Trigonometry: A Unit Circle Approach Student Solutions Manual

STUDENT SOLUTIONS MANUAL KEVIN BODDEN RANDY GALLAHER LEWIS AND CLARK COMMUNITY COLLEGE ��e6ra & 8 �ri8onometry MIC...

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STUDENT SOLUTIONS MANUAL KEVIN BODDEN

RANDY GALLAHER

LEWIS AND CLARK COMMUNITY COLLEGE

��e6ra &

8 �ri8onometry

MICHAEL SULLIVAN PEARSON

----

Prentice Hall

Upper Saddle River, NJ

07458

Vice President and Editorial Director, Mathematics: Christine Hoag Senior Editor: Adam Jaworski Editorial AssistantlPrint Supplements Editor: Christine Whitlock Senior Managing Editor: Linda Behrens Assistant Managing Editor: Lynn Savino Wendel Production Editor: Ashley M. Booth Supplement Cover Manager: Paul Gourhan Supplement Cover Designer: Victoria Colotta Manufacturing Buyer: Ilene Kahn Associate Director of Operations Alexis Heydt-Long

PEARSON

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Prentice Hall

© 2008 Pearson Education, Inc. Pearson Prentice Hall Pearson Education, Inc.

Upper Saddle River, NJ 07458

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Printed in the United States of America

10

9

8

ISBN 13: ISBN 10:

7

6

5

4

3

2

978-0-13-232124-2 0-13-232124-6

Pearson Education Ltd., London

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Table of Contents Preface Chapter R

Review

Real Numbers ............................................................................................................................... 1 R.l ... . . . .. . . .. .. . . .. . .. . . 3 R.2 Algebra Essentials . .... R.3 Geometry Essentials ..................................................................................................................... 6 R.4 Polynomials .................................................................................................................................. 8 .. . .. . .. . . . . . . . II R.5 Factoring Polynomials . R.6 Synthetic Division ...................................................................................................................... 14 R.7 Rational expressions .....................: ............................................................................................. 15 R.8 nth Roots; Rational Exponents ................................................................................................... 20 Chapter Review ..................................................................................................................................... 24 Chapter Test .......................................................................................................................................... 28 . .

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Chapter 1 Equations and Inequalities Linear Equations ......................................................................................................................... 30 1.1 Quadratic Equations.................................................................................................................... 39 1.2 Complex Numbers; Quadratic Equations in the Complex Number System ............................... 48 1.3 Radical Equations; Equations Quadratic in Form; Factorable Equations ................................... 50 1.4 Solving Inequalities .................................................................................................................... 59 1.5 Equations and Inequalities Involving Absolute Value ................................................................ 65 1.6 Problem Solving: Interest, Mixture, Uniform Motion, and Constant Rate Job Applications ..... 69 1.7 Chapter Review ..................................................................................................................................... 73 Chapter Test .......................................................................................................................................... 80 Chapter 2

Graphs

The Distance and Midpoint Formulas......................................................................................... 82 2.1 Graphs of Equations in Two Variables; Intercepts; Symmetry .................................................. 87 2.2 Lines 93 2.3 2.4 Circles ....................................................................................................................................... 101 Variation ................................................................................................................................... 106 2.5 Chapter Review ................................................................................................................................... 108 Chapter Test ........................................................................................................................................ 114 Cumulative Review ............................................................................................................................. 116 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Chapter 3 Functions and Their Graphs 3.1 Functions................................................................................................................................... 118 The Graph of a Function ........................................................................................................... 125 3.2 Properties of Functions ............................................................................................................. 129 3.3 Library of Functions; Piecewise-defined Functions ................................................................. 136 3.4 3.5 Graphing Techniques: Transformations ................................................................................... 142 Mathematical Models: Building Functions ............................................................................... 150 3.6 Chapter Review................................................................................................................................... 153 Chapter Test ........................................................................................................................................ 159 Cumulative Review ............................................................................................................................. 163

© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 4

Linear and Quadratic Functions

Linear Functions and Their Properties . . . 165 4.1 Building Linear Functions from Data ....................................................................................... 170 4.2 Linear and Quadratic Functions . . . . . . 172 4.3 4.4 Properties of Quadratic Functions ............................................................................................ 182 Inequalities Involving Quadratic Functions . . 185 4.5 Chapter Review . .. . . . . . 195 Chapter Test 201 Cumulative Review ................................................................................................ :............................ 203 ...... ............................. . .. ......................................... ....

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Chapter 5

Polynomial and Rational Functions

Polynomial Functions and Models . . :.................................................................. 205 5.1 Properties of Rational Functions . . . . . . . . .. 215 5.2 The Graph of a Rational Function . . . . . .. . 219 5.3 Polynomial and Rational Inequalities . . . . . . 242 5.4 The Real Zeros of a Polynomial Function . . . . . 249 5.5 Complex Zeros; Fundamental Theorem of Algebra . . . . 270 5.6 Chapter Review . . . . : ................................................................... 273 Chapter Test . . . .. . . . . . . 289 Cumulative Review . . . . . . . . 293 .... ............. ......

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Chapter 6

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Exponential and Logarithmic Functions

Composite Functions ................................................................................................................ 296 One-to-One Functions; Inverse Functions . . . .. . . . . 303 Exponential Functions . . . . . . . . 313 Logarithmic Functions .............................................................................................................. 321 Properties of Logarithms .......................................................................................................... 329 Logarithmic and Exponential Equations . ... . . 333 Compound Interest . . . . . . . . . 341 Exponential Growth and Decay Models; Newton's Law; Logistic Growth and Decay Models . . . . . . 345 Building Exponential, Logarithmic, and Logistic Models from Data....................................... 348 6.9 Chapter Review . . . . .. 351 Chapter Test . . . . 361 Cumulative Review . . . . . . .. . . . 364 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8

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Chapter 7

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Trigonometric Functions

Angles and Their Measure . . . . . . . 367 7.1 Right Triangle Trigonometry . . . . . 371 7.2 Computing the Values of Trigonometric Functions of Acute Angles . . . 378 7.3 Trigonometric Functions of General Angles ............................................................................ 384 7.4 Unit Circle Approach: Properties of the Trigonometric Functions . . . 391 7.5 7.6 Graphs of the Sine and Cosine Functions ... . . .. . . 394 Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions.. . .. . 404 7.7 Phase Shift; Sinusoidal Curve Fitting . . . . . 407 7.8 Chapter Review . . . . 413 Chapter Test . . . . . . . . . . . 421 Cumulative Review . . . . .. . . . .. . . . . . . . 425 .. ........ ......... ...........................................

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Chapter 8

Analytic Trigonometry

The Inverse Sine, Cosine, and Tangent Functions . 8.1 The Inverse Trigonometric Functions (continued) 8.2 Trigonometric Identities 8.3 Sum and Difference Formulas 8.4 Double-Angle and Half-Angle Formulas 8.5 Product-to-Sum and Sum-to-Product Formulas 8.6 Trigonometric Equations I 8.7 Trigonometric Equations 11. 8.8 Chapter Review Chapter Test Cumulative Review

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Chapter 9 Applications of Trigonometric Functions Applications Involving Right Triangles . 9.1 The Law of Sines . . . . . . . 9.2 The Law of Cosines . . .. . . 9.3 9.4 Area of a Triangle .. . .. . Simple Harmonic Motion; Damped Motion; Combining Waves 9.5 Chapter Review . . . . . . .. Chapter Test .. .. . . . Cumulative Review . . .

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Chapter 10 Polar Coordinates; Vectors 10.1 Polar Coordinates . 10.2 Polar Equations and Graphs 10.3 The Complex Plane; DeMoivre's Theorem 10.4 Vectors 10.5 The Dot Product. Chapter Review . . Chapter Test . Cumulative Review

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Chapter 1 1

503 506 513 517 521 524 529 532

535 538 550 555 558 562 569 572

Analytic Geometry

. 11.2 The Parabola 11.3 The Ellipse . . . . .. . . 11.4 The Hyperbola 11.5 Rotation of Axes; General Form of a Conic . . . . . . . 11.6 Polar Equations of Conics .. . . . 11.7 Plane Curves and Parametric Equations Chapter Review . .. . . .. Chapter Test .. . . . . .. . . . Cumulative Review . .

