

267

5, the area is A(5)

=

3 ( 5 ) + � V36 - ( 5 )2 4

�

19.15 square feet

If the length of the playpen is 5 feet, its area is 19.15 square feet. �I (d) See Figure 63. The maximum area is about 19.82 square feet, obtained when x � is about 5.58 feet. •

3.6 Assess Your Understanding Appl ications a n d Extensions

Let P = (x, y) be a point on the graph of y = x2 8. (a) Express the distance d from P to the origin as a function of x. (b) What is d if x = O? (c) What is d if x 1 ? / (d) Use a graphing utility to graph d = d(x). (e) For what values of x is d smallest? 2. Let P = (x, y) be a point on the graph of y = x2 8. (a) Express the distance d from P to the point (0, - 1 ) as a function of x. (b) What is d if x = O? (c) What is d if x = - 1 ? " � (d) Use a graphing utility to graph d d(x). (c) For what values of x is d smallest? 3. Let P (x, y) be a point on the graph of y Vi . ( a ) Express the distance d from P t o the point ( 1 , 0) a s a function of x. } (b) Use a graphing utility to graph d = d(x). ( c) For what values of x is d smallest? . 1 4. Let P = (x, y) be a pomt on the graph of y = - . 1.

-

=

third on the positive x-axis at (x, 0). Express the area A of the triangle as a function of x. 7. A rectangle has one corner in quadrant I on the graph of y = 16 - x2, another at the origin, a third on the positive y axis, and the fourth on the positive x-axis. See the figure.

-

=

=

=

;� 8.

y

x

(a) Express the distance d from P to the origin as a func tion of x. tIl (b) Use a graphing utility to graph d d(x). ( c ) For what values of x is d smallest? 5. A right triangle has one vertex on the graph of y = x3, X > 0, at (x, y), another at the origin, and the third on the positive y-axis at (0, y), as shown in the figure. Ex press the area A of the triangle as a function of x.

(a) Express the area A of the rectangle as a function of x. (b) What is the domain of A ? (c) Graph A = A(x). For what value of x is A largest? A rectangle is inscribed in a semicircle of radius 2. See the figure. Let P = ( x, y) be the point in quadrant I that is a vertex of the rectangle and is on the circle.

y = .J4

x2 x

=

•

9.

(a) Express the area A of the rectangle as a function of x. (b) Express the perimeter p of the rectangle as a function of x. ( c ) Graph A = A (x). For what value of x is A largest? (d) Graph p = p( x). For what value of x is p largest? A rectangle is inscribed in a circle of radius 2. See the figure. Let P = (x, y) be the point in quadrant I that is a vertex of the rectangle and is on the circle.

x -

6. A right triangle has one vertex on the graph of y = 9 - x2, X > 0, at (x, y), another at the origin, and the

-

2

268

Ij 10.

C H A PTER 3


(a) Express the area A of the rectangle as a function of x. (b) Express the perimeter p of the rectangle as a function of x. (c) Graph A = A (x ) . For what value of x is A largest? (d) Graph p = p( x ) . For what value of x is p largest? A circle of radius r is inscribed in a square. See the figure.

16.

x2 [Hint: First show that r2 = 3 ']

17.

1 1.

(a) Express the area A of the square as a function of the radius r of the circle. (b) Express the perimeter p of the square as a function of r. A wire 1 0 meters long is to be cut into two pieces. One piece will be shaped as a square, and the other piece will be shaped as a circle. See the figure.

I 1

10

An equilateral triangle is inscribed in a circle of radius r. See the figure in Problem 16. Express the area A within the cir cle, but outside the triangle, as a function of the length x of a side of the triangle. Two cars leave an intersection at the same time. One is head ed south at a constant speed of 30 miles per hour, and the other is headed west at a constant speed of 40 miles per hour (see the figure). Express the distance d between the cars as a function of the time t. [Hint: At t = 0, the cars leave the intersection.]

o

m

10

18.

[Q] x

4x

An equilateral triangle is inscribed in a circle of radius r. See the figure. Express the circumference C of the circle as a func tion of the length x of a side of the triangle.

-

4x

(a) Express the total area A enclosed by the pieces of wire as a function of the length x of a side of the square. (b) What is the domain of A ? .r (c) Graph A = A ( x ) . For what value of x is A smallest? 12. A wire 10 meters long is to be cut into two pieces. One piece will be shaped as an equilateral triangle, and the other piece will be shaped as a circle. (a) Express the total area A enclosed by the pieces of wire as a function of the length x of a side of the equilateral triangle. (b) What is the domain of A ? , (e) Graph A = A ( x ) . For what value of x is A smallest? 13. A wire of length x is bent into the shape of a circle. (a) Express the circumference of the circle as a function of x. (b) Express the area of the circle as a function of x. 14. A wire of length x is bent into the shape of a square. (a) Express the perimeter of the square as a function of x. (b) Express the area of the square as a function of x. 15. A semicircle of radius r is inscribed in a rectangle so that the diameter of the semicircle is the length of the rectangle. See the figure.

19.

1� 20.

Two cars are approaching an intersection. One is 2 miles south of the intersection and is moving at a constant speed of 30 miles per hour. At the same time, the other car is 3 miles east of the intersection and is moving at a constant speed of 40 miles per hour. (a) Express the distance d between the cars as a function of time t. [ Hint: At t = 0, the cars are 2 miles south and 3 miles east of the intersection, respectively.] (b) Use a graphing utility to graph d = d(t). For what value of t is d smallest? Inscribing a Cylinder in a Sphere Inscribe a right circular cylinder of height h and radius r in a sphere af fixed radius R. See the illustration. Express the volume V of the cylinder as a function of h. [Hint: V = 7Tr2h. Note also the right triangle.] r

T h

(a) Express the area A of the rectangle as a function of the radius r of the semicircle. (b) Express the perimeter p of the rectangle as a function of r.

Sphere

1

Chapter Review

Inscribe a right circular cylinder of height h and radius r in a cone of fixed radius R and fixed height H. See the illustration. Express the volume V of the cylinder as a function of r. [Hint: V = 7Tr2h. Note also the similar triangles.)

21. Inscribing a Cylinder in a Cone

An island is 2 miles from the nearest point P on a straight shoreline. A town is 12 miles down the shore from P. See the illustration.

23. Time Required to Go from an Island to a Town

'E N

I t

\R

Cone

MetroMedia Cable is asked to provide service to a customer whose house is located 2 miles from the road along which the cable is buried. The nearest con nection box for the cable is located 5 miles down the road. See the figure.

269

"

12-x�x ", ".,,-" 1 2Jni f(c). The average rate of change of f from a to b is -

tl y tlx

feb) - f(a) b-a

a"* b

Objectives --------� Section

You should be a ble to ...

Review Exercises

3.1

Determine whether a relation represents a function (p. 208) Find the value of a fun ction (p. 212) Find the domain of a function (p. 215) Form the sum, difference, product, and quotient of two functions (p. 217)

1,2 3-8, 23, 24, 7 1 , 72, 73 9-16 17-22

Identify the graph of a function (p. 223) Obtain information from or about the graph of a function (p. 224)

47-50 25(a)-(e), 26(a)-(e), 27(a), 27(d) 27(f), 28(a), 28(d), 28(f)

Determine even and odd functions from a graph (p. 231 ) Iden tify even and odd functions from the equation (p. 232) Use a graph to determine where a function is increasing, decreasing, or constan t (p. 233) Use a graph to locate local maxi ma and local minima (p. 234) Use a graphing utility to approximate local maxima and local minima and to determjne where a function is increasing or decreasing (p. 235) Find the average rate of change of a function (p. 236)

27(e), 28(e) 29-36 27(b), 28(b) 27(c), 28(c)

Graph the functions listed in the library of functions (p. 244) G raph piecewise-defined functions (p. 247)

5 1 -54 67-70

Graph functions using vertical and horizontal shifts (p. 252) Graph functions using compressions and stretches (p. 255) Graph functions using reflections about the x-axis and the y-axis (p. 258)

25(f), 26(f), 26(g) 55, 56, 59-66 25(g), 26(h), 57, 58, 65, 66 25(h), 57, 61, 62, 66

Build and an alyze function s (p. 264)

74-76

2

3

4

3.2 2

3.3 2

3

�

4 5

6

3.4 2

3.5 2

3

3.6

37-40, 75(e), 76(b) 41-46

Review Exercises

In Problems 1 and 2, determine whether each relation represents a function. For each function, state the domain and range. 2. {( 4, - 1 ), (2, 1 ) , (4, 2) } 1. { ( -1, 0), (2, 3 ) , (4, 0)} In Problems 3-8, find the following for each function: (c) f( -x ) (b) f( - 2 ) (a) f(2) x2 3x 4. f(x) = 3. f(x) = 1 x +1 r- 1

--

(d) -f(x)

(e) f(x

-

2)

5 . f(x) =

(f) f(2x)

�

272

CHAPTE R 3

6. f(x )

= Ix2 - 41

In Problems

9. f(x ) 13. h(x )


=

=

7. f(x )

20. f(x )

=

=

x2 - 4 x2

find the domain of each function. 3x2 x 10. f(x ) = -2x 2 x -9

9-16,

=� x 15. f(x ) = x2 + 2x

_

Ixl 14. g(x ) = �

\IX

Ixl

+

2 - x; g(x ) 3x; g(x )

8. f(x )

11. f(x)

-

In Problems 1 7-22, find f

17. f(x)

=

= x2 x3 9 _

12.

_

f(x)

16. F(x )

3

= Vx+2 =

= 3x + 1

= 2x - 1; g(x ) = 2x + 1 1 x+ 1 21. f(x) --; g(x ) = x x- 1

= 1 + x + x2

=

f(x

+

25. Using the graph of the function f shown: y 4

19. f(x)

�;

-

=

-

X

+

4

26. Using the graph of the function g shown: ( -5,1) .

y 3 (-1,1) ...

-5

x

5

(a) (b) (c) (d) (e) (f)

h) - f(x ) h "* 0 h 24. f(x ) 3x2 - 2x

(3,3)

x

5 (3, -3)

-4

Fi nd the domain and the range of f . List the intercepts. Find f( -2). For what value of x does f(x ) -3? Solve f(x) > O. Graph y f(x - 3).

=

= (g) Graph y = f G x ) . (h) Graph y = -f(x) .

(a) (b) (c) (d) (e) (f) (g) (h)

Find the domain and the range of g. Find g(- 1 ) . List the intercepts. For what value of x does g(x ) -3? Solve g(x ) > O. Graph y g(x - 2). Graph y g(x ) + l. Graph y 2g(x ) .

===

=

In Problems 27 and 28, use the graph of the fitnction f to find: (a) The domain and the range of f . (b) The intervals on which f is increasing, decreasing, or constant. (c) The local minima and local maxima. (d) Whether the graph is symmetric with respect to the x-axis, the y-axis, o r the origin. (e) Whether the function is even, odd, or neither. (j) The intercepts, if any. 27. 28. y (-1,1 ) -5

3

(4,0)

5 x

(3, -3)

-

4

= 3x2 + X + 1 ; g(x ) = 3x 3 1 22. f(x) = -- g(x ) = x - .)

18. f(x )

= -2x2 + X + 1

(-4, -3)

_

g, f - g, f . g, and L for each pair of functions. State the domain of each one. g

In Problems 23 and 24, find the difference quotient of each function f ; that is, flnd

23. f( x)

x2

1 3x

6 x

Chapter Review

In Problems 29-36, determine (algebraically) whether the given function is even, odd, or neitha 4 + x2 1 1 29. f(x) = x3 - 4x 30. g(x) = --4 31. hex) = 4 + 2" + 1 1 +x x x

33. G ( x) = 1 - x + x3 ;,�

34. H (x) = I + x + x2

35. f(x)

32. F(x)= � 1 + x2 3 36. g(x) = -X

x + x2

_ = _

l

273

In Problems 37-40, use a graphing utility to graph each function over the indicated interval. Approximate any local maxima and local minima. Determine where the function is increasing and where it is decreasing. 37. f(x) = 2x3 - Sx + I (-3,3) 38. f(x) = -x3 + 3x - 5 (-3,3)

2X4 - Sx3

2x + 1 (-2,3)

40. f(x)

=

-x4 + 3x3 - 4x

In Problems 41 and 42, find the average rate of change of f: ( c) From 2 to 4 (b) From ° to 1 (a) From 1 to 2 = 2 x 8x 41. f(x) 42. f(x)

=

2x3 + X

39. f(x)

=

+

+

3 ( -2, 3 )

In Problems 43-46, find the average rate of change from 2 to 3 for each function f. Be sure to simplify. 43. f(x) = 2 - Sx 44. f(x) = 2x + 7 46. f(x) = x2 - 3x + 2 45. f(x) = 3x - 4x2

2

In Problems 47-50, is the graph shown the graph of a function? 47. 48. y y

x

49.

50.

y

x

x

x

In Problems 51-54, sketch the graph of each function. Be sure to label at least three points. 51. f(x) = Ixl 52. f(x) = \YX 53. f(x) = �

54. f(x) = ..!.. x

In Problems 55-66, graph each function using the techniques of shifting, compressing or stretching, and reflections. Identify any intercepts on the graph. State the domain and, based on the graph, find the range. I 57. g(x) = -21xl 55. F(x) = Ixl - 4 56. f(x) = Ixl + 4 58. g(x) = 21xl

60. hex) = vx - 1

63. hex) = (x - If

+

64. h e x ) = (x + 2f -

2

In Problems 67-70, (a) Find the domain of each function. ( c ) Graph each functiol1. if -2 < x :s; 1 67. f(x) = if x > 1 x+ 1

{3X

69. f(x) =

{

1 3x \:

if -4 :s; x < 0 if x = 0 if x > 0

71. A function f is defined by

If f(l) = 4, find A .

Ax + 5 f(x) = -6x - 2

3

61. f(x)

=

62. f (x) = -Vx + 3

�

66. g(x) = -2(x + 2)3 - 8

65. g(x) = 3(x - 1)3 + 1

(b) Locate any intercepts. ( d ) Based on the graph, find the range. i f -3 < x < 0 x-1 68. f ( x) = if x :2: 0 3x - 1

{

70. f(x) =

{X2

2x - 1

if -2 :s; x :s; 2 if x > 2

72. A function g is defined by g(x)

If g( -1) = 0, fi nd A.

A

=

x

8 +? r

274

C H A PTER 3

73. Spheres


The volume

V of a

r

sphere of radius is V

r

=

�'7Tr3;

the surface area S of this sphere is S 4'7T 2 . If the radius doubles, how does the volume change? How does the surface area change? 1 74. Page Design A page with dimensions of 8 inches by 2" 11 inches has a border of uniform width x surrounding the printed matter of the page, as shown in the figure. =

8� in.

Cd) Graph the function A = A(x). (c) Use TRACE to determine what margin should be used to obtain areas of 70 square inches and 50 square inches. 75. Material Needed to Make a Drum A steel drum in the shape of a right circular cylinder is required to have a volume of 1 00 cubic feet. (a) Express the amount A of material required to make the drum as a function of the radius of the cylinder. (b) How much material is required if the drum is of radius 3 feet? (c) How much material is required if the drum is of radius 4 feet? (d) How much material is required if the drum is of radius 5 feet? ?� (e) Graph A = A ( ). For what value of is A smallest? Hint: The volume V and surface area S of a right circular cylinder are S = 2m.2 + :,�

r

r

where

(a) Express the area A of the printed part of the page as a function of the width x of the border. (b) Give the domain and the range of A . (c) Find the area of printed page for borders of widths 1 inch, 1.2 inches, and 1 .5 inches.

r

h is the height and r is the radius

2'7Trh

76. A rectangle has one vertex in quadrant I on the graph of

y = 10 x2, another at the origin, one on the positive x-axis, and one on the positive y-axis. (a) Express the area A of the rectangle as a function of x. (b ) Find the largest area A that can be enclosed by the rectangle. -

'��

CHAPTER TEST 1. Determine whether each relation represents a function. For

each function, state the domain and range. (a) {(2 , 5 ), (4, 6), (6, 7), (8, 8) } (b) { (1, 3), (4, 2) ( -3 5) ( 1, 7 )} (c) -

,

,

,

In Problems 2-4, find the domain of each function and evaluate each function at x = - 1 . 2. f(x) � x + 2 3. g (x) x+2 I-I x-4 4. hex) = 2 X + 5x 36 5. Using the graph of the function f: =

=

--- -

y

-4

-2

4 (1 , 3)

y

(d)

6

( - 5 , -3)

2 -4

-2

2 -2

4

x

(a) (b) (c) (d) (e)

-4

(3, - 3)

Find the domain and the range of f. List the intercepts. Find f ( l). For what value(s) of x does f(x) = -3 ? Solve f(x) < O.

Cumulative Review

LJ

6. Use

a graphing utility to graph the function f(x) -x4 + 2x3 + 4x2 - 2 on the interval ( -5, 5). Ap proximate any local maxima and local minima rounded to two decimal places. Determine where the f unction is in creasing and where it is decreasing. =

{

. 2X + 1 . 7. Consider the function g( x) = x-4

8. 9.

if x if x

6

9. 14x

+

11 :2! 7

= (-2, -3) to P 2 = (3, -5). (b) What is the midpoint of the line segment from P1 to P ? 2 (c) What is the slope of the line containing the points PI to P2 ?

10. (a) Find the distance from PI

In Problems 11-14, graph each equation. 11. 3x - 2y = 12 13. x2

+ (y - 3)2

=

16

15. For the equa tion 3x2 - 4 y = 12, find the intercepts and check for symmetry. In Problems 17-19, graph each function. 1 18. f(x) = 17. f(x) (x + 2)2 - 3 x =

12.

x

14. y

=i =

V;

16. Find the slope-intercept form of the equa tion of the line con taining the points ( - 2,4) and (6, 8).

19. f(x) =

{2 - x Ixl

2

if x ::5 if x > 2

276

CHAPTER 3


CHAPTER PROJECTS I.

This month, as you were paying your bills, you noticed tha t your cell phone service contract had expired. Since cell phone numbers are now portable, you decided to investigate different companies. Since you regularly travel outside your local calling area, you decide to look only at plans with no roaming charges. Here are your findings: Cell Phone Service

Anytime Minutes Included

Charge for Each Extra Minute

Mobile-to-Mobile Minutes

National Long Distance

Nights and Weekends

Plan A 1 : $39.99

450 with rollover

$0.45

Unlimited

Included

5000 (9 PM-6 AM)

Plan A2: $59.99

900 with rollover

$0.40

Unlimited

Included

Unlimited (9 PM-6 AM)

600

$0.40

None

Included


1 000

$0.40

None

Included


55 1 -1050, $5 for each

Unlimited

Included


Unlimited

Included


Company A:

Company B: Plan 81: $39.99 Plan 82: $49.99

Company C: Plan ( 1 : $59.99

550

50 minutes; Above 1 050, $0. 1 0 each minute

Plan C2: $69.99

800

80 1 - 1 300, $5 for each 50 minutes; Above 1 300, $0. 1 0 each minute

Based on rates from the Web sites of the companies Cingular, T Mobile and Sprint PCS for the zip code 76201 on May 20, 2006. (www.cingular.com , www. t-mobile. com, www.sprintpcs.com)

Source:

Each plan requires a 2-year contract. 1.

Determine the total cost of each plan for the life of the con tract, assuming that you stay within the allotted anytime min utes provided by each contract. 2. If you expect to use 400 anytime, 200 mobile-to-mobile, and 4500 night and weekend minutes per month, which plan pro vides the best deal? If you expect to use 400 anytime, 200 mo bile-to-mobile, and 5500 night and weekend minutes per month, which plan provides the best deal? If you expect to use 500 anytime, and 1 000 mobile-to-mobile, and 2000 night and weekend minutes per month, which plan provides the best deal? 3. Ignoring any mobile-to-mobile and night and weekend usage, if you expect to use 850 anytime minutes each month, which option provides the best deal? What if you use 1050 anytime minutes per month?

4. Each monthly charge includes a specific number of peak time

minutes in the monthly fee. Write a function for each option, where C is the monthly cost and x is the number of anytime minutes used. 5. Graph each function from part 4. 6. For each of the companies A, B, and C, determine the aver age price per minute for each plan, based on no extra min utes used. For each company, which plan is better? 7. Looking at the three plans that you found to be the best for compa nies A, B, and C, in part 6, which of these three seems to be the best deal? 8. Based on your own cell phone usage, which plan would be the best for you?

The following projects are available on the Instructor's Resource Center (IRC): Wireless Internet Service Use functions and their graphs to analyze the total cost of various wireless In ternet service plans. III. Cost of Cable When government regulations and customer preference influence the path of a new cable line, the Pythagorean Theorem can be used to assess the cost of installation. IV. Oil Spill Functions are used to analyze the size and spread of an oil spill from a leaking tanker.

II. Project at Motorola:

Linear and Quadratic Functions Valuing Hewlett-Packard 17 MAY 2006-With the market focused on Hewlett-Packard Co. (HPQ) taking market share from Dell Inc. (DELL), the casual observer might be fooled into mistaking it as a computer hardware company. It's not; while the Compaq acquisition of 2002 expanded the company's hardware business, 53.4 percent of H PQ's profits still come from its imaging and printing division. When we value stocks, we use the capital asset pricing model to price shares, which ex amines the volatility of the stock through beta and then projects how much earnings growth would be needed to justify the current stock price. Historically, HPQ has a beta of 2.05. The computer peripherals industry average, though, is 1.98. But the industry average beta for the computer hardware companies is 1 .58. A weighted average beta for the two industries seems appropriate for a company that walks the line between them. In this case, a beta weighted between the computer peripher als and computer hardware industries by the percentages of HPQ's profits yields an expected beta of 1 .8. If we use this beta in a capital-asset-pricing model, assuming that Hewlett-Packard will maintain a price-to-earnings to-growth ratio of 1 .3 and that full fiscal year 2006 earnings per share will be at the low end of management's guidance, $2.04, then a new position taken at the current price would require an annualized, per-share earnings growth rate of 9.9% over the next five years to break even. The average long-term growth rate among the 12 analysts who provide one is 12.3 % for Hewlett-Packard; this is well above what our analysis indicates that we require, implying a strong undervaluation even after the price spike. Source: Adapted from Paul DeMartino, "Valuing Hewlett-Packard," May 17, 2006, www.reuters.com . A dapted with permission.


A Look Back Up to now, o u r d i scussion has focused on g raphs of equations and the idea of a function. We have learned how to graph equations using the point-plotting method, intercepts, and the tests for symm etry. In addition, we have learned what a function is and how to identi fy whether a relation represents a function. We also in troduced properties of functions, such as domain/range, increasing/decreasing, even/odd, and average rate of change.

A Look Ahead Going forward, we w i l l l ook at classes of functions. In this chapter, we focus on linear and quad ratic functions and their properties and appl ications.

Outline

4.1 linear Functions and Their Properties 4.2 Building linear Functions from Data

4.3 Quadratic Functions and Their Properties 4.4 Quadratic Models; Building Quadratic Functions from Data

4.5 Inequalities Involving Quadratic Functions

Chapter Review Chapter Test Cumulative Review Chapter Projects

277

278

CHAPTER 4

Linear and Quadratic Functions

4.1 Linear Functions and Their Properties PREPARING FOR THIS SECTION •

• •


Lines (Section 2.3, pp. 173-185) Graphs of Equations in Two Variables; Intercepts; Symmetry (Section 2.2, pp. 163-171 ) Linear Equations (Section 1.1, pp. 86-93)

•

Now Work the 'Are You Prepared?' problems on page 284.

•

Functions (Section 3.1, pp. 208-219) The Graph of a Function (Section 3.2, pp. 222-226) Properties of Functions (Section 3.3, pp. 231-238)

OBJECTIVES 1 Gra ph Linear Functions (p. 278)

2 Use Average Rate of Change to Identify Linear Functions (p. 278) 3 Determ ine Whether a Linear Function is Increasing, Decreasing,

o r Constant (p. 281)

4 Work with Applications of Linear Functions (p. 282)

1

Graph Linear Functions

In Section 2.3 we discussed lines. In particular, we developed the slope-intercept form of the equation of a line y = mx + b. When we write the slope-intercept form of a line using function notation, we have a linear function.

DEFINITION

A

linear function

is a function of the form

f(x) = mx + b The graph of a linear function is a line with slope m and y-intercept b. Its domain is the set of all real numbers.

E XAM P L E 1 Figure

1 Y

G raphing a Linear Function Graph the linear function:

(0 , 7)

.J

f(x) = -3x + 7

This is a linear function with slope m = -3 and y-intercept b = 7. To graph this function, we start by plotting the point (0, 7), the y-intercept, and use the slope to find an additional point by moving right 1 unit and down 3 units. See Figure 1 .

Solution

•

Alternatively, we could have found an additional point by evaluating the func tion at some x =F- O. For x = 1 , we find f(l) = -3 ( 1 ) + 7 = 4 and obtain the point ( 1 , 4 ) on the graph.

5 x

I
2

Now Work

P R O B L EMS 1 3 (a ) AND (b )

Use Average Rate of Change to Identify Linear Functions

Look at Table 1, which shows certain values of the independent variable x and cor responding values of the dependent variable y for the function f(x) = -3x + 7. Notice that as the value of the independent variable, x, increases by 1 the value of the dependent variable y decreases by 3. That is, the average rate of change of y with respect to x is a constant, -3.

SECTION 4.1

Table

1 x

y=f(x)=-3x+7

-2

Linear Functions and Their P roperties

279

!l.y Average Rate of Change =!l.x

13 10- 13

-----:-----,--

-1 -( - 2)

-3 1

= - =

-3

10

-1

7

- 10

-3

--- = - =

0-(-1) o

1

-3

7

-3 4

-3 2

-3 -2

3

It is not a coincidence that the average rate of change of the linear function

� f(x) = - 3x + 7 is the slope of the linear function. That is, y = m = -3. The fol �x lowing theorem states this fact. THEOREM

Average Rate of Change of a Linear Function

Linear functions have a constant average rate of change. That is, the average rate of change of a linear function f(x) = mx + b is

I

�y =m �x

_.J

� -------'---

Proof

The average rate of change of f(x) = mx + b from X l to x2 , X l

�y �x

=1=

X2 , is

(mx2 + b) - (mxl + b) ! ( X2 ) - ! ( Xl) X 2 - Xl X2 Xl mX2 - mXl m(x2 - X l) = nl x2 - Xl x2 - X l

•

B ased on the theorem just proved, the average rate of change of the function

2 2 g(x) = - - x + 5 is - -. 5

:!I!l

Now Work

5

PROBLEM

1 3(c)

As it turns out, only linear functions have a constant average rate of change. Because of this, we can use the average rate of change to determine whether a par ticular function is linear or not. Functions that are not linear are said to be nonlinear.

E XA M P L E 2

U sing the Average Rate of Change to Identify Linear Functions (a) A strain of E-coli Beu 397-recA441 is placed into a Petri dish at 30° Celsius and allowed to grow. The data shown in Table 2 are collected. The population is measured in grams and the time in hours. Plot the ordered pairs (x, y) in the Cartesian plane and use the average rate of change to determine whether the function is linear.

280

CHAPTER 4


(b) The data in Table 3 represent the maximum number of heartbeats that a healthy individual should have during a IS-second interval of time while exercising for different ages. Plot the ordered pairs (x, y ) in the Cartesian plane and use the average rate of change to determine whether the function is linear. Table 3

Table 2

r

j�

l�

Age, x

Maximum Number of Heart Beats, y

(x,y)

(0,0.09)

20

50

(20,50)

(1,0.12)

30

47.5

(30,47.5)

Time (hours), x

Population (grams), y

(x, y)

0

0.09 0.12

�

2

0.16

(2,0.16)

40

45

(40,45)

3

0.22

(3,0.22)

50

42.5

(50,42.5)

4

0.29

(4,0.29)

60

40

(60,40)

5

0.39

(5,0.39)

70

37.5

(70,37.5)

Source: American Heart Association

We compute the average rate of change of each function. If the average rate of change is constant, the function is linear. If the average rate of change is not constant, the function is nonlinear.

Solution

(a) Figure 2 shows the points listed in Table 2 plotted in the Cartesian plane. Notice that it is impossible to draw a straight line that contains all the points. Table 4 displays the average rate of change of population. Table 4

Figure 2 >-

y

Time (hours), x

a

c::

�

:::J Cl.

a 0..

•

0.3 0.2 0.1

Ily

Average Rate of Change =

e;;- OA E � 9

Population (grams), y

•

0

•

•

2

•

0

Ilx

0.09 0.12 - 0.09

•

1

-

0

=

0.03

0. 1 2 0.04

3

4

Time (hou rs), x

5

X

2

0. 1 6 0.06

3

0.22 0.07

4

0.29 0. 1 0

5

0.39

Because the average rate of change is not constant, we know that the func tion is not linear. In fact, because the average rate of change is increasing as the value of the independent variable increases, we say that the function is in creasing at an increasing rate. So not only is the population increasing over time, but it is also growing more rapidly as time passes. (b) Figure 3 shows the points listed in Table 3 plotted in the Cartesian plane. We can see that the data in Figure 3 lie on a straight line. Table 5 contains the av erage rate of change of the maximum number of heartbeats. The average rate of change of the heartbeat data is constant, -0.25 beats per year, so the func tion is linear.

SECTION 4.1

Figure 3

Table

y

Linear Functions an d Their Properties

5

50

Age, x

Maximum Number of Heartbeats, y

20

50

281

Average Rate of !!.y Change !!.x =-

(/)

� Q)

co

47.5 - 50

45 � Q)

30 - 20

I

30

47.5

40

45

50

42.5

60

40

70

37.5

=

-0.25

-0.25

40 20

30

40

50

60

x

70

-0.25

Age

-0.25

-0.25

�

3

=--

Now Work

• PROBLEM 21

Determine Whether a Linear Function Is Increasing, Decreasing, or Constant

Look back at the Seeing the Concept on page 176. We know that when the slope m of a linear function is positive ( m > 0 ) the line slants upward from left to right. When the slope m of a linear function is negative ( m < 0 ) , the line slants downward from left to right. When the slope m of a linear function is zero ( m = 0 ) , the line is horizontal. Based on these results, we have the following theorem. THEOREM

Increasing, Decreasing, and Constant Linear Functions

A linear function f (x) = mx + b is increasing over its domain if its slope, m, is positive. It is decreasing over its domain if its slope, m, is negative. It is constant over its domain if its slope, m, is zero.

--1

E XAM P LE 3

Determining Whether a Linear F unction I s Increasing, Decreasing, or Constant Determine whether the following linear functions are increasing, decreasing, or constant. (a) f ( x )

5x - 2 3 (c) set ) = - t - 4 4

Solution

(b) g (x) = -2x + 8

=

(d) h ( z)

=

7

(a) For the linear function f(x) = 5x - 2, the slope is 5, which is positive. The function f is increasing on the interval ( -00, (0 ) .

(b) For the linear function g( x) = -2x + 8 , the slope is -2, which is negative. The function g is decreasing on the interval ( -00,00 ) .

- 4 , the slope is 2, which is positive. The func4 4 tion s is increasing on the interval ( 00, (0 ) .

( c) For the linear function s ( t) =

2t

-

(d) We can write the linear function h as h ( z) = Oz + 7. Because the slope is 0, the function h is constant on the interval ( -00, 00 ) . ,.

1_

Now Work

PROBLEM 1

3(d)

•

282

CHAPTER 4


4

Work with Applications of Linear Functions

There are many applications of linear functions. Let's look at one from accounting.

E XA M P LE 4

Straight-line Depreciation Book value is the value of an asset that a company uses to create its balance sheet. Some companies depreciate their assets using straight-line depreciation so that the value of the asset declines by a fixed amount each year. The amount of the decline depends on the useful life that the company places on the asset. Suppose that a com pany just purchased a fleet of new cars for its sales force at a cost of $28,000 per car. The company chooses to depreciate each vehicle using the straight-line method over $28,000 7 years. This means that each car will depreciate by = $4000 per year. 7 (a) Write a linear function that expresses the book value V of each car as a function of its age, x. (b) Graph the linear function. (c) What is the book value of each car after 3 years? (d) Interpret the slope. (e) When will the book value of each car be $8000? [Hint: Solve the equation Vex) = 8000.]

Solution

(a) If we let Vex) represent the value of each car after x years, then V(O) represents the original value of each car, so V (O ) = $28,000. The y-intercept of the linear function is $28,000. Because each car depreciates by $4000 per year, the slope of the linear function is -4000. The linear function that represents the book value V of each car after x years is Vex) = -4000x + 28,000 (b) Figure 4 shows the graph of V.

Figure 4

(c) The book value of each car after 3 years is 28,000

V ( 3 ) = -4000 ( 3 ) + 28,000

24,000 _

�

�

= $16,000

20,000

(d) Since the slope of V (x) = -4000x + 28,000 is -4000, the average rate of change of book value is -$4000/year. So for each additional year that passes the book value of the car decreases by $4000.

16,000

g 12,000 en

(e) To find when the book value is $8000, we solve the equation

8000

Vex)

4000 2 3 4 5 6 7 Age of vehicle (years)

=

8000

-4000x + 28,000 = 8000 -4000x = -20,000 x =

-20,000 = 5 -4000

Su btract 28,000 from each side. Divide by -4000.

The car will have a book value of $8000 when it is 5 years old. ""=>-

Now Work

•

P R O B LE M 4 7

Next we look at an application from economics.

E XA M P L E 5

Supply and Demand The quantity supplied of a good is the amount of a product that a company is will ing to make available for sale at a given price. The quantity demanded of a good is the amount of a product that consumers are willing to purchase at a given price.

SECTION 4.1

Linear Functions an d Their Properties

283

Suppose that the quantity supplied, S, and quantity demanded, D, of cellular tele phones each month are given by the following functions: S(p) D(p)

60p - 900

=

=

- 1 5p + 2850

where p is the price (in dollars) of the telephone. (a) The equilibrium price of a product is defined as the price at which quantity sup plied equals quantity demanded. That is, the equilibrium price is the price at which S(p) = D(p). Find the equilibrium price of cellular telephones. What is the equil ibrium quantity, the amount demanded (or supplied) at the equilibrium price? (b) Determine the prices for which quantity supplied is greater than quantity de manded. That is, solve the inequality S(p) > D ( p ) . (c) Graph S = S(p), D = D(p) and label the equilibrium price.

Solution

(a) To find the equilibrium price, we solve the equation S(p) = D ( p ) . 60p - 900

=

- 1 5p + 2850

60p

=

- 1 5p + 3750

S(p) = 60p - 900; D(p) = -15p + 2850 Add 900 to each side.

75 p = 3750 p

=

Add

15p to each side.

Divide each side by 75.

50

The equilibrium price is $50 per cellular phone. To find the equilibrium quan tity, we evaluate either S(P) or D(P) at p = 50. S(50)

=

60(50) - 900

=

2100

The equilibrium quantity is 2100 cellular phones. At a price of $50 per phone, the company will produce and sell 2100 phones each month and have no short ages or excess inventory.

(b) We solve the inequality S(p) > D ( p ) .

60p - 900 > - 15p + 2850 60p > -15p + 3750 75p > 3750

S(p) > D(p) Add 900 to each side. Add

15p to each side.


p > 50

If the company charges more than $50 per phone, quantity supplied will exceed quantity demanded. In this case the company will have excess phones in in ventory. (c) Figure 5 shows the graphs of S point labeled. Figure

S, D

5 _"0 "0 '" '" "0 :.= C '"

C. C.E (/)0 .2:'»

::::l '"

:.;::::::;-� c � '" C ::::l '" a::::l a

=

S(p) and D = D(p) with the equilibrium

S= S(p)

3000 2000 1 000

50 Price ($) ==�

Now Work

P R O B L EM 4

1

•

284

CHAPTER 4


4. 1 Assess Your Understa n ding JAre You Prepared?' Answers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red.

1. Graph y

2 x - 3. (pp. 1 73- ( 85) 2. Find the slope of the line joining the points (2, 5) and ( -1, 3 ) . (pp. 1 73-1 74) =

3. Find the average rate of change of f(x)

=

2 to 4. (pp. 23 1 -238)

3x2 - 2, from

=

4. Solve: 60x - 900 5. If f(x) 6.

=

- 1 5x +

2850. (pp. 86-93)

x2 - 4, find f( -2). (pp. 2 08-2 1 9)

True or False The graph of the function f(x) creasing on the interval (0, oc ) . (pp. 23 1-238)

=

x2 is in


7. For the graph of the linear function f(x)

and b is the

=

8. For the graph of the linear function H ( z )

slope is 9.

__

and the y-intercept is

__

.

mx + b, m is the =

-4z + 3, the

If the slope m of the graph of a linear function is function is increasing over its domain.

__

, the

True or False The slope of a line is the average rate of change of the linear function. 11. True or False If the average rate of change of a linear func-

10.

tion 12.

iS�, then if

True or False

y

increases by 3 , x will increase by 2.

The average rate of change of f(x)

is 8.

2x + 8

=

Skill Building

In Problems 13-20, a linear fitnction is given. (a) Determine the slope and y-intercept of each function. (b) Use the slope and y-intercept to graph the linear function. (c) Determine the average rate of change of each function. (d) Determine whether the linear function is increasing, decreasing, or constant. 15. hex) = 14. g(x) = 5x - 4 13. f(x) = 2 x + 3 17. f( X )

=

�

X

18. hex)

-3

=

2 --x 3

+

4

19. F(x)

=

- 3x +

16. p( x)

4

20. G(x)

4

=

=

-x

+ 6

-2

In Problems 21 -28, determine whether the given function is linear or nonlinear. If it is linear, determine the slope. 21.

x

22.

x

y =f (x) 1 /4

-2

4

-2

0

-2

0

2

-8

-1

25.

y =f (x)

x

-1

-5

y =f (x)

2

26.

x

-2

-26

-2

0

2

0

-1

2

-4 -2

-10

-1

2

23.

2

4

y = f (x) -4

-3.5

-3

- 2.5

-2

-8

-2

-4

0

0

4

-2 0 2

27.

y =f (x)

y = f (x)

-1

1 /2

x -2

-1

0 2

24.

x

x

-1

-3

2

0

y =f (x)

28.

x

0 8

12

y = f (x)

8

-2

0

8

0

4

2

16

8

8

8

-1

9

Applications and Extensions = 4x - 1 and g(x) = - 2 x + 5. (a) So lve f(x) = O . (b) Solve f(x) > O. (c) Sol ve f(x) = g(x). (d) Solve f(x) :s g(x). (e) Graph y = f(x) and y = g(x) and label the point that represents the solution to the equation f(x) = g(x).

29. Suppose that f(x)

= 3 x + 5 and g ( x ) = - 2 x + 15. (a) Solve f(x) = O . (b) Solve f(x) < O. (c) Solve f(x) = g(x). (d) Solve f(x) 2: g(x). (e) Graph y = f(x) and y = g(x) and label the point that represents the solution to the equation f(x) = g(x).

30. Suppose that f(x)

SECTION 4.1

31. In parts (a)-(f), use the following figure.

Linea r Functions an d Their Properties

(a) Solve the equation: f(x) = g(x). (b) Solve the inequality: g(x) :5 f(x)

Y

50. (e) Solve f(x) :5 80. (f) Solve 0 < f(x) < 80. 32. In parts (a)-(f), use the following figure. =

x

=

Y

Y= g (x)

Y= g (x)

Y= f(x)

(a) Solve the equation: f(x) = g(x). (b) Solve the inequality: g(x) < f(x)

:5

hex).

The cost C, in dollars, of renting a moving truck for a day is given by the function C(x) = 0.25x + 35, where x is the number of miles driven. (a) What is the cost if you drive x 40 miles? (b) If the cost of renting the moving truck is $80, how many miles did you drive? (c) Suppose that you want the cost to be no more than $100. What is the maximum number of miles that you can drive?

37. Car Rentals

=

x

(a) Solve g(x) 20. (b) Solve g(x) = 60. (c) Solve g(x) = O. (d) Solve g(x) > 20. (e) Solve g(x) :5 60. (f) Solve 0 < g(x) < 60. 33. In parts (a) and (b) use the following figure. =

Y = f(x)

Y

tional calls on a certain cellular phone plan is given by the function C(x) = 0.38x + 5, where x is the number of min utes used.

x

(a) Solve the equation: f(x) = g(x). (b) Solve the inequality: f(x) > g(x). 34. In parts (a) and (b), use the following figure. Y

38. Phone Charges The monthly cost C, in dollars, for interna

(a) What is the cost if you talk on the phone for x = 50 minutes? (b) Suppose that your monthly bill is $29.32. How many minutes did you use the phone? (c) Suppose that you budget yourself $60 per month for the phone. What is the maximum number of minutes that you can talk? The average monthly benefit B, in dol lars, for individuals on disability is given by the function B(t) = 19.25t + 585.72 where t is the number of years since January 1 , 1 990. (a) What was the average monthly benefit in 2000 (t = 10)? (b) In what year will the average monthly benefit be $893.72? [Hint: use a table with 6t = 1 .] (c) In what year will the average monthly benefit exceed $1000?

39. Disability Benefits

Y= f(x)

x

(a) Solve the equation: f(x) = g(x). (b) Solve the inequality: f(x) :5 g(x). 35. In parts (a) and (b), use the following figure.

40. Health Expenditures The total private health expenditures

Y = f(x)

Y= h (x)

x Y= g (x)

H, in billions of dollars, is given by the function H ( t ) = 26t + 411 where t is the number of years since January 1 , 1 990. (a) What were total private health expenditures in 2000 (t = 1O) ? (b) In what year will total private health expenditures be $879 billion? [Hint: use a table with 6t = 1.] (c) In what year will total private health expenditures exceed $1 trillion ($1,000 billion)?

286

CHAPTER 4


41. Supply and Demand Suppose that the quantity supplied S

and quantity demanded D of T-shirts at a concert are given by the following functions: S(p) = -200 + SOp D(p) = 1 000 - 25p where p is the price of a T-shirt. (a) Find the equilibrium price for T-shirts at this concert. What is the equilibrium quantity? (b) Determine the prices for which quantity demanded is greater than quantity supplied. (c) What do you think will eventually happen to the price of T-shirts if quantity demanded is greater than quan tity supplied? 42. Supply and Demand Suppose that the quantity supplied S and quantity demanded D of hot dogs at a baseball game are given by the following functions: S(p) = -2000 + 3000p D(p) = 10,000 - 1 000p where p is the price of a hot dog. (a) Find the equilibrium price for hot dogs at the baseball game. What is the equilibrium quantity? (b) Determine the prices for which quantity demanded is less than quantity supplied. (c) What do you think will eventually happen to the price of hot dogs if quantity demanded is less than quantity supplied? 43. Taxes The function T(x) = 0.15(x - 7300) + 730 repre sents the tax bill T of a single person whose adjusted gross in come is x dollars for income between $7300 and $29,700, inclusive in 2005. Source: Internal Revenue Service (a) What is the domain of this linear function? (b) What is a single filer's tax bill if adjusted gross income is $18,000? (c) Which variable is independent and which is dependent? (d) Graph the linear function over the domain specified in part (a). (e) What is a single filer's adjusted gross income if the tax bill is $2860? 44. Luxury Tax In 2002, major league baseball signed a labor agreement with the players. In this agreement, any team whose payroll exceeds $128 million in 2005 will have to pay a luxury tax of 22.5% (for first-time offenses). The linear func tion T(p) = 0.225( p 128) describes the luxury tax T of a team whose payroll is p (in millions of dollars). (a) What is the implied domain of this linear function? (b) What is the luxury tax for a team whose payroll is $160 million? (c) Graph the linear function. (d) What is the payroll of a team that pays a lUxury tax of $11.7 million? -

The point at which a company 's profits equal zero is called the company 's break-even point. For Problems 45 and 46, let R represent a company 's revenue, let C represent the company's costs, and let x represent the number of units produced and sold each day. (a) Find the flrm 's break-even point; that is,find x so that R = C. (b) Find the values ofx such that R (x) > C(x). This represents the number of units that the company must sell to earn a profit.

= 8x C(x) = 4.5x + 1 7,500 46. R(x) = 1 2x C(x) = lOx + 15,000

45. R(x)

Suppose that a company has just purchased a new computer for $3000. The company chooses to depreciate the computer using the straight-line method over 3 years. (a) Write a linear function that expresses the book value V of the computer as a function of its age x. (b) Graph the linear function. (c) What is the book value of the computer after 2 years? (d) When will the computer have a book value of $2000?

47. Straight-line Depreciation

Suppose that a company has just purchased a new machine for its manufacturing facility for $ 120,000. The company chooses to depreciate the machine using the straight-line method over 10 years. (a) Write a linear function that expresses the book value V of the machine as a function of its age x. (b) Graph the linear function. (c) What is the book value of the machine after 4 years? (d) When will the machine have a book value of $72,000?

48. Straight-line Depreciation

The simplest cost function is the linear cost function, C( x) = mx + b, where the y-intercept b represents the fixed costs of operating a business and the slope m rep resents the cost of each item produced. Suppose that a small bicycle manufacturer has daily fixed costs of $1800 and each bicycle costs $90 to manufacture. (a) Write a linear function that expresses the cost C of man ufacturing x bicycles in a day. (b) Graph the linear function. (c) What is the cost of manufacturing 14 bicycles in a day? (d) How many bicycles could be manufactured for $3780?

49. Cost Function

Refer to Problem 49. Suppose that the land lord of the building increases the bicycle manufacturer's rent by $100 per month. (a) Assuming that the manufacturer is open for business 20 days per month, what are the new daily fixed costs? (b) Write a linear function that expresses the cost C of man ufacturing x bicycles in a day with the higher rent. (c) Graph the linear function. (d) What is the cost of manufacturing 14 bicycles in a day? (e) How many bicycles could be manufactured for $3780?

50. Cost Function

51. Truck Rentals A truck rental company rents a truck for one

day by charging $29 plus $0.07 per mile. (a) Write a linear function that relates the cost C, in dollars, of renting the truck to the number x of miles driven. (b) What is the cost of renting the truck if the truck is dri ven 110 miles? 230 miles? A phone company offers a domestic long distance package by charging $5 plus $0.05 per minute. (a) Write a linear function that relates the cost C, in dol lars, of talking x minutes. (b) What is the cost of talking 1 05 minutes? 180 minutes?

52. Long Dishll1ce

SECTION 4.2

Bui lding Linear Functions from Data

287

Discussion and Writing 53.

Which of the following functions might have the graph shown? (More than one answer is possible.) Y (a) f ( x ) = 2x - 7 (b) g ( x ) = -3x + 4 (c) H ( x ) = 5 (d) F ( x ) = 3x + 4

shown? (More than one answer is possible.) (a) f ( x ) = 3x + 1 Y (b) g ( x ) = -2x + 3 (c) H ( x ) = 3 (d) F ( x ) = -4x - 1 2 (e) G( ,x: ) = - - x + 3

x

1 (e) G ( x ) = 2 x + 2

55.

54. Which of the following functions might have the graph

3

Under what circumstances is a linear function f ( x ) = mx + b odd? Can a linear function ever be even?

tAre You Prepared?, Answers

1.

2.

Y

.3 2

4. {50}

3. 36

5. 0

6. True

2 x

4.2 Building Linear Functions from Data 1 Draw and Interpret Scatter Diagrams (p. 287)

OBJECTIVES

2

Disting u i s h between Linear and Nonl inear Relations (p. 288)

• 3 Use a G raph ing Util ity to Find the Line of Best Fit (p. 289)

1

Draw and Interpret Scatter Diagrams

In Section 4.1 , we built linear functions from verbal descriptions. Linear functions can also be constructed by fitting a linear function to data. The first step is to plot the ordered pairs using rectangular coordinates. The resulting graph is called a scatter diagram.

E XA M P L E 1

Drawing and I nterpreting a Scatter Diagram The data listed in Table 6 represent the apparent temperature versus the relative humidity in a room whose actual temperature is 72° Fahrenheit.

Table 6

Relative Humidity (%), x

Apparent Temperature of, y

(x, y)

°

64

(0, 64)

10

65

( 1 0, 65)

20

67

(20, 67)

30

68

(30, 68)

40

70

(40, 70)

50

71

(50, 7 1 )

60

72

(60, 72)

70

73

(70, 73)

80

74

(80, 74)

90

75

(90, 75)

76

( 1 00, 76)

1 00

288

CHAPTER 4


�

(a) Draw a scatter diagram by hand treating relative humidity as the independent variable. (b) Use a graphing utility to draw a scatter diagram.* (c) Describe what happens to the apparent temperature as the relative humidity increases. (a) To draw a scatter diagram by hand, we plot the ordered pairs listed in Table 6, with relative humidity as the x-coordinate and apparent temperature as the y-coordinate. See Figure 6(a). Notice that the points in a scatter diagram are not connected.

Solution

(b) Figure 6(b) shows a scatter diagram using a graphing utility. 80 ,..----, 78 r----------------------1 ---� � 75 r-----·--��--- 1 - ---------� 74 �---.

Figure 6

78

m

E 72 r----------.---�-------1

� 70 �--------·-------1 58 • � 56 f_-------------------J · � 54 �----------------------1 5 c

�

�f-?------�

-10 .

o 1 0 20 30 40 50 60 70 80 90 1 00 Relative Humidity (a)

a

a

a

a

a

a

a

a

a

a

110

52 (b)

(c) We see from the scatter diagrams that, as the relative humidity increases, the apparent temperature increases. 'I"

2

w, """'""

Now Work

PROBLEM 9

•

(a )

Distinguish between Linear and N onlinear Relations

Scatter diagrams are used to help us see the type of relation that exists between two variables. In this text, we will discuss a variety of different relations that may exist between two variables. For now, we concentrate on distinguishing between linear and nonlinear relations. See Figure 7. Figure 7

. .

.

(a) Linear Y = mx + b, m > 0

E XA M P L E 2

(b) Linear Y = mx + b, m < 0

(c) Nonlinear

.

' .

'

.

.

(e) Nonlinear

(d) Nonlinear

Distinguishing between Linear and Nonlinear Relations Determine whether the relation between the two variables in Figure 8 is linear or nonlinear.

Figure 8 ' .

.

(a)

(b)

' .

(c)

'" Consult your owner's manual for the proper keystrokes.

.

' .

.

.

.

.

(d)

SECTION 4.2

(a) Linear

Solution

(b) Nonlinear

�- - N o w Work

B ui l d i ng Linear Functions from Data

( c) Nonlinear

289

(d) Nonlinear •

PROBLEM 3

In this section we will study data whose scatter diagrams imply that a linear relation exists between the two variables. Suppose that the scatter diagram of a set of data appears to be linearly related as in Figure 7(a) or (b). We might wish to find an equation of a line that relates the two variables. One way to obtain an equation for such data is to draw a line through two points on the scatter diagram and determine the equation of the line.

EXAM P L E 3

Finding an Equation for Linearly Related Data Using the data in Table 6 from Example 1 , (a) Select two points and find an equation of the line containing the points. (b) Graph the line on the scatter diagram obtained in Example l (a). (a) Select two points, say ( 1 0, 65) and (70, 73). The slope of the line joining the points (10, 65) and (70, 73) is

Solution

n1 =

The equation o f the line with slope

Figure 9

.a Q)

�

E

� C

�

�

0,

for a

=

1, a

=

Figure 14

y

2 l / f(X) = X2 f(x) � X2

f(x)

t

=

Figure 1 5

3x

y

4

,

=

-4

4

-4

x

\

-4

'",- f(x)

f(x)

Figure 1 6 Graphs of a quadratic fu nction , f( x) = a>? + bx + c, a -:f- 0

Axis of symmetry

Vertex is highest point

Vertex is lowest point

Axis of symmetry

(a) Opens up a>O

(b) Opens down

a ° and down if a < 0. The axis of sym metry is the vertical line x = h.

SECTION 4.3

f

For example, compare equation ( 1 ) with the solution given in Example (x) = 2 ( x + 2 )2 - 3 2 ( x - ( -2) ? + ( - 3 ) = a(x h ) 2 + Ie =

-

We conclude that a = 2, so the graph opens up. Also, we find that Ie = -3, so its vertex is at ( -2, - 3 ) . 2

297

Q u a d ratic Functions an d Their Properties

h

1.

-2 and

=

Identify the Vertex and Axis of Symmetry of a Quadratic Function

We do not need to complete the square to obtain the vertex. In almost every case, it is easier to obtain the vertex of a quadratic function f by remembering that its

-�. f (- �) .

x-coordinate is

h

=

- �. That is' Ie

=

.

2a

2a

The y-coordinate Ie can then be found by evaluating

f

at

2a

We summarize these remarks as follows:

f (-�,f(-�))

Properties of the Graph of a Quadratic Function

( x)

Vertex

=

2a

=

2a

ax2 + bx +

c

a *" O

Axis of symmetry: the line x =

-� 2a

(2)

Parabola opens up if a >

0; the vertex is a minimum point. Parabola opens down if a < 0; the vertex is a maximum point. E XAM P L E 2

Locating the Vertex without G raphing

f

Without graphing, locate the vertex and axis of symmetry of the parabola defined by ( x ) = -3x2 + 6x + 1 . Does it open up or down? For this quadratic function, a = -3, b vertex is b h = --

Solution

2a

=

6, and

c

=

1.

The x-coordinate of the

= -- = l

6 -6

The y-coordinate of the vertex is Ie

=f

( : ) f(l) -

a

=

=

-3 + 6 + 1

=

4

The vertex is located at the point ( 1 , 4). The axis of symmetry is the line x Because a = - 3 < 0, the parabola opens down. 3

=

1. •

Graph a Quadratic Function Using Its Vertex, Axis, and Intercepts

f

The information we gathered in Example 2, together with the location of the inter cepts, usually provides enough information to graph ( x ) = ax2 + bx + c, a *" O.

f

The y-intercept is the value of at x

=

f

0; that is, the y-intercept is ( O )

=

c.

The x-intercepts, if there are any, are found by solving the quadratic equation ax2 + bx + c = 0

298

CHAPTER 4


This equation has two, one, or no real solutions, depending on whether the discrim inant b2 - 4ac is positive, 0, or negative. Depending on the value of the discrimi nant, the graph of 1 has x-intercepts, as follows: The x-Intercepts of a Quadratic Function

ax2 + bx + c has places. ax2 + bx + c has

- 4ac > 0, the graph of I(x) = two distinct x-intercepts so it crosses the x-axis in two 2. If the discriminant b2 - 4ac 0, the graph of 1 ( x ) one x-intercept so it touches the x-axis at its vertex. 3. If the discriminant b2 - 4ac < 0, the graph of I ( x ) = no x-intercept so it does not cross or touch the x-axis. 1. If the discriminant b2

=

=

a x2

+

bx

+

c has

Figure 18 illustrates these possibilities for parabolas that open up. ((x)

Figure 1 8

=

ax2 + bx + c, a >

0

Axis of symmetry x = - JL

Axis of symmetry x = _JL

Axis of symmetry x = _ JL

2a

2a

2a

\)

x-intercept (_ JL f (- JL))

2a 2a ' (a) bL 4ac > 0

(c) bL 4ac < 0

(b) bL 4ac = 0

No x-intercepts

One x-intercept

Two x-intercepts

E XAM P L E 3

x

b 0) (- 2a '

G raphing a Quadratic F unction Using Its Vertex, Axis, and I ntercepts (a) Use the information from Example 2 and the locations of the intercepts to graph I ( x ) = -3x2 + 6x + 1 . (b) Determine the domain and the range o f f (c) Determine where 1 is increasing and where it is decreasing.

Solution

(a) In Example 2, we found the vertex to be at (1, 4) and the axis of symmetry to be x = 1. The y-intercept is found by letting x = O. The y-intercept is 1 (0) = 1 . The x-intercepts are found b y solving the equation I ( x ) = O. This results in the equation -3x2 + 6x +

1

= a

a =

- 3,

b

=

6,

-6

+

C =

1

The discriminant b2 - 4ac = ( 6 ? - 4( - 3) ( 1 ) = 3 6 + 12 = 48 > 0, so the equation has two real solutions and the graph has two x-intercepts. Using the quadratic formula, we find that x = Figure 1 9

Axis of symmetry x =l

2a

-6 + V48 -6

4 \1'3 ;:::::: -0.15 -6

and

y

x = (2, 1 )

�4

-b + Yb2 - 4ac

x

(2.1 5, 0)

-b - Yb2 - 4ac

2a

-6 - V48 -6

-6 - 4 \1'3 ;:::::: 2.15 -6

The x-intercepts are approximately -0.15 and 2.15. The graph is illustrated in Figure 19. Notice how we used the y-intercept and the axis of symmetry, x = 1 , to obtain the additional point (2, 1 ) on the graph.

SECTION 4.3

Quadratic Functions and Their Properties

299

(b) The domain of f is the set of all real numbers. B ased on the graph, the range of f is the interval ( - 00 , 4]. (c) The function f is increasing on the interval ( - 00, 1) and decreasing on the interval ( 1 , 00 ) . •

'I,j

=___

_'l!>: = ="- '-

Graph the function in Example

3 by completing the square and using

transformations. Which method do you prefer?


If the graph of a quadratic function has only one x-intercept or no x-intercepts, it is usually necessary to plot an additional point to obtain the graph.

EXAM P L E 4

G raphing a Quadratic F unction Using Its Vertex, Axis, and I ntercepts (a) Graph f(x) = x2 - 6x + 9 by determining whether the graph opens up or down and by finding its vertex, axis of symmetry, y-intercept, and x-intercepts, if any. (b) Determine the domain and the range of f (c) Determine where f is increasing and where it is decreasing.

Solution

-6 b h = - - = - -- = 3 2( 1 ) 2a

Figure 20

y

(a) For f(x) = x2 - 6x + 9, we have a = 1 , b = -6, and c = 9. Since a = 1 > 0, the parabola opens up. The x-coordinate of the vertex is

Axis of symmetry

x= 3

The y-coordinate of the vertex is

k = f(3) = ( 3 ? - 6(3)

+

9 =0

So the vertex is at (3, 0). The axis of symmetry is the line x = 3. The y-intercept is f(O) = 9. Since the vertex (3, 0) lies on the x-axis, the graph touches the x-axis at the x-intercept. By using the axis of symmetry and the y-intercept at (0, 9), we can locate the additional point (6, 9) on the graph. See Figure 20. (b) The domain of f is the set of all real numbers. B ased on the graph, the range of f is the interval [0, 00 ) . (3, 0)

x

(c) The function f i s decreasing o n the interval ( - 00, 3 ) and increasing on the interval (3, 00 ) . •

Now Wor k P R O B L E M 4 3

E XA M P L E 5

G raphing a Quad ratic F unction Using Its Vertex, Axis, and I ntercepts (a) Graph f(x) = 2x2 + X + 1 by determining whether the graph opens up or down and by finding its vertex, axis of symmetry, y-intercept, and x-intercepts, if any. (b) Determine the domain and the range of f (c) Determine where f is increasing and where it is decreasing.

Solution

(a) For f(x) = 2 x2 + X + 1, we have a = 2, b = 1, and c = l . Since a = 2 > 0, the parabola opens up. The x-coordinate of the vertex is

b 1 17. = - - = - 4 2a

300

CHAPTER 4


The y-coordinate of the vertex is NGrE In Example 5, since the vertex is above the x-axis and the parabola opens up, we can conclude that the graph of the quadratic function will have no • x-intercepts.

So the vertex is at

(-±, i}

The axis of symmetry is the line x

-

=

-

±. The

y-intercept is f(O) = 1. The x-intercept(s), if any, obey the equation 2x2 + x + 1 = 0. Since the discriminant b2 4ac = ( 1 ? - 4(2) ( 1 ) = -7 < 0, this equation has no real solutions, and therefore the graph has no x-intercepts.

Figure 2 1

We use the point (0, 1) and the axis of symmetry x = tional point

(-�, ) [i, ) (-±, )

_

1:. to locate the addi4

on the graph. See Figure 2 1 .

1

(b) The domain of f is the set of all real numbers. Based on the graph, the range 00

of f is the interval

.

(c) The function f is decreasing on the interval -1

the interval

x

00

(-oo,-±)

and is increasing on

.

• 1l!I!lI _ _ ",,- '-


Given the vertex (h, k) and one additional point on the graph of a qua dratic function f(x) = ax2 + bx + c , a *' 0, we can use

f(x) = a(x - h ) 2

+

(3 )

k

to obtain the quadratic function.

E XA M P L E 6

Finding the Quadratic F unction G iven Its Vertex and One Other Point

-

Determine the quadratic function whose vertex is ( 1 , -5) and whose y-intercept is 3. The graph of the parabola is shown in Figure 22.

Solution Figure 22

The vertex is ( 1 , - 5 ) , so h = 1 and k = -5. Substitute these values into equation (3).

f(x) = a(x

-

h? f(x) = a(x - 1 ) 2

y

-

+

k

Equation (3)

5

h

=

1, k

=

To determine the value of a, we use the fact that f(O) =

f(x) = a(x

-8

-3

(the y-intercept).

- 1) 2 - 5

-3 = a(O - 1 ? -3 = a - 5 (1 , -5 )

-5

-

5 x=

G, y

= f( G ) =

-3

a=2 The quadratic function whose graph is shown in Figure 22 is

f(x)

=

a(x

- h?

+

k = 2(x - 1 ?

- 5 = 2x2 - 4x - 3 •

� =:::I> '-

Now Work

PROBlEM 53

SECTION 4.3

SUMMA RY Option STEP STEP

Option

1

Steps for Graphing a Quadratic Function f(x)

+

bx +

c,

a

2

STEP

2: Determine the vertex

STEP

3: Determine the axis of symmetry, x = -

(-:a ' ( : ) ) f -

a

.

(a

>

0) or down

(a
0, the graph of the quadratic function has two x-intercepts, which are found by solving the equation + bx + c = O. (b) If b2 - 4 c = 0, the vertex is the x-intercept.

(c) If b2 -

STEP 5: STEP

a>?

1: Complete the square in x to write the quadratic function in the form f(x)

STEP

STEP

=

Q u a d ratic Functions and Their Properties

a4ac

0

. This vertex is the highest point on

0 and the lowest point on the graph if

highest point lowest point

ax2

a

> O. If the vertex is the

0), then f

is the

maximum value

of f If the vertex is the

0), then f

is the

minimum value

of f

F i nding the M aximum or M i n i m u m Value of a Quad ratic Function Determine whether the quadratic function f(x)

=

x2 - 4x - 5

has a maximum or minimum value. Then find the maximum or minimum value.

Solution

We compare f(x) = x2 - 4x - 5 to f(x) = ax2 + bx + c . We conclude that a = 1, b = -4, and c = -5. Since a > 0, the graph of f opens up, so the vertex is a minimum point. The minimum value occurs at x

-4 = - - = - -- = - = 2 a = 1, b

The minimum value is f

& g(x), that is, the region below f and above g 69. f(x) = 2x - 1; g(x) = x2 - 4 70. f(x) = - 2x - 1 ; g(x) = x2 9 2 = 2 71. f(x) = - x + 4; g(x) - 2x + 1 72. f(x) = -x + 9; g(x) = 2x + 1 73. f(x) = -x2 + 5x; g(x) = x2 + 3x - 4 74. f(x) = -x2 + 7x - 6; g(x) = x2 + X - 6 -

Answer Problems 75 and 76 using the following: A quadratic function of the form f(x) = ax2 + bx + c with b2 - 4ac > a may a/so be written in the form f(x) = a(x - r1 ) (x - r2 ) , where r1 and r2 are the x-intercepts of the graph of the quadratic function. 76. (a) Find a quadratic function whose x-intercepts are -5 and 75. (a) Find a quadratic function whose x-intercepts are -3 and 3 with a = 1; a = 2; a = -2; a = 5. 1 with a = 1; a = 2; a = -2; a = 5. (b) How does the value of a affect the intercepts? (b) How does the value of a affect the intercepts? (c) How does the value of a affect the axis of sym (c) How does the value of a affect the axis of sym metry? metry? (d) How does the value of a affect the vertex? (d) How does the value of a affect the vertex? (e) Compare the x-coordinate of the vertex with the mid (e) Compare the x-coordinate of the vertex with the mid point of the x-intercepts. What might you conclude? point of the x-intercepts. What might you conclude?

304

CHAPTER 4


77. Suppose that f(x) = x2 + 4x - 21

(a) What is the vertex of f? (b) What are the x-intercepts of the graph of f? (c) Solve f(x) = -21 for x. What points are on the graph of f? (d) Use the information obtained in parts (a)-(c) to graph f(x) = x2 + 4x - 2l. 78. Suppose that f(x) = x2 + 2x - 8 (a) What is the vertex of f? (b) What are the x-intercepts of the graph of f? (c) Solve f(x) = -8 for x. What points are on the graph off? (d) Use the information obtained in parts (a)-(c) to graph f(x) = x2 + 2x - 8. 79. Find the point on the line y = x that is closest to the point (3, 1 ) . [Hint: Express the distance d from the point to the line as a function of x, and then find the minimum value of [d(x)f 80. Find the point on the line y = x + 1 that is closest to the point (4, 1 ) . 81. Maximizing Revenue Suppose that the manufacturer of a gas clothes dryer has found that, when the unit price is p dollars, the revenue R (in dollars) is R(p) -4p2 + 4000p What unit price should be established for the dryer to maxi mize revenue? What is the maximum revenue? 82. Maximizing Revenue The John Deere company has found that the revcnue, in dollars, from sales of heavy-duty tractors is a function of the unit price p, in dollars, that it charges. If the revenue R is 1 R(p) - Z p2 + 1900p what unit price p should be charged to maximize revenue? What is the maximum revenue? 83. Minimizing Marginal Cost The marginal cost of a product can be thought of as the cost of producing one additional unit of output. For example, if the marginal cost of producing the 50th product is $6.20, it cost $6.20 to increase production from 49 to 50 units of output. Suppose the marginal cost C (in dollars) to produce x thousand mp3 players is given by the function C(x) = x2 - 140x + 7400 (a) How many players should be produced to minimize the marginal cost? (b) What is the minimum marginal cost? 84. Minimizing Marginal Cost (See Problem 83.) The margin al cost C (in dollars) of manufacturing x cell phones (in thou sands) is given by C(x) = 5x2 - 200x + 4000. (a) How many cell phones should be manufactured to min imize the marginal cost? (b) What is the minimum marginal cost? 85. Hunting The function H(x) = -3.24x2 + 242.1x - 738.4 models the number of individuals whose age is x and engage in hunting activities. (a) What is the age at which there are the most hunters? Approximately how many hunters are this age? =

=

(b) Are the number of hunters increasing or decreasing for individuals who are between 40 and 45 years of age? Source: National Sporting Goods Association 86. Advanced Degrees The function P(x) -0.008x2 + 0.815x - 9.983 models the percentage of the U.S. population whose age is x that have earned an advanced degree (more than a bache lor's degree) in March 2003. (a) What is the age for which the highest percentage of Americans have earned an advanced degree? What is the highest percentage? (b) Is the percentage of Americans who have earned an ad vanced degree increasing or decreasing for individuals between the ages of 40 and 50? Source: Us. Census Bureau 87. Male Murder V ictims The function M(x) = l.00x2 - 136.74x + 4764.89 models the number of male murder victims who are x years of age (20 :=; x < 90). (a) Use the model to approximate the number of male murder victims who are x = 23 years of age. (b) At what age is the number of male murder victims 1456? Describe what happens to the number of male murder (c) victims as age increases from 20 to 65. Source: Federal Bureau of Investigation 88. Health Care Expenditures The function H(x) 0.004x2 - 0.197x + 5.406 models the percentage of total income that an individual who is x years of age spends on health care. (a) Use the model to approximate the percentage of total income an individual who is x 45 years of age spends on health care. (b) At what age is the percentage of income spent on health care 1O%? ;;r . . (c) Using a graphing utility, graph H = H(x). (d) Based on the graph drawn in part (c), describe what happens to the percentage of income spent on health care as individuals age. Source: Bureau of Labor Statistics 89. Business The monthly revenue R achieved by selling x wristwatches is figured to be R(x) = 75x - 0.2x2. The monthly cost C of selling x wristwatches is C(x) 32x + 1750. (a) How many wristwatches must the firm sell to maximize revenue? What is the maximum revenue? (b) Profit is given as P(x) = R(x) - C(x). What is the profit function? (c) How many wristwatches must the firm sell to maximize profit? What is the maximum profit? (d) Provide a reasonable explanation as to why the answers found in parts (a) and (c) differ. Explain why a qua dratic function is a reasonable model for revenue. 90. Business The daily revenue R achieved by selling x boxes of candy is figured to be R(x) = 9.5x - 0.04x2. The daily cost C of selling x boxes of candy is C (x) 1.25x + 250. =

=

=

=

=

SECTION 4.4

(a) How many boxes of candy must the firm sell to maxi mize revenue? What is the maximum revenue? (b) Profit is given as P(x) = R(x) - C(x). What is the profit function? (c) How many boxes of candy must the firm sell to maxi mize profit? What is the maximum profit?

91.

Quadratic Models; Building Q u a d ratic Functions from Data

305

(d) Provide a reasonable explanation as to why the answers found in parts (a) and (c) differ. Explain why a qua dratic function is a reasonable model for revenue. Let f(x ) = ax2 + bx + c, where a, b, and c are odd integers. If x is an integer, show that f(x) must be an odd integer. [Hint: x is either an even integer or an odd integer.]


93.

94.

95.

Make up a quadratic function that opens down and has only one x-intercept. Compare yours with others in the class. What are the similarities? What are the differences? On one set of coordinate axes, graph the family of parabolas f(x) = x2 + 2x + c for c = -3, c = 0, and c = 1. Describe the characteristics of a member of this family. On one set of coordinate axes, graph the family of parabolas f(x) = x2 + bx + 1 for b = -4, b = 0, and b = 4. De scribe the general characteristics of this family.

96. 97. 98.

State the circumstances that cause the graph of a quadratic function f(x) = ax2 + bx + c to have no x-intercepts. Why does the graph of a quadratic function open up if a > 0 and down if a < O? Can a quadratic function have a range of ( - 00, oo )? Justify your answer. What are the possibilities for the number of times the graphs of two different quadratic functions intersect?


1. (0, -9), ( -3, 0), (3, 0)

2.

{-4, �}

25 4

3. -

4.

right; 4

4.4 Quadratic Models; Build ing Quadratic Functions from Data PREPARING FOR THIS SECTION •


Problem Solving (Section 1.7, pp. 139-145)

•

Building Linear Functions from Data (Section 4.2, pp. 287-290)

'\. N o w Work the 'Are You Prepared?' problems on page 3 1 0. OBJECTIVES

1 Solve Applied Problems I nvolving Quad ratic Fu nctions (p. 305)

� 2

Use a G ra p h i ng Util ity to Find the Quadratic Function of Best Fit (p. 309)

In this section we will first discuss models that lead to a quadratic function when a verbal description of the problem is given. We end the section by fitting a quadratic function to data, which is another form of modeling. When a mathematical model leads to a quadratic function, the properties of the graph of the quadratic function can provide important information about the model. In particular, we can use the quadratic function to determine the maximum or minimum value of the function. The fact that the graph of a quadratic function has a maximum or minimum value enables us to answer questions involving optim ization, that is, finding the maximum or minimum values in models involving qua dratic functions. 1

S olve Applied Problems Involving Quadratic Functions

In economics, revenue R, in dollars, is defined as the amount of money received from the sale of an item and is equal to the unit selling price p, in dollars, of the item times the number x of units actually sold. That is,

R

=

xp

In economics, the Law of Demand states that p and x are related: As one increases, the other decreases. The equation that relates p and x is called the demand

equation.

306

CHAPTER 4


EXAM P L E 1

M aximizing Revenue The marketing department at Texas Instruments has found that, when certain cal culators are sold at a price of p dollars per unit, the number x of calculators sold is given by the demand equation x = 21,000 - 150p (a) (b) (c) (d) (e)

Solution

Express the revenue R as a function of the price p. What urnt price should be used to maximize revenue? If this price is charged, what is the maximum revenue? How many units are sold at this price? Graph R.

(a) The revenue R is R R

=

xp

=

=

xp, where x = 21,000 - 150p. (21 ,000 - 150p) p

=

- 150p2

+

21,000p

(b) The function R is a quadratic function with a = - 1 50, b = 21 ,000, and c = O. Because a < 0, the vertex is the highest point on the parabola. The revenue R is therefore a maximum when the price p is

_ _ � _ _ 21,000 _ _ 21 ,000 - $ 70.00 _

p -

2a

2( - 150)

-300

i

a = - 1 50, b = 21,000

( c) The maximum revenue R is R (70) = - 150(70) 2 + 21,000(70)

=

$735,000

(d) The number of calculators sold is given by the demand equation x = 21,000 - 150p. At a price of p = $70, x calculators are sold.

=

21,000 - 150(70)

=

10,500

(e) See Figure 23 for the graph of R. R 800 , 000 I ·

Figure 23

··· ············································································:�c···cc·�·�·�·�·:·········........................................,

700,000 ... _

�

i o

�

�

600,000 .... . . 500 '000 ... 400,000 ... . ... 300,000 ... . .. 200 , 000 ... . . . ................................ . .. .

14

28

42

....... . . ... ... . .. .... . .. .... ...

56

Now Work

E XA M P LE 2

70

84

p

98

Price p per calculator (dollars)

•

PROBLEM 3

M aximizing the Area E nclosed by a F ence A farmer has 2000 yards of fence to enclose a rectangular field. What are the di mensions of the rectangle that encloses the most area?

Solution

Figure 24 illustrates the situation. The available fence represents the perimeter of the rectangle. If x is the length and w is the width, then 2x

+

2w = 2000

The area A of the rectangle is

A

=

xw

(1)

SECTION 4.4

Q uad ratic Models; Building Q uadratic Fu nctions from Data

307

To express A in terms of a single variable, we solve equation (1) for w and substi tute the result in A = xw. Then A involves only the variable x. [You could also solve equation (1) for x and express A in terms of w alone. Try it!]

Figure 24

x

2x

w

2w 2w

+

=

=

w

x

=

2000 2000 - 2x 2000 - 2x 2

Equation (1)

=

Solve for w.

1000 - x

Then the area A is

A

xw

=

=

x( lOOO - x)

=

-x2 + 1000x

Now, A is a quadratic function of x. =

A ( x)

1000x

+

a = - 1, b = 1000, C = 0

Figure 25 shows a graph of A(x) = -x2 + 1000x. Since a < 0, the vertex is a maximum point on the graph of A . The maximum value occurs at

Figure 2S A

-x2

(500, 250000)

x (1 000, 0) -'--'---'*- ---'--�,..:.... 500 1 000 x

=

b

-2a

=

1000 - -2(-1)

=

500

The maximum value of A is

( :a)

A -

=

A (500)

=

-5002 + 1000(500)

=

-250,000 + 500,000

=

250,000

The largest rectangle that can be enclosed by 2000 yards of fence has an area of 250,000 square yards. Its dimensions are 500 yards by 500 yards.

•

= = =-

E XA M P L E 3


Analyzing the Motion of a Projectil e A projectile i s fired from a cliff 500 feet above t h e water a t a n inclination o f 45° to the horizontal, with a muzzle velocity of 400 feet per second. In physics, it is estab lished that the height h of the proj ectile above the water is given by

h(x)

=

-32x2 + x + 500 (400?

where x is the horizontal distance of the proj ectile from the base of the cliff. See Figure 26. Figure 26

h (x)

2500 2000 1 500 /' 1 000 5° V4 500

-

"9-

t-.....

....... -

1 000 2000

""

�

3000 4000 5000

x

(a) Find the maximum height of the projectile. (b) How far from the base of the cliff will the projectile strike the water?

Solution

(a) The height of the projectile is given by a quadratic function.

h (x)

=

-32x2 + X + 500 (400) -

-7

=

-1 --x2 + X + 500 5000

308

CHAPTER 4


We are looking for the maximum value of h. Since a < 0, the maximum value is obtained at the vertex. We compute x = - !!..- = 2a

2

(

)

=

=

- 1250 + 2500 + 500

1

1 __ 5000

5000 = 2500 2

The maximum height of the projectile is h(2500) =

-1 (2500) 2 + 2500 5000

+

500

=

1750 ft

(b) The projectile will strike the water when the height is zero. To find the distance x traveled, we need to solve the equation hex) =

1/ Seeing the Concept

We find the discriminant first.

Graph

h (x) = o :s; Use

-1 --

5000

x2 +

x :s; 5 5 00

MAXIMUM

X

-1 x2 + X + 500 = 0 5000

b2 - 4ac

+ 500

= 12 - 4

( )

2 ( 500) = 1 . 4 5000

Then

to find the maxi m u m

height o f t h e projectile, a n d u s e ROOT o r

x =

ZERO t o find t h e distance from t h e base

-b

±

Vb2

- 4ac

2a

(

-1

±

2 -

of the cliff to where it strikes the water. Compare you r results with those ob

ViA

1 5000

)

�

{

-458 5458

We discard the negative solution and find that the projectile will strike the water at a distance of about 5458 feet from the base of the cliff.

tained in Example 3.

• �

EXAM P L E 4


The Golden Gate Bridge The Golden Gate Bridge, a suspension bridge, spans the entrance to San Francisco Bay. Its 746-foot-tall towers are 4200 feet apart. The bridge is suspended from two huge cables more than 3 feet in diameter; the 90-foot-wide roadway is 220 feet above the water. The cables are parabolic in shape':' and touch the road surface at the center of the bridge. Find the height of the cable above the road at a distance of 1000 feet from the center.

Solution

Figure 27

See Figure 27. We begin by choosing the placement of the coordinate axes so that the x-axis coincides with the road surface and the origin coincides with the center of the bridge. As a result, the twin towers will be vertical (height 746 - 220 = 526 feet above the road) and located 2100 feet from the center. Also, the cable, which has the shape of a parabola, will extend from the towers, open up, and have its vertex at (0, 0). The choice of placement of the axes enables us to identify the equation of the parabola as y = ax2 , a > O. We can also see that the points ( -2100, 526) and (2100, 526) are on the graph. ( - 21 00, 526)

y

(21 00, 526)

':' A cable suspended from two towers is in the shape of a catenary, but when a horizontal roadway is

suspended from the cable, the cable takes the shape of a parabola.

SECTION 4.4

Quadratic Models; B u i l d i n g Quadratic Functions from Data

309

B ased on these facts, we can find the value of a in y = ax2 . y =

526

ax2

=

a(2100) 2

x = 2100, y = 526

526 (2100) 2

a=

---

The equation of the parabola is therefore

526 2 x (2100 ) 2

=

Y

The height of the cable when x = 1000 is

526 ( 1000) 2 (2100) 2

y =

�

1 19.3 feet

The cable is 1 19.3 feet above the road at a distance of 1000 feet from the center of the bridge. • Q) =:;:>-

2


Use a Graphing Utility to Find the Qua dratic Function of Best Fit

In Section 4.2 we found the line of best fit for data that appeared to be linearly related. It was noted that data may also follow a nonlinear relation. Figures 28(a) and (b) show scatter diagrams of data that follow a quadratic relation. Figure 28 ..

.

.

.

. •

.

•

"0 • ° 0

• ° 0 .

0 •

.

•

• . .

y

E XA M P L E 5

= ax2

+

bx + (a)

c,

a>0

y

.

° 0

•

:' •

.

' .

. .

.

'

"0 "

.

.

= ax2 + bx + (b)

c,

a=:::;

31.5 pounds of fertilizer per 100 square feet

(d) We evaluate the function Y(x) for x = 31.5.

Y ( 3 1 .5 )

=

-0.0171 (31.5)2 + 1.0765 (31.5) + 3.8939

>=:::;

20.8 bushels

If we apply 31.5 pounds of fertilizer per 100 square feet, the crop yield will be 20.8 bushels according to the quadratic function of best fit. (e) Figure 31 shows the graph of the quadratic function found in part (b) drawn

Figure 3 1

on the scatter diagram.

25

.y� /.�

Look again at Figure 30. Notice that the output given by the graphing calculator does not include r, the correlation coefficient. Recall that the correlation coefficient is a measure of the strength of a linear relation that exists between two variables. The graphing calculator does not provide an indication of how well the function fits the data in terms of r since a quadratic function cannot be expressed as a linear function.

a ...a.-i-iI--

11'__�

...d

}r /a

-5

•

45 a

i.lm�


4.4 Assess You r U nderstan d i n g 'Are You Prepared?' A nswers are given a t the end of these exercises. If you get a wrong answer, read the pages listed in red.

1. Translate the following sentence to a mathematical equation: The total revenue R from selling x hot dogs is $3 times the number

� 2.

of hot dogs sold. (pp. 139-1 45) Use a graphing utility to find the line of best fit for the following data: (pp. 287-290)

x y

I

3 10

5 13

5 12

6 15

7 16

8 19

Applications and Extensions . 3.

The price p (in dollars) and the quan tity x sold of a certain product obey the demand equation

Demand E(luation

P =

1 6

- -x

+ 100

a :S X :S 600

(a) Express the revenue R as a function of x. (Remember, R = xp.) (b) What is the revenue if 200 units are sold?

(c) What quantity x maximizes revenue? What is the max imum revenue? (d) What price should the company charge to maximize revenue? 4. Demand Equation The price p (in dollars) and the quantity x sold of a certain product obey the demand equation 1 P = - - x + 1 00 a :S X :S 300 3

SECTION 4.4

(a) Express the revenue R as a function of x. (b) What is the revenue if 100 units are sold? (c) What quantity x maximizes revenue? What is the max imum revenue? (d) What price should the company charge to maximize revenue? s. Demand Equation The price p (in dollars) and the quanti ty x sold of a certain product obey the demand equation x = -5p + 100, 0 :5 P :5 20 (a) Express the revenue R as a function of x. (b) What is the revenue if 15 units are sold? (c) What quantity x maximizes revenue? What is the max imum revenue? (d) What price should the company charge to maximize revenue? 6. Demand Equation The price p (in dollars) and the quanti ty x sold of a certain product obey the demand equation x = -20p + 500, 0 :5 P :5 25 (a) Express the revenue R as a function of x. (b) What is the revenue if 20 units are sold? (c) What quantity x maximizes revenue? What is the max imum revenue? (d) What price should the company charge to maximize revenue? 7. Enclosing a Rectangular Field D avid has 400 yards of fenc ing and wishes to enclose a rectangular area. (a) Express the area A of the rectangle as a function of the width w of the rectangle. (b) For what value of w is the area largest? (c) What is the maximum area? 8. Enclosing a Rectangular Field Beth has 3000 feet of fenc ing available to enclose a rectangular field. (a) Express the area A of the rectangle as a function of x, where x is the length of the rectangle. (b) For what value of x is the area largest? (c) What is the maximum area? 9. Enclosing the Most Area with a Fence A farmer with

4000 meters of fencing wants to enclose a rectangular plot that borders on a river. If the farmer does not fence the side along the river, what is the largest area that can be enclosed? (See the figure.)

x

A farmer with 2000 meters of fencing wants to enclose a rectangular plot that bor ders on a straight highway. If the farmer does not fence the side along the highway, what is the largest area that can be enclosed? 11. Anal)'zing the Motion of a Projectile A projectile is fired from a cliff 200 feet above the water at an inclination of 45° to the horizontal, with a muzzle velocity of 50 feet per second. The height h of the projectile above the water is given by -32x2 hex) = -, + x + 200 (50)-

10. Enclosing the Most Area with a Fence

Quadratic Models; Building Quad ratic Functions from Data

31 1

where x is the horizontal distance of the projectile from the face of the cliff. (a) At what horizontal distance from the face of the cliff is the height of the projectile a maximum? (b) Find the maximum height of the projectile. (c) At what horizontal distance from the face of the cliff will the projectile strike the water? � (d) Using a graphing utility, graph the function h, o :5 X :5 200. (e) Use a graphing utility to verify the solutions found in parts (b) and (c). (f) When the height of the projectile is 100 feet above the water, how far is it from the cliff? 12. Analyzing the Motion of a Projectile A projectile is fired at an inclination of 45° to the horizontal, with a muzzle velocity of 100 feet per second. The height h of the projectile is given by -32x2 hex) = + x ( 100) 2 where x is the horizontal distance of the projectile from the firing point. (a) At what horizontal distance from the firing point is the height of the projectile a maximum? (b) Find the maximum height of the projectile. (c) At what horizontal distance from the firing point will the projectile strike the ground? � (d) Using a graphing utility, graph the function h, o :5 X :5 350. (e) Use a graphing utility to verify the results obtained in parts (b) and (c). (f) When the height of the projectile is 50 feet above the ground, how far has it traveled horizontally? 13. Suspension Bridge A suspension bridge with weight uni formly distributed along its length has twin towers that extend 75 meters above the road surface and are 400 meters apart. The cables are parabolic in shape and are suspended from the tops of the towers. The cables touch the road surface at the center of the bridge. Find the height of the cables at a point 100 meters from the center. (Assume that the road is level.) 14. Architecture A parabolic arch has a span of 120 feet and a maximum height of 25 feet. Choose suitable rectangular co ordinate axes and find the equation of the parabola. Then calculate the height of the arch at points 10 feet, 20 feet, and 40 feet from the center. 15. Constructing Rain Gutters A rain gutter is to be made of aluminum sheets that are 12 inches wide by turning up the edges 90°. See the illustration. What depth will provide maximum cross-sectional area and hence allow the most water to flow?

31 2

CHAPTER 4


A Norman window has the shape of a rectangle surmounted by a semicircle of diameter equal to the width of the rectangle. See the figure. If the perimeter of the window is 20 feet, what dimensions will admit the most light (maximize the area)? [Hint: Circumference of a circle = 2 7fr ; area of a circle = 7f r2, where r is the radius of the circle.]

16. Norman Windows

catastrophes in U.S. history (as of 2005). For the United States Automobile Association (USAA) and its affiliates, the total cost of claims for catastrophic losses, in millions, can be approximated by C(x) = 34.87x2 - 98.1x + 258.3, where x = 0 for 1999, x = 1 for 2000, x = 2 for 2001, and so on. (a) Estimate the total cost of claims for the year 2003. (b) According to the model, during which year were cata strophic loss claims at a minimum? ( c) Would C(x) be useful for predicting total catastrophic loss claims for the year 2015? Why or why not? Source: USAA Report to Members 2005 A self-catalytic chemical reaction re sults in the formation of a compound that causes the forma tion ratio to increase. If the reaction rate V is given by

21. Chemical Reactions

Vex) A track and field playing area is in the shape of a rectangle with semicircles at each end. See the figure. The inside perimeter of the track is to be 1500 meters. What should the dimensions of the rectangle be so that the area of the rectangle is a maximum?

17. Constructing a Stadium

=

kx(a - x),

where k is a positive constant, a is the initial amount of the compound, and x is the variable amount of the compound, for what value of x is the reaction rate a maximum?

f,. 22. Calculus: Siml)SOn'S Rule The figure shows the graph of y = ax2 + bx + c. Suppose that the points ( -h, Yo) , (0, Yl ) , and ( h , Y2 ) are o n the graph. It can b e shown that the area en closed by the parabola, the x-axis, and the lines x = -h and x = h is h Area = (2ah2 + 6c) 3

-

Show that this area may also be given by

y

A special window has the shape of a rectan gle surmounted by an equilateral triangle. See the figure. If the perimeter of the window is 16 feet, what dimensions will admit the most light? [Hint: Area of an equilateral triangle = x2, where x is the length of a side of the triangle.]

18. Architecture

(�)

x

f,. 23. Use the result obtained in Problem 22 to find the area en closed by f (x) = 5x2 + 8, the x-axis, and the lines x = -1 and x = 1 .

-

An accepted relationship between stopping distance, d (in feet), and the speed of a car, v (in mph), is d = 1 . 1 v + 0.06v2 on dry, level concrete. (a) How many feet will it take a car traveling 45 mph to stop on dry, level concrete? (b) If an accident occurs 200 feet ahead of you, what is the maximum speed you should be traveling to avoid being involved? (c) What might the term 1.1v represent? Source: www2.nsta.orgIEnergylfn_braking.html

19. Stopping Distance

The years 1999 to 2005 were particularly costly for insurance companies, with 7 of the 10 most costly

20. Insurance Claims

f,. 24. Use the result obtained in Problem 2 2 t o find the area en closed by f (x) = 2x2 + 8, the x-axis, and the lines x = - 2 and x = 2.

f,. 25. Use the result obtained in Problem 22 to find the area en closed by f(x) = x2 + 3x + 5, the x-axis, and the lines x = -4 and x = 4.

f,. 26. Use the result obtained in Problem 22 to find the area en closed by f (x) = -x2 + X + 4, the x-axis, and the lines x = -1 and x = 1 . A n individual's income varies with his or her age. The following table shows the median in come I of individuals of different age groups within the United States for 2003. For each age group, let the class midpoint represent the independent variable x. For the class

27. Life Cycle Hypothesis

SECTION 4.4

"65 years and older," we will assume that the class midpoint is 69.5. Age (yearsl 1 5-24

Class Midpoint, x

Median Income (Sl, I

1 9.5

8,614

25-34

29.5

26,21 2

35-44

39.5

30,9 14

45-54

49.5

32,583

55-64

59.5

28,068

65 a n d older

69.5

1 4,664

"�

29.

Quadratic Models; Building Quadratic Functions from Data

(d) Use a graphing utility to verify that the function given in part (b) is the quadratic function of best fit. (e) With a graphing utility, draw a scatter diagram of the data and then graph the quadratic function of best fit on the scatter diagram. Height of a Ball A shot-putter throws a ball at an inclina tion of 45° to the horizontal. The following data represent the height h of the ball at the instant that it has traveled x feet horizontally.

Distance, x

Source: U.S. Census Bureau, 2003 Annual Social and Economic Supplement

(a) Draw a scatter diagram of the data. Comment on the type of relation that may exist between the two variables. (b) The quadratic function of best fit to these data is I ( x ) = -34.3x2 + 3157x - 39,115

Use this function to determine the age at which an in dividual can expect to earn the most income. (c) Use the function to predict the peak income earned. � (d) Use a graphing utility to verify that the function given in part (b) is the quadratic function of best fit. (e) With a graphing utility, draw a scatter diagram of the data and then graph the quadratic function of best fit on the scatter diagram. 28. Life Cycle Hypothesis An individual's income varies with his or her age. The following table shows the median income I of individuals of different age groups within the United States for 2004. For each age group, let the class mid point represent the independent variable x. For the class "65 years and older," we will assume that the class midpoint is 69.5.

Age (yearsl 1 5-24 25-34

Class Midpoint, x

Median Income (Sl, I

1 9.5

8,782

29.5

26,642

�

Height, h

20

25

40

40

60

55

80

65

1 00

71

1 20

77

1 40

77

1 60

75

1 80

71

200

64

(a) Draw a scatter diagram of the data. Comment on the type of relation that may exist between the two variables. (b) The quadratic function of best fit to these data is hex) = -O.0037x2 + 1 .03x + 5.7 Use this function to determine the horizontal distance the ball will travel before it reaches its maximum height. (c) Use the function to find the maximum height of the ball. (d) Use a graphing utility to verify that the function given in part (b) is the quadratic function of best fit. (e) With a graphing utility, draw a scatter diagram of the data and then graph the quadratic function of best fit on the scatter diagram. An engineer collects data showing the speed s of a Ford Taurus and its average miles per gallon, M. See the table.

30. Miles per Gallon

35-44

39.5

31 ,629

45-54

49.5

32,908

Speed, s

Miles per Gal lon, M

55-64

59.5

28,5 1 8

30

18

6 5 a n d older

69.5

1 5, 1 93

35

20

40

23

40

25

45

25

50

28

Source: U.S. Census B ureau, 2004 Annual Social and Economic Supplement

(a) Draw a scatter diagram of the data. Comment on the type of relation that may exist between the two variables. (b) The quadratic function of best fit to these data is I ( x ) = -34.5x2 + 3186x - 39,335 Use this function to determine the age at which an in dividual can expect to earn the most income. (c) Use the function to predict the peak income earned.

31 3

55

30

60

29

65

26

65

25

70

25

31 4

CHAPTER 4


(c) Use the function to predict miles per gallon for a speed of 63 miles per hour. ;,; (d) Use a graphing utility to verify that the function given in part (b) is the quadratic function of best fit. (e) With a graphing utility, draw a scatter diagram of the data and then graph the quadratic function of best fit on the scatter diagram.

(a) Draw a scatter diagram of the data. Comment on the type of relation that may exist between the two variables. (b) The quadratic function of best fit to these data is M(s) = -0.017s2 + 1 .93s - 25.34

Use this function to determine the speed that maximizes miles per gallon. Discussion and Writing 31.

Refer to Example 1 on page 306. Notice that if the price charged for the calculators is $0 or $140 the revenue is $0. It

is easy to explain why revenue would be $0 if the price charged is $0, but how can revenue be $0 if the price charged is $140?


1. R

=

2.

3x

Y =

l.7826x + 4.0652

4.5 Ineq ualities Involving Quadratic Functions PREPARING FOR THIS SECTION •


Solve Inequalities (Section 1 .5, p. 131)

Use Interval Notation (Section 1 .5, pp. 125-126)

Now Work the 'Are You Prepared? , problem on page 3 1 6.

OBJECTIVE

1

1

Solve Inequal ities I nvolving a Q u a d ratic Function (p. 314)

Solve Inequalities Involving a Quadratic Function

In this section we solve inequalities that involve quadratic functions. We will accom plish this by using their graphs. For example, to solve the inequality

a *- O ax2 + bx + c > 0 2 we graph the function f( x ) = ax + bx + c and, from the graph, determine where it is above the x-axis, that is, where f( x ) > O. To solve the inequality ax2 + bx + c < 0, a *- 0, we graph the function f(x) = ax2 + bx + c and deter mine where the graph is below the x-axis. If the inequality is not strict, we include the x-intercepts in the solution. Let's look at an example.

EXA M P L E 1

Solving an I n equal ity

Solve the inequality x2 - 4x - 12 :5 0 and graph the solution set. Solution

We graph the function f(x) = x2 - 4x - 12. The intercepts are y-intercept:

x-intercepts ( if any ) :

f(O) = - 12

x2 - 4x - 12 = 0

( x - 6 ) (x + 2 ) x - 6 = 0 or x + 2 x = 6 or x

=

=

=

0 0 -2

Eva luate f at O. Solve f(x) = O. Factor. Apply the Zero-Product Property.

The y-intercept is - 12; the x-intercepts are -2 and 6. The vertex is at x = -

b

2a

S e e Figure 32 for the graph.

=

-4

- 2 = 2. Since f(2) = - 16, the vertex is (2, -16).

SECTION 4.5

I [ -4 -2

0

2

4

1 I I 6 8

•

x

31 5

y

Figure 32

Figure 33

I nequalities Involving Quadratic Functions

The graph is below the x-axis for -2 < x < 6. Since the original inequality is not strict, we include the x-intercepts. The solution set is { x 1 - 2 ::5 X ::5 6} or, us ing interval notation, [-2, 6]. See Figure 33 for the graph of the solution set. • I
EXAM P L E 2


Solving an I neq uality

Solve the inequality 2X2 < x + 10 and graph the solution set. Solution

Method 1

We rearrange the inequality so that 0 is on the right side. 2x2 < x + 10 2x2 - X - 10 < 0

Subtract x +

10 from both sides.

This inequality is equivalent to the one that we wish to solve. Next we graph the function f(x) = 2x2 - X - 10. The intercepts are

f(O)

y-intercept:

=

-10

Eva luate f at O.

x-intercepts (if any) : 2x2 - x - 1 0 = 0 ( 2x - 5 ) (x + 2 ) 0

Solve f(x) = O. Factor.

=

2x - 5

=

Figure 34

y 4

x

0 or x 5

=

-

2

or

x

+ 2 =

=

0

Apply the Zero-Product Property.

-2

5 . . - 10 ; th e x-lI1tercepts IS are - 2 an d 2" Th e y-mtercept ' TIle vertex

(�,

)

.

IS

at x

=

-

b 2a

=

-1 -4

=

()

l 1 S mce ' f 4 4'

=

-10 . 125, the vertex is

- 10.125 . See Figure 34 for the graph. The graph is below the x-axis between x

is strict, the solution set is

{

x1 -2 < x

Q2
0 is an integer.

(2)

326 t r r

r

CHAPTER 5

Polynomial and Rational Functions

Examples of power functions are

In Words

single monomial.

f(x) = -5x2

f(x) = 3x

A power function is defined by a

f(x) = -5x4

f(x) = 8x3 degree 3

degree 2

degree 1

degree 4

The graph of a power function of degree 1 , f(x) = ax, is a straight line, with slope a, that passes through the origin. The graph of a power function of degree 2, f(x) = ax2, is a parabola, with vertex at the origin, that opens up if a > 0 and down if a < O. If we know how to graph a power function of the form f(x) = x", a compres sion or stretch and, perhaps, a reflection about the x-axis will enable us to obtain the graph of g(x) = ax". Consequently, we shall concentrate on graphing power func tions of the form f (x) = x". We begin with power functions of even degree of the form f (x) = XII, n 2:: 2 and n even. The domain of f is the set of all real numbers, and the range is the set of nonnegative real numbers. Such a power function is an even function (do you see why?) , so its graph is symmetric with respect to the y-axis. Its graph always contains the origin and the points ( - 1 , 1) and ( 1 , 1). If n = 2, the graph is the familiar parabola y = x2 that opens up, with vertex at the origin. If n 2:: 4, the graph of f(x) = x", n even, will be closer to the x-axis than the parabola y = x2 if - 1 < x < 1 and farther from the x-axis than the parabola y = x2 if x < -1 or if x > 1 . Figure 2(a) illustrates this conclusion. Figure 2(b) shows the graphs of y = x4 and y = x8 for comparison. f(x) = x" n2 4 n even

Figure 2

Y=

x8 Y

4 Y= X

4

2

(-1 ,1)

-3

3

x

-3

(1, 1 ) (0,0)

(a)

3

x

(b)

From Figure 2, we can see tha t as n increases the graph of f (x) = XII, n 2:: 2 and n even, tends to flatten out near the origin and to increase very rapidly when x is far from O. For large n, it may appear that the graph coincides with the x-axis near the origin, but it does not; the graph actually touches the x-axis only at the origin (see Table 2). Also, for large n, it may appear that for x < -1 or for x > 1 the graph is vertical, but it is not; it is only increasing very rapidly in these intervals. If the graphs were enlarged many times, these distinctions would be clear. Table 2

x =

{ (x)

=

{ (x)

=

{ (x)

=

x =

0.1

x8 x20

1 0-8 1 0-20

x40

1 0-40

0.3

x =

0.5

0.0000656

0.0039063

3.487 ' 1 0-'1 1 .2 1 6. 1 0-21

0.000001 9.095' 1 0-13

Seeing the Concept Graph Y, X4, Y2 = xB, and Y3 X'2 usi n g the viewing rectangle -2 :5 X :5 2, -4 :5 Y :5 1 6. Then g raph each again usi n g the viewing rectangle -1 :5 x :5 1 ,0 :5 Y :5 1 . See Figure 3.TRACE along one =

=

of the graphs to confirm that for x close to 0 the graph is above the x-axis a n d that for x is i ncreasi ng.

>

0 the graph

SECTION 5.1

Polynomial Functions and Models

327

Figure 3

'\..11' II'

... III -2

�

/1)

. j.' .., '

�II \,\.

2 -1

-4

�1��

�

I�

..�... . ... .

,.//{

..

_

o

(a)

..

(b)

Properties of Power Functions, ((xl

=

X',

n Is an Even Integer

1. f is an even function, so its graph is symmetric with respect to the y-axis. 2. The domain is the set of all real numbers. The range is the set of non negative real numbers. 3. The graph always contains the points ( - 1 , 1), (0,0), and ( 1 , 1). 4. As the exponent n increases in magnitude, the graph becomes more ver tical when x < - l or x > 1 ; but for x near the origin, the graph tends to flatten out and lie closer to the x-axis. Now we consider power functions of odd degree of the form f(x) = XII, n 2: 3 and n odd. The domain and the range of f are the set of real numbers. Such a power function is an odd function (do you see why?), so its graph is symmetric with respect to the origin. Its graph always contains the origin and the points ( - 1 , - 1 ) and (1, 1). The graph of f(x) = x" when n = 3 has been shown several times and is repeated in Figure 4. If n 2: 5, the graph of f(x) = x", n odd, will be closer to the x-axis than that of y = x3 if - 1 < x < 1 and farther from the x-axis than that of y = x 3 if x < - 1 or if x > 1. Figure 4 also illustrates this conclusion. Figure 5 shows the graph of y = x S and the graph of y = x 9 for further comparison. n

Figure 5

Y= x n? 5 n odd

Figure 4

-3

3 x

-3

3 x

It appears that each graph coincides with the x-axis near the origin, but it does not; each graph actually crosses the x-axis at the origin. Also, it appears that as x increases the graph becomes vertical, but it does not; each graph is increasing very rapidly.

328

CHAPTER 5

Polynomial and Ratio nal Functions

�I Seeing the Concept eJ Graph Yl = x3, Y2 = x7, a n d Y3 = Xll using the viewing rectangle -2 :oS X :oS

2, - 1 6 :oS Y :oS 1 6. Then graph each again using the viewing rectangle -1 :oS x :oS 1, -1 :oS Y :oS 1 . See Figure 6. TRACE along one of the graphs to confirm that the graph is increasing and crosses the x-axis at the origin.

Figure 6

-1

-16

(b)

(a)

To summarize: Properties of Power Functions, ((x)

=

x",

n Is an Odd Integer

1. f is an odd function, so its graph is symmetric with respect to the origin. The domain and the range are the set of all real numbers. 3. The graph always contains the points (-1 , -1), (0, 0), and (1,1). 4. As the exponent n increases in magnitude, the graph becomes more vertical when x < -1 or x > 1; but for x near the origin, the graph tends to flatten out and lie closer to the x-axis.

2.

2

Graph Polynomial Functions Using Transformations

The methods of shifting, compression, stretching, and reflection studied in Section 3.5, when used with the facts just presented, will enable us to graph polyno mial functions that are transformations of power functions. E XA M P L E 2

Graphing Polynomial Functions Using Transformations

Graph: f(x) Solution

=

1 - x5

Figure 7 shows the required stages.

Figure 7

2

-2

x

(-1,-1) -2 Multi ply by -1; ref lect about x-a xis

(a) y = XS

EXAM P L E 3

Add 1 ;

shi ft up

(b)

s Y = -x

1 unit

(c)

y = -x s + 1

Graphing Polynomial Functions Using Transformatio n s

Graph: f(x)

=

1

2 (x - 1)4

= 1 -xs

•

SECTION 5.1

329

Figure 8 shows the required stages.

Solution Figu re


8

Y 2

-2 -2

-2

-2

(a) Y= x4 � == ....

3

x

(1 , 0) 2

Replace x by x - 1; shift right 1 unit

�

(b) y=(x-1)4

Now Work

PRO B

L

-2

�;

Multiply by co mpression by a factor o f

(c) Y= � (X_1)4

9

EMS 2 3 A N D 2

•

Identify the Real Zeros of a Polynomial Function and Their Multiplicity

Figure 9 shows the graph of a polynomial function with four x-intercepts. Notice that at the x-intercepts the graph must either cross the x-axis or touch the x-axis. Consequently, between consecutive x-intercepts the graph is either above the x-axis or below the x-axis. We will make use of this property of the graph of a polynomial function shortly. Figure 9

Y - Above---+ x-axis

x-axis ___

Below x-axis

x-axis

__

x-axis

Below x-axis

If a polynomial function f is factored completely, it is easy to solve the equation f(x) = 0 using the Zero-Product Property and locate the x-intercepts of the graph . For example, if f(x) = (x - 1)2(x + 3) , then the solutions of the equation f(x)

=

(x - l)\x + 3)

=

0

are identified as 1 and - 3. Based on this result, we make the following observations: DEFINITION

If f is a function and r is a real number for which f(r) a real zero of f·

=

0, then r is called

.J

As a consequence of this definition, the following statements are equivalent. 1. r is a real zero of a polynomial function f 2. r is an x-intercept of the graph of f 3. x - r is a factor of f So the real zeros of a polynomial function are the x-intercepts of its graph, and they are found by solving the equation f(x) = O.

330

CHAPTER 5


F i nding a Polynom ial from Its Zeros

E XA M P L E 4

Solution Figure

10 40

.

.. .---.

l

,/'

"

l 6

- 50

(a) Find a polynomial of degree 3 whose zeros are -3,2, and 5. (b) Use a graphing utility to graph the polynomial found in part (a) to verify your result. (a) If r is a real zero of a polynomial [, then x - r is a factor of f. This means that x - ( -3) = x + 3, x - 2, and x - 5 are factors of f. As a result, any polynomial of the form [(x) = a(x + 3)(x - 2)(x - 5) where a is any nonzero real number, qualifies. The value of a causes a stretch, compression, or reflection, but does not affect the x-intercepts. Do you know why? We choose to graph [ with a = 1. Then [(x) = (x + 3)(x - 2)(x - 5) = x3 - 4x2 - llx + 3 0 Figure 10 shows the graph of f. Notice that the x-intercepts are -3, 2, and 5.

I (b)

•

Seeing the Concept Graph the function found in Example 4 for

a=

2 and

of f? How does the value of a affect the graph of f?

lJ'l!l:=iOlIII ""II: .- Now Work

a=

-1. Does the value of

a affect the

zeros

PRO B L E M 3 7

If the same factor x -

occurs more than once,

r

r

is called a repeated, or

multiple, zero of f. More precisely, we have the following definition.

If (x - r ) 111 is a factor of a polynomial [ and ( x then r is called a zero of multiplicity m off· *

DEFINITION E XA M P L E 5

-

r

) I11+1 is not a factor of [,

.J

Identifyi ng Zeros and Thei r M u lti p l icities

For the polynomial [(x)

=

( �y

5 (x - 2)(x + 3? X -

2 is a zero of multiplicity 1 because the exponent on the factor x - 2 is 1. -3 is a zero of multiplicity 2 because the exponent on the factor x + 3 is 2.

�

is a zero of multiplicity 4 because the exponent on the factor x -

�

Now Work

PRO B L EM 4 5 (a)

�

is 4. •

Suppose that it is possible to factor completely a polynomial function and, as a result, locate all the x-intercepts of its graph (the real zeros of the function). These x-intercepts then divide the x-axis into open intervals and, on each such interval, the graph of the polynomial will be either above or below the x-axis. Let's look at an example. E XA M P L E 6

Graphing a Polynomial Using Its x-Intercepts

For the polynomial: [(x)

=

x2(x - 2)

(a) Find the x- and y-intercepts of the graph of f. (b) Use the x-intercepts to find the intervals on which the graph of [ is above the x-axis and the intervals on which the graph of [ is below the x-axis. ':' Some books use the terms lIIultiple root and root of lIIultiplicity

m.

SECTION 5.1


331

(c) Locate other points on the graph and connect all the points plotted with a smooth, continuous curve. Solution

(a) The y-intercept is f (O)

=

02(0 - 2) = O. The x-intercepts satisfy the equation f(x) =x2(x - 2) = 0

from which we find x2=0 or x - 2=0 x =O or x=2 The x-intercepts are 0 and 2. (b) The two x-intercepts divide the x-axis into three intervals: (0, 2)

( - 00, 0)

( 2,

(0

)

Since the graph of f crosses or touches the x-axis only at x = 0 and x = 2, it follows that the graph of f is either above the x-axis [I(x) > 0] or below the x-axis [I(x) < 0] on each of these three intervals. To see where the graph lies, we only need to pick a number in each interval, evaluate f there, and see whether the value is positive (above the x-axis) or negative (below the x-axis). See Table 3. (c) In constructing Table 3, we obtained three additional points on the graph: ( - 1 , -3), (1, -1), and (3, 9). Figure 11 illustrates these points, the intercepts, and a smooth, continuous curve (the graph of f) connecting them. Table 3

Figure

.2

.0 Interval Number chosen

-1

Value of f

f(-1)

Location of graph

Below x-axis

Point on graph

=

-3

(-1,-3)

f(1)

3

f(3)

= -1

Below x-axis

(1, -1)

(3,9)

(2,00)

(0,2)

(-00,0)

• x

11

=9

Above x-axis

(3,9)

-2

4

x

•

Look again at Table 3. Since the graph of f(x ) = x2 (x - 2) is below the x-axis on both sides of 0, the graph of f touches the x-axis at x = 0, a zero of multiplicity 2. Since the graph of f is below the x-axis for x < 2 and above the x-axis for x > 2, the graph of f crosses the x-axis at x = 2, a zero of multip licity 1. This suggests the following results: If r Is a Zero of Even Multiplicity

The sign of f (x) does not change from one side to the other side of r. If r Is a Zero of Odd Multiplicity

The sign of f(x) changes from one side to the other side of r. 'I!=>-

Now Work

PRO B L EM 4 5 (b)

The graph of f touches the x-axis at r.

The graph of f crosses the x-axis at r.

332

C H A PTER 5


Behavior Near a Zero

We have just learned how the multiplicity of a zero can be used to determine whether the graph of a function touches or crosses the x-axis at the zero. However, we can learn more about the behavior of the graph near its zeros than just whether the graph crosses or touches the x-axis. Consider the function f(x) = x2(x - 2) whose graph is drawn in Figure 1 1 . The zeros of f are 0 and 2. Table 4 shows the values of f(x) = x2(x - 2) and y = -2x2 for x near O. Figure 12 shows the points ( -0.1 , -0.021) , ( -0.05, -0.0051), and so on, that are on the graph of f(x) = x2(x - 2) along with the graph of y = _2x2 on the same Cartesian plane. From the table and graph, we can see that the points on the graph of f(x) = x2(x - 2) and the points on the graph of y = -2x2 are indistinguishable near x = O. So y = _2x2 describes the behavior of the graph of f(x) = x2(x - 2) near x = O. Table 4

Figure

fIx) = �(x - 2)

x

y=-�

-0.1

-0.021

-0.02

-0.05

-0.005 1 25

-0.005

-0.03

-0.00 1 827

-0.00 1 8

-0.01

-0.000201

-0.0002

0

0

12 Y 0.005 x

0

0.01

-0.0001 99

-0.0002

0.03

-0.00 1 773

-0.001 8

0.05

-0.004875

-0.005

0.1

-0.01 9

-0.02

But how did we know that the function f(x) = x\x - 2) behaves like -2x2 when x is close to O? In other words, where did y = -2x2 come from? Because the zero, 0, comes from the factor x2, we evaluate all factors in the function f at 0 with the exception of x2. y =

f(x)

=

�

=

x2(x - 2) x2(0 - 2) _2x2

The factor the factor

J'- g ives rise to the J'- a nd let x = 0 in

zero, so we kee p the rema i ni ng

factors to fi nd the behavio r near O.

This tells us that the graph of f(x) = x2(x - 2) will behave like the graph of y = _2x2 near x = O. Now let's discuss the behavior of f(x) = x2(x - 2) near x = 2, the other zero. Because the zero, 2, comes from the factor x - 2, we evaluate all factors of the func tion f at 2 with the exception of x - 2. f(x)

=

�

=

x2(x - 2) 22(x - 2) 4(x - 2)

T h e factor

x

-

2 gives rise t o the zero, so we

keep the factor x

-

2 a nd let x = 2 in the

rema i n ing factors to find the behavior near 2.

So the graph of f(x) = x2(x - 2) will behave like the graph of y = 4(x - 2) near x = 2. Table 5 verifies that f(x) = x2(x - 2) and y = 4(x - 2 ) have similar values for x near 2. Figure 13 shows the points ( 1 .9, -0.361 ) , ( 1 99 -0.0396 ) , and so on, that are on the graph of f(x) = x2(x - 2) along with the graph of y = 4( x - 2) on the same Cartesian plane. We can see that the points on the graph of f(x) = x2(x - 2) and the points on the graph of y = 4(x - 2) are indistin guishable near x = 2. So Y = 4(x - 2 ) , a line with slope 4, describes the behavior of the graph of f(x) = x2(x - 2) near x = 2. .

,

SECTION 5.1

Table 5

f(x)

x

1.9

Figure

14

=

xl(x - 2)

y

=

4(x - 2)

-0.361

-0.4

1.99

-0.0396

-0.04

1.999

-0.003996

-0.004

2

0

0

2.001

0.004004

0.004

2.01

0.0404

0.04

2.1

0.441

0.4


Figure

13

333

y •

x

2.2

Figure 14 illustrates how we would use this information to begin to graph f(x) = x2(x - 2). (2,0)

1

Behaves 1 ke -2 y=-2x near x= 0 -4

3 x

Behaves l i ke Y= 4(x- 2) near x= 2

By determining the multiplicity of a real zero, we determine whether the graph crosses or touches the x-axis at the zero. By determining the behavior of the graph near the real zero, we determine how the graph touches or crosses the x-axis. =:>-

Now Work

PRO B l EM

45( c )

Turning Points

Look again at Figure 11. We cannot be sure just how low the graph actually goes between x = 0 and x = 2. But we do know that somewhere in the interval (0, 2) the graph of f must change direction (from decreasing to increasing). The points at which a graph changes direction are called turning points. In calculus, such points are called local maxima or local minima, and techniques for locating them are given. So we shall not ask for the location of turning points in our graphs. Instead, we will use the following result from calculus, which tells us the maximum number of turn ing points that the graph of a polynomial function can have. THEOREM

Turning Points

If f is a polynomial function of degree n, then f has at most n - 1 turning points. If the graph of a polynomial function f has n - 1 turning points, the degree of f is at least n.

.J

For example, the graph of f(x) = x\x - 2) shown in Figure 11 is the graph of a polynomial of degree 3 and has 3 - 1 = 2 turning points: one at (0, 0) and the other somewhere between x = 0 and x = 2.

l�.·1 Exploration lM A g raphing utility can be used to locate the turning points of a g raph. Graph MINIMUM to find the location of the turning point for 0

Figure 15

=- Now Work

PRO B L EM

4 5(d)

? + x? -

1 2x. Compa re what you see with Figure 1 9(b). Use MAXIMUM/MINIMUM to locate the two turning points.

L.'l'l:=::::o:- Now Work P R O B L E M 6 7 E XA M P L E 1 0

Analyzing the Graph of a Polynom ial F u n ction

Follow the instructions of Example 9 for the following polynomial: f(x) (a) The y-intercept is f(O)

Solution

=

=

x2(x - 4 ) ( x

+

1)

O. The x-intercepts satisfy the equation

f(x)

=

x2(x - 4 ) ( x

+

1) = 0

So x2 0 or x - 4 x x = 0 or =

(b) (c) (d) (e)

=

=

=

0 or x + 1 0 x = 1 4 or =

-

The x-intercepts are - 1 , 0, and 4. The intercept 0 is a zero of multiplicity 2, so the graph of f will touch the x-axis at 0; 4 and - 1 are zeros of multiplicity 1, so the graph of f will cross the x-axis at 4 and - l . End behavior: the graph of [" resembles that of the power function y = X4 for large values of I x l . The graph of f will contain at most three turning points. The three x-intercepts are - 1 , 0, and 4.

Near - I : f(x) = x2( x - 4 ) ( x + 1) Near 0: f(x) x2( x - 4)(x + 1) Near 4: f(x)

=

x2( x - 4 ) ( x + 1 )

� � �

( - I f( - I - 4) (x + 1)

=

-5(x

x2(0 - 4 ) (0 + 1) = -4x2 42(x - 4 ) (4 + 1 ) = 80(x - 4)

+

1)

A l ine with slope - 5 A parabola opening down A l in e with slope

80

(f) Figure 20(a) illustrates the information obtained from parts (a), (b), (c), and (e). 1

The graph of f is given in Figure 20(b) . Notice that we evaluated f at -2' - 2' 2, and 5 to help establish the scale on the y-axis.

338

CHAPTER 5


Figure 20

y

I

End behavior: 4 Resembles y = x

Near x = 0 the graph behaves like y = -4} a parabola opening down

End behavior: 4 Resembles y = x

\

Near x = 4 the graph behaves like y = 80(x - 4) a line with slope 80 (4, 0)

2

-2

5

3

x

Near x = -1 the graph behaves like y = -5(x + 1 ) a line with slope -5

x

- 40

(b)

(a)

Exploration Graph Yl = x2 (x

- 4)(x + 1 ). Compare

•

what you see with F i g u re 20(b). Use MAXIMUM/MINIMUM to

locate the two turning points besides (0, 0).

SUMMARY

Steps for Ana lyzing the Graph of a Polynomial

To analyze the graph of a polynomial function

y

=

1 (x), follow these steps:

1: (a) Find the y-intercept by letting x = 0 and finding the value of 1(0). (b) Find the x-intercepts, if any, by solving the equation I(x) = O. STEP 2: Determine whether the graph of 1 crosses or touches the x-axis at each x-intercept. STEP 3: End behavior: find the power function that the graph of 1 resembles for large values of I x i . STEP 4: Determine the maximum number of turning points on the graph of I. STEP 5: Determine the behavior of the graph of 1 near each x-intercept. STEP 6: Put all the information together to obtain the graph of f. This usually requires finding additional points on the graph. STEP

'1'1

,-

Now Work PRO B L E M 7 7

For polynomial functions that have non integer coefficients and for polynomials that are not easily factored, we utilize the graphing utility early in the analysis of the graph. This is because the amount of information that can be obtained from algebraic analysis is limited.

II

E XA M P L E 1 1

Using a Graphing Utility to Analyze the Graph of a Polynomial F unction

For the polynomial I(x)

=

x3 + 2.48x2 - 4.3155x + 2.484406:

(a) Find the degree of the polynomial. Determine the end behavior; that is, find the power function that the graph of 1 resembles for large values of I x l . (b) Graph 1 using a graphing utility.

SECTION 5 . 1

Polynomial Functio ns and Models

339

(c) Find the x- and y-intercepts of the graph. (d) Use a TABLE to find points on the graph around each x-intercept. Determine on which intervals the graph is above and below the x-axis. ( e) Determine the local maxima and local minima, if any exist, rounded to two dec imal places. That is, locate any turning points. (f) Use the information obtained in parts (a) to (e) to draw a complete graph of I by hand. Be sure to label the intercepts, turning points, and the points ob tained in part (d). (g) Find the domain of f. Use the graph to find the range of I. (h) Use the graph to determine where I is increasing and decreasing. Solution Figure 2 1 15

Table 6 ::-:: � -3.D "3:.6 -3.'1

-�.2 -3 -2.8

X= - 4

V1 -'1.S;''1 -.177S �.S(lS 6.S219 D.!'-

•

Now Work PRO B L E M S 4 1 A N D 4 3

F i n d i n g Horizontal or O b l i q u e Asymptotes

E XA M P L E 8

Find the horizontal or oblique asymptotes, if any, of the graph of 2x5 - x3 + 2 G ( x) = ----x3 - 1 Since the degree of the numerator, S, is larger than the degree of the denominator, 3, the rational function G is improper. To find any horizontal or oblique asymptotes, we use long division.

Solution

x3 - 1 hx5 - x3 + 2x5 - 2x2 -x3 + 2x2 + + -x3 2x2 +

2 2 1 1

As a result, G( x) =

2x5 - x3 + 2 x' - 1 0

=

2x2 - 1

+

2x2 + 1 x3 - 1

---

Then, as x --i> - 00 or as x --i> 00 , 2X2 + 1 2x2 2 ::::; 0 = - --i> 0 x x3 - 1 x '

---

-

As x --i> - 00 or as x --i> 00, we have G (x) --i> 2x2 - 1. We conclude that, for large values of lxi, the graph of G approaches the graph of y = 2x2 - 1 . That is, the graph of G will look like the graph of y = 2x2 - 1 as x --i> - 00 or x --i> 00 . Since y = 2X2 - 1 is not a linear function, G has no horizontal or oblique asymptotes. •

We now summarize the procedure for finding horizontal and oblique asymptotes. SUMMARY

Finding Horizo nta l and Obl ique Asym ptotes of a Rational Function R

Consider the rational function R(x) =

p (x) q(x)

=

x" + a,,_ l x"- 1 + . . . + a l x + ao " bm xm + bm_ 1 ,·'C'I/-l + . . . + b1x + bo a

in which the degree of the numerator is n. and the degree of the denominator is m . 1. If the degree of the numerator is less than the degree of the denominator, R is a proper rational function, and the graph of R will have the horizontal asymptote y = 0 (the x-axis) .

352

Pol y n omia l and Rational Functions

CHAPTER 5

2. If the degree of the numerator is greater than or equal to the degree of the denominator, then R is improper.

Here long division is used. (a) If the degree of the numerator equals the degree of the denominator, the quotient obtained will be the an

number - , and the line bm

y

an

= - is a horizontal asymptote. bm

(b) If the degree of the numerator is one more than the degree of the denominator, the quotient obtained is of the form a x + b (a polynomial of degree 1), and the line y = ax + b is an oblique asymptote. (c) If the degree of the numerator is two or more than the degree of the denominator, the quotient obtained is a polynomial of degree 2 or higher, and R has neither a horizontal nor an oblique asymptote. In this case, for Ixl unbounded, the graph of R will behave like the graph of the quotient. Note: The graph of a rational function either has one horizontal or one oblique asymptote or else has no horizontal and no oblique asymptote. It cannot have both a horizontal asymptote and an oblique asymptote.

5.2 Assess You r U ndersta nding 'Are You P repared?, Answers are given a t the end of these exercises. If you get a wrong answer, read the pages listed in red.

1. True or False The quotient of two polynomial expressions is a rational expression. (pp. 6 1-68) 2. What is the quotient and remainder when 3x4 - x2 is divid ed by x3 - x2 + 1 . ( pp. 44-47)

3. Graph y =

1 -

x

. (pp. 1 70-171 )

4. Graph y = 2(x + 1) 2 - 3 using transformations.

(pp. 2S2-260)

Concepts and Voca bulary 5•

· The hne

6. The line 7.

__

.

IS

__

x3 - 1 3+ 1

.

a honzontal asymptote of R ( x ) = --- ' X

is a vertical asymptote of R(x) =

: : �.

9. True or False If an asymptote is neither horizontal nor ver tical, it is called oblique.

10.

For a rational function R, if the degree of the numerator is less than the degree of the denominator, then R is . __

True or False If the degree of the numerator of a rational function equals the degree of the denominator, then the ratio of the leading coefficients gives rise to the horizontal asymptote.

8. True or False The domain of every rational function is the set of all real numbers. Skill B u i l d i n g

In Problems 11-22, find the domain of each rational function. 4x Sx2 11. R(x) = 12 R( x ) = x-3 3+x . --

14. G(x) =

6

(x + 3) (4 - x) x

17. R(x) = -- X3 - 8

20. G(x) =

x - 3 + 1

4 X

3x(x - 1 ) 15. F ( x) = ---'-----'--2x2 - Sx - 3 x

13.

H(x) =

-4x2 (x - 2)(x + 4) -x( 1 - x)

16. Q ( x) = ---:-------'---

3x2 + Sx - 2 3x2 + X x2 + 4

18. R(x) = -4- x - I

19. H(x) =

3(x2 - X - 6) 21. R(x) = -----'4(x2 - 9)

22. F (x) = ---'-----'--

-

2 ( x2 - 4)

3(x2 + 4x + 4)

SECTION 5.2

In Problems 23-28, use the graph shown to find: (a) The domain and range ofeach function (d) Vertical asymptotes, if any y

23.

(b) The intercepts, if any (e) Oblique asymptotes, if any

4 x

-4

25.

�!

-- l�

y

3 x

I I I I I I

-3

35 3

(c) Horizontal asymptotes, if any

24.

4

Properties of Rational Functions

-3

1 -3

-4

I

26.

27. y

) � : l y

:I

3

I

I I

I I

I I

_ _ _ -.-l _ _ _ _ -3

3 x

In Problems

29-40, 1

35. R(x ) = 38. F(x ) =

3 x

2

=

33. H ( x ) =

-

x2

+

-

L___ 3 x

-3

30. Q(x )

x

=

_ _ _

x

graph each rational function using transformations.

29. F(x ) = 2 + 32. R(x )

28.

-1 4x

+

1

1

x +

4

36. R(x ) = 39. R (x )

=

1

3+? x-

31. R(x )

-2

=

3x

x

+

42. R(x )

4 +

-x2 1 44. G(x ) = x2 _ 5x + 6 47. Q(x )

=

50. R(x ) =

5 - x2

=

45. T(x ) =

�

48. F(x ) =

6x2 + X + 1 2 3x2 - 5x - 2

51. G ( x )

=

2

x+ 1

34. G(x ) =

1 + 1 x-I

37. G(x ) = 1 +

x?- - 4 x2

40. R(x ) =

--

In Problems 41-52, find the vertical, horizontal, and oblique asymptotes, if any, of each rational function.

41. R(x ) =

1 (x - I f (x

+

2f 2

(x _ 3 ) 2

x -4 x

3x + 5 x -6

43. H ( x ) =

x3 - 8 2 x _ 5x + 6

x3 _ 1

46. P(x ) =

4x5 x3 _

x4

-2X2

2x 3

+

1

+ 4x2

x3 - 1

x - x2

49. R(x )

=

52. F(x ) =

3x4

x3

1

+

4

+ 3x

x - I

x

_

x3

354

C H APTER 5



53.

(a) Let Rl = 10 ohms, and graph Rtot as a function of R2 . (b) Find and interpret any asymptotes of the graph obtained in part (a). (c) If R2 = 2�, what value of R I will yield an Rtot of 1 7 ohms?

In physics, it is established that the acceleration due Gravity to gravity, g (in meters/sec2 ), at a height h meters above sea level is given by 3.99

g(h) = (6.374

X

10 1 4 106 + h ) 2

X

where 6.374 X 106 is the radius of Earth in meters. (a) What is the acceleration due to gravity at sea level? (b) The Sears Tower in Chicago, Illinois, is 443 meters tall. What is the acceleration due to gravity at the top of the Sears Tower? (c) The peak of Mount Everest is 8848 meters above sea level. What is the acceleration due to gravity on the peak of Mount Everest? (d) Find the horizontal asymptote of g( h ) . ( e ) Solve g ( h ) O . How d o you interpret your answer? 54. Population Model A rare species of insect was discovered in the Amazon Rain Forest. To protect the species, environ mentalists declare the insect endangered and transplant the insect into a protected area. The population P of the insect ( months after being transplanted is

Source: en. wikipedia.org/wiki/SeriesJmd_paralleCcircuits

gs, 56.

2

is an efficient method for finding the x-intercepts (or real zeros) of a function, such as p(x). The steps below outline Newton's Method STEP 1 : Select an initial value Xo that is somewhat close to the x-intercept being sought. STEP 2: Find values for x using the relation p(X,,) X,,+ I = X" - ' - n = 1 , 2, . . . p ( x" )

50( 1 + 0.5t) (2 + O.Olt)

until you get two consecutive values Xk and Xk+ 1 that agree to whatever decimal place accuracy you desire. STEP 3: The approximate zero will be Xk+ l ' Consider the polynomial p(x) = x3 7x - 40. (a) Evaluate p (5) and p( -3). (b) What might we conclude about a zero of p . Explain. (c) Use Newton's Method to approximate an x-intercept, 1', -3 < r < 5, of p(x) to four decimal places. .- (d) Use a graphing utility to graph p(x) and verify your an swer in part (c). (e) Using a graphing utility, evaluate per) to verify your result.

(a) How many insects were discovered? In other words,

what was the population when t = O? (b) What will the population be after 5 years? (c) Determine the horizontal asymptote of p e t ) . What is the largest population that the protected area can sustain? 55. R esistance in Parallel Circuits From Ohm's law for circuits, it follows that the total resistance R tot of two components hooked in parallel is given by the equation Rtot

=

RI R 2 RI + R

"

Newton's Method

=

pet ) =

I n calculus you will learn that, if p(x) = a"x + a,,_lx,,- I + . . . + ajX + ao is a polynomial, then the derivative of p(x) is p ' (x) = na"x,,- I + ( n - 1 )a,,_jx,,-2 + . . . + 2a x + al

Newton's Method

-

---

2

where R l and R 2 are the individual resistances. Discussion a n d Writing

57. If the graph of a rational function R has the vertical asymp tote x = 4, the factor x 4 must be present in the denomi

59. Can the graph of a rational function have both a horizontal

and an oblique asymptote? Explain.

-

nator of R. Explain why. 58. If the graph of a rational function R has the horizontal asymptote y = 2, the degree of the numerator of R equals the degree of the denominator of R . Explain why.

60. Make up a rational function that has y

2x + 1 as an oblique asymptote. Explain the methodology that you used. =

'Are You Prepa red ?' An swers 1.

True

2. Quotient: 3x + 3; Remainder: 2x2 - 3x - 3

3.

4.

y 3

-

3 x

3 (0, - 1 )

SECTION 5.3

The Graph of a Rational Function

355

5.3 The Graph of a Rational Function PREPARING FOR THIS SECTION •


Intercepts (Section 2.2, pp. 1 65-166) Now Work the 'Are You

Prepared?' problems on page 366.

OBJECTIVES 1 Ana lyze the G raph

of a

Rational Fu nction (p. 355)

2 Solve Appl ied Problems I nvolving Rational F u n ctions (p. 365)

1

Analyze the Graph of a Rational Function

We commented earlier that calculus provides the tools required to graph a polyno mial function accurately. The same holds true for rational functions. However, we can gather together quite a bit of information about their graphs to get an idea of the general shape and position of the graph. In the examples that follow, we will analyze the graph of a rational function by applying the following steps:

Analyzing the Graph of a Rational Function R

STEP 1: Factor the numerator and denominator of R and find its domain. If o is in the domain, find the y-intercept, R(O), and plot it.

STEP 2: Write R in lowest terms as

STEP 3:

STEP 4:

STEP 5:

STEP 6: STEP

E XA M P L E 1

7:

�i��

and find the real zeros of the numer

ator; that is, find the real solutions of the equation p(x) 0, if any. These are the x-intercepts of the graph. D etermine the behavior of the graph of R near each x-intercept, using the same procedure as for polynomial functions. Plot each x-intercept and indicate the behavior of the graph near it. ( x) With R written in lowest terms as p , find the real zeros of the deq( x) nominator; that is, find the real solutions of the equation q(x) = 0, if any. These determine the vertical asymptotes of the graph. Graph each vertical asymptote using a dashed line. Locate any horizontal or oblique asymptotes using the procedure given in the previous section. Graph the asymptotes using a dashed line. D etermine the points, if any, at which the graph of R intersects these asymptotes. Plot any such points. Using the real zeros of the numerator and the denominator of the given equation for R, divide the x-axis into intervals and determine where the graph is above the x-axis and where it is below the x-axis by choosing a number in each interval and evaluating R there. Plot the points found. Analyze the behavior of the graph of R near each asymptote and in dicate this behavior on the graph. Put all the information together to obtain the graph of R. =

A n alyzi ng the G raph of a Ration al Functio n

Analyze the graph of the rational function:

R(x)

=

x - I x- - 4

-?-

356

CHAPTER 5

Polynomial a nd Rational Functions

Solution

STEP 1: We factor the numerator and denominator of R, obtaining

R(x ) The domain of R is { x i x

"*

x - 1 (x + 2) (x - 2)

- ----

-2, x

"*

2 } . The y-intercept is

-1 1 R(O) = - = 4 -4

( �) .

Plot the point 0,

STEP 2: R is in lowest terms. The real zero of the numerator satisfies the equation

x - 1 = O. The only x-intercept is 1 . Near 1 :

x - 1 (x + 2) (x - 2)

R(x ) =

�

x - 1 1 = - 3(x - 1 ) (1 + 2) (1 - 2)

Plot the point (1, 0) and indicate a line with slope -

�

there.

STEP 3: R is in lowest terms. The real zeros of the denominator are the real solutions

of the equation (x + 2) (x - 2) = 0, that is, -2 and 2. The graph of R has two vertical asymptotes: the lines x = -2 and x = 2. Graph each of these asymptotes using dashed lines. STEP 4: The degree of the numerator is less than the degree of the denominator, so R is proper and the line y = 0 (the x-axis) is a horizontal asymptote of the graph. Indicate this line by graphing y = 0 using a dashed line. To determine if the graph of R intersects the horizontal asymptote, we solve the equation R( x) = o. x - 1 -= 0 x2 - 4

Multiply both sides by

x - 1 = 0 x = 1

:l- 4.

The only solution is x = 1 , so the graph of R intersects the horizontal asymptote at ( 1 , 0). We have already plotted this point. STEP 5: The zero of the numerator, 1, and the zeros of the denominator, -2 and 2, divide the x-axis into four intervals:

( -00, -2)

(2,

( 1 , 2)

( -2, 1 )

(0

)

Now construct Table 8.

-----------. ----- ----- .2 -----

Table 8 r-

_-:2

�

Interval

( -00, -2)

(-2, 1)

Number chosen

-3

0

Value of R

R(-3)

Location of graph

Below x-axis

Point on graph

( - 3, -0.8)

=

-0.8

R(O)

-::-

(2, 00)

(1, 2) 3

=

� 4

Above x-axis

( O,�)

---,

, x

3

2

R(%) � =

-

Below x-axis

e- -�) 2'

7

R(3)

=

0.4

Above x-axis (3, 0.4)

Plot the points from Table 8. You should now have Figure 27(a). STEP 6: Next, we determine the behavior of the graph near the asymptotes. • Since the x-axis is a horizontal asymptote and the graph lies below the x-axis for x < -2, we can sketch a portion of the graph by placing a small arrow to the far left and under the x-axis.

357

SECTION 5.3 The Graph of a Rational Function

Since the line x = -2 is a vertical asymptote and the graph lies below the x-axis for x < -2, we continue by placing an arrow well below the x-axis and approaching the line x = - 2 on the left. • Since the graph is above the x-axis for -2 < x < 1 and x = -2 is a ver tical asymptote, the graph will continue on the right of x = -2 at the top. Similar explanations account for the other arrows shown in Figure 27(b). STEP 7: Figure 27(c) shows the complete graph. •

Figure 27

X=-2

Y

x=2 I I I I I I I I

3

-3 (-3, -0.8)

(0, �)

•

(1,0) : •

(�2' _f) 7

X=-2

i\

(3,0.4) •

3

xY=

0 -.... -3 (-3,-0.8) •

I I I I I I I I I I I I I

I I I I I

(1,0) :

(0, �)

•

I

(�,-n :

(a)

(3,0.4) 3 XY=0

1,

.-

1-3, -0.8)

�

li I

(b)

�. lim

x=2

3

I I I

-3

Y

X= -2

it

3

V

-3

x=2

Y

.Jl 3

l

0

X Y=

-3 (c)

•

Exploration x-l Graph R(x) = � x-

-4

Result The analysis just completed in Example 1 helps us to set the viewing rectangle to obtain a com x-l -2-- in connected mode, and Figure 28(b) shows x -4 it in dot mode. Notice in Figure 28(a) that the g ra ph has vertical lines at x = -2 and x = 2. This is due plete g ra ph. Figure 28(a) shows the graph of R(x) =

to the fact that, when the g ra phing utility is in connected mode, it will connect the dots between con secutive pixels. We know that the graph of R does not cross the lines x = -2 and x = 2, since R is not defined at x = -2 or x = 2. So, when graphing rational functions, dot mode should be used to avoid extraneous vertical lines that are not part of the g raph. See Figure 28(b).

Figure 28

-4

l

---,\

I

4

4 '\

"--.,-,. 4

4 -4

-4

Dot mode

Connected mode

(b)

(a)

"'1'1

EXAM P L E 2

¥ 5 .-

Now Work

PRO B L E M

7

A n alyzi ng the G raph of a Rational Function

Analyze the graph of the rational function: Solution

STEP 1: The domain of R is { x i x

=1=

R(x)

x2

-

1

= --

x

o} . B ecause x cannot equal 0, there is no y-intercept. Now factor R to obtain (x + l ) ( x - 1) R(x) x STEP 2: R is in lowest terms.

358

CHAPTER 5

Polynomial a n d Rational Functions

Solving the equation R(x) = 0, we find the graph has two x-intercepts: - 1 and 1 . (x + l ) (x - 1) (x + 1)( - 1 - 1 ) � = 2(x + 1) Near -1: R(x) = x -1 (x + 1) (x - 1) (1 + l) (x - 1) � = 2 ( x - 1) Near 1 : R(x) = 1 x

Plot the point ( - 1, 0) and indicate a line with slope 2 there. Plot the point ( 1, 0) and indicate a line with slope 2 there. STEP 3: R is in lowest terms, so the graph of R has the line x = 0 (the y-axis) as a vertical asymptote. Graph x = 0 using a dashed line. STEP 4: The rational function R is improper, since the degree of the numerator, 2, is larger than the degree of the denominator, 1. To find any horizontal or oblique asymptotes, we use long division. x

-1 x i s a n oblique asymptote o f the graph.

The quotient is x , s o the line y = Graph y = x using a dashed line. To determine whether the graph of R intersects the asymptote y we solve the equation R(x) = X. R(x)

=

x2 - 1 -X

=

=

x,

x

x2 - 1 = x2 -1 = 0

Im possible

---

x2 - 1 . ' We conclud e th at t h e equatIOn = x h as no soi utlOn, so t he x graph of R does not intersect the line y = X. STEP 5: The zeros of the numerator are -1 and 1; the zero of the denominator is O. We use these values to divide the x-axis into four intervals: ( - 1, 0)

( - 00 , -1) Table 9

(0, 1)

( 1, 00 )

Now we construct Table 9.

.

-1

Interval

(-00, - 1)

Number chosen

-2

Value of R

R(-2)

Location of graph

Below x-axis

Point on graph

( -1, 0) -

=

( D 2 - ,-

3

--

2

.0

'x

(0, 1)

(1,00)

-

2

1

1

-

2

2

( -D = % (_ � i)

G) = - %

R

R

Above x-axis

Below x-axis

U _ i)

2'2

2' 2

R(2)

=%

Above x-axis

(D 2

,

Plot the points from Table 9. You should now have Figure 29(a) . STEP 6: • Since the graph of R is below the x-axis for x < - 1 and is above the x-axis for x > 1, and since the graph of R does not intersect the oblique asymptote y = x, the graph of R will approach the line y = x as shown in Figure 29(b). • Since the graph of R is above the x-axis for - 1 < x < 0, the graph of R will approach the vertical asymptote x 0 at the top to the left of x = O. =

SECTION 5.3 The Graph of a Figure 29

x= O

Rational Function

x= O

y

x=O

y 3 /

/ //

//

/ // e

e

y

y =x / //

y= x /

( 2 3) ,2 3

359

/ // x

/

/

//

/

/

( 2, 23)

e

x

3

-3

x

(12' _;1)2

-3

(a)

.[� ;.;I �

(b)

Seeing the Concept x2 - 1 Graph R(x) = -- and compare what x you see with Figu re 29(c). Could you have predicted from the g raph that y = x is an oblique asym ptote? Graph y = x and ZOOM-OUT. What do you observe?

EXA M P L E 3

(c)

Since the graph of R is below the x-axis for 0 < x < 1 , the graph of R will approach the vertical asymptote x = 0 at the bottom to the right of x = O. See Figure 29 (b ) . STEP 7: The complete graph is given in Figure 29 ( c) . •

•

Iii-

:;.>-

Now Work

PROBlEM 1 5

Analyzing the G raph of a Rational Function


R( x )

=

4 + 1 -- X2

X

STEP 1: R is completely factored. The domain of R is

y-intercept.

"*

O } . There is no

4

+ 1 = 0 has no real solutions, there are no x-intercepts. STEP 3: R is in lowest terms, so x = 0 (the y-axis) is a vertical asymptote of R. Graph the line x = 0 using dashes. STEP 4: The rational function R is improper. To find any horizontal or oblique asymptotes, we use long division.

STEP 2: R is in lowest terms. Since

X

{x i x

1 The quotient is x2 , so the graph has no horizontal or oblique asymptotes. However, the graph of R will approach the graph of y = x2 as x � -00 and as x � 00. The graph of R does not intersect y = x2 . Graph y = x2 using dashes. STEP 5: The numerator has no zeros, and the denominator has one zero at O. We divide the x-axis into the two intervals

( -00, 0) and construct Table 10.

( 0, 00 )

360

CHAPTER 5 Polynomial and Rational Functions o

Table 10

.

. )(

Interval

(-00,0)

(0,00)

Number chosen

-1

1

Value of R

R(-l)

Location of graph

Above x-axis

Above x-axis

Point on graph

(-1, 2)

( 1 , 2)

R( 1 )

2

=

=

2

Plot the points ( - 1 ,2 ) and ( 1 , 2 ) . STEP 6: • Since the graph of R is above the x-axis and does not intersect y = x2, we place arrows above y = x2 as shown in Figure30(a). • Also,since the graph of R is above the x-axis,it will approach the verti cal asymptote x = 0 at the top to the left of x = 0 and at the top to the right of x = O. See Figure30(a) . STEP 7: Figure30(b) shows the complete graph. Figure 30 x=O

x=O

Y

Y

6

\

Graph R(x)

=

\

\ \

I �I Seeing the Concept

I\lm

t

X4 + 1

-2

- and compare what

x you see with Figure 30(b). Use MINIMUM to find the two turning points. Enter Y2 = x2 and ZOOM-OUT. What do you see?

\

\

,

.

\

,

-3

/

/

.

(1, 2)

(-1,2) \

,

6

2 Y= x

,

/

/

/

/

1/

I I

I

/ / / /

/ /

,

(-1,2)

/

x

3

\

\

EXAM P L E 4

Now Work

"

3

x

(b)

P R O BLE M 1

•

3

A n alyzi ng the Graph of a Ration al Function


,

-3

(a)

""= -

2 Y= x

I I

R(x )

=

3x2 -3x x2 + X - 12

STEP 1: We factor R to get

R(x)

=

3x(x - 1 ) (x + 4 ) ( x - 3)

The domain of R is { x i x -=I- -4, x -=I- 3}. The y-intercept is R(O) = O. Plot the point (0,0). STEP 2: R is in lowest terms. Since the real solutions of the equation3x( x - I ) = 0 are x = 0 and x = 1, the graph has two x-intercepts, 0 and 1. We determine the behavior of the graph of R near each x-intercept. Near 0:

R( x )

=

3x( x - 1 ) (x + 4 ) ( x - 3)

-,----'� -,-----'

Near 1:

R(x)

=

3x(x - 1 ) ( x + 4 ) ( x - 3)

�

3x(0 - 1 ) (0 + 4) (0 - 3)

=

1 -x 4

3( 1 ) (x - 1 ) (1 + 4) ( 1 - 3)

=

-

3 (x 10

1)

SECTION 5.3


36 1

Plot the point (0, 0) and show a line with slope � there. 4 3 Plot the point (1, 0) and show a line with slope there. 10 STEP 3: R is in lowest terms. Since the real solutions of the equation (x + 4 ) ( x - 3) = 0 are x = -4 and x = 3, the graph of R has two verti cal asymptotes, the lines x = -4 and x 3. Plot these lines using dashes. STEP 4: Since the degree of the numerator equals the degree of the denominator, the graph has a horizontal asymptote. To find it, we either use long division or form the quotient of the leading coefficient of the numerator, 3, and the leading coefficient of the denominator, 1. The graph of R has the horizon tal asymptote y = 3. To find out whether the graph of R intersects the asymptote, we solve the equation R(x) = 3. =

3x2 - 3x ==3 x2 + x - 12 3x2 - 3x = 3x2 + 3x - 36 -6x = -36 x 6

R(x)

=

The graph intersects the line y = 3 only at x 6, and (6, 3) is a point on the graph of R. Plot the point (6, 3 ) and the line y = 3 using dashes. STEP 5: The zeros of the numerator, 0 and 1 , and the zeros of the denominator, -4 and 3, divide the x-axis into five intervals: ( -00, -4) (-4, 0 ) (0, 1 ) ( 1 , 3 ) (3, (0 ) =

Construct Table 1 1 .

Table 11

-4

.0

.

.3

Interval

(-00, -4)

( -4, 0)

Number chosen

-5

-2

Value of R

R(-5) = 1 1 .25

R(-2) = -1.8

R

Location of graph

Above x-axis

Below x-axis

Above x-axis

Point on graph

( - 5, 1 1 .25)

(-2, - 1 .8)

(0, 1 ) 2

G)

=

1 15

G �) '1

• x

( 1 , 3)

(3, 00)

2

4

R(2) = - 1

R(4) = 4.5

Below x-axis

Above x-axis

(2, - 1 )

(4, 4.5)

Plot the points from Table 1 1 . Figure 31(a) shows the graph we have so far. STEP 6: • Since the graph of R is above the x-axis for x < -4 and only crosses the line y = 3 at (6, 3 ) , as x approaches -00 the graph of R will approach the horizontal asymptote y = 3 from above. • The graph of R will approach the vertical asymptote x = -4 at the top to the left of x = -4 and at the bottom to the right of x -4. • The graph of R will approach the vertical asymptote x = 3 at the bot tom to the left of x 3 and at the top to the right of x 3. • We do not know whether the graph of R crosses or touches the line y 3 at (6, 3 ) . To see whether the graph, in fact, crosses or touches the line y = 3, we plot an additional point to the right of (6, 3 ) . We use x = 7 =

=

=

=

to find R(7)

=

�� < 3. The graph crosses

y =

3 at x

=

6. Because (6, 3)

is the only point where the graph of R intersects the asymptote y = 3, the graph must approach the line y = 3 from below as x � 00. See Figure 31(b). STEP 7: The complete graph is shown in Figure 3 1 (c).

362

CHAPTER 5


Figure 31

x= -4 (-5,11.25)-

I I I I

10

I

-----...1 - - (0,6)" -5

I

_

1(-2,-1.8) I I I I I

x=-4 t (-5,11.25)-

x=3

Y

1.2'15.1..) I

-( --t------ -I \ I

(4,4.5) _ (6,3)

-(2 -1) : (1,0) I

I

-----

(a)

'i

I I

:It

2'15 I

.1..)1 (4,4.5) _ (6,3)

( 6)

\

_

I

if

.

-(2,-1) I

(1,0)

I

Ii

(7,�) 22

5

Y= 3

x

+

-10 (b)

ii� '

x= 3

Y 10

"

:

-

---------:: - - (0, 0) -5

10

1(-2,-1.8)

x = -4

(-5 11 25)

--

I

-5

I I I

-10

x= 3

___ ...1 6:- " -I-(1 -t--------

Y=3

x

5

Y

I

11 45) ( 2' 15):4 :

---1-- - - -- (6,3) - Y=3 -

,

I

(1,0)

(7�)

I 5 ,-1) ' 22 :

x

-10 (c)

II

Exploration

= -,2:-----3 x 2 - 3x

x +x - 1 2

Result Figure 32 shows the graph in connected mode, and Figure 33(a) shows it in dot mode. Neither graph displays clearly the behavior between the two x-intercepts, 0 and 1. Nor do they clearly display the fact that the graph crosses the horizonta l asymptote at (6, 3).To see these parts better, we graph R for -1 oS x oS 2 and - 1 oS Y oS 0.5 [Figure 33(b)] and for 4 oS X oS 60 and 2.5 oS Y oS 3.5. [Figure 34(b)].

Figure 32

Figure 33

10

-----/

�

-10

Graph R(x)

•

/. I

l

-10

\

�

10

:

'

---""",/

10

-10

0.5

.

-1 10

.'"

\\

-1

-10

Connected mode

�

/

/,..

Dot mode

---

10

/

-10

(b)

(a)

Figure 34

3.5

'

.

,

�

10

-10 Dot mode

(a)

2

-

4. 2.5

.

.

.

(b)

.

y= 3

60

SECTION 5.3 The Graph of a Rational Function

363

The new g raphs reflect the behavior produced by the analysis. Furthermore, we observe two turning points, one between 0 and 1 and the other to the right of 4. Rounded to two decimal places, these turn ing points a re (0.52,0.07) and ( 1 1 .48,2.75).

E XA M P L E 5

A n alyzi ng the G raph of a Ratio nal Function with a Hole

R( x )

Analyze the graph of the rational function: Soluti o n

2x2 - 5x + 2 x2 - 4

=

STEP 1: We factor R and obtain

R(x) The domain of R is { x i x

( �) .

=

=1=

( 2x - l ) (x - 2 ) ( x + 2 ) (x - 2 ) -2, x

=1=

2 } . The y-intercept i s R(O)

=

-�.

-

Plot the point 0, STEP 2: In lowest terms,

R(x) =

2x - 1 x + 2

x

=1=

-2

1

.

The graph has one x-mtercept: "2' Near

Plot the point

1 i

(�, ) °

R(x) =

2x - 1 x + 2

�

2x - 1 2 = - (2x - 1 ) 5 1 - + 2 2

showing a line with slope

�.

STEP 3: Look at R in lowest terms. The graph has one vertical asymptote, x = -2, since x + 2 is the only factor of the denominator of R(x) in lowest terms.

Remember though, the rational function is undefined at both x 2 and x = -2. Graph the line x = -2 using dashes. STEP 4: Since the degree of the numerator equals the degree of the denominator, the graph has a horizontal asymptote. To find it, we either use long division or form the quotient of the leading coefficient of the numerator, 2, and the leading coefficient of the denominator, 1 . The graph of R has the horizon tal asymptote y 2. Graph the line y = 2 using dashes. To find out whether the graph of R intersects the horizontal asymptote y = 2, we solve the equation R ( x ) = 2. =

=

R(x)

=

2x - 1 = 2 x + 2 2x - 1 = 2 ( x + 2 ) 2x - 1 = 2x + 4 -1 = 4

Im possible

The graph does not intersect the line y = 2. STEP 5: Look at the given expression for R. The zeros of the numerator and denominator, -2, .!, and 2, divide the x-axis into four intervals: 2

( -00, -2) Construct Table 12.

(2, (0)

364

CHAPTER 5 Polynomial and Rational Functions

.

Table 12

-2

.

( l)

(-00,-2)

Number chosen

-3

Value of R

R(-3)

Location of graph

Above x-axis

Below x-axis

Point on graph

(-3,7)

(-1, -3)

-1

R(-1)

=

G)

• x

2 '

-2,

Interval

=7

.2

1/2

R(l)

-3

(2,00) 3 1

=-

R(3)

3

=1

Above x-axis

Above x-axis

(1'�)

(3,1)

Plot these points. STEP 6: • From Table 12 we know the graph of R is above the x-axis for x < -2. From Step 4 we know the graph of R does not intersect the asymptote y = 2. Therefore, the graph of R will approach y = 2 from above as x � -00 and will approach the vertical asymptote x = -2 at the top from the left. • Since the graph of R is below the x-axis for -2 < x < �, the graph of R 2 will approach x - 2 at the bottom from the right. =

•

Finally, since the graph of R is above the x-axis for x >

�

and does not

intersect the horizontal asymptote y = 2, the graph of R will approach 2 from below as x � 00. See Figure35 ( a) . STEP 7: See Figure35 ( b ) for the complete graph. Since R is not defined at 2, there y =

( %).

is a hole at the point 2, Figure 35

(-3,7).

/1II x=-2 I I

I I I I I

y 8 6

:

I

4

I

I I I I

x=-2

Y 8

6

4

(11) I 1 2 ----------- i io-:::: l)- 2 \---- - (3.�- y 2 -----------,----I (0, -�) �, -4 -3 -2 -1 -4 -3 \1 2 3 x -2 0,0) /

:

' 3

=

"

2

Hole at

(1 1) (2, �)

3

x

1(-1,\-3)-4

(a)

�

Exploration

Graph R ( x )

=

2X2

( %}

the hole at 2,

-

2

x

5x + 2 -

4

.

(b)

•

NOTE The coordinates of the hole were obta ined by evaluating R in lowest terms at 2. R in lowest Do you see

TRACE along the graph.

Did you obtain an ERROR at x = 2 ? Are you convinced that an algebraic analysis of a rational function is required in order to accurately interpret the graph obtained with a g ra phing utility?

terms is

--

2x - 1 , x+2

which, at

x = 2,

As Example 5 shows,the

is

2(2) - 1 3 = -. 2+2 4

real zeros of the denominator of a rational function

give rise to either vertical asymptotes or holes on the graph. m>:: -==:a

Now Work

•

PROBLEM

33

We now discuss the problem of finding a rational function from its graph.

SECTION 5.3 EXAM P L E 6


365

Constructing a Rational Function from Its G raph

Find a rational function that might have the graph shown in Figure 36. Figure 36

x= -5 I 1 1

y 10

x= 2 I I I I I 1 1 1

:

5 1 --------1---1

-15

T

---------

-10

y= 2

15 x

10

-5

-10

The numerator of a rational function R( x)

Solution

=

��;?

in lowest terms determines the

x-intercepts of its graph. The graph shown in Figure 36 has x-intercepts -2 (even multiplicity; graph touches the x-axis) and 5 (odd multiplicity; graph crosses the x-axis). So one possibility for the numerator is p(x) = (x + 2)2(x - 5 ) . The denominator o f a rational function i n lowest terms determines the vertical asymptotes of its graph. The vertical asymptotes of the graph are x = -5 and x = 2. Since R( x) approaches 00 to the left of x = -5 and R(x) approaches - 00 to the right of x = -5, then x + 5 is a factor of odd multiplicity in q(x). Also, because R(x) approaches -00 on both sides of x = 2, then x - 2 is a factor of even mul tiplicity in q(x ) . A possibility for the denominator is q(x) = (x + 5 ) (x - 2 f So far w e have R(x)

-15

(x

+

2)2(x - 5 )

(x + 5 ) (x - 2 ) 2

'

The horizontal asymptote o f the graph given i n Figure 36 i s y = 2, s o w e know that the degree of the numerator must equal the degree of the denominator and the

Figure 37

J

=

quotient of leading coefficients must be

5

I(

I

-5

,------

R( x ) -

10

I

- -

2 ( x + 2)2(x - 5 )

----'--

----'-'--

---'--

( x + 5 ) (x - 2 ) 2

•

R on a g ra p h i n g util ity. S i n ce Fig u re 37 looks s i m i l a r to Fig u re 36, we have fou n d a rational fu nction R for the g ra p h in Fig u re 36.

Check: Figure 37 shows the g ra p h of r.i!i�

2 EXAM P L E 7

i. This leads to

Now Work

PROBLEM

45

Solve Applied Problems Involving Rational Functions F ind ing the Least Cost of a Can

Reynolds M etal Company manufactures aluminum cans in the shape of a right circular cylinder with a capacity of 500 cubic centimeters

(� )

liter ' The top and

bottom of the can are made of a special aluminum alloy that costs 0.05¢ per square centimeter. The sides of the can are made of material that costs 0.02¢ per square centimeter.

366

CHAPTER 5


\?!

(a) Express the cost C of material for the can as a function of the radius r of the can. (b) Find any vertical asymptotes. Discuss the cost of the can near any vertical asymptotes. C (r ). (c) Use a graphing utility to graph the function C (d) What value of r will result in the least cost? (e) What is this least cost? =

(a) Figure 38 illustrates the components of a can in the shape of a right circular cylinder. Notice that the material required to produce a cylindrical can of height h and radius r consists of a rectangle of area 27Trh and two circles, each of area 7Tr 2 . The total cost C (in cents) of manufacturing the can is therefore

Solution

Figure 38

rd r

.

.

Top Area

h h

=

Tir

2

C

La_ t--:eral Surface Area = 2Tirh

=

----'

L__

Area

=

=

Tir2

Cost of the top and bottom + Cost of the side 2(7Tr2) (0.05) + (2 'ITI'h) (0.02) "---v---'

�

"---v---'

Total area

Cost/unit

of top and

area

=

Cost/unit

area of

area

side

bottom

Bottom

�

Total

O.lO'TTr2 + O.04'TTrh

But we have the additional restriction that the height h and radius r must be chosen so that the volume V of the can is 500 cubic centimeters. Since V = 7Tr 2 h, we have 500 500 = 7TI, 2h so h = -

7Tr 2

Substituting this expression for h, the cost C , in cents, as a function of the radius r is

500 20 0.107Tr 3 + 20 0.047Tr-) = 0. 107Tr 2 + - = ----r r 7Tr (b) The only vertical asymptote is r = O. As the radius r of the can gets closer to 0, the cost C of the can gets higher. (c) See Figure 39 for the graph of C = C ( r ) . (d) Using the M INI M UM command, the cost is least for a radius of about 3.17 cen

Figure 39

C (r )

=

0.107Tr2

+

timeters. (c) The least cost is C (3.17 ) I'll

- Now Work

'

�

9.47¢.

•

PROBLEM S5


'Are You Prepared?' The answer is given at the end of these exercises. If you get a wrong answel; read the pages listed in red. 1.

The intercepts of the equation y =

- 1 -)--

x are x- - 4

____

. (pp. 1 65- 1 66)

Concepts and Vocabulary 2. If the numerator and the denominator of a rational function

have no common factors, the rational function is

____

3. True or False

The graph of a polynomial function some times has a hole.

4. True or False

5. True or False

The graph of a rational function sometimes in tersects a vertical asymptote.

6. True or False

The graph of a rational function sometimes

has a hole.

The graph of a rational function never inter sects a horizontal asymptote.

Skill Building In Problems 7. R(x)

=

7-44,

follow Steps

x + 1 x( A.v + 4)

1

through

7

on page 355 to analyze the graph of each function. 8.

R(x) =

x ( x - 1)( x + 2)

9.

R(x) =

3x + 3 2" _, + 4

SECTION 5.3 2x + 4 x - I

10. R(x)

=

13. P( x)

=

16. G(x)

=

-

19. G( x)

=

- -

22. R (x)

=

--

25. F(x )

=

--

28. R ( x )

-

31. R(x )

_

=

4

X + x2 + 1 ----:=--" \: 2 - 1

=

40. f(x)

=

43. f (x)

=

45.

=

20. G( x)

-4

(x + 1 ) ( x2 - 9)

x2 - 3x x + 2

---

4-

2

3 -

- -

X

4

12. R ( x )

4

X - 1 -? x- - 4

,

. 15. H ( x )

23. H (x)

=

=

3x ? x- - 1

- -

6 r- x - 6

=

18. R ( x )

=

21. R ( x )

=

x3? - 1 r- 9

- -

X - 12 x2 - 4

x2 +

3 ? (x - 1 ) ( r - 4)

x2 - 1

4 - X

- 16

x2 + X - 12 x - 4

=

x2 + 3x + 2 x-I

27. R ( x )

=

29. F(x)

=

x2 + X - 12 x + 2

30. G ( x )

=

---

32. R ( x )

=

(x - 1 ) (x + 2 ) ( x - 3 ) ? x(x - 4)-

33. R ( x )

=

x2 + X - 12 ? x- - x - 6

x;: :-2 + 3 x l0 ? r + 8x + 15

35. R(x )

=

6x2 - 7x - 3 2r - 7x + 6

36 . R ( x ) -

8x2 + 26x + 15 2x2 - x - 1 5

x2 + 5x + 6 x +j

3 8. R( x

39. f ( x )

=

1 x +x

9 2x + x

41. f ( x )

42. f(x)

=

x2 - X - 1 2 x + ) _

x(x - 1)2 ,

(x + 3 ) "

,.,

x +

1 x

3"

45-48,

)

=

=

367

-?:---

=

26. F ( x)

_

-=? ------

x2 + x

X

+

- 30 6

1 ? r +x

44. f(x)

)[-': " l

In Problems

14. Q(x)

�v3 + 1

-

37. R(x)

=

"

x- + 2x x X2 - 4

_ _ _ _ _ _ _ _

34. R ( x ) -

11. R ( x )

The Graph of a Rationa[ Function

=

2x +

x2 - x - 12 x + 1

2x2 +

16

X

9 x

3"

: '�1

find a rational function that might have the given graph. (More than one answer might be possible.)

'�-

46.

---�---- ---f--- y=1 -3 3 x

x 3

x= -3

48.

47.

I

Y

10 8

6

-4

-3

4

-2 x=2

-15

-10

-5

x= 4 I

[ [ [ [ [ [

�-------------y=3 [ [

[

-2 [

[5 [ [

[ -8

[

[

[ -4 [ [ -6

[ [ [ [

1

0

15

20 x

368

CHAPTER 5



tf0, lLYx

The concentration C of a certain drug in a patient's bloodstream I hours after injection is given by

49. Drug Concentration

C( I )

=

212

1

+ 1

(a) Find the horizontal asymptote of C ( I ) . What happens to the concentration of the drug as I increases? c, (b) Using your graphing utility, graph C ( t ) . , (c) Determine the time at which the concentration is highest. 50. Drug Concentration The concentration C of a certain drug in a patient's bloodstream t minutes after injection is given by C( I )

fil

=

SOt 25

+ 12

(a) Find the horizontal asymptote of C ( I ) . What happens to the concentration of the drug as t increases? (b) Using our graphing utility, graph C ( t ) . . . � (c) Determme the time at which the concentration IS highest.

A rectangular area adjacent to a river is to be fenced in; no fence is needed on the river side. The en closed area is to be 1000 square feet. Fencing for the side par allel to the river is $5 per linear foot, and fencing for the other two sides is $8 per linear foot; the four corner posts are $25 apiece. Let x be the length of one of the sides perpendicular to the river. (a) Write a function C(x ) that describes the cost of the project. (b) What is the domain of C? 6tl (c) Use a graphing utility to graph C(x) . (d) Find the dimensions of the cheapest enclosure. Source: www.uncwil. edulcourseslmalhlllhbIPandRirationa/1 rational.hlml 52. Doppler Effect The Doppler effect (named after Christian Doppler) is the change in the pitch (frequency) of the sound from a source ( s ) as heard by an observer ( 0 ) when one or both are in motion. If we assume both the source and the ob server are moving in the same direction, the relationship is

x

(a) Express the surface area S of the box as a function of x. 'J; (11) Using a graphing utility, graph the function found in part (a). (c) What is the minimum amount of cardboard that can be used to construct the box? (d) What are the dimensions of the box that minimize the surface area? (e) Why might UPS be interested in designing a box that minimizes the surface area? United Parcel Service has con tracted you to design an open box with a square base that has a volume of 5000 cubic inches. See the illustration.

54. Minimizing Surface Area

r' lLYx

51. Minimum Cost

f' f' fa

where

v va

Vs

= = = = =

=

fa

C = ::)

perceived pitch by the observer actual pitch of the source speed of sound in air (assume 772.4 mph) speed of the observer speed of the source

Suppose you are traveling down the road at 45 mph and you hear an ambulance (with siren) coming toward you from the rear. The actual pitch of the siren is 600 hertz (Hz). (a) Write a function f'(v\) that describes this scenario. (b) If f' 620 Hz, find the speed of the ambulance. ;'1 (c) Use a graphing utility to graph the function. (d) Verify your answer from part (b). Source: www. kettering. edul-drusselll 53. Minimizing Surface Area United Parcel Service has con tracted you to design a closed box with a square base that has a volume of 10,000 cubic inches. See the illustration. =

x

(a) Express the surface area S of the box as a function of x. Using a graphing utility, graph the function found in part (a). (c) What is the minimum amount of cardboard that can be used to construct the box? (d) What are the dimensions of the box that minimize the surface area? (e) Why might UPS be interested in designing a box that minimizes the surface area?

lMl (b)

A can in the shape of a right circular cylinder is required to have a volume of 500 cubic centimeters. The top and bottom are made of material that costs 6¢ per square centimeter, while the sides are made of material that costs 4¢ per square centimeter. (a) Express the total cost C of the material as a function of the radius r of the cylinder. ( Refer to Figure 38.) C ( r ) . For what value of r is the cost C a i ,;? (b) Graph C " minimum?

55. Cost of a Can

=

A steel drum in the shape of a right circular cylinder is required to have a vol ume of 100 cubic feet.

56. Material Needed to Make a Drum

SECTION 5.4

(a) Express the amount A of material required to make the drum as a function of the radius r of the cylinder. (b) How much material is required if the drum's radius is 3 feet? (c) How much material is required if the drum's radius is 4 feet?

Polynomial and Rational Inequalities

369

(d) How much material is required if the drum's radius is 5 feet? 'g� (e) Graph A = A(r). For what value of r is A smallest?

Discussion and Writing 57. Graph each of the following functions:

x2-1 x-I X4 1 y=- x-I y=

x=

-

y= y=

60. Create a rational function that has the following characteris

tics: crosses the x-axis at 2; touches the x-axis at -1; one ver tical asymptote at x = -5 and another at x = 6; and one horizontal asymptote, y= 3. Compare your function to a fellow classmate's. How do they differ? What are their similarities?

x3- 1

x -I x5 - 1

-- x-I

1 a vertical asymptote? Why not? What is happening x" - 1 . for x I ? What do you conjecture about y= , x-I n 2: 1 an integer, for x = I ? Is

---

=

58. Graph each of the following functions: y=

- x-I x2

y=

- x-I X4

x6

y=-

x-I

y=

xS -

61. Create a rational function that has the following characteris

tics: crosses the x-axis at 3; touches the x-axis at - 2; one ver tical asymptote, x = 1; and one horizontal asymptote, y= 2. Give your rational function to a fellow classmate and ask for a written critique of your rational function. 62. Create a rational function with the following characteristics:

x-I

What similarities do you see? What differences? 59. Write a few paragraphs that provide a general strategy for

graphing a rational function. Be sure to mention the follow ing: proper, improper, intercepts, and asymptotes.

three real zeros, one of multiplicity 2;y-intercept 1; vertical as ymptotes x = - 2 and x = 3; oblique asymptote y = 2x + l. Is this rational function unique? Compare your function with those of other students. What will be the same as everyone else's? Add some more characteristics, such as symmetry or naming the real zeros. How does this modify the rational function?

'Are You Prepared?' Answers 1.

( �), 0,

(1, 0)

5.4 Polynomial and Rational Inequalities PREPARING FOR THIS SECTION •


Solving Inequalities (Section 1 .5, pp. 124--131) Now Work

the 'Are You Prepared?' problem o n page 373.

OBJECTIVES

1 Solve Polynomial Inequal ities (p. 370)

2 Solve Rational I neq u a l ities (p. 372)

In this section we solve inequalities that involve polynomials of degree 3 and higher, as well as some that involve rational expressions. To solve such inequalities, we use the information obtained in the previous three sections about the graph of polyno mial and rational functions. The general idea follows: Suppose that the polynomial or rational inequality is in one of the forms

f(x) < 0

f(x) > 0

f(x)

:5

0

f(x)

2:

0

Locate the zeros of f if f is a polynomial function, and locate the zeros of the nu merator and the denominator if f is a rational function. If we use these zeros to di vide the real number line into intervals, we know that on each interval the graph of f is either above the x-axis [f(x) > 0 ] or below the x-axis [ f(x) < 0]. In other words, we have found the solution of the inequality.

370

CHAPTER 5


The following steps provide more detail.

Steps for Solving Polynomial and Rational Inequalities STEP 1: Write the inequality so that a polynomial or rational expression f is on

the left side and zero is on the right side in one of the following forms:

f(x) > 0

2:

f(x)

f(x) < 0

0

$

f(x)

0

For rational expressions, be sure that the left side is written as a single quotient and find its domain. STEP 2: Determine the real numbers at which the expression f on the left side equals zero and, if the expression is rational, the real numbers at which the expression f on the left side is undefined. STEP 3: Use the numbers found in Step 2 to separate the real number line into intervals. STEP 4: Select a number in each interval and evaluate f at the number. (a) If the value of f is positive, then f (x) > 0 for all numbers x in the interval. (b) If the value of f is negative, then f (x) < 0 for all numbers x in the interval. If the inequality is not strict, include the solutions of f (x) 0 in the solution set. =

1 EXAM P L E 1

Solve Polynomial Ineq ualities Solving a Polynomial Inequality

Solve the inequality x4 Solution

$

4x2 , and graph the solution set .

STEP 1: Rearrange the inequality s o that 0 is o n the right side.

x4 $ 4x2 X4 - 4x2 $ 0

Subtract 41 from both sides of the inequality.

This inequality is equivalent to the one that we wish to solve. STEP 2: Find the real zeros of

x4 - 4x2

x

=

=

f(x)

o.

X4 - 4x2 x2 ( x2 - 4) x2 (x + 2 ) ( x - 2 ) 0 or x -2 or x =

=

=

=

=

=

x4 - 4x2 by solving the equation 0 0

Factor. Factor.

0 2

Use the Zero-Product Property and solve.

STEP 3: We use these zeros of f to separate the real number line into four intervals:

( -00, -2)

( -2, 0)

(0, 2 )

(2,

(0

STEP 4: Select a number in each interval and evaluate f(x) mine if f(x) is positive or negative. See Table 13. • -2

Table 1 3 Interval

( -00, -2)

Number chosen

-3

Value of f

f(-3)

Conclusion

Positive

=

45

• 0

( -2,0) -1

f(-1)

=

Negative

• 2 (0, 2) 1

-3

f(1)

=

x (2, 00) 3

=

)

-3

Negative

f(3) = 45 Positive

X4 - 4x2 to deter

SECTION 5.4

I [

-2

1

2

0

4

6

8

•

x

•

� !==-

E XA M P L E 2

371

B ased on Table 13, we know that f(x) < 0 for all x in the intervals ( -2, 0 ) U (0, 2 ) , that is, for all x such that -2 < x < 0 or 0 < x < 2. How ever, because the original inequality is not strict, numbers x that satisfy the equation f(x) = x4 - 4x2 = 0 are also solutions of the inequality X4 :::; 4x2 . Thus, we include -2, 0, and 2. The solution set of the given inequality is { x l -2 :::; x :::; 2} or, using interval notation, [ -2, 2 J . Figure 40 shows the graph of the solution set.

Figure 40 -4

Polynomial and Rational Inequal ities

Now Work

5 AND 1 3

PROBlEMS

Solving a Po lynomial I nequality

Solve the inequality X4 > x, and graph the solution set. Solution

STEP 1: Rearrange the inequality so that 0 is on the right side.

X4 > X 4 x - X > 0

S ubtract x from both sides of the inequa lity.

This inequality is equivalent to the one that we wish to solve. STEP 2: Find the real zeros of f(x) = X4 - x by solving X4 - x = O. X4 - X = 0 x(x3 - 1 ) = 0

Factor out x.

x( x - I ) (x2 + X + 1 ) = 0

Factor the difference of two cu bes.

x = 0

or x - I = 0 or x2 + x + 1 = 0

x = 0

or

x = 1

Set each factor equal to zero a n d solve. The equation x2 + x + 1 0 ha s no real solutions. (Do you see why?) =

STEP 3: We use these zeros 0 and 1 to separate the real number line into three

intervals: ( 1 , 00 )

(0, 1 )

( - 00 , 0 )

STEP 4: We choose a number in each interval and evaluate f(x) = x4 - x to deter

mine if f(x) is positive or negative. See Table 14. Table 1 4

.0 (- 00,0)

(0, 1 )

(1,00)

Number chosen

-1

-

2

Value of f

f(- l )

Conclusion

Positive

Interval

Figure 41 I

I

-2 -1

) ( 0

1

• x

1

2

=

f

2

G)

=

Negative

-

7 16

f(2)

=

14

Positive

B ased on Table 14, we know that f(x) > 0 for all x in ( - 00 , 0) U ( 1 , 00 ) , that is, for all numbers x for which x < 0 or x > 1 . Because the orig inal inequality is strict, the solution set of the given inequality is {x i x < 0 or x > I j or, using interval notation, ( - 00 , 0) U ( 1 , 00 ) . Figure 41 shows the graph of the solution set. •

2

� - Now Work

.

PROBLEM

17

372

CHAPTER 5


2

Solve Rational Inequalities Solving a Rational I nequal ity

E XA M P L E 3

(x + 3 ) ( 2 - x) > 0, and graph the solution set. (x - 1 ) 2

Solve the inequality Solution

STEP 1: The domain of the variable x i s { x I x

form with 0 on the right side.

=f.

I } . The inequality is already in a

(x + 3 ) ( 2 - x)

2 . The real zeros of the numerator of f are -3 (x - 1 ) and 2 ; the real zero of the denominator is l . STEP 3: We use the zeros -3, 1, and 2 , found i n Step 2, to separate the real number line into four intervals:

STEP 2 : Let f(x)

=

(

-

00

,

-3)

( -3, 1 )

(2,

(1, 2)

(0

STEP 4: Select a number in each interval and evaluate f(x)

)

=

(x + 3 ) (2 - x) (x - I f

determine if f(x) is positive or negative. See Table 15. Table

Figure 42 (

-4 -3

2 -1

-

.

lS

@)

a 1 2 3

Interval

(-00, -3)

( -3, 1 )

( 1 , 2)

(2, (0)

Number chosen

-4

0

-

3 2

3

Value of f

f( -4)

Conclusion

Negative

6

- -

flO)

25

=

f(V =

6

Positive

Positive

9

f(3)

=

)(

3

- -

2

Negative

=

•

Now Work

PROBLEM 2 1

Solving a Rational Inequality

Solve the inequality Solution

=

.

Based on Table 15, we know that f(x) > O for all x in ( -3, 1 ) U ( 1 , 2), that is, for all x such that -3 < x < 1 or 1 < x < 2. Because the original inequality is strict, the solution set of the given inequality is { x l -3 < x < 2, x =f. I } or, using interval notation, ( -3, 1 ) U ( 1 , 2 ) . Figure 42 shows the graph of the solution set. Notice the hole at x 1 to indicate that 1 is to be excluded. m=

E XA M P L E 4

.2

-3

to

4x + 5 x + 2

:2:

3, and graph the solution set.

STEP 1: The domain of the variable x is { x i x

=f.

- 2 } . Rearrange the inequality so that 0 is on the right side. Then express the left side as a single quotient.

--( --) -------

4x + 5 - 3 :2: 0 x + 2

4x + 5 X + 2 --- - 3 x + 2 x + 2

:2: 0

4x + 5 - 3x - 6 :2: 0 x + 2 x - I -- :2: 0 x + 2

Su btract 3 from both sides of the inequality. Multiply -3 by

--. x + 2 x + 2

Write as a single quotient. Combine like terms.

The domain of the variable is { x I x

=f.

-2}.

SECTION 5.4

Polynomial and Rational Inequal ities

373

x - I --. The zero of the numerator of f is 1 , and the zero of the x + 2 denominator is -2.

STEP 2: Let f(x)

=

STEP 3: We use the zeros found in Step 2, -2 and

into three intervals: ( - 00 , -2)

1,

( -2, 1 )

to separate the real number line (1,

00

)

STEP 4: Select a number in each interval and evaluate f(x)

determine if f( x ) is positive or negative. See Table 16. Table

16

-2

• x

.

Interval

( -00,-2)

Number chosen

-3

Value of f

f(-3)

Conclusion

Positive

=

( 1 , (0 )

( - 2, 1 )

0

4

4x + 5 - 3 to x + 2

2

f(O) =

1

f(2)

- -

2

1

= 4"

Positive

Negative

B ased on Table 16, we know that f ( x ) > 0 for all x in ( - 00 , -2) U ( 1 , 00 ) , that is, for all x such that x < -2 or x > 1 . Because the original x - I inequality is not strict, numbers x that satisfy the equation f(x) = -- = 0 x + 2 x - I = 0 only if x are also solutions of the inequality. Since 1, we x + 2 conclude that the solution set is { x i x < - 2 or x 2: I } or, using interval notation, ( - 00 , -2) U [ 1 , 00 ) . =

Figure 43 I.. I I ) -4

-2

I

[

I

0

! ); ,

2

4

Figure 43 shows the graph of the solution set. = :::cD1l'l"i:

Now Work

•

PROBLEM 29


'Are You Prepared?' The answer is given at the end of these exercises. If you get a wrong answel; read the pages listed in red. 1. Solve the inequality: 3 - 4x > 5. Graph the solution set. ( pp. 124- 1 3 1 )

Concepts and Vocabulary 2.

True

or

False

The first step in solving the inequality x2 + 4x

2::

-4 is to factor the expression x2 + 4x.

Skill Building In Problems 3-40, solve each inequality. 3. (x - 5)2(x + 2) < 0 4. (x - 5 ) ( x + 2f > 0 7. x3 - 9x :::; 0

11. (x - 1 ) ( x2 +

X

+ 4)

2::

0

12. (x + 2) (x2

(x - 2)2 x2 - I

2::

-

X4

< 9x2

+ 1)

-

26 .

(x + 5)2 x2 - 4

2::

0

. 13. (x

2::

23.

0

- l ) (x - 2 ) ( x - 3) :::; 0

16. x3 + 2x2 - 3x > 0 19.

X 3 22. -- > 0 x + 1

0

X

15. x3 - 2x2 - 3x > 0 18.

25.

10. 3x3 < -15x2

9. 2x3 > -8x2

14. (x + l ) (x + 2 ) ( x + 3) :::; 0

x + 1 21. -- > 0 x - I

6. x3 + 8x2 < 0

5. x3 - 4x2 > 0

X4

20. x3 >

> 1

(x - 1 ) (x + 1 ) x 6 x

27. 6 x - ) < �

:::; 0

24.

1

(x - 3 ) (x + 2) x - I

12 28. x + - < 7 x

:::; 0

374

CHAPTER 5


x +2 � 1 x - 4 5 3 34. -- > -x - 3 x + 1

x +4 � 1 x - 2 2 1 33. -- < -x -2 3x - 9 29.

37.

x2( 3 + x ) ( x + 4) (x + 5 ) (x - 1 )

�

0

38.

x -4 1 2x + 4 1 3 36. -- > -x + 2 x + 1

3x - 5 �2 x + 2 2x + 5 x + 1 35. -- > -x + 1 x - I

30.

31.

x(x2 + l ) (x - 2) (x - l ) (x + 1 )

�

0

39.

(3 - x)3(2x + 1 ) 3

x - I

32.

--- �

(2 - x )3 (3x - 2)

40. -----:3:--- < 0

< 0

X + 1

Application and Extensions 41. For what positive numbers will the cube of a number exceed

four times its square? 42. For what positive numbers will the cube of a number be less than the number? 43. What is the domain of the function f(x) = � ? 44. What is the domain of the function f(x) = 45. What is the domain of the function f(x) = 46. What is the domain of the function f ( x ) =

V� - 3�?

) )

x x x x

+ +

2 ? 4 1 ? 4

In Problems 47-50, determine where the graph of f is below the graph of g by solving the inequality f ( x ) � g ( x ) . Graph f and g togethel: 47. f(x) = X4 - 1 48. f ( x) = X4 - 1 g(x)

=

_2x2 + 2

g(x) = x - I

49. f(x) = X4 - 4

50. f ( x ) = X4

g(x) = 3x2

g(x)

=2

- x2

Suppose that the daily cost C of manufac turing bicycles is given by C(x) = 80x + 5000. Then the 80x + 5000 -. . . average dally cost C IS given by C(x) = . How x many bicycles must be produced each day for the average cost to be no more than $100? 52. Average Cost See Problem 5 1 . Suppose that the govern ment imposes a $1000 per day tax on the bicycle manufac turer so that the daily cost C of manufacturing x bicycles is now given by C(x) = 80x + 6000. Now the average daily 80x + 6000 . How many bicycles cost C is given by C(x) = x must be produced each day for the average cost to be no more than $100? 53. Bungee "umping Originating on Pentecost Island in the Pa cific, the practice of a person j umping from a high place har nessed to a flexible attachment was introduced to Western culture in 1 979 by the Oxford University Dangerous Sport 51. Average Cost

Club. One important parameter to know before attempting a bungee j ump is the amount the cord will stretch at the bot tom of the fall. The stiffness of the cord is related to the amount of stretch by the equation K

where

=

2W(S + L) -'---:,--: "":" S2

-

W = weight of the jumper (pounds) K = cord's stiffness

(pounds per foot)

L = free length of the cord (feet)

S = stretch (feet) (a) A ISO-pound person plans to j ump off a ledge attached to a cord of length 42 feet. If the stiffness of the cord is no less than 16 pounds per foot, how much will the cord stretch? (b) If safety requirements will not permit the jumper to get any closer than 3 feet to the ground, what is the mini mum height required for the ledge in part (a)?

Source: American

Institute of Physics, Physics News Update, No. 150, November 5, 1993

According to Newton's Law of uni versal gravitation, the attractive force F between two bodies is given by m 1m F = G .2 2 , where m 1 , m2 are the masses of the two bodies

54. Gravitational Force

r

=

distance between the Two bodies

G = gravitational constant 6.6742 newtons meter2 kilogram- 2

X

10- 1 1

Suppose an object is traveling directly from Earth to the moon. The mass of the Earth is 5.9742 X 1024 kilograms, the mass of the moon is 7.349 X 1022 kilograms and the mean dis tance from Earth to the moon is 384,400 kilometers. For an object between Earth and the moon, how far from Earth is the force on the object due to the moon greater than the force on the object due to Earth? Source: www.solarviews. com;en. wikipedia. org

Discussion and Writing 55. Make up an inequality that has no solution. Make up one that

has exactly one solution. 56. The inequality X4 + 1 < -5 has no solution. Explain why.

. . x +4 57. A student attempted to solve the Inequality -- � 0 by x -3 multiplying both sides of the inequality by x - 3 to get

1} "

'Are You Prepared?, 1.

{i

x x < - 2"

-2

x + 4 � O. This led to a solution of {xix � -4J. Is the stu dent correct? Explain. 58. Write

a rational inequality whose solution set is {xl -3 < x � 5 J .

SECTION 5.5

The Real Zeros of a Polynomial Function

375

5.5 The Real Zeros of a Polynomial Function PREPARING FOR THIS SECTION •

• •


Evaluating Functions (Section 3 . 1 , pp. 212-214) Factoring Polynomials (Chapter R, Section R.5, pp. 49-55) Synthetic Division (Chapter R, Section R.6, pp. 57-60) Now Work

the 'Are You Prepared?' problems on page

OBJECTIVES

• •

Polynomial Division (Chapter R, Section R.4, pp. 44-47) Quadratic Formula (Section 1 .2, pp. 102-104)

386.

1 Use the Remainder and Factor Theorems (p. 375) 2

Use Descartes' Rule of Signs to Determine the N u mber of Positive and the N u m ber of Negative Rea l Zeros of a Polynomia l Fu nction (p. 378)

3 U se the Rational Zeros Theorem to List the Potential Rational Zeros of

a Polynomial Fu nction (p. 379)

4 Find the Real Zeros of a Polynom i a l Function (p. 38 1 ) 5 Solve Polynomial Equations (p. 382)

6 Use the Theorem for Bounds on Zeros (p. 383) 7

Use the I ntermed iate Va l u e Theorem (p. 384)

In this section, we discuss techniques that can be used to find the real zeros of a polynomial function. Recall that if r is a real zero of a polynomial function f then f(r) = 0, r is an x-intercept of the graph of f, and r is a solution of the equation f(x) = 0. For polynomial and rational functions, we have seen the importance of the real zeros for graphing. In most cases, however, the real zeros of a polynomial function are difficult to find using algebraic methods. No nice formulas like the quadratic formula are available to help us find zeros for polynomials of degree 3 or higher. Formulas do exist for solving any third- or fourth-degree polynomial equation, but they are some what complicated. No general formulas exist for polynomial equations of degree 5 or higher. Refer to the Historical Feature at the end of this section for more information. 1

Use the Remainder and Factor Theorems

When we divide one polynomial (the dividend) by another (the divisor), we obtain a quotient polynomial and a remainder, the remainder being either the zero poly nomial or a polynomial whose degree is less than the degree of the divisor. To check our work, we verify that ( Quotient) (Divisor) + Remainder

=

Dividend

This checking routine is the basis for a famous theorem called the division algo which we now state without proof.

rithm* for polynomials,

THEOREM

Division Algorithm for Polynomials

If f(x) and g(x) denote polynomial functions and if g(x) is not the zero poly nomial, there are unique polynomial functions q(x) and r(x) such that

r(x) f(x) - q (x) + or f(x) g(x) g(x)

-

-

i

=

q(x)g(x) + r(x) i

i

dividend quotient divisor

i

(1)

remainder

where r( x ) is either the zero polynomial or a polynomial of degree less than that of g(x).

--.J

,', A systematic process in which certain steps are repeated a fin i te number of times is called an algorithm.

For example, long division is an algorithm.

376

CHAPTER 5

Polynomial a n d Rational Functions

In equation (1), f(x) is the dividend, g(x) is the divisor, q(x) is the and r ex ) is the remainder. If the divisor g( x) is a first-degree polynomial of the form

quotient,

c a real number

g(x) = x - c,

then the remainder r ex) is either the zero polynomial or a polynomial of degree O. As a result, for such divisors, the remainder is some number, say R, and we may write

f(x)

=

(x - c) q (x) + R

(2)

This equation is an identity in x and is true for all real numbers x. Suppose that x = c. Then equation (2) becomes

f(c) = (c - c)q(c) + R f(c) = R Substitute f(c) for R in equation (2) to obtain

f(x) We have now proved the REMAINDER THEOREM

E XA M P L E 1

=

(x - c) q (x) + f(c)

Remainder Theorem.

Let f be a polynomial function. If f(x) is divided by x - c, then the remainder is f( c).

--1

U s i ng the Remainder Theorem

Find the remainder if f(x) = x3 - 4x2 (a) x - 3 Solutio n

( 3)

-

5 is divided by

(b) x + 2

(a) We could use long division or synthetic division, but it is easier to use the Re mainder Theorem, which says that the remainder is f(3).

f(3)

=

(3)3 - 4(3) 2 - 5

=

27 - 3 6 - 5 = - 14

The remainder is -14. (b) To find the remainder when f(x) is divided by x ate f( -2).

f(-2) = ( -2) 3 - 4( -2) 2 - 5

=

-8

-

+

2

=

16 - 5

x - ( -2), we evalu =

-29

The remainder is -29.

•

Compare the method used in Example l (a) with the method used in Example 1 of Chapter R, Section R.6. Which method do you prefer? Give reasons.

� COM MENT

A graphing utility provides another way to find the value of a function using the eVALUEate feature. Consult your manual for details. Then check the results of Exam ple 1. •

An important and useful consequence of the Remainder Theorem is the

Factor

Theorem.

FACTOR THEOREM

Let f be a polynomial function. Then x - c is a factor of f (x) if and only if f(c) = o.

--1

The Factor Theorem actually consists of two separate statements: 1. If f(c)

=

0, then x - c is a factor of f(x). 2 . If x - c i s a factor o f f(x), then f(c) = o .

SECTION 5.5

The Real Zeros o f a Polynomial Function

377

The proof requires two parts. Proof

1. Suppose that

2.

f(c) = O. Then, by equation (3), we have

f(x) = (x - c) q (x) for some polynomial q (x). That is, x - c is a factor o f f(x). Suppose that x - c is a factor of f(x). Then there is a polynomial function q such that

f(x) = (x - c) q (x) Replacing x by c, we find that

f(c) = (c - c)q(c) = O · q (c) = 0 •

This completes the proof.

One use of the Factor Theorem is to determine whether a polynomial has a par ticular factor. EXAM P L E 2

U s i n g the Factor Theorem

Use the Factor Theorem to determine whether the function

f (x) = 2x3 - x2 + 2x - 3 has the factor (a) x - I Soluti o n

(b) x + 3

The Factor Theorem states that if f(c) = 0 then x - c is a factor. (a) Because x - I is of the form x - c with c choose to use substitution.

f( l )

=

=

1, we find the value of f( l ) . We

2 ( 1 ) 3 - ( 1 )2 + 2 ( 1 ) - 3 = 2 - 1 + 2 - 3 = 0

By the Factor Theorem, x - I is a factor of f(x). (b) To test the factor x + 3, w e first need t o write i t in the form x - c. Since x + 3 = x - ( - 3 ) , we find the value of f( - 3 ) . We choose to use synthetic division. -3 )2 - 1 2 -6 21 2 - 7 23

-3 -69 -72

Because f( -3) = -72 * 0, we conclude from the Factor Theorem that x - ( -3) x + 3 is not a factor of f(x). =

..m:.: = =-

Now Work

•

PROBLEM

1 1

In Example 2(a), we found that x - I was a factor of f. To write f in factored form, we use long division or synthetic division. Using synthetic division, we find that 1 )2 - 1 2

2

The quotient is q (x) 2x2 write f in factored form as =

+

1

2 1 3

-3 3 0

X + 3 with a remainder of 0, as expected. We can

f (x) = 2 x3 - x2 + 2x - 3 = (x - 1 ) ( 2x2 +

X

+ 3)

378

CHAPTER 5


The next theorem concerns the number of real zeros that a polynomial function may have. In counting the zeros of a polynomial, we count each zero as many times as its multiplicity. THEOREM

N u m ber of Real Zeros A

polynomial function cannot have more real zeros than its degree.

-.J

Proof The proof is based on the Factor Theorem. If I' is a real zero of a polynomial function f , then fe r ) a and, hence, x - I' is a factor of f(x). Each real zero cor responds to a factor of degree Because f cannot have more first-degree factors • than its degree, the result follows.

1.

=

2

Use Descartes' Rule of Signs to Determine the Number of Positive and the Number of Negative Real Zeros of a Polynomial Function

Descartes' Rule of Signs provides information about the number and location of the real zeros of a polynomial function written in standard form (descending powers of x). It requires that we count the number of variations in the sign of the coeffi cients of f(x) and f( -x). For example, the following polynomial function has two variations in the signs of the coefficients.

f(x)

=

=

-3x7 + 4X4 + 3x2 - 2x - 1 -3x7 + Ox6 + Ox5 + 4X4 + Ox3 + 3x2 - 2x -

�

�

1

+ to -

- to +

Notice that we ignored the zero coefficients in Ox6, Ox5, and Ox3 in counting the number of variations in the sign of f(x). Replacing x by -x, we get f( -x )

=

=

- 3( -x ) 7 + 4( -X )4 + 3( -X )2 - 2( -x ) 3x7 + 4X4 + 3x2 + 2x - 1 �

-1

+ to -

which has one variation in sign. THEOREM

Descartes' Rule of S i gns

Let f denote a polynomial function written in standard form. The number of positive real zeros of f either equals the number of variations in the sign of the nonzero coefficients of f(x) or else equals that number less an even integer. The number of negative real zeros of f either equals the number of variations in the sign of the nonzero coefficients of f( -x) or else equals that number less an even integer. We shall not prove Descartes' Rule of Signs. Let's see how it is used. EXAM P L E 3

U s i n g the Number of Real Zeros Theorem and Descartes' Rule of Signs

Discuss the real zeros of f (x) Solution

=

3x6 - 4x4 + 3x3 + 2x2 - X - 3.

Because the polynomial is of degree 6, by the Number of Real Zeros Theorem there are at most six real zeros. Since there are three variations in the sign of the nonzero

SECTION 5.5

The Real Zeros of a Polynomial Function

379

coefficients of f ( x ) , by Descartes' Rule of Signs we expect either three or one pos itive real zeros. To continue, we look at f( -x ) . f ( -x )

=

3x6 - 4X4 - 3x3 + 2x2 + X - 3

There are three variations in sign, so we expect either three or one negative real zeros. Equivalently, we now know that the graph of f has either three or one pos itive x-intercepts and three or one negative x-intercepts.

•

1i::!l!:O: :::; ;Z: =--=

3

Now Work

21

PROBLEM

Use the Rational Zeros Theorem to List the Potential Rational Zeros of a Polynomial Function

The next result, called the Rational Zeros Theorem, provides information about the rational zeros of a polynomial with integer coefficients. THEOREM

Rational Zeros Theorem

Let f be a polynomial function of degree 1 or higher of the form an

=I- 0,

ao

=I- °

where each coefficient is an integer. If E , in lowest terms, is a rational zero of

q

f, then p must be a factor of ao , and q must be a factor of EXA M P L E 4

an '

.J

L isti ng Potential Rational Zeros

List the potential rational zeros of

f (x) Solution

± 1 , ±2, ±3, ± 6 ±1, ±2

q:

r

r r

r

r

r

2x3 + llx2 - 7 x - 6

Because f has integer coefficients, we may use the Rational Zeros Theorem. First, we list all the integers p that are factors of the constant term ao = -6 and all the integers q that are factors of the leading coefficient a3 = 2. p:

r

=

Factors of - 6

Factors of 2

Now we form all possible ratios E .

q

In Words

For the polynomial function f(x) = 2>1' + 11>-


f

Remember, if is a one-to-one function, it has an inverse function. We use the symbol r l to denote the inverse of f. Figure 1 1 illustrates this definition. Based on Figure 11 , two facts are now apparent about a one-to-one function and its inverse rl.

f

Domain of

1

,1

1 Domain of,-

Range of,-

f = Range of r1

Range of f = Domain of rI

f,

Look again at Figure 1 1 to visualize the relationship. If we start with x, apply and then applyrI, we get x back again. If we start with x, apply rI, and then apply f, we get the number x back again. To put it simply, what does, rl undoes, and vice versa. See the illustration that follows.

f

WARNING Be ca reful!

f-1

is a symbol

for the inverse function of f. The -1 used i n

f-1

is not a n exponent. That is,

f-1 does not mea n the reCiprocal of f; 1 f-\x) is not equal to • f(x)'

•

I

Inpu t x

I

I Input x I

Apply f

-1

Apply f

)

-1

Apply f

I f(x) I

) rl(x)

I

I

Apply f

) rl(f(x) )

I

)

=

xI

I f(f-l ( X) ) = x I

41 4

CHAPTER 6

Exponential and Logarithmic Functions

In other words,

= x where x is in the domain of f

rl(f(X»

f(f-l(x» = x where x is in the domain of f -1

Consider the one-to-one function f(x) = 2x, which multiplies the argument x by 2. The inverse function f -1 undoes whatever f does. So the inverse function of f

is rl(x) = %x, which divides the argument by 2. For example, f(3) = 2(3) = 6 Figure

and r I ( 6) = "2(6) = 3, so r1 undoes what f did. We can verify this by showing that

1

12

f

X

r\f"(x) = rl(2x) =

� f(X)=2X

j

% (2X) = x

and f(r1 (x))

=f

f-\x) = -x 2

r-\2x) = � ( 2x)= x

See Figure 12.

2

X

(� ) =j (� ) = X

x

f(x) = 2x

1

Verifyi ng I nverse Functions

EXAMPLE 5

(a) We verify that the inverse of g(x)

g-l(g(x))

g(g-I(X))

= x3 is g-l(x)

= g-1(x3) = # = x = g(\YX) = (\YX? = x

1 = r1(2x+ 3) = "2[(2x + 3)

.nrl(x»

=f

(

-

.!. . ( X-3» = 2 1:. (x - 3) 2 2

) [

]

\YX by showing that

for all x in the domain of g

for all x in the domain of g-I

(b) We verify that the inverse of f(x) = 2x that

r1(f (x»

=

+

3 is rl(x) = .!..(x-3) by showing 2

1

3J = "2(2x) +

3

for all x in the domain of f

=x

for all x in the domain of r1

= (x - 3)+ 3 = x

•

Verifying I nverse F unctions

EXAMPLE 6

_ is rl(x) = 1:.x

Verify that the inverse of f(x) = _1

x -I

r1(f(x» = x? For what values of x is f(r1(x» The domain of f is {xix

Solution

rl(f(x» f(r1(x»)

"'m

Figure 1 3

y =x

Y b

3

t

= rl

(

_ l _ x-I

(� )

a

b

x

I}

1. For what values of x is

= x?

and the domain of r1 is {xix

) = _ 11_ + 1 = x - I + 1 =

=f +1 =

x-I

1 1 -+ 1-1 x

1 =-=x 1 x

x

O}. Now

oj.

provided x

provided x

oj.

oj.

1

0 •

i - Now Work P R O B L E M 3 1

O bta i n the Gra p h of the Inverse Function fro m the Gra p h of t h e Function

a

oj.

+

Suppose that ( a, b) is a point on the graph of a one-to-one function f defined by y = f(x). Then b = f( a ) . This means that a = rl(b), so (b, a ) is a point on the graph of the inverse function ri. The relationship between the point ( a, b) on f and the point (b, a ) on rl is shown in Figure 13. The line segment containing ( a, b) and

SECTION 6.2

One-to-One Functions; Inverse Functions

41 5

(b, a ) is perpendicular to the line y = x and is bisected by the line y = x. (Do you see why?) It follows that the point (b, a ) on r1 is the reflection about the line y = x of the point ( a, b) on f. THEOREM

The graph of a function j and the graph of its inverse j-l are symmetric with respect to the line y = x.

-.J

Figure 14 illustrates this result. Notice that, once the graph of j is known, the graph of r1 may be obtained by reflecting the graph of j about the line y = x. Figure

14

Y = f(x)

Y

Seeing the Concept

= X, Y2 =

3 x , and Y3 = Vx on a square screen with 3 � x � 3. What do 3 you observe about the graphs of Y2 = x , its inverse Y3 = Vx, and the line Y1 = x? Repeat this Simultaneously graph Y1

-

experiment by simultaneously graphing Y1 with

-6 � x � 3. = x?

=

X,

Y2

=

2x + 3,

and Y3

=

1

"2(x

-

3) on a square screen

Do you see the symmetry of the graph of Y2 and its inverse Y3 with respect to the

line Y1

EXAMPLE 7

Solution

G raphi ng the I nverse Function

The graph in Figure lS(a) is that of a one-to-one function y = j ( x). Draw the graph of its inverse.

We begin by adding the graph of y = x to Figure lS(a). Since the points ( -2, - 1 ) , ( - 1 , 0 ) , and (2, 1 ) are on the graph of j, we know that the points (-1 , -2), (0, - 1 ) , and ( 1 , 2) must be on the graph of rl. Keeping in mil1d that the graph of rJ is the reflection about the line y = x of the graph of j, we can draw ri. See Figure lS(b). Figu re

15

Y 3

Y 3 (1 , 2) Y = f(x)

(2, 1 ) 3 x

-3

( -2 - 1 ) ,

Y=x

(2, 1 ) 3 x

-3

(-2, - 1 )

-3

(a)

1 Y = f- (x)

(b) •

==".. -


416

CHAPTER 6

Exponential a n d Logarithmic Functions

4

F i n d the Inverse of a Function Defi ned by an Equation

The fact that the graphs of a one-to-one function I and its inverse function 1-1 are symmetric with respect to the line y = x tells us more. It says that we can obtain 1-1 by interchanging the roles of x and y in I. Look again at Figure 14. If I is defined by the equation

y = I(x) then r1 is defined by the equation

x = I(y) The equation x = I(y) defines r1 implicitly. If we can solve this equation for y , we will have the explicit form of rl , that is, y = rl(x) Let's use this procedure t o find the inverse o f I(x) = 2x+ 3. (Since I i s a lin

ear function and is increasing, we know that I is one-to-one and so has an inverse function.) E X A M P LE 8

F i n d i ng the I nverse Fu nction

Find the inverse of I(x) = 2x+ 3. Also find the domain and range of I and Graph I and I-Ion the same coordinate axes.

Solution

r1 .

y = 2x + 3 , interchange the variables x and y. The result, x = 2y+ 3 that defines the inverse r1 implicitly. To find the explicit form, we

In the equation is an equation solve for y.

2y+ 3 = x 2y = x - 3 y = "21 (x - 3) The explicit form of the inverse r1 is therefore 1 rJ (x) = "2 (x - 3) which w e verified i n Example 5 (b). Next we find

= Range of I =

Domain ofl The graphs of I(x) Figure

Figure 16

=( Domain of r1 = Range ofr1

= 2x+ 3 and

its inverse

- 00 , 00

)

( -00,00 )

rl (x)

=

� (x - 3)

are shown in

16. Note the symmetry of the graphs with respect to the line y = x. y

•

SECTION 6.2


41 7

We now outline the steps to follow for finding the inverse of a one-to-one function. Procedure for Finding the Inverse of a One-to-One Function STEP

1: In y= f(x), interchange the variables x and y to obtain

x= f(y)

STEP

STEP

This equation defines the inverse function r1 implicitly.

2: If possible, solve the implicit equation for y in terms of x to obtain

the explicit form of rl:

= r1(x)

3: Check the result by showing that

r1(f(x))

EXA M P LE 9

y

=x

and

f(r1(x))= x

Finding the I nverse Function

The function

f(x)

+1 = 2xx-I

x =t- 1

is one-to-one. Find its inverse and check the result. Solut i o n

STEP

1: Interchange the variables x and y in

y =

2x + 1 x-I

x=

2y + 1 y- 1

to obtain

STEP

2: Solve for y.

2y + 1 y- 1 x(y -1) 2y + 1 xy -x 2y + 1 xy - 2y = x + 1 (x -2)y x + 1 x+1 y x-2 x=

= =

= =

--

The inverse is

f-l(X) STEP

3:

M u ltiply both sides by

-

1.

Apply the Distributive Property. Subtract

2y from both sides; add x to both sides.

Factor. Divide by x

-

2.

+1 _X X =t- 2 x-2

Replace

y by f-1 (x).

Check:

2x+1 +1 2X + 1 x-I r1(f(x))= rl = x -I 2x+l _2 x -I

(

f(r1(x))

y

)

= f ( :� � �) =

--

X+ 2 --1 +1 -2 + 1 - 1 x-2

(�r ) --

3x 2x + 1+x-I =x = 2x+I-2(x-l) = 3 2(x + 1)+ x -2 x + 1 - (x-2)

3x ==x 3

x =t-l

x=t-2 •

41 8

CHAPTER 6


Exploration

In Example 9, we found that, if {(x) = --, then (-'(x) = --, Compare the vertical and hori2x+ 1

x+ 1

x- 1

x- 2

zontal asymptotes of { and {-'. What did you find? Are you su rprised?

Result You should have found that the vertical asymptote of { is x = 1, and the horizontal asymptote , is y = 2. The vertical asymptote of (- is x = 2, and the horizontal asymptote is y = 1 . "",\::::

: : >- Now Work

P R O B L E M 49

We said in Chapter 3 that finding the range of a function f is not easy. However, if f is one-to-one, we can find its range by finding the domain of the inverse func tion r1. EXAM P LE 10

Finding the Range of a Function

Find the domain and range of

f(x) Solution

1 = 2xx+ - 1

The domain of f is {xix *' 1 }. To find the range of f, we first find the inverse rl . B ased on Example 9, we have

The domain of r1 is {xix *' 2}, so the range of f is {yly "!IIi

>-

*'

2}.

•


If a function is not one-to-one, it has no inverse function. Sometimes, though, an appropriate restriction on the domain of such a function will yield a new function that is one-to-one. Then the function defined on the restricted domain has an inverse function. Let's look at an example of this common practice. EXAMPLE 1 1

Finding the I nverse of a Domain-restricted Function

Find the inverse of y

Solution

=

STEP

Y

2

f(x)

= x2 if x 2: 0.

The function y x2 is not one-to-one. [Refer to Example 2(a).] However, if we restrict the domain of this function to x 2: 0, as indicated, we have a new function that is increasing and therefore is one-to-one. As a result, the function defined by y f(x) x2, X 2: 0, has an inverse function, rl . We follow the steps given previously to find rl.

=

Figure

=

=

1: In the equation y

=

x2, X

2:

0, interchange the variables x and y. The result is

x= l

This equation defines (implicitly) the inverse function.

17

f(x) = x2, X2: 0

STEP

Y= x

x

2: We solve for y to get the explicit form of the inverse. Since y 2:

solution for y is obtained: y

= Yx. So rl (x)

=

Yx.

0, only one

= rl(x2) # Ixl x since x 2: ° = f(Yx) = (Yxf x Figure 17 illustrates the graphs of f(x) = x2, X 2: 0, and r1(x) Yx. STEP

2

y 2: 0

3:

Check:

r1(f(x) ) f(rl(x) )

=

=

=

=

=

•

SECTION 6.2


41 9

SUMMARY 1. If a function I is one-to-one, then it has an inverse function 2. Domain of I

=

Range of rl; Range of I

=

1-1.

Domain of rl.

r1(f(x) ) = x for every x in the domain of I and l(r1(x) ) = x for every x in the domain of rl . 4. The graphs of I and r1 are symmetric with respect to the line y = x. 5. To find the range of a one-to-one function I, find the domain of its inverse function ri. 3. To verify that r1 is the inverse of I, show that

6.2 Assess Your Understanding 'Are You Prepared?' Answers are given at the end of tfuse exercises. If you get a wrong answel; read the pages listed in red. x + 5 1. Is the set of ordered pairs { ( I,3 ) ,( 2,3 ) ,( - 1,2)}a function? ?(pp. 215-216) 3. What is the domain of f ( x ) x2 + 3x - 18 =

Why or why not? (pp. 208-210)

2. Where is the function f(x) decreasing? (pp. 233-234 )

=

x2 increasing? Where is it


4. If every horizontal line intersects the graph of a function f at no more than one point,f is a(n) function.

7.

True orFalse If f and g are inverse functions,the domain of f is the same as the domain of g.

8.

True orFalse If f and g are i nverse functions,their graphs are symmetric with respect to the line y = x.

__

5. If rl denotes the inverse of a function f, then the graphs of f and rl are symmetric with respect to the line . __

6. If the domain of a one-to-one function f is [4,00 ) , the range of its inverse,f -I, is . __

Skill Building In Problems 9-16, determine whether the function is one-to-one.

9.

11.

Domain

Range

20 Hou rs

$200

25 Hou rs

$300

30 Hours

$350

40 Hou rs

$425

Domain

Range

20 Hours

10.

Domain Bob

Range -� Karla

t---

-

Dave -�

� Debra

John

12.

Dawn

Chuck

Phoebe

Domain

Range

...j.- $200

Bob

25 Hours -

Karla

Dave

30 Hours

$350

John

40 Hours

$425

Chuck

Debra ....j.- Phoebe r-

13. { ( 2,6 ) ,( - 3,6) , ( 4,9 ) ,( 1,10)}

14. {( - 2,5 ) ,( - 1,3 ) , (3,7 ) ,( 4,12)}

15. {C O,0),( 1,1), ( 2,1 6 ) ,(3 , 8 1 ) }

16. { ( I,2) ,( 2,8) , (3,18) ,( 4,3 2)}

In Problems 17-22, the graph of a function f is given. Use the horizontal-line test to determine whether f is one-to-one.

17.

18.

19.

y

3

3 x

-3

-3

3 x

-3

-3

3 x

-3

-3

420

CHAPTER 6

20.


21.

Y

x

"

22.

-3

Y 3

3 x

-3

In Problems 23-30, find the inverse of each one-to-one function. State the domain and the range of each inverse function. 23.

Annual Rainfall (inches)

Location Mt Waialeale, Hawaii

24.

460.00

Star Wars

$461

Star Wars: Episode One - The Phantom Menace

$431

E. T. the Extra Terrestrial

$400

Jurassic Park

$357

Forrest Gump

$330

Monrovia, Liberia

202 01

Pago Pago, American Samoa

196.46

Moulmein, Bu rma

191.02

Lae, Papua New Guinea

182.87

Source: Information Please Almanac

25.

Domestic Gross (in millions)

Title

Source: Information Please Almanac

Age

Monthly Cost of Life Insurance

30

$7.09

26.

$8.40

40

$11.29

45

State

Unemployment Rate

Virginia

11%

Nevada

5.5%

Tennessee

5.1%

Texas

6.3%

Source: United

Source: eterm.com

.27. {( -3, 5), (-2, 9), (-1, 2), ( 0, ll), (1, -5)}

States Statistical Abstract

28. {( -2, 2), ( - 1 ,6), (0, 8), (1, -3), (2, 9)}

29. { (-2, 1), (-3,2), ( - 10,0) , (1, 9), (2, 4)}

30. {( -2, -8), ( - 1 , -1), ( 0,0) , (1, 1), (2, 8)}

In Problems 31-40, verify that the functions f and g are inverses of each other by showing that f (g (x)) = x and g (f (x)) = x. Give any values of x that need to be excluded. 1 l .31. f (x) = 3x + 4; g (x) = 3 (x - 4) 32. f (x) = 3 - 2x; g (x) = _ (x - 3) 2 x

33. f (x)

= 4x - 8; g (x) = "4

35. f(x)

= x3 - 8; g (x) = �

1 37. f (x) = - ; x 39. f (x)

=

g (x)

2x+ 3 x+4

-- ;

=

+

2

1

-

x

4x - 3 g (x) = -2 - x

1 2

34. f (x)

= 2x

36. f (x)

= (x - 2j2, x

38. f (x)

= x; g (x) = x

40. f (x)

=

+

6; g (x) = -x - 3 2:

2; g (x) = vX + 2

x -5 ; g (x) 2x + 3

--

3x + 5 = -1 - 2x

In Problems 41-46, the graph of a one-to-one function f is given. Draw the graph of the inverse function rl. For convenience (and as a hint), the graph of y = x is also given. 41.

Y 3

Y= x

42.

Y 3

y=X

43.

y 3

(1,2 )

Y=X

(2,1) 3 x

-3

3 x

-3

3 x

-3 ( -1, -1)

-3

SECTION 6.2

44.

Y =X

Y 3

45.


421

46.

Y=X

( - 2,1) 3 x (1 , -1)

-3

3 x

-3

-3

-3

In Problems 47-58, the [unction [ is one-to-one. F ind its inverse and check your answeJ: State the domain and the range o[ [ and rl . Graph [, [-I , and y = x on the same coordinate axes. 49. [(x) = 4x+ 2 48. [ (x) = -4x 47. [(x) = 3x 50. [(x)

= 1 - 3x

51. [(x)

= x3 - 1

52. [( x)

=

x3+1

53. [( x)

= x2

54. [( x)

= x2+ 9 x � o

55. [ (x)

=

-

56. [(x)

=

57. [(x)

=

58. [ (x)

=

--

+

4 x�o

3 x

- -

1 x-2

--

4 x

4 x+2

In Problems 59-70, the [unction [ is one-to-one. F ind its inverse and check your answeJ: State the domain o[[ and find its range using rl . 4 3x 2 .,9. [(x) = 61. [ (x) = 60. [(x) 3+x x+2 2-x _

--

62. [(x)

=

65. [(x)

=

68. [(x)

=

=

2x x - I

'-,, 63. [(x)

- -

=

-

--

2x .)x -

64. [(x)

=

- --

67. [(x)

=

2x+ 3 x+2

70. [(x)

=

-

1

3x+4 2x - 3

66. [(x)

=

2x - 3 x+4

-3x - 4 x-2

69. [(x)

=

x2 - 4 x> 2x-,

--

3x+ 1 x

� -

°

x2 + 3 x> 3x-,

°


71. Use the graph of y the following:

(a) (b) (c) (d)

[( - 1) [(1) rl (l) rl (2)

=

[ (x) given in Problem 41 to eval uate

79. A function y = [(x) is increasing o n the interval (0, 5). What conclusions can you draw about the graph of y = rl (x) ?

= [ (x) given in Problem 42 to evaluate

(c) rl (O) (d) rl ( - l )

= 13 and [ is one-to-one, what is r l ( B)?

74. If g (-5)

=

80. A function y [(x) is decreasing on the interval (0,5). What conclusions can you draw about the graph of y = rl (x) ? =

72. Use the graph of y the following: (a) [(2) (b) [(1)

73. If [(7)

78. The domain of a one-to-one function g is [0, 15] , and its range is (0, 8). State the domain and the range of g-I.

3 and g is one-to-one,what is g-I (3)?

75. The domain of a one-to-one function [is [5,00 ) , and its range is [-2, 00). State the domain and the range of rl . 76. The domain of a one-to-one function [ is [0, 00 ) ,and its range is [5, 00 ) . State the domain and the range of r l .

77. The domain o f a one-to-one function g is the set o f all real n umbers, and its range is [0,00). State the domain and the range of g-I .

81. Find the inverse of the linear function

[(x) = mx+b

In "* °

82. Find the inverse of the function

[(x) = Vr2 - x2 O :=;

X

:=;

r

83. A function [ has an inverse function. If the graph of [ lies in quadrant I, in which q uadrant does the graph of r 1 lie? 84. A function [ has an inverse function. If the graph of [ lies in quadrant II, in which quadrant does the graph of r 1 lie?

85. The function [(x) = I x l is not one-to-one. Find a suitable re striction on the domain of [ so that the new function that re sults is one-to-one. Then find the inverse off. 86. The function [(x) = x4 is not one-to-one. Find a suitable re striction on the domain of [ so that the new function that results is one-to-one. Then find the inverse of f.

422

C HAPTER 6


In applications, the symbols used for the independent and dependent variables are often based on common usage. So, rather than using y = f(x ) to represent a function, an applied problem might use C C(q) to represent the cost C of manufacturing q units of a good since, in economics, q is used for output. Because of this, the inverse notation f-1 used in a pure mathematics problem is not used when finding in verses of applied problems. Rathel; the inverse of a function such as C C( q) will be q = q( C). So C = C( q) is a function that represents the cost C as a function of the output q, while q = q( C) is a jil.l1ction that represents the output q as a function of the cost C. Problems 87-90 illustrate this idea. =

=

87. Vehicle Stopping Distance Taking into account reaction time, the distance d (in feet) that a car requires to come to a complete stop while traveling r miles per hour is given by the function d ( r)

= 6.97r - 90.39

(a) Express the speed r at which the car is traveling as a function of the distance d required to come to a com plete stop. (b) Verify that r = r e d ) is the inverse of d = d ( r ) by show ing that r ( d ( r ) ) r and d ( r ( d) ) = d. (c) Predict the speed that a car was traveling if the distance required to stop was 300 feet. =

88. Height and Head Circumference The head circumference C of a child is related to the height H of the child (both in inches) through the function

H (C) = 2.15C - 1 0. 53 (a) Express the head circumference C as a function of height H. (b) Verify that C = C ( H ) is the inverse of H H(C) by showing that H ( C ( H ) ) = H and C ( H ( C ) ) C. (c) Predict the head circumference of a child who is 26 inches tall. =

=

89. Ideal Body Weight The ideal body weight W for men (in kilograms) as a function of height h (in inches) is given by the function W(h) SO + 2.3(h - 60 ) =

(a) What is the ideal weight of a 6-foot male? (b) Express the height h as a function of weight W. (c) Verify that h = h e W ) is the inverse of W = W ( h ) by W. showing that h ( W ( h ) ) = h and W ( h ( W ) ) (d) What is the height of a male who is at his ideal weight of 80 kilograms? =

Note: The ideal body weight W for women (in kilograms)

as a function of height h (in inches) is given by W ( h) 45.5 + 2.3 ( h - 60) . =

9 90. Temperature Con"ersion The function F(C) = s C + 32 converts a temperature from C degrees Celsius to F degrees Fahrenheit. (a) Express the temperature in degrees Celsius C as a func tion of the temperature in degrees Fahrenheit F. (b) Verify that C C ( F ) is the inverse of F F ( C) by showing that C ( F(C) ) = C and F ( C ( F ) ) F. (c) What is the temperature in degrees Celsius if it is 70 de grees Fahrenheit? =

91. Income Taxes

The function

T ( g ) = 4220 + 0.25(g - 30,650 ) represents the 2006 federal income tax T (in dollars) due for a "single" filer whose adjusted gross income is g dollars, where 30,650 � g � 74,200. (a) What is the domain of the function T? (b) Given that the tax due T is an increasing linear func tion of adjusted gross income g, find the range of the function T. (c) Find adjusted gross income g as a function of federal in come tax T. What are the domain and the range of this function? 92. Income Taxes

The function

T ( g ) = 1510 + 0. 15(g - 15,100 ) represents the 2006 federal income tax T (in dollars) due for a "married filing jointly" filer whose adjusted gross income is g dollars, where 15,100 � g � 61 ,300. (a) What is the domain of the function T? (b) Given that the tax due T is an increasing linear function of adjusted gross income g, find the range of the func tion T. (c) Find adjusted gross income g as a function of federal in come Lax T. What are the domain and the range of this function?

93. Gra"ity on Earth If a rock falls from a height of 100 meters on Earth, the height H (in meters) after t seconds is approx imately H ( t ) = 100 - 4.9t2

(a) In general, quadratic functions are not one-to-one. How ever, the function H(t) is one-to-one. Why? (b) Find the inverse of H and verify your result. ( c) How long will it take a rock to fall 80 meters? 94. Period of a Pendululll The period T (in seconds) of a sim ple pendulum as a function of its length I (in feet) is given by

�32.2

T ( l ) = 27T

(a) Express the length I as a function of the period T. (b) How long is a pendulum whose period is 3 seconds? 95. Given f(x) =

=

=

I

find r1 ( x ) . If c is f r l ?

oF

ax + b cx + d

0, under what conditions on a, b, c, and d

=


96. Can a one-to-one function and its inverse be equal? What must be true about the graph of f for this to happen? Give some examples to support your conclusion.

97. Draw the graph o f a one-to-one function that contains the points ( -2, -3 ) , (0, 0), and ( 1 , 5 ) . Now draw the graph of its inverse. Compare your graph to those of other students. Dis cuss any similarities. What differences do you see?

SECTION 6.3

98. Give an example of a function whose domain is the set of real numbers and that is neither increasing nor decreasing on its domain, but is one-to-one. [Hint: Use a piecewise-defined function.]

Exponential Functions

423

99. Is every odd function one-to-one? Explain. 100. Suppose C(g) represents the cost C, in dollars, of manufac turing g cars. Explain what C-1 (800,000) represents.


3. {xix

1. Yes; for each input x there is one output y .

2. Increasing on (0, 00 ) ; decreasing on ( - 00 , 0)

*"

-

6, x

3}

*"

6.3 Exponentia l Functions PREPARING FOR THIS SECT ION

• •


Exponents (Section R.2, pp. 21-24, and Section R.8, pp. 75-76) Graphing Techniques: Transformations (Section 3.5, pp. 252-260) Now Work the 'Are You Prepared?, problems on page

OBJECT IVES 1

• •

Average Rate of Change (Section 3.3, pp. 236-238) Solving Equations (Section 1 . 1 , pp. 86-92) Horizontal Asymptotes (Section 5.2, pp. 346-352)

432.

Evaluate Exponential Functions (p. 423)

2 Graph Exponential Functions (p. 425) 3 Defin e the Number e (p.429) 4 Solve Exponential Equations (p.431)

1

Eva l ua te Expo nential Fu nctions

In Chapter R, Section R.8, we give a definition for raising a real number a to a ratio nal power. Based on that discussion, we gave meaning to expressions of the form

a

r

where the base a is a positive real number and the exponent r is a rational number. B ut what is the meaning of aX, where the b ase a is a positive real number and the exponent x is an irrational number? Although a rigorous definition requires methods discussed in calculus, the basis for the definition is easy to follow: Select a rational number r that is formed by truncating (removing) all but a finite number of digits from the irrational number x. Then it is reasonable to expect that For example, take the irrational number tion to a7T is

7T =

3.141 5 9 . . . . Then an approxima

where the digits after the hundredths position have been removed from the value for 7T. A better approximation would be

a7T

:::::

3 14159 a .

where the digits after the hundred-thousandths position have been removed. Con . tinuing in this way, we can obtain approximations to aT to any desired degree of accuracy. Most calculators have an key or a caret key for working with expo

[2]

nents. To evaluate expressions of the form key (or the

[] key), enter the exponent

X

a ,

x,

�I

enter the base

and press

a,

then press the

I 1 (or I enter I)· =

[2]

424

CHAPTER 6

Exponential and Logarithmic Functions Using a Calcu l ator to Evaluate Powers of 2

E XA M P LE 1

Using a calculator, evaluate: Solution

(a) 21 .4 � 2.639015822 ( c) 2 1 .41 4 � 2.66474965 (e) 2° � 2.665144143 �j ! =---

(e) 20

(d) 21 .4142

(c) 21 .414

(b) 2 1 .41

(a) 2 1 .4

(b) 2 1.41 � 2.657371628 (d) 2 1.4142 � 2.665119089

•


It can be shown that the familiar laws for rational exponents hold for real exponents. THEOREM

Laws of Exponents

If

s,

t, a, and b are real numbers with a

> 0 and

b

> 0, then

(aby

a-S = � as

1S = 1

=

=

as . bS

(!a ) S

( 1)

�------�

�

We are now ready for the following definition: DEFINITION

An

exponential function

is a function of the form

f(x) where a is a positive real number set of all real numbers.

WARNING It is important to distinguish a power function, g(x)

=

x",

n �

2, an

integer, from an exponential function, f(x)

=

aX, a *- 1, a

>

O.

In

a power

function, the base i s a variable and the exponent is a constant. In a n exponen tial function, the base i s a constant and the exponent is a variable.

_

(a

=

aX

> 0) and

a

1 . The domain of

*-

f is the

We exclude the base a = 1 because this function is simply the constant func tion f (x) = 1x = 1. We also need to exclude bases that are negative; otherwise, we

�

�.

would have to exclude many values of x from the domain, such as x = and x = [Recall that ( _2 ) 1/2 = vC2, ( _3 )3/4 = �, and s o on, are not defined in the set of real numbers.] Some examples of exponential functions are =

f(x)

=

�

F (x)

2x

=

(3)X 1

Notice for each function that the base is a constant and the exponent is a variable. You may wonder what role the base a plays in the exponential function f(x) = aX. We use the following Exploration to find out.

Exploration

(a) Evaluate ((x) (b) Eval u ate g(x)

=

=

2X at x

=

-

2, - 1 , 0, 1, 2, and 3.

3x + 2 at x

=

-

2, - 1, 0, 1, 2, and 3.

(c) Comment on the pattern that exists in the val ues of ( and g. Result

(a) Table 1 shows the values of ((x) (b) Table 2 shows the val ues of g(x)

= =

2x for x

=

-

2, - 1, 0, 1 , 2, and 3.

3x + 2 for x

=

-

2, - 1 , 0, 1 , 2, and 3.

SECTION 6.3

Table

x

1

f(x)

2

f(- 2) =

1

=

r2

Table 2

r

x

1

1

2

2

4

3

8

=

3x +

425

2

g(- 2) = 3 ( - 2) + 2 = -4

-1

2

0

g(x)

-2

1 1 = - = 4 22

Exponential Functions

-1

0

2

1

5

2

8

3

11

(c) In Table 1 we notice that each va lue of the exponential function f(x) = aX = Y could be found by m Ultiplying the previous value of the function by the base, a = 2. For example, f( - 1 ) = 2 · f( - 2) = 2 '

1 1 = 2' "4

f(O) = 2 · f( - 1 ) = 2 · 2 = 1 , 1

f(1 ) = 2 · f(O) = 2 · 1 = 2

and so on. Put another way, the ratio of consecutive outputs is constant for u n it increases in the input. The constant equals the value of the base a of the exponential function. For example, for the function f(x) = 2x, we notice that 1

2 f(- 1 ) -- = - = 2' f(- 2) 1

---

f( 1 ) 2 - = - = 2' f(O) 1

f(x + 1 ) f(x)

y+ 1 = - = 2 Y

4

and so on. From Table 2 we see that the function g(x) = 3x + 2 does not have the ratio of consecutive outputs that a re constant because it is not exponential. For example, --

g(- 1 )

g(-2)

g(1 ) 5 -1 1 = - = - i' - = -4 4 g(O) 2

Instead, because g(x) = 3x + 2 is a linear function, for unit increases in the input, the outputs increase by a fixed amount equal to the value of the slope, 3.

The results of the Exploration lead to the following result. THEOREM r

r

r

r

r

In Words

For

f(x)

an

=

For an exponential function f(x) then

=

f(x

exponential

X a , a

+

1)

f(x)

'----

> 0,

a

I

= a

-----------.l..J

f u n ction

aX, for l-unit c h anges in

Proof

the input x, the ratio of consecu

f(x

tive outputs is the constant a.

+

1)

•

f(x) �1iiIi = ill:lO "'"

2

EXAM P L E 2


G r a p h Exponential Fu nctions

First, we graph the exponential function f(x)

=

2x.

G raphing an Exponential Function

Graph the exponential function: Solution

=F 1, if x is any real number,

f ( x ) = 2x

The domain of f(x) = 2x consists of all real numbers. We begin by locating some points on the graph of f(x) = 2x, as listed in Table 3.

426

CHAPTER 6


Table 3

Figure 1 8

x r 1 0 '" 0.00098

-10 -3

r3

=

�

-2

r2

=

�

-1

r1

=

�

o

20

=

1

21

=

2

2

22

=

4

3

23

=

8

10

2 10

=

1 024

8

4 2

Since 2x > ° for all x, the range of f i s the interval (0, 00 ) . From this, we con clude that the graph has no x-intercepts, and, in fact, the graph will lie above the x-axis. As Table 3 indicates, the y-intercept is Table 3 also indicates that as x � - 00 the value of f (x) = 2x gets closer and closer to 0. We conclude that the x-axis is a horizontal asymptote to the graph as x � - 00. This gives us the end behavior of the graph for x large and negative. To determine the end behavior for x large and positive, look again at Table 3. As x � 00 , f(x) 2x grows very quickly, causing the graph of f(x) = 2x to rise very rapidly. It is apparent that f is an increasing function and so is one-to-one. Using all this information, we plot some of the points from Table 3 and con nect them with a smooth, continuous curve, as shown in Figure 18.

l.

=

•

As we shall see, graphs that look like the one in Figure 18 occur very frequently in a variety of situations. For example, look at the graph in Figure 19, which illus trates the number of cellular telephone subscribers at the end of each year from 1 985 to 2005. We might conclude from this graph that the number of cellular tele phone subscribers is growing exponentially. Figure

Number of Cellular Phone Subscribers at Year End

19 0 and is closer to the x-axis when x < O .

l�1 I!m

Seeing the Concept

Gra p h Y, 2x and compare what you see to Figure 1 8. Clear the screen and g raph Y, 3x and Y2 = 6x = = a n d compare what you see to Figure 20. Clear the screen a n d graph Y, 1 0x and Y2 1 00x. What = viewing recta ngle seems to work best? =

-3

Y=0

The following display summarizes the information that we have about f(x) = > 1.

aX, a

Properties of the Exponential Function f(x)

((x)

4. f(x)

=

aX, a

= 0) i s a horizontal asymptote a s x � - (X) .

> 1, i s an increasing function and is one-to-one.

( �} -1,

5. The graph of f contains the points y= o

x

E XA M P L E 3

Solution

f(x)

-2 -1 o

2 3 10

=

(1"2)-10

(�r

=

1 024

Gr3 Gr2 Gr' GY GY = � GY � GY � G) =

8

= 4 =

=

2

a).

See Figure 21.

aX

when 0

1

x

a" ,

1. The domain is the set of all real numbers; the range is the set of positive

Figure 2 1 =

=

The domain of f(x)

=

(�r

f(x) =

1.

(l)X "2

consists of all real numbers. As before, we locate some

points on the graph by creating Table 4. Since

(�r

> 0 for all x, the range of f is

the interval ( 0, (0 ) . The graph lies above the x-axis and so has no x-intercepts. The

y-intercept is L As x � - (X) , f(x) =

(l)X "2

grows very quickly. As x �

00 ,

the values

of f(x) approach O . The x-axis (y = 0) is a horizontal asymptote as x � 00 . It is apparent that f is a decreasing function and so is one-to-one. Figure 22 illustrates the graph. Figure 22

1

=

=

10 ""

0.0009 8

-3

3

x Y=0 •

428

CHAPTER 6


We could have obtained the graph of y transformations. The graph of y = the graph of y

=

(�r

=

=

(�r

from the graph of y

TX is a reflection

=

2x using

about the y-axis of

2x (replace x by -x). See Figures 23(a) and 23(b).

Figure 23

(a) y = 2 X

l� mlI

Replace x by Reflect about the y-axis

Yj

y=

GY

(b) Yj

W

(-1 , 6) y=

G) x

-3

=

3

x y= 0

Y = 2-x =

=

(�Y,

for

a>

=

=

=

(�Y

(�r

in Figure 22 is typical of all exponential functions

that have a base between ° and 1. Such functions are decreasing and one-to-one. Their graphs lie above the x-axis and pass through the point (0, 1 ) . The graphs rise rapidly as x -7 - 00 . As x -7 00 , the x-axis (y = 0) is a horizontal asymptote. There are no vertical asymptotes. Finally, the graphs are smooth and continuous, with no corners or gaps. Figure 24 illustrates the graphs of two more exponential functions whose bases are between ° and 1. Notice that the smaller base results in a graph that is steeper when x < 0. When x > 0, the graph of the equation with the smaller base is closer to the x-axis. The following display summarizes the information that we have about the func tion f(x) = aX, ° < a < 1 .

1.

Figure 25

6x, Y2

0, i s the reflection about the y-axis of the g raph of

Properties o f the Exponential Function f(x)

{(x)

(W

«.

The graph of f(x) =

X y

(b)

Using a g raphing utility, simultaneously g raph:

Conclude that the graph of Y2

24

x;

Seeing the Concept (a) Yj = 3x, Y2 =

Figure

-

=

tr,

0 < a < 1

The domain is the set of all real numbers; the range is the set of positive real numbers.

2. There are no x-intercepts; the y-intercept is 1 .

aX, 0 < a < 1

3. The x-axis ( y 4. f(x)

=

=

0 ) is a horizontal asymptote as x -7 00 .

aX, ° < a
I;;Ii -

3


Defi n e the N u mber

e

As we shall see shortly, many problems that occur in nature require the use of an exponential function whose base is a certain irrational number, symbolized by the letter e . Let's look at one way of arriving at this important number e .

DEFINITION

The number

e

is defined as the number that the expression

( I )"

( 2)

1 +n

approaches as

n �

00 . In calculus, this is expressed using limit notation as

e

=

lim

,, --> 00

(

n

1 + I

)

"

I

-.J

Table 5, on page 430, illustrates what happens to the defining expression (2) as takes on increasingly large values. The last number in the right column in the table is correct to nine decimal places and is the same as the entry given for e on your cal culator (if expressed correctly to nine decimal places). The exponential function f(x) = eX, whose base is the number e , occurs with such frequency in applications that it is usually referred to as the exponential function. Indeed, most calculators have the key or exp(x) which may be used to

n

[2]

1

evaluate the exponential function for a given value of x . *

I,

':' I f your calculator does n o t have one of these keys, refer t o y o u r Owner's Manual.

430

CHAPTER 6

Table 5


n

(

+ n 1

n 2

1

)

-2

0.1 4

-1

0.37

0.5

1 .5

2.25

0.2

1 .2

2.48832

10

0.1

1 .1

2.59374246

0.01

1 .01

2.70481 3829

a

2.72

0.001

1 .001

2.7 1 6923932

1 0,000

0.0001

1 .0001

2.7 1 8 1 45927

1 00,000

0.00001

1 .00001

2.71 8268237

0.000001

1 .000001

2.7 1 8280469

1 0- 9

1 + 1 0- 9

2.7 1 8281 827

1 ,000,000 1 ,000,000,000

y

e"

x

2

5

1 ,000

Table 6

"

2

1 00

Figure 27 y = e"

+ n

1

2

eX

7.39

Now use your calculator to approximate for x -2, x = - 1 , x = 0, x = 1 , and x = 2 , as we have done to create Table 6 . The graph of the exponential function f(x) = is given in Figure 27. Since 2 < < 3, the graph of y = lies between the graphs of y = 2 and y Y. Do you see why? (Refer to Figures 18 and 20.)

(2, 7.39)

eX

I� . --:· I �

x

e

=

=

eX

Seeing the Concept

Gra ph Y1 = e" and compare what you see to Fig u re 27. Use eVALUEate or TABLE to ver ify the points on the g raph shown in Figure 27. Now graph Y2 = 2x and Y3 = 3x on the same screen as Y1 e". Notice that the graph of Y1 = eX l ies between these two graphs. =

Y= 0

3

(-2, 0.1 4)

x

E XA M P L E 5

G raphing Exponential F unctions Using Transformations

Graph f(x) of f. Solution

=

_ex-3

and determine the domain, range, and horizontal asymptote

We begin with the graph of

Figure 28

y

y

=

eX.

Figure 28 shows the stages. y

y= o

3

(2, 7.39) ( - 2 , - 0 . 1 4) ( -1, -0.37)

6

x

Y= 0

x

-3

3

-I--..J---,---,I

..-1_

_ _ __

Y= 0

-6 (2, -7.39 )

__

3

(-2, 0.1 4 )

(a) y = eX

x

Multiply by -1; Reflect about x-axis.

(b) y = - ex

Replace x by x - 3; Shift right 3 units.

(c) Y = _ ex-3

_ex-3

As Figure 28(c) illustrates, the domain of f(x) = is the interval ( - 00 , 00 ) , and the range is the interval ( - 00 , 0). The horizontal asymptote is the line y = O . rr �

-

•


SECTION 6.3

4 r

r , r r

In Words

E XA M P L E 6

If

a

U

=

V a ,

then

u

=

v

( 3)

Property (3) is a consequence of the fact that exponential functions are one-to one. To use property (3), each side of the equality must be written with the same base. Solving an Exponential Eq uation

Solve: Solution

431

Equations that involve terms of the form aX , a > 0, a "* 1 , are often referred to as exponential equations. Such equations can sometimes be solved by appropriately applying the Laws of Exponents and property (3):

sions with the sa m e base are equal.

Functions

So lve Exponential Equations

When two exponential expres equal, then their exponents are

Exponential

3x +1 = 81

Since 81 = 34 , we can write the equation as 3x + 1 = 81 = 34 Now we have the same base, 3, on each side, so we can set the exponents equal to each other to obtain x + 1 = 4 x = 3 The solution set is (3). L"!J1!

E XA M P L E 7


Solving a n Exponential Equation

Solve: Solution

•

e-x

2

=

(ex) 2 .

�e�

We use the Laws of Exponents first to get the base e on the right side. (ex ) 2 . As a result,

�e� = e2x • e-3 = e2x -3

e-X2 = e2x -3 -x2 = 2x - 3

x2 + 2x - 3 = 0 (x + 3 ) (x - 1 ) = 0 x = -3 or x = 1

A p ply Property ( 3 ) . Place th e quadratic equation i n standard form . Factor. U se the Zero-Prod uct Property.

The solution set is { -3, 1 } .

•

Many applications involve the exponential function. Let's look at one. E XA M P L E 8

Exponential Probabil ity

B etween 9:00 PM and 10:00 PM cars arrive at Burger King's drive-thru at the rate of 12 cars per hour (0.2 car per minute ) . The following formula from statistics can be used to determine the probability that a car will arrive within t minutes of 9:00 PM.

F(t)

=

1 - e-O.2t

( a) Determine the probability that a car will arrive within 5 minutes of 9 PM ( that is, before 9:05

PM).

( b ) Determine the probability that a car will arrive within 30 minutes of 9 ( before 9:30 PM).

PM

432

CHAPTER 6


I:l

(c) What value does F approach as t becomes unbounded in the positive direction? (d ) Graph F using your graphing utility.

5

(a) The probability that a car will arrive within F(t) at t = 5 .

Solution

F( 5 )

=

1

- e-O. 2(5)

;::::

i

minutes is found by evaluating

0.63212

U se a ca lc ulator.

We conclude that there is a 63 % probability that a car will arrive within 5 minutes. (b) The probability that a car will arrive within 30 minutes is found by evaluating F(t) at t = 30.

F(30)

=

1 - e-O. 2(30)

Figure 29

;::::

i

0.9975

Use a ca lcu lator.

1

There is a 99.75 % probability that a car will arrive within 30 minutes. (c) As time passes, the probability that a car will arrive increases. The value that

1

t ---? 00. Since e-O.2t = 02' it follows that e.t e-O . 2t ---? 0 as t ---? 00. We conclude that F approaches 1 as t gets large.

approaches can be found by letting

I:l

O �======� 30

o

(d)

S e e Figure 2 9 for the graph o f F.

�-� ..... - Now Work

SUMMARY f(x)

X a ,

=

a

X a ,

=

•

P R O B L EM 1 0 1

Properties of the Exponential Function

>

1

Domain: the interval ( - 00, 00 ) ; range: the interval x-intercepts: none; y-intercept: 1 Horizontal asymptote: x-axis ( y

f (x)

F

a

°
0, a * 1 , are equivalent. (b) If eb = 9, then b = loge 9.

loga 24.

•


C hanging Logarith m i c Expressions to Exponential Expressions

E XA M P L E 3

Change each logarithmic expression to an equivalent expression involving an exponent. (a) log" 4

=

5

(a) If loga 4 =

Solution

(c)

=

5, then as = 4. If log3 5 = c, then 3c = 5.

UJ1':I = ::::r.tl � .,..

2

(b) loge b

(c) log3 5

=

c

(b) If loge b

=

-3, then e-3

-3

=

b. •


Eva l u ate Loga rithmic Expressi o n s

To find the exact value of a logarithm, we write the logarithm in exponential nota tion and use the fact that if a" = aV then u = v.

SECTION 6.4

E XA M P L E 4

Logarithmic Functions

439

F i nd i ng the Exact Value of a Logarit h m i c Expression

Find the exact value of:

1 b (b) louo J 27 1 (b) y = log3 27

( a) log2 1 6 (a)

Solution

y= 2Y = 2Y = Y

=

log2 1 6 16 24 4

Ch ang e to expone ntial form .

16

= 24

3Y

Equate expon ents.

Therefore, log2 16

1

=27

1

T3

= y=

3Y

= 4.

C h ange to exponential form .

27

-

-3

3

-

Equate exponents.

Therefore, log3 �I"

1

= 33 = 3-3

1 27

=

-3. •

, or-


Determine the Doma i n of a Logarit h m ic Fu nction

The logarithmic function y = loga x has been defined as the inverse of the expo nential function y = aX. That is, if f(x) = aX, then rl (x) = loga x. Based on the discussion given in Section 6.2 on inverse functions, for a function f and its inverse r\ we have Domain ofr1

= Range of f

and

Range ofr1

=

Domain of f

Consequently, it follows that Domain of the logarithmic function Range of the logarithmic function

=

=

Range of the exponential function

Domain of the exponential function

= (0, 00 ) = ( - 00, 00 )

In the next box, we summarize some properties of the logarithmic function:

y = loga x Domain:

(defining equation:

° < x
0 Solving this inequality, we find that the domain of g consists of all x between - 1 and 1 , that is, -1 < x < 1 Of, using interval notation, ( -1 , 1 ) .

( c) Since Ixl > 0, provided that x "* 0, the domain of h consists of all real num bers except zero or, using interval notation, ( - 00 , 0) U (0, 00 ). "'Ill!

•


440

CHAPTER 6


4

G ra p h Log a rith mic Fu nctions

Since exponential functions and logarithmic functions are inverses of each other, the graph of the logarithmic function y = logo x is the reflection about the line y = x of the graph of the exponential function y = a X , as shown in Figure 30.

Figure 30

Y Y = aX

Y = aX y = x

Y

y= x

3 x

-3

-3

-3

-3

(b) a > 1

(a) a < a < 1

For example, to graph y

=

log2 x, graph y

y = x. See Figure 31. To graph y the line y = x. See Figure 32.

=

=

2x

[Ogl/3 x, graph Figure 3 2

Figure 3 1

Y = (�

r

( - 1 , 3)

and reflect it about the line

y =

(l)X 3

and reflect it about

y=x

Y

-3

-3

�====> - Now Work P R O B L E M 5 9

Properties of the Logarithmic Function f(x) 1.

loga x

The domain is the set of positive real numbers; the range is the set of all real numbers.

2. The x-intercept of the graph is

3. The y-axis (x 4.

=

1 . There is no y-intercept.

0) is a vertical asymptote of the graph. A logarithmic function is decreasing if 0 < a < 1 and increasing if a =

5. The graph of f contains the points

( 1 , 0), (a, 1 ), and

(�, -1).

>

1.

6. The graph is smooth and continuous, with n o corners or gaps.

If the base of a logarithmic function is the number e, then we have the natural function occurs so frequently in applications that it is given a special symbol, In (from the Latin, logarithmus naturalis). That is,

logarithm function. This

y

=

In x if and only if x

=

eY

( 1)

Since y = In x and the exponential function y = eX are inverse functions, we can obtain the graph of y = In x by reflecting the graph of y = eX about the line y = x. See Figure 33.

SECTION 6.4

Using a calculator with an f ( x ) = In x. See Table 7. Figure 3 3


� key, we can obtain other points on the graph of Table 7

Y

5

y= e x

y=x

In x

x 2

� �

441

-0.69

2

0.69

3

1 .1 0

Seeing the Concept Graph Y1 = eX and Y2

I n x on the same square screen. Use eVALUEate to verify the points on the graph given in Figure 33. Do you see the symmetry of the two g raphs with respect to the line y = x? =

E XA M P L E 6

X= 0

Figure 34

I I I

G raph i ng a Logarithmic F u n ction and Its I nverse

(a) (b) ( c) (d) (e) (f) Solution

3 x

y = 0 -3

Find the domain of the logarithmic function f ( x ) = -In ( x - 2 ) . Graph f From the graph, determine the range and vertical asymptote of f Find f - 1 , the inverse of f Use rl to find the range of f Graph f - l .

(a) The domain of f consists of all x for which x - 2 > 0 or, equivalently, x > 2. (b) To obtain the graph of y = -In ( x - 2 ) , we begin with the graph of y = In x and use transformations. See Figure 34.

Y

Y 3

3

x= O

Y 3

x- 2

x= o

x

-1

-1

-1

(a) y = In x

Multi ply by - 1 ; reflect about x-axis

Replace x by x - 2; shift right 2 u n its.

(b) y = - I n x

(c) y = - l n (x - 2)

(c) The range of f(x) = - I n( x - 2 ) is the set of all real numbers. The vertical as ymptote is x = 2. [Do you see why? The original asymptote ( x = 0) is shifted to the right 2 units.] (d) We begin with y - In ( x - 2 ) . The inverse function is defined (implicitly) by the equation =

x = -In ( y - 2 ) We proceed to solve for y. -x = In(y - 2 ) e-x = y - 2 y e -x + 2 =

Isolate the l ogarith m . Change to an exponential expression . Solve for y.

The inverse of f is f -l ( x ) = e-x

+

2.

442

CHAPTER 6


(e) The range of jis the domain of j -I, the set of all real numbers, confirming what we found from the graph of f. (f) To graph rI, we use the graph of j in Figure 34 (c) and reflect it about the line y = x. See Figure 35.

Figure 3S

Y 5

x= 2 Y= X

-1

I I

fix) = : I n (x

-1

Figure 36

I I I

-

-

2)

y =x

• 1.l!J ! �

Y = log

4

x


If the base of a logarithmic function is the number 10, then we have the common If the base a of the logarithmic function is not indicated, it is understood to be 10. That is,

logarithm function.

y=

x

log x if and only if

x=

lOY

Since y = log x and the exponential function y = lOX are inverse functions, we can obtain the graph of y = log x by reflecting the graph of y = lOX about the line y = x. See Figure 36.

-2 EXAM PLE 7

G raphing a Logarith m i c F un ction and Its I nverse

(a) (b) ( c) (d) ( e) (f)

Find the domain of the logarithmic function j (x) = 3 log (x - 1 ) . Graph f From the graph, determine the range and vertical asymptote of f Find j -l, the inverse of f Use j -1 to find the range of f Graph j -1 .

( a) The domain of j consists of all x for which x - 1 > a or, equivalently, x > l . (b) To obtain the graph of y = 3 loge x - 1 ) , we begin with the graph of y = log x and use transformations. See Figure 37.

Solution

Figure 3 7

2 -2 -2

I I I I

Y X= 1

Y X= 0

2

(10, 1 ) ( 1 , 0) ._-----1 2 (fo, - 1 )

4

6

8

10

1 2 x -2

Y X= 1

(1 1 , 1 ) : (2, 0) ._-----6

10

1 2 x -2

-2

-2 ----+-

Replace x by x - 1 ; horizontal shift right 1 unit (a) Y = log x

8

2

M u ltiply by 3; vertical stretch by a factor of 3.

(b) y = log (x - 1 )

(c) y = 3 log (x - 1 )

SECTION 6.4


443

(c) The range of f(x) = 3 10g(x - 1 ) is the set of all real numbers. The vertical as ymptote is x = 1 . (d) We begin with y = 3 10g(x - 1 ) . The inverse function i s defined (implicitly) by the equation x

=

3 log(y - 1 )

We proceed to solve for y. x '3 = log (y - 1 ) lOx/3

Isolate th e logarith m .

- 1 Change to an exponential expression . 3 Solve for y. Y = lOx/ + 1 3 The inverse of f is f -1 ( x) = lOx/ + 1 . (e) The range of fis the domain of f -l, the set of all real numbers, confirming what we found from the graph of f. (f) To graph f -1 , we use the graph of f in Figure 37(c) and reflect it about the line y = x. See Figure 38. Figure 38

=

Y

y

•

�==> .5

E XA M P L E 8

So lve Loga rith mic Equations

Equations that contain logarithms are called logarithmic equations. Care must be taken when solving logarithmic equations algebraically. In the expression logo M , remember that a and M are positive and a *' 1 . B e sure to check each apparent solution in the original equation and discard any that are extraneous. Some logarithmic equations can be solved by changing from a logarithmic expression to an exponential expression. Solving a Logarith m i c E q u ation

Solve: S o l ution


(a) log3 (4x - 7)

=

2

(b) logx 64

=

2

(a) We can obtain an exact solution by changing the logarithm to exponential form. log3 (4x - 7) 4x - 7

=

=

4x - 7

=

4x

=

x

=

2

32

9 16 4

C h a ng e to an exponential expression .

444

CHAPTER 6

Exponential and Logarithmic Fu nctions

Check: log3 ( 4x - 7 )

log3 ( 1 6

=

-

7)

=

i og3 9

=

2

32

= 9

The solution set is (4J. (b) We can obtain an exact solution by changing the logarithm to exponential form. logx 64 x2

=

2

=

64

X = ±

V64

Change to an exponentia l expression .

=

±8

Square Root Method

The base of a logarithm is always positive. As a result, we discard -8. We check the solution 8.

Check: logs 64

=

2

82

=

64

The solution set is (8J.

EXAM PLE 9

U s i n g Logarithms t o Solve Exponential E quations

Solve: Solution

•

e2x =

5

We can obtain an exact solution by changing the exponential equation to logarithmic form. e2 x 5 =

In 5

x

The solution set 1"

EXAM P L E 1 0

E .....

=

= �

2x

In 5 -2 0.805

Change to a loga rith m ic expression using the

fact that if eY = x then y = In

x.

Exact solution A pproxi mate sol ution

. { In 5 }

IS

2 .

•


Alcohol and D rivi ng

The blood alcohol concentration (BAC) is the amount of alcohol in a person's blood stream. A BAC of 0.04% means that a person has 4 parts alcohol per 10,000 parts blood in the body. Relative risk is defined as the likelihood of one event occurring divided by the likelihood of a second event occurring. For example, if an individ ual with a BAC of 0.02% is 1 .4 times as likely to have a car accident as an indi vidual that has not been drinking, the relative risk of an accident with a BAC of 0.02% is 1 .4. Recent medical research suggests that the relative risk R of having an accident while driving a car can be modeled by the equation R = e

kx

where x is the percent of concentration of alcohol in the bloodstream and k is a constant. (a) Medical research indicates that the relative risk of a person having an accident with a BAC of 0.02% is 1 .4. Find the constant k in the equation. (b) Using this value of k, what is the relative risk if the concentration is 0. 17%? (c) Using this same value of k , what BAC corresponds to a relative risk of 100? (d) If the law asserts that anyone with a relative risk of 5 or more should not have driving privileges, at what concentration of alcohol in the bloodstream should a driver be arrested and charged with a DUI (driving under the influence)? Solution

(a) For a concentration of alcohol in the blood of 0.02% and a relative risk of 1 .4, we let x = 0.02 and R = 1 .4 in the equation and solve for k. R

1.4

k e x k (O.02) = e

=

R

= 1.4; x =

0.02

S E CTION 6.4

0.02k

=

In 1.4 In 1.4 --

445

Change to a logarithm ic expression. �

16.82 Solve for k. 0.02 For a concentration of 0.17 % , we have x = 0.17. Using k = 16.82 in the equa tion, we find the relative risk R to be R = ekx e(16.82)(O.17) = 17.5 For a concentration of alcohol in the blood of 0.17%, the relative risk of an ac cident is about 17.5. That is, a person with a BAC of 0.17% is 17.5 times as likely k

(b)

=


=

to have a car accident as a person with no alcohol in the bloodstream. risk of 100, we have R = 100. Using k = 16.82 in the equation ( c) For a relative x ek R , we find the concentration x of alcohol in the blood obeys =

NOTE A BAC of 0.30 results in a loss of consciousness in most people. •

100 = e16.82x 16.82x = In 100 In 100 x= 16.82

R =

ekx, R

=

100; k = 1 6.82

Change to a logarithmic expression. �

0.27

Solve for x.

For a concentration of alcohol in the blood of 0.27 % the relative risk of an accident is 100. (d) For a relative risk of 5, we have R 5. Using k = 16.82 in the equation R = ekx, we find the concentration x of alcohol in the bloodstream obeys =

5 = e16.82x 16.82x = In 5 In 5 � 0.096 x 16.82 =

NOTE Most states use 0.08 or 0.10 as the blood alcohol content at which a • DUI citation is g iven.

SUMMARY =

BAC

of 0.096 % or more should be arrested and charged •

Properties of the Logarithmic Function

logax, a> 1 (y = loga x means x aY)

f(x)

A

driver with a with DUI.

=

loga x, 0 < a < 1 (y = loga x means x = aY)

f(x) =

Domain: the interval (0, 00 ); range: the interval ( -00 , 00) x-intercept: 1; y-intercept: none; vertical asymptote: x 0 (y-axis); increasing; one-to-one See Figure 39(a) for a typical graph. Domain: the interval (0, ) range: the interval ( -00, (0) x-intercept: 1; y-intercept: none; vertical asymptote: x = 0 (y-axis); decreasing; one-to-one See Figure 39(b) for a typical graph. =

(0 ;

Figure 39

y 3

Y 3

: X= 0

-3 -3 (a)

-3

I

: X= 0

a>

1

(b)

0
0 (pp. 314-316) Concepts and Vocabulary 4. The domain of the logarithmic function f(x) = log" x is

7. True or False 8. True or False

5. The graph of every logarithmic function f(x) = log" x,

a >

0,

y

=

log"

x, then y

The graph of

f(x)

=

aX.

= log"

x, a >

has an x-intercept equal to1 and no y-intercept.

*- 1, passes through three points: __, __, and

a

If

6. If the graph of a logarithmic function f(x) = log" x, 0,

a >

than

a

*-

1,

is increasing, then its base must be larger

Skill Building

In Problems 9-16, change each exponential expression to an equivalent expression in volving a logarithm. 9. 9

13. 2x

=

32

=

10. 16

=

42

11.

14. 3x = 4. 6

7. 2

a2

12. a3 = 2.1

1 .6

=

15. eX = 8

16. e2.2 = M

In Problems 1 7-24, change each logarithmic expression to an equivalent expression involving an exponent.

{�)

17. log28 = 3

18.10g

21. log3 2 = x

22. log 6

2

=

= -2

19. log" 3 = 6

20.10gb 4

x

23. In 4

24. In x = 4

=

x

In Problems 25-36, find the exact value of each logarithm without using a calculatOl:

=

2

(i)

25. log21

26. logs 8

27. logs 25

28.10g

29. log'!216 33. logv'2 4

30. log'!39

31. 1 0g lO ViO

32.10gsV2s

34.10gy'39

3

36. In e

35.1nVe

3

In Problems 37-48, find the domain of each function. 37. f(x) 40. H(x)

=

I n ( x- 3)

38. g ( x) = In (x - 1)

logs

41. f(x)

=

3- 210g4

44. g(x)

=

l_ In _ x- 5

47·f(x)

=

=

x3

(x + ) C� )

43. f(x) = In _l_ 1

46. hex)

=

IOg3

1

(

[1 ] ) - 5

�

8 + 5 In(2x + 3)

42. g(x)

=

45. g(x)

=

48. g(x)

=-

log5

In

l

(X-+x- )

x

In Problems 49-56, use a calculator to evaluate each expression. Round your answer to three decimal places.

49.

5 In]'

53.

----

In 4 + In 2

log4 + log 2

10

50

•

54.

In-

�

51.

--=----=

55.

3

log 15+ log 20 In15+ In 20

57. Find a so that the graph of f( x)

=

log"

x contains the point ( 2, 2).

58. Find a so that the graph of f(x)

=

log"

x contains the point

3

52.

0 . 04

21n 5+ log 50

----=-

56.

log4 - In 2

(�, )

-4 .

In Problems 59-62, graph each fun ction and its inverse on the same Cartesian plane. 59. f(x) = 3'\ r' (x) = log3 x

60. f(x) = 4\rl (x)

=

log4

x

1

2 In3 -0 .1

310g80 - In 5

-=-----.,--

log 5 + In

-

20

0,

a

*- 1,

SECTION 6.4

61. f(x) =

(1)X "2

;r 1(x) = loglx

(l)X;rl(x)

62. f(x) =

� J

=

447


logp: '

In Problems 63-70, the graph of a logarithmic function is given. Match each graph to on e of the following functions: A. E.

Y = log3X

Y = IOg3x-I

F.

3 rx x -3 Y

63.

=

0

x=O3 -3

= IOg3(X- 1)

y

31- :x -3 Yt

64.

y

= -log3x

D.

G. Y = log3(1 - x)

I I

1

67.

c.

Y = IOg3(-x)

B.

YI

65. = 1

H.

66.

_�o

68.

----'----

69.

5x

-3 1-

-1

-1

y

1

= - log3x

3 �x=O L t � ytx=3 1 Yt

,..

5

-1

-1

Y

Y =-log3(-x)

3 rx= -3 Y

-3 �

x

-3 70.

0

-3

-1

In Problems 71-86, use the given function f to: (a) Find the domain of! (b) Graph! (c) From the graph, determine the range and any asymptotes of! (d) Find r1, the in verse of! (e) Use f-1 to find the range of! (f) Graph rl.

73. f(x) = 2 +Inx

74. f(x)

76. f(x) =-21n (x +1)

77. f(x) = log(x- 4) +2

78. f(x) =- Iogx- 5 2

80. f(x) = loge -2x)

81. f(x) = 3 + log3(X

82. f(x)

84. f(x) = 3ex +2

3 85. f(x) = 2 x/ + 4

86. f(x) =-3 x+1

89. IOg (2x + 1) = 3 2

90. log3(3x- 2) = 2

93.In e = 5

94.In e-2x = 8

71. f(x) = In(x + 4)

72. f(x)

75. f(x) = In(2x)- 3 79. f(x) =

1

"2

log(2x)

83. f(x) = ex+2- 3

=

In(x- 3)

In Prob lems 87-11 0, solve each equation. 87. log3x = 2

88. log5 x = 3

91. log 4 x

92. IOg

2

=

103. log3(X2+ 107. 5eO.2x

=

1)

(�)

=

3

96. logs625 = x

95. IOg464 = x

99. e3x = 10

x

100. e-2x = =2

111. SupposethatG(x)

+

108. 8·102x-7

=

IOg3(2x +

(a) What is thedomain of G?

x + 4) = 2 =

3

1).

(b) What i s G(40)? What point is on the graph o f G? (c) If G(x) = 2, what is x? What point is on the graph of G?

97. log3243 = 2x + 101. e2x+S

3

104. logs(x 2

7

.!.

X

=

=

-In(-x)

=

98. log636 102. e- 2x +1

8

105. IOg 8x = -3 2 109.2 _102-x

1

+ 2)

=

5

1

2- IOg3(X + 1)

= =

106. log3 3x = 110. 4 ex+1

=

5x +3

13 -1

5

112. Supposethat F(x) = IOg (X+ 1)- 3. 2 (a) What is thedomain of F ?

(b) What is F(7)? What point is on the graph of f? (c) If F(x) = -1, what is x? What point is graph of F ?

on

the

448

CHAPTER 6

{ { -In x


{ {ln x

In Problems 113-116, graph each function. Based on the graph, state the domain and the range and find any intercepts. if x :5 - 1 ln (- X) In (-X) if x < 0 114. f (x ) = 113. f(x) = - l n ( -x ) if - 1 < x < 0 In x if x > 0 115. f ( x)

=

In x

if 0 < x < 1 if x 2: 1

116. f (x)

=

- In x

if 0 < x < 1 if x 2: 1


The pH of a chemical solution is given by the

117. Chemistry formula

where [H +] is the concentration of hydrogen ions in moles per liter. Values of pH range from 0 (acidic) to 14 (alkaline). (a) What is the pH of a solution for which [H+] is 0.1? (b) What is the pH of a solution for which [H+] is 0.01? (c) What is the pH of a solution for which [H+] is 0.001? (d) What happens to pH as the hydrogen ion concentration decreases? (e) Determine the hydrogen ion concentration of an orange (pH 3.5 ) . (f) Determine the hydrogen i o n concentration o f human blood (pH = 7.4 ). =

118. Diversity Index Shannon's diversity index is a measure of the diversity of a population. The diversity index is given by the formula H

=

- (PI log PI

+ P2 10g P2 + . . . +

Pl1log PH )

where PI is the proportion of the population that is species 1 , P 2 i s the proportion o f the population that is species 2 , and so on. (a) According to the U.S. Census B ureau, the distribution of race in the United States in 2000 was as follows: Proportion

Race American Indian or Native

0.014

Alaskan

0.041

Asian Black or African American

0.128

Hispanic

0.124

Native Hawaiian or Pacific Islander

0.003

White

0.690

Source: u.s.

Census Bureau

Compute the diversity index of the United States in 2000. (b) The largest value of the diversity index is given by logeS) , where S is the number of categories of Hmax race. Compute Hmax' =

(c) The evenness ratio is given by Ef{

=

H

Hmax

--

, where

0:5 EN:5 1 . If Ef{ = 1 , there is complete evenness. Compute the evenness ratio for the United States. (d) Obtain the distribution of race for the United States in 1 990 from the Census B ureau. Compute Shannon's di versity index. Is the United States becoming more di verse? Why?

119. Atmospheric Pressure TIle atmospheric pressure p on a bal loon or an aircraft decreases with increasing height. This pres sure, measured in millimeters of mercury, is related to the height h (in kilometers) above sea level by the formula O 45h P 760e- .1 (a) Find the height of an aircraft if the atmospheric pres sure is 320 millimeters of mercury. (b) Find the height of a mountain if the atmospheric pres sure is 667 millimeters of mercury. =

120. Healing of Wounds The normal healing of wounds can be modeled by an exponential function. If Ao represents the original area of the wound and if A equals the area of the wound, then the formula A Aoe-0.3511 =

describes the area of a wound after n days following an injury when no infection is present to retard the healing. Suppose that a wound initially had an area of 100 square millimeters. (a) If healing is taking place, after how many days will the wound be one-half its original size? (b) How long before the wound is 10% of its original size?

121. Exponential Probability Between 1 2:00 PM and 1 :00 PM, cars arrive at Citibank's drive-thru at the rate of 6 cars per hour (0.1 car per minute). The following formula from statistics can be used to determine the probability that a car will arrive within t minutes of 12:00 PM. F(t ) 1 e-O.II =

-

(a) Determine how many minutes are needed for the prob ability to reach 50%. (b) Determine how many minutes are needed for the prob ability to reach 80% . (c) I s it possible for the probability t o equal 100% ? Explain.

122. Exponential Probability Between 5 :00 PM and 6:00 PM, cars arrive at Jiffy Lube at the rate of 9 cars per hour (0.15 car per minute). The following formula from statistics can be used to determine the probability that a car will arrive within t minutes of 5:00 PM. F(t ) = 1 - e-O.151 (a) Determine how many minutes are needed for the prob ability to reach 50 % . (b) Determine how many minutes are needed for the prob ability to reach 80% . 123. Drug Medication The formula D

=

5e-0.4h

can be used to find the number of milligrams D of a certain drug that is in a patient's bloodstream h hours after the drug was administered. When the number of milligrams reaches 2, the drug is to be administered again. What is the time between injections?

SECTION 6.4

124. Spreading of Rumors A model for the number N of people in a college community who have heard a certain rumor is N

=

126. Learning Curve

P ( l - e-0.J5d)

449

Psychologists sometimes use the function L(t)

=

A( l - e -kl)

to measure the amount L learned at time t. The number A represents the amount to be learned, and the number k mea sures the rate of learning. Suppose that a student has an amount A of 200 vocabulary words to learn. A psychologist determines that the student learned 20 vocabulary words after 5 minutes. (a) Determine the rate of learning k. (b) Approximately how many words will the student have learned after 10 minutes? (c) After 15 minutes? (d) How long does i t take for the student to learn 1 80 words?

where P is the total population of the community and d is the number of days that have elapsed since the rumor began. In a community of 1000 students, how many days will elapse before 450 students have heard the rumor? 125. Current in a RL Circuit The equation governing the amount of current 1 (in amperes) after time t (in seconds) in a simple RL circuit consisting of a resistance R (in ohms), an induc tance L (in henrys), and an electromotive force E (in volts) is I =


.£ [ 1 - e -(R/L)I] R

If E = 12 volts, R = 10 ohms, and L = 5 henrys, how long does it take to obtain a current of 0.5 ampere? Of 1.0 am pere? Graph the equation.

-

Problems 127-130 use the following discussion: The loudness L( x), measured in decibels, of a sound of intensity x, x measured in watts per square metel; is defined as L ( x) = 10 log , where 10 = 10- 1 2 watt per square meter is the least intense sound

Loudness of Sound

10

that a human ear can detect. Determine the loudness, in decibels, of each of the following sounds. 3 127. Normal conversation: intensity of x = 10- 7 watt per square 129. Heavy city traffic: intensity of x = 10- watt per square meter. meter. 130. Diesel truck traveling 40 miles per hour 50 feet away: inten 128. Amplified rock music: intensity of 1 0- 1 watt per square meter. sity 10 times that of a passenger car traveling 50 miles per hour 50 feet away whose loudness is 70 decibels.

The Richter Scale Problems 131 and 132 use the following discussion: The Richter scale is one way of converting seismographic readings into numbers the Richter scale that provide an easy reference for measuring the magnitude M of an earthquake. All earthquakes are com pared to a zero-level earthquake whose seismographic reading measures 0.001 millimeter at a distance of 1 00 kilometers from the epicenteJ: An earthquake whose seismogra phic reading measures x millimeters has magnitude M( x), given by M( x)

= 10g

where Xo = 10- 3 is the reading of a zero-level earthquake the same distance from its epicenter. In Problems 131 and 132, determine the magnitude of each earthquake.

131. Magnitude of an Earthquake Mexico City in 1985: seismo graphic reading of 125,892 millimeters 100 kilometers from the center. 132. Magnitude of an Earthquake San Francisco in 1 906: seis mographic reading of 7943 millimeters 100 kilometers from the center. 133. Alcohol and Driving The concentration of alcohol in a per son's bloodstream is measurable. Suppose that the relative risk R of having an accident while driving a car can be mod eled by the equation

where x is the percent of concentration of alcohol in the bloodstream and k is a constant.

( �) Xo

(a) Suppose that a concentration of alcohol in the blood stream of 0.03 percent results in a relative risk of an ac cident of 1 .4. Find the constant k in the equation. (b) Using this value of k, what is the relative risk if the con centration is 0.17 percent? (c) Using the same value of k, what concentration of alco hol corresponds to a relative risk of 100? (d) If the law asserts that anyone with a relative risk of hav ing an accident of 5 or more should not have driving privileges, at what concentration of alcohol in the blood stream should a driver be arrested and charged with a DUI? (e) Compare this situation with that of Example 10. If you were a lawmaker, which situation would you support? Give your reasons.

450

CHAPTER 6



134. Is there any function of the form y x G', O < a < 1 , that in creases more slowly than a logarithmic function whose base is greater than I? Explain. =

Age in Years

2

3

4

5

$32,400

$28,750

$25,400

$21 ,200

New

135. In the definition of the logarithmic function, the base a is not

$38,000

$36,600

allowed to equal 1. Why?

Use the formula New = Old (eRt) to find R, the annual de preciation rate, for a specific time t. When might be the best time to trade in the car? Consult the NADA ("blue") book and compare two like models that you are interested in. Which has the better depreciation rate?

136. Critical Thinking In buying a new car, one consideration might be how well the price of the car holds up over time. Different makes of cars have different depreciation rates. One way to compute a depreciation rate for a car is given here. Suppose that the current prices of a certain Mercedes automobile are as follows: 'Are You Prepared?' Answers

1. x ::; 3 '�'-

3. x < -4 or x > 1

2. x < -2 or x > 3

� ..

6.5 Properties of Logarithms OBJECTIVES 1 Work with the Properties of Logarithms (p.450)

2 Write a Logarithmic Expression as a Sum or Difference of Logarithms (p.452)

3

Write a Logarithmic Expression as a Single Logarithm (p.453)

4 Evaluate Logarithms Whose Base Is Neither 10 nor e (p.454)

1

E XA M PLE 1

Work with the Properties of Logarithms

Logarithms have some very useful properties that can be derived directly from the definition and the laws of exponents. Establishing Properties of Logarithms

(a) Show that loga 1 = o. (b) Show that loga a 1. (a) This fact was established when we graphed y = loga (see Figure 30). To show the result algebraically, let y loga 1 . Then y = loga 1 =

Solution

x

=

Change to an exponential expression. aO = 1

y =o

(b) Let y

=

loga 1 = 0 loga a. Then y = loga a

Solve for y.

y

aY = a aY = a 1 y= 1 loga a = 1

=

loga 1

Change to an exponential expression. a a1 =

Solve for y.

y

=

logaa

To summarize: loga 1

=

0

loga a = 1

•

SECTION 6.S

THEOREM

45 1

Properties of Logarithms

Properties of Logarithms =1=

r

In the properties given next, M and a are positive real numbers, a 1, and is any real number. The number loga M is the exponent to which a must be raised to obtain M. That is, (1)

I

The logarithm to the base a of a raised to a power equals that power. That is, logaaf (2) I --

�----------

= r

----

--------

y

=

--

--

.�

--

�

----

y =

The proof uses the fact that aX and loga x are inverses. Proof of Property (1) For inverse functions, .f(r'(x)) x for all x in the domain of r' Using f(x) aX and r1(x) toga we find f(r'(x)) a'ogax x for x > 0 Now letx M to obtain a'og{/I'vl M, where M > o. Proof of Property (2) For inverse functions, rl(f(x)) x for all x in the domain of f Using f(x) aXand rl(x) !og"x, we find r' (f(x)) logac["' x for all real numbers x Now let x to obtain loga af where r is any real number. =

=

=

X,

=

=

=

=

•

=

=

=

=

= r

E XA M PLE 2

=

= r,

•

Using Properties (1) and (2) 2'og21T

(a) L'Ji

-

=

1T

(b) log0 2 o.rV2

=

-

v2

(c) In

ekl =

kt

•

Now Work PRO B L E M 9

Other useful properties of logarithms are given next. THEOREM


In the following properties, M, N, and a are positive real numbers, a is any real number.

r

=1= 1,

and

The Log of a Product Equals the S um of the Logs

(3) The Log of a Quotient Equals the Difference of the Logs

(4)

452

CHAPTER 6


The Log of a Power Equals the Prod uct of the Power and the Log

logaMr = r logaM

(5)

I

--------------------------------��

�

We shall derive properties (3) and (5) and leave the derivation of property (4) as an exercise (see Problem 103). Proof of Property (3) Let A = logaM and let B = logaN. These expressions are equivalent to the exponential expressions A a = M and aB = N Now loga(MN) = loga( Aa ) = loga aA+B Law of Exponents a

B

= A + B = logaM

Proof of Property (5)

Now

+ logaN Let A = logaM. This expression is equivalent to A a =M

loga M r = loga (aA) r = loga arA

= rA = r logaM

� == » -

2

Property (2) of logarithms •

Law of Exponents Property (2) of logarithms •


Write a Logarithmic Expression as a Sum or Difference of Logarithms

Logarithms can be used to transform products into sums, quotients into differences, and powers into factors. Such transformations prove useful in certain types of cal culus problems. EXAMPLE 3

Solution

Writing a Logarithmic Expression as a Sum of Logarithms

Write loga (x�i) ,x> 0, as a sum of logarithms. Express all powers as factors. loga (xP+1) = logax + loga P+1 logA . N) logaM + logaN = logax + loga(x2 + 1) 1 /2 1 = logax + "2 loga(x- + 1) loga loga =

M

?

Mr

= r

M

•

EXAMPLE 4

Writing a Logarithmic Expression as a Difference of Logarithms

Write

x2--:: x> 1 I n -(x 1)3 as a difference of logarithms. Express all powers as factors. -

Properties o f Logarithms

SECTION 6.5

453

In (x x2- 1) 3 Inx2 - In(x - 1)3 2 1nx - 3 1n(x -1)

Solution

=

i

i

=

loga

(�) =

loga M - loga N

loga M'

= r

logaM •

Writing a Logarithmic Expression as a Su m and Difference of Logarithms

E XA M P L E 5

Write as a sum and difference of logarithms. Express all powers as factors. , � loga xyQ"+l 3 (x + 1)4 loga V + 1 -loga[x3 (x + 1)4]

WARNING

I n using properties (3) through (5), be careful about the values that the variab le may assume. For example, the domain of the variable for loga x is x > 0 and for loga(x - 1) it is x > 1 . If we add these func tions, the domain is x > 1. T hat is, the equality loga x + loga(x - 1) = 10gA x(x - 1)]

is true only for x > 1 .

= =

=

3

Solution

=::;;:;;:>

�

•

E XAMPLE 6

Xk

=

Solution

-

Property (4)

loga yQ"+l - [logax3 + loga(x + 1)4] loga(x2 + 1)1/2 - logax3 - log(,(x + 1)4 1 2 loga(x2 + 1) - 3 loga - 4 loga( + 1) X

X

Property (3)

Property (5) •


Write a Logarithmic Expression as a Single Logarithm

Another use of properties (3) through (5) is to write sums and/or differences of log arithms with the same base as a single logarithm. This skill will be needed to solve certain logarithmic equations discussed in the next section. Writing Expressions as a Single Logarithm

Write each of the following as a single logarithm. (a) loga 7 + 4 logo 3 (b) "32 1n 8 - In(34 - 8) (c) logax + loga 9 + loga(x2 + 1) - loga 5 (a) loga 7 + 4 logo 3 logo 7 + loga 34 logaM logaM' logo 7 + loga 81 logo (7' 81) logo 567 (b) "32 In 8 - In(34 - 8) In 82/3 - In( 81 - 8) rlogaM logaM' In 4 - In 73 In ( 7�) r

=

=

= =

=

=

=

=

=

454

CHAPTER 6


• WARNING

A common error made by some students is to express the logarithm of a sum as the sum of logarithms. loga( M

Correct statement

is not equal to

+ N)

loga( MN)

=

loga M

+

loga M

+

loga N

Property (3)

loga N

Another common error is to express the d ifference of logarithms as the quotient of logarithms.

Correct statement

M-

loga M

loga N

loga

loga N

is not equal to =

(�)

IOga

loga M loga N

Property (4)

A third common error is to express a logarithm ra ised to a power as the product of the power times the logarithm. is not equal to

(Ioga MY Correct statement

�:;;:::::.

-

loga Mr

= r loga M

r

loga M

•

Property (5)


Two other properties of logarithms that we need to know are consequences of the fact that the logarithmic function loga x is a one-to-one function. y =

THEOREM

Properties of Logarithms =f.

In the following properties, M, N, and a are positive real numbers, a 1. If M N, then loga M loga N. (6) If loga M loga N , then M N . (7) =

=

=

=

�------�� =

When property (6) is used, we start with the equation M N and say "take the logarithm of both sides" to obtain loga M loga N . Properties (6) and (7) are useful for solving exponential and logarithmic equa tions, a topic discussed in the next section. =

4

Evaluate Logarithms Whose Base Is Neither 1 0 nor

e

Logarithms to the base 10, common logarithms, were used to facilitate arithmetic computations before the widespread use of calculators. (See the Historical Feature at the end of this section.) Natural logarithms, that is, logarithms whose base is the number e, remain very important because they arise frequently in the study of nat ural phenomena. Common logarithms are usually abbreviated by writing with the base understood to be 10, just as natural logarithms are abbreviated by with the base understood to be e. Most calculators have both � and � keys to calculate the common loga rithm and natural logarithm of a number. Let's look at an example to see how to approximate logarithms having a base other than 10 or e. log,

In,

S ECTION 6.S

EXAMPLE 7 Solution

Properties o f Logarithms

Approximating a Logarithm Whose Base Is Neither 1 0 nor

455

e

Approximate log2 7. Round the answer to four decimal places. Let log2 7. Then 2Y 7, so 2Y 7 In 2Y In 7 Property (6) y in 2 In 7 Property (5) In 7 Exact va lue In 2 2.8074 Approximate val ue rounded to four y=

=

=

=

=

y= y

�

decima l places

•

Example 7 shows how to approximate a logarithm whose base is 2 by changing to logarithms involving the base In general, we use the e.

THEOREM

Change-or-Base Formula.

Change-of-Base Formula

If a i= 1, b i= 1, and M are positive real numbers, then (8)

I

�

�----------------------------------�

Proof

y=

We derive this formula as follows: Let logoM. Then aY M 10gb aY 10gbM Property (6) 10gb a 10gbM Property (5) 10gbM -- Solve for 10gb a 10gaM 10gbM loga M 10gb a =

=

=

y

y=

y.

=

y =

•

Since calculators have keys only for � and � , in practice, the Change-of Base Formula uses either b 10 or b That is, =

= e.

M and log M In M logaM log -1-og In a =

E XAMPLE 8

Solution

a

a

Using the C hange-of-Base Form ula

Approximate: (a) logs 89 (b ) 10gy'2 Vs Round answers to four decimal places. log 89 1.949390007 2.7889 ( a ) logs 89 log 5 0.6989700043 or 4.48863637 2 .7889 log-) 89 InIn895 1.609437912 =

=

�

�

�

�

__

(9)

456

CHAPTER 6


,r;

b logyz v5

( )

1 log 5 2.3219 log V 5 -2 log 5 12 log V 2 l. log 2 og 2 ,

=

r; ;:::.

,

=

-- =

�

or logyz V 5

,r;

C'l:: == _ -

COMMENT

1 In 5 In V 5 = 2 In 5 2.3219 ;:::. In In V 2 l. In 2 2 2 , r;

=

,

--

=

-

�

Now Work P R O B L E M S , 7 AND 6 5

To graph logarithmic functions when the base is different from

Change-of-Base Formula. For exam ple, to g raph y = log2

0�

C'l::==" -

SUMMARY

•

x,

e

or 10 requi res the In x we would instead g raph y = hl'

•



In the list that follows, a, b, M, N, and are real numbers. Also, a> 0, a *- 1, b> 0, b *- 1, M > ° and N > 0. aY y = loga means loga 1 0; toga a 1 logaMr = r logaM 1a 0gaM = M; loga ar = r If M N, then logaM logaN. 10ga(MN) = logaM logaN If logaM = logaN, then M N. lOga( �) logaM -logaN M 1-10gaM -10gb ogb a r

x

Definition

Properties of logarithms

x

=

=

=

=

=

+

=

=

=

Change-of-Base Formula

I-Hs:torical Feature

L John

Napier (1550-1617)

ogarithms were invented about 1 590 by

tion 6.4). Napier's tables, p u blished in 1 6 1 4, listed what would now

John Napier (1 550- 1 6 1 7) and Joost Burgi

be ca lled natural logarithms of sines and were rather difficult to use. A

(1 552- 1 632),

independently.

London professor, Henry Briggs, became interested in the tables and vis

N a pier, whose work had the greater influence,

ited Napier. I n their conversations, they developed the idea of common

working

was a Scottish lord,a secretive man whose neigh

logarithms, which were published in 1 61 7. Their importance for calcula

bors were inclined to believe him to be in leag ue

tion was immediately recognized, and by 1 650 they were being printed

with the devil. His approach to logarithms was

as far away as China. They remained an important calculation tool until

very different from ou rs; it was based on the re-

the advent of the inexpensive handheld calculator about 1972,which has

lationship between arithmetic and geometric

decreased their calculational, but not their theoretical, importance.

sequences, discussed in a later cha pter, and not on the inverse function relationship of logarithms to exponential functions (described in Sec-

A side effect of the invention of logarithms was the popularization

of the decimal system of notation for real numbers.

457


SECTION 6.5

6.S Assess Yo u r Understanding Concepts and Vocabulary 1. The logarithm of a product equals the logarithms.

logs 7

-- ' 1 ogs 8

2. Hlogs M = 3. loga Mr

=

__

__

of the

5. True or False log2(3x4)

then M =

6. True or False logz16

.

In(x + 3)

In(x + 3) - In(2x) =

4. True or False

=

In(2x)

4 log2 (3x)

In 16

= --

In

2

Skill Building

In Prob lems 7-22, use p rop eriies of logarithms to find the exact value of each expression. Do not use a calculatO/: 13 -4 · 15 8. log2 T 7. log 3371 ', 9. In e 10. In eV" 12. e1n 15. log6 18 - log63

s

16. logs 16 - logs 2

In Problems 23-30, suppose that

In 2

=

a and

In 3 =

',13. logs 2 + logs 4

14. log69 + log64

.,17. log2 6 ·log64

18. log3 8 . logs

9

b . Use properties of logarithms to write each logarithm in terms of a and b.

2

23. I n 6

24. In '3

25. In 1.5

26. In 0 .5

27. In 8

28. In 27

29. In

30. ln

'#6

�

In Prob lems 31-50, write each expression as a sum and/or differen ce of logarithms. Express po wers as factors. x

31. logs(25x)

32. log 3'9

35. In(ex)

36. In�

x

40. IOg2

43. log2 47. In

49. In

[

(--) x

3

x-

3

X2 - x - 2 2 ( x + 4)

5 x V1+3x (x

- 4)

x> 3

J

3

3 1/

44. logs

x e'

38.lnr

37. In( xeX)

(; ) ( ' 3�)

2

41. In(x �)

a> O,b > 0

2

V r+1 2

x -I

x> 1

'

. 45.

[ [ [

log

X(X + 2)

(x

2

48. In -'---2----'-- 5O. In

x>4

JJ

+ 3)

(x - 4)

x>2

0

x -I

<x 0

2

2/3

5X2� ? 4(x + 1 )-

[

X3�

( x - 2)

x>4

J

x> 0

42. In x

2

J

x>2

O<x

=F

0, a 1

For example, to solve the equation log2 (1 2x) = 3, we use the equivalent expo nential expression 1 2x = 23 and solve for x. log2( 1 2x) = 3 1 2x = 23 Change to an exponential expression. -

-

-

-

-2x = 7 X=

Simplify.

7 2

--

Solve.

You should check this solution for yourself. For most logarithmic equations, some manipulation of the equation (usually using properties of logarithms) is required to obtain a solution. Also, to avoid extra neous solutions with logarithmic equations, we determine the domain of the vari able first. We begin with an example of a logarithmic equation that requires using the fact that a logarithmic function is a one-to-one function. If logaM logaN, thenM N M, N, and a are positive and a 1 =

EXAMPLE 1 Solution

=F

=

Solvi ng a Logarithmic E quation

Solve: 2 logs x = logs 9 The domain of the variable in this equation is x> O. Because each logarithm is to the same base, 5, we can obtain an exact solution as follows: 2 logs x = logs 9 loga loga logs x2 = logs 9 x2 = 9 If loga M loga N, then N. x = 3 or � Reca ll that the domain of the variable is x > Mr = r =

M

M =

o.

Therefore, -3 is extraneous and we discard it.

460

CHAPTER 6


Check:

2 logs 3

J,

logs 9 logs 32 J, logs 9 logs 9 logs 9 The solution set is {3}.

r

loga M = loga Mr

=

mt=- -

•

Now Work PR O B L E M 1 3

In the next example we employ the log of a product property to solve a loga rithmic equation. EXAMPLE 2

Solving a Logarithmic E quation +

Solution

=

Solve: logs(x + 6) logs(x + 2) 1 The domain of the variable requires that x + 6 > 0 and x + 2 > 0 so x > -6 and x > -2. This means any solution must satisfy x > -2. To obtain an exact solution, we need to express the left side as a single logarithm. Then we will change the expression to exponential form. logs(x + 6) + logs(x + 2) 1 logs[(x + 6)(x + 2)J 1 loga M + loga N = loga (MN) (x + 6)(x + 2) Sl S Change to an exponentia l expression. x2 + 8x 12 S Simplify. x2 + 8x + 7 0 P lace the quad ratic equation in standard form. (x + 7)(x + 1 ) 0 Factor. x -7 or x -1 Zero-Product Property Only x -1 satisfies the restriction that x > -2, so x -7 is extraneous. The so lution set is {-1}, which you should check. = =

=

+

=

= =

WARNING

A negative sol ution is not a utomatica l ly extra neous. You m u st determine whether the potential solu tion causes the a rg u ment of any loga rithmic expression in the equation to be negative. _

=

=

=

Solution

=

•

,"'l!l;: =� �

EXA M P L E 3

=

Now Work PR O B L E M 2 1

Solving a Logarithmic E quation =

Solve: In x + In(x - 4) In(x + 6) The domain of the variable requires that x > 0, x 4 > 0, and x + 6 > O. As a result, the domain of the variable here is x > 4. We begin the solution using the log of a product property. In x + In(x 4) In(x + 6) In[ x(x - 4)J In(x + 6) I n M + I n N I n (MN) x(x - 4) x + 6 If In M In N, then M N. x2 - 4x x 6 Si m pl ify. x2 - Sx - 6 0 P lace the quad ratic equation in sta nda rd form. (x - 6)(x + 1) 0 Factor. 6 or x - 1 Zero-Prod uct Property Since the domain of the variable is x > 4, we discard -1 as extraneous. The solu tion set is 161, which you should check. -

-

=

=

=

=

=

=

X =

+

=

=

=

=

��"-=�- Now Work PR O B L E M 3 1

•

SECTION 6.6

2

Logarithmic and Exponential Equations

461

Solve Exponential Equations

In Sections 6.3 and 6.4, we solved exponential equations algebraically by expressing each side of the equation using the same base. That is, we used the one-to-one prop erty of the exponential function: If aU aV, then a > O, a *- 1 =

u

= v

For example, to solve the exponential equation 42x+1 16, we notice that 16 42 and apply the property above to obtain 2x + 1 2, from which we find �. For most exponential equations, we cannot express each side of the equation using the same base. In such cases, algebraic techniques can sometimes be used to obtain exact solutions. When algebraic techniques cannot be used, we use a graph ing utility to obtain approximate solutions. You should pay particular attention to the form of equations for which exact solutions are obtained. In the next example we solve two exponential equations by changing the expo nential expression to a logarithmic expression. =

=

EXAMPLE 4

x =

Solving an Exponential Equation =

Solve: (a) 2x 5 (b) 8 · 3x 5 (a) Since 5 cannot be written as an integral power of 2, we write the exponential equation as the equivalent logarithmic equation. 21' 5 x log 5 2 Now use the Change of Base Formula (9) on page 455. In 5 Exact solution x = In 2 =

Solution

=

=

=

In 5- } . . set IS. { The solutIOn In 2 (b) 8 · 3x 5 . 5 3"

:::; 2.322

Approximate sol ution

=

= -

8

Solve for Y.

Exact solution

In 3 :::; -0.428

Approximate solution

{ (�) }.

In The solution set is In 3

•

",,,

EXAMPLE 5

;:.. - Now Work P R O B L E

M

35

Solving an Exponential E quation

Solve:

y-2

=

33x+ 2

462

CHAPTER 6


Because the bases are different, we first apply Property (6), Section 6.5 (take the natural logarithm of each side), and then use a property of logarithms. The result is an equation in x that we can solve.

Solution

5x -2 5x - 2

=

33x +2

+

In 33x 2 In ( 2) In 5 (3x + 2) In 3 (ln 5)x - 2 1n 5 = (3 1n 3)x + 2 1n 3 (In 5)x - (3 In 3)x 2 In 3 + 2 in 5 (In 5 3 1n 3)x = 2(1n 3 In 5) 3 + In 5 ) x = 2(in In S - 3 1n 3 -3.212 3 + In 5 ) } The solution set is { 2(In In 5 - 3 In 3 . x

=

=

-

=

+

-

�

If M = N, In

M

= I n N.

In Mr = r In M

Distribute. Place terms i nvolving

x

on the left.

Factor. Exact solution Approxi mate solution

•

"""' - Now Work P R O B L E M 4 5

'l'l

The next example deals with an exponential equation that is quadratic in form. E XA M P L E 6

Solving an Exponential Equation That Is Quadratic in Form

Solve: 4x - 2x - 12 0 We note that 4'" (22 ) X 2(2x) = (2x /, so the equation is quadratic in form, and we can rewrite it as (2x) 2 - 2x - 12 0 Let u = then u2 - u - = Now we can factor as usual. (2X - 4)(2X + 3) 0 (u - 4)(u + 3) = 0 2X - 4 0 or 2x + 3 = 0 u - 4 = 0 or u + 3 = 0 2x 4 2x -3 u= =4 u= = -3 The equation on the left has the solution x 2, since 2x 4 22; the equation on the right has no solution, since 2x 0 for all x. The only solution is 2. The solution set is {2}. =

=

Solution

=

=

12

2x;

o.

=

=

>

.� "'=

3

2X

=

=

=

2x

=

=

- - Now Work P R O B L E M S 3

•

Solve Logarithmic and Exponential Equations Using a G raphing Utility

The algebraic techniques introduced in this section to obtain exact solutions apply only to certain types of logarithmic and exponential equations. Solutions for other types are usually studied in calculus, using numerical methods. For such types, we can use a graphing utility to approximate the solution. EXAMPLE 7

Solving Equations Using a G raphing Utility

Solve: x + eX = 2 Express the solution(s) rounded to two decimal places.

4

- -- ---------- �- .--.. -- . -

=

1

•

.

�---- "'--'"

= X + eX

1

The solution is found by graphing Y and Y 2 2. Since Y is an increasing function (do you know why?), there is only one point of intersection for Y 1 and Y2 . Figure 40 shows the graphs of Y1 and Y 2 . Using the INTERSECT command, the solution is 0.44 rounded to two decimal places.

Solution

Figure 40

463

Logarithmic and Exponential Equations

SECTION 6.6

'rr = ..__ -

Ir.tti:t"�.;:·:ti_:· ... o :-:=.'1'1;?;B�'1'1 �'(=;?; o


6.6 Assess You r U nde rsta nd i n g 'Are You Prepared?' A nswers are giv en at the end of thes e exercises. I f yo u get a wrong answel; read the pages Lis ted in red. 1.

Solve x2 - 7x - 30 O. (pp. 97-106) 2. Solve ( x + 3 ) 2 - 4 ( x + 3 ) + 3 = O. (pp. 11 9-1 2 1 )

6 3. Approximate the solution(s) to x3

=

= x2 - 5 using a graphing

utility. (pp. AS-AIO) � 4. Approximate the solution(s) to x3 - 2x + 2 graphing utility. (pp. AS-A lO)

0

=

using a

Skill Building

In Pro blems 5-32, s olve each logarithmic eq uation. Express irrarional sol utions in exact form and as a decimal rou nded to 3 decimal places.

8. IOg3 ( 3 x - 1 ) 11.

1

- 100, x 2 bJ

14. 2 IOg5 x

=

=

6. log (x + 6) 9. IOg4( X + 2)

2

2 2 100, bJ

12. -2 IOg4 x

17. log x + log(x + 1 5 )

=

=

. 21. log2 (x

1 - logs ( x + 4 )

26. In (x + 1 ) - I n x

=

24. log5 (x

31. log{[ (x - 1 ) - log{[ (x + 6 )

=

=

+

7)

+

=

=

2

+

19. log(2x 1

=

IOg2 (X + S)

=

32. logll x

2

1 ) = 1 + log(x - 2)

22. log6(X + 4) + log6 (X + 3 ) 25. In x + In(x

+

2)

=

4

28. IOg2 (X + 1 ) + IOg2 (X + 7)

2

30. IOg4(X2 - 9 ) - lo�(x + 3 )

log{[ (x - 2 ) - loga(x + 3 )

=

16. 2 log3 (x + 4) - IOg3 9

5

3 ) = 1 - IOg5 (X - 1 )

-1

= log5 3

. 13. 31og2 X = - log2 27

IOg4 9

27. log3(X + 1 ) + IOg3(X + 4 )

2

29. IOg l/3(X2 + x) - IOg1 /3 (X2 - x)

+

4

10. IOg5(2x + 3)

IOg4 S

18. log x + log (x - 21 )

2

20. log(2x) - log(x - 3 ) = 1 23. logs(x + 6)

=

=

=

7. log2 ( 5 x)

1

15. 3 IOg2 ( X - 1) + IOg2 4

3 IOg5 4

=

=

+ 1 0ga C x

- 2)

=

=

=

1

=3

3

loga(x + 4)

In Problems 33-60, s olve each exponential eq uation. Express irrational sol utions in exact form and as a decimal rounded to 3 decimal places. . 35. 2x = 10 36. 3x = 14 34. 5 - x = 25 33. 2x - 5 = S

37. S-x

=

41. 3 1 - 2x

1.2 =

38.

4x

TX

=

42. 2x + 1

45. 1 .2x (0.5 fx 2 49. 2 x + 2x - 12

1 .5

=

=

46. 0.3 1 + x 50. 32x

0

. 53. 1 6x + 4x+ 1 - 3 = 0 57. 3 · 4x + 4 · 2x + S

=

+ 3"

(_.)"')o )X

=

7 ' -x

44.

47. 7T 1 - x = eX 51. 3 2x + 3x + I - 4

.

58. 2 · 49x + 1 1 · 7x + 5

0

43.

1 72x - 1

- 2 = 0 54. 9x - 3x + 1 + 1 = 0

=

=

5 1 - 2x

40. 0.3 ( 402X )

55. 25x - S · 5x =

0

=

59. 4x - 10 · 4-x

=

=

),-x G3

=

=

5x

48. ex + = 7Tx 2 52. 22x + 2x + - 1 2

0

63. eX

=

67. In x 71.

+

1 ) - IOg4(X - 2 )

56. 36x - 6 · 6x

3

60. 3"' - 14 · Tx

-x

eX + In x

=

62. IOg2 (X - 1 ) - IOg6(X + 2 )

1

64. e2x

-x =

=

=

4

=

2

x + 2

68. I n (2x)

=

72. eX - In x

-x + 2

69. In x

=

73. e-x

4

= =

x3 - 1

In x

=

-16

!.{ In Pro blems 61-74, use a graphing utility to s olve each eq uation. Express yo ur answer rounded to two decimal places.

61. logs(x

0 .2

70. ln x 74. e-x

= =

-x2

- In x

=

=

-9 5

0

464

C H A PTER 6



In Prob lems 75-86, solv e each eq uation. Express irrational sol utions in exact form and as a decimal ro unded to 3 decimal places.

75. log2(x

+

1 ) - log4

X =

76. l og2 (3 x + 2) - log4 X

1

[Hint: Change log4 x to base 2.]

78. log9 x + 3 log3 X 81.

eX

+

e-x

2

=

14

79.

eX - e-x 2

=

3

77. IOg16 x + log4 X + log2 X

=

7

2X2

= 1

[Hint: Multiply each side by eX .] 84.

( � ) 2- X

=

=

85. log5 x + log3 X

-2

=

86. log2 x

1

+

log6 x

=

3

[Hint: Use the Change-of-Base Formula and factor out In x. ]

87. f ( x ) = l og2 (x + 3 ) and g(x ) log2 (3 x + 1 ) . (a) Solve f(x ) = 3. What point is on the graph of f? (b) Solve g(x) = 4. What point is on the graph of g? (c) Solve f(x ) = g(x ) . Do the graphs of f and g intersect? If so, where? (d) Solve f(x ) + g(x) = 7. (e) Solve f(x ) - g(x ) = 2 . =

88. [ ( x) = log3(x + 5 ) and g(x ) = log3 (x - 1 ) . (a) Solve f(x) = 2. What point is on the graph of f? (b) Solve g(x ) = 3. What point is on the graph of g? (c) Solve f(x) = g(x ) . Do the graphs of f and g intersect? If so, where? (d) Solve f(x) + g(x ) = 3 . ( e ) Solve f(x ) - g(x ) = 2.

89. (a) If f(x) = y+1 and g(x) = 2x + 2 , graph f and g on the same Cartesian plane. (b) Find the point(s) of intersection of the graphs off and g by solvin g f ( x ) = g(x) .Round answers to three decimal places. Label this point on the graph drawn in part (a). (c) Based on the graph, solve f(x ) > g(x ) .

90. ( a ) If f(x) = 5x- 1 a n d g(x ) = 2x+ l , graph f a nd g o n the same Cartesian plane. (b) Find the point(s) of intersection of the graphs of f and g by solving f(x) = g(x ) .Label this point on the graph drawn in part (a). (c) Based on the graph, solve f(x ) > g(x ) .

91. ( a) Graph f (x ) = 3 x and g (x ) = 1 0 o n the same Cartesian plane. (b) Shade the region bounded by the y-axis, f(x) = 3" and g(x ) = 10 on the graph drawn in part (a). (c) Solve f(x ) = g(x) and label the point of intersection on the graph drawn in part (b).

92. (a) Graph f(x) = 2x and g(x) = 1 2 on the same Cartesian plane. (b) Shade the region bounded by the y-axis,/ ( x) = 2-', and g(x) = 12 on the graph drawn in part (a). (c) Solve f(x ) = g(x ) and label the point of intersection on the graph drawn in part (b) . 93. (a) Graph f(x ) = 2x + 1 and g(x) Cartesian plane.

=

Tx+ 2 on the same

(b) Shade the region bounded by the y-axis, f(x ) = 2x+ \ and g(x) = Tx+2 o n the graph draw i n part (a). (c) Solve f(x) = g(x) and label the point of intersection on the graph drawn in part (b). 2 94. (a) Graph f(x) = rx+ 1 and g(x) = y - on the same Cartesian plane. (b) Shade the region bounded by the y-axis,f ( x) = rr + l , and g(x) = 3x - 2 on the graph draw in part (a). (c) Solve f(x ) = g(x) and label the point of intersection on the graph drawn in part (b). 95. (a) Graph I (x) = 2x - 4. (b) Based on the graph, solve f(x ) < O.

96. (a) Graph g(x) = 3x - 9. (b) Based on the graph, solve g(x) > O. 97. A Population Model The resident popUlation of the United States in 2006 was 298 million people and was growing at a rate of 0.9% per year. Assuming that this growth rate con tinues, the model p e t) = 298 ( 1 .009)' - 2006 represents the population P (in millions of people) in year t. (a) According to this model, when will the population of the United States be 310 million people? (b) According to this model, when will the population of the United States be 360 million people? Source:

2006.

S tati stical Ab stract of t he U ni ted States, 1 25th ed.,

SECTION 6.7

98. A Population Model The population of the world in 2006 was 6.53 billion people and was growing at a rate of 1.14% per year. Assuming that this growth rate continues, the model p e t ) = 6.53 ( 1 .0 1 1 4y-2oo6 represents the population P (in billions of people) in year t. (a) According to this model, when will the population of the world be 9.25 billion people? (b) According to this model, when will the population of the world be 1 1 .75 billion people?

Source:

465

100. Depreciation The value V of a Dodge Stratus that is t years old can be modeled by V e t ) = 19,282 (0.84t (a) According to the model, when will the car be worth $ 1 5 ,000? (b) According to the model, when will the car be worth $8000? (c) According to the model, when will the car be worth $2000?

Census Bu reau.

99. Depreciation The value V of a Chevy Cobalt that is t years old can be modeled by V e t ) = 14,5 12(0.82 t (a) According to the model, when will the car be worth $9000? (b) According to the model, when will the car be worth $4000? (c) According to the model , when will the car be worth $2000? Source: u.s.

Compound Interest

Source: Kelley B lu e Book

Kelley Blu e Book


101. Fill in reasons for each step in the following two solutions. Solve:

log3 ( x - 1 )2

=2

Solution A

Solution B

log3 ( x - 1 ) 2 = 2 (x - 1 )2 = 32 = 9 (x - 1 ) x - I

=

=

±3

log3( x - 1 )2 = 2 2 1 0g 3 ( x - 1 ) = 2

__

log3 ( x - 1) = 1

__

-3 orx -

1

=3

x -

x = -2 orx = 4

l

__

__

= 3i = 3

x = 4

Both solutions given in Solution A check. Explain what caused the solution x

=

-2 to be lost in Solution B.


1. { - 3, 10}

2. { -2, O}

3. { - 1 .43}

4. { - 1.77}

6.7 Compound Interest PREPARI NG FOR THIS SECTION •


Simple Interest (Section 1 .7, pp. 1 41-142) Now Work the A re You Prepared?' problems on page 472. '

OBJECTIVES 1 Determine the Future Value of a Lump Sum of Money (p.465)

2 Calculate Effective Rates of Return (p.469)

3

Determine the Present Value of a Lump Sum of Money (p.470)

4 Determine the Rate of Interest or Time Required to Double a Lump

Sum of Money (p.47 l )

1

Determine the Future Val u e of a Lump S u m of Money

Interest is money paid for the use of money. The total amount borrowed (whether by an individual from a bank in the form of a loan or by a bank from an individual in the form of a savings account) is called the The principal.

rate of interest,

466

CHAPTER 6


expressed as a percent, is the amount charged for the use of the principal for a given period of time, usually on a yearly (that is, per annum) basis. THEOREM

Simple Interest Formula

If a principal of P dollars is borrowed for a period of t years at a per annum interest rate r, expressed as a decimal, the interest I charged is (1)

1 = Prt

Interest charged according to formula (1) is called

simple interest.

In working with problems involving interest, we define the term as follows: Once per year 12 times per year 365 times per year* Twice per year Four times per year When the interest due at the end of a payment period is added to the principal so that the interest computed at the end of the next payment period is based on this new principal amount ( old principal interest ) , the interest is said to have been is interest paid on principal and previously earned interest. payment

period

Annually:

Monthly:

Semiannually:

Daily:

Quarterly:

+

compounded. Compound interest

EXAMPLE 1

Solution

Computing Compound I nterest

A credit union pays interest of S% per annum compounded quarterly on a certain savings plan. If $1000 is deposited in such a plan and the interest is left to accu mulate, how much is in the account after 1 year? We use the simple interest formula, I = Prt. The principal P is $1000 and the rate of interest is S% = O.OS. After the first quarter of a year, the time t is ± year, so the interest earned is 1 = Prt = ($1000)(0.oS) (±) $20 =

+

The new principal is P I = $1000 $20 = $1020. At the end of the second quar ter, the interest on this principal is +

I =

($1020)(0.OS) (±) = $20.40

At the end of the third quarter, the interest on the new principal of $1020 $20.40 = $1040.40 is +

I =

($1040.40)(0.os -


future value

of the

468

CHAPTER 6


EXAMPLE 2

Comparing I nvestments Using Different Compounding Periods

Investing $1000 at an annual rate of 10% compounded annually, semiannually, quar terly, monthly, and daily will yield the following amounts after 1 year: A p . (1 + ) Annual compounding (n 1): = ($1000)(1 + 0.10) = $1100.00 Semiannual compounding ( = 2): A p . (1 + � ) 2 = ($1000)(1 + 0. 0 5? $1102.50 Quarterly compounding (n 4): A = p . (1 + �) 4 = ($1000)(1 + 0.025) 4 = $1103.81 Monthly compounding (n = 12): A = p . ( 1 + ;2 ) 1 2 0 10 12 = ($1000) ( 1 + --t2 ) = $1104.71 r

=

=

n

=

=

=

Daily compounding (n 365): =

From Example 2, we can see that the effect of compounding more frequently is that the amount after 1 year is higher: $1000 compounded 4 times a year at 10% results in $1103.81, $1000 compounded 12 times a year at 10% results in $1104.71, and $1000 compounded 365 times a year at 10% results in $1105.16. This leads to the following question: What would happen to the amount after 1 year if the num ber of times that the interest is compounded were increased without bound? Let's find the answer. Suppose that P is the principal, is the per annum inter est rate, and n is the number of times that the interest is compounded each year. The amount after 1 year is ) A = P ' (1 + � Rewrite this expression as follows: r

r

11

n h = � r

Now suppose that the number n of times that the interest is compounded per year gets larger and larger; that is, suppose that n Then h = '!.. and the ex pression in brackets in Equation (3) equals e. That is, A Per . Table 8 compares (1 + !....)", for large values of n, to er for = 0.05, 0.10, = 0.15, and = T� e larger that n gets, the closer (1 + � )" gets to cr. No matter how frequent the compounding, the amount after 1 year has the definite ceiling Per . ---7 00 .

---7

r

---7 00 ,

r

r

=

r

r

1.