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Chapter 12

12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8

Systems of Equations and Inequalities ... . ...... . .

. . Systems of Linear Equations: Determinants . . . . . . Matrix Algebra. . . . . . Partial Fraction Decomposition . . .. . .. . . Systems of Linear Equations: Matrices

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Systems of Nonlinear Equations Systems of Inequalities Linear Programming

Chapter Review Chapter Test

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Arithmetic Sequences

Mathematical Induction The Binomial Theorem

Chapter Review

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Chapter 14 Counting and Probability .. . . 14.1 Sets and Counting . . 14.2 Permutations and Combinations 14.3 Probability . . . . . .. . . . . Chapter Review . . . . . . . .. . . . Chapter Test .. . . . . Cumulative Review . .. . .

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Graphing Utilities

Section I The Viewing Rectangle

. .. . . 2 Using a Graphing Utility to Graph Equations .. . Section 3 Using a Graphing Utility to Locate Intercepts and Check for Symmetry Section 5 Square Screens Section

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Sequences; Induction; the Binomial Theorem

Sequences

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Chapter 13

Chapter Test

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Cumulative Review

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Systems of Linear Equations: Substitution and Elimination

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Preface This solution manual accompanies Algebra & Trigonometry, 8e by Michael Sullivan.

The Instructor Solutions Manual (ISM) contains detailed solutions to all exercises in the text and the chapter projects (both in the text and those posted on the internet). The Student Solutions Manual (SSM) contains detailed solutions to all odd exercises in the text and all solutions to chapter tests. In both manuals, some TI-84 Plus graphing calculator screenshots have been included to demonstrate how technology can be used to solve problems and check solutions. A concerted effort has been made to make this manual as user-friendly and error free as possible. Please feel free to send us any suggestions or corrections. We would like to extend our thanks to Dawn Murrin, Christine Whitlock and Bob Walters from Prentice Hall for all their help with manuscript pages and logistics. Thanks for everything! We would also like to thank our wives (Angie and Karen) and our children (Annie, Ben, Ethan,Logan, Payton, and Shawn) for their patient support and for enduring many late evemngs.

Kevin Bodden and Randy Gallaher Department of Mathematics Lewis and Clark Community College 5800 Godfrey Road Godfrey,IL 62035 [email protected]

[email protected]

Chapter R Review

Section R.t

23. a.

I. rational

b.

3. Distributive

c.

5. True

II. 13.

15. 1 7.

1 9.

{O,I}

{0,1'21 '3I '4I }

d. None

7. False; 6 is the Greatest Common Factor of 1 2 and 1 8 . The Least Common Multiple i s the smallest value that both numbers will divide evenly. The LCM for 1 2 and 1 8 is 36. 9.

{I}

e.

2 5 . a.

AuB = {I, 3, 4,5, 9} u{ 2, 4,6,7,8} = {I, 2,3,4, 5, 6, 7,8, 9} AnB = { l, 3, 4, 5 ;9} n{2, 4,6, 7,8} = {4}

None

b.

None

c.

None

d.

(AuB) nC = ({I, 3, 4,5, 9} u{2,4,6, 7,8})n{I,3,4,6} = {l,2,3,4,5,6, 7,8,9} n{1,3,4,6} = {I, 3,4,6}

{0,1'21 '3I '4I }

e.

{�,Jr,�+1,Jr + �} {�,Jr,�+1,Jr+�}

27. a.

1 8 .953

b.

1 8.952

28 .653

b.

28.653

A = { 0, 2, 6, 7, 8}

29. a. 3I. a.

0.063

b.

0.062

AnB = {I, 3, 4, 5, 9} Il {2, 4, 6, 7, 8} = {4} = {O, 1, 2, 3, 5, 6, 7, 8, 9}

33. a.

9.999

b.

9.998

35. a.

0.429

b.

0.428

37. a.

34.733

b.

34.733

Au B = { 0, 2, 6, 7, 8} u { 0, 1, 3, 5, 9}

39.

3+2 = 5

{2,5}

4I.

x+2 = 3·4

b. {-6,2,5}

43.

3y = 1+ 2

45.

x-2 = 6

47.

�=6 2

49.

9 - 4+2 = 5+2 = 7

5I.

-6+ 4 . 3 = -6+ 1 2 = 6

53.

4+5 - 8 = 9 - 8 = 1

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

2I. a.

c.

{-6,� ,- 1 .333 . .. = - 1 .3,2,5 }

d. {Jr} . e.

{-6,� ,-1 .333 ... = -1 .3,Jr,2,5 } 1

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Chapter R: Review

55

. 4+ .!.3 = 12+13 = .!.i3 6-[3 .5+2.(3-2)J = 6-[15+2 . (1)J = 6-17 = -11 2·(3-5)+8 . 2-1 = 2 . (-2)+16-1 = -4+16-1 = 12-1 = 11 10-[6-2·2 +(8-3)}2 = 10-[6-4+5]· 2 = 10-[2+5] .2 = 10-[7] . 2 = 10-14 = -4 (5-3) � = ( 2) � = 1

81.

57.

83.

59.

85.

61.

87.

89. �1 .

63.

93.

67. 6 9.

73.

75. 77. 79.

3 10 3·2·5 ·2· 2 5'21 = 5·3·7 = tt . t't7 = 7 6 10 2·3·5·2 2· · ·2 4 25 ' 27 = 5·5·3·9 = t t. 5 . tt . 9 = 45

95. 97.

99.

5 9 25+54 =-79 -+-= 6 5 -30 30 5 1 10+3 13 18 + 12 = � = 36 1 7 3-35 32 16 ---=-30 18 90 = --90 = --45 3 2 9-8 ---=--= 20 15 60 60

101.

103.

1 05.

107.

5 C�8 ) = 185 '1127 = 5 . 9 . 3 = 5 . , . 3 = 15 9·2·11 ,. 2 . 11 22 ) U 1 3 7 3 7 -3+7 =-= 10 1 -·_+-=-+-= 2 3 3 2 3 3 6 3 6 . -+2 3 2·-+-=4 8 1 . -+-=-+-=4 8 4 8 42 8 12 3 12+3 = -15 =-+-=-8 8 8 8 6 ( x + 4) = 6x + 24 x(X-4) = X2 -4x 2 (�X4 .!.2 ) = 2 . �x4 2 . .!.2 = 2·2·23x 3.2 t ·3x 2 =-x-1 3 =---t·2 2 2 (x+2)(x+4) = x2 +4x+2x+8 = x2 +6x+8 (x -2)( x + 1 ) = x2 + X -2x -2 =x2 -x-2 ( x -8)( x -2) = x2 -2x -8x + 16 =x2 -10x+16 2x+3x = 2·x+3·x = (2+3)·x = ( 5)· x = 5x 2 ( 3·4) = 2 ( 12) = 24 (2 . 3) . ( 2 . 4) = ( 6) (8) = 48 Subtraction is not commutative; for example: 2 -3 = -1 * 1 = 3 -2 . Division is not commutative; for example: 2 *"23 ' "3 The Symmetric Property implies that if 2 x, then x 2 . 5

10

-

10

10

10

_

10

_

=

=

2

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Section R.2: Algebra Essentials

1 09. There are no real numbers that are both rational

29. Graph on the number line:

1

and irrational, since an irrational number, by definition, is a number that cannot be expressed as the ratio of two integers; that is, not a rational number

1

1 1 3 . Answers will vary.

Section R.2 1 . variable

X

1 . 2345678 103

7 . True

1 -I

-2.5

13.

1 0 -> 2

1 5.

-1> - 2

1 7.

7t> 3.14

19.

1 = 0. 5 '2

21.

3' < 0.67

23.

x>O

25.

x -l

0

I.

1

�

1

•

37.

d(A,E) = d( -3 , 3 ) = 1 3 - ( -3 ) 1 = 161 =6

39.

x+2y = - 2 +2· 3 = -2+6 = 4

41.

5xy +2 = 5 ( -2)( 3 ) + 2 = -30+2 = - 28

43.