S ECTION 6.7

Ta ble

8 n

r r r r

= = = =

=

1 00

(1 � �)" n

=

1 000

n

=

Compound Interest

1 0,000

469

e'

0.05

1 .05 1 2580

1 .05 1 2698

1 .0 5 1 271

0.1 0

1 . 1 05 1 1 57

1 . 1 05 1 654

1 . 1 05 1 704

1 . 1 05 1 709

0.1 5

1 . 1 6 1 7037

1 . 1 6 1 82 1 2

1 . 1 6 1 8329

1 . 1 6 1 8342

1

2.7048 1 38

2 .7 1 69239

2 .7 1 8 1 459

2 .7 1 828 1 8

1 .05 1 27 1 1

When interest is compounded so that the amount after 1 year is Per, we say the interest is compounded continuously.

THEOREM

Continuous Compounding

The amount A after t years due to a principal P invested at an annual inter est rate compounded continuously is A Pert (4) I r

=

�

�----------------------------------�

EXAM P L E 3

Using Continuous Compounding

The amount A that results from investing a principal P of $1000 at an annual rate of 10% compounded continuously for a time t of 1 year is A $1000eO. 1 0 ($1000) (1.10517) $1105.17

r

=

w= -

2

=

=

•


Calculate Effective Rates of Return

The is the equivalent annual simple rate of interest that would yield the same amount as compounding after 1 year. For example, based on Example 3, a principal of $1000 will result in $1105.17 at a rate of 10% compounded continuously. To get this same amount using a simple rate of interest would require that interest of $1105.17 - $1000.00 $105.17 be earned on the principal. Since $105.17 is 10.517% of $1000, a simple rate of interest of 10.517% is needed to equal 10% compounded continuously. The effective rate of interest of 10% compounded continuously is 10.517%. Based on the results of Examples 2 and 3, we find the following comparisons: effective rate of interest

=

Annual Rate

EXAM P L E 4

Effective Rate

Annual compounding

1 0%

1 0%

Semia n n u a l compounding

1 0%

1 0 .25%

Quarterly compounding

1 0%

1 0 .38 1 %

Monthly compou nding

1 0%

1 0.47 1 %

Daily compou nding

1 0%

1 0 .5 1 6%

Conti nuous compou nding

1 0%

1 0 .51 7%

Computing the Effective Rate of I nterest

On January 2, 2007, $2000 is placed in an Individual Retirement Account (IRA) that will pay interest of 7% per annum compounded continuously. (a) What will the IRA be worth on January 1, 2027? (b) What is the effective annual rate of interest?

470

CHAPTER 6


(a) On January, 1, 2027, the initial principal of $2000 will have earned interest of 7% compounded continuously for 20 years. The amount A after 20 years is

Solution

A = Perr = $2000e ( O.07 ) ( 20 ) = $8110 . 40

(b) First, we compute the interest earned on $2000 at r = 7 % compounded con tinuously for 1 year. A = $2000eO.07( 1) = $2145 . 02

So the interest earned is $2145 .02 - $2000.00 = $145 .02. To find the effective rate of interest R, we use the simple interest formula = P Rt, with I $145.02, P = $2000, and t = 1 . 1

=

� 001

Exploration

$145.02 = $2000 · R · l $145.02 R= = 0.07251 $2000

4, how until A = $4000? $6000? Y, = 2000eO.0 7x and Y2 =

For the IRA described in Exa mple

[Hint: Graph 4000. Use I NTERS ECT to find x.I long will it be

Solve for

s.

The effective annual rate of interest R is 7.251 %. """

3

1 = PRt

N*"> -

•


Determine the Present Val u e of a Lump Sum of Money

When people engaged in finance speak of the "time value of money," they are usu ally referring to the present value of money. The of A dollars to be received at a future date is the principal that you would need to invest now so that it will grow to A dollars in the specified time period. The present value of money to be received at a future date is always less than the amount to be received, since the amount to be received will equal the present value (money invested now) plus the interest accrued over the time period. We use the compound interest formula (2) to get a formula for present value. If P is the present value of A dollars to be received after t years at a per annum inter est rate r compounded n times per year, then, by formula (2), present value

( �rl To solve for P, we divide both sides by ( 1 + r ) . The result is A = P or P = A · ( 1 + -;; ) r ( 1 + -n) A = P' 1 +

-;;

r

fll

THEOREM

1ll

-11I

Present Value Formulas

The present value P of A dollars to be received after t years, assuming a per annum interest rate r compounded n times per year, is

( ) r

P = A · 1 + -;;

-fll

(5)

If the interest is compounded continuously, P = Ae-rr

(6)

I�

� -----------------�

To prove (6), solve formula (4) for P.

SECTION 6.7

Compound Interest

47 1

Computing the Value of a Zero-coupon Bond

E XA M P L E 5

A zero-coupon (noninterest-bearing) bond can be redeemed in 10 years for $1000. How much should you be willing to pay for it now if you want a return of (a) 8 % compounded monthly? (b) 7% compounded continuously? (a) We are seeking the present value of $1000. We use formula (5) with A = $1000, n = 12, r = 0.08, and t = 10. ( r )-11 = $1000( 1 + -t0 08 )- 1 2 (10) = $450.52 P = A· 1 +

Solution

-;;

For a return of 8% compounded monthly, you should pay $450.52 for the bond. (b) Here we use formula (6) with A = $1000, r = 0.07, and t = 10. P = Ae-rt = $1000e- ( 007 )( l O ) = $496.59 For a return of 7% compounded continuously, you should pay $496.59 for the bond. ,, = = = -

4 E XA M P L E 6

Solution

•


Determine the Rate of Interest or Time Required to Double a Lump Sum of Money

Rate of I nterest Required to Double an Investment

What annual rate of interest compounded annually should you seek if you want to double your investment in 5 years? If P is the principal and we want P to double, the amount A will be 2P. We use the compound interest formula with n = 1 and t = 5 to find r.

(

A = p . 1 + �) 2P = P ' ( l + r ) 5 2 (1 + r ) 5

1/1

=

1 +

r =

V'2 V'2

A = 2 P, n = 1 , t = 5 Cancel the Ps. Ta ke the fifth root of each side.

1 :::::: 1.148698 - 1 = 0.148698 The annual rate of interest needed to double the principal in 5 years is 14.87% r =

I.C 'JT: ==- -

E XA M P L E 7

Solution

-


.

•

Time Required to Double or Triple an I nvestment

(a) How long will it take for an investment to double in value if it earns 5% com pounded continuously? (b) How long will it take to triple at this rate? (a) If P is the initial investment and we want P to double, the amount A will be 2P. We use formula (4) for continuously compounded interest with r = 0.05.

472

CHAPTER 6


Then A

2P

2

=

=

=

=

Pert Peo.OS! eO.OS!

A = 2P,

r =

0.05

Cancel the P's.

O.OSt In 2 Rewrite as a logarith m . In 2 t 0.05 13.86 Solve for It will take about 14 years to double the investment. (b) To triple the investment, we set A 3P in formula (4). t.

;:::::

=

=

A

= =

Perl PeO.OSl eO.OS!

A

r

=

3P 3P, 0.05 3 Cancel the P's. O.OSt In 3 Rewrite as a logarithm . In 3 t 0.05 21.97 Solve for It will take about 22 years to triple the investment. =

=

=

,, =

-- Now Work P R O B

t.

;:::::

=

l

EM

35

6.7 Assess You r U n de rsta n d i n g 'Are You Prepared?' Answers are given a t the end of these exercises. If you get a wrong answer, read the pages listed in red.

1. What is the interest due if $500 is borrowed for 6 months at a simple interest rate of 6% per annum? (pp. 1 41-142) Skill Building

2. If you borrow $5000 and, after 9 months, pay off the loan in the amount of $5500, what per annum rate of interest was charged? (pp. 141-142)

In Problems 3-12, find the amount that results from each investment. . 3. $100 invested at 4% compounded quarterly after a period of 2 years

4. $50 invested at 6% compounded monthly after a period of

5. $500 invested at 8% compounded quarterly after a period of 1 2 '2 years

6. $300 invested at 12% compounded monthly after a period of 1 1 years '2

7. $600 invested at 5 % compounded daily after a period of 3 years

8. $700 invested at 6% compounded daily after a period of 2 years

9. $10 invested at 11 % compounded continuously after a period of 2 years

10. $40 invested at 7% compounded continuously after a period of 3 years

11.

$ 1 00 invested at 1 0 % compounded continuously after a period of 2

� years

3 years

12. $100 invested at 12% compounded continuously after a pe

%

riod of 3 years

In Problems 13-22, find the principal needed now to get each amount; that is, find the present value. 13. To get $100 after 2 years at 6% compounded monthly

14. To get $75 after 3 years at 8% compounded quarterly

15. To get $1000 after 2 years at 6% compounded daily

16. To get $800 after 3

17. To get $600 after 2 years at 4% compounded quarterly

18. To get $300 after 4 years at 3% compounded daily

19. To get $80 after 3 years at 9% compounded continuously

20. To get $800 after 2 years at 8% compounded continuously

21. To get $400 after 1 year at 10% compounded continuously

22. To get $ 1000 after 1 year at 12% compounded continuously

�

�

1 years at 7% compounded monthly '2

�

SECTION 6.7

Compound Interest

473

In Problem s 23-26, which of the two rates wo uld yield the larger am ount in 1 y ear? [Hint: Start with a principal of$10, 000 i n each i nstance. ]

1 23. 6% compounded quarterly or 6 "4 % compounded annually

1 24. 9% compounded quarterly or 9 "4 % compounded annually

25. 9% compounded monthly or 8.8% compounded daily

26. 8% compounded semiannually or 7.9% compounded daily

In Problem s 27-30, find the effective rate of i nterest.

27. For 5 % compounded quarterly 28. For 6% compounded monthly 29. For 5% compounded continuously 30. For 6% compounded continuously 31. What rate of interest compounded annually is required to double an investment in 3 years? 32. What rate of interest compounded annually is required to double an investment in 6 years? 33. What rate of interest compounded annually is required to triple an investment in 5 years? 34. What rate of interest compounded annually is required to triple an investment in 10 years?

. 35. (a) How long does it take for an investment to double in value if it is invested at 8% compounded monthly? (b) How long does it take if the interest is compounded con tinuously? 36. (a) How long does it take for an investment to triple in value if it is invested at 6 % compounded monthly? (b) How long does it take if the interest is compounded con tinuously? 37. What rate of interest compounded quarterly will yield an effective interest rate of 7 % ? 38. What rate of interest compounded continuously will yield an effective interest rate of 6 % ?


39. Time Required to Reach a Goal I f Tanisha has $100 to in vest at 8% per annum compounded monthly, how long will it be before she has $ 1 50? If the compounding is continuous, how long will it be? 40. Time Required to Reach a Goal If Angela has $100 to in vest at 10% per annum compounded monthly, how long will it be before she has $175? If the compounding is continuous, how long will it be? 41. Time Required to Reach a Goal How many years will it take for an initial investment of $ 10,000 to grow to $25,000? Assume a rate of interest of 6% compounded continuously. 42. Time Required to Reach a Goal How many years will it take for an initial investment of $25,000 to grow to $80,000? Assume a rate of interest of 7% compounded continuously. 43. Price Appreciation of Homes What will a $90,000 house cost 5 years from now if the price appreciation for homes over that period averages 3% compounded annually? 44.

Credit Card Interest Sears charges 1 .25 % per month on the unpaid balance for customers with charge accounts (interest is compounded monthly). A customer charges $200 and does not pay her bill for 6 months. What is the bill at that time?

45. Saving for a Car Jerome will be buying a used car for $15,000 in 3 years. How much money should he ask his par ents for now so that, if he invests it at 5 % compounded con tinuously, he will have enough to buy the car? 46. Paying off a Loan John will require $3000 in 6 months to pay off a loan that has no prepayment privileges. If he has the $3000 now, how much of it should he save in an account paying 3% compounded monthly so that in 6 months he will have exactly $3000?

47. Return o n a Stock George is contemplating the purchase of 100 shares of a stock selling for $15 per share. The stock pays no dividends. The history of the stock indicates that it should grow at an annual rate of 1 5 % per year. How much will the 100 shares of stock be worth in 5 years? 48. Return on an Investment A business purchased for $650,000 in 2001 is sold i n 2004 for $850,000. What is the annual rate of return for this investment? 49. Comparing Savings P lans Jim places $1000 in a bank account that pays 5 . 6 % compounded continuously. After 1 year, will he have enough money to buy a computer system that costs $1060? If another bank will pay Jim 5.9% com pounded monthly, is this a better deal? 50. Savings Plans On January 1, Kim places $1000 in a certifi cate of deposit that pays 6.8% compounded continuously and matures in 3 months. Then Kim places the $1000 and the in terest in a passbook account that pays 5.25% compounded monthly. How much does Kim have in the passbook account on May I? 51. Comparing IRA Investments Will invests $2000 in his IRA in a bond trust that pays 9% interest compounded semiannu ally. His friend Henry invests $2000 in his IRA in a certificate 1 of deposit that pays 8 "2 % compounded continuously. Who has more money after 20 years, Will or Henry? 52. Comparing Two Alternatives Suppose that April has access to an investment that will pay 10% interest compounded con tinuously. Which is better: to be given $1000 now so that she can take advantage of this investment opportunity or to be given $1325 after 3 years?

474

CHAPTER 6


53. College Costs The average cost of college at 4-year private colleges was $29,026 in 2005. This was a 5.5% increase from the previous year. Source: Trends in College Pricing 2005, College Board

� �% interest compounded continuously

(b) 11 % interest compounded monthly (c)

11

Which option is best; that is, which results in the least inter est on the loan?

(a) I f the cost of college increases by 5.5% each year, what will be the average cost of college at a 4-year private college in 2015?

55. Federal Deficit At the end of fiscal year 2005, the federal budget deficit was $319 billion. At that time, 20-year Series EE bonds had a fixed rate of 3 .2% compounded semiannu ally. If the federal government financed this deficit through EE bonds, how much would it have to pay back in 2025? Source: Us. Treasury Department

(b) College savings plans, such as a 529 plan, allow individ uals to put money aside now to help pay for college later. If one such plan offers a rate of 4% compounded continuously, how much should be put in a college sav ings plan in 2005 to pay for 1 year of the cost of college at a 4-year private college in 2015?

56. Federal Deficit On February 6, 2006, President B ush pro posed the fiscal year 2007 federal budget. The proposal projected a fiscal year 2006 deficit of $423 billion and a fiscal year 2007 deficit of $354 billion. Assuming the deficit de creases at the same rate each year, when will the deficit be cut to $100 billion? Source: Office of Management and Budget

54. Analyzing Interest Rates on a Mortgage Colleen and Bill have j ust purchased a house for $650,000, with the seller holding a second mortgage of $ 1 00,000. They promise to pay the seller $100,000 plus all accrued interest 5 years from now. The seller offers them three interest options on the second mortgage: (a) Simple interest at 1 2 % per annum

Inflation Problems 57-62 require the following discussion. Inflation is a term used to describe the erosion of the purchasing power of money. For example, suppose the annual inflation rate is 3 %. Then $1000 worth ofpurchasing power now will have only $970 worth of purchasing power in one year because 3 % of the original $1 000 (0. 03 X 1000 30) has been eroded due to inflation. In general, if the rate of inflation averages r% over n years, the amount A that $P will purchase after n years is =

A=

p.

where r is expressed as a decimal.

(1

- r)"

57. Intlation If the inflation rate averages 3 % , how much will $ 1 000 purchase in 2 years?

60. Inflation If the amount that $1000 will purchase is only $930 after 2 years, what was the average inflation rate?

59. lnDation If the amount that $1000 will purchase is only $950 after 2 years, what was the average inflation rate?

62. Inflation If the average inflation rate is 4 % , how long is it until purchasing power is cut in half?

58. Inflation If the inflation rate averages 2 % , how much will $ 1000 purchase in 3 years?

61. InDation I f the average inflation rate is 2 %, how long is it until purchasing power is cut in half?

Problems 63-66 involve zero-coupon bonds. A zero- coupon bond is a bond that is sold now at a discount and will pay its face value at the lime when it matures; no interest payments are made. 63. Zero-Coupon Bonds A zero-coupon bond can be redeemed in 20 years for $ 10,000. How much should you be willing to pay for it now if you want a return of: (a) 10% compounded monthly? (b) 10% compounded continuously? 64. Zero-Coupon Bonds A child's grandparents are consider ing buying a $40,000 face value zero-coupon bond at birth so that she will have enough money for her college education 1 7 years later. If they want a rate of return o f 8 % compounded annually, what should they pay for the bond? 65. Zero-Coupon Bonds How much should a $ 10,000 face value zero-coupon bond, maturing in 10 years, be sold for now if its rate of return is to be 8% compounded annually?

66. Zero-Coupon Bonds If Pat pays $ 12,485.52 for a $25 ,000 face value zero-coupon bond that matures in 8 years, what is his annual rate of return? 67. Time to Double or Triple an I nvestment The formula In m t = n In 1 + n

----( �)

can be used to find the number of years t required to multi ply an investment m times when r is the per annum interest rate compounded n times a year. (a) How many years will it take to double the value of an I RA that compounds annually at the rate of 12%? (b) How many years will it take to triple the value of a sav ings account that compounds quarterly at an annual rate of 6 % ? (c) Give a derivation o f this formula.

-

68. Time to Reach an Investment Goal The formula In A In P t = ----r can be used to find the number of years t required for an in vestment P to grow to a value A when compounded contin uously at an annual rate r. (a) How long will it take to increase an initial investment of $1 000 to $8000 at an annual rate of 1 O % ? ( b ) What annual rate i s required t o increase t h e value o f a $2000 IRA to $30,000 in 35 years? (c) Give a derivation of this formula.

SECTION 6.8

Exponential Growth and Decay Models; Newton's Law; Logistic Growth and Decay Models

475

Problems 69-72, require the following discussion. The Consumer Price Index (CPI) indicates the relative change in price over time for a fixed basket of goods and services. It is a cost of living index that helps measure the effect of inflation on the cost ofgoods and services. The CPI Llses the base period 1982-1984 for comparison (the CPI for this period is ZOO). The CPI for January 2006 was ]98.3. This means that $100 in the period 1 982-1984 had the same purchasing power as $198.30 in January 2006. In general, if the rate of inflation averages r% over n years, then the CPI index after n years is

CPI

=

( + 1�0 )"

CPIo 1

where CPlo is the CPI index at the beginning of the n-year period.

69. Consumer Price Index

(a) The cpr was 1 52.4 for 1 995 and 1 95 .3 for 2005. Assum ing that annual inflation remained constant for this time period, determine the average annual inflation rate. (b) Using the inflation rate from part (a), in what year will the cpr reach 300?

70. Consumer Price Index If the current c p r is 234.2 and the average annual inflation rate is 2.8 % , what will be the cpr in 5 years?

Source: u.s.

Bureau of Labor Statistics

71. Consumer Price Index I f the average annual inflation rate is 3.1 %, how long will it take for the cpr index to double?

72. Consumer Price Index The base period for the CPI changed in 1 998. Under the previous weight and item structure, the cpr for 1 995 was 456.5. If the average annual inflation rate was 5 .57 % , what year was used as the base period for the CPI?


73. Explain in your own words what the term compound interest means. What does continuous compounding mean? 74. Explain in your own words the meaning of present value. 75. Critical Thinking You have j ust contracted to buy a house and will seek financing in the amount of $1 00,000. You go to several banks. B ank 1 will lend you $ 1 00,000 at the rate of 8.75 % amortized over 30 years with a loan origination fee of 1 . 75 % . Bank 2 will lend you $ 100,000 at the rate of 8.375 % amortized over 15 years with a loan origination fee of 1 .5 % . Bank 3 will lend you $100,000 a t the rate o f 9.125% amor tized over 30 years with no loan origination fee. Bank 4 will lend you $ 1 00,000 at the rate of 8.625 % amortized over 1 5 years with no loan origination fee. Which loan would you take? Why? Be sure to have sound reasons for your choice.

Use the information in the table to assist you. If the amollnt of the monthly payment does not matter to you, which loan would you take? Again, have sound reasons for your choice. Compare your final decision with others in the class. Discllss.

Monthly Payment

Loan Origination Fee

$786.70

$1 ,750.00

Bank 2

$977.42

$1 ,500.00

Bank 3

$813.63

$0.00

Bank 4

$990.68

$0.00

Bank 1


$15

6.8 Exponential Growth and Decay Models; Newton's Law; Logistic Growth and Decay Models OBJECTIVES 1 Find Equations of Populations That Obey the Law of Uninhibited Growth (p. 475)

2 Find Equations of Populations That Obey the Law of Decay (p. 478)

3

Use Newton's Law of Cooling (p. 479)

4 Use Logistic Models (p. 48 1 )

1

Find Equations of Populations That Obey the Law of U nin hibited Growth

Many natural phenomena have been found to follow the law that an amount A varies with time t according to the function (1)

Here Ao is the original amount (t = 0) and k *" 0 is a constant.

476

CHAPTER 6


If k > 0, then equation (1) states that the amount A is increasing over time; if k < 0, the amount A is decreasing over time. In either case, when an amount A

varies over time according to equation (1), it is said to follow the the ( k > 0) (Ie < 0). See Figure 41.

exponential law

law of uninhibited growth

or

or decay

Fig u re 41

A

For example, we saw in Section 6.7 that continuously compounded interest fol lows the law of uninhibited growth. In this section we shall look at three additional phenomena that follow the exponential law. Cell division is the growth process of many living organisms, such as amoebas, plants, and human skin cells. Based on an ideal situation in which no cells die and no by-products are produced, the number of cells present at a given time follows the law of uninhibited growth. Actually, however, after enough time has passed, growth at an exponential rate will cease due to the influence of factors such as lack of liv ing space and dwindling food supply. The law of uninhibited growth accurately reflects only the early stages of the cell division process. The cell division process begins with a culture containing No cells. Each cell in the culture grows for a certain period of time and then divides into two identical cells. We assume that the time needed for each cell to divide in two is constant and does not change as the number of cells increases. These new cells then grow, and eventually each divides in two, and so on. U n inhibited Growth of Cells

A model that gives the number N of cells in a culture after a time t has passed (in the early stages of growth) is 1e > 0

(2)

where No is the initial number of cells and k is a positive constant that rep resents the growth rate of the cells. In using formula (2) to model the growth of cells, we are using a function that yields positive real numbers, even though we are counting the number of cells, which must be an integer. This is a common practice in many applications. E XA M P L E 1

Bacterial G rowth

A colony of bacteria that grows according to the law of uninhibited growth is mod eled by the function N (t) = 100eO.045r, where N is measured in grams and t is mea sured in days. (a) Determine the initial amount of bacteria. (b) What is the growth rate of the bacteria? ( c) What is the population after 5 days?

SECTION 6.8


477

(d) How long will it take for the population to reach grams? (e) What is the doubling time for the population? (a) The initial amount of bacteria, is obtained when t = so ( ) grams to ( t ) The value of k, . 4 , indicates a (b) Compare N ( t ) = growth rate of 4.5 % . 125.2 grams. (c) The population after 5 days is ( 5 ) (d) To find how long it takes for the population to reach 140 grams, we solve the equation ( t ) 4 . 140

Solution

No ,

No

N O

=

=

oo 100e . 45t

1 00eO.045(O)

N

=

N

N

=

eO. 045t

=

1

100eO.045(5)

00 5

=

Divide both sides of the equation by 100.

00 5

Rewrite as a loga rith m. Divide both sides of the equation by 0.045.

--

0.045 7.5 days

�

(e) The population doubles when by solving the equation 200

In 2

1 00

100ekt .

140

.4 . 4 t = In 1.4 In 1.4 t=

2

=

=

1 0

=

oo 100e . 45t

200

0,

=

N(t)

=

grams, so we find the doubling time for t.

200

100eO .045t

=

oo 1 00e . 45t

=

eO.045 t

Divide both sides of the equation by 100.

=

0.045t

Rewrite as a loga rith m.

In

2

t = -0.045 � 15 . 4 days

Divide both sides of the equation by 0.045.

The population doubles approximately every 15.4 days.

•

�-��> - Now Work P R O B L E M

E XA M P LE 2

Bacterial G rowth

A colony of bacteria increases according to the law of uninhibited growth. (a) If is the number of cells and t is the time in hours, express N as a function of t. (b) If the number of bacteria doubles in 3 hours, find the function that gives the number of cells in the culture. (c) How long will it take for the size of the colony to triple? (d) How long will it take for the population to double a second time (that is, in crease four times)? (a) Using formula (2), the number of cells at a time t is N

Solution

N

N (t)

=

No ekt

where is the initial number of bacteria present and k is a positive number. (b) We seek the number k. The number of cells doubles in 3 hours, so we have No

But

N (3)

=

No ek(3),

so

N o ek(3)

=

2N o

e3k

=

2

3k = k

=

In 1 3 ln

N(

3)

= 2N o

Divide both sides by No .

2

Write the ex ponentia l equation as a logarithm. 2

478

C H A PTEH 6


The function that models this growth process is therefore N(t)

=

Noe (�ln2}

c The time t needed for the size of the colony to triple requires that N We substitute 3N for N to get

( )

0

3No 3

(� In } 2

t

=

=

=

=

3N o.

Noe(�l n 2} e G l n 2}

In 3 --2

3 1n 3

:0:::

In 4.755 hours It will take about 4.755 hours or 4 hours, 45 minutes for the size of the colony to triple. ( d ) If a population doubles in 3 hours, it will double a second time in 3 more hours, for a total time of 6 hours. =

•

2

Find Equations of Populations That Obey the Law of Decay

Radioactive materials follow the law of uninhibited decay. U n inhi bited Radioactive Decay

The amount A of a radioactive material present at time t is given by kCO

(b)

y=

c

482

C H A PT E R 6


Based on the figures, we have the following properties of logistic growth functions. Properties of the Logistic Growth Function, Equation (6)

c

,

The domain is the set of all real numbers. The range is the interval (0, ) where is the carrying capacity. 2. There are no x-intercepts; the y-intercept is P(O). 3 . There are two horizontal asymptotes: y = 0 and y = 4 . P(t ) is an increasing function if b > 0 and a decreasing function if b < O. 5. There is an where P(t) equals "21 of the carrying capacity. The inflection point is the point on the graph where the graph changes from being curved upward to curved downward for growth functions and the point where the graph changes from being curved downward to curved upward for decay functions. 6. The graph is smooth and continuous, with no corners or gaps. 1.

c

c.

inflection point

EXAM P L E 5

F ruit Fly Population

Fruit flies are placed in a half-pint milk bottle with a banana (for food) and yeast plants (for food and to provide a stimulus to lay eggs). Suppose that the fruit fly population after t days is given by 23_0 P(t) = 1 + 56.5 e-037r (a) State the carrying capacity and the growth rate. (b) Determine the initial population. (c) What is the population after 5 days? (d) How long does it take for the population to reach 180? I (e) Use a graphing utility to determine how long it takes for the population to reach one-half of the carrying capacity by graphing Y1 = pet) and Y = 1 15 and us ing INTERSECT. e-O.37r (a) As t 0 and P ( t ) 230 . The carrying capacity of the half-pint Solution 1 bottle is 230 fruit flies. The growth rate is Ibl = 1 0 .371 = 37% per day. (b) To find the initial number of fruit flies in the half-pint bottle, we evaluate P(O). 230 P(O) = 1 + 56.5 e-0 37 ( O ) 230 1 + 56.5 _ _

_ _

2

--,> 00 ,

--,>

--,>

=4

So, initially, there were 4 fruit flies in the half-pint bottle. (c) To find the number of fruit flies in the half-pint bottle after 5 days, we evalu ate P ( 5 ) . 230 23 fruit flies P(5) - 1 56.5 e-O.37( 5 ) After 5 days, there are approximately 23 fruit flies in the bottle. +

:::;:j

SECTION 6.8

Exponential G rowth and Decay Models; Newton's Law; Logistic Growth and Decay Models

483

(d) To determine when the population of fruit flies will be 180, we solve the equa tion p et ) 180. 230 3 -:-:-- = 180 ---1 + 56.5e-O. 71 230 = 180(1 56.5e-0371 ) 1.2778 1 + 56.5e-0371 Divide both sides by 180. 0.2778 56.5e-0371 Subtract 1 from both sides. 0.0049 = e-0371 Divide both sides by 56.5. In(0.0049) -0.37t Rewrite as a logarithmic expression. t 14.4 days Divide both sides by - 0.37. It will take approximately 14.4 days (14 days, 10 hours) for the population to reach 180 fruit flies. One-half of the carrying capacity is 115 fruit flies. We solve P ( t ) = 115 by graph 230 3 and Y = 115 and using INTERSECT. See Figure 43. ing 1 + 56.5e-0 7I The population will reach one-half of the carrying capacity in about 10.9 days (10 days, 22 hours). =

+

=

=

=

::::;

y1 =

2

.

Figure 43

25 0

Y, =

1 +

230 56.5e 0.37/

- 50

lQ.J b9

Look back at Figure 43. Notice the point where the graph reaches 115 fruit flies ;{ � (one-half of the carrying capacity): the graph changes from being curved upward to

Exploration On

Y,

the same viewing recta n gle, gra p h

=

1 +

•

500 24e-o.03t

and

Y2 =

500

1 + 24e-o.08t

----

What effect does the growth rate have on the logistic growth function?

E XA M P L E 6

Ibl

being curved downward. Using the language of calculus, we say the graph changes from increasing at an increasing rate to increasing at a decreasing rate. For any logis tic growth function, when the population reaches one-half the carrying capacity, the population growth starts to slow down. �==""' Now Work P R O B L E M 2 3

Wood Products

The EFISCEN wood product model classifies wood products according to their life span. There are four classifications: short (1 year), medium short (4 years), medium long (16 years), and long (50 years). Based on data obtained from the European Forest Institute, the percentage of remaining wood products after t years for wood products with long life-spans (such as those used in the building industry) is given by 100.3952 pe t ) 1 + 0.0316eo.058lt (a) What is the decay rate? (b) What is the percentage of remaining wood products after 10 years? ( c) How long does it take for the percentage of remaining wood products to reach 50%? (d) Explain why the numerator given in the model is reasonable.

484

CI-IAPTER 6


Solution

(a) The decay rate is Ibl (b) Evaluate P ( 1 0 ) .

1 -0.0581 1

P ( 10 )

=

1 +

=

5.81%.

1 00.3952 :;:::j 95 . 0 . 0.031 6 eo . o58 J ( 10 )

So 95 % of long-life-span wood products remains after 1 0 years. (c) Solve the equation p e t ) 50. =

1

+

100.3952 0.031 6eo o5811

=

50

1 00.3 952

=

2.00 79

=

1 .0079

=

3 1 .8957

=

0 . 031 6 eo o58 11 ) 50( 1 1 0.03 1 6eo.05811 0.031 6 eo.o581 1 eO.058 1 I

I n ( 3 1 .8957)

=

0.05S1 t

+

+

Divide both sides by 50. Subtract 1 from both sides. Divide both sides by 0.031 6 . Rewrite as a loga rithmic expression .

Divide both sides by 0.0581 . years It will take approximately 59. 6 years for the percentage of long-life-span wood products remaining to reach 50%. (eI ) The numerator of 100.3952 is reasonable because the maximum percentage of wood products remaining that is possible is 1 00 % .

t :;:::j 59.6

•

6.8 Assess You r Understandi ng Applications and Extensions

. 1. Growth of an Insect Population The size P of a certain in sect popula tion at time t (in days) obeys the function P(t) = 500eo o21. (a) Determine the number of insects at t = 0 days. (b) What is the growth rate of the insect population? (c) What is the population after 10 days? (d) When will the insect population reach 800? (e) When will the insect population double? 2. Growth of Bacteria The number N of bacteria present in a culture at time t (in hours) obeys the law of uninhibited growth N ( t ) = 1 000eo.oll• (a) Determine the number of bacteria at t = 0 hours. (b) What is the growth rate of the bacteria? (c) What is the population after 4 hours? (d) When will the number of bacteria reach 1 700? (e) When will the number of bacteria double?

3. Radioactive Decay Strontium 90 is a radioactive material that decays according to the function A ( t ) = A e-o.o2441, o where A is the initial amount present and A is the amount o present at time t (in years). Assume that a scientist has a sam ple of 500 grams of strontium 90. (a) What is the decay rate of strontium 90? (b) How much strontium 90 is left after 10 years? (c) When will 400 grams of strontium 90 be left? (d) What is the half-life of strontiulll 90?

4. Radioactive Decay Iodine 131 is a radioactive material that decays according to the function A ( I ) = A e-o.oS?I, where A o o is the initial amount present and A is the amount present at time 1 (in days). Assume that a scientist has a sample of 100 grams of iodine 1 3 l . (a) What is the decay rate of iodine 131? (b) H ow much iodine 131 is left after 9 days? (c) When will 70 grams of iodine 1 3 1 be left? (d) What is the half-life of iodine 131? 5. Growth of a Colony of Mosquitoes The population of a colony of Illosquitoes obeys the law of uninhibited growth. (a) If N is the population of the colony and t is the time in days, express N as a function of I. (b)

If there are 1 000 mosquitoes initially and there are 1 800 after 1 day, what is the size of the colony after 3 days?

(c) How long is it until there are 1 0,000 mosquitoes? 6. Bacterial Growth A culture of bacteria obeys the law of uninhibited growth. (a) If N is the number of bacteria in the culture and 1 is the time in hours, express N as a function of t.

(b) If 500 bacteria are present initially and there are 800 after 1 hour, how many will be present in the culture after 5 hours? (c) How long is it until there are 20,000 bacteria?

SECTION 6.8


7. Population Growth The population of a southern city fol lows the exponential law. (a) If N is the population of the city and t is the time in years, express N as a function of t. (b) If the population doubled in size over an 18-month pe riod and the current population is 10,000, what will the population be 2 years from now? 8. Population Decline The population of a midwestern city follows the exponential law. (a) If N is the population of the city and t is the time in years, express N as a function of t. (b) If the population decreased from 900,000 to 800,000 from 2003 to 2005, what will the population be in 2007? 9. Radioactive Decay The half-life of radium is 1690 years. If 10 grams are present now, how much will be present in 50 years? 10. Radioactive Decay The half-life of radioactive potassium is 1 .3 billion years. If 1 0 grams are present now, how much will be present in 1 00 years? In 1 000 years? 11. Estimating the Age of a Tree

A piece of charcoal is found to contain 30% of the carbon 14 that it origin al ly had. When did the tree die from which the charcoal came? Use 5600 years as the half-life of carbon 14.

12. Estimating the Age of a Fossil A fossilized leaf contains 70% of its normal amount of carbon 1 4. How old is the fossil? 13. Cooling Time of a Pizza Pan A pizza pan is removed at 5:00 PM from a n oven whose temperature is fixed at 450°F into a room that is a constant 70°F. After 5 minutes, the pan is at 300°F. (a) At what time is the temperature of the pan 135°F? (b) Determine the time that needs to elapse before the pan is 1 60°F. (c) What do you notice about the temperature as time passes?

485

the thermometer reads 15°C after 3 minutes, what will it read after being in the room for 5 minutes? For 10 minutes? [Hint: You need to construct a formula similar to equation (4).] 16. Warming Time of a Beerstein A beerstein has a tempera ture of 28°F. It is placed in a room with a constant tempera ture of 70°F. After 10 minutes, the temperature of the stein has risen to 35°F. What will the temperature of the stein be after 30 minutes? How long will it take the stein to reach a temperature of 45°F? (See the hint given for Problem 15.) 17. Decomposition of Chlorine in a Pool Under certain water conditions, the free chlorine (hypochlorous acid, HOCl) in a swimming pool decomposes according to the law of unin hibited decay. After shocking his pool, Ben tested the water and found the amount of free chlorine to be 2.5 parts per mil lion (ppm) . Twenty-four hours l ater, Ben tested the water again and found the amount of free chlorine to be 2.2 ppm. What will be the reading after 3 days (that is, n hours)? When the chlorine level reaches 1 .0 ppm, Ben must shock the pool again. How long can Ben go before he must shock the pool again? 18. Decomposition of Dinitrogen Pentoxide At 45°C, dinitro gen pentoxide (N205) decomposes into nitrous dioxide (N02) and oxygen (02) according to the law of uninhibited decay. An initial amount of 0.25 M of dinitrogen pentoxide decom poses to 0.15 M in 17 minutes. How much dinitrogen pen toxide will remain after 30 minutes? How long wil l it take until 0.01 M of dinitrogen pentoxide remains? 19. Decomposition of Sucrose Reacting with water in an acidic solution at 35°C, sucrose (C12H2201 1 ) decomposes into glu cose (C6H 1 z06) and fructose (C6H 1 206)* according to the law of uninhibited decay. An initial amount of 0.40 M of sucrose decomposes to 0.36 M in 30 minutes. How much sucrose will remain after 2 hours? How long will it take until 0.10 M of sucrose remains? 20. Decomposition of Salt in Water Salt (NaCl) decomposes in water into sodium (Na+) and chloride (Cl-) ions according to the law of uninhibited decay. If the initial amount of salt is 25 kilograms and, after 10 hours, 15 kilograms of salt is left, how much salt is left after 1 day? How long does it take until

� kilogram of salt is left?

21. Radioactivit)1 from Chernobyl After the release of ra dioactive material into the atmosphere from a nuclear power plant at Chernobyl (Ukraine) in 1986, the hay in Austria was contaminated by iodine 131 (half-life 8 days). If it is safe to feed the hay to cows when 10% of the iodine 131 remains, how long did the farmers need to wait to use this hay? 14. Newton's Law of Cooling A thermometer reading nOF is placed in a refrigerator where the temperature is a constant 38°F. (a) If the thermometer reads 60°F after 2 minutes, what will it read after 7 minutes? (b) How long will it take before the thermometer reads 39°F? (c) Determine the time needed to elapse before the ther mometer reads 45°F. (d) What do you notice about the temperature as time passes? 15. Newton's Law of Heating A thermometer reading 8°C is brought into a room with a constant temperature of 35°C. If

22. Pig Roasts The hotel Bora-Bora is having a pig roast. At noon, the chef put the pig in a large earthen oven. The pig's original temperature was 75°F. At 2:00 PM the chef checked the pig's temperature and was upset because it had reached

':' Author's Note: Surprisingly, the chemical formulas for glucose and fruc tose are the same. This is not a typo.

486

CH A PTER 6


expect the population to grow according to the model

only lOO°F. If the oven's temperature remains a constant 325°F, at what time may the hotel serve its guests, assuming that pork is done when it reaches 1 75°F?

_

P(I) -

23. Proportion of the Population That Owns a DVD Player The logistic growth model P ( t)

=

1 +

24. Market Penetration of Intel's Coprocessor growth model =

1 +

The logistic

0.90 3 .5 e-O.3391

relates the proportion of new personal computers (pes) sold at Best Buy that have Intel's latest coprocessor t months after it has been introduced. (a) Determine the maximum proportion of pes sold at Best Buy that will have Intel's latest coprocessor. (b) What proportion of computers sold at Best Buy will have I ntel's latest coprocessor when it is first introduced (t = OJ? (c) What proportion of pes sold will have Intel's latest coprocessor t = 4 months after it is introduced? (d) When will 0.75 (75 % ) of pes sold at Best B uy have Intel's latest coprocessor? (e) How long will it be before 0.45 (45 % ) of the pes sold by Best Buy have Intel's latest coprocessor? 25. Population of a Bacteria Culture The logistic growth model pet)

where t is measured in years.

0.9 6e-O.321

relates the proportion of U.S. households that own a DVD player to the year. Let I = 0 represent 2000, I = 1 represent 200 1 , and so on. (a) Determine the maximum proportion of households that will own a DVD player. (b) What proportion of households owned a DVD player in 2000 (t = OJ? (c) What proportion of households owned a DVD player in 2005 (t = 5 ) ? (d) When will 0.8 (80 % ) o f U.S. households own a DVD player?

pet)

500 1 + 83 . 33e-O. 1 621

1000

= --+---- 91 0.-:4:037" 3 2. 33e-7 1

(a) (b) (c) (d) (e)

Determine the carrying capacity of the environment. What is the growth rate of the bald eagle? What is the population after 3 years? When will the population be 3 00 eagles? How long does it take for the population to reach one half of the carrying capacity? 27. The Challenger Disaster After the Challenger disaster in 1 986, a study was made of the 23 launches that preceded the fatal flight. A mathematical model was developed involving the relationship between the Fahrenheit temperature x around the O-rings and the number y of eroded or leaky primary O-rings. The model stated that y =

[i;TI

1

+

6 (5.085 - 0.l l56x ) e

where the number 6 indicates the 6 primary O-rings on the spacecraft. (a) What is the predicted number of eroded or leaky primary O-rings at a temperature of 1 00°F? (b) What is the predicted number of eroded or leaky primary O-rings at a temperature of 60°F? (c) What is the predicted number of eroded or leaky primary O-rings at a temperature of 30°F? (d) Graph the equation using a graphing utility. At what temperature is the predicted number of eroded or leaky O-rings I? 3 ? 5?

represents the population (in grams) of a bacterium after I hours. (a) Determine the carrying capacity of the environment. (b) What is the growth rate of the bacteria? (c) Determine the initial population size. (d) What is the population after 9 hours? (e) When will the population be 700 grams? (E) How long does it take for the population to reach one half the carrying capacity? 26. Population of an Endangered Species Often environmen talists capture an endangered species and transport the species to a controlled environment where the species can produce offspring and regenerate its population. Suppose that six American bald eagles are captured, transported to Montana, and set free. B ased on experience, the environmentalists

Linda Tappin, "Analyzing Data Relating to the Challenger Disaster," Mathematics Teacher, Vol. 87, No. 6, September 1 994, pp. 423-426. Source:


PREPARING FOR THI S SECTION •

•

Building Linear Functions from Data (Section 4.2, pp. 287-290)

�1 �2 kfj 3

OBJECTIVES

487

Building Exponential, Logarithmic, and Logistic Models from Data

SECTION 6.9

B uilding Quadratic Functions from Data (Section 4.4, pp. 309-310)

Use a Graphing Utility to Fit a n Exponential Fu nction to Data (p. 487) Use a Graphing Utility to Fit a Logarithmic Fu nction to Data (p.489) Use a Graphing Utility to Fit a Logistic Function to Data (p. 490)

+

In Section 4.2, we discussed how to find the linear function of best fit ( y ax b), and in Section 4.4, we discussed how to find the quadratic function of best fit ( y ax2 + bx c). In this section we will discuss how to use a graphing utility to find equations of best fit that describe the relation between two variables when the relation is thought to be exponential ( y abX) , logarithmic ( y a + b In ) or logistic (y 1 + ae . ) . As before, we draw a scatter diagram of the data to help to determine the appropriate model to use. Figure 44 shows scatter diagrams that will typically be observed for the three models. Below each scatter diagram are any restrictions on the values of the parameters. =

+

=

=

c

=

Figure 44

'

.

abx, a > 0, b > 1

x

y

y

y

x

Exponential

• 0 ° ' : .: •

.

'

•

y=

-b

y

Y

x ,

=

y=

°

0 , . '

•

..

..

' '

abx, 0 0, c > 0

Logistic

Most graphing utilities have REGression options that fit data to a specific type of curve. Once the data have been entered and a scatter diagram obtained, the type of curve that you want to fit to the data is selected. Then that REGression option is used to obtain the curve of best fit of the type selected. The correlation coefficient will appear only if the model can be written as a linear expression. As it turns out, r will appear for the linear, power, exponential, and logarithmic models, since these models can be written as a linear expression. Remember, the closer Irl is to 1 , the better the fit. Let's look at some examples. r

� !:lIE!. '

1

Use a G raphing Utility to Fit an Exponential Function to Data

We saw in Section 6 .7 that the future value of money behaves exponentially, and we saw in Section 6 .8 that growth and decay models also behave exponentially. The next example shows how data can lead to an exponential model.

488

CHAPTER 6


Fitting an Exponential Function to Data

E XA M P L E 1 Ta ble 9

Number of Subscribers (in millionsl, y

r, x 1 985 (x = 1 1

0.34

1 986 (x = 21

0.68

1 987 (x = 31

1 .23

1 988 (x = 41

2.07

= 51

3.51

1 990 (x = 61

5.28

1 991 (x = 7 1

7.56

1 992 (x = 81

1 1 .03

1 993 (x = 91

1 6.01

1 994 (x = 1 0 1

24. 1 3

1 995 (x = 1 1 1

33.76

1 996 (x = 1 21

44.04

1 989 (x

1 997 (x = 131

55.31

1 998 (x = 1 41

69.21

1 999 (x = 1 5 1

86.05

2000 (x = 161

1 09.48

2001 (x = 1 7 1

1 28.37

2002 (x = 1 8 1

140.77

2003 (x = 1 9 1

1 58.72

2004 (x = 201

1 82.14

2005 (x = 2 1 1

207.90

Kathleen is interested in finding a function that explains the growth of cell phone usage in the United States. She gathers data on the number (in millions) of U.S. cell phone subscribers from 1985 through 2005. The data are shown in Table 9. (a) Using a graphing utility, draw a scatter diagram with year as the independent variable. (b) Using a graphing utility, fit an exponential function to the data. (c) Express the function found in part (b) in the form A = Aoela. (d) Graph the exponential function found in part (b) or (c) on the scatter diagram. (e) Using the solution to part (b) or (c), predict the number of U.S. cell phone subscribers in 2009. (f) Interpret the value of k found in part (c). Solution

(a) Enter the data into the graphing utility, letting 1 represent 1985, 2 represent 1986, and so on. We obtain the scatter diagram shown in Figure 45. (b) A graphing utility fits the data in Figure 45 to an exponential function of the form y = abx using the EXPonential REGression option. From Figure 46 we find that y = abx = 0.66703(1.3621)--

=

663.0 = 661 . 6 1 + 71 .6e-O.5470(1 9 )


•

6.9 Assess You r U nde rsta n d i ng Applications and Extensions

1. Biology A strain of E-coli B eu 397-recA441 is placed into a nutrient broth at 30°Celsius and allowed to grow. The fol lowing data are collected. Theory states that the number of bacteria in the petri dish will initially grow according to the law of uninhibited growth. The population is measured using an optical device in which the amount of light that passes through the petri dish is measured.

�------------� TIme (hours), Population, y o

x

0.09

following data are collected. Theory states that the number of bacteria in the petri dish will initially grow according to the law of uninhibited growth. The population is measured using an optical device in which the amount of light that passes through the petri dish is measured.

� � :�' 1

Domain: the interval ( - 00 , Range: the interval (0, 00 )

00

)

x.

x-intercepts: none; y-intercept: 1 Horizontal asymptote: x-axis (y

=

0) as x -> - 00

Increasing; one-to-one; smooth; continuous

f(x)

=

See Figure 21 for a typical graph. aX , 0 < a < 1

Domain: the interval ( - 00 , Range: the in terval (0,

00

)

00

)

x-intercepts: none; y-intercept: 1 Horizontal asymptote: x-axis (y

=

0) as x -> 00

Decreasing; one-to-one; smooth; continuous See Figure 25 for a typical graph.

Chapter Review

Number

e

(p. 429)

Property of ex(wnents (p. 431) Properties of the logarithmic function (p. 440)

(

Value approached by the expression 1 If a" = aV, then u = v.

f(x) = log" x,

a

>

(y= log" x means x

1 =

+

1

n

)

"

as n � 00; that is, lim

1/---",00

1

495

"

( 1 + -) = e n

Domain: the interval (0,00) Range: the interval (-00,00) x-intercept: 1;y-intercept;none

aY )

Vertical asymptote: x= 0 (y-axis) Increasing; one-to-one; smooth; continuous See Figure 30(b) for a typical graph. f(x) = log" x,

0

O. (e) Find r1(x) . Graph r1 on the same Cartesian plane asf

84.

=4

4.1 - 14,4-.1 =5 Suppose that f(x) =lOg 3(X + 1) - 4. (a) Graphf (b) What isf(8)? What point is on the graph of f? (c) Solve f(x) = -3. What point is on the graph off? (d) Based on the graph drawn in part (a), solve f(x) < O. (e) Find r 1(x). Graph r1 on the same Cartesian plane asf

In Problems 85 and 86, use the foLLowing result: If x is the atmospheric pressure (measured in miLLimeters of mercury), then the formula for the altitude hex) (measured in meters above sea level) is hex) =(30T

+

8000) 10g

(:0)

where T is the temperature (in degrees Celsius) and Po is the atmospheric pressure at sea level, which is approximately 760 miLLimeters of mercury. 86. Finding the Height of a Mountain How high is a mountain 85. Finding the Altitude of an Airplane At what height is a Piper if instruments placed on its peak record a temperature of SoC Cub whose instruments record an outside temperature of O°C and a barometric pressure of 500 millimeters of mercury? and a barometric pressure of 300 millimeters of mercury?

An amplifier's power output P (in watts) is related to its decibel voltage gain d by the formula P =25eo1d . .

87. Amplifying Sound

(a) Find the power output for a decibel voltage gain of 4 decibels. (b) For a power output of 50 watts, what is the decibel voltage gain? A telescope is limited in its usefulness by the brightness of the star that it is aimed at and by the diameter of its lens. One measure of a star's bright ness is its magnitude; the dimmer the star, the larger its mag nitude. A formula for the limiting magnitude L of a telescope,

88. Limiting Magnitude of a Telescope

Chapter Review

that is, the magnitude of the dimmest star that it can be used to view, is given by L = 9 + 5.1 log d

The annual growth rate of the world's population in 2005 was k 1.15% = 0.0115. The popUlation of the world in 2005 was 6,451,058,790. Letting t 0 repre sent 2005, use the uninhibited growth model to predict the world's population in the year 2015. Source: u.s. Census Bureau

95. World Population

=

=

where d is the diameter (in inches) of the lens. (a) What is the limiting magnitude of a 3.5-inch telescope? (b) What diameter is required to view a star of magnitude 14?

The half-life of radioactive cobalt is 5.27 years. If 100 grams of radioactive cobalt is present now, how much will be present in 20 years? In 40 years? 97. Federal Deficit In fiscal year 2005, the federal deficit was $319 billion. At that time, 10-year treasury notes were paying 4.25% interest per annum. If the federal government financed this deficit through lO-year notes, how much would if have to pay back in 2015? Source: u.s. Treasury Department

96. Radioactive Decay

The number of years n for a piece of ma chinery to depreciate to a known salvage value can be found using the formula log s log i 11. = 10g(1 d)

89. Salvage Value

-

-

where s is the salvage value of the machinery, i is its initial value, and d is the annual rate of depreciation. (a) How many years will it take for a piece of machinery to decline in value from $90,000 to $10,000 if the annual rate of depreciation is 0.20 (20%)? (b) How many years will it take for a piece of machinery to lose half of its value if the annual rate of deprecia tion is 15%? 90. Funding a College Education A child's grandparents pur chase a $10,000 bond fund that matures in 18 years to be used for her college education. The bond fund pays 4% interest compounded semiannually. How much will the bond fund be worth at maturity? What is the effective rate of interest? How long will it take the bond to double in value under these terms? 91. Funding a College Education A child's grandparents wish to purchase a bond that matures in 18 years to be used for her college education. The bond pays 4% interest compounded semiannually. How much should they pay so that the bond will be worth $85,000 at maturity? 92. Funding an IRA First Colonial Bankshares Corporation advertised the following IRA investment plans.

98. Logistic Growth

The logistic growth model pet)

OJ: •

1 +

0.8 1.67e

O . 161

-

The following data were collected by placing a temperature probe in a portable heater, removing the probe, and then recording temperature over time.

99. CBL Experiment

Time (sec.)

0

Temperature (OF)

165.07 164.77

For each $5000 Maturity Value Desired At a Term of:

$620.17

20 Years

$1045.02

15 Years

$1760.92

10 Years

$2967.26

5 Years

(a) Assuming continuous compounding, what annual rate of interest did they offer? (b) First Colonial Bankshares claims that $4000 invested today will have a value of over $32,000 in 20 years. Use the answer found in part (a) to find the actual value of $4000 in 20 years. Assume continuous compounding. The bones of a prehistoric man found in the desert of New Mexico contain approximately 5% of the original amount of carbon 14. If the half-life of carbon 14 is 5600 years, approximately how long ago did the man die?

93. Estimating the Date That a P rehistoric Man Died

A skillet is removed from an oven whose temperature is 450°F and placed in a room whose tem perature is 70°F After 5 minutes, the temperature of the skil let is 400°F. How long will it be until its temperature is 150°F?

94. Temperature of a Skillet

=

represents the proportion of new cars with a global posi tioning system (GPS). Let t = 0 represent 2006, t = 1 rep resent 2007 , and so on. (a) What proportion of new cars in 2006 had a GPS? (b) Determine the maximum proportion of new cars that have a GPS. (c) Using a graphing utility, graph pet) . (d) When will 75% of new cars have a GPS?

Target IRA Plans

Deposit:

499

2 3 4 5

163.99 163.22 162.82 161.96

6

161.20

7

160.45

8

159.35

9

158.61

10

157.89

12

156.11

11

156.83

13

155.08

14

154.40

15

153.72

According to Newton's Law of Cooling, these data should follow an exponential model. (a) Using a graphing utility, draw a scatter diagram for the data. (b) Using a graphing utility, fit an exponential function to the data.

U1Q51 CUll.

data. (c) Using a graphing utility, draw the logarithmic function found in part (b) on the scatter diagram. (d) Use the function found in part (b) to predict the wind chill factor if the air temperature is 15°F and the wind speed is 23 mph.

(d) According to the function found in part (b), what is the maximum number of people who will catch the cold? In reality, what is the maximum number of people who could catch the cold? (e) Sometime between the second and third day, 10 people in the town had a cold. According to the model found in part (b), when did 10 people have a cold? (f) How long will it take for 46 people to catch the cold?

Jack and Diane live in a small town of 50 people. Unfortunately, both Jack and Diane have a cold.

101. Spreading of a Disease

CHAPTER TEST 1.

x + 2 and g(x) = 2x x - 2 (a) fog and state its domain (b) (g f )(-2) (c) (f g)(-2)

Given f(x)

=

--

+

In Problems 12 and 13, use the given function f to: (a) Find the domain of ! (b) Graph! (c) From the graph, determine the range and any asymptotes of! (d) Findr\ the inverse of! (e) Use r 1 to find the range of! (f) Graph rl.

. 5, fmd:

°

°

2.

Determine whether the function is one-to-one. (a) y = 4 x2 + 3 (b) y

3.

4.

=

Vx+3

-

5

Find the inverse of f(x)

=

2

---

and check your answer.

3x 5 State the domain and the range of f and rl.

If the point (3, 5 ) is on the graph of a one-to-one function f, what point must be on the graph of rl?

-

In Problems calculator. 5. Y

=

2 43

5-7,

find the unknown value without using a 6.

10gb 16

=

2

7.

logs x

=

4

In Problems 8-11, use a calculator to evaluate each expression. Round your answer to three decimal places. 8. e 3 + 2 9. log 20 10.

log3 2 1

11.

In 133

12. f(x)

=

4-'+1 - 2

13. f(x)

=

1 - logs(x - 2)

In Problems 14-19, solve each equation. 14. 5x+ 2 = 125 15. log(x + 9) = 2 2 17. log(x + 3) log(x + 6) 16. 8 2e-x 4 3 18. y+ = eX 19. log2(x - 4) + log2(x + 4)

-

=

=

=

3

4x3 as the sum and/or difference of x 2 - 3x 18 logarithms. Express powers as factors.

(

)

20.

Write log2

21.

A 50-mg sample of a radioactive substance decays to 34 mg after 30 days. How long will it take for there to be 2 mg remaining?

-

Chapter Review

that is, the magnitude of the dimmest star that it can be used to view, is given by L 9 + 5. 1 log d

The annual growth rate of the world's population in 2005 was k= 1.15%= 0. 0115. The population of the world in 2005 was 6,451,058,790. Letting t = 0 repre sent 2005, use the uninhibited growth model to predict the world's population in the year 2015. Source: Us. Census Bureau

95. World Population

=

where d is the diameter (in inches) of the lens. (a) What is the limiting magnitude of a 3. 5-inch telescope? (b) What diameter is required to view a star of magnitude 14?

The half-life of radioactive cobalt is 5.27 years. If 100 grams of radioactive cobalt is present now, how much will be present in 20 years? In 40 years? 97. Federal Deficit In fiscal year 2005, the federal deficit was $319 billion. At that time, 10-year treasury notes were paying 4.25% interest per annum. If the federal government financed this deficit through 10-year notes, how much would if have to pay back in 2015? Source: Us. Treasury Department 96. Radioactive Decay

The number of years n for a piece of ma chinery to depreciate to a known salvage value can be found using the formula log s - log i n= 10g(1 - d)

89. Salvage Value

where s is the salvage value of the machinery, i is its initial value, and d is the annual rate of depreciation. (a) How many years will it take for a piece of machinery to decline in value from $90,000 to $10,000 if the annual rate of depreciation is 0.20 (20%)? (b) How many years will it take for a piece of machinery to lose half of its value if the annual rate of deprecia tion is 15%? A child's grandparents pur chase a $10,000 bond fund that matures in 18 years to be used for her college education. The bond fund pays 4% interest compounded semiannually. How much will the bond fund be worth at maturity? What is the effective rate of interest? How long will it take the bond to double in value under these terms? 91. Funding a College Education A child's grandparents wish to purchase a bond that matures in 18 years to be used for her college education. The bond pays 4% interest compounded semiannually. How much should they pay so that the bond will be worth $85,000 at maturity? 92. Funding an IRA First Colonial Bankshares Corporation advertised the following IRA investment plans.

98. Logistic Growth

The logistic growth model P( t ) =

90. Funding a College Education

�

+

1

0.8 .67e

O. 16t

-

Time (sec.)

0

Temperature (OF)

165.07 164.77

For each $5000 Maturity Value Desired At a Term of:

$620.17

20 Years

$1045.02

15 Years

$1760.92

10 Years

$2967.26

5 Years

(a) Assuming continuous compounding, what annual rate of interest did they offer? (b) First Colonial Bankshares claims that $4000 invested today will have a value of over $32,000 in 20 years. Use the answer found in part (a) to find the actual value of $4000 in 20 years. Assume continuous compounding. The bones of a prehistoric man found in the desert of New Mexico contain approximately 5% of the original amount of carbon 14. If the half-life of carbon 14 is 5600 years, approximately how long ago did the man die?

93. Estimating the Date That a P rehistoric Man Died

A skillet is removed from an oven whose temperature is 450°F and placed in a room whose tem perature is 70°F. After 5 minutes, the temperature of the skil let is 400°F. How long will it be until its temperature is 150°F?

94. Temperature of a Skillet

1

represents the proportion of new cars with a global posi tioning system (GPS). Let t = 0 represent 2006, t = 1 rep resent 2007, and so on. (a) What proportion of new cars in 2006 had a GPS? (b) Determine the maximum proportion of new cars that have a GPS. � (c) Using a graphing utility, graph pet) . (d) When will 75% of new cars have a GPS? 99. CB L Experiment The following data were collected by placing a temperature probe in a portable heater, removing the probe, and then recording temperature over time.

Target IRA Plans

Deposit:

499

2 3

163.99

163.22

4

162.82

5

161.96

6

161.20

7

160.45

8

159.35

9

158.61

10

157.89

11

156.83

12

156.11

13

155.08

14

154.40

15

153.72

According to Newton's Law of Cooling, these data should follow an exponential model. (a) Using a graphing utility, draw a scatter diagram for the data. (b) Using a graphing utility, fit an exponential function to the data.

500

CHAPTER 6


Those who come in contact with someone who has this cold will themselves catch the cold. The following data represent the number of people in the small town who have caught the cold after t days.

(c) Graph the exponential function found in part (b) on the scatter diagram. (d) Predict how long it will take for the probe to reach a temperature of llO°F. The following data represent the wind speed (mph) and wind chill factor at an air temperature of 15°F.

100. Wind Chill Factor

�:.

J

J. � l.f Days. t 0

Wind Speed (mph)

2

Wind Chill Factor (OF)

7

5

15

0

20

-2

30

-5

35

-7

8

5

30

7

42

8

22

37

44

(a) Using a graphing utility, draw a scatter diagram of the data. Comment on the type of relation that appears to exist between the days and number of people with a cold. (b) Using a graphing utility, fit a logistic function to the data. (c) Graph the function found in part (b) on the scatter diagram. (d) According to the function found in part (b), what is the maximum number of people who will catch the cold? In reality, what is the maximum number of people who could catch the cold? (e) Sometime between the second and third day, 10 people in the town had a cold. According to the model found in part (b), when did 10 people have a cold? (f) How long will it take for 46 people to catch the cold?

Source: U.S. National Weather Service

(a) Using a graphing utility, draw a scatter diagram with wind speed as the independent variable. (b) Using a graphing utility, fit a logarithmic function to the data. (c) Using a graphing utility, draw the logarithmic function found in part (b) on the scatter diagram. (d) Use the function found in part (b) to predict the wind chill factor if the air temperature is 15°F and the wind speed is 23 mph. ,J 101. Spreading of a Disease

Jack and Diane live in a small town of 50 people. Unfortunately, both Jack and Diane have a cold.

'"

4

14

6

-4

25

2

3

4

3

10

Number of People with Cold. C

CHAPTER TEST 1.

2.

x + 2 and g(x) = 2x x-2 (a) fog and state its domain (b) (g ° f)(- 2) (c) (f ° g)(- 2)

Given f(x)

4.

=

Vx+3

-

In Problems 12 and 13, use the given function f to: (a) Find the domain of ! (b) Graph! (c) From the graph, determine the range and any asymptotes of! (d) Findr\ the inverse of! (e) Use rl to find the range off (f) Graphrl.

5, find:

5

2 and check your answer. 3x - 5 State the domain and the range of f and rl.

. . Fmd the mverse of f(x)

=

---

If the point (3, - 5) is on the graph of a one-to-one function f, what point must be on the graph of rl?

In Problems calculator. S.

+

Determine whether the function is one -to-one. (a) y = 4 x 2 + 3 (b) y

3.

=

3x

=

243

5-7,

find the unknown value without using a 6.

10gb 16

=

2

7.

logs x

=

4

In Problems 8-11, use a calculator to evaluate each expression. Round your answer to three decimal places. 8. e3 + 10.

2

9.

log 20

log 321

11.

In 133

12. f(x)

=

4x+1

13. f(x)

=

1 - logs(x - 2)

- 2

In Problems 14-19, solve each equation. 2 125 15. lo g(x + 9) = 2 14. 5x+ 17. log(x2 + 3) = log(x 16. 8 - 2e x = 4 =

18. y+3

=

-

eX

19.

log 2(x - 4)

+

log 2(x

+

+

6) 4)

=

3

4x 3 as the sum and/or difference of x- - 3x - 18 logarithms. Express powers as factors.

(

7

)

20.

Write log 2

21.

A 50-mg sample of a radioactive substance decays to 34 mg after 30 days. How long will it take for there to be 2 mg remaining?

Cumulative Review

22.

23.

(a) If $1000 is invested at 5% compounded monthly, how much is there after 8 months? (b) If you want to have $1000 in 9 months, how much do you need now to place in a savings account now that pays 5% compounded quarterly? (c) How long does it take to double your money if you can invest it at 6% compounded annually? The decibel level,

D

10

=

10 10g

(:0),

(a) If the shout of a single person measures 80 decibels, how loud will the sound be if two people shout at the same time? That is, how loud would the sound be if the in tensity doubled? (b) The pain threshold for sound is 1 25 decibels. If the Athens Olympic Stadium 2004 (Olympiako Stadio Athinas 'Spyros Louis') can seat 74,400 people, how many people in the crowd need to shout at the same time for the resulting sound level to meet or exceed the pain threshold? (Ignore any possible sound dampen ing.)

D, of sound is given by the equation

where I is the intensity of the sound and

= 10-12 watt per square meter.

CUMULATIVE REVIEW 1.

9.

Is the following graph the graph of a function? If it is, is the function one-to-one?

y

10.

/

-4

2. 3.

4

x 11.

For the function [(x) = 2X2

- 3x + 1 , find the following: (b) [(-x) (c) [(x + h)

(a) [(3)

Determine which of the following points are on the graph of x2 + l 1.

U.

13.

=

4.

5. 6.

7.

Solve the equation 3(x

-

2) = 4(x

Graph the line 2x - 4y = 16.

+ 5).

(a) Graph the quadratic function [(x) = -x2 + 2x - 3 by determining whether its graph opens up or down and by finding its vertex, axis of symmetry, y-intercept, and x-intercept(s), if any. (b) Solve [(x) ::; O.

Determine the quadratic function whose graph is given in the figure. y

-2

8.

-10

Graph [(x) =

(4, -8)

Vertex:

3(x + 1 )3

-

8

x

2 using transformations.

SOl

14. •

;.

15.

Given that [(x) = x2

+ 2 and g(x) =

and state its domain. What is [(g(5»?

x

= 3' find [(g(x»

For the polynomial function [(x) = 4x3 + 9x2 - 30x - 8: (a) Find the real zeros of f. (b) Determine the intercepts of the graph of f. (c) Use a graphing utility to approximate the local maxima and local minima. (d) Draw a complete graph of f. Be sure to label the in tercepts and turning points.

For the function g(x) = Y + 2: (a) Graph g using transformations. State the domain, range, and horizontal asymptote of g. (b) Determine the inverse of g. State the domain, range, and vertical asymptote of g-l. (c) On the same graph as g, graph g-l. Solve the equation 4x 3 = 82x.

-

+ 1) + log3( 2x - 3) = log99 Suppose that [(x) + 2). Solve: (a) [(x) = O. (b) [(x) > O. (c) [(x) = 3. Solve the equation:

log3(x = log3(x

Data Analysis The following data represent the percent of all drivers by age that have been stopped by the police for any reason within the past year. The median age represents the midpoint of the upper and lower limit for the age range.

Age Range

Median Age,x

Percentage Stopped, y

16- 1 9

1 7.5

1 8.2

20-29

24.5

1 6.8

30-39

34.5

1 1 .3

40-49

44.5

9.4

50-59

54.5

7.7

2::60

69.5

3.8

(a) Using your graphing utility, draw a scatter diagram of the data treating median age, x, as the independent variable. (b) Determine a model that you feel best describes the rela tion between median age and percentage stopped. You may choose from among linear, quadratic, cubic, power, exponential, logarithmic, or logistic models. (c) Provide a justification for the model that you selected in part (b).

502

CHAPTER 6


CHAPTER PROJECTS 2.

3.

The TempControl Company has a container that reduces the temperature of a liquid from 200° to 110°F in 25 min utes by maintaining a constant temperature of 60°F. The Hot'n'Cold Company has a container that reduces the temperature of a liquid from 200° to 1 20°F in 20 minutes by maintaining a constant temperature of 65°F.

You need to recommend which container the restaurant should purchase. (a) Use Newton's Law of Cooling to find a function relat ing the temperature of the liquid over time for each container. I.

(b) How long does it take each container to lower the coffee temperature from 200° to 1 30°F? Hot Co ffee A fast-food restaurant wants a special container to hold coffee. The restaurant wishes the container to quickly cool the coffee from 200° to 130°F and keep the liquid be tween 1 10° and 130°F as long as possible. The restaurant has three containers to select from . 1.

The CentiKeeper Company has a container that reduces the temperature of a liquid from 200° to 100°F in 30 min utes by maintaining a constant temperature of 70°F.

(c) How long will the coffee temperature remain between 110° and 130°F? This temperature is considered the opti mal drinking temperature. (d) Graph each function using a graphing utility. (e) Which company would you recommend to the restau rant? Why? (f) How might the cost of the container affect your decision ?

The following projects are available on the Instructor's Resource Center (IRC): II. Project at Motorola Thermal Fatigue of Solder Connections

Product reliability is a major concern of a manufacturer. Here a logarithmic transformation is used to simplify the analysis of a cell phone's ability to withstand temperature change.

III. Depreciation of a New Car

Resale value is a factor to con sider when purchasing a car, and exponential functions provide a way to compare the depreciation rates of different makes and models.

Trigonometric Functions Surf's Up: Using Models To Predict Huge Waves

GALVESTON, Feb. 15, 2005

Estimated Mean Circulation Field 2-m depth 24-Aug-2006

.' R

Texas

•

30' are a ship captain and there might be Reliability Index: 7 50-foot waves headed your way, you would appreciate some information (1 low 10 high) 29'N about them, right? Y':. That's the idea behind a wave model system a Texas A&M University at ' ...'>r(.�....� '"" - , 30' Galveston professor has developed. His detailed wave prediction system is .. :.. ,� --....... � "'\1 .. currently in use in the Gulf of Mexico and the Gulf of Maine. .... , ..... '.....- � 28"N • Vijay Panchang, head of the Department of Maritime Systems Engi " "' N 't\J'o neering, doesn't make waves; he predicts what they will do, when they will 30' do it, and how high they will get. Using data provided daily from NOAA and his own complex mathemati cal models, Panchang and research engineer Doncheng Li provide daily wave model predictions for much of the Texas coast, the Gulf of Mexico, and the Gulf of Maine. Their simulations, updated every 12 hours, provide a forecast for two days ahead. Because the models use wind data, tsunamis that are created by under sea earthquakes cannot be predicted. But that is not to say his modeling sys 30 tem does not come up with some big waves. 30' 30 30' 96'W 93'W 95'W 94'W His wave model predicted big waves in November 2003 in the Gulf of Maine, and it was accurate; waves as high as 30 feet were recorded during one storm even in coastal regions. Last summer during H urricane Ivan, a buoy located 60 miles south of the Alabama coast recorded a whopping 60-foot wave. "There may have been higher waves because right after recording the 60-foot wave, the buoy snapped and stopped functioning," he says.

- If you

Source: Science Daily www.sciencedaily.comlreleasesI20051021050222193810.htm. Posted February 23, 2005,


A Look Back

In C h a pter 3, we began ou r discussion of functions. We defined domain and range

and independent and d ependent vari a bles; we found the value of a function and g ra p h ed functions. We continued our stu d y of functions by l isting properties that a function might have, like being even or odd, and we created a l i brary of functions, na ming key functions and l isting their properties, including the g ra p h .

Outline

7.1

Angles and Their Measure

7.2 Right Triangle Trigonometry

7.3 Computing the Values ofTrigonometric

Functions ofAcute Angles

7.4 Trigonometric Functions of

General Angles

A Look Ahead

In this c h a pter we define the trigon o m etric functions, six functions that have w i d e

a p p l i cation. We s h a l l t a l k a bout their d om a in a n d range, s e e how t o find values, g ra p h them, and d evelop a l i st of their properties. There a re two widely accepted a pp roaches to the development of the trigon o metric functions: one uses right trian g les; the other uses circles, especia l l y the unit circle. In this book, we d evelop the trigonometric functions using right trian g l es. In

7.5 Unit Circle Approach; Properties ofthe

Trigonometric Functions

7.6 Graphs ofthe Sine and Cosine Functions 7.7 Graphs ofthe Tangent, Cotangent,

Cosecant, and Secant Functions

7.8 Phase Shift; Sinusoidal Curve Fit ting

Chapter Review

Chapter Test

Section 7.5, we introduce trigonometric functions using the unit circle and show

Cumulative Review

that this a pproach leads to the d efin ition using right triang les.

Chapter Projects

503

30'

504

CHAPTER 7




Circumference and Area of a Circle (Chapter R, Review, Section R . 3, pp. 3 1-32)

"NOW Work


OBJECTIVES 1 Convert between Deci mals a n d Deg rees, M i n utes, Seconds Forms for A n g l es (p. 506)

2 Find the Arc Length of a Circle (p. 507)

3 Convert from Deg rees to R a d i a n s and from Radians to Degrees (p. 508)

4 Find the Area of a Sector of a C i rcle (p. 5 1 1 )

5 Find the Linea r Speed of a n Object Trave l i n g i n Circ u l a r Motion (p. 5 1 2)

A ray, or half-line, is that portion of a line that starts at a point Von the line and extends indefinitely in one direction. The starting point Vof a ray is called its vertex. See Figure 1 . If two rays are drawn with a common vertex, they form an angle. We call one of the rays of an angle the initial side and the other the terminal side. The angle formed is identified by showing the direction and amount of rotation from the initial side to the terminal side. If the rotation is in the counterclockwise direction, the angle is positive; if the rotation is clockwise, the angle is negative. See Figure 2.

Figure 1

----.-----.� Line V Ray

Figure

2

Clockwise rotation

Counterclockwise rotation

Positive angle

Negative angle

Positive angle

(a)

(b)

Counterclockwise rotation

(c)

Lowercase Greek letters, such as a (alpha), f3 (beta), 'Y (gamma), and e (theta), will often be used to denote angles. Notice in Figure 2(a) that the angle a is positive because the direction of the rotation from the initial side to the terminal side is counterclockwise. The angle f3 in Figure 2(b) is negative because the rotation is clockwise. The angle 'Y in Figure 2( c) is positive. Notice that the angle a in Figure 2( a) and the angle 'Yin Figure 2( c) have the same initial side and the same terminal side. However, a and 'Yare unequal, because the amount of rotation required to go from the initial side to the terminal side is greater for angle 'Y than for angle a. An angle e is said to be in standard position if its vertex is at the origin of a rec tangular coordinate system and its initial side coincides with the positive x-axis. See Figure 3. Figure

y

3

x

(a)

e is in standard position; e is positive

I nitial side

(b)

e is in standard position; e is negative

x

SECTION 7.1


50S

When an angle e is in standard position, the terminal side will lie either in a quadrant, in which case we say that e lies in that quadrant, or e will lie on the x-axis or the y-axis, in which case we say that e is a quadrantal angle. For example, the angle e in Figure 4(a) lies in quadrant II, the angle e in Figure 4(b) lies in quadrant IV, and the angle e in Figure 4(c) is a quadrantal angle. Figure 4

y

y

x (a)

x

a lies in quadrant II

(b)

x

a lies in quadrant IV

(c) a is a quad rantal angle

We measure angles by determining the amount of rotation needed for the ini tial side to become coincident with the terminal side. The two commonly used mea sures for angles are degrees and radians. Degrees 360° due to the Baby lonian year, which had 360 days in it . • HISTORICAL NOTE

One counterclock

wise rotation is

The angle formed by rotating the initial side exactly once in the counterclockwise direction until it coincides with itself (1 revolution) is said to measure 360 degrees, 1 abbreviated 360°. One degree, r, is -.., - revolution. A right angle is an angle that .)60 1 1 measures 90°, or '4 revolution; a straight angle is an angle that measures 180°, or 2: revolution. See Figure 5. As Figure 5(b) shows, it is customary to indicate a right angle by using the symbol �.

Figure 5

erminal side

Vertex

(a)

e

E XA M P L E 1

T

TInitial side

1 revolution

'

;�

"

L

�Vertex

Initial side

Terminal side Vertex Initial side

(c) straight angle, � revolution

(b) right angle, revolution counter-clockwise, 90°

counterclockwise, 3600

*

counter-clockwise, 180°

It is also customary to refer to an angle that measures degrees.

e

degrees as an angle of

D rawing an Angle

Draw each angle. (b) -90° Solutio n

(a) An angle of 45° is

�

(c) 225° of a right

(d) 405° (b) An angle of -90° is

� revolution in

angle. See Figure 6.

the clockwise direction. See Figure 7.

Figure

Figure 7

6

Vertex Terminal side

Initial side

n,:-- l� -"u

506

CHAPTER 7


( c) An angle of 225° consists of a ro tation through 180° followed by a rotation through 45°. See Figure 8. Figure

(d) An angle of 405° consists of 1 rev olution (3 60°) followed by a ro tation through 45°. See Figure 9. Figure 9

8 Initial side

==:::>-

1

Now Work PROBLEM

• 11

Convert between Decimals and Degrees, Minutes, Seconds Forms for Angles

Although subdivisions of a degree may be obtained by using decimals, we also may use the notion of minutes and seconds. One minute, denoted by 1', is defined as

1 degree. 60

1 minute, or equivalently, 1 degree. 60 3 600 An angle of, say, 30 degrees, 40 minutes, 10 seconds is written compactly as 30°40 ' 10".

One second, denoted by

1", is defined as

To summarize:

1 counterclockwise revolution l' = 60" 1° = 60 '

=

3 60°

(1)

It is sometimes necessary to convert from the degree, minute, second notation ( DOM'S") to a decimal form, and vice versa. Check your calculator; it should be capable of doing the conversion for you. Before getting started, though, you must set the mode to degrees because there are two common ways to measure angles: degree mode and radian mode. (We will define radians shortly.) Usually, a menu is used to change from one mode to another. Check your owner's manual to find out how your particular calculator works. Now let's see how to convert from the degree, minute, second notation (DOM' S") to a decimal form, and vice versa, by looking at some examples:

15°30 '

32.25°

= 32°15 '

= 15 .5°

because

because 0 .25° =

30 '

()

1 ° 4

=

=

30"l'

1 "1° 4

=

r

l'

=

=

( )

1 ° 60

=

0.5°

(610Y

1 ( 60 ' ) 4

r 60' 1°

E X A M P LE 2

30"

= 15 '

=

Converting between Degrees, M i nutes, Seconds F orm and Deci mal Form

(a) Convert 50° 6' 21" to a decimal in degrees. Round the answer to four decimal places. (b) Convert 21.25 6° to the DOM'S" form. Round the answer to the nearest second.

SECTION 7.1

(a) Because

Solutio n

l'

=

(1) 60

0

and 1" =

50°6'21"

60 60

60

+

6'

50°

+

6·

"'" 50°

+

0.1°

=

507

( 1 ) ' = ( 1 ' 1 ) °' we convert as follows:

50°

=


+

21"

(1)

= 50. 1058°

60 +

0

+

21·

( 1 ' 1 )° 60 60

0 . 0058°

(b) We proceed as follows: 21.256° = 21° = 21 °

= 21° = 21° = 21°

+

+

+

+

+

0.256° (0.256) (60 ' ) 15 . 36' 15' + 0 . 36'

15' 21° + 15' "'" 21°15'22"

=

+

+

(0.36) (60") 21 . 6"

Convert fraction o f degree t o minutes;

Convert fraction of minute to seconds; l' Round to the nearest second.

Now Work PROB L EMS 2 3 AND 2 9

Ql!l:==;> -

1° = 60'. = 60". •

In many applications, such as describing the exact location of a star or the pre cise position of a ship at sea, angles measured in degrees, minutes, and even seconds are used. For calculation purposes, these are transformed to decimal fonn. In other applications, especially those in calculus, angles are measured using radians. Radians

A central angle is a positive angle whose vertex is at the center of a circle. The rays of a central angle subtend (intersect) an arc on the circle. If the radius of the circle is r and the length of the arc subtended by the central angle is also r, then the mea sure of the angle is 1 radian. See Figure 10(a). For a circle of radius 1, the rays of a central angle with measure 1 radian would subtend an arc of length 1. For a circle of radius 3, the rays of a central angle with measure 1 radian would subtend an arc of length 3. See Figure 10(b). Figure 10

(a)

2

(b)

Find the Arc Length of a Circle

Now consider a circle of radius r and two central angles, e and e1, measured in radi ans. Suppose that these central angles subtend arcs of lengths sand sl, respectively,

508

C HA PTER 7

Figure 1 1

e

e,


s

as shown in Figure 1 1 . From geometry, we know that the ratio of the measures of the angles equals the ratio of the corresponding lengths of the arcs subtended by these angles; that is, 8 (2) 81

s,

Suppose that 81 = 1 radian. Refer again to Figure 10(a). The length SI of the arc subtended by the central angle 81 = 1 radian equals the radius r of the circle. Then SI = r, so equation (2) reduces to S

8 1

THEOREM

or

r

S

=

(3)

r8

Arc Length

For a circle of radius r, a central angle of 8 radians subtends an arc whose length sis

I �----------------------------------�� S

=

(4)

r8

NOTE Formulas must be consistent with regard to the units used. In equation (4), we write s = re

To see the units, however, we must go back to equation (3) and write

r length units

s length units

e radians 1 radian

e radians s length u nits = r length units --1 radian

Since the radians cancel, we are left with s length units

=

(r length units)e

s

=

re

where e appears to be "dimensionless" but, in fact, is measured in radians. So, in using the formula s = re, the dimension for e is radians, and any convenient unit of length (such as inches or me ters) may be used for s and r. •

EXAM PLE 3

Finding the Length of an Arc of a C i rcle

Find the length of the arc of a circle of radius 2 meters subtended by a central an gle of 0.25 radian. Solution

We use equation (4) with r

=

2 meters and 8 = 0.25. The length s of the arc is

s = r8 �====> - NowWork P R O B L E M

3 Figure 1 2

1 revolution

=

2 7T radians

=

2 (0.25 )

=

0.5 meter

•

71

Convert from Degrees to Radians and from Radians to Degrees

Next we discuss the relationship between angles measured in degrees and angles measured in radians. Consider a circle of radius r . A central angle of 1 revolution will subtend an arc equal to the circumference of the circle (Figure 12). Because the circumference of a circle equals 21Tr, we use s = 21Tr in equation (4) to find that, for an angle 8 of 1 revolution, s = r8

21Tr 8

=

=

r8 21T radians

e = 1 revolutio n; s Solve for e.

=

27Tr

SECTION 7.1


509

From this we have,

1 revolution Since 1 revolution = 360°, we have

= 27r radians

(5)

360° = 27r radians D ividing both sides by 2 yields

180°

(6)

= 7r radians

Divide both sides of equation (6) by 180 . Then

7r rad'13n 180

1 degree = Divide both sides of (6) by 7r, Then

180 7r

- degrees

' = 1 radIan

We have the following two conversion formulas:

1 degree =

E XA M P L E 4

7r rad'Ian 180

1 rad'Ian

=

7r

(7)

Converting from Degrees to Radians

Convert each angle in degrees to radians, (a) 60° Solution

180

- d egrees

(a) 60°

(b) 150°

= 60·1 degree

(b) 150°

(e) lOr

(c) -45°

;

= 60· radian 1 0

= 150·1° = 150·� radian =

=

; radians

57r radians 6

180 7r , = - 7r c -45° = -45 . - radIan () - radian 4 180 7r 7r , (d) 90° = 90 . rad13n = 2' radians 180 7r (e) 107° = 107· radian ;::j 1.868 radians 180

•

Example 4, parts (a)-(d), illustrates that angles that are "nice" fractions of a revolution are expressed in radian measure as fractional multiples of 7r, rather than as decimals, For example, a right angle, as in Example 4 (d), is left in the

7r

form 2' radians, which is exact, rather than using the approximation

7r 2'

3,1416

, ' ' I ;::j --2- = 1 ,5708 rad'lans, When th e fractlOns are not "l11ce," we use th e d eClIna

approximation of the angle, as in Example 4 (e), � -

-

Now Work PRO B L EM S 3 5 AND 6 1

510

C H A PTER 7


EXAM P LE 5

Converting Radians to Degrees

Convert each angle in radians to degrees.

7T

3 7T

.

(d) Solution

7

; radians

( e) 3 radians

7T 7T . 7T 180 =- degrees = 30° 6 7T 6 6 3 7T 180 3 7T - radians = - degrees = 270° 2 2 7T 3 7T . 3 7T 180 -135° - - radIans - - · - dearees 4 b 4 7T 7 7T 180 7 7T . - radIans - - degrees = 420° 3 3 7T 180 3 radians = 3 · degrees:::::; 171.89° 7T

(a) - radian = - 1 radIan .

(b) (c) (d)

( e)

3 7T . ( C) - - radIans 4

. (b) 2 radians

(a) "6 radIan

�.,.",.,...

.

-.

=

=

=

.


•

47

Table 1 lists the degree and radian measures of some commonly encountered angles. You should learn to feel equally comfortable using degree or radian measure for these angles. Table 1

Degrees

00

Radians

0

300 7T 6 2100 77T 6

Degrees Radians

EXAMP LE 6

Figure 13

450 7T 4 2250 -

57T 4

600 7T 3 2400

-

471" 3

900 7T 2 2700

-

371" 2

13 5 0

1200 271" 3 3000 571" 3

1800

371" 4 3150

1500 571" 6 3300

7T

771" 4

117T 6

27T

3600

Finding the D istance between Two Cities

See Figure 13(a). The latitude of a location L is the angle formed by a ray drawn from the center of Earth to the Equator and a ray drawn from the center of Earth to L. See Figure 13(b ) . Glasgow, Montana, is due north of Albuquerque, New Mexico. Find the distance between Glasgow (48°9' north latitude) and Albuquerque (35°5 ' north latitude). Assume that the radius of Earth is 39 60 miles.

, North Pole

, North Pole

Equator

Equator

(a)

(b)

SECTION 7.1

51 1


The measure of the central angle between the two cities is 48°9' - 35°5' = 13°4' . We use equation ( 4) , s = r8, but first we must convert the angle of 13°4' to radians.

Solution

8 = 13°4' ;:::: 13.0667° i 4° 4' =-

=

7T 13.0667· - radian;:::: 0.228 radian 180

60

We use 8 = 0.228 radian and r = 3960 miles in equation ( 4 ) . The distance between the two cities is s = r8 = 3960·0 .228 ;:::: 903 miles

•

NOTE If the measure of an angle is given as 5, it is understood to mean 5 radians; if the measure of an angle is given as 5°, it mean s 5 degrees. •

4

Figure 1 4

.

Figure 15

() ()l

A Al

- = -

When an angle is measured in degrees, the degree symbol will always be shown. However, when an angle is measured in radians, we will follow the usual practice and omit the word radians. So, if the measure of an angle is given as

7T . stood to mean (5 radlan. &l' 1==::!P-

7T ' it is under6

Now Work P R O B L E M 1 0 1

Find the Area of a Sector of a Circle

Consider a circle of radius r. Suppose that 8, measured in radians, is a central angle of this circle. See Figure 14. We seek a formula for the area A of the sector (shown in blue ) formed by the angle 8. Now consider a circle of radius r and two central angles 8 and 81, both measured in radians. See Figure 15. From geometry, we know the ratio of the measures of the angles equals the ratio of the corresponding areas of the sectors formed by these angles. That is, A 8 81 Al Suppose that 81 for A, we find

=

2 7T radians . Then Al = area of the circle = 7Tr2. Solving 1 2

8 2 7T

8 81 i

A = AI- = 7Tr2- = -r28 AI

()1

THEOREM

= 7iT2 =

27f

Area of a Sector

The area A of the sector of a circle of radius 8 radians is

1 2

r

formed by a central angle of

7

A = - r8

(8)

EXA M P LE 7

I

��

L-________________________________

Finding the Area of a Sector of a Circle

Find the area of the sector of a circle of radius 2 feet formed by an angle of 30°. Round the answer to two decimal places. Solution

7T

We use equation (8) with r = 2 feet and 8 = 30° = - radian . [ Remember, in 6 equation (8 ) , 8 must be in radians. ]

1 2

1 2

7T 6

7T 3

A = - r28 = - ( 2 ? - = The area A of the sector is 1.05 square feet, rounded to two decimal places. __ •w4·

,,> -


79

.... .

_

.

512

CHAPTER 7


5

Find the Linear Speed of an Object Traveling in Circular Motion

We have already defined the average speed of an object as the distance traveled divided by the elapsed time. Figure 16

v

5

DEFINITION

=t

Suppose that an object moves around a circle of radius r at a constant speed. If s is the distance traveled in time t around this circle, then the linear speed v of the object is defined as v

s =t

(9)

I

�--------------------------------��

As this object travels around the circle, suppose that 8 (measured in radians) is the central angle swept out in time t. See Figure 16.

DEFINITION

The angular speed w (the Greek letter omega) of this object is the angle 8 (measured in radians) swept out, divided by the elapsed time t; that is,

8 w=t

(10)

I

�----------------------------------��

Angular speed is the way the turning rate of an engine is described. For exam ple, an engine idling at 900 rpm (revolutions per minute) is one that rotates at an angular speed of revolutions � radians radians = 900 . 27T = 1 8007T minute minute minute -wvetuttmf There is an important relationship between linear speed and angular speed:

---

900

linear speed =

v

= I

(9)

� = rt8 = r ( �) = r' w I s=

So, v

=

rw

rl3

i (10)

(11)

where w is measured in radians per unit time. When using equation (11), remember that v = !.. (the linear speed) has the t dimensions of length per unit of time (such as feet per second or miles per hour), r (the radius of the circular motion) has the same length dimension as s, and w (the angular speed) has the dimensions of radians per unit of time. If the angular speed is given in terms of revolutions per unit of time (as is often the case), be sure to convert it to radians per unit of time before attempting to use equation (11). Remember, 1 revolution = 2 7T radians. E XA M P L E 8

Finding Linear Speed

A child is spinning a rock at the end of a 2-foot rope at the rate of 1 80 revolutions per minute (rpm). Find the linear speed of the rock when it is released. Sol ution

Look at Figure 17. The rock is moving around a circle of radius r = 2 feet. The angular speed w of the rock is

w = 1 80

� radians radians revolutions =1 80 ·27T = 3607T minute minute ..revehrtiOn minute

---

SECTION 7.1

Figure

513


From equation (1 1), the linear speed v of the rock is

17

v = rw = 2 feet· 3607T

radians minute

=

n07T

feet . mmute

>:::::

feet 2262 -. mmute

The linear speed of the rock when it is released is 2262 ft/min &1'l

-


-

>:::::

25.7 mi/h r.

-' _.

_

__

97

l--iis:torical Feature

T

(1514-1576). Rhaeticus's

rigonometry was developed by Greek astronomers, who regarded

Rhaeticus

the sky as the inside of a sphere, so it was natural that triangles

trigonometric functions as ratios of sides of triangles, although he did

on a sphere were investigated early (by Menelaus of Alexandria

about AD

100)

and that triangles in the plane were studied much later.

book was the first to define the six

not give the functions their present names. Credit for this is due to Thomas Finck

(1583), but

Finck's notation was by no means universally

The first book containing a systematic treatment of plane and spherical

accepted at the time.The notation was finally stabilized by the textbooks

trigonometry was written by the Persian astronomer Nasir Eddin (about

of Leonhard Euler

AD

1250).

(1707-1783).

Trigonometry has since evolved from its use by surveyors, naviga is the person most responsible for

tors, and engineers to present applications involving ocean tides, the

moving trigonometry from astronomy into mathematics. His work

rise and fall of food supplies in certain ecologies, brain wave patterns,

Regiomontanus

(1436-1476)

was improved by Copernicus

(1473-1543)

and Copernicus's student

and many other phenomena.

7.1 Assess Your Understanding 'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red. 1.

What is the formula for the circumference C of a circle of ra dius r? (pp. 3 1-32)

2. What is the formula for the area A of a circle of radius r? (pp. 31-32)


3. An angle I} is in

__ __

if its vertex is at the origin of a rectangular coordinate system and its initial side coincides with the positive x-axis.

4. On a circle of radius r, a central angle of I} radians subtends an arc of length s = ; the area of the sector formed by 5.

this angle I} is A

6.True o r False 7f = 180. 7. Trite o r False 180° 7f radians . =

8.

__

=

__ .

An object travels around a circle of radius r with constant speed . If s is the distance traveled in time t around the circle and I} is the central angle (in radians) swept out in time t, then the linear speed of the object is v = and the angular speed of the object is w = . __

__

9.

On the unit circle, if s is the lengtb of the arc subtended by a central angle I}, measured in radians, then True or False

s = I}.

Trite or False

The area A of the sector of a circle of radius r

formed by a central angle of I} degrees is A

10.

�

= r21}.

True or False For circular motion on a circle of radius r, linear speed equals angular speed divided by r.

Skill Building

In Problems 11-22, draw each angle. 12. 60° 1 1. 30° 3 7f 47f 17 18. . 4 3

13.135°

-�

19.

14.- 120° 2 7f 20.- 3

15.450° 167f 21. 3

16.540° 2 17f 22. 4 -

In Problems 23-28, convert each angle to a decimal in degrees. Round your answer to two decimal places. 40°10'25 " 24.61°42'21" 25. 1°2'3 " 26.73°40'40 " 27.9°9'9"

'\.. 23.

In Problems 29-34, convert each angle to 29. 40.32° 30. 6 1.24°

D OM'S"

form. Round your answer to the nearest second. 31. 18 .255° 32. 29.411° 33.19.99°

34. 44.01°

514

C H A PTER 7


In Problems 35-46, convert each angle in degrees to radians. Express your answer as a multiple of 71". 35. 30° 36. 120 ° 37. 240° 38. 330° 39. -60° 41.

180°

42.

2 70°

43.

- 135°

In Problems 47-58, convert each angle in radians to degrees. 5 71" 5 71" 6 47. � 49 . -48. 3 4

'

71" 12

71"

5 71" 12

53. -

5 4. -

55 -•

2

44.

50. 56.

-225 °

45.

271"

-3

51.

40.

-90°

71"

"2 71" 6

-71"

57. --

-30°

46.

- 180°

52.

471"

58.

3 71"

-4

In Problems 59-64, convert each angle in degrees to radians. Express your answer in decimal form, rounded to two decimal places. 60. 73 ° 59. 17° 64. 350° 61. -40° 62. -51° 63. 125° In Problems 65-70, convert each angle in radians to degrees. Express your answer in decimal form, rounded to two decimal places. 65. 3 . 14 66. 0.75 67. 2 68. 3 69. 6 .32 70. V2

In Problems 71-78, s denotes the length of the arc of a circle of radius r subtended by the central angle 8. Find the missing quantity. Round answers to three decimal places. 1 . s= ? 71. r 10 meters, 8 = "2 radian, 72. r 6 feet, 8 = 2 radians, s = ? =

=

1

73.

8

= :3 ra d'Ian, s = 2 feet, r

75.

r

=

5 miles, s

77.

r

=

2 inches, 8 = 30°, s

79.

r

=

81.

8

= radian,

83.

r

85.

r

=

=

?

3 miles, 8 = ? =

?

�

74.

8

= radian, s

76.

r

=

6 meters, s = 8 meters, 8

78.

r

=

3 meters, 8 = 120°, s

80.

r

=

6 feet, 8 = 2 radians, A

82.

8

=

�

=

r =?

6 centimeters,

=

=

?

?

In Problems 79-86, A denotes the area of the sector of a circle of radius r formed by the central angle 8. Find the missing quantity. Round answers to three decimal places. 10 meters, 8

�

=

A A

5 miles,

C!3 =

=

�

A =?

radian,

= 2 square feet, r

=

?

= 3 square miles, 8 = ?

2 inches, 8 = 30°,

A

=

?

8

84. 86.

A

"3 2ft 1t

88·

1t

6

A S 4m

89.

?

radian,

A

= 6 square centimeters, r = ?

r = 6 meters,

A

= 8 square meters,

r

=

3 meters, 8

In Problems 87-90, find the length s and area A. Round answers to three decimal places. 87.

=

=

120° ,

A

=

?

90.

8

=?

(3 50° gem A

S


Movement of a Minute Hand The minute hand of a clock is 6 inches long. How far does the tip of the minute hand move in 15 minutes? How far does it move in 25 minutes? Round answers to two decimal places.

92.

Movement of a Pendulum A pendulum swings through an angle of 20° each second . If the pendulum is 40 inches long, how far does its tip move each second? Round answers to two decimal places.

93.

Area of a Sector Find the area of the sector of a circle of ra dius 4 meters formed by an angle of 45 °. Round the answer to two decimal places.

94.

Area of a Sector Find the area of the sector of a circle of ra dius 3 centimeters formed by an angle of 60°. Round the an swer to two decimal places.

SECTION 7.1

95.

96.

" 97.

Watering a Lawn A water sprinkler sprays water over a dis tance of 30 feet while rotating through an angle of 135°. What area of lawn receives water?

of

�

Distance between Cities Memphis, Tennessee, is due north of New Orleans, Louisiana. Find the distance be tween Memphis (35°9' north latitude) and New Orleans ( 29°57' north latitude). Assume that the radius of Earth is 3960 miles.

102.

Distance between Cities Charleston, West Virginia, is due north of Jacksonville, Florida. Find the distance be tween Charleston (38°21 ' north latitude) and Jacksonville (30°20' north l a titude) . Assume that the radius of Earth is 3960 miles.

103.

Linear Speed on Earth Earth rotates on an axis through its poles. The distance from the axis to a location on Earth 30° north latitude is about 3429.5 miles. Therefore, a location on Earth at 30° north latitude is spinning on a circle of radius 3429.5 miles. Compute the linear speed on the surface of Earth at 30° north latitude.

104.

Linear Speed on Earth Earth rotates on an axis through its poles. The distance from the axis to a location on Earth 40° north latitude is about 3033.5 miles. Therefore, a location on Earth at 40° north latitude is spinning on a circle of radius 3033.5 miles. Compute the linear speed on the surface of Earth at 40° north latitude.

105.

Speed of the Moon The mean distance of the Moon from Earth is 2.39 X 10 5 miles. Assuming that the orbit of the Moon around Earth is circular and that 1 revolution takes 27.3 days, find the linear speed of the Moon. Express your answer in miles per hour.

106.

Speed of Earth The mean distance of Earth from the Sun is 9.29 X 10 7 miles. Assuming that the orbit of Earth around the Sun is circular and that 1 revolution takes 365 days, find the linear speed of Earth. Express your answer in miles per hour.

107.

Pulleys Two pulleys, one with radius 2 inches and the other with radius 8 inches, are connected by a belt. (See the figure.) If the 2-inch pulley is caused to rotate at 3 revolutions per minute, determine the revolutions per minute of the 8-inch pulley.

radian is swept out, what is the angular speed of the

object? What is its linear speed? 9S.

Motion on a Circle An object is traveling around a cir cle with a radius of 2 meters. If in 20 seconds the object travels 5 meters, what is its angular speed? What is its linear speed?

99.

Bicycle Wheels The diameter of each wheel of a bicycle is 26 inches. If you are traveling at a speed of 35 miles per hour on this bicycle, through how many revolutions per minute are the wheels turning?

100.

51 5

101.

Designing a Water Sprinkler An engineer is asked to de sign a water sprinkler that will cover a field of 100 square yards that is in the shape of a sector of a circle of radius 50 yards. Through what angle should the sprinkler rotate? Motion on a Circle An object is traveling around a circle with a radius of 5 centimeters. If in 20 seconds a central angle


[ Hint: The linear speeds of the pulleys are the same; both equal the speed of the belt.]

Car Wheels The radius of each wheel of a car is 15 inches. If the wheels are turning at the rate of 3 revolutions per second, how fast is the car moving? Express your answer in inches per second and in mjles per hour.

In Problems 101-104, the latitude of a location L is the angle formed by a ray drawn from the center of Earth to the Equator and a ray drawn from the center of Earth to L. See the figure. North Pole

lOS.

Ferris Wheels A neighborhood carnival has a Ferris wheel whose radius is 30 feet. You measure the time it takes for one revolution to be 70 seconds. What is the linear speed (in feet per second) of this Ferris wheel? What is the angular speed in radians per second?

109.

Computing the Speed of a RiYer Current To approximate the speed of the current of a river, a circular paddle wheel with radius 4 feet is lowered into the water. If the current causes the wheel to rotate at a speed of 10 revolutions

Equator

South Pole

516

CHAPTER 7


per minute, what is the speed of the current? Express your answer in m iles per hour.

,

110.

111.

The Cable Cars of San Francisco At the Cable Cm' Museum you can see the four cable lines that are used to pull cable cars up and down the hills of San Francisco. Each cable trav els at a speed of 9.55 miles per hour, caused by a rotating wheel whose diameter is 8.5 feet. How fast is the wheel ro tating? Express your answer in revolutions per minute.

112.

Difference in Time of Sunrise Naples, Florida, is approxi mately 90 m iles due west of Ft. Lauderdale. How m uch sooner would a person in Ft. Lauderdale first see the rising Sun than a person in Naples? See the h int. [Hint: Consult the figure. When a person at Q sees the first rays of the Sun, a person at P is still in the dark. The person at P sees the first rays after Earth has rotated so that P is at the location Q. Now use the fact that at the latitude of Ft. Lauderdale in 24 hours a length of arc of 21/ (3559) miles is subtended.] 90 m i les

p!a '

/

-

-

- _

----_

'�) ;of:�"" Earth � ,,

W+E � S

Y

--

a

,

-

--

Fort

\\.. ()Lau N p les P ,,- • 113.

115.

116.

- -

Nautical Miles A nautical mile equals the length of arc sub tended by a central angle of 1 minute on a great circle* on the surface of Earth. (See the figure.) If the radius of Earth is taken as 3960 miles, express 1 nautical mile in terms of ordi nary, or statute, miles.

4 ft

SI)in Balancing Tires A spin balancer rotates the wheel of a car at 480 revolutions per m inute. If the diameter of the wheel is 26 inches, what road speed is being tested? Express your answer in miles per hour. At how many revolutions per minute should the balancer be set to test a road speed of 80 miles per hour?

3miles 559

114.

erdale,

Sun

a

Keeping UI) with the Sun How fast would you have to travel on the surface of Earth at the equator to keep up with the Sun (that is, so that the Sun would appear to remain in the same position in the sky)?

Approximating the Circumference of Earth Eratosthenes of Cyrene (276-194 Be) was a Greek scholar who lived and worked in Cyrene and Alexandria. One day while visiting in Syene he noticed that the Sun's rays shone directly down a well. On this date 1 year later, in Alexandria, which is 500 miles due north of Syene he measured the angle of the Sun to be about 7.2 degrees. See the figure. Use this infor mation to approximate the radius and circumference of Earth.

Designing a Little League Field For a 60-foot Little League Baseball field, the distance from home base to the nearest fence (or other obstruction) on fair territory should be a min imum of 200 feet. The commissioner of parks and recreation is making plans for a new 60-foot field. Because of limited ground availability, he will use the minimum required dis tance to the outfield fence. To increase safety, however, he plans to include a 10-foot wide warning track on the inside of the fence. To further increase safety, the fence and warning track will extend both directions into foul territory. In total the arc formed by the outfield fence (including the exten sions into the foul territories) will be subtencled by a central angle at home plate measuring 96°, as illustrated. (a) Determine the length of the outfield fence. (b) Determine the area of the warning track.

'" Any circle drawn on the surface of Earth that divides Earth into two equal hemispheres.

SECTION 7.2

Right Triangle Trigonometry

51 7

[Note: There is a 900 angle between the two foul lines. Then there are two 3°angles between the foul lines and the dotted lines shown. The angle between the two dotted lines outside the 200 foot foul lines is 96°.] 117.

Pulleys Two pulleys, one with radius rl and the other with radius r2 , are connected by a belt. The pulley with radius r 1 rotates at W I revolutions per minute, whereas the pulley with radius r2 rotates at W2 revolutions per minute. Show that rl

r2

Source:

W2 WI

www.littleleague.org


Do you prefer to measure angles using degrees or radians? Provide justification and a rationale for your choice.

119.

What is 1 radian?

UO.

Which angle has the larger measure: 1 degree or 1 radian? Or are they equal?

U1.

Explain the difference between linear speed and angular speed.

U2.

For a circle of radius r, a central angle of (J degrees subtends an arc whose length s is s

;

123.

124.

125.

r(J. Discuss whether this is a 1 0 true or false statement. Give reasons to defend your position. =

Discuss why ships and airplanes use nautical miles to measure distance. Explain the difference between a nautical mile and a statute mile. Investigate the way that speed bicycles work. In particular, ex plain the differences and similarities between 5-speed and 9-speed derailleurs. Be sure to include a discussion of linear speed and angular speed. In Example 6, we found that the distance between Albu querque, New Mexico, and Glasgow, Montana, is approxi mately 903 miles. According to mapquest.com, the distance is approximately 1 300 miles. What might account for the difference?

'Are You Prepared?' Answers 1. C

=

2 7T r

7.2 Right Triangle Trigonometry PREPARI NG FOR THIS SECTION •


Geometry Essentials (Chapter R, Review, Section R.3, pp. 30-35)

'\.Now Work

•

Functions (Section 3.1, pp. 208-218)


OBJECTIVES 1 Find the Va l ues of Trigonometric Fu nctions of Acute Ang les (p. 5 1 7) 2 Use the Fundamenta l Identities (p. 5 1 9)

3 Find the Va lues of the Rema i n i n g Trigonometric Functions, Given the

Val u e of One of Them (p. 5 2 1 )

4 Use the Complementary A n g l e Theorem (p. 5 2 3)

1 Figure 1 8

b a

Find the Values of Trigonometric Functions of Acute Angles

A triangle in which one angle is a right angle (90°) is called a right triangle. Recall that the side opposite the right angle is called the hypotenuse, and the remaining two sides are called the legs of the triangle. In Figure 18 we have labeled the hypotenuse as c to indicate that its length is c units, .a nd, in a like manner, we have labeled the legs as a and b. Because the triangle is a right triangle, the Pythagorean Theorem tells us that

518

C H A PTER 7


Now, suppose that e is an acute angle; that is, 0°

degrees) and 0

< e
--

2

= V2 . 1 = V2 •

Now Work P R O B L E M S 5 AND 1 7

F ind the Exact Values of the Trigonometric Functions '1T

6

of EXAMP LE 3

1 1

--

=

'1T

30 ° and 3

=

60°

Finding the Exact Val ues of the Trigonometric F u n ctions of

TT

6

-

=

3 0 ° and

TT

3

-

=

60°

1T

1T

Find the exact values of the six trigonometric functions of 6 = 30° and "3 = 60°. Solution

1T

Form a right triangle in which one of the angles is 6 = 30°. It then follows that the 1T

third angle is "3 = 60°. Figure 28(a) illustrates such a triangle with hypotenuse of length 2. Our problem is to determine a and b. We begin by placing next to the triangle in Figure 28(a) another triangle con gruent to the first, as shown in Figure 28(b). Notice that we now have a triangle whose angles are each 60°. This triangle is therefore equilateral, so each side is of length 2. In particular, the base is 2a = 2, so a = 1 . By the Pythagorean Theorem, b satisfies the equation a2 + b2 = c2 , so we have a2 12

+ +

b2 = c2 b2 = 22 b2 = 4 - 1 = 3 b =

Vi

a =

1,

C=

2

SECTION 7.3

sin 7T

-

6

b

=

sin 30° =

7T = cos 30° cos -

6

a

(a)

opposite hypotenuse

2

1

cos 7T

adj acent hypotenuse

V3 2

. 7T = SIn 60° = -SIn

-=--=-

=

-----

sin 30° cos 30°

6

=

tan 30°

=

7T csc -

=

csc 30°

= --- =

---

1

1

a

-l....L _ _ _ _ _ _ _ •

sin 30°

(b)

6

cot 7T

6

=

=

7T = 30°, 7T = -

6

(e)

4

1

=

cot 30°

Table 3

a 1

1 cos 30°

= ---

7T = sec 30° sec -

a

tan 30°

3

1

2

V3

V3 2

=

V3 3

V3

1

_ 1_

V3 3

=

=

2

V3 2

cot 7T

-

3

=

cot 60°

V3 = --

7T sec 3

=

sec 60°

=2

7T csc -

=

csc 60° =

3

_3_ = V3 V3

1

cos 60° = -

.

3

=2

1

-

2

_ _

1

-

2 V3 2

7T tan -

6

L-

= 60° are com-

Using the triangle in Figure 28( c) and the fact that 7T = 30° and 7T 3 6 plementary angles, we find

Figure 28

53 1

Computing the Values of Trigonometric Functions of Acute Angles

7T

tan

3

3

2V3 3

--

= tan 60° = V3 •

summanzes the information j ust derived for the angles . . . 7T 60°. Rather than memonze the entnes In Tabl e 3 , you 45°, and -

3

=

can draw the appropriate triangle to determine the values given in the table. Table 3

(J (Radians)

(J (Degrees)

sin (J

cos (J

tan (J

30°

2

1

V3 2

V3

V2

V2

V3 2

2

7T

-

6 7T

45°

4 7T

60°

3

E XA M P L E 4

2

esc (J

3

sec (J

2 V3 3

2

V2

2

V3

cot (J

V3

V2

2 V3

V3

2

3

3

F in d i n g the Exact Value of a Trigo n ometric Expression

Find the exact value of each expression.

Soluti o n

?

7T - sin 7T (b) tan -

( a) sin 45° cos 30°

.

3

4

v2

7T (c) tan- -

6

+

. 7T SIn2 -

4

V3 v'6 2 4 V3 2 - V3 1 -2 = 2

(a) SIn 45° cos 30° = -- . -- = --

2

. 7T 7T - SIn (b) tan "4

3

(c) tan2 7T

6

11'I!i: =: :m. _

+

=

sin2 7T

4

( ) ( v2 )2

= V3 3

2

+


2

=

9 A ND 19

1:. + 1:. = � 3

2

6

•

532

C HA PTER 7


7T

7T

The exact values of the trigonometric functions for the angles {5 = 30°,

= 450, and

7T

= 60° are relatively easy to calculate, because the triangles that 3 4 contain such angles have "nice" geometric features. For most other angles, we can only approximate the value of each trigonometric function. To do this, we will need a calculator.

3

Use a Calculator to Approximate the Values of the Trigonometric Functions of Acute Angles

Before getting started, you must first decide whether to enter the angle in the cal culator using radians or degrees and then set the calculator to the correct MODE. (Check your instruction manual to find out how your calculator handles degrees and radians. ) Your calculator has the keys marked 1 sin I, 1 cos I, and 1 tan I. To find the values of the remaining three trigonometric functions (secant, cosecant, and cotan gent), we use the reciprocal identities. sec e =

1 cos e

-

1 csc e = - sin e

cot e =

1 tan e

-

Using a Calculator to Approximate the Value of Trigonometric Functions

E XA M P L E 5

Use a calculator to find the approximate value of: (a) cos 48°

�1g1. (c) tan

(b) csc 21 °

�

,7T

12

Express your answer rounded to two decimal places. Solution

(a) First, we set the MODE to receive degrees. Rounded to two decimal places, cos 48° = 0.67

Figure 29

tan ( n ....· 1 2 ) . 267949 1 924

(b) Most calculators do not have a csc key. The manufacturers assume the user knows some trigonometry. To find the value of csc 21°, we use the fact that 1 . csc 21 ° = ---:--- ' Rounded to two deCImal places, csc 21 ° = 2.79 . sm 21° . � (c) Set the MODE to receive radians. Figure 29 shows the solution using a TI-84 Plus graphing calculator. Rounded to two decimal places, tan " j l �

4

7T 12

= 0.27

•


Model and Solve Applied Problems Involving Right Triangles

Right triangles can be used to model many types of situations, such as the optimal design of a rain gutter. * '" l n applied problems, it is important that answers be reported with both justifiable accuracy and ap propriate significant figures. We shall assume that the problem data are accurate to the number of sig n i ficant digits, resulting in sides being rounded to two decimal places and angles being rounded to one decimal p l ace.

SECTION 7.3

E XA M P L E 6 Figure 30

1,

-

4 in

A rain gutter is to be constructed of aluminum sheets 12 inches wide. After mark ing off a length of 4 inches from each edge, this length is bent up at an angle 8. See Figure 30.

4 in

•

b

I:fl Solution Figure 3 1

533

Constructing a Rain Gutter

1 2 in ----1--, 4 in

•

Computing the Values of Trigonometric Functions of Acute Angles

(a) Express the area A of the opening as a function of 8 . [Hint: Let denote the vertical height of the bend.] (b) Find the area A of the opening for 8 = 30°, 8 = 45°, 8 = 60°, and 8 = 75°. (c) Graph A = A(8). Find the angle 8 that makes A largest. (This bend will allow the most water to flow through the gutter.) (a) Look again at Figure 30. The area A of the opening is the sum of the areas of two congruent right triangles and one rectangle. Look at Figure 31, showing one of the triangles in Figure 30 redrawn. We see that cos 8

a

= 4" so

a=

.

4 cos 8

SIn 8

The area of the triangle is area of triangle =

�

(baSe) ( height) =

b

b

= "4 so

�ab � b, =

= 4 sin 8

(4 COS 8 ) (4 sin 8)

= 8 sin 8 cos 8

So the area of the two congruent triangles is 16 sin 8 cos 8. The rectangle has length 4 and height so its area is area of rectangle

b

= 4 = 4(4 sin 8)

= 16 sin 8

The area A of the opening is A = area of the two triangles + area of the rectangle A( 8) = 16 sin 8 cos 8 + 16 sin 8 = 16 sin 8( cos 8 + 1 )

(b) For 8

= 30°:

A( 300) = 16 sin 300(cos 30° = 16

+

(�) (� ) +

1)

=

1

4 V3

+

8

�

14.9

The area of the opening for 8 = 30° is about 14.9 square inches. A ( 45°) = 16 sin 45°( cos 45°

+

+

1

= 16 The area of the opening for 8 A( 60°)

22

The area of the opening for 8

--

....-----*'.. --- -

0°

//'"

�.H -

3

Now Work

PROBLEM

21

Determine the Signs of the Trigonometric Fu nctions of a n Angle i n a Given Quadra nt If () is not a quadrantal angle, then it will lie in a particular quadrant. In such a case, the signs of the x-coordinate and y-coordinate of a point (a, b) on the terminal side

Figure

51

of () are known. Because r = a2 + b2 > 0, it follows that the signs of the trigono metric functions of an angle () can be found if we know in which quadrant () lies. For example, if () lies in quadrant II, as shown in Figure 51, then a point (a, b) on the terminal side of () has a negative x-coordinate and a positive y-coordinate. Then,

y

() i n quadrant II, a

< 0, b

y

>

0, r

>

0

b r r csc () = b .

SID

x

-

() =

> °

a r r sec () = a

cos () =

> °

-

< °

tan () =

< °

b a a

-

< °

cot () = b < °

Table 5 lists the signs of the six trigonometric functions for each quadrant. Figure 52 provides two illustrations. Table

5

Quadrant of (J

sin (J, esc (J

II

III

IV

Figure

52 II

(-, +)

sin e > 0, csc e > ° others negative

y

Positive

Positive

Positive

Negative

Negative

Negative

Negative

Positive

Negative

Positive

Negative

+-

positive

IV (+,-)

111(-, -)

Positive

1(+, +)

All

+y +y

x

-

cos e > 0 , sec e > ° others negative

tan e > 0 , cot e > 0 others negative

(a)

Solution

+

+

+

E XA M PL E 4

tan (J, cot (J

cos (J, sec (J

+

-

sine cosecant

x

cosine secant

x

tangent cotangent

(b)

Finding the Quadrant in Which an Angle Lies

If sin () < ° and cos () < 0, name the quadrant in which the angle () lies.

If sin () < 0, then () lies in quadrant III or IV. If cos () or III. Therefore, () lies in quadrant III. == � ,...

Now Work

PROBLEM

33

--4V17 17

V17 4

Now Work

cose

=

sece

=

V17

D V17

tane

=

cote

= --

-4 1 4

•

PROBLEM 99

7.4 Assess Your Understanding Concepts and Vocabulary 1.

2. 3. 4. 5.

For an angle e that l ies i n quadrant III, the trigonometric functions and are positive. Two angles in s tandard position that have the s ame terminal side are T he reference angle of 2 40° is . True or False sin 182 ° = cos 2 ° . 7T True or False tan "2 is not defined. __

6. The reference angle is always an acute angle. What is the reference angl e of 6 00 ° ? 8. n which quadrants is the cosin e function positive? True or False

7.

__

I

If 0 :5 e < 27T, for what angles e, if any, is tan e undefi ned?

9.

__

1o

. What

IS '

nIT

the reference angle of - - ? 3

Skill Building In Problems

"11.

(-3, 4)

16.

(2, -2)

11-20,

a point on the terminal side of an angle e is given. Find the exact value of each of the six trigonometric functions of e.

12. ( 5, - 12)

13.

17.

18.

( \132 '!)2

(2 , -3)

14.

(_!2 ' \132 )

19.

( -1, -2)

( V2

_

2 '

V2 2

15. ( -3, -3) 20. ( V22 '

)

_

In Problems 21-32, use a coterminal angle to find the exact value of each expression. Do not use a calculator.

22. cos 42 0° 28. sec 42 0°

. 21. sin 4 05° 27. cot 3 9 0°

In Problems

"33. sin e 36. cos e 39.

>

>

33-40,

23. tan 4 05° 337T 29. cos 4

name the quadrant in which the angle

0, cos e < 0 0, tan e

sec e < 0, tan e

>

>

0 0

34. s in e < 37. cos e

>

24. sin 3 9 0° 30. . 4'97T SlI1

e lies.

0, cos e

>

0

0, cot e < 0

40. csc e

>

25. csc 4 500 31. tan(217T) 35. s in e < 38. s in e

0

)

549

SECTION 7.4 Trigonometric Functions of General Angles In Problems 41-58, find the reference angle of each angle.

41.

-30°

42.

4

53.

60°

43.

-3

120°

55.

300°

44.

871' 4 9. 3

571' 48. 6 771' 54. -6

571' . 74 271'

771'

'4

50

440°

490°

56.

45.

210°

46.

330°

51.

-135°

52.

-240°

57.

1571' -4

58.

1971' - 6

In Problems 59-88, use the reference angle to find the exact value of each expression. Do not use a calculator. 59.

sin 150°

60.

cos 210°

61.

cos 315°

62.

sin 120°

63.

sin 510°

64.

cos 600°

65.

cos( -45° )

66.

sine - 240°)

67.

sec 240°

68.

csc 300°

69.

cot 330°

70.

tan 225°

71.

. 371' SIn

72.

271' cos 3

73.

771' cot6

74.

771' csc4

75.

1371' cos 4

76.

871' tan 3

77.

sin

78.

cot

79.

1471' tan - 3

80.

1171' sec4

81.

csc ( -315°)

82.

sec( -225°)

83.

sin(871')

84.

cos( -271')

85.

tan(h)

86.

cot(571')

87.

sec ( -371')

88.

csc

4

( 2;) _

(-�)

In Problems 89-106, find the exact value of each of the remaining trigonometric functions of

��,

89.

sin 8

92.

sin 8 = - � 8 in Quadrant III 13'

93.

95.

1 cos 8 = - 3"

96.

9S.

1 cos 8 = - 4"' tan 8

101. 104.

=

3 tan 8 = 4"'

8 in Quadrant I I

90.

180° < 8 < 270° >

0

99.

sin 8 < 0

sec e = -2, tan e

>

0

Find the exact v alue of tan 40°

)

71' .

109.

If

112.

IfG (e) = cot 8 = - 2, findG (8

+

Find the exact v alue of sin 1° + sin 2° + sin 3° + . . .

sin 358°

115.

f(8

+

+

cos 8

=

94.

cos e

=

97.

. SIn e

0

=

-

�,

_

8 in Quadrant III

5' 270° < e < 360° 4

3" 2

tan e < 0

csc e = 3, cot e < 0 1 - 3'

sin e

> 0

103.

tan 8

=

106.

cot 8

=

111.

If F (8) = tan e = 3, find F(e

114.

2 . If cos 8 = 3', fmd sec (8

'

-2, sec 8

>

0

sin 220° + sin 310°.

tan 140°.


(8) = sin 8 = 0.2, find

>

91.

100.

cos f) < 0

csc 8 = -2, tan e

lOS.

271" or if

t

-

Table 8 x

1T 2

21T 3

1T

41T 3 31T 2

51T 3

21T

PROBLEM 45

u s

I N G T R A N 5 F O R M AT I O N 5

The Graph of the Cosine Fu nction y

= C05 X

0

1T 3

Now Work

•

The cosine function also has period 21T. We proceed as we did with the sine function by constructing Table 8, which lists some points on the graph of y = cos x, o s x S 21T. As the table shows, the graph of y = cos x, 0 s x S 21T, begins at the

(x, y)

(0, 1 ) 1

2

0 2

-1 2

0 2

( �, �)

(f, o)

- 1 ; as x increases from 1T to

1T 3 '-� 2

C ) ( �)

(1T,

- 1)

41T 3 '

_

3;

point (0, 1 ) . As x increases from 0 to

2

; to 1T, the value of y decreases from 1 to 0 to

to 21T, the value of y increases from - 1 to 0 to 1 . As

before, we plot the points in Table 8 to get one period or cycle of the graph. See Figure 78. Y = cos x, 0 Figure 78

::;

x

::;

21T

y

( 21T , 1 )

(0, 1 )

e;, o) 1T 3 '� 2

e )

(21T, 1 )

(1T, - 1 ) A more complete graph of y = cos x is obtained by continuing the graph in each direction, as shown in Figure 79. Y = COS x, - 00 < Figure 79

x < 00

y

(21T, 1 ) x

(-1T, -1 )

(1T, -1 )

The graph of y = cos x illustrates some of the facts that we already know about the cosine function.

SECTION 7.6 Graphs of t h e Sine a n d Cosine Functions

563

Properties of the Cosine Function

1. The domain is the set of all real numbers. 2. The range consists of all real numbers from - 1 to 1, inclusive. 3. 4. 5.

6. 2

The cosine function is an even function, as the symmetry of the graph with respect to the y-axis indicates. The cosine function is periodic, with period 2'TT . 3'TT 5'TT ' . . . - 3'TT 'TT 'TT 2 ' 2 . ; the y-mtercept IS The x-mtercepts are . . , 2 ' "2' "2 ' The maximum value i s 1 and occurs at x = . , -2'TT , 0 , 2'TT , 4'TT , 6'TT , . . . ; the minimum value is - 1 and occurs at x = . , -'TT , 'TT , 3'TT , 5'TT , . . . .

..

.

..

l.

..

G ra p h Fu nctions of the Form y = A cos(wx) Using Tra nsformations Graphing Functions of the Form y = A cos(wx)

EXAM P L E 3

Using Transformations

Graph y = 2 cos ( 3x ) using transformations. Solution Figure

Figure 80 shows the steps.

80

y

y

( 2'TT, 1) (-'IT, -1)

('IT, -1 ) (a) y cos x =

Multiply by 2; Vertical stretch by a factor of 2

(b) Y =

2 cos x

Replace x by 3x; Horizontal compression by a factor of +

(-¥, -2)

(c) y =

( ¥ , -2) 2 cos ( 3x)

Notice in Figure 80 ( c) that the period of the function y = 2 cos ( 3x ) is to the compression of the original period 2'TT by a factor of I0I'l"

=--

Now Work

PROBLEM 5 3

S i n u soidal G raphs

�.

2;

•

due

U 5 I N G T R A N 5 F O R M AT I O N 5

'TT

Shift the graph of y = cos x to the right "2 units to obtain the graph of

( �) . ( �) .

y = cos x -

See Figure 81 ( a ) . Now look at the graph of y = sin x in

Figure 81 ( b ) . We see that the graph of y = sin x is the same as the graph of y = cos x Figure 81

y

y

(a) y = cos x y = cos ( x ¥ ) -

(b)

Y sin x =

564

CHAPTER 7 Trigonometric Functions

B ased on Figure 81, we conjecture that

�

Seeing the Concept

Graph Y,

( ;)

sin x = cos x -

( - %).

= sin x and Y2 = cos x

How many graphs do you see?

(We shall prove this fact in Chapter 8.) Because of this relationship, the graphs of functions of the form y = A sin( wx) or y = A cos( wx) are referred to as sinusoidal graphs.

Let's look at some general properties of sinusoidal graphs. 3

Determine the Amplitude a n d Period of Sinusoidal Fu nctions In Figure 82(b) we show the graph of y = 2 cos x. Notice that the values of y = 2 cos x lie between -2 and 2, inclusive.

Figure 82 y

(2'IT, 1 )

Multiply by 2; Vertical stretch by a factor of 2 ( -'IT, -2) (a) y =

('IT, -2)

(b) Y = 2 cos x

cos x

A

In general, the values of the functions y = A sin x and y = A cos x, where 0, will always satisfy the inequalities

=F

- [A [

:s

A sin x

:s

[A[ and -[A[ :S A cos x

respectively. The number [A[ is called the amplitude of y See Figure 83. Figure 83

=

:s

[A[

A sin x or y = A cos x.

y

y = A sin x, A > 0 Period = 21T

In Figure 84(b), we show the graph of y = cos(3x ) . Notice that the period of 21T this function is -, due to the horizontal compression of the original period 21T by 3 1 a factor of 3' Figure 84

y

(2'IT, 1 )

(a) y =

cos x

Replace x by 3x; Horizontal compression by a factor of .1.3

(b) Y = cos (3x)

In general, if w > 0, the functions y = sin( wx) and y = cos( wx) will have 27T period T = . To see why, recall that the graph of y = sin( wx) is obtained from the w

SECTION 7.6 Graphs of the Sine and Cosine Functions

565

1 graph of y = sin x by performing a horizontal compression or stretch by a factor -. w This horizontal compression replaces the interval [0, 21T) , which contains one peri2 which contains one period of od of the graph of y sin x, by the interval 0,

: ],

[

=

the graph of y = sin e wx) . The period of the functions y = sine wx) and 21T y = cos ( wx ) , w > 0, is -. w For example, for the function y = cos (3x) , graphed in Figure 84(b), w = 3, so 21T 21T the period is - = - . w 3 One period of the graph of y sine wx) or y = cos( wx) is called a cycle. Figure 85 illustrates the general situation. The blue portion of the graph is one cycle. =

Figure 85

y

y A sin (wx), A > 0, W > 0 Period = 'iH =

If w < ° in y = sin(wx) or y = cos ( wx ) , we use the Even-Odd Properties of the sine and cosine functions as follows: sine -wx) = -sine wx)

and

cos( - wx ) = cos( wx)

This gives us an equivalent form in which the coefficient of x in the argument is positive. For example, sin( -2x) = -sin(2x)

B ecause of this, we can assume w > 0.

and COS ( -1TX)

=

COS(1TX)

If w > 0, the amplitude and period of y = A sin(wx) and y = A cos(wx) are given by

THEOREM

Amplitude = IAI

21T . Penod = T = w

(1)

I

�

� -----------------�

EXAM P L E 4

Finding the Amplitude and Period of a Sinusoidal F unction D etermine the amplitude and period of y = 3 sine 4x) . Comparing y = 3 sin(4x) to y = A sin( wx ) , we find that A = 3 and w = 4 . From equation ( 1 ) ,

Solution

Amplitude = I AI = 3 �

4

Now Work

1T 21T 21T Period = T = - = - = w 4 2

•

PROBLEM 23

G ra p h Sinusoidal Functions Using Key Points So far, we have graphed functions of the form y = A sine wx) or y = A cos( wx) using transformations. We now introduce another method that can be used to graph these functions.

566


Figure 86 shows one cycle of the graphs of y = sin x and y = cos x on the inter val [0, 2'7T] . Notice that each graph consists of four parts corresponding to the four subintervals:

'7T Each subinterval is of length 2 (the period 2'7T divided by 4, the number of parts), '7T 3'7T and the endpoints of these intervals x = 0, x = , X = '7T, X = , X = 2'7T give rise 2 2 to five key points on each graph: For y = sin x:

(0, 0),

(; )

, 1 , ('7T , 0),

( 3; )

, - 1 , (2'7T, 0)

For y = cos x: Look again at Figure 86. Fig ure 86

y

y 1

(21T, 1 )

(0, 1 )

x

(0, 0)

-1

(�, -1 )

-1

(1T, - 1 )

(a) y = sin x

(b) y = cos x

Steps for Graphing a Sinusoidal Function of the Form y = A sin(wx) or y = A cos(wx) Using Key Points STEP 1:

Use the amplitude A to determine the maximum and minimum val ues of the function. This sets the scale for the y-axis. 2 2 STEP 2: Use the period and divide the interval 0, into four subintervals

[ :]

:

of the same length. STEP 3: Use the endpoints of these subintervals to obtain five key points on the graph. STEP 4: Connect these points with a sinusoidal graph to obtain the graph of one cycle and extend the graph in each direction to make it complete.

E XA M P LE 5

G raphing a Sinusoidal Function Using Key Points Graph:

Solution

y = 3 sin(4x)

'7T Refer to Example 4. For y = 3 sine 4x), the amplitude is 3 and the period is 2 ' Because

the amplitude is 3, the graph of y = 3 sin( 4x ) will lie between -3 and 3 on the y-axis.

; one cycle will begin at x = and end at x = ; . We divide the interval [ 0, ; ] into four subintervals, each of length ; -;- 4 = �,

Because the period is

°

,

by finding the following values: °

initial value

'7T '7T = 0+8 8 2nd value

'7T '7T '7T =8 +8 4 3rd value

31T '7T '7T =8 8 4 +4th value

1T 31T '7T =8 + 8 2 final va lue


NOTE We could a lso obta i n the five key

( )

points by eva l uati n g y = 3 si n 4x

each va l u e of x,

567

These values of x determine the x-coordinates of the five key points on the graph, To obtain the y-coordinates of the five key points of y = 3 sine 4x ) , we mUltiply the y-coordinates of the five key points for y = sin x in Figure 86(a) by A = 3, The five key points are

at

•

We plot these five points and fill in the graph of the sine curve as shown in Fig ure 87(a). We extend the graph in either direction to obtain the complete graph shown in Figure 87(b). y

Figure 87

�

�

(�, 3 )

( "f, 3 )

(0, 0) COM M ENT To graph a s i n u soid a l fu nc tion of the form

y=

A

cas ( wx )

y=

A

sin ( wx )

or

using a g ra p h i n g uti l

ity, we u s e the a m plitude t o set Ym i n

-3

a n d Ymax a n d u s e t h e period t o set

_

Xm i n and Xmax,

(a)

("f, - 3 )

er, - 3 )

(b) Y = 3 sin (4x)

Check: Graph y = 3 sin(4x) using transformations. Which graphing method do you

prefer?

IJ.'I II

E XA M P L E 6

""-

Now Work

•

PROBlEM 45

U5ING

KEY POINT5

Finding the Amplitude and Period of a Sinusoidal F unction and G raphing It Using Key Points

( ;)

Determine the amplitude and period of y = 2 sin -

Solution

x , and graph the function.

Since the sine function is odd, we can use the e quivalent form:

Comparing y = -2 sin The amphtude .

. IAI

IS

(; )

y = -2 sin

= 2, and the penod .

(; ) x

x

A sin ( wx ) , we find

x to y =

The graph of y = -2 sin

(; )

.

IS

T = - =

2n w

that A = -2 and w

=

%.

2n = 4. n 2

-

will lie between -2 and 2 on the y-axis. One

cycle will begin at x = 0 and end at x = 4. We divide the interval [0, 4] into four subintervals, each of length 4 + 4 = 1 , by finding the following values: 0

0 + 1 = 1

i n itia l va l ue

Since y = -2 sin

1 st va l ue

(; )

1 + 1 = 2 2nd va lu e

2 + 1 = 3

3 + 1 =4

3rd va l ue

final va l u e

x , we multiply the y-coordinates of the five key points in

Figure 86(a) by A = -2. The five key points on the graph are (0, 0) ,

( 1 , -2 ) ,

(2, 0 ) ,

(3, 2 ) ,

(4, 0)

We plot these five points and fill in the graph of the sine function as shown in Figure 88(a) on page 568. Extending the graph in each direction, we obtain Figure 88(b) .

568

CHAPTER 7 Trigonometric Functions Figure 88

Y

(4, 0 ) x

( 5 , - 2)

(1 , - 2 ) (a)

Check: Graph y =

2 (;)

sin do you prefer?

� = = ",..

Now Work

x

(b) Y = 2 sin

( -� X)

using transformations. Which graphing method •

U

P RO B LEM 49

S I NG KEY P O I NT5

If the function to be graphed is of the form y = A sine wx ) + B [or y = A cos( wx ) + B], first graph y = A sin(wx) [or y = A cos(wx) ] and then use a vertical shift.

E XA M P LE 7

Finding the Amplitude and Period of a Sinusoidal F unction and G raphing It Using Key Points Determine the amplitude and period of y = -4 cos( 7Tx )

Solution

-

2,

and graph the function.

We begin by graphing the function y = -4 COS( 7TX ) . Comparing y = -4 COS(7TX) with y = A cos(wx ) , we find that A = -4 and w = 7T. The amplitude is 27T 7T . . IS T = = 2. I A I = I -4 I = 4, and the penod w 7T The graph of y = -4 cos( 7TX) will lie between -4 and 4 on the y-axis. One cycle will begin at x = ° and end at x = We divide the interval into four

=2 2.

-

subintervals, each of length °

initial value

0+-=1

1

2 2 1st value

2

--'-- 4 =

1

-

2

+

�,

-

[0, 2J

by finding the following values:

1

= 1 2 2nd value -

+

1

1

3

= 2 2 3rd value -

-

3

-

2

1

=2 2 final value +

-

Since y = -4 cos( 7TX) , we multiply the y-coordinates of the five key points of y = cos x shown in Figure 86(b) by A = -4 to obtain the five key points on the graph of y = -4 COS(7TX):

( 0, - 4 ) ,

(�, o),

( 1 , 4) ,

(%, 0), ( 2, -4)

We plot these five points and fill in the graph of the cosine function as shown in Figure 89( a). Extending the graph in each direction, we obtain Figure 89(b), the graph of y = -4 cos( 7TX) . A vertical shift down 2 units gives the graph of y = -4 cas ( 7T x) - 2, as shown in Figure 89( c). Figure 89

Y

(1 , 4)

Y

(1 , 2)

x x

x

(2, -4)

(a) 1t!'II'= = � _

Now Work

(b) Y = -4 cos (1TX) PROBLEM

59

�

Subtract 2; vertical Sllift down 2 unit (c) Y = -4 cos (1Tx) - 2

•

SECTION 7.6

5

Graphs of the Sine and Cosine Functions

569

Find a n Equation for a Sinusoidal Gra p h We can also use the ideas of amplitude and period t o identify a sinusoidal function when its graph is given.

E XA M P L E 8

Finding an Equation for a Sinusoidal G raph Find an equation for the graph shown in Figure 90.

Figure 90

y 3

•

�--- Period

1

The graph has the characteristics of a cosine function. Do you see why? So we view the equation as a cosine function y = A cos( wx) with A = 3 and period T = 1 . 2 '7T = 1 , s o w = 2'7T. The cosine function whose graph i s given in Figure 90 is Then w y = A cos(wx) = 3 cos( 2'7Tx)

Solution

Check: G raph Y1 = 3 cos ( 2'7Tx ) and compare the result with Fig u re 90.

E XA M P L E 9

•

Finding an Equation for a Sinusoidal G raph Find an equation for the graph shown in Figure 91.

Figure 91

y

�--- Period

.1

2'7T The graph is sinusoidal, with amplitude IAI = 2. The period is 4, so - = 4 or w . Since the graph passes through the origin, it is easiest to view the equation w =

Solution

;

as a sine function,* but notice that the graph is actually the reflection of a sine function about the x-axis (since the graph is decreasing near the origin). This requires that A = -2. The sine function whose graph is given in Figure 91 is y = A sin (wx) = -2 sin

[I

Check: G raph Y1

=

L.=== _ Now Work _

-2 sin

(; ) x

PROBLEMS

(; ) x

•

a n d com pa re the result with Figu re 9 1 .

67

AND 7 1

" The equation could also be viewed as a cosine function with a horizontal shift, but viewing it as a sine function i s easier.

570


7.6 Assess Your Understanding 'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in 1. Use transformations to graph y = 3x2 (pp. 252-260)

2. Use transformations to graph y = -x2 . (pp. 252-260)

Concepts and Vocabulary 3. The maximum value of y = sin x, 0 :S

and occurs at x = 4. The function y = A sine wx ) , A riod 2; then A = and w ___

>

=

X

:S 27r, is

6.

_ _ _

7.

0, has amplitude 3 and pe.

___

5. The function y = 3 cos( 6x) has amplitude

___

red.

True or False The graphs of y = sin x and y = cos x are identical except for a horizontal shift. True or False

8.

and period

7r

period is 2 .

For y = 2 sin( 1TX ) , the amplitude is 2 and the

True or False The graph o f the sine function has infinitely many x-intercepts.

Skill Building In Problems 9-18, if necessary, refer to a graph to answer each question. 9.

, 11.

What is the y-intercept of y = sin x?

10. What is the y-intercept of y = cos x?

For what numbers x, -7r :S increasing?

12. For what numbers x, -7r :S decreasing?

X

:S 7r, is the graph of y = sin x

13. What is the largest value of y = sin x?

15. For what numbers x, 0

:S X :S

27r, does sin x

1 9-28,

=

16. For what numbers x, 0 :S

O?

Y

X

:S 27r, does cos x = O?

18. For what numbers x, -27r :S Where does cos x = - 1 ?

determine the amplitude and period of each function withou.t graphing.

19. y = 2 sin x 23.

:S 7r, is the graph of y = cos x

14. What is the smallest value of y = cos x?

17. For what numbers x, -27r :S x :S 27r, does sin x = 1 ? Where does sin x = - 1 ? In Problems

X

20. y = 3 cos x 24.

= 6 sin(7rx)

Y

21. y = -4 cos ( 2x )

22.

= -3 cos(3x)

X :S

Y

27r, does cos x = 1 ?

G) � G)

= -sin

26. y =

sin

x

x

In Problems 29-38, match the given fu.nction to one of the graphs (A)-(J). y 2

y

y

x

x -2

(A)

(D)

(B)

(E)

(e)

(F)

y

x -2

(G)

(H)

(I)

571


�

2 COS(� X) 33. y = -3 sin(2x) 36. Y = -2 COS(�X)

29. Y = 2 Sin(�X) 32. Y = 3 cos(2x) 35. Y = -2 COSGX) 38. y = -2 SinGX)

(J)

31. Y = 2 COsG x) 34. y = 2 SinGx) 37. Y = 3 sin(2x)

30. Y =

In Problems 39-42, match the given function to one of the graphs (A)-(D).

3

o lr

3

\

\

I ' 8'0

\.j

-3

(A)

39. Y = 3 SinG x )

01

-1/\1 //'�\\ \j

-3 40. Y =

3

/

\j

(8)

-3 sin(2x)

2 '0

0

3

;.------:=:---/--:=:/'---,\ \. /

-3

.

I 2 '0

o

41. y = 3 sin(2x)

=

44. y = 3 sin x ' 45. Y = -4 sin x 48. y = sin(3x) 49. Y = sin(-2x) .:13. Y = - "21 cos(2x) 52. y = 2 cos (� x) 56_ Y = cos 57. Y = 5 COS(7TX) 7T 2 61. Y = 5 - 3 sin(2x) 60. y = -3 cos ( 4 x ) 37T x ) 64. Y = 59 cos ( - 2 65. Y - "23 cos (47T x ) + "2 �

3

x

+ 2

3

+

1

=

\\./ \jl

69. Amplitude: 3 Period: 2 72.

73.

74. -3

y

y

\'1 8'0

(D)

46. Y = -3 cos x 50. Y = cas ( -2x) 54. Y = -4 SinGx) 58. y = 4 Sin(� ) - 2 62. y = 2 - 4 cos(3x) 66. y = _1:.2 sin (�8 r) 2� x

"

In Problems 67-70, write the equation of a sine function that has the given characteristics.

67. Amplitude: 3 68. Amplitude: 2 Period: 7T Period: 47T In Problems 71-84, find an equation for each graph. 71.

\ I

42. Y = -3 SinGx)

In Problems 43-66, graph each function . Be sure to label key points and show at least two cycles.

43. y = 4 cos x 47. Y = cos(4x) 51. y = 2 sinG x) 55. Y = 2 sin x + 3 " 59. Y -6 sin(� x) + 4 27T x ) 63. y = 35 sm. (- 3

I� I

-3

(C)

/' l

0\ \

1

70.

Amplitude: 4 Period: 1

+

572


75.

y

76.

77.

y

78.

79.

y 2

80.

y

x , 81.

82.

3 -2

1\

.(\\ If\ I

I

\j-3 V

6

4

83.

2

'IT

2

\.

-4

Applications and Extensions In Problems

85-88,

85. f( x) = sin x

In Problems

89-92,

89. f(x)

=

sin x

g(x)

=

4x

93.

find the average rate of change off from 0 to 86. f(x) find (f

0

g) (x) and (g

=

cos x

87. f( x) = sin

(�)

f) (x) and graph each of these functions.

90. f(x) = cos x 1 g(x) -x: 2 0

�.

=

-

91. f(x)

=

- 2x

g(x)

=

cos x

92. f(x) = - 3x g(x) = sin x

(c) I f a resistance of R = 10 ohms is present, what is the current /? [Hint: Use Ohm's Law, V = IR.] (d) What is the amplitude and period of the current /? (e) Graph J over two periods, beginning at t = O.

The current I, in amperes, flowing through an ac (alternating current) circuit at time t in seconds, is I � 0 1 ( / ) = 220 sin(601TI)

Alternating Current (ae) Circuits

What is the period? What is the amplitude? Graph this function over two periods. 94. Alternating Current (ae) Circuits The current /, in amperes, flowing through an ac (alternating current) circuit at time I in seconds, is l e t ) = 1 20 sin ( 301T/) I � 0 What is the period? What is the amplitude? Graph this func tion over two periods. 95. Alternating C u rrent (ae) Generators The voltage V, in volts, produced by an ac generator at time I, in seconds, is V ( / ) = 220 sin( 1201T/ ) (a) What is the amplitude? What is the period? (b) Graph V over two periods, beginning at I = O.

88. f(x) = cos(2x)

96.

Alternating Current (ae) Generators The voltage V, in volts, produced by an ac generator at time I, in seconds, is

V ( / ) = 120 sin ( 1 201T/)

(a) What is the amplitude? What is the period? (b) Graph V over two periods, beginning at t O. (c) If a resistance of R = 20 ohms is present, what is the current /? [Hint: Use Ohm ' s Law, V = JR.] (d) What is the amplitude and period of the current I? (e) Graph J over two periods, beginning at t = O. =


97.

(a) Find an equation for the sine curve that fits the opening. Place the origin at the left end of the sine curve. (b) If the road is 1 4 feet wide with 7 foot shoulders on each side, what is the height of the tunnel at the edge of the road.

Alternating Current (ac) Generators The voltage V pro duced by an ac generator is sinusoidal. As a function of time, the voltage V is V t)

e

= Vo sin(27Tft)

where f is the frequency, the number of complete oscilla tions (cycles) per second. [In the United States and Canada, f is 60 hertz (Hz).] The power P delivered to a resistance R at any time t is defined as

[ V (t ) f e ) = -R-

Sources: en. wikipedia. orglwikillnterstate_Highway_standards

99.

and

Ohio Revised Code

In the theory of biorhythms, a sine function of

Biorhythms

the form

p t)

e

p t

(a) Show that

p t)

e

(b) The graph o f P i s shown in the figure. Express sinusoidal function.

P

as a

1

1

3

if

ff

1

7

Power in an ac generator

(c) Deduce that sin2(27Tft) 98.

=

+

50

Physical potential: period of 23 days

Emotional potential: period of 28 days

� 4f

= 50 sin( wt )

is used to measure the percent P of a person's potential at time t, where t is measured in days and t = 0 is the person's birthday. Three characteristics are commonly measured:

V

? = Ii6 sin-(27Tft ).

573

Intellectual potential: period of 33 days (a) Find w for each characteristic. (b) Using a graphing utility, graph all three functions on the same screen. (c) Is there a time t when all three characteristics have 100% potential? When is it? (d) Suppose that you are 20 years old today (t = 7305 days). Describe your physical, emotional, and intellectual po tential for the next 30 days.

� [ 1 - COS ( 47Tft ) ]

Bridge Clearance A one-lane highway runs through a tun nel in the shape of one-half a sine curve cycle. The opening is 28 feet wide at road level and is 15 feet tall at its highest point.

100. Graph

101. Graph

y y

= Icos xl, -27T :5 x :5 27T. Isin x l , -27T :5 x :5 27T.

=

Discussion and Writing 1 02.

Explain how you would scale the x-axis and y-axis before graphing y 3 COS(7TX). =

103. Explain the term amplitude as it relates to the graph of a sinusoidal function.

104. Explain how the amplitude and period of a sinusoidal graph are used to establish the scale on each coordinate axis.

105. Find an application in your major field that leads to a sinu soidal graph. Write a paper about your findings.

'Are You Prepared?' Answers 1. Vertical stretch by a factor of 3

2. Reflection about the x-axis y 2

2 x

(0, 0)

5

x

574


7.7 Graphs of the Ta ngent, Cotangent, Cosecant, and Seca nt Functions PREPARING FOR THIS SECTION •


Vertical Asymptotes (Section 5.2, pp. 346-348)

Now Work the 'Are You Prepared?' problems on page 579. 1

OBJECTIVES

G raph Functions of the Form y = a n d y = A cot (wx) + B (p. 576)

2 Gra ph Fu n ctions of the Form y = a n d y = A sec (wx) + B (p. 578)

A tan (wx)

+ B

A csc (wx)

+ B

The G ra p h of the Ta ngent Fu nction Because the tangent function has period 7T, we only need to determine the graph over some interval of length 7T. The rest of the graph will consist of repetitions of that graph. Because the tangent function is not defined at 37T 7T 7T 37T 7T 7T we will concentrate on the interval - 2 -2 2 '2 ' · · · ' 2' ' ' 2' · · · ' of length 7T, and construct Table 9, which lists some points 011 the graph of 7T 7T Y = tan x, 2 < x < 2 We plot the points in the table and connect them with a · smooth curve. See Figure 92 for a partial graph of y = tan x, where - ::; x ::; .

(

-

Table 9

x 7r

3

7r

4

7r

-6

0

7r

6

y

=

n

ta x

- \13 "" - 1 .73 -1 -

-\13 -\13 3

"" -0.58

o

3

7r

3

Figure 92

Y = ta n x, - - ::; x ::;

(x, y)

(-f' - \13) (- �, - 1 ) ( � \133 ) _

6'

7r

3

;

7r

3

Y

_

(0, 0) "" 0.58

7r

4

;

)

\13 "" 1 .73

(� \133 ) (�, 1 ) (�, \13) 6'

To complete one period of the graph of y tan x, we need to investigate the 7T 7T - - and - . We must be careful, though, behavior of the function as x approaches . 2 2 because y = tan x is not defined at these numbers. To determine this behavior, we use the identity =

tan x = See Table 10. If x is close to

7T 2

�

- sin x cos x

1 .5708, but remains less than

7T 2

, then sin x will

be close to 1 and cos x will be positive and close to O. (To see this, refer back to the

SECTION 7.7 Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions

graphs of the sine function and the cosine function.) So the ratio

(�

x Table 1 0

=

x

sin x

7r

v3

- "" 1 .05 3 1 .5 1 .57 1 .5707

2

)

_tan x = (0 . In other words, the vertical line

� is a vertical asymptote to the graph of x

sin x will be pas cas x

�, the closer sin x gets to 1 and cos x gets

itive and large. In fact, the closer x gets to to 0, so tan x approaches 00 l

575

y =

tan x. y = tan x

cos x

2 0.9975 0.9999 0.9999

v3 "" 1 .73

2 0.0707 7.96E-4 9.6E-s

7r

- "" 1 .5708 2

1 4.1 1 25 5.8 1 0,381 Undefi ned

0

then sin x will be close to 1 2 �x and cos x will be positive and close to O. The ratio -- approaches cos x - 00 li � tan x = (0 . In other words, the vertical line x = - 7T is also a vertical 2 x-f asymptote to the graph. With these observations, we can complete one period of the graph. We obtain the complete graph of y = tan x by repeating this period, as shown in Figure 93. If x is close to

-

(

Y

tan x, -00 < x < 00, x not equal to odd

multiples of

7r

2

2

-

, - 00 < y < 00

but remains greater than

-

7T ,

-

)

371 x = -2

5 71 X = -2

Figure 93

=

7T ,

x = -.:!!:. 2

x = 571 2

-

x

Check: G raph Y, = ta n x a n d compare the result with Fig u re 93. Use TRACE to see

7T 7T . Be s u re to set the what h appens as x gets c l ose to -, but is less than 2 2 WIN DOW accord i n g ly a n d to use DOT mode.

The graph of y

=

tan x in Figure 93 illustrates the following properties.

Properties of the Tangent Function

1. The domain is the set of all real numbers, except odd multiples of 2. The range is the set of all real numbers. 3.

�.

The tangent function is an odd function, as the symmetry of the graph with respect to the origin indicates.

(continued)

576


4. The tangent function is periodic, with period 11".

5. The x-intercepts are . . . , -211", -11", 0, 11", 211", 311", . . . ; the y-intercept is O. 6. Vertical asymptotes occur at x = �j� e-

1

Now Work

311" 11" 11" 311" . . . , - 2 ' - 2' 2 ' 2' . . . .

PROBLEMS 7 AND

1 5

Gra p h Fu nctions of the Form y = A tan (wx)

+

8

For tangent functions, there is no concept of amplitude since the range of the tan gent function is (-00, 00 ) . The role of A in y = A tan (wx) + B is to provide the magnitude of the vertical stretch. The period of y = tan x is 11", so the period of 11" y = A tan(wx) + B is -, caused by the horizontal compression of the graph by a w 1 factor of -. Finally, the presence of B indicates a vertical shift is required. w

. EX-AMPLE-1 ml Solution Figure 94

x = -..:!I.2

G raphing Functions of the Form y = A tan (wx) + B Graph:

y = 2 tan x - 1

Figure 94 shows the steps using transformations.

x = 3211

x = .TI.2

/: �� h i\ /: (- � ,

� 1)1 l-

I

(a) y =

I

EXAM P L E 2

UI

Solution Figure 95

11 X = -2

/:

y

tan x

X = -1i:.2

x = 1i:.2

�� I(�, / 2)

X

1

I I

�

I I I I I I I

'� /

( - �, 2 ) Multiply by 2; Vertical stretch by a factor of 2

I

1

(b) Y =

x = _1I.2 Y j

317

X= 2

j j

j

j j j j

j

j

j I j

I I

I I I I I I I I I (-�, - 3)

I

L-. : x I I I I I ,

Subtract 1 ; vertically shift down one unit

2 tan x

x = 37i 2

x =�

j

I I I I I X I (11', -I1) I I I I I (c) y =

2 tan x -1 •

G raphing F unctions of the Form y = A tan (wx) + B Graph:

y = 3 tan(2x)

Figure 9S shows the steps using transformations.

x = 1I.2

31T X =T

X = -2

/: I I I I

X

il (a) y =

y

11

tan x

x = 311 I 2 I I I

3

I I I I

I X I I I I I I

I I I I

I I

(- � , -3)

Multiply by 3 ; Vertical strech by a factor of 3

I

(b) Y =

3 tan x

y x = _ 1I.4 : 3 I I I I :(0, 0) I I I I I 17 I (-8 , -3)

I x = .:;f I I I

Replace x by 2x ; (c) y = 3 tan ( 2x) Horizontal compression by a factor of t

37T

�=4 ( �, 3 ) x

•


577

'iT

Notice in Figure 95(c) that the period of y = 3 tan(2x) is 2 due to the compres-

sion of the original period 'iT by a factor of l. Also notice that the asymptotes are 2 'iT 'iT 3 'iT x = 4' x 4' x = 4' and so on, also due to the compression.

-

'I'

=

IZ-

-

Now Work

PROBLEM 2 1

The G ra ph of the Cotangent Fu nction We obtain the graph of y cot x as we did the graph of y = tan x. The period of y = cot x is 'iT . Because the cotangent function is not defined for integer multiples of 'iT, we will concentrate on the interval (0, 7T ) . Table 11 lists some points on the graph of y = cot x, 0 < x < 'iT . As x approaches 0, but remains greater than 0, the value of cos x will be close to 1 and the value of sin x will be positive and close to O. c�s x Hence, the ratio = cot x will be positive and large; so as x approaches 0, with Sill x x > 0, cot x approaches 00 ( Iim +cot x = 00 ) . Similarly, as x approaches 7T, but =

Table 1 1 x 'iT

6

'iT

4

'iT

3

'iT

2

2 'iT

3

3 'iT 4 5 'iT 6

y

=

\13 ( �, \13) (�, 1 ) \13 ( 3 ' \133 ) 3 cot x

0

(x, y)

-,-

�

(%, 0 )

\133 ( 3 '

-1

X-'J>O

27T

_

\13 3

e: )

-

remains less than 7T , the value of cos x will be close to 1 and the value of sin x will cos x ' be positive and close to O. So the ratio = cot x wIlI be negatIve and will approach - 00 as x approaches

)

, -1

7T

Sill X

( lim _cot x = X � '1T

,

,

- 00 ) . Figure 96 shows the graph.

cot x, -00 < x < 00, x not equal to integral mUltiples of 'iT, -00 < Y < 00 ,

Figure 96

Y =

X = -2'iT

X= O Y

x = -'iT

x = 2'iT

- \13 e:, - \13)

\1

ifCheck: Graph Y,

= cot x and compa re the res u lt with Figure 96. U se TRACE to see what happens when x is close to 0,

The graph of y =

A cot (wx)

+ B has similar characteristics to those of the tan

7T , The role of w A is to provide the magnitude of the vertical stretch; the presence of B indicates a vertical shift is required, gent function. The cotangent function y =

�=--

Now Work

A cot (wx)

+ B has period

PROBLEM 23

The Graph of the Cosecant Function a n d the Secant Function The cosecant and secant functions, sometimes referred to as reciprocal functions, are graphed by making use of the reciprocal identities csc x =

1 -,x Sill

and

sec x =

1 - cos x

578


For example, the value of the cosecant function y = csc x at a given number x equals the reciprocal of the corresponding value of the sine function, provided that the value of the sine function is not O. If the value of sin x is 0, then x is an integer multiple of 'IT . At such numbers, the cosecant function is not defined. In fact, the graph of the cosecant function has vertical asymptotes at integer multiples of 'IT . Figure 97 shows the graph. Y = esc x, Figure 97

-

00 < x < 00, x not equal to integer m ultiples of 7T, x

- 'IT

X= 0

X

Y

iyi

'IT

2': 1

x = 2'IT

j\

I I I I

x

\j Using the idea of reciprocals, we can similarly obtain the graph of See Figure 98. Y

sec x, - 00 < x < 00, x not equal to odd mu ltiples of

Figure 98 =

x = _ 37T 2

J

� 2

,

iy i

y=

sec x.

2': 1

x = -1I 2

I

x -1

2

G ra p h Fu nctions of the Form y = A csc (wx) a n d y = A sec (wx) + 8

+

8

The role of A in these functions is to set the range. The range of y = csc x is {Yi iyi 2: I } ; the range of y = A csc x is {Yi iyi 2: iAi } ' due to the vertical stretch of the graph by a factor of iAi. Just as with the sine and cosine functions, the period of 2 y csc(wx) and y = sec(wx) becomes w'IT , due to the horizontal compression of the graph by a factor of 1... The presence of B indicates a vertical shift is required. w We shall graph these functions in two ways: using transformations and using the reciprocal function. =

E XA M P L E 3

G raphing Functions of the Form y = A csc (wx) + B Graph:

y = 2 csc x

- I

579


Figure 99 shows the required steps. .

Solution Using Transfo rmations Figure 99

x= 0

x = -'IT Y I I I I I I I I

x= O X = - 'IT n X = 'IT x = 2'IT

x = 2'IT I I I I I I I I I

x

I

f\2

x= 0 x = - 'IT n x = 'IT X = 2'IT

( ¥, 1)

x

I I I I

-1

: ( 31T , -2) I I I I I I

Multiply by 2; �

Vertical stretch by a factor of 2

(a) y = csc x

(b) Y =

I I I I I I I

I I I I I I

x

( 3; , - 3) Subtract 1 ; Vertical shift down 1 unit

2 csc x

(e) y =

I I I I I I

2 csc x -1

We graph y = 2 csc x - I by first graphing the reciprocal function y = 2 sin x - I and then filling in the graph of y = 2 csc x - I, using the idea of reciprocals. See Figure 100.

Solution Using the Reciprocal F unction Figure 1 00

X= y

2

- 'iT

x = 21T

X = 'IT

x= o y

(¥, 1 ) Iy = I I

51T X T

2 sin x - 1

� (�, - 3 ) (a) y =

2 sin x - 1

",I'

-

(b) Y =

Now Work

2 csc x - 1

•

PROBLEM 29

7.7 Assess Yo ur Understanding 'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 3x 6 . . It . ?. 2. True or False A function f has at most one vertical asymptote. 1. Th e grap h a f y = has a vertlca I asymptote. What IS x 4 (pp. 346-348) ( pp. 346-348) -

---

Concepts and Vocabulary 3. TIle graph of y = tan x is symmetric with respect to the and has vertical asymptotes at .

__

__

4. The graph of y sec x is symmetric with respect to the and has vertical asymptotes at . =

__

__

5. It is easiest to graph y of 6.

=

sec x by first sketching the graph

The graphs of y = tan x, y = cot x, Y sec x, and y = csc x each have infinitely many vertical asymptotes.

Tnle or False

=

580



7-1 6,

if necessary, refer to the graphs to answer each question.

'- 7. What is the y-intercept of y = tan x? 9. What is the y-intercept of y = sec x? 11.

13. " 15.

8. What is the y-intercept of y = cot x? 10. What is the y-intercept of y = csc x? 12. For what numbers x, -27T 27T, does csc x = I? For what numbers x does csc x = - I ? 14. For what numbers x , - 2 7T 27T, does t h e graph of y = csc x have vertical asymptotes? 16. For what numbers x, - 27T 27T, does the graph of

For what numbers x, -27T ::; X ::; 27T, does sec x = I? For what numbers x does sec x = - I ? For what numbers x , -27T ::; X ::; 27T, does the graph of y = sec x have vertical asymptotes? For what numbers x, -27T ::; X ::; 27T, does the graph of y = tan x have vertical asymptotes?

In Problems

1 7-40,

::; X ::;

::; X ::; ::; X ::;

y = cot x have vertical asymptotes?

graph each function. Be sure to label key points and show at least two cycles.

17. y = 3 tan x "'\.. 21. y = tan ( �x ) 2S. Y = 2 sec x ' 29. Y = 4 sec (�x ) 33. y = tan ( �x ) 1 37. Y = � tan Gx ) - 2

18. y = -2 tan x 22. y = tan Gx ) 26. Y "21 csc x 30. y = "21 csc(2x) 34. Y = 2 cot x - I 38. y = 3 cotGx ) - 2

19. y = 4 cot x 23. y = cot Gx ) 27. Y -3 csc x 31. y = -2 CSC(7TX) 3S. Y = sec ( 2;x ) 39. y = 2 csc Gx )

=

+

20. Y = 3 cot x 24. Y = co{%x ) 28. y = -4 sec x 32. y = -3 sec (�x ) 36. Y = csc C;x ) 40. Y = 3 sec (�x ) + 1 -

=

+

2

-

1

Applications and Extensions In Problems

41-44,

find the average rate of change offfmm 0 to

41. f(x) = tan x In Problems

4S. f(x)

=

45-48,

tan x

g ( x ) = 4x

49.

42. f(x) = sec x find (f

0

g ) (x) and ( g

46.

6. 7T

43. f(x) = tan (2x)

f ) (x) and graph each of these functions. f(x) = 2 sec x f(x) = 2x 0

g(x) = � x 2

Two hallways, one of width 3 feet, the other of width 4 feet, meet at a right angle. See the illustration.

Carrying a Ladder around a Corner

47.

SO.

-

g(x) = cot x

44. f(x) = sec(2x) 48. f(x) = "21 x g(x) = 2 csc x

A Rotating Beacon Suppose that a fire truck is parked in front of a building as shown in the figure.

(a) Show that the length L of the line segment shown as a function of the angle 0 is � (b) Graph

L

L ( O ) = 3 sec O 7T = L(O), O < 0 < "2 .

+

4 csc O

(c) For what value o f 0 i s L the least? (d) What is the length of the longest ladder that can be car ried around the corner? Why is this also the least value of L?

The beacon light on top of the fire truck is located 10 feet from the wall and has a light on each side. If the beacon light rotates 1 revolution every 2 seconds, then a model for deter mining the distance d that the beacon of light is from point A on the wall after t seconds is given by d(t) = 1 10 tan(7Tt) 1 (a) Graph d(t) = 1 10 tan(7Tt) 1 for 0 ::; t ::; 2.

SECTION 7.8

51.

Exploration

1 0 tan(1T t)

d(O. I ) - d(O) d ( O .2) - d(O.I ) and so on, ' 0.2 - 0 . 1 0.1 - 0 for each consecutive value o f t . These are called first

(d) Compute

Graph

y = tan x

t =

581

(e) Interpret the first differences found in part (d). What is happening to the speed of the beam of light as d in creases?

(b) For what values of t is the function undefined? Explain what this means in terms of the beam of light on the wall. (c) Fill in the following table.

d(t)

Phase Shift; Sinusoidal Curve Fitting

and

( �) ( �)

y = -cot

Do you think that tan x = -cot

'

x

+

x

+

?

differences.


x

= 4

2. False

OBJECTIVES

1

Graph Sin usoidal Fu nctions of the Form (p. 5 8 1 )

y = A sin(wx - 1» +

8

2 Find a Sin u soidal Function from Data (p. 585)

1 Figure 1 0 1

One cycle y

= A sin(wx), A > 0, w > °

NOTE We c a n a lso find the beg i n n i ng

and end of the period by solving the

o :s

(/J cP w

-

:s

:S

We have seen that the graph of y = A sin(wx), w > 0, has amplitude [A[ and period 211 . 211 . T = - . One cycle can be drawn as x vanes from 0 to - or, eqUIvalently, as wx w

w

x

equal ity:

G ra p h Sinusoidal Fu nctions of the Form y = A sin(wx - 4» + B

wx - cP :s 211 wx :s 211 + cP cP 211 x :S + w w -

-

in

varies from 0 to 211. See Figure 10l. We now want to discuss the graph of y = A sin( wx - 1» which may also be written as

where w > 0 and 1> (the Greek letter phi) are real numbers. The graph will be a sine curve with amplitude [A[. As wx - 1> varies from 0 to 211, one period will be traced out. This period will begin when 1> or x = wx - 1> = 0 w

and will end when •

wx - 1> = 211

1> 211 or x = - +

See Figure 102. O n e cycle y

Figure 1 02

= A sin(wx - cP), A > 0, w > 0, cP > 0

w

w

582


[ ( �) ]

We see that the graph of y = A sin e wx - ¢) = A sin w x -

is the same

as the graph of y = A sin(wx), except that it has been shifted t units (to the right

t

if ¢ > 0 and to the left if ¢ < 0). This number is called th phase shift of the w graph of y = A sin(wx ¢). _

�

For the graphs of y = A sin(wx - ¢) or y = A cos(wx - ¢ ) , w > 0, Amplitude = IA I

Period =

T

=

271

Phase shift =

W

The phase shift is to the left if ¢ < 0 and to the right if ¢ > O.

t w

F i nding the Amplitude, Period, and Phase Shift of a Sinusoidal Function and G raphing It

EXAM P L E 1

Find the amplitude, period, and phase shift of y = 3 sin(2x - 71) and graph the function.

Solution

[ ( �) ] [ ( �) ]

Comparing

y = 3 sin (2x - 71) = 3 sin 2 x to

-

y = A sin(wx - ¢) = A sin w x -

N OTE W e also c a n f i n d t h e i nterval defin

ing one cycle by solving the inequal ity

o $ 2x

-

71

$

-

271.

Then 71 $ 2x $ 371 71

- $ x $

2

371

-

2

we find that A = 3, w = 2, and ¢ = 71. The graph is a sine curve with amplitude ¢ 71 271 271 . I A I = 3, penod T = - = - = 71, and phase shift = - = -. w w 2 2 The graph of y = 3 sin(2x - 71 ) will lie between -3 and 3 on the y-axis. One ¢ 71 ¢ 271 71 371 . . cycle WIll begm at x = - = - and end at x = + - = - + 71 = -. To find w w w 2 2 2

•

[ ]

71 371 into four subintervals, each of , the five key points, we divide the interval 2 2 71 length 71 -7- 4 = 4' by finding the following values of x: 71 2

-+-=-

- + - = 71

571 71 71 + - = 4 4

-+-=-

i n itia l va l u e

2nd va l ue

3rd va lue

4th va lue

fi na l va lue

71 2

71 4

371 4

371 4

71 4

571 4

71 4

371 2

Use these values of x to determine the five key points on the graph:

We plot these five points and fill in the graph of the sine function as shown in Figure 103 ( a). Extending the graph in each direction, we obtain Figure 103(b). y

(�, 3)

( gr, 3)

x

( If , -3 )

(�, -3)

(b)

(�, -3)

The graph of y

=

3 sin (2x - 7T)

[ ( ;)]

3 sin 2 X

=

583


SECTION 7.S

-

may also be obtained us

ing transformations. See Figure 104. Figure 1 04

(a) y

=

Multiply by 3; Vertical stretch by a factor of 3

sin x

Replace x by 2x; Horizontal compression by a factor of (c) y = (b) Y = 3 sin x

!

3

Replace x by x Shift right � units sin (2x )

-

�; (d ) Y = =

[2 (x ¥)]

3 sin 3 sin

'TT)

-

( 2x

-

•

To graph a sinusoidal function of the form y A sin (wx - ¢) + B, we first graph the function y = A sine wx - ¢) and then apply a vertical shift. =

Finding the Amplitude, Period, and P hase Shift of a Sinusoidal Function and G raphing I t

E XA M P L E 2

Find the amplitude, period, and phase shift of y the function. We begin by graphing y

Solution

=

y

=

2 cos(4x +

7T

2 cos(4x + 37T)

NOTE We can a lso find the interva l defi n

ing one cyc le b y solving the inequ a l ity

Then

o

:::;

4x

+

37T :::; 27T

-37T :::; 4x :::; - 7T 7T 37T - - :::; x :::; - 4 4

A

i n itia l va l ue

= [ ( ;)] �) ] A [ ( 2 COS 4 X +

- ¢)

cos w x

=

3

-

2, w = 4, and (p -37T. The graph is a cosine curve with ampji¢ 27T 27T 7T 37T . tude I A I 2 , period T and phase shift = - = - - . w w 4 2' 4 The graph of y = 2 cos( 4x + 37T) will lie between -2 and 2 on the y-axis. One ¢ 27T ¢ 37T 7T 7T 37T + + = . cycle will beain at x = - = - - and end at x = =b w w W 4 2 4 4 3 To find the five key points, we divide the interval , into four subintervals,

=

we see that

= = =

=

=

-

-

-

-

•

each of the length 37T 4

A cas( wx

2 cos( 4x + 37T) + 1 and graph

) . Comparing

to y =

=

37T 7T -- + - = 8 4

57T 8

- -

2 n d va l u e

;

[ ; :]

-

-

-

-

-:- 4

-

-

=

; by finding the following values. ,

57T 7T 7T + = -8 8 2

The five key points on the graph of y

(_ ) (_ ) (_ 57T 0 8 "

7T 2

-- +

-

3rd va lue

37T 2 4 "

-

=

7T 8

-

37T 8

= --

37T 7T -- + 8 8

-

7T 4

= --

fi n a l va l u e

4th va l ue

2 cos(4x + 7T) are

) (_ ) ( )

7T -2 , 2'

37T 0 8 "

_

7T 2 4'

We plot these five points and fill in the graph of the cosine function as shown in Figure l05(a). Extending the graph in each direction, we obtain Figure l05(b), the graph of y = 2 cos( 4x + 7T). A vertical shift up 1 unit gives the final graph. See Fig ure l05(c).

584

CHAPTER 7 Trigonometric Functions Figure

1 05

Y

(- � , 2)

( - � , 2)

2

'IT "8

-:!! 8

y

(� , 2 )

X

x

-2

( - -¥ ' -2)

Add 1 ;

Vertical sh ift up 1 unit

(b) Y = 2 cos (4x + 37T)

(a)

(c) y = 2 cos (4x + 37T) + 1

2 COS[ ( ;) ]

2

4 x+ 3

cos ( 4x + 7T) + 1 = 3 obtained using transformations. See Figure 106. The graph of y

Figure

y

=

+ 1 may also be

1 06 ( -,¥, 2)

( � , 2)

( - � , 2)

x

(a) y

=

2 cos x

( - t , -2)

( � , - 2)

('" -2) Replace

x by 4x;

(b) Y

=

2 cos (4x)

Replace

t

Horizontal compression by a factor of

Shift left

(e) y =

x by x + �; � units

=

Add

2 cos [4 (x +�)l 2 c o s (4x + 3 .. )

1; Vertical shift up 1 unit

( -,¥, 3)

I

y

x -2

�;,.... SUMMARY

(d) Y

Now Work

27T

=

Determine the starting point of one cycle of the graph,

STEP 3:

Determine the ending point of one cycle of the graph,

[t , t 27T]

A sin(wx

t. t 27T .

STEP 2:

-

f/J ) + B or y

=

w

W

+

W

27T

-7- 4. + into four subintervals, each of length w w w w STEP 5: Use the endpoints of the subintervals to find the five key points on the graph. STEP 6: Fill in one cycle of the graph. STEP 7: Extend the graph in each direction to make it complete. STEP 8: If B =F 0, apply a vertical shift.

Divide the interval

2 cos (4x + 37T)+ 1

PROBLEM 3

Steps for Graphing Sinusoidal Fu nctions y . STEP 1: Determine the amplitude I A I and penod T = - . w

STEP 4:

=

A cos(wx

-

f/J) + B

•

SECTION 7.8

2


585

Find a S i n usoidal Fu nction from Data Scatter diagrams of data sometimes take the form of a sinusoidal function. Let's look at an example. The data given in Table 12 represent the average monthly temperatures in Denver, Colorado. Since the data represent average monthly temperatures collected over many years, the data will not vary much from year to year and so will essen tially repeat each year. In other words, the data are periodic. Figure 107 shows the scatter diagram of these data repeated over 2 years, where x 1 represents January, x = 2 represents February, and so on. =

Table 1 2

Figure 1 07 y

Average Monthly Temperature, of

Month, x J a n u a ry, 1

29.7

Febru a ry, 2

33.4

M a rch, 3

39.0

April, 4

48.2

M a y, 5

57.2

June, 6

66.9

J u ly, 7

73.5

August, 8

7 1 .4

September, 9

62.3

October, 1 0

51 .4

November, 1 1

39.0

D e cem ber, 1 2

3 1 .0

75

•

•

•

•

•

• •

•

• • • •

•

•

•

• • •

•

•

0

25

x

SOURC[ U.S. National Oceanic and Atmos p heric Administration

Notice that the scatter diagram looks like the graph of a sinusoidal function. We choose to fit the data to a sine function of the form y = A sin ( wx - cP ) + B

where A, B, w, and cP are constants.

EXAM P L E 3

Finding a Sinusoidal Function from Temperatu re Data Fit a sine function to the data in Table 12. We begin with a scatter diagram of the data for one year. See Figure 108. The data will be fitted to a sine function of the form STEP 1:

y

=

A sin( wx - cP ) + B

To find the amplitude A, we compute largest data value - smallest data value 2 73.5 - 29.7 = 21 . 9 = 2

Amplitude =

To see the remaining steps in this process, we superimpose the graph of the function y = 21.9 sin x, where x represents months, on the scatter diagram.


586

Figure 1 09

Figure 109 shows the two graphs. To fit the data, the graph needs to be shifted vertically, shifted horizontally, and stretched horizontally. STEP 2: We determine the vertical shift by finding the average of the highest and lowest data values. 73.5 + 29.7 51.6 Vertical shift 2 Now we superimpose the graph of y 21.9 sin + 51.6 on the scatter dia gram. See Figure 110 . We see that the graph needs to be shifted horizontally and stretched horizontally. STEP 3: It is easier to find the horizontal stretch factor first. Since the temperatures repeat every 12 months, the period of the function is T = 12. Since 27T T = - = 12, we find

y 75

•

•

=

• • • •

•

•

• •

•

•

w

-25

=

=

w = 12 = 6 27T

x

7T

Now we superimpose the graph of

=

y

21.9 sin

(� x)

+ 51.6 on the

scatter diagram. See Figure 1 1 1 . We see that the graph still needs to be shifted horizontally. Figure 1 1 0

Figure 1 1 1

Y

··v

25

6 STEP 4:

75

25 0

x

12

6

To determine the horizontal shift, we use the period T interval [0, 12] into four subintervals of length 12 -;- 4 [0, 3],

[3, 6],

[6, 9],

=

=

12

x

12 and divide the 3:

[9, 12]

x=

The sine curve is increasing on the interval (0, 3) and is decreasing on the in terval (3, 9 ) , so a local maximum occurs at 3. The data indicate that a maximum occurs at 7 (corresponding to July's temperature), so we must shift the graph of the function 4 units to the right by replacing by 4. Doing this, we obtain

x=

=

Figure 1 1 2

wx

y

=

21.9 Sin

(� (x ) - 4)

The graph of y

=

21.9 sin

(� x ;) (� x ;)

21.9 sin

-

-

of the data are shown in Figure 1 12.

25 6

12

x

2

2

-

+ 51.6

Multiplying out, we find that a sine y A sin( - cP) + B that fits the data is y =

x x

function of the form

+ 51.6

+ 51.6 and the scatter diagram •

SECTION 7.8 Phase Shift; Sinusoidal Curve Fitting

587

The steps to fit a sine function y

= A sin (wx - 4» + B

to sinusoidal data follow: Steps for fitting Data to a Sine Function y = A sin( wx STEP 1:

-

4» +

8

Determine A , the amplitude of the function. largest data value - smallest data value Amplitude = 2 STEP 2: Determine B, the vertical shift of the function. largest data value + smallest data value Vertical shift = 2 STEP 3: Determine w. Since the period T, the time it takes for the data to re27T' . peat, IS T = -, we have w 27T' w = T

STEP 4:

Determine the horizontal shift of the function by using the period of the data. Divide the period into four subintervals of equal length. De termine the x-coordinate for the maximum of the sine function and the x-coordinate for the maximum value of the data. Use this information to determine the value of the phase shift, 1:. w

k'!!l;;

;;':> -

Now Work

PROBLEM 29

(a )

-

(c)

Let's look at another example. Since the number of hours of sunlight in a day cycles annually, the number of hours of sunlight in a day for a given location can be modeled by a sinusoidal function. The longest day of the year (in terms of hours of sunlight) occurs on the day of the summer solstice. For locations in the northern hemisphere, the summer solstice is the time when the sun is farthest north. In 2005, the summer solstice occurred on June 21 (the 172nd day of the year) at 2:46 AM EDT. The shortest day of the year occurs on the day of the winter solstice. The winter solstice is the time when the Sun is farthest south (again, for locations in the northern hemisphere). In 2005, the win ter solstice occurred on December 21 (the 355th day of the year) at 1:35 PM (EST).

EXAM P L E 4

F i nd i ng a Sinusoidal F unction for Hours of Daylight According to the Old Farmer's Almanac, the number of hours of sunlight in Boston on the summer solstice is 1 5 .30 and the number of hours of sunlight on the winter solstice is 9.08.

(a) Find a sinusoidal function of the form y = A sin (wx - 4» + B that fits the data. (b) Use the function found in part (a) to predict the number of hours of sunlight on April 1, the 91st day of the year. (c) Draw a graph of the function found in part (a). (d) Look up the number of hours of sunlight for April 1 in the Old Farmer's A lmanac and compare it to the results found in part (b). Source: The Old Farmer's Almanac, www.almanac.comlrise

Solution

(a) STEP 1: Amplitude =

largest data value - smallest data value 2 15 .30 - 9.08 = 3.11 2 =


588

largest data value + smallest data value 2 15.30 + 9.08 = 12 . 19 = 2

STEP 2:

Vertical shift =

STEP 3:

The data repeat every 365 days. Since T = W

So far, we have y = 3 . 1 1 sin STEP 4:

=

27T = 365, we find w

27T 365

(�:S ) x

-

¢ + 12.19.

To determine the horizontal shift, we use the period T = 365 and divide the interval [0, 365 J into four subintervals of length 365 -7- 4 = 91 .25: [ 0, 91 .25 J ,

[91.25, 182.5 J ,

[182.5, 273.75 J ,

[273.75, 365 J

The sine curve is increasing on the interval (0, 91 .25 ) and is decreas ing on the interval ( 91 .25, 273.75 ) , so a local maximum occurs at x = 91 .25. Since the maximum occurs on the summer solstice at x = 172, we must shift the graph of the function 172 - 91.25 = 80.75 units to the right by replacing x by x - 80.75. Doing this, we obtain

y = 3 . 1 1 sin

(�:S

(x - 80.75 )

)

+ 12.19

Multiplying out, we find that a sine function of the form y = A sin( wx - ¢) + B that fits the data is

Y = 3.11 sin

(

27T 3237T x 365 730

)

+ 12.19

(b) To predict the number of hours of daylight on April 1 , we let x = 91 in the function found in part (a) and obtain

(

. 27T . 91 - 323 7T Y = 3.11 Slll 365 730 ::::; 12.74

Figure 1 1 3

16

/....... .

/.

!

o

./ .." ....

8

\

...

'"'\

...

..

...,.......

400

)

+ 12.19

So we predict that there will be about 12.74 hours = 12 hours, 44 minutes of sunlight on April 1 in Boston. (c) The graph of the function found in part (a) is given in Figure 1 1 3 . (d) According to the Old Farmer's Almanac, there will be 12 hours 45 minutes of sunlight on April 1 in Boston. 'i"11 "

w >-

Now Work

III

PROBLEM 3 5

Certain graphing utilities (such as a TI-83, TI-84 Plus, and TI-86) have the capa bility of finding the sine function of best fit for sinusoidal data. At least four data points are required for this process.

E XA M P L E 5

Finding the Sine Function of Best Fit Use a graphing utility to find the sine function of best fit for the data in Table 12. Graph this function with the scatter diagram of the data.

Solution

Enter the data from Table 12 and execute the SINe REGression program. The result is shown in Figure 1 14.


SECTION 7.8

589

The output that the utility provides shows the equation y

=

a

sin ( bx + c) + d

The sinusoidal function of best fit is y

=

21.15 sin ( O.55x - 2.35 ) + 51.19

where x represents the month and y represents the average temperature. Figure 115 shows the graph of the sinusoidal function of best fit on the scatter diagram. Figure

75

S i nRe9 '::I=o3*'s i n ( b>(+c. ) +d 03=2 1 1 46827'36 b= . 549459 1 1 '39 c.= - 2 . 35007307 d=5 1 . 1 9288889 •

a

� ==-

115

Figure

1 14

Now Work

PROBLEMS 2

9(d)

,.

.A"\

�

J \

25

.

.

.

.

.

.

.

.

.

13

•

AND ( e )

7.8 Assess Your Understanding Concepts and Vocabulary 1. For the graph of y the

=

A sin( wx - -

1 + tan u 1 + cot u

=

tan u

1 + tan u 1 + tan u 1 tan u + 1 1 + - tan u tan u tan u� = tan Ll ..tafl--tt4=1

Now Work

•

P R O B L E M S 2 3 AND 2 7

When sums or differences of quotients appear, it is usually best to rewrite them as a single quotient, especially if the other side of the identity consists of only one term. E XA M P L E 6

E stab l i s h i n g an I dentity

sin () 1 + cos () . Establish the identity: ---- + 1 + cos () SIll () Solution

=

2 csc (}

The left side is more complicated, so we start with it and proceed to add. sin () I + cos ()

---- +

sin2 () + ( 1 + cos (})2 (1 + cos (}) (sin ())

Add the quotients.

sin2 () + 1 + 2 cos () + cos2 () ( 1 + cos ()) (sin ())

Remove parentheses in the numerator.

(sin2 () + cos2 () ) + 1 + 2 cos () ( 1 + cos ()) (sin ())

Regroup.

2 + 2 cos () ( 1 + cos ()) (sin ())

Pythagorean Identity

2 (� �(sin (})

Factor a nd cancel.

1 + cos () sin ()

----

2 sin () = 2 csc () il!l!l: = = "" '-

No w Work

Reci proca l Identity

•

PROBLEM 49

Sometimes it helps to write one side in terms of sine and cosine functions only. E XA M P L E 7

Estab l i s h i ng an Identity

tan v + cot v Establish the identity: ----sec v csc v

=

1

624

CHAPTER 8

Analytic Trigonometry

Solution

tan v + cot v sec v esc v

sin v cos v -- + - cos v sin v 1 1 cos v sin v

Cha nge to sines and cosines.

r

sin2 v + cos2 V cos v sin v 1 cos v sin v

Add the quotients in the n umerator.

1 ---. cos v sin v = 1 1 cos v sin v

Divide the quotients; 2 2 sin v + cos V = 1 . 1.. "I!l: = = :'- '-

NowWork

•

PROBLEM 69

Sometimes, multiplying the numerator and denominator by an appropriate fac tor will result in a simplification. E XA M P L E 8

E stab li s h i ng an I dentity

Establish the identity: Solution

1 - sin e cos e

cos e 1 + sin e

We start with the left side and multiply the numerator and the denominator by 1 + sin e. ( Alternatively, we could multiply the numerator and denominator of the right side by 1 - sin e.) 1 - sin e cos e

1 - sin e 1 + sin e 1 + sin e cos e cos e(l + sin e) cos2 e cos e( l + sin e) cos e 1 + sin e

L "l'];;= = :a- -

NowWork

M u ltiply the numerator and denom inator by 1 + sin e.

Cancel.

•

PROBLEM 5 3

Although a lot of practice is the only real way to learn how to establish identi ties, the following guidelines should prove helpful. WARN ING Be carefu l not to handle

identities to be established as if they were conditional equations. You cannot establish an identity by such methods as adding the same expression to each side and obtaining a true statement. Th is practice is not allowed, because the original statement is precisely the one that you are trying to esta blish. You do not know until it has been estab lished that it is, in fact, true. _

G u idelines for Esta blishing Identities 1.

It is almost always preferable to start with the side containing the more complicated expression . 2. Rewrite sums or differences of quotients as a single quotient. 3. Sometimes rewriting one side in terms of sine and cosine functions only will help. 4. Always keep your goal in mind . As you manipulate one side of the ex pression, you must keep in mind the form of the expression on the other side.

8.3 Asssess You r Understa n d i n g 'Are You Prepared?' Answers are given a t the end of these exercises. If you get a wrong answel; read the pages listed in red. 1.

True or False

sin2 e

=

1 - cos2 e. (p. 520)

2. True or False

sin ( -e)

+

cos ( -e)

=

cos e - sin e. (p. 556)

625

SECTION 8.3 Trigonometric Identities

Concepts and Voc a b u l a ry

sin( -e) + sin e a for any value of e. 7. True or False In establishing an identity, it is often easiest to j ust multiply both sides by a well-chosen nonzero expression involving the variable. 8. True or False tan e ' cos e sin e for any e i= (2k +

3. Suppose that f and g are two functions with the same do main. If f(x) = g ( x ) for every x in the domain, the equation is called a(n) . Otherwise, it is called a(n) equa tion. __

4. tan2 e - sec2 e

=

5. cos( -e) - cos e

=

6. True or False

__

=

1)�.

=

__

.

Skill B u i l d i n g

In Problems 9.

11.

simplify each trigonometric expression by following the indicated direction.

9-18,

Rewrite in terms of sine and cosine functions: cos e .

Multiply

SIll

1 -

1 1

by

e

+

+

tan e · csc e.

sin e ' . SIll e

. 12. MultIply

3 sin2 e

17. Factor and simplify: In Problems 1 9-98,

1 9.

csc e . cos e

25. tan

u

=

+

34. tan2 e cos2 e

+

37. sec u - tan

u =

+

tan

v

46.

csc e - 1 cot e

49.

1 - sin v cos v 1

55.

cos e 1 - tan e

cos e) - 1

sin e

2 sin e

+

+ 1

( tan e

16. Multiply and simplify:

1

18.

1)

20. sec e . sin e

=

Factor and simplify:

+

+

1) - sec2 e

tan e)

=

sec e

+

1)

1 ) ( tan e tan e

cos2 e - 1 cos2 e - cos e

, 23. cos e(tan e

+

=

29. ( sec e

cot2 e

1 = 1

cos u sin u 5

+

sec2 e

=

tan4 e

47.

2 sec v

cos e

sin e 1 - cot e

----

=

sin e

+

cos e

cos2 e + sin e

csc v - I csc v + 1

1

+

sin e 1 - sin e cos v sin v

50.

1 +

53.

1

1

- sin e . S Ill e

+

+ =

+

=

1

1

+

cot e ) ( csc e - cot e)

33. ( sin e

+

cos e)2

39. 3 sin2 e

1

+ 4

sin2 e 1 - cos e

-

1 - sin v 1 + sin v

4;,.

--

csc e + 1 csc e - 1

48.

cos e + 1 cos e - 1

51.

sin e sin e - cos e

54.

1 - cos e 1 + cos e

+ sin cos v

v

�

=

2 sec v

( sec e - tan e)2

56.

+

cos2 e

42.

1

tan2 e

30. ( csc e

sin e

=

=

=

cot e 1 - tan e

+

sec e csc e

tan e 1 - cot e

----

=

+

1 +

sin e cos e

--

= =

=

3

+

cos2 e

-cos e 2 tan e

1 + sec e 1 - sec e

=

1

1 - cot e ?

( csc e - cot e)-

tan e

+

1

(sin e - cos e)2

tan2 e

sin u = --- I + cos u

- cot u

1

=

+

27. ( sec e - 1 ) ( sec e

tan e ) ( sec e - tan e)

35. sec4 e - sec2 e

41. 1 -

24. sin e( cot e

sin2 u

cot2 e)

cot e csc e + 1 =

=

csc e

+

38. csc u 4

+

=

32. ( 1 - cos2 e) ( l

44.

cos v 1 - sin v

cot e)

26. sin u csc u - cos2 U

1 +

=

tan e

sin2 Ll

cot2 e sin2 e

---- =

+

+ 4

+

+

sin e cos e

cat v + 1 cat v - 1

sin2 e + cos e

52.

-

+

=

=

40. 9 sec2 e - 5 tan2 e 1 +

cos e ) ( sin e

cot e

tan2 e)

1 - tan v

+

establish each identity.

28. (csc e - 1 ) ( csc e

43.

sin2 e

cot u - cos2 U

31. cos2 e ( 1

cos e - sin e sin e

( sin e

15. Multiply and simplify:

1 - cos e si n e b y --cos e 1 - cos e

I +

14. Rewrite over a common denominator: 1 1 1 - cos v 1 + cos v

13. Rewrite over a common denominator:

sin e + cos e cos e

cot e · sec e.

10. Rewrite in terms of sine and cosine functions:

cot e

=

2

626

CHAPTER 8

57. tan 0

60.


cos 0 1 + sin 0

+

=

sin 0 - cos 0 + 1 sin 0 + cos 0 - 1

-----

63.

tan u - cot u tan u + cot u

66.

sec 0 1 + sec 0

69.

sec 0 - csc 0 sec 0 esc 0

+

=

1

.

S i ll

78.

80.

1

sec 0 - sin 0

+

tan 0 - cot 0 . = SIIl2 0 - cos2 0 tan 0 + cot 0

62.

2 Sill ' 2

64.

tan u - cot u + 2 cos2 u = 1 tan u + cot u

65.

sec 0 + tan 0 = tan 0 sec 0 cot 0 + cos 0

67.

1 tan2 0 1 + tan2 0

68.

1 - cot2 0 + 2 cos2 0 = 1 1 + cot2 0

70.

sin2 0 - tan 0 cos-? 0 - cot 0

73.

1 1 + 1 - sin 0 1 + sin 0

76.

1 + sin 0 ? = (sec 0 + tan 0)1 - sin 0

U

0 - cos 0

tan v

84.

cos 0 + sin 0 - sin3 0 sin 0

88.

1 - 2 cos2 0

sin 0 cos 0

= sin v + cos v

=

+

1

=

=

cot 0 + cos-? 0

tan 2 0

+

=

? 2 sec- 0

74.

77.

sin 0 + cos 0 cos 0

-----

1 + sin 0

1 - sin 0

1 + cos2 0

sin 0 - cos 0 sin 0

=

-----

(2 cos2 0 - 1 ) 2

cos4 0 - sin4 0

sec 0 csc 0

? = 1 - 2 sin- 0

1 + sin 0 + cos 0 1 + sin e - cos e

------

1 + cos e sin e

89. (a sin e + b cos e)2 + (a cos e - b si.n e)2 = a2 + b2 +

tan a cot a =

tan f3 cot f3

+

=

tan a tan f3

0

93. (sin a + cos (3 )2 + (cos f3 + sin a ) ( cos f3 - sin a) = 2 cos f3(sin a + cos (3 )

95. In /sec 0 / 97. Inll

+

=

101 . f(e) =

(cos f3 + sin a ) ( cos f3 - sin a )

+

99-102,

=

- 2 cos f3(sin a - cos (3 ) 96. In/tan el = Inlsin el - Inlcos e l

- In /cos e/

cos 01

In Problems 99. f(x)

=

+

Inl1 - cos e l = 2 In lsin e l

98. In lsec 0 + tan el + Inlsec e - tan el

show that the functions f and g are identically equal.

sin x ' tan x g (x) = sec x - cos x 1 - sin e cos e

--

cos 0 g(e) 1 + sin e

sec 0

(sec v - tan v)2 + 1 = 2 tan v esc v(sec v - tan v)

cos2 0 - sin2 0 = cos2 0 1 - tan2 0

tan (3 ) ( 1 - cot a cot (3 ) + (cot a + cot (3 ) ( 1 - tan a tan (3 )

94. (sin a - cos f3 ?

+

1 - sin 0 --.- = 4 tan 0 sec 0 1 + Sill 0

---

83.

91. 92. ( tan a

tan 0

71. sec 0 - cos 0 = sin 0 tan 0

sin3 0 + cos3 0 = 1 - sin 0 cos 0 . S i ll 0 + cos 0

87.

1 + cos 0 + sin 0 = sec O + tan e 1 + cos e - sin e

sec 0 - cos 0 sec 0 + cos 0

----

81.

85.

= tan 0 - cot 0

59.

2 cos2 0

79.

sec 0 - sin 0 tan 0 - 1

sin3 0 + cos3 0 1 - 2 cos2 0

86.

-

-----

82.

1

COS-

cos 0 - sin 0 = sec 0 csc 0 cos e

sin 0 + cos 0 sin 0

---

=

61.

1 + sin 0 cos3 0

sec2 v - tan2 v sec v

tan 0 + sec 0 - 1 tan 0 - sec 0 + 1

sin 0 + 1 cos 0

72. tan 0 + cot 0 = sec 0 csc 0

75.

tan 0 - tan2 0

58.

1 - cos 0 sin2 0 =

sin 0 cos 0 ? 0 - sin 2 0

sec 0

=

0

100. f(x)

=

cos X · cot x g(x)

102. f( e)

=

tan e + sec e g( e)

=

=

0

esc x - sin x

=

1

cos 0 .

- SIIl

e

SECTION 8.4

Sum and Difference Formulas

627


103.

Searchlights A searchlight at the grand opening of a new car dealership casts a spot of light on a wall located 75 meters from the searchlight. The acceleration r of the spot of light is found to be r = 1200 sec e(2sec2 e - 1 ) . Show that this is 1 + sin2 e . . eqUivalent to r = 1200 cos3 e Source: Adapted from Hibbeler, Engineering Mechanics: Dynamics, 1 0th ed. © 2004.

(

104.

)

Opti cal Measurement Optical methods of measurement often rely on the interference of two light waves. If two light waves, identical except for a phase lag, are mixed together, the resulting intensity, or irradiance, is given by (csc e - l ) (sec e + tan e) [t = 4A 2 . Show that this is equivacsc e sec e lent to It = (2A cos ef

Source: Experimental Techniques, July/August 2002

Discussion a n d Writing

105. Write a few paragraphs outlining your strategy for estab lishing identities. 106. Write down the three Pythagorean Identities.

107. Why do you think it is usually preferable to start with the side containing the more complicated expression when establish ing an identity? 108. Make up an identity that is not a Fundamental Identity.


1. True

2. True

8.4 Sum and Difference Form ulas Before getting started, review the following:


•

•

Distance Formula (Section 2 . 1 , p. 157) Values of the Trigonometric Functions (Section 7.3, pp. 529-532 and Section 7.4, pp. 540-547)

"NOW Work the 'Are You Prepared?' problems on page

Finding Exact Values Given the Value of a Trigonometric Function and the Quadrant of the Angle (Section 7.4, pp. 546-548)

634.

OBJECTIVES 1 Use S u m a n d Difference Formulas to F i n d Exact Va lues (p. 628) 2

3

Use S u m and Difference Formu las to Esta b l i s h I dentities (p. 629) Use Sum a n d Difference Formu l a s I nvolvi n g I nverse Trigonometric Functions (p. 633)

In this section, we continue our derivation of trigonometric identities by obtaining formulas that involve the sum or difference of two angles, such as cos( O' + (3 ) , cos(O' - (3 ) , or sin( O' + (3 ) . These formulas are referred to as the sum and difference formulas. We begin with the formulas for cos( 0' + (3 ) and cos( 0' - (3 ) . THEOREM

('

('

In

Words

(' Formula (1) states that the r cosine of the sum of two angles r equals the cosine of the first angle times the cosine of the r second angle minus the sine of (' the first angle times the sine (' of the second a ng le.

Sum and Difference Formulas for the Cosine Function

cos(O' + (3) = cos 0' cos {3 - sin 0' sin (3

(1)

cos(O' - (3 )

(2)

=

cos 0' cos {3 + sin

0'

sin {3

�------�� Proof

We will prove formula (2) first. Although this formula is true for all numbers 0' and {3, we shall assume in our proof that 0 < {3 < 0' < 27T . We begin with the unit circle and place the angles 0' and (3 in standard position, as shown in Figure 20(a). The point Pl lies on the terminal side of {3, so its coordinates are ( cos (3, sin (3 ) ; and the point P2 lies on the terminal side of 0' , so its coordinates are ( cos 0' , sin 0' ) .

628

CHAPTER 8


P2

Figure 20

=

(cos Ci, sin Ci )

�, �

P1

=

P3

=

(cos( Ci - (3), sin( Ci - (3)) Y

1

(cos [3, sin (3)

I x

-1

\,

A

-1

=

( 1 , 0) x

-1

-1

(b)

(a)

Now place the angle Cl' - (3 in standard position, as shown in Figure 20(b). The point A has coordinates ( 1 , 0), and the point P3 is on the terminal side of the angle Cl' - {3, so its coordinates are (cos(Cl' - (3 ) , sin(Cl' - (3 ) ) . Looking a t triangle OP1 P in Figure 20(a) and triangle OAP3 in Figure 20(b), we 2 see that these triangles are congruent. (Do you see why? We have SAS: two sides and the included angle, Cl' - (3, are equal. ) As a result, the unknown side of each tri angle must be equal; that is, Using the distance formula, we find that

V[ cos( Cl' - (3 ) - I F + [ sin ( Cl' - (3) - Of = V( cos Cl' - cos (3)2 + (sin Cl' - sin (3)2

[ cos( Cl' - (3 ) - I f + sin2(Cl' - (3 ) = ( COS Cl' - cos (3 )2 + ( sin Cl' - sin {3 ? cos2(Cl' - (3) - 2 cos( Cl' - (3) + 1 + sin2( Cl' - (3) = cos2 Cl' - 2 cos Cl' cos {3 + cos2 {3 + sin2 Cl' - 2 sin Cl' sin {3 + sin2 (3 2 - 2 cos( Cl' - (3 ) = 2 - 2 cos Cl' cos {3 - 2 sin Cl' sin {3 -2 cos( Cl' - (3 ) = -2 cos Cl' cos {3 - 2 sin Cl' sin (3 cos( Cl' - (3 ) = cos Cl' cos {3 + sin Cl' sin {3

d(A, P3)

=

d(P" P2)

Square both sides. M u ltiply out the squared terms. Apply a Pythagorean Identity (3 times). Su btract 2 from each side. Divide each side by - 2.

•

This is formula (2).

The proof of formula ( 1 ) follows from formula (2) and the Even-Odd Identities. We use the fact that Cl' + {3 = Cl' - ( -(3 ) . Then cos( Cl' + (3) = cos[Cl' - ( - (3 ) ] = cos Cl' cos( - (3 ) + sin Cl' sine -(3 ) = cos Cl' cos {3 - sin Cl' sin (3 1

EXAM P L E 1

Use form u la (2). Even-Odd Identities

•

Use Sum and Difference Formulas to Find Exact Values

One use of formulas (1) and (2) is to obtain the exact value of the cosine of an angle that can be expressed as the sum or difference of angles whose sine and cosine are known exactly. Using the Sum Formula to F i n d Exact Values

Find the exact value of cos 75°. Solution

Since 75° = 45° + 30°, we use formula ( 1 ) to obtain cos 75°

=

cos( 45° + 30° )

=

i

cos 45°cos 30° - sin 45°sin 30°

Form ula (1) =

v'2 . V3 2

2

_

v'2 . 1. 2

2

=

1.4 ( V6 v'2) _

•

SECTION 8.4

E XA M P L E 2

Sum

629

and Difference Form ulas

U s i n g the Diffe rence F o r m u l a to F in d Exact Values

Find the exact value of cos � . 12

Sol ution

cos

)

(

4

= ""I

2

(

7T 37T 27T 7T 7T = cos 12 - 12 = cos 4 - 6 12 7T 7T . 7T . 7T = cos - cos + sm - sm 4

6

-

V2 . � = �

NowWork

PROBLEM 1 1

.:> -

2

2

Use formu la (2).

6

V2 . v3 + 2

)

4

2

(V6 + V2)

•

Use Sum and Difference Formulas to Establish Identities

Another use of formulas ( 1 ) and (2) is to establish other identities. Two important identities we conjectured earlier in Section 7.6 are given next.

I

Seeing the Concept Graph Y1

=

cos

(� e) -

and Y2

=

sin e

- e

= sin e

(3a)

sin

-e

= cos e

(3b)

on the same screen. Does this demon strate the result 3 (a)? How wou ld you demonstrate the result 3(b)?

(; ) (; )

cos

Proof To prove formula (3a), we use the formula for cos (a - f3 ) with a = f3 = e.

cos

(; ) - e

= cos

;

cos e + sin

% sin e

;

and

= o · cos e + 1 . sin e

= sin e To prove formula (3b), we make use of the identity (3a) just established. sin

(; ) t [; (; ) ] cos

- e

-

- e

= cos e •

Use (3a).

Formulas (3a) and (3b) should look familiar. They are the basis for the theorem stated in Chapter 7: Cofunctions of complementary angles are equal. Also, since cos

(; ) - e

[ ( ;) ] r ( ; )

= cos - e -

cos e -

Even Property of Cosine

and since cos

( ;)

(; ) t - e

3(a)

sin e

( ;)

it fOHows that cos e = sin e . The graphs of y = cos e and y = sin e . . are IdentICal. Having established the identities in formulas (3a) and (3b) , we now can derive the sum and difference formulas for sin (a + f3 ) and sin(a - f3 ) .

630

CHAPTER 8


Proof

sin(a + f3 )

=

=

r r

r

In

=

Words

Form ula (4) states that the sine of the sum of two angles equals r the sine of the first angle times the cosine of the second angle r plus the cosine of the first angle r times the sine of the second r angle. r

cos

[;

cos

(� )

- (a + f3 )

]

Form u la (3a)

- a cos f3 + sin

(� )

- a sin f3

sin a cos f3 + cos a sin f3

Formula (2) Formulas (3a) and (3b)

sin(a - f3 ) = sin[a + ( -f3 ) ] =

sin a cos( -f3 ) + cos a sine - f3 )

Use the sum formula for sine just obtained.

=

sin a cos f3 + cos a( -sin f3 )

Even-Odd Identities.

=

•

sin a cos f3 - cos a sin f3

Sum and Difference Formulas for the Sine Fu nction

THEOREM

=

sin(a + f3 )

=

sin (a - f3 )

(4)

sin a cos f3 + cos a sin f3

(5)

sin a cos f3 - cos a sin f3

�

�------�

EXA M P L E 3

Using the S u m Formula to F i nd E xact Val ues

Find the exact value of sin Solution

7

71 sin 12

3

) (

4 17 . 12 = sm

17 17 '4 + '3

=

17 sin 12 +

=

17 . 17 17 + cos sm - cos sm -

=

.

17 4

V2 . ! + 2

==> '(Jl!!';

EXAM P L E 4

(

�;.

2

4

3

V2 . V3

NowWork

2

2

)

3

=

! ( V2 + 4

Formula (4)

V6)

•

PROBLEM 1 7

U s i n g the Difference Formula to F i n d Exact Val ues

Find the exact value of sin 80° cos 20° - cos 80° sin 20° . Solution

The form of the expression sin 80° cos 20° - cos 80° sin 20° is that of the right side of the formula (5) for sin(a - f3) with a = 80° and f3 = 20°. That is, sin 80° cos 20° - cos 80° sin 20° i!l!

EXAM P L E 5

-

NowWork

=

sin(80° - 20° )

=

sin 600

If it is known that sin a 31

=

�, ; < a
- NowWork

•

PROBLEM 3 1

(d )

E stablishing an I dentity

Prove the identity: Solution

tan a + tan 13 tan(a + 13 ) = -----'----1 - tan a tan 13

tan ( ()

+

7T' ) =

tan( () +

7T'

) = tan ()

tan () + tan 7T' tan () + 0 = = tan () 1 - tan () tan 7T' 1 - tan () 0 0

n

The result obtained in Example 7 verifies that the tangent function is periodic with period 7T' , a fact that we discussed earlier.

SECTION 804

EXAM P L E 8

Solution

( ;)

tan e +

= -cot e

We cannot use formu la (6), since tan

( %) ( ;)

WARNING Be ca reful when using for mulas (6) and (7). These formu las can be used only for angles a and f3 for which tan a and tan f3 are defined, that is, all angles except odd integer 'iT'

633

Establish ing an Identity

Prove the i dentity:

multiples of 2'


sin e +

sin e cos

cos e +

cos e cos

E XA M P L E 9

- sin e sin

Trigonometric F unctions

( �

sin cos -1

a ::;

7T'

�)

+ sin -1

We seek the sine of the sum of two angles,

a =

0' =

V1

- cos?- 0' =

cos f3 =

V1

? - s nr f3

.

Sll1

Sll1 .

(

II

cos- 2'

+

.

3 ) = s .m (

s .m-1 5'


a

an d cos f3. Since sin a

25 = = 'V� 1 - 25 = 'V[16

a +

2::

0 an d

V3 'VrI(3 1 - "4 = 'V "4 = 2

f3 ) = =

C" ,� = :;::c__

3

1

cos -1 2' an df3 = sin -15" The n

3 7T' 7T' an d sin f3 = - -- ::; f3 ::; 2 2 5

We use Pythagorean I dentities to obta in sin cos f3 2:: 0 ( do you know why ?), we fin d

As a result,

Solution

•

Finding the Exact Value of an Expression Involving Inverse

1

EXA M P L E 10

% %


cos a = - 0::; 2 NOTE In Example 9, we could also find a 1 sina by using cosa = - = - , so a = 1 r 2 and r = 2. Then b = v3 and b v3 sin 0' = - = -. We could find cos f3 r 2 • in a similar fashion.

+ cos e sin

Use Sum and Difference Formulas Involving Inverse

Fin d the exact va lue of: Solution

% %

(sin e) (0) + (cos e) (1) cos e ------ = ----- = -cot e (cos e) ( o) - (sin e) ( l) -sin e

•

3

; is not defin e d. Instea d, we procee d as follows:

Sill

.

+

0' cos f3

V3 4 + 1

---0-

2

5

4

5'

cos 0' Sll1 f3 .

3 = 4V3 + 3

_0_

2 5

_______

10

73

•

Writing a Trigonometric Expression as an Algebraic Expression

Write sin (sin -l u + cos -1 v ) as an algebra ic expression containing u an d v (that is, wit hout any trigonometric functions). Give the restrictions on u an d v. First, for sin -1 u, we have ::; u ::; an d for cos -1 v, we have ::; v::; 1. Now 1 let 0' = sin - u an df3 = cos -1 v. Then

-1

sin 0' = cos f3 =

u

v

1

7T'

-1

7T'

-- ::; 0' ::; - -l::;u::;l 2 2 0::; f3 ::; 7T' - 1 ::; v ::;

1

634

CHAPTER 8

Analytic Trigonometry 71

.

71

Smce - - ::; Q' ::; -, we know that cos Q' 2 2

Similarly, since 0

::;

f3

As a result,

::;

71,

2:

O. As a result,

we know that sin f3

sin f3 =

VI

- cos2 f3 =

2:

O. Then

�

sin(sin- 1 u + cos- 1 v) = sin(Q' + f3) = sin Q' cos f3 + cos Q' sin f3

��

= uv + "II

S U M MARY

-- Now Work P R O B L E M


cos(eX + f3) = cos Q' cos f3 - sin Q' sin f3 sin(Q' + f3) = sin Q' cos f3 + cos Q' sin f3 tan Q' + tan f3 1 - tan Q' tan f3

tan(Q' + f3)

83

cos(Q' - f3) = cos Q' cos f3

+

sin Q' sin f3

sin( Q' - f3) = sin Q' cos f3 - cos Q' sin f3

tan(Q' - f3)

tan Q' - tan f3 1 + tan Q' tan f3

8.4 Assess Your Understanding 'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red.

1. The distance d from the point (2, -3) to the point (5,1) is . (p.l57)

3. (a) sin

__

�.

cos

�=

__

(b) tan � - sin �6 = 4

4 2. If sin 8 = "5 and 8 is in quadrant II, then cos 8 = __ . (pp.54 -6 548)

. (pp.529-532)

__

.(pp. 529-532)

Concepts and Vocab�lary

4. cos(O' + (3 ) = cos a cos {3

5 . sin ( a - (3 ) = sinO'cos{3

__

__

sin a sin (3 .

cosO'sin{3.

6. True or False sine a + (3 ) = sin

a

+ sin {3 + 2 sin a sin (3

7. True or False tan 75° = tan 300 + tan 450

(

8. True or False cos � - 8

)=

cos 8

Skill Building

In Problems 9.

. 57T sll112

15. tan 15°

9-20,

find the exact value of each expression. . 7T 12

1o .

Sll1

77T 11. cos 12

16.

tan 1950

177T 1 7. sinU

In Problems 21-30, find the exact value of each expression.

77T 12. tan 12

197T 18. tan u

13.

cos 1650

19.

sec -

( �)

21. sin 200 cos 100 + cos 20° sin 100

22. sin 20° cos 800 - cos 200 sin 80°

tan 20° + tan 250 25. ------1 - tan 20° tan 250

26.

23. cos 70° cos 200 - sin 700 sin 200 7T 77T 7T 77T 27. sin -cos - - cos -sin 12 12 12 12 7T 57T . 57T. 7T 29. cos - cos - + Sll1 - S111 12 12 12 12

24. cos 40° cos 10° + sin 400 sin 10° tan 400 - tan 100 1 + tan 40° tan 100

57T 77T . 57T 77T 28. cos - cos - - S111 - sin 12 12 12 12 7T . 57T . 7T 57T 30. Sll1 -cos - + cos - S111 18 18 18 18

14. sin 1050

•

SECTION 8.4

In Problems

3 1-36,

(a) sin(a + f3)

, 31. sma

=

33. tan a

=

,

(b) cos(a + f3)

3

7r

- , 0 < 0' < - ; cosf3 = 5

2

4 7r - - - < a < 7r'' 3' 2

=

(a) cos 8

38. If cos 8

=

�,

cosf3

2Vs 5

--

=

1

(c) sin(a - f3)

(d) tan(a -f3)

7r

, - - < f3 < 0 2

7r - , 0 < f3 < 2 2

tanf3

=

-

,V3 J;; '

7r < f3 < 7r "2

(b) sin(8

+�)

=

5' 0 < a < "2; smf3

=

34. tan a

=

12,7r < a

=

·

- sin 15° and then apply formula (lOa) .

- cos 30°

2 /1 - V3/2 _ )2 - V3 \j 2 4 =

=

_

'1/2 - V3 2

•

It is interesting to compare the answer found in Example 5 (a) with the answer to Example of Section 8.4. There we calculated

2

cos!!.. 12

=

cos 15°

= .!4

(V6 + vz)

B ased on this and the result of Example 5 (a), we conclude that

�(V6 + VZ)

and

are equal. (Since each expression is positive, you can verify this equality by squar ing each expression.) Two very different looking, yet correct, answers can be obtained, depending on the approach taken to solve a problem. oJ'l'===- -


642

CHAPTER 8


E XA M P L E 6

F inding E xact Values Using H alf-angle Formulas

3 If cos a = -"5' 1T .

a

�:dr:�t

F q

a

31T 2' find the exact value of:

< a

a

(c) tan 2"

(b) cos 2

(a) sm 2 Solution

14. csc 8 = -Vs cos 8 < 0

0

1 9-28,

18. cot 8 = 3

17 . tan 8 = -3 sin 8 < 0

csc 8 < 0

sec 8 < 0

15. cot 8 = -2

cos 8 < 0

use the Haltangle Formulas to find the exact value of each expression.

" 19. sin 22.5"

•

7� S 7� 26. csc S 21 . tan

20. cos 22.So

l S� 25 . sec - 8

24 . sin 1 9So

22. tan

( �)

y

+

y2

=

x2

5

+

y2

x

35. g(2a)

36. f (2a)

� � cos(28) � cos(

'41. Show that sin4 8 =

+

-

=

cos x, and hex) = tan x.

=

1

x

32. f 37. f

( S3� )

28. cos -

y

x2

30. g (28)

23. cos 16So

27 . sin -

In Problems 29-40, use the figures to evaluate each jill1.ction given that f ( x ) = sin x, g ( x)

29. f(28)

9� S

(� )

(%)

33. h (28)

(%)

34.

h

40.

h ( 2a )

42. Show that sine 48) = (cos 8 ) ( 4 sin 8 - 8 sin3 8).

48) .

43. Develop a formula for cos(38) as a third-degree polynomial

44. Develop a formula for cos( 48) as a fourth-degree polynomial

45. Fi nd an expression for sin(S8) as a fifth-degree polynomial i n

46. Find an expression for cos(S8) as a fifth-degree polynomial

in the variable cos 8.

in the variable cos 8,

the variable sin 8. In Problems

47-68,

in the variable cos 8.

establish each identity,

47. cos 4 8 - sin4 8 = cos(28)

48.

1

cot 8 - tan 8 = cos(28) cot 8 + tan 8

cot2 8 - 1 2 co t 8

49. cot( 28) = - - - - - -

1 52. csc(28) = 2 sec 8 csc 8

sec2 8 2 - sec-1 8

50. cot(28) = 2 (cot 8 - tan 8 )

51. sec(28)

53. cos2 (2u ) - sin2 ( 2u )

54. (4 sin U cos u ) ( l - 2 sin2 u ) = sin(4u )

=

cos( 4u)

=

8 57 . sec-1 - = 2

sec v + 1 59. cot2 = 2 sec v-I

64.

_

+

+

cos 3 8 cos 8

cos 8 + sin 8 cos 8 - sin 8 . . = 2 tan( 28 ) cos 8 - s m 8 cos 8 + s m 8

66. tan 8

+

68. I n\cos 81

tan(8 =

1

+

1 20°) + tan(8 + 240°) = 3 tan(38)

2(ln\1

+

2

cos(28) 1 - I n 2)

63. '\"

-

18

1

cot 8 cot 8 +

2 - cos 8

---

8 - tan2 2 61. cos 8 = 8 1 + tan2 2 1

60. tan 2 = csc v - cot v

SiI 13 8 1. sin(28) = . sm 8 2

I +

58. csc- - =

---

v

�

62. 1

2 1 + cos 8

cos(28 ) '---sin(28)

55. -

----

sin(38) cos(38) = 2 SII1 8 cos 8

,-

,, 65. tan(38) =

--

3 tan 8 - tan3 8 1 - 3 tan2 8

67. I nlsi n 8 1 = .( l nl1 - cos(28 ) 1 - In 2 )

1 2

1

1

SECTION 8.5

In Problems

69-80,

find the exact value of each expression.

( �) tan [ 2 cos- 1 ( - �) ]

[ �] tan ( 2 t an-1 �)

( �) �) sin ( 2 sec ( 2 tan �)

69. Sin 2 Sin-l

70. Sin 2 s in-1

71. cos 2 sin-I

73.

74.

75.

79.

Double-angle a nd Half-angle Formulas

645

( �) cos [ 2 tan-1 ( - �) ] csc [ 2 sin-1 ( - D]

72. cos 2 cos-1

COS-I

76.

-]

80.


81.

of v arying periods and amp litudes. A first approxim ation to the s awtooth curve is given by

Laser Projection In a laser projection system, the optical or scanning angle e is related to the throw distance D from the scanner to the screen and the projected image width W by

y =

!W 2 the equat ion D = csc e - cot e ( a) S how that the projected image width is given by e W = 2D t an i (b) Find the optical angle if t he t hrow distance is 15 feet and t he projected image w id th is 6 .5 feet.

� s in (27Tx) � Sin( 47TX) +

S how that y = sin(27Tx) COS2 ( 7TX) .

_____

Source: Pangolin Laser Systems, Inc. 82.

Product of Inertia The product of inertia for an area about inclined axes is given by the formula II/v = Ir sin e cos e - Iy sin e cos e + lry( cos 2 e - sin2 e ) . Show that this i s equivalent to

Ir - Iy

85.

sin 2e + lxy cos 2e. 2 Source: Adapted from Hibbeler, Engineering Mechanics: Statics, 10th ed., Prentice H all © 2004. 83. Projectile Motion An object is propelled upw ard at an angle e , 45° < e < 90°, to the horizontal with an initi al ve locity of Va feet per second from the b ase of a pl ane that makes an angle of 45°; with the horizontal . See t he illustra tion. If air resistance is ignored, t he dist ance R t hat it travels up the inclined plane is given by t he function I"v =

R(e) =

v

2

, .... �-'4iI

( a) S how that

\

2

v2 [sin(2e) - cos(2e) - 1 ) � v

, (b) Grap h R = R(e). (Use va = 3 2 feet per second.) (c) W hat v alue of e m akes R the l argest? (Use va = 32 feet per second.) 84. Sawtooth Curve An oscilloscope often displ ays a sawtooth curve. This curve can be approximated by sinusoidal curves "

See the illustrat ion. The height h bisects the angle e and is the perpendicular bisector of the b ase.]

[Hint:

v2 � cos e(sin e - cos e) I f f I

R(e) =

Area of an Isosceles Triangle S how t hat the are a A of an isosceles triangle whose equal sides are of length sand e is the angle between them is 1 2 . S Sil l e Z

86.

Geometry A rectangle is inscribed in a semicircle of radius l. See t he illustration.

l-x�1 1...-1 --1

( a) Express t he are a A of the rectangle as a function of the angle e shown in the illustration. (b) S how t hat A ( e ) = sin(2e). (c) Find the angle e that results in the l argest area A. (d) Find t he dimensions of this l argest rectangle. 87. If x = 2 tan e, express sin(2e ) as a function of x. 88. If x = 2 t an e, express cos(2e) as a function of x.

646

CHAPTER 8

Ana lytic Trigonometry

95. Use the fact that

89. Find the value of the number C: 1 1 - sin2 x + C - - cos(2x ) 4 2 90. Find the value of the number C:

cos

=

1

- cos2 X + C = - cos(2x) 4 2 2z a . tan -, show that sm a = -- ' 2 1 + Z2 1

If j

91. If z

=

to find sin

a

§S. 92. If z = tan -, show that cos a 2

=

7T . 7T and use it to find s m and cos ' 16 16 97. Show that

-

1 - cos(2x ) . for 0 s m2 x = . 2 ' USll1g transf ormatlOns.

93. Graph f (x)

=

94. R epeat Problem 93 for g(x)

=

7T 7T and cos . 24 24

96. Show that

Z2 --2' 1 + Z 1

� = �(V6 + Yz)

cos2 X.

�

x

�

sin3 {/ + sin3( {/ + 120°) + sin3( {/ + 240°)

2 7T by

98. I f tan {/

= a

= -

3 "4 sin(3{/)

e e. tan 3"' express tan 3" m terms 0f a.


99. Go to the library and research Chebyshev polynomials. Write a report on your findings.

OBJECTIVES

1

1

Express Products as Sums (p. 64 6)

2 Express Sums as Products (p. 64 7)

Express Products as Sums

Sum and d ifference formulas can be used to d erive formulas for writing the prod ucts of sines and /or cosines as sums or d ifferences.These id entities are usually called the Product-to-Sum Formulas.

THEOREM

Product-to-Sum Formulas

sina sin13= � [ cos(a -13) - cos(a + 13)] cosa cos13= 2"1 [ cos(a - 13)+ cos(a + 13)] sina cos13= � [sin(a + 13) + sin(a -13)]

(1)

(2) (3)

�------��

These formulas d o not have to be memorized . Instead , y ou should remember how they are d erived . Then, when y ou want to use them, either look them up or d erive them, as need ed . To d erive formulas( 1 ) and (2), write d own the sum and d ifference formulas for the cosine: cos(a -13) = cosa cos13+ sina sinf3 (4) cos(a + 13) cosa cos13- sina sin13 (5) Subtract equation (5 ) from equation (4) to get cos(a -13) - cos(a + 13) = 2 sina sin13 from which sina sin13= � [cos(a -13) - cos(a + 13)] =

SECTION 8.6

Product-to-S u m and Sum-to-Product Formulas

647

Now add equations (4) and (5) to get cos(o'. - 13) + cos(o'. + 13) = 2 cos a cos13

from which

1 cos a cos13 = 2" [cos(a - 13) + cos(a + 13)] To derive Product-to-Sum Formula (3), use the sum and difference formulas for sine in a similar way. (You are asked to do this in Problem 43 .) EXAM P L E 1

Express i n g P roducts as Sums

Express each of the following products as a sum containing only sines or only cosines. (a) sin(68) sin(48)

(b) cos(38) cos 8

(a) We use formula (1) to get

Solution

sine 68) sine 48) =

(c) sin(3 8) cos(58)

� [cos(68 - 48) - cos(68 + 48)] 1

= 2" [cos(2 8) - cos(108)] (b) We use formula (2 ) to get 1 cos(3 8) cos 8 = 2" [cos(3 8 - 8) + cos(3 8 + 8)] 1

= 2"[cos(2 8) + cos(48)] (c) We use formula (3 ) to get 1 sin(3 8) cos(58) = 2"[sin(3 8 + 58) + sin(3 8 - 58)]

= �I!=> -

2

THEOREM

� [sin(S8) + sine - 2 8)] = �[sin(S8) - sin(28)]


•

1

Express Sums as Products

The Sum-to-Product Formulas are given next. Sum-to-Product Formulas

0'.-13 0'. + 13 sin a + sin 13 = 2 sin - - cos -2 2 0'.-13 a + f3 sin a - sin13 = 2 sin - - cos - 2 2 a + f3 a-f3 cos a + cos13 = 2 cos - - cos - 2 2 a + f3 a -f3 cos a - cos13 = -2 sin -- sin 2 2

--

(6) (7) (8) (9)

�------�

�

We will derive formula (6) and leave the derivations of fonnulas (7) through (9) as exercises (see Problems 44 through 46).

648

CHAPTER 8


Proof

2 sin

:

a f3

COs

; � 2 . �[ sin (a:f3

a f3

+

; ) + sin (a:f3

; )]

a f3

a f3

_

Product-to-Su m Formula (3)

. 2a

. 213

.

.

= Sill 2 + Sill 2 = SI\1 a + Sll1 13 E XA M PLE 2

•

Expressi n g Sums (or Differe n ces) as a Product

Express each sum or difference as a product of sines and/or cosines. (b) cos(38)

(a) sin(58) - sin(38) Solution

+

cos(28)

(a) We use formula (7) to get .

58 + 38 . 58 - 38 cos 2 2 = 2 sin8 cos(48)

.

Sll1(58) - S1l1 (38) = 2 SI\1

(b) cos(38) + cos(28) = 2 cos

38 + 28 38 - 28 cos 2 2 58 2

Formula (8)

8 2

= 2 coscos-

�=� - Now Work P R O B L E M

•

11

8.6 Assess Your Understanding Skill Building

In Problems

1 - 1 0,

express each product as a sum containing only sines or only cosines.

1 . sin(48) sin(Z8)

2. cos(48) cos(Z8)

3. sine 48) cos(Z8)

4. sin(38) sin(58)

6. sin(48) cos(68)

7. sin 8 sin(Z8)

8. cos(38) cos(48)

9.

In Problems n.

11-18,

12.

15. sin 8 + sin(38)

19. 22.

19-36,

sin(48) + sin(Z8)

16. cos 8 + cos(38)

2. Sill (2) 8 8 - cos(38)

sin(38) - sin 8

=

cos 8

= tan(28)

20. 23.

COS

29

sine 48) + sin(S8) cos(48) + cos(S8)

= tan(68)

sin(48) + sin(S8)

tan(68) = --. sin(48) - sin(S8) tan(Z8)

sin 0' + sin f3 ex + f3 ex - f3 = tan -- cot -31. . . Sll1 0' - Sll1 f3 2 Z

8 + cos(38)

10.

.

S1l1

8

58

"2 cos 2

13. cos(28) + cos(48)

14. cos(58) - cos(38)

8 38 17. cos- - cos2 2

18.

cos 8

21.

8 - cos(38) . = tan 8 8 + SIl1 (38)

24.

2cos(2) 8 COS

.

S1l1

25. sin 8[sin 8 + sin(38)J = cos 8[cos 8 - cos(38)J 27.

8

.

8 2

. 38 2

S1l1 - - Sill -

establish each identity.

sin 8 + sin(38) COS

2 cos "2

express each sum or difference as a product of sines and/or cosines.

sin(48) - sin(28)

In Problems

. 38

S1l1

5. cos(38) cos(58)

=

sine 48) + sin(28) cos(48) + cos(Z8) cos 8 - cos(58) . . S1l1 8 + S1l1 (58 )

= tan(38)

= tan(Z8)

26. sin 8[sin(38) + sin(58)J = cos 8[cos(38) - cos(58)J 28.

30. 32.

sin(48) - sin(S8) cos(48) - cos(S8) cos(48) - cos(S8) cos(4) 8 + cos(S) 8

= - cot(68)

= tan(28) tan(68)

0' + f3 ex - f3 cos 0' + cos f3 = -cot -- cot -2 cos ex - cos f3 2

SECTION 8.7

33.

sin a + sin f3 cos a + cos f3

35.

1 +

=

a + f3 tan-2

cos(2e) + cost 4e)

+

34.

cos t 6(1)

4 cos e cos(2e) cos(3e)

=

Trigonometric Equations (I)

649

sin a - sin f3 a + f3 = - cot -2 cos a - cos f3 =

36. 1 - cos(2e) + cos t 4e) - cost 68)

4 sin 8 cos(28) sin(38)


37.

On a Touch-Ton e phone, each button p roduces a unique sound. Th e sound produced is th e sum of two ton es, given by

to a s et of rotated axes. Th es e moments are given by th e equations

Touch-Tone Phones

y =

sin(27Tlt) and

y =

wh ere I and h are the low and h igh frequencies (cycles p er second) shown on the illustration. For example, if you touch 7, the low frequency is I 852 cycles p er second and the high frequency is h = 1 209 cycl es p er s econd. The sound emitted by touching 7 is =

y

=

III = Ix

Iv = It

sin(27Tht)

sin[27T(852 ) tJ + sin[27T( 1209) tJ

Touch-Tone phone

cos2 8 +

38.

LTI 39.

2Ity 2Ity

sin 8 cos 8 sin 8 cos e

Use product-to-sum formulas to show that Ix + Iy It - I y . III = - - + -- - cos 28 - Ity S1l1 28 2 2 and It + I y It - Iy . Iv = - - - - - - cos 28 + Ity sll1 28 2 2 Source: Adapt ed from Hibbel er, Engineering Mechanics: Statics, 10th ed. , Prentice Hall © 2004. 40. Projectile Motion The range of a proj ectile propelled down ward from the top of an inclined plane at an angle 8 to the in clined plane is given by 2 V6 sin 8 cos t e - ¢) -::-'- -'g cos2 ¢

---'---

-

when the projecti l e hits the inclined plane. Here va is the ini tial velocity of the proj ecti l e, ¢ is the angle th e plane makes with respect to the horizontal, and g is a ccel eration due to gravity. (a) Show that for fix ed va and ¢ the maximum range down

1 4 77

cyc l es/sec

1 336 cycles/sec

•

sin2 e -

sin2 8 +Iy cos2 8 +

R (8) =

1 209 cycles/sec

Iy

th e II1clin e is given by Rmax

(a) Write this sound as a product of sines and/or cosines. (b) D etermine th e maximum valu e of y. (c) Graph the sound emitted by touching 7. Touch-Tone Phones

(a) Write the sound emitted by touching th e # k ey as a product of sines and/or cosin es. (b) D etermine the maximum value of y. (c) Graph the sound emitted by touching the # k ey.

The moment of inertia I of an obj ect is a m easure of how easy it is to rotate the obj ect about som e fix ed point. In engineering m echanics, it is sometimes n ecessary to compute moments of i n ertia with r espect Moment of Inertia

.

=

(

v

2

0

g 1 - SIl1 ¢ )

(b) D etermine the maximum range if the proj ectil e has an initial velocity of 50 meters/second, the angle of th e plane is ¢ = 35°, and g = 9.8 m eters/second2 . 41. If a + f3 + Y = 7T, show that sin(2a) + sin(2f3 ) + sin(2y) 42. If a + f3 +

Y =

=

4 sin a sin f3 sin y

7T, show that

tan a + tan f3 + tan y

=

tan a tan f3 tan y

43. D erive formula (3). 44. D erive formula (7). 45. D erive formula (8). 46. D erive formula (9).

.

8.7 Trigonometric Equations (I) PREPARING FOR THIS SECTION •

Solving Equations (Section

"NOW Work

1.1,


pp. 86-92)


OBJECTIVE

•

Values of the Trigonometric Functions (Section 7.3 , pp. 529-532 and S ection 7.4, pp. 540-548)

1 Solve Equations Involving a Single Trigonometric Function (p. 650)

650

CHAPTER 8


1

E XA M P L E 1

Solve Equations Involving a Single Trigonometric Function

The previous four sections of this chapter were devoted to trigonometric identities, that is, equations involving trigonometric functions that are satisfied by every value in the domain of the variable. In the remaining two sections, we discuss trigonometric equations, that is, equations involving trigonometric functions that are satisfied only by some values of the variable (or, possibly, are not satisfied by any values of the variable) . The values that satisfy the equation are called solutions of the equation. Checking Whether a Given Number Is a Solution of a Trigonometri c Equation

1 1T . . Determine whether e = "4 IS a solution of the equation sm e = 2"' Is e solution? Solution

Replace e by

1T

= "6

a

: in the given equation. The result is .

1T

sm- = 4

V2

--

2

1 -=1- 2

: is not a solution. Next replace by : in the equation. The result is

We conclude that

e

1 . 1T SIn 6 = 2 -

1T We conclude that "6 is a solution of the given equation.

:

•

The equation given in Example 1 has other solutions besides e = . For exam51T . 131T ' . p Ie, e = 6 1S a Iso a soIutlOn, as IS e = -6- ' (You should check this for yourself. ) In fact, the equation has an infinite number of solutions due to the periodicity of the sine function, as can be seen in Figure 2 5.

Figure 25

y

Unless the domain of the variable is restricted, we need to find all the solutions of a trigonometric equation. As the next example illustrates, finding all the solu tions can be accomplished by first finding solutions over an interval whose length equals the period of the function and then adding multiples of that period to the solutions found. Let's look at some examples.

E XA M P L E 2

F i n d i ng A l l the Solutions of a Trigo nometric Equati o n

Solve the equation:

1 cos e = 2"

Give a general formula for all the solutions. List eight of the solutions.

SECTION 8.7

Sol ution

Figure 26

y 2

-2

-2

-

'IF 3

The period of the cosine function is 2 1T. In the interval [0, 2 1T) , there are two angles 1 1T 51T . 8 for WhICh cos 8 = i 8 = 3 and 8 = 3' See Figure 2 6. Because the cosine

1T 8 = 3 + 2 k1T 2 x

51T 8 = 3 + 2 k1T

or

( 1 , - b)

1T 3 '

1T 3'

k = -1

51T 3 '

71T 3 '

ll1T 3 '

=

cos X

13 1T 3 '

171T 3

k = 2

=

•

cos x and Y2

1

= - to determine

2 where the g ra p h s i ntersect. (Be s u re to g r a p h i n rad i a n mode.) See Fig u re 27.

The g ra p h of Y1 i ntersects the g ra p h of Y2 at 7.33 mn ==--

-1

any integer

k = l

k = 0

Check: We can verify the sol utions by g ra p h i n g Y1

Figure 27

k

Eight of the solutions are 51T 3 '

Yl

65 1

�� has period 2 1T, all the solutions of cos 8 = � may be given by the general

(1 , b) e


( ) 7 1T

;:::; 3

, a n d 1 1 .52

(

l l 1T

)

x=

1 .05

( ;) (;:::; 5;), ;:::;

, 5 .24

;:::; -- , rou nded to two deci m a l places. 3


3 1

In most of our work, we shall be interested only in finding solutions of trigono metric equations for ° ::; 8 < 2 1T. E XA M P L E 3

Solving a Linear Trigonometri c Equati o n


2 sin 8 +

V3 = 0,

0 ::; 8

" -


}

•

7

When the argument of the trigonometric function in an equation is a multiple of 8, the general formula must be used to solve the equation. E XA M P L E 4

Solvi n g a Trigo nometri c Equation


1 sin ( 2 8) = "2 '

0 ::; 8

� on the interval

[-27T, 47T] .

56. [(x) = 2 cos x.

(a) Find the x-intercepts of the graph of [ on the interval [-27T, 47T].

(b) Graph [(x) = 2 cos x on the interval [- 27T, 47T] . (c) Solve [(x) = -\13 on the interval [-27T, 47T]. What points are on the graph of f? Label these points on the graph drawn in part (b). (d) Use the graph drawn in part (b) along with the results of part ( c) to determine the values of x such that [(x) < - \13 on the interval [-27T, 47T] . 57. [(x) = 4 tan x. (a) Solve [(x) = -4. (b) For what values of x is [(x) < -4 on the interval

58.

(-�,�> [(x) = cot x.

(a) Solve [(x) = - \13. (b) For what values of x is f(x) (0, 7T ) ?

=

>

\13 2

- -

f:J 2

40. tan -

=

-1

=


53. What are the x-intercepts of the graph of f ( x) on the interval [0, 27T] ?

=

e < 2 7T . Round answers to two decimal places. 44. cot e = 2 43. tan e 5 48. csc f:J = - 3 47. sec e = - 4 52. 4 cos e + 3 = ° 51. 3 sin e - 2 = °

In Problems 41-52, use a calculator to solve each equation on the interval

g(x) on the interval [0, 7T]. (d) Shade the region bounded by [(x) = 3 sin(2x) + 2 and 7 g(x) = 2' between the two points found in part (b) on =

the graph drawn i n part (a). x !fi 60. (a) Graph [(x) = 2 cos 2 + 3 and g(x) = 4 on the same Cartesian plane for the interval [0, 47T] .

=

(b) Solve [( x) g( x) on the interval [0, 47T] and label the points of intersection on the graph drawn in part (b). (c) Solve f(x) < g(x) on the interval [0, 47T] . x (d) Shade the region bounded by [(x) = 2 cos 2 + 3 and

=

g( x) 4 between the two points found in part (b) on the graph drawn in part (a).

!fi 61. (a) Graph f(x) = -4 cos x and g(x) = 2 cos x + 3 on the same Cartesian plane for the interval [0, 27T] . (b) Solve [(x) g(x) on the interval [0, 27T] and label the points of intersection on the graph drawn in part (b). (c) Solve [(x) > g(x) on the interval [0, 27T] . (d) Shade the region bounded by [(x) = -4 cos x and g( x) 2 cos x + 3 between the two points found in part (b) on the graph drawn i n part (a).

=

=

!fi 62. (a) Graph [(x) = 2 sin x and g(x) = -2 sin x + 2 on the same Cartesian plane for the interval [0, 27T] . (b) Solve [( x) = g( x) on the interval [0, 27T] and label the points of intersection on the graph drawn in part (b). (c) Solve [(x) > g(x) on the interval [0, 27T] . (d) Shade the region bounded by [(x) = 2 sin x and g(x) = -2 sin x + 2 between the two points found in part (b) on the graph drawn in part (a). 63. The Ferris Wheel In 1893, George Ferris engineered the Ferris Wheel. It was 250 feet in diameter. If the wheel makes 1 revolution every 40 seconds, then the function h (t )

=

125 sin

( 0.157t - �) + 125

represents the height h, in feet, of a seat on the wheel as a function of time t, where t is measured in seconds. The ride be gins when t = 0. (a) During the first 40 seconds of the ride, at what time t is an individual on the Ferris Wheel exactly 125 feet above the ground? (b) During the first 80 seconds of the ride, at what time t is an i ndividual on the Ferris Wheel exactly 250 feet above the ground? (c) During the first 40 seconds of the ride, over what in terval of time t is an individual on the Ferris Wheel more than 125 feet above the ground? 64. Tire Rotation The P215/65R15 Cobra Radial G/T tire has a diameter of exactly 26 i nches. Suppose that a car ' s wheel is making 2 revolutions per second (the car is travel ing a little less than 10 miles per hour). Then h ( t )

(

13 sin 47Tt -

� ) + 13 represents the height

=

h (in inches)

of a point on the tire as a function of time t (in seconds). The car starts to move when t = 0.

SECTION 8.7

(a) During the first second that the car is moving, at what time t is the point on the tire exactly 13 inches above the ground? (b) During the first second that the car is moving, at what time t is the point on the tire exactly 6.5 inches above the ground? (c) During the first second that the car is moving, at what time t is the point on the tire more than 13 inches above the ground?

655

(c) During the first 20 minutes after the plane enters the holding pattern, at what time x is the plane more than 100 miles from the airport? (d) While the plane is in the holding pattern, will it ever be within 70 miles of the airport? Why? 66. Projectile Motion A golfer hits a golf ball with an initial velocity of 100 miles per hour. The range R of the ball as a function of the angle () to the horizontal is given by R( () ) = 672 sin(2()), where R is measured in feet. (a) At what angle () should the ball be hit if the golfer wants the ball to travel 450 feet (150 yards)? (b) At what angle () should the ball be hit if the golfer wants the ball to travel 540 feet (180 yards)? (c) At what angle () should the ball be hit if the golfer wants the ball to travel at least 480 feet (160 yards)? (d) Can the golfer hit the ball 720 feet (240 yards)?

Source: Cobra Tire 65.


An airplane is asked to stay within a hold ing pattern near Chicago 's O'Hare International Airport. The function d ( x) = 70 sin( 0.65x) + 150 represents the distance d, in miles, that the airplane is from the airport at time x, in minutes. (a) When the plane enters the holding pattern, x = 0, how far is it from O ' Hare? (b) During the first 20 minutes after the plane enters the holding pattern, at what time x is the plane exactly 100 miles from the airport? Holding Pattern

IX The following discussion of Snell's Law of Refraction * (named after Willebrord Snell, 1580-1626) is needed for Problems 67-73. Light, T sound, and other waves travel at different speeds, depending on the media (ail; wate}; wood, and so on) through which they pass. Suppose

that light travels from a point A in one medium, where its speed is V I , to a point B in another medium, where its speed is V2 ' Refer to the figure, where the angle ii i is called the angle of incidence and the angle 1i2 is the angle of refraction. Snell's Law, which can be proved using calculus, states that

sin ()l sin ()2

VI V2

The ratio � is called the index of refraction. Some values are given in the following table. V2

Some Indexes of Refraction

Angle of incidence

A

Index of Refraction t

Medium I ncident ray, speed

v1

water is 1 .33. If the angle of incidence is 400, determine the angle of refraction. 68. The index of refraction of light in passing from a vacuum into dense flint glass is 1 .66. If the angle of incidence is 500, de termine the angle of refraction. 69. Ptolemy, who lived in the city of Alexandria in Egypt during the second century A D, gave the measured values in the table following for the angle of incidence () j and the angle of re fraction ()2 for a light beam passing from air into water. Do *

Because this law was also deduced by Rene Descartes in France, it is

1 .33

Ethyl alcohol (20°C)

1 .36

Carbon d i s u lfide

1 .63

Ai r ( 1 atm an d O°C)

1 .00029

Dia mond

2.42

Fused q u artz

1 .46

Glass, crown

1 .5 2

G l as s , d e n s e flint

1 .66

Sodium chloride

1 .54

these values agree with Snell 's Law? If so, what index of re fraction results? (These data are interesting as the oldest recorded physical measurements.) t

67. The index of refraction of light in passing from a vacuum into

also known as Descartes's Law.

Water

8,

t

82

8,

82

1 00

80

500

35°0'

200

1 5° 3 0 '

60°

40°30'

30°

22°30'

70°

45°30'

400

2900'

80°

5000'

For light of wavelength 589 n an ometers, measured with respect to a

vacuum. The index with respect to air is negligibly d ifferent in most cases.

656

CHAPTER 8


The speed of yellow sodium light (wave length of 589 nanometers) in a certain liquid is measured to be 1.92 x 1 08 meters per second. What is the index of re fraction of this liquid, with respect to air, for sodium light?'" [Hint: The speed of light in air is approximately 2.998 x 1 08 meters per second.] 71. Bending Light A beam of light with a wavelength of 589 nanometers traveling in air makes an angle of incidence of 40° on a slab of transparent material, and the refracted beam makes an angle of refraction of 26°. Find the index of refrac tion of the materia!.'" 72. Bending Light A light ray with a wavelength of 589 nanometers (produced by a sodium lamp) traveling through air makes an angle of incidence of 30° on a smooth, flat slab of crown glass. Find the angle of refraction.'"

70.

Bending Light

'" Adapted from H a l liday and Resnick,

73. A light beam passes through a thick slab of material whose

index of refraction is n2 ' Show that the emerging beam is par allel to the incident beam.* 74. Brewster's Law If the angle of incidence and the angle of refraction are complementary angles, the angle of incidence is referred to as the Brewster angle () B' The Brewster angle is related to the index of refractions of the two media, n l and /12, by the equation /11 sin ()B = /12 cos () B, where 111 is the index of refraction of the incident medium and n2 is the index of refraction of the refractive medium. Determine the Brewster angle for a light beam traveling through water (at 20°e) that makes an angle of incidence with a smooth, flat slab of crown glass.

Fundamentals of Physics, 7th ed . , 2005, John Wiley

& Sons.


75. Explain in your own words how you would use your calculator to solve the equation cos x = -0.6, 0 modify your approach to solve the equation cot x = 5, 0 < x < 27T?

::;

x < 27T. How would you


{ ':?'2 }

2 . v2 . - �

2 '

2

PREPARING FOR THIS SECTION • •


Solving Quadratic Equations by Factoring (Section 1.2, pp. 98-99) The Quadratic Formula (Section 1 .2, pp. 102-104) .NOW Work the 'Are You

•

Solving Equations Quadratic in Form (Section 1.4, pp. 1 19-121) Using a Graphing Utility to Solve Equations (Appendix, Section 4, pp. A8-A10)

Prepared?' problems on page 66 1 .

OBJECTIVES 1 Solve Trigonometric Equations Quadratic i n Form (p. 656) 2 Solve Trigonometric Equations Using Identities (p. 657)

3 Solve Trigonometric Equations Linear i n Sine and Cosine (p. 659)

�4 1

Solve Trigonometric Equations Using a G ra p h i n g Util ity (p. 66 1 )

Solve Trigonometric Equations Quadratic in Form

In this section we continue our study of trigonometric equations. Many trigonomet ric equations can be solved by applying techniques that we already know, such as applying the quadratic formula (if the equation is a second-degree polynomial) or factoring. E XA M P L E 1

Solving a Trigonometric Equation Quadrati c i n Form

Solve the equation: S olution

2 sin2 e - 3 sin e + 1 = 0,

0 :::; e

0 is measured in radians.

45. Geometry

The distance from home plate to the fence in dead center at the Oak Lawn Little League field is 280 feet. How far is it from the fence in dead center to third base?

42. Little League Baseball

[Hint: The distance between the bases in Little League is 60 feet.]

Clint is building a wooden swing set for his children. Each supporting end of the swing set is to be an A-frame constructed with two 10-foot-long 4 by 4s joined at a 45° angle. To prevent the swing set from tipping over, Clint wants to secure the base of each A-frame to con crete footings. How far apart should the footings for each A-frame be?

46. For any triangle, show that

cos f = 2

43. Building a Swing Set

Rod OA rotates about the fixed point 0 so that point A travels on a circle of radius r. Connected to point A is another rod AB of length L > 2,. , and point B is connected to a piston. See the figure. Show that the dis tance x between point 0 and point B is given by

where

x = ,. cos (j +

L2

+

-

1

s = 2 (a + b + c).

[Hint: Use a Half-angle Formula and the Law of Cosines.] 47. For any triangle show that

44. Rods and Pistons

yr2 cos2 (j

� "V�

wheres

=

. C Sln - = 2

)r:"(s - -a-:--.)(:-s---b-:-:-) ab

1

2(a + b + c).

48. Use the Law of Cosines to prove the identity

cos A

a

r2

--

+

cos B

b

--

+

cos C

c

--

a2 + b2 + c2

= ----��

Discussion and Writing 49. What do you do first if you are asked to solve a triangle and

51. Make up an applied problem that requires using the Law of

are given two sides and the included angle?

Cosines.

50. What do you do first if you are asked to solve a triangle and

52. Write down your strategy for solving an oblique triangle.

are given three sides? 'Are You Prepared?' Answers

2. (j = 45°

9.4 Area of a Triangle PREPARING FOR THIS SECTION •


Geometry Essentials (Section R.3, pp. 30-35) Now Work

the 'Are You Prepared?' problem on page

OBJECTIVES

1

2

694.

Find the Area of SAS Triangles (p. 692) Find the Area of SSS Tria n g les (p. 693)

692

CHAPTER 9

Applications ofTrigonometric Functions

In this section, we will derive several formulas for calculating the area of a triangle. The most familiar of these is the following: The area K of a triangle is

THEOREM

�b

K

h

= �_________________________________________

where b is the base and h is an altitude drawn to that base.

�

-.J

Proof The derivation of this formula is rather easy once a rectangle of base band height h is constructed around the triangle. See Figures 29 and 30. Triangles 1 and 2 in Figure 30 are equal in area, as are triangles 3 and 4. Conse quently, the area of the triangle with base band altitude h is exactly half the area of the rectangle, which is bh. Figure 29

Figure 30

r------------

I I I I I I

b

1 Figure 31

I I I I I I I

b

•

Find the Area of SAS Triangles

If the base band altitude h to that base are known, then we can find the area of such a triangle using formula ( 1). Usually, though, the information required to use for mula ( 1) is not given. Suppose, for example, that we know two sides aand band the included angle C. See Figure 3 1. Then the altitude h can be found by noting that - =

sin C

=

asin C

h

b

----------1

a

so that

h

Using this fact in formula ( 1) produces K

=

� bh � b(asin C) � absin C =

=

We now have the formula K

=

� absin C

(2)

By dropping altitudes from the other two vertices of the triangle, we obtain the following corresponding formulas: K K

=

=

2bc S111 A

(3)

2ac S1l1 B

(4)

1

.

1

.

It is easiest to remember these formulas using the following wording: THEOREM

The area K of a triangle equals one-half the product of two of its sides times the sine of their included angle.

-.J

Area of a Triangle

SECTION 9.4

EXAMPLE 1

Finding the Area of a SAS Triangle

Find the area K of the triangle for which See Figure

Solution

Figure 32

8

THEOREM

=

1i.EII-=r:_ - Now Work 2

=

6, and C

=

=

30°.

32. We use formula (2) to get �ab sin C �. 8 . 6 . sin 30° 12 square units 2 2 K

� �

a 8, b

693

=

=

•

PROB l E M 5

Find the Area of SSS Triangles

If the three sides of a triangle are known, another formula, called Heron's Formula (named after Heron of Alexandria), can be used to find the area of a triangle. Heron's Formula

The area K of a triangle with sides

a, b, and c is ys(s - o)(s - b)(s - c) 1 b c). "2(a K

where

s =

(5)

=

+

+

A proof of Heron's Formula is given at the end of this section. EXAMPLE 2

Finding the Area of a SSS Triangle

4,5, and 1 5 -(4

7.

Find the area of a triangle whose sides are Solution

a 4, b

7.

c Then s -(a 2 b ) 2 Heron's Formula then gives the area as ys(s - a)(s - b)(s - c) V8·4·3·1

We let

=

=

5, and =

K

1

=

+

+

c

K

=

+

=

=

1l!I!l:lI=� - Now Work

+

=

7)

=

v96

8 =

4V6 square units

PROB l E M 1 1

•

Proof of Heron's Formula The proof that we shall give uses the Law of Cosines and is quite different from the proof given by Heron. From the Law of Cosines, = + cos C and the Half-angle Formula C 1 + cos C cosz- =

c2 a2 bZ - 2ab 2

2 aZ b2 - c21 ---2ab C 1 cos C cos2 2 2 2 a2 2ab b2 - cZ (a b)2 - cZ 4ab 4ab (a b c)(a b c) 2(s - c)·2s s(s 4ab 4ab ab i

we find that

=

i

Factor

+

+

+

_

+

+

= -------

+

+

+

a

+

+ b

- c = a + b + c - 2c = 2s - 2c = 2(s - c)

c

)

(6)

694

CHAPTER 9

Applications ofTrigonometric Functions . ? S'imi'1arIy, usmg sm"2

C

.

=

1 - cos C

' we find that 2 C (s - a)(s - b) . 2 Sin -= 2 ab Now we use formula (2) for the area. K

=

1

.

2absm C

=

� ab'2sin % cos %

=

ab

=

[ (%)] = 2sin % cos %

sin e sin 2 =

(s - ) (s - b) 'V� Use equations (6) and (7). I ---;;;;-ab 'V a

Vs(s -

a)(s - b)(s - c )

I-li�torica! Feature

1-1

(7)

•

book Metrica, on making such devices, has survived and was discovered

eron's Formula (also known as Hero's Formula) is due to Heron

of Alexandria (first century AD), who had, besides his mathe

in

1896 in the city of Constantinople.

matical talents, a good deal of engineering skills. In various tem

Heron's Formulas for the area of a triangle caused some mild dis

ples his mechanical devices produced effects that seemed supernatural,

comfort in Greek mathematics, because a product with two factors was

and visitors presumably were thus influenced to generosity. Heron's

an area, while one with three factors was a volume, but four factors seemed contradictory in Heron's time.

9.4 Assess Your Understanding 'Are You Prepared?' Answer as given at the end of these exercises. If you get a wrong answer, read the page listed in red. 1. The area of a triangle whose base is b and whose height is h is

__

(p. 31)

Concepts and Vocabulary 2. If three sides of a triangle are given,

___

Formula is used

Given two sides and the included angle, there is a formula that can be used to find the area of the triangle.

4. True or False

to find the area of the triangle.

3. True or False

No formula exists for finding the area of a triangle when only three sides are given.

Skill Building

6' � 3

In Problems 5-12, find the area of each triangle. Round answers to two decimal places. .5.

2�b / WA � 4

8.

2�

---2QOA -------

a

BW 4

9.

5

12.

11.

4

7.

? �3

� � c

10.

SECTION 9.4

Area of a Triangle

695

In Problems 13-24, find the area of each triangle. Round answers to two decimal places. a

= 3,

b = 4,

C = 40°

14. a = 2,

c = 1,

B

=

100

15. b = 1 ,

c = 3,

A = 8 00

16. a = 6,

b = 4,

C = 600

17. a = 3,

c = 2,

B

=

1 10°

18. b = 4,

c = 1,

A = 120°

c = 5

20. a = 4,

b = 5,

c = 3

21.

= 2,

b = 2,

c

23. a = 5,

b = 8,

c

24. a = 4,

b = 3,

c = 6

13.

19. a

=

b = 1 3,

12,

22. a = 3,

b = 3,

c = 2

= 9

a

= 2


If two angles and the included side are given, the third angle is easy to find. Use the Law of Sines to show that the area K of a triangle with side a and angles A, B, and C is a 2 sin B sin C K = ----2 sin A

25. Area of an ASA Triangle

Prove the two other forms of the for mula given in Problem 25. b 2 sin A sin C c 2 sin A sin B K= and K = �:...--..2 sin B 2 sin C

26. Area of a Triangle

In Problems 27-32, use the results of Problem 25 or 26 to find the area of each triangle. Round answers to two decimal places. 27. A = 400,

B = 200,

a = 2

28. A = 5 00,

C = 200,

a = 3

29. B = 700,

C = 10°,

b= 5

Find the area of the segment (shaded in blue in the figure) of a circle whose radius is 8 feet, formed by a central angle of 7 0°.

33. Area of a Segment

P �--��--� R 10

[Hint: Subtract the area of the triangle from the area of the sector to obtain the area of the segment.]

Consult the figure, which shows a circle of radius r with center at O. Find the area Kof the shaded region as a function of the central angle e.

39. Geometry

Find the area of the segment of a circle whose radius is 5 inches, formed by a central angle of 400.


The dimensions of a triangular lot are 100 feet by 50 feet by 75 feet. If the price of such land is $3 per square foot, how much does the lot cost?

35. Cost of a Triangular Lot

A cone-shaped tent is made from a circular piece of canvas 24 feet in diameter by removing a sector with central angle 1 00° and connecting the ends. What is the surface area of the tent?

36. Amount of Material to Make a Tent

40. Approximating the Area of a Lake To approximate the area

of a lake, a surveyor walks around the perimeter of the lake, taking the measurements shown in the illustration. Using this technique, what is the approximate area of the lake? [Hint: Use the Law of Cosines on the three triangles shown and then find the sum of their areas.]

The dimensions of home plate at any major league baseball stadium are shown. Find the area of home plate.

37. Dimensions of Home Plate

• .-

�.

8.5 in.

8.5 in. 17 in.

Find the area of the shaded region en closed in a semicircle of diameter 10 inches. The length of the chord PQ is 8 inches.

38. Computing Areas

[Hint: Triangle PQR is a right triangle.]

Completed in 1902 in New York City, the Flatiron Building is triangular shaped and bounded by 22nd Street, B roadway, and 5 th Avenue. The building measures approximately 87 feet on the 22nd Street side, 1 90 feet on the Broadway side, and 173 feet on the 5th Avenue side. Approximate the ground area covered by the building.

41. The Flatiron Building

696

Applications ofTrigonometric Functions

CHAPTER 9

to show that 1
0 are constants, moves with simple harmonic motion. The motion has amplitude lal and period

271". w

The frequency f of an object in simple harmonic motion is the number of oscil lations per unit time. Since the period is the time required for one oscillation, it fol lows that the frequency is the reciprocal of the period; that is,

f=�

271"

Figure 36

EXAMPLE 1

Finding an E q u ation for an O bject in H armonic Motion

Suppose that an object attached to a coiled spring is pulled down a distance of 5 inches from its rest position and then released. If the time for one oscillation is 3 seconds, write an equation that relates the displacement d of the object from its rest position after time t (in seconds). Assume no friction.

d Solution

5

o

The motion of the object is simple harmonic. See Figure 36. When the object is released (t = 0 ) , the displacement of the object from the rest position is -5 units (since the object was pulled down). Because d = -5 when t = 0, it is easier to use the cosine function*

d Rest position

1= 0

=

a cos(wt)

to describe the motion. Now the amplitude is I - 5 1

a -5

w > O

=

-5

and

271"

- = period w

" No phase shift is required if a cosine function is used.

=

=

3,

5 and the period is 3, so

271"

w = 3

700

CHAPTER 9

Applications of Trigonometric Functions

NOfE In the solution to Exa m ple 1, we let a

=

An equation of the motion of the object is

-5, since the i n itial motion is

II

down. If t he i n itial direction were u p, we wou ld let a

=

5.

=

d W �.,....

2 E XA M P L E 2

Now Work

-5 cos

[ 2; t ]

P RO B L E M 5

•

Analyze Simple Harmonic Motion Analyzing the M otion of an Object

Suppose that the displacement d (in meters) of an object at time t (in seconds) sat isfies the equation

d (a) (b) (c) (d) Solution

=

10 sin(5t)

Describe the motion of the object. What is the maximum displacement from its resting position? What is the time required for one oscillation? What is the frequency?

We observe that the given equation is of the form where a

=

d 10 and w = 5 .

=

d

a sine wt)

=

1 0 s i n (5t)

(a) The motion i s simple harmonic. (b) The maximum displacement of the object from its resting position is the am plitude: lal = 10 meters. (c) The time required for one oscillation is the period: Period

=

27T w

-

=

27T seconds 5

-

(d) The frequency is the reciprocal of the period. Thus, Frequency i:l'I ' \

3

�-

Now Work

=

f

=

5 oscillations per second 27T

-

P RO B L E M 1 3

•

Analyze an Object in Damped Motion

Most physical phenomena are affected by friction or other resistive forces. These forces remove energy from a moving system and thereby damp its motion. For example, when a mass hanging from a spring is pulled down a distance a and released, the friction in the spring causes the distance that the mass moves from its at-rest position to decrease over time. As a result, the amplitude of any real oscil lating spring or swinging pendulum decreases with time due to air resistance, fric tion, and so forth. See Figure 37. Figure 37

a

-a

SECTION 9.5

701

Simple Harmonic Motion; Damped Motion; Combining Waves

A function that describes this phenomenon maintains a sinusoidal component, but the amplitude of this component will decrease with time to account for the damping effect. In addition, the period of the oscillating component will be affected by the damping. The next result, from physics, describes damped motion. THEOREM

Damped Motion

The displacement d of an oscillating object from its at-rest position at time t is given by

where b is the damping factor or damping coefficient and In is the mass of the

oscillating object. Here l al is the displacement at t = ° and - is the period W under simple harmonic motion (no damping) . 27T

..J

Notice for b = ° (zero damping) that we have the formula for simple harmonic 27T

motion with amplitude l al and period - . W

EXAMPLE 3

Analyzin g a Damped Vibration C u rve

Analyze the damped vibration curve

2:

d(t) = e-I/7T cos t, t Solution

°

The displacement d is the product of y = e-I/r. and y = cos t. Using properties of absolute value and the fact that Icos t l :S 1 , we find that

I d ( t ) 1 = le-I/". cos t l = le-I/7Tl lcos t l

:S

1 e -1/7T1 = e-I/7T i e- r/ r.

> 0

As a result,

This means that the graph of d will lie between the graphs of y = e-I/7T and bounding curves of d. Also, the graph of d will touch these graphs when Icos t l = 1 , that is, when t = 0, 7T, 27T, and so on. The x-intercepts of the graph of d occur when cos t = 0,

y = - e-I/r., the

that is, at

Ta b l e 1

7T 37T 57T

2' 2' 2'

t cos t

d(t) =

and so on. See Table 1 .

o

e-t/r. cos t

Point on graph of d

7T 2

3 7T 2

e- 1/2

e- 1

e-3/2

0

1 - e-

0

0

(0, 1)

17

( %, 0)

-1

(7T , - e- 1 )

0

e; , o )

2 17

e-2 e-2

(27T,

e-2)

702

CHAPTER 9


We graph y

=

cos t,

Y = e-t/'",

y

= -e-t/",

= e-t/7T

and d ( t )

cos t in Figure 38.

Figure 38

d 1

-1

•

Exploration Graph Y,

=

e-x/7f cos x, along with Y2

=

e-x/7f, and Y3

=

_ e-x/7f, for 0

,0; x ,0;

27T.

Determine where Y,

has its first turning point (local mini mum). Compare this to where Y, intersects Y3.

Result Figure

39 shows the graphs of Y, e-x/7f cos x, Y2 = e-x/7f, and Y3 = _ e-x/7f• X "" 2.83; Y, I NT ERSECT S Y3 at x = 7T "" 3.14. =

Using MINIMUM,

the first turning point occurs at

Figure 39

�"l!l"".> """' - Now Work 4

P ROB L EM 2 1

Graph the Sum of Two Functions

Many physical and biological applications require the graph of the sum of two func tions, such as f(x)

=

x

+

sin x

or

g(x)

=

sin x

+

cos( 2x )

For example, i f two tones are emitted, the sound produced i s the sum of the waves produced by the two tones. See Problem 51 for an explanation of Touch-Tone phones. To graph the sum of two (or more) functions, we can use the method of adding y-coordinates described next. E XA M P L E 4

G raph ing the Sum of Two F un ctions

Use the method of adding y-coordinates to graph f(x) Solution

First, we graph the component functions, y

=

h(x)

=

=

x

+

sin x.

sin x

in the same coordinate system. See Figure 40(a). Now, select several values of x, 3 71' 71' say, x = 0, x = 2 ' x = 71', X = 2 ' and x = 2 71' , at w h'lCh we compute f(x)

=

h (x)

+

h ( x ) . Table 2 shows the computation. We plot these points and

connect them to get the graph, as shown in Figure 40(b).

SECTION 9.5

Table 2

Y = f1 (x) = x

°

f(x) = x + si n x

°

Point on graph of f

Figure 40

7T 2

7T 2

(0, 0)

°

'"

7T 3 2

7T

-

- + 1

2 7T

2

2

°

Y = f2(x) = si n x

37T

7T

0

X

703

Simple Harmonic Motion; Damped Motion; Combi ning Waves

2.57

7T

(�, ) 2.57

27T

°

-1

-- 1 3 7T

'"

2

3

.71

e ;, )

(7T, 7T )

3

.71

27T

(27T, 27T)

Y

(b)

(a)

y

[�I � . .

=

In Figure 40(b), notice that the graph of f(x) = x + sin x intersects the line x whenever sin x = O. Also, notice that the graph of f is not periodic.

•

= X, Y2 = s i n x, a n d Y3 = X + s i n x and compare the result with Fig u re 40(b). Use I NTERSECT to verify that the gra phs of Y1 and Y3 intersect when sin x = O.

Check: Gra ph Yl

The next example shows a periodic graph. G raph i ng the S u m of Two S i nusoidal F un ctions

EXAMP L E 5

Use the method of adding y-coordinates to graph f(x) Solution

Ta ble 3 x

7T 2

-1

f(x) = sin x + cos(2x)

-2

Point on graph of f

sin x

+

cos(2x)

Table 3 shows the steps for computing several points on the graph of .f. Figure 41 on page 704 illustrates the graphs of the component functions, y = fl (X) = sin x and y = fz(x) = cos(2x), and the graph of f(x) = sin x + cos(2x ) , which is shown in red.

y = f1 (x) = sin x

y = f2(x) = cos(2x)

=

-1

o °

(0, 1 )

7T 2

-1

°

37T

7T

27T

2

°

°

-1 -1

(

-2 3 7T - -2 2 '

)

(27T, 1)

704

CHAPTER 9


Figure 4 1 Y 2

Notice that f is periodic, with period 27T.

•

9.5 Assess You r Understa n d i n g 'Are You Prepared?' Answer given a t the end of these exercises. If you get a wrong answel; read the pages listed i n red. 1. The amplitude A and period T of f(x)

= 5 sin (4x) are

__

and

__

. (pp. 563-569)

Concepts a nd Voca bulary

2. The motion of an object obeys the equation d = 4 cos(6t ) . Such

motion i s described a s cal led the

__ __

. The n umber 4 is

If the distance d of an object from its rest po sition at time t is given by a sinusoidal graph, the motion of the object is simple harmonic motion.

4. True or False

3. When a mass hanging from a spring is pulled down and then

released, the motion is called

tional force to retard the motion, and the motion is called if there is friction.

__ __

if there is no fric-

Skill Building

In Problems 5-8, an object attached to a coiled spring is pulled down a distance a fi'om its rest position and then released. Assuming that the motion is simple harmonic with period T, write an equation that relates the displacement d of the object from its rest position after t seconds. Also assume that the positive direction of the motion is up. 5. a

= 5; T = 2 seconds

6. a

= 10; T = 3 seconds

7. a

= 6; T = 7T' seconds

8. a

= 4; T = 2 seconds

9. Rework Problem 5 under the same conditions except that,

at time t down. 11.

=

0, the object is at its resting position and moving

Rework Problem 7 under the same conditions except that, at time t 0, the object is at its resting position and moving down. =

7T'

10. Rework Problem 6 under the same conditions except that, at

time t down.

= 0, the object is at its resting position and moving =

12. Rework Problem 8 under the same conditions except that, at

time t down.

0, the object is at its resting position and moving

In Problems 13-20, the displacement d (in meters) of an object at time t (in seconds) is given. (a) Describe the motion of the object. (b) What is the maximum displacement from its resting position? (c) What is the time required for one oscillation? (d) What is the frequency? 13. d

=

17. d

= -3 sin

5 sin(3t)

G) t

(� )

14. d

= 4 sin(2t)

15. d

= 6 cos(7T't)

16. d

= 5 cos

18. d

= - 2 cos(2t)

19. d

= 6 + 2 cos (27T't)

20. d

= 4 + 3 sin (7T't)

t

SECTION 9.5

In Problems 21-24, graph each damped vibration curve for 0 21. d(t) = e-I/-rr cos(2t )

:S

t

22. d ( t ) = e-I/2r. cos (2t)

:S

Simple Harmonic Motion; Damped Motion; Combi ning Waves

70S

2 71.

23. del) = e-I/2r. cos t

In Problems 25-32, use the method of adding y-coordinates to graph each function.

24. d(t) = e-I/4-rr cos

25. f (x) = x + cos x

26. f( x) = x + cos(2x)

27. f (x) = x - sin x

28. f ( x) = x - cos x

29. f (x) = sin x + cos x

30. f (x) = sin (2x) + cos x

[

32. g ( x ) = cos (2x) + cos x

31. g( x) = sin x + sin(2x)


In Problems 33-38, an object of mass m (in grams) altached to a coiled spring with damping factor b (in grams per second) is pulled down a distance a (in centimeters) from its rest position and then released. Assume that the positive direction of the motion is up and the period is T (in seconds) under simple harmonic motion. (a) Write an equation that relates the distance d of the object from its rest position after t seconds. ",.� (b) Graph the equation found in part (a) for 5 oscillations using a graphing utility. 33. m

=

25,

35. m = 30, 37. m = 10,

a = 10, b = 0.7, T = 5

34. m = 20,

a

a

36. m = 15,

a

38. m = 10,

a

=

18,

b = 0.6,

T = 4

a = 5, b = 0.8, T

=

3

=

=

=

15,

b = 0.75, T = 6

16,

b = 0.65,

5, b = 0.7,

T

T

=

5

=3

In Problems 39-44, the distance d (in meters) of the bob of a pendulum of mass m (in kilograms) from its rest position at time t (in seconds) is given. The bob is released from the left of its rest position and represents a negative direction. (a) Describe the motion of the object. Be sure to give the mass and damping factor. (b) What is the initial displacement of the bob? That is, what is the displacement at t = O? l] (c) Graph the motion using a graphing utility. (d) What is the displacement of the bob at the start of the second oscillation? (e) What happens to the displacement of the bob as time increases without bound? 39. d = _ 20e- 0.71/40 cos 41 . d =

-

30e-0.61/80 cos

( )( ) (( )

271 2

5

-

0.49 t 1 600

0 ·3 6 2 71 2 t 7 640 0

) )

A loudspeaker diaphragm is oscil l ating in simple harmonic motion described by the equation d = a cos(w t) with a frequency of 520 hertz (cycles per secon d) and a maximum displacement of 0.80 millimeter. Find w and then determine the equation that describes the movement of the diaphragm.

45. Loudspeaker

40. d = _ 20e-0.81/40 cos

42. d =

-

30e-o ')/- /70 cos

44. d =

-

1 0e

The end of a tuning fork moves in simple barmonic motion described by tbe equation d = a sin(w t). If a tuning fork for tbe note A above middle C on an even tempered scale (A4 ' the tone by which an orcbestra tunes itself) b as a frequency of 440 b ertz (cycles per second) , find w. If the maximum displacement of the end of the tun ing fork is 0.01 millimeter, determine the equation that describes the movement of the tuning fork. David Lapp. Physics of Music and Musical Instru ments. Medford, MA: Tufts U niversity, 2003

Source:

0 .sl so

/ cos

271 2

5

_ 71, 2

2

-

) ) )

0.64 t 1 600

_ 0 .25 t 49 0 0

271 2 _ 0.64 t 2500 3

Added to Six Flags St. Louis in 1986, the Colossus is a giant Ferris wheel. Its diameter is 1 65 feet, it rotates at a rate of about 1 .6 revolutions per minute, and the bottom of the wheel is 1 5 feet above the ground. Determine an equation that relates a rider's height above the ground at time t. As sume the passenger begins the ride at the bottom of the wheel.

46. Colossus

SOl/rce:

47. Tuning Fork

-

( )( ) (() (( )

Six Flags Theme Parks, Inc.

The end of a tuning fork moves in simple harmonic motion described by the equation d a sin (wt). I f a tuning fork for the note E above middle C on an even tempered scale (E4) has a frequency of approximately 329.63 hertz (cycles per second), find w. If the maximum dis placement of the end of the tuning fork is 0.025 millimeter, determine the equation that describes the movement of the tuning fork.

48. Tuning Fork

=

David Lapp. Physics of Music and Mu.sical Instru ments. Medford, MA: Tufts University, 2003 Source:

706

49.


CHAPTER 9

Charging a Capacitor See the illustration. If a charged ca pacitor is connected to a coil by closing a switch, energy is transferred to the coil and then back to the capacitor in an oscillatory motion. The voltage V (in volts) across the capac itor will gradually diminish to 0 with time t (in seconds). (a) Graph the function relating V and t: Vet)

=

e-I/3 COS(7Tt),

O :s:: t :s:: 3

o

� 51.

of y = - e I/3 ? (c) When will the voltage V be between -0.4 and 0.4 volt?

Switch

i,

50.

,.

sin(27Tlt)

y =

and

sin(27Tht)

where I and h are the low and high frequencies (cycles per second) shown in the illustration. For example, if you touch 7, the low frequency is I = 852 cycles per second and the high frequency is h = 1209 cycles per second. The sound emitted by touching 7 is

-

Capacitor

Touch-Tone Phones On a Touch-Tone phone, each button produces a unique sound. The sound produced is the sum of two tones, given by y =

(b) At what times t will the graph of V touch the graph of 3 y = e-I/ ? When does the graph of V touch the graph

+

Use a graphing utility to graph this function for :s:: x :s:: 4 and compare the result to the graphs ob tained in parts (a) and (b). (d) What do you think the next approximation to the saw tooth curve is?

y =

sin[27T(852) t ) + sin[27T( 1209)t)

Use a graphing utility to graph the sound emitted by touching 7.

Coil

Touch-Tone phone

The Sawtooth Curve An oscilloscope often displays a sawtooth curve. This curve can be approximated by sinusoidal curves of varying periods and amplitudes. (a) Use a graphing utility to graph the following function, which can be used to approximate the sawtooth curve. f(x)

=

�

sin( 27Tx) +

�

sin(47Tx ),

697 cycles/sec

O :s:: x :s:: 4

(b) A better approximation to the sawtooth curve is given by f(x)

=

�

sin( 27Tx) +

�

sin(47Tx) +

�

sin(87Tx)

1 209 cycles/sec

o

Use a graphing utility to graph this function for :s:: x :s:: 4 and compare the result to the graph ob tained in part (a). (c) A third and even better approximation to the sawtooth curve is given by f(x)

=

�

sin (27Tx)

+

�

sin(47Tx) +

�

sin(87Tx) +

�

6

sin(167Tx)

1336 cycles/sec

lliTI 52. �.,. 53.

I 54.


fir 55. \hll 56. r.

graphing utility to graph the function . sm x f(x) = -- , x > O. Based on the graph, what do you conx sin x jecture about the value of -- for x close to O? x Use a graphing utility to graph y = x sin x, y = x2 sin x, and 3 y = x sin x for x > O. What patterns do you observe? Use

a

'Are You Prepared?' Answer 1. A =

5', T

= -

7T 2

1477 cycles/sec

Use a graphing utility to graph the sound emitted by the ,;, key on a Touch-Tone phone. See Problem 5 l . CBL Experiment Pendulum motion is analyzed to esti mate simple harmonic motion. A plot is generated with the position of the pendulum over time. The graph is used to find a sinusoidal curve of the form y = A cos [ B ( x - C) ) + D. Determine the amplitude, period, and frequency. (Activity 16, Real-World Math with the CBL System.) CBL Experiment The sound from a tuning fork is collected over time. Determine the amplitude, frequency, and period of the graph. A model of the form y = A cos [ B ( x - C ) ) is fitted to the data. (Activity 23, Real-World Math with the CBL System.)

- U se a grap h'mg utI·t·Ity to graph y = � :J7. •

y =

� x

I . 1 . sm x, y = -2 sm x, and x x sin x for x > O. What patterns do you observe? -

58. How would you explain to a friend what simple harmonic

motion is? How would you explain damped motion?

Chapter Review

707

C HAPTER REVIEW Things t o Know

Formulas Law of Sines (p. 676)

sin A a

Law of Cosines (p. 686)

c2 b2 a2

= =

=

K=

Area of a triangle (pp. 692-693)

sin B b

sin C c

a2 + b2 - 2ab cos C a2 + c2 - 2ac cos B b2 + c2 - 2bc cos l "2 bh

A

I

K = ys(s

. K = "2 a b Sin C K=

1 . "2 bc Sin

- a)(s - b)(s - c),

where

1

. B A K = "2 ac Sin

s

=

�

(a + b + c)

Objectives --------, Section


Review Exercises

9. 1

2

Solve right triangles (p. 670) Solve applied problems (p. 67 1 )

1-4 35, 36, 45-47

2

Solve SAA or ASA triangles (p. 676) Solve SSA triangles (p. 677) Solve applied problems (p. 679)

5, 6, 22 7-10, 12, 17, 18, 21 37-39

2

Solve SAS triangles (p. 686) Solve SSS triangles (p. 687) Solve applied problems (p. 688)

1 1 , 15, 16, 23, 24 13, 14, 19, 20 40, 41

2

Find the area of SAS triangles (p. 692) Find the area of SSS triangles (p. 693)

25-28, 43, 44 29-32, 42

2

Find an equation for an object in simple harmonic motion (p. 697) Analyze simple harmonic motion (p. 700) Analyze an object in damped motion (p. 700) Graph the sum of two functions (p. 702)

48, 49

1

9.2

3

9.3

3 1

9.4

1

9.5

3

4

50-53 54-57 58, 59

Review Exercises

In Problems 1-4, solve each triangle. 1.

1 0� A �o r1 b

2.

a

n B

35°

c

5

3·

b

� a

4·

2

�1 3

In Problems 5-24, find the remaining anglers) and siders) of each triangle, if it (they) exists. If no triangle exists, say "No triangle. " S.

A=

B=

50°,

a

30°,

=

1

6.

A =

10°,

C=

40°,

c

=

2

7.

A=

=

a

100°,

=

c

5,

2

8. a

=

2,

c

=

5,

A=

60°

9. a

=

3,

c

=

1,

C=

1 1 0°

10. a

=

3,

c

=

1,

C=

20°

1 1. a

=

3,

c

=

1,

B=

100°

12. a

=

3,

b

=

5,

B=

80°

13. a

=

2,

b

=

3,

c

1

14. a

=

1 0,

=

8

15. a

=

1,

b

=

3,

C=

40°

16. a

=

4,

b

=

1,

C=

17. a

=

5,

b = 3,

A

=

80°

18. a = 2,

b

=

3,

A=

20°

19. a

=

1,

b

= "2 '

1

4 c = 3

20. a

=

3,

b

=

2,

c

b

=4

22. a

=

4,

A=

c

=

5,

b

=

4,

A=

23.

b

=

c

7,

=

2 70°

21. a

=

3,

A=

10°,

24. a

=

1,

b

=

2,

C=

60°

20°,

=

100°

B=

100°

708

CHAPTER 9


In Problems 25-34, find the area of each triangle. a

=

2,

b

28. a = 2,

b

25.

31.

a

=

4,

b

=

=

=

3,

C

= 40°

1,

C

=

2,

C =

1 00°

26. b

=

5,

c

=

5,

A

29. a

=

4,

b

=

3,

c

5

=

=

20°

27. b

5

30. a = 10,

32. a = 3,

A straight trail with a uniform inclination leads from a hotel, elevation 5000 feet, to a lake in a valley, elevation 4100 feet. The length of the trail is 4100 feet. What is the inclination (grade) of the trail?

35. Finding the Grade of a Mountain Trail

b

=

2,

C

=

=

4,

C =

b

1 0,

=

7,

A = 70° C =

8

2

decide to go around the bay. The illustration shows the path that they decide on and the measurements taken. What is the length of highway needed to go around the bay?

The hypotenuse of a right triangle is 12 feet. If one leg is 8 feet, find the degree measure of each angle.

36. Geometry

Two observers simulta neously measure the angle of elevation of a helicopter. One angle is measured as 25°, the other as 40° (see the figure). If the observers are 1 00 feet apart and the helicopter lies over the line joining them, how high is the helicopter?

37. Finding the Height of a Helicopter

,

,

,

A sailboat leaves St. Thomas bound for an island in the British West Indies, 200 miles away. M aintaining a constant speed of 18 miles per hour, but encountering heavy crosswinds and strong currents, the crew finds after 4 hours that the sailboat is off course by 15°. (a) How far is the sailboat from the island at this time? (b) Through what angle should the sailboat turn to correct its course? (c) How much time has been added to the trip because of this? (Assume that the speed remains at 18 miles per hour.)

40. Correcting a Navigational Error , ,

,

,

,

,

"

'>

ti

Rebecca, the navigator of a ship at sea, spots two lighthouses that she knows to be 2 miles apart along a straight shoreline. She determines that the an gles formed between two line-of-sight observations of the lighthouses and the line from the ship directly to shore are 12° and 30°. See the illustration. (a) How far is the ship from lighthouse Ll? (b) How far is the ship from lighthouse L2 ? ( c) How far is the ship from shore?

38. D et ermining Distances at Sea

Two homes are located on opposite sides of a small hill. See the illustration. To measure the distance be tween them, a surveyor walks a distance of 50 feet from house P to point R, uses a transit to measure L. PRQ, which is found to be 80°, and then walks to house Q, a distance of 60 feet. How far apart are the houses?

41. Surveying

A highway whose primary direc tions are north-south is being constructed along the west coast of Florida. Near Naples, a bay obstructs the straight path of the road. Since the cost of a bridge is prohibitive, engineers

39. Constructing a Highway

R

Chapter Review

42. Approximating the Area of a Lake To approximate the area

of a lake, Cindy walks around the perimeter of the lake, tak ing the measurements shown in the illustration. Using this technique, what is the approximate area of the lake?

709

speed of 10 knots. After 1 hour, the ship turns 90° toward the southwest. After 2 hours at an average speed of 20 knots, what is the bearing of the ship from Boston?

[Hint: Use the Law of Cosines on the three triangles shown and then find the sum of their areas.]

46. Drive Wheels of an Engine

The irregular parcel of land shown in the figure is being sold for $ 100 per square foot. What is the cost of this parcel?

47. Rework Problem 46 if the belt is crossed, as shown in the

43. Calculating the Cost of Land

TIle drive wheel of an engine is 13 inches in diameter, and the pulley on the rotary pump is 5 inches in diameter. If the shafts of the drive wheel and the pulley are 2 feet apart, what length of belt is required to join them as shown in the figure?

figure.

20 ft

�

50

40°

1 00 ft

Find the area of the segment of a circle whose radius is 6 inches formed by a central angle of 50°.


The Majesty leaves the Port at Boston for Bermuda with a bearing of S800E at an average

45. Finding the Bearing of a Ship

In Problems 48 and 49, an object auached to a coiled spring is pulled down a distance a from its rest position and then released. As suming that the motion is simple harmonic with period T, write an equation that relales the displacement d of the object from its resl po silion after I seconds. Also assume that the positive direction of the motion is up. 48. a = 3;

T

= 4

49. a = 5;

seconds

T

=

6 seconds

In Problems 50-53, the distance d (in feet) that an object travels in time t (in seconds) is given. (a) Describe the motion of the object. (b) What is the maximum displacement fi'om its rest position? (c) What is Ihe time required for one oscillation ? (d) What is the frequency?

50.

d

=

6 sin ( 2t )

51. d

=

2 cos(4t )

52. d = -2 cos('1Tt)

53. d =

-3

sin

[% ] t

In Problems 54 and 55, an object of mass m attached to a coiled spring with damping factor b is pulled down a distance a from its rest position and then released. Assume that the positive direction of the motion is up and the period is T under simple harmonic motion. (a) Write an equation that relates the distance d of the object from its rest position after t seconds. U (/J) Graph Ihe equation found in part (a) for 5 oscillations. 54.

m =

40 grams; a

55. m = 25 grams;

a

=

=

15 centimeters;

b

13 centimeters; b

=

=

0.75 gram/second; T 0.65 gram/second; T

=

=

5 seconds 4 seconds

In Problems 56 and 57, the distance d (in meters) of the bob of a pendulum of mass 111. (in kilograms) from its rest position at time t (in seconds) is given. (a) Describe the motion of the object. (b) What is the initial displacement of the bob? That is, what is the displacement at t = O? :.: (c) Graph the motion using a graphing utility. (d) What is the displacement of the bob at the start of the second oscillation ? (e) What happens to the displacement of the bob as time increases withoUl bound? 56. d

=

_ 1 5e-0 6t/4o

cos

(

C; Y - �� ) t

710

CH A PTER 9


In Problems 58 and 59, use the method of adding y-coordinates to graph each function.

59. y = 2 cos(2x) + sm 2

58. y = 2 sin x + cos(2x)

. x

C HAPTER TEST 1. A 12-foot ladder leans against a building. The top of the lad

2. A hot-air balloon is flying at a height of 600 feet and is di

der leans against the wall 1 0.5 feet from the ground. What is the angle formed by the ground and the ladder?

rectly above the Marshall Space Flight Center in Huntsville, Alabama. The pilot of the balloon looks down at the airport that is known to be 5 miles from the Marshall Space Flight Center. What is the angle of depression from the balloon to the airport?

�

In Problems 3-5, use the given information to determine the three remaining parts of each triangle.

3.

4.

b�

5.

/� � c

In Problems 6-8, solve each triangle. 6. A

=

55°,

C

=

20°,

a

=

4

7. a = 3 ,

b = 7,

9. Find the area of the triangle described in Problem 8.

10. Find the area of the triangle described in Problem 5 . 11.

Find the area of the shaded region enclosed in a semicircle of diameter 8 centimeters. The length of the chord A B is 6 cen timeters.

[Hint: Triangle A B C is a right triangle.]

10

A = 40°

8. a

=

8,

b = 4,

C = 70°

distance first. Highway 20 goes right past the boat ramp and County Road 3 goes to the lodge. The two roads intersect at point ( C) , 4.2 miles from the ramp and 3.5 miles from the lodge. Madison uses a transit to measure the angle of inter section of the two roads to be 32°. How far will she need to swim? 14. Given that 60A B is an isosceles triangle and the shaded sec

tor is a semicircle, find the area of the entire region. Express your answer as a decimal rounded to two places. A L O.

CHAPTER PROJ ECTS

II. The Lewis and Clark Expedition

+

4.

s.

Use the Pythagorean Theorem to find another value for OQ2 - CQ2 and OP2 - C P2 . Now solve for cos c . Replacing the ratios in part (4) by the cosines of the sides of the spherical triangle, you should now have the Law of Cosines for spherical triangles: cos c

=

cos a cos b + sin

a

sin b cos C

Source: For the spherical Law of Cosines; see Mathematics from the Birth of Numbers by Jan Gullberg. W. W. Norton & Co., Publishers, 1 996, pp. 491-494.

we must account for the curvature of Earth when computing the distance that they traveled. Assume that the radius of Earth is 3960 miles. 1. Great Falls is at approximately 47.soN and 1 l 1 .3°W. Lemhi

is at approximately 4S.0oN and

I 1 3.S°W.

( We will assume

712

CHAPTER 9


2. From Lemhi, they went up the Bitteroot River and the

3.

4. that the rivers flow straight from Great Falls to Lemhi on the surface of Earth.) This line is called a geodesic line. Apply the Law of Cosines for a spherical triangle to find the angle between Great Falls and Lemhi. (The central an gles are found by using the differences in the latitudes and longitudes of the towns. See the diagram.) Then find the length of the arc j oining the two towns. (Recall s = re.) Diagram ii

Snake River to what is now Lewiston and Clarkston on the border of Idaho and Washington. Although this is not really a side to a triangle, we will make a side that goes from Lemhi to Lewiston and Clarkston. If Lewiston and Clarkston are at about 46SN 1 1 7 .00W, find the distance from Lemhi using the Law of Cosines for a spherical tri angle and the arc length. How far did the explorers travel just to get that far? Draw a plane triangle connecting the three towns. If the distance from Lewiston to Great Falls is 282 miles and the angle at Great Falls is 42° and the angle at Lewiston is 48.5°, find the distance from Great Falls to Lemhi and from Lemhi to Lewiston. How do these distances compare with the ones computed in parts (a) and (b)? For Lewis and Clark Expedition: American Jour ney: The Quest for Liberty to 1877, Texas Edition. Prentice Hall, 1992, p. 345.

Source:

Source: For map coordinates: National Geographic Atlas of the World, published by National Geographic Society, 1981, pp. 74-75.

South

The following projects are available at the Instructor's Resource Center (IRC): III.

Project at Motorola: How Can You Build or Analyze a Vibration Profile? Fourier functions are not only important to analyze vibrations, but they are also what a mathematician would call interesting. Complete the project to see why.

IV.

Leaning Tower of Pisa

V.

Locating Lost Treasure

Trigonometry is used to analyze the apparent height and tilt of the Leaning Tower of Pisa. Clever treasure seekers who know the Law of Sines are able to efficiently find a buried treasure.

V I. Jacob's Field Angles of elevation and the Law of Sines are used to determine the height of the stadium wall and the distance from home plate to the top of the wall.

Polar Coordinates; Vectors How Do Airplanes Fly? Have you ever watched a big jetliner lumber into position on the runway for takeoff and wonder, "How does that thing ever get off the ground?" You know it's because of the wing that it stays up i n " the air, b u t how does it really work? f' When air flows around a wing, it creates lift. The way it creates lift is based on the wing's movement through the air and the air pres ,.. , , _ . "�,I sure created around the wing. An airplane's wing, in varying degrees ,' depending on the type and design of the airplane, is curved over II the top of the wing and straighter underneath the wing. As air hits the wing, it is "split in two," with air moving both over and under the wing. Since the top of the wing has more curve than the bottom of the wing, the air moving over the top of the wing has farther to travel, and thus must move faster than the air moving underneath the wing. The air moving over the top of the wing now exerts less air pressure on the wing than the slower-moving air under the wing. Lift is created. The difference in air pressure is the primary force creating lift on a wing, but one other force exerted on the wing also helps to pro duce lift. This is the force of deflection. Air moving along the underside of the wing is deflected downward. Remember the New tonian principle: For every action, there is an equal and opposite reaction. The air that is deflected downward (action) helps to push the wing upward (reaction), producing more lift. These two natural forces on the wing, pressure and deflection, produce lift. The faster the wing moves through the air, the greater the forces become, and the greater the lift. 0, the point is on the terminal side of e, and r 0, Vx2 + l. Since . e y cos e x we have sin e x cos e If < 0, the point e) can be represented as + e), where -r > 0. Since sine + e) -sin e L -cos e + e) we have sin e x cos e

Figure 1 2

Y

x

Pr =

= d( P) =

=-r SIn =-r y=r =r

r

P = (r, X COS(7T = = -r =r y =r

7T

E XA M P L E 4

Solution

y x

(-r, 7T = = -r

-

Converting from Polar Coordinates to Rectangular Coordinates

Find the rectangular coordinates of the points with the following polar coordinates:

( :) We use formula (1): x = r cos e and y = r sin e. (a) Figure 13(a) shows (6, �) plotted. Notice that (6, �) lies in quadrant I of the rectangular coordinate system. So we expect both the x-coordinate and the y-coordinate to be positive. With r = 6 and e = �, we have x r cos e = 6 cos 7T = 6· Y3 = 3 Y3 6 2 y = r sm. e = 6· sm 7T6 = 61 .-2 = 3 The rectangular coordinates of the point (6, �) are ( 3Y3, 3 ) , which lies in quadrant I, as expected. (b) Figure 13(b) shows ( :) plotted. Notice that (- 4, - :) lies in quadrant II of the rectangular coordinate system. With r = - 4 and e = 7T '4' we have (b)

Figure 1 3

717

-4, -

=

(a)

-

x

- 4, -

-

(b)

= r cos e cos (- ) = - 4 . V22 = - 2V-2 Y = rsin e = -4 sin ( : ) = ( �) = 2V2 The rectangular coordinates of the point ( ) are ( -2\12,2\12), which lies in quadrant II, as expected. x

COMMENT

M ost calculators have the capability of converting from polar coor dinates to rectangular coordinates. Consult your owner's manual for the proper key strokes. Since this proce dure is often tedious, you will find that _ using formula (1) is faster.

=

7T

-4

4

-

-4 -

- 4,- 7T 4

•

= ... 'l!lO ;r;::; I:>o ;;o -

Now Work PRO B l EMS 3 9 AND 5 1

718

CHAPTER 10

Polar Coordinates; Vectors

3

EXA M P LE 5

Find polar coordinates of a point whose rectangular coordinates are (0, 3). See Figure 14. The point (0, 3) lies on the y-axis a distance of 3 units from the origin 71 ( pole ) , so r = 3. A ray with vertex at the pole through (0, 3) forms an angle e =

Figure 1 4

(

with the polar axis. Polar coordinates for this point can be given by 3, �).

y =

Converting from rectangular coordinates ( x, y) to polar coordinates ( r, e) is a little more complicated. Notice that we begin each example by plotting the given rectan gular coordinates. Converting from Rectangular Coordinates to Polar Coordinates

Solution

(x, y)

Convert fro m Recta n g u l a r Coord i n ates to Po l a r Coord i n ates

(0, 3)

1T

3

2

2

•

COMMENT

Most calculators have the capability of converting from rectangular coordinates to _ polar coordinates. Consult your owner's manual for the proper keystrokes.

x

Figure 15 shows polar coordinates of points that lie on either the x-axis or the y-axis. In each illustration, a > 0. y (x, y) = (0, a) (r, 0) = (a, ¥)

y

Figure 1 5

(x, y) (r, 8)

(a, D)

= (a, 0) =

a

a

y

1T

(x, y) (r, 0)

2

x

x

y

= (-a, D) =

(a, 71)

x

a

(r, 8)

(x, y) (a) (x, y)

=

(b) (x, y)

(a, 0), a> 0

=

(0, a), a> 0

(c) (x, y)

=

(

== (a, 3f)

(d) (x, y)

-a, 0), a> 0

x

(0, -a)

=

(0, -a), a> 0

L�==�� - NowWork PROB LEM 5 5

EXAM P L E 6

Converting from Rectangular Coordinates to Polar Coordinates

Find polar coordinates of a point whose rectangular coordinates are: (a) ( b ) ( -1, -\13)

(2, -2)

Solution

( a)

See Figure 16 ( a) . The distance r from the origin to the point

(2, -2) is

Figure 1 6

y

x

-1 -1 (a)

(x, y)

=

( 2,

- 2)

-,

-, 2

2

. e by reca 11'mg that tan e = Y so e = tan- Y -71 71 We fmd < e < -. x x Since lies in quadrant IV, we know that � < e < 0. As a result,

(2,-2)

e

=

. (-2-2 )

tan-1 -yx tan- 1 =

-

=

1

tan-1 (-1)

7T

= -4

(2V2' - :) . Other possible re (2V2' ) (-2V2, 3;).

A set of polar coordinates for this point is 7; and presentations include

SECTION 10.1 Figure 1 6

r=

-

V(-I? + (- v3y = V4 = 2

To find e, we use e = tan-lx.x , - 2'iT < e < �. 2 Since the point ( -1, -v3) lies in quadrant III and the inverse tangent function gives an angle in quadrant I, we an add 'iT to the result to obtain ) angle in quadrant III. V3 'iT = "3 4'iT = 'iT + tan- 1 v;:::3 = 'iT + 3 e = 'iT + tan-1 -

x

=

71 9

(b) See Figure 16(b). The distance r from the origin to the point ( -1, v3) is

y o

(x,y)

Polar Coordinates

(-1, --J3)

( ---=l

(b)

'

( )

A set of polar coordinates for this point is 2, 4; . Other possible represen 2 ;. tations include -2, ; and

(2, )

( )

_

•

Figure 17 shows how to find polar coordinates of a point that lies in a quadrant when its rectangular coordinates (x, y) are given. Figure 1 7

(x,y)

(x,y)

y

y

y

y

o

x

x

x

x

(x,y)

(x,y) (a)

(b) r= .x) 2 +l

r= .x) 2 + l 0= tan-1 y..

e

x

=

'iT +

tan- 1 y.. x

(c)

(d)

r= .x) 2 + l = + tan-1 y..

e

'iT

x

r= .)x2 + l = tan-1 y..

e

x

Based on the preceding discussion, we have the formulas tan e =

y x

-

if x "* 0

(2)

To use formula (2) effectively, follow these steps: Steps for Converting from Recta n g u l a r to Pol a r Coordinates

STEP 1: STEP 2:

Always plot the point (x, y) first, as we did in Examples 5 and 6. If x = ° or y = 0, use your illustration to find (r, e). See Figure STEP 3: If x "* ° and y "* 0, then r = Vx2 + i.

15.

To find e, first determine the quadrant in which the point lies. y Quadrant I or IV: e = tan-IXx. Quadrant II or III: e = 'iT + tan-Ix See Figure 17. STEP 4:

'-'l!l:==_ Now Work 4

PRO B L EM 5 9

Transfo rm Equations from Po l a r to Rectang u l a r Form

Formulas (1) and (2) may also be used to transform equations from polar form to rectangular form, and vice-versa. Two common techniques for transforming an equation from polar form to rectangular form are 1. multiplying both sides of the equation by r 2. squaring both sides of the equation.

720

CHAPTER 10 Polar Coordinates; Vectors

Transforming an Equation from Polar to Rectangular Form

E XA M P L E 7

Transform the equation r 6 cos 8 from polar coordinates to rectangular coordi nates, and identify the graph. If we multiply each side by r, it will be easier to apply formulas (1) and (2). r 6 cos8 r2 6r cos 8 M ultiply each side by r. =

Solution

=

,2

=

x2 + l

=

6x

=

J' +

j; x= r cos 0

This is the equation of a circle, so we proceed to complete the square to obtain the standard form of the equation. x2 + l 2 (x - 6x) + l = 2 (x - 6x + 9) + l (x - 3 ? + l = =

=

6x 0 9 9

General form Complete the square in

x.

Factor.

This is the standard form of the equation of a circle with center (3 , 0) and radius 3.

•

-===--

E XA M P L E 8

Now Work PRO B LEM 7 5

Transforming an Equation from Rectangular to Polar Form

Transform the equation 4xy 9 from rectangular coordinates to polar coordinates. We use formula (1): x = r cos8 and y = r sin 8. 4xy 9 rcosO,y= rsinO 4(rcos8)(rsin8) = 9 2 4r cos 8 sin 8 = 9 This is the polar form of the equation. It can be simplified as shown next: 2r2 (2 sin 8 cos8) = 9 Factor out 2,2. 2r2 sin(28) = 9 Double-angle Formula =

Solution

=

x=

•

�.

Now Work PRO B LEM 6 9

10.1 Assess Your Understanding 'Are You Prepared?'

Answers are given al Ihe end of Ihese exercises. If you gel a wrong answer, read Ihe pages listed in red.

1. Plot the point whose rectangular coordinates are (3, (p. 156)

2.

To complete the square of x2 + 6x, add __ . (pp.

-

1)

.

99-1 OJ)

3. If P

(a,

b)

is a point on the terminal side of the angle 0 at

a distance rfrom the origin, then sin 0 =

4.

=

tan-1( - 1 )

=

__

Concepts a n d Voca bulary 5.

In polar coordinates, the origin is called the __ and the

positive x-axis is referred to as the ____

.

6. Another representation in polar coordinates for the point (

7.

2,

�)

is

, (_ 4;).

The polar coordinates gular coordinates by

(

-2,

(__ ,

�)

8. True or False 9.

True or False

unique.

10. True or False are represented in rectan

).

__

. (pp. 608-610)

__

.

(pp. 540-542)

The polar coordinates of a point are unique .

The rectangular coordinates of a point are

In (r, 0), the number r can be negative.

SECTION 10.1

Polar Coordinates

721

Skill Building In Problems 1 1-18, match each point in polar coordinates with either A, B, C, or D on the graph. 11.

( _�) 1 7T

2,

12.

(

-2 , -

�)

13.

17.

( �) ( ) -2,

14.

77T -2'6

18.

( ;) ( ) 2,

7

1 17T 2' - 6

In Problems 19-30, plot each point gi ven in polar coordinates. 19. (3, 90°) 23.

27.

( �) ( �)

20. (4, 270° )

6,

24.

-1, -

28.

( ;) ( :) 5,

5

-3, -

3

21. (-2, 0 )

22. ( -3, 7T)

25. (-2, 135°)

26. (-3,120°)

29. ( -2, -7T)

30.

(

, � -32

)

In Problems 31-38, plot each poin t gi ven in polar coordinates, and find other polar coordinates (r, e ) of the poin t for which: (a) r

31.

>

0,

( ;) 5,

- 27T:S e < 0

2

( b) r < 0,

32.

O:s e < 27T

( :) 4,

3

(c) r

>

0,

27T:S e < 47T

34. (-3, 47T)

33. ( -2, 37T)

36. (2, 7T)

37.

(

7T -3'- "4

)

38.

(

27T -2 ' - 3

)

In Problems 39-54, the polar coordin ates of a point are gi ven. Find the rectangular coordinates of each point. 39.

( �) 3,

40.

43. (6, 150° )

47.

(

-1'-

�)

51. (7.5, 1 1 0°)

( ;) 4,

3

44. ( 5 , 300° )

48.

(

42. ( -3, 7T)

41. (- 2, 0)

:)

3 _3, _

52. ( -3 .1, 182° )

45.

(

-2

37T , 4

)

46.

(

27T -2 ' 3

)

49. (-2, -180°)

50. ( -3, -90° )

53. (6.3, 3.8)

54. (8. 1 , 5.2)

In Problems 55-66, the rectangular coordinates of a point are gi ven. Find polar coordinates for each point. 55. (3, 0)

56. (0, 2)

57. ( - 1 , 0 )

59. (1, -1)

60. ( -3, 3 )

61.

63. ( 1 .3, -2.1)

64. ( -0.8, -2.1)

65. (8.3, 4.2 )

(V3, 1)

58. (0, -2) 62. (-2,

-2V3)

66. ( -2.3, 0.2)

In Problems 67-74, the letters x and y represent rectangular coordinates. Write each equation using polar coordinates (r, e) . 67. 2x2 + 2/ = 3

68. x2 + /= x

71. 2xy= 1

69. x2= 4y

70. /= 2x

73. x= 4

74. Y= -3

77. r2= cos e

78. r = sin e - cos e

4 81. r=--1 - cos e

82. r=

In Problems 75-82, the letters r an d e represent polar coordinates. Write each equation using rectangular coordinates ( x, y).

" 75.

r = cos e

79. r= 2

76. r

=

sin e + 1

80. r= 4

3 --3 - cos e

Applications a n d Extensions 83. Chicago I n Chicago, the road system is set up like a Cartesian plane, where streets are indicated by the number of blocks they are from Madison Street and State Street. For example, Wrigley Field in Chicago is located at 1060 West Addison, which is

10 blocks west of State Street and 36 blocks north of Madison Street. Treat the intersection of Madison Street and State Street as the origin of a coordinate system, with East being the posi tive x-axis.

722

CHAPTER 10

Polar Coordinates; Vectors City of Chicago, Illinois

(a) Write the location of Wrigley Field using rectangular coordinates. (b) Write the location of Wrigley Field using polar coordi nates. Use the East direction for the polar axis. Express f) in degrees. (c) U.S. Cellular Field, home of the White Sox, is located at 35th and Princeton, which is 3 blocks west of State Street and 35 blocks south of Madison. Write the location of U.S. Cellular Field using rectangular coordinates. (d) Write the location of U.S. Cellular Field using polar coordinates. Use the East direction for the polar axis. Express f) in degrees. 84. Show that the formula for the distance d between two points PI= ( 1'1 , f)I)and P2 = h, f)2)is d=

VrT + I'�

-

21'11'2 COS(f)2 - f)1 )

Discussion a n d Writing 85. In converting from polar coordinates to rectangular coordi nates, what formulas will you use? 86. Explain how you proceed to convert from rectangular coor dinates to polar coordinates. 'Are You Prepa red?' Answers 1.

y 2 -2

-2

2. 9

3.

� r

4.

-

87. Is the street system in your town based on a rectangular co ordinate system, a polar coordinate system, or some other system? Explain.

�

4

2 • 4 x (3, -1 )

10.2 Polar Equations and Graphs PREPARING FOR THIS SECTION •

•

•


Symmetry (Section 2.2, pp. 1 67-168) Circles (Section 2.4, pp. 189-193) Even-Odd Properties of Trigonometric Functions (Section 7.5, pp. 556-557)

• •

Difference Formulas for Sine and Cosine (Section 8.4, pp. 627 and 630) Value of the Sine and Cosine Functions at Certain Angles (Section 7.3, pp. 5 29-532 and Section 7.4, pp. 540-547)

Now Work the 'Are You Prepared?' problems on page 735. Equations (p. 723)

OBJECTIVES 1 Graph and Identify Polar Equations by Converting to Rectangular 2 Test Polar Equations for Symmetry (p. 727)

3 Graph Polar Equations by Plotting Points (p.

728)

Just as a rectangular grid may be used to plot points given by rectangular coordinates, as in Figure 18(a), we can use a grid consisting of concentric circles (with centers at the pole) and rays (with vertices at the pole) to plot points given by polar coordinates, as shown in Figure 18(b). We shall use such polar grids to graph polar equations.

SECTION 10.2 Polar Equations and Graphs

723

Figure 1 8

_�_I_L-L-L.·

B _:tIT - 2

'---'

__

(b) Polar grid

(a) Rectangular grid

An equation whose variables are polar coordinates is called a polar equation. The graph of a polar equation consists of all points whose polar coordinates satisfy the equation. .-J

DEFINITION

1

G ra ph and Identify Po l a r Equations by Converting to Rectang u l a r Equations

One method that we can use to graph a polar equation is to convert the equation to rectangular coordinates. In the discussion that follows, ( x, y) represent the rectan gular coordinates of a point P, and (r, e) represent polar coordinates of the point P. E XA M P L E 1 Solution

Identifying and Graphing a Polar E qu ation (Circle)

Identify and graph the equation: r = 3 We convert the polar equation to a rectangular equation. r 3 r2 = 9 Square both sides . x2 +l = 9 1- = >1-+1 The graph of r = 3 is a circle, with center at the pole and radius 3. See Figure 1 9 . =

x2

Figure 1 9

r

=3

or

+ .; = 9

•

�;:�� - NowWork PRO B L EM 1 3

E XA M P L E 2

Identifying and Graphing a Polar Equation (Line)

Identify and graph the equation:

e

7T

=4

-

724

CHAPTER 10


We convert the polar equation to a rectangular equation.

Solution

e

=

7T 4

e

tan = tan

7T

4

=1

Take the ta ngent of both sides. tan e = �

I =l x y=x

x

The graph of = : is a line passing through the pole making an angle Of: with the polar axis. See Figure 20. e

Figure 20

e

=

7T

4

-

ory = x

=

e !tIT 4

"' "' �-

8 =� 2

•

Now Work PRO B L EM 1 5

Identifying and Graphing a Polar Equation (Horizontal Line)

E XA M P L E 3

e

Identify and graph the equation: sin = 2 Since y = r sin we can write the equation as y= 2 We concl the graph of sin = 2 is a horizontal line 2 units above the pole. See Figureude2 that . r

e,

Solution

1

2 or Y

Figure 21

r sin

e

=

e

r

yt

e='IT

=2

�2 .

e --/-' !�/ i

\/\.-- ---!--,-.;

A.

\

__

-

\,-�!

e= !tIT 4

�I

COMMENT

O= �2

'IT

4

X I �

e= 0

0= err 4

•

A gra phing utility can be used to graph polar equations. Read

ity to Graph a Polar Equation

. h F'Igure 21. resu ItWit

in the Appendix. Section

8.

Then graph

r =

Using a Graphing Util2

.-

sin e

and compare the

•


E XA M P L E 4

Identifying and Graphing a Polar E qu ation (Vertical Line)

e =

Identify and graph the equation: cos -3 Since cos we can write the equation as -3 Wethe polconcle. uSeede Figure that the graph of cos -3 is a vertical line 3 units to the left of r

Solution

725

x =r

e,

x =

e =

r

22.