2x 2( - 2 ) = -4 = 4 = -x - y --2- 3 -5 5

45.

3x+ 2y = 3( - 2 ) + 2( 3 ) = -6+ 6 = .2. = 0 2+ 3 5 5 2+y

47.

Ix + y l = 1 3 + ( - 2 ) 1 = 1 1 1 = 1

49.

Ix l+IYI = 1 3 1+1 - 2 1 = 3 +2 = 5

.

53.

55.

2

-

( 1

1

d(D,E) = d( 1 , 3 ) = 1 3 -ti = 1 2 1 =2

51

1

1

1

35.

9. False; a number in scientific notation is

expressed as the product of a number, x , 1 :::; x < 10 or - 1 0 < x :::; - 1 , and a power of 10.

1

x �-2

d(C ,D) = d( 0 , 1 ) = II - 0 1 = 1 1 1 = 1

3. strict

0.25

0

1

33.

1 1 1 . Answers will vary.

11.

-2

1

3 1 . Graph on the number line:

Every real number is either a rational number or an irrational number, since the decimal form of a real number either involves an infinitely repeating pattern of digits or an infinite, non repeating string of digits.

5.

liE 1

L:l J�_Li =1 3 3 x 14x - 5 y I = 14( 3 ) - 5( - 2) I = 1 1 2 + 1 01 = 1 221 = 22 11 4x l - 1 5y II = 114( 3 ) 1 - 1 5( - 2) II = 11 121 - 1 - 1 0 11 = 1 12- 1 0 1 = 121 =2 x

Part (c) must be excluded. The value x = 0 must be excluded from the domain because it causes division by O.

3

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Chapter R: Review

59.

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

Part (a) must be excluded. The values and must be excluded from the domain because they cause division by o. 61.

87.

89.

1

None of the given values are excluded. The domain is all real numbers. 63.

x2 +5x-10 x2 +5x-10 x3 -x x(x-l)(x+l) x = 0, x = 1, x = -1

from the domain because they cause division by

67.

x x+4 x = -4

Y

-I

1

-

93 .

must be exluded b ecause it makes the denominator equal O. Domain

-I

Y

Y

O.

4 xx-5= 5

-I

-I

Y

Parts (b), (c), and (d) must be excluded. The values and must be excluded

65.

(x2 y )2 = (x2 )2 .(y )2 =x4 y-2 = 7X4 x2 y34 x2-1 3-4 = x1 y = -X --= xy -8x4iz2 9x/z -8 4-1 2 -3 z2 -1 =-x 9 -8 3 z = -x 9 8x3 z 9y [�)-2 = ( 3y ) 2 = ( 4X )2 42 x2 = 1 6x2 4x 3y 32 i 9i 4y-1

= { xl x 5} "#

95.

--

--

2(2) 2xy-1 = 2x = = -4 y

(-1)

must b e excluded sine it makes the denominator equal O. Domain =

69.

71. 73.

{

xl x -4} "#

= �9 (F - 32) = �9 (32 - 32) = �9 (O) = = �9 (F - 32) = �9 (77 - 32) = �9 (45) = 25°e (_ 4) 2 = (-4)(-4) = 16 C

101.

ooe

C

77.

3-6. 34 = 3

79.

(T2

-6

+4 =

1 05.

1 07.

J;z =

Ixl = 1 21 = 2

x -_ 2-1 -_ -21 = 2, 2x3 - 3x2 + 5x -4 = 2 . 23 -3 . 22 + 5 2 - 4 = 16-12+10-4 = 10 Ifx = 1, 2x3 -3x2 +5x-4 2 13 -3 . 12 +5·1-4 = 2 - 3 +5 - 4 =0 y

If x

.

T2

= 1= .!.. 32

r' = 3(-2)(-1) = 32 = 9

9

=

83.

�( _4)2 = 1 -41 = 4

.

4

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Section R.2: Algebra Essentials

1 09.

1 1 1.

1 13 . 1 1 5. 1 1 7. 1 1 9. 121. 1 23. 1 25. 1 27. 1 29. 131 . 133. 135. 137.

139.

141.

1 43.

(666) 4 = ( 666 = 34 = 81 (222)4 222 J (8. 2)6"" 304,006. 671 (6.1)-3 "" 0.004 (-2. 8)6"" 481. 890 (-8 . 11r4 "" 0.000 454.2 = 4. 542 x 102 0. 0 13 = 1. 3 x 10-2 32,155 = 3. 2 155x104 0.000423 = 4. 23x 10-4 6.15 x 104 = 61,500 l.2 14x10-3 = 0. 001214 1.1 x 108 = 11 0, 000, 000 8 . 1x10-2 = 0. 0 81 A =lw C=Trd A = ,J34 x2 4 r3 V =-Tr 3 V = x3 If x = 1000, C = 4000+2x = 4000 + 2 (1000) = 4000+2000 = $6000 The cost of producing 1000 watches is $6000.

If x = 2000, C = 4000+2x = 4000 + 2(2000) = 4000+4000 = $8000 The cost of producing 2000 watches is $8000. We want the difference between x and 4 to be at least 6 units. Since we don't care whether the value for x is larger or smaller than 4, we take b.

147.

the absolute value of the difference. We want the inequality to be non-strict since we are dealing with an 'at least' situation. Thus, we have

IX -41 6 Ix-110 1 = II08-110 1 = 1 -2 1 =2�5 108 volts is acceptable. I x -11 0 I = 1 104 - 1 1 0 I= 1 -6 1 = 6 5 104 volts is acceptable. Ix -3 1 = 1 2. 999 -3 1 = 1 -0 .001 1 = 0. 001 0.0 1 radius of2. 9 99 centimeters is acceptable. Ix-3 1 = 1 2 . 89-3 1 = 1 -0. 1 1 1 = 0. 11{'0. 0 1 radius of2.89 centimeters is acceptable. �

1 49. a.

b.

>

not

1 5 1 . a.

�

A

b.

not

A

1 53 .

1 55.

1 45. a.

1 57.

The distance from Earth to the Moon is about 4 x 108 = 400,000,000 meters. The wavelength of visible light is about 5xl0-7 = 0 .0000005 meters. The smallest commercial copper wire has a diameter of about 0 . 0005 5 x 10-4 inches. =

5

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Chapter R: Review

1 59.

186,000·60·60·24·365 =(1.86 X 105)( 6 x 101 )2 ( 2.4 x 101)( 3.65 X 102)

1 5.

a= 7, b= 24, c2 = a2 + b2 =72 + 242 =49 + 576 =625 :::> c= 25

1 7.

52 =32 + 42 25 =9 + 1 6 25 = 25

= 586.5696 xl 010 = 5.865696 x 1 012 There are about 5.9 x 1 012 miles in one light

year.

161.

1 63 .

1 65.

1

'3= 0.333333

>

... 0.333 � is larger by approximately 0.0003333 ...

The given triangle is a right triangle. The hypotenuse is 5.

No. For any positive number a, the value � is smaller and therefore closer to O. Answers will vary.

19.

The given triangle is not a right triangle.

Section R.3 1. 3. 5.

7.

9. 11.

13.

21.

right; hypotenuse C

=

True. False; the volume of a sphere of radius is given by V=34 7fr3

23.

r

True. Two corresponding angles are equal.

25.

a=5, b=1 2, c2 = a2 + b2 = 52 + 122 =25 + 144 = 1 69 :::> c= 1 3

27.

29.

a=1 0, b=24, c2 = a2 + b2 = 1 02 + 242

31.

62 =32 + 42 36=9 + 1 6 36= 25 false

The given triangle is not a right triangle. A=I·w =4·2=8 in2

}-

}-

A= b .h= (2)(4)=4in2 A=nr2 =n (5)2 = 25n m2 C= 2nr= 2n (5)= I On m V =1 w h=8 . 4 . 7 = 224 ft3 S = 2/w + 21h + 2wh = 2 (8 )( 4 ) + 2 (8 )( 7 )+ 2 ( 4 )( 7 )

= 64+112+56

= 100+576 :::>

252 = 7 2 + 242 625=49 + 576 625= 625

The given triangle is a right triangle. The hypotenuse is 25.