Figure 22 r

cos e

=

-

3 or x

=

-

3

e=� 4

e=I1T 4

e =� 2

•

es 3 and left asBased exercionses.Exampl See Probl ems 75weandare76.)led to the following results. (The proofs are Let be a nonzero real number. Then the graph of the equation sin isif a horizontal line units above the pole if and units below the pole The graph of the equation cos iles fta ofvertithecalpollinee ifaaunits to the right of the pole if and lal units to the 4,

THEOREM

a

e =a

r

a

a < O.

a>0

e =a

r

a >0

< O.

t.l'I!::==o--

EXA M P L E 5

.J


Identifying and Graphing a Polar E quation (Circle)

e

Identify and graph the equation: sin To transform the equation to rectangular coordinates, we multiply each side by sin Now we use the facts that + and sin Then r=4

Solution

lal

r2 = 4r

,.2 = x2

x2 + i x2 + (i - 4y) x2 + (i - 4y + 4) x2 + (y - 2?

= = = =

i

4y 0 4 4

e

y= r

e.

Com plete the square

in y.

Factor.

Thi s is theandstandard dinates radius equati See Fiognureof 23.a circle with center at 2.

r.

(0,2)

in rectangular coor

726

CHAPTER 10


Figure 23

( = 4 sin () or

x2 + (y - 2)2

=4

•

E XA M P L E 6 Solution

Identifying and Graphing a Polar E quation (Circle)

Identify and graph the equation: We proceed as in Example 5.

cos -2r cos

r = -2

r2 =

x2 + i = -2x x2 + 2x + i = 0 (x2 + 2x + 1) + i = 1 (x + 1)2 + i = 1

e

e Mu ltiply both sides by r. r2

=

? +

I;

x

= r

cos ()

Complete the squa re in

x.

Factor.

Thicoordis isnates the standard on of a circle with center at (-1,0) rectangular and radiusequati 1. See Figure 111

24.

( = -2 co s () or (x + 1)2 + i = 1 Figure 24

S=� 4

•

Exploration = sine,

= 2 si ne, and (3

the scree n a nd graph (1 = - si n e, r2 = -2 sine, a nd (3 = -3 sin e. Do you see the patter n? Clear the scree n and graphrl cose,r2 = 2 cose, and r3 = 3cose. Do yo u see the patter n? Clear the scree n a nd graph r1 = -COS(},r2 = -2cos(J, a ndr3 = -3co se. Do yo u see the pattern? U si ng a square screen, graph

(1

(2

=

3 si n (J. Do you see the p attern? Cle ar

=

THEOREM

on Exampl s 5 andare6 and precedises.nSee g ExplProbloratiemson,77-80. we are) led to the folloBased wing resul ts. (Theeproofs left astheexerci Let be a positive real number. Then Description Equation (a) r sin Circle: radius center at (0, in rectangular coordinates (b) r -2a sin Circle: radius a; center at (0, in rectangular coordinates a

= 2a

=

e

e

a;

a)

-a)


=

727

(c) r 2a cos 8 Circle: radius center at 0) in rectangular coordinates (d) r 2a cos 8 Circle: radius center at 0) in rectangular coordinates Each circle passes through the pole. ..J =

a; a;

-

(a, (-a,

Ij!� - Now Work P R O B L E M 2 1

The method ofgraph converti ng ala wpolaysarhelpful, equationor n toisanitialdenti fianecessary. ble rectangulUsualar equa tion to obtai n the i s not w ays lymay , we setbe possi up a btablle toe thatreduceliststheseveral poi n ts on the graph. By checki n g for symmetry, it number of points needed to draw the graph. 2

Test Po l a r Equations for Sym m et ry

pointsSee(1',Figure 8) and25(r(a)., -8)Thearepoisymmetri with(1respect polIn polar aaxir coordi s (andntoates,thethex-axis). nts (1', 8)c and ', 8)to arethe symmetric with respect to the line 8 (the y-axis). See Figure 25(b). The points (r, 8) and (-r,8) are symmetric with respect to the pole (the origin). See Figure 25(c). 1T

1T

= "2

9=

'IT

++-+-!-=t;

9=0

9= :tIT

9=:tIT

2

2

(b)

-

Points symmetric with respect to the line 9=

(c)

T

Points symmetric with respect to the pole

The following tests are a consequence of these observations. THEOREM

Tests for Symmetry

Symmetry with Respect to the Polar Axis (x-Axis)

In a polisasymmetri r equatiocn,wireplth arespect ce 8 byto-8.theIfpolanaequival graph r axis. ent equation results, the Symmetry with Respect to the Line (y-Axis) In a polar equation, replace 8 by 8. If an equivalent equation results, the graph is symmetric with respect to the line 8 ; . Symmetry with Respect to the Pole (Origin) Ingrapha polar equation,c wireplth arespect ce r byto-r.theIfpolan eequivalent equation results, the is symmetri . ..J for symmetry hereis,areansufficient for symmetry, but The they three are nottestsnecessary conditigions.venThat equationcondi may tfaiionsl these tests and stil have a graph that is symmetric with respect to the polar axis, the line 8 2 or the pole. For example, the graph of r sin (28) turns out to be symmetric with respect to the polar axis, the line 8 2 and the pole, but all three tests given here fail. See also Problems 81-83. ()

=

1T

-

TT

2"

=

=

=

1T,

=

1T,

728

CHAPTER 10


3

EXAMPLE 7 Solution

G ra p h Pol a r Equations by Plott i n g Poi nts

Graphing a Polar Equation (Cardioid)

Graph the equation: r = 1- sine We check for symmetry first. Replacee by -e. The result is r = 1 - sine-e) = 1 sine aXITheS. test fails, so the graph may or may not be symmetric with respect to the polar ; Replace e by 7T- e. The result is r = 1- sin(7T- e) = 1- (sin7Tcose - COS7Tsin e) = 1- [o'cose - (-l) sineJ 1- sine The test is satisfied, so the graph is symmetric with respect to the line e ; by -rmay . Then the result is -r = 1- sine, so r -1 sine. The test faiRepl ls, soacether graph or may not be symmetric with respect to the pole. Nextthewecorrespondi identify poinngtsvalues on theofgraph by assigning values to the anglee and cal culating r. Due to the symmetry with respect to the line e 7T we only need to aSSI'gn values toe from 7T to 7T as gI. ven Tablel. Now we plot the points (r, e) from Table 1and trace out the graph, beginning at the point (2, - ;) and ending7Tat the point ( ;) . Then we reflect this portion ofFigtheure graph about the line e -2 (the y-axis) to obtain the complete graph. See 26. Polar Axis:

+

The Line (J

Table 1 r =

(J

1

-

-

7T

1 - (-1) =

-

7T

1- -

-

2

-

3

7T

1

6 0

-

1

sin (J

2

( �) ( �) %

0

� 1.87

-

= 1

=

1 1 1--=-

7T

6

-

-

1-

7T

-

3

2

v3 2

1-1=

-2 7T

2

"" 0.13

0

=

sin(-e)

=

-sin e

:

=

=

The Pole:

=

= 2'

-

2

2'

.

+

.

m

0,

=

Figure 26 e =:tIT 4

� Exploration mI Graph = 1 + si n e.

a nd gr aph r1 = 1 - cos e. C lear the scree n a nd graph r1 = 1 + cos e. Do you see a pattern? r1

8= 'IT 4

8=� 4

Clear the scree n

DEFINITION

/

•

The curve in Figure 26 is an example of a cardioid (a heart-shaped curve). Cardioids are characterized by equations of the form r=a cose) r a(l sine) r a(1 sin e) r a(l - cose) where a 0. The graph of a cardioid passes through the pole. (1 +

=

>

==t;!'I!C


=

+

=

-

.J

SECTION 10.2

EXA M P L E 8

Graph the equation: r = cos We check for symmetry first. Replace by The result is r= 2 cos( = cos The test is satisfied, so the graph is symmetric with respect to the polar axis. Replace8 by The result is r= cos( = 2( cos cos sin sin = cose The test fails, so the graph mayor maynot be symmetric with respect to the line e Repl a ce r by -r. The test fails, so the graph mayor maynot be sym metric with respect to the pole. Nextthewe correspondi identify poinntsg onvaluthees graph by assigning values to the angle and cal culpolarating of r . Due to the symmetry with respect to the axis,weweplonlotytheneedpoitontsassi(r,ge)n valfrom ues toTablefromand0 totraceasoutgiventheingraph, Table begin Now nintheg atgraph the poiabout nt (5, the0) andpolaendi at thex-axi poins)t to1( , obtaiThen refleecttethigraph. s portiSeeon ofFigure r axinsg (the n thewecompl Polar Axis:

8

r =

(J

The Line

3 + 2 cos (J

7T

6

3 + 2

7T

3 + 2

3 7T

(�) G)

3 + 2(0)

2 27T

�4.73

=

=3

(-D = 2(-�)

3 57T

6

3 +

7T

3 + 2(-1)

=

=

8

3 +

-8)

'TT

2:

7T

2

7T

2

cas( -8) = cos 8

8

- 8.

- e)

3 + 3 -

e +

7T

2

7T

e)

7T

2

-.

The Pole:

4

3 + 2

=

3 +

3 + 2(1) = 5

0

0

2

-8.

3 +

Table 2

729

Graphing a Polar Equation (Lima�on without an Inner Loop) 3 +

Solution

Polar Equations and Graphs

e

2 �

e

1.27

1

2

2.

7T,

7T

).

27.

Figure 27

�

8

Exploration

� Graph = 3 r1

- 2 cos e. Clear the sc ree n and graph (1 3 + 2 si n e. Cle a r t he screen and g raph (1 3 - 2 sin e. Do

= �4

=

=

you see a p attern?

= :t!T ' 2

•

The curve in Figure is an example of a limar;on (the French word for snail)

without an inner loop.

DEFINITION

o 27

Lima',;ons without an inner loop

are characterized by equations of the form r=a cose, r = a sin r=a cose, r = a sin where a 0, 0, and a The graph of a l i m a

0= =-::::::;:; >-

b >

> b.


+ b - b

8

e

730

Polar Coordi nates; Vectors

CHAPTER 10

EXA M P L E 9

Solution

Graphing a Polar Equation (Lima�on with an Inner Loop)

=

Graph the equation: r 1 + 2 cos 8 First, we check for symmetry. Replace 8 by -8. The result is r 1 + 2 cos( -8) 1 + 2 cos 8 The test is satisfied, so the graph is symmetric with respect to the polar axis. �: Replace 8 by 8. The result is r 1 + 2 cos( 8) 1 + 2( cos cos 8 + sin sin 8) 1 - 2 cos 8 The test fails, so the graph may or may not be symmetric with respect to the line 8 2' ace rtobythe-r.polThe metric withRepl respect e. test fails, so the graph may or may not be sym Nexte 8weandidenticalcfuly apoitinngtstheoncorrespondi the graph ofngrvalu1es+of2 cosDue8 byto assi gsymmetry ning valueswitoth therespect angl the only need to assign values to 8 from to as given in TablNoweto3.wetheplpolot ather axipois,nwets (r,8) beginonin nofg theat (3,0) andaboutendithengpoat (l-ar 1,axis (theSeex-axi Figures) to28(a).obtaiFinnalthely, compl wefromrefleTabl etectgraph. thie 3,s porti graph See Figure 28(b).

Polar Axis:

=

The Line

Table 3 r

(J

=

0

1 + 2(1)

7T 6

1 + 2

1 + 2

7T -

1 + 2(0)

2

27T 3 57T

6

7T

1 + 2

=

3

( �) G)

7T 3

1 + 2 cos (J

=

=

�2.73

1 + 2 -

=

1T -

=

1T -

= =

1T

1T

1T

=-

The Pole:

2

1

( - �) ( �)

()

=

=

= 0

� -0.73

1 + 2(-1) = -1

1T

r.

°

1T,

).

Figure 28

x

I� Exploration

Graph (1 = 1 - 2 cos fi. C lear the scree n and gr ap h (1 = 1 + 2 s i n fi. Clear t he screen and g rap h (1 = 1 - 2 sin fi. Do you see a patter n?

DEFINITION

2 (a)

e=:tIT

2

e = :tIT

(b) r = 1

+ 2 cos e

•

The curve in Figure 28(b) is an example of a limar,;on with an innerloop. Lima-;ons with an inner loop are characterized by equations of the form r a cos 8, r a + sin 8 r a cos 8, r a sin 8 0, the0, and wilwherel passa through poleatwice. The graph of a lima

==>tr.:!llli:

b >

+ b

=

b

- b

=

- b

< b.


SECTION 10.2

E XA M P L E 1 0

Polar Equatio ns and Graphs

731

Graphing a Polar Equation (Rose)

Graph the equation: 2 cos(28) We check for symmetry. If we replace 8 by -8, the result is 2 cos[2(-8)] 2 cos(28) The test is satisfied, so the graph is symmetric with respect to the polar axis. ; If we replace 8 by 7T - 8, we obtain 2 cos[2(7T - 8)] 2 COS(27T - 28) 2 cos(28) The test is satisfied, so the graph is symmetric with respect to the line 8 7T Since the graph is symmetric with respect to both the polar axis and the line 8 ; , it must be symmetric with respect to the pole. Do you see why? Next, we construct Table Due to the symmetry with respect to the polar axis, the line 8 and the pole, we consider only values of 8 from 0to pointsfiristn Fiabout gure the29(a).polFiarnalaxilys, because we reflWeeplctothit ands porticonnect o7Tn of these the graph (the x-axiofs)symmetry, and then about the line 8 (the y-axis) to obtain the complete graph. See Figure 29(b). r =

Solution

Polar Axis:

=

r =

The Line (J

:

=

r =

Table 4 (J

r =

0

2(1) = 2

2

7r

-

6 7r

2 cos(2(J)

G)

4

2

7r

-

3

= 1

( -D

2(-1)

7r

2

-

=

=

= "2'

The Pole:

=

2(0) = 0

-

=

=

-1

-2

4.

TI

�

= "2'

"2'

= "2

Figure 29

r�

Exploration

rl = 2 cos ( 48 ) ; clear the scree n and graph rl = 2 cos ( 68 ) . How ma ny petals did each of these graphs have?

tlii Graph

clear screen, rl = 2 cos ( 38 ) , 2 cos ( 58 ) , an d r1 = 2 cos ( 78 ) . What do you notice a bout the n um ber of petals? Clear the screen a n d graph, in or der, each

on

a

r1 =

DEFINITION

•

The curve in Figure 29(b) is called a with four petals. Rose curves are characterized by equations of the form cos(n8), sin (n8), n2:2 and is odd,havethegraphs rose hasthatn arepetalroses. shaped. If is even, the rose has 2n petals; if n..J rose

r = a

a *- 0,

r = a

n

1.ll!iZ 1ISIII; _� -


732

CHAPTER 10


EXA M P L E 11

Solution Table 5 (J 0 7T

6 7T -

4

7T -

3

7T -

2

,2 =

4(0) = 0 4

(�)

4( 1) = 4 4

,

4 sin(2(J)

(�)

4(0) = 0

=

0

2\13

2\13

Graph the equation: sin(28) We leave it to you to verify that the graph is symmetric with respect to the pole. Table 5 lists points on the graph for values of 8 0 through 8 Note that there are no points on the graph for ; 8 (quadrant II), since sin(28) 0 for such values. Thepoipointsntsonfrom Tableare5, obtai wherened by0,usiareng symmetry. plotted inFiFiggureure30(b) 30(a).shows The remaining the graph the final graph.

± 1 .9 0

7T

= 2'

=

0

Yt

Yt

Typical graph

I

center on the line 8 = �, 2 radius a

I

= ± 2 ay,

r = ±2asin 8,

a> 0

-

a> 0

a> 0

Yt

....'---� .

--""

-

X

-

tangent to the polar axis,

x2 +

a> 0

-+ +---1I----l�

--

Passing through the pole,

x

Other Equations Name

Cardioid

Polar equations

r = a ± a cos Ii, r =

Lima 0

a> 0

r = a ± b cos 8, r = a ± b sin 8,

Yt

LimaO

a> 0

---�"":;""----1� x

r = a ± b sin 8,

Yt

0 < a 0

r = a cos(21i),

a> 0

Yt

---1� ------3I4E---+

X

a> 0

a> 0

--;-----:;JE---1I-�

x

S ketching Quickly

a polarofequati on inbyvolmaki ves onlngyusea siofneTabl (or ecosi7, nperie) functi aIfsketch its graph odicity,on,andyoua can shortquitablckley. obtain E XA M P L E 1 3

Sketch in g the Graph of a Polar Equation Quickly

Graph the equation: r 2+ 2 sine recogni z e the pol a r equati o n: Its graph is a cardi o i d . The period of si n is 21T, We formofaatablcardie usioidngase varie es 21T, theepointsand(rFi,e),gureand32. sketch theso wegraph fromcompute to 21T.r,SeeplotTabl =

Solution

0 ::5

::5

0

8

e

SECTION 10.2


735

Figure 32

Table 8 (J

r =

0

2 + 2(0) = 2

71"

2 + 2 sin (J

2 + 2(1) = 4

2 71"

2 + 2(0) = 2

371"

2 + 2( -1) = 0

2

2 + 2(0) = 2

271"

•

For those of polyouarwhoequatiareonsplainnis innorder. g to study calculus, a com mentInabout one i m portant rol e of r coordiof naates, theon.equati onit requires + l 1, whose graph is the unit cigraph rcle, ofisrectangul notthe theunitagraph functi In fact, two functions to obtain the circle: Calculus Comment

x2

Yl

=

�

Y2

Upper semicircle

=

=

=

-

�

Lower semicircle

In polar coordionn.ates,Thattheis, forequatieachonchoice1,ofwhosetheregraph iys aloneso correspondi the unit circlnge,valdoesue defi n e a functi i s onl ofopportuni that ist,y to express l . Since many problems in calculus require the use of functions, the nonfunctiy useful ons in.rectangular coordinates as functions in polar coordiNotenatesalsbecomes extremel o thatntheates.vertical-line test for functions is valid only for equations in rectangular coordi r

r,

r

e

=

l n5;torical Feature ....

P

olar coordinates see m to have bee n in ve nted by Jakob Be rnoulli (1654-1705) in

a bout 1691, although, a s with most such ideas, earlier trace s of the notion exist. Early users

mid-1800s, applied mathe maticia ns realized the tremendous simplif ica tio n that pola r coordinates make possible in the descr iption of objects with ci rc ular or cyl i n dr ical sym metry. From then o n their use beca me widespread.

of calc ul us remained committed to rectang ular

Jakob Bernoulli

(7654-1 705)

coordinates, a n d polar coordinates did not be come w i dely used until the early 1800s. Eve n then, it was mostly geometers who used them fo r descri bing o d d c urves. Final ly, a bout the

10.2 Assess Your Understanding 'Are You Prepa red?' Answers are given at the en d of these exercises. If you get a wrong answel; read the pages listed in red. 1. If the rectangular coordinates of a point are (4, - 6 ) , the point symmetric to i t with respect to the origin is . (pp. 1 67- L 68)

4. Is the sine function even, odd, or neither? ( pp. 556-557)

__

2. The difference formula for cosine is cos ( A - B) = . (p. 627) 3. The standard equation of a circle with center at ( -2, 5 ) and radius 3 is . ( pp. 1 89- 1 93) __

__

5.

. Slll 4= 571"

271" 6. cos 3

=

. (pp. )40-547) _

__

. (pp. 540-547)

__

736

CHAPTER 10


Concepts a n d Vocabulary 7. An equation whose variables are polar coordinates is called a(n) . 8. Using polar coordinates ( r,e), the circle x2 + l = 2x takes the form 9. A polar equation is symmetric with respect to the pole if an equivalent equation results when r is replaced by . __ __

10. True o r False T h e tests for symmetry in polar coordinates are necessary, but not sufficient. 11. True or False the pole.

The graph of a cardioid never passes through

12. True or False All polar equations have a symmetric feature.

__

Skill B u i l d i n g In Problems 1 3-28, transform each polar equation to an equation in rectangular coordinates. Then identify and graph the equation. 13. r

=

4

14.

r

17. r sin 13 = 4

18. r cos 13

2 1. r = 2 cos 13

22. r

25. r sec 13

=

4

7T

= 2

=

=4

7T

15. 13 = 3

16. 13 = - 4

19.

20. r sin 13

r cos 13 = -2

=

-2

2 sin 13

23. r = -4 sin 13

24. r = -4 cos 13

26. r csc e = 8

27. r csc 13 = - 2

28. r sec 13 = -4

In Problems 29-36, match each of the graphs (A) through ( H) to one of the following polar equations. 29. r = 2

30. 13 = '!!.. 4

31. r = 2 cos 13

33. r

34. r

35. 13

=

1 + cos 13

=

2 sin 13

( A)

(6)

(E)

(F)

=

37T 4

32. r cos 13 36.

-

r

sin 13

=2 =

2

(C)

(D)

(G)

(H)

In Problems 37-60, identify and graph each polar equation. 37. r = 2

+

2 cos 13

38. r = 1 + sin 13

39. r = 3 - 3 sin 13

41. r = 2

+

sin 13

42. r

=

2 - cos 13

43.

46. r

=

1 - 2 sin 13

47. r

45. r

=

1 + 2 sin 13

r

= 4 - 2 cos 13 =

2 - 3 cos 13

49. r = 3 cos( 2e)

50. r = 2 sin(3e)

51. r = 4 sin(Se)

53. r2

54. r2

sin(2e)

55. r

=

3 + cos 13

59. r

=

=

9 cos(2e)

57. r = 1 - cos 13

58. r

=

=

44. r = 4 + 2 sin 13 48. r

=

2

+

4 cos 13

52. r = 3 cos( 4(3) 56. r = 3°

2° 1

40. r = 2 - 2 cos 13

- 3 cos 13

60. r

=

4 cos (3e)

SECTION 10.2


73 7


62.

61.

x

9-=0 8 = 5TI 4

9 = 32TI

64.

63.

8 = 32TI

In Problems 65-74, graph each polar equation. 2 (parabola) 65. r = 1 - cos e ---

66.

r = 1 - 22 cos e (hyperbola) ----

67.

r = 3 - 21 cos e (ellipse)

6S.

r = 1 - 1cos e (parabola)

69.

r = e,

70.

r = e3 (reciprocal spiral)

71.

r = csc e - 2,

72.

r = sin e tan e (cissoid)

73.

r = tan e,

74.

r = cos 2e

----

e

2:

(spiral ofArchimedes)

0

0 < e < 7T

-- < e
=

E XA M P LE 2

=

+

+

+

=

+

=

+

= V3 +

•


Find Prod u cts and Quotients of Com p l ex N u m bers in Po l a r Form

The polarandformquotiof eantscompl ex number provides an alternative method for finding products of compl ex numbers. Let z] rThen l ( cosel i sineJ) and Z2 r2(cose2 i sin (2) be two complex numbers. +

=

+

=

(5)

If Z2 0, then *'

(6)

r

I n Word s The magnitude of a complex numr ber z is r and its argument is e, r so when r r z = r (cos e + i sin e) r the magnitude of the product r (quotient) of two complex numbers equals the product (quor tient) of their magnitudes; the r argument of the product (quor tient) of two complex num bers is r determined by the s u m (differr ence) of their arguments. r

E XA M P L E 3

�

�------�

(see ProblWeemwi66).l prove formula (5). The proof of formula (6) is left as an exercise Z,Z2 [rl(COSel isine])][r2(cose2 isin (2)] r]r2[(cosel sin (1 )(cose2 i sin (2)] r,r2[(coseJ cose2 - sinel sin (2) i(sine, cose2 cose1 sin (2)] rlr2[cOS(el (2) i sin Ce] (2)] Let's look at an example of how this theorem can be used.

Proof

+

=

=

+

+ i

+

=

+

+

=

+

+

+

•

Finding P roducts and Quotients of Complex N u m bers in Polar Form

=

If(leave z your 3(cosanswers 20° i isinnpol20°)ar form): and HI 5(cos 100° i sin 100°), find the following (a) ZHI (b) +

+

=

!:.. HI

SECTION 10.3 The Complex Plane; De Moivre's Theorem

(a) zw = [3(cos 20° i sin 200)J[5(cos 100° i sin 1000)J = (3 ' 5)[cos(20° 100°) i sin (20° 100°) J 15(cos 120° i sin 120°) (b) wz 5(3(coscos100°20° ii sisinn 20°) 100°) 3 [cos(20° - 100°) sin(20° - 100°) J = "53 [cos(-80°) i sine -80°) J 2( 5 cas 280° i sin 280°)

Solution

+

+

+

=

741

+

+

+

Apply equation (5)

+ +

=

= "5

Apply equation (6)

+ i

+

=

1-


he Babylo nians, Greeks, and Ara bs co nsid ered square roots of negat ive quantities to be impossi ble and equations with com plex solutio ns to be unso lvable. The fi rst hint

T

worked with complex nu m be rs without much belief in their actual ex istence. In 1 673, John Wallis appears to have bee n t he first to suggest the graphical rep rese ntatio n of co mplex nu m be rs, a truly significant idea that was not pu rsued fu rther u n t il about 1 800. Severa l people, includ

that there was some co n nection between rea l

ing Kar l Friedrich Gauss (1 777- 1 855), then rediscovered t he idea, and graphical rep rese ntatio n helped to establish complex nu mbe rs as equal me m be rs of the nu m be r family. In practical appl icatio ns, co mp lex num

solutions o f equations and co mplex nu m be rs came when Giro lamo Card a no ( 1 50 1 - 1 576) and Tar t aglia ( 1 499- 1 5 57) fou nd real roots of cubic

John Wallis

equations by t a ki n g cube roots of complex quan tit ies. For ce nturies thereafter, mathematicians

H i sto rical Pro b l e m s

bers have found their g reatest uses i n the study of alte rnati ng cu rre nt, whe re they are a co m mo nplace tool, and in the fie ld of subatomic physics.

1 . The quadratic for mula will work perfectly well if the coefficie nts a re co mplex nu mbers. Solve t he fol lowing usi ng De Moivre's Theore m w he re

necessary.

[Hint: The answers a re "nice."]

(a) Z 2

- (2 + 5i)z - 3 + 5i

=

( b)

0

Z2

- (1 + i)z - 2 - i = 0

10.3 Assess your Understanding

Answers are given at the end of these exercises. If you. get a wrong answel; read the pages listed in red. 3. The sum formula for the cosine functi o n is cos (A + 8) The conjugate of -4 - 3i is . (pp. 1 09-114) . (p. 627) The sum formula for the sine function is sin ( A + 8) 4. sin 12 0° ; cos 240° . (pp. 540-547) . (p. 630)

'Are You Prepared?' 1. 2.

__

__

=

__

__

=

__

Concepts a n d Voca b u l a ry 5. When a complex number z is written in the polar form z = r(cos e + i sin e), the nonnegative number r is the or of z, and the angle e, 0 s; e < 271", is the of z .

__

__

6.

__

Theorem can b e used t o raise a complex number t o a power.

__

7. Every non-zero complex n umber will have exactly cube roots.

__

8. True or False De Moivre's Theorem is useful for raising a complex number to a positive i n teger power. 9. True or False Using De Moivre's Theorem, the square of a complex number will have two answers. 10. True or False unique.

The polar form of a complex number is

S kill Building

[n Problems 1 1-22, plot each complex nu.mber in the complex plane and write it in polar form. Express the argument in degrees. 12. -1 + i ' 1 1. 1 + i 14. 1 - V3i 13. V3 - i 15.

-3i

16. -2

19.

3 - 4i

20. 2 +

V3i

17.

4 - 4i

18.

9V3 + 9i

21.

-2 + 3i

22.

Vs

In Problems 23-32, write each complex number in rectangular form. 24. 3( cos 210° + i sin 210°) 23. 2( cos 120° + i sin 120°)

-i

( 771" + 771")

25. 4 cos -=!

i sin 4

SECTION 10.3

27. 3 ( 3 7T

( 57T i . 657T)

26. 2 cos 6 +

Sin

29. 0.2( cos 100° + 31.

2

( 7T cos

18

+

cos T

+

37T)

( 7T

28. 4 cos "2 +

i sin T

i sin 100°)

30. 0.4( cos 200°

. 7T )

i Sin 1 8

32.

3 ( 7T cos

10

+

+

36.

39.

=

z = 2 ( cos 80° + i sin 80° ) w = 6( cos 200° + i sin 200° )

( i = 2 ( 7T

z=2

37.

W

cos cos

-

10

+ +

z = 2 + 2i w = V3 - i

i sin

=

1

=

1

( 37T i . 837T) ( 1697T i 1697T )

4 cos 8 +

sin

+

sin

=2

cos

-i - V3i

[ ( 7T

43. 2 cos

. ]6 [ ,v3;:: ( 1857T i 1857T ) +

Z =

W

42. [3(

cos

z = 3 ( cos 130° + i sin 1300) w = 4( cos 270° + i sin 270°)

38.

In Problems 41-52, write each expression in the standard form a + bi. cos 80° + i sin 80° ) ] 3

48.

7T )

i sin 10 z w

7T)

i sin "2

i sin 200° )

35.

i) 7T )

40.

745

i sin 1 0

In Problems 33-40, find zw and �w . Leave your answers in polar form. 34. z = cos 1 20° + i sin 1 20° 33. 2 ( cos 40° + i sin 40°) w cos 100° + i sin 100° w = 4( cos 20° + i sin 20° ) z =

The Complex Plane; De Moivre's Theorem

Sin

49.

(1

10

+

7T ) J5

i sin 1 0

- i)5

In Problems 53-60, find all the complex roots. Leave your answers in polar form with the argument in degrees. 54. TIle complex fourth roots of V3 - i 53. TIle complex cube roots of 1 + i 55. The complex fourth roots of 4 - 4 V3i

56. The complex cube roots of -8

57. The complex fourth roots of -16i

58. The complex cube roots of -8

59. The complex fifth roots of i

60. The complex fifth roots of

- 8i

-i

Applications a n d Exten sions 61.

Find the four complex fourth roots of unity them.

62. Find the six complex sixth roots of unity them.

( 1 ) and plot ( 1 ) and plot

63. Show that each complex nth root of a nonzero complex num ber w has the same magnitude.

64. Use the result of Problem 63 to draw the conclusion that each complex nth root lies on a circle with center at the origin. What is the radius of this circle? 65. Refer to Problem 64. Show that the complex nth roots of a nonzero complex number w are equally spaced on the circle. 66. Prove formula (6).

'Are You Prepa red?' An swers 1.

-4 +

3i

2. sin A cos B

+

cos A sin B

3. cos A cos B - sin

A

sin B

4.

V3

2 '

1

-' -

2

746

CHAPTER 10


10.4 Vectors

OBJECTIVES 1 Graph Vectors (p. 748)

2 Find a Position Vector (p.

749)

3 Add and S u btract Vectors Alge b ra i cally (p. 750)

4 Find a Scalar Multiple and the Magnitude of a Vector (p.

5 Find a Unit Vector (p. 75 1 )

75 1 )

6 Find a Vector from Its Direction and Magnitude (p. 752) 7 Analyze Objects in Static Equilib rium (p. 753)

Inquanti simplety thatterms,has abothvectormagni(deritudevedandfromdirecti theoLati nis customary meaning "to carry"a vector ) is a n. It to represent byandusithengarrowhead an arrow. Theindilceatesngththeof direction the arrowofrepresents the magnitude of the vector, the vector. icann physics can be represented bythatvectors. Fortheexampl e, theof velmovement; oMany city of quantities antheaircraft be represented by an arrow poi n ts in direction lengthif theof theaircraft arrowchanges represents speed. If theintroduce aircraftanspeeds up,in thewe lnew engthen the arrow; di r ecti o n, we arrow See Filginuree segments 39 . Based on this representation, it is not surprising that vectorsdirecti andodin.rected are somehow related. vehere,

Figure 39

Geo metric Vectors P

Q

Ifbothandandare two distinct points in the xy-plane, there is exactly one line containing [Figure 40 ( a)] . The points on that part of the line that joins to incl formproceed what is calfromled theto linewesegment [Figure 40 ( b If we order theto upoidiornngatsgeometric soandthat they have a directed line segment from vector, which we denote by PQ. In a directed line segment PQ, we call the initial point and the terminal point, as indicated in Figure 40(c) . P

Q,

P

Q

Q,

P

Q

P

Q,

PQ

Figure 40

/

P (a) Line containing P and a

( b) Line seg ment PO

)].

P

/ �� /'

Q,

P

O.

I n itial point

inal

(c) Directed line segment PO P

P

ofis,theit isditherectedlength line ofsegment PQ is the distance from the poin,t tofromtheThepoitomagni nt Iftudethat the l i n e segment. The direction of is a vector v':, has the same magnitude and the same direction as the directed line segment PQ, we write Thegnedvector v whose magnitude is 0 is called the zero vector, The zero vector is assiTwo no di vectors rectiandon.ware equal, written v=w if theyForhave theesame magnivectors tude andshown the isame directi41 have on. the same magnitude and exampl , the three n Figure the same direction, so they are equal, even though they have different initial points P

Q.

Q;

PQ

v = PQ

-->

Figure 41

O.

v

'" Boldface letters will be used to denote vectors, to distinguish them from numbers. For handwritten work, an arrow is placed over the letter to signify a vector.

SECTION 10.4

Vectors

747

andasdifanferent terminal poiinnmind ts. Asthata result, we find(vectors) it usefulareto think ofif they a vector simthe plysame arrow, keepi n g two arrows equal have direction and the same magnitude (length). The of twopoivectors isv coidefinncieddesaswifolthlothews: iWenitiapositi on ofthew,vectors v and wFigure sosum thatvtheThewterminal n t of l point as shown in vector v + w is then the unique vector whose initial point coincides wiofw.th the initial point of v and whose terminal point coincides with the terminal point Vector addition is commutative. That is, if v and w are any two vectors, then Ad d i n g Vectors +

Figure 42 , Terminal point of' w

...}--;: "" , /( k Initial point of v

42.

v+w=w+v Figure 43

Figureway43of sayi il ustrates s fact.te si(Observe that the commutative another ng that thiopposi des of a parallelogram are equal andproperty parallel.is) Vector addition is also associative. That is, if u, v, and w are vectors, then u + v + w) = u + v + w Figure il ustrates the associative property for vectors. The zero vector 0 has the property that (

Figure 44

(u + v) +

w =

u + (v +

w

)

(

)

44

v+O=O+v=v

for anyIf v vector v. is a vector, then -v is the vector having the same magnitude as v, but whose direction i s opposi t e to v, as shown in Figure Furthermore, 45.

Figure 45

v + ( - v) = O

If v and w are two vectors, we define the difference v w as v - w v + (-w) Figure illustrates the relationships among v, w, v + w, and v - w. When dealing wityh magni vectors,tude.we Exampl refer toereal numbers asofscalars. Scalarstieares arequanti ties that have onl s from physi c s scal a r quanti tem perature, speed, and time. We now define how to multiply a vector by a scalar. If is a scalar and v is a vector, the scalar multiple av is defined as follows: If 0, av is the vector whose magnitude is times the magnitude of v and whose direction is the same as v. Ifv and0' whose 0, av is the vector whose magnitude is 10'1 times the magnitude of direction is opposite that of v. or if If = v 0, then av = O. .J -

=

Figure 46

46

M u ltiplying Vectors by N u m bers

DEFINITION

a

1.

a

O. Similarly, because the dis tance from V to the directrix D is also a and, because D must be perpendicular to the x-axis (since the x-axis is the axis of symmetry), the equation of the directrix D must be x = -a. Now, if P = (x, y) is any point on the parabola, P must obey equation (1):

Figure 4

0: X= -a

1y

d(F, P)

=

d(P, D)

So we have x

vex - a)2 + l ( x - a) 2

x2 - 2ax

+

=

l

=

a2 + l

=

l

=

+

Ix

+

al

Use the Distance Formula.

Square both sides. (x + a ? x2 + 2ax + a2 Remove parentheses.

4ax

Simplify.

CHAPTER 11

774

Analytic Geometry

THEOREM

Equation of a Parabola: Vertex at (0, 0), Focus at (a, 0), a > 0

The equation of a parabola with vertex at (0, 0), focus at ( a, 0), and directrix x = -a, a > 0, is l

=

4ax

(2)

I�

�----------------------------------�

F in d in g the E q uation of a Parabola and G raphing It

E XA M P L E 1

Find an equation of the parabola with vertex at (0, 0) and focus at (3, 0). Graph the equation. Solution

Figure 5

D:x= -3

The distance from the vertex (0, 0) to the focus (3, 0) is a equation (2), the equation of this parabola is l l

Y 6

= =

4ax 12x a

=

=

3. Based on

3

To graph this parabola, it is helpful to plot the points on the graph directly above or below the focus. To locate these two points, we let x = 3. Then v (0,0)

-6

6x

l y

=

=

12x ±6

=

12(3)

=

36 Solve for y.

The points on the parabola directly above or below the focus are (3, 6) and (3, -6). These points help in graphing the parabola because they determine the "opening." See Figure S.

•

-6

In general, the points on a parabola l = 4ax that lie directly above or below the focus ( a, 0) are each at a distance 2a from the focus. This follows from the fact that if x = a then l 4ax = 4a2 , so y = ±2a. The line segment joining these two points is called the latus rectum; its length is 4a. =

C O M M E NT

tions ""

Yl

=

To graph the parabola y 2 = 12x discussed in Example 1, we need to g raph the two func • V12x and Y2 = V12x Do this and compare what you see with Figure 5.

!,--

-

Now Work

.

PROB L EM 1 9

By reversing the steps we used to obtain equation (2), it follows that the graph of an equation of the form of equation (2), l = 4ax, is a parabola; its vertex is at (0, 0 ) , its focus is at (a, 0), its directrix is the line x = - a, and its axis of symmetry is the x-axis. For the remainder of this section, the direction "Analyze the equation" will mean to find the vertex, focus, and directrix of the parabola and graph it. EXAM PLE 2

Analyzing the E q uation of a Parabola

Analyze the equation: l

Figure 6

D: x= - 2

Sol ution Y

5

-5

5

-5

=

8x

The equation l = 8x is of the form l = 4ax, where 4a = 8, so that a = 2. Consequently, the graph of the equation is a parabola with vertex at (0, 0) and focus on the positive x-axis at (a, 0) = (2, 0 ) . The directrix is the vertical line x = -2. The two points that determine the latus rectum are obtained by letting x = 2. Then l = 16, so y = ±4. The points (2, -4) and (2, 4) determine the latus rectum. See Figure 6 for the graph.

•

Recall that we arrived at equation (2) after placing the focus on the positive x-axis. If the focus is placed on the negative x-axis, positive y-axis, or negative y-axis, a different form of the equation for the parabola results. The four forms of the equa tion of a parabola with vertex at (0, 0) and focus on a coordinate axis a distance a

SECTION 1 1 .2

The Parabola

775

from (0,0 ) are given in Table 1 , and their graphs are given in Figure 7. Notice that each graph is symmetric with respect to its axis of symmetry. Table 1

Equations of a Parabola: Vertex at (0,0); Focus on an Axis; a > 0 Focus

Vertex

(0,0) (0,0)

Directrix

(a, 0)

x = -a

(-a,O)

x = a

(0, a)

(0,0)

y = -a

(0, -a)

(0,0)

y=a

Equation

Description

1= 4ax 1=

Axis of symmetry is the x-axis, opens right

- 4ax

Axis of symmetry is the x-axis, opens left

x2 = 4ay x2 = -4ay

Axis of symmetry is the y-axis, opens up Axis of symmetry is the y-axis, opens down

Figure 7 D: x = a

D: x = -a y

y.l.

y

F= (0, a )

v

D: y = a

x

x

x D:y = -a

F= (0, -a)

l = 4 ax

(a)

(b)

E X A M PLE 3

2 (c) x = 4ay

- 4 ax

2 (d) x = -4ay

Analyz i n g the E q uation of a Parabola

Analyze the equation: x2 = - 1 2 y

Figure 8

S o l ut i o n

y 6

l=

D:y = 3

V

The equation x2 = - 1 2y is of the form x2 = -4ay, with a = 3. Consequently, the graph of the equation is a parabola with vertex at (0,0 ), focus at (0, -3), and directrix the line y = 3. The parabola opens down, and its axis of symmetry is the y-axis. To obtain the points defining the latus rectum, let y = -3. Then x2 = 36, so x = ± 6. The points (-6,-3 ) and ( 6,-3) determine the latus rectum. See Figure 8 for the graph.

•

"-'l!C:=m&- Now Work E X A M PLE 4

PROB L EM 3 9

F i n d i n g the E q uati o n of a Parabola

Find the equation of the parabola with focus at (0,4 ) and directrix the line y = -4. Graph the equation. S o l ut i o n

Figure 9

Y 10

A parabola whose focus is at (0,4 ) and whose directrix is the horizontal line y = -4 will have its vertex at (0,0). (Do you see why? The vertex is midway between the focus and the directrix.) Since the focus is on the positive y-axis at (0,4), the equation of this parabola is of the form x2 = 4ay, with a = 4; that is, x2

x

-6

D: y = -4

=

4ay

=

i

4(4)y = 16y

a=4

The points ( 8,4) and (-8, 4) determine the latus rectum. Figure 9 shows the graph of x2 = 16y.

•

776

CHAPTER 1 1

Analytic Geometry

F in d in g the E q uation of a Parabol a

E XA M P L E 5

Find the equation of a parabola with vertex at (0, 0 ) if its axis of symmetry is the

( )

x-axis and its graph contains the point - �, 2 . Find its focus and directrix, and 2 graph the equation. The vertex is at the origin, the axis of symmetry is the x-axis, and the graph contains a point in the second quadrant, so the parabola opens to the left. We see from Table 1 that the form of the equation is

Solutio n

i Because the point

( �, 2) -

=

0: X = 2

4ax

is on the parabola, the coordinates x

��t satisfy the equation. Substituting x

Figure 10

-

=

-

� and

y =

=

-

�,y

=

2

2 into the equation, we find

1 . 2 = -4ax', x = - - y = 2 Y 2' a =

(0, 0)

5

2

The equation of the parabola is

x

i

=

-4(2)x

=

-8x

y

The focus is at ( -2,0) and the directrix is the line x = 2. Letting x = -2, we find i = 1 6, so = ±4. The points ( -2, 4 ) and ( -2, -4) determine the latus rectum. See Figure 10. • 'i'E=--

2

Now Work

PROB L EM 27

Analyze Parabolas with Vertex at (h, k)

If a parabola with vertex at the origin and axis of symmetry along a coordinate axis is shifted horizontally h units and then vertically k units, the result is a parabola with vertex at (h, k) and axis of symmetry parallel to a coordinate axis. The equations of such parabolas have the same forms as those in Table 1, but with x replaced by x - h ( the horizontal shift ) and replaced by - k ( the vertical shift ) . Table 2 gives the forms of the equations of such parabolas. Figures l1 ( a) -( d ) illustrate the graphs for h > 0, k > O.

y

Table 2

y

Equations of a Parabola: Vertex at (h, k); Axis of Symmetry Parallel to a Coordinate Axis; a > 0 Vertex

Focus

Description

Equation

Directrix

(h,k)

(h + o,k)

x = h

-

0

(y-k)2 = 40(x-h)

(h,k)

(h-Q,k)

x = h +

o

(y

(h,k)

(h,k + 0)

y= k-o

(x-h)2 = 40(y-k)

Axis of symmetry is parallel to the y-axis, opens up

(h,k)

(h,k - 0)

Y =k + 0

(x-h)2 =

Axis of symmetry is parallel to the y-axis, opens down

- k)2

=

- 40(x-h)

- 40

(y - k)

Axis of symmetry is parallel to the x-axis, opens right Axis of symmetry is parallel to the x-axis, opens left

The Parabola

SECTION 1 1 .2

Figure 11

y Axis of symmetry y=k

D: X = !

/7 - a

x= /7 + a

D:

x

x

2 (b) (y - k ) = - 4a(x - h)

(a) (y - k ) = 4a(x - h)

2

Axis of symmetry X=h

Y

Axis of symmetry X=h

F=(h, k + a)

y D:y = k + a

V = (h, k)

V=(h, k)

x

x D:

'\

F=(h, k - a)

y= k - a

2

(c) (x - h) = 4a(y - k)

E X A M PLE 6

777

2 (d) (x - /7) = -4a(y - k)

F i n d i n g the E q uation of a Parabola, Vertex Not at the Origin

Find an equation of the parabola with vertex at ( -2 , 3 ) and focus at (0, 3 ) . Graph the equation. S o l ution Figure 12 D:

X= - 4

Axis of symmetry y= 3

V=( - 2 , 3)

The vertex (-2,3) and focus (0, 3 ) both lie on the horizontal line y = 3 ( the axis of symmetry ) . The distance a from the vertex ( -2,3) to the focus (0, 3) is a = 2. Also, because the focus lies to the right of the vertex, we know that the parabola opens to the right. Consequently, the form of the equation is (y - k ) 2 = 4a(x - h ) where ( h , Ie ) = (-2, 3 ) and

(y - 3? = 8(x + 2 ) 6x

-4

= 2 . Therefore, the equation is

(y - 3f = 4'2[x - ( -2 )]

F = (0, 3) -6

a

I f x = 0 , then (y - 3 )2 = 16. Then y - 3 = ±4, s o Y = - lor y = 7. The points (0, -1) and (0, 7) determine the latus rectum; the line x = -4 is the directrix. See Figure 12. •

�'l!a _ __

- Now Work

PROB L EM 2 9

Polynomial equations define parabolas whenever they involve two variables that are quadratic in one variable and linear in the other. To analyze this type of equation, we first complete the square of the variable that is quadratic. E XAM P L E 7

Analyz i n g the E q uati o n of a Parabola

Analyze the equation: x2 + 4x - 4y =

°

778

CHAPTER 11

Analytic Geometry

To analyze the equation x2 + 4x - 4y variable x. x2 + 4x - 4y = 0

Solution Figure 13

Axis of symmetry x = -2

y

x2 + 4x

4

x2 + 4x + 4 (x + 2) 2

=

0, we complete the square involving the

=

4y

Isolate the terms involving x on the left side.

=

4y + 4

Complete the square on the left side.

4(y + 1) Factor. This equation is of the form ( x - h ) 2 = 4a( y - k ) , with h -2, k -1, and a = 1. The graph is a parabola with vertex at (h, k ) = ( -2, -1) that opens up. The focus is at ( -2, 0), and the directrix is the line y = -2. See Figure 13.

4 x

=

=

_3D:y=-2

k � 1U-

3

Now Work

=

•

PROB L EM 47

Solve Applied Problems Involving Parabolas

Parabolas find their way into many applications. For example, as we discussed in Section 4.4, suspension bridges have cables in the shape of a parabola. Another property of parabolas that is used in applications is their reflecting property. Suppose that a mirror is shaped like a paraboloid of revolution, a surface formed by rotating a parabola about its axis of symmetry. If a light (or any other emitting source) is placed at the focus of the parabola, all the rays emanating from the light will reflect off the mirror in lines parallel to the axis of symmetry. This prin ciple is used in the design of searchlights, flashlights, certain automobile headlights, and other such devices. See Figure 14. Conversely, suppose that rays of light (or other signals) emanate from a distant source so that they are essentially parallel. When these rays strike the surface of a parabolic mirror whose axis of symmetry is parallel to these rays, they are reflected to a single point at the focus. This principle is used in the design of some solar ener gy devices, satellite dishes, and the mirrors used in some types of telescopes. See Figure 15.

EXAMPLE 8

Figure 14

Figure 1 5

Search Light

Telescope

Satel l ite Dish

A satellite dish is shaped like a paraboloid of revolution. The signals that emanate from a satellite strike the surface of the dish and are reflected to a single point, where the receiver is located. If the dish is 8 feet across at its opening and 3 feet deep at its center, at what position should the receiver be placed? That is, where is the focus? S o l ut i o n

Figure 16(a) shows the satellite dish. We draw the parabola used to form the dish on a rectangular coordinate system so that the vertex of the parabola is at the origin and its focus is on the positive y-axis. See Figure 16(b).

SECTION 11.2

The Parabola

779

F igure 16 y

1-+-- 8'

(- 4,3)

T 3'

-

4

-

3 2 -

-

1

t

2 3 4 X

0 (b)

(a)

The form of the equation of the parabola is x2 = 4ay and its focus is at (0, a ) . Since (4, 3) is a point on the graph, we have 42

4a(3) x = 4, Y = 3 4 a=Solve for a. 3 =

�h;�:�;��er should be located 13 feet from the base of the dish, along its axis of 1

�'IJ!::=::::;:;� ;z Now Work

• PROB L EM 6 3

11.2 Assess Your Understanding 'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red. 1.

The formula for the distance d from PI P2 = (X2 ,)'2 ) is d = . (p. 157)

=

(Xl, yd to

4.

__

2. 3.

To complete the square of x2 - 4x, add

The point that is symmetric with respect to the x-axis to the point (-2, 5) is . (pp. 167-169) To graph y = (x - 3 ) 2 + 1, shift the graph of y = x2 to the right units and then 1 unit. (pp. 252-260) __

. (pp. 99-101) Use the Square Root Method to find the real solutions of (x + 4) 2 = 9. (pp. 98-99) __

5.

__

__

Concepts and Vocabula ry 6.

7.

A(n) is the collection of all points in the plane such that the distance from each point to a fixed point equals its distance to a fixed line. __

The surface formed by rotating a parabola about its axis of symmetry is called a .

If a light is placed at the focus of a parabola, all the rays reflected off the parabola will be parallel to the axis of symmetry.

9. True or False

10. True or False

parabola.

______

The vertex of a parabola is a point on the parabola that also is on its axis of symmetry.

The graph of a quadratic function is a

8. True or False


(a) i (b) x2 11.

=

=

J 1-18,

4x 4y y

the graph of a parabola is given. Match each graph to its equation.

(c) i (d) x2

=

=

(e) (y - 1)2

-4x -4y

(f) (x +

13.

12.

If y 2

( 1,

1)

-2 -2 -2

2

x

-2 -2

= =

4(x - 1) 4(y + 1)

(g) ( y - 1)2 (h) (x + 1)2 14.

y 2

=

=

-4(x - 1) -4(y + 1)

780

CHAPTER 1 1

Analytic Geometry

16.

y 2

15.

2 X

-2

In Problems

19-36, find

y 2

2 X

2 X

-2

the equation of the paraboLa described. Find the two points that define the Latus rectum, and graph the equation.

. 19. Focus at (4, 0); vertex at (0, 0)

20. Focus at (0, 2); vertex at (0, 0) 22. Focus at (-4, 0); vertex at (0, 0)

21. Focus at (0, -3); vertex at (0, 0) 23. Focus at ( - 2, 0); directrix the line x 25. Directrix the line y

18.

17.

2

=

2

=

24. Focus at (O,- I ); directrix the line y

1

- "2 ; vertex at (0,0)

26. Directrix the line x

=

=

1

-�; vertex at (0, 0)

27. Vertex at (0, 0); axis of symmetry the y-axis; containing the

28. Vertex at (0, 0); axis of symmetry the x-axis; containing the

29. Vertex at (2, -3); focus at (2, -5 )

30. Vertex at (4, -2); focus at (6, -2)

31. Vertex at ( - 1 , -2); focus at (0, -2)

32. Vertex at (3, 0); focus at (3, -2)

point (2,3)

33. Focus at (-3,4); directrix the line y

2

=

35. Focus at (-3,-2); clirectrix the line x

point (2, 3)

34. Focus at (2,4); directrix the line x

36. Focus at (-4,4); directrix the line y

=1

In ProbLems 3 7-54, find the vertex, focus, and directrix of each paraboLa. Graph the equation. �" 39. y-? = -16x 37. x-? = 4y 38. y2 = 8x

41. (y - 2) 2

=

8(x + 1 )

42. (x + 4)2

=

3f

=

8(x - 2)

46. (x - 2) 2

=

45. ( y

+

49. x2 + 8x 53. x2 - 4x In Problems

=

50. l - 2Y

4y - 8

=

Y +

55-62,

=

=

16(y + 2)

43. (x - 3) 2 ',, 47.

4(y - 3) 8x

-

1

?

y-

- 4y

=

-(y + 1 )

+ 4x

51. l + 2Y - x

4

54. l + 1 2y

=

57.

y

40. x2

+4

=

=

-4 =

-2

=

-4y

44. (y + 1 )2

48. x2 + 6x - 4Y + 1

0

52. x2 - 4x

0

y

56.

2y

=

y

58.

2

(0, 1 ) X

-2

X

-2

-2

- 2 f-

-2

;�') � (0, 1 ) I

-2

0

-x + 1

2

-2

=

write an equation for each parabola.

55.

59.

-4(x - 2)

=

I

2

X

60.

-2

61.

�t -2

(1, -1)

-2

x

(0, - 1 )

y

62.

2

2 -2

2

-2

�t

x

-2

�

X

fl1 O)

x

x

-2 -2

SECTION 11.2

The Parabola

781

Applications a n d Extensions 63.

64.

65.

66.

67.

Satellite Dish A satellite dish is shaped like a paraboloid of revolution. The signals that emanate from a satellite strike the surface of the dish and are reflected to a single point, where the receiver is located. If the dish is 10 feet across at its opening and 4 feet deep at its center, at what position should the receiver be placed? Constructing a TV Dish A cable T V receiving dish is in the shape of a paraboloid of revolution. Find the location of the receiver, which is placed at the focus, if the dish is 6 feet across at its opening and 2 feet deep. Constructing a Flashlight The reflector of a flashlight is in the shape of a paraboloid of revolution. Its diameter is 4 inch es and its depth is 1 inch. How far from the vertex should the light bulb be placed so that the rays will be reflected parallel to the axis? Constructing a Headlight A sealed-beam headlight is in the shape of a paraboloid of revolution. The bulb, which is placed at the focus, is 1 inch from the vertex. If the depth is to be 2 inches, what is the diameter of the headlight at its opening? Suspension Bridge The cables of a suspension bridge are in the shape of a parabola, as shown in the figure. The towers supporting the cable are 600 feet apart and 80 feet high. If the cables touch the road surface midway between the tow ers, what is the height of the cable from the road at a point 1 50 feet from the center of the bridge?

72.

73.

74.

75.

600 ft 68.

69.

70.

71.

Suspension Bridge The cables of a suspension bridge are in the shape of a parabola. The towers supporting the cable are 400 feet apart and 100 feet high. If the cables are at a height of 10 feet midway between the towers, what is the height of the cable at a point 50 feet from the center of the bridge? Searchlight A searchlight is shaped like a paraboloid of rev olution. If the light source is located 2 feet from the base along the axis of symmetry and the opening is 5 feet across, how deep should the searchlight be? Searchlight A searchlight is shaped like a paraboloid of rev olution. If the light source is located 2 feet from the base along the axis of symmetry and the depth of the searchlight is 4 feet, what should the width of the opening be? Solar Heat A mirror is shaped like a paraboloid of revolu tion and will be used to concentrate the rays of the sun at its focus, creating a heat source. See the figure. If the mirror is 20 feet across at its opening and is 6 feet deep, where will the heat source be concentrated?

Sun's

Reflecting Telescope A reflecting telescope contains a mir ror shaped like a paraboloid of revolution. If the mirror is 4 inches across at its opening and is 3 inches deep, where will the collected light be concentrated? Parabolic Arch Bridge A bridge is built in the shape of a parabolic arch. The bridge has a span of 120 feet and a max imum height of 25 feet. See the illustration. Choose a suit able rectangular coordinate system and find the height of the arch at distances of 10, 30, and 50 feet from the center.

Parabolic Arch Bridge A bridge is to be built in the shape of a parabolic arch and is to have a span of 100 feet. The height of the arch a distance of 40 feet from the center is to be 10 feet. Find the height of the arch at its center. Gateway Arch The Gateway Arch in St. Louis is often mis taken to be parabolic is shape. In fact, it is a catenary, which has a more complicated formula than a parabola. The Arch is 625 feet high and 598 feet wide at its base. (a) Find the equation of a parabola with the same dimen sions. Let x equal the horizontal distance from the cen ter of the arc. (b) The table below gives the height of the Arch at various widths; find the corresponding heights for the parabola found in (a). Height (tt)

Width (tt)

1 00

567

3 1 2.5

478

525

308

(c) Do the data support the notion that the Arch is in the shape of a parabola?

Source:

76.

Wikipedia, the free encyclopedia

Show that an equation of the form AX2 + Ey

77.

A "* 0, E "* 0

0,

is the equation of a parabola with vertex at (0, 0) and axis of symmetry the y-axis. Find its focus and directrix. Show that an equation of the form

cl

78.

=

+

Dx

=

c "* 0, D "* 0

0,

is the equation of a parabola with vertex at (0, 0 ) and axis of symmetry the x-axis. Find its focus and directrix. Show that the graph of an equation of the form Ax2 + Dx + E y + F

(a) (b) (c) (d)

=

0,

A "* 0

Is a parabola if E "* O. Is a vertical line if E 0 and D2 4AF = O. Is two vertical lines if E = 0 and D2 4AF > O. Contains no points if E 0 and D2 4A F < O. =

-

-

=

-

782 79.

CHAPTER 11

Analytic Geometry

Show that the graph of an equation of the form =

cl + Dx + Ey + F (a) (b) (c) (d)

Is a parabola if D *- O. 4CF = O. Is a horizontal line if D = 0 and E2 Is two horizontal lines if D = 0 and £2 4CF > Contains no points if D = 0 and E2 - 4CF < O.

0,

c *- o

-

-

O.

'Are You Prepared?' Answers 2. 4

3. x + 4

=

±3; { -7,

-

I}

4.

( -2, - 5 )

s.

3 ; up

11.3 The Ellipse PREPARING FOR THIS SECTION • • • •


Distance Formula (Section 2.1 , p. 157) Completing the Square (Section 1 .2, pp. 99-101 ) Intercepts (Section 2.2, pp. 1 65-166) Symmetry (Section 2.2, pp. 1 67-169) Now Work the 'Are You

• •

Circles (Section 2.4, pp. 189-193) Graphing Techniques: Transformations (Section 3.5, pp. 252-260)

Prepared?' prob lems on page 789.

OBJECTIVES 1

DEFINITION

Ana lyze Elli pses with Center at the Origin (p. 782)

2

Ana lyze Elli pses with Center at (h, k) (p. 786)

3

Solve Appl ied Problems I nvolving E l l i pses (p. 788)

An ellipse is the collection of all points in the plane, the sum of whose distances from two fixed points, called the foci, is a constant.

-.I

The definition contains within it a physical means for drawing an ellipse. Find a piece of string (the length of this string is the constant referred to in the definjtion) . Then take two thumbtacks (the foci) and stick them into a piece of cardboard so that the distance between them is less than the length of the string. Now attach the ends of the string to the thumbtacks and, using the point of a pencil, pull the string taut. See Figure 17. Keeping the string taut, rotate the pencil around the two thumb tacks. The pencil traces out an ellipse, as shown in Figure 17. In Figure 17, the foci are labeled F 1 and F2 ' The line containing the foci is called the major axis. The midpoint of the line segment joining the foci is the center of the ellipse. The line through the center and perpendicular to the major axis is the minor

Figure 17

axis.

The two points of intersection of the ellipse and the major axis are the vertices, and V2 , of the ellipse. The distance from one vertex to the other is the length of the major axis. The ellipse is symmetric with respect to its major axis, with respect to its minor axis, and with respect to its center. V1

Figure 18

y p = (x, y)

1 x

Analyze Ellipses with Center at the Origin

With these ideas in mind, we are now ready to find the equation of an ellipse in a rectangular coordinate system. First, we place the center of the ellipse at the origin. Second, we position the ellipse so that its major axis coincides with a coordinate axis. Suppose that the major axis coincides with the x-axis, as shown in Figure 18. If c is the distance from the center to a focus, one focus will be at F 1 = ( -c, 0) and the other at F2 = (c, 0 ) . As we shall see, it is convenient to let 2a denote the constant

SECTION 11.3

The

Ellipse

783

distance referred to in the definition. Then, if P ( x, y) is any point on the ellipse, we have d ( F 1 , P) + d( F2 , P) 2a Sum of the distances from P =

=

V(x + e ? + l + V( x - e ? + l V( x + e ? + l (x + e ?

+

l

=

2a

=

2a -

=

x2 + 2ex + e2 + l

=

4ex - 4a2

=

ex - a2 (ex - a2) 2 ( e2 - a2)x2 - a2 1 ( a2 - e2)x2 + a2 1

to the foci equals a constant, 2a. Use the Distance Formula.

V(x - e ? + l 4a2 - 4aV(x - e) 2 + l

Isolate one radical. Square both sides.

+ (x - e)2 + l

4a2 - 4aV(x - e)2 + l Remove pa rentheses. + x2 - 2ex + e2 + l -4aV(x - e ?

+

l

Simplify; isolate the radical.

=

-aV(x - e ? + l a2[ (x - e ? + lJ

Square both sides again.

=

a2e2 - a4

Rearrange the terms.

=

=

a2(a2 - e2)


Multiply each side by - 1 ; factor a2 on the right side.

(1) To obtain points on the ellipse off the x-axis, it must be that a > e. To see why, look again at Figure 18. Then d ( Fj , P) + d( F2 , P ) > d( Fj , F2 ) The sum of the lengths of two sides of a triangle is greater than the length of the third side.

2a > 2e d(F" p) + d(F2' p) = 2a, d(F" F2) = 2c a>e Since a > e, we also have a2 > e2, so a2 - e2 > O. Let b2 a2 - e2, b > O. Then a > b and equation (1) can be written as b2x2 + a2 1 = a2b2 =

x2 l Divide each side by a2/l. -+- 1 a2 b2 As you can verify, this equation is symmetric with respect to the x-axis, y-axis, and origin. Because the major axis is the x-axis, we find the vertices of this ellipse by 2 letting y = O. The vertices satisfy the equation \ 1, the solutions of which are a x = ±a. Consequently, the vertices of this ellipse are V I ( -a, 0) and V = (a, O). 2 The y-intercepts of the ellipse, found by letting x = 0, have coordinates (0, -b) and (0, b). These four intercepts, (a, 0), ( -a, 0), (0, b), and (0, -b), are used to graph the ellipse. =

=

=

THEOREM

Figure 1 9

y

Equation of an Ellipse: Center at (0, 0); Major Axis along the x-Axis

An equation of the ellipse with center at (0, 0 ) , foci at ( - e, 0) and (e, 0), and vertices at ( -a, 0) and (a, 0) is

(0, b) V = (-a, I

0)

V = 2

(a, 0) x

(0, - b)

x2 l + 2" a b

2"

=

1,

where a > b > 0 and b2

The major axis is the x-axis. See Figure 19.

=

a2 - e2

(2)

784

CHAPTER 11

Analytic Geometry

Notice in Figure 19 the right triangle formed with the points (0, 0), (c, 0), and (0, b ) . Because b2 = a2 - c2 (or b2 + c2 = a2 ), the distance from the focus at ( c, 0) to the point (0, b) is a. This can be seen another way. Look at the two right triangles in Figure 19. They are congruent. Do you see why? Because the sum of the distances from the foci to a point on the ellipse is 2a, it follows that the distance from (c, 0) to (0, b) is a. F i n d i n g an Equ ation of an El l i pse

E XA M P L E 1

Find an equation of the ellipse with center at the origin, one focus at (3, 0), and a vertex at ( -4, 0). Graph the equation. Solution Figure 20 x2

16

y2

7

- + - =

1

The ellipse has its center at the origin and, since the given focus and vertex lie on the x-axis, the major axis is the x-axis. The distance from the center, (0, 0), to one of the foci, (3, 0), is c = 3. The distance from the center, (0, 0), to one of the vertices, ( -4, 0 ) , is a = 4. From equation (2), it follows that b2 = a2 - c2 = 16 - 9 = 7

Y 5

so an equation of the ellipse is x2 l -+-=1 16 7

\5 \

(0, - ,f[ )

V.

2

=

x

(4

'

-5

0)

Figure 20 shows the graph. '"

Notice in Figure 20 how we used the intercepts of the equation to graph the ellipse. Following this practice will make it easier for you to obtain an accurate graph of an ellipse when graphing. COM M E NT The intercepts of the ellipse also provide information about how to set the viewing rec tangle for g raphing an ellipse. To graph the ellipse

x2 -

i

+ - = 1 7 16

discussed in Example 1, we set the viewing rectangle using a square screen that includes the intercepts, perhaps -4.5 :=; x :=; 4.5, -3 :=; Y :=; 3. Then we proceed to solve the equation for y: x2

y2

- + - = 1 7 16

y2

Subtract - from each side.

y2 = 7 1

Multiply both sides by 7.

( - �;) ) ( �)

Figure 2 1

y = ±

I

-..... ..\

-4.5 1-----+--'-- 4 . 5

\.

�

x2

- = 1 - 7 16

7 1

-

16

Take the square root of each side.

Now graph the two functions

/ Figure

21

shows the result.

!"� .

Now Work

PROB L EM 2 7

II

SECTION 11.3

The Elli pse

785

An equation of the form of equation (2), with a > b, is the equation of an ellipse with center at the origin, foci on the x-axis at (-c, 0) and (c, 0 ) , where c2 = a2 - b2, and major axis along the x-axis. For the remainder of this section, the direction "Analyze the equation" will mean to find the center, major axis, foci, and vertices of the ellipse and graph it. EXAM P LE 2

A nalyz i n g the E q u ation of an E l l i pse

2 . Analyze the equatIOn: �� 25 Solution

+

i

9= 1

The given equation is of the form of equation (2), with a2 = 25 and b2 = 9. The equation is that of an ellipse with center (0, 0 ) and major axis along the x-axis. The vertices are at ( ±a, 0 ) ( ±5, 0 ) . Because b2 = a2 - c2, we find that =

c2 = a2 - b2 = 25 - 9

The foci are at ( ±c, 0)

=

=

16

( ±4, 0). Figure 22 shows the graph. y

Figure 22

6

(0, 3) V2 = (5, 0)

/.

6 x

-6

(0, -3)

L. 'I'i: ::Z::;: lIIiOil:::o-

Now Work

• PROB L EM

17

If the major axis of an ellipse with center at (0, 0 ) lies on the y-axis, the foci are at (0, -c) and (0, c). Using the same steps as before, the definition of an ellipse leads to the following result: THEOREM

Equation of an Ellipse: Center at (0, 0); Major Axis along the y-Axis

An equation of the ellipse with center at (0, 0 ) , foci at (0, -c) and (0, c), and vertices at (0, -a) and (0, a) is x2 i + 2" b a

2"

Figure 23

y

V2 =

(0, a )

1,

where a > b > ° and b2 = a2 - c2

(3)

The major axis is the y-axis.

( b, O) ( - b, O)

x

V1 =

=

(0, - a)

E XA M P L E 3

Figure 23 illustrates the graph of such an ellipse. Again, notice the right triangle with the points at (0, 0), (b, 0), and (0, c ) . Look closely a t equations ( 2 ) and (3) . Although they may look alike, there is a difference! In equation (2) , the larger number, a2, is in the denominator of the x2-term, so the major axis of the ellipse is along the x-axis. In equation (3) , the larger number, a2, is in the denominator of the i -term, so the major axis is along the y-axis. Analyz i n g the Eq uation of an E l l i pse

Analyze the equation: 9x2 +

i

= 9

786

C HAPTER 11

Analytic Geometry

To put the equation in proper form, we divide each side by 9.

Solution

Figure 24

y 3 V2

=

x2 +

(0, 3)

( 1 , 0)

3

x

=

1

The larger number, 9 , is in the denominator of the l-term so, based on equation (3), this is the equation of an ellipse with center at the origin and major axis along the y-axis. Also, we conclude that a2 = 9, b2 = 1, and c2 = a2 - b2 = 9 - 1 = 8. The vertices are at (0, ±a) = (0, ±3) , and the foci are at (0, ±c) = (0, ±2 v'2). Figure 24 shows the graph. £!,,� >-

EXAMPLE 4

l 9

-

Now Work

'"

PROB L EM 2 1

F i n d i n g an Equation of an E ll ipse

Find an equation of the ellipse having one focus at (0, 2 ) and vertices at (0, -3 ) and (0, 3 ) . Graph the equation. Figure 25

( ,g -

,

0)

Because the vertices are at (0, -3) and (0, 3 ) , the center of this ellipse is at their midpoint, the origin. Also, its major axis lies on the y-axis. The distance from the center, (0, 0), to one of the foci, (0, 2 ) , is c = 2. The distance from the center, (0, 0 ) , to one o f the vertices, (0, 3 ) , i s a = 3. S o b l = al - cl = 9 - 4 = 5. The form of the equation of this ellipse is given by equation (3).

Solution

({S, O)

-3

3

x

-3 V1 = (0, -3)

Figure 25 shows the graph.

..

The circle may be considered a special kind of ellipse. To see why, let a equation (2) or (3). Then

=

b in

x2 l + - = 1 al al xl + l = al

-

This is the equation of a circle with center at the origin and radius a. The value of c is cl

=

a2 - bl

=

i

°

a = b

We conclude that the closer the two foci of an ellipse are to the center, the more the ellipse will look like a circle. 2

Analyze Ellipses with Center at (h, k)

If an ellipse with center at the origin and major axis coinciding with a coordinate axis is shifted horizontally h units and then vertically k units, the result is an ellipse with center at (h, k) and major axis parallel to a coordinate axis. The equations of such ellipses have the same forms as those given in equations (2) and (3), except that x is replaced by x - h (the horizontal shift) and y is replaced by y - k (the vertical shift). Table 3 gives the forms of the equations of such ellipses, and Figure 26 shows their graphs.

SECTION 11.3

The Ellipse

787

Equations of an Ellipse: Center at (h, k); Major Axis Parallel to a Coordinate Axis

Table 3

Center

(h, k)

Major Axis

Foci

Vertices

Parallel to the x-axis

(h + e, k)

(h + o, k)

(h - e, k)

(h - o, k)

(h, k + e)

(h, k + 0)

(h, k - e)

(h, k

Parallel to the y-axis

(h, k)

Figure 26

-

0)

Equation (x - W (y - W + --- = 1 , -b2 o2o > b and b2 = 02 - e2

(y - k) 2 + --- = 1 , b2 02 o > b and b2 02 - e2 (x - h) 2

--

=

y y

Major axis

-

( h - a, k)

(h, k - c)

x

x

(b)

h) 2 (y - k) 2 + -- = 1 a2 b2

(x -

--

F i n d i n g an E q u ation of an E l l i pse, Cente r N ot at the O ri gi n

E XA M P L E 5

Find an equation for the ellipse with center at one vertex at (5, -3). Graph the equation. Solution

(2, -3), one focus at (3, -3), and

The center is at (h, k) (2, -3), so h = 2 and k -3. Since the center, focus, and vertex all lie on the line y -3, the major axis is parallel to the x-axis. The distance from the center (2, -3) to a focus (3, -3) is c = 1; the distance from the center (2, -3) to a vertex (5, -3) is a = 3. Then b2 = a2 - c2 = 9 - 1 = 8. The form of the equation is =

=

=

- hf (y - k) 2 -'-(x-----=- '- + a2 b2 2 (x - 2) -'--- -------' + (y + 3) 2 8 -9

-2

V1

=

(-1 ,

1, where h

=

2' k

=

-3" a

=

3b

=

2'V�L

= 1

To graph the equation, we use the center (h, k) = (2, -3 ) to locate the ver tices. The major axis is parallel to the x-axis, so the vertices are a = 3 units left and right of the center (2, -3) . Therefore, the vertices are

Figure 27 y 2

=

(2,

-3

I

\

VI

2 ,J2)

=

(2 - 3, - 3)

= ( -1,

- 3) and V2

=

(2 + 3, -3)

= (5, -3)

x

Since c = 1 and the major axis is parallel to the x-axis, the foci are 1 unit left and right of the center. Therefore, the foci are

V2 = ( 5 , -3 )

(1, -3) and F2 = (2 + 1 , -3) = (3, -3) Finally, we use the value of b = 2V2 to find the two points above and below the

6

-3 )

(2,

+

Fl

=

(2

- 1, -3)

=

center.

-3

(2, -3 - 2\12) and (2, -3 + 2\12)

- 2 ,J2)

Figure 27 shows the graph. � = =-- -

Now Work

PROB L EM 5 5

•

788

CHAPTER 11

Analytic Geometry

EXAMPLE 6

Analyzi n g the Eq uation of an E l l ipse

Analyze the equation: 4x2 + l - 8x

+

4y + 4 = 0

We proceed to complete the squares in x and in y.

Solution

4x2 + l - 8x + 4y + 4

=

0

4x2 - 8x + l + 4y = -4

Group like variables; place the constant on the right side. Factor out 4 from the first two terms.

4(x2 - 2x) + (l + 4y) = -4 4 ( x2 - 2x + 1 ) + (l + 4y + 4) = -4

Figure 28

y

4(x - 1 ) 2 + (y + 2 ) 2 (1 , 0)

( x - 1 )2 +

x

(1 , - 2 + -[:3)

=

4

(y + 2 ) 2 = 1 4

+

4 + 4 Complete each square. Factor. Divide each side by 4.

This is the equation of an ellipse with center at ( 1 , -2) and major axis parallel to the y-axis. Since a2 = 4 and b2 = 1, we have c2 = a2 - b2 = 4 - 1 = 3. The ver tices are at (h, k ± a) = ( 1 , 2 ± 2) or (1, 0) and ( 1 , -4 ) . The foci are at (h, k ± c) = ( 1 , -2 ± \13) or ( 1 , -2 - \13) and ( 1 , -2 + \13). Figure 28 shows the graph.

(2, -2) (1 , - 2 - -[:3)

-

(1 , -4)

�i\

3

Mp- Now Work

•

P R OB l EM 47

Solve Applied Problems Involving Ellipses

Ellipses are found in many applications in science and engineering. For example, the orbits of the planets around the Sun are elliptical, with the Sun's position at a focus. See Figure 29.

Figure 29

Stone and concrete bridges are often shaped as semielliptical arches. Elliptical gears are used in machinery when a variable rate of motion is required. Ellipses also have an interesting reflection property. If a source of light (or sound) is placed at one focus, the waves transmitted by the source will reflect off the ellipse and concentrate at the other focus. This is the principle behind whispering galleries, which are rooms designed with elliptical ceilings. A person standing at one focus of the ellipse can whisper and be heard by a person standing at the other focus, because all the sound waves that reach the ceiling are reflected to the other person .

SECTION 11.3

The Ellipse

789

A W h i speri n g Gallery

EXAMPLE 7

The whispering gallery in the Museum of Science and Industry in Chicago is 47.3 feet long. The distance from the center of the room to the foci is 20.3 feet. Find an equa tion that describes the shape of the room. How high is the room at its center? Source:

Chicago Museum of Science and Industry Web site; www.msichicago. org

We set up a rectangular coordinate system so that the center of the ellipse is at the origin and the major axis is along the x-axis. The equation of the ellipse is

S o lution

Since the length of the room is 47.3 feet, the distance from the center of the room

.

47.3 2 = 23.65 feet; so a = 23.65 feet. The distance from the center of the room to each focus is = 20.3 feet. See Figure 30. Since b2 = a2 - c2 , we find b2 = 23.652 - 20.32 = 147.2325. An equation that

to each vertex (the end of the room) wlil be --

c

describes the shape of the room is given by x2

l

=1 147.2325 The height of the room at its center is b = \1147.2325 ;::; 12.1 feet. --

23.65 2

+

y 15

Figure 30

( - 23.65, 0)

\

/ 25 x (20.3, 0)

-25 \ ( - 20.3 , 0)

"1 ! �

Now Work

• PROB L EM 7 1

11 .3 Assess Your Understanding 'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red. 1.

The distance d from Pl . (p.157)

=

(2, - 5 ) to P2

2. To complete the square of x2 - 3x, add

=

(4, -2 ) is d

=

5.

__

__

3. 4.

To graph y = (x + I f - 4, shift the graph of y = x2 to the (left/right) unit(s) and then (up/down) unit(s). (pp. 252-260) The standard equation of a circle with center at (2, -3) and radius1 is . (pp.189-193)

__

. (pp. 99-101)

Find the intercepts of the equation l (pp. 165-1 66)

=

16 - 4x2 .

6.

__

__

The point that is symmetric with respect to the y-axis to the point ( -2, 5 ) is . (pp.167-169) __

Concepts and Vocabulary 7. 8. 9.

A(n) is the collection of all points in the plane the sum of whose distances from two fixed points is a constant. For an ellipse, the foci lie on a line called the axis. __

For the elliPse and

2

:2 + �5

__

= 1,

the vertices are the points

__

True or False The foci, vertices,and center of an ellipse lie on a line called the axis of symmetry. 11. True or False If the center of an ellipse is at the origin and the foci lie on the y-axis, the ellipse is symmetric with respect to the x-axis, the y-axis, and the origin.

10.

12.

True or False

A circle is a certain type of ellipse.

790

CHAPTER 11

Analytic Geometry


13-16,

x2 ? (a) "4 + y-

=

Y 4

13.

the graph of a n ellipse is given. Match each graph to its equation.

l (b) x2 + "4

1

=

1

(c) Y

14.

l x2 + "4 16

-4

x

4

=

l x2 (d) "4 + 16 15. Y 3

1

-4 In Problems

17.

=

-

21. 4x2 + i 25. x2 + l

Y 3

16.

3X

-3

x

-3

-3

1 7-26,

l � + 25 4

-

1

3

-3

x

=

l � + 9 4

1

18.

16

22. x2 + 9 l

=

=

find the vertices and foci of each ellipse. Graph each equation. -

-

=

=

l � + 9 25

19.

1

-

-

23. 4l + x 2

18

26. x2 + l

16

l 16

=

1

20. x2 +

=

8

24. 4l + 9x2

=

=

1 =

36

4

In Problems 2 7-38, find an equation for each ellipse. Graph the equation.

27. Center at (0, 0 ) ; focus at (3, 0); vertex at ( 5 , 0)

28. Center at (0, 0); focus at ( - 1 , 0); vertex at ( 3, 0)

29. Center at (0, 0 ) ; focus at (0, -4); vertex at (0, 5 )

30. Center at (0, 0); focus at (0, 1 ) ; vertex at (0, - 2 )

31. Foci a t ( ±2 , 0 ) ; length of the major axis is 6

32. Foci at ( 0 , ± 2 ) ; length o f t h e major axis i s 8

33. Focus at ( -4, 0); vertices at ( ±5, 0)

34. Focus at (0, -4); vertices at (0, ±8)

35. Foci at (0, ± 3 ) ; x-intercepts are ±2

36. Vertices at ( ±4, 0); y-intercepts are ± 1

37. Center at (0, 0); vertex at (0, 4 ) ; b

=

38. Vertices a t ( ±5, 0); c

1

In Problems 39-42, write an equation for each ellipse. Y 3

39.

Y 3

40.

3

Y 3

41.

3

x

43.

43-54,

x

3

-3

x

-3

analyze each equation; that is, find the centel; foci, and vertices of each ellipse. Graph each equation.

46. 9(x - 3f + (y + 2 )2

= =

44.

1

( x + 4)2 9

+

(y + 2)2 =

4

=

0

50. 4x2 + 3l + 8x - 6y

52. x2 + 9i + 6x - 18y + 9

=

0

53. 4x2 + l + 4y

=

0

=

45. (x + 5 )2 + 4(y - 4)2

1

" 47. x2 + 4x + 4l - 8y + 4

18

49. 2x2 + 3l - 8x + 6y + 5

55-64,

Y 3

-3

( X - 3)2 ( y + l )2 + 4 9

In Problems

2

42.

3

-3

x

-3 In Problems

=

=

5

0

48. x2 + 31 - 12y + 9

=

=

16

0

51. 9x2 + 4l - 18x + 1 6y - 1 1 54. 9x2 + l - 1 8x

=

=

0

0

find an equation for each ellipse. Graph the equation.

55. Center at (2, - 2 ) ; vertex at (7, - 2 ) ; focus at (4, -2)

56. Center at ( -3, 1 ) ; vertex at ( -3, 3 ) ; focus at ( -3, 0)

57. Vertices at (4, 3 ) and (4, 9 ) ; focus a t (4, 8)

58. Foci at ( 1 , 2 ) and ( -3, 2 ) ; vertex a t ( -4, 2 )

59. Foci at (5, 1) and ( - 1 , 1 ) ; length of the major axis is 8

60. Vertices a t (2, 5 ) and (2, - 1 ) ; c

61. Center at ( 1 , 2 ) ; focus a t (4, 2 ) ; contains the point ( 1 , 3)

62. Center at ( 1 , 2 ) ; focus a t ( 1 , 4); contains the point (2, 2 )

63. Center at ( I , 2 ) ; vertex at (4, 2 ) ; contains the point ( I , 3)

64. Center at ( 1 , 2 ) ; vertex a t ( 1 , 4 ) ; contains the point ( 2 , 2 )

=

2

SECTION 11.3

The Ellipse

79 1

In Problems 65-68, graph each function. Be sure to label all the intercepts. [ Hint: Notice that each function is half an ellipse.]

65. f(x)

=

V16 - 4x2

66. f(x)

=

V9 - 9x2

67. f(x)

68. f(x)

Applications a n d Extensions 69. Semielliptical Arch Bridge

An arch in the shape o f t h e upper half o f a n ellipse is used to support a bridge that i s to span a river 20 meters wide. TIle center of the arch is 6 me ters above the center of the river. See the figure. Write an equation for the ellipse in which the x-axis coincides with the water level and the y-axis passes through the center of the arch.

74. Semielliptical Arch Bridge A bridge is to be built in the

shape of a semielliptical arch and is to have a span of 100 feet. The height of the arch, at a distance of 40 feet from the cen ter, is to be 10 feet. Find the height of the arch at its center.

75. Racetrack Design Consult the figure. A racetrack is in the shape of an ellipse, 100 feet long and 50 feet wide. What is the width 10 feet from a vertex?

76. Semielliptical Arch Bridge An arch for a bridge over a high

70. Semielliptical Arch Bridge The arch of a bridge is a semi ellipse with a horizontal major axis. The span is 30 feet, and the top of the arch is 10 feet above the major axis. The road way is horizontal and is 2 feet above the top of the arch. Find the vertical distance from the roadway to the arch at 5-foot intervals along the roadway.

71. Whispering Gallery A h a l l 1 00 feet in length is to be designed as a whispering gallery. If the foci are located 25 feet from the center, how high will the ceiling be at the center?

72. Whispering Gallery Jim, standing at one focus of a whis

pering gallery, is 6 feet from the nearest wall. His friend is standing at the other focus, 100 feet away. What is the length of this whispering gallery? How high is its elliptical ceiling at the center?

73. Semielliptical Arch Bridge A bridge is built in the shape of a semielliptical arch. The bridge has a span of 120 feet and a maximum height of 25 feet. Choose a suitable rectangular co ordinate system and find the height of the arch at distances of 10, 30, and 50 feet from the center.

way is in the form of half an ellipse. The top of the arch is 20 feet above the ground level (the major axis). The highway has four lanes, each 12 feet wide; a center safety strip 8 feet wide; and two side strips, each 4 feet wide. What should the span of the bridge be (the length of its major axis) if the height 28 feet from the center is to be 13 feet?

77. Installing a Vent Pipe A homeowner is putting in a fire place that has a 4-inch-radius vent pipe. He needs to cut an elliptical hole in his roof to accommodate the pipe. If the pitch of his roof is of the hole? Source:

lion miles. If the aphelion of Earth is 94.5 million miles, what is the perihelion? Write an equation for the orbit of Earth around the Sun . 80. Mars TIle mean distance of Mars from the Sun is 142 mil lion miles. If the perihelion of M ars is 1 28.5 million miles, what is the aphelion? Write an equation for the orbit of Mars about the Sun. 81. Jupiter The aphelion of Jupiter is 507 million miles. If the distance from the center of its elliptical orbit to the Sun is

(a rise of 5, run of 4) what are the dimensions

www.pen. k12.va.us

78. Volume of a Football A football is in the shape of a prolate spheroid, which is simply a solid obtained by rotating an y = 1 about its major axis. An inflated NFL + ellipse abfootball averages 1 1 . 125 inches in length and 28.25 inches in center circumference. If the volume of a prolate spheroid is 4 2 37Tab , how much air does the football contain? (Neglect material thickness)? Source: www.answerbagcom

(x.: : )

In Problems 79-82, use the fact that the orbit of a planet about the Sun is an ellipse, with the Sun at one focus. The aphelion of a planet is its greatest distance from the Sun, and the perihelion is its shortest distance. The mean distance of a planet from the Sun is the length of the semimajor axis of the elliptical orbit. See the illustration.

79. Earth The mean distance of Earth from the Sun is 93 mil

%,

I

-

f

Mean di stance

�.c:::P:A helion � Major axl s 0 -Center Sun

�--

23.2 million miles, what is the perihelion? What is the mean distance? Write an equation for the orbit of Jupiter around the Sun.

82. Pluto The perihelion of Pluto is 4551 million miles, and the distance from the center of its elliptical orbit to the Sun is 897.5 million miles. Find the aphelion of Pluto. What is the mean distance of Pluto from the Sun? Write an equation for the orbit of Pluto about the Sun.

792 83.

CH APTE R 11

Analytic Geometry

2 2 (a) I s an e ll'Ipse I'f D + £ 4A 4C

Show that an equation of the form Ax2 +

cl + F

=

0,

A =f. 0 , C =f. 0 , F =f.

°

2 2 (b) Is a point if D + £ 4A 4C

where A and C are of the same sign and F is of opposite sign, (a) Is the equation of an ellipse with center at (0, 0) if A =f. C. (b) Is the eq uation of a circle with center (0, 0) if A = C. 84.

A =f. 0 , C =f.

I F ' t1e . IS

- F =

.

same sign as A .

0.

. . . . 2 2 (c) Con tams ' no pomts I f D + £ - F IS ' 0f opposite sign 4A 4C to A .

Show that the graph of an equation of the form

A X2 + c l + Dx + y + F = 0 , £ where A and C are of the same sign,

-

°


c

The eccentricity e of an ellipse is defined as the number - , where a and c are the numbers given in equation (2). Because a a

>

c,

it

follows that e < 1 . Write a brief paragraph about the general shape of each of the following ellipses. Be sure to justify your conclusions. (a) Eccentricity close to ° (b) Eccentricity = 0.5 (c) Eccentricity close to 1


Vi3

2.

2. 4

3.

( -2, 0), (2, 0 ) , (0, - 4 ) , (0, 4 )

4.

(2, 5 )

5.

left; 1 ; down: 4

6. ( x - 2 ) 2 + (y + 3 ) 2

=

1

11.4 The Hyperbola PREPARING FOR THIS SECTION • • • •


Distance Formula (Section 2 . 1 , p. 157) Completing the Square (Section 1 .2, pp. 99-101) Intercepts (Section 2.2, pp. 1 65-166) Symmetry (Section 2.2, pp. 167-169) Now Work the 'Are You

2

DEFINITION

•

Asymptotes (Section 5 .2, pp. 346-352) Graphing Techniques: Transformations (Section 3.5, pp. 252-260)

•

Square Root Method (Section 1 .2, pp. 98-99)

Prepared ?' pro blems on page 80 1 .

OBJECTIVES 1

Figure 31

•

Analyze Hyperbolas with Center at the Origin (p. 792) Find the Asymptotes of a Hyperbola (p.797)

3

Ana lyze Hyperbolas with Center at (h, k) (p. 798)

4

Solve Applied Problems I nvolving Hyperbolas (p. 800)

A hyperbola is the collection of all points in the plane, the difference of whose distances from two fixed points, called the foci, is a constant.

-.J

Conjugate

Figure 31 illustrates a hyperbola with foci FI and Fz . The line containing the foci is called the transverse axis. The midpoint of the line segment joining the foci is the center of the hyperbola. The line through the center and perpendicular to the transverse axis is the conjugate axis. The hyperbola consists of two separate curves, called branches, that are symmetric with respect to the transverse axis, conjugate axis, and center. The two points of intersection of the hyperbola and the transverse axis are the vertices, VI and V2 , of the hyperbola. 1

Analyze Hyperbolas with Center at the Origin

With these ideas in mind, we are now ready to find the equation of a hyperbola in the rectangular coordinate system. First, we place the center at the origin. Next, we

SECTION 1 1.4

Figure 32

d(F1 , P)

- d(F2< P)

= ±2a

The Hyperbola

793

position the hyperbola so that its transverse axis coincides with a coordinate axis. Suppose that the transverse axis coincides with the x-axis, as shown in Figure 32. If e is the distance from the center to a focus, one focus will be at F 1 = ( - e, 0) and the other at F2 = (e, 0). Now we let the constant difference of the distances from any point P = (x, y) on the hyperbola to the foci Fl and F 2 be denoted by ±2a. ( If P is on the right branch, the + sign is used; if P is on the left branch, the - sign is used. ) The coordinates of P must satisfy the equation

d(F1 , P) - d(F2 , P) = ±2a Y(x + e ? + i Y(x - e ? + i -

Difference of the d ista nces from

=

P to the

±2a

Use the Distance Form u la .

y( x e) 2 + i = ±2a + y(x - e) 2 + i (x + e) 2 + i = 4a2 ± 4ay(x e) 2 + i +

Isolate one radical.

-

+

(x

foci equals ± 2a.

Squ a re both sides.

- e? + i

Next we remove the parentheses.

e2 + i 4ex 4a2

=

ex - a2 (ex - a2 ) 2 e2 x2 - 2e a2x + a4 e2 x2 + a4 (e2 - a2 )x2 - a2 y2

=

x2

+

2ex

+

-

=

4a2 ± 4ay(x - e ) 2 + i + x2 - 2ex + e2 + i ±4ay(x e) 2 + i Si m pl ify; isolate the radical . -

±ay(x e) 2 + i = a2 [ (x - e ? + i J a2 (x2 - 2ex + e2 + i ) a2 x2 + a2 e2 + a2 y2 = a2 e2 - a4 -

=

=

Divide each side by 4. Square both s ides. Simplify. Remove pa rentheses and si m p lify. Rearrange terms. Factor

a2 on

To obtain points on the hyperbola off the x-axis, it must be that a why, look again at Figure 32.

d(F[ , P) d(Fl ' P ) - d(F2 ' P) 2a

O. Let

b2

=

e2 - a2 , b >

O.

a2 b2 2

1

Divide eac h side by a li.

To find the vertices of the hyperbola defined by this equation, let y O. x = 1, the solutions of which are x = ±a. The vertices satisfy the equation a Consequently, the vertices of the hyperbola are VI = ( - a, O ) and V2 = (a, O). Notice that the distance from the center (0, 0) to either vertex is a.

�

=

794

CHAPTER 1 1

Analytic Geometry

THEOREM

Equation of a Hyperbola: Center at (0, 0); Transverse Axis along the x-Axis

An equation of the hyperbola with center at and vertices at ( -a, 0) and ( a, 0) is

(0, 0), foci at ( -c, 0) and ( c, 0), (2)

y

The transverse axis is the x-axis. See Figure 33. As you can verify, the hyperbola defined by equation (2) is sym metric with respect to the x-axis, y-axis, and origin. To find the y-intercepts, if any, let y? x = 0 in equation (2). This results in the equation : = -1, which has no real solu

( - a, 0) Transverse '\. 1 aXIs '\. F , ( - c, 0) v, =

b

=

tion. We conclude that the hyperbola defined by equation (2) has no y-intercepts. 2 2 x? In fact, since \ - 1 = Y ;::: 0, it follows that � ;::: 1. There are no points on the a b2 a graph for -a < x < a. E XA M P L E 1

F i nd i n g and G raph i n g an Eq u ation of a Hyperbola

Find an equation of the hyperbola with center at the origin, one focus at (3, 0), and one vertex at ( -2, 0). Graph the equation. Solution

The hyperbola has its center at the origin, and the transverse axis coincides with the x-axis. One focus is at ( c , 0) = (3, 0), so c = 3. One vertex is at ( -a, 0) = ( -2, 0), so a = 2. From equation (2), it follows that b2 c2 - a2 = 9 - 4 = 5, so an equation of the hyperbola is =

x2 - i=1 4 5 To graph a hyperbola, it is helpful to locate and plot other points on the graph. For example, to find the points above and below the foci, we let x = ± 3. Then x2 i

---=1 4 5 ( ±3 )2 i ---=1 4 5 l -49 - 5 =1 i 5 5 4 25 y2 = 4 5 Y = ±2

Figure 34

x = ±3

( %) ( - %).

The points above and below the foci are ± 3,

and ±3,

termine the "opening" of the hyperbola. See Figure 34. x2

COMM ENI 10 graph the hyperbola functions Fig u re

34.

Y, =

V5 \j�1 4"

4

im!l::==-- Now Work

-

4

and Y2

-

=

These points de •

y2 - = 1 discu ssed in Exa mple 1, we need to graph the two

5

-

V5 \j�1 .

PRO B L EM 1 7

4"

4

-

Do this and compa re what you see with

•

SECTION 11.4 The Hyperbola

795

An equation of the form of equation (2) is the equation of a hyperbola with center at the origin, foci on the x-axis at ( -c, 0) and (c, 0), where c2 = a2 + b2, and transverse axis along the x-axis. For the next two examples, the direction " Analyze the equation" will mean to find the center, transverse axis, vertices, and foci of the hyperbola and graph it. EXAM P LE 2

Analyzing the E q u ation of a Hyperbola

Analyze the equation: Solution

x2

i

16 - 4"" =

1

The given equation is of the form of equation (2), with a2 = 16 and b2 = 4. The graph of the equation is a hyperbola with center at (0, 0) and transverse axis along the x-axis. Also, we know that c2 = a2 + b2 = 16 + 4 = 20. The vertices are at ( ±a, O) = ( ±4, 0), and the foci are at ( ±c, 0) = ( ±2V5, 0). To locate the points o n the graph above and below the foci, we let x = ±2 V5 . Then x2 i

16 - 4 = 1 ( ±2VS) 2 i -'--- 16---' - 4 = 1 -

-

-

-

-

i

20

16 4 = i -45 - 4 = i 4 y=

-

Figure 35 y

4 (-2 -15, 1 ) I V1 (- 4 , 0) =

./

(2 -15, 1 )

(

V2 (4 , 0) I \

=

-

-

1

-

1

±2Vs

1 4 ±1

The points above and below the foci are Figure 35.

-4

x =

( ±2V5 , 1 ) and ( ±2V5, - 1 ) . See •

The next result gives the form of the equation of a hyperbola with center at the origin and transverse axis along the y-axis. THEOREM

Equation of a Hyperbola: Center at (0, 0); Transverse Axis along the y-Axis

An equation of the hyperbola with center at and vertices at (0, -a) and (0, a) is

(0, 0), foci at (0, -c) and (0, c), (3)

The transverse axis is the y-axis.

..J

Figure 36 shows the graph of a typical hyperbola defined by equation (3). x2 y2 An equation of the form of equation (2), - ? 1, i s the equation of a

y

V2

=

( 0,

a2

a) x

V1 = (0,

F,

=

- a)

(0, - c)

b-

=

hyperbola with center at the origin, foci on the x-axis at ( -c, 0) and (c, 0), where c2 = a2 + b2 , and transverse axis along the x-axis. 2 x? An equation of the form of equation (3), Y - � = 1, is the equation of a a2 bhyperbola with center at the origin, foci on the y-axis at (0, -c) and (0, c), where c2 = a2 + b2 , and transverse axis along the y-axis.

796

CHAPTER 11

Analytic Geometry

Notice the difference in the forms of equations (2) and (3). When the i -term is subtracted from the x2-term, the transverse axis is along the x-axis. When the x2 -term is subtracted from the i-term, the transverse axis is along the y-axis. E XA M P L E 3

Analyzi n g the Eq uation of a Hyperbola

Analyze the equation: / - 4x2 Solution

-5

4

To put the equation in proper form, we divide each side by 4:

Since the x2-term is subtracted from the i-term, the equation is that of a hyperbola with center at the origin and transverse axis along the y-axis. Also, comparing the above equation to equation (3), we find a2 = 4, b2 = 1, and c 2 = a2 + b2 = 5. The vertices are at (0, ±a) = (0, ±2), and the foci are at (0, ±c) = (0, ± v's). To locate other points o n the graph, w e let x = ±2. Then

Figure 3 7

(- 2, 2 "5 )

=

/ - 4x2 / - 4( ± 2 ?

(2, 2 "5 )

5 x

(- 2, - 2 "5 )

=

4

=

4

x = ±2

=4 / = 20 Y = ±2Vs Four other points on the graph are ( ±2, 2v's) and ( ±2, -2v's). See Figure 37 . /

-

16

•

E XA M P L E 4

F i nd i ng an E quation of a Hyperbola

Find an equation of the hyperbola having one vertex at and (0, 3). Graph the equation. Figure 38

Solution

(0, 2) and foci at (0, -3)

Since the foci are at (0, -3) and (0, 3), the center of the hyperbola, which is at their midpoint, is the origin. Also, the transverse axis is along the y-axis. The given information also reveals that c = 3, a = 2, and b2 = c 2 - a2 = 9 - 4 = 5. The form of the equation of the hyperbola is given by equation (3): / a2

x2 --=

1

/ 4

1

-

Let y L' II

b2

-

x2 5

-

=

= ±3 to obtain points on the graph across from the foci. See Figure 38 .

£' --

Now Work

• PROB L EM

1 9

Look at the equations of the hyperbolas in Examples 2 and 4. For the hyperbola in Example 2, a2 = 16 and b2 = 4, so a > b; for the hyperbola in Example 4, a2 = 4 and b2 = 5, so a < b. We conclude that, for hyperbolas, there are no requirements involving the relative sizes of a and b. Contrast this situation to the case of an ellipse, in which the relative sizes of a and b dictate which axis is the major axis. Hyperbolas have another feature to distinguish them from ellipses and parabolas: Hyperbolas have asymptotes.

SECTION 11.4

2

The Hyperbola

797

Find the Asymptotes of a Hyperbola

Recall from Section 5.2 that a horizontal or oblique asymptote of a graph is a line with the property that the distance from the line to points on the graph approaches 0 as x ---,) - 00 or as x ---,) 00 . The asymptotes provide information about the end behavior of the graph of a hyperbola. THEOREM

Asym ptotes of a Hyperbola

x2 l - 2" = 1 has the two oblique asymptotes a- b

The hyperbola ?

I

b b y = -x (4) a and y = - -x a �---------------------------------�� Proof

We begin by solving for y in the equation of the hyperbola.

x2 l ---= a2 b2

1

l x2 -=-1 b2 a2 y- = b?- x- a2

(? )

?

Since x

=1=

1

0, we can rearrange the right side in the form

Y= Now, as x ---,) - 00 or as x ---,)

00,

bx a

±-

R 2 x2

1 --

a2 x

the term 2" approaches 0, so the expression under the

bx

radical approaches 1. So, as x ---,) - 00 or as x ---,) 00 , the value of y approaches ± �; a that is, the graph of the hyperbola approaches the lines b y = ba x and y = -x a - -

These lines are oblique asymptotes of the hyperbola.

•

The asymptotes of a hyperbola are not part of the hyperbola, but they do serve as a guide for graphing a hyperbola. For example, suppose that we want to graph the equation y

We begin by plotting the vertices ( -a, 0) and (a, 0 ) . Then we plot the points (0, -b) and (0, b) and use these four points to construct a rectangle, as shown in Figure 39. b b . The dIagonals . of thIS . rectangI e have slopes - an d - - , an d t h ell' ' extensIOns are th e a a b b asymptotes y = - x and y = - -x of the hyperbola. If we graph the asymptotes, we

a

a

can use them to establish the "opening" of the hyperbola and avoid plotting other points.

798

CHAPTER 11

Analytic Geometry

THEOREM

Asym ptotes of a Hyperbola

2 2 Y The hyperbola 2 a b-

\ - 1 has the two oblique asymptotes =

�

�

�

I

)I = x' and Y = - X (5 ) ____________________________ __ _ _________________��

You are asked to prove this result in Problem 74. For the remainder of this section, the direction " Analyze the equation" will mean to find the center, transverse axis, vertices, foci, and asymptotes of the hyper bola and graph it. EXAM PLE 5

Analyz i n g the Equation of a Hyperbola

Analyze the equation:

/ - - x2 =

1

4

Since the x2 -tenn is subtracted from the i -term, the equation is of the form of equation (3) and is a hyperbola with center at the origin and transverse axis along the y-axis. A lso, comparing this equation to equation (3), we find that a2 = 4, b2 = 1, and c2 = a2 + b2 = 5. The vertices are at (0, ±a) = (0, ±2), and the foci are at (0, ±c) = (0, ± VS). Using equation (5) with a = 2 and b = 1, the

Solution

a

a

asymptotes are the lines y = b x = 2x and y = - b x = -2x. Form the rectangle

containing the points (0, ±a) = (0, ±2) and ( ±b, 0) = (±1, 0). The extensions of the diagonals of this rectangle are the asymptotes. Now graph the rectangle, the asymptotes, and the hyperbola. See Figure 40. 11\

EXAM P L E 6

Analyzing the E quation of a Hyperbola

9x2

36 Divide each side of the equation by 36 to put the equation in proper form.

Analyze the equation: Solution

- 4/

=

/ x2 - - - = 4 9

1

We now proceed to analyze the equation. The center of the hyperbola is the origin. Since the x2 -term is first in the equation, we know that the transverse axis is along the x-axis and the vertices and foci will lie on the x-axis. Using equation (2), we find a2 = 4, b2 = 9, and c2 = a2 + b2 = 13. The vertices are a = 2 units left and right of the center at ( ±a, 0) = ( ±2, 0), the foci are c = VI3 units left and right of the center at ( ±c, 0) = ( ± VI3, 0), and the asymptotes have the equations b 3 y = -x = -x

a

2

and

b 3 y = - -x = - -x

a

2

To graph the hyperbola, form the rectangle containing the points ( ±a, 0) and and (0, 3). The extensions of the diagonals of this rectangle are the asymptotes. See Figure 41 for the graph.

(0, ±b), that is, ( -2, 0), (2, 0), (0, -3), C!lO :' III:IC:> -

3

Now Work

•

PROB L EM 2 9

Analyze Hyperbolas with Center at (h, k)

If a hyperbola with center at the origin and transverse axis coinciding with a coor dinate axis is shifted horizontally h units and then vertically k units, the result is a hyperbola with center at (h, k. ) and transverse axis parallel to a coordinate axis. The

SECTION 11.4

The Hyperbola

799

equations of such hyperbolas have the same forms as those given in equations (2) and (3), except that x is replaced by x - h (the horizontal shift) and y is replaced by y - k (the vertical shift). Table 4 gives the forms of the equations of such hyperbo las. See Figure 42 for typical graphs. Equation of a Hyperbola: Center at (h, k); Transverse Axis Parallel to a Coordinate Axis

Table 4

Center

Transverse Axis

(h, k)

(h, k)

Foci

Vertices

Parallel to the x-axis

(h ± (, k)

(h ± a, k)

Parallel to the y-axis

(h, k ± ()

(h, k ± a)

Figure 42

Asymptotes

Equation

(x - h}2 a2

(y - k} 2

b a

= 1'

b2

=

(2

_

a2

y - k = ±- (x - h)

--- - --- = 1 '

b2

=

(2

_

a2

y - k = ± - (x - h)

---

-

---

b2

(x - h)2 b2

(y - W

a2

y

a b

Y

Transverse axis

e -':---'�---4ij'--+-""--=-_ Transve rs.=... axis x

f

EXAM PLE 7

F i n d i n g an E q uation of a Hyperbola, C e nter N ot at the O rigi n

Find an equation for the hyperbola with center at ( 1 , - 2 ) , one focus at (4, -2) , and one vertex at (3, -2 ) . Graph the equation. Solution

Figure 43 Y 6 ,

,

,

,

V1 = ( - 1 ' -2)

,

"

The center is at ( h , k ) = ( 1 , -2 ) , so h = 1 and k = -2. Since the center, focus, and vertex all lie on the line y = -2, the transverse axis is parallel to the x-axis. The distance from the center ( 1 , -2) to the focus (4, -2) is c = 3; the distance from the center ( 1 , -2) to the vertex (3, -2) is a = 2. Then, b2 = c2 - a2 = 9 - 4 = 5. The equation is

,

-6

Transverse axis

(x - 1 ) 2

4

( y + 2 )2 = 1 5

---

See Figure 43. 1
-

=

1

Refer again to Figure 46. Since the lightning strikes due north of the individual at the point A = (2640, 0), we let x = 2640 and solve the resulting equation.

l 2640-2 - -""""'--- = 1 5502 6,667,100 Y

2

6,667,100 = -22.04 l = 146,942,884 y = 12,122

Subtract

2 2640 2 550

--

from both sides.

Mult i ply both sides by

- 6,667,1 00.

y > 0

since the lightning strike occured in quadrant I

Check: The difference between the distance from

(2640, 12, 122) to the person at the point B = ( -2640, 0) and the distance from (2640, 12, 122) to the person at the point A (2640, 0) should be 1 100. Using the distance formula, we find the difference in the distances is =

V[2640 - ( -2640)F + ( 12,122 - 0)2 - V(2640 - 2640? + ( 12,122 - 0) 2 = 1100 as required. The lightning strike is I;;!I!';: = :: = -

Now Work

12,122 feet north of the person standing at point A .

•

PROB L E M 6 5

11 .4 Assess Your Understanding 'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red. 1.

The distance d from PI ( p. 1 57)

=

(3, -4) to P

To complete the square of x2 + 5x, add

__.

2. 3.

Find the intercepts of the ( pp. 1 65- 1 66)

2=

( -2, 1 ) is d

__

equation

=

5.

To graph y = (x - 5 ) 3 - 4, shift the graph of y (Jeftlright) unites) and then (up/down) ( pp. 252-260) __

. ( pp. 99-10 1 ) l = 9 + 4x2.

6.

=

x3 to the unites).

__

Find the vertical asymptotes, if any, and the horizontal or -9 oblique asymptotes, if any, of y = . ( pp. 346-352) x- - 4

a, it follows that e > 1. Describe the general

Source:

71.

c

69. Rutherford's Experiment In May 1 9 1 1 , Ernest Rutherford

published a paper in Philosophical Magazine. In this article, he described the motion of alpha particles as they are shot at a piece of gold foil 0.00004 cm thick. Before conducting this experiment, Rutherford expected that the alpha particles would shoot through the foil j ust as a bullet would shoot through snow. Instead, a small fraction of the alpha particles bounced off the foil. This led to the conclusion that the nu cleus of an atom is dense, while the remainder of the atom is sparse. Only the density of the nucleus could cause the alpha particles to deviate from their path. TIle figure shows a dia gram from Rutherford's paper that indicates that the de flected alpha particles follow the path of one branch of a hyperbola. (a) Find an equation of the asymptotes under this scenario. (b) If the vertex of the path of the alpha particles is 10 cm from the center of the hyperbola, find an equation that describes the path of the particle.

�2 - �� ?

is located. If the equation of the hyperbola is

shape of a hyperbola whose eccentricity is close to 1. What is the shape if e is very large?

a = b is called an equilateral hyper bola. Find the eccentricity e of an equilateral hyperbola.

72. A hyperbola for which

[Note: The eccentricity of a hyperbola is defined in Prob lem 71.]

73. Two hyperbolas that have the same set of asymptotes are called conjugate. Show that the hyperbolas

x2 4

--l=

1

and l -

;r2 4 =1

�

are conjugate. Graph each hyperbola on the same set of coordinate axes. 74.

Prove that the hyperbola

l

-

a2

-

x2 =1 b2

-

804

CHAPTER 11

Analytic Geometry

where A and

has the two oblique asymptotes a

y = bx

and

a

y = - bx

(a)

75. Show that the graph of an equation of the form

C are of opposite sign, D2 £2 Is a hyperbola if 4A + 4C - F

D2 £2 4C - F = O 4A + -

at (0, 0) .

Show that the graph of an equation of the form

AX2 + cl + Dx + £y + F = 0,

A

"1=

0, C

"1=

O.

(b) Is two intersecting lines if

Ax2 + Cl + F = 0, A "1= 0, C "1= 0, F "1= 0 where A and C are of opposite sign, is a hyperbola with center

76.

"1=

0

'Are You Prepared?' Answers 1. 5 V2

2

. 254

-

3. (0, - 3 ) , (0, 3)

4.

True

5.

right; 5; down; 4

6.

Vertical: x

=

-2, x

= 2; horizontal: y = 1

1 1.5 Rotation of Axes; General Form of a Conic PREPARING FOR THIS SECTION •

•


Sum Formu las for Sine and Cosine (Section 8.4, pp. 627 and 630) Half-angle Formulas for Sine and Cosine (Section 8.5, p. 640) Now Work

Double-angle Formulas for Sine and Cosine (Section 8.5, p. 637)


OBJECTIVES 1 Identify a Conic (p. 804) 2

Use a Rotation of Axes to Tra nsform Eq u ations (p. 805)

3

Ana lyze an Eq uation Using a Rotation of Axes (p. 808)

4

I dentify Con ics without a Rotation of Axes (p. 8 1 0)

In this section, we show that the graph of a general second-degree polynomial con taining two variables x and y, that is, an equation of the form

AX2

+ Bxy

+ cl +

Dx

+

Ey + F

=

0

(1)

where A, B, and C are not simultaneously 0, is a conic. We shall not concern our selves here with the degenerate cases of equation (1), such as x2 + l = 0, whose graph is a single point (0, 0 ) ; or x2 + 3l + 3 = 0, whose graph contains no points; or x2 - 4l = 0, whose graph is two lines, x - 2y = 0 and x + 2y = O. We begin with the case where B = O. In this case, the term containing xy is not present, so equation (1) has the form

AX2 + Cl where either A *, O or C 1

+

Dx + E y + F

=

0

*' O.

Identify a Conic

We have already discussed the procedure for identifying the graph of this kind of equation; we complete the squares of the quadratic expressions in x or y, or both. Once this has been done, the conic can be identified by comparing it to one of the forms studied in Sections 1 1 .2 through 1 1 .4. In fact, though, we can identify the conic directly from the equation without completing the squares.

SECTION 11.5

THEOREM

Rotation of Axes; General Form of a Co nic

805

Identifying Conics without Completing the Squares

Excluding degenerate cases, the equation

AX2

+

cl

+

Dx

+

Ey

+ F

=°

(2)

where A and C cannot both equal zero:

( a ) Defines a parabola if A C = 0. ( b ) Defines an ellipse ( or a circle ) if A C > 0. ( c) Defines a hyperbola if AC < 0. Proof

( a ) If AC

= 0, then either A tion (2) is either

=

AX2 + Dx or

cl

+

° or +

C = 0, but not both, so the form of equa

Ey + F

Dx + Ey

+ F

=

0,

A *, O

= 0,

c *' o

Using the results of Problems 78 and 79 in Exercise 1 1 .2, it follows that, except for the degenerate cases, the equation is a parabola. (b ) If A C > 0, then A and C are of the same sign. Using the results of Problem 84 in Exercise 1 1 .3, except for the degenerate cases, the equation is an ellipse if A *' C or a circle if A = C.

( c ) If A C

< 0, then A and C are of opposite sign. Using the results of Problem 76 in Exercise 1 1 .4, except for the degenerate cases, the equation is a hyperbola. •

We will not be concerned with the degenerate cases of equation (2). However, in practice, you should be alert to the possibility of degeneracy. Identifying a Conic without Complet i n g t h e Squares

E XA M P L E 1

Identify each equation without completing the squares. ( a ) 3x2 + 61 + 6x - 12y = ° ( b ) 2x2 - 3 1 + 6y + 4 = °

( c) y2 - 2x + 4

=

°

( a ) We compare the given equation to equation (2) and conclude that A = 3 and C = 6. Since A C 1 8 > 0, the equation is an ellipse. ( b ) Here A = 2 and C = -3, so A C = -6 < 0. The equation is a hyperbola. ( c ) Here A = ° and C = 1, so A C = 0. The equation is a parabola.

Solution

=

•

t>=_ -

Now Work

PROB L EM

1 1

Although we can now identify the type of conic represented by any equation of the form of equation (2) without completing the squares, we will still need to complete the squares if we desire additional information about the conic, such as its graph. Now we turn our attention to equations of the form of equation ( 1 ) , where B *' 0. To discuss this case, we first need to investigate a new procedure: rotation of axes. 2

Use a Rotation of Axes to Transfo rm Equations

In a rotation of axes, the origin remains fixed while the x-axis and y-axis are rotated through an angle e to a new position; the new positions of the x-axis and the y-axis are denoted by x' and y ' , respectively, as shown in Figure 47 ( a ) .

806

CHAPTER 1 1

Analytic Geometry

Now look at Figure 47(b). There the point P has the coordinates (x, y) relative to the xy-plane, while the same point P has coordinates ( x ' , y ' ) relative to the x ' y' -plane. We seek relationships that will enable us to express x and y in terms of x ' , y ' , and e. As Figure 47(b) shows, r denotes the distance from the origin 0 to the point P, and a denotes the angle between the positive x ' -axis and the ray from 0 through P. Then , using the definitions of sine and cosine, we have

Figure 47 y'

y e

x' e x

y'

x' = r cos a

(3)

r sin a

y = r sin (e

x = r cas( e + a ) (a)

=

+

(4)

a)

Now x = r cos (e =

+

a) Apply the Sum Formula for sine.

r( cos e cos a - sin e sin a)

= (r cos a) (cos e ) - ( r sin a) (sin e ) = x ' cos e - y' sin e

By equation (3)

Similarly, y

=

r sin(e + a )

= r( sin e cos a

(b)

=

THEOREM

+

Apply the S u m Formula for sine.

cos e s i n a )

By equation (3)

x ' sin e + y' cos e

Rotation Formulas

If the x- and y-axes are rotated through an angle e, the coordinates (x, y) of a point P relative to the xy-plane and the coordinates ( x ' , y ' ) of the same point relative to the new x' - and y' -axes are related by the formulas x = x' cos e

-

y' sin e

y

=

x ' sin e +

y

'

cos e

(5)

I�

� -----------------�

E XA M P LE 2

Rotating Axes

Express the equation xy = 1 in terms of new x' y' -coordinates by rotating the axes through a 45° angle. Discuss the new equation. Solution

Let e = 4SO in equation ( 5 ) . Then x = x' cos 45° - y' y = x'

.

Sill

.

Sill

V2

V2

V2

V2

V2

V2

45° = x ' -- - y ' -- = -- (x ' - y ' )

45° + y' cos 45°

2

=

x ' --

2

2

+

y' --

2

2

=

-- (x ' + y ' )

2

Substituting these expressions for x and y in xy = 1 gives Figure 48

This is the equation of a hyperbola with center at (0, 0 ) and transverse axis along the x' -axis. The vertices are at ( ± V2, 0) on the x' -axis; the asymptotes are y' = x' and y' = -x' (which correspond to the original x- and y-axes) . See Figure 48 for the graph. •

SECTION 11.5

807

Rotation of Axes; General Form of a Conic

As Example 2 illustrates, a rotation of axes through an appropriate angle can transform a second-degree equation in x and y containing an xy-term into one in x' and y ' in which no x'y '-term appears. In fact, we will show that a rotation of axes through an appropriate angle will transform any equation of the form of equa tion (1) into an equation in x ' and y' without an x 'y '-term. To find the formula for choosing an appropriate angle 8through which to rotate the axes, we begin with equation ( 1 ) , B*O AX2 + Bxy + cl + Dx + E y + F = 0 , Next we rotate through an angle 8using rotation formulas (5). A(X ' cos 8 - y 'sin 8) 2 + B(X ' cos 8 - y 'sin 8) (X 'sin 8 + y ' cos 8) + C(x ' sin 8 + y ' cos 8) 2 + D(x ' cos 8 - y' sin 8) + E(X ' sin 8 + y ' cos 8) + F = 0

By expanding and collecting like terms, we obtain 2 (Acos 8 + B sin 8cos 8 + Csin2 8)x '2 + [B(cos2 8 - sin2 8) + 2(C - A) (sin 8cos 8) ]x'y ' + (Asin2 8 - B sin 8cos 8 + Ccos2 8)y '2 + (D cos 8 + E sin 8) x '

+ (- D sin 8 + E cos 8) y ' + F

=

(6)

0

In equation (6), the coefficient of x 'y ' is B(cos2 8 - sin2 8) + 2(C - A) (sin 8cos 8)

Since we want to eliminate the x 'y' -term, we select an angle 8 so that this coeffi cient is O. B(cos2 8 - sin 2 8) + 2 (C - A) (sin 8cos 8) = 0 B cos(28) + (C - A) sin(28) = 0 B cos(28)

cot(28)

THEOREM

=

=

(A - C) sin(28) A- C B

To transform the equation AX2 + Bxy + cl + Dx + Ey + F =

0

Dou ble-a ng Ie Formulas

B*O

B*O

into an equation in x ' and y ' without an x 'y '-term, rotate the axes through an angle 8 that satisfies the equation cot(28)

=

A- C B

(7)

--

�------------------

1

----�.�

--------

Equation (7) has an infinite number of solutions for 8. We shall adopt the con vention of choosing the acute angle 8 that satisfies (7). Then we have the follow ing two possibilities:

If cot(28) If cot(28)

2:
°

=

a cos

t

y =

a

sin t

is a constant. Graph this curve, indicating its orientation.

820

CHAPTER 11

Ana lytic Geometry

Solution

cos t

we find that

Figure 57

(0, a)

(-a, 0)

The presence of sines and cosines in the parametric equations suggests that we use a Pythagorean Identity. In fact, since =

x a

. smt

x2 +

l

-

y a

=

y

(a, 0) x

=

a2

The curve is a circle with center at (0,0) and radius a. As the parameter t increases, 7T say from t = 0 [the point (a, O)] to t = [the point (0, a)] to t = 7T [the point 2 (-a, 0)], we see that the corresponding points are traced in a counterclockwise di rection around the circle. The orientation is as indicated in Figure

57.

� == -

Now Work P R O B L E M S 7

•

AND 1 9

Let's analyze the curve in Example 2 further. The domain of each parametric equation is -(X) < t < 00. This means, the graph in Figure is actually being repeated each time that t increases by 27T. If we wanted the curve to consist of exactly 1 revolution in the counterclockwise direction, we could write

57

x = a cos t,

y = a sin t,

o ::5 t ::5 27T

This curve starts at t = 0 [the point ( a,O)] and, proceeding counterclockwise around the circle, ends at t = 27T [also the point ( a, 0)]. If we wanted the curve to consist of exactly three revolutions in the counter clockwise direction, we could write x

or

x

or E XA M P L E 3

= =

a cos t,

y = a sin t,

-27T ::5 t ::5 47T

a cos t,

y =

o ::5 t ::5 67T

x = a cos t,

a sin t,

y = a sin t,

27T ::5 t ::5 87T

Describing Parametric E q u ations

Find rectangular equations for the following curves defined by parametric equa tions. Graph each curve.

Soluti o n

( a ) x = a cos t, Y = a sin t, 0 ::5 t ::5 7T, a> 0 (b) x = -a sin t, y = -a cost, 0 ::5 t ::5 7T, a> 0

(a) We eliminate the parameter t using a Pythagorean Identity. cos2 t + sin2 t = 1 x2 + l = a2 The curve defined by these parametric equations is a circle, with radius a and center at (0, 0). The circle begins at the point ( a, 0), t = 0; passes through the point ( 0,a),

t =

; and ends at the point (-a, 0),t ;

=

7T.

SECTION 11.7 Figure 58

(0, a)

(a, 0)

(-a, O) y

(b) We eliminate the parameter t using a Pythagorean Identity. sin2 t + cos2 t = 1

x

(0, a)

x2 +

point ( -a, 0 ) , t

(0, -a)

a

Seeing the Concept

=

x=

x=

-sin t, y

=

=

�; and ends at the point (0, a ) , t

= 7T.

See Figure 59. The parametric equations define a left semicircle of radius with a clockwise orientation. The rectangular equation is x = - Va2 - l, -a ::; y ::; a

Example 3 illustrates the versatility of parametric equations for replacing com plicated rectangular equations, while providing additional information about orien tation. These characteristics make parametric equations very useful in applications, such as projectile motion.

sin t for 0 :s t :s 7r. Compare to

Figure 58. Graph

= a2

•

cos t, y = sin t, 0 :s t :s 27r. Compare to Figure 57. Graph x = cos t,

Y

l

The curve defined by these parametric equations is a circle, with radius a and center at (0, 0). The circle begins at the point ( 0, -a), t = 0; passes through the

x

(-a, O)

Graph

821

See Figure 58. The parametric equations define an upper semicircle of radius a with a counterclockwise orientation. The rectangular equation is - a ::; x ::; a y = Va2 - x2 ,

y

Figure 59

Plane Curves and Parametric Equations

-cos t

for 0 :s t :s 7r. Compare to Figure 59.

3

Use Time as a Parameter in Parametric Equations

If we think of the parameter t as time, the parametric equations x

=

f(t) and

y = get) of a curve specify how the x- and y-coordinates of a moving point vary

with time. For example, we can use parametric equations to describe the motion of an object, sometimes referred to as curvilinear motion. Using parametric equations, we can specify not only where the object travels, that is, its location (x, y), but also when it gets there, that is, the time t. When an object is propelled upward at an inclination e to the horizontal with initial speed vo, the resulting motion is called projectile motion. See Figure 60(a). In calculus it is shown that the parametric equations of the path of a projectile fired at an inclination e to the horizontal, with an initial speed vo, from a height h above the horizontal are x = (vo cos e)t

1

Y = - - gt2 + (Vosine)t + h 2

(1)

where t is the time and g is the constant acceleration due to gravity (approximately 32 ft/sec/sec or 9.8 m/sec/sec). See Figure 60(b). Figure 60

y "

(a)

"

, ....

....

.,,----- .... "" h

(b)

.., ....

(x(t), y(t))

.... ....

.... .... ,, ,

x

822

CHAPTER 11

Analytic Geometry

P rojectile Motion

E XAMP L E 4

Suppose that Adam hit a golf ball with an initial velocity of 150 feet per second at an angle of 30° to the horizontal. See Figure 61.

Figure 61

�J Solution

(a) Find parametric equations that describe the position of the ball as a function of time. (b) How long is the golf ball in the air? (c) When is the ball at its maximum height? Determine the maximum height of the ball. (d) Determine the horizontal distance that the ball traveled. (e) Using a graphing utility, simulate the motion of the golf ball by simultaneously graphing the equations found in part (a).

(a) We have va = 150 ft/sec,8 = 3 0 °, = 0 (the ball is on the ground), and g = 32 ft/sec2 (since the units are in feet and seconds). Substituting these values into equations (1), we find that

h

x = (va cos 8) t

y =

�

_ gt2

= - 16t2

+

+

=

( 150 cos 300)t = 75 V3t

(vasin8)t

+

h

75t

�

= - (32)t2

+

(150 sin 300)t

+

0

(b) To determine the length of time that the ball is in the air, we solve the equa tion y = O. - 16 P + 75t = 0 t( - 16t + 75) = 0 75 t = 0 sec or t = = 4.6875 sec 16

The ball will strike the ground after 4.6875 seconds. (c) Notice that the height y of the ball is a quadratic function of t,so the maximum height of the ball can be found by determining the vertex of y = - 16r2 + 75t. The value of t at the vertex is t=

Maximum height

-156

-75 = 2.34375 sec -32

--

=

- 16(2.34375?

+

(75 )2.34375

�

87.89 feet

(d) Since the ball is in the air for 4.6875 seconds,the horizontal distance that the ball travels is

246

------.... -....-----.---

2a

=

The ball is at its maximum height after 2.34375 seconds. The maximum height of the ball is found by evaluating the function y at t = 2.34375 seconds.

Figure 62

o

-b

-

610

x = (75 V3)4.6875

�

608.92 feet x = 75\13 t

� (e) We enter the equations from part (a) into a graphing utility with

Tmin = 0, Tmax = 4.7, and Tstep = 0.1. We use ZOOM-SQUARE to avoid any distortion to the angle of elevation. See Figure 62.

•

Exploration

Simulate the motion of a ball thrown straight up with an initial speed of 100 feet per

second from a height of 5 feet above the ground. Use PARametric mode with Tmin Tstep

= 0.1, Xmin

= 0,Xmax

=

= 0, Tmax =

6.5,

5, Ymin = 0, and Ymax = 180. What happens to the speed with

which the graph is drawn as the ball goes up and then comes back down? How do you interpret this physically? Repeat the experiment using other values for Tstep. How does this affect the experiment?

[Hint: In the projectile motion equations, let () = 900,Vo = 100, h

instead of

x=

° to see the vertical motion better.]

=

5, and 9 = 32. Use x

=

3

SECTION 11.7

Plane Cu rves and Parametric Equation s

823

Result See Figure 63. In Figure 63(a) the ball is going up. In Figure 63(b) the ball is near its highest point. Final l y, in Figure 63(c) the ball is coming back down. Notice that, as the ball goes up, its speed decreases, u ntil at the highest poi n t it is ze ro. Then the speed increases as the ball comes back down.

Figure 63

1 80

1 80

"

0 �======�====� 5 o

(I = 0.7) (a) nbS:=

�I

180

0 �'====��====� 5

0 �======� 5 o ( 1 = 3)

o

(1 = 4) (c)

(b)

:

�


A graphing utility can be used to simulate other kinds of motion as well. Let's work Example 5 from Section 1.7 again. E X A M PLE 5

Figure 64

S i m u lati n g M otion

Tanya, who is a long distance runner, runs at an average velocity of 8 miles per hour. Two hours after Tanya leaves your house, you leave in your Honda and follow the same route. If your average velocity is 40 miles per hour, how long will it be before you catch up to Tanya? See Figure 64. Use a simulation of the two motions to ver ify the answer. -+------ Time

2 hr

t --------+-

----- :r --- � t= 2 � . -.�

YfDF

:1

t=2 S o l u ti o n

--+ .

We begin with two sets of parametric equations: one to describe Tanya's motion, the other to describe the motion of the Honda. We choose time t = 0 to be when Tanya leaves the house. We choose Yl = 2 as Tanya's path and Y2 = 4 as the parallel path of the Honda in order to more easily see the two motions. The horizontal distances traversed in time t ( Distance = Velocity X Time ) are Tanya:

xl

=

8t

Honda: X2 The Honda catches u p to Tanya when Xl = X2 ' 8t 8f -32t

=

40( t - 2 )

=

40( t - 2 ) = 40t - 80 = -80 -80 t= = 2.5 -32 The Honda catches up to Tanya 2.5 hours after Tanya leaves the house. In PARametric mode with Tstep = 0.01 , we simultaneously graph Tanya: Xl = 81 Yl = 2

-

Honda: X2 Y2

= =

40( t - 2 ) 4

for 0 ::; t ::; 3. Figure 65 shows the relative position of Tanya and the Honda for t = 0, t = 2, t = 2.25, t = 2.5, and t = 2.75.

824

CHAPTER 11

Ana lytic Geometry

5

Figure 65

5

5

P

r------e

(73&

=:t

o

�=======:=.J 40

0

40 0

o

5

�===============.J 40 o

4

0

40 0

t= 2

t= 0

o

�

t = 2.25

5

o

t = 2.5

�:=:::=======� 40 o

t = 2.75

•

Find Parametric Equations for Curves Defined by Rectang ular Equations

We now take up the question of how to find parametric equations of a given curve. If a curve is defined by the equation y = J (x), where J is a function, one way of finding parametric equations is to let x = t. Then y = J(t) and x

=

t,

Y

t in the domain of J

= J (t),

are parametric equations of the curve. E XA M P L E 6

S o l u ti o n

F i nd i n g Parametric E quations for a Curve Defined by a Rectangular Equatio n

Find parametric equations for the equation

y

Let x = t. Then the parametric equations are x = t,

Y

= t2

- 4,

= x2 - (X)

- 4. < t

0

47. x= 4 sin t - 2 sin( 2t ) y

"

=

46. x= sin t

+

cos t,

Y

=

sin t - cos t

48. x= 4 sin t + 2 sin(2t)

y= 4 cos t + 2 cos(2t )

4 cos t - 2 cos ( 2t )


49.

Projectile Motion Bob throws a ball straight up with an in itial speed of 50 feet per secon d from a height of 6 feet. (a) Fin d parametric equation s that describe the motion of the ball as a fun ction of time. (b) How lon g is the ball in the air? (c) When is the ball at its maximum height? Determin e the maximum height of the ball. [j. ( d ) Simulate the motion of the ball by graphin g the equation s foun d in part (a).

50.

•

51.

Projectile Motion Alice throws a ball straight up with an in itial speed of 40 feet per secon d from a height of 5 feet. (a) Fin d parametric equation s that describe the motion of the ball as a fun ction of time. (b) How lon g is the ball in the air? (c) When is the ball at its maximum height? Determin e the maximum height of the ball. (d) Simulate the motion of the ball by graphin g the equa tion s foun d in part (a).

B ill's train leaves at 8:06 AM an d acceler ates at the rate of 2 meters per secon d per secon d. Bill, who can run 5 meters per secon d, arrives at the train station 5 secon ds after the train has left an d run s for the train . (a) Fin d parametric equation s that describe the motion s of the train an d Bill as a fun ction of time. [Hint: The position s at time t of an object havin g Catching a Train

( a ) Fin d parametric equation s that describe the motion s of the bus an d Jodi as a fun ction of time. [Hint: The position s at time t of an object havin g

.

53.

Catching a B u s Jodi's bus leaves at 5 :30 PM an d accelerates at the rate of 3 meters per secon d per secon d. Jodi, who can run 5 meters per secon d, arrives at the bus station 2 secon ds after the bus has left an d run s for the bus.

1

Projectile Motion Ichiro throws a baseball with an in itial speed of 145 feet per secon d at an an gle of 200 to the hori zon tal. The ball leaves Ichiro's han d at a height of 5 feet.

(a) Fin d parametric equation s that describe the position of the ball as a fun ction of time. (b) How lon g is the ball in the air? (c) Determin e the horizon tal distan ce that the ball traveled. (d) When is the ball at its maximum height? Determin e the maximum height of the ball. "

•

54.

1

52.

=

z at2. ] (b) Determin e algebraically whether Jo di will catch the bus. H so, when ? - (c) Simulate the motion of the bus an d Jodi by simultan e ously graphin g the equation s foun d in part (a).

acceleration a is s= at2 ] z (b) Determin e algebraically whether Bill will catch the train . If so, when ? - (c) Simulate the motion of the train an d Bill by simultan e ously graphin g the equation s foun d in part (a).

.

acceleratIOn a IS s

•

55.

(e) Usin g a graphin g utility, simultan eously graph the equation s foun d in part (a). Projectile Motion Barry Bonds hit a baseball with an in itial speed of 1 25 feet per secon d at an an gle of 400 to the hori zon tal. The ball was hit at a height of 3 feet off the groun d. (a) Fin d parametric equation s that describe the position of the ball as a fun ction of time. (b) How lon g is the ball in the air? (c) Determin e the horizon tal distan ce that the ball traveled. (d) When is the ball at its maximum height? Determin e the maximum height of the ball. (e) Usin g a graphin g utility, simultan eously graph the equa tion s foun d in part (a). Projectile Motion Suppose that Adam hits a golf ball off a cliff 300 meters high with an in itial speed of 40 meters per secon d at an an gle of 450 to the horizon tal.

SECTION 11.7

• ·

56.

•

57.

-

1

.:

59.

(a) Find parametric equations that describe the motion of the Paseo and B onneville. (b) Find a formula for the distance between the cars as a function of time. (c) Graph the function in part (b) using a graphing utility. (d) What is the minimum distance between the cars? When are the cars closest? (e) Simulate the motion of the cars by simultaneously graphing the equations found in part (a). A Cessna (heading south at 1 20 mph) and a Boeing 747 (heading west at 600 mph) are flying toward the same point at the same altitude. The Cessna is 100 miles from the point where the flight patterns intersect, and the 747 is 550 miles from this intersection point. See the figure. ( a) Find parametric equations that describe the motion of the Cessna and the 747. Uniform Motion

(b) Find a formula for the distance between the planes as a function of time.

600

I

55 0 mi

829

mph

��

)I

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

10(

60. 30 m p h

I

t1 l hr

1 00 mi

Uniform Motion A Toyota Paseo (traveling east at 40 mph) and a Pontiac B o nneville ( traveling north at 30 mph) are heading toward the same intersection. The Paseo is 5 miles from the intersection when the Bonneville is 4 miles from the intersection. See the figure.

4 mi

58.

�mph

(a) Find parametric equations that describe the position of the ball as a function of time. (b) How long is the ball in the air? (c) Determine the horizontal distance that the ball traveled. (d) When is the ball at its maximum height? Determine the maximum height of the ball. (e) Using a graphing utility, simultaneously graph the equations found in part (a). Projectile Motion Suppose that Karla hits a golf ball off a cliff 300 meters high with an initial speed of 40 meters per second at an angle of 45° to the horizontal on the Moon (grav ity on the Moon is one-sixth of that on Earth). (a) Find parametric equations that describe the position of the ball as a function of time. (b) How long is the ball in the air? (c) Determine the horizontal distance that the ball traveled. (d) When is the ball at its maximum height? Determine the maximum height of the ball. (e) Using a graphing utility, simultaneously graph the equa tions found in part (a).

Plane Curves and Parametric Equations

( c) Graph the function in part (b) using a graphing utility. (d) What is the minimum distance between the planes? When are the planes closest? (c) Simulate the motion of the planes by simultaneously graphing the equations found in part (a). The left field wall at Fenway Park is 310 feet from home plate; the waH itself (affectionately named The Green Monster) is 37 feet high. A batted ball must clear the wall to be a home run . Suppose a ball leaves the bat 3 feet off the ground, at an angle of 45°. Use g = 32 ft/sec2 as the ac celeration due to gravity and ignore any air resistance. (a) Find parametric equations that describe the position of the ball as a function of time. (b) What is the maximum height of the ball if it leaves the bat with a speed of 90 miles per hour? Give answer in feet.

T h e Green Monster

(c) How far is the ball from home plate at its maximum height? Give answer in feet. (d) If the ball is hit straight down the left field wall, will it clear the " Green Monster"? If it does, by how much does it clear the wall? Source: The Boston Red Sox Projectile Motion The position of a projectile fired with an initial velocity va feet per second and at an angle fJ to the horizontal at the end of t seconds is given by the parametric equations x = (va cos fJ)t y = (va sin fJ )t - 1 6t2

See the following illustration.

A�

.,. ' - - - - - - - .... . ....

" '�

�------ R --------�

(a) Obtain the rectangular equation of the trajectory and identify the curve. (b) Show that the projectile hits the ground 1 . t = 16 va sm fJ.

(y = 0) when

(c) How far has the proj ectile traveled (horizontally) when it strikes the ground? In other words, find the range R. (d) Find the time t when x = y. Then find the horizontal distance x and the vertical distance y traveled by the projectile in this time. Then compute � . This is the distance R, the range, that the projectile travels up a plane inclined at 45° to the horizontal (x = y).

83 0

C HAPTER 11

Ana lytic Geometry

x (X2 - Xt)t + XI Y = (Y2 - Yl )t

83 in

See the following illustration. (See also Problem Section

8.5.)

=

- 00 < t < 00

+ Yl ,

What is the orientation of this line?

62.

x(t)

� (a)

, yet)

=

t,

21T

Graph the hypocycloid using a graphing utility. (b) Find rectangular equations of the hypocycloid.

61. Show that the parametric equations for a line passing through the points

Hypocycloid The hypocycloid is a curve defined by the parametric equations cos3 t = sin3 O :s t :s

(XI , Y1) and (X2 , Y2 ) are


�

62,

63. In Problem we graphed the hypocycloid. Now graph the rectangular equations of the hypocycloid. Did you obtain a complete graph? If not, experiment until you do.

hypocycloid

epicycloid.

64. Look up the curves called and Write a report on what you find. Be sure to draw comparisons with the cycloid.

'Are You Prepared?' Answers 1T

1. 3 ; "2

CHAPTER REVIEW Things to Know Equations

773-778) Ellipse (pp. 782-788) Hyperbola (pp. 792-800)

1 and 2 (pp. 775 and 776). See Table 3 (p. 787). See Table 4 (p. 799). Ax2 + Bxy + ci + Dx Ey + F See Tables

Parabola (pp.

General equation of a conic (p.

810)

+

=

0

-

B2 4AC 0 Elli pse (or circle) if B2 4A C 0 Hyperbola if B2 4AC 0

Parabola if

=

-

-

Polar equations of a conic with focus at the pole (pp.

812-816)

Parametric equations of a curve (p.

818)

>

,

{

2x + y - z = -2 (1) = -7 (2) -x + Y 0 = -22 (3)

Equation (3) has no solution so the system is inconsistent. 7

•

Express the Solution of a System of Dependent Equations Containing Three Variables

EXAMPLE 1 1

Solve: S o l u ti o n

- 2y - z = 8 (1) 2x - 3y + z = 23 (2) 4x - 5y + 5z = 53 (3)

Our plan is to eliminate x from equations (2) and (3). Multiply each side of equa tion ( 1 ) by -2 and add the result to equation (2). Also, multiply each side of equa tion ( 1 ) by and add the result to equation (3).

-4

x - 2y - z = 8 2x - 3y + z = 23

(1 ) Multiply by - 2 (2)

x - 2y - z = 8 (1) - 5y + 5z = 53 (3)

4x

{X

Solvin g a System of Dependent E quati o n s

Multiply by -4.

-2x + 4y + 2z = - 16 (1 ) 2x - 3y + z = 23 (2) Y + 3z = 7 Add.

______

{X

-4x + 8y + 4z = - 32 (1) 4x - 5y + 5z = 53 � 3y + 9z = 21 Add.

- 2y -

z

= 8

(1)

Y + 3z = 7 (2) 3y + 9z = 21 (3)

Treat equations (2) and (3) as a system of two equations containing two variables, and eliminate the variable y by multiplying both sides of equation (2) by -3 and adding the result to equation (3). y + 3z = 7 3y + 9z = 21

Multiply by -3.

-3y - 9z = -21 3y + 9z = 21 a

=

Add.

a ------->

{X

- 2y - z = 8 (1 ) Y + 3z = 7 (2) a = a (3)

The original system is equivalent to a system containing two equations, so the equations are dependent and the system has infinitely many solutions. If we solve

846

CHAPTER 12

Systems of Equations and Inequalities

equation (2) for y, we can express y in terms of z as y = -3z + 7. Substitute this expression into equation (1) to determine x in terms of z. x - 2y - z x - 2( -3z + 7 ) - z x + 6z - 14 - z X + 5z

X

= 8 = 8

(1)

Substitute y

= 8

(1 ).

Remove parentheses.

= 22 = -5z + 22

{X

= - 3z + 7 in

Com bine l ike terms. Solve for x.

We will write the solution to the system as y

= -5z + 22 = -3z + 7

where z can be any real number. This way of writing the solution makes it easier to find specific solutions of the system. To find specific solutions, choose any value of z and use the equations x = -5z + 22 and y = -3z + 7 to determine x and y. For example, if z = 0, then x = 22 and y = 7, and if z = 1, then x = 17 and y = 4. Using ordered triplets, the solution is {(x, y, z ) l x = -5z + 22, y = -3z + 7, z any real number}

Li!l!O

-

•


Two distinct points in the Cartesian plane determine a unique line. Given three noncollinear points, we can find the (unique) quadratic function whose graph con tains these three points. E XA M P L E 1 2

Curve F i ttin g

Find real numbers a, b, and c so that the graph of the quadratic function y = ax2 + bx + c contains the points ( - 1 , -4), ( 1 , 6 ) , and (3, 0 ) .

Soluti o n

y

= ax2 + bx + c.

For the point ( - 1, -4) we have: -4 = a( - l? + b( - l ) + c -4 = a - b + c 6 = a+b+c For the point ( 1 , 6) we have: 6 = a( 1 ? + b( l ) + c For the point (3, 0 ) we have: o = 9a + 3b + c o = a(3? + b(3 ) + c

Figure 6

-4

We require that the three points satisfy the equation

{

We wish to determine a, b, and c so that each equation is satisfied. That is, we want to solve the following system of three equations containing three variables:

-2

4

(-1 , -4) 5

x

a - b + c = -4 (1) a + b + c = 6 (2) 9a + 3b + c = 0 (3)

Solving this system of equations, we obtain a = -2, b = 5, and c = 3. So the qua dratic function whose graph contains the points ( - 1 , -4 ) , ( 1 , 6 ) , and (3, 0 ) is y = _2x2 + 5x + 3 y = aXZ + bx + c, a = -2, b = 5, C = 3 Figure 6 shows the graph of the function along with the three points.

"'_ ::::0: _ ::11 ....


•

SECTION 12.1 1 2.1

Systems of Linear Equations: Su bstitution and Elimi nation

847

Assess Your Understanding

'Are You Prepared?'

1. Solve the equation:

Answers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red. 3x + 4 = 8 - x. (pp. 86-94) 2. (a) Graph the line: 3x + 4y = 12. (b) What is the slope of a line parallel to this line? (pp. 173-185)

Concepts and Vocabulary 3. If

5. True or False

a system of equations h as no solution, it is said to be

A system of two linear equations containing two variables always has at least one solution.

6 . True or False

4. If a system of equations has one or more solutions, the system is said to be

A solution of a system of equations consists of values for the variables that are solutions of each equa tion of the system.

Skill Building

In Problems 7-16, verify that the values of the variables listed are solutions of the system of equations. { 3x - 4y = 4 {2X = 5 {3X 2y = 2 9 7. 8. . . 1 x - 3y = 1 5x 2y 8 x - 7y = -30 2 -2 x = 2, y = - ; ( 2 , - ) x = -2, y = 4; (-2, 4) x = 2, = L2" (2 1:.) 2

-+ Y

+

=

1

1

{X - y = 3 3 �x x = 4, y = (4, 1 )

{ 2X + �y 0 3x - 4y = - 192 x = - 1:.2 ' y = 2,' (-1:.2' 2 ) =

10.

)1

+ YY +

{ +Y Y -3,+

{ 3x 3 2z = 4 4x - z = 7 14. 8x 5 - z = 0 13. - z= 0 2y - 3z = -8 -x - y 5z = 6 x = 2, = z = 1; x = 1, y = -1, z = 2; (2, -3 , 1 ) (1, -1, 2) { 3x 3y 2z = 4 { 4x - 5z = 6 x - 3y = 10 5y - z = -17 5x - 2y - 3z = 8 -x - 6y 5z = 24 x 2, = -2, z = 2; (2, -2, 2) x = 4, y = -3, z = 2; (4, -3, 2) In Problems solve each system of equations. If the system has no solution, say that it is inconsistent. {x 2y = 5 {X = 8 { x 3y = 5 = 13 '- ] 9. {5X 20. 8. 1 7. 2x = -8 x 2x 3 = 12 -y=4 3 { 2x 4y 2 3x = 24 {4X 5y = -3 {3X 6y 2 { 21 . 22. 23. 24. x 2y = 0 - 2y = -4 5x 4y = 3x - 5y = -10 11.

+Y

1 5.

=

=

1;

+ ++ Y +Y

x-y=3 12. { -3x y = 1 x = -2, = -5; (-2, -5)

+Y

16.

Z

1 7-54,

x

25.

' 29.

33.

+ {2\: + Y = 1 4x 2y { x + 2y = 4 2x + 4y = 8 �

= J�

+

{ 2X + 3y = 6 x - = 21

Y

x

++ Y +

1

26. 0 3 .

34

.

+

- YY

�

=

{ x- y=5 -3x + 3y = 2

-Y { �Xx +- 2yy � ·28 {3X = 7 9x - 3y = 21 =

+

27.

31.

+ {2X - y = O 3x + 2y = 7 { 2x - 3y lOx + Y =

= -::;J

1

=

1

35.

=

28.

-1

11

1

32.

,

{ 21 x + -::;- y = "4 x - "32 y = - l J

-3+ Y + {'X+3Y � -1 4x + Y 8 { 3 2y = 0 5x + lOy = 4

J

36.

=

X

�

-::;J

':'"

{ �X + � -5 3 1 = -x 3 4 + -y

11

848

37.

CHAPTER 12

{I I

Systems of Equations and Inequa lities

{ 15x3x -+ 5y5y == 213

x y 39. -x3 - -y5 = 0 . Let = -1 and = 1 [ x y

�

u.

Hmt:

and y

6 { X- y 2x - 3z = 16 2y + Z = 4 x- y-z=1 45. 2x + 3y + = 2 3x + 2y = 0 { 2x - 2y + 3z = 6 49. 4x - 3y + 2z = 0 -2x + 3y - 7z = 1 { X + 2y - z = -3 53. 2x - 4 + Z = -7 -2x + 2y - 3z = 4 =

41.

{

Z

{ -x4 - -y3 = 0 40. 3 =2 -x6 + 2y

-+-=8

-1

{ 2X - � y = 38. x+ y=

{ 2X + = -4 -2y + 4z = 0 3x - 2z = -11 { 2x - 3y - z = 0 46. -x + 2y + = 5 3x - 4y - z = 1 { 3x - 2y + 2z = 6 7x - 3y + 2z = - 1 2x - 3y + 4z = 0 Y

42.

Z

SO.

=

v

- ,

and solve for

u.

and

v.

Then

x

1

=u.

1

-.J v

{ X - 2y + 3z = 7 2x + + Z = 4 -3x + 2y - 2z = -10 { x- y- z= 1 47. -x + 2y - 3z = -4 3x - 2y - 7z = 0 { x+ y- z= 6 51. 3x - 2y + z -5 x + 3y - 2z = 14 { X + 4y - 3z -8 54. 3x - y + 3z = 12 x + y + 6z 1

43.

Y

=

{ 2x + - 3z 0 -2x + 2y + = -7 3x - 4y - 3z = 7 { 2X - 3y - z = 0 48. 3x + 2y + 2z = 2 x + 5y + 3z 2 { X - y + Z = -4 52. 2x - 3y + 4z = -15 5x + y - 2z 12 Y

44.

=

Z

=

=

=

Y

=


90

55. The perimeter of a rectangular floor is feet. Find the dimensions of the floor if the length is twice the width.

(a) How much should b e invested i n each t o realize exactly

56. The length of fence required to enclose a rectangular field is meters. What are the dimensions of the field if it is known that the difference between its length and width is meters?

(b) I f, after years, the couple requires per year in income, how should they reallocate their investment to achieve the new amount?

3000 50

57.

2005

$12,000? 2

61.

55

Orbital Launches In there was a total of commer cial and noncommercial orbital launches worldwide. In ad dition, the number of noncommercial orbital launches was one more than twice the number of commercia l orbital launches. Determine the number of commercial and non commercial orbital launches in Source: Federal A viation Administration

Computing Wind Speed With a tail wind, a small Piper air craft can fly miles in hours. Against this same wind, the Piper can fly the same distance in hours. Find the average wind speed and the average airspeed of the Piper.

600

2005.

58.

325 $2495.

59.

30

60.

$7.00

$1.50 $3.00

$12,000

A recently retired couple needs per year to supplement their Social Security. They have to invest to obtain this income. They have decided on two investment options: AA bonds yielding per annum and a Bank Certificate yielding Financial Planning

$150,000

5%.

10%

3

62.

63.

64.

4

hours

......

I

$5.00

A store sells cashews for per pound and peanuts for per pound. The manager decides to mix pounds of peanuts with some cashews and sell the mixture for per pound. How many pounds of cashews should be mixed with the peanuts so that the mixture will produce the same revenue as would selling the nuts separately? Mixing Nuts

3

$9.00

Movie Theater Tickets A movie theater charges for adults and for senior citizens. On a day when people paid an admission, the total receipts were How many who paid were adults? How many were seniors?

$14,000

-'-��----

600

mi.

..

--------.·

I

Computing Wind Speed TIle average airspeed of a single engine aircraft is miles per hour. If the aircraft flew the same distance in hours with the wind as it flew in hours against the wind, what was the wind speed?

150 2

3

Restaurant Management A restaurant manager wants to purchase sets of dishes. One design costs per set, while another costs per set. If she only has to spend, how many of each design should be ordered?

200

$45

$25 $7400

10

One group of people purchased hot dogs and soft drinks at a cost of A second bought hot dogs and soft drinks at a cost of What is the cost of a single hot dog? A single soft drink?

Cost of Fast Food

7

5

4

$35.00. $25.25.

SECTION 12.1 We paid $35.00. How much is one hot dog? How much is one soda?

We paid $25.25. How much is one hot dog? How much is one soda?

72.

Systems of Linear Equations: Substitution and Elimi nation

849

IS-LM Model in Economics In economics, the IS curve is a linear equation that represents all combinations of income Y and interest rates that maintain an equilibrium in the mar ket for goods in the economy. The LM curve is a linear equa tion that represents all combinations of income Y and interest rates r that maintain an equilibrium in the market for money in the economy. In an economy, suppose the equilibrium level of income (in millions of dollars) and interest rates satisfy the system of equations

r

{

0.05Y - 1000r = 10 0.05Y + 800r = 100

Find the equilibrium level of income and interest rates.

73. 65.

66.

67.

Computing a Refund The grocery store we use does not mark prices on its goods. My wife went to this store, bought three I -pound packages of bacon and two cartons of eggs, and paid a total of $ 13 .45 . Not knowing that she went to the store, I also went to the same store, purchased two I -pound packages of bacon and three cartons of eggs, and paid a total of $ 1 1 .45. Now we want to return two I -pound packages of bacon and two cartons of eggs. How much will be refunded?

1 2 , and 1 3 ,

A doctor's prescription calls for a daily intake containing 40 mg of vitamin C and 30 mg of vitamin D. Your pharmacy stocks two liquids that can be used: one contains 20% vitamin C and 30% vitamin D, the other 40% vitamin C and 20% vitamin D. How many milligrams of each compound should be mixed to fill the prescription?

'3

-

'1 -

5V +

3n

10V

Pharmacy

Pharmacy A doctor's prescription calls for the creation of pills that contain 12 units of vitamin B I 2 and 1 2 units of vita min E. Your pharmacy stocks two powders that can be used to make these pills: one contains 20% vitamin B 1 2 and 30% vitamin E, the other 40% vitamin B I 2 and 20% vitamin E. How many units of each powder should be mixed in each pill?

69.