27fr

= 676

6 2 = 4 2 + 52 36 = 1 6 + 25 36= 4 1 false

c= 26

= 232 ft2 33.

4 3 =-n·4 4 3 =-ncm 256 3 V=-nr 3 3 3 S = 4nr2 = 4n·42 = 64n cm2

6

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Section R.3: Geometry Essentials

35.

37.

39.

= 1t r2 h = 1t(9i (8) = 6481t in3 = 27ir2 + 27irh = 27i ( 9)2 + 27i ( 9) ( 8) = 1627i + 1447i = 3067i in2 The diameter of the circle is 2, so its radius is 1 . A = 1t r2 = 1t(1) 2 = 1t square units

45.

V S

Total Distance

47.

49.

The diameter of the circle is the length of the diagonal of the square .

43.

4C

=

647t '" 20 1 . 1 inches ", 1 6 . 8 feet

=

4( 7t d)

=

47t · 1 6

Area of the border area ofEFGH - area of ABCD = 102 _62 = 100-36 = 64 ft 2 Area of the window area of the rectangle + area of the semicircle. =

=

= (6)(4)+ "21 . 1t . 2 2 = 24+21t '" 30.28 ft2 Perimeter of the window 2 heights + width one-half the circumference. = 2(6)+4+ 21 "1t(4) = 12+4+21t = 16 + 21t '" 22 .2 8 feet =

+

P

51.

We can form similar triangles using the Great Pyramid's height/shadow and Thales' height/shadow: h

Since the triangles are similar, the lengths of corresponding sides are proportional. Therefore, we get

126

240

1 14

2�

This allows us to write

8 x 4 2 8·2 -4 = x 4=x In addition, corresponding angles must have the same angle measure. Therefore, we have A = 90° = 60° , and = 30° . , B

=

A

d 2 = 2 2 +22 = 4+4 =8 d = .J8 = 2.fi r = !!...2 = 2.fi2 = .fi The area of the circle is: A = 1t r2 = 1t ( .fi t = 21t square units

41.

The total distance traveled is 4 times the circumference of the wheel.

3

h =-2 -240 3 h = 2·240 3 = 160

53.

C

Since the triangles are similar, the lengths of corresponding sides are proportional. Therefore, we get

The height of the Great Pyramid is 160 paces. Convert 20 feet to miles, and solve the Pythagorean Theorem to find the distance: 1 mile� = 0. 003788 nules. 20 feet = 20 feet · 5280 eet

d 2 = (3960 + 0 . 003788) 2 -39602 = 30 '" 5 .477 miles

30 x 20 45 30·45 = x -20 135 = x or x = 67. 5 2

d

sq. miles

In addition, corresponding angles must have the same angle measure. Therefore, we have A = 60° , = 95° , and C = 25° . B

7

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Chapter R: Review

55.

Convert 100 feet to miles, and solve the Pythagorean Theorem to find the distance: . 1 milefeet = 0. 0 18939 rrules 100 feet = 100 feet · 5280

d 2 = (3960 + 0. 0 18939)2 -39602 150 miles Convert 150 feet to miles, and solve the Pythagorean Theorem to find the distance: I mile 150 feet = 150 feet · 5280 feet = 0. 028409 miles d 2 = (3960 + 0.028409) 2 -39602 225 d 15 . 0 miles Let 1 length of the rectangle and w width of the rectangle. Notice that (l + W) 2 - (I - w) 2 = [(l + w) + (/-w)][(l + w) -(/-w)] = (2/)(2w) = 4/w = 4A So A = ± [(l+ w) 2 _(l-W) 2 ] Since (I -W) 2 0, the largest area will occur when 1 - w 0 or 1 w; that is, when the rectangle is a square. But 1000 = 21 + 2 w = 2(/ + w) 500 = 1 +w= 21 250 = 1 = w The largest possible area is 2502 = 62500 sq ft. A circular pool with circumference 1000 feet yields the equation: 2 7'Cr = 1 000 r = 500 7'C The area enclosed by the circular pool is: A = 7'C r2 = 7'C ( 5�0 J = 5�2 79577.47 ft2 d

""

�

12.2

sq. miles

�

1 1.

1 3.

=

=

1 7. 19. 21.

�

=

x = 8x- 1 k Not a monomial; when written in the form ax , the variable has a negative exponent. -2xi Monomial; Variable: x,y ; Coefficient: Degree: 3 8x = 8xy-l Not a monomial; when written in the form axn , the exponent on the variable y is negative. !

-2 ;

y

y

sq. miles

�

57.

9.

=

23.

25.

=

�

2 7.

�

29.

Thus, a circular pool will enclose the most area.

Not a monomial; the expression contains more than one term. This expression is a binomial. 3x2 -5 Polynomial; Degree: 2 5 Polynomial; Degree: 0

3x2 x5

--

Not a polynomial; the variable in the denominator results in an exponent that is not a nonnegative integer. 2/ - J2 Polynomial; Degree: 3

x2 +5 Not a polynomial; the polynomial in x3 -1 the denominator has a degree greater than O. (x2 +4x+5)+ (3x-3) = x2 +(4x+3x)+(5-3) = x2 +7x+2 (x3 _2X2 +5x+10)-(2x2 -4x+3) = x3 _2X2 +5x+ 1O-2x2 +4x-3 = x 3 + ( 2X 2 _

Section R.4

_

2X 2 ) + (5X + 4X) + (1 0 3) -

= x 3 - 4x2 + 9x + 7

1.

4; 3

3 1.

5.

True

33.

7.

m

3 Monomial; Variable: x ; 2xCoefficient: 2; Degree: 3

( 6x5 +X3 +X ) + ( 5X4 _X3 +3x2 ) = 6x5 +5x4 +3X2 +x (x2 -3x+1)+2(3x2 +x-4) = x2 -3x + 1 + 6x2 + 2x -8 = 7x2 -x-7

8

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Section R.4: Polynomials

35.

37.

39.

41. 43. 45.

47.

49.

51.

53.

55.

57.

59.

61.

6(x3 +x2 -3)-4(2x3 -3x2) = 6x3 +6x2 -18-8x3 +12x2 = -2x3 + 18x2 -18 ( X2 -X + 2 ) + ( 2X2 -3x + 5 ) - ( x2 + 1 ) = x2 -x+2+2� -3x+5-X2 _1 = 2X2 -4x+6 9 (/ -3y+4 ) -6 ( 1-/ ) =9/ -27y+36-6+6/ =15y2 -27y+30 x(x2 +x-4) = x3 +X2 -4x _2x2 (4x3 + 5) = -8x5 -I Ox2 (x+l)(x2 +2x-4) = x(x2 +2x-4)+I(x2 +2x-4) =x3 +2X2 _4X+X2 +2x-4 = x3 + 3x2 -2x -4 (x+ 2)(x+4) = x2 +4x+ 2x+ 8 = x2 +6x+8 (2x+5)(x+2) = 2X2 +4x+5x+l0 = 2X2 +9x+1O (x-4)(x+2) = x2 + 2x -4x-8 = x2 -2x-8 (x -3)(x -2) = x2 -2x -3x + 6 = x2 -5x+6 (2x+3)(x-2) = 2X2 -4x+3x-6 = 2X2 -x-6 (-2x + 3)(x -4) = _2X2 + 8x + 3x -12 = -2X2 + llx -12 (-x-2)(-2x-4) 2X2 +4x+4x+8 = 2x2 +8x+8 (x-2y)(x+ y) = x2 + xy -2xy -2/ = X2 - xy - 2Y2

63.

(- 2x - 3y)(3x + 2y) -6x2 - 4xy - 9xy - 6/ =

=

65. 67.

73. 75. 77. 79.

83.

85.

87.

89.

-6x2 - 1 3xy - 6/

(x-7)(x+7) = x2 _72 = x2 -49 (2x+3)(2x-3) = (2X)2 _32 = 4x2 -9 (3x+4)(3x-4) = (3X)2 _42 = 9x2 -16 (2x-3)2 = (2X)2 _2(2X)(3)+32 =4x2 -12x+9 (x+y)(x_ Y) = (X)2 _(y)2 =x2 _/ (3x+y)(3x-y) =(3x)2 _(y)2 =9x2 -/ (X_2y)2 = x2 +2(x . (-2Y ))+(2y)2 = x2 -4xy+4y2 (X-2)3 = x3 -3.x2 ·2+3·x . 22 _23 = x3 -6x2 + 12x -8 (2x + 1)3 = (2X)3 + 3(2x)2 (1) + 3(2x)· e + e = 8x3 +12x2 +6x+l 4x 2 - l lx + 23 x + 2 4x 3 - 3x 2 + X + 1

)

4x 3

+

8x 2

- l lx 2

+

X

- l lx 2 - 22x 23x + 1 23x + 46 - 45

Check:

(x + 2)(4x2 -llx+23)+(-45) = 4x3 -llx2 + 23x + 8x2 -22x + 46 -45 = 4x3 -3x2 +x+l The quotient is 4x 2 - l lx 23 ; the remainder is --45.

=

+

9

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Chapter R: Review

91.

4x - 3 X 2 4x 3 - 3x 2 + x +

)

97.

)

2X 2 + x + 1 2x4 - 3x 3 + Ox 2 + x + 1

4x 3

2X4 + x 3 + x 2 - 4x 3 - x 2 + X

-4x3 - 2X2 - 2 x x 2 + 3x + 1

x+ 1

Check:

(x 2 )(4x - 3) + (x + 1) = 4x 3 - 3x 2 + x + 1 The quotient is 4x - 3 ; the remainder is x + 1 . 93.

Check:

( 2X2 + X + 1 ) ( x2 - 2x + !) + � x + ! = 2X4 - 4x 3 + x 2 + X 3 _ 2X 2 + ! x

)

x 2 + 2 5x4 + Ox 3 - 3x 2 + X + 1 5x4

1 1 x2 + - x + 2 2 5 1 -x + 2 2

+ 1 0x 2

+ x 2 _ 2x + .12 + 2. 2 x + .1

2

- 1 3x 2 + x + 1 - 26 x + 27

= 2X4 - 3x 3 + X + 1 The quotient is x 2 - 2x + t ; the remainder is 5

Check:

1

"2 x + "2 .

( X2 + 2 )( 5x2 - 1 3 ) + ( x + 27 ) = 5x4 + 1 0x 2 - 1 3x 2 - 26 + x + 27

99.

= 5x4 - 3x 2 + X + 1 The quotient is 5x 2 - 1 3 ; the remainder is x + 27 .

- 4x 2 - 3x - 3 x - 1 - 4x 3 + x 2 + Ox - 4

)

-4x 3 + 4x 2 - 3x 2 -3x 2 + 3x - 3x - 4 -3x + 3 -7

Check:

Check:

( 2x3 - 1 )( 2x 2 ) + ( _x 2 + X + 1 )

(x - 1)(-4x 2 - 3x - 3) + (- 7)

= - 4x 3 - 3x 2 - 3x + 4x 2 + 3x + 3 - 7

= 4x5 _ 2X 2 _ x 2 + x + 1 = 4x5 - 3x 2 + x + 1 The quotient is 2x2 ; the remainder is _x 2 + x + 1 .

= - 4x 3 + x 2 - 4

The quotient is - 4x 2 - 3x - 3 ; the remainder is

-7.

10

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Section R.S: Factoring Polynomials

101.

1 05.

)

x 2 + x + l x4 + 0x 3 - x 2 + 0x + l x4 + x 3 + x 2

PI

_ x2 + x + 1 _x 2 - x - l 2x + 2

Check:

1 07.

(x 2 + x + l)(x 2 - x - l) + 2x + 2

= x4 + x 3 + X 2 _ x 3 _ x 2 - x - x 2 - x - 1 + 2x + 2 = x4 _ x 2 + 1

The quotient is

1 09.

x2 - x - 1 ;

2x + 2 .

1 03 .

When we multiply polynomials PI ( x ) and P2 ( x ) , each term of PI ( x ) will be multiplied by each term of P2 ( x ) . So when the highest powered term of ( x ) multiplies by the highest powered term of P2 ( x ) , the exponents on the variables in those terms will add according to the basic rules of exponents. Therefore, the highest powered term of the product polynomial will have degree equal to the sum of the degrees of PI ( x ) and P2 ( x ) . When we add two polynomials PI ( x ) and P2 ( x ) , where the degree of PI ( x ) = the degree of P2 ( x ) , the new polynomial will have degree :; the degree of PI ( x ) and P2 ( x ) . Answers will vary.

the remainder is

Section R.S

)

x - a x3 + ox2 + Ox - a 3 x3 _ ax2 ax2 ax2 - a 2 x a2 x _ a 3 a2 x - a 3 o

1. 3.

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

True; x2 + 4 is prime over the set of real numbers.

5.

3x + 6 = 3(x + 2)

7.

ax2 + a = a(x2 + 1)

Check:

(x - a)(x2 + ax + a2 ) + O = x3 + ax2 + a2 x - ax2 _ a2 x a3 = x3 _ a 3 The quotient is x2 + ax + a2 ; the remainder is O.

_

11.

2X2 - 2x = 2x(x - l)

1 3.

3x2 y - 6xy2 + 1 2xy = 3xy(x - 2y + 4)

1 5.

x2 - 1 = x2 _ 12 = (x - 1)(x + 1)

1 7.

4x2 - 1 = (2X)2 _ 1 2 = (2x - 1)(2x + 1)

19.

x2 - 1 6 = x2 _ 42 = (x - 4)(x + 4)

21.

25x2 - 4 = (5x - 2)(5x + 2)

23.

x2 + 2x + l = (x + l)2

11

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Chapter R: Review

67. 3x2 + l Ox - S = (3x - 2)(x + 4)

25. x2 + 4x + 4 = (x + 2)2 27.

x2 - l Ox + 25 = (x _ 5)2

69. x2 - 36 = (x - 6)(x + 6)

29.

4x2 + 4x + 1 = (2x + 1)2

71. 2 - Sx2 = 2(1 - 4x2 ) = 2 1 - 2x 1 + 2x

31.

1 6x2 + Sx + l = (4x + l)2

73.

33.

x3 - 27 = x3 - 33 = (x - 3)(x2 + 3x + 9)

75. x2 - 1 0x + 2 1 = x - 7 x - 3

35. x3 + 27 = x3 + 3 3 = (x + 3)(x2 - 3x + 9)

77.

37.

Sx3 + 27 = (2X) 3 + 33 = (2x + 3)(4x2 - 6x + 9)

79.

39.

x2 + 5x + 6 = (x + 2)(x + 3)

(

)( )

x2 + l 1x + l0 = (x + l)(x + l 0)

( )( ) 4x2 - Sx + 32 = 4 ( x2 - 2x + S )

is prime over the reals because there are no factors of 1 6 whose sum is 4. x2 + 4x + 1 6

81. 1 5 + 2x - x2 = - (x2 - 2x - 1 5) = -(x - 5)(x + 3) 83. 3x2 - 1 2x - 36 = 3(x2 - 4x - 1 2) = 3(x - 6)(x + 2)

4 1 . x2 + 7x + 6 = (x + 6)(x + l)

45. x2 - 1 0x + 1 6 = (x - 2)(x - S)

85. / + l l i + 30 / = / ( / + l ly + 30) = y2 (y + 5)(y + 6)

47. x2 - 7x - S = (x + l)(x - S)

87. 4x2 + 1 2x + 9 = (2x + 3)2

43.

49.

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

89.

x2 + 7x - S = (x + S)(x - l)

51. 2X2 + 4x + 3x + 6 = 2x(x + 2) + 3(x + 2) = (x + 2)(2x + 3)

91.