Curve Fitting

y

c

& 3rd ed., by Serway. 1990 Brooks/Cole, a division of Thomson Learning.

Source:

74.

©

Electricity: }(jrchhoff's Rules An application of Kirchhoff's Rules to the circuit shown results in the following system of equations:

c

c

/2 , and 1 3 , an

'1 �

Curve Fitting Find real numbers a , b , and so that the graph of the function = ax2 + bx + contains the points ( 1 - 2 ) , ( 1 , - 4 ) , and (2, 4 ) .

y

Physicsfor Scientists Engineers,

Find the currents 1 1 ,

Find real numbers a, b, and so that the graph of the function = ax2 + bx + contains the points ( - 1 , 4 ) , (2, 3 ) , and (0, 1 ) .

-

71.

Find the currents 1 1 ,

Finding the Current of a Stream Pamela requires 3 hours to swim 15 miles downstream on the I llinois River. The return trip upstream takes 5 hours. Find Pamela's average speed in still water. How fast is the current? (Assume that Pamela's speed is the same i n each direction.)

68.

70.

An application of Kirchhoff's Rules to the circuit shown results in the fol lowing system of equations: Electricity: Kirchhoff's Rules

c

4V

,

IS-L M Model i n Economics I n economics, the I S curve i s a linear equation that represents all combinations of income Y and interest rates that maintain an equilibrium in the mar ket for goods in the economy. The LM curve is a linear equa tion that represents all combinations of income Y and interest rates r that maintain an equilibrium in the market for money in the economy. In an economy, suppose the equilibrium level of income (in millions of dollars) and interest rates satisfy the system of equations

r

{

0.06Y - 5000r 0.06Y + 6000r

= 240 = 900

Find the equilibrium level of income and interest rates.

& 3rd ed., by Serway. 1 990 Brooks/Cole, a division of Thomson Learning.

Source:

75.

©

Physics for Scientists Engineers,

A B roadway theater has 500 seats, di vided into orchestra, main, and balcony seating. Orchestra seats sell for $50, main seats for $35, and balcony seats for $25 . I f all the seats are sold, the gross revenue to the theater is $ 1 7 , 1 00. If all the main and balcony seats are sold, but only half the orchestra seats are sold, the gross revenue is $14,600. How many are there of each kind of seat?

Theater Revenues

850

76.

CHAPTER 12

Systems of Equations and I nequalities

A movie theater charges $8. 0 0 for adults, $4.50 for children, and $6.00 for senior citizens. One day the theater sold 405 tickets and collected $2320 in receipts. There were twice as many children's tickets sold as adult tickets. How many adults, children, and senior citizens went to the theater that day?

77. Nutrition

A dietitian wishes a patient to have a meal that has grams of protein, grams of carbohydrates, and milligrams of calcium . The hospital food service tells the dietitian that the dinner for today is chicken, corn, and milk. Each serving of chicken has grams of protein, grams of carbohydrates, and milligrams of calcium. Each serving of corn has 3 grams of protein, grams of carbohydrates, and milligrams of calcium. Each glass of 2 milk has grams of protein, grams of carbohydrates, and milligrams of calcium. How many servings of each food should the dietit ian provide for the patient?

910

66

10

30 16

35

%

13

80.

2% 9

300

$20,000 5%

Kelly has to invest. As her financial planner, you recommend that she diversify i nto three invest ments: Treasury bills that yield simple interest, Treasury bonds that yield simple interest, and corporate bonds that yield simple interest. Kelly wishes to earn per year in income. Also, Kelly wants her investment in Treasury bills to be $3000 more than her investment i n corporate bonds. How much money should Kelly place in each investment? Investments

7%

10%

79.

$2.25,

94.5

200

78.

sufficient information to determine the price of each food item? If not, construct a table showing the various possibili ties. Assume that the h a mburgers cost between and the fries between and and the colas between and

Theater Revenues

P rices of Fast Food

81.

$0.60

$0.90.

$0.75

$1.00,

$1.75

Use the information given in Prob lem Suppose that a third group purchased deluxe ham burgers, large fries, and large colas for Now is there sufficient information to determine the price of each food item? If so, determine each price. Prices of Fast Food

79.

2

4

3 $10.95.

Painting a House Three painters, Beth, Bill, and Edie, working together, can paint the exterior of a home in hours. Bill and Edie together have painted a similar house in hours. One day, all three worked on this same kind of house for hours, after which Edie left. Beth and Bill re quired 8 more hours to finish. Assuming no gain or loss in efficiency, how long should it take each person to complete such a job alone?

10

15

4

$1390

One group of customers bought

8 deluxe hamburgers, orders of large fries, and large colas

$26.10.

6

10

6 $31.60.

for A second group ordered deluxe hamburgers, large fries, and 8 large colas and paid Is there

6


Make up a system of three linear equations containing three variables that has: (a) No solution (b) Exactly one solution (c) Infinitely many solutions

83. Write a brief paragraph outlining your strategy for solving a system of two linear equations containing two variables. 84. Do you prefer the method of substitution or the method of elimination for solving a system of two linear equations con taining two variables? Give reasons.

Give the three systems to a friend to solve and critique. 'Are You Prepared?' Answers

1. { I }

2.

(a)

(b)

y

3 4

-2 12.2 Systems of Linear Equations: Matrices OBJECTIVES

1 Write the Augmented Matrix of a System of Linea r Equations (p. 851 ) 2 Write the System of Equations from the Aug mented Matrix (p. 852) 3 Perform Row Operations on a Matrix (p. 853) 4 Solve a System of Linea r Equations U s i n g Matrices (p. 854)

SECTION 12.2

Systems of Linear Equations: Matrices

851

The systematic approach of the method of elimination for solving a system of linear equations provides another method of solution that involves a simplified notation. Consider the following system of linear equations:

{ X 2y4y 14 3x

=

+

= 0

-

If we choose not to write the symbols used for the variables, we can represent this system as

1� ]

4 -2

where it is understood that the first column represents the coefficients of the vari able x, the second column the coefficients of and the third column the constants on the right side of the equal signs. The vertical line serves as a reminder of the equal signs. The large square brackets are used to denote a in algebra.

y,

matrix

A

DEFINITION

matrix

is defined as a rectangular array of numbers, Col u m n 1

Row 1

Colum n 2

Col u m n )

Col u m n n

Row 2

al l

a21

al 2

a22

a2j

a lj

al n

Row i

ail

ail

aij

ain

Row m

amI

am2

a,nj

amn

a2n

(1)

-.J

Each number aU of the matrix has two indexes: the row index i and the column The matrix shown in display has rows and n columns. The numbers aij are usually referred to as the entries of the matrix. For example, a23 refers to the entry in the second row, third column.

(1)

index j.

1

m

Write the Augmented Matrix of a System of Linear Equations

Now we will use matrix notation to represent a system of linear equations. The matrix used to represent a system of linear equations is called an augmented matrix. In writing the augmented matrix of a system, the variables of each equation must be on the left side of the equal sign and the constants on the right side. A variable that does not appear in an equation has a coefficient of O. E XA M P L E 1

{ 2x

Writin g the Augmented M atrix of a System of L i n ear E q u ations

{ 4y 2x - y

Write the augmented matrix of each system of equations. (a) Solution

3x

-

3

=

=

6 -5

-

(1)

(b)

(2)

(a) The augmented matrix is

[�

-

x + x +

y+Z=0 1 0 2y 8 0 z

-

=

-

=

(1 ) (2)

(3)

-4 -3

(b) Care must be taken that the system be written so that the coefficients of all variables are present (if any variable is missing, its coefficient is 0). Also, all

852

CHAPTER 12

Systems of Equations and I nequalities

constants must be to the right of the equal sign. We need to rearrange the given system as follows: (1)

{ 2X - Y + Z = 0 x+ Z - 1 =0 x + 2y - 8 = 0

{

(2)

(3)

2x - Y + z = 0 (1) x + o · y + Z = 1 (2) x + 2y + o · z = 8 (3)

The augmented matrix is

•

If we do not include the constants to the right of the equal sign, that is, to the right of the vertical bar in the augmented matrix of a system of equations, the result ing matrix is called the coefficient matrix of the system. For the systems discussed in Example 1, the coefficient matrices are

[ � =�J

"'I'

2 EXAM P LE 2

=-


and

[; -� r

Write the System of Equations from the Augmented Matrix Writing the System of Li near E q u ations from the Augmented M atrix

Write the system of linear equations corresponding to each augmented matrix. [3 -1 -1 (b) 2 0 2 0 1 1

13J -10 Solution

(a) The matrix has two rows and so represents a system of two equations. The two columns to the left of the vertical bar indicate that the system has two variables. If x and y are used to denote these variables, the system of equations is

{

5x + 2y = 13 - 3x + Y = - 10

(1)

(2)

(b) Since the augmented matrix has three rows, it represents a system of three equations. Since there are three columns to the left of the vertical bar, the sys tem contains three variables. If x, y, and z are the three variables, the system of equations is

{

3x - Y - z = 7 (1) + 2z = 8 (2) 2x Y + z = 0 (3)

•

SECTION U.2 3

853


Perform Row Operations on a Matrix Row operations on a matrix are used to solve systems of equations when the system is written as an augmented matrix. There are three basic row operations.

Row Operations

1. Interchange any two rows.

2. Replace a row by a nonzero multiple of that row. 3.

Replace a row by the sum of that row and a constant nonzero multiple of some other row.

These three row operations correspond to the three rules given earlier for obtaining an equivalent system of equations. When a row operation is performed on a matrix, the resulting matrix represents a system of equations equivalent to the sys tem represented by the original matrix. For example, consider the augmented matrix 2 -1

[ �J

Suppose that w e want t o apply a row operation t o this matrix that results i n a matrix whose entry in row 2, column 1 is a O. The row operation to use is Multiply each entry in row 1 by -4 and add the result to the corresponding entries in row 2.

(2)

If we use R2 to represent the new entries in row 2 and we use r1 and r2 to rep resent the original entries in rows 1 and 2, respectively, we can represent the row operation in statement (2) by Then

-1

2

3 2

J [ T

1 2 -4( 1 ) + 4 -4(2) + ( - 1 )

R2 = -4r1

+

r2

As desired, we now have the entry 0 in row 2, column E XA M P LE 3

2 -9

3 -4(3)

-1

[ �]

1.

Applying a Row Operation to a n Augmented M atrix

Apply the row operation R2 = -3r1 + r2 to the augmented matrix

[�

Solution

D

-2 -5

-2 -5

[ �J

-3

The row operation R2 = -3r1 + r2 tells us that the entries in row 2 are to be replaced by the entries obtained after multiplying each entry in row 1 by and adding the result to the corresponding entries in row 2.

n T [ -3 ( �)

+ 3

R2 = -3r, + r2

(-3) (

-2 -2) + ( -5 )


2 -3(2) +

9

J [�

-2 1

[ �J

•

854

CHAPTER

12

Systems of Equations and I nequa l ities

F i n d i n g a Particular Row Operatio n

E XA M P L E 4

Find a row operation that will result in the augmented matrix having a 0 in row 1, column 2.

We want a 0 in row 1, column 2. This result can be accomplished by multiplying row 2 by 2 and adding the result to row 1 . That is, we apply the row operation R l = 2r2 + rJ .

Solution

[�

-2 1

2 J ---;. [2(0) 0 3

+

r R, = 2r2 + r,

1 2( 1 ) 1

+

( -2 )

[� � I �J

•

A word about the notation that we have introduced. A row operation such as 2r2 + rl changes the entries in row 1 . Note also that for this type of row oper ation we change the entries in a given row by multiplying the entries in some other row by an appropriate nonzero number and adding the results to the original entries of the row to be changed. Rl

4

=

Solve a System of Linear Equations Using Matrices

To solve a system of linear equations using matrices, we use row operations on the augmented matrix of the system to obtain a matrix that is in ro w echelon form. DEFINITION

A matrix is in row echelon form when the following conditions are met:

The entry in row 1, column 1 is a 1, and only O's appear below it. 2. The first nonzero entry in each row after the first row is a 1, only O's appear below it, and the 1 appears to the right of the first nonzero entry in any row above. 3. Any rows that contain all O's to the left of the vertical bar appear at the bottom . 1.

[1 b

For example, for a system of three equations containing three variables with a unique solution, the augmented matrix is in row echelon form if it is of the form a

o 1 c 0 0 1

where a , c , e, and f are real numbers. The last row of this augmented matrix states that z = We can then determine the value of y using back-substitution with z = f, since row 2 represents the equation y + cz = e. Finally, x is determined using back-substitution again. Two advantages of solving a system of equations by writing the augmented matrix in row echelon form are the following:

b, d,f.

1.

The process is algorithmic; that is, it consists of repetitive steps that can be pro grammed on a computer. 2. The process works on any system of linear equations, no matter how many equa tions or variables are present. The next example shows how to write a matrix in row echelon form.

SECTION 12.2

E XA M P L E 5


855

Solving a System of Li near E q u ations U s i n g M atrices ( Row Echelon Form)

Solve: Sol ution

=

6 (1)

{2Xx ++ 2YY + = 1 3x + 4y - z = 13 Z

(2)

(3)

[ 1 21 01

First, we write the augmented matrix that represents this system. 2

3 4 -1 1 1,

The first step requires getting the entry i n row column A n interchange of rows and is the easiest way to do this. [Note that this is equivalent to inter changing equations and (2) of the system. ]

1.

[3� 4 -1� 13 � �] -2r0 l + r2 1-3rl + 0r} 3, 1. [3I 42I -10 13I ] 1 [ 00I 011 -4-21 I n

1 2 (1)

Next, we want a in row 2, column and a in row column We use the row operations R2 = and R} = to accomplish this. Notice that row is unchanged using these row operations. Also, do you see that performing these row operations simultaneously is the same as doing one followed by the other? 1

2

Now we want the entry complish this.

6

�

R2 = - 2r1 + r2 R3 = -3r1 + r3

1

1

in row 2, column

Interchanging rows

2. 2 [ 1 01 -21 ! ] � [ � � -! 1�] 1 -4 10 0 0 -2 1 3, 1 . R} = - r}. [ � 0 -41 1041 ] [100 011 -411 -21�] �

and 3 will ac

o o

Finally, w e want a 2"

i n row

The result IS

column 3. To obtain it, w e use the row operation

�

o

-2

R3

I

=

1 - - r3 2

This matrix is the row echelon form of the augmented matrix. The third row of this matrix represents the equation -2. Using we back-substitute into the equation = ( from the second row ) and obtain

y - 4z 10

z = z = -2, y 4z 10 y - 4( Y == 210 -2 )

=

z =

-2

Solve for y.

856

CHAPTER

12


Finally, we back-substitute y = 2 and z = -2 into the equation x + y + z = 1 (from the first row) and obtain x + y + z = 1 x + 2 + ( -2 ) = 1 x = 1

y

= 2, Z

=

-2

Solve for x.

The solution of the system is x = 1 , y = 2, z = -2 or, using ordered triplets, ( 1 , 2, -2 ) .

•

The steps that we used to solve the system of linear equations in Example 5 can be summarized as follows: Matrix Method for Solving a System of Linear Equations ( Row Echelon Form)

r

r

r r

r

In Words

To obtai n an augmented matrix in row echelon form: •

Add rows, excha nge rows, or multiply the row by a nonzero r consta nt r • Work from top to bottom a nd r left to right r • Get 1's i n the main diagonal r with O's below the 1's.

STEP 1: STEP 2: STEP 3:

STEP 4:

STEP 5: STEP 6:

Write the augmented matrix that represents the system. Perform row operations that place the entry 1 in row 1 , column Perform row operations that leave the entry 1 in row 1 , column 1 unchanged, while causing O's to appear below it in column Perform row operations that place the entry 1 in row 2, column 2, but leave the entries in columns to the left unchanged. If it is impossible to place a 1 in row 2, column 2, proceed to place a 1 in row 2, column 3. Once a 1 is in place, perform row operations to place O's below it. [Place any rows that contain only O's on the left side of the vertical bar, at the bottom of the matrix.] Now repeat Step 4, placing a 1 in the next row, but one column to the right. Continue until the bottom row or the vertical bar is reached. The matrix that results is the row echelon form of the augmented matrix. Analyze the system of equations corresponding to it to solve the original system.

l.

l.

In the next example, we solve a system of linear equations using these steps. E XA M P L E 6

( Row Echelon Form)

Solve: Solution

STEP 1:

STEP 2: STEP 3:

r

r

r

r

I n Words

Notice we use m u ltiples of the entry in row 1, col u m n 1 to obtai n O's i n t h e entries below t h e 1 .

{

Solving a System of Li near E q u ations Using M atrices

x - y + z � 8 2x + 3y - z = -2 3x - 2y - 9z = 9

(1 )

(2)

(3)

The augmented matrix of the system is

U

-1 3 -2

1 -1 -9

-

n

Because the entry 1 is already present in row 1 , column 1 , we can go to step 3. Perform the row operations R2 = - 2r1 + r2 and R3 = -3r1 + r3 ' Each of these leaves the entry 1 in row 1, column 1 unchanged, while causing O's to appear under it.

[�

-1 3 -2

1 -1 -9

-�hD

-1 5 1

R2 -2 r, + r2 R3 = -3r, + r3 =

1 -3 - 12

�

-1 -15

]

STEP 4:

12

SECTION 12.2

1

2

1

2

by 5' but this introduces fractions) .

[� -� -1� -1-18�] 1 2, 1 8 ] [1 -1 1 1 -12 -1�] -12 -18 rO O

To get a 0 under the R3 = -5r 2 + r 3 '

-3

5

o

in row 2, column

-)

-15

-3

R3

=

perform the row operation

0

57

57

-5r2 + r3

1 in row 3, column 3 by using R3 517

[� -� -�� -:�h U -� -1� -In

Continuing, we obtain a

=

R3

STEP 6:

2,

857

The easiest way to obtain the entry in row column without altering column is to interchange rows and 3. (Another way would be to multiply row

STEP 5 :


=

{X - - 12z

r3'

1 -r3 57

The matrix on the right is the row echelon form of the augmented matrix. The system of equations represented by the matrix in row echelon form is y +

z =

Y

=

z

=

=

8 -15 1

z 1, we back-substitute to get {X - - 12 -158 (1)

Using

y +

Y

1

=

(1 )

=

(2)

Simplify.

(1) (2)

{X (3)

We get y = -3, and back-substituting into x The solution of the system is x = 4, Y = - 3,

-

z

(4, - 3 , 1 ) .

y =

y =

7 -3

(1 ) (2)

y =

= 1

7, we find that x = 4. or, using ordered triplets, •

Sometimes it is advantageous to write a matrix in reduced row echelon form. In this form, row operations are used to obtain entries that are 0 above (as well as below) the leading in a row. For example, the row echelon form obtained in the solution to Example 6 is

1

-1 1 1 -12 o 1

-1:] r U

-In

-1� ] [I 1 r� n

To write this matrix in reduced row echelon fonn, we proceed as follows:

[�

-1 1

0

1 -12 1

RI

=

0

-11 1 -12 0

r2 +rl

1

-7

RI R2

=

=

0

0

0

1

0

1 1 r3 + rl 1 2r3 + r2

858

CHAPTER 12

Systems of Equations a n d Inequalities

The matrix is now written in reduced row echelon form. The advantage of writing the matrix in this form is that the solution to the system, = 4, Y = Z = is readily found, without the need to back-substitute. Another advantage will be seen in Section where the inverse of a matrix is discussed. The methodology used to write a matrix in reduced row echelon form is called Gauss-Jordan

x

-3,

1,

12.4,

elimination. Now Work P R O B LE M S 3 7 A N D 4 7

�

The matrix method for solving a system of linear equations also identifies systems that have infinitely many solutions and systems that are inconsistent. Let's see how.

{ -12x -+ 2yy -+ 2z

Solving a Dependent System of Linear E quations Using M atrices

E X A MPL E 7

Z

6x

Solve:

5x

Solution

+

Z

y-

=

=

=

4

(1)

-8 3

(2)

(3)

We start with the augmented matrix of the system and proceed to obtain a column with O's below.

[

1[ -�h - � +

1 -1 -1 2 2 1 -1

-1�

=

1

-1r3

2,

in row

1,

-�]3 �r [� 1 -n 0

0

-2 2 2 1 -1

R1

1

-2 -22 2 1 -1

0

R2 R3

r1

2

=

=

12r1 + r2 -5r1 + r3

1

Obtaining a in row column without altering column can be accom. r2 or by R3 = il r3 and mterchanging rows and or by phshed by R2 =

- 221

1

.

2

3

R2 = il r3 + r2' We shall use the flrst of these.

23

u

.

0

-2 -22 2 -1 11

�]�r:

-2 I

0

R2

=

:l �r:

0

-2 1 111 11 -1

- 12 -2 I

1

- 22 r 2

R3

{ yX--12y

0

=

0

1 - 112

-2 1 111 0

-11r2

+

0

0

r3

l

This matrix is in row echelon form. Because the bottom row consists entirely of O's, the system actually consists of only two equations. =

1 2 11 Z = - 11

(1) (2)

x

y

To make it easier to write down some of the solutions, we express both and in terms of -. Now back-substitute this solution From the second equation, Z for mto . the f'lrst equatIOn to get

z.

y 111 - 112 x 2y + 1 = 2(1ilz - il2) + 1 li2 Z + 117 =

y

-

.

=

=

{ x = -Z112 + 117 SECTION 12.2


859

The original system is equivalent to the system - (1)

liz - 112 1

Y =

(2)

where Z can be any real number. Let's look at the situation. The original system of three equations is equivalent to a system containing two equations. This means that any values of that satisfy both

x 112 +-117 , = "x = -1, x = li' y = - li =

-z

will be solutions. For example z and z

3

5

0

x, y, Z

y = li1 � 112 711' y = --'11'2 Z = 1 , x = -11' y 1 x, y,

and

-

7

-

9

-

=

=

0f

are some

11'

the solutIOns of the ongll1aI system. ' .

.

{ x = 112 + -117

There are, in fact, infinitely many values of and zfor which the two equations are satisfied. That is, the original system has infinitely many solutions. We will write the solution of the original system as -z

y = liZ - li2 1

where

z

can be any real number, or, using ordered triplets, as

({ x, y,z) lx = 121z + 117 ' y = 1 2

}

, zany real number . z ll -ll

[I -2l - � - 1 ]

•

We can also find the solution by writing the augmented matrix in reduced row echelon form. Starting with the row echelon form, we have

o o

0

I�

li 0

{x - 2

� R,

1 112 : � 11 0

1

-

=

0

711 211

-

0

2r2 + r,

The matrix on the right is in reduced row echelon form. The corresponding system of equations is

711 y - li1 Z = -li2 x 112 + 117 y liZ1 - li2 -z = 11

or, equivalently,

where 'L.''1

z

{

=

=

can be any real number. -

Now Work P R O B L E M S 3

-z

- (1) (2)

- (1) (2)

860

CHAPTER 12

Systems of Equations a n d Ineq u a l ities

E XA M P L E

8

{X+2x - YY +Z- z == 63 x + 2y + 2z 0 1� -3 -31 -� [� 1 11 -:] [� 11 11 O�]r�O[ 1 1 -6] r 0 -3 -3 r 0 0 0 OxOy+ Oz+ -27 Solving an Inconsistent System of Linear Equations

Using M atrices

Solve:

Solution

We proceed as follows, beginning with the augmented matrix. 1

�

R2 = -2r1 + r2 R3 = -1r1 + r3

�

Intercha nge rows 2 and 3. 9

R3

-

=

3r2 + r3

6-6] -27

This matrix is in row echelon form. The bottom row is equivalent to the equation =

which has no solution. The original system is inconsistent. 1!E

•


The matrix method is especially effective for systems of equations for which the number of equations and the number of variables are unequal. Here, too, such a sys tem is either inconsistent or consistent. If it is consistent, it will have either exactly one solution or infinitely many solutions. Let's look at a system of four equations containing three variables.

EXAMPLE

9

{ 2Xx -2y+ 2y +-3zz == -30 z Y -x + 4 y 2z == 13 -2 -31 -30l l0l -26 -51 -3 I0 -21 -11 -l � 41 -12 -113 r 00 21 -13 -113°l rI00 62 -53 -2 1 -2 IrOI O0 -151 -1l�l � I:I0 00 -10 -ll + °

Solving a System of Linear Equations Using Matrices

(1 )

Solve:

-

+

Solution

1

(2)

(3) (4)

We proceed as follows, beginning with the augmented matrix.

2

�

�

R2 = -2r1 + r2 R4 = r1 + r4

�

o

o

1

R3 = -6r2 r3 -2r2 + r4 R4 =

1

--3� 13l

Interchange rows 2 and 3.

1

R4 = -5r3 + r4

1

SECTION 12.2

86 1


I

ll

We could stop here, since the matrix is in row echelon form, and back-substitute z = 3 to find x and y. Or we can continue to obtain the reduced row echelon form.

O O rO

1 -2 1 -1 1 0 0 0

l�

-1 3

JI --i>

Rl

=

0 -1 0 1 -1 1 0 0 0 0 2r2

-2 -1 3 o

--i>

r

Rl R2

+ rl

1 0 0 o

=

=

0 1 0 0

0 0 1 0

r3 + rl r3 + r2

�l

The matrix is now in reduced row echelon form, and we can see that the solution is x = 1, y = 2, z = 3 , or using ordered triplets, (1,2, 3 ). - Now Work P R O B L E M 6 9

.""

E X A M P L E 10

•

Penalties in the 2006 Fifa World Cup

Italy and France combined for a total of 46 penalties during the 2006 Fifa World Cup. The penalties were a combination of fouls, yellow cards (cautions), and red cards (expulsions). There was one less red card than half the number of yellow cards and one more foul than 8 times the total number of cards. How many of each type of penalty were there during the match?

Source: f�fa worldcup. com Solution

Letf�y, and r represent the number of fouls, yellow cards, and red cards, respectively. There was a total of 46 penalties, which is found by adding the number of fouls, yellow cards, and red cards. The first equation is f + Y + r 46. There was one less red card than half the number of yellow cards, so, the number of red cards equals 1 . . . "2 the number of yellow cards, mmus 1. ThIs statement leads to the second equatIOn: =

r =

�y - 1. We also know that there was one more foul than 8 times the total

number of cards, so, the number of fouls equals 1 plus the product of 8 and the sum of the number of yellow cards and the number of red cards. This statement gives the third equation: f = 1 + 8 (y + r). Putting each equation in standard form, we have the following system of equations:

r

+

1

--

�

y+ r y

r

+

=

=

f - y - 8r =

46

(1)

-1 1

(2)

(3)

l -I �[�

We begin with the augmented matrix and proceed as follows:

l:

1 1 2 -8

1

1

-8

�:J�r l: -i 1

0

[I

R3

=

-rl + r3

1 0 1

rO O

--i>

R3

-9

=

9r2 +

1

1 -9

1 -2 -27 r3

46

-4 46 2 -27

r

] [II 5

R2

=

R3

0 1

=

46

1 -2 -9

�

-4

-2r2

rO O

--i>

1 1 -9

1

1 -2 1

--r3 27

4

n

]

862

CHAPTER 12

Systems of Equations and Ineq u a l ities

{f

The matrix is now in echelon form. The final matrix represents the system + Y+ Y

I' = =

- 21' '

I

46

(1)

2

(2)

1

=

( 3)

From equation (3), we determine that 1 red card was given. Back-substitute I' = 1 into equation (2) to find that y = 4, so 4 yellow cards were given. Back-substitute these values into equation ( 1 ) and to find that = 41, so 41 fouls were called.

i f4,!.····.·.··.·j �

f

COMMENT Most g raphing utilities have the capabil ity to put an aug mented matrix into row echelon form (ref) and also red uced row echelon form (rref). See the Appendix, Section 7, for a _ discussion.

1 2 .2 Assess Your Understanding An m by n rectangular array of numbers is called a (n )

Concepts and Vocabulary 1.

3. True or False

The augmented matrix of a system of two equations containing three variables has two rows and four columns.

In Problems

Skill Building 5-16,

.:l.

9.

{O.OlX - 0.03y : 0.06 O .13x+0.10y - 0.20

13.

1[ 1

The matrix used t o represent a system o f linear equations is called a(n) matrix. __

4. True or False

The matrix

form.

3

6.

X+Y-Z= 2 3x - 2y = 2 5x+3y - Z = 1

{

{3X+4y = 7 4x - 2y = 5

7.

a

11.

14.

2X+3y - 4z = a { x - 5z+ 2 = a x+2y - 3z = -2

15.

{ 2X+3y - 6 = a 4x - 6y+2 = a

{

12.

2:: 2� -3x+4y = 4x - 5y+ Z =

�:

: ��

5

a

�

a

8.

X - y+ Z = 10 { 3x + 3y = 5 x+Y+2z = 2

-2 ] is in row echelon

a

write the augmenced matrix of the given system of equ.ations.

{ x - 5y = 5 4x+3y = 6

_

2.

16.

{

{

9x - y = a 3x - Y - 4= a

5x - y - z = a x+ y = 5 2x - 3z = 2

X - y + 2z- w = 5 { x + 3y- 4z+2w = 2 3x- y- 5z- w= -l

In Problems 1 7-24, write the system of equations corresponding to each augmented matrix. Then pelform each row operation on the given au.gmented matrix.

,17. 19.

21

·

23

D

-3 -5

-3 -5 3

I -�J 4 6 4

R2 = -21'] +1'2

[ � �J U [� 1 -5

-4

-3 -5 -6

-3 -5

2 3 4 1

6 4

-6 -4] 6

-2 -2] 6

(a) R2= -31'] +1'2 (b) R3 = 5 1'1 + 1'3

(a) R2 =-21'] +1'2 (b) R3 = 31'] +1'3

(a) R] =-21'2+1'] (b) R[ = 1'3 +1']

18.

20.

22.

24.

[�

-3 -5

[ -!

-3

-3 -5 2

I

-3 -4J

3 -3 4

-3 -4 -5 6 4 -

U 1 U

-3 -5 -6

-1

2 4

R2 =-21'1 +r 2

-5 -5] 6

-6 -6] 6

n

(a) R2 = 4r 1 + 1'2 (b) R3 = 31'] + r3

(a) R2 =-6r1 +1'2 (b) R3 = 1'] + r3

(a) R[ =-r 2+1'1 (b) R1 = 1'3+1'1

SECTION 12.2


863

In Problems 25-36, the reduced row echelon form of a system of linear equations is given. Write the system of equations corresponding {o {he given matrix. Use x, y; or x, y, z; or XI, X20 X3, X4 as variables. De{ermine whether {he system is consistent or inconsistent. If it is consistent, give the solution. 25.

28.

[

1

1

[' [I [I

0

34.

1

0

0

0

31.

I -n

0

n

0

0

0

0

0

0

0

1

0

0

0

0

1

0

a

a

0

1

0

0

o

0

1

]

2

2

26 .

29.

[

1

0

0

1

[' [I

2 1 -4 0 0 0

0

o

n �j

32.

0

l'

1

0

o

35.

I -�J

0

0 0 1

"

-1 -2

] n

27.

30.

0

2 3

0

0

a

0

1

0

0

0

1

1 2 -1

0

0

0

0

-�l

33.

['

0

0

0

1

0

o

0

0

[I

4 3

1

o

0

0

0

0

4 3

0

0

0 0

[I

II

0 0

36.

0

a

1

1

a

a

0

1

0

o

0

o

0

1

0

iJ �]

n

il

0 0 1

In Problems 3 7- 72, solve each sysrem of equations using matrices (row operations). If the system has no solution, say that it is inconsistent. 37.

40.

43.

46.

49.

52.

55.

58.

61 .

{ X +Y = 8 X- Y= 4

{

38.

3x + .Jy = .J 8 4x+2y = 3 �

41 .

�

{2X +3y = 6 1 x - y ="2 {2X - y x

-

-I--y =

2 1

44.

=

�

2

X - 2y+3z = 7 { 2x+ y+ Z = 4 -3x+2y - 2z = -10 2x - 3y - z = 0 { -x+2y+Z = 5 3x - 4y - z = 1

2x - 2y+3z = 6 4x - 3y+2z = 0 -2x+3y - 7z = 1

{

X- y+ Z = -4 { 2x- 3Y+4z = -15 5x+ y - 2z = 12 3x +

47.

50.

53.

" + 4r

z=l 8 2y =3

{ x+2y = 4 2x + 4y = 8 {�x

-I-

Y =

59.

-2

x- 2y = 8

45.

x- y= 6 { 2x - 3z = 1 6 2y+ z = 4

48.

2x+ Y - 3z = 0 { -2x+2Y+ Z =-7 3x - 4y - 3z = 7 -x+ Y+ Z = - 1 { � x+2y - 3z=-4 .JX - 2y - 7z = 0

62.

X + 2y- z = -3 2x - 4y+ Z = -7 -2x+2y - 3z = 4

{

{

39.

42.

3x - 2y+2z = 6 { 5. 6 7x- 3y+2z=-1 2x - 3y+4z = 0

2 y- z=3

2x- y +

{x+2y = 5 x+ Y = 3

+

Z=

�

" +2y+Z = 3

\:

= -2

=

3

{3X - Y = 7 9x- 3y = 21

{ 3x- 5y = 3 15x + 5y = 21

2x + Y = -4 { -2y+4z = 0 3x- 2z = - 1 1

2x- 2y - 2z 2 { 51.2x+3y+ Z = 2 3x+2y = 0 =

54.

57.

60.

x+y=l

2x- y

{ 2X - 4Y 3x+2y

63.

2X- 3y - z = 0 { 3x + 2y+2z = 2 x + 5y+3z = 2

x+ y- z= 6 { 3x- 2y + Z =-5 x+3y - 2z = 14 X + 4y - 3z = -8 { 3x - y+3z = 12 x + y+6z = 1

{

x+y+z+w= 4 2x - Y+Z = 0 3x+2y+Z - w = 6 x - 2y - 2z+2w = -1

864

64.

67.

70.

{ {

CHAPTER 12


x+y+z+w= 4 -x+2y + Z = o 2x + 3y+ Z - w= 6 -2x+ y - 2z+2w=-1

65.

x- y+ z=5 3x+2y - 2z=0

68.

X - 3y + Z = 1 { 2x - y - 4z=0 x - 3y+ 2Z =1 x - 2y =5

71.

{X + 2y+ Z =1 2x - y+2z=2 3x + Y+3z=3

{

66.

2x+Y - z=4 -x+Y+n=1

69.

�

{4X+ y+ z - w=4 x - y + 2z+3w=3

2 Curve Fitting Find the function y ax +bx + c whose graph contains the points (1,2), (-2, -7), and (2, -3). 2 Curve Fitting Find the function y = ax +bx + c whose graph contains the points (1, -1),(3, -1), and (-2, 14). 3 Curve Fitting Find the function f(x)=ax +bx2 + cx+d for which f(-3)= -11 2,f (-1)=-2,f (1)=4, and f(2)=13. 2 Curve Fitting Find the function f(x) =ax 3 +bx +

72.

{X+2y - z=3 2x - y + 2z=6 x - 3y+3z=4

{

{

2x + 3y - z=3 x- y - z=O -x + Y + Z =0 x+ Y+3z=5

-4x+y=5 2x - y+ Z - w=5 z+w = 4


74. 75. 76.

77.

78.

79.

=

cx +d for which f(-2)=-10,f(-1)= 3,f(1)=5, and f(3)=15.

Nutrition A dietitian at Palos Community Hospital wants a patient to have a meal that has 78 grams of protein,59 grams of carbohydrates, and 75 milligrams of vitamin A. The hospi tal food service tells the dietitian that the dinner for today is salmon steak, baked eggs,and acorn squash. Each serving of salmon steak has 30 grams of protein, 20 grams of caJ'bohy drates,and 2 milligrams of vitamin A. Each serving of baked eggs contains 15 grams of protein, 2 grams of carbohydrates, and 20 milligrams of vitamin A. Each serving of acorn squash contains 3 grams of protein, 25 grams of carbohydra tes, and 32 milligrams of vitamin A. How many servings of each food should the dietitian provide for the patient? Nutrition A dietitian at General Hospital wants a patient to have a meal that has 47 grams of protein, 58 grams of carbo hydrates, and 630 milligrams of calcium. The hospital food service tells the dietitian that the dinner for today is pork chops, corn on the cob, and 2% milk. Each serving of pork chops has 23 grams of protein,0 grams of carbohydrates, and 10 milligrams of calcium. Each serving of corn on the cob contains 3 grams of protein, 16 grams of carbohydrates, and 10 milligrams of calcium. Each glass of 2% milk contains 9 grams of protein, 13 grams of carbohydrates, and 300 mil ligrams of calcium. How many servings of each food should the dietitian provide for the patient?

Financial Planning Carletta has $10,000 to invest. As her financial consultant, you recommend that she invest in Treasury bills that yield 6%,Treasury bonds that yield 7%, and corporate bonds that yield 8%. Cadetta wants to have an annual income of $680, and the amount invested in corporate bonds must be half that invested in Treasury bills. Find the amount in each investment.

80.

81.

82.

83.

Landscaping A landscape company is hired to plant trees in three new subdivisions. The company charges the devel oper for each tree planted,an hourly rate to plant the trees, and a fixed delivery charge. In one subdivision it took 166 labor hours to plant 250 trees for a cost of $7520. In a second subdivision it took 124 labor hours to plant 200 trees for a cost of $5945. In the final subdivision it took 200 labor hours to plant 300 trees for a cost of $8985. Determine the cost for each tree, the hourly labor charge, and the fixed delivery charge. Sources: gurney. com; www.bx.org

Production To manufacture an automobile requires paint ing, drying, and polishing. Epsilon Motor Company produces three types of cars: the Delta, the Beta, and the Sigma. Each Delta requires 10 hours for painting, 3 hours for drying, and 2 hours for polishing. A Beta requires 16 hours for painting, 5 hours for drying, and 3 hours for polishing, and a Sigma re quires 8 hours for painting, 2 hours for drying,and 1 hour for polishing. If the company has 240 hours for painting, 69 hours for drying, and 41 hours for polishing per month, how many of each type of car are produced? Production A Florida juice company completes the prepa ration of its products by sterilizing, filling, and labeling bot tles. Each case of orange juice requires 9 minutes for sterilizing, 6 minutes for filling, and 1 minute for labeling. Each case of grapefruit juice requires 10 minutes for steriliz ing,4 minutes for filling,and 2 minutes for labeling. Each case of tomato juice requires 12 minutes for sterilizing,4 minutes for filling, and 1 minute for labeling. If the company runs the sterilizing machine for 398 minutes, the filling machine for 164 minutes, and the labeling machine for 58 minutes, how many cases of each type of juice are prepared?

{

Electricity: Kirchhoff's Rules An application of Kirchhoff's Rules to the circuit shown results in the following system of equations:

-4+8 - 212=0 8=514+11 4=313+11 13+14=11

Find the currents 11, [2, 13 , and 14,

SECTION U.3

3D

2D

Source:

84.

Based on Raymond Serway, Physics, 3rd ed. (Philadel phia: Saunders, 1990), Prob. 34, p. 790. Electricity: Kirchhoff's Rules An application of Kirchhoff's Rules to the circuit shown results in the following system of equations:

{

86.

11=13 +12 24 - 61, - 3 1 3 = 0 12 +24 - 61, - 612 0

Find the currents

=

11,

1 2 , and 1 3 , 87.

24 V 3D

88.

85.

Source:

865

that yield 9%, and some money in junk bonds that yield 1 1%. Prepare a table for each couple showing the various ways that their goals can be achieved: (a) If the first couple has $20,000 to invest. (b) If the second couple has $25,000 to invest. (c) If the third couple has $30,000 to invest. (d) What advice would you give each couple regarding the amount to invest and the choices available? [Hint: Higher yields generally carry more risk.] Financial Planning A young couple has $25,000 to invest. As their financial consultant, you recommend that they invest some money in Treasury bills that yield 7%, some money in corporate bonds that yield 9% , and some money in junk bonds that yield 1 1 %. Prepare a table showing the various ways that this couple can achieve the following goals: (a) The couple wants $1500 per year in income. (b) The couple wants $2000 per year in income. (c) The couple wants $2500 per year in income. (d) What advice would you give this couple regarding the income that they require and the choices available? [Hint: Higher yields generally carry more risk.]

12 V 5D

Systems of Linear Equations: Determ i nants

Ibid., Prob. 38, p. 791 .

Financial Planning Three retired couples each require an additional annual income of $2000 per year. As their financial consultant, you recommend that they invest some money in Treasury bills that yield 7%, some money in corporate bonds

Pharmacy A doctor's prescription calls for a daily intake of a supplement containing 40 mg of vitamin C and 30 mg of vi tamin D. Your pharmacy stocks three supplements that can be used: one contains 20% vitamin C and 30% vitamin D; a second, 40% vitamin C and 20% vitamin D; and a third, 30% vitamin C and 50% vitamin D. Create a table showing the pos sible combinations that could be used to fill the prescription. Pharmacy A doctor's prescription calls for the creation of pills that contain 12 units of vitamin B1 2 and 12 units of vita min E. Your pharmacy stocks three powders that can be used to make these pills: one contains 20% vitamin B' and 30% 2 vitamin E; a second, 40% vitamin B12 and 20% vitamin E; and a third, 30% vitamin B1 and 40% vitamin E. Create a 2 table showing the possible combinations of each powder that could be mixed in each pill.


90.

Write a brief paragraph or two that outline your strategy for solving a system of linear equations using matrices.

91.

When solving a system of linear equations using matrices, do you prefer to p lace the augmented matrix in row echelon form or in reduced row echelon form? Give reasons for your choice.

Make up a system of three linear equations containing three variables that has: (a) No solution (b) Exactly one solution (c) Infinitely many solutions Give the three systems to a friend to solve and critique.

12.3 Systems of Linear Equations: Determinants OBJECTIVES

1 Eva l uate 2 by 2 Determinants (p. 866)

2 Use Cramer's R u l e to Solve a System of Two Eq u ation s Conta i n i ng Two Varia bles (p. 866)

3 Eva l uate 3 by 3 Determ inants (p. 869)

4 5

Use Cram er's R u l e to Solve a System of Three Equations Conta i ning Three Va riables (p. 870) Know Properties of Determina nts (p. 872)

866

CHAPTER 12


In the preceding section, we described a method of using matrices to solve a system of linear equations. This section deals with yet another method for solving systems of linear equations; however, it can be used only when the number of equations equals the number of variables. Although the method will work for any system ( pro vided that the number of equations equals the number of variables) , it is most often used for systems of two equations containing two variables or three equations con taining three variables. This method, called is based on the concept of a

Cramer'sRule,

determinant.

1

Evalu ate 2 by 2 Determi na nts

DEFINITION

If

a, b, e, d and

are four real numbers, the symbol

is called a 2 by 2 determinant. Its value is the number D

COMMENT A

=

lea db l ad - be =

ad - be;

that is,

(1)

I

�

�----------------------------------�

A

The following device may be helpful for remembering the value of a 2 by

matrix is an array of numbers; it has no value. deter minant represents a num ber. •

2 determinant:

/

be =

ad

-

be

Evaluating a 2 x 2 Determinant

EXAMPLE 1

Evaluate:

1

Solution

3 6

-2 1

1

= = =

,m

2

1# >-

1

-2 6 1

3

1

(3 ) ( 1 ) - ( 6 ) ( -2) 3 - (- 1 2 ) 15

•

Now Work P R O B l E M 7

Use Cra mer's R ule to Solve a System of Two Equations Conta i n i ng Two Varia bles Let's now see the role that a 2 by 2 determinant plays in the solution of a system of two equations containing two variables. Consider the system

{ exax dbyy s =

+

=

+

d

t

b { adXbex bdy bdy tbsd

(1)

2 ( )

(2)

We shall use the method of elimination to solve this system. Provided =f. 0 and =f. 0, this system is equivalent to the system + +

=

=

(1) (2)

M ultiply by d.

Multiply b y b.

SECTION U.3

Systems of Linear Equations: Determinants

867

Subtracting the second equation from the first equation, we get

(ad - be) x Y= sd - tb lae db Ix = Its db I = I� � I = ad - be x x= I� �I�I I� �I I� +

0·

which can be rewritten using determinant notation as

If

D

"*

0, we can solve for

to get

D

Return now to the original system (2). Provided that tem is equivalent to

bey == eats { aeaexx ady x (ad - be)y = at - es I; �IY = I; �I = I; � I = ad - be y +

+

(1)

(2)

+

(3)

a

"* ° and

e

"*

0, the sys

Multiply by c. Multiply by a.

Subtracting the first equation from the second equation, we get °

.

which can be rewritten using determinant notation as

If

D

"*

0, we can solve for

to get

(4)

Equations (3) and (4) lead us to the following result, called Cramer's Rule.

THEOREM

C ra m e r's Rule for Two Equations Conta i n i n g Two Variabl es

The solution to the system of equations

+

{ ea xx dyby == ts +

is given by

x= provided that D

(1)

(2)

(5)

I � I; y= a b le dl

( 6)

= lea db l = ad - be

"* °

868

CHAPTER 12

Systems of Equations and I n eq u a l ities

b,

In this derivation for Cramer's Rule, we assumed that none of the numbers a, and was O. In Problem you will be asked to complete the proof under the less stringent condition that D *" O. Now look carefully at the pattern in Cramer's Rule. The denominator in the solution is the determinant of the coefficients of the variables.

e, d

61= ad - be { exax dyby ==

(6)

+

+

x,

D

S

t

= 1 ea db1

In the solution for the numerator is the determinant, denoted by Dx, formed by replacing the entries in the first column ( the coefficients of of D by the constants on the right side of the equal sign.

= Is bl

Dx

y,

x)

d

t

In the solution for the numerator is the determinant, denoted by Dy, formed by replacing the entries in the second column ( the coefficients of of D by the con stants on the right side of the equal sign.

Cramer's Rule then states that, if

D *"

y)

0, Dy

x=- y= Dx D

E X A MP L E

2

D

(7)

Solving a System of Linear Equations Using Determinants

Use Cramer's Rule, if applicable, to solve the system

Solution

The determinant

D

{ 3X6x - 2y == 134 +

D *"

Y

of the coefficients of the variables is D

Because

(1 )

(2)

= 136 -211 = (3)(1) - (6)(-2) = 15

0, Cramer's Rule (7) can be used.

1134 -211 x = - = 15 (4)(1) - (13)(-2) 3015 15 =2 x = 2, y = 1, Dx D

The solution is

I� 1�1 - - 15 (3)(13) - (6)(4) 1515 15 = (2, 1). Dy

Y--D

1

or, using ordered pairs

•

In attempting to use Cramer's Rule, if the determinant D of the coefficients of the variables equals 0 ( so that Cramer's Rule is not applicable ) , the system either is inconsistent or has infinitely many solutions. I!J!!

-


SECTION U.3

3

869

Systems o f Linear Equations: Determinants

Eva l u ate 3 by 3 Determinants

To use Cramer's Rule to solve a system of three equations containing three vari ables, we need to define a 3 by 3 determinant. A 3 by 3 determinant is symbolized by au a12 a l 3 (8) a21 a22 a23 J a3 a32 a33 in which al l , a12 ,"" are real numbers. As with matrices, we use a double subscript to identify an entry by indicating its row and column numbers. For example, the entry a23 is in row 2, column 3. The value of a 3 by 3 determinant may be defined in terms of 2 by 2 determi nants by the following formula: al l a2 l a31

I

a13 an a23 = all -a32 a33

a l2 a22 a3 2

11 l

Minus

a23 a - an 2 1 a33 a31 l'

2 by 2

l1 l Plus

l'

+

a13

a" a31

a22 a32 l'

1

(9)

by 2 2 by 2 determ i na nt determina nt left after left after removing the row removing the row and col u m n and colum n containing a'2 conta ining a'3

determinant left after removing the row and colu m n contai n ing an

2

a?o _J a33

The 2 by 2 determinants shown in formula ( 9 ) are called minors of the 3 by 3 deter minant. For an n by n determinant, the minor Mij of entry aij is the determinant resulting from removing the ith row and jth column. EXAMPLE

3

Finding Minors of a

For the determinant A Solution

3

by

2 -2 o

=

3

Determinant

-1 5 6

3 1 , find: -9

(a) M12

(b) M23

(a) MI2 is the determinant that results from removing the first row and second column from A .

�

A= -

-� � -�

�

Ml2

=

1

-2 0

1

-9

1

= (-2) ( - 9 ) - ( 0)(1 )

=

18

( b ) M23 i s the determinant that results from removing the second row and third column from A . -1

A= -

0

6

'

M23

-

=

I � -� I

=

(2)(6) - (0)(- 1 )

=

12 •

Referring back to formula ( 9 ) , we see that each element aij is multiplied by its minor, but sometimes this term is added and other times, subtracted. To determine whether to add or subtract a term, we must consider the cofactor.

DEFINITION

For an n by given by

n

determinant A , the cofactor of entry aij ' denoted by Aij , is

where Mij is the minor of entry aij '

870

CHAPTER 12

+


i The exponent of ( -I) +j is the sum of the row and column of the entry ai , so if j i i j is even, ( - I ) +j will equal 1, and if i + j is odd, ( -I /+j will equal -1. To find the value of a determinant, multiply each entry in any row or column by its cofactor and sum the results. This process is referred to as expanding across a row or column. For example, the value of the 3 by 3 determinant in formula (9) was found by expanding across row 1 . I f we choose t o expand down column 2, we obtain an a2 1 a31

an a2 3 a33

al2 a22 a32

r

=

!

Expand down col um n 2. If we

al l a21 a31

!

!

a a23 al3 + ( - I ) 2+2 a22 ll a33 a31 a33

a ( - 1 ) 1+2 aI2 21 a3 l

!

+

a l l an ( - I ) 3+2 a32 a21 a23

1

+

a l a12 ( - I ?+ 3 a33 l a21 a22

choose to expand across row 3, we obtain

al2 al3 a22 a23 a32 a33

=

r

!

I

1

a a ( -I) 3+ l a31 12 a13 + ( - I ?+2 a32 n a21 a22 a23

Exp and across row 3.

a1 3 a23

!

!

I

!

It can be shown that the value of a determinant does not depend on the choice of the row or column used in the expansion. However, expanding across a row or column that has an element equal to 0 reduces the amount of work needed to com pute the value of the determinant.

EXAMPLE

E valuating a

4

3

x

3

Determinant

Find the value of the 3 by 3 determinant:

Solution

3 4 8

0 6 -2

-1 2 3

Because of the 0, it is easier to expand across row 1 or down row 2 . We choose to expand across row 1 . 0 6 -2

-1 2 "

.)

= == +

( - 1 ) 1+1 .3.

=

3(18

�.

+

1 �I 6 -2

+

( - 1 )1+2 .0.

1 : �I

4) - 0 + ( - 1 ) ( -8 - 48)

+

( - 1 )1+ 3.( - 1 )·

1 : -�I

3 (22 ) + ( - 1 ) (-56) 66

4

3 4 8

56

=

122

..

- Now Work P R O B L E M 1 1

Use Cra mer's Rule to Solve a System of Three Equations

{ Xx ++ ++ = x+ + =

Conta i n i ng Three Va riables

Consider the following system of three equations containing three variables. al l a2l a31 .

anY a22Y a32Y

a l 3Z a2 3 Z a33Z

=

Cl c2 c3

(10)

SECTION 1203

Systems o f Linear Equations: Determinants

871

It can be shown that if the determinant D of the coefficients of the variables is not 0; that is, if all al2 [/21 an

D =

al3 [/23

*-

0

the unique solution of system (10) is given by Cramer's Rule for Three Equations Contai n i ng Three Variables D

D

x x =D

where Dx

y y =-

CI = C2 C3

a12 [/22 a32

a13 an a33

D

Z

D

y

all Cl = a2l C2 a31 C3

=

Dz D

a 13 an a33

D_< =

a ll a2 l a31

a12 CI a22 C2 a32 C3

Do you see the similarity of this pattern and the pattern observed earlier for a system of two equations containing two variables? Using Cramer's Rule

EXAMPLE 5

{+

Use Cramer's Rule, if applicable, to solve the following system: y - z = 3 2x -x + 2y + 4z = -3 x - 2y - 3z = 4

The value of the determinant

Solution D=

2 -1 1

1 2 -2

\

=

2(2) - 1 ( - 1 )

= 4 + 1 = 5 Because Dx

,..,

3 =

-oJ

4

1 2 -2

D *-

Dy

3 -3 4

\

Dz

1 2 -2

�\

-oJ

+

4 -3

2 ( -7) - 3 ( - 1 ) +

3

( _ 1 ) 1+20 1 0

+ 1 =

1

+

\

+

( _ 1 ) 1+20 1 0

+

=

15

(-1) 1+2030

\

\

-1 1

�\

-oJ

-

!I

+ ( - 1 ) 1+20 1 0

3(0) = 5

(_1 ) 1+3 ( - 1 )

and

Dzo

\ -�\ -1 1

�\

+

( _ 1 ) 1+ 3 ( _ 1 )

\ -�\

�\

+

(_1 ) 1+3 ( _ 1 )

\ -!\

-oJ

-1 1

-oJ

1 � -!I -

+

Dr, D , y

-3 4

( - 1 ) (-1 )

-10

3 2 -3 = ( - 1 ) 1+1020 -2 4 = 2 ( 2 ) - 1 (-1 )

+

( - 1 ) ( -2 )

\! \

-1 4 = ( - 1 )1 +1020 -3 = -14

(3)

of the coefficients of the variables is

(-1)(0)

+

-1 2 4 = ( - 1 ) 1 +1030 -2 -3

=

2 = -1 1

- oJ

+

(2)

0, we proceed to find the values of

= 3(2) - 1 ( -7) 2 = -1 1

�\

-1 2 4 = ( - I ) l+1020 -2 -3

D

(1)

-3 4

-1 1

1 � -�I

+ ( - 1 ) 1 +3030

872

CHAPTER 1 2

Systems o f Equations a n d Inequal ities

As a result,

x

The solution is

DDy = = -2 z = -' = - = l = = -2, z -2, 3,

y

-10 5

Y

=

=

D7

'

-

D

5 5

1, or, using ordered triplets, (3,

1).

u

If the determinant of the coefficients of the variables of a system of three linear equations containing three variables is 0, Cramer's Rule is not applicable. In such a case, the system either is inconsistent or has infinitely many solutions. To solve such a system, use elimination or row reduction on the augmented matrix. t;i!J!l;; = =- -

5


Know Properties of Determ i n a nts

Determinants have several properties that are sometimes helpful for obtaining their value. We list some of them here. THEOREM

The value of a determinant changes sign if any two rows (or any two columns) are interchanged. (11).J Proof for 2 by 2 Determ ina nts

I� �I

E XA M P L E 6

THEOREM

ad - be

I � �I

and

=2

=

be - ad

=

- ( ad - be)

Demonstrating Theorem (1 1 )

I � �I

If all the entries determinant is O.

111

=

6 - 4

I � !I

=

4 - 6

=

•

-2

•

any row (or any column) equal 0, the value of the

(12).J

Expand across the row (or down the column) containing the O's.

Proof

THEOREM

=

•

If any two rows (or any two columns) of a determinant have corresponding entries that are equal, the value of the determinant is O. (13).J You are asked to prove this result for a 3 by 3 determinant in which the entries in column 1 equal the entries in column 3 in Problem 64.

E XA M P L E 7

1 22

Demonstrating Theorem (1 3) 1

4

5

3 3 6

( - 1 ) 1+1 0 1 0 1 ( -3) -

THEOREM

1� �I

= 2(

=

-6)

+

+

( - 1 ) 1 +2 0 2 0

3( -3)

=

-3

I! �I

+

+

12 - 9

( - 1 ) 1+ 3 0 30 =

0

I! �I

•

If any row (or any column) of a determinant is multiplied by a nonzero number k, the value of the determinant is also changed by a factor of k. (14).J

SECTION U.3

You are asked to prove this result for a 3 by 3 determinant using row Problem 63.

EXAM P L E 8

- -2

2

873

Systems of Linear Equations: Determinants

in

Demonstrating Theorem (1 4)

I� �I I� �I

=

2

=

6

8

=

6k - 8k

=

-

2

k

k ( - 2)

=

=

k

I� �I

•

If the entries of any row (or any column) of a determinant are multiplied by a nonzero number k and the result is added to the corresponding entries of another row (or column), the value of the determinant remains unchanged.

THEOREM

(15).J

1 2.

In Problem 65, you are asked to prove this result for a 3 by 3 determinant using rows and

-14

Demonstrating Theorem (1 5)

EXAM P L E 9

I� � I

=

Multiply row 2 by

-2 and add to row 1 .

•

1 2.3 Assess Your Understanding Concepts and Vocabulary 1.

to solve a system of linear

Cramer's Rule uses equations.

2. D

= I : !I = -

3 . True or False

4. True or False

A

by determinant can never equal O.

3 3

The value of a determinant rema ins un changed if any two rows or any two columns are inter changed.

Skill Building 5-14,

In Problems find the value of each determinant. S.

I ! �I 1-4-5 � I

6.

I� �I

7.

I -� �I 1

8.

I : -�I

9.

I -! -� I -9

3 44 4 2 3 -2 3 4 52 -5 1 -3 1 -3 4 1 2 -2 2 3 In Problems solve each system of equations using Cramer's Rule if it is applicable. If Cramer's Rule is not applicable, say so. {2xX - "y3y = 5 {5X2x - "y = 13 {Xx - 2y = {Xx - y = 4 12 =3 {2X 4y {3X - 6y 24 {4X -2y5y = -3-4 { 3x = 24 3x - 5y = 5x 4y 12 x 2y 10.

15.

19.

+ Y

+

n.

-1

15-42,

8

16.

12.

8

+

=

=

0

20.

6

+

Y

5

=

-1

1

13.

1 7.

21.

+

�

Y

-1

1 4.

0

18.

=

22.

1 8

+

�

=

=

+

6

+

0 1

= =

8

16

-9

CHAPTER 12

874

23.

27.

31.

35.

39.

Systems of Equations a n d I nequalities

{ 3X - 2y=4 6x - 4y = 0

24.

{ 2x - 3y = -1 lOx + l Oy= 5

{

28.

{ 3x - 5y = 3 15x+ 5y = 21

46.

I � �I 3 1 0

2 x 1

3X - 2y =0 5x + l Oy = 4

29.

. 33.

{ x+ 4y - 3, � C 3x - y+ 3z =0 x+ Y+6z=0

40.

In Problems 43-48, solve for x. 43.

25.

{ x + 4y - 3F -8 3x - y+ 3z = 12 x+ Y+ 6z= 1

36.

x + 2y - , � O 2x - 4y+ Z =0 -2x + 2y - 3z=0

-x+2y = 5 4x - 8y=6

r x - y � -1 3 1 ,r + -y=2 2

32.

{ x+2y - , � -3 2x - 4y + Z = - 7 -2x + 2y - 3z = 4

{

{

= 5

44.

4 5 =0 -2

I � �I

37.

41.

{ 2X - 4y=-2 3x+2y= 3 { 2X+ 3y = 6 1 x - y= "2

47.

30.

{ x+ y - z= 6 3x - 2y+ Z =-5 x + 3y - 2z = 14 { X - 2y+3z = 1 3x+ y - 2z=0 2x - 4y+6z = 2

{ X - 2y + 3z = 0 3x+ y - 2z=0 2x - 4y+6z=0

=-2

45.

2 3 0 = 7 1 x 6 1 -2

x

26.

48.

34.

38.

42.

�

{ 3x + 3y = {

4x , +2y = 3

�

X+ Y=-2 x - 2y =

8

{X - y+ Z= -4 2x - 3y + 4z=-15 5x+ Y - 2z = 12

5 { X - y+2z = 3x + 2y= 4 -2x + 2y - 4z=- 10 X - y + 2z=0 3x+2y=0 -2x+2y - 4z = 0

{

x 1 1 4 3 2 =2 -1 2 5

x 1 2 1 x 3 =-4x 0 1 2

In Problems 49-56, use properties of determinants to find the value of each determinant if it is known that

49.

53.

1

u

x

2 v y

1 x-3 2u

3

x

y v 2 4

u

50.

w

z

2 3 y-6 z - 9 2w 2v

x

u

54.

1

Applicati ons and Extensi ons 57.

y

v

2

x y Z u v w =4 1 2 3

z

w

51.

6

u

Z -x

w - u

y YI yz

1 1

55.

2

Geometry: Equation of a Line An eq uation of the line con taining the two points (x, , )' , ) and ( X2 , Y2) may be expressed as the determinant

x x,

59.

=0

1 X2 Prove this result by expanding the determinant and compar i ng the result to the two-point form of the equation of a line. 58.

Geometry: Collinear Points Using the result obtained i n Problem 57, show that three distinct points (XI , Y I ), (xz , yz), and (X3 , Y3) are collinear ( lie on the same l ine ) if and only if X,

X2 x3

y, Y2 Y3

1

1 =0 1

x -3

60.

61.

y -6

1 2x u -1

v

z -9

52.

w

2 2y v -2

1 x-u u

3 2z w - 3

56.

x+3 3u - 1 1

2 y-v v

3 z-

w

w

y+6 z+9 3v - 2 3 w - 3 2 3

Geometry: Area of a Triangle A triangle has vertices (Xl , )' , ), (xz, Y 2 ) , and (X3, Y3)' S how that the area of the triangle is given by the absolute value of D , where x , X2 x3 1 D = - YI Yz Y3 ' Use this formula to find the area of a 2 1 1 1 triangle with vertices (2, 3), (5, 2), and (6, 5).

x2 x 1 Show that i y 1 =(y - z)(x - y ) ( x - z). Z2 Z 1

Complete the proof of Cramer's Rule for two equations con taining two variables. [Hint: I n system (5), page 867, if a =0, then b *" 0 and c *" 0, since D = - be *" O. Now show that equation (6) provides a

SECTION 12.4

62.

63.

solution of the system when a = 0. There are then three remaining cases: b = 0, C = 0, and d = 0.] Interchange columns 1 and 3 of a 3 by 3 determinant. Show that the value of the new determinant is -1 times the value of the original determinant. Multiply each entry in row 2 of a 3 by 3 determinant by the number k, k =F 0. Show that the value of the new determi nant is k times the value of the original determinant.

64. 65.

Matrix Algebra

875

Prove that a 3 by 3 determinant in which the entries in col umn 1 equal those in column 3 has the value 0. Prove that, if row 2 of a 3 by 3 determinant is multiplied by k, k =F 0, and the result is added to the entries in row 1, there is no change in the value of the determinant.

1 2.4 Matrix Algebra OBJECTIVES 1 Find the Sum and Difference of Two Matrices (p. 876)

2 Find Sca l a r M u lti ples of a Matrix (p. 878) 3 Find the Product of Two Matrices (p. 879)

4

5

Find the I nverse of a Matrix (p. 884) Solve a System of Linea r Equations Using an I nverse Matrix (p. 887)

In Section 12.2, we defined a matrix as a rectangular array of real numbers and used an augmented matrix to represent a system of linear equations. There is, however, a branch of mathematics, called linear algebra, that deals with matrices in such a way that an algebra of matrices is permitted. In this section, we provide a survey of how this matrix algebra is developed. Before getting started, we restate the definition of a matrix.

DEFINITION

A matrix is defined as a rectangular array of numbers: Column 1

Column 2

Col u m n )

Colu m n n

Row 2

a1l

a2l

al2

alj

aln

Row i

ai l

ai2

aij

ain

Row m

aml

am2

amj

amn.

Row 1

a22

a2j

a2n

-.J

Each number aij of the matrix has two indexes: the row index i and the column index j. The matrix shown here has m rows and n columns. The numbers aij are usu ally referred to as the entries of the matrix. For example, a23 refers to the entry in

the second row, third column. Let's begin with an example that illustrates how matrices can be used to conve niently represent an array of information.

EXAM P L E 1

Arranging Data in a Matrix In a survey of 900 people, the following information was obtained: 200 males 150 males 45 males 315 females 125 females 65 females

Thought federal defense spending Thought federal defense spending Had no opinion Thought federal defense spending Thought federal defense spending Had no opinion

was too high was too low was too high was too low

876

CHAPTER 12

Systems of E q u a t i o n s a n d I n e q u a l ities

We can arrange these data in a rectangular array as follows: Too High

Too Low

No Opinion

Male

200

1 50

45

Female

315

1 25

65

[

or as the matrix

23015 115205 4655] 1 2 3=6

This matrix has two rows (representing male and female) and three columns (rep resenting "too high," "too low," and "no opinion").

•

The matrix we developed in Example has rows and 3 columns. In general, a matrix with m rows and n columns is called an m by n matrix. The matrix we devel oped in Example 1 is a 2 by 3 matrix and contains 2 . entries. An m by n matrix will contain m . n entries. If an m by n matrix has the same number of rows as columns; that is, if m = n, then the matrix is referred to as a square matrix.

5-6 O1J [468 � 4�]

Examples of Matrices

EXAM P L E 2

(a)

(c)

1

[

A 2 by 2 square matrix

(b) [

1 0 3]

A 1 by 3 matrix

-2

A 3 by 3 square matrix •

F i n d the S um and Difference of Two Matrices

We begin our discussion of matrix algebra by first defining what is meant by two matrices being equal and then defining the operations of addition and subtraction. It is important to note that these definitions require each matrix to have the same number of rows and the same number of columns as a condition for equality and for addition and subtraction. We usually represent matrices by capital letters, such as and C.

DEFINITION

A B A=B A A [o � : ] � [ � _: ] [� � J * [� �J [� � �J * [� � � ! J

Two matrices

and

are said to be equal, written as

A, B,

provided that and have the same number of rows and the same number of columns and each entry aij in is equal to the corresponding entry bij in

B

For example,

and

-

B'--1

1 �J

v4

Because the entries in row 1, col u m n 2 a re not equal

Because the matrix on the left has 3 colu mns and the matrix on the rig ht has 4 colum ns.

SECTION 12.4

A B B. B

Matrix Algebra

877

Suppose that and represent two m by n matrices. We define their sum to be the In by n matrix formed by adding the corresponding entries aij of and bij of The difference A - B is defined as the In by n matrix formed by sub tracting the entries bij in from the corresponding entries aij in Addition and subtraction of matrices are allowed only for matrices having the same number In of rows anel the same number n of columns. For example, a by matrix and a by matrix cannot be added or subtracted.

A + B

2 3

A.

A = [20 41 28 - n B = [ -36 48 20 A-8 A+B A + B = [� 41 28 -n + [-� 48 20 �J = [20 ++ 6(-3) 4 ++ 48 28 ++ 02 -33 ++ 0 = [-16 89 48 -32J A - B = [� � � -n - [ - � : � �J = [2 -- 6 41 -- 48 28 - 20 -33 - 0

�J

Adding and Subtracting Matrices

EXAM P L E 3

Suppose that

and

Find: (a)

Solution

(a)

(b)

1

1

(b)

o

( -3 )

1

-

_

Figure 7

[A] + [S] [ [ -1 8 [6 9 [A] - [S] [ [5 0 [ -6 -7

8 -2 ] 4 3 ] ] 8 -4] 0 3 ] ]

l�.1 �

J

J

A

2 4

A.dd correspondi ng entries.

Subtract correspond ing entries.

•

Seeing the Concept Graphing utilities can m a ke the sometimes ted i o u s process of matrix algebra easy. I n fact, most g raph i n g calculators can handle matrices as large as 9 by 9, some even larger ones. Enter the matrices into a g ra p h i n g utility. N a m e them [AJ a n d [8J. Fig u re 7 shows the results of a d d i n g a n d su btracting

[AJ and [8J.

;z=


Many of the algebraic properties of sums of real numbers are also true for sums of matrices. Suppose that and C are m by n matrices. Then matrix addition is commutative. That is,

A, B,

Commutative Property of Matrix Addition

A+B=8+A

Matrix addition is also associative. That is, Associative Property of Matrix Addition

(A + B) + = A + (B + C

C)

878

CHAPTER 12

Systems o f Equations and Inequalities

Although we shall not prove these results, the proofs, as the following example illustrates, are based on the commutative and associative properties for real numbers.

03 -�J + [-� -32 41J == [[42+(-1+ +-21 ) 203+2+ 3(-3) -1 ++ 41 J + J -3 0 4 4 [ -� -32 !J + [! 03 -�J 0 [0 0 0] [0 00 O0J [0 O0J 0 A+ O = O +A=A O. A k A k A A k. A. kA

Demonstrating the Commutative Property

EXAM PLE 4

[!

5

7

+ +

5

l + (-l) + 7

=

A matrix whose entries are all equal to lowing matrices is a zero matrix. 2 by 2 square zero matnx

o

o

n

is called a zero matrix. Each of the fol 1 by 3 zero

2 by 3 zero

matrix

matnx

Zero matrices have properties similar to the real number m atrix and is an In by n zero matrix, then

If

is an

In

by

n

In other words, the zero matrix is the additive identity in matrix algebra. 2

F i n d Scalar M u ltip les of a Matrix

We can also multiply a matrix by a real number. If is a real number and is an In by n matrix, the matrix is the In by n matrix formed by multiplying each entry aij in by The number k is sometimes referred to as a scalar, and the matrix is called a scalar multiple of

EXA M P LE 5

Operations Using Matrices Suppose that

Solution

4A 1:3.c 3A -2B 4A { -23 01 6J - [4(4'-32) 44·. 01 44 .· 6J = [ 12 40 2204J O ' 0 9 3 O [ � [ � ' J [ 9 3 3 -3 6 � 1:.3 ( -3) -·31 6J � -1 2J

Find:

(a)

(a)

=

(b) ! C � !

(b)

5

(c)

5

-8

( c ) 3 A - 2B

[ [ =[ 9 [9 =

-

3

3 1 5 -2 ° 6

3·3 3 ( -2 )

-

J

3·1 3·0

3 0

-6

[ OJ J [ OJ J [ ]

15 18

3·5 3·6

-

- 8 3 - 2 -6 - 16 0 - 2

= [ � � �! -2

-

4 1 8 1

- 2

J

-3

2 '4 2·8

_

SECTION 12.4

8 2 16 2

2·1 2·1

2·° 2( - 3 )

Matrix Algebra

879

J

-6

15 - 0 1 8 - ( -6 )

1

•

Check: Enter the matrices [AJ, [BJ, a n d [C] i nto a graphing uti l it y. Then fi nd 4A, - C, 3

a n d 3A - 2B.

k=_=_ Now Wor k

PROBLEM 1 1

We list next some of the algebraic properties of scalar multiplication. Let h and k be real numbers, and let A and B be In by n matrices. Then Properties of Sca lar Multiplication

k ( hA) = ( k h ) A ( k + h ) A kA + h A k(A + B) kA + k B =

=

3

F i n d the Product of Two Matrices

Unlike the straightforward definition for adding two matrices, the definition for multiplying two matrices is not what we might expect. In preparation for this defin ition, we need the following definitions: A row vector R is a 1 by n matrix R = [ r)

r

2

rll ]

A column vector C is an n by 1 matrix

DEFINITION

The product RC of R times C is defined as the number

880

CHAPTER 12


Notice that a row vector and a column vector can be multiplied only if they contain the same number of entries.

EXA M P L E

6

[ :l

The Product of a Row Vector by a Column Vector If R � [ 3

RC � [3

-5

-5

2 ] and c �

2

{ -n

then

_

� 3·3

+

( -5 )4

+

2( -5 ) � 9 - 20 - 10 � - 21 ..

Let's look at an application of the product of a row vector by a column vector.

EXA M P LE

7

Using Matrices to Compute Revenue A clothing store sells men's shirts for $40, silk ties for $20, and wool suits for $400. Last month, the store had sales consisting of 100 shirts, 200 ties, and 50 suits. What was the total revenue due to these sales?

Solution

We set up a row vector R to represent the prices of each item and a column vector C to represent the corresponding number of items sold. Then Prices Shirts Ties Suits R =

[ ��l

N u m ber sold

[40 20 400 l,

C

=

1 00 2

J

Shirts Ties Suits

The total revenue obtained is the product R C. That is,

[;�]

RC � [40 20 400] =

40 . 1 00

Shirt revenue �

+

20 · 200 �

Tie revenue

+

400 · 50 '-v------'

Suit revenue

=

$28,000 '-v------'

Total revenue

..

The definition for multiplying two matrices is based on the definition of a row vector times a column vector.

DEFINITION

matrix and let B denote an r by n matrix. The m by n matrix whose entry in row i, column j is the product of the ith row of A and the jth column of B . Let A denote an

m

by

r

product A B is defined as the

.J

The definition of the product A B of two matrices A and B, in this order, requires that the number of columns of A equal the number of rows of B; otherwise, no product is defined.

I

m

A by r t

r

M ust be sa me for

AB to be defined AB is m by n.

t

An example will help to clarify the definition.

B by n

I

SECTION 12.4

M u ltiplying Two Matrices

EXAM P L E 8

Find the product

�[ - I J

AB if =

A

4

48

Matrix Algebra

881

= [ 4 581 1 -14] 4, 2

and

0

B

_

0 -2

3

6

First, we note that is 2 by 3 and is 3 by so the product is defined and will be a 2 by matrix. Suppose that we want the entry in row 2, column 3 of To find it, we find the product of the row vector from row 2 of and the column vector from column 3 of

Solution

A

B

AB

AB.

A

1 [ [5 8 _� ] = 5 ' 1 + 8 · 0 +0(-2) = 5

B.

Colu m n 3 of B

Row 2 of A

0]

So far, we have

[

-+-

Colu m n 3

AB � = =

=

]

4 4 l 4 [2 4 -1 [ � = 2' 4 +4 ' 6 +( - 1)(-1 ) = 3 5 1 4l 4 [ = [5 8 -�J 4 81 -1 1 1 1 1 4 1+ 4· 4 + ( - 1 )( 5 + 4· 8 + (-1 ) 1 44· ( [5· 2 + 8 · 4 + 8 8 8 · 6 5· 4 5· 5 · + 5 + = [!� :� � ��J Now, to find the entry in row 1 , column of

A and column

B.

of

'1.


I

:

I-H�torical Feature

M

Arthur Cayley (1821-1895)

atrices

were

Arth u r Cayley

(1821-1895) 1857 invented

in

I

by

as a way of

efficiently computing

the

result of

of their lives ela borating the theory. The torch was then passed to Georg Froben i u s

(1849-1917),

rather

to the surprise of physicists, it was fou n d that matrices (with complex

Historical Problem

n u m bers i n them) were exactly the right tool for descri bing the behav

3).

The res u l t i n g system had

i n cred i b l e richness, i n the sense that a wide vari

ior of atomic systems. Today, matrices are used in a wide variety of

ety of mathematical systems could be m i m icked

appl ications.

by the matrices. Cayley a n d his friend James J. Sylvester

(1814-1897)

spent m u c h of the rest

1 . Matrices a n d Complex Numbers Froben i u s e m p h asized i n h i s re search how matrices could be used to m i m i c other mathematical

3 . Cayley's Definition of Matrix Multiplication Cayley invented

{ uv arcr + dsbs {X kumu ++ Ivnv s u 2 2

matrix m u ltiplication to s i m p l ify the fol l owing problem:

=

systems. Here, we m i m i c the behavior of comp lex n u m bers using matrices. Mathematici a n s call such a relatio n s h i p a n

Com plex n u m ber

a bi

+

Matrix

isomorphism.

[ -ba baJ

matrix. Thus,

2 + 3i

[ -32 23 J

and

(b) M u ltiply the two matrices.

and

+

i n part (b).

2 - 5i 1 + 3i. and

The result s h o u l d be the same as

that fou n d i n part (c).

The process a l so works for a d d ition and s u btraction. Try it for

(a bi)(a - bi) +

a n d y i n terms of r a n d

=

by su bstituti n g

(b) Use the result of part (a) to find the

and

v

from the

by

matrix A in

(c) Now look at the following way to d o it. Write the equations in matrix form.

[�J [: :J[:J [;J [� �][�J =

(c) F i n d the corresp o n d i n g com plex n u mber for the matrix fou n d

(d) Mu ltiply

x

y =

first system of equations into the second system of equations.

[42 -24J 4 - 2i 2 - 5i 1 3i.

(a) Find the matrices correspo n d i n g to

+

=

(a) Find

Note that the comp lex n u mber can be read off the top l i n e of the

2. Compute

1924,

su bstituting one li near system i nto a nother (see

H i sto rica l Pro b l e m s

yourself.

whose deep investigations established

a central place for matrices in modern mathematics. In

=

So

[;J [� �][ : :][:J =

Do you see how Cayley defined matrix m u ltiplication? u s i n g matrices. I nterpret t h e result.

SECTION 12.4

Matrix Algebra

889

1 2.4 Assess Your Understanding A matrix B, for which A B = 'II ' the identity matrix, is called of A . the

Concepts and Vocabulary 1. 2.

I n the algebra of matrices, the matrix that h as properties similar to the number 1 is called the matrix.

In Problems

'\

..

A

+

7-22,

A=

B

8.

1 1 . 3A - 2B 15.

12.

CA

16.

1 9. AC - 31 2

In Problems ' 23 .

20.

23-28,

find the producl.

[�

3 2

A - B

-5

2A + 4B

[ -� �][ -1

2 8 3 6

-1

0

B=

J

6 '

Any pair of matrices can be multiplied.

CA +

J

1 3

4 -2

[

17.

21.

513

27.

-�J

9. 4A

1 3.

CB

[� -�J [� -� � �J 1

26.

6. True or False

use the following matrices to compute the given expression.

Skill Building

7.

Every square matrix h as an inverse.

5 . True o r False

A matrix that has the same number of rows as columns is called a(n) matrix. __

3.

4. True or False

c

�

AC

34.

[� �J [� �J

30.

b *- 0

35.

0 2 4 3 6

-1 -2 -3

In Problems 39-58, use the inverses found in Problems 39.

43.

{ 2X +Y= 8 x +Y= 5

{ 6X + 5y = 7 2x + 2y 2

40.

44.

=

47.

51.

{ 2X ax X {

+

+

-

Y = -3 a *- O ay = -a

+ Z= 0 -2y + Z = - 1 -2x - 3y = - 5 y

48.

52.

{ 3x - y = 8 -2x + y = 4

31.

36.

18.

CA - CB

22.

28.

[� �J

[ -; n

29-38

37.

0

[ -: -�J

45.

{bX + Y 3 = 2b + 3 b *- 0 bx + 2y = 2b + 2 X + 2z = 6 -x + 2y + 3z = - 5 6 x - y

49.

53.

+Y

+Y

= 0 = 5

[: -i] 1 2 1

y= 2 a a *- O ax + ay = 5

{

X- y + Z -2y + z

=

2 2

- 2"r - 3y = � =

2

1

50.

2

0

·

46.

{2X +

A C + BC

38

42.

{6X + 5y= 13 2x + 2y = 5

( A + B)C

33.

to solve each system of equations. {2X x

BC

[ � -� � ][� -�] -1

32.

0 2 -1

41.

=

10. -3B 14.

8

{ -4X + Y= 0 6x - 2y = 14

{

-2

[ 1 �]1 [� -lJ�l

[-� -�J

U �I

[:n

C(A + B )

In Problems 29-38, each matrix is nonsingular. Find the inverse of each matrix. 29.

Matrix multiplication is commutative.

[� �]

U

3 2

-1

a *- O

1] 1

1

{ 3x - y = 4 -2x +Y= 5

{-4X + Y= 5 6x - 2y = -9

{bX + 3y = 1 4 b *- 0 bx + 2y = 10

890

55.

CHAPTER 12

{

Systems of Equations a n d I n eq u a l ities

x + y + z = 9 3x + 2y - z = 8 3x + Y + 2z = 1

56.

{

3x + 3y + Z = 8 x + 2Y + Z = 5 2x - Y + Z = 4

58.

[n Problems 59-64, show that each matrix has no inverse. 59.

62.

b;l

[

[

4 2 2 1

J

61

OJ

-3 4 0

•

64.

[�

[

15 10

�J

[ � -� -5

7

3X + 3y + Z = 1 { x + 2y + Z = 0 2x - y + Z = 4

� - ] 1

� [�

In Problems 65-68, use a graphing utility to find the inverse, if it exists, of each matrix. Round answers to two decimal places. 65.

[ 25 61 1 8 -2 8 35

{

-12] 4 21

66.

[ 18 6 10

-3 4] - 20 14 25 -15

67.

21 -8

��

1 2 -12 4 -16 4 9

68.

{

2 5

_�� -!

5 8 20 8 27 15 -3 -10

J

, In Problems 69-72, use the idea behind Example 1 5 with a graphing utility to solve the following systems of equations. Round answers to two decimal places. 69.

25x + 61y - 12z = 10 18x - 12y + 7y = -9 3x + 4y - z = 12

70.

{ 25X + 61y - 12z = 15 18x 1 2y + 7z = -3 3x + 4y z = 12 -

71.

{ 25X + 61y - 12z = 21 18x - 12y + 7z = 7 3x + 4y z = -2

72.

25x + 61y - 12z = 25 18x - 12y + 7z = 10 3x + 4y - z = -4


College Tuition Nikki and Joe take classes at a community college LCCC and a local university SlUE. The number of credit hours taken and the cost per credit hour ( 2006-2007 academic year, tuition only) are as follows: LCCC

SlUE

Cost p e r Credit Hour

N i kki

6

9

LCCC

Joe

3

12

SlUE

74.

75.

$ 7 1 .00 $ 1 S8.60

(a) Write a matrix A for the credit hours taken by each stu dent and a matrix B for the cost per credit hour. (b) Compute A B and interpret the results.

Sources:

www. lc.edu. www.siue.edu

School Loan Interest Jamal and Stephanie each have school loans issued from the same two banks. The amounts borrowed and the monthly interest rates are given next (interest is com pounded monthly): Lender 1

Lender 2

Jamal

$4000

$3000

Lender 1

0.0 1 1 ( 1 . 1 %)

Ste p h a n i e

$2500

$3800

Lender 2

0.006 (0.6%)

Monthly Interest Rate

76.

(a) Write a matrix A for the amounts borrowed by each student and a matrix B for the monthly interest rates. (b) Compute A B and interpret the results. [ J (c) Let C � . Compute A (C + B) and interpret the =

results.

Computing the Cost of Production The Acme Steel Company is a producer of stainless steel and aluminum con tainers. On a certain day, the following stainless steel containers were manufactured: 500 with l O-gallon capacity, 350 with 5-gallon capacity, and 400 with 1-gallon capacity. On the same day, the following aluminum containers were manufactured: 700 with l O-gallon capacity, 500 with 5-gallon capacity, and 850 with I-gallon capacity. (a) Find a 2 by 3 matrix representing these data. Find a 3 by 2 matrix to represent the same data. (b) If the amount of material used in the 10-gallon con tainers is 15 pounds, the amount used in the 5-gallon containers is 8 pounds, and the amount used in the 1-gallon containers is 3 pounds, find a 3 by 1 matrix rep resenting the amount of material used. (c) Multiply the 2 by 3 matrix found in part (a) and the 3 by 1 matrix found in part (b) to get a 2 by 1 matrix showing the day's usage of material. (d) If stainless steel costs Acme $0.10 per pound and alu minum costs $0.05 per pound, find a 1 by 2 matrix rep resenting cost. (e) Multiply the matrices found in parts (c) and (d) to de termine the total cost of the day's production. Computing Pro tit Rizza Ford has two locations, one in the city and the other in the suburbs. In January, the city location sold 400 subcompacts, 250 intermediate-size cars, and 50 SUVs; in February, it sold 350 subcompacts, 1 00 interme diates, and 30 SUVs. At the suburban location in January, 450 subcompacts, 200 intermediates, and 140 SUVs were sold. I n February, the suburban location sold 350 subcompacts, 300 intermediates, and 100 SUVs.

SECTION 12.5

77.

(a) Find 2 by 3 matrices that summarize the sales data for each location for January and February (one matrix for each month). (b) Use matrix addition to obtain total sales for the two month period. (c) The profit on each kind of car is $100 per subcompact, $150 per intermediate, and $200 per SUY. Find a 3 by 1 matrix representing this profit. (d) Multiply the matrices found in parts (b) and (c) to get a 2 by 1 matrix showing the profit at each location.

Cryptography One method of encryption is to use a matrix to encrypt the message and then use the corresponding in verse matrix to decode the message. The encrypted matrix, E, is obtained by mUltiplying the message matrix, M, by a key matrix, K. The original message can be retrieved by multi plying the encrypted matrix by the inverse of the key matrix. That is, E = M . K and M = E · K- I •

( , ) G;,e" ,he key mot';' K

�

[: : n

78.

_[

79.

[

]

Child's income

M H

[: �J

Consider the 2 by 2 square matrix A =

]

L

is called a left stochastic transition matrix. For example, the entry P2 1 = 0.5 means that 50% of the children of low rela tive income parents will transition to the medium level income. The diagonal entry Pi, i represents the percent of chil dren who remain in the same income level as their parents. Assuming the transition matrix is valid from one generation to the next, compute and interpret p2 . Source: Understanding Mobility in America, April 2006

';"d ;t; ;,w",e,

If D

(b) Use your result from part (a) to decode the encrypted 47 34 33 matrix E = 44 36 27 . 47 41 20 (c) Each entry in your result for part (b) represents the po sition of a letter in the English alphabet (A = 1 , B = 2, C 3, and so on). What is the original message?

=

ad - bc *' 0, show that A is nonsingular and that A-I

=

�[ D

d

-b

-c

a.

J

=

Source:

goldenmuseum.com


Create a situation different from any found in the text that can be represented by a matrix.

1 2.5 Partial Fraction Decomposition PREPARING FOR THIS S ECTION • •


Identity (Section 1 . 1 , p. 86) Proper and Improper Rational Functions (Section 5.2, p. 349)

• •

Factoring Polynomials (Review, Section R.5, pp. 49-55) Complex Zeros; Fundamental Theorem of Algebra (Section 5.6, pp. 389-393)

Now Work the 'Are You Prepared?' problems on page 897. OBJECTIVES 1 Decom pose

�, Where Q Has O n ly Q

N o n re peated Linea r Factors (p. 892)

p

.

2 Decom pose -, Where Q Has Repeated Linear Factors (p. 894) Q 3 Decompose

(p. 896) 4

Decompose (p. 897)

891

Economic Mobility The relative income of a child ( low, medium, or high) generally depends on the relative income of the child's parents. The matrix P, given by

Parent's I ncome L M H 0.4 0.2 0.1 P - 0.5 0.6 0.5 0.1 0.2 0.4

(Note: This key matrix is known as the Q� Fibonacci encryption matrix.)

K- I •

Partial Fraction Decomposition

�, Where Q Has a Non repeated I rred ucible Quadratic Factor Q

�, Where Q Has a Q

Repeated I rred ucible Quadratic Factor

2

Consider the problem of adding two rational expressions: 3 x + 4

and

x

-

3

892

CHAPTER 12

Systems of Equations a n d I ne q u a l ities

The result is 2 3 + -x + 4 x - 3

--

3 (x - 3 ) + 2 ( x + 4) (x + 4)(x - 3)

-'--'------'---'-= ----

5x - 1 x2 + X - 1 2

5x - 1 --- The reverse procedure, of starting with the rational expression x2 + X - 12 3 and writing it as the sum (or difference) of the two simpler fractions -- and x + 4 2 -- , is referred to as partial fraction decomposition, and the two simpler fractions x - 3 are called partial fractions. Decomposing a rational expression into a sum of partial fractions is important in solving certain types of calculus problems. This section presents a systematic way to decompose rational expressions. We begin by recalling that a rational expression is the ratio of two polynomials, say, P and Q =1= O. We assume that P and Q have no common factors. Recall also that a rational expression

p

is called proper if the degree of the polynomial in the Q numerator is less than the degree of the polynomial in the denominator. Otherwise, the rational expression is termed improper. Because any improper rational expression can be reduced by long division to a mixed form consisting of the sum of a polynomial and a proper rational expression, we shall restrict the discussion that follows to proper rational expressions. The partial fraction decomposition of the rational expression

� depends on the

factors of the denominator Q. Recall from Section 5.6 that any polynomial whose coefficients are real numbers can be factored (over the real numbers) into products of linear and/or irreducible quadratic factors. This means that the denominator Q of p

the rational expression - will contain only factors of one or both of the following Q types: 1.

2.

Linear factors of the form x - a, where a is a real number. Irreducible quadratic factors of the form ax2 + bx + c, where a, b, and c are real numbers, a =1= 0, and b2 - 4ac < 0 (which guarantees that ax2 + bx + c cannot be written as the product of two linear factors with real coefficients).

As it turns out, there are four cases to be examined. We begin with the case for which Q h as only nonrepeated linear factors. 1

Decom pose

Case 1 :

p

Q

' Where Q Has Only Non repeated Linear Factors

Q has only non repeated linear factors.

Under the assumption that Q has only nonrepeated linear factors, the poly nomial Q has the form

where none of the numbers a l , a2 , . . . , an is equal. In this case, the partial fraction decomposition of P(x) --

Q(x)

=

� is of the form

Al A2 A ll + --+ . . . + --x - an X - al x - a2

---

where the numbers A ] , A2 , . . . , An are to be determined.

(1)

SECTION 12.5

Parti a l Fraction Decomposition

893

We show how to find these numbers i n the example that follows.

EXAM P L E 1

Nonrepeated Linear Factors x Write the partial fraction decomposition of 2 . x - 5x + 6

Solution

First, we factor the denominator, x2 - 5x + 6 = (x - 2 ) ( x - 3) and conclude that the denominator contains only nonrepeated linear factors. Then we decompose the rational expression according to equation ( 1 ) : A x B ---- = -- + -2 x - 2 x - 3 x - 5x + 6

(2)

where A and B are to be determined. To find A and B, we clear the fractions by m ultiplying each side by (x - 2) (x - 3 ) = x2 - 5x + 6. The result is x = A ( x - 3 ) + B ( x - 2)

(3)

or x = ( A + B)x + ( -3A - 2B) TillS equation is an identity in x. We equate the coefficients of like powers of x to get

{

1 = A + B o = -3A - 2B

Equate coefficients of x: 1x Equate the constants: 0

=

=

(A

-3A

+

-

B)x. 2B.

This system of two equations containing two variables, A and B , can be solved us ing whatever method you wish. Solving it, we get A = -2

B = 3

From equation (2), the partial fraction decomposition is x 3 -2 ---: :- --- = -- + -x - 2 x - 3 x2 - 5x + 6 Check: The decomposition can be checked by adding the rational expressions.

-2(x - 3 ) + 3 ( x - 2 ) 3 -2 -- + -- = ------x - 2 x - 3 (x - 2)(x - 3) x 2 x - 5x + 6

x (x - 2)(x - 3) •

The numbers to be found in the partial fraction decomposition can sometimes be found more readily by using suitable choices for x (which may include complex numbers ) in the identity obtained after fractions h ave been cleared. In Example 1 , the identity after clearing fractions i s equation (3): x = A(x - 3 ) + B ( x - 2) If we let x = 2 in this expression, the term containing B drops out, leaving 2 = A ( - 1 ) , or A = -2. Similarly, if we let x = 3, the term containing A drops out, leaving 3 = B. As before, A = -2 and B = 3 . ..:>il

-


894

CHAPTER 12

Systems of Equations and I n eq u a l ities

2

P Where Q Has Repeate d Lmea · r Factors Decom pose -, Q

Case 2:

Q has repeated linear factors.

If the polynomial Q has a repeated linear factor, say (x - a) n , n 2:: 2 an integer,

then, in the partial fraction decomposition of

p

Q

' we allow for the terms

An A Al 2 -+ + . . . + ----'''X - a (x - a),' (x - a)2

--

where the numbers A I , A , . . . , An are to be determined.

2

EXA M P L E 2

Repeated Linear Factors Write the partial fraction decomposition of

Solution

x+2 . X 3 - 2x2 + x

First, we factor the denominator,

x3 - 2x2 +

X

=

X

(x2 - 2x + 1 )

=

x (x - 1 f

and find that the denominator has the nonrepeated linear factor x and the repeated

A x

linear factor (x - 1 f By Case 1, we must allow for the term - in the decomposition; B and, by Case 2, we must allow for the terms -- +

x - I

We write

x+2 x3 - 2x2 + x

:- -:--- = ---:-

A x

-+

B x- I

-- +

e . .. 111 the decomposItIon. (x - 1 ) 2 e (x - l )2

----=-

Again, we clear fractions by multiplying each side by x3 - 2x2 + The result is the identity

x+2

=

A(x - I ? + Bx(x - 1 ) + ex

(4) X

=

x(x - I f (5)

=

0 in this expression, the terms containing B and e drop out, leaving = 2. Similarly, if we let x = 1, the terms containing A and B drop out, leaving 3 = C. Then equation (5) becomes

If we let x

2

=

A ( -1 )2, or A

x+2

Now let x

=

=

2(x - I ? + Bx(x - 1 ) + 3x

2 (any choice other than 0 or 1 will work as well). The result is 4= 4 = 2B = B =

2( 1 ? + B(2 ) ( I ) + 3 ( 2 ) 2 + 2B + 6 -4 -2

We have A 2, B -2, and e = 3. From equation (4), the partial fraction decomposition is =

=

x+2 x3 - 2x2 +

---: :::---

E XAM P L E 3

X

2

= X

+

-2

-- +

x - I

Repeated Linear Factors Write the partial fraction decomposition of

3

----=-

(x - 1 f

x3 - 8 3 . X2 (x - 1 )

•

SECTION 12.5

Solution


895

The denominator contains the repeated linear factor x2 and the repeated linear factor (x - 1 )3. The partial fraction decomposition takes the form

+ + + +

x3 - 8 x- (x - l ) � 0

?

=

A x

-

B x-

+ ?

+

D C + x - I ( x - l ) ?-

--

E ( x - l )3

(6)

As before, we clear fractions and obtain the identity x3 - 8

=

Ax(x - 1 )3

B ( x - I ? + Cx2( x - I ?

Dx2( x - 1 )

Ex2

(7)

Let x = O. (Do you see why this choice was made?) Then -8 = B ( - l ) B = 8 Now let x

=

1 in equation (7). Then -7 = E

Use B = 8 and E = -7 in equation (7) and collect like terms.

+ + + + + ++ + + + + + x3 - 8 = Ax(x - 1 )3

+

8(x - 1 ) 3

Cx2( x - I ?

7 x2 = Ax(x - 1 )3

x3 - 8 - 8 ( x3 - 3x2 + 3x - 1 ) -7x� o

Dx2(x - 1 ) - 7x2

Cx2 (x - 1 )2

31x- - 24x = x ( x - l ) [ A (x - 1 ) 2

24) = x ( x - l ) [ A (x - I ?

-7x + 24 = A ( x - I f

-

1)

Cx( x - 1 ) + Dx]

?

x(x - 1 ) ( -7x

Dx2(x

Cx( x - 1 )

Cx( x - 1 )

Dx

Dx]

(8)

We now work with equation (8). Let x = O. Then 24 = A Now let x = 1 in equation (8). Then

+

17 = D

+

Use A = 24 and D = 1 7 in equation (8) and collect like terms. -7 x Now let x = 2. Then

24 = 24 (x - I ? + C x (x -14

-

+ + + 24 = 24 -48 = 2C -24 = C

C( 2 )

1)

17 x

34

We now know all the numbers A , B, C, D , and E, so, from equation (6), we have the decomposition x3 - 8 x2( x - l )3

-:----:-

24

=

X

+ -+ + + ---,8 x2

-24 x - I

--

17 ( x - l )2

-7 ( x - l )3

-

•

� I=e! �- Now Work Example 3 by solving the system of five equations containing five variables that the expansion of equation (7) leads to. 1d'J!i: =m: =--


++ + +

The final two cases involve irreducible quadratic factors. A quadratic factor is irreducible if it cannot be factored into linear factors with real coefficients. A qua dratic expression ax2 bx c is irreducible whenever b2 - 4ac < O. For example, x2 x 1 and x2 + 4 are irreducible.

896

CHAPTER 12


3

Decom pose

P

Q

' Where Q Has a No n repeated I rreducible

Quadratic Factor Case 3:

Q contains a nonrepeated i rreducible quadratic factor

If Q contains a nonrepeated irreducible quadratic factor of the form ax2 + bx + the term

c,

then, in the partial fraction decomposition of Ax + B 2 ax + bx +

P,

Q

allow for

C

where the numbers A and B are to be determined.

EXAM P L E 4

Nonrepeated I rreducible Quadratic Factor 3x - 5 . . . . ' d ecompOSltion 0f 3 W nte · t Ile partIaI fractIon x - I

Solution

We factor the denominator, x3 - 1 = (x - 1 ) (x2 +

X

+ 1)

and find that i t has a nonrepeated linear factor x - I and a nonrepeated irreducible quadratic factor x2 + x + 1. We allow for the term � by Case 1, and we allow x - I Bx + C for the term 2 by Case 3. We write x + x + 1 Bx + C 3x - 5 A --= -- + --::3 2 x I x + X + 1 x - 1

----: :-

(9)

We clear fractions by multiplying each side of equation (9) by x3 - 1 = ( x - 1 ) (x2 + X + 1 ) to obtain the identity 3x - 5 = A ( x2 +

X

+ 1 ) + ( Bx + C ) ( x - 1 )

(10)

Expand the identity in (10) to obtain 3x - 5 = (A + B)x2 + (A - B + C)x + (A - C)

{� � �

This identity leads to the system of equations

A

The solution of this system is A see that

=

� g�

+ C - C = -5

:

(3 )

13 - �, B = �, C = . Then from equation (9), we 3 3 3

2

2 13 -x + 3x - 5 3 3 3 = -- + --::--2 x - I x3 - 1 + x + l x 'I"

-- Now Work Example 4 using equation (10) and assigning values to x

. ..=>-


•

SECTION 12.5

4

Decompose

P


897

' Where Q Has a Repeated I rred ucible

Q

Q u a d ratic Factor Case 4:

Q contains a repeated irreducible quadratic factor.

( a x2 bx

If the polynomial Q contains a repeated irreducible quadratic factor + + C ) " , n ?: 2, n an integer, then, in the partial fraction decomposiP . tlon of ' allow for the terms Q

A] , B1 , A2 , B2 , . . , AI" BIl x3(x2 x42)2" 2 ( x 4/, x(x32 x4)22 Axx2 4B (Cxx2 4)2 x3 x2 (Ax B) (x2 4) Cx x" x-= Ax3 (4A C)x 4B = A B { 4A 4BC === l A = x"B =x- C==x-4, = -4-4.x -4 (x2 4)2 x2 4 (x2 4)2 are to be determined.

where the numbers

EXAM P LE 5

Repeated I rreducible Quad ratic Factor

+

Write the partial fraction decomposition of

Solution

+ The denominator contains the repeated irreducible quadratic factor we write + + D ---:: - + ----::-

+

+

We clear fractions to obtain +

+

=

+

+

Collecting like terms yields the identity ,

+

7

+

Rc 7

+

Equating coefficients, we arrive at the system

so

(11)

+

+

+

+ D

+

+

+ D

1

+

0 0

D +

The solution is

1,

,

+

+

From equation ( 1 1) ,

D

1,

7

+ 1 + +

+

Now Wor k P R O B l E M 3 5

w== =

•

1 2.5 Assess Your Understanding 'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red. The equation (x - I ? example of an identity. (p. 86)

1. True or False

2. True or False

(p. 349)

1

=

5x2 The rational expression ,

x(x

-

1

x" + 1

- 2) is an

is proper.

3.

Factor completely: 3x4

4. True or False

+

6x3 + 3x2. (pp. 49-55)

Every polynomial with real numbers as coefficients can be factored into products of linear and/or irreducible quadratic factors. (pp. 389-393)

898

CHAPTER 12


Skill Building In Problems 5-12, tell whether the given rational expression is proper or improper. If improper, rewrite it as the sUin of a polynomial and a proper rational expression. 5.

9.

x x2 - I

6.

-

5x3

� 2x - 1

10.

x- - 4

+ 2 , x' - 1 3x4 + x2 - 2 5

7.

:

X

11.

3 + 8

3x2 - 2

x2 + 5

8. ----;:-2-

2 X

x - I

- 4

x(x - 1 )

(x + 4)(x - 3 )

----

2X(x2 + 4 )

12 •

x2 + 1

.

In Problems 13-46, write the partial fraction decomposition of each rational expression. . 13.

17. '

4

14.

x(x - 1 )

29.

(x - 1 ) (x - 2)

x - 3 (x + 2 ) ( x + 1 )

X 2

x

41.

45.

+ 2x + 3

2

( x + 1 ) ( x2 + 2x + 4 )

33.

37.

2

( x + 2 ) ( x - 4)

7x + 3

30.

38.

x3 - 2x2 - 3x x3 '

42.

(x2 + 16)3 2x + 3 4

26.

34.

+ 2x - 3

46.

x - 9x

2

19.

18. -,-----,-------,-

22.

x

15.

3x

x ----

21.

25.

3x ( x + 2 ) (x - 1 )

2x + 4

3 -

23.

x - I

2

'x

+ x

27.

( x + 2 ) ( x - 1 )2 x2 - 1 1x - 1 8

31.

x ( x2 + 3x + 3 ) x2 - x - 8 ( x + 1 ) ( x- + 5x + 6 ) 7

x5 + 1 x6 X

35.

39.

4 X

1

7-=-

x (x- + 1 ) x2

----;c -

-

(x - I f( x + 1 ) x2

(x - 1 ) 2 ( x + 1 ) 2 x + 4 x (----=x2 + 4)

----=2

x ( 3x - 2) (2x + 1 ) x2 + 2x + 3 (x2 + 4 ) 2 2 X

x3 - 4x2 + 5x - 2 4

2

43. ---:2:---

2x - 5x - 3

( x2 + 4)3

16.

20.

24.

28.

32.

36.

1 (x + l ) (x2 + 4) x + 1 x2( x - 2 ) x + 1 x2 ( x - 2 ) 2 10x2 + 2x (x - 1 )- ( x2 + 2 ) 7

1 (2x + 3 ) (4x -

1)

-----

x3 + 1 (x2 + 1 6 ) 2

40. X

3

x2 + 1 +

7 x-

- 5x + 3

4x

44. ---:7:---

2x- + 3x - 2

x2 + 9 7 4 x - 2x- - 8

'Are You Prepared?, Answers 1.

True

2.

True

4.

True

1 2.6 Systems of Nonlinear Equations PREPARING FOR TH I S S ECTION • • •


Lines (Section 2.3, pp. 173-185) Circles (Section 2.4, pp. 189-193) Parabolas (Section 1 1 .2, pp. 773-779)

• •

Ellipses (Section 1 1 .3, pp. 782-789) Hyperbolas (Section 1 1 .4, pp. 792-801)

Now Wor k the 'Are You Prepared?' problems on page 904. OBJECTIVES

1 Solve a System of Nonlinear Equations Using Su bstitution (p. 898)

2 Solve a System of Nonli nea r Equations Using Elimination (p. 900)

1

Solve a System of No nl i near Equations Using Su bstitution

In Section 1 2 . 1 , we observed that the solution to a system of linear equations could be found geometrically by determining the point(s) of intersection (if any) of the equations in the system. Similarly, when solving systems of nonlinear equations, the solution(s) also represents the point(s) of intersection (if any) of the graphs of the equations.

SECTION U.6

Systems of Nonlinear Equations

899

There is no general methodology for solving a system of nonlinear equations. At times substitution is best; other times, elimination is best; and sometimes neither of these methods works. Experience and a certain degree of imagination are your allies here. B efore we begin, two comments are in order. 1.

2.

EXAM P L E 1

If the system contains two variables and if the equations in the system are easy to graph, then graph them. By graphing each equation in the system, you can get an idea of how many solutions a system has and approximately where they are located. Extraneous solutions can creep in when solving nonlinear systems, so it is im perative that all apparent solutions be checked.

Solving a System of Nonlinear Equations

{

Solve the following system of equations:

Solution Using Substitution Figure 9

2X2 - Y = 0 (y = 2X 2 )

-6

3x- y = -2 (y = 3x + 2) (2, 8 )

6 x

3X - y = -2 2x2 - Y = 0

(1)

(2)

+

First, we notice that the system contains two variables and that we know how to graph each equation. Equation ( 1 ) is the line y = 3x 2 and equation (2) is the parabola y = 2x2 . See Figure 9. The system apparently has two solutions. We will use substitution to solve the system. We choose to solve equation ( 1 ) for y . Equation (1) 3x y = -2 y = 3x + 2

-

We substitute this expression for y in equation (2). The result is an equation con taining j ust the variable x, which we can then solve. 2X2 Y = 0 Equation (2)

-

2x

+

+

2x2 - (3x 2 ) = 0 2x2 - 3x - 2 = 0

(2x

1 = 0 X

-

=

+

1 ) (x - 2) = 0 x - 2 = 0

or

1 2

( -�)

The apparent solutions are x Check: For x =

-

1 1 2 y =2 '

Remove pa rentheses. Factor. Apply the Zero-Prod uct Property.

x = 2

or

Using these values for x in y = 3x

y = 3

Substitute 3x + 2 for y.

+

=

+

2 =

2, we find

�

or

y = 3 (2 )

+

1 1 - 2 ' y = 2 and x = 2, y = 8.

'

1 2 1 2 For x = 2,

y = 8,

2 = 8

{

6 - 8 = -2 3 (2 ) - 8 = 2 (2 ? - 8 = 2 ( 4 ) - 8 = 0

-2

(1)

o

(2)

(1)

(2)

900

CHAPTER 12


)

(

Each solution checks. Now we know that the graphs in Figure 9 intersect at the 1 1 . POll1ts - '2' 2' and (2, 8 ) . Check: Gra ph

3x

-

y

= -2 ( Y1 =

3x

•

+ 2) a n d 2X2 - Y = 0 ( Y2 = 2X2 ) and

compare what you see with Fig u re 9. Use I NTERS ECT (twice) to find the two poi nts of i ntersection.

1;l!I!;Z: = = --

2

Now Work P R O B l E M 1 5 U 5 I N G S U B S T I T U T I 0 N

Solve a System of Nonli near Equations Using E l i m i nation Our next example illustrates how the method of elimination works for nonlinear systems.

Solve:

Solution Using Elimination Fig ure 1 0

y

;c

+ l = 13 - y = 7

(1 )

(2)

+

{X2x2 - y

x

+ l = 13 7 6 l + Y

y2 = 1 3 6

x2

Equation (1) is a circle and equation (2) is the parabola y = - 7. We graph each equation, as shown in Figure 10. Based on the graph, we expect four solutions. By subtracting equation (2) from equation ( 1 ) , the variable can b e eliminated.

x2 - Y = 7 (y = x2 - 7)

x2 -6

{x�

Solving a System of Nonlinear Equations

EXA M P L E 2

=

Subtract

=

This quadratic equation in y can be solved by factoring.

x

l + y - 6 = 0

( y + 3 ) (y - 2 ) = 0

-8

x2 x2

x

We use these values for y in equation (2) to find If y = 2, then

x

= y + 7 = 9, so

If y = -3, then

x

x.

or y = 2

Y = -3

= y + 7 = 4, so

x

= 3 or -3.

x

x

= 2 or -2.

= -3, Y = 2; We have four solutions: = 3, Y = 2; = 2, Y = -3; and = -2, y -3. You should verify that, in fact, these four solutions also satisfy equation ( 1 ), so all four are solutions of the system. The four points, ( 3 2 ) ( -3, 2 ) , (2, -3), and ( -2, - 3 ) , are the points of intersection of the graphs. Look again at Figure 10 . =

,

�

Check: G ra p h X2 + l =

13

•

a n d X2 - Y

= 7. ( R e m e m b e r that to g ra p h

X2 + l =

13

�= :::z,_

Now Work P R O B l E M 1 3 U 5 I N G E I I M I N A T I O N

req u i res two fu nctions: Y 1 I NTERSECT to find the points of i ntersection.

EXAM P L E 3

{x2

=

VB

- X2 a n d Y2 =

Solving a System of Nonlinear Equations Solve:

x2

- l = 4 Y =

(1)

(2)

,

- VB

- X2.) Use

SECTION 12.6 Figure 1 1

Solution

Systems of N o n l i near Equations

901

Equation ( 1 ) is a hyperbola and equation (2) is a parabola. See Figure 11 . It appears the system has no solution. We verify this using substitution. Replace x2 by y in equation ( 1 ) . The result is x2 - / = 4 y - / = 4

/ - y + 4 = 0

Equation (1 )

y = .;z. Place

sta ndard form .

in

This is a quadratic eq uation. Its discriminant is ( - 1 ) 2 - 4 · 1 . 4 1 - 1 6 = 1 5 < O. The equation has no real solutions, so the system is inconsistent. The graphs of these two equations do not intersect. =

-

•

-5

EXAM P L E 4

Solve:

Solution Using E l i m i nation

{

Solving a System of Nonlinear Equations x2 +

+ / - 3 + 2 = 0 y- - Y x + 1 + = 0 x

�

X

--

(1) (2)

First, we multiply equation (2) by x to eliminate the fraction. The result is an equivalent system because x cannot be O. [Look at equation (2) to see why.]

{

x2 + x + / - 3y + 2 = 0 x2 + x + / - y = 0

(1)

(2)

x

*- 0

Now subtract equation (2) from equation ( 1 ) to eliminate x. The result is -2y + 2 = 0 y = 1

Solve for y.

To find x, we back-substitute y = 1 in equation ( 1 ) : x2 + x + / - 3 y + 2 = 0 x2 + x + 1 - 3 + 2 = 0 x2 + x = 0

x = 0

x(x + 1 ) = 0

or x = - 1

Equatio n (1 )

S ubstitute 1 for y

in ( 1 ) .

Si m plify. Factor.

Apply the Zero-Prod uct Property.

B ecause x cannot be 0, the value x = 0 is extraneous, and we discard it.

{ (-I?

Check: We now check x = - 1 , y = 1 :

The solution i s x tions is ( - 1 , 1).

+ ( - 1 ) + 12 - 3 ( 1 ) + 2 = 1 - 1 + 1 - 3 + 2 = 0 0 12 - 1 0 + = 0 -1 + 1 + -1 -1 --

=

=

-

(1)

(2)

- 1 , y = 1 . The point o f intersection o f the graphs o f the equa •

In Problem 55 you are asked to graph the equations given in Example 4. Be sure to show holes in the graph of equation (2) for x = O. = =1i'Ir::

EXAM P L E 5

{


Solving a System of Nonlinear E quations Solve:

3XY - 2 / = -2 10 9x2 + 4/ =

(1) (2)

902

CHAPTER 12

Systems of Equations and I ne q u a l ities

Solution

(1) 2 (2) l { 69x2XY - == 10 9x23x2 6xy 2xy == 62 x 0 y Y = 2 -2x3x2 X 0 y (2) 9x2 3\': = 10 9x2 (? 2x , 2 )2 = 10 9x2 - 12x2x- 9x4 = 10 9x4 - 12x2 9x4 = 10x2 18x4 - 22x2 = 0 9x4 - llx2 2 = 0 x2) (9x2 - 2)(x2 - 1) = 0 9x2 - 2 = 0 x2 - 1 = 0 x2 = -92 x2 = 1 x = ±J% = ± � x = ±1 y, (3): - .? 2 2 2 2 3x V x = 3 . y = 2x � 2Yz Yz 2( ) 2- 3 x = - -V23-: y = 2 -3x2 2x + � ) -2Yz x= y = 2 -2x3x2 = 2 - 23(1? - --21 Ifx = - I: y = 2 -2x3x2 = 2 - -23(-1)2 21 (V2,3 V2), (- V2,3 -V2), (I, -.!2.) , (-I"!').2

We multiply equation terms.

Since

*

by and add the result to equation

++ +

41

-4

(1 )

(2)

41

Add.


(do you see why?) , we can solve for

+ + + ++ + ++

Now substitute for

(3)

of the system.

41

Equation (2)

S u bstitute y =

_

-

4

2

-

3';-

2x

in (2) .

Expand and simplify

?

M ultiply both sides by ';-.

4

This quadratic equation (in

in this equation to get

*

in equation

4

to eliminate the

4

Subtract 10';- from both sides. Divide both sides by 2.

can be factored: or or

or

To find

we use equation

4

3_

If

__

�

.?

If

If

__

>-

� � -Yz 4

l:

The system has four solutions: Check them for yourself. £""

�


•

The next example illustrates an imaginative solution to a system of nonlinear equations.

SECTION 12.6

EXAM P L E

Systems of Nonlinear Equations

903

Running a Long Distance Race

6

In a 50-mile race, the winner crosses the finish line 1 mile ahead of the second-place runner and 4 miles ahead of the third place runner. Assuming that each runner maintains a constant speed throughout the race, by how many miles does the second place runner beat the third-place runner?

11....[------ 3 miles ------.+-I.� 1 mile �I

Solution

Let VI , V2 , and V3 denote the speeds of the first-, secondo, and third-place runners, respectively. Let tl and 1 2 denote the times (in hours) required for the first-place runner and second-place runner to finish the race. Then we have the system of equations 50 = V I I I 49 = V2 1 1 { 46 = v3t l 50 = v2t2

(1 )

First-place run ner goes 5 0 miles in t, hours.

(2)

Second-place ru n ner goes 49 m iles in t, hours. Th ird-place runner goes 46 miles in t , hours.

(3)

Second-place run ner goes 50 miles in t 2 hours.

(4)

We seek the distance d of the third-place runner from the finish at time t? At time t2 , the third-place runner has gone a distance of V3t2 miles, so the distal�ce d remaining is 50 - v3t2 ' Now d

50

-

= 50

-

=

V3t2 V3

C .�) I

t? = 50 - (V3tl ) ' -= tl 50 V? = 50 - 46 · --=-50 Vj V = 50 - 46 . , V2 50 = 50 - 46 · 49 ::::: 3.06 miles

1

From (3), v3 t , From (4) , t 2 From (1), t,

=

=

=

46

50

-

V2 50

v,

From the quotient of (1) and (2) . •

f-lisJorical Feature

I

n the beg i n n i n g of this section, we said that i m a g i nation a n d experi

of the equations i nvolved. Th i s conjecture was i n d eed made by Etienne

ence are i m portant in solving systems of n o n l i near equations. I n d eed,

Bezout ( 1 730- 1 783), but working out the deta ils took a bout 1 50 years.

these k i n d s of problems lead into some of the deepest and most dif

It turns out that, to a rrive at the correct n u m ber of i n tersections, we must

ficult parts of modern mathematics. Look a g a i n at the g raphs i n Exam

count not o n ly the complex n u m ber intersections, but a l so those inter

ples 1 and 2 of this section (Figures 9 a n d 1 0) . We see that Exa m ple 1

sections that, in a certa i n sense, lie at infin ity. For example, a parabola

ture that the n u m ber of solutions is e q u a l to the product of the degrees

at infin ity. T h i s t o p i c is p a r t o f the study of a l gebraic geometry.

has two solutions, a n d Exa m ple 2 has four sol utions. We might conjec

H i storical Problem

A p a p y r u s d a t i n g b a c k t o 1 95 0

BC

c o n ta i n s t h e fo l l ow i n g p ro b l e m :

" A g iven s u rface a rea o f 1 00 u n its o f a rea s h a l l be represented a s t h e s u m o f two s q u a res wh ose s i d e s a re to e a c h other a s l ' �

. 4'

"

and a l i n e lying on the axis of the parabola i n tersect at the vertex a n d

{X2

Solve for the sides by solving the system of equations +

I

x

=

=

�oo

.�?

904

CHAPTER 12

Systems of Equations and I nequa lities

1 2.6 Assess Your Understanding

'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1.

2.

Graph the equation: Graph the equation:

In Problems

Skill Building

5.

5-24,

{ y = x2 + 1 y x + 1

y = 3x + 2 (pp. 173-1 85 ) y + 4 = x2 (pp. 773-779)

13.

6.

{ y = vX y= 2 - x

17.

21.

{

x2 + i = 4 x2 + 2x + i = 0

{x2 + i = 4 i- x = 4

{

x2 + i= 4 y i" x2- 9

In Problems 25.

28.

31.

34.

37.

40•

43.

25-54,

10.

14.

18.

22.

{ y = x2 + 1 y = 4x + 1

{ X2 - 4i = 16 2y - x 2

{ y = vX y = 6- x

{

x2 + i = 8 x2 + i + 4y = 0

{x2 + i = 16 ? x-- 2y = 8

{ Xy = 1 y = 2x + 1

4X2 - 3xy + 9i = 15 2x + 3y = 5

26.

29.

=

{ 3X2 - 2i + 5 = 0 2x2 - i + 2 = 0 { x 2 + 2xy = 10 3x2 - xy = 2

{ i- x 2 + 4= 0 2x2 + 3i = 6

{ { �' �,

�- � + 3= 0 x2 / 3 1 - + -= 7 x2 i

32.

35.

38.

41.

44.

-

46.

,

49.

52.

55.

1 - + -= 4 X4 l �

/ +Y + X2- X- 2= 0 { x- 2 y + 1 + -- = 0 y

2 )= 3 { IOgxC Y l ogxC 4y ) = 2

7.

11.

] 5.

19 •

23.

solve each system. Use any method you wish.

{ 2x2 + i = 18 x y= 4

{

Graph t h e equation:

4.

Graph the equation:

i = x2 - 1 (pp. 792-801 )

x2 + 4i = 4 (pp. 782-789)

graph each equation of the system. Then solve the system to find the points of intersection.

=

9.

3.

47.

50.

53 .

Graph the equations given in Example 4.

{ y = V36- x2 Y= 8 - x {x = 2y x = i - 2y

{

{

20.

{ y = x2 - 4 y 6x - 13

24.

=

x + y + 1= 0 x2 + i + 6y - x =-5

{ 7x2 - 3/ + 5= 0 ? ? 3x- + 5y-= 1 2

27.

30.

33.

36.

Y + 1 3 i + 3 6= 0 { 5X xy + 7y 2 = 6

39.

{ x2 + 2i = 16 4x-? - Y2 = 24

42.

2 3 -- - + 1 = 0 x2 i 6 7 -- - + 2= 0 x2 i

45.

{ x 2 - 3xy + 2i = 0 x2 + xy = 6

48.

3 x - 2x2 + i + 3 y - 4= 0 { i- y x- 2 + --= 0 x2

{

16.

xy = 4 x2 + i = 8

{ 2i - 3xy + 6y + 2x + 4= 0 2x - 3y + 4= 0

{

12.

y = 3x - 5 x2 + i = 5

{ x 2 - i = 21 x +Y = 7

{

8.

In x = 4 1n y log3 x = 2 + 2 log3 y

56.

5 1.

54.

{

{y = � y = 2x + 4

{ y = x- I y = x2 - 6x + 9 { X2 + i = 10 y =x + 2

{ x2 = y xy = 1

{x 2 + i = 1 0 xy = 3

y = 2x + 1 ? 2x 2 + y- = 1

{ 2x2 - xy + i = 8 xy = 4

{

x 2 - 41 + 7 = 0 3x 2 + y2 = 3 1

{ x2 - 3 i + 1 = 0 2x 2- 7 i + 5= 0

{

{

2x2 + i = 2 ? x 2 - 2y- + 8 = 0

4x2 + 3 i= 4 2x2- 6i= -3

I{

6 - + -= 6 x4 l

� - �= 1 9 x4 l

{ X2 - xy - 2i = 0 xy + x + 6= 0 { {

logxY= 3 logxC 4y ) = 5

In x = 5 In y log x = 3 + 2 log Y 2 2

Graph the equations given in Problem 49.

SECTION 12.6

[n Problems

905

graph each equation and find the poinc(s) of intersection, if any. y 2 (y If (x 2y 2 (y f (x l - 2y (y - 2 (x l- 1)2 y - x y x2 l y [n Problems use a graphing utility to solve each system of equations. Express the solution(s) rounded to two decimal places. {x3 x2l 2 {x2 x'!3y {yy ex3/2 {yy ex2/3 {X2 ly {Xy x {X4 xyl {X4 xy2 122 5 7-62,

= 0 and the circle (x - 1 ) + -

57. The line x + 2

59. The circle

the parabola

61.

:�

Systems of Nonli near Equations

63.

67.

=

4 x -

3

--

58. The line x +

the circle

5

+ (y + 2f = 4 and + 1 = 0

60. The circle

the parabola

. ? and the circle x- - 6x + Y2 + 1

=

62.

0

=

64.

-.\

+ l

=

68.

=

=

=

=

-.\

+

=

6

69.

= 1

Y

=

=

+ 2 )2 +

1) - x -5

4 . and the CIrcle x + 2

+

65.

=

+ 6 = 0 and + 1) + + I

--

63-70,

=

=

+ 4

=

2 4

2 In

=

=

=

5

4 and 0

+ 4x +

66.

+

70.

+

- 4

=

Y =

=

=

=

0

4

4 In x

Applications and Extensions 82. Geometry

The altitude of an isosceles triangle drawn to its base is 3 centimeters, and its perimeter is 1 8 centimeters. Find the length of its base. 83. The Tortoise and the Hare In a 2 1 -meter race between a tortoise and a hare, the tortoise leaves 9 minutes before the hare. The hare, by running at an average speed of 0.5 meter per hour faster than the tortoise, crosses the finish line 3 min utes before the tortoise. What are the average speeds of the tortoise and the hare?

71. The difference of two numbers is 2 and the sum of their

squares is 1 0. Find the numbers.

72. The sum of two numbers is 7 and the difference of their

squares is 2 1 . Find the numbers.

73. TIle product of two numbers is 4 and the sum of their squares

is 8. Find the numbers.

74. The product of two n umbers is 1 0 and the difference of their

squares is

21.

Find the numbers.

75. The difference of two numbers is the same as their product,

and the sum of their reciprocals is 5. Find the numbers.

76. The sum of two numbers is the same as their product, and

the difference of their reciprocals is 3. Find the numbers. 2 77. The ratio of to is '3' The sum of and is What is the ratio of

a a b a - ab +

to

78. TIle ratio of to

ratio of

79. Geometry

b b-a ba b?

to

?

a b 10. a b

is 4: 3. The sum of and +

�----

is 1 4. What is the

84. Running a Race

I n a I-mile race, the winner crosses the fin ish line 1 0 feet ahead of the second-place runner and 20 feet ahead of the third-place runner. Assuming that each runner maintains a constant speed throughout the race, by how many feet does the second-place runner beat the third place runner?

The perimeter of a rectangle is 1 6 inches and its area is 15 square inches. What are its dimensions?

80. Geometry

21 meters ----�

An area of 52 square feet is to be enclosed by two squares whose sides are in the ratio of 3. Find the sides of the squares.

2:

81. Geometry

Two circles have circumferences that add up to 127T centimeters and areas that add up to 207T square cen timeters. Find the radius of each circle.

85. Constructing a Box

A rectangular piece of cardboard , whose area is 2 1 6 square centimeters, is made into an open box by cutting a 2-cen timeter square from each corner and turning up the sides. See the figure . If the box is to have a volume of 224 cubic centinleters, what size cardboard should you start with?

�----------_L ___________

l

I

CHAPTER 12

906


86.

90. 91.

Constructing a Cylindrical Tube A rectangular piece of cardboard, whose area is 216 square centimeters, is made into a cylindrical tube by j oining together two sides of the rec tangle. See the figure . If the tube is to have a volume of 224 cubic centimeters, what size cardboard should you start with?

87.

Geometry Find formulas for the base b and one of the equal sides L of an isosceles triangle in terms of its altitude hand perimeter P. Descartes's Method of Equal Roots Descartes's method for finding tangents depends on the idea that, for many graphs, the tangent line at a given point is the line that inter sects the graph at that point only. We will apply his method to find an equation of the tangent line to the parabola at the point (2, 4 ) . See the figure.

u.nique

y=x2

Fencing A farmer has 300 feet of fence available to enclose a 4500-square-foot region in the shape of adjoining squares, with sides of length and See the figure . Find and

x y.

x y.

First, we know that the equation of the tangent line must be in the form b. Using the fact that the point (2, 4 ) i s o n the line, w e can solve for b i n terms o f a n d get the equation (4 Now we want (2, 4) to be the solution to the system

y=mx+ m y = mx + 2m). unique {� : :::x+ m x2 - mx+ m = x= m "'-1m2 x,m2 m, 4- 2

x

From this system, we get using the quadratic formula, we get

88. 89.

±

Bending Wire A wire 60 feet long is cut into two pieces. Is it possible to bend one piece into the shape of a square and the other into the shape of a circle so that the total area enclosed by the two pieces is 100 square feet? If this is possible, find the length of the side of the square and the radius of the circle.

(2

- 4)

O. By

- 4(2m - 4 )

2

To obtain a unique solution for the two roots must be equal; in other words, the discriminant - 4(2m - 4) must be O. Complete the work to get and write an equation of the tangent line.

L P. In Problems use Descartes's method fi·om ProbLem to find the equation of the Line tangent to each graph at the given point. 92.x2+1=10; 93.y=x2+2; 94.x2+y=5; 96. 3x2+l 95. 2X2+31 = 97.x2 - l = 3; (2,1) 98.21 - x2= 99. r2 ax2+ c= b a Geometry Find formulas for the length and width rectangle in terms of its area A and perimeter 92-98,

IV

of a

91

at ( 1 , 3 )

14;

1 4;

If

rl

and

at ( 1 , 3 )

at ( 1 , 2 )

=

7;

at (2, 3 )

are two solutions o f a quadratic equation

at ( -2, 1 )

at ( -1 , 2 )

bx +

at

0 , i t can b e shown that

and

Solve this system of equations for /"1 and

/"2.

Discussion and Writing 100. A

circle and a line i ntersect at most twice. A circle and a parabola intersect at most four times. Deduce that a circle and the graph of a polynomial of degree intersect at most six times. What do you conjecture about a polynomial of de gree 4? What about a polynomial of degree Can you ex plain your conclusions using an algebraic argument?

3

101.

n?

Suppose that you are the manager of a sheet metal shop. A customer asks you to manufacture 10,000 boxes, each box being open on top. The boxes are required to have a square

base and a 9-cubic-foot capacity. You construct the boxes by cutting out a square from each corner of a square piece of sheet metal and folding along the edges. (a) What are the dimensions of the square to be cut if the area of the square piece of sheet metal is 1 00 square feet? (b) Could you make the box using a smaller piece of sheet metal? Make a list of the dimensions of the box for various pieces of sheet metal .

SECTION

Systems of Inequalities

U.7

907


2.

3.

y

4.

y 5

y 5

(0, 1)

-2

2

( - 1, -1)

x

-5

5

x

( - 2,0)

(2,0)

-5

5 (0, - 1)

-5

x

-5

12.7 Systems of Inequalities PREPARING FOR THIS SECTION •

•


Solving Linear Inequalities (Section 1 .5 , pp. 124-13 1 )

•

Lines (Section 2.3, pp. 1 73-185)

Circles (Section 2.4, pp. 1 89-193) Graphing Teclmiques: Transformations (Section 3.5, pp. 252-260)

Now Work the 'Are You Prepared?' problems on page 912. OBJ ECTIVES

1 Graph an Inequality (p. 907) 2 Gra ph a System of I neq u a l ities (p. 909)

In Section 1 .5 , we discussed inequalities in one variable. In this section, we discuss inequalities in two variables. EXAM PLE 1

Exam ples of I nequal ities in Two Variables

(a) 3x

1

E XA M P L E 2

+

y

:S

6

l

< 4

(c)

l

>

x

•

G raph a n Ineq ual ity

An inequality in two variables x and y is satisfied by an ordered pair (a, b) if, when x is replaced by a and y by b, a true statement results. The graph of an inequality in two variables x and y consists of all points (x, y) whose coordinates satisfy the inequality. Let's look at an example. G raphing an I n eq uality

Graph the linear inequality: Solution

+

(b) x2

3x

+

y :S 6

We begin by graphing the equation

3x

+ Y

= 6

formed by replacing (for now) the :s symbol with an = sign. The graph of the equa tion is a line. See Figure 12(a) on page 9 08. This line is part of the graph of the in equality that we seek because the inequality is nonstrict. (Do you see why? We are seeking points for which 3x + y is less than or equal to 6.)

908

CHAPTER

U


Figure 12

Y

Y

e (5, 5)

e (5,5)

( - 1, 2)e

-6

( 1. 2) e

6 e (4, -1)

( - 2, - 2) e

x

(a) 3x+ Y= 6

-6

6 e (4, -1)

2.

x

(b) Graph of 3x+ Y::; 6

Now let's test a few randomly selected points to see whether they belong to the graph of the inequality. 3x+ys6

(4, -1)

3(4)+(-1) = 11 > 6 3(5)+5 = 20 > 6

(5,5) (-1,2) (-2, -2)

3(-1)+2 = -1 :::; 6 3(-2)+(-2) = - 8:::; 6

Conclusion

Does not belong to the graph Does not belong to the graph Belongs to the graph Belongs to the graph

Look again at Figure 12(a). Notice that the two points that belong to the graph both lie on the same side of the line, and the two points that do not belong to the graph lie on the opposite side. As it turns out,this is always the case. The graph we seek consists of all paints that lie on the same side of the line as (-1, 2) and (-2, -2) and is shown as the shaded region in Figure 12(b). -- Now Work PRO B L E M 1 5

"""

NOTE The strict inequalities are and >. The nonstrict inequalities • are :=; and 2::.
4 Does not belong to the graph

All the points inside and on the circle satisfy the inequality. See Figure 13. 'I!! e'

3

-

!

-

909


•

Linear Ineq u a l ities

-3

Linear inequalities are inequalities in one of the forms

Figure 14

Ax

y

+

By < C

Ax+By > C

Ax +By :S C

Ax

+

By 2: C

where A and B are not both zero. The graph of the corresponding equation of a linear inequality is a line that sep arates the xy-plane into two regions, called half-planes. See Figure 14. As shown, Ax + By = C is the equation of the boundary line and it divides the plane into two half-planes: one for which Ax + By < C and the other for which Ax +By > C. Because of this, for linear inequalities, only one test point is required. EXAM P L E 4

G raphing Li near I nequal ities

Graph:

(b) y 2: 2x

(a) y < 2 (a) The graph of the equation y = 2 is a horizontal line and is not part of the graph of the inequality. Since (0, 0) satisfies the inequality, the graph consists of the half-plane below the line y = 2. See Figure 15.

Solution

Figure 16

Figure 15

y Graph of y

:S

2-

(15,5) (20,5) (25,0)

,�-��::-.;-.. y =5

1= 1.65

•

SECTION 12.8

DEFINITION

Linear Prog ramming

91 7

A solution to a linear programming problem consists of a feasible point that maximizes (or minimizes) the objective function, together with the corresponding value of the objective function. One condition for a linear programming problem in two variables to have a solution is that the graph of the feasible points be bounded. (Refer to page 912.) If none of the feasible points maximizes (or minimizes) the objective function or if there are no feasible points, the linear programming problem has no solution. Consider the linear programming problem stated in Example 2, and look again at Figure 24. The feasible points are the points that lie in the shaded region. For example, (20,3) is a feasible point, as are ( 15,5), (20,S), (18, 4), and so on. To find the solution of the problem requires that we find a feasible point (x, y) that makes I = 0.06x + 0 . 09y as large as possible. Notice that as I increases in value from I = 0 to I = 0.9 to I = 1.35 to I = 1.65 to I = 1.8 we obtain a collection of parallel lines. Further, notice that the largest value of I that can be obtained using feasible points is I = 1.65,which corresponds to the line 1.65 = 0.06x + 0.09y . Any larger value of I results in a line that does not pass through any feasible points. Finally, notice that the feasible point that yields I = 1.65 is the point (20, 5),a corner point. These obser vations form the basis of the following result, which we state without proof.

THEOREM

Location of the Solution of a Linear Progra m m i ng Problem

If a linear programming problem has a solution, it is located at a corner point of the graph of the feasible points. If a linear programming problem has multiple solutions, at least one of them is located at a corner point of the graph of the feasible points. In either case, the corresponding value of the objective function is unique. We shall not consider here linear programming problems that have no solution. As a result, we can outline the procedure for solving a linear programming problem as follows: Procedure for Solving a Linear Programming Problem

STEP 1: Write an expression for the quantity to be maximized (or minimized).

STEP 2: Write all the constraints as a system of linear inequalities and graph

This expression is the objective function.

STEP 3: List the corner points of the graph of the feasible points. STEP 4: List the corresponding values of the objective function at each cor

the system.

ner point. The largest (or smallest) of these is the solution. EXAM PLE 1

Solvi ng a M i n i m u m Linear Progra m m i ng P roblem

Minimize the expression

z

=

2x + 3y

subject to the constraints

2':0, Y2':0 x:::=;6 x + Y2':2, function is z = 2x + 3 y. We seek the smallest value of z

y:::=;5, Solution

X

The objective occur if x and y are solutions of the system of linear inequalities y:::=; 5

x:::=;6 x + y2':2 2':0 y2':O X

that can

918

CHAPTER 12


The graph of this system (the set of feasible points) is shown as the shaded region in Figure 25 . We have also plotted the corner points. Table 1 lists the corner points and the corresponding values of the objective function. From the table, we can see that the minimum value of z is 4, and it occurs at the point (2, 0) . Table 1

x=5

Figure 25 y

Value of the Objective Function

Corner Point

��

5 �� __� y = (5, 5)

______

B

z =

(x,y)

x

2x + 3y

(0, 2)

z

= 2(0) + 3(2) = 6

(0, 5)

z

= 2(0) + 3(5)

15

(6, 5)

z

= 2(6) + 3(5) = 27

(6, 0)

z

= 2(6) + 3(0) = 1 2

(2, 0)

z =

2(2)

+

3(0) =

x+y=2

4 •

1.i!E:::z!lOOiiiIiiI

E XA M P L E 4

=

Now Work P R O B L E M S 5 A N D 1 1

M ax i m i zing P rofit

At the end of every month, after filling orders for its regular customers, a coffee company has some pure Colombian coffee and some special-blend coffee remain ing. The practice of the company has been to package a mixture of the two coffees into I-pound packages as follows: a low-grade mixture containing 4 ounces of Colombian coffee and 12 ounces of special-blend coffee and a high-grade mixture containing 8 ounces of Colombian and 8 ounces of special-blend coffee. A profit of $0.30 per package is made on the low-grade mixture, whereas a profit of $0 . 40 per package is made on the high-grade mixture. This month, 120 pounds of special-blend coffee and 100 pounds of pure Colombian coffee remain. How many packages of each mixture should be prepared to achieve a maximum profit? Assume that all packages prepared can be sold. Solution

We begin by assigning symbols for the two variables.

x = Number of packages of the low-grade mixture

y = Number of packages of the high-grade mixture

If P denotes the profit, then

P

=

$0.30x + $0.40y

This expression is the objective function. We seek to maximize P subject to certain constraints on x and y. Because x and y represent numbers of packages, the only mean ingful values for x and y are nonnegative integers. So we have the two constraints X

2:: 0,

Y 2:: 0

Nonnegative constraints

We also have only so much of each type of coffee available. For example, the total amount of Colombian coffee used in the two mixtures cannot exceed 100 pounds, or 1600 ounces. Because we use 4 ounces in each low-grade package and 8 ounces in each high-grade package, we are led to the constraint

4x + 8y :s 1600 Colombian coffee constraint Similarly, the supply of 120 pounds, or 1920 ounces, of special-blend coffee leads to the constraint

12x + 8y :s 1920

Special-blend coffee constraint

The linear programming problem may be stated as Maximize

P

=

0.3x + O.4y

SECTION 12.8

Linear Programming

919

subject to the constraints

4x + 8y :s; 1600,

y :::::: 0,

x:::::: 0,

12x+8y :s; 1920

The graph of the constraints (the feasible points) is illustrated in Figure 26 . We list the corner points and evaluate the objective function at each. In Table 2, we can see that the maximum profit, $84, is achieved with 40 packages of the low-grade mixture and 180 packages of the high-grade mixture. Table 2

Figure 26

Value of Profit

Corner Point (x,y) P =

(0,0)

P

P

=

O.3x + OAy

0

= 0.3(0) + 0.4(200) = $80 P = 0.3(40) + 0.4(180) = $84

(0,200) (40,180)

P

(160,0)

= 0.3(160)

+

0.4(0)

4x+ 8y = 1600

12x+ 8y = 1920 &:!In::==-

=

$48

•


12.8 Assess Your Understanding Concepts and Vocabulary 1. A linear programming problem requires that a linear expression, called the , be maximized or minimized .

2.

____

True or False If a linear programming problem has a solu tion, it is located at a corner point of the graph of the feasi ble points.

Skill Building In Problems 3-8, find the maximum and minimum value of the given objective function of a linear programming problem. The figure illustrates the graph of the feasible points. 3. z =

5. z =

7.

z =

4. z =

x+ Y x + lOy

Y 8 (0,6)

6. z = lOx + Y 8. z = 7x + Sy

Sx + 7y

-4

In Problems

9-18,

9. Maximize 10.

Maximize

U.

Minimize

n.

13.

14.

15.

Minimize Maxi mize Maximize Minimize

16. Minimize

17. Maximize 18. Maximize

z

= 2x + y x + 3y

z = =

z = z =

z

=

z = z

8

-1

solve each linear programming problem.

z =

z

(5,6)

2x + 3y

=

z =

subject to

x 2:: 0,

subject to

x 2:: 0,

subject to

x 2:: 0,

3x + 4y

subject to

x 2:: 0,

+

+

Sy subject to 3y

subject to

Sx + 4y

subject to

2x

subject to

Sx

+

3y

Sx + 2y

subject to

2x + 4y

subject to

x + Y ::s 6, x + Y 2:: 3,

x + Y 2:: 1 x::s 5, y::s 7

Y 2:: 0, x + Y 2:: 2, x::s 5, y::s 3 Y 2:: 0, 2x + 3y 2:: 6, x + y::s 8 x 2:: 0, Y 2:: 0, x + Y 2:: 2, 2x + 3y::s 12,

2x + Sy 3x

Y 2:: 0, Y 2:: 0,

3x

+

2y::s 12

x 2:: 0, Y 2:: 0, x + Y 2:: 2, x + y::s 8, 2x + y::s 10 x 2:: 0, Y 2:: 0, x + y 2:: 2, 2x + 3y::s 12, 3x + y::s 12 x 2:: 0, Y 2:: 0, x + y 2:: 3, x + y::s 9, x + 3y 2:: 6 x 2:: 0, x 2:: 0,

Y 2:: 0, x + y::s 10, 2x + y 2:: 10, Y 2:: 0, 2x + Y 2:: 4, x + y ::s 9

x

+

2y 2:: 10

x

920

CHAPTER U



'- 19.

Maximizing Profit A manufacturer of skis produces two types: downhill and cross-country. Use the following table to determine how many of each kind of ski should be produced to achieve a maximum profit. What is the maximum profit? What would the maximum profit be if the time available for manufacturing is increased to 48 hours? Cross country

Downhill

Time Available

Man ufacturing time per ski

2

hours

1

hour

40

Finishing time per ski

1

hour

1

hour

32 hours

Profit per ski

$70

hours

$50

20. Farm Management A farmer has 70 acres of land available for planting either soybeans or wheat. The cost of preparing the soil, the workdays required, and the expected profit per acre planted for each type of crop are given in the following table:

Preparation cost per acre Workdays req u i red per acre P rofit per acre

Soybeans

Wheat

$60

$30

3

4

$180

$100

(b) How much should the broker recommend that the client place in each investment to maximize income if the client insists that the amount invested in T-bills must not exceed the amount placed in j unk bonds?

24. Production Scheduling In a factory, machine 1 produces 8-inch pliers at the rate of 60 units per hour and 6-inch pliers at the rate of 70 units per hour. Machine 2 produces 8-inch pli ers at the rate of 40 units per hour and 6-inch pliers at the rate of 20 units per hour. It costs $50 per hour to operate ma chine 1 , and machine 2 costs $30 per hour to operate . The production schedule requires that at least 240 units of 8-inch pliers and at least 1 40 units of 6-inch pliers be produced dur ing each lO-hour day. Which combination of machines will cost the least money to operate? 25. Managing a Meat Market A meat market combines ground beef and ground pork in a single package for meat loaf. The ground beef is 7 5% lean (75% beef, 25% fat) and costs the market $0.75 per pound. The ground pork is 60% lean and costs the market $0.45 per pound. The meat loaf must be at least 70% lean. If the market wants to use at least 50 lb of its available pork, but no more than 200 lb of its available ground beef, how much ground beef should be mixed with ground pork so that the cost is minimized? 60 % lean ground pork

The farmer cannot spend more than $1800 in preparation costs nor use more than a total of 120 workdays. How many acres of each crop should �e planted to maximize the profit? What is the maximum profit? What is the maximum profit if the farmer is willing to spend no more than $2400 on preparation? 21. Banquet Seating A banquet hall offers two types of tables for rent: 6-person rectangular tables at a cost of $28 each and lO-person round tables at a cost of $52 each. Kathleen would like to rent the hall for a wedding banquet and needs tables for 250 people. The room can have a maximum of 35 tables and the hall only has 1 5 rectangular tables available. How many of each type of table should be rented to minimize cost and what is the minimum cost?

26. Ice Cream The Mom and Pop Ice Cream Company makes two kinds of chocolate ice cream: regular and premium. The properties of 1 gallon of each type are shown in the table: Regular

Flavoring

Source: facilities.princeton. edu

22. Spring Break The student activities department of a com munity college plans to rent buses and vans for a spring-break trip. Each bus has 40 regular seats and 1 handicapped seat; each van has 8 regular seats and 3 handicapped seats. The rental cost is $350 for each van and $975 for each bus. If 320 regular and 36 handicapped seats are required for the trip, how many vehicles of each type should be rented to minimize cost? Source:

www. busrates. com

23. Return on Investment An investment broker is instructed by her client to invest up to $20,000, some in a junk bond yielding 9% per annum and some in Treasury bills yielding 7% per annum. The client wants to invest at least $8000 in T-bills and no more than $12,000 in the j unk bond.

(a) How much should the broker recommend that the client place in each investment to maximize income if the client insists that the amount invested in T-bills must equal or exceed the amount placed in j unk bonds?

Milk-fat products Shipping weigh t Profit

24 0z

12 o z 5 1 bs

$0.75

Premium 20 0z

20 0z

6 1bs

$0.90

In addition, current commitments require the company to make at least 1 gallon of premium for every 4 gallons of regular. Each day, the company has available 725 pounds of flavoring and 425 pounds of milk-fat products. If the company can ship no more than 3000 pounds of product per day, how many gallons of each type should be produced daily to maximize profit? Source:

www.scitoys. com/ingredients/iceJream. html

27. Maximizing Profit on Ice Skates A factory manufactures two kinds of ice skates: racing skates and figure skates. The racing skates require 6 work-hours in the fabrication depart ment, whereas the figure skates require 4 work-hours there. The racing skates require 1 work-hour in the finishing de partment, whereas the figure skates require 2 work-hours there. The fabricating department has available at most 1 20 work-hours per day, and the finishing department has no more

Cha pter Review

than 40 work-hours per day available. If the profit on each racing skate is $ 1 0 and the profit on each figure skate is $12, how many of each should be manufactured each day to max imize profit? (Assume that all skates made are sold.) 28. Financial Planning A retired couple has up to $50,000 to place in fixed-income securities. Their financial adviser sug gests two securities to them: one is an AAA bond that yields 8 % per annum; the other is a certificate of deposit (CD) that yields 4 % . After careful consideration of the alternatives, the couple decides to place at most $20,000 in the AAA bond and at least $ 1 5,000 in the CD. They also instruct the finan cial adviser to place at least as much in the CD as in the AAA bond. How should the financial adviser proceed to maximize the return on their investment? 29. Product Design An entrepreneur is having a design group produce at least six samples of a new kind of fastener that he wants to market. It costs $9.00 to produce each metal fasten er and $4 . 00 to produce each plastic fastener. He wants to have at least two of each version of the fastener and needs to have all the samples 24 hours from now. It takes 4 hours to produce each metal sample and 2 hours to produce each plas tic sample. To minimize the cost of the samples, how many of each kind should the entrepreneur order? What will be the cost of the samples?

921

30. Animal Nutrition Kevin's dog Amadeus likes two kinds of canned dog food. Gourmet Dog costs 40 cents a can and has 20 units of a vitamin complex; the calorie content is 75 calo ries. Chow Hound costs 32 cents a can and has 35 units of vi tamins and 50 calories. Kevin likes Amadeus to have at least 1 1 75 units of vitamins a month and at least 2375 calories dur ing the same time period. Kevin has space to store only 60 cans of dog food at a time. How much of each kind of dog food should Kevin buy each month to mini mize his cost? 31. Airline Revenue An airline has two classes of service: first class and coach. Management's experience has been that each aircraft should have at least 8 but no more than 16 first-class seats and at least 80 but not more than 120 coach seats. (a) If management decides that the ratio of first class to coach seats should never exceed 1 : 1 2 , with how many of each type of seat should an aircraft be configured to maximize revenue?

(b) If management decides that the ratio of first class to coach seats should never exceed 1 :8, with how many of each type of seat should an aircraft be configured to maximize revenue? (c) If you were management, what would you do? [Hint: Assume that the airline charges $C for a coach seat and $F for a first-class seat; C > 0, F > C.]

Discussion and Writing 32. Explain in your own words what a linear programming problem is and how it can be solved.

C HA PTER R EVI EW Things to Know Systems of equations (pp. 836-846) Systems with no solutions are inconsistent.

Systems with a solution are consistent.

Consistent systems of l inear equations have either a unique solution or an infirute number of solutions. Matrix (p. 851)

Rectangular array of numbers, called entries

Augmented matrix (p. 851 ) Row operations (p. 853) Row echelon form (p. 854) Determinants and Cramer's Rule (pp. 866, 867, and 871) Matrix (p. 875) m

by n matrix (p. 876)

M atrix with In rows and n columns

Identity matrix I" (p. 883)

An n by n square matrix whose diagonal entries are l 's, while all other entries are O's

Inverse of a matrix (pp. 884-885)

A- I

Nonsingular matrix (p. 884)

A square matrix that has an inverse

is the inverse of A if AA - I

= A- I A =

I"

Linear programming problem (p. 916)

Maximize (or minimize) a l inear objective function, Z = A x + By, subject to certain conditions, or constraints, expressible as linear inequalities in x and y. A feasible point (x, y) is a point that satisfies the constraints of a linear programming problem. Location of solution (p. 917)

If a l inear programming problem has a solution, it is located at a corner point of the graph of the feasible points. If a linear program ming problem has multiple solutions, at least one of them is located at a corner point of the graph of the feasible points. In either case, the corresponding value of the objective function is unique.

922

CHAPTER 1 2

Objectives


--------,

Review Exercises

Section


12. 1

Solve systems of equations by substitution (p. Solve systems of equations by elimination (p. Identify inconsistent systems of equations containing two variables (p. Express the solution of a system of dependent equations containing two variables (p. Solve systems of three equations containing three variables (p. Identify inconsistent systems of equations containing three variables (p. Express the solution of a system of dependent equations containing three variables (p.

2 3

4 5

6

7

12.2

2

3

12.3

4 2 3

4

12.4

5 2 3

4

5

12.5

12.6 12.7 12.8

838) 840)

841)

842)

843)

844)

845)

851) 852)

Write the augmented matrix of a system of linear equations (p. Write the system of equations from the augmented matrix (p. Perform row operations on a matrix (p. Solve a system of linear equations using matrices (p.

853)

2 2 866) 3 3 870)

854)

866) 869) 872) 876) 878) 884) 879)

Evaluate by determinants (p. Use Cramer's Rule to solve a system of two equations containing two variables (p. Evaluate by determinants (p. Use Cramer's Rule to solve a system of three equations containing three variables (p. Know properties of determinants (p. Find the sum and difference of two matrices (p. Find scalar multiples of a matrix (p. Find the product of two matrices (p. Find the inverse of a matrix (p.

Solve a system of linear equations using an inverse matrix (p.

1

Decompose

2

Decompose

3

Decompose

4

Decompose

2

1-14,101,102,105-107 1-14,101,102,105-107 9,10,13,98 14,97 15-18,99,100,103 18 17 35-44 19,20 35-44 35-44 45,46 51-54 47-50 55,56 57,58 21,22 23,24 25-28 29-34 35-44 59,60 61,62 63,64,67,68 65,66 69-78 69-78 79-82 83-92,1 04 108,1 09 93-96,1 08,109

�, p

Q

where Q h as only nonrepeated linear factors (p.

. ' where Q has repeated lmear factors (p.

�, �

887) 892)

894) 896) 897)

where Q has a nonrepeated irreducible quadratic factor (p.

, where Q has a repeated irreducible quadratic factor (p.

Solve a system of nonlinear equations using substitution (p. Solve a system of nonlinear equations using elimination (p. Graph an inequality (p.

907)

2

Graph a system of inequalities (p.

2

Solve a linear programming problem (p.

898) 900)

909) 915) 916)

Set up a linear programming problem (p.

Review Exercises

In Problems 1-18,itsolve eachstent. system of equations using the method of substitution or the method of elimination. If the system has no soluti on, say that is inconsi { 2X + y = 0 { 3X - 4y = 4 {2X + 3y = 2 {2X - y = 5 5x + 2y = 8 7x - y = 3 5x - 4y = � x - 3y = -2 {X = 5y + 2 {yx == 2x3y -+ 54 { X - 2y - 4 = 0 { 2xX +- 3y3y +- 55 == 00 y = 5x + 2 3x + 2y - 4 = 0 1.

2.

5.

6.

3.

7.

1

4.

8.