( ) = 2 ( 3x + 1 ) ( x + 1 ) x 4 - S 1 = ( X2 r - 92 = (x2 - 9)(x2 + 9)

6x2 + Sx + 2 = 2 3x2 + 4x + l

= (x - 3)(x + 3)(X2 + 9)

53. 2X2 - 4x + x - 2 = 2x(x - 2) + I(x - 2) = (x - 2)(2x + 1)

93.

x6 - 2x3 + 1 = (x3 _ 1) 2

[

= (x - l)(x2 + x + l)

55. 6x2 + 9x + 4x + 6 = 3x(2x + 3) + 2(2x + 3) = (2x + 3)(3x + 2)

Y

= (x _ l)2 (x2 + x + l)2

57. 3x2 + 4x + l = (3x + l)(x + l)

95.

x7 _ x5 = X5 (x2 - 1) = x \ x - l)(x + 1)

2z2 + 5z + 3 = (2z + 3)(z + l)

97.

1 6x2 + 24x + 9 = 4x + 3 2

59.

61. 3x2 + 2x - S = ( 3x - 4)(x + 2)

99.

63. 3x2 - 2x - S = (3x + 4)(x - 2) 101.

65. 3x2 + 14x + S = (3x + 2)(x + 4)

(

)

5 + 1 6x - 1 6x2 = -(1 6x2 - 1 6x - 5) = -(4x - 5)(4x + l) 4y2 - 1 6y + 1 5 = (2y - 5)(2y - 3)

12

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Section R.5: Factoring Polynomials

103. 1-8x2 -9x4 = -(9x4 +8x2 -1) = -(9x2 -1)(x2 + 1) = -(3x -1)(3x + 1)(x2 + 1) 105. x(x +3) -6(x + 3) = (x + 3)(x -6) 107. (X+2)2 -5(x+2) = (x+2)[(x+2)-5] = (x+2)(x-3) 109. (3x-2)3 _27 = (3x -2)3 _ 33 = [(3x -2) -3 J[(3x -2)2 + 3(3x -2) + 9] = (3x-5)(9x2 -12x+4+9x-6+9) = (3x -5)(9x2 -3x + 7) 111. 3(X2 +lOx+25 )-4(x+5) =3(x+5)2 -4(x+5) =(x+5)[3(x+5)-4J =(x+5)(3x+15-4) = (x + 5)(3x + 11) 113. x3 +2x2 -x-2 = x2(x+2)-I(x+2) = (x+2)(x2 -1) = (x + 2)(x -1)(x + 1) 115. X4 _x3 +x-l =x\x-l) +I(x-l) = (x -1)(x3 + 1) = (x-l)(x+l)(x2 -x+l) 117. 2(3x+4)2 +(2x+3) . 2(3x+4) . 3 = 2(3x+ 4)((3x + 4) + (2x + 3)·3) = 2(3x+ 4)(3x+4 + 6x+9) = 2(3x+4)(9x+13)

119. 2x(2x+5)+x2 . 2 = 2x((2x+5)+x) = 2x(2x+5 +x) = 2x(3x+5) 121. 2(x+3)(x-2)3 +(X+3)2 . 3(x-2)2 = (x+3)(x-2)2 (2(x-2)+(x+3) . 3) = (x+3)(x -2)2 (2x -4 +3x+9) = (x+3)(x -2)2 (5x +5) = 5(x+3)(x-2)2 (x+ 1) 123. (4x-3)2 +x . 2(4x-3) . 4 = (4x-3)(( 4x-3)+8x) = (4x-3)( 4x-3+8x) = (4x-3)(12x-3) =3(4x-3)(4x-l) 125. 2(3x-5) . 3 (2x+l)3 +(3x-5)2 . 3(2x+l/ . 2 = 6(3x-5)(2x+ 1)2 ((2x+ 1)+(3x-5)) = 6(3x -5)(2x + 1)2 (2x + 1 + 3x -5) = 6(3x-5)(2x+l)2 (5x-4) 127. FactorsSum:of4: 1,5 4 2,4 2 -1,-5-4 -2, -2 None of the sums of the factors is 0, so x2 + 4 is prime. Alternatively, the possibilities are ( x ± 1)( x ± 4) = x2 ± 5x + 4 or (x±2 )(x±2) = x2 ±4x+4 , none of which equals x2 + 4 . 129. Answers will vary. -4

13

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Chapter R: Review

Section R.6 1. 3. 5.

17.

quotient; divisor; remainder True 2)1 -1 2 4

2)4 -3 -8 4 8 10 4 4 5 2 8 Remainder 8 of O. Therefore, x -2 is not a factor of 4x3 -3x2 -8x + 4 . 2)3 -6 0 -5 10 6 0 0 -10 3 0 0 -5 0 Remainder O. Therefore, x -2 is a factor of 3x4 -6x3 -5x+10 . -3)3 0 0 82 0 0 27 - 9 27 -81 -3 9 -27 3 - 9 27 1 -3 9 o Remainder O. Therefore, x + 3 is a factor of 3x6 + 82x3 + 27 . -4)4 0 -64 0 1 0 -15 -16 64 0 0 -4 16 4 -16 0 0 - 4 Remainder 1 O . Therefore, x + 3 is not a factor of 4 x -64x4 +X2 -15 . -i )2 -1 0 2 -1 0 2 002 0 Remainder 0; therefore x -1 is a factor of 2X4 _x3 +2x-1 . -2)1 -2 3 5 -2 8 -22 -4 11 -17 3----17 x -2X2 + 3x + 5 -- = x2 - 4x+ 11 + -x+2 + b+c + = 1- 4 + 11-17 = -9 =

2 2 8 4 12 Quotient: x2 +x+4 Remainder: 12 3)3 2 -1 3 9 33 9 6 3 11 32 99 Quotient: 3x2 +l1x+32 Remainder: 99 0 9. -3)1 0 -4 0 -3 9 -15 45 -138 1 -3 5 -15 46 -138 Quotient: X4 -3x3 + 5x2 -15x + 46 Remainder: -138 1)4 0 -3 0 0 5 2 2 4 4 4 4 2 2 7 Quotient: 4x5 +4X4 +X3 +X2 +2x+2 Remainder: 7 -1.1)0.1 0 0. 2 0 -0. 11 0.121 -0. 3 531 0. 1 -0. 11 0. 3 21 -0. 3 531 Quotient: 0 . lx2 -0.l1x+0. 3 21 Remainder: -0 . 3 531 1)1 0 0 0 0 -1 1 o Quotient: X4 + x3 + x2 + X + 1 Remainder: 0

1 9.

=

7.

21.

=

23.

=

6

11.

2 5.

of

o

=

1 3.

27.

15.

a

d

x+2

14

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Section R. 7: Rational Expressions

Section R.7

1.

1 9.

lowest tenns

3 -4x 2X ( X2 -2 ) 2xx-2 x-2 3x+9 3(x + 3) 3 x2 -9 (x-3)(x+3) x-3 x2 -2x x(x-2) = -x 3x-6 3(x -2) 3 24x2 24x2 4x 12x2 -6x 6x(2x -1) 2x-I y2 -25 (y+5)(y-5) 2y2 -8y-IO 2 ( / -4y-5 ) (y+5 )(y-5) 2(y-5)(y + I) y+5 2(y + 1) x2 +4x-5 (x+5)(x-I) x+5 x2 -2x+I (x-I)(x-I) x-I x2 +5x-I4 _ (x+7)(x-2) 2-x 2-x x + 7)(x-2) - ( -I(x-2) = -(x+7) = -x-7 3x+6 x 3(x+2) x 5x2 x2 -4 5x2 (x-2)(x+2) 3 5x(x -2)

3. True;

5.

7.

9.

1 1.

13.

1 5.

21.

_

-

23.

25.

_

17.

-- . _-

27.

4x2 x3 -64 x2 -16 2x 4x2 (X_4) ( X2 +4x+I6) -4 (x )(x + 4) 2x 2x . 2x(x-4) ( x2 +4x+I6 ) 2x(x -4)(x + 4) 2x ( x2 +4x+I6) x+4 4x-8 12 4(x-2) 12 -3x 12-6x -3x 6(2-x) 4(x-2) 2 -3x (-I)(x-2) 8 3x x2 -3x-1O x2 +4x-21 x2 + 2x -35 x2 + 9x + 14 (x-5)(x+2) ' (x+7)(x-3) (x+7)(x-5) (x+7)(x+2) x-3 x+ 7 6x x23x-9-4 = � . 2x+4 3x-9 2x+ 4 x2 -4 6x 2(x+2) (x-2)(x+2) 3(x-3) 4x (x-2)(x-3) 8x x2�-1 = 8x ' x+I x+ 1 x2 -1 lOx 8x x+I (x -1)( x + 1) lOx 4 5(x -I) -- ' -

-- ' --

15

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Chapter R: Review

29.

31.

33.

35.

37.

39.

41 .

43.

4-x 4-x x2 -16 4+x 4x 4+x 4x x2 -16 4-x ( x+4) ( x-4) 4+x 4x ( 4-x ) ( x-4 ) 4x ( x _4 )2 4x x2 + 7x+12 -x -12 x2 _ 7 x + 12 x2 + 7x + 12 -x2 --x2 + x -12 x2 -7 x + 12 x2 + x -12 x2 -x-12 (x + 3)(x + 4) . (x -4)(x + 3) (x-3)(x-4) (x+4)(x-3) (x +3)2 (X_3) 2 2X2 -x-28 3x2 -x-2 2x2 -x - 28 3x2 + l lx + 6 4x2 +16x+7 3x2 -x-2 4x2 +16x+7 3x2 +llx+6 (2x+7)(x-4) (3x + 2)(x + 3) (3x + 2)(x -1) (2x + 7)(2x + 1) (x -4)(x + 3) (x -1)(2x + 1) -x2 + -25 = x+52 ( x -2 ) x2 -4 ( x + 2 )-'-:x2 4 = -2x-3 --2x-3 2x-3 = -'----:-'2x-3 -"x + 1 2x -3 x + 1 + 2x -3 3x -2 x-3 + x-3 = x-3 x-3 3x+5 2x-4 (3x+5)-(2x-4) 2x -1 2x -1 2x -1 3x+5-2x+4 2x-l x+9 2x-l 4 + x = 4 - x = 4-x x-2 2-x x-2 x-2 x-2

45.

-_ . _-

47.

49.

51.

--

--

--

--

-

-

53.

----

-

--

--

--

--

--

--

or

--

--

4 2 = 4(x+2) 2(x-l)x-l) --x-I x+2 (x-l)(x+2) (x+2)( 4x+8-2x+2 (x + 2)(x -1) 2x+l0 (x+2)(x-l) 2(x+5) (x+2)(x-l) x + -2x-3 = x(x-l) + -'--(2x-3)(x+l) -'--'--'x+l x-I (x+l)(x-l) (x-l)(x+l) x2 -x+2x2 -x-3 (x-l)(x+l) 3X2 -2x-3 (x-l)(x+l) x-3 x+4 (x-3)(x-2) _ (x+4)(x+2) x+2 ---= x-2 (x+2)(x-2) (x-2)(x+2) x2 -5x+6-(x2 +6x+8) (x + 2)(x -2) x2 -5x+ 6-x2 -6x-8 (x+2)(x-2) -(llx + 2) -llx-2 (x+2)(x-2) (x + 2)(x -2) x -1 = x2 +X2 -4) --+ x2 -4 x x ( x2 -4 2X2 -4 X ( X2 -4) ) 2 ( x2 -2 x ( x-2 ) ( x+2 ) x2 -4 = ( x+2 ) ( x-2 ) x2 -x -2 = ( x + 1) (x -2 ) LCM ( x 2 ) ( x - 2) ( x + 1 ) . x3 -x = X ( X2 -1 ) = x ( x+l ) ( x-l ) x2 -x = x ( x -1 ) LCM x ( x+l ) ( x-l ) .

55.

Therefore,

=

Therefore,

=

+

-

16

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57.

4x3 -4x2 +x = x ( 4x2 -4x+I ) = x ( 2x -1)( 2x -1 ) 2x3 -x2 = x2 (2x -1) x3 LCM = x3 ( 2x _1) 2 . x3 -x = x ( x2 -I ) = x (x+I)(x-1) x3 -2X2 +x = x ( x2 -2x+1 ) = x(x-I)2 x3 -1 = (x -1) ( X2 + X + 1 ) LCM x ( x 1)( x_1)2 ( x2 + X + 1 ) . x x 2 x2 - 7x + 6 x -2x -24 x x (x -6)(x -1) (x - 6)(x + 4) x(x+4) x(x-1) (x -6)(x -I)(x + 4) (x -6)(x + 4)(x -1) x2 +4x-x2 +x 5x (x -6)(x + 4)(x -1) (x - 6)(x + 4)(x -1) 4x 2 x2 -4 x2 +x- 6 4x 2 (x-2)(x+2) (x+3)(x-2) 2(x+2) ) = (x -24x)(x(x+3 + 2)(x + 3) (x + 3)(x -2)(x + 2) 4x2 +I2x-2x-4 (x -2)(x + 2)(x + 3) 4x2 +lOx-4 (x -2)(x + 2)(x + 3) 2(2x2 + 5x -2) (x -2)(x + 2)(x + 3) 3 + 2 (x-I)2 (x+I) (x-1)(x+1)2 3(x+I)+2 (x-1) (x_I)2 (x+I)2 3x+3+2x-2 (x_I)2 (x+1)2 5x+I

67.

4)(x

Therefore, 59.

Therefore,

61.

63.

65.

=

x+4 2x+3 2x -x-2 x2 +2x-8 x+4 2x+3 (x -2)(x + 1) (x + 4)(x -2) (2x 3)(x + (x + + (x -2)(x + I)(x + 4) (x + 4)(x -2)(x + 1) 2 +5x+3) _- x2 +8x+I6-(2x -2)(x + l)(x + 4) (x _x2 +3x+13 (x -2)(x + l)(x + 4) 1 2 3 -x x 2 +x + x3 _x2 1 2 3 =-X X ( X + 1) + x2 ( x-I) -2x ( x-I) 3 ( x + 1 ) _- x ( x + 1) ( x-I) x2 ( X + 1)( x-I) x ( x2 -1 ) - 2X2 + 2x + 3x + 3 x2 (x + 1)( x-I) x3 -X-2X2 +5x+3 x2 (x + 1)( x -1 ) x3 -2X2 +4x+3 x2 (x + 1) (x -1)

69.

+

--

+

4)

1)

-

---

+

73.

hx(x+h) -1 x(x+h) (� + �) ( X; I ) x+1 x x+I (� _ � ) = ( X � I ) = -x . x-I = -;=}

17

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Chapter R: Review

75.

77.

79.

x+Ix 2x-x-I -2xx --x 3(x+I) +-x-I 3x+3+x-I x+I x+I x+I x-I x-I . x+I = x 2 (2x + I) _ = 4xx_ + 2 x+I (x-I)(x+I ) 2x(2x + 1) x+4 x-3 x-2x+Ix+I X+4)(X+I) (X-3)(X-2) ) ( (x-2)(x+I) (x+I)(x-2) x+I -SX+6) ) ( X2 +SX+4-(x2 (x -2)(x + 1) x+I lOx-2 1 2(Sx - 1) (x-2)(x+I) x+I (x-2)(x+I)2 x-2 x-I --+-x+2 x+I 2x-3 x-- -x+I x X-2)(X+I) + (X-I)(X+2) ) ( (x+2)(x+I) (x+I)(x+2) ( (x +X2I)(x) (2X-3)(X+I) x(x + 1) J -2 + x2 + X -2 ( 2x2 -4 ( x2 -X(x+2)(x+I) (x+2)(x+I) J J ( X2 _ ��:2+�; _3) ) ( -:�::I; 3 ) 2(x2 -2) . x(x+I) (x+2)(x+I) _(x2 -x-3) 2X(X2 -2) -(x + 2)(x2 -x-3) -2X(X2 -2) (x 2)(x2 -x-3)

2- x+Ix x-I 3+x+I

81 .