_

Chapter Review

0

{ X - 3y + 4 { x + !y4 = 2 {4X5x ++ 6y5y = 4221 {2X + 3x3y -- 2y13 == !2 x �2 y + i3 = Y + 4x + 2 = {3X - 22Y = 8 { 2xX +- 2yy -+ 3zz == -13 {2X4x lOy5y = 20 x - "3Y = 12 3x - 2y + 3z = -16 { -3xX +- 4yy -+ 3z5z == -515 {2Xx -+ 4Y2y +- 4zz = -1527 { 2xX ++ 5yY -+ z = 27 x - Y + 2z = 5x - 6y - 2z -3 -7x - 5y - 9z = In Problems and write the system of equations corresponding to the given augmented matrix. [� ! I -:] w. [ � � -� -n In Problems use the following matrices to compute each expression. [ 3 -4 = [ � �], B = [� -� -�J C � � ] 2 -4B +C -C BC CB BA AB In Problems 29-34, find the inverse, if there is one, of each matrix. If there is not an inverse, say that the matrix is singular. [1 32 3 ] [ -3 [� �J 2 [ -3 2 J [4 �[ � -: ] In Problems solve each system of equations using matrices. If the system has no solution, say that it is inconsistent. 3X + 2y = 6 { 4x5x -- 7y6y -- 3z2z -3 { 3xOx -+ lOy2y == 5 l 3x + y - 7z = x - y = - "2 X + 2y - z = 2 { 2xX +- 2z = { 2X4x +- yy +- 3zz == 5 { 2x 2y + 3 y = -3 4x - 3y - 4z = 3 6x + 4y + 3z = 5 8x Y - = 5 x= { 4x2x -+ 3y4y -+ 3z5z == { x - xY -- 5zy -+ z = O 2x yY -- zz -+ 2tt 3 2x - 2y + 6x + 2y + = 3xx -- 42yy +- 2z - 3t5t == -3 X - 3y + 3z - t = 4 xx ++ 2y3z -+ 2tz == -3 x + Y + 5z 3 In Problems find the value of each determinant. 4 2 I -� � I I � �I 4 3 -2 2 -3 23 24 23 5 2 6 3 2 =

0

9.

_

10.

0

13.

= 10

+ +

14.

0

U.

0

11.

=

6

15.

=

Z

16.

17.

19

18.

=

=

11

10

20,

19.

_

21-28,

A

=

-1

21. A

22. A

23. 6A

24.

25.

26.

27.

28.

1

30 .

29.

3L

33.

35-44,

1

35.

36.

{

z

+

41.

44.

{

Z

6

=

0

1

=

0

=

=

1

40.

0

42.

Z

0

0

43.

48.

-1

1

10

5

{

Z =

1

49.

0

1

0

1

47.

-1

50.

1

-1

1

1

-1

=

0

1

6

1

+

6

46.

0

=

Z

45-50,

45.

-6

37.

39.

1

1

-1 1

34.

-1

1

38.

1 1

31.

6

0

+

0

923

924

Systems of Eq uations and Inequalities

CHAPTER 12

In Problems use Cramer's Rule, if applicable, to solve each system. {2X + 3x3y -- 2y13 == { 2xx +- 3y3y == -55 { 3xX -+ 2y2y == 44 { 2xX +- 2yy -+ 3zz == -136 { 2xX -+ 3Yy +- ZZ == -2 {3X5x +- 2y4y +- 126 = 3x - y - 9z = 9 3x - 2y + 3z = -16 In Problems and use properties of determinants find the value of each determinant if it is known that I � � I = 51-56,

51.

°

54.

=

57.

1 �� I

57

°

52.

53.

55.

56.

10

58,

58 ·

°

°

8

1 � �I

8

.

In Problems write the partial fractix on decomposition of each rational expression. x x-4 6 2x - 6 ( x + 2)( x - 3) (r + 9)(x + 1) x(x - 4) (x - 2)-(x - 1) r(x - 1) 2 3+1 2 4 3x 2 x x 2(X + 16) (x2 + 4)(x2 - 1) (x - 2)(x + 1) (x2 + l )(x2 In Problems solve each system of 2equations. { 7x2 -3X2y-2?--l5 == { 2xx -+ yl2 16 {2XY {2X +X2Y++ y-3? == 5 23y -+ xyl = 102 2 + 2xy - 2l = 6 {2Xx22 ++ Yl2 == 99 {3X22 ++ 4xy + 5l = { 3x { X2 + xl2 == 6y3y 2 xy - 2)'- + 4 = x 3xy + 2y = { X2 + X + l = 2Y -+ Y2 { x2x-- -3xx+ l + Y -2 -x + 1 = - y +y+1= x In Problems graph each inequality. 2x - 3y 2: 6 3x + 4y :S 12 x 2: l In Problems graph each system of inequalities. Tell whether the graph is bounded or unbounded, and label the corner points. X {-2Xx ++ yY 2::S 22 {X2x-+2yy 2::S 26 x2x ++ 3yYy :S:S 46 X 2: X 2:2: X 2:2: y 2: 0 Y Y 3x + y 2: 6 2xx ++ 2yy 2::S 2 3x + y :S 9 2x + Y 2: 2 2x + 3y 2: 6 In Problems graph each system of inequalities. 2 22 + l2 2: 1 {xx ++ ly :S2: 216 { xyy :S:Sx42 {l :S x - I {X x + y :S 4 x - y :S 3 In Problems solve each linear programming problem. x 2: 0, Y 2: 0,3x + 2y 2: 6, x + Y :S z = 3x + 4y z = 2x + 4y x 2: 0, Y 2: 0, X + Y :S 6, x 2: 2 z = 3x + 5y x 2: 0, Y 2: 0, X + Y 2: 1, 3x + 2y :S 12, x + 3y :S 12 z = 3x + y x 2: 0, Y 2: 0, X :S y :S 6,2x + Y 4 2 + bx + c y = ax (0, 1), (1,0), (-2, 1). {2X4x ++ 5yOy == 5 (0, 1), (1,0), (-2, 1). l 97 x2 + l + Dx + Ey + = 0.] 59-68,

60.

59.

64.

69-78,

69.

°

61. --=?--

62.

66.

67.

=

70.

=

-

71.

8

=

?

77.

? -

68.

1)

=

1

°

72.

8

°

75.

74.

73.

63.

?

76.

?

°

78.

°

79-82

80.

79.

82.

83-88,

83.

86.

{

84.

°

87.

85.

{

° °

88.

8

{ {

2: ° 2: °

°

°

89-92,

8�

9�

91.

92.

93-96,

93.

Maximize

subject to

94.

Maximize

subject to

8

95.

Minimize

96.

Minimize

subject to

97.

Find A so that the system of equations has infinitely many solutions.

8,

subject to

2:

99.

100.

A

98.

Find

A

so that the system in Problem

is inconsistent.

Curve Fitting Find the quadratic function that passes through the three points

and

Curve Fitting Find the general equation of the circle that passes through the three points and [Hint: The general equation of a circle is F

925

Chapter Test

101.

Blending Coffee A coffee distributor is blending a new cof fee that will cost $6.90 per pound. It will consist of a blend of $6.00 per pound coffee and $9 . 00 per pound coffee. What amounts of each type of coffee should be mixed to achieve the desired blend?

Flotel Orellana. As I watched the Amazon unfold, I wondered how fast the speedboat was going and how fast the current of the white-water Aguarico River was. I timed the trip downstream at 2.5 hours and the return trip at 3 hours. What were the two speeds?

[Hint: Assume that the weight of the blended coffee is 100 pounds.]

102.

Farming A 1000-acre farm in Illinois is used to grow corn and soybeans. The cost per acre for raising corn is $65, and the cost per acre for soybeans is $45. If $54,325 has been bud geted for costs and all the acreage is to be used, how many acres should be allocated for each crop?

103.

Cookie Orders A cookie company m akes three kinds of cookies, oatmeal raisin, chocolate chip, and shortbread, pack aged in small, medium, and large boxes. The small box con tains 1 dozen oatmeal raisin and 1 dozen chocolate chip; the medium box has 2 dozen oatmeal raisin , 1 dozen chocolate chip, and 1 dozen shortbread; the large box contains 2 dozen oatmeal raisin, 2 dozen chocolate chip, and 3 dozen short bread . If you require exactly 1 5 dozen oatmeal raisin, 10 dozen chocolate chip, and 1 1 dozen shortbread, how many of each size box should you buy?

104.

Mixed Nuts A store that specializes in selling nuts has avail able 72 pounds of cashews and 1 20 pounds of peanuts. These are to be mixed in 1 2-ounce packages as follows: a lower priced package containing 8 ounces of peanuts and 4 ounces of cashews and a quality package containing 6 ounces of peanuts and 6 ounces of cashews. (a) Use x to denote the number of lower-priced packages and use to denote the number of quality packages. Write a system of linear inequalities that describes the possible number of each kind of package. (b) Graph the system and label the corner points.

106.

Finding the Speed of the Jet Stream On a flight between M idway Airport in Chicago and Ft. Lauderdale, Florida, a Boeing 737 jet maintains an airspeed of 475 miles per hour. If the trip from Chicago to Ft. Lauderdale takes 2 hours, 30 minutes and the return flight takes 2 hours, 50 minutes, what is the speed of the jet stream? (Assume that the speed of the jet stream remains constant at the various altitudes of the plane and that the plane flies with the jet stream one way and against it the other way.)

107.

Constant Rate Jobs If Bruce and Bryce work together for 1 hour and 20 minutes, they will finish a certain job. If Bryce and Marty work together for 1 hour and 36 minutes, the same job can be finished. If M arty and Bruce work together, they can complete this job in 2 hours and 40 minutes. How long will it take each of them working alone to finish the job?

108.

Maximizing Profit on Figurines A factory manufactures two kinds of ceramic figurines: a dancing girl and a mermaid. Each requires three processes: molding, painting, and glazing. The daily labor available for molding is no more than 90 work-hours, labor available for painting does not exceed 1 20 work-hours, and labor available for glazing is no more than 60 work-hours. The dancing girl requires 3 work-hours for molding, 6 work hours for painting, and 2 work-hours for glazing. The mermaid requires 3 work-hours for molding, 4 work-hours for paint ing, and 3 work-hours for glazing. If the profit on each figurine is $25 for dancing girls and $30 for mermaids, how many of each should be produced each day to maximize profit? If man agement decides to produce the number of each figurine that maximizes profit, determine which of these processes has work-hours assigned to it that are not used.

109.

Minimizing Production Cost A factory produces gasoline engines and diesel engines. Each week the factory is obligated to deliver at least 20 gasoline engines and at least 15 diesel engines. Due to physical limitations, however, the factory can not m ake m ore than 60 gasoline engines nor more than 40 diesel engines in any given week. Finally, to prevent lay offs, a total of at least 50 engines must be produced. If gaso line engines cost $450 each to produce and diesel engines cost $550 each to produce, how many of each should be produced per week to minimize the cost? What is the excess capacity of the factory; that is, how many of each kind of engine is being produced in excess of the number that the factory is obligated to deliver?

110.

Describe four ways of solving a system of three linear equa tions containing three variables. Which method do you pre fer? Why?

y

105.

Determining the Speed of the Current of the Aguarico River On a recent trip to the Cuyabeno Wildlife Reserve in the Amazon region of Ecuador, I took a 1 00-kilometer trip by speedboat down the Aguarico River from Chiritza to the

C H A PTER TEST

{

InlutiProblems eachstent.system of equations using the method ofsubstitution or the method of elimination. If the system has no so on, say that1-4,it issolveinconsi {I { X yy { .?: 2y �y 1.

-2X + Y 4x + 3y

=

=

-7 9

2.

-

x

-

2y

3 5x - 30y

=

=

1

18

+ 2z 3x + 4 - z 5x + + 3z -

3.

=

=

=

5 -2 8

4.

3x + 2y - 8z x

-

6x

-

-

y +

=

Z =

+ 15z

=

-3 1

8

926 5.

{ -4xx -- Syy ++ 6Z == -19.0

Write the augmented matrix corresponding to the system of equations:

6.


CHAPTER 12

2

[ � �1 3:

Inexpression. Problems A

7.

"

-

A

�l

- 11

21

.

B "

21.

[: -23 S1 J C = [ -1; -�]8

9. AC

23.

=

2

3C -

25,

24.

25.

26.

Graph the system o f inequalities. Tell whether the graph is bounded or unbounded, and label all corner points.

{ Xy 2:2: 00

10. BA

12,

1 1. A

12. B =

2

13-16,

2 -2

13.

2xX +-2y3y 2:8 x 2: 0,2x + y 8, x - 3y z-3.= Sx + 8y 4 $90.00. $42.2S0, 1 3 $62.00. 2:

2

27.

Maximize

subject to

s

28.

14.

=

+

22.

Graph the system of inequalities: 24

In Problems and find the inverse of each nonsingular matrix. [ 1 -1S = [� !J 2 3 eachsaysysttehmatofit equations IfIn theProblems system has nosolve solution, is inconsisusitent.ng matrices. { X + 41 Y = 7 {6X2x -+ 3yy == 1 8x + 2y = S6 { 2xX + 2y7Y ++ lSz4z = -3-12 { 2Xx -+ 2Yy +- 2z3Z == S8 4x + 7y + 13z -10 3x + Sy - 8z = -2 In Problems and find the value of each determinant. 6 -4 1 42 -40 15.

?

22,

Inof each Problems and owrin. te the partial fraction decomposition rational expressi 4x2 - 3 3x + 7 (x + 3)2 x(x2 + 3)2

,

8. A

C

11

2

20.

IS

use the given matrices to compute each

7-1 0,

[� =n

2 +

20,

19.

Write the system of equations corresponding to the

-2

19

=

x + Sy - Sz = 10

augmented matrix:

In Problems each system. and use Cramer's Rule, if possible, to solve { -4xx -+ 3YY +- 2z3z = -IS {4X3x -+ Sy3y == 19-23 Sx - Sy + 2z = 18 In Problems and solve each system of equations. {3X2 + y-l = 9x1 {2yl -- 3x2x == S1

s

and

Megan went clothes shopping and bought 2 pairs of flare jeans, 2 camisoles, and T-shirts for At the same store, Paige bought one pair of flare jeans and 3 T-shirts for while Kara bought pair of flare jeans, camisoles, and T shirts for Determine the price of each clothing item.

16.

=

17

18,

2

18.

-1

C U M U LATIVE R EVIEW

In Problems solve each equation. 2x2 - X 0 = 1-6,

1. 4.

7.

Y

S.

Determine whether the function odd, or neither. Is the graph of the x-axis, y-axis, or origin?

8.

2.

= 9x + L

x2 + l - 2x + 4y - 11 =

g

�

log3 ( x

g(x) = x 2x3+ 1 4

-

-

=4 - 1) +

is even,

+ 1) = 9.

symmetric with respect to

Find the center and radius of the circle O. Graph the circle.

2x3 - 3x2 - 8x - 3 = 0 =e f(x) = 3x-2 + 1 f f(x) = x + 2 3.

log3 (2x

10.

2

6. Y

Graph using transformations. What is the domain, range, and horizontal asymptote of ? The function

S_

_

is one-to-one. Find r l . Find the

domain and the range of f and the domain and the range of r L .

Chapter Projects

11.

Graph each equation. (b) x2 + 1 = 4 1 (d) y = x (f) Y = (h) 2x2 + 51 = 1 (j) x2 - 2x - 4Y + 1 = 0

(a) y = 3x + 6

(C) y = x3

(e) y = Vx

Y

=

t..

eX

(g) y = In x (i) x2 - 31 = (k)

lli:l u.

sin x

1

(I)

Y

13.

927

f(x) = x3 - 3x + 5 (a) Using a graphing utility, graph f and approximate the zeroes) of f. (b) Using a graphing utility, approximate the local maxima and local minima. (c) Determine the intervals on which is increasing. 2 Solve: 2 sin x = 3 cos x

f

= 3 sin(2x)

C HA PTER P ROJ ECTS I.

Markov Ch ains A Markov chain (or process) is one in which future outcomes are determined by a current state. Fu ture outcomes are based on probabilities. The probability of moving to a certain state depends only on the state previ ously occupied and does not vary with time. An example of a Markov chain is the maxim u m education achieved by chil dren based on the highest education attained by their par ents, where the states are ( 1 ) earned college degree, (2) high school diploma only, (3) elementary school only. If Pi} is the probability of moving from state to state the transition matrix is the m X m matrix

p =

[

i

Pl l :

P l2

Pml

Pm2

j,

4.

The table represents the probabilities of the highest educa tiona I level of children based on the highest educational level of their parents. For example, the table shows that the prob ability P2 1 is 40% that parents with a high-school education (row 2) will have children with a college education (column 1).

i

5.

6.

Highest Educational Maximum Education That Children Achieve Level of College High School Elementary Parents College High school Elementary

80%

40% 20%

1 8%

50%

60%

1.

Convert the percentages to decimals.

2.

What is the transition matrix?

3.

2%

1 0%

20%

7.

Sum across the rows. What do you notice? Why do you think that you obtained this result?

If P is the transition matrix of a Markov chain, the ( i, j) lh entry of pn (nth power of P) gives the probability of passing from state to state j in stages. What is the probability that a grandchild of a college graduate is a college graduate?

n

What is the probability that the grandchild of a high school graduate finishes college?

The row vector v(O) = [0.277 0.575 0. 148J represents the proportion of the U.S. population 25 years or older that has college, high school, and elementary school, re spectively, as the highest educational level in 2004.';' In a Markov chain the probability distribution V( k ) after k stages is V( k ) = V(O)pk , where pk is the kth power of the transition matrix. What will be the distribution of highest educational attainment of the grandchildren of the cur rent population? Calculate p3, p4, p5, . . Continue until the matrix does not change. This is called the long-run distribution . What is the long-run distribution of highest educational attain ment of the population? .

*Source: Us.

.

Census Bureau.

The following projects are available at the Instructor's Resource Center (IRC): II. Project at Motorola: Error Control Coding

The high powered engineering needed to assure that wireless communications are transmitted correctly is analyzed using matrices to control coding errors.

Ill. Using Matrices to Find the Line of Best Fit

Have you wondered how our calculators get a line of best fit? See how to find the

line by solving a matrix equation.

IV. CBL Experiment

Simulate two people walking toward each other at a constant rate. Then solve the resulting system of equations to determine when and where they will meet.

Sequences; I nduction; the Binomial Theorem The Future of the World Population New York, 24 Febru a ry - World

population is expected to increase by billion over the next years, from billion today to billion in Almost all growth will take place in the less developed regions, where today's billion population is expected to swell to 7.8 billion in B y contrast, the population of the more developed regions wiLl remain mostly unchanged, at billion. Future population growth is highly dependent on the path that fu ture fertility takes. In the medium variant, fertility declines from children per woman today to slightly over children per woman in If fertility were to remain about half a child above the levels pro jected in the medium variant, world population would reach bil lion by A fertility path half a child below the medium would lead to a population of 7.7 billion by mid-century. That is, at the world level, continued population growth until is inevitable even if the de cline of fertility accelerates. Because of its low and declining rate of population growth, the pop ulation of developed countries as a whole is expected to remain vir tually unchanged between and at about billion. In con trast, the population of the least developed countries is projected to more than double, passing from 0.8 billion in to 1.7 billion in projected to be robust, though less rapid, with its population rising from

2.62050. 2050.

45

6.5

9. 1

5.3

1.2

2.6

2

2050.

10.6

2050.

2050

200550 2050,

Source: UN.

1.2 2005

Press Release POP1918, February 24, 2005

2050. 4.5

Growth in the rest of the developing world is also billion to billion between and

6. 1

2005 2050.

- See the Chapter Project-

A Look Back, A Look Ahead

Th is cha pter may be d ivided i nto th ree i nd ependent parts: Sections 1 3 . 1 - 1 3 .3, Section 1 3 .4, and Section 1 3.5. In Chapter 3, we d efi ned a fu nction and its domain, which was usually some set of real n u m bers. I n Sections 1 3 . 1 - 1 3 .3 , we d iscuss a sequence, which is a fu nction whose domain is the set of positive i ntegers. Throughout this text, where it seemed a ppropriate, we have g iven proofs of many of the results. In Section 1 3.4, a tec h n iq u e for proving theorems i nvolving natural n u m bers is discussed. I n Chapter R, Section R.4, there are form ulas for expanding (x + a)2 and (x + a)3. I n Section 1 3 .5, we d i scuss the B i n o m i a l Theorem, a formula for the expa nsion of (x + a)n, where n is any positive i nteger. The topics i ntrod uced in this cha pter a re covered in more detai l i n courses titled Discrete Mathematics. Appl ications of these topics can be fou n d in the fields of computer science, engi neering, b u s i n ess a n d economics, the social sciences, and the physical and biological sciences.

Outline 1 3. 1 Sequences

1 3.2 Arithmetic Sequences 1 3.3 Geometric Sequences; Geometric Series

1 3.4 Mathematical Ind uction 1 3.5 The Binomial Theorem Chapter Review Chapter Test Cumulative Review Chapter Projects

929

930 ,

CHAPTER 13

Seq uences; Induction; the Binomial Theorem

. .

1 3 . 1 Seq uences PREPARING FOR THIS SECTION •

Functions (Section

"NOW Work

3.1, pp.

Before getting started, review the following concept:

208-219)


OBJECTIVES

1 Write the Fi rst Several Terms 2 Write the Terms

of a Sequence (p. 930)

of a Seq uence Defined by a Recursive Form u la (p. 933)

3 Use S u m mation Notation (p. 934) 4 Find the Sum

of a Sequence (p. 935)

A sequence is a function whose domain is the set of positive integers.

DEFINITION

Because a sequence is a function, it will have a graph. In Figure l ( a ) , we show the graph of the function f (x)

=

�, x > O. If all the points on this graph were removed

(2,�), (3, �),

except those whose x-coordinates are positive integers, that is, if all points were

removed except ( 1 , 1 ) ,

the graph of the sequence f en)

=

and so on, the remaining points would be

1..n , as shown in Figure l (b ) . Notice that we use n

to represent the independent variable in a sequence. This serves to remind us that n is a positive integer. f(n)

Figure 1

3

2

2

( 1, 1)

·

(a) f(x) 1

2 =

3

-i 0

4

i2 , �) (3, D (4 , �) 2 3 4 n •

x (b) f(n)

=

-11, n a positive i nteger

Write the Fi rst Several Terms of a Sequ ence

A sequence is usually represented by listing its values in order. For example, the sequence whose graph is given in Figure l (b ) might be represented as

f ( l ) , f (2),f (3), f ( 4), . . .

or

1 1 1 1 ' 2' 3' 4' . . .

The list never ends, as the ellipsis indicates. The numbers in this ordered list are called the terms of the sequence . In dealing with sequences, we usually use subscripted letters, such as a l , to rep resent the first term, a2 for the second term, a3 for the third term, and so on. For the sequence f en)

al = f ( l ) =

1,

=

1

-

n

.

, we wnte

1 1 1 1 a2 = f (2) = 2 ' a3 = f (3) = 3 ' a4 = f (4) = 4 ' · · · ' an = f e n) = -;{ , . . .

SECTION 13.1

Sequences

931

In other words, we usually do not use the traditional function notation f(n) for sequences. For this particular sequence, we have a rule for the nth term, which is 1 . . a = , so It IS easy to fi nd any term of the sequence. n n When a formula for the nth term (sometimes called the general term) of a sequence is known, rather than write out the terms of the sequence, we usually rep resent the entire sequence by placing braces around the formula for the nth term.

-

For example, the sequence whose nth term is

{ bn } = or by

b1 = E XA M P L E 1

-1 2

'

()

1 n

bn = "2

{ (�r }

may be represented as

1 1 b-? = 4 ' b3 = 8" ' " -

Writing the F i rst Several Terms o f a Sequence

Write down the first six terms of the following sequence and graph it.

{a } n Figure 2

1 .0 0.8 0. 6 0.4 0.2

Solution

(5 ( 3, '3 ) , � ) (4 �) (6 , 6) (2 , �) (1 , 0 ) 2

•

•

•

•

' 4

•

1 2 3 4 5 6

{n - 1} -n-

The first six terms of the sequence are al

5

n

=

See Figure

= 0,

a?-

-

= 21 '

2 for the graph.

a,�

=

2

3'

-

a-

�

4

= 5' -

•

COMMENT

Graphing utilities can be used to write the terms of a sequence and graph them. Figure 3 shows the sequence given in Example 1 generated on a TI-84 Plus graphing calculator. We can see the first few terms of the sequence on the viewing window. You need to press the right arrow key to scroll right to see the remaining terms of the sequence. Figure 4 shows a graph of the sequence. Notice that the first term of the sequence is not visible since it lies on the x-axis. TRACEing the graph will allow you to see the terms of the sequence. The TABLE feature can also be used to generate the terms of the sequence. See Table 1 . Figure 4

Figure 3

se� « X- l ) /X , X , I . 6, 1) {O 5 66666666 . .. Ans � Frac {O 1 /2 2/3 3/4 ...

. .

Table 1

1

0 �============� 7 o

'.!' •

EXAM P L E 2

' R"" -


Writi ng the F i rst Several Terms of a Sequence

Write down the first six terms of the following sequence and graph it.

•

932

Sequences; Ind uction; the B i nomial Theorem

CHAPTER 13

Figure 5

Solution

bn

2

The first six terms of the sequence are

. ( 1 . 2)

See Figure 5 for the graph.

(3. n (5 2) '5 •

2

-1

Notice in the sequence { bn } in Example 2 that the signs of the terms alternate_ When this occurs, we use factors such as ( - 1 ) n + 1, which equals 1 if n is odd and -1 if n is even,o r ( - 1 ) /1, which equals -1 i f n is odd and 1 if n i s even.

•

34 5 6 · (4 . - �) (6 · -D •

•

n

· (2. -1)

Writing the F i rst Several Terms of a Sequence

EXAM P L E 3

{

}

Write down the first six terms of the following sequence and graph it. Figure 6 _

•

6

5 4 3

�

C

{ /1 } -

(4. 4) (6. 6) (1�1 ) ·

6

. 1'f n 1s odd

The first six terms of the sequence are

Solution

(2. 2)

2

if n is even

1 n

•

(3. �) ( . D 5 3 4 5

n

n

C 1 = 1, c2 = 2, See Figure 6 for the graph.

:=;;o.

= ..mr=

-

•


Sometimes a sequence is indicated by an observed pattern in the first few terms that makes it possible to infer the makeup of the nth term. In the example that fol lows,a sufficient number of terms of the sequence is given so that a natural choice for the nth term is suggested.

EXAM P L E 4

Determ i ni n g a Sequence from a Pattern

en a/1 = n

e2 e3 e4 (a) e' 2 ' 3 ' 4 " " 1 1 1 (b) 1' 3 9 " " 27 '

bll

'

1

(d)

1, 3, 5, 7 , . . . 1 4 9 16 25 . . .

= n-1 3 cn = 2n - 1 2 dll = n

(e)

1 1 1 1 1'- 2' 3' - 4' 5" "

en = ( -1 )" + 1

(c)

,

,

,

1oi!J11: = : =-

,

,

(�)

•


The Factorial Sym bol DEFINITION

If n

;:::: 0 is an integer, the factorial symbol n! is defined as follows: I! = 1 n ! = n(n - 1 ) · . . . · 3 - 2 - 1 O! =

�

1

if n ;:::: 2

I

�

-----I

SECTION 13.1

Table 2

For example, 2! = 2 · 1 = 2, 3 ! = 3 · 2 · 1 = 6, 4! = 4 · 3 · 2 · 1 Table 2 lists the values of n ! for 0 :::; n :::; 6. Because n! = n �n - 1 )(n - 2) . ·3·2·1

n!

1-------1 n

0

2

2

3

6

4

24

5

1 20

6

720

Sequences

=

933

24, and so on.

r----�

(n - l)!

we can use the formula

=

n!

III Exploration

n(n - 1 ) !

to find successive factorials. For example, because 6 !

You r calculator has a factoria l key. Use it

to see how fast factorials in crease i n value. Find t h e va lue o f 69!. What hap

7! and 8!

pens when you try to fi n d 70!? In fact, 70! is larger than 1 0 1 00 (a googol).

tl!l!!: : =--= =

=

=

7 · 6!

=

= 720, we have

7 ( 720) = 5040

=

8 · 7! = 8 (5040)


40,320

2 Write the Terms of a Sequence Defi ned by a Recu rsive Form ul a

A second way of defining a sequence is to assign a value to the first (or the first few) term(s) and specify the nth term by a formula or equation that involves one or more of the terms preceding it. Sequences defined this way are said to be defined recurs· ively, and the rule or formula is called a recursive formula. EXAM PLE 5

Writi ng the Terms of a Recursively Defi ned Seq uence

Write down the first five terms of the following recursively defined sequence. S1

Solution

=

1,

Sn

=

The first term is given as Sl = 1 . To get the second term, we use n = 2 in the formula Sn = nSn - l to get S2 = 2s1 = 2 · 1 = 2. To get the third term, we use n = 3 in the formula to get S3 = 3s2 = 3 . 2 = 6. To get a new term requires that we know the value of the preceding term. The first five terms are Sl

S2 S3

S4

Ss

=

1

=

3·2

= 2·1 =

2

= 4·6 =

24

=

=

5 . 24

6

=

Do you recognize this sequence? Sn = n !

E XA M P L E 6

nSn - l

120 •

Writing the Terms o f a Recursively Defi ned Sequ e n ce

Write down the first five terms of the following recursively defined sequence. Solution

1,

Lin + Lln + l We are given the first two terms. To get the third term requires that we know both of the previous two terms. That is, Ul = 1 Ul

=

Ll2

LiZ

U3

Ll4

Us

= =

= =

1

Li t

=

Lln + 2

1,

+ u2

Ll2 + Ll3

u3 + U4

= =

=

=

1 + 1 1 + 2 2 + 3

=

2

=

5

=

3 •

934

CHAPTER 13

Sequences; I n d uction; the Bi nomial Theorem

The sequence defined in Example 6 is called the Fibonacci sequence, and the terms of this sequence are called Fibonacci numbers. These numbers appear in a wide variety of applications (see Problems 85-88). Now Work PRO B L E M S 3 5 A N D 4 3

�

3

Use S u m mation Notation

It is often important to be able to find the sum of the first { an } , that is, al

+ a2 + a3 + . . . +

n terms of a sequence

an

Rather than write down all these terms, we introduce a more concise way to express the sum, called summation notation. Using summation notation, we write the sum as al

+ a2 + a3

11

+

. . + an = L ak .

k=l

The symbol 2: (the Greek letter sigma, which is an S in our alphabet) is simply an instruction to sum, or add up, the terms. The integer k is called the index of the sum; it tells you where to start the sum and where to end it. The expression 11

L ak

k=1

i s a n instruction t o add the terms ak o f the sequence { an } starting with k = 1 and ending with k = n . We read the expression as "the sum of ak from k = 1 to k = n." EXAM P L E 7

Expan d i n g Summation Notation

Write out each sum. (a) Solution

(a)

n

1

k�k

n

L k!

k=1 11

n

1 1 1 1 L -k = 1 + -2 + -3 + . . . + -n

(b)

k=l

1Oi!l!I:= = >-

E XA M P L E 8

(b)

L k! = I ! + 2! + . . . + n!

k=l

•

Now Work PRO B L E M S 1

Writi ng a S u m i n Summation N otatio n

Express each sum using summation notation.

1 + 2"1 + 41 + "81 + . . . + 2 n1- 1 The sum 1 2 + 22 + 3 2 + . . . + 92 has 9 terms, each of the form k2 , and starts at k = 1 and ends at k = 9: (b)

Solution

(a)

1 2 + 22 + 3 2 + . . . +

2 9

=

9

L k2

k=1

(b) The sum

1 1 + -21 + -41 + -81 + . . . + -2,, - 1 has

n terms, each of the form 2 k1- 1 ' and starts at k = 1 and ends at k

=

n:

11

1 = "' 1 1 + -21 + -1 + -1 + · · · + � 2k - 1 2n - 1 4

8

•

SECTION 13.1

The index of summation need not always begin at could have expressed the sum in Example 8(b) as

1 or end at

n;

Sequences

93 5

for example, we

11 - 1

1 1 1 1 "' - = 1 + - + - + · · · + 1 2 4 2" -

Letters other than

� 2k

k may be used as the index. For example, '" . , L,. J . 11

'" L,. l. ., i=l 11

and

j=l each represent the same sum as the one given in Example 7(b). =(.m;::=

4


F i n d the S u m of a Seq ue nce

Next we list some properties of sequences using summation notation. These prop erties are useful for adding the terms of a sequence. THEOREM

Properties of Sequences

If

{ an } and { bn } are two sequences and c is a real number, then: n n 2: ( cak ) = cal + ca2 + . . . + can = c (al + a2 + . . . + an ) = C 2: ak k=l k=l n n n 2: (ak + bk ) = 2: ak + 2: bk k=l k=l k=l n n n = a bk 2: 2: (ak bd 2: k k=l k=l k=l n 11 j 2: ak 2: ak - 2: ak , where 0 < j < =

k =j+ l

k=l

n

k=l

(1) (2) (3) ( 4)

.J

The proof of property (1) follows from the distributive property of real num bers. The proofs of properties 2 and 3 are based on the commutative and associative properties of real numbers. Property (4) states that the sum from j + 1 to n equals the sum from 1 to n minus the sum from 1 to j. It can be helpful to employ this prop erty when the index of summation begins at a number larger than l . Next we give some formulas for finding the sums o f certain sequences. THEOREM

Formulas for Sums of Sequences II

2: c = c + c + · · · + c

k=l

\

terms

)

� k2 L,.

k=l

= 12 +

?

2-

n 2: k3 = 13+2

k=l

c is a real number

(5)

2

(6 )

n(n + 1 ) n = --n2 n( n + 6(2n + 1 ) + 32 + [n(n + 1 ) ]2 3 + 33 + . . . + n3 = n

n 2: k = 1 + 2 + 3 + . . . +

k=l

cn

. . .+

1)

= ---''--'_ ":'" -

2

(7) (8)

.J

The proof of formula (5) follows from the definition of summation notation . You are asked to prove formula (6) in Problem 92. The proofs of formulas (7) and (8) require mathematical induction, which is discussed in Section 1 3 . 4 . Notice the difference between formulas (5) and (6) . In (5), the constant c is being summed from 1 to while in (6), the index of summation k is being summed from 1 to

n.

n,

936

C H A PTER 13

Seq uences; Ind uction; the Binomial Theorem

EXAM P L E 9

F i n d i n g the Sum of a Sequence

Find the sum of each sequence. 5

� (3k) ( c) � (k2 - 7 k + 2) (a)

� (k3 + 1) (d) �(4k2) 10

(b)

k=l

k=l

24

20

k=l

Solution

5

k=6

5

� (3k) 3 � k Property (1) 3 C(5 : 1 ) ) Formula (6) 3(15) 45 (b) � (k3 + 1) = � k3 + � 1 Property (2) (10(10 + 1 ) ) 2 + 1(10) Formulas (8) and (5) 2 3025 + 10 3035 24 ( c) � (k2 - 7 k + 2) = � k2 - � (7 k) + � 2 Properties (2) and = � k2 - 7 � k + � 2 Property (1) 24(24+1)(2·24 + 1) - 7 (24(24 + 1)) + 2(24) Formulas (7), (6), (5) 6 2 4900 - 2100 + 48 2848 (d) Notice that the index of summation starts at 6. We use property (4) as follows: � (4k2 ) 4 � k-? 4 [ � ? - � k 4 [ 20(21)(41) - 5(6)(11) J 6 r r Jr 6 (a)

=

k= l

k=1

=

=

=

10

k=l

10

10

k=1

k=1

=

=

=

24

24

24

k=l

k=1

k=1

24

24

24

k=1

k=l

k=l

(3)

k=l

=

=

=

L"

k=6

=

L"

k=6

=

Property (1) Property (4) =

� = �-

L" Ie

k=l

L"

k=l

2

=

--'---'--'-'---

Formula (7)

4[2870 - 55] 11,260 =

•


1 3 . 1 Assess Your U nderstanding 1.

For the function

(pp.208-219)

red.

Answersx -areI given at the end of these exercises. If you get a wrong answer, read the pages listed in f(x) x , f(2) y x (pp. 208-219)

'Are You Prepared?'

=

find

and f ( 3 ) .

A function is a relation between two sets D and R so that each element in the first set D is related to ex actly one element i n the second set R.

2 . True

01'

False

Sequences

SECTION 13.1

937


3. A(n) integers.

__

is a function whose domain is the set of positive

= 4. For the sequence - I}, the first term is and the fourth term is S4 =

{sn } {4n

__

__

4

.

Sl

=

Sequences are sometimes defined recUTsively.

6.

True or False

7.

True or False

8.

True or False

A sequence is a function.

2

Lk

k=1

= _.

5. L (2k )

k= l

=

3

Skill Building

In Problems

evaluate each factorial expression. 9! 12! 5! 8! 3!4!7! 9! In Problems write down the first five terms of each sequence. (snl { n } (sil {n2 + I } (bil { 2n2n I } ( al l { _ n +n 2 }} (dil l {(-1)"- 1 ( 2n n_ )} (sil { n 21 t( ill { (n + (-l )(nl ) n + 2) } (al l {-;;} (cl l {�I�} (bill {;�, } In Problems the given pattern continues. Write down the nth term of a sequence (anl suggested by the pattern. 24 8 1 12 4 2 ' 3 < 4' < 5" " .1 2 ' 2 . 3 ' 3 . 4 ' 4 . 5 " " 4 -8, 10, ... In Problems a sequence is defined recursively. Write down the first five terms. al al 4 - al - l al 2; an + an -I a] -2; an n + al -l al 5; al 2al - 1 al an n - an-I al 2; al -an- l al an ann-I al al -2; al n + 3al - l al al al -l + d a] al ran - I , r 0 al v2; al - )al2- 1 a1 \12 ; al V2 + an-l In Problems write out eachnsum. n n 1 k2 Lk L + 1) 2 L (2k + L + 2) Lk=l k=1 k= 1 2 n ( _ 1)k+12k nL-1 11-1 1L L L (2 + 1) L k+l C2 Y In Problems express each sum using summation notation. 1 + 23 + + 83 2 + . . 20 32 + 4 + .. . + + + 5 + 7 + . . . + [2(12) 2 1 2 4 28 - . .. + (- 1 ) 12(32 ) 11 1 - - + . . . 6( ) 3 '9 + 9 27 7 9-14,

9. 1O!

"

e ,

10.

1 1.

12.

6!

14. 3!

13. -

1O !

1 5-26,

=

15.

=

23.

16.

=

20.

=

24.

=

I

3

1l

2 7-34,

3

27.

35-48,

=

38.

= 1;

41.

=

54.

3;

1

1

36.

=

=

39.

=

= -

42.

=

45.

=

=

35.

49.

1

=

3

3;

=

21.

=

25.

=

3

+

33. 1,

-2, 3,

=

=

40.

=

. 43.

=

=

=

46.

48.

=

,

;:::.

50.

(k

55.

k=O

59. 1 +

63. 1

2,

-

, 6,

=

=

1;

= A;

'1=

=

49-58,

11

1

=

+ --

_

1)

k=o 3

1

52.

51.

+ 3

. +

3

+

1 - - +

3

-

k=O

57.

k

60. 3

13 3 + 1

1

+

(-1 )

k=1 11

56.

59-68,

6 1.

26.

34.

-4, 5, -6, . . .

37.

=

=

16 30. 3 ' 9 ' ' 27 81 " "

=

A;

18.

1

1 1 29. 1 ' 2 ' 4 ' 8 " "

1 1 1 1 32. 1 ' 2 , 3 ' 4 , 5 ' 6 , 7 ' 8 " "

31. 1, - 1 , 1 , - 1, 1, - 1 , . . .

47.

1

28.

1 7.

33

62. 1

1

36

-

64.

3

-

k=2 +

(k

53.

( - l ) k ln k

58.

...

-

1J

1

k=o 3 k= 3

938

CHAPTER 13

Sequences; Ind uction; the Binomial Theorem

65. 3 322 333 3nn 67. a (a d) (a 2d) In Problems find the sum 69. 2: 5 . 73. 2: (5k 3) 77. 2: (2k) +-+-

+ ... + -

+

+

40

+

+

+

69-80,

k=l 20

k=l

+

60

k = lO

66. -e1 e22 e33 68. a ar ar2

en" arn-J

+ - + - + ... + -

+

(a nd) each sequence. 70. 2: 8 74. 2:26 (3k - 7) 78. 2: (-3k)

... of

+

+

+

+ ... +

40

50

24

71. 2:k 75. 2: (k2 4) 79. 2: k3

72. 2: (-k) 76. 2: (k2 - 4) 80. 2:2 k3

k= l

k=l

16

k=1

14

+

k=1

40

k= l

k=O

20

4

k=5

k=8

k=4


81.

$3000

Credit Card Debt John has a balance of on his Dis cover card that charges interest per month on any un paid balance. John can afford to pay toward the balance each month. His balance each month after making a payment is given by the recursively defined sequence

1%

$100 $100 Bo $3000, Bn 1.0 1Bn- J - 100 Bj• 82. 200020 86. n 3% Po 2000, Pn l.03pn- l 20 P2 83. 0.5 % $18,500 $100$434.47 87. Bo $18,500, Bn 1. 005Bn_1 - 534.47 Bl 84. 250 10% =

=

Determine John ' s balance after making the first payment. That is, determine

Trout Population A pond currently has trout i n it. A fish hatchery decides to add an additional trout each month. In addition, it is known that the trout population is growing per month. The size of the population after months is given by the recursively defined sequence =

[Hint: A Fibonacci sequence models this colony. Do you see why?] 1 mature

pair 1 mature pair mature pairs 3 mature pairs

2

How many trout are in the pond after two months? That is, what is ?

Car Loans Phil bought a car by taking out a loan for at interest per month. Phil's normal monthly payment is per month, but he decides that he can afford to pay extra toward the balance each month. His balance each month is given by the recursively defined sequence =

Determine Phil 's balance after making the first payment. That is, determine .

n (1 vs)" (1 vs) Un 2n VS U j 1 U2 Un+2 Un+l u n

n Po 250, Pn O.9Pn-l 15 =

85.

+

=

2

Determine the amount of pollutant in the lake after years. That is, determine P2 ' Growth of a Rabbit Colony A colony o f rabbits begins with one pair of mature rabbits, which will produce a pair of offspring (one male, one female) each month. Assume that all rabbits mature in month and produce a pair of offspring (one male, one female) after months. If no rabbits

1

2

+

=

define the nth term of a sequence. (a) Show that = and = 1. + UI1 " = (b) Show that

(c) Draw the conclusion that { } is a Fibonacci sequence. Pascal's Triangle Divide the triangular array shown (called Pascal 's triangle) using diagonal lines as indica ted. Find the sum of the numbers in each diagonal row. Do you recognize this sequence?

/ �

Environmental Control The Environmental Protection Agency (EPA) determines that Maple Lake has tons of pollutant as a result of industrial waste and that of the pollutant present is neutralized by solar oxidation every year. The EPA imposes new pollution control laws that result in tons of new pollutant entering the lake each year. The amount of pollutant in the lake after years is given by the recursively defined sequence

15

Let

Fibonacci Sequence

+

=

=

ever die, how many pairs of mature rabbits are there after months?

7

ffi, 1

� �

1

5

6

88.

Fibonacci Sequence following problems:

(a) Write the first

11

10

15

15

5

6

86 Un+l Un

Use the result of Problem

to do the

terms of the Fibonacci sequence.

(b) Write down the first

n

10

20

10

terms of the ratio --.

(c) As gets large, what number does the ratio approach? This number is referred to as the golden ratio. Rectangles whose sides are in this ratio were considered pleasing to

S E CTION 13.2

the eye by the Greeks. For example, the facade of the Parthenon was constructed using the golden ratio. Ull

(d) Write down the first 10 terms of the ratio --.

(c)

UIl+ 1

(e) As n gets large, what number does the ratio approach? This number is also referred to as the conjugate golden ratio. This ratio is believed to have been used in the con struction of the Great Pyramid in Egypt. The ratio equals the sum of the areas of the four face triangles di vided by the total surface area of the Great Pyramid. (/. 89.

' e

In calculus, it can be shown that 00 xk [(x) = eX = �

Approximating I(x) =

k=o k !

We can approximate the value of [(x) = the following sum [(x) = eX

(/.

(d)

""

X e

for any x using

= 0.4,

(f)

:ck � ! k=O k Il

�

for some n. (a) Approximate [ ( 1 .3) with n = 4 (b) Approximate [ (1 .3) with n = 7. (c) Use a calculator to approximate [ (1 .3). ,� ' (d) Using trial and error along with a graphing utility 'S SEQuence mode, determine the value of n required to approximate [ (1 .3) correct to eight decimal places. Refer to Problem 89. 90. Approximating I(x) = e' (a) Approximate [( - 2.4) with n = 3. (b) Approximate [( - 2.4) with n = 6. (c) Use a calculator to approximate [ ( - 2.4) . ' - : (d) Using trial and error along with a graphing utility S SEQuence mode, determine the value of n required to approximate [( - 2.4) correct to eight decimal places. 91. Bode's Law In 1772, Johann Bode published the following formula for predicting the mean distances, in astronomical units (AU), of the planets from the sun: a1

(e)

{ a, J = { 0.4 + 0.3 ' 21l-2 } , n � 2

where n is the number of the planet from the sun. (a) Determine the first eight terms of this sequence. (b) At the time of Bode 's publication, the known planets were Mercury (0.39 AU), Venus (0.72 AU), Earth (1 AU),

Arithmetic Sequences

939

Mars (1 .52 AU), Jupiter (5.20 AU), and Saturn (9.54 AU). How do the actual distances compare to the terms of the sequence? The planet Uranus was discovered in 1781 and the aster oid Ceres was discovered in 1801 . The mean orbital dis tances from the sun to Uranus and Ceres';' are 19.2 AU and 2.77 AU, respectively. How well do these values fit within the sequence? Determine the ninth and tenth terms of Bode's sequence. The planets Neptune and Pluto* were discovered in 1 846 and 1930, respectively. Their mean orbital distances from the sun are 30.07 AU and 39.44 AU, respectively. How do these actual distances compare to the terms of the sequence? On July 29, 2005, NASA announced the discovery of a tenth planet* (n = 1 1 ) , which has temporarily been named 2003 UB313* until a permanent name is decided on. Use Bode 's Law to predict the mean orbital distance of 2003 UB313 from the sun. Its actual mean distance is not yet known, but 2003 UB313 is currently about 97 astronomical units from the sun.

Sources: NASA. 92.

Show that

n(n + 1 ) 1 + 2 + . . . + ( n - 1 ) + n = ---2

[Hint: Let

s S

( n - 1) + n = n + ( n - 1 ) + (n - 2 ) + . . . + 1 = 1 + 2 + ... +

Add these equations. Then 25

=

[1 + nJ + [2 n

+

(n

terms

in

1)] + . . . + [n + 1] brackets

Now complete the derivation.] "Ceres, Pluto, and 2003 UB3 1 3 are now referred to as dwarf planets.


Investigate various applications that lead to a Fibonacci sequence, such as art, architecture, or financial markets. Write an essay on these applications.


[(2)

=

1 2 '2 ; [(3) = :3

2.

True

OBJECTIVES

1 Determine If a Seq uence I s Arithmetic (p. 940) 2 Find a Formula for an Arithmetic Sequence (p. 941) 3 Find the Sum of a n Arith metic Sequence (p. 942)

940

CHAPTER 13

Sequences; I n d u ction; the Bi nomial Theorem

1

Dete r mi n e If a Seq uence I s Arithmetic

When the difference between successive terms of a sequence is always the same number, the sequence is called arithmetic. DEFINITION

An arithmetic sequence* may be defined recursively as a1 or as

an

=

=

a, an - an- l

=

d,

(1)

an- l + d

where a1 = a and d are real numbers. The number a is the first term, and the number d is called the common difference.

The terms of an arithmetic sequence with first term a l and common difference d follow the pattern

E XA M P L E 1

Determ i n i n g If a Sequence Is Arithmetic

The sequence

4, 6, 8,

10, . . .

is arithmetic since the difference of successive terms is and the common difference is d = 2. EXAM P L E 2

2. The first term is al

=

4, •

Determ i n i ng If a Sequence Is Arith metic

Show that the following sequence is arithmetic. Find the first term and the common difference.

{Sn} Solution

= 3·1

The first term is SI sequence { sn } are

Sn

=

3n

+

+ 5 5

=

and

=

{3n + 5 }

8. The nth term and the ( n - 1 )st term of the

sn- 1

=

3(n

-

1) + 5

=

3n + 2

Their difference d is

d

=

Sn - Sn- 1

=

(3n + 5 ) - (3n +

2)

=

5

-

2

=

3

Since the difference of any two successive terms is the constant 3, the sequence is arithmetic and the common difference is 3.

•

EXAM PLE 3

Determ i n i ng If a Sequence Is Arith metic

Show that the sequence {tn} common difference. Solution

The first term is t1 tIl

= =

=

{4 - n } is arithmetic. Find the first term and the

4 - 1 3. The nth term and the (n - l )st term are 4 - n and t,, - l = 4 - (n - 1 ) = 5 - n =

Their difference d is

d

=

tn - t,, - l

= (4

-

':' Sometimes called an arithmetic progression.

n) - (5 - n )

=

4

-

5

=

-1

SECTION 13.2

Arithmetic Seq uences

941

Since the difference of any two successive terms is the constant - 1 ; {tn} is an arith metic sequence whose common difference is - 1 . t;:!J!l1I -= = o-

2

•

Now Work PRO B L E M S

F i n d a Form ul a for a n Arithmetic Sequence

Suppose that a is the first term of an arithmetic sequence whose common difference is d. We seek a formula for the nth term, all ' To see the pattern, we write down the first few terms.

= = =

al

+

a2

+ d

= a3

as

=

a4

all

=

al l - l

at a2 a3

a4

a

d

=

+ 1 .

al

=

(al

+ d

=

( al

+ d

=

+

( al

+

d

+

d =

2 · d)

d

= al

d

= al

d)

+

+

3 · d) +

d = [ al

+

+

+

at

(n - 2)d]

+

2·d + +

d

3'd 4·d = al

+

( n - l )d

We are led to the following result:

THEOREM

nth Term of an Arithmetic Sequence

For an arithmetic sequence {an } whose first term is a l and whose common difference is d, the nth term is determined by the formula an = a l

+

(2)

( n - l )d

I

�----------------------------------�.� E XA M P L E 4

F i n d i ng a Parti cular Term of an Arith metic Sequ en ce

Find the forty-first term of the arithmetic sequence: S o l ution

The first term of this arithmetic sequence is d = 4. By formula (2), the nth term is an

= 2

+

(n - 1 )4

an

=

a,

al

=

2, 6, 10, 14, 18, . . .

2, and the common difference is

+ (n - l)d; a, =

2, d = 4

The forty-first term is a4l

E XA M P L E 5

= 2

+

40 · 4

=

162

F i n d ing a Recu rsive Form u l a for an Arithm etic Seq uence

The eighth term of an arithmetic sequence is 75, and the twentieth term is 39. (a) Find the first term and the common difference. (b) Give a recursive formula for the sequence. (c) What is the nth term of the sequence? Solution

(a) By formula (2), we know that

an

as

=

{

a20

= al

= al

al

+

(n - l )d. As a result,

+ 7d = 75 + 19d = 39

•

942

CHAPTER 13

Sequences; I nduction; the Binomial Theorem

TI1is is a system of two linear equations containing two variables, al and d, which we can solve by elimination. Subtracting the second equation from the fIrst, we get

� Exploration

1M Graph

-12d d recu rsive

formula

=

With d = -3, we use al + 7d = 75 and flnd that al = 75 - 7d = 75 The fIrst term is al = 96, and the common difference is d = -3. (b) Using formula (1), a recursive formula for this sequence is

Exam ple 5 , 0 1 = 96, on = °0- 1 - 3, using a graphing util ity. Conclude that the

36 -3

=

from

7(-3) = 96.

the g ra p h of the rec u rs ive fo r m u l a behaves l i ke the g ra p h of a l i near func

(c) Using formula

tion How i s d, the common d iffe rence, . related to m, the slope of a line?

an 'J"

3

;:; ;...'>-

=

al

(2), a formula for the nth term of the sequence { a,J is + (n - l )d = 96 + ( n - 1 ) ( -3) = 99 - 3n

•

Now Work PRO B L E M S 2 1 A N D 2 7

Find t h e Sum of a n Arith m etic Sequence

The next result gives two formulas for fInding the sum of the first n terms of an arith metic sequence. THEOREM

Sum of the Fi rst n Terms of an Arith metic Sequence

Let {an} be an arithmetic sequence with fIrst term al and common difference d. The sum Sn of the fIrst n terms of {an } may be found in two ways: Sn = al+a2+a3 + . . . + an n n = L [ al + (k - l )d] = - [2al

2

k=l

Sn = al n =

(3)

+ ( n - l )d ]

+ a2 + a3 + . . . + an

� [al + (k - l )d]

=

n '2 (al

+ an)

(4)

�------�

�

Proof

Sn

=

= =

+ a2 + a3 + .. . + an al + (al + d) + (al + 2d) + . . . + [al + (n - l )d] (al + al + ... + al) + [d + 2d + ... + (n - l )d] al

\

IWI+

=

Form u la (2) Rearrange terms

I

n terms

d[l + 2 + . . . + (n - 1)] n l ) n] = nal+d [ ( � 6, = nal + � (n - l)d

=

S u m of first n terms

� [2a] + ( - l )d] n

Formula

Scetion 1 3.1

Factor out

�;

this is Form u la (3).

Use Formula (2) ; this is Formula (4). •

There are two ways to find the sum of the first n terms of an arithmetic sequence. Notice that formula ( 3) involves the first term and common difference, whereas formula (4) involves the first term and the nth term. Use whichever form is easier.

SECTION 13.2

E XA M P L E 6

943

Arithmetic Sequences

F i n d i n g the S u m of an Arith metic Seq uence

Find the sum 5n of the first n terms of the sequence { a,. } = { 3n + 5 } ; that is, find n 8 + 1 1 + 14+. . . + ( 3n + 5 ) = � (3k + 5) k= l

Solution

The sequence {an } = {3n + 5 } is an arithmetic sequence with first term and the nth term an = 3n + 5 . To find the sum 5n , we use formula (4) . n n n 5n = � (3k + 5) = - [ 8 + (3n+5 ) ] = - ( 3n+ 13) 2 2 k=] i n Sn = 2 (a, + a n) 1J!l!: """ = =

EXAM P L E 7


=

8

•

F i nd i n g the S u m of an Arith metic Seq uence

Find the sum: Solution

al

60 + 64 + 68 + 72 + . . . + 120

This is the sum 5n of an arithmetic sequence {an I whose first term is a] 60 and = 120. We use formula (2) whose common difference is d = 4. The nth term is an to find n. an = at + (n - l)d Formula (2) 120 = 60 + (n - 1) ' 4 an = 120, a, = 60, d = 4 =

60 = 15 = n = Now we use formula

4(n - 1) Simplify n - 1 Simplify Solve for n 16 (4) to find the sum 51 6 16 6 0 + 6 4 + 6 8 + . . . + 120 = 5 1 6 = - (60 + 120) 2 i n Sn = 2 (a, + an)

Q>1!: ==- -

E XA M P L E 8


=

1440 •

C reating a F l oo r Design

A ceramic tile floor is designed in the shape of a trapezoid 20 feet wide at the base and 10 feet wide at the top. See Figure 7. The tiles, 12 inches by 12 inches, are to be placed so that each successive row contains one less tile than the preceding row. How many tiles will be required? Figure 7

Solution

The bottom row requires 20 tiles and the top row, 10 tiles. Since each successive row requires one less tile, the total number of tiles required is 5

=

20 + 19 + 18 + . . . + 1 1 + 10

944

CHAPTER 13

Sequences; Induction; the Binomial Theorem

This is the sum of an arithmetic sequence; the common difference is - 1 . The num ber of terms to be added is n = 11 , with the first term a1 = 20 and the last term a l l = 10. The sum S is n 11 S = "2 (a1 + a l l ) = 2 (20+10) = 165 In all, 165 tiles will be required.

•


True or False In an arithmetic sequence the sum of the first and last terms equals twice the sum of all the terms.

2.

I n a(n) sequence, the difference between successive terms is a constant. __


show that each sequence is arithmetic. Find the common difference and write out the first four terms.

3-12,

3. (sill

=

{n + 4}

4. ( sill

8 . ( all)

=

{ 4 - 2n }

9 ( t 11 I •

=

{n - 5}

{2

!

=

-

! n 3

5. ( aliI

}

10. ( till

=

=

6. {bill

{ 2n - 5 }

{ 32 4" } +

n

11. ( sill

=

=

7. ( clll

{ 3n + I } 11 { In 3 }

=

=

12. ( sill

{ 6 - 2n } {e

1nll }

In Problems 13-20, find the nth term of the arithmetic sequence ( alII whose initial term a and common difference d are given. What is the fifty-first term ?

13. al 17. al

=

=

2;

d

3

14. al

O',

d = -

18. al

In Problems

=

1

2

21-26,

= =

d = 4

-2; 1;

d

=

15. a 1

-3 1

19. a 1

= =

5;

d = -3

v2 ;

d

=

v2

16. a 1

=

6;

d = -2

20. aJ

=

0;

d = 7T

find the indicated term in each arithmetic sequence.

2 1 . 100th term of 2, 4, 6, . . .

22. 80th term of - 1 , 1 , 3, . . .

24. 80th term of 5 , 0, -5, . . .

7 5 25. 80th term of 2 , 3 , 2' 2' . . .

23. 90th term of 1 , -2, - 5 , . . . 26. 70th term of 2 Vs , 4 Vs , 6 Vs , . . .

In Problems 2 7-34, find the first term and the common difference of the arithmetic sequence described. Give a recursive formula for the sequence. Find a formula for the nth term.

27. 8th term is 8;

20th term is 44

30. 8th term is 4;

1 8th term is - 96 33. 1 4th term is - 1 ; 1 8th term is - 9

In Problems 35-52, find each sum. 35. 1 + 3 + 5 + . . . + ( 2n - 1 )

44. 7 + 1 - 5 47.

80

-

+

2: (2 n - 5)

11 = 1

11

- · · · -

34. 1 2th term is 4;

299

20th term is 35 40th term is - 50

29. 9th term is -5;

32. 5th term is -2;

1 8th term is 28

+

37. 7 + 12

45. 4 + 4.5+5+5.5 + . . . + 100

46. 8

90

48. 2: (3 - 2 n ) lI = i

49.

1 5th term is 31

13th term is 30

+

17 + . . . + (2 + 5 n ) 40. 1 + 3 + 5 + . . . + 59 43. 73 + 78 + 83+88 + . . . + 558

4 + 6 + . . . + 2n 39. 2 + 4 + 6 + . . . + 70 42. 2 + 5 + 8 + . . . + 41

36. 2

. . . + ( 4n - 5 ) 41. 5 + 9 + 13 + . . . + 49 38. - 1 + 3 + 7

28. 4th term is 3;

31. 1 5th term is 0;

JOO

(

2: 6

11 = 1

1 2

- -n

)

+

1

1

3

8 - + 8- + 8- + 9 + . . . + 50 4 2 4

(

80 1 50. 2: - n 11 = 1

3

1 2

+ -

)

51. The sum of the first 1 20 terms of the sequence 1 4, 16, 18, 20, . . . .

52. The sum of the first 46 terms of the sequence 2, - 1 , -4, -7, . . . . Applications and Extensions

53. Find x so that x + 3, 2x + 1 , and 5x + 2 are consecutive terms of an arithmetic sequence.

54. Find x so that 2x, 3x + 2, and 5x + 3 are consecutive terms of an arithmetic sequence.

55. Drury Lane Theater The Drury Lane Theater h as 25 seats in the first row and 30 rows i n all. Each successive row contains one additional seat. How many seats are in the thea ter?

SECTION 13.3

56.

57.

Football Stadium The corner section of a football stadium has 15 seats in the first row and 40 rows in all. Each succes sive row contains two additional seats. How many seats are in this section?

20 '

...

'

T 'II'

... '

.

945

58.

Constructing a Brick Staircase A brick staircase has a total of 30 steps. The bottom step requires 1 00 bricks. Each suc cessive step requires two less bricks than the prior step. (a) How many bricks are required for the top step? (b) How many bricks are required to build the staircase?

59.

Cooling Air As a parcel of air rises (for example, as it is pushed over a mountain), it cools at the dry adiabatic lapse rate of 5.5° F per 1000 feet until it reaches its dew point. If the ground temperature is 67° F, write a formula for the sequence of temperatures, {Til), of a parcel of air that has risen n. thou sand feet. What is the temperature of a parcel of air if it has risen 5000 feet? Source: National Aeronautics and Space Administration

60.

Citrus Ladders Ladders used by fruit pickers are typically tapered with a wide bottom for stability and a narrow top for ease of picking. Suppose the bottom rung of such a ladder is 49 inches wide and the top rung is 24 inches wide. How many rungs does the ladder have if each rung is 2.5 inches shorter than the one below it? How much material would be needed to make the rungs for the ladder described? Source: www.slokesladclers. com

61.

Seats in an Aml)hitheater An outdoor amphitheater has 35 seats in the first row, 37 in the second row, 39 in the third row, and so on. There are 27 rows altogether. How many can the amphitheater seat?

62.

Stadium Construction How many rows are in the corner section of a stadium containing 2040 seats if the first row has 10 seats and each successive row has 4 additional seats?

63.

Salary Suppose that you j ust received a job offer with a starting salary of $35,000 per year and a guaranteed raise of $ 1 400 per year. How many years will it take before your aggregate salary is $280,000?

Creating a Mosaic A mosaic is designed in the shape of an equilateral triangle, 20 feet on each side. Each tile in the mo saic is in the shape of an equilateral triangle, 12 inches to a side. The tiles are to alternate in color as shown in the illus tration. How many tiles of each color will be required?

/

Geometric Sequences; Geometric Series

, 20 '

[ Hint: Your aggregate salary $35,000 + ( $35,000 + $ 1 400) .]

'T T T ...T ·...T ...T TTTTTT T TT TT TTTT 20 '

after

2

years

is


Make up an arithmetic sequence. Give it to a friend and ask for its twentieth term.

65.

Describe the similarities and differences between arithmetic sequences and linear functions.

1 3 .3 Geometric Seq uences; Geometric Series PREPARING FOR THIS SECTION •


Compound Interest (Section 6.7, pp. 465-472) Now Work the 'Are You Prepared?' problems on page 954. OBJECTIVES

1 Determ i n e If a Sequence I s Geometric (p. 946) 2 Find a Form ula for a Geometric Sequence (p. 947) 3 Find the S u m of a Geometric Sequence (p. 948) 4 Determine whether a Geometric Series Converges or Diverges (p. 949) 5 Solve A n n u ity Problems (p. 952)

946

CHAPTER 13


1

Dete rmine If a Sequence I s G eometric

When the ratio of successive terms of a sequence is always the same nonzero num ber, the sequence is called geometric. DEFINITION

A geometric sequence* may be defined recursively as al an. =

=

an a, -- = r ' or as al1-1 (1)

r al1 - 1

where al = a and r t:- 0 are real numbers. The number al is the first term, and the nonzero number r is called the common ratio.

..J

al

The terms of a geometric sequence with first term low the pattern

E XA M P L E 1

and common ratio r fol

Determ i n i n g If a Sequence Is G eometric

The sequence 2, 6, 18,

54,

1 62, . . .

=

. . . . . . IS geometnc smce t h e ratIO af succeSSIve terms IS 3 ; first term is ai 2, and the common ratio is 3. E XA M P L E 2

(

6

- =

2

"6 18

=

54 '8 1

= =) . . .

3 . Th e •

Determ i n i ng If a Sequence Is Geometric

=

Show that the following sequence is geometric. { SI1 }

T"

Find the first term and the common ratio. Solution

The fir�t term is s l = { SI1 } are Their ratio is

T

1

=

1

2' The nth term and the ( n

�= sn - i

--- = 2

2-

-( 11 - 1 )

2

-11 + ( 11 - 1 )

=

�.

-

2 1

1 )st

=

1

-

2

Because the ratio of successive terms is the nonzero constant is geometric with common ratio

E XA M P L E 3

Determi n ing I f a Sequence I s Geometric

Show that the following sequence is geometric.

Find the first term and the common ratio. *

Sometimes called a geometric progression.

term of the sequence

�,

the sequence {sn } •

SECTION 13.3

Solution

The first term is t1

=

41


947

The nth term and the ( n - l ) st term are n tl = 4 and t,, - l = 411- 1 1

4

=

.

Their ratio is t' 11_ _ t l1 - 1

=

411

__ = 411 - 1

411 - ( 11 - 1 )

=

4

The sequence, { tn } , is a geometric sequence with common ratio = =.,.. �

4

.

•

Now Work PRO B l E M 1 1

2 F i n d a For m u l a for a Geometric Sequence

Suppose that a 1 is the first term of a geometric sequence with common ratio r #- O. We seek a formula for the nth term al1 . To see the pattern, we write down the first few terms: al = al 1 = alro •

a

2

a3 a4 as a"

al r

l

=

ra l

=

ra = r ( a]r ) = a1r2 2 ra3 r ( alr2 ) = alr3 ra4 r ( alr3 ) = a 1 r4

=

=

=

=

=

=

ra n - I .

=

r e a'1 rl1-2 )

=

a 1 r" - \

We are led to the following result:

THEOREM

nth Term of a Geometric Sequence

For a geometric sequence { al1 } whose first term is al and whose common ratio is r, the nth term is determined by the formula

r #- 0 (2) �----------------------------------�.�

I

E XA M P L E 4

F i n d i n g a Particular Term of a Geometric Sequence

( a ) Find the nth term of the geometric sequence: 10, 9, (b ) Find the ninth term of this sequence . ( c) Find a recursive formula for this sequence. Solution

( a ) The first term of this geometric sequence is a1 81 9 9 10 IS ' (Use or 9 . 10 10

=

nth term is

an

=

10

=

81 729 10 ' 100 . . . .

10 and the common ratio

9 or any two succeSSIve . terms. ) Then by fOlmula (2), the 10

( ),, - 1 9 10

an

=

a1

r n-1•' a1

=

10 , r

9 = -

10

948

CHAPTER 13

Sequences; I nduction; the Binomial Theorem

i� Exploration

(b) The ninth term is a9 = l O

Use a g raphing utility to fi n d the ninth term of the seq uence given i n Example 4. Use it to fi n d the twentieth and fiftieth terms. Now use a g ra p h i n g util ity to

(: Y O

-l

= lO

(: Y O

�

4.3046721

(c) The first term in the sequence is 10 and the common ratio is r

g raph the recursive fo rm u la fou n d i n Example 4(c). Conclude that t h e graph of

formula ( 1 ), the recursive formula is a j

the recursive fo rmula behaves like the

=

10, an

=

graph of a n expon e ntial function. How i s r, t h e co mmon ratio, related t o a , t h e base

of the exponential function y

=

aX?

"'''I!::==� -

3

Now Work PRO B L E M S 3 3

I

41

I

:0 an - I '

:0 ' Using •

AND 49

Find the S u m of a G eom etric Sequence

The next result gives us a formula for finding the sum of the first geometric sequence.

THEOREM

=

n

terms of a

S u m of the First n Terms of a Geometric Sequence

Let { a,,} be a geometric sequence with first term a l and common ratio r, where r *" 0, r *" 1 . The sum S" of the first n terms of {an } is (3 )

r *" 0, 1

�

�------�

n The sum Sn of the first n terms of { an } = { a l r - 1 } is n S" = al + a I r + . . . + a j r - l

Proof

(4)

MUltiply each side by r to obtain

(5)

Now, subtract (5) from (4). The result is n Sn - rSn = a l - alr ( 1 - r ) Sn = al ( l - r") Since

r

*" 1 , we can solve for Sw SII = a l "

E XA M P L E 5

-

•

F i n d i n g the Sum of the F i rst n Terms of a Geometric Sequence

Find the sum Sn of the first

Sol uti o n

n r --1 - r 1

The sequence

{ ( l)n } 2

n

{ (�)" } ; ) L - (-

terms of the sequence

n l

that is, find

l k-1

k=1 2 2 is a geometric sequence with al

2 and r = 2 ' So we use 1

=

1

formula

(3 ) 5

SECTION 13.3

to get 11

=

� 2:1 ( 2: l

11

)

k-l

=

2: 1

"41

+

+

"81


+

.. () +

.

2:1

11

Formula (3); a,

....

= = I;!!!r; ;:

II

=

1 1 2' r = "2

•


U s i ng a G raph ing Util ity to F i n d the S u m

E XA M P L E 6

o f a Geometric Sequence

Use � graphing utility to find the sum of the first 1 5 terms of the sequence that IS, find

1:. 1:. 3 +

Figure 8

949

S o l ution

9

+

� 27

(1:.) 15 � 1:. (1:.) k-l 3 k =1 3 3 { (�)" }

+ . . . +

{ (�)" } ;

=

Figure 8 shows the result obtained using a TI-84 Plus graphing calculator. The sum of the first 15 terms of the sequence ",

4

is 0.4999999652.

•

- Now Work PRO B L E M 6 1

Determine whether a G eometric Series Co nverges or Dive rges

DEFINITION

An infinite sum of the form

with first term a1 and common ratio r, is called an infinite geometric series and is denoted by

Based on formula 511

(3), =

the sum

al

11

of the first

-- = -1 - rl1

'

5

l - r

al

l - r

-

n

terms of a geometric series is

a I r" l - r

-

(6)

950

CHAPTER 13


If this finite sum S/1 approaches a number L as n �

NOTE In calculus, we use limit notation and write L =

n --+ 00

lim

Sn =

n

00

lim 2: a,rk-' n--+ 00 = 1 k

=

2: a,rk-' k= 1

•

series

CXl

2: a l rk - 1

k=l we write

00 ,

we say the infinite geometric

converges. We call L the sum of the infinite geometric series. and CXl

L =

2: a l rk - 1

k=l

If a series does not converge, i t i s called a divergent series.

THEOREM

Convergence of an I nfi n ite Geometric Series

If

Irl

< 1 , the infinite geometric series

CXl

2: a l rk - 1 converges. Its sum is

k=l

I

a = 1 k=l �----------------------------------�� CXl

Intu itive Proof

(7)

1_

2: al rk - 1

_ _ -

r

Since I r I < 1 , it follows that I r " l approaches a as n � 00 . Then, ��

--

based on formula (6), the term -- approaches 0, so the sum S" approaches 1 - r 1 - r as n � 00 . • E XA M P L E 7

�

Determ i n i n g whether a Geometric Series C o nverges o r Diverges

() 3

Determine if the geometric series CXl

2 k-l

2: 2 -

k=l

() 3

=

2 +

3

4

-

8

9

+

-

+

.

.

·

converges or diverges. If it converges, find its sum. Sol uti o n

CXl

Comparing

2: 2

k=l

common ratio is r find its sum:

2 k- l to =

3

CXl

k=l

E XA M P L E 8

Solution

.-

k=l

2 -. Since

2: 2

t;. 'I!llI :sr&I ____

CXl

2: a l rk - \ we

() 3

2 k- l -

Irl

=

see that the first term is a l

= 2, and the

< 1 , the series converges. We use formula (7) to

2+

3

4

-

+

8

9

+. .

-

.

2

= --- =

3

2 1 - -


6 •

Repeating Decimals

Show that the repeating decimal 0.999 . . . equals 1 .

. = 0 .9+0.09+0 .009+ .

The decimal 0 999 . .

.

. .

=

10

100

�+� + CXl

9 _ _+ 1000

. .

. is an

infinite geometric series. We will write it in the form 2: alrk - 1 so we can use k=l formula (7) . CXl CXl CXl 9 9 9 9 9 + . 9 1 k- l . = = 0 999 . . = + + 2: - Ok 1000 100 10 . 10 . 1 0k - l k = 1 1 0 10

.

.

'&d

�

( )

� alr L.J

. . N ow we can compare thIS senes to Since

Irl

k

and conclude that al

k=1

The repeating decimal

Figure 9

1

=

9 10

=

1

_

9 10 = - = 9 � 10 10

1 10

= -.

1 •

0.999 . . equals 1. .

P e n d u l u m Swings

Initially, a pendulum swings through an arc of 18 inches. See Figure cessive swing, the length of the arc is 0 . 98 of the previous length. (a) (b) (c) (d) Solution

9

- and r 10

< 1 , the series converges and its sum is

0. 999 . . .

EXAM PLE 9

951


SECTION 13.3

What is the length of the arc of the 1 0th swing? On which swing is the length of the arc first less than 12 inches? After 15 swings, what total distance will the pendulum have swung? When it stops, what total distance will the pendulum have swung?

(a) The The The The

length length length length

of the of the of the of the

first swing is 18 inches. second swing is 0 . 98(18) inches. third swing is 0.98(0 . 98) (18) = 0.982 (18) inches. arc of the 10th swing is

(0.98) 9 (18)

(b)

9 . On each suc

15.007 inches The length of the arc of the nth swing is (0.98),,- 1 (18) . For this to be exactly �

12 inches requires that

(0 . 98) ,, -1(18) = (0.98) ,, - 1 = n

-

12

� 18 = 3 12

1 = I Og0 98 n

=1

+


(�) (�)

Express as a logarithm.

In In

0.98

�

1

+

20 . 07

=

2 1 . 07

Solve for n; use the Change of Base Formula.

The length of the arc of the pendulum exceeds 1 2 inches on the 21st swing and is first less than 12 inches on the 22nd swing. (c) After 15 swings, the pendulum will have swung the following total distance L =

18

+

0 . 98(18) 2nd

1st

+

(0 . 98) 2 (18) 3rd

+

(0.98?(18)

+ ... +

(0.98)14(18) 15th

4th

This is the sum of a geometric sequence. The common ratio is term is 18 . The sum has 15 terms, so L

= 18 ·

1

1

-

_

0�9815 . 98

�

18(13 . 07)

L:

�

0.98; the first

235.3 inches

The pendulum will have swung through 235 . 3 inches after

15 swings.

(d) When the pendulum stops, it will have swung the following total distance T: T

=

18 + 0 . 98(18)

+

(0 . 98?(18)

+

(0.98) 3 (18)

+ ...

952

CHAPTER 13


This is the sum of an infinite geometric series. The common ratio is r = 0.98; the first term is a l = 18 . Since I r I < 1, the series converges. Its sum is al 18 = = 900 = r T 1 1 0.98 The pendulum will have swung a total of 900 inches when it finally stops. _

_

•

i -

5


Solve A n n u ity Problems

In Section 6 . 7 we developed the compound interest formula that gives the future value when a fixed amount of money is deposited in an account that pays interest compounded periodically. Often, though, money is invested in small amounts at periodic intervals. An annuity is a sequence of equal periodic deposits. The periodic deposits may be made annually, quarterly, monthly, or daily. When deposits are made at the same time that the interest is credited, the annu ity is called ordinary. We will only deal with ordinary annuities here. The amount of an annuity is the sum of all deposits made plus all interest paid. Suppose the interest rate that an account earns is i percent per payment period (expressed as a decimal). For example, if an account pays 12 % compounded monthly 0.12 (12 times a year), then i = 12 = 0.01 . If an account pays 8 % compounded quar-

0.08 . terly (4 tImes a year) , then i = 4- = 0.02.

To develop a formula for the amount of an annuity, suppose that $P is deposited each payment period for n payment periods in an account that earns i percent per payment period . When the last deposit is made at the nth payment period, the first deposit of $P has earned interest compounded for n 1 payment periods, the sec ond deposit of $P has earned interest compounded for n 2 payment periods, and so on. Table 3 shows the value of each deposit after n deposits have been made. -

-

Table 3

Deposit Amount

1

P(l + i)n- 1

2 P(1

+

i)n-2

P(l + i)n- 3 3

. . .

.

.

.

n

n

1

P(l + i) -

P

The amount A of the annuity is the sum of the amounts shown in Table 3; that is, p.

p.

( 1 + i)" - 2 + . . . + = P [ l + ( 1 + i) + . . . + ( 1 + i)" - I J

A =

( 1 + i)"-I +

p.

( 1 + i) + P

The expression in brackets is the sum of a geometric sequence with n terms and a common ratio of ( 1 + i ) . As a result,

A = P [ l + (1 + i) + . . + (1 + i)"- 2 + (1 + it- I ] .

= P

1 - ( 1 + i)n 1 - ( 1 + i)" ( 1 + i)" - 1 = P = P 1 - ( 1 + i) i -i

----

We have established the following result: THEOREM

Amount of a n Annuity

Suppose P is the deposit in dollars made at the end of each payment period for an annuity paying i percent interest per payment period. The amount A of the annuity after n deposits is

�----

A = P

( 1 + i)" - 1 i

----

(8)

I

------------------------��

---

SECTION 13.3


953

NOTE: I n using formula (8), remember that when the nth deposit is made, the first deposit has earned interest for n 1 compounding periods. -

E X A M P L E 10

Determi n i n g the Amount of an A n n u ity

To save for retirement, Brett decides to place $2000 into an Individual Retirement Account (IRA) each year for the next 30 years. What will the value of the IRA be when Brett makes his 30th deposit? Assume that the rate of return of the IRA is 1 0 % per annum compounded annually. Solution

n

=

. . . mterest per payment penod I S i

=

0.10 -1

30 deposits is

A = 2000

EXAM PLE 1 1

30 annual deposits of P = $2000. The rate of

This is an ordinary annuity with

( 1 + 0 . 10) 30 - 1 0.10

=

0.10. The amount A of the annuity after

= 2000( 164.494023 )

= $328,988 . 05

•

Determ i n i n g the Amount of an A n n u ity

To save for her daughter's college education, Ms. Miranda decides to put $50 aside every month in a credit union account paying 10% interest compounded monthly. She begins this savings program when her daughter is 3 years old. How much will she have saved by the time she makes the 180th deposit? How old is her daughter at this time? Sol ution

This is an annuity with P

A = 50

(1

+

=

$50,

n

yso

0 . 10 12 0.10 12

=

_

180, and i 1

=

=

O��O . The amount A saved is

50(414 . 47035 ) = $20,723.52

Since there are 12 deposits per year, when the 180th deposit is made 1 80

12 =

15 years have passed and Ms. Miranda's daughter is 18 years old .

..m!l:==> -


�i�torical Feature

�

equences a re a mong the oldest objects of mathematical investigation, havi ng been studied for over 3S00 years. After the i n itial

steps, however, little progress was made u ntil Arithmetic and geometric sequences appear in the Rhind papyrus, a mathematical text con taining 85 problems copied around 1 650

(about AD 75) and Diophantus (about AD 250). One problem, again mod ified s l ig htly, is sti l l with u s in the fa miliar puzzle rhyme "As I was going

to st. I ves . . . " (see Historical Problem 2).

The Rhind papyrus i n d icates that the Egyptia n s knew how to add up the terms of an arithmetic o r geometric sequence, as did the Babyloni

about 1 600. Fibon.acci

•

BC

by

the Egyptian scribe Ah mes from an earlier work

(see Historical Problem 1 ). Fibonacci (AD 1 220) wrote about problems

similar to those found i n the R h i n d papyrus, lead i n g one to suspect that

ans. The rule for s u m m i n g up a geometric sequence is fou n d in Euclid's

Elements (Book IX, 35, 36), where, l i ke a l l Euclid's algebra, it is presented

in a geometric form.

I nvestigations of other kinds of sequences began i n the 1 500s, when algebra became sufficiently developed to handle the more complicated problems. The development of ca lculus in the 1 600s added a powerful

Fibonacci may have had material available that is now lost. Th is mater

new tool, especially for fi n d i n g the s u m of i n fi n ite series, a n d the sub

ial wou l d have been in the non-Euclidean Greek tradition of Heron

ject conti nues to fl ourish today.

(Con tinued)

954

CHAPTER 13


H i storical Pro b l e m s

1 . Arithmetic sequence problem from the Rhind papyrus (statement mod

Each wife had seven sacks

ified slightly for clarity) One h u n d red loaves of bread a re to be d i

Each sack had seven ca�

vided among five people so that the amou nts that they receive form

Each cat had seven kits [kittens]

an arithmetic seq uence. The first two together receive one-seventh

Kits, cats, sacks, wives

of what the last t h ree receive. How many loaves does each receive?

How many were going to St. lves?

[Partial answer: First person receives 1

'3 loaves.] 2

(a) Ass u m i n g that the speaker and the cat fa nciers met by travel ing in opposite d i rections, what is the a nswer?

2. The following old E n g l i s h c h i l d ren's rhyme resem bles one of the

(b) How many kittens a re being transported?

(e) Kits, cats, sacks, wives; how many?

Rhind papyrus p roblems.

As I was going to St. lves I met a man with seven wives

, 3.3 Assess Your U nderstanding 'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in 1. If $ 1 000 is invested at 4% per annum compounded semian nually, how m uch is in the account after two years?

red.

2. How much do you need to invest now at 5 % per annum com pounded monthly so that in 1 year you will have $ 1 0, 000?

(pp. 465-472)

(pp. 465-472)

Concepts and Vocabulary 3. In a(n) a constant.

____

sequence the ratio of successive terms is

6. True or False

A geometric sequence may be defined re

cursively.

4. If 11'1 < 1 , the sum of the geometric series

co

2: ark - I is

k= 1

7.

True or False In a geometric sequence the common ratio is always a positive number.

8.

True or False For a geometric sequence with first term al and common ratio 1' , where I' of. 0, I' of. 1 , the sum of the first 1 - I'" n terms is SI1 = al ' -- . 1 - I'

5. A sequence of equal periodic deposits is called a(n)


9-18,

a geometric sequence is given. Find the common ratio and write out the first four terms.

9. {Sill = { 3" } 14. { dill =

10.

{} 9 3 11

{Sill = { ( -5 ) " }

11. { a"l =

15. { e" l = { 21l/3 }

16.

{ -3G)"}

21 {f n l = { 3 1 }

12. { bill = 17. { tl1l =

{G),, }

{ } 1

� 3n -

13. { clll = 18. { unl =

{ } { } 2" - 1 -4 2"

3 'I - l

In Problems 1 9-32, determine whether the given sequence is arithmetic, geometric, or neither. If the sequence is arithmetic, find the com mon difference; if it is geometric, find the common ratio. 19. { n + 2 }

24.

20.

{8 - � }

21.

25. 1 , 3, 6, 10, . . .

n

29. - 1 , -2, -4, -8, . . . In Problems

{ 2n - 5 }

30.

1 , 1 , 2, 3, 5, 8, . . .

{4n2 }

22. { 5 n2 + I }

23.

{ (�)"}

26. 2, 4, 6, 8, . . .

27.

31. {3 n/2 }

32. { ( - I )" }

28.

{ 3 - �n }

{(%)"}

33. al = 2;

find the fifth term and the nth term of the geometric sequence whose initial term al and common ratio 1' = 3 34. a1 = -2; 1' = 4 35. a1 = 5 ; 1' = - 1 36. al = 6; 1' = -2

37. a l = 0;

.! r = .

33-40,

2

38. a l = 1 ;

r = -

.!.

3

39. al = V2 ;

r = V2

40. a l = 0;

1' =

1.

.

w

I'

are given.

SECTION 13.3

In Problems

41-46,


find the indicated term of each geometric sequence.

.

1 1 41. 7th term of 1'"2'"4""

42. 8th term of 1 , 3, 9, . .

43. 9th term of 1,-1, 1, .. .

44. 10th term of -1,2,-4, ...

45. 8th term of 0.4, 0.04, 0.004,. . .

46. 7th term o f 0.1, 1. 0, 10. 0, .. .

In Problems

955

find the nth term all of each geometric sequence. When given, r is the common ratio. 1 1 48. 5, 1 0, 20, 40,. . . 47. 7, 14, 28, 56, . . . 49. -3'9" 47-54,

3

51. a6 = 24 ;

In Problems

1

I'

= -3

55-60,

I' =3

53. (/2

=

7;

a4

=

1575

3 3 2 33

find each sum.

--

(�)k

2 22 23 2 11-1 . 55. - + -+- +- + . . . + 4 4 4 4 4

311 56. -+ - + - + ...+ 9 9 9 9

57.

11 58. 2.: 4·

59. -1 - 2 - 4 - 8 - . . . - (2"-1 )

3 11-1 6 18 60. 2 + - +- + . . . + 2 5 25 5

k-1 k=l 3

, For Problems

0'

1

52. a2 = 7;

-3,1,

61-66,

use a graphing utility to find the sum of each geometric sequence.

. -

1 2 2 14 23 61. -+- + - + - + .. . + 4 4 4 4 4 15 64. 2.: 4 .311-1 22

1/=1

k=1i

()

3

()

32 3 315 33 62. -+- +- + . . + 9 9 9 9

15 2 11 63. 2.: 11=1 3

65. - 1 - 2 - 4 - 8 - . . . - 214

3 15 6 18 66. 2 + -+- + . .. + 2 5 25 5

()

In Problems 67-82, determine whether each infinite geometric series con verges or diverges. If it converges, find its sum .

3

. 3 3 (l)k-l 2.: k=l ( l)k k=l2.:

1 1 67. 1 +-+-+ ... 9

8 4 68 2 +-+ - + ... 9

1 1 1 71. 2 --+ - --+ . . . 2 8

27 9 72. 1 --+---+ 4 64 16

(l)k-l k=1 ( )k-l 2 2.: 3 k=1 00

75. 2.: 5 4 79.

00

32

6 --

76. 80.

00

00

2 70. 6 + 2 +3 + . . .

73. 8 + 1 2 + 18 + 27 + ...

64 74. 9 + 12 + 16 + 3 + ...

1 77. 2.: - 3k-1

78. 2.: 3 2

-1

81.

(3)k-1 k=1 ( )k 3 2.: k=l 00

00

8 3

4 -2

69. 8 + 4 + 2 + ...

k=12(2)k �3 3 00

00

82.

2 4


83. Find x so that x, x + 2, and a geometric sequence.

x

+

3

are consecutive terms of

84. Find x so that x - 1, x, and x + 2 are consecutive terms of a geometric sequence. 85.

86.

87.

Salary Increases Suppose that you have j ust been hired at an annual salary of $18,000 and expect to receive annual in creases of 5%.What will your salary be when you begin your fifth year?

A new piece of equipment cost a company $ 1 5,000. Each year, for tax purposes, the company depreciates the value by 15%.What value should the com pany give the equipment after 5 years? Equipment Depreciation

88.

Bouncing Balls A ball is dropped from a height of 30 feet. Each time it strikes the ground, it bounces up to 0.8 of the previous height.

-11\\ 30'

1 // 1

0 /1

I I

/\\

1\

1 1 \ 1\ /19.2'1 \ ?t \I // 24' I / I I / I / I /1 I /

I

1/ 1/ 1/ 1/

-

/ 1// I / I 1 / 1/ 11//1I 1 / 1/ 1/ I 1/ I

I

Initially, a pendulum swings through an arc of 2 feet. On each successive swing, the length of the arc is 0.9 of the previous length. (a) What is the length of the arc of the 10th swing? (b) On which swing is the length of the arc first less than 1 foot? (c) After 15 swings, what total length will the pendulum have swung?

(a) What height will the ball bounce up to after it strikes the ground for the third time? (b) How high will it bounce after it strikes the ground for the nth time? (c) How many times does the ball need to strike the ground before its bounce is less than 6 inches?

(d) When it stops, what total length will the pendulum have swung?

(d) What total distance does the ball travel before it stops bouncing?

Pendulum Swings

956

89.

CHAPTER 1 3

Sequences; Induction; t h e Binomial Theorem

Retirement Christine contributes $100 each month to her 401 (k). What will be the value of Christine's 401 (k) after the 360th deposit (30 years) if the per annum rate of return is as sumed to be 12% compounded monthly?

Jolene wants to purchase a new home. Suppose that she invests $400 per month into a mutual fund. If the per annum rate of return of the mutual fund is assumed to be 10% compounded monthly, how much will Jolene have for a down payment after the 36th deposit (3 years)?

90. SaYing for a Home

Don contributes $500 at the end of each quarter to a tax sheltered annuity (TSA). What will the value of the TSA be after the 80th deposit (20 years) if the per annum rate of return is assumed to be 8 % compounded quarterly?

.91. Tax Sheltered Annuity

97.

Ray contributes $1 000 to an Individual Re tirement Account (IRA) semiannually. What will the value of the IRA be when Ray makes his 30th deposit (after 15 years) if the per annum rate of return is assumed to be 10% com pounded semiannually?

92. Retirement

Scott and Alice want to purchase a vacation home in 10 years and need $50,000 for a down payment. How much should they place in a savings account each month if the per annum rate of return is assumed to be 6% compounded monthly?

93. Sinking Fund

94.

95.

Sinking Fund For a child born in 1 9 96, a 4-year college edu cation at a public university is projected to be $150,000. As slllning an 8% per annum rate of return compounded monthly, how much must be contributed to a college fund every month to have $150,000 in 18 years when the child begins college?

In an old fable, a com moner who had saved the king's life was told he could ask the king for any j ust reward. Being a shrewd man, the com moner said, "A simple wish, sire. Place one grain of wheat on the first square of a chessboard, two grains on the second square, four grains on the third square, continuing until you have filled the board. This is all I seek." Compute the total number of grains needed to do this to see why the request, seemingly simple, could not be granted. (A chessboard con sists of 8 X 8 64 squares.) Grains of Wheat on a Chess Board

=

Multiplier Suppose that, throughout the U.S. economy, in dividuals spend 90% of every additional dollar that they earn. Economists would say that an individual's marginal propen sity to consume is 0. 90. For example, if Jane earns an addi tional dollar, she will spend 0. 9(1 ) = $0. 90 of it. The individual that earns $0. 90 (from Jane) will spend 90% of it or $0.81. This process of spending continues and results in an infinite geometric series as follows: 1, 0. 90, 0. 902, 0. 903, 0. 904, . . .

The sum of this infinite geometric series is called the multi plier. What is the multiplier if individuals spend 90% of every additional dollar that they earn?

98.

Multiplier Refer to Problem 97. Suppose that the marginal propensity to consume throughout the U.S. economy is 0. 95. What is the multiplier for the U.S. economy?

99.

Stock Price One method of pricing a stock is to discount the stream of future dividends of the stock. Suppose that a stock pays $P per year in dividends and, historically, the div idend has been increased i% per year. If you desire an annual rate of return of r%, this method of pricing a stock states that the price that you should pay is the present value of an infi nite stream of payments: 1 +i + 1 +i 3 . . . Price = P + P +P +P + . 1 +, l+r l +r

-- (1--i)2 ( --)

The price of the stock is the sum of an infinite geometric se ries. Suppose that a stock pays an annual dividend of $4.00 and, historically, the dividend has been increased 3% per year. You desire an annual rate of return of 9 % . What is the most you should pay for the stock? Refer to Problem 9 9. Suppose that a stock pays an annual dividend of $2.50 and, historically, the dividend has increased 4 % per year. You desire an annual rate of return of 11 % . What is the most that you should pay for the stock?

100. Stock Price

A rich man promises to give you $1000 on September 1 , 2007. Each day thereafter he will give

101. A Rich Man's Promise

you

96.

Look at the figure. What fraction of the square is eventually shaded if the indicated shading process continues indefinitely?

:

of what he gave you the previous day. What is the 0 first date on which the amount you receive is less than 1¢? How much have you received when this happens?

Discussion and Writing 102. Critical Thinking

You are interviewing for a job and receive

two offers:

A: B:

$20,000 to start, with guaranteed annual increases of 6 % for the first 5 years $22,000 to start, with guaranteed annual increases of 3% for the first 5 years

Which offer is best if your goal is to be making as much as possible after 5 years? Which is best if your goal is to make as much money as possible over the contract (5 years)? Which of the following choices, A or B, results in more money?

103. Critical Thinking

S E CTION 13.4

A: B:

104.

To receive $1000 on day 1, $ 9 9 9 on day 2, $ 9 98 on day 3, with the process to end after 1 000 days To receive $1 on day 1, $2 on day 2, $4 on day 3, for 1 9 days

Critical Thinking You have just signed a 7-year professional football league contract with a beginning salary of $2,000,000 per year. Management gives you the following options with regard to your salary over the 7 years. 1. A bonus of $ 1 00,000 each year 2. An annual increase of 4.5 % per year beginning after 1 year

957

be 1 penny. On the second day your pay would be two pennies; the third day 4 pennies. Your pay would double on each successive workday. There are 22 workdays in the month. There will be no sick days. If you miss a day of work, there is no pay or pay increase. How much would you get paid if you work all 22 days? How much do you get paid for the 22nd workday? What risks do you run if you take this job offer? Would you take the job?

106. Can a sequence be both arithmetic and geometric? Give rea sons for your answer. 107. Make up a geometric sequence. Give it to a friend and ask for its 20th term.

An annual increase of $ 95,000 per year beginning after 1 year Which option provides the most money over the 7-year pe riod? Which the least? Which would you choose? Why?

108. Make up two infinite geometric series, one that has a sum and

Critical Thinking Suppose you were offered a j ob in which you would work 8 hours per day for 5 workdays per week for 1 month at hard manual labor. Your pay the first day would

109. Describe the similarities and differences between geometric sequences and exponential functions.

3.

105.

Mathematical lnduction

one that does not. Give them to a friend and ask for the sum of each series.


1. $1082.43

2. $ 95 13.28

OBJECTIVE 1 Prove Statements Using M athematical Induction (p. 957) 1

Prove Statements Using Mathematical Induction

Mathematical induction is a method for proving that statements involving natural numbers are true for all natural numbers.* For example, the statement "2n is always an even integer" can be proved for all natural numbers n by using mathematical induction. Also, the statement "the sum of the first n positive odd integers equals n2," that is, 1 + 3 + 5 + ... + ( 2n - 1 ) = n2

(1)

can be proved for all natural numbers n by using mathematical induction. Before stating the method of mathematical induction, let's try to gain a sense of the power of the method. We shall use the statement in equation ( 1 ) for this purpose by restating it for various values of n = 1 , 2, 3, .... n = 1

n = 2

The sum of the first positive odd integer is 12; 1 = 12. The sum of the first 2 positive odd integers is 22; 1 + 3 = 4 = 22.

n = 3

The sum of the first 3 positive odd integers is 32; 1 + 3 + 5 = 9 = 32.

n = 4

The sum of the first 4 positive odd integers is 42; 1 + 3 + 5 + 7 = 16 = 42.

Although from this pattern we might conjecture that statement (1) is true for any choice of n, can we really be sure that it does not fail for some choice of n? The method of proof by mathematical induction will, in fact, prove that the statement is true for all n. * Recall that the natural numbers are the numbers 1,2,3,4,

bel'S and posilive ilUegers are synonymous.

. . . .

I n other words, the terms naturalnum

958

CHAPTER 1 3


THEOREM

The Principle o f Mathematical Induction

Suppose that the following two conditions are satisfied with regard to a state ment about natural numbers: CONDITION I:

The statement is true for the natural number 1 .

CONDITION II:

If the statement is true for some natural number k, it is also true for the next natural number k + 1.

Then the statement is true for all natural numbers.

Figure

We shall not prove this principle. However, we can provide a physical interpre tation that will help us to see why the principle works. Think of a collection of nat ural numbers obeying a statement as a collection of infinitely many dominoes. See Figure 10. Now, suppose that we are told two facts:

10

1. The first domino is pushed over. 2. If one domino falls over, say the kth domino, so will the next one, the (k + l)st

domino. Is it safe to conclude that all the dominoes fall over? The answer is yes, because if the first one falls (Condition I), the second one does also (by Condition II); and if the second one falls, so does the third (by Condition II); and so on. Now let's prove some statements about natural numbers using mathematical induction.

EXAM P L E 1

Using Mathematical Induction Show that the following statement is true for all natural numbers n.

Solution

1 + 3 + 5 + . . . + (2n - 1 ) = n2

(2)

1 + 3 + .. . + (2k - 1) = k2

(3)

We need to show first that statement (2) holds for n = 1. Because 1 = 12, statement (2) is true for n = 1 . Condition I holds. Next, we need to show that Condition II holds. Suppose that we know for some k that We wish to show that, based on equation (3), statement (2) holds for k + 1. We look at the sum of the first k + 1 positive odd integers to determine whether this sum equals (k + If

1 + 3 + . . . + (2k - 1) + [2(k + 1)-1]

=

[1 + 3 + . . . + (2k

\

=

-

!?- by equation (3) ,

1)] + (2k + 1) I

= k2 + (2k + 1) k2 + 2k + 1 = (k + 1)2

=

Conditions I and II are satisfied; by the Principle of Mathematical Induction, state ment (2) is true for all natural numbers n. •

E XA M P LE 2

Using M athematical Induction Show that the following statement is true for all natural numbers n. 211 > n

Solution

First, we show that the statement 211 > n holds when n = 1. Because 21 = 2 > 1, the inequality is true for n = 1. Condition I holds.

SECTION 13.4

Mathematica l l nduction

959

k kk. We wish to show that k, 2 k+ 1; 2 +1 > k+ 1. Now 2k+1 = 2· 2k > 2·k = k+ k 2: k+ 1

Next, we assume, for some natural number that > the formula holds for that is, we wish to show that i

We know that

2 k k,

2k+1 k+ 1,

2k > k.

i

k 2:: 1

If > then > so Condition II of the Principle of Mathematical Induction is satisfied. The statement > n is true for all natural numbers n. •

EXAM P L E 3

Using M athematical Induction

Show that the following formula is true for all natural numbers n.

1 +

Solution

211

2 + 3 +. . . + n = n ( n 2+ 1)

First, we show that formula (4) is true when n = 1. Because

(4)

1(1+ 1) = 1(2) = 1 2 2

Condition I of the Principle of Mathematical Induction holds. Next, we assume that formula (4) holds for some and we determine whether the formula then holds for We assume that

k, k+ 1. k(k + 1) for some k 1 + 2 + 3 +. .. + k = 2

(5)

Now we need to show that

(k+ l )[ (k+ 1)+-1] --=:' (k+ l)(k+ 2) 1+ 2 + 3 +. .. + k+ (k+ 1) = �-�'----'--2 2 We do this as follows: 1+ 2+ 3+ . .+ k+ (k + 1) P + 2+ 3+' y " + kl,+ (k + 1) 2 1) k(k + + (k+ 1) = 2 k2 + k+2 2k+ 2 2 + 23k + 2 - (k+l)(k2 +2) - k =

_

k(k + 1)

by equation (5)

_

Condition II also holds. As a result, formula (4) is true for all natural numbers n . •

....

� = ;;;;;m;


Now Work

PRO B L E M

1

Using Mathematical Induction

3" - 1 is divisible by 2 for all natural numbers n. First, we show that the statement is true when n = 1. Because 3 1 -1 = 3 - 1 = 2 is divisible by 2, the statement is true when n = 1. Condition I is satisfied . Next, we assume that the statement holds for some k, and we determine whether the statement then holds for k+ 1. We assume that 3 k -1 is divisible by 2 for some k. We need to show that 3 k+1 -1 is divisible by 2. Now 3k 3 k+1 -1 = 3kk+1 -3 k+ 3 kk-1 k k = 3 (3 - 1) + (3 - 1) = 3 • 2 + (3 - 1) Show that

Subtract and add

960

CHAPTER 13


2

2

2,

Because 3k• is divisible by and 3k - 1 is divisible by it follows that k k+1 3" . + ( 3 - 1) = 3 - 1 is divisible by Condition II is also satisfied. As a re sult, the statement "3" - 1 is divisible by is true for all natural numbers n . -

2

WARNING

2"2.

both Conditions I and II of the Principle of Mathematical Induction have been

The conclusion that a statement involving natural numbers is true for all n atural num

bers is made only after

satisfted. Problem 28 demonstrates a statement for which only Condition I holds, but the state ment is not true for all natural numbers. Problem 29 demonstrates a statement for which only

_

Condition II holds, but the statement is not true for any natural n umber.

13.4 Assess Your Understanding Skill Building

In Problems 1-22, use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers 2. 1 +5 + 9 + . . . + (4n - 3) = n (2n - 1 ) 1. 2 + 4 + 6 + . . . + 2n = n(n + 1 ) 4. 3 + 5 + 7 + . . . + (2n +1 ) = n(n + 2)

1 3. 3 +4 +5 + . . . + (n + 2) = 2 n(n + 5) 5. 2 + 5 + 8 + . . . +(3n - 1 )

=

1 6. 1 + 4 + 7 + . . . + (3n - 2 ) =2 n(3n - 1 )

1 2 n(3n + 1 )

8. 1 + 3 + 32 + . .. + 3n-1 = .!.(3// - 1 ) 2

7. 1 + 2 + 2 2 + . . . + 2//-1 = 2// - 1

�

10. 1 +5 + 52 + . . . + 5//-1

9. 1 + 4 + 42 + . .. + 4//-1 = (4// - 1 )

1 1 1 1 11. - + - + - + . . .+--,------,1 ·2 2·3 3·4 n(n + l ) 2

? ? 13. 1 +2- +3- + . . . +n

2

=

n n + l

1 "6n ( n + 1 ) (2n + 1 )

1 15. 4 + 3 +2 + . . . + (5 - n) =2 n(9 - n) 17. 1 · 2 + 2 · 3 + 3·4 + .. . +n(n + 1 )

n.

=

=

�

(5n - 1 )

1 1 1 1 12. - + + - + ... + (2n - 1) (2n + 1) 1·3 3·5 5· 7

-------

-

14. 13 + 23 + 33 + . . . + n3

=

n 2n + 1

1 "4 n2 (n + 1 )2

16. -2 - 3 - 4 - . . . - (n + 1 )

=

1 -2 n(n + 3)

1 n ( n + l ) ( n + 2) -;:; .J

1 18. 1,2 + 3·4 + 5 · 6 + . . . + (2n - 1 ) (2n) = -;:; n(n + 1 ) (4n - 1)

.J

19. n2 + n is divisible by 2.

20. n3 + 2n is divisible by 3.

22. n(n + 1 ) (n +2) is divisible by 6 .

21. n2 - n + 2 is divisible by 2. Applications and Extensions

In Problems 23-27, prove each statement. 23. If x > 1, then xn > 1 . 24. If 0 < x < 1, then 0 < xn < 1. 25. a - b is a factor of an - bn. [Hint: ak+1 - bk+1 = a(ak - bk ) + bk (a - b)] 26. a + b is a factor of a2n+1 + b2//+I. 27. (1 + a)" � 1 + na, for a > 0 28. Show that the statement n2 - n +41 is a prime number" is true for n =1, but is not true for n = 41 .

30. Use mathematical induction to prove that if r =f. 1 then

1 - r// a + ar + ar2 + . . . + arn- 1 = a I - r

--

31. Use mathematical induction to prove that

a + (a + d) + (a + 2d)

n(n - 1 ) + . . . + [a + (n - l) d] = n a + d --'-- --'2 -

"

29. Show that the formula

2 +4 + 6 + . . . + 2n = n2 +

n

+ 2

obeys Condition II of the Principle of Mathematical Induc tion. That is, show that if the formula is true for some k it is also true for k + 1. Then show that the formula is false for n 1 (or for any other choice of n). =

32.

Extended Principle of Mathematical Induction The Ex tended Principle of Mathematical Induction states that if Conditions I and II hold, that is,

(I) A statement is true for a natural number j. (II) If the statement is true for some natural number k � j, then it is also true for the next natural number

k+1.

SECTION 13.5

then the statement is true for all natural numbers "2 j. Use the Extended Principle of Mathematical Induction t o show that the number o f diagonals i n a convex polygon o f n 1 . .IS "2 sides n(n 3) .

961

[Hint: Begin by showing that the result is true when n (Condition I ) . ]

33.

-

The B i n om i al Theorem

=

4

Geometry Use the Extended Principle of Mathematical Induction to show that the sum of the interior angles of a convex polygon of n sides equals (n 2) 180°.

- .


34. How would you explain the Principle of Mathematical Induction to a friend?

13.5 The Binomial Theorem OBJECTIVES 1 Eva l uate

(�)

(p.961)

2 Use the Binomial Theorem (p. 963)

+ +

2

Formulas have been given for expanding (x a)" for n = and n 3. The Binomial Theorem'" is a formula for the expansion of (x a)" for any positive integer n. If n = and 4, the expansion of (x a)" is straightforward.

(x

1,2,3, + a)l = x +a

+

=

1 Two terms, begi n ning with x and 1 endin g with a 2 Three terms, beginnin g with x 2 and ending with a Four terms, beginning with x 3 and ending with a

3

4 Five terms, beginning with x and 4 ending with a

+

Notice that each expansion of (x a)" begins with x" and ends with a". As you read from left to right, the powers of x are decreasing by while the powers of a are increasing by Also, the number of terms equals n Notice, too, that the degree of each monomial in the expansion equals n. For example, in the expansion of (x a?, each monomial (x3, 3ax2, 3a2x, a3) is of degree 3. As a result, we might conjecture that the expansion of (x a)" would look like this: (x a)" = x" axn-1 a 2x,,-2 an-Ix + a"

+

1.

+

+

__

+ +

+ 1. 1,

+... +

__

__

where the blanks are numbers to be found. This is, in fact, the case, as we shall see shortly. Before we can fill in the blanks, we need to introduce the symbol 11. . j

� .' �

COMMENT the symbol the key

1 On a graphing calculator,

(J)

�.

may be denoted by -

DEFINITION

Eva l u ate

(;)

We define the symbol

()

(�).

read

"11.

taken j at a time, " as follows:

If j and n are integers with 0 ::; j ::;

11.,

the symbol

(;)

is defined as

(�) I �--------------------------------�� n! j! ( n - j)!

'" The name binomial is derived from the fact that

x + a

(1)

is a binomial; that is, it contains two terms.

962

CHAPTER 13

Sequences; I n d uction; the B i nomial Theorem

E XA M P L E 1 Evaluating

(;)

Find: (a)

Solution

Figure 11

65 nCr 1 5 2. 073746998E1 4

G)

(b)

(�)

(c)

(�)

I

(d)

(��)

3·2·1 (13 ) = 1! (3 3!- I)! = I!3!2! = 1(2 . 1) = "26 = 3 4! = 4! = 4·3·2 ·1 24 = 6 4 (b) ( ) = 2 2! (4 - 2)! 2! 2! (2· 1) (2. 1) = 4 8! = 8! = 8·;rrl =8 = 8 8 (c) ( ) = 7 7! (8 - 7)! 7! 1! ;rr. ! "1 8! =r 8·7! l;. (d) Figure 11 shows the solution using a TI-84 Plus graphing calculator. So (��) 2.073746 998 X 1014 (a)

::::::

"""

:>-

Now Work

PRO B L E M

•

5

Four useful formulas involving the symbol

(�.)

are

( �) =1 ( � ) =n ( n:l) = n ( :) = 1 ( n) = Olenn!- O)! =O!ntnt =1"1=1 (1n) =l! (nn!-1)! = (n -n! I)! =n�! �! = n You are asked to show the remaining two formulas in Problem 45. Suppose that we arrange the values of the symbol (� ) in a triangular display, ] as shown next and in Figure 12. Proof

o

•

(�) (�) C) (�) (�) (�) (�) G) (�) G)

(�) (�) (�) (;) (:) (�) G) (�) (�) (!) G)

SECTION 13.5

Figure 12

/

Pascal trian g l e

n

=

2 -----+-

4 5

2

3

6

10

The Binomial Theorem

j=0

/

3

j=1

/

j=2

/

)

4

10

963

=

3

/

)=4

5

/

j=5

This display is called the Pascal triangle, named after Blaise Pascal (1623-1 662), a French mathematician. The Pascal triangle has 1 's down the sides. To get any other entry, add the two nearest entries in the row above it. The shaded triangles in Figure 1 2 illustrate this feature of the Pascal triangle. Based on this feature, the row corresponding to n = 6 is found as follows: 1 5 10 10 5 1 VVVVV 6 15 20 15 6

n=5-? n =

6-?

Later we shall prove that this addition always works (see the theorem on page 965). Although the Pascal triangle provides an interesting and organized display of n the symbol , in practice it is not all that helpful. For example, if you wanted to 12 ] know the value of , you would need to produce 13 rows of the triangle before 5 seeing the answer. It is much faster to use the definition ( 1 ) .

()

2

THEOREM

( )

Use the Binomial Theorem

Now we are ready to state the

Binomial Theorem.

Binomial Theorem

Let x and

a be real numbers. For any positive integer n, we have

()

(2)

± � xn-jaj j=O )

�

�------�

Now you know why we needed to introduce the symbol

(� }

the numerical coefficients that appear in the expansion of (x the symbol

E XAM P LE 2

(� )

is called a

binomial coefficient.

Expanding a Binomial Use the Binomial Theorem to expand

(x

+ 2)5.

these symbols are

+a)". Because of this,

964

C H A PTER 13

Sequences; I n d uction; the Binomial Theorem

Solution

2 and n = 5. Then (x 2)5 = (�) x5 + G) 2x4 + G) 22x3 + G) 2 3x2+ (!) 2 4X+ G) 2 5

In the Binomial Theorem, let a = +

i

1·x5 +5·2X4 + 10·4x3+ 10·8x2+ 5'16x+ 1·32

Use equation (2).

=

�

se row

=

EXAM P L E 3

n = 5 of the

Pascal triangle or formula (1) for

x5 + 10x4 + 40x3+ 80x2+ 80x+ 32

(� )

.

•

Expanding a Binomial

(2y - 3)4 using the Binomial Theorem. First, we rewrite the expression (2y - 3)4 as [2y + (-3)t Now we use the Binomial Theorem with n = 4, x = 2y, and -3. [2y + (-3)J4 = (�}2y)4 + (�) (-3)(2Y? + (�) (-3?(2Y? +G}-3)3(2y)+ G}-3)4 = 1'16/ + 4(-3)8l + 6'9·4/ + 4(-2 7)2y + 1·81 Expand

Solution

a =

i

Use row

=

n = 4 of the Pascal triangle or formula (1)

16/ - 96l + 216/ -216y + 81

for

(� )

.

In this expansion, note that the signs alternate due to the fact that a = �,


"'--

-3 < •0.


F inding a Particular Coefficient i n a Binomial Expansion

(2y + 3)10 . We write out the expansion using the Binomial Theorem. (2y + 3)10 COO}2Y)1O + C2}2y)9(3)1+ C20}2y)8(3)2+ C30}2Y?(3)3 0 + C�}2y)6(3)4 + ... + C ) (2Y)(3)9 + G�) (3)10 9

Find the coefficient of l in the expansion of

=

From the third term in the expansion, the coefficient of

l is

10·9. .sr. 2 8. 9 = 103 680 8 . = 2 9 (102 ) (2)8(3? = �. 2! 8! 2·sr. '

•

As this solution demonstrates, we can use the B inomial Theorem to find a par ticular term in an expansion without writing the entire expansion.

SECTION 13.5

Based on the expansion of

965

The Bi nomial Theorem

(x+ a)", the term containing xj is

we can solve Example 4 by using formula (3) with 10, aexample, n = For 3, x 2y, = = and j = 8. Then the term containing l is

EXAM PLE 5

F inding a Particular Term in a Binomial E xpansion Find the sixth term in the expansion of

Solution A

(x+ 2)9.

We expand using the Binomial Theorem until the sixth term is reached.

+( 59) x·2 4 5 +. .. The sixth term is

. ( 59) 2 5 = _ 5!94!_' . x4 . 32 = 4032x4 ;"(4

•

•

Solution B

(x+ 2)9,

The sixth term in the expansion of which has 10 terms total, contains (Do you see why?) By formula (3), the sixth term is

&;i']!l;; = ==

Now Work

PRO B L E MS

2 9 AND

X4.

• 35

triangular addition feature of the Pascal triangle illus

12

Next we show that the trated in Figure always works.

j

THEOREM

If n and are integers with

1 :5 j :5 n, then (4)

I

�--------------------------------�� c � ) +(;) 1

Proof

_ l-)!-[n-n-�_(J'- )] + j! (nn� j)! n! n! ----+ (j - l)! (n - j+ I)! j! (n -j)! = j(j - l)! (njn!- j + I)!+ j!(n (n-j-+j+1)n! l)(n - j)!

=

-( ].

1

!

-

----

------

.

j

� and J n- j + 1 the second term by -� n-j + 1 Multiply the first term by

966

CHAPTER 13


( n - j+1)n! jn! - --'-----+�--'--- - '--j! ( n - j+1)! j! ( n - j+ 1)!

Now the denominators are equal.

jn! + ( n - j+ 1)n! j! ( n - j+1)! n! (j+ n - j+1) j! ( n - j+1)! n! ( n + 1) j! ( n - j + 1)!

( n +1)! j![ (n +1) - j]!

•

�i�torical Feature

T Omar Khayyam (1050-1123)

he case n

(a + b)2,

=2

i of the Binomial Theorem,

The heart of the Binomial Theorem is the form u l a for the nu merical

BC,

coefficients, and, as we saw, they can be written i n a symmetric trian

was known to Euclid in 300

but the general law seems to have been

g u l a r form. The Pascal tria ngle appears first i n the books of Yang Hui

discovered by the Persian mathematician and as

(about 1 270) a nd Chu Shih-chieh (1 303). Pascal's name is attached to

tronomer Omar Khayyam ( 1 050- 1 1 23), who is

the triangle beca use of the many applications he made of it, especia l ly

also well known as the a uthor of the Rubdiydt, a

to counting and probability. In esta blishing these results, he was one of

collection of fou r-line poems making observa

the earliest users of mathematical induction.

tions on the human condition. Omar Khayyam

Many people worked on the proof of the Binomial Theorem, which

did not state the Binomial Theorem explicitly, but

was finally com pleted for a l l n (incl uding complex n u m bers) by Niels

he claimed to have a method for extracting third, fourth, fifth roots, and

Abel (1 802-1 829).

so on. A little study shows that one must know the Binomial Theorem to create such a method.


2.

The coefficients.

__ __

is a triangular display of the binomial

(�) =_

3.

True or False

4.

The like (2x + 3 )6.

(J ) = n

__ __

p)

(n - J ! n!

can be used t o expand expressions

Skill Building

In Problems 5.

9.

13.

C) (!�) CD

In Problems 17. 21. 25.

5-16,

evaluate eac h expression. 6.

10.

14.

17-28,

(x + 1)5

(3x + 1)4

(\IX + \12)6

C) (19°8°)

7.

11.

G�)

15.

expand each expressio n using the Binomial Theo rem . 18. 22. 26.

(x - 1)5

(2x + 3 )5 (\IX - v'3t

1 9. 23. 27.

G) ( ) 1 000 1000

G�) (x - 2)6

(x2 + l)5 (ax + by) 5

8.

12.

16.

20. 24. 28.

(�) ( ) ° (��) 1000

5 (x + 3) (x2 - l)6

(ax - by) 4

In Problems 29-42, use the Binomial Theo rem to find the ind icated coefficient or term. 10 3 10 30. The coefficient of x in the expansion of (x - 3) '-29. The coefficient of x6 in the expansion of (x + 3 ) 7 1 3 12 1) 2 32. The coefficient of x in the expansion of (2x + 31. The coefficient of x in the expansion of (2x 9 7 9 2 33. The coefficient of x in the expansion of (2x + 3) 34. The coefficient of x in the expansion of (2x - 3 ) -

1)

967

Chapter Review

"

35.

The fifth term in the expansion of (x + 3)7

37.

The third term in the expansion of (3x - 2)9

39.

The coefficient of xo in the expansion of

41.

The coefficient of

X4

( + �) ( � yo 12

2 x

in the expansion of x

_

36.

The third term in the expansion of (x - 3)7

38.

The sixth term in the expansion of (3x + 2)8

40.

1 . . of x° .Jl1 the expansIOn The coeffi Clent of x - 2

42.

The coefficient of

48.

If n is a positive integer,show that

(

2 x

in the expansion of

(

x

Vx +

)9

:rxy


Use the Binomial Theorem to find the numerical value of (1.001)5 correct to five decimal places. 5 (1 + 1 0-3) ] [Hint: (1.001)5

(:) = (z) (�) G) (�)GY G)(�)4G) G)(�YGY C)GYGY + C)G)GY G)(�Y

=

44.

-

Use the Binomial Theorem to find the numerical value of (0.998)6 correct to five decimal places.

C: )=

Show that

46.

Show that if n and j are integers with a

1

n and

( :) =

45.

+

49.

1. j

�

n then

SO.

+

Stirling's Formula

is given by n!

If n is a positive integer, show that

[Hint:

21/

a

+

+

�

Conclude that the Pascal triangle is symmetric with respect to a vertical line drawn from the topmost entry. 47.

- . . . + (-1)''

+

�

=

An approximation for n!, when n is large,

v:2;;;(; !2)n(l + e

1 _ _ 12n - 1

)

Calculate 12!, 20!, and 25! on your calculator. Then use Stirling's formula to approximate 12!, 20!,and 25!.

= ( 1 + 1)" ; now use the Binomial Theorem.]

CHAPTER REVIEW Things to Know

Sequence (p. 930)

A function whose domain is the set of positive integers.

Factorials (p. 932)

a!

Arithmetic sequence (pp. 940 and 941) Sum of the first n terms of an arithmetic sequence (p. 942) Geometric sequence (pp. 946 and 947) Sum of the first n terms of a geometric sequence (p. 948) Infinite geometric series (p. 949)

?

a1

a, an

all

S

= 1,1! = 1,n! = =

Il

=

=

al all

= =

al -l /

n(n - 1)' ... ·3·2·1 if n

2" [ 2a1 + (n - l)d] a,

all

l -1 / ajr ,

=

rall-1 , r

a

ll

1 -r , 1 -r

alr

+

I r I < 1,

..

i= a r

= 2"n (

where

aj

2:

2 is an integer

= = first term,d = common difference

+ d, where al

+ (n - l)d

al

n

Sn = l --+ . + al

=

a

aj

+

an

)

= = first term, = common ratio a

r

i= 0, 1

ll-1 a1r

+

... = 2: 00

k=l

k 1 ajr -

Sum of a convergent infinite geometric series (p. 950)

If

Principle of Mathematical Induction (p. 958)

Suppose the following two conditions are satisfied. Condition I : TIle statement is true for the natural number 1. Condition II: If the statement is true for some natural number k, it is also true for k + 1. TIlen the statement is true for all natural numbers.

968

CHAPTER 1 3


() n j

Binomial coefficient (p. 961)

n! = j! ( n - j) !

See Figure 12.

Pascal triangle (p. 963)

(x + a) "

Binomial Theorem (p. 963)

= (n° ) x" + (n1 ) axil-I + . . . + ( nj ) ai x"-i + ... + ( n ) an = � ( j ) x"-iai .

.

II

.

11.

11.

.

.

Objectives -------, Section

You should be able to ...

Review Exercises

13.1

Write the first several terms of a sequence (p. 930) Write the terms of a sequence defined by a recursive formula (p. 933) Use summation notation (p. 934)

1-4

2 3

Determine if a sequence is arithmetic (p. 940)

13-24

Find a formula for an arithmetic sequence (p. 941) Find the sum of an arithmetic sequence (p. 942)

31, 32,35, 37-40, 65 13, 14,19, 20,27,28, 66

2 3 4 5

Determine if a sequence is geometric (p. 946) Find a formula for a geometric sequence (p. 947) Find the sum of a geometric sequence (p. 948) Determine whether a geometric series converges or diverges (p. 949) Solve annuity problems (p. 952)

13-24 22, 33,34,36, 70 17, 18,21, 22, 29, 30, 67(a)-(c), 70

2 3 4 Find the sum of a sequence (p. 935)

13.2

13.3

(� )

5-8 9-12 25, 26

41-48,67(d) 68, 69

13.4

Prove statements using mathematical induction (p. 957)

49-54

13.5

Evaluate

55,56

2

(p. 961)

57-64

Use the Binomial Theorem (p. 963)

Review Exercises

In Problems 1. ( all) 5.

al

=

=

write down the first five terms of each sequence.

{( ( : : D} -1)"

3;

In Problems 9.

1-8,

4

2. (bll)

6. al 9

and

10,

= {( -1)1l+1(2n + 3)}

= 4;

10.

k=i

11

and

7. al

write out each sum.

2,: (4k+2)

In Problems

3. (cll)

12,

=

= 2;

{ } 21l

"""1 n-

all

4.

= 2 - an-l

8. al

= { -;;en

= -3;

} a

n

= 4 + an-l

3

2,: (3 - k2)

k=i

express each sum using summation notation.

1 1 1 1 11. 1 --+ --- + ··· + 2 3 4 13

( dn)

--

22 23 2n+1 12. 2 + - + - + ··· + 3" 3 32

In Problems 13-24, determine whether the given sequence is arithmetic, geometric, or neither. If the sequence is arithmetic, find the com mon difference and the sum of the first n terms. If the sequence is geometric, find the common ratio and the sum of the first n terms. 16. ( dll) { 2 n2 - 1} 15. (cll) { 2 n3 } 1 4. ( bn) { 4 n + 3} 13. (all) { n + 5}

=

=

=

19. 0, 4, 8, 12, . . . 2 3 4 5 23. "3'"4'5'6' ...

=

20.1, -3, -7, -11, . . .

C hapter Review

In Problems 25-30, find each sum.

26. � k2 k=l

0 4 28. � ( -2k k=l

+

7 29 . � k=\

8)

31-36,

27. � (3k - 9) k=l

)k ( l -::;-'

10

k 30. � (_2) k=l

find the indicated term in each sequence. [Hint: Find the

31. 9th term of 3, 7,11,15,...

In Problems

30

30

50

25. � (3k) k=l

32. 8th term of

4, 2, 8, ... 34. 11th term of 1,

35. 9th term of

1, -1,

-3 ,

general term first.]

-5,...

33. 11th term of 1,

Yz, 2Yz, 3Yz, ...

36. 9th term of

In Problems 3 7-40, find a general formula for each arithmetic sequence. 38. 8th term is -20; 17th term is

39. 10th term is 0; 18th term is 8

40. 12th term is 30; 22nd term is 50

1

3

24

49. 3

9

+.

-

8

49-54,

.

00 ( )k-l 5 46. � 5· -4 k=l

.

47.

�4 00

( )k-l 3 48. �3 -k=l 4

"2

use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers.

+

6

+ 9+. . .+3n

2 +

6

+

3n

= 2 ( n+ 1 )

18+. . .+2. 3 11- J

In Problems 55 and

=

50. 2

3 11 - 1

+ 6+10+. . .+ (4n -

52. 3+6

+

12+. ..

56,

+

=

=

2112

3(2/1 - 1)

·5+. . .+n ( n + 2)

= 6(11 + 1 ) ( 2n + 7) n

evaluate each binomial coefficient.

G)

56.

57. (x+2)5

G)

58.

(x -

(x+2t

62. Find the coefficient of x3 in the expansion of

(x -

63.

(2x+ 1)7.

2 Find the coefficient of x in the expansion of 6 Find the coefficient of x in the expansion of

65. Constructing a B rick Staircase

3)

8.

(2x+ 1

)8.

A brick staircase has a total

of 25 steps .The bottom step requires 80 bricks. Each succes

sive step requires three less bricks than the prior step. (a) How many bricks are required for the top step?

(b) How many bricks are required to build the staircase? 66. Creating a Floor Design

A mosaic tile floor is designed in

the shape of a trapezoid 30 feet wide at the base and 15 feet

wide at the top.The tiles, 12 inches by 12 inches, are to be

placed so that each successive row contains one less tile than

the row below.How many tiles will be required?

A ball is dropped from a height of 20 feet.

Each time it strikes the ground, it bounces up to three

quarters of the previous height.

(a) What height will the ball bounce up to after it strikes the ground for the third time?

60. (3x4)4

59. (2x+3)5

3)4

61. Find the coefficient of x7 in the expansion of

67. Bouncing Balls

2)

3 .211-1

In Problems 5 7-60, expand each expression using the Binomial Theorem.

64.

, . . •

00

( l )k-l

54.1·3+ · 4 2 +3

55.

10' 100""

,V�2 , 2, 20'/2

41-48,

+- +

-

In Problems

51.

1

determine whether each infinite geometric series converges or diverges. If it converges, find its sum. 1 1 1 1 8 16 1 1 3+ 1+"3+"9+. . . 42. 2 + 1+"2+4"+. . . 43. 2 - 1 + "2 - 4" + . .. 44. 64 + "3 - 9+. . .

In Problems

45.

1

-47

37. 7th term is 31; 20th term is 96

41.

969

(b) How high will it bounce after it strikes the ground for the nth time?

(c) How many times does the ball need to strike the ground before its bounce is less than 6 inches?

(d) What total distance does the ball travel before it stops bouncing?

68. Retirement Planning

Chris gets paid once a month and con

tributes $200 each pay period into 401(k). his If Chris plans

on retiring in 20 years, what will be the value of 401(k) his if the per annum rate of return of the 401(k) is 10% com pounded monthly?

69. Retirement Planning

Jacky contributes $500 every quarter

to an IRA. If Jacky plans on retiring in 30 years, what will be

the value of the IRA if the per annum rate of return of the IRA is 8% compounded quarterly?

70. Salary Increases

Your friend has just been hired at an an

nual salary of $20,000. If she expects to receive annual in

creases4%, of what will be her salary as she begins her fifth year?

970

CHAPTER 1 3

Sequences; Induction; the B i n o m i a l Theorem

CHAPTER TEST In Problems sequence.

and

1

write down the first five terms of each

{ : :; } � ( _l )k+l( k ; )

1. {sn l =

4,

5. Write the following sum using summation notation.

l

11 2 3 4 --+- - -+. . .+5 6 7 14

2. a = 4, an = 3all- 1 + 2

2

In Problems 3 and 3.

2,

write out each sum. Evaluate each sum. 1

� [(�y k ] -

4.

In Problems 6-11, determine whether the given sequence is arithmetic, geometric, or neither. If the sequence is arithmetic, find the common difference and the sum of the first n terms. If the sequence is geometric, find the common ratio and the sum of the first n terms.

8. -2, - 10,

6. 6, 12, 36, 144, . . . 8 10. 25, 10, 4, 5"' . . .

U. Determine whether the infinite geometric series 256 - 64+ 16 - 4 + . . .

11.

-

{ -- }

18 -26, . . . ,

2n - 3 2n+1

15. A 2004 Dodge Durango sold for $31 ,000. If the vehicle loses 15% of its value each year, how much will it be worth after 10 years?

converges or diverges. If it converges, find its sum.

13. Expand (3m+ 2)5 using the Binomial Theorem. 14. Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers.

16. A weightlifter begins his routine by benching 100 pounds and increases the weight by 30 pounds for each set. If he does 10 repetitions in each set, what is the total weight lifted after 5 sets?

CUMU LATIVE REVIEW (g) The function g-I and its domain

1. Find all the solutions, real and complex, of the equation

I x21 = 9 2. (a) Graph the circle x2+i 100 2

and the parabola y = 3x .

=

7.

(b) Solve the system of equations: (c) Where do the circle and the parabola intersect?

3. Solve the equation 2ex

=

5.

4. Find an equation of the line with slope 5 and x-intercept 2. 5. Find the standard equation of the circle whose center is the point ( - 1 , 2) if (3, 5 ) is a point on the circle. 6. f( x )

=

Find: (a) (f

3x x - 2

-- ,

0

g ) (2)

g ( x ) = 2x+ 1 (b) (g

(d) The domain of (f (e) (g

0

f)(x)

0

0

f) (4)

(h) The function rl and its domain

Find the equation of an ellipse with center at the origin , a focus at (0, 3 ) , and a vertex at (0, 4 ) .

8. Find the equation o f a parabola with vertex a t ( - 1 , 2) and focus at ( - 1 , 3 ) . 9. Find the polar equation o f a circle with center a t (0, 4) that passes through the pole. What is the rectangular equation? 10. Solve the equation 2 sin2 x - sin x - 3

g) (x)

(f) The domain of (g

0

0

g) (x)

f )( x )

0, 0 :::; x < 27T

11. Find the exact value of COS-I ( -0.5 ) .

12. I f sin 8 = (c) (f

=

(a) cos 8

�

(c) sin (28) (e) sin

and 8 i s i n the second quadrant, find:

(� ) 8

(b) tan 8 (d) cos(28)

Chapter Projects

971

CHAPTER PROJECTS population, while death rates are given a s the number of deaths per 100,000 population. Each must be computed as the number of births (deaths) per individual. For exam ple, in 2000, the birth rate was 14.7 per 1000 and the death 14 .7 = 0.0147, while rate was 873 . 1 per 100,000, so b = 1000 873.1 d = 0 . 00873 1 . 1 00,000 = Next, using data from the Immigration and Naturaliza tion Service wWlv,fedstats.gov, determine the net immi gration to the United States for the same year used to ob tain b and d in Problem 1 .

2. Determine the value o f r, the growth rate o f the popula tion.

3. Find a recursive formula for the population of the United States.

I.

Population Growth The size of the population of the United States essentially depends on its current population, the birth and death rates of the population, and immigration. Suppose that b represents the birth rate of the US. population and d represents its death rate. Then r = b - d represents the growth rate of the population, where r varies from year to year. The US. population after n years can be modeled using the recursive function

PII = (1 + r) P n - 1 + I

where I represents net immigration into the United States.

1. Using data from the National Center for Health Statistics wwwfedstats.gov, determine the birth and death rates for all races for the most recent year that data are available. Birth rates are given as the number of live births per 1000

4. Use the recursive formula to predict the population of the United States in the following year. In other words, if data are available up to the year 2005, predict the US. popu lation in 2006. 5. Compare your prediction to actual data. 6. Repeat Problems 1-5 for Uganda using the CIA World Factbook at www.cia.govlcialpllblicatiol1slfactbooklil1dex.html (in 2000, the birth rate was 48.04 per 1000 and the death rate was 18.44 per 1000). 7.

Do your results for the United States (a developed coun try) and Uganda (a developing country) seem in line with the article in the chapter opener? Explain.

S. Do you think the recursive formula found in Problem 3

will be useful in predicting future populations? Why or why not?

The following projects are available at the Instructor's Resource Center (IRC):

II.

Project at Motorola Digital Wireless Commu1licatio1l Cell phones take speech and change it into digital code using only zeros and ones. See how the code length can be modeled using a mathematical sequence.

III. Economics IV.

Economists use the current price of a good and a recursive model to predict future consumer demand and to de termine future production. Standardized Tests

matical sequence.

Many tests of intelligence, aptitude, and achievement contain questions asking for the terms of a mathe

Counting a nd Probability Deal o r No D eal

By LYNN ELBER, AP Television Writer-LOS ANGELES - The promise of an easy million bucks, a stage crowded with sexy models and the smoothly calibrated charm of host Howie Mandel made " Deal or No Deal" an unexpected hit in television's December dead zone. B ased on a series that debuted in Holland in 2002 and became an interna tional hit, "Deal or No D e al" is about luck and playing the odds. Con testants are faced with 26 briefcases held by 26 models, each case with a hidden value ranging from a penny to the top prize that will esca late by week's end to $3 million. As the game progresses and cases are eliminated, a contestant weighs the chance of snaring a big prize against lesser but still tempting offers made by the show's "bank," rep resented by an anonymous, silhouetted figure.

Source: Adapted from Lynn

© 2006 Associated Press.

Elber, "'Deal or No Deal' back with bigger

prizes, " Associated Press, February 24, 2006.

- See the Chapter Project-

A Look Back

We first introduced sets in Chapter R. We have been using sets to represent solutions of equations and inequalities and to rep resent the domain and range of functions.

A Look Ahead

Here we discuss methods for counting the number of elements in a set and the role of sets in probabil ity.

Outline 1 4.1 1 4.2 1 4.3

Counting

Permutations and Combinations Probability Chapter Review Chapter Test Cumulative Review Chapter Projects

973

974

CHAPTER 14

1 4. 1

Counting a n d Probability

Counting



Sets (Review, Section R.1, pp. 1-3)

Now Work the 'Are You Prepared?, problems on page OBJECTIVES

978.

1 Find All the Subsets

of a

Set (p. 974)

2 Count the Number of Elements in a Set (p. 974) 3 Solve Counting Problems U sing the Multiplication Principle (p. 976)

Counting plays a major role in many diverse areas, such as probability, statistics, and computer science; counting techniques are a part of a branch of mathematics called combinatorics.

1

F i n d A l l the S u bsets of a Set

AA BB . A AA BB, A A B.=I' B, A.A B,

A A

We begin by reviewing the ways that two sets can be compared. If two sets and have precisely the same elements, we say that and B are equal and write If each element of a set is also an element of a set we say that is a subset of B and write � If � and we say that is a proper subset of and write C If � every element in set A is also in set but B may or may not have additional elements. If C every element in is also in and B has at least one element not found in Finally, w e agree that the empty set i s a subset o f every set; that is,

B,

=

A

o�

A

A

for any set

B,

B,

B

A B.

A

F i nding All the Subsets of a Set

EXAM P L E 1

Write down all the subsets of the set

Solution

{ a , b , c} .

To organize our work, we write down all the subsets with no elements, then those with one element, then those with two elements, and finally those with three elements. These will give us all the subsets. Do you see why? o Elements o

1 Element

2 Elements

{a} , {b}, {c}

{ a , b}, {b, c}, { a , c}

3 Elements

{ a , b , c}

•

��==:;> - Now Work P R O B L E M 7

2

Count the N u mber of Elements i n a Set

As you count the number of students in a classroom or the number of pennies in your pocket, what you are really doing is matching, on a one-to-one basis, each object to be counted with the set of counting numbers, 1, 2, 3, . . . , 11, for some number If a set A matched up in this fashion with the set { I, 2, . . . , 25 } , you would conclude that there are 25 elements in the set We use the notation 25 to indicate that there are 25 elements in the set Because the empty set has no elements, we write = 0

n.

(" r In Words (" We use the notation n(A) to r mean the number of elements in set A.

n(0 )

AA..

n(A)

=

If the number of elements in a set is a nonnegative integer, we say that the set is finite. Otherwise, it is infinite. We shall concern ourselves only with finite sets. Look again at Example A set with 3 elements has 23 = 8 subsets. This result can be generalized.

1.

SECTION 14.1

Counting

2/l subsets. d, e} has 25 = 32 subsets.

975

If A is a set with n elements, A has For example, the set {a, b, c ,

EXAM PLE 2

Analyzing Survey Data

35

In a survey of 100 college students, were registered in College Algebra, registered in Computer Science I, and 18 were registered in both courses.

52 were

(a) How many students were registered in College Algebra or Computer Science I? (b) How many were registered in neither course?

Solution

(a) First, let A = set of students in College Algebra B = set of students in Computer Science I Then the given information tells us that n(A) =

Figure 1 U n iversal set

�

VJ

nCB) =

52,

n ( A n B ) = 18,

Refer to Figure 1. Since n ( A n B) = 18, we know that the common part of the circles representing set A and set B has 18 elements. In addition, we know that the remaining portion of the circle representing set A will have 18 = 17 elements. Similarly, we know that the remaining portion of the circle repre senting set B has 18 = elements. We conclude that 17 18 + = 69 students were registered in College Algebra or Computer Science 1. (b) Since 100 students were surveyed, it follows that 100 69 = 31 were registered in neither course. •

52

31

35,

34

-

+

35

-

34

-

1b. "1" =Z>-

Now Work

PRO B L E M S 1 5

2

AND

25

The solution to Example contains the basis for a general counting formula. If we count the elements in each of two sets A and B, we necessarily count twice any elements that are in both A and B, that is, those elements in A n B. To count cor rectly the elements that are in A or B, that is, to find n(A U B ) , we need to subtract those in A n B from n ( A ) nCB).

+

THEOREM

Counting Form ula

If A and B are finite sets,

n ( A U B) = n ( A ) + n C B) - n ( A n B)

(1)

I�

� -----------------�

2. Using (1), we have n ( A U B) = n ( A ) + n C B ) - n ( A = 35+ 52 1 8

Refer back to Example

= 69

-

n B)

There are 69 students registered in College Algebra or Computer Science 1. A special case of the counting formula (1) occurs if A and B have no elements in common. In this case, A n B = 0, so n ( A n B) = O. THEOREM

Addition Principle of Counting

If two sets A and B have no elements in common, that is, if A n B = 0, then n ( A U B) = n ( A )

+n C B )

�----------------------------------�I� (2)

976

CHAPTER 1 4

Counting and Probabil ity

We can generalize formula (2). THEOREM

General Addition Principle of Counting

If, for n sets A1 , A2 , . . . , An , no two have elements in common, �---� ---�

Counting

E XA M P L E 3

In the year 2004, US. universities awarded 42,155 doctoral degrees. Table 1 lists the number of doctorates conferred by broad fields of study. Table 1 Broad Field of Study

Number of Doctorates

6049 5776 8819 6795 5467 6635 261 4

Physical s c i e n c e s Engineering Life s c i e n c e s Soci al sci ences H u m a n ities E d u c ation P rofessional/oth e r fi elds

Source: NSF/NIH/USED/NEH/USDA/NASA, 2004 Survey of Earned Doctorates

(a) How many doctorates were awarded by US. universities in physical sciences or life sciences? (b) How many doctorates were awarded by US. universities in physical sciences, life sciences, or engineering?

Solution

Let A represent the set of physical science doctorates, B represent the set of life science doctorates, and C represent the set of engineering doctorates. No two of the sets A , B, and C have elements in common since a single degree cannot be classified into more than one broad field of study. Then n( A)

=

6049

nCB) = 8819

n(C) = 5776

(a) Using formula (2), we have n ( A U B)

=

n (A ) + nCB)

=

6049 + 8819

=

14,868

There were 14,868 doctorates awarded in physical sciences or life sciences. (b) Using formula (3), we have n(A U B U C) = n(A) + nCB) + n(C) = 6049 + 8819 + 5776 = 20,644 There were 20,644 doctorates awarded in physical sciences, life sciences, or engineering. • urr;:== > .-

3


Solve Cou nting Problems Using the M u ltipl ication Principle

We begin with an example that illustrates the multiplication principle.

SECTION 14.1

EXA M P L E 4

Counting

977

Counting the Number of Possible M eals The fixed-price dinner at Mabenka Restaurant provides the following choices: Appetizer: soup or salad Entree: baked chicken, broiled beef patty, baby beef liver, or roast beef au jus Dessert: ice cream or cheese cake How many different meals can be ordered?

Solution

Ordering such a meal requires three separate decisions: Choose an Ap petizer

Choose an Entree

Choose a Dessert

2 choices

4 choices

2 choices

Look at the tree diagram in Figure 2. We see that, for each choice of appetizer, there are 4 choices of entrees. And for each of these 2 · 4 8 choices, there are 2 choices for dessert. A total of =

2·4·2 =

16

different meals can be ordered.

Figure 2

Appetizer

Entree

Dessert I ce cr eam Ch e es e cake I ce cr ea m Ch ee se cake I ce cr eam Ch eese cake I ce cr eam Ch ees e cake I ce cr eam Ch e es e cake I ce cr eam Ch e es e cake I ce cr eam Ch ees e c ake I ce cr eam Ch ees e cake

S o u p , chicken, ice cream Soup, chicken, cheese cake Soup, patty, ice cream Soup, patty, cheese cake Soup, liver, ice cream S o u p , liver, cheese cake Soup, beef, ice cream Soup, beef, cheese cake Salad, chicken, ice cream Salad, chicken, cheese cake Salad, patty, ice cream Salad, patty, cheese cake Salad, liver, ice cream Salad, liver, cheese cake Salad, beef, ice cream

•

Salad, beef, cheese cake

Example 4 demonstrates a general principle of counting. THEOREM

Multiplication Principle of Counting

If a task consists of a sequence of choices in which there are p selections for the first choice, q selections for the second choice, r selections for the third choice, and so on, the task of making these selections can be done in p'q'r'

different ways.

. . .

..J

978

CHAPTE R 1 4


Forming Codes

E XA M P L E 5

How many two-symbol code words can be formed if the first symbol is an upper case letter and the second symbol is a digit?

Solution

It sometimes helps to begin by listing some of the possibilities. The code consists of an uppercase letter followed by a digit, so some possibilities are AI, A2, B3, XO, and so on. The task consists of making two selections: the first selection requires choosing an uppercase letter (26 choices) and the second task requires choosing a digit (10 choices). By the Multiplication Principle, there are 26 · 10 = 260

different codewords of the type described. �= =-

Now Work

•

PRO B l E M 2 1

14. 1 Assess Your Understanding IAre You Prepared r Answers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red.

of A and B consists of all elements in either A or The B or both. (p. 2) of A with B consists of all elements in both A 2. The and B. ( p. 2)

1.

3. True or False The intersection of two sets is always a sub set of their union. (p. 2)

4. True or False If A is a set, the complement of A is the set of aLl the elements in the universal set that are not in A. (p. 3 )


5. The Counting Formula states that if A and B are two finite sets then

n(A U B) =

6 . True o r False The Multiplication Principle states that i f A and B are two finite sets then

n(A U B) = n(A) · nCB)

Skill Building 7.

8. Write down all the subsets of {a, b, c, d, e } .

Write down all the subsets of {a, b, c, d } .

9. If neAl 15, nCB) find n(A U B) . =

11. If n(A U B) find n e A l .

=

=

20, and n ( A n B)

50, n ( A n B)

=

=

10. I f n e A l = 30, n C B ) find n ( A n B ) .

10,

1 0, and nCB)

=

20,

13-20, use the information given in the figure. 13. How many are in set A ? 14. How many are in set B?

17.

How many are in A but not C?

19. How many are in A and B and C?

40, and n ( A U B ) = 45,

12. If n ( A U B) = 60, n ( A n B) find n e A l .

In Problems

15. How many are i n A o r B?

=

=

40, and n e A l

=

nCB),

u

16. How many are in A and B? 18. How many are not in A? 20. How many are in A or B or C?


21.

S h irts and Ties A man has 5 shirts and 3 ties. How many different shirt and tie arrangements can he wear?

22.

Blouses and Skirts A woman has 5 blouses and 8 skirts. H ow many different outfits can she wear?

23.

Four-digit Numbers How many four-digit numbers can be formed using the digits 0, 1 , 2, 3, 4, 5, 6, 7, 8, and 9 if the first digit cannot be o? Repeated digits are allowed.

24.

Five-digit Numbers How many five-digit numbers can be formed using the digits 0, 1 , 2, 3, 4, 5 , 6, 7, 8, and 9 if the first digit cannot be ° or 1? Repeated digits are allowed.

25.

Analyzing Survey Data In a consumer survey of 500 people, 200 indicated that they would be buying a major appliance within the next month, 150 indicated that they would buy a car, and 25 said that they would purchase both a major ap pliance and a car. How many will purchase neither? How many will purchase only a car?

26.

Analyzing S urvey Data In a student survey, 200 indicated that they would attend Summer Session I and 150 indicated Summer Session II. If 75 students plan to attend both summer sessions and 275 indicated that they would attend neither ses sion, how many students participated in the survey?

SECTION 14.2

27.

28.

. 29.

b----------,

In a survey of 100 investors in the stock market, 50 owned shares in I B M 40 owned shares in AT&T 45 owned shares in GE

Analyzing Survey Data

20 owned shares i n both IBM and GE 15 owned shares in both AT&T and GE 20 owned shares in both IBM and AT&T 5 owned shares in all three (a) How many of the investors surveyed did not have shares in any of the three companies? (b) How many owned just IBM shares? (c) How many owned just GE shares? (d) How many owned neither I B M nor GE? (e) How many owned either IBM or AT&T but no GE?

�

Marital Status M a rrie d

Widowed D ivorced Never m a rried

979

Number (in thousands)

62,486 2,643 8,954 29,561

Source: Current Population Survey 30.

The following data represent the marital sta tus of females 18 years old and older in 2004. (a) Determine the number of females 18 years old and older who are widowed or divorced.

Demographics

Marital Status

Human blood is classified as ei ther Rh + or Rh - . Blood is also classified by type: A, if it con tains an A antigen but not a B antigen; B, if it contains a B antigen but not an A antigen;AB, if it contains both A and B antigens; and 0, if it contains neither antigen. Draw a Venn diagram illustrating the various blood types. B ased on this classification, how many different kinds of blood are there? Classifying Blood Types

Demographics The following data represent the marital sta tus of males 18 years old and older in 2004. (a) Determine the number of males 18 years old and older who are widowed or divorced. (b) Determine the number of males 18 years old and older who are married, widowed, or divorced.

Permutations and Combinations

M a rried Widowed Divorced Never m a rried

Number (in thousands)

64,829 1 1 ,1 40 1 2,803 23,654

Source: Current Population Survey

(b) Determine the number of females 18 years old and older who are married, widowed, or divorced.

31.

As a financial planner, you are asked to select one stock e ach from the following groups: 8 DOW stocks, 15 NASDAQ stocks, and 4 global stocks. How many different portfolios are possible?

Stock Portfolios


32. Make up a problem different from any found in the text that requires the addition principle of counting to solve. Give it to a friend to solve and critique.

33. Investigate the notion of counting as it relates to infinite sets. Write an essay on your findings.


3. True

2. intersection

1. union

4. True

14.2 Permutations and Combinations PREPARING FOR THIS SECTION •


Factorial (Section 1 3 . 1 , pp. 931-933) . N o w Work the 'Are You Prepared?, problems on page OBJ ECTIVES

985.

1 Solve Counting Problems Using Permutations I nvolving

Disti nct Objects (p. 979)

2 Solve Counting Problems Using Combi nations (p. 982) 3 Solve Counting Problems Using Permutations I nvolving

Objects (p. 984)

1

n n Nond isti nct

Sol ve Co u nting Problems Using Pe rm utations I nvolving n D i stinct Obj ects

We begin with a definition. DEFINITION

A

permutation

is an ordered arrangement of r objects chosen from n objects ..J .

980

CHAPTER 14

Counting and Probability

We discuss three types of permutations:

1. The n objects are distinct (different), and repetition is allowed in the selection

of r of them. [Distinct, with repetition] 2. The n objects are distinct (different), and repetition is not allowed in the se lection of r of them, where r ::; n. [Distinct, without repetition] 3. The n objects are not distinct, and we use all of them in the arrangement. [Not distinct] We take up the first two types here and deal with the third type at the end of this section. The first type of permutation is handled using the Multiplication Principle.

EXA M P LE 1

Counting Airport Codes [Permutation: Distinct, with Repetition] The International Airline Transportation Association (lATA) assigns three-letter codes to represent airport locations. For example, the airport code for Ft. Lauderdale, Florida, is FLL. Notice that repetition is allowed in forming this code. How many airport codes are possible?

Solution

We are choosing 3 letters from 26 letters and arranging them in order. In the ordered arrangement a letter may be repeated. This is an example of a permutation with repetition in which 3 objects are chosen from 26 distinct objects. The task of counting the number of such arrangements consists of making three selections. Each selection requires choosing a letter of the alphabet (26 choices). By the Multiplication Principle, there are 26 · 26 · 26 = 263 = 17,576 possible airport codes.

•

The solution given to Example 1 can be generalized. THEOREM

Permutations: Distinct Objects with Repetition

The number of ordered arrangements of r objects chosen from n objects, in which the n objects are distinct and repetition is allowed, is n ' . !;O m

-.... Now Work

.J

PRO B l E M 3 3

We begin the discussion of permutations in which the objects are distinct and repetition is not allowed with an example.

EXAM P L E 2

Forming Codes [Permutation: Distinct, without Repetition] Suppose that we wish to establish a three-letter code using any of the 26 uppercase letters of the alphabet, but we require that no letter be used more than once. How many different three-letter codes are there?

Solution

Some of the possibilities are ABC, ABD, ABZ, ACB, CBA, and so on. The task consists of making three selections. The first selection requires choosing from 26 letters. Because no letter can be used more than once, the second selection requires choosing from 25 letters. The third selection requires choosing from 24 letters. (Do you see why?) By the Multiplication Principle, there are 26 · 25 · 24 = 15,600

different three-letter codes with no letter repeated.

•

SECTION 14.2

Perm utations and Combinations

981

For the second type of permutation, we introduce the following notation. The notation P(n, r) represents the number of ordered arrangements of r ob j ects chosen from n distinct objects, where r :s n and repetition is not allowed. For example, the question posed in Example 2 asks for the number of ways that the 26 letters of the alphabet can be arranged in order using three nonrepeated let ters. The answer is P ( 26, 3 ) = 26 · 25 · 24 = 15,600

E XA M P L E 3

Li n i ng Up People In how many ways can 5 people be lined up?

Solution

The 5 people are distinct. Once a person is in line, that person will not be repeated elsewhere in the line; and, in lining up people, order is important. We have a permutation of 5 objects taken 5 at a time. We can line up 5 people in P(5, 5 ) = 5 · 4 ·

3

· 2 · 1 = 120 ways

"-----v----' 5 factors

= = L]!!;: -

•


To arrive at a formula for P ( n, r ) , we note that the task of obtaining an ordered arrangement of n objects ill which only r :s n of them are used, without repeating any of them, requires making r selections. For the first selection, there are n choices; for the second selection, there are n - 1 choices; for the third selection, there are n - 2 choices; . . . ; for the rth selection, there are n - (r - 1 ) choices. By the Multiplication Principle, we have 1st

3rd

2nd

rth

P(n, r ) = n ' ( n - l ) · (n - 2 ) · . . ' [n - (r - 1 ) ] = n ' (n - l ) ' (n - 2 ) · . . ' (11 - r + 1 ) ·

·

This formula for P ( I1, r ) can be compactly written using factorial notation."' P ( n, r ) = n ' (n - 1 ) ' ( 11 - 2 ) · . . ' ( n - r + 1 ) ·

3

. .2. 1 (11 - r) ' -= n ' (n - 1) . (n - 2) . . . . . (n - r + 1) . """-----'-'3(n - r) ' ' ' 2'1 •

THEOREM

Permutations of r Objects Chosen from

without Repetition

n

. . .

n! ( 11 - r) !

Distinct Objects

The number of arrangements of n objects using r :s n of them, in which 1. the

11

objects are distinct, 2. once an object is used it cannot be repeated, and 3. order is important, is given by the formula n! ( 11 - r ) !

P(n, r)

L-

1, 1!

1 , 2!

2 ' 1 , . . , 17 !

/7 ( /7

- 1) · . . 3 2 1

(1)

'" Recall that

O!

=

=

=

=

·

·

·

·

I

��

__ __ __ __ __ __ __ __ __ __ __ __ __ __ __ __

.

982

CHA PTER 14

Co unt i ng and Probability

EXAM P L E 4

Computing Permutations Evaluate:

Solution

(a) P(7, 3 )

(b) P(6, 1 )

We shall work parts (a) and (b) in two ways.

I] (c) P(52, 5 )

(a) P(7, 3) = 7 · 6 · 5 = 210 '---v-----' 3 factors

or

=(7 7 ! 3 ) ! =74!! =7 · 6 M'· 5 · M' =210 (b) P(6, 1) 6 6 == P(7, 3 )

_

y 1 factor

Figure 3

52 nPt� 5 3 1 1 875200

or

=( 6 6 ! I ) ! = 516 ! = 6 st· st = 6

P( 6, 1 )

_

Figure 3 shows the solution using a TI-84 Plus graphing calculator. So P(52, 5 ) 3 1 1 ,875,200

=

(1!1! ! .."

....- Now Work

•

PROBLEM 7

The Birthday P roblem

E XAM P LE 5

All we know about Shannon, Patrick, and Ryan is that they have different birth days. If we listed all the possible ways this could occur, how many would there be? Assume that there are 365 days in a year.

Solution

This is an example of a permutation in which 3 birthdays are selected from a possible 365 days, and no birthday may repeat itself. The number of ways that this can occur is P ( 365, 3 )

=(365365 ! 3 ) ! =365 . 364 . 363 _

.36Zt

.

36-2t

=365 · 364 · 363 = 48,228,180

There are 48,228,180 ways in a group of three people for each to have a different birthday. • '11\

2

'"

,...


Solve Counting Problems Using Combinations

In a permutation, order is important. For example, the arrangements ABC, CAB, BAC, . . are considered different arrangements of the letters A , B, and C. In many situations, though, order is unimportant. For example, in the card game of poker, the order in which the cards are received does not matter; it is the combination of the cards that matters. .

DEFINITION

E XA M P L E 6

A combination is an arrangement, without regard to order, of r objects selected from n distinct objects without repetition, where r :5 n. The notation C(n, r) represents the number of combinations of n distinct objects using r of them . .-J

Listing Combinations List all the combinations of the 4 objects a, b, c, d taken 2 at a time. What is C (4, 2 ) ?

Solution

One combination of a , b , c , d taken 2 a t a time is ab

SECTION 14.2

Permutations a n d Combinations

983

We exclude ba from the list because order is not important in a combination (this means that we do not distinguish ab from ba). The list of all combinations of a, b, e, d taken 2 at a time is

ab, ae, ad, be, bd, ed

so

C (4, 2 ) = 6

•

We can find a formula for C(n, r ) by noting that the only difference between a permutation of type 2 (distinct, without repetition) and a combination is that we dis regard order in combinations. To determine C ( n , r ) , we need only eliminate from the formula for P(n, r ) the number of permutations that were simply rearrange ments of a given set of r objects. This can be determined from the formula for P(n, r ) by calculating P ( r, r ) r ! . So, if we divide P(n, r ) by r!, we will have the desired formula for C ( n , r ) : =

n! P ( n, r ) (n - r)! C( n, r ) = ----'r! l' r! Use formula

(1).

n! (n - r) !r!

We have proved the following result: THEOREM

Num ber of Combinations of

n

Distinct Objects Taken

r at a Time

The number of arrangements of n objects using r ::5 n of them, in which 1. the n objects are distinct, 2. once an object is used, it cannot be repeated, and 3. order is not important,

is given by the formula

�----------------------------------�I� () C(n, r ) =

n! (n - r) !r!

--

(2)

n r for the binomial coefficients are, in fact, the same. The Pascal triangle (see Section 13.5) can be used to find the value of C( n, r ) . However, because it is more practical and convenient, we will use formula (2) instead. . Based on formula (2), we discover that the symbol C( n, r ) and the symbol

EXAM P L E 7

Using Formula (2) Use formula (2) to find the value of each expression.

Solution

(a) C(3, 1 )

(b) C(6, 3 )

(a) C ( 3 , 1 ) =

3! (3 - 1 ) ! 1 !

(b) C (6, 3 ) =

=

(c) C ( n , n )

( d) C(n, O )

3! 3·2· 1 = = 3 2. 1 . 1 2!1!

6 · 5 · 4 · .M 6! S'5.4 = 20 = = 3 ! · .M S 3) !3! (6 _

1 .m' n! (c) C ( n, n ) = =-= -= 1 (n - n)!n! O!. nt 1

� (e) C ( 52, 5 )

984

CHAPTER 14

Figure 4

52 nCr 5


_

2598960

(d) C ( n, O ) -

n! _ .nt - 1. 1 .mO ! - 1 ( n - O) !O!

II (e) Figure 4 shows the solution using a TI-84 Plus graphing calculator. So C ( 52, 5 )

cm= ::::> =

E XA M P L E 8

Now Work

=

2,598,960

•

PRO B L E M 1 5

Forming Committees How many different committees of 3 people can be formed from a pool of 7 people? The 7 people are distinct. More important, though, is the observation that the order of being selected for a committee is not significant. The problem asks for the number of combinations of 7 objects taken 3 at a time.

Solution

C ( 7, 3 ) =

7! 4!3!

=

=7 · ffff · 5

7 · 6 · 5 · At At3 !

=

35 •

Thirty-five different committees can be formed.

Forming Committees

E XA M P L E 9

In how many ways can a committee consisting of 2 faculty members and 3 students be formed if 6 faculty members and 10 students are eligible to serve on the com mittee?

Solution

The problem can be separated into two parts: the number of ways that the faculty members can be chosen, C ( 6, 2 ) , and the number of ways that the student members can be chosen, C ( 1 0 , 3 ) . By the Multiplication Principle, the committee can be formed in C ( 6, 2 ) · C ( 1 0, 3 ) =

6! 10! . 4!2! 7 ! 3 !

=

6 · 5 . At 1 0 · 9 · 8 · .:tr ' At2! .:tr3 !

720 =230 ' 6 =1 800 ways = � :::::;;;... -

3

•


Solve Cou nting Problems Using Permutations I nvo lving n Nond istinct Objects

We begin with an example.

E XAM P L E 1 0

Forming Different Words How many different words (real or imaginary) can be formed using all the letters in the word REARRANGE?

Solution

Each word formed will have 9 letters: 3 R 's, 2 A's, 2 E's, 1 N, and 1 G. To construct each word, we need to fill in 9 positions with the 9 letters:

I 2 3 4 5 6 7 8

9

The process of forming a word consists of five tasks: Task 1 : Choose the positions for the 3 R 's. Task 2: Choose the positions for the 2 A's. Task 3: Choose the positions for the 2 E's.

SECTION 14.2


985

Task 4: Choose the position for the 1 N. Task 5: Choose the position for the 1 G. Task 1 can be done in C ( 9, 3 ) ways. There then remain 6 positions to be filled, so Task 2 can be done in C( 6, 2 ) ways. There remain 4 positions to be filled, so Task 3 can be done in C( 4, 2 ) ways. There remain 2 positions to be filled, so Task 4 can be done in C(2, 1 ) ways. The last position can be filled in C ( 1 , 1 ) way. Using the Multiplication Principle, the number of possible words that can be formed is C ( 9, 3 ) · C ( 6 , 2 ) ' C(4, 2 ) ' C(2, 1 ) ' C ( 1 , 1 )

=

9! .6-1' � 2t -U 3 ! · .&r 2 ! ' � 2 ! · 2t 1 ! · 'B. 0! ' 1 !

---- . ---- . ---- . ---- . ----

9! = 1),120 3 .1 . 2 .1 . 2 1. . 1 I. . 1 .I _

•

15,120 possible words can be formed.

The form of the expression before the answer to Example 10 is suggestive of a general result. Had the letters in REARRANGE each been different, there would have been P ( 9, 9 ) 9 ! possible words formed. This is the numerator of the answer. The presence of 3 R's, 2 A's, and 2 E's reduces the number of different words, as the entries in the denominator illustrate. We are led to the following result: =

THEOREM

Permutations Involving

n

Objects That Are Not Distinct

The number of permutations of n objects of which n1 are of one kind, n2 are of a second kind, . . . , and nk are of a kth kind is given by (3)

n!

where n

E XA M P L E 1 1

=

nl + /12 + . . . + nk '

.J

Arranging Flags How many different vertical arrangements are there of 8 flags if 4 are white, 3 are blue, and 1 is red?

Solution

We seek the number of permutations of 8 objects, of which 4 are of one kind, 3 are of a second kind, and 1 is of a third kind. Using formula (3) , we find that there are 8! 4! . 3 ! ' 1 ! �---

Now Work

8 · 7 · 6 · 5 · At . ----- = 280 different arrangements At · 3! ' 1 !

•

PROBLEM S1

14.2 Assess Your Understanding 'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.

1. O!

=

__

; I! =

__

. (p. 932)

2. True or False

n!

=

(n + I ) !

n

. (pp. 932-933)


3. A(n) is an ordered arrangement of r objects chosen from n objects. 4. A(n) is an arrangement of r objects chosen from n dis tinct objects, without repetition and without regard to order. __

__

5. True or False In a combination problem, order is not im pOI·tant. 6. True or False In some permutation problems, once an ob ject is used, it cannot be repeated.

986

CHAPTER

14


Skill Building

In Problems

'.

7-1 4,

find the value of each permutation.

. 7. P(6, 2 )

8. P(7, 2 )

9. P(4, 4)

10. P(8, 8 )

1 1 . P(7, 0)

12. P(9, 0)

13. P(8, 4)

14. P(8, 3 )

In Problems 15-22, use formula (2) to find the value of each combination. 17. C(7, 4) 15. C(8, 2 ) 16. C(8, 6 )

18. C(6, 2)

1 9. C( 1 5 , 1 5 )

20. C ( 1 8, 1 )

21. C(26, 1 3 )

22. C(18, 9)


23. List all the ordered arrangements of 5 objects a, b, c, d, and e choosing 3 at a time without repetition. What is P(5, 3 ) ? 24. List all the ordered arrangements o f 5 objects a, b, c , d , and e choosing 2 at a time without repetition. What is P(5, 2 ) ? 25. List a l l the ordered arrangements of 4 objects 1 , 2, 3, and 4 choosing 3 at a time without repetition. What is P( 4, 3 ) ? 26. List all the ordered arrangements o f 6 objects 1 , 2, 3, 4, 5, and 6 choosing 3 at a time without repetition. What is P(6, 3 ) ? 27. List all the combinations o f 5 objects a , b , c , d , and e taken 3 at a time. What is C(5, 3 ) ? 28. List a l l the combinations of 5 objects a, b, c, d, a n d e taken 2 at a time. What is C(5, 2)? 29. List all the combinations of 4 objects 1 , 2, 3, and 4 taken 3 at a time. What is C( 4, 3 ) ? 30. List all the combinations o f 6 objects 1 , 2, 3, 4, 5, and 6 taken 3 at a time. What is C ( 6, 3 ) ? 31. Forming Codes H o w many two-letter codes c a n be formed using the letters A, B, C, and D? Repeated letters are allowed. 32. Forming Codes How many two-letter codes can be formed using the letters A , B, C, D, and E? Repeated letters are allowed. 33. Forming Numbers How many three-digit numbers can be formed using the digits 0 and I ? Repeated digits are allowed. 34. Forming Numbers How many three-digit numbers can be formed using the digits 0, 1 , 2, 3, 4, 5, 6, 7, 8, and 9? Repeated digits are allowed. 35. Lining Up People In how many ways can 4 people be lined up? 36. Stacking Boxes In how many ways can 5 different boxes be stacked? 37. Forming Codes How many different three-letter codes are .. there if only the letters A, B, C, D, and E can be used and no letter can be used more than once? 38. Forming Codes How many different four-letter codes are there if only the letters A, B, C, D, E, and F can be used and no letter can be used more than once? 39. Stocks on the NYSE Companies whose stocks are listed on the New York Stock Exchange (NYSE) have their company name represented by either 1 , 2, or 3 letters (repetition of let ters is allowed). What is the maximum number of companies that can be listed on the NYSE? 40. Stocks on the NASDAQ Companies whose stocks are listed on the NASDAQ stock exchange have their company name represented by either 4 or 5 letters (repetition of letters is allowed). What is the maximum number of companies that can be listed on the NASDAQ?

41.

Establishing Committees In how many ways can a com mittee of 4 students be formed from a pool of 7 students?

42.

Establishing Committees In how many ways can a com mittee of 3 professors be formed from a department having 8 professors?

43.

Possible A nswers on a n·ue/False Test

44.

Possible Answers on a Multil>le-choice Test How many arrangements of answers are possible in a multiple-choice test with 5 questions, each of which has 4 possible answers?

45.

Arranging Books Five different mathematics books are to be arranged on a student's desk. How many arrangements are possible?

46.

Forming License Plate Numbers How many different li cense plate numbers can be made using 2 letters followed by 4 digits selected from the digits 0 through 9, if (a) letters and digits may be repeated? (b) letters may be repeated, but digits may not be repeated? (c) neither letters nor digits may be repeated?

47.

Birthday Problem In how many ways can 2 people each have different birthdays? Assume that there are 365 days in a year.

48.

Birthday Problem In how many ways can 5 people each have different birthdays? Assume that there are 365 days in a year.

49.

Forming a Committee A student dance committee is to be formed consisting of 2 boys and 3 girls. If the membership is to be chosen from 4 boys and 8 girls, how many different com mittees are possible?

50.

The student relations committee of a college consists of 2 administrators, 3 faculty members, and 5 students. Four administrators, 8 faculty members, and 20 students are eligible to serve. How many different com mittees are possible?

How many arrange ments of answers are possible for a true/false test with 10 questions?

Forming a Committee

SECTION 14.2

"

51. Forming Words

58. Baseball

52. Forming Words

59. Baseball Teams

How many different 9-letter words (real or imaginary) can be formed from the letters in the word ECONOMICS? How many different 1 1 -letter words (real or imaginary) can be formed from the letters in the word MATHEMATICS?

54. Selecting Objects

55.

An urn contains 15 red balls and 10 white balls. Five balls are selected. In how many ways can the 5 balls be drawn from the total of 25 balls: (a) If all 5 balls are red? (b) If 3 balls are red and 2 are white? (c) If at least 4 are red balls?

Senate Committees The U.S. Senate has 100 members. Sup pose that it is desired to place each senator on exactly 1 of 7 possible committees. The first committee has 22 members, the second has 1 3 , the third has 10, the fourth has 5, the fifth has 16, and the sixth and seventh have 17 apiece. In how many ways can these committees be formed?

A baseball team has 15 members. Four of the players are pitchers, and the remaining 1 1 members can play any position. How many different teams of 9 players can be formed?

60. World Series

61.

62.

In the World Series the American League team ( A ) and the National League team ( N ) play until one team wins four games. If the sequence of winners is designated by letters (for example, NAAAA means that the National League team won the first game and the American League won the next four), how many different sequences are possible? Basketball Teams A basketball team has 6 players who play guard (2 of 5 starting positions). How many different teams are possible, assuming that the remaining 3 positions are filled and it is not possible to distinguish a left guard from a right guard? Basketball Teams On a basketball team of 12 players, 2 only play center, 3 only play guard, and the rest play forward (5 players on a team: 2 forwards, 2 guards, and 1 center). How many different teams are possible, assuming that it is not pos sible to distinguish left and right guards and left and right forwards?

63. Combination Locks

A combination lock displays 50 num bers. To open it, you turn to a number, then rotate clockwise to a second number, and then counterclockwise to the third number. (a) How many different lock combinations are there? (b) Comment on the description of such a lock as a comb ination lock.

56. Football Teams

A defensive football squad consists of 25 players. Of these, 10 are linemen, 10 are linebackers, and 5 are safeties. How many different teams of 5 linemen, 3 line backers, and 3 safeties can be formed?

57.

987

In the National Baseball League, the pitcher usu ally bats ninth. If this is the case, how many batting orders is it possible for a manager to use?

53. Selecting Objects

An urn contains 7 white balls and 3 red balls. Three balls are selected. In how many ways can the 3 balls be drawn from the total of 10 balls: (a) If 2 balls are white and 1 is red? (b) If all 3 balls are white? ( c) If all 3 balls are red?


Baseball In the American B aseball League, a designated hitter may be used. How many batting orders is it possible for a manager to use? (There are 9 regular players on a team.)


64. Create a problem different from any found in the text that re

quires a permutation to solve. Give it to a friend to solve and critique. 65. Create a problem different from any found in the text that re quires a combination to solve. Give it to a friend to solve and critique.


1; 1

2. False

66.

Explain the difference between a permutation and a combi nation. Give an example to illustrate your explanation.

988

CHAPTER 14

Cou nting a n d Proba bility

1 4.3 Probability OBJECTIVES

1 Construct Proba bility Models (p. 988)

2 Compute Proba bilities of E q u a l ly Li kely Outcomes (p. 990) 3 Find Probabi lities of the U nion of Two Events (p. 992) 4 Use the Complement Rule to Find Probabilities (p . 993)

is an area of mathematics that deals with experiments that yield random results, yet admit a certain regularity. Such experiments do not always produce the same result or outcome, so the result of any one observation is not predictable. However, the results of the experiment over a long period do produce regular pat terns that enable us to predict with remarkable accuracy. Probability

Tossing a Fair Coin

E XA M P LE 1

In tossing a fair coin, we know that the outcome is either a head or a tail. On any particular throw, we cannot predict what will happen, but, if we toss the coin many times, we observe that the number of times that a head comes up is approximately equal to the number of times that we get a tail. It seems reasonable, therefore, to assign a probability of comes up. 1

� that a head comes up and a probability of � that a tail

•

Construct Probabil ity Models

The discussion in Example 1 constitutes the construction of a probability model for the experiment of tossing a fair coin once. A probability model has two components: a sample space and an assignment of probabilities. A sample space S is a set whose elements represent all the possibilities that can occur as a result of the experiment. Each element of S is called an outcome. To each outcome, we assign a number, called the probability of that outcome, which has two properties: 1. The probability assigned to each outcome is nonnegative. 2. The sum of all the probabilities equals 1 .

DEFINITION

A

probability model

with the sample space

where el , e2 , · · · , en are the possible outcomes and P (el), P ( e2), . . . , P (en) are the respective probabilities of these outcomes, requires that

+

+. ..+

= P(el ) P (en) = 1 P ( e2) (2) i=l �---� ---�

n

L P( ei)

EXAM P L E 2

Determ ining P robability Models In a bag of M&Ms,nvl the candies are colored red, green, blue, brown, yellow, and orange. Suppose that a candy is drawn from the bag and the color is recorded. The sample space of this experiment is { red, green, blue, brown, yellow, orange } . Determine which o f the following are probability models.

SECTION 14.3

(a)

(c)

Outcome

0.3

red

0.1

0.1 5

g reen

0.1

blue

0

blue

0.1

brown

0.1 5

brown

0.4

yellow

0.2

yellow

0.2

orange

0.2

orange

0.3

Outcome

(d )

Probability 0.3 -0.3

Outcome

989

Probability

red

red

Probability

red

0

g reen

0

blue

0.2

blue

0

brown

0.4

brown

0

yellow

0.2

yellow

orange

0.2

orange

0

(a) This model is a probability model since all the outcomes have probabilities that are nonnegative and the sum of the probabilities is 1 . (b) This model is not a probability model because the sum of the probabilities is not 1 . (c ) This model is not a probability model because P (green) i s less than O . Recall, all probabilities must be nonnegative. (d) This model is a probability model because all the outcomes have probabilities that are nonnegative, and the sum of the probabilities is 1 . Notice that P (yellow) = 1, meaning that this outcome will occur with 100% certainty each time that the experiment is repeated. This means that the bag of M&MsTM has only yellow candies. • "' 'I!l: = �-

E XA M P L E 3

Outcome

g reen

g reen

Solution

(b)

Probability

Probabil ity

Now Work P R O B l E M 7

C onstructing a Probability Model An experiment consists of rolling a fair die once. A die is a cube with each face having either 1 , 2, 3, 4, 5, or 6 dots on it. See Figure 5 . Construct a probability model for this experiment.

Figure 5

Solution

A sample space S consists of all the possibilities that can occur. Because rolling the die will result in one of six faces showing, the sample space S consists of S = { 1 , 2, 3, 4, 5, 6}

Because the die is fair, one face is no more likely to occur than another. As a result, our assignment of probabilities is P(l) P(3) P(5)

1 -

6 1 6

1 -

6

P(2) P(4) P(6)

1 -

6 1 6 1 6

•

990

CHAPTER 14


Now suppose that a die is loaded (weighted) so that the probability assignments are P ( l ) = 0,

P ( 2) = 0 ,

P(3) =

�,

P ( 4) =

�,

pe S ) = 0,

P( 6 ) = °

This assignment would be made if the die were loaded so that only a 3 or 4 could occur and the 4 is twice as likely as the 3 to occur. This assignment is consistent with the definition, since each assignment is nonnegative and the sum of all the proba bility assignments equals 1 . �==>- Now Work P R O B L E M 2 3

E XA M P LE 4

Constructing a Probability Model An experiment consists of tossing a coin. The coin is weighted so that heads (H) is three times as likely to occur as tails (T). Construct a probability model for this ex periment. The sample space S is S = {H, T} . If x denotes the probability that a tail occurs,

Solution

peT) = x

and

P ( H ) = 3x

Since the sum of the probabilities of the possible outcomes must equal 1 , we have peT)

+ P ( H ) = x + 3x = 1

4x = 1 1 x =4

We assign the probabilities peT) = (i'I!Il --= = ;:;" '-

Now Work

�,

P(H) =

�

•

PROBLEM 2 7

In working with probability models, the term event is used to describe a set of possible outcomes of the experiment. An event E is some subset of the sample space S. The probability of an event E, E *- 0, denoted by p e E ) , is defined as the sum of the probabilities of the outcomes in E. We can also think of the probability of an event E as the likelihood that the event E occurs. If E = 0, then p e E ) = 0; if E = S, then p e E ) = p e S ) = 1 . 2

THEOREM

Co m p ute Proba b i l ities of Equally Likely Outcomes

When the same probability is assigned to each outcome of the sample space, the experiment is said to have equally likely outcomes. Proba bility for Equally Likely Outcomes

If an experiment has n equally likely outcomes and if the number of ways that an event E can occur is m, then the probability of E is peE) =

Number of ways that E can occur m =n Number of possible outcomes

(3)

If S is the sample space of this experiment, peE) =

neE)

(4)

I

�----------------------------------�� n(S)

SECTION 14.3

EXAM P LE 5

Probability

991

Calculating P robabilities of Events I nvolving Equally Likely Outcomes Calculate the probability that in a 3-child family there are 2 boys and 1 girl. As sume equally likely outcomes.

Solution Figure 6

We begin by constructing a tree diagram to help in listing the possible outcomes of the experiment. See Figure 6, where B stands for boy and G for girl. The sample space 5 of this experiment is 5 = {BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG }

1 st child 2 n d child 3rd child B

B B

BBB BB

B

B B

so n ( 5 ) = 8. We wish to know the probability of the event E: "having two boys and one girl." From Figure 6, we conclude that E = {BBG, BGB, GBB } , so n ( E ) = 3. Since the outcomes are equally likely, the probability of E is

B

P(E) =

1.�'''''3

/

/

/ _/

-

3

-10

•

2 Exercises In Problems 1-16, graph each equation using the following viewing windows: (a) Xmin =-S Xmax = S

Xmax =

Xsci = 1

Ymin = -4

-8 Ymax = 8 Yscl = 1

Yscl = 1 l. y=x+2 5. y= 2x

+

2

9. y=x2+ 2 13. 3x+ 2y= 6

10 Xsci = 2 Ymin -8 Ymax = 8

10

Xsci = 1 Ymax= 4

(c) Xmin = -10

(b) Xmin = -10

Xmax =

Ymin =

(d) Xmin = -S Xmax = S Xscl = 1 Ymin = -20

=

Ymax = 20

Yscl = 2

Yscl = S

2. Y = x - 2

3. Y = -x+ 2

6. y = 2x - 2

7. Y =-2x+ 2

-2

11. y=-x2+2

10. y = x2

14. 3x - 2y= 6

15. -3x+2y

17-32. For each of the preceding equations 1-16, create a table, -3 ::::; x ::::;

=

3, and list

4. Y =-x - 2 8. y=-2x - 2 12. y=-x2 - 2 6

16. -3x - 2y= 6

points on the graph.

A6

APPENDlX

Graphing Utilities

3 Using·a Graphing Utility to Locate Intercepts and Check for Symmetry ,

Value and Zero (or Root) Most graphing utilities have an eVALUEate feature that, given a value of x, deter mines the value of y for an equation. We can use this feature to evaluate an equa tion at x = ° to determine the y-intercept. Most graphing utilities also have a ZERO (or ROOT) feature that can be used to determine the x-intercept(s) of an equation. Finding Intercepts Using a Graphing Utility

EXAMPLE 1

Solution Figure

Use a graphing utility to find the intercepts of the equation y Figure lO(a) shows the graph of y = x3 - 8.

10

10

10

-5

l

1-....

/---

.I

I l

5 -5

"

I

�/

r--'

X=Q

-20

(a)

I

'1'= -&

=

x3 - 8.

10

i

1

5 -5

/ f---'"'

Z�t·1) M=;::

.II

5

'i=(1

-20

-20

(b)

(c)

The eVALUEate feature of a TI-84 Plus graphing calculator accepts as input a value of x and determines the value of y. If we let x = 0, we find that the y-intercept is -8. See Figure lO(b). The ZERO feature of a TI-84 Plus is used to find the x-intercept (s). See Fig ure 10(c). The x-intercept is 2.

•

Trace Most graphing utilities allow you t o move from point t o point along the graph, dis playing on the screen the coordinates of each point. This feature is called TRACE.

EXAMPLE 2

Using TRACE to Locate Intercepts

x3 - 8. Use TRACE to locate the intercepts. Figure 11 shows the graph of y = x3 - 8. Graph the equation y

Solution Figure

11

=

10

/

-5

1-_/

5

f--

.1

-20

Activate the TRACE feature. As you move the cursor along the graph, you will see the coordinates of each point displayed. When the cursor is on the y-axis, we find that the y-intercept is -8. See Figure 12.

SECTION 3

Figure

10

12 -5

l ...-�f,-j

i/

x=(-

Using a Graphing Utility to Locate Intercepts and Check for Symmetry

1

,/

A7

5

'I'=-B

-20

Continue moving the cursor along the graph. Just before you get to the x-axis, the display will look like the one in Figure 13(a). (Due to differences in graphing utilities, y our display may be slightly different from the one shown here.) Figure

13

1

l / l'

1 · 1· ·r'

" ,

-

�:=1.!l1'1B!l36

1

j

�J'l

/I�

'i= - !l7B'13'1� .

�=;;:.(';;:1;;:766

(a)

'r-"jl

\'=.;;:S:B(I'1'1!1�

(b)

In Figure 13(a), the negative value of the y-coordinate indicates that we are still be low the x-axis. The next position of the cursor is shown in Figure 13(b). The posi tive value of the y-coordinate indicates that we are now above the x-axis. This means that between these two points the x-axis was crossed. The x-intercept lies between 1.9 148936 and 2.0212766.

•

EXAMPLE 3

Graphing the Equation y =

.! x

1 � With the viewing window set as

Graph the equation:

y

=

Xmin Xmax Xscl

Solution

�"

4

�

J\

.-...----

.

-3

=

-3 3 1

Ymin Ymax Yscl

= =

=

-4 4 1

use TRACE to infer information about intercepts and symmetry.

Figure 14

Y1 =

= =

---.l..--...

"\

3

Figure 14 illustrates the graph. We infer from the graph that there are no intercepts; we may also infer that symmetry with respect to the origin is a possibility. The TRACE feature on a graphing utility can provide further evidence of symmetry with respect to the origin. Using TRACE, we observe that for any ordered pair (x, y) the ordered pair ( -x, -y) is also a point on the graph. For example, the points (0.95744681, 1.0444444) and ( -0.95744681, -1.0444444) both lie on the graph.

•

-4

3 Exercises /11 Problems 1-6, use ZERO (or ROOT) to approximate the smaller of the two x-intercepts of each equation. Express the answer rounded

to two decimal places.

1. y = x

2

+ 4x + 2

2 4. Y = 3x + Sx + 1

2. Y = x 5. Y =

2

+ 4x - 3

2x2

- 3x - 1

3. Y = 2X 6. y

=

2

+ 4x + 1

2 2x - 4x - 1

A8

APPENDIX

Graphing Utilities

In Problems 7-14, use ZERO (or ROOT) to approximate the positive x-intercepts of each equation. Express each answer rounded to two decimal places. 7. y =

x3

9. y =

X

4

8. y =

+ 3.2x2 - 16.83x - 5.31 - 1.4x3 - 33.71x2 + 23.94x + 292.41

11. y =

7TX3 - (8.887T

+

1)x2 - (42.0667T - 8.88)x

12. y =

7TX3 - (5.637T

+

2)x2 - (108.3927T - 11.26)x

13. y =

x3

+

19.5x2 - 1021x

+

x3 4

10. y =

X

14. Y =

x3

+ 3.2x2 - 7.25x - 6.3 +

1.2x3 - 7.46x2 - 4.692x

+ 15.2881

42.066 +

216.784

+ 1000.5

+ 14.2x2 - 4.8x - 12.4

4 Using a Graphing Utility to Solve Equations For m any equations, there are no algebraic techniques that lead to a solution. For such equations, a graphing utility can often be used to investigate possible solutions. When a graphing utility is used to solve an equation, usually approximate solutions are obtained. Unless otherwise stated, we shall follow the practice of giving approx imate solutions rounded to two decimal places. The ZERO (or ROOT) feature of a graphing utility can be used to find the solutions of an equation when one side of the equation is O. In using this feature to solve equations, we m ake use of the fact that the x-intercepts (or zeros) of the graph of an equation are found by letting y = 0 and solvi ng the equation for x. Solving an equation for x when one side of the equation is 0 is equivalent to finding where the graph of the corresponding equation crosses or touches the x-axis. Using ZERO (or ROOT) to Approximate Solutions

EXAMPLE 1

of an Equation

Find the solution(s) of the equation x 2 - 6x + 7 imal places.

O. Round answers to two dec

The solutions of the equation x 2 - 6x + 7 = 0 are the same as the x-intercepts of the graph of Y1 = x 2 - 6x + 7. We begin by graphing Y 1. See Figure 15(a). From the graph there appear to be two x-intercepts (solutions to the equation): one between 1 and 2, the other between 4 and 5.

Solution

15

Figure

8

1

-1

=

�� \

\."

-2

,I /1/ .f .....""-"J".

(a)

8

8

1

/ /

\

7

-1

\ \...\ 2�I'(O

, .-

X=1.�B�(,B6't _'1'=0

-2

(b)

\

/ /

.)1

�\

7

-1

2�I'(O

�\

/

�

l

7

X='t.'t1't

1 3. C 217,. 4. similar 5. T 6. T 7. 8. T 9. T 10. F 11. 13. 26 15. 25 17. 1. right; hypotenuse 2. A "2bh triangle; 5 19. Not a right triangle 21. Right triangle; 25 23. Not a right triangle 25. 8 in 2 27. 4 in 2 29. A = 2517 m2 ; C 1017 m 256 cmJ;, 6417 cm- 35. V = 6487T in.J; 30617 in. - 37. 17 square units 39. 217 square units 31. = 224 ft3; 232 ft2 33. = -7T 3 41. x = 4 units; A = 90°; 60°; C 30° 43. x 67.5 units; A 60°; 95°; C = 25° 45. About 16.8 ft 47. 64 ft2 49. 24 + 27T "" 30.28 ft2; 16 217 "" 22.28 ft 51. 160 paces 53. About 5.477 mi 55. 100 ft: 12.2 mi; From 150 ft: 15.0 mi =

F

=

13

Right

=

V

S

V

=

B

S

=

=

7

=

=

+

R.4 Assess Yo u r U nderstand i n g

(page 4 7)

'

=

B

S

7

=

=

From

1. 4; 3 2. - 16 3. x3 - 8 4. 5. T 6. 7. Monomial; variable: x; coefficient: degree: 3 9. Not a monomial; the exponent of the variable is not a nonnegative integer 11. Monomial; variables: x, y; coefficient: -2; degree: 3 13. Not a monomial; the exponent of one of the variables is not a nonnegative integer 15. Not a monomial; it has more than one term 17. Yes; 19. Yes; 0 X4

F

F

2;

2

AN1

ANSWERS Section RA

AN2

21. No; the variable of one of the terms is not a nonnegative integer 23. Yes; 3 25. No; the polynomial of the denominator has a degree greater than 0 27. x2 + 7x + 2 29. x3 - 4x2 + 9x + 7 31. 6x5 + 5x4 + 3x2 + x 33. 7x2 - x - 7 35. -2x3 + 18x2 - 1 8 37. 2x2 - 4x + 6 39. 1 5i - 27y + 3 0 41. x 3 + x 2 - 4x 43. -8xs - 10x2 45. x3 + 3x2 - 2x - 4 47. x2 + 6x + 8 49. 2X2 + 9x + 1 0 51. x2 - 2x - 8 53. x2 - 5 x + 6 55. 2X2 - x - 6 57. -2x2 + 1 1 x - 12 59. 2X2 + 8x + 8 61. x2 - xy - 2i 63. -6x2 - 1 3xy - 6i 65. x2 - 49 67. 4x2 - 9 69. x2 + 8x + 1 6 71. x2 - 8x + 1 6 73. 9x2 - 16 75. 4x2 - 1 2x + 9 77. x2 - i 79. 9x2 - i 81. x2 + 2xy + i 83. x2 - 4xy + 4i 85. x3 - 6x2 + 12x - 8 87. 8x3 + 12x2 + 6x + 1 89. 4x2, - 1 1 x +1 23; remainder5 -451 91. 4x - 3; remainder x + 1 93. 5x2 - 13; remainder x + 27 95. 2x2; remainder -x2 + X + 1 97. - 2x + 2; remal.llder 2 x + 2 99. -4x2 - 3x - remalllder -7 101. x2 - x - I ; remainder 2x + 2 103. x2 + ax + a-;, remainder 0 .:l;

r

�

R.S Assess You r Unde rstanding

.

(page 56)

1. 3x(x - 2 ) ( x + 2 ) 2. Prime 3. T 4. F 5. 3(x + 2) 7. a(x2 + 1 ) 9. x(x2 + X + 1 ) 11. 2x(x - 1 ) 13. 3xy(x - 2y + 4) 15. (x + 1 ) (x - 1) 17. (2x + 1 ) (2x - 1) 19. (x + 4 ) ( x - 4) 21. (5x + 2) (5x - 2) 23. (x + 1 )2 25. (x + 2)2 27. (x - 5)2 29. (2x + 1 )2 31. (4x + 1 )2 33. (x - 3 ) (x2 + 3x + 9) 35. (x + 3 ) (x2 - 3x + 9) 37. (2x + 3 ) (4x2 - 6x + 9) 39. (x + 2 ) ( x + 3 ) 41. ( x + 6)(x + 1 ) 43. (x + 5 ) ( x + 2) 45. (x - 8)(x - 2) 47. (x - 8 ) ( x + 1 ) 49. (x + 8)(x - 1 ) 51. (x + 2) (2x 3 ) 53. (x - 2) (2x + 1 ) 55. (2x + 3 ) (3x + 2 ) 57. (3x + 1 ) (x + 1 ) 59. (z + 1 ) (2z + 3 ) 61. (x + 2) (3x - 4 ) 63. (x - 2) (3x + 4 ) 65. ( x + 4)(3x + 2 ) 67. (x + 4) ( 3x - 2) 69. (x + 6)(x - 6) 71. 2 ( 1 + 2x) ( 1 - 2x) 73. (x + 1 ) (x + 10) 75. (x - 7)(x - 3 ) 77. 4(x2 - 2x + 8) 79. Prime 81. -(x - 5)(x + 3 ) 83. 3(x + 2)(x - 6) 85. i C y + 5)(y + 6) 87. (2x + 3 )2 89. 2 (3x + l ) (x + 1 ) 91. (x - 3 ) ( x + 3 ) (x2 + 9 ) 93. (x - 1 )2(x2 + X + I f 95. xs(x - 1 ) (x + 1 ) 97. (4x + 3)2 99. - (4x - 5 ) (4x + 1 ) 101. (2y - 5 ) (2y - 3 ) 103. - (3x - 1 ) (3x + l ) ( x2 + 1 ) 105. (x + 3)(x - 6) 107. (x + 2)(x - 3 ) 109. (3x - 5 ) (9x2 - 3x + 7) 111. (x + 5) (3x + 1 1 ) 113. (x - 1 ) (x + l ) (x + 2) 115. (x - 1 ) (x + 1 ) ( x2 - X + 1) 117. 2(3x + 4) (9x + 13) 119. 2x(3x + 5 ) 121. 5(x + 3 ) ( x - 2f(x + 1 ) 123. 3 (4x - 3 ) (4x - 1 ) 125. 6(3x - 5 ) (2x + 1 f(5x - 4) 127. The possibilities are (x l ) (x 4) x2 ± 5x + 4 or (x 2 ) ( x 2) x2 ± 4x + 4, none of which equals x2 + 4. +

±

±

±

±

=

=

R.6 Assess Your U n derstanding

(page 60)

R. 7 Assess Your Understanding

(page 69)

quotient; divisor; remainder 2. -3)2 0 -5 1 3. T 4. T 5. x2 + X + 4; remainder 12 7. 3x2 + I l x + 32; remainder 99 9. X4 - 3x3 + 5x2 - 1 5x + 46; remainder - 1 38 11. 4xs + 4X4 + x3 + x2 + 2x + 2; remainder 7 13. 0.1x2 - O.l1x + 0.321; remainder -0.3531 15. X4 + x3 + x2 + X + 1; remainder 0 17. No 19. Yes 21. Yes 23. No 25. Yes 27. -9 1.

1. lowest terms 2. least common multiple 3. T 4. F 5. x -3 3 7. -x3 9. 2x4x- l 2(yy ++ 51 ) xx +- 5I 15. -(x + 7) 3f 17. 5x(x3 - 2) 19. 2X(X2 x+ +4x4 + 16) 21. -3x8 23. xx +- 73 25. (x - 24x) (x - 3) 27. 5(x 4 1) 29. (x 4x- 4)2 31. (x(x -+ 3)2 -:: 3x - 2 x+9 33. (x(x -- l4)) (2x( x ++ 3l)) 35. x +2 5 37. (x -2x2 )-( x3+ 2) 39. -41. -43. 4x -- x2 45. (x -2(xl )+(x5+) 2) x-3 2x - l - 2) x + 2) 53. (x - 2)(x + 2 ) ( x + 1 ) 55. x(x - l ) (x + 1 ) 57. x3(2x - 1 )2 51. x(x2(x2 47. (x3x2+ -l )2x(x -- 31 ) 49. (x- ( ll2)(x - 2) - 2)(x + 2) + 5x - 2) 5x + l 59. x(x - 1 )2 (x + l ) ( x2 + x + 1) 61. (x 6)(x 5x 1 ) (x + 4) 63. (x -2(2x2 65. --:- --::2)(x + 2)(x + 3) (x - 1 )2(x + 1 )2 (x - l ) (x + 1 ) x+l x3 - 2x2 + 4x + 3 -x2 + 3x + 1 3 - 71. -X-'--(�l-'-177. ( x 2(Sx2)(x- +l ) 1 )2 67. -----,---- 69. --::. ) 73. x - I 75. 2x(2x + 1 ) + 17-:( - 2 ) ( x + 1 ) (x + 4) x2(x + l ) (x - 1 ) x(3x + 2) (x + 1-:-) ( x - I ) - 2) 19 91. (x +(x32 )+(3x1 )-2 1 ) 81. x --1 I 83. 23 xx -+ 1 85. ----:- 87. __-- ,--__,--_ 89. 79. (x +-2x(x2 (3x - 5 ) 2 (x2 + 1 ) 2 (3x + 1 ) 2 2 ) ( x-, - x - 3 ) 93 f ( - 1RI) ( R• R,I + R ) ' -152 m --

--

-c--:-:---,-

--

11.

13.

-

_

--

--

--

__ -

-

+

-

x

-

--

.

=

n

-

2

-

_

_

-

_

1

--

'

R.8 Assess Y o u r U n derstanding

(page 77)

3. index 4. T 5. cube root 6. F 7. 3 9. -2 11. 2 \12 13. -2x� 15. x3i 17. x2y 19. 6vX 21. 6xvX 23. lSV3 25. 12V3 27. 7\12 29. \12 31. 2V3 33. -'V2 35. x 2vX + 1 37. (2x - l ) � 39. (2x - 1 5 ) Th ViS ( 5 + \12 ) V3 5 V4 2 x + h - 2 Vx2 - xh \12 - 47. - 53. 49. 8Vs41- 19 51. 45. -41. - (x + 5y) � 43. 2 2 23 h 5 + 2) 27\12 27 \12 8x5/4 3x + 2 67. X7/12 69. xi 71. x2/3y 73. y3/4 75. ( 1 + X) I/2 77. x(3x2 55. 4 57. - 3 59. 64 61. 63. 65. 32 32 ( x2 + 1 ) 1/2 -

1 2'7

ANSWERS Section 1 . 1

AN3

5 1 " 89. -()X + 2 ) (x 1 )1/2 81. 2( 12 ++ xxy'/2 83. (x4+-4)x3/'- 85. r(x79. l O Vx22x- 5+� " I- 1 ) I/,- 87. 2 VXIx (-1 3x2 2 + r)91. 2X /2( 3x - 4)(x + 1) 93. (x-, + 4) 1/3 (l1r' + 1 2 ) 95. (3x + 5) '(2x + 3) 1/'-( 1 7x + 27) 97. 3(x2x+l/22) 99. 1 .4 1 101. 1 .59 7rV36- '" 0.91 sec 103. 4.89 105. 2 . 1 5 107. (a) 15,660.4 gal (b) 390.7 gal 109. 2 '" 8.89 sec 111. -� == =� == =

+

_

II'

I

,� V 27r

Review Exercises

(page 8 7)

1. { 1 , 3, 5, 6, 7, 8} 3. {3, 7 } 5. {2, 4, 6, 8, 9 } 7. { 1 , 2, 4, 5, 6, 9} 9. (a) None (b) { - 1 0} { -1O, 0.65 , 1 .343434 · · , i } 21. 5 23. - 1 0 (d) { v/} (e) { - 1 0, 0.65, 1 .343434 . . . , v/,i} 11. 14 13. % 15. 3 17. 4x - 12 19. 3 25. -49 27. 5 29. {xix 6} 31. x 3 \12 units; A 90°; 45°; C 45° 33. 5.0625 35. Coefficients: 3,4, -2, 0, 5, - 1 2; degree: 52 37. 2X4 - 2x3 + x2 + 5x 3 39. 6x2 - txy - 5l 41. 1 6x2 - 1 43. x3 - 7x - 6 45. 3x2 + Sx + 25; remainder 79 47. -3x + 4; remainder -2 49. X4 - x3 + x2 - X + 1 ; remainder 0 51. (x + 7)(x - 2) 53. (3x + 2 )(2x - 3) 55. 3(x + 2)(x - 7) 2x + 7 57. (2x + 1)(4x2 - 2x + 1) 59. (2x + 3 ) (x - 1 ) (x + 1) 61. (5x - 2)(5x + 2) 63. Prime 65. (x + 4)2 67. --:;:-=-z 1 25 69. (x +3(3x3)(3x- 1 +) 1 ) 71. (x + 14x)(x - 1 ) 73. (xx2-+2 )1(7xx ++ 2)-2 , 75. xx -_ I 77. 4\12 79. -2 2 81. 2 V2 83. x1 y 85. -,x-y 2 ( 1 + x2) x(3x + 1 6) 4 v'5- 93. -2(1 + \12) 95. - 3 + v'5 87. )-3xy2 89. 3xy vx 91. -- 97. 5 2 (2 + x-) 1/2 99. 2(x + 4)'' /2 101. x2vx2 - 1 107. Yes 109. $0.35 per share 111. 167r '" 50.27 ft2; 1 0 7r '" 31.42 103. (x2 + 4)1 /3(llx2 + 12) 105. 2.8142 1906 (c)

II

1=

+

=

B

=

=

=

,1 � v

'l.

�

4 .3 ,

,�

,

x

Chapter Test

10 8

ft

(page 84)

1. (a) {7} (b) { 0, 7 } { O, 1 .2, 7, � } (d) { V2 , 7r} (e) { O, 1.2, V2 , 7, �, 7r} 2. (a) - 1 6 (b) 18 3. x 2;2 A 40°; 95°; C 45° 4. -8x3 1 1 x2 - 14x + 1 1 (b) -lOx2 + 1 9x - 6 5. (a) (x - 4)(x - 2) (b) ( 2x - 5)(2x 5 ) (3x 1 )( 2x - 7) 6. x2 - X + 6 ; remainder 2 ,, 10. 0; b 0 7. (a) 3 3 (b) -2 (c) I"x (d) 64x2 8. 1 5 -223 V3 9. -3( 11. 2 1 6 square feet; 84 feet of fence (c)

=

B

=

s

=

=

+

. r:. V

(a)

(c )

+

(c) 5

(d) - 174

+

a

y

-

=

=

CHAPTER 1 Equations a n d I n e q u a l ities 1 .1 Assess You r U nderstan d i n g

(page 94)

4. F 5. identity 6. linear; first degree 7. 8. T 9. {3} {-5} 13. a} 15. {%} {-2} 19. {3} 21. {-1 } 23. {-2} 25. {-IS} 27. { -4} 29. {-%} 31. {-20} 33. {2} 35. { 0.5} 37. g� } 39. { 2 } 41. {8} 43. {2} 45. {-I} 47. {3} 49. No solution 51. No solution 53. {-6} 55. {34} 57. {-��} 59. {-I} 61. { _ 161} 63. {-6} 65. {5.9 1 } 67. {0.41 } 69. x = b : c x ! 73. x mv2 81. ----- 83. $ 1 1 ,500 wiLl be invested in bonds and $S500 in CDs. 85. Yahoo! was 75. = 3 77. 79. F used for 1 .52 billion searches;2 Google was used for 2.05 billion searches. 87. The regular hourly rate is $S.50. 89. Brooke needs a score of 85. 91. The original price was $500,000; purchasing the model saves $75,000. 93. The bookstore paid $68 . 1 5 for the book. 95. There were 2 187 adults. 97. The length is 19 ft; the width is ft. 99. There were about 694.06 million Internet users worldwide. F

17.

11.

a

R

=

R I R,-

RI

+

R

R

=

r =

S

S

71.

a

=

a

a

C

=

b

a2

11

H i storical Problems

(page 7 06)

1. The area of each shaded square is 9, so the larger square will have area 85 + 4( 9 ) 1 2 1 . The area of the larger square is also given by the expression (x + 6)2, so (x + 6)2 121 . Taking the positive square root of each side, x 6 1 1 or x = 5. 2. Let -6, so Z2 12z - 85 = -121. We get the equation u2 - 12 1 0 or u2 Thus u = ± 1 1 , so x = ±11 - 6. x -1 7 or x = 5. z =

+

=

=

=

=

121.

+

=

=

AN 4

3.

ANSWERS Section 1 .2

( x + �r = ( Yb22: 4ac r ( x + :a ) 2 _ ( Yb22: 4aC ) Z 0 --_ ( X + �2a _ Yr-bZcO;2a--4a-c ) ( X + �2a + Ybz2a- 4aC ) 0 ( X + b - Y2ab2 - 4aC ) ( X + b + Y2ab2 - 4aC ) 0 - 4ac or x -b - Yb2 - 4ac x -b Ybz 2a 2a =

=

=

+

=

=

(page 1 06)

5. add; ( % r 21 6. discriminant;negative 7. 8. 9. {0, 9} 11. {-5, 5} 13. {-3, 2} 15. {- �, 3} 17. {-4, 4} 19. {2, 6} 21. a} 23. {-�, �} 25. {-�,�} 27. {-%,2} 29. {-5, 5} 31. {-1, 3} 33. {-3, 0} 35. 16 37. 116 39. i- 41. {-7, 3} 43. { - 4'I 43 } 45. { -l -6 V7 ' -1 +6 V7 } 47. {2 - \1'2, 2 + \1'2} 49. {2 - \IS, 2 + \IS} 51. { 1, 23 } 53. No real solution 55. { -I � \IS, - l : \IS } 57. {o,n 59. U} 61. {-�, l} 63. 3 -1� , 3 +1� 65. - 2 -2 V1O , - 2 +2 V1O 67. { 1 - 8\133' 1 + 8\133 } 69. f1 9 - 2v73 ' 9 + 2v73 71. {0.63, 3.47} 73. {-2.80, 1.07} 75. {-0.85, 1.17} 77. {-8.16, -0.22} 79. {-\IS, \IS} 81. { 41 } 83. { - S3 ' 25 } 85. { - 2'1 32 } 87. { -\1'22 2 ' -\1'22 - 2 } 89. { -I -2 VI? ' -1 +2 VI? } 91. {5} 93. No real solution 95. Repeated real solution 97. Two unequal real solutions 99. Academic year 2009-2010 101. The dimensions are 11 ft by 13 ft. 103. The dimensions are 5 m by 8 m. 105. TI1e dimensions should be 4 ft by 4 ft. 107. (3) TI1e ball strikes the ground after 6 sec. (b) The ball passes the top of the building on its way down after 5 sec. 109. The dimensions should be 11.55 cm by 6.55 cm by 3 cm. 111. The border will be 2.71 ft wide. 113. TI1e border will be 2.56 ft wide. 115. The screen of a 37-inch TV in 4:3 format has an area of 657.12 square inches; the screen of a 37-inch TV in 16 : 9 format has an area of 584.97 square inches. The traditional TV has a larger screen. 117. 36 consecutive integers must be added. 1 1 - 4ac + -b - Ybz - 4ac = -2b = - -b 121. k -or 119. -b + Ybz 2a 2a 2a a 2 k - -2 - 4ac 125. (b) 123. ax-. + bx + c = 0'"r = - b ± Y2ab2 - 4ac .' ax2 - bx + c 0, x b Y�(--2ab-)""z -"""--4-a-c b Y2ab2 - 4ac -b ± Ybz 2a 1 .2 Assess You r U nderstand i n g

F

=

F

)

)

) \

1

+

=

?

±

=

1 .3 Assess You r U nderstanding

(page 1 1 6)

=

±

= -------

4. real; imaginary; imaginary unit 5. {-2i, 2i} 6. F 7. T 8. 9. 8 + 5i 11. -7 + 6i -6 - 11i 15. 6 - 18i 6 + 4i V3 19. 10 - 5i 21. 37 23. S6 + S8i 25. 1 - 2i 27. 25 - 2i7 29. - 21 + T i 31. 2i 33. -i 35. i 37. -6 39. -10i 41. -2 + 2i 43. 0 45. 0 47. 2i 49. 5i 51. 5i 53. {-2i, 2i} 55. {-4, 4} 57. {3 - 2i, 3 + 2i} 59. {3 - i, 3 + i} 61. { "4I - "4l i' "4l + "4I i } V3 V3.} 67. {2, -1 - V31,. -1 + V31.} 69. {-2, 2, -2i, 2i} 71. {-3i, -2i, 2i, 3i} i, - 21 + Tl 63. { SI - S2i, S1 S2 i } 65. { - 2I - T 73. Two complex solutions that are conjugates of each other 75. Two unequal real solutions 77. A repeated real solution 79. 2 - 3i 81. 6 83. 25 85. 2 + 3i ohms 87. z + (a + bi) + (a - bi) 2a; z - (a + bi) - (a - bi) 2bi 89. z + w (a + bi) + (c + di) = (a + c) + (b + d)i (a + c) - (b + d)i (a - bi) + (c - di) + F

13.

17.

+

=

z

=

1 .4 Assess Y o u r U nderstan d i n g

=

=

z

=

=

=

=

(page 1 22)

z

W

4. extraneous 5. quadratic in form 6. T 7. {1} 9. No real solution 11. {-13} 13. {4} 15. {-I} 17. {0, 64} 19. {3} 21. {2} 23. {- �} 25. {8} 27. {-1,3} 29. {1, 5} 31. {I} 33. {5} 35. {2} 37. {-4, 4} 39. {0, 3} 41. {-2, -1, 1, 2} 43. {-I, I} 45. {-2, I} 47. {-6, -5} 49. { - 3I } 51. { - 2'3 2 } 53. { 0, 16I } 55. {16} 57. {I} :::9. {( 9 - 8VI? )4, ( 9 +;m)4} 61. {\1'2, V3} 63. {-4, 1} 65. {-2, -�} 67. {-�,�} 69. { - l 27 } 71. {-2' -�} 73. {-3,0,3} 75. {o, %} 77. {-5,0,4} 79. {-I, I} 81. {-2,2, 3} 83. {-2, �, 2} 85. H} 87. {o, %, 3} 89. {0.34, 11.66} 91. {-1.03, 1.03} 93. {-1.85, 0.17} 95. {l.5 , 5} 97. The depth of the well is 229.94 ft. 99. 220.7 ft �

Section 1 . 6

ANSWERS

1 . 5 Assess You r U nderstand i n g

AN 5

(page 13 2)

3. negative 4. closed interval 5. Multiplication Properties 6. T 7. T 8. T 9. F lO. T 11. [0, 2]; 0 oS x oS 2 13. [2, (0); x ;2: 2 15. [0, 3); 17. 19. 7 > 0 -1 > -8 12 > -9 (d) -8 < 6 21. 2x + 4 < 5 29. ( -00, -4) 27. [4, (0) 23. 25. o oS x < 3 (a) 6 < 8 (b) -2 < 0 (c) 9 < 1 5 (d) -6 > -10 (b) 2x - 4 < -3 (c) 6x + 3 < 6 (d) -4x - 2 > -4 [0, 4) [4, 6 ) [ ) • 1 . [ 6 4 4 0 -3 < x < -2 2 oS x oS 5 [ 1 , ) . f

33.

31.

5

2

39. < 41. > 43. ;2: 45. 53. {xix < 4} or (-00, 4)

47. 49. > 51. ;2: 55. {xlx ;2: - I } or [- 1 , (0) oS

4

71. { x 1 32

1

•

5

3

77. {xix < -5} or (

2 '3

-00,

oS

x oS 3

59. {xIX ;2: 2} or [2, (0) [ 2

65. {xix < -20} or (-00, -20) �--+)----20

} or [ ] 73. { 1 11

1 •

2 3' 3

I } or ( 11 1 ) 75.

x - 2" < x < "2 ) I

3

"2

-

.

2

' "2

4

3"

{xl-6 < x < O} ) o

-6

79. { xIX ;2: - I } or [- 1 , (0)

-5)

-4

-3

4

2

69. {xl3 oS x oS 5} or [3, 5)

[

3

1

'3

37. x < -3

----�f--�

63. {_+ oS � } or (-oo,�]

-7

)

4

35. x ;2: 4

-I

61. {xix > -7} or (-7, 00)

[

57. {xix > 3 } or (3,00)

[

)

(a)

(c)

(b)

-2

-3

3} or (3, (0) 10

3

"3

89. a 3, b 5 91. a - 12, -8 93. a 3, b 11 95. a �, b 1 97. a = 4, b 16 99. { xlx ;2: -2} 101. 21 < Age < 30 103. Male ;2: 79.66 years Female ;2: 83.58 years A female can expect to live 3.92 years longer. 105. The agent's commission ranges From $45,000 to $95,000, inclusive. As a percent of selling price, the commission ranges from 5% to 8.6%, inclusive. 107. The amount withheld varies from $98.30 to $148.30, inclusive. 109. The usage varies from 675.4 1 kW hr to 2500.91 kW hr, inclusive. 111. The dealer's cost varies from $ 1 5,254.24 to $ 1 6,07 1.43, inclusive. 113. You need at least a 74 on the fifth test. You need at least a 77 on the fifth test. a+ b a+ b a+ b 2b - a - b b- a a b a b - 2a b- a -- > O' therefore, b > -115. -- a -- > O' therefore, a < --. b 2 2 2 2 ' 2 2 2 2 ' 117. ( VI.ib)2 - a2 ab - a2 = a(b - a) > 0; thus ( VI.ib)2 > a2 and VI.ib > a. - ( VI.ib? b2 - ab b(b - a) > 0; thus b2 > ( VI.ib? and b > VI.ib. b(b - a) a(b - a) 2ab b2 - ab 2ab ab - a2 - a > 0; thus < b. > 0; thus > a. b 119. - a -b - -- = --- = a + b a + b a + b a + b a + b a + b a-b a b 1 1 1 1 121. Sl.I1ce 0 < a < b, then a - b < 0 and -< O. So - - - < 0, or - - - < O. Therefore, - < -. And 0 < - because b > O. ab ab ab b a b a b =

(a)

=

=

=

=

=

.

.

(a)

+

=

=

=

(c)

(b)

+

(b)

=

=

-- =

=

Ii h

b

=

=

=

h

II

= --- =

=

h

=

1

1 .6 Assess Yo u r U nderstan d i n g

(page 1 3 7)

3. {-5, 5 } 4. {xl-5 < x < 5 } 5. T 6. T 7. {-3, 3 } 9. {-4, I } 11. { - I , n 13. {-4, 4} 15. {2} 17. { _ 272 ' 272 } 19. { _ 356 , 254 } 21. No solution 23. {-�, �} 25. {-3, 3 } 27. {- I , 3 } 29. {-2, -I , 0, I } 31. H' 4 } 33. {-�, O } 35. { xl-4 < x < 4 } ; (-4, 4) 39. {xiI < x < 3 } ; ( 1 , 3 ) 37. { xix < -4 0r x > 4 } ; (-00, -4) U (4, 00) f

-4

)

4

,

)

-4

43. { XIX � or x oS

[ 2

-3

J '

2

f ·,

4

1 I

2:

;2:

% } ; ( �] U [ %, )

•,

-00,

00

) ,

3

45. { x l - 1 < x < �} ; ( -1 , � ) f -]

AN6

ANSWERS Section 1 . 6

47. {xix < -l orx > 2}; (-00, -1) or (2, 00) 49. No solution 2. 55. No solution 53. {xl-l :5 x :5 2}; [-1,2] 2 -1 59. {xJ- � < x < %}; (-� , �) 61. All real numbers; (- 00, (

-1

[

.

o

1 •

) � 4

3

2:

57. All real numbers; (- 00, (0) I o

o

,

••

(0

I.

)

J.

o

63. Ix - 98.61 :2 1.5 ; x :5 97.1 or :2 100.1 65. Ix - 13.41 < 1.35; between 12.05 and 14.75 books per year are read. 67. Ix - 31 2;1 25 < x < 27 69. Ix + 31 > 2; x < -5 or x > -1 71. a = 2, b = 8 73. a = -15, b = -7 75. a = -1, b = - ]51 b - a = ( Vb - Va) ( Vb + Va). Since Vb - Va > 0, Va > 0, Vb > 0, then b - a > 0, so a < b. 79. ( + b)2 a2 2ab + b2 la l2 + 2 1al l bl + I bl2 = ( Ia l + Ibl )2; thus, la bl :5 la l + Ib l . 81. x2 - a < 0; (x - Va)(x + Va) 0; therefore,-Va < x < Va. 83. {xl-l < x 2} 91. {-1,5} x

O }; (0, 00) All real numbers -x i f x :5 0 -r if -l :5 x :5 0 43. le x) = { , 41. f(x) = { 1 -x + 2 if 0 < x :5 2 (Other answers are possible.) z x if 0 < x :5 2 (Other answers are possible. ) 45. 2 3 -4 47. (a) $35 $61 $35.40 49. $67.43 51. For schedule $477.04 O.lOx if 0 < x :5 7550 , {1.15955 + 9. 4 5 if 0 x :5 50 X 755 + 0.15(x 7550) if 7550 < x :5 30,650 C 0. 9 1025x + 21.915 if x 50 4220 0. 2 5(x 30,650) 650 < x :5 74,200 (d) le x) = 15,107.50 + 0.28(x - 74,200) ifif 30, C 74, 2 00 < x 154,800 (100, 112.94) 37, 6 75. 5 0 + 0. 3 3(x 8 00) if 154, 8 00 < x :5 336,550 154, ., 100 97,653.00 + 0.35(x - 336,550) if x 336,550 (d)

�

(3) (3)

(b)

(c)

(b)

(b) (c)

:5

_

>

2!l

� =§ -::: '0 6 '-'

50 (0, 9.45)

:2

X:

+

:5

>

"' �

u�

(c)

(50,67.43) x L...J.-'--'-'---5 .. -'-0---'-L...J.--'-.. .I. 0""10-Gas Usage (therms)

53. (a)

{X

y 300

'?

,§o

(960, 270)

� 1 50

8

y

1 00 50

Distance (miles)

=

C 50 O.4(x - 100) C 170 0.25(x - 400)

(b)

+

=

�

+

"

AN1 7

:s:

70 i3 60 � � 50 � � 40 � � 30 § � 20

- �::-'--::-"':-:-x ��'--="�-'-4"" S00 960 0-' """ 60:-:: 0::0--'--:-:

(c)

Section 3.5

0 x < 10 1O :s: x < 500 30 i f 500 :s: x < 1000 50 if 1000 :s: x < 1500 70 if 1500 :s: x if

10 if

55. f(x) =

(SOD, 270)

250 200

ANSWERS

10

x

1 00 500 Bill (dollars)

10

1000

1 500

(d) -4°e (e) The wind chill is equal to the air temperature. 20 m/sec, the wind chill factor depends only on the air temperature. 59. 0.39 O < x :s: 1 0.63 < x :s: 2 0.87 2 < x :s: 3 3<x 4 1.1 1 '§ l.35 4 < x :s: 5 1.59 5 < x :s: 6 C( x ) 1 .83 6 < x :s: 7 8 2.07 7 < x :s: 8 2.3 1 8 < x :s: 9 2.55 9 < x :s: 10 2.79 10 < x :S: 3.03 < x :S: 12 3.27 12 < x :S: 1 3 61. Each graph is that of y x2, but shifted horizontally. If y (x - k )2, k > 0, the shift is right k units; if y (x + k ) 2, k > 0, the shift is left k units. 63. The graph of y f( x) is the reflection a bout the x-axis of the graph of y f (x). 65. Yes. The graph of y (x - 1)3 + 2 is the graph of y x3 shifted right 1 unit and up 2 units. 67. They all have the same general shape. All three go through the points (-1, -1), (0,0), and ( 1 , 1 ). As the exponent increases, the steepness of the curve increases (except near x 0).

100e

57. (a)

(b)

4°e

(c)

-3°e

(I) At wind speed greater than

c

l

" �n

:s:

:: ::

=

o--e

o--e

o--e

o--e

o--e

o--e

o--e

o--e

1 .35

11

=

=

o--e

3.27 3.03 2.79 2.55 2 .3 1 2.07 I .S3 1 .59

11

o--e

1.11 0.S7 0.63 0.3

o--e

0

2

=

o--e

4

6

S

x

10

12

14

Weight (ounces)

=

=

=

-

=

3.5 Assess You r U nderstand i n g

(page 26 1)

H

y 3. -5; -2; 2 4. T 5. F 6. T 7. B 9. 11. I 13. L 15. F 23. -x3 25. 4x3 27. ( l ) y v'X + 2; (2) y -(v'X + 2); (3) -(� + 2) 29. (1) - v'X ; (2) y -v'X + 2; (3) y - Vx+3 + 2 31. (c) 33. (c) 1. horizontal; right 2. Y =

Y =

35.

Y =

=

=

y

=

=

37.

2

Y =

x

x

45.

- 4)3

21. Y =

y

41.

-5

---'--_--'--_+---'_---;';---* x

S

y 10

47.

(4 , S )

(0, 0 )

19. Y = ( x

-5

(1. 4)

-2 _ 1

G

-2

5

-2

y 9

17.

y 5

39.

(0, - I )

43.

=

( -8, 2le - l

- 10

,1

49.

x

o 1 , - 1 )( 8, - 2 )

y 10

(-8, 2) (-1 , 1 ) -1 0

x - 10

-10

x3 + 4

AN 1 8

Section 3.5

ANSWERS

55.

53.

51.

(1 , 5)

-5

x

x

5

-2 59.

x

5

-5

5

( 1 , 0)

x

F(x) f(x) + 3

(b)

=

f": -5

H, ,"

,

,

�'

G(x) f(x 2)

, , , , i

5

( , 0)

2 x

-7 ( -6,

3

-2)

( e)

Q(x) �f(x)

(0

=

Y 5

�

(2, J )

�I (4, 0) 5

x

..

�t

( - 'IT, 3 )

(e)

'iT

_'!!:

2

-2)

-5

G(x) = f(x + 2 )

(e)

2

(o r) '

11'

2

-1

1 .

y

2

H , '- �) -2

(

"!! 1 2' 2

)

x

x

(1, 0) 5

(3, -2)

�

(-5, -4) _ 5

x (-2, -2)

P (x) = -f(x)

(d)

''IT

�

, - 1) -2 i ( � (g)

l1(x) = f(2x) Y

2, (� , 1)

2

x

( � , - 1)

H(x) f(x + 1 ) - 2 =

Y

Y.j.

Y

-2

x

H , I) ,

g(x) = f(-x) (_ � , I)

( - 1, 0) -5

h(x) = f(2x)

H - 2, - 1 ) -2 (I)

Q(x) = 2f (x)

Y 3

2

( 'IT, 3 )

x

��

H (x) = f(x + 1) - 2

-5 I

_ ( � , 4)

(-f2 ) - 'IT

,

-5

(d)

=

-5 (b)

-5

P(x) -f(x)

(- 4 2)

(g)

x

F(x) = f (x) + 3

x

,

Y 5

(4,

67. (a)

"

x

Y 5

- 5 ( - 4, 0)

-5

"

-5

g(x) = f( -x )

(-2, 2)

-1

,x

T

-5

-3

8

(-4,

0) I �' '!b.1 -5 (- 1, ) ( 0 , - 2)

Y

(e)

+

=

Y 5

Y 7

(4 , 3 )

-2

Y 5

-5

-5 65. (a)

(2

I I, V I I I I

63.

61.

Y 5

l� �

57.

- 11'

�

_�

H , -I)

-2

�

11'

-2'IT

-'IT

(-'IT - L - 2(-1 lT

,

-2)

27T ( 'IT - I ,

x

-2)

ANSWERS

69. (a)

-7 and l

(b)

-3 and S

(c)

(d) - 3 and 5 71. (a) ( -3, 3 )

-S and 3

73. (a)

(-2, 1) (- 1 , 1)

��

(1 , 1 )

-J 79 [(x)

.

2

(-1, I) (-2, 0)

(2, 0 ) � 2

8 1 . [(x)

�t

(�, �) 2

x

(2, 3)

Xl

T

0-

E

�

80 76 72 68 64 60 56 o

-)

3)

5

(4, - 15)

Y

IlfI

85.

I I 5

-

c =

x

fh

-3 -8) ' ' (-1, -8) (c)

-6

� AM

PM

A M,

T,

80 76 72 68 64 60 56

G:' i!.... � 3 '" v

0.

5

I-

4

8

12

1 6 20 24 28

o

4

8

12

16

20 24 28

Time (hours after midnight)

288 256 224 192 160 128 96 64 32

F

OF I

�1 ;'0 2'0'3'0 410 510 6'0 7'0 810 910 100

91. (a )

288 256 224 192 160 128 96 64 32 o

C

(c)

o

2000 -15,000

(373, 212)

330 350 370

10% tax Y1 is the graph of p(x) shifted down vertically 10,000 units. Y2 is the graph of p(x) vertically compressed by a factor of 0.9. (d) 1 0% tax

(b)

Y

-2

TIle time at which the temperature adjusts between the daytime and overnight settings is moved to 1 hI' sooner. It begins warming up at S:OO instead of 6:00 and it begins cooling down at 8:00 instead of 9:00

Time (hours after midnight)

89.

-2

( -2, - 5 )

-5

The temperature decreases by 2° to 70°F during the day and 63°F overnight.

:J

" .

�I I I

-

I I I I IX

= -3(x + 2)2 - 5

83. [ (x)

(3, I )x

-5

I

t:7[10. 0) -5 ,

= 2(x - 3)2 + 1

87. (a) nOF; 6soF

�

I I I

-

Y

- 51 - ? n

2

-I

e"

Y+

-2

-2

E

(c)

( 1 , 1)

-2

= ( .r. + -21 Y + �4� Yl

(- � , � ) (- � , � )

(b)

(4,10) Decreasing on (-l, S) (d) Decreasing on (-S, I ) 75. [ (x) = (x + 1)2 - 1 77. [ (x) = (x 4) 2 15

(b) Y+

(b)

AN 1 9

Section 3.5

K

PM.

AN20

ANSWERS

Section 3.6

3 . 6 Assess Your U nderstan d i n g 1.

(page 2 6 7)

d(x) = Vx2 - X + 1 (b) 2 ....----------,

d(x) = Vx4 - 15x2 + 64 (b) d(O) = 8 d(l) = V50 "" 7. 0 7 (a)

3. (a)

(c)

40

(d)

- 10

lLL -5

(c)

10

d is smallest when x "" -2. 7 4 or x "" 2.74. 9. (a) A(x) = 4x� A is largest when x "" 1.41. 10 (e)

1 A(x) = -x4 2

5.

A(x) = x(16 - x2) (b) Domain: {xiO < x < 4} The area is largest when x "" 2.3 1. 30

(a)

(c)

�

2 Oo d is smallest when x = 0.5. A(x) = x2 + 25 - 20x + 4x2 (d) Domain: {xiO < x < 2.5} A is smallest when x "" 1.40 m.

4x + 4V4 - x2 (d) p is largest when x "" 1.4 1. 12 (b) p(x) =

(c)

7.

11. (a)

r.

(c)

8

� !� ....-.

.. - _-. .

o

o

A(x) = :: 15. (a) A(r) = 21'2 19. (a) d(l) V250012 - 3601 13 d is smallest when "" 0.07 hI'.

13. ( a)

C( x ) = x (b)

(b)

p(r) = 61'

+

=

A(x) = (� - � ) x2 r.J-/ ( R - 1')1' 2 --=--- --'----21 . V(r) R 12 - x '\0+4 23. (a) T(x) = 5 + 3 {xi O :5 x :5 12} 3.09 hr (d) 3. 5 5 hr -

-

(b)

5

(c)

Review Exerc i ses

(page 2 7 1)

3(x - 2) 6x 3x (d) Function; domain { -I, 2, 4}, range {O, 3} 3 . (a) 2 -2 ,..2 1 (e) x2 - 4x + 3 4x2 1 x2 - 4 (d) x2 - 4 x(x - 4 ) Vx2 - 4x 2� (a) 0 (b) 0 -v?""--=-4 (d) -Vx2 - 4 (x 2)2 x2 x2 9. {xi x *- -3, x *- 3} 11. {xIX :5 2} 13. {xix > O} 15. {xi x *- -3, x I} 19. (f + g)(x) = 3x2 + 4x 1 ; Domain: all real numbers (f + g)(x) 2x + 3; Domain: all real numbers (f - g)(x) 3x2 - 2x + 1; Domain: all real numbers (f - g) (x) -4x + 1; Domain: all real numbers (f . g )(x) 9x3 + 3x2 + 3x; Domain: aU real numbers (f·g)(x) -3x2 + 5x + 2; Domain: all real numbers 1 2 x (-fg ) (x) = 3t2- + r + l ; Domain: {xix *- O} (Lg ) (x) = 3x + ; Domain: {xix --3 } x2 + 2x - 1 ; Domalll:. { x I x *- 0, } 23. -4x - 2h 21 . (f + g)(x) = x ( x - I) (f - g)(x) x(�x2 +- 1I ) ; Domain: {xi x *- 0, I } (f . g)(x) = . (xx + 11 ) ; Domain: {xix *- 0, I} ( fg ) ( ) x(xx -+ I1 ) ; Domalll:. {xix *- 0, I } (b)

1.

(c )

(I)

(e )

(c )

A

=

=

=

x

=

+

=

*-

-

.n �

J

x

(e)

*-

=

1

-

_

(I)

_

(c )

7.

=

17.

-

25

17.

=

{

(b)

o

+ 1

_

_

5. (a)

0

(b)

0

ANSWERS

25. (a) Domain: ro

{x l -4 :5 x :5 3 } ; Range: {y l -3 :5 y :5 3} w y 4

(b)

(e)

(0.0) y 4

-1

-4

(d)

39.

I\ / U

(I) x-intercepts:

40

-2

(-4, 3 )

37.

-3

=

5

41. (a) 43. - 5 47. No

55.

y

(4, 4)

(1, 1)

x

-5

(0, 0 )

(4, 2) 5

61.

(-4, 0), (4, 0), (0, -4) {yly -4} [-4, 00 )

( 1 , 0)

-5

-3

63.

(0, 1 )

-5

5

(1,0) Domain: {xix I } or [1, 00 ) Range: {yl y 2: O} o r [0, 00 )

x

x

-5

(e)

Intercept: (0,3) (0, 1 ), (1,0) Domain: {xi x :5 I} or ( - 00 , 1] Domain: all real numbers Range: {yl y O} or [0, 00) Range: {yl y 2} or [2, 00 ) (0,0) 69. {xi x -41 or [-4, 00) (0, 1) 2:

2:

67. (a) { x i x > -21 0r ( -2, 00 )

(b)

y 5

(a)

2:

(e)

(b)

y

(2, 3 ) -5

(-2 , -6)

(d) { yl y >

5

x

5

-5

x

( -4, -4) -5

-6} or (-6, 00 )

(d)

x

-8

(0, 0) {y l y O}

Intercept: Domain: all real numbers :5 or ( - 00, 0] Range: y

-5

{y l -4 y < 0 or y > O} o r [-4, 0) (0, 00) :5

U

(0, -2)

-)

-5

Intercepts:

2:

y 2

65.

-5

Intercept:

47

x

x

y 5 ( -3, 2)

(e)

7 - 17

(b) 45. 49. Yes

(0, 0)

Intercepts: Domain: all real numbers Range: 2: or y 5

23

57.

y 5

-5

-5

59.

3

(-0.91, 4.04) (0.9 1, -2.04) ( -3, -0.91); (0.9 1,3) (-0.91, 0.91)

Local maximum: Local minimum: Increasing: Decreasing:

-20

53.

y

x

(3, -3)

#

(0.4 1, 1.53) (-0.34, 0.54); (1.80, -3.56) (-0.34, 0.4 1); ( 1.80,3) (-2, -0.34); (0.41, 1 .80)

3

4

20

Local maximum: Local minima: Increasing: Decreasing:

I

y

-4 29. O d d 31. Even 33. Neither 35. Odd

AN 2 1

{xiO < x :5 3}

-2, 0,4; y-intercept: 0

-4

51.

(h)

5

(d) No symmetry (e) Neither

(e)

x

10

-3)-4 Domain: {xl-4 :5 x :5 4} or [-4, 4] Range: {y l -3 :5 Y :5 I } or [-3, 1 ] Increasing on (-4, - 1 ) and (3,4); Decreasing on ( - 1 , 3 ) Local maximum is 1 and occurs at x -1. Local minimum i s -3 and occurs a t x 3. =

(e)

(6, 3 )

x

-5 (- 1 ,

27. (a)

(b)

Chapter 3 Review Exercises

5

x

(0, -2), (1 - �, 0) or about (0.3,0)

Intercepts:

Domain: all real numbers Range: all real numbers

AN22

Chapter 3 Review Exercises

ANSWERS

= 11 73. If the radius doubles, the volume of the new sphere is 8 times as large as the original sphere, and the surface area of the new sphere is 4 times as large as the original sphere. 200 75. (a) = 27T1·2 + 71.

A

A

r

(b) 123.22 ft2

(c) 1 50.53 ft2

(d) 1 97.08 ft2

(e) 500

r.:".....--,

\�//1 ..

oo

A is smallest when

r

10

"" 2.52 ft.

(page 2 74)

C h apter Test

{

1. (a) Function; domain: {2, 4, 6, 8 } ; range: {5, 6, 7, 8}

range: { y l y ;;o: 2}

2. Domain:

x i x :5

�};f(-I)

=

(b) Not a function

3. Domain: { x i x *" -2 } ; g( - I )

3

(d) Function; domain; all real n umbers;

(c) Not a function =

1

4. Domain: { x i x *" -9, x *" 4 } ; h ( - I )

=

�

5. (a) Domain: { x l - 5 :5 x :5 5 } ; range: { y l -3 :5 y :5 3 } (b) (0, 2), ( -2, 0), and (2, 0) (c) [( 1 ) = 3 (d) x = -S and x = 3 -2 or 2 x :5 S } or [-5, -2) U (2, S] (e) { x l -S :5 x 6. Local maxima: [( -0.8S) "" -0.86; [(2.3S) "" IS.5S; Local minima: [(0) = -2; the function is increasing on the intervals ( -S, -0.8S) and (0, 2.3S) and decreasing on the intervals ( -0.8S, 0) and (2.3S, S). (b) (0, -4), (4, 0) 8. 19 7. (a) y y = x - 4, X 2 - 1 9. (a) - g) = 2x2 - 3x 3 3 (c) g(-S) = - 9

3 } ; ( - 00 , -4) or (3, 00 ) 19. N o real solution

21.

(b)

25. (a) { - 1 , 1 }

{ I -j x x

(e) { - 1 , 4}

{-I}

- 00 ,

{ i}

(b) -

29. ( a ) { -2, 2}

(b)

(c) { -4, 0 }

{ -2, 2 }

or

00

- ,3

17. { x l x < - 1 0r x > 8 } ; ( - 00, - 1 ) or (8, 00 )

{l

(d) { x l - 1 < x < 1 } ; ( -1 , 1 ) (e)

(c) { -2, 2 }

x x :5 -

i} (

or - oo, -

VU] [

+ \,t"U} (

�]

)

( I) { x i x
2 } ; ( - 00 , - 1 ) or (2, 00 )

4 } ; ( - 00 , -4] o r [4, 00 )

(e) { x I X :5 - I } o r ( - 00 , - 1 ]

(d) { x l x < -2 0r x > 2 } ; ( - 00, -2) or (2, 00 ) (g)

2:

23. { x i x :5 - 4 o r x

{ x l x :5 - V2 o r x 2:V2}; ( - 00 , - V2] o r [ V2, 00 )

(g)

( 0 { x i x < -2 0r x > 2 } ; ( - 00 , -2) or (2, 00 )

{I 1

< X < 3 ;

(d) { x i x < - l or x > l } ; ( - oo , - l ) or ( l, oo )

(I) { x i x < - 1 or x > 4 } ; ( - 00 , - 1 ) o r ( 4 , 00 )

27. (a) { - 1 , 1 }

x -

15.

31. (a) { - 1 , 2}

(b ) { -2, 1 }

(c) ( 01

O J ; ( - 00 , 0 )

1 1 + VU . VU or x ; - 00, or 33. (a) 5 sec (b) The ball IS more than 96 ft above the ground , 00 2 2 2 2 from time I between 2 and 3 sec, 2 < ( < 3. 35. (a) $0, $1000 (b) The revenue is more than $800,000 for prices between $276.39 and $723.61, $276.39 < p < $ 723.61 . 37. ( a ) {c I 0.1 1 2 < c < 81 .907} ; (0.112, 81.907) (b) It is possible to hit a target 75 km away if c = 0.651 or c = 1.536.

(g)

x x :5

2: 1

-

Review Exercises 1. (a)

(b)

m =

2; b

=

(page 3 1 8) 3 . (a)

-5

(b)

y 2

x

-5

111 =

.

4 -' b

5'

=

-6

5. (a)

(b)

y 2

In =

0; b

=

7. Linear; Slope: 5

4

y 5

(0, 4)

(-3, 4 )

-2 (5 , -2 )

(4, 4)

-5

5

x

(0 , - 6 ) -8 ( c ) Increasing

(c) Increasing 9.

11.

y

(c) Constant

-5

13.

y 5

15. (a) y

(4, 0) -2

-2

x

8 -2

-2

x

-5 -4

-1

4

(b)

-2

I I I

Domain: ( - 00, 00 ) Range: [2, 00 ) (c) Decreasing: ( - 00, 2) Increasing: (2, 00)

x

8

AN28

ANSWERS Cha pter 4 Review Exercises 23. (a)

21. (a)

19. (a)

17. (a)

x

x

-2

-

1

8 (0, - 16)

(b) Domain: ( - 00 , (0

Range: [ - 16,

(0

(b) Domain: ( - 00, (0

)

(0,

(0

(e) Increasing:

33.

{

x ix

(e)

:=;

-�

or x �

0 :=; x < 350

41. (a) 63 clubs

5}; (

-�] [5,

- 00 ,

$151.90

39 38 37

43.

1

=

12

2

S

B 36 � 35

34 0

•

Range:

31. { x l - S

2}; ( -00, 1 ) U ( 2, (0 ) ) (

3

1 2

o.

..

43. { xix < - 4 o r 2 < x < 4 or x> 6 } ; ( - 00 , -4) U (2, 4) U (6, (0 ) •

( )(

)

-4

2 46

45. R = 10; g is not a factor off. 47. R = O; g is a factor off. 49. [(4)

=

47,105

� % ±�, � � � �

51. 4, 2, or 0 positive; 2 or 0 negative 53. ± 1 , ±3 , ± , ± ,

( )

1

x + 2 ; x-intercept: -2 x + 1

x

37. { xl-3 < x:5 3 } ; ( -3, 3]

.

--

6-7.

•

( - 1 , 2)

••

2 3

2

•

( -2, - 1 )

2 41. { xiI :5 x :5 2 or x> 3}; [1, 2] U (3, 00 ) (

-1

1, not i ntersected

( - 1 , 2)

)

f 1

=

•

35. {xix < -20r -l < x < 2 } ; -1 . ) ( -2

2. In lowest terms, G(x) =

4. Horizontal asymptote: y

•

I

x

(2, 3 2)

-2

Location of Graph Above x-axis

U

32

Above x-axis

+ 2 ) (x - 2) ; domam: . {x I x "" - 1 , x "" 2 } ; y-mtercept: 2 ' + 1 ) (x - 2 )

3. Vertical asymptote: x

( - 00, -2)

=

R(2)

=

Above x-axis

_2, 3

Point on Graph

m�

R

Location of Graph Above x-axis

1 2

± ,± ,± ,±

55. -2, 1 , 4; [(x) = (x

+ 2 ) ( x - 1 ) ( x - 4)

I

37

{

1

. . .

}

57. z,multiplicity 2; -2; [(x) = 4 x - Z (x + 2 ) 59. 2, multIplicIty 2 ; 61. { -3, 2 } 63. - 3 , - 1 , - Z ' 1 65. 5 67. Z 69. [(0) = - 1 ; [ ( 1 ) = 1 71. [(0) = - 1 ; [( 1 ) = 1 73. 1.52 75. 0.93 77. 4 - i; [(x) x3 - 14x2 + 65x - 102 79. -i, 1 - i; [(x) = X4 - 2x3 + 3x2 -

2x + 2

85. 2 (multiplicity 2), -Vsi, Vsi; [(x)

V; V;

87. -3, 2, -

i

=

(

89. (a) A(r) = 27rr2 +

-

)( )( ) { V; )( V; )

81. -2, 1, 4; [(x) = (x + 2 ) (x - 1 ) ( x - 4) 83. -2,

x + Vsi x - Vsi x - 2 2

i;[(X) = 2(x + 3 ) ( x - 2 500

=

r +

(b) 223.22 cm2

r

-

i

X-

2 (e) 257.08 cm

y 7

f(x)

=

(x-3)4

(2,-1) -2

-3

(3, -2)

(4, -1) 8

x

2

( �y

(multiplicity 2); [(x) = 4(x + 2) x -

i

1000 r;-:---------,

(d)

L�

A is smallest when r "" 3.41 cm.

o L::::::==:::J 8 o

Chapter Test (page 398)

1.

�

2. (a) 3 (b) Every zero of g lies between - 15 and 15. p

(i)

1 3 5 15 (e) q: ± Z ' ±1, ± Z, ± Z , ±3, ±5, ±Z , ±15 1 (d) -5, - Z' 3; g(x) = (x + 5 ) (2x + 1 ) ( x - 3 )

(e) y-intercept: -15; x-intercepts: -5, 1 (0 Crosses at -5, - 3 (g) y = 2x' Z' (h) Near -5: g(x) "" 72(x + 5 ) 1 63 Near -z: g(x) "" - 4 (2x + 1 ) Near 3: g(x) "" 56(x - 3 ) _

-�,

3 -50

AN42

ANSWERS

Chapter 5 Test

{

5 - V61 5 + V61 6 5. Domain: { xlx,.. - 1 0, x,..4}; asymptotes: x = - 10 , y = 2 6 6. Domain: { xix'" - l } ;asymptotes: x = - l , y = x + 1 3. 4, -5i, 5i

4.

],

}

'

8. Answers may vary. One possibility is [(x) =

7.

X4 - 4x3 -

2x2 + 20x.

2(x - 9 ) ( x - 1 )

. (x - 4) ( x - 9 ) -36 Since [(0) = 8 > 0 and [(4) = -36 < 0, the I ntermediate Value Theorem guarantees that there is at least one real zero between 0 and 4.

9. Answers may vary. One possibility i s rex) = 1 0. f(O) = 8;f (4 ) x

11. { xix

=

< 30r x > 8}; ( -00, 3 ) U (8,00)

C u m u lative Review (page 398)

1. V26

os 0 or x 2: 1 } ; (-00, 0] or [ 1 , 00)

2. { xix

] a

4. [(x) = -3x + 1

5.

3. { xl-l < x < 4}; ( - 1 , 4) , (

-1

t»

Y = 2x - 1

6.

y 6

x

x

5

-5

5

-5 7.

y = x3

(3,5)

l(x)= -3x+1 -5

-8

x

Not a function; 3 has two images.

8. {0, 2, 4} 9.

o

I

1

[I

2

J I t

3

2

17

12. y = -3x + 3 13. Not a function; it fails the Vertical Line Test.

5 ( 1 , 1) x 2

(-2,-2)

'I

1 1 . x-intercepts: -3, 0, 3; y-intercept: 0; symmetric with respect to the origin

1 0. Center: (-2,1); radius: 3 y

{x I X 2: %} ; %,00)

8 -8

-4

(-2, 4)

) .

4

(d)�; (�, )

14. (a) 22 (b) x2 - 5x - 2 (c) -x2 - 5x + 2 (d) 9x2 + 15x - 2 (e) 2x + h + 5 15. (a) {xix,.. I} (b) No; (2, 7) is on the graph. (c) 4; (3, 4) is on the graph. the graph.

9 is on

-3

17.

16.

y

18. 6; y = 6x - 1 19. (a) x-intercepts: -5, - 1 , 5; y-intercept: -3

x=1

(b) No symmetry

(c) Neither

(d) Increasing: ( -00, - 3 ) and (2, 00);

7

-]

3

-1

-1

(1 - 't. 0 )-2

x 3

21. (a) Domain: { xix > -3 } or ( -3,00) 1 . . (b) x-lI1tercept: -"2; y-lI1tercept: 1 (c) y 5

-4 -5 ( -3, -5) (d) Range: {yly

\

< 5} or ( - 00,5)

(Lg ) (x)

x2 - 5x + 1; domain: x x ", -4x - 7 1 24. (a) R(x) = - x2 + 1 50x 10 (b) $ 14,000 (c) 750; $ 56,250 (d) $ 75

(b)

x

5 , - 2)

(e)

_2}

23. (a) (f + g) (x) = x 2 - 9x - 6; domain: all real numbers

22.

(2, 5)

x

decreasing: ( -3, 2 ) Local maximum i s 5 and occurs at x = -3. (f) Local minimum is -6 and occllrs at x = 2. 20. Odd

3

=

{i

4

ANSWERS

AN43

Section 6.2

CHAPTER 6 Exponential a n d logarithmic Functions 6.1 Assess You r U nderstand i ng (page 407)

(g 0 n(x) 5. F 6. F 7. (a) -1 (b) - 1 8 0 (e) 8 (f) -7 9. (a) 4 (b) 5 - 1 -2 11. (a) 98 (b) 49 (c) 4 3 163 (c) 1 --3 15. (a) 2 v 2 (b) 2 v 2 1 (d) 0 17. (a) 1 (b) -1 (c) 1 -1 19. (a) � 4 (a) 97 (b) -2 2 2 17 5 v4 + 1 % 0 21. {xix *- 0, x *- 2 } 23. {xix *- -4, x *- O} 25. {xix ;;;, -&} 27. {xix ;;;, I } 29. (a) (f 0 g)(x) = 6x + 3; (b) 1 all real numbers (b) (g n(x) = 6x + 9; all real n u mbers (f n(x) = 4x + 9; all real numbers (g g) (x) 9x; a l l real n umbers 31. (a) (f g)(x) 3x2 + 1; all real numbers (b) (g n(x) = 9x2 + 6x + 1 ; all real numbers (f n(x) 9x + 4; all real n umbers (g g) (x) = x4; all real numbers 33. (a (f 0 g)(x) = X4 + 8x2 + 16; all real numbers (b) (g n(x) = X4 4; all real numbers (c (f 0 n(x) = x ; all real numbers (g g)(x) = x + 8x- + 20; all real numbers 35. (a) (f 0 g)(x) = 2 3x x ; {x I x *- 0, x *- 2} 2 (x - 1 ) ; {xix *- I } (c) (f n(x) = 3(x - 1 ) ; {x i x *- 1, x *- 4 } (g g)(x) x; {xix *- O} (b) (g n(x) = 4 _ x -4(x - 1) 3 4 37. (a) (f 0 g)(x) 4 + x ; {xix *- -4, x O} (b) (g n(x) = x ; {xix *- 0, x *- I} (f n(x) = x; {xi x *- I} (d) (g g)(x) x; {xi x *- O } 39. a) (f 0 g) (x) = V2x+3; {xix ;;;, -&} (b) (g n(x) = 2vx + 3; {xi x ;;;, O } (f n(x) = \YX; {xi x ;;;, O} (g g)(x) = 4x + 9; all real n umbers 41. a (f 0 g)(x) x; {xix ;;;, -I} - 1 ; {xi x ;;;, 2 } (b) (g 0 n(x) = Ixl; all real numbers (f 0 n(x) = X4 + 2x2 + 2; all real numbers (g 0 g)(x) = yry-',x='=-=1�(c)

4.

13.

(d)

(d)

(c)

(d)

, f:::

, f:::

0

(c)

=

-

(c)

(d)

)

4

(d)

(c)

4

,

--

(d)

if;

(

=

0

+

=

0

(c)

0

0

0

(d)

0

=

_

0

=

0

=

0

0

0

0

(d)

0

0

)

0

(c)

(d)

(d)

0

(d)

(c)

(c)

( )

0

=

(f g)(x) ��,-::.. 117 ; {xix 3; x *- H (b) (g f)(x) = -�� : �; {xix *- -4; x *- -I} 2x + 5 3x - 4 { I ll } c) (f n(x) = x 2 ; {xix *- -1; x *- 2} (g 0 g)(x) = - 2x - 11 ; x x 2 ; x *- 3 45. (f 0 g)(x) = f(g(x)) = fGx) = 2 G x ) = x; (g n(x) = g(f(x)) = g(2x) = �(2X) x 47. (f g) (x) = f(g(x)) = f(�) = ( �)3 = x; (g n(x) g(f(x)) = g(x3 ) = V? = x 49. (f g) (x) = f(g(x)) = fG(x + 6 ) ) 2 [ � ( X 6)] - 6 = x + 6 - 6 = x; 1 1 (g n(x) = g(f(x)) = g(2x - 6) = -(2x 2 - 6 + 6 ) = -2 (2x) = x 51. (f 0 g)(x) = f(g(x)) = fG(x - b) ) = a [�(x - b) ] + b = x; (g n(x) = g(f(x)) = g(ax + b) = �(ax b - b) x 53. f(x) x ; g(x) 2x + 3 (Other answers are possible.) 55. f(x) vx ; g(x) = x2 + 1 (Other answers are possible.) 57. f(x) = Ixl; g(x) 2x 1 (Other answers are possible.) 59. (f g)(x) = 11; (g n(x) 2 61. -3, 3 63. a (f 0 g)(x) = aex + ad + b f o g = g f when ad + b = be + d (b) (g f) (x) = aex + be + d (c The domains of both f o g and g f are all real numbers 2Y100 p + 600, 0 s; p s; 100 71. VCr) = 2m3 16 65. SCI) = 9' 7T16 67. C(l) = 15,000 + 800,000t - 40,000t2 69. C(p) = 25 73. (a) f(x) = 0.8382x (b) g(x) 140. 9 687x (c) g(f(x)) = g(0.8382x) = 118.l5996x 118,159.96 yen -x) = f(g( -x)) f( -g(x)) -f(g(x)) = - (f g) (x). So 75. f and g are odd functions, so f( -x) -f(x) and g( - x) -g(x). Then (f g)( odd. o f g is also 43. (a) (

0

=

if;

-

0

'-

0

(d)

--

(d)

if; -

=

0

0

0

0

=

=

+

0

+

0

4

=

=

=

0

=

+

0

0

)

=

(d)

0

0

=

( )

-

(d)

=

=

=

0

=

=

0

6.2 Assess You r U nderstand i ng (page 4 1 9)

x [4, 00 )

4. one-to-one 5. y 7. F 8. T 9. one-to-one 11. Not one-to-one 13. Not one-to-one 15. one-to-one 17. one-la-one 6. 19. Not one-to-one 21. one-to-one =

23.

Annual Rainfall

460.00 202.01 196.46 191.02 182.87

Location Mt Waialeale, Hawaii Monrovia, Liberia Pago Pago, American Samoa Moulmein, B urma Lae, Papua New G uinea

{460.00, 202.01, 196.46, 191.02, 182.87}

Domain: Range: { Mt Waialeale, Monrovia, Pago Pago, Moulmein, Lae}

25. Monthly Cost of Life Insurance

$7.09 $8.40

$ 1 1 .29

Age

30 40 45

$8.04, $ 1 1 .29} {30, 40, 45}

Domain: { $ 7.09, Range:

AN44

27.

31.

ANSWERS

Section 6.2

{(S, -3), (9, -2 ) ,(2 , -1 ) , ( 1 1 , 0 ) , (-S, I ) } Domain: {S, 9, 2, 11 , -S } Range: { -3, -2, - 1 , 0, I } f(g(x))

=

G

f (x - 4)

{(1, -2), (2, -3), (0, - 1 0 ) , ( 9, 1 ) , (4, 2 ) } Domain: { I 2,0, 9, 4} Range: { -2, -3, - 10, 2} X + 2 = 4 "4X + 2 33. f(g(x)) = f "4 8 = (x ,

) 3[�(X - 4) ] + 4 =

= (x - 4) + 4 = x g(f(x))

=

f(g(x)) g(f(x))

=

=

g(3x + 4)

1

3

[(3x + 4 ) - 4]

=

1

3 (3x)

=

f(�) = ( �)3 - 8 = (x + 8) = g(x3 - 8) = V(x3 - 8) + 8 = Vx3 = x -3 4X__ 2 _ 2-x +3 3 4X 39. f(g(x)) = f ___ = ----_ -'.- _---'-_ -2-x 4x - 3 -- + 4 2-x 2(4x - 3) + 3(2 - x) Sx = S =x 4x - 3 + 4(2 - x) = 35.

(

g(f(x) )

=

+

3

=

/ =

x - 8

=

x

37.

,

2

-2 1

x 3 f(r l(x)) = f x = 3

G ) G X) = x 1

= x r l(f(x) ) = r l(3x) = -(3x 3 ) Domain f = range f -I = all real numbers Range f = domain f-I = all real numbers

g(4x

f(g(x)) = f

2

-2

Vx+l =

8) - 8

.

/

/

=

g

(1 ) �

+ =

=x

2=x

G) 1

_- . x

/y = x

/ (2, 1 ) ? f- 1 / x / 2 Z l , 0)

�;::-2}(0, -1) ( -2, _ 2)? 1 f/

/

/

/

45.

/ / /y = X /

Y

3

//

/

f- 1 /,�, x -3 3 // / / // / -3 / x 1 49. r et ) = . 4 2 .f er I (x)) = f "4x - 2'1 = 4 "4x - 2'1 + 2 = (x - 2 ) + 2 = x 1 1 4x + 2 1 x - --=x r(f.(x) ) = .r- 1(4x + 2) = 4 2= +2 2 Domain f = range f -I = all real n umbers Range f = domain f-I = all real numbers f(x) = 4x + 2 = x /y -

( ) ( ) -

-- -

S

=

x; g(f(x))

=

Y

/,

f(rl (x) ) = f( Vx + 1 ) = (Vx+l )3 - 1 = x rl (f(x) ) = r l(x3 - 1 ) = V(x3 - 1 ) + 1 = x Domain f range f-I = all real n u mbers Range f = domain f - I = all real n u mbers

51. r l(x)

1

�

+

4x - 8 + 2 == (x - 2 ) 4-

(1) = ) G

�t f

-S

8)

-

//

+ -

X

=

( ) [ ]

41.

4

x+4 4(2x + 3 ) - 3(x + 4) Sx 2(x + 4) - (2x + 3) - S -x Y (1 ,2) /y = x 2 � / 1- 1/ / //

47. r l(x )

g(f(x) )

(-��t-:-1) - 3 g ( �) 2 - 2x 3 2X

43.

/ //

)

) (

1,

29.

53.

//

/

=" / / f-l(x) 4

-

( )

- 1. 2

(Vx'=4)2

r l(x) = �, x � 4 f(r l (x) ) = f(�) = + 4 = x r l(f(x)) = r l(x + 4 ) = V(x2 + 4) - 4 = \I? = x,x � 0 Domain f = range r l = {xix � O} or [0,00) Range f domai n r l = {xix � 4} or [4, 00

2

=

x -2

/

/

/ �2

8

x

)

ANSWERS

55.

rl(x) = -x4

l(r1(x)) = 1

(4) = 4 _ . (�) �

- x

( ) (�)

4 rl(f(x)) = r1 � = 4 =

Domain 1 range I�l Range 1 domain I�l =

Yf

/,/ ,

-5

//

/

_t.

(

r1(f(x))

=

(

?

_

=

_

2 3+x Domain 1 = all real numbers except -3 Range 1 = domain r1 = all real numbers except 0

61.

2

=

-

2

=

x

r I (x) = -2x x 3

-6x ( ) (�) -2x 3(+-2x) 2(x - 3) = -6 = x -2 ( x 3: 2 ) ( -2(3x) -6x r1 x + 2 ) � 3x - 3(x + 2) -6 -3

_

3

= -2x x-3 +2

1(f�I(X)) = I -2X� x - .) r I (f(x)) =

�

--

� .

=

= - = r

---: _ -:-' -----'-

x+2

=

Domain 1 = all real numbers except -2 Range 1 = domain I�l all real numbers except 3

63.

rl(x) = _x_ 3x ·- 2

_

IW1 (x)) = I

( 3x - 2

-x'

C

)

2 rl(f(x)) r x � 1 =

Domain 1 Range 1

=

( J ( )

2 3. ----:-_X----: -� '_ X_ 1 3 _ 3x - 2 2x _

�)1

) = 3 (-3x -

= all real numbers except :31

= domain r1

=

1)

-

1

= all real numbers except 2 = all real numbers except 0

x=2 I

2x 2x = x 2 =)=3 + � 3x +--2 - 3x 2 . 3 x ) 2 - 3( 2 ) 2(3 + x) - 3 ' 2 2x 3+x 2 _

rl 3 + x

=

-

1 1

x -2

Domain 1 range I�l Range 1 = domain I�l

5

1

-2

2x 2x = -= x 3x - (3x - 2) 2 2x 2x -x -,--- -,-= -= 6x - 2(3x - 1 ) 2 2

all real numbers except :3

AN45

) = "2x + - 2 (2x + x - 2x = X x ) 2 (-- + ( 2 + (x - 2) x-2 =x r1 (f(x)) = r 1 x 2 ) = ---

_

Y =x

rl(x) = 2 -x 3x l(r1(x)) = 1 2 -_t, 3X

(

1

/

59.

+1 r l (x) = 2x-x+1 I W 1(X)) = 1 2X-x' 1

/'(x ) = l � l (x ) = x4 , , , x /

•

_

= all real numbers except 0 = all real numbers except 0

5�� /'\H .

57.

Section 6.2

AN46

ANSWERS

Section 6.2

3x + 4 2x - 3 ( 2x3X - 43 ) + 4 3 ( 3 X + 4 ) l (f- I (x)) = j 2x - 3 = 2 ( 3X + 4 ) - 3 2x - 3 3(3x + 4) + 4(2x - 3) 2(3x + 4) - 3(2x - 3 ) = 17x = x

65. rl (x)

= --

r l (f (.>.: ) )

+

=

.

= Domain . j' Range I

U

-2x + 3 x-2 ( -2xX-+2 3 ) + 3 2 ( -2X + 3 ) 1 (r I (X)) = 1 x-2 -2x + 3 + 2 x-2 2(-2x + 3) 3 ( x - 2) -x = -2x + 3 2(x - 2) = -=1 = x ( 2X + 3 ) 3 -2 2X + 3 ) r l (f( x ) ) = r l ( = 2x� 3 x + 2 x+2 -2 -2(2x + 3 ) + 3 (x + 2) = -=1 -x = 2x + 3 - 2(x + 2 ) Domain I = all real numbers except -2

69. I-I

=

(x )

=

=

3(3x + 4) + 4(2x - 3) = 1 7x = x 2(3x + 4) - 3(2x - 3) 17 3 all real numbers except 2 -

domain I- I

=

2

v I - 2x

--

8x x r l (f(x )) = r l ( X2 - 4 ) =8=

+

�

0

domain I- I

2

=

0

all real numbers except

1

one-to-one; r l (x)

=

X, X 2': 0

�1 2 ( X2 -, 4 ) = � = v? 2x- \j � 2

2

_

x, since x > 0 Domain I = {x i x > O} or (0, 00 )

2

Range l

=

In

=

domai n r

=

{ i < �} (-oo,D x x

or

[-2, 00 ) range of r [5, 00) 77. Domain of g-L [0, 00); range of g-L all real numbers - b), 0 83. Quadrant I 85. Possible answer: f(x ) lxi, x 0, is

71. (a) (b) (e) (d) 73. 7 75. Domain o f r L � (x 79. Increasing on the interval ( f(0), /(5)) 81. r l

(x )

=

=

x

=

%

- . �

_

--

Range I

all real n umbers except

4 -4 4 - 4(1 - 2x ) ) 1 2X 1 (rI (X) ) = / ( = 2·4 4 V1="2x 2._ _ 1 - 2x

+

+

+

=

2

=

+

+

--

--

67. r l (x)

rl

--

( 2x3X - 34 ) 3 ( 3X2x +- 43 ) 4 2 ( 3X2x -+ 34 ) - 3

;

L

=

In ""

2':

1'( d) = d +6.990.7 39 6.971' - 90.39 + 90.39 = 6.971' r (b) ( d ( )) = 6.97 6.97 ( d + 90.39 ) d(r(d) ) 6. 9 7 6.97 - 90.39 = d + 90.39 - 90.39 d 56 miles per hour 2.3h = h h(W(h)) = 50 2.3 (h2.-3 60) + 88 = 89. (a) 77. 6 kg 2.3 88 ) W ( W 50 W + 88 � (b) heW) = � + 60 = � W(h( W )) 50 + 2.3 2 3 - 60 = 50 W + 88 - 138 = W 73 inches -dx + b ; f = r l. lf. a = -d 95. r l (x) 93. (a) { represents time, so t 2': O. 91. (a) { g I 30, 6 50 :s g :s 74,200} ex - a � H - 100 � 100 - H ( ) (b) {TI4220 :s T :s 15,107. 5 0} 99. No (b) t H = -4.9 4.9 4220 2.02 seconds g eT) = 0.25 30,6)0 Domain: {T14220 :s T :s 15,l07. 5 0} Range: {gI30,650 :s g 74,200} 87. (a)

I'

I'

--

=

=

=

(e)

+

(e)

=

+

(d)

=

=

(e)

T

+

(e)

_

:s

6.3 Assess You r U n derstand i n g (page 432) 6.

4 9. F lO. F 11. (a) 11.2 12 (b) 11.587 11.664 11.665 8.825 1 5. (a) 21. 2 17 (b) 22.217 22.440 22.459 17. 3.320 19. 0.427 21. Not exponential = 2 27. Not exponential 29. B 31. D 33. A 35.

(d) a

( -1, �), (0, 1), (1,

a)

7. 1 8.

(e)

(e)

(d)

E

(d)

13. (a)

8.815

(b)

23. Exponential; a

8.821 8.824 = 4 25. Exponential; (e)

ANSWERS

37.

39.

Y

9

41.

Y

4

-3

Domain: A l l real n umbers

8

-2 {yly > 2}

or

Range:

{yly

Range:

>

y 8 (0, 7.39) ( - 1, e)

-5

2}

or

(2, (0 ) =2

(_ 3, 1 )( -2, 1)

-2

y=0 5

x

e

-5

y 5

57.

- - -y = 2 x

5

-2

y=0 5

Domain: All real numbers

Domain: A l l real n umbers

{yly > O} or (0, (0 ) Horizontal asymptote: y = 0

O} or (0, (0 ) Horizontal asymptote: y 0

Range:

Horizontal asymptote: y

55.

{ y l y > -2} or ( -2, 00 ) = -2

51.

3

-2

5


r��


(2, (0 ) = 2

8

(0, (0 ) = 0

x

-3

x

-}

-5

( -2, 7.39) __

x

Y

or

(2, 5)


53.

{yly > O}

( 1, )

.�-=-=-=-'.

_ _ _ _


49.

8

Domain: A l l real numbers Range:

Range:

L

y = -2

x


Y

____ _ _ _

6

3

=

Domain : All real numbers

{yly > O} or (0, (0 ) =0

(2, 6) ( 1 , 3) \ Y y=2 �=:4::"t:.

-4

-3


47.

Y

(-l, �)

(0, -1)

x

)

-1

Range:

Y

10

Domain: A l l real numbers

{yly > I } or ( 1, (0 ) Horizontal asymptote: y = 1

45.

1)

(0, � 3

3 Range:

43.

(2, 3) (1,

AN47

Section 6.3

{3} 59. { -4 } 61. {2} 63.

Range:

{yly

x

>

=

U}

65.

{-\12, 0, \12}

67.

{6}

{ - I, 7 } 71. { -4, 2} 73. { -4} 75. {1, 2} 1 77. (a) 16; (4, 16) (b) -4; -4' 1 6 79. (a) (b) 3; (3, 66) 81. (a) 10; ( -2, 10) (b) 3 ; 3, 1 1 1 . . 83. 49 85. 4' 87. f( x) 3 ' 89. f ( x) = -6' 69.

(

)

�; ( -l,�) ( :) _

-5

=

Domain: All real numbers Range:


{yly < 5 } or ( -00, 5) =5

91.

{yly < 2} or ( - 00 , 2) =2

Range:



93.

0.5

-3

3

x

74% (b) 47% 97. (a) $12,123 (b) $6443 99. 3.35 mg; 0.45 mg 101. (a) 0.632 (b) 0.982 (e) 1 103. (a) 0.0516 (b) 0.0888 y = 1 - e -O.lr (d) 105. (a) 70.95% (b) 72.62% 1 I (e) 1 00%

95. (a)

Y

x

(0, - 1 )

-2 (-00, 00 ) Range: [1, (0 ) Intercept: (0, 1)

Domain:

Domain: ( - 00 , (0 ) Range:

[- 1, 0 ) (0, - 1 )

In tercept:

o

(e) About 7 min

AN48

ANSWERS

Section 6.3

5.41 amp, 7.59 amp, 10.38 amp (b) 12 amp 3.34 amp, 5.31 amp, 9.44 amp (e) 24 amp 222024 161418 1210 - ------------(1.0, 10.376)-----21) 86 (0.3, 5.414)(0.5, 7.585) (1.0, 9.443) II (I) 12(1 -e 42 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 2.0 3.0 109. 4: 2.7083; 6: 2.7181; 8: 2.7182788; 10: 2.7182818 ax+h - aX aXah - aX aX (ah - 1) f(x + h ) - f(x) . 1 1 111. 1 13. fe -x) a-I a h h h h " f( x ) �ee-x - eX) -�(eX - e-X) -f(x) 117. f ( l) 5, [(2 ) 17, [(3) 257, [ (4) 65,537 115. (a) f e - x ) � ee-x - e- (-x) f(5 ) 4,294,967,297 = 641 6,700,417 (b) 35 �(eX - e-x) 107. (a)

(c), (t)

(d)

y

- - - - - - --- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

=

o

11 =

n =

n =

=

11 =

= ---

=

=

=

= -;: =

=

=

=

=

=

=

x

=

lY

I.

-6

/

V

=

=

6

-3.5

6.4 Assess Your U nderstan d i n g (page 446)

(0, 00 ) 17. 23 8 19. a6 3 4. { x i x > O} or

=

=

except O; { x l x ,c O}

47. { x i x

�

1 } ; [1, 00

5. 21.

G, -1 ) . (1, 0), 2 2 1.6 2 eX 4 0 2 -4 1 4 1 ( a, 1 ) 6.

3'

=

23.

(10, 00 ) 49. 0. 5 11 51. 30.099

)

25.

=

41. { x i x > 1 O } ;

7. F 8. T 9.

27.

53.

2.303

(a) Domain: ( I

x=

:

-5

I I I I

-4

-4, 00 )

( c) Range:

-53.991

4 1

x

(0

-3

( - 00 , 00 )

eX - 4

log"

=

)

=

x

y

4 = -4

15. x

=

In

8

U

B D A E

l og.!, x 2

73. (a) Domain: (b)

(c) Range: ( - 00, 00 )

(0, 00 )

5

(d) r l (x) =

(e)

x

-5

x=O

ex-2

Vertical asymptote: x

Y

-1

-4

log2 7.2

3

f- (x)

5

=

3}; (3, 00 ) 39. A l l real numbers (0, 00 ) 45. { x l x < - l or x > O} ; ( - oo , - l )

57. v2

(2, -1) I -4

13. x

35. 2" 37. { x i x >

63. 65. 67. 69.

Y

-5

=

( t )'

Vertical asymptote: x (d) r l (x) = (e) Range off : ( - 00, 00

Y

-4

=

11.

log3 9

31. 2" 33.

29.

55.

61. f (x)

(b)

=

43. { x l x > - l } ; ( - l, oo )

59.

71.

1

(0

=

0

Range off: ( - oo , oo )

Y

8

y

=

(2,...L -L..l---L1) ';:'x 0 -3�:f=,I,=j=:lL..L..L 7 -2

ANSWERS

75. (a) Domain: (b)

(0, 00 )

( d) r I (x)

2 (e) Range o f f : (-00, 00)

Y

2

-2

= -1 e'. + 0'

(1,

-8

.. .L3-7 ..J..d=-.L...J.+-,-'y _-4.1=


79. (a) Domain:

=

(e)

Y

( t)

2

x

-2 x=O

1 = "2 , 102,

(b)

y7

(-1, 3) -5

U, 5 ) 0, n ( y = 0 -5-""'= 5 =I=!:±::±::,!,:...LL.l..--L-L;.:x -3

(e)

( t)

rl(x) = In(x + Range off:

(-3, yS

3) - 2

x 5

-S

-5

O

1 13.

91.

{2}

y3

115.

x

O} (-1, 0), (1,0)

Domain: {xix if' Range: (- 00, 00 ) I n tercepts:

95.

{3}

97.

fn31 O } 101. fn 82- 5 } (b) 4; (40 , 4) (c) 4; (4, 2)

{2}

99.

3

Y

x -+--""'--1---'---'-+

-�2---L -1 -3

O} {yly O}

D o m a i n : {xix > Range: � I ntercept: ( 1 , 0)

y 6

y = -2

(d) r l (x)

7 (4, 00) 103.

-----

-4

-2

=

( t)

=

3 10g2 (x

Range off:

y8

(4, 00 )

-

4)

x

-2 x

Horizontal asymptote:

{ x i x > -H; ( -�, oo)

111. (a)

-3

{5}

Range off: (- 00, 00)

x

(6, 8)

(O, S) y = 4 ---(c) Range:

r l (x) = y - 3 - 2

(e)

-3

y = -3 93.

(t)

5

10y

(b)

x = -3

Horizontal asymptote:

{9} 89. G} 109. { 2 - l g % }

S

(-2, -2)

(-3, 00 )

(e)

(1, 4)

85. (a) Domain: (-00, 00)

00)

x

4 (d)


(b)

87.

I I I I I

5

( c ) Range: (-00, 00)

(d)

=

5

=0

83. (a) Domain: (- 00, 00 )

(c) Range:

x = -2 Y I

4

-4

81. (a) Domain: ( -2, 00 )

Range off: (-00, 00)

(2, 5) ----y =4

-4


r l (x)

y6

-�=-=-�-=-=1:::=- -

x

10

( c ) Range: (- 00, 00)

(d)


i i�

( t)

-2

0

(0, 00)

(c) Range: (- 00 , 00 )

(e) Range off : (-00, 00)

x

0

-3

(c) Range: (- 00 , 00 )

(4, 00 ) x=4 I

x=O

(b)

(b)

(t)

x

-2.3 ) 8 " -3 U' )

77. (a) Domain:

AN49

Section 6 . 4

4

y=4

{ -2Vz, 2Vz} 105.

{-I}

107.

{ S in n

1 (b) 2 (c) 3 (d) It increases (e) 0.000316 3.981 10-8 S.97 km (b) 0.90 km 1 21. (a) 6.93 min (b) 16.09 min (c) No, since F(r) can never equal l 123. Ii 2.29, so the time between injections is about 2 hr, 17 min. ( t)

117. (a) 119. (a)

""

X

ANSO

ANSWERS

Section 6.4

8.1

2.0

125. 0.2695 sec

� 1.2 � 0.8 1

1 .6

on

0.

x

21

1

0.4 0.8 1 .2

k=

0.4 1

1 .6 2.0

4. F

5

6.5 Assess You r U n derstand i n g (page 457)

1. sum 2. 7 3.

r

2 + l og5 x 33. 3 10g2 z log" M

3 10g2 x - log2(x - 3)

5. F 6. T 7. 7 1

31.

35. 1 + ln x 37. ln x + x 39. +

45.

-

X

"3

57.

1r=== -2

-2

tF== l/

-2

25. b

-2 In(x - 1 )

_

4

a+b

2 10g" u + 3 log" v 41. 2 In x + "21 ln(1 - x) 1 47. "3ln(x - 2) + "31 l n( x + 1 ) - 2 ln(x + 4)

9. -4 11. 7 13. 1 15. 1 17. 2 19. 4 21. 4 23.

log x + log(x 2) - 2 10g(x + 3) 49. In 5 + In x + �In(l + 3x) - 3 ln(x - 4 ) 51. log5 u3v4 53. log3 ( !/2 ) 55. IOg4 [ (xx +- 11) 4 ] ( x + 1 )2 ] 65. 2.771 67. -3.880 69. 5.615 71. 0.874 25x6 61. log" ( � ) 63. log [ 2 (x + 3)(x 1) log x log(x + 2) 73. Y = -75. Y = log 4 log 2 3 2 1 29' S(a + b)

-1

true for x "" 1

(0.8959, 1 )

Seconds

43.

(d)

127. 50 decibels (dB) 129. 90 dB 131. 133. (a) 1 1 . 6 (b) 6.73 (c) % 0.14% 135. Because y = log l means F = = x, which cannot be

0.8959 sec

77.

Y =

59.

log(x + 1 )

-a

3a

log2[x(3x - 2) 4]

log(x - 1 )

op 4

5

27.

5

x; {xix is any real number} or ( - 00 , ) (b) (g fle x) = x; { x i x > O } or ( 0, 00 ) ( c) 5 (d) (f h )(x) = In x2 ; (xl x "" O) or (- oo, O) U (O, (0) (e) 2 '. . *(2x + 1)1/6 91. 3 93. 1 81. y = Cx 83. y = Cx(x + 1 ) 85. Y = Ceo' 87. y = Ce-4 , + 3 89. y = (x + 4 ) 1/9 95. l og, ( x + �) + log,, ( x - �) = log,,[(x + �)(x - �)l = log,,[x2 - (x2 - 1 )] = log" 1 = 0 97. In(l + e2x) = In [e 2X(e-2X + 1 )] = In e2x + In(e-2X 1 ) = 2x + In(1 + e-2x ) 99. y = f(x) = log" x; a Y = x implies aY = (�rY = x, so -y = logl/" x = -lex). 79. (a) (f

0

g)(x) =

-4

(0

0

0

+

1 0g,,� = log" 1 - log" x = -lex ) M " ' ' rl es a -log" N- ' -- I og" ( M ' N- I ) - I 0b" M + I og" N - I - I 0b'" M - I og" N , sll1ce 103 . Iog" alog" W ' -- N - I Imp - N , I.e., I og" N - -I og" N-I . N

101. f(x )

= l og " X ; f

(� )

=

a

a

{ 28 } 21. {-6} 23. { 2 } 25. {-I + V1+ii} "" { 6.456} I 2 35. {log2 1O} = f�n�O } {3.322} 37. {-logs 1.2} { _ � l� } "" { -0.088} n

6.6 Assess You r U nderstand i n g (page 463)

{-5 \3 Vs} "" ( 0.854 )

5. { 1 6} 7.

{�6 }

2

9. { 6 } 11. { 1 6 } 13.

U}

8

5

In S8 In 3 39. h IOg2 n = "" {0.226} 41. C In 3 + In 4 } {0.307 } 3 In 2 1 } 49. {:��} "" { l .585 } 51. (0) 53. {IOg4(-2 + \17)} "" (-0.315) 55. 27.

1

29. { } 31.

G}

1

15. { 3 } 17. { } 19. 33. { }

""

-

{ In 0.6In+7 I n 7 } "" { 1.356} ""

43.

{ logs 4}

"" { 0. 86 1 }

57.

=

45. {OJ 47.

No real solutioll

59.

{ 1 +In I n 7r

7r

} ""

{ 0.

534}

{ log4 5 } "" { 1. l 6 1 }

ANSWERS

61.

{2.79}

83.

{

63.

{-0.57}

{-0.70}

65.

', {2 + v's)) " (I444 )

85.

{0.57}

{ : ,: } " (I.92I) 67.

'"

e

'

{0.39, 1.00}

x

-3

-3 -4 3 (b) (0. 7 10,6.541) (c) {xix > 0. 7 10) or (0. 7 10, 00 )

(c)

97. (a)

95. (a)

f(x) = 2x_ 4 -5

(b)

s

{1.3 2}

{ -1, n {O} ( 5 ), (5, 4) (,) (I), ye,,' (1,2) ( 5 ) (e) {- I\}

73.

(b)

22y

91. (a), (b)

g

71.

( 5 ). (5 , 3)

87. ( ,)

"

f (x) = 3X + l (x) = 2x + 2

89. (a)

69.

{1.3 1}

75.

{I}

77.

{16}

79.

81.

(d)

f(x) = 3X

3

-4

AN 5 1

Section 6.8

93. (a), (b)

5

-s

x

x

-s

( I g 10, 10) 0 3

2010

(b)

2027

99. (a)

After 2.4 yr

(b)

After 6.5 yr (c) After 10 yr

x

-5

{x i x 2} or ( - 00, 2 )
�}; (�, 00 31. {xix l or x > 2}; (-00 , 1) U (2, 00) 33. -3 1 25 35. 2 37. OA 39. log3 u + 2 g3 - log3 41. 2 g x + "2 log(XO + 1 ) 43. In x + ;:;- In(r + 1) In(x - 3) 45. - log4 x 47. -2 In(x + 1) .) 4 4 x3 ) 51. 2.124 53. 49. log( 3 [(x + 3)(x - 2)]1/2 Domain of f = range of [-I = all real numbers except Range of f = domain of [-I all real numbers except =

z

,V;;:.

10

V

'

1

10

W

or Horizontal sym ptote: = = l o g (x e) Range off: 4

1. (a) f o g 2. (a)

4.

'" { - 1 .366, 0.366}

77. { - I }


3.

}

'" {4.301 }

a

o

153

(b) Y

dB

(e)

(e)

_ _ _ _ _

a.

2

(b) { 3 } , (6, 3 ) (e) { 10}, (10, 4) (d)

99. (a) 1 65 1

"

1 + V3

Y 13

87. (a) 37.3 W (b) 6.9

r.� �

-

83. (a),

-3

-5

85. 3229.5

'

(-2 + V7)

81.

(0, - 2 )

-3 -

V3

2

5} '" { -0.609}

Y

-5

-

69.

( -(X), (0

(f)

1

73. { 83 } 75.

(0,0.55)5

-5

-

e} U} Ln ��n 3 } n, } In { IOg 3 }={ 'n (-I�: } (e) f-l(x)=

67. (4, 0.97 )

65.

y

-2

ANSWERS Section 7.1

13. (a)

(b)

Domain off: {xix > 2 } or (2, (0) y

Y

7 (0, 7)

(3, 1 ) x

(c)

14.

=

(I)

6

I I II x=2 Range off: { y l - oo < y < oo } o r ( -00, (0); Vertical symptote: x = 2

18.

7

-3

Yes; no 2. (a) 10 (b) 2X2

+

3x +

Y

10

5.

- 10 (0, -4)

-

if 31,623 people shouted at the same time.

(c)

1

x

-3

C u m u l ative Review (page 5 0 1 )

1.

19. { 2V6 } "" {4.899} 20. 2 + 3 log2 X log2(x - 6) - log2(x + 3 ) 21. About 250.39 days 22. (a) $1033.82 (b) $963.42 (c) 1 1.9 yr 23. (a) About 83 dB (b) The pain threshold will be exceeded

(1, 3)

y = 2 -- - -

+"VD } {1 {/��7 } "" { -6.172}

{ I } 15. {91} 16. { - In 2} "" { -0.693} - Vi3 1 17. "" { - 1 .303, 2.303} 2 ' 2

rl (x) 5 1-x + 2 Range of f: ( -00, (0) (e)

(d)

2x2

+

4xh + 2/72 - 3x - 3/7 + 1 3. Y

10

6 . (a)

(�, �) is on the graph. 7. 8.

(8, 0) . 10

4.

{ -26}

[(x) = 2(x - 4? - 8 = 2x2 - 16x + 24

x 5

- 10 (b) 9.

[(g(x» =

10. (a) (b) (c) (d)

4 + 2; domam: . {xix (x - 3) 2

{xl-oo

79. ]

81. v2

2

O.

7.4 Assess You r U n derstand i n g (page 548) 1. tangent; cotangent

tan e

=

4

4 3 . 7T' 7j }jj 3. 60° 4. F 5. T 6. T 7. 60° 8. I, IV 9. 2' 2 10. '3 11. sm e = '5 ; cos e = -'5 ; 5 3 3V13 2 V13 3 V13 V13 . -"3; cot e = -4' 13. sm e = ---u-; cos e = 13; tan e = -'2; csc e = ---; sec e = - ; cot e 3 2

2. coterminal

5

-"3; csc e = 4'; sec e

=

-

=

2

-"3

ANs8

V2 sin e = - 2 '' cos e

15

·

17

ANSWERS

·

19.

sin IJ

=

�, cos IJ

=

-

V2

,

tan e

2 '

l ,' csc e

=

V2

29. -- 31. 0 33. 2

,

- V2: sec IJ = - V2:, cot e =

=

csc e 2', sec e 2 ', tan IJ V3, 3 '

V3 =

=

=

�;cos e = �; tan e

2'

sin e = -

Section 7.4

II

=

-1; csc e

2 V3 , cot e 3 '

=

- V2; sec II

=

=

1

V3

=

V2; cot II

- 1 21.

=

�

23.

1 25. 1 27. Y3

11' 11' 1 11' 11' 35. IV 37. IV 39. I I I 41. 30° 43. 60° 45. 30° 47. '4 49. '3 51. 45° 53. '3 55. 80° 57. '4 59. 2

V3 V2 V2 1 V2 V2 65. - 67. -2 69. - .vr::3 71. - 73. .vr::3 75. -- 77. -- 79. - .vr::3 81. V2 83. 0 85. 0 87. - 1 61' 2 63. 2 2 2 3 3 4 12 13 , 89. cos e = -- tan e = --' csc e = -' sec e = -- cot e = -- 91. sm e = --' tan e = -' esc e = --' sec e = --' cot e = ' 12' ' 12 4' 3' 4' 3 13 ' 2V2 12 13 13 12 3 V2 , tan e = 2 V2', esc e = 93. cos e = --' tan IJ = --' csc e = -'' sec e = -- cot e = -- 95. sin e = , sec e = -3', cot e 12' 3 ' 4 ' 13 ' 12' Vs Vs 3 3 Vs 2 Vs 97. cos e = --' tan e = - --' csc e -' sec e = - --' cot e = -' 2 3 ' 5 '

5'

5 5

2 13 ' 5 '

5

2' - Y3', esc e

sin e

101.

=

sin e

-Y32 ', cos e �2'' tan e -�5 ', cos e _.:£5 ', esc e =

=

=

=

5

- .?3 ''sec e

=

, e = -2; 1 cos e -2; Y3 tan e sm 117. (a) Approximately 1 6,6 ft (b) 105.

=

=

=

=

_

2Y3 , cot e =

=

5

=

2Y3

:!. 103.

sin e

3

---; cot e 3

=

=

V2

4

Y3 3

_

3 '

?, cot e -.4'

V3 --; sec e 3 20

5

5'

5

=

99.

2

5

Y3

=

VlQ , cos e 10 '

=

-

3 VlQ ,

10 '

esc e

=

107. 0 109. -0,2 111. 3 113.

VlQ', sec e

-5

=

-

VlQ

, cot e

3 '

=

-3

115. 0

(c) 67,SO

7.5 Assess You r U nderstan d i n g (page 557)

4. 211'; 11' 5.

3 , cot t -Y3 Y3, esc t -2'" sec t 2 Y 3 " 3 ' 2 ' tan t -V3, V2 - V2 , cos = - 2 ', tan , esc t - V2', sec - V2', cot t 1 2 ' Vs Vs 3 3 Vs 2 Vs 2 - ' cos t. -' tan t 3 2' 5 ' cot 2 5 '' esc t -'sec 3' 4 5 3 --4 '' cos e -3 ' tan e - -' esc e - -5 ' sec -' cot 5' 3' 4' 3' 4 5 3 V13 3 V13' sec - V13 ' cot ' cos e = - 2V13 3 ' tan e - -' csc

sin t = -L eos (

2'

9.

sin t

=

, t 13. sin

=

11

11' All real numbers, except odd multiples of "2 6. All real numbers from -1 to 1 inclusive 7. -0,2; 0,2 8. T

·

,

IJ

= --

17.

S111

19.

sin II

,

'

-

I'

�; cos e -�; tan e 1; esc � � � 1

=

'

=

II

2'

=

=

( = -

t = --'

=

--

'

( =

=

=

=

=

=

= --

=

13

=

=

t =

=

=

S111

=

t

e

15.

=

II

=

e

II

=

II

= --'

3

- V2; sec II

= --

=

=

--

2 '

- V2; cot e

�

-1

=

1

II

2 = --3 21.

�

23. 1 25.

1 27.

�

Y3

29.

�

31. 0

2 V2 35. 37. 39. 41. 2 43. 45. -1 47. 49. 0 51. - V2 53. 55. - 1 57. -2 - V2 11' 11' 59. --- 61. All real numbers 63. At odd multiples of "2 65. At odd multiples of "2 67. [-1 , 1] 69. (-00, 00) 71. (-oo, -l] or [l , oo) 2

33.

73.

2

Odd; yes; origin 75. Odd; yes; origin 77. Even; yes; y-axis 79. 0,9 81. 9 83. (a)

1

1

-� 1 (b)

85. (a) -2 (b) 6 87. (a) -4 (b) - 1 2

Using graph: sin 1 "" 0,8, cos "" D,S , tan 1 "" 1 . 6, esc "" 1.3, sec 1 "" 2 , cot 1 "" 0,6; Using calculator: sin 1 "" 0,8, cos 1 "" D,S, tan 1 "" 1.6, esc 1 "" 1.2, sec 1 "" 1.9, cot 1 "" 0,6 "" 0.4, tan 5.1 "" -2.3, csc 5,1 "" -1.1, sec cot "" -0.4; (b) Using graph: sin , 1 "" -0 , 9, cos Using calculator: sin "" - 0,9, cos , 1 "" 0.4, tan 5 , 1 "" -2.4, esc "" - 1 . 1 , sec "" 2 , 6, cot "" -0.4 91. Let P = ( x, y) be the point on the unit circle that corresponds to t, Consider the equation tan t � = a, Then y x 89. (a)

5

5.1

5,1

5

5.1

5, 1 "" 2 ,5 , 5.1 5.1 5,1

h and y = �; that is, for any real number 1 =

But x2 + i = 1, so x 2

+ a2x2

unit circle for which tan t

=

a,

=

1 . Thus x

=

±

+

a-

±

1 + a2

a,

=

ax.

there is a point P

=

( x,

In other words, -00 < tan t < 00, and the range of the tangent function is the set of all real numbers,

y) on the

ANSWERS

27T, for which sin(O + p) = s i n 0 for a l l O. If 0 0, then sin(O + p ) = s i n p = sin = 0; 7T 37T ) = -1 = Sin. ( 27T ) = This. IS. Impossible. 7T . . Therefore, the . 7T . 7T. Thus Sin. ( 2 so p = 7T. If 0 = 2' then Sin ( 2 p ) = Sin ( 2 ) . But p smallest positive number p for which sin (O + p ) = sin 0 for all 0 is p = 27T. 1 . . 27T, so does sec O. 95. sec 0 = -- ; Since cos 0 has penod cos 0

93. Suppose that there i s a number p, ° +

< P
-5

x

53. Vertex:(2, -8); focus:(2, -341 }

51. Vertex: (-1,-1); focus:( -�, -1}

49. Vertex: (-4, -2); focus: (-4, -1);

(0, 2)

-3

55. (y - If = x 57. (y - 1)2 = -(x - 2) 59. x2 = 4(y - 1) 61. l = .!.2 (x + 2) 63. 1.5625 ft from the base of the dish,along the axis of symmetry 65. 1 in. from the vertex,along the axis of symmetry 67. 20 ft 69. 0.78125 ft 71. 4.17 ft from the base, along the axis of symmetry 625 r + 625 (b) 567 ft: 63.12 ft; 478 ft:225.67 ft; 308 ft:459.2 ft (c) No 73. 24.31 ft, 18. 75 ft,7.64 ft 75. (3) = - (299) 2 Y

77. cl + Dx = 0, C*"0, D*"O cl = -Dx 79. ci + Dx + £y + F = 0, C*"O cl + £y = -Dx - F

( (

£

D -F --x C C � 2 y + 2C = _ll... C x _ !.C+£ 4C2 2 D £2 4CF £ Y + 2C = -Cx + 4C2 i + -y C

) )

=

?

lllis is the equation of a parabola with vertex at (0, 0) and axis of symmetry the x-axis. The focus is

--§

>

( -:c' o} the directrix is the line :c. The parabola opens to the right if

0 and to the left if --§ < O.

x

=

If D *" 0, then the equation may be written as � 2 _ll... x _ £2 - 4CF y+ 2C C 4CD · £ . vertex at £2 - 4CF ,-Th·IS IS . the equatIOn · of a parabola wIth 4CD 2C and axis of symmetry parallel to the x-axis. (b)-(d) If D = 0, the graph of the equation contains no points if £2 - 4CF < 0, is a single horizontal line if £ 2 - 4CF 0, and is two horizontal lines if £2 - 4CF > O. (3)

(

)

=

(

)

(

=

)

ANSWERS Section 11.3

AN 89

11.3 Assess Your Understanding (page 789)

7. ellipse 8. major 9. (0, -5); (0,5) 10. F 11. T 12. T 13. C 15. B 21. -x24 +-16y2 = 1 17. Vertices: (-5,0),(5,0) 19. Vertices: (0,-5), (0,5) Foci: (-vn, 0), (vn, 0)

Vertices: (0, -4 ), (0, 4) Foci: (0, -2V3), ( 0, 2V3)

Foci: (0, -4), (0,4)

Y

5

(0,-2) -5

.

Vertices: (-4,0), (4,0), (0, -4), (0,4) Focus: (0,0)

31. 9+Y5- = 1 ? r

x2 i -+9 25 = 1

(-3,0) -5 (0,-4)

(5,0) 5x

(-5,0) -5 -5 (0,-4) x2 -= l 1 35. -+ 4 13

x2 i 33. -+ 25 9 = 1

-3 ?

y 5

y (0,5)

y 5 (4,0) x 5

(-4,0) -5

3 (0,.[2)

-5 _

Y

Y

(2,0) x 5

79

5

Foci: (-ifG, 0), (ifG, 0 )

5

-5 (0, -2{3)

25. 16x2 + 16l = 1

Vertices: (-2Vz,0),(2Vz,0)

Y

(0,2{3) (3,0) x 5

23. "8x2 +2l = 1

(3,0) x 5

(-2,0) (-3,0)

(0,15)

-5

-5

(0,-5)

i 1 37. -r2+-= 16 (o,m)

(-2,0) -5

(2,0) x 5

(1,0) 4

(0, -"13) -5

39. (x+4 1)2 + (y - 1)2 = 1 43.

41.

Center: (3,-1); vertices: (3, -4), (3,2);

foci: (3, - 1 - v's),(3, -1+ v's)

(0, -m) (0, -4) ?

(x - 1)2 +:- = 1

45.

(x+5)2 + (y - 4)2 =1 16 4 Center: ( -5,4); vertices: (-9,4), (-1,4); foci: (-5 - 2V3,4),(-5 +2V3,4)

Y

5

-5

(3, 2) (3, -1+;/5) • x

(1,-1)

-5

x

-3

47.

(x+2f + (y - 1)2 = 1 4 Center: (-2,1); vertices: ( -4 ,1), (0, 1); foci: (-2 - V3,1),(-2+V3,1)

(-5 +2{3,4)

(-2 -,13,1) (-2,2)

(-1,4) x

(-4,1) -5

Y

3 (-2+,13,1)

(-2,0) -2

AN 9 0 49 .

+

(x - 2? (y + 1 )2 = 1 2 3 Center: (2, -1); vertices: (2 - V3 , -1 ) ; (2

+

51.

V3 , - I ); foci: ( 1 , -I), (3, - 1 )

( x - 2) 2 25

(y + 2) 2 21

+

=

-5 (- 1 , -2) ( 1 , -2)

(2, -2 -.J2i )

- 1)2 + (y - 2? = 1 10

(-2, 2) (1 M 2)

67.

(1, 3) •

(1, 2)

( 1 , 1)

-I

(-2, 0) 2

(1, -2 -/5)

(4+$ , 6) (4, 3)

-1

7

( x - 1)2 9

63.

(2, 0)

(1- 2-./2 , 2)

(4, 2) 4

x

83. (a) Ax2 + cl + F = ° If A =

C,

-3

x2

1 00

+

(1 , 3) (1, 1 )

(y - 1) 2 = 7

1

(5 , 1 ) (6,1 )

(2, 1 -17 )

7 x

65. 5

( 1 + 2fi, 2) (4, 2)

4

x

(0, -4)

5 (2, 1 +17 )

Y

. ( 1 , 2)

-1

Y

+

V3 )

x

(- 2 , 0) -3

(0, 4)

-1

(2, 0\ 3

73. 24.65 ft, 21. 65 ft, 1 3 .82 ft 75. 30 It 77. The elliptical hole will have a y-' . . . x2 major axis of length 2Y41 in . and a minor axis of length 8 in . 79. 9 1.5 mIlhon ml; -- + - 1 (93) 2 8 646 .7) l =1 . x2 . . ' 81. en' heI'Ion: 460.6 ml'11'I on 1m;. mean d Istance: 48 3. 8 mil lIOn ml; ---, +

69. 5

-2

x

-5

-5

l=

36

1

71.

43.3 ft

_

(483.8) -

P

(b)

(-1 , 1 ) (-2, 1 )

x

+ (y - 2? = 1

5

(-2, 2)

(1 +M,2)

y

59. (x -1 6 2)2

Y

,

Y

-5

-3

- V3), (0, - 2

( 1 , -5)

(4 -,15, 6)

7 (7, -2)

-2

foci: (0, -2

(0, -2 -f3 )

Y 9

x

y 5

x 5 (3 , -2)

+

Center: (0, -2); v ertices: (0, -4), (0, 0);

-2

(y - 6)2 9- = 1 57. (X -5 4)2 + -

1

(2, -2 +.J2i)

-7

(1, 1 )

( 1 , -2 -t;f5)

(2, - 1 +-./2 ) x (2 +-G, -1) (3, - 1 ) (2, -1 -fi)

3

-

()I + 2)2 53. x2 + --=1 4

Y

5

y

x

- VS), ( 1 , -2 + VS)

foci: ( 1 , -2

Y

-5 (2 - /J, -I) ( 1, -1) -5

61. (

(y - 2? (x - 1 )2 + - 9- = 1 4 Center: ( 1 , - 2); vertices: (1, -5 ), ( 1 , 1 );

5 (2, -I)

55.

+


(:�) + (�f)

I f A and ?

C a re

the equa tion may be written as x2

+

233,524.2

of the same sign and F is of opposite sign, then the equa tion takes the form ?

=

1 , where

=

-5 and -f are positive. This is the equation of an ellipse with center at (0, 0).

l = --:4'F

This is the equa tion of a circle with center at (0, 0) anci radius equal to

g.


11.4 Assess Your Understanding (page 801)

hyperbola 3 9. = -x; y 2 12. F 13. B

8. transverse axis 3 x 10. F 1 1. T 2

7.

Y

1 7.

--

=

15.

x2

_

Y

A

-5

21.

X2 - 1'...2 . = "9 16

Y

36

x2

"9 =

SX

,

y-

x-

?

?

1

25.

F2 = (0,3 $)

x Y=-4, I

10

8-8

1

=

y=- x,

V2 =(3,0)

VI =(-3, 0)

-L.....L.

10

-10

x

(0, -4)

27.

1'...2 .

F2 = (0,6)

y =_2,f)

Y= 2 fix (0,2fi) V2 =(1,0) F2 =(3.0) 5 (0,-2-/2)

23 .

1

19. 16Y - -20x2 = 1

i=1

8

? r

x2 i = 1 25 - 9 Centel': (0,0) Transverse axis: x-axis Vertices: (-5,0), (5,0) Foci: ( -\/34, 0 ) , (\/34, 0 ) Asymptotes: y = ±s3 x

?

y-

i - .1'2 = 1

1 4 16 Center: (0,0) Transverse axis: x-axis Vertices: ( -2,0), (2,0) Foci: (-2Vs, 0), (2Vs, 0 ) Asymptotes: y = ±2x

29. - - - =

-

31. "9

Center: (0,0) Transverse axis: y-axis Vertices: (0,-3), (0,3) Foci: ( 0,- vTO), ( 0, v'iO) Asymptotes: y = ±3x y;;;; 3x I F2 =(0,.J16)

y=-2x

Y

10

-4

y2 25

-

x2 25

-

\

= 1

Center: (0,0) Transverse axis: y-axis Vertices: (0,-5), (0,5) Foci: ( 0, -5Yz ) , ( 0,5Yz ) Asymptotes: y = ±x 39 .

(x - 4)2 4

y

(y + 1)2 5

V2 = (0, 5)

1

41.

=

i

1

x2

37. - - - = 1 36 9

(0,-5-/2)

(y+4 )2 4

.,

(x+3)2 12

- ---

'.>J;?>.-..L I

=

1

y 6

I.I

43.

.J} (x +3)

y+4 = �

I J..v:J�I

'1

.

I I I I (� X

8

I

�' -:[ j

AN1 04


19.

23.

21.

25.

x+y=2

LJ

-5

29.

27.

5

5

-5

-5 Y 5

x

35.

39.

37.

5 -5

y =x-2 x 5

x

x+y=3

x

-2

-5

49. Bounded; corner points (2, 0 ) , ( 5 , 0 ) ,

y 16

x+y=8

57. (a)

os; 50,000 � 35,000 os; 1 0 ,000 �O �O

(b)

( �)

53.

( 1 , 0 ) , (10, 0 ), ( 0, 5 ), 0 ,

y

x + 2y =

55.

10

x y 80,000

24 12 7' 7

2r + y = 4

51. B ounded; corner points

(2, 6 ), (0, 8) , (0, 2 )

x = 35,000

y = 1 0,000 --+---"'4----�x - 20,000 80,000 -20,000 x + y = 50,000 ( 35,000, 0 ) (50,000, 0 )

59. (a)

{

x y x + 2y 3x + 2y

�0

>O

-5

x

�

-5

47. Bounded; corner points (2 , 0), (4, 0 ) ,

(

( 2 , 0 ), ( 0, 4)

y

+Y x y x y

5 -5

45. Unbounded; corner points

( 0 , 0), ( 3, 0 ) , (2, 2 ), ( 0 , 3 )

{X

-5

-5

43. Bounded; corner points

( 2, 0)

y 5

3x + 2y = 6

2x - 3y = 0 _ 5

-2

33. N o solution

31.

os; 300 os; 480

(b)

{

)

'

(0, 4), (0, 2 )

x x+y x y

{X

os; os; � �

4 6 0 O

os; 20 y � 15 x + Y os; 50 x - y os; O x�0

y

400

61 . (a)

{ 23X 2y 01601 0 + x + 3y x

s:

s:

2

ANSWERS Chapter 12 Review Exercises

AN1 05

(b)

5

y 0 2

1 2.8 Assess Your Understanding (page 919)

1. objective function 2. T 3. Maximum value is 1 1; m inimum value is 3. 5. Maximum value is 65; minimum value is 4. 7. Maximum value is 67; minimum value is 20. 9. The maximum value of is 1 2, and it occurs at the point (6,0). 1 1 . The minimum value of Z is 4, and it occurs at the point (2,0). 13. The maximum value of is 20, and i t occurs at the point (0,4). 15. The minimum value of z is 8, and it occurs at the point (0,2). 17. TIle maximum value of z is 50, and it occurs at the point (10,0). 19. 8 downhill, 24 cross-country; $1760; $1 920 21. Rent 15 rectangular tables and 16 round tables for a minimum cost of $1252.00. 23. (a) $10,000 in a junk bond and $10,000 in Treasury bills (b) $12,000 in j unk bonds and $8000 in Treasury bills 25. 100 Ib of ground beef and 50 Ib of pork 27. 1 0 racing skates, 15 figure skates 29. 2 metal samples, 4 plastic samples; $34 31. (a) 10 first class, 120 coach (b) 1 5 (irst class, 120 coach Z

Z

1 1 y = -� or ( -:-1 1 , --::-3 ) 9. Inconsistent 1. x = 2, y - 1 or (2, -1) 3. x = 2, y = 21 0r ( 2' 21 ) 5. x = 2, y = -1 or (2,-1) 7. x = -:-, ) ) ) ) 39 y = 89 Z + 8' 69 Z .IS any real n um ber or 11. x = 2, Y = 3 or (2, 3) 13. I nconsisten t IS. x = -1, Y = 2, z = - 3 or (-1 , 2, -3) 17. x = 4"7 z + 4' Review Exercises (page 922)

�

=

[� -�l [ 1 � l [

l

69 Z IS. any real number 19. { 3X + 2y = 8 21. 39 y = 89 Z + 8' { (x, y, z,) ! x = 4"7 z + 4' } x + 4y = - 1 4 4 23. -6 212� 25. -21 � =�5 --4� 5 -9 -3 7 7 7 -13 1 31. -7 7 27 33. Singular 35. x = 25' y = 101 or e5' 10:l ) 29. 27. : 2 22 -13 -4 3 1 4 7 7 -7 37. 9, y = Z = � or (9, ;1 , 1 ) 39. x = -�, y = -�, z = -� or ( -�, -�, -�) 41. z = -1, x = Y + Y is any real n umber or {(X, y, Z)!X = Y + l, Z = - l , Y iS any real n umber } 43. x = 4, y = 2, z = 3, 1 = -2 or (4,2, 3,-2) 45. 5 47. 108 49. - 100 3 -3 -51. x = 2, y = - 1 0r (2, 1 ) 53. x = 2, y = 3 0r (2, 3 ) 55. x = -l,y = 2, z = -3 or ( - 1 , 2, -3) 57. 16 59. �2 + x 2 4 1 1x+ 9 1 1 -3 3 4 10 x -4x 2 4 4 2 11 lO 10 + 67. -+ -- + -- 69. x = - - ' Y = - ; X = -2, Y = 1 or 65. -+ - + - 63. -- + 61. -x-1 x+1 5 ) x2 + J x + l x-1 x x2 x 2 + 4 (x2 + 4) 2 x2 + 9 1 (-�, 5 ) (-2, 71. x = 2Yz,y = Yz; x = -2Yz, y = -Yz or (2Yz, Yz), (-2Yz, -Yz) 73. x = O, y = O; x = -3, y = 3; .� = 3, y = 3 0r (0, 0), ( -3, 3 ) , (3, 3 ) 75. x = Yz, y = -Yz; x = -Yz, y = Yz; x = j Yz, y = -� Yz; = - j Yz, y = � Yz or ( Yz, -Yz), ( -Yz, Yz), (jYz, - �Yz), ( - j Yz, �Yz) 77. x = Y = -1 or (1, -1) 85. Bounded 83. Unbounded 81. 79. Y 5 Y 8 3x + 4y 12

[

x

-l�l [ -t -�] 1;, ; _1 1 )

-

1,

=

1

-

_ _

'

1,

x

s:

5x

-5 -

5 r-----'

2x

+

3)'

=6

AN1 06

ANSWERS Chapter 12 Review Exercises

87. Bounded

y

91.

89. -5

x

9

93. The maximum value is 3 2 when x = 0 and y = 8. 95. The minimum value is 3 when x = 1 and y = O. 97. 1 0 99. y = -3I ? - 32 x + 1 101. 70 pounds of $6.00 coffee and 30 pounds of $9.00 coffee 103. 1 small, S medium, 2 large 105. Speedboat: 36.67 km/hr; Aguarico River: 3.33 km/hr 107. Bruce: 4 hours; Bryce: 2 hours; Marty: 8 hours 109. 35 gasoline engines, 15 diesel engines; 15 gasoline engines, 0 diesel engines r


1 17 18 y = z - 7' where z can be any real number. 4. x = 3' y 1. x = 3, y = - 1 2. Inconsistent. 3. x = -z + 7'

5.

[: -

-5 -1

5

1 6 -5

] {

: 6.

-1 10

3 x + 2, + 4 , - 6 Ix + Oy + 8z = 2 or -2x + ly + 3z = - 11 �

{

9. TIle operation cannot beperformed. 10. [ 1 � - :; ] 11. I

3 x + 2y + 4 , -6 x + 8z = 2 -2x + y + 3z = - 11 �

2

H

-1

1 [

3

� 12. =:

7

[: :] 8. [-1 1 :] -1 12

3 -2 -5

-3 6

=

-1

-2, z

=

0

-22

13. x = 2'1 y = 3 14. The system is dependent

and therefore has an infinite number of solutions. Any ordered pair satisfying the equation x = -.!. y + 7 or y = -4x + 28 is a solution to the system. 4 x 1 , y = -2, Z = 0 Inconsistent -29 x = -2, y = -5 x = 1, Y = - 1 , Z = 4 ( 1 , -3) and ( 1 , 3)

15. 16. 22. (3, 4) and (1 , 2) 23. =

x2 + i = 100

17.

18. -12 19.

3 24. -+3

Y 12

X

1

20.

+

21.

-2 (x + 3 ) ?-

27. The maximum value of z is 64,

and it occurs at the point (0, 8).

1

-x

25. � + (x23+ 3 ) + (x2 5x+ 3)2 28. Flare jeans cost $24.50,

8 6 4

camisoles cost $8.50, and T-shirts cost $6.00 .

X

- 12

12 x

26. The graph is unbounded. TIle corner points are (4, 2) and (8, 0): y 8

4x - 3y = 0

- 12

Cumulative Review (page 926) 1.

{o,�} 2. { 5 } 3. { - 1 , -� , 3 } 4. { -2 }

8. Center: ( 1 , -2); radius = 4 Y 5

5.

{%} 6. { In\ } 7. Odd; symmetric with respect to origin

9. Domain: all real numbers

Range: {yly > I } Horizontal asymptote: y = 1 y 5 (2, 2) -+---¥"'- - - - - x -5 5 -5

10. [ 1 (x) = -x5 - 2 _

Domain off: {xix 0/= -2} Range off: {yly 0/= O } Domain of rL { xix 0/= O} Range of rL {yly 0/= -2}

AN1 07

ANSW ERS Section 13.1 11, ( . )

2

_2

,

� 2

L

-2

-I ' �

-2

- 1)

Y

'

(2, 0)

2

x

I - T '° I

-x

"

1

\

y

2

;

�

�I

•

-2

(0

1 °, Vs,I

I

, I

_

3

(2, 1 )

yI 1

2

/I I ".. 2 -1 � ( - 2, , -I ) 1 (2 , - I )

.x

(k )

x

2

--...... ( -2, 1 ) "0.. ( - 1 , 0)

/'I' \ 11,01

( 0, -,%) - 1

yl

0, 774-75 vertex 01', 295, 773 Paraboloids of revolution, 771 , 778-79 Parallel lines, 1 82-83 Parallel vectors, 760 Parameter, 818 time as, 821 -24 Parametric equations, 8 1 8-30 for curves defined by rectangular equations, 824-27 applications to mechanics, 826-27 cycloid, 825-26 defined, 818 describing, 820-2 1 graphing, 8 1 9 rectangular equation for curve defined parametrically, 8 1 9-21 time as parameter in, 821 -24 Parentheses, order of operations and, 8 Partial fraction decomposition, 835, 89 1 -98 defined, 892 where denominator has nonrepeated irreducible quadratic factor, 896 where denominator has only nonrepeated linear factors, 892-93 where denominator has repeated irreducible quadratic factors, 897 where denominator has repeated linear factors, 894-95 Partial fractions, 892 Participation rate, 222 Pascal, Blaise, 826, 963, 994 Pascal triangle, 963, 966 Payment period, 466 Peano, G iuseppe, 995 Pendulum period 01', 79, 199 simple, 199 Perfect cubes, 44 Perfect roots, 72 Perfect squares, 43, 5 1-52 Perfect triangle, 696 Perihelion, 791 , 8 1 7 , 834 Perimeter, formulas for, 3 1 Period fundamental, 555 of pendulum, 79, 199 of simple harmonic motion, 698 of sinusoidal functions, 564, 567-68, 582-84 Periodic functions, 555

Index Periodic properties, 555-56 Permutations, 979-82 computing, 982 defined, 979 distinct objects without repetition, 980-82 distinct objects with repetition, 980 involving n nondistincl objects, 984-85 Perpendicular line, 760n Phase angle, 527 Phase shift, 581 -84 to graph y = A sin(wx -

Trigonometry: A Unit Circle Approach

Trigonometry: A Unit Circle Approach

Trigonometry: A Unit Circle Approach, 9th Edition