1 1- --1 1- --= 1- 1.x x-I = I- � x-I x-I-x x-I -1 x-I 2 ( x-I ) +3 -_ _x-I2_ +3 - _x-I2_ + 3(x-I) x-I 2(x-I) 3 3(x-I rl +2 � x-I +2 _x-I_ + x-I 2+3(x-I) x-I 3+2(x-I) x-I 2+3(x-I) ---;x-I--:x-I 3+2(x-I) 2+3(x-I) 2+3x-3 3+2(x-I) 3+2x-2 3x-I = -2x+I ( 2x + 3) . 3 -( 3x - S ) . 2 --6x + 9 -6x + 10(3X-S)2 (3x-S)2 19 x

-1

83.

85.

87.

_

_

-

--;:--

x . 2x-(x2 +1 ) . 1 2x2 -x2 -I (X2 + 1t (X2 + 1 f x2 -1 (X2 +It (x-I)(x+I) (X2 +I t

+

18

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89.

(3x + l ) . 2x - x2 · 3 6x 2 + 2x - 3x2 (3x + l) 2 (3x + 1) 2 3x 2 + 2x (3x + 1) 2 = x(3x + 2)2 (3x + l) ( x2 + 1 ) . 3 -(3x + 4) 2x 3x 2 + 3 - 6x 2 - 8x ( x2 + 1)2 (X 2 + 1 t -3x2 - 8x + 3 ( X 2 + It - ( 3x2 + 8x -3 ) (X2 + 1t ( 3x -1) ( x + 3 ) ( x2 + 1 )2

95.

.

91.

93.

1 + -1 ) -1 = (n -l) (R, R2 I l. = (n - l) ( RR,2 +· RR, ) I 2 R, · R -I-2 = (n - l) ( R2 + R, ) I I R, · R2 (n - l) (R2 + R, ) R, · R2 1 = (n - 1)(R 2 + R, ) 0.1(0.2) I - (1.5 - 1)(0.2 + 0. 1 ) 0.02 2 meters 0.02 = -= 0.5(0.3) 0.15 = 15

x + l =:> a = 1, b = 1, c = 0 1 + -x1 = -x 1 1 = 1 + -x 1 + --� = 1 + --+l x X;I ( ) 1+ x + l + x = -2x + l --x+l x+l =:> a = 2,b = l,c = 1 x+l 1 + _1 = 1 + ( 2X1+ l = 1 +-) 2x + l 1+ _ l + x l + lx 2x + l + x + l = -3x + 2 2x + l 2x + l =:> a = 3,b = 2,c = 1 1+ = 1 + ( 3X 1+ 2 ) = 1 + 3x2x ++21 1 1+ 1 2x + l 1 + __1 1 + -x 3x + 2 + 2x + l 5x +3 3x + 2 3x +2 =:> a = 5,b = 3,c = 2 If we continue this process, the values of a, b and c produce the following sequences: 1, 2,3,5, 8,l3, 21, .. b : 1,1, 2,3,5,8, l3, 21, .. c : 0, 1, 1, 2,3, 5, 8, l3, 21, .... . In each case we have a Fibonacci Sequence, where the next value in the list is obtained from the sum of the previous 2 values in the list. Answers will vary. a :

. .

. . .

__

97.

19

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Chapter R: Review

Section R.8

1.

33.

9; -9

3. index

root m=W=3 r-s = �(_2)3 = -2

5. cube 7. 9.

35.

37.

39.

41.

�3x2 ..l12x = �36x2 . X = 6xvlx

21.

( .j5 �f = ( .j5 f ( � t

23.

= 5 · W = 5 Wi. = 5 · 3 � = 15 �

43.

45.

+

27.

3J2 4J2 = (3 + 4)J2 = 7J2

29.

-.J18 + 2../8 = -J9:2 + 2#2

47.

= -3J2 + 4J2 = (-3 + 4)J2 = J2

( vIx - 1 f = (vlxf - 2v1x + 1 = x - 2v1x + 1 �16x4 - mx = �8x3 · 2x -mx = 2x mx -mx = (2x - 1)mx �8X3 -3..l50x = �4x 2 . 2x -3..l25 · 2x = 2x£ -15£ = ( 2x -15)£

� - 3x�2xy + 5 �-2xl = �8X3 . 2xy _ 3x �2xy + 5 �r-_--y '---:3 . -2-- x-y = 2x�2xy -3x �2xy - 5 y �2xy = ( 2x - 3x - 5y )�2xy = ( -x - 5 y )�2xy or - (x + 5 y )�2xy 1 1 J2 = J2 J2 = J2 ' J2 7:

.j5 = --J15-J3 -J3 = ' .j5 .j5 .j5 5 J3 = J3 5 + J2 5 -J2 5 - J2 ' 5 + J2 J3 ( 5+J2)

25 - 2 J3(5 +J2) 5J3 + ./6 or 23 23

( J3 + 3)( J3 - 1) = ( J3 t + 3J3 - J3 - 3 = 3 + 2J3 - 3 = 2J3

31.

5ifi -2� = 5ifi - 2 · 3ifi = 5 ifi - 6 ifi = (5 - 6) ifi = -ifi

20

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exist. No portion of thi s material may be reproduced, in any form or by any means , without permission in writing from the publisher.

/

Section R. B: n th Roots; Rational Exponents

49.

2 -.[5 2 - .[5 2 -3.[5 2 + 3v5 2 + 3.[5 . 2 - 3.[5 4 - 2.[5 - 6.[5 + 15 4 - 45 - 8.[5 8-15 41 -41 if,t 5 if,t 5 5 if,t lfi lfi . 2r;

19

51.

53.

69.

19

=

( -VSY

(-27t 3 -V-27 -3

=

=

73.

=

=

=

=

= _

=

=

=

=

2

3

=

Y

=

---'-

-'----,:.

-'---

_

27 27 27 .fi 8 · 2.fi 16J2 16.fi · .fi 27.fi 32 33 2 3 (.fi t 27 . .fi 16.fi .fi

xy

=

=

22 4

9-3/ 2 _311_2 _1_ �3 271 9 ( J9 t 3 =

=

=

75.

61.

1 13

( X2 y (yt3 (xt3 (i t 3 X2 l 3 y2 / 3 X2 / 3 / / 3 X2 / 3 y 4 / 3 X2 / 3 y 2 / 3 X2 / 3+ 2 / 3-2 / 3 y I l 3+4 / 3-2 / 3 X2 l 3 yl x2 / 3 ( 16x2 y-1 /3 )3/ 4 163/ 4 (X2 t 4 ( y-1 /3 )3/4 X l / 4 ( y2 )1 / 4 (� t x3/ 2 y-1 / 4 x1 l4/ 1 2 2 3 X 3/ 2 -1 / 4 y -1 / 4 -1 / 2 8x5 / 4 y -3/ 4 8x5/ 4 /1 4 1 + X )1 / 2 x + 2 (I-+ xt 2 ( x -,-,- + 2 ( 1 + x ) 1 / 2 -(1 + X) 1 / 2 (1 + x) 1 I 2 - x(1+ +2 (Ix)+1I 2x) x + 2 + 2x (1 + X) 1 / 2 3x + 2 (I + X) 1I 2 =

( x + h) - 2)x ( x + h ) + x ( x + h) - x x + h - 2�X2 + xh + x x+h-x 2x + h - 2 �r-X-:-2-+-x-h h

57.

=

6

=

=

.Jx+h - Fx ..Jx+h - Fx ..Jx+h - Fx + h + vxr Jx + h + Fx · ..Jx+h - Fx vx�

8 2 /3

6

l

=

55.

( x 3 y ) 1 / 3 ( x3 ) 1 / 3 ( y )

27 27 8 · 2J2 16.fi 27J2

=

=

----n-

21

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exist. No portion of this material may be reproduced, i n any form or by any means, without permi ssion in wri ting from the publi sher.

Chapter R: Review

77.

79.

2x ( x2 + 1 )1/2 + x2 2"1 ( x 2 + 1 )-1/2 . 2x = 2x ( X2 + 1 )112 + ( 2 x3 1/2 X + 1) 2x ( x2 + 1 )112 . ( x2 + 1 )1/2 + x3 ( X 2 + 1 )1/2 2x ( x2 + l )1/2+1/2 + x3 2x ( x 2 + 1 Y + x3 ( X 2 + 1 )1/2 ( X 2 + 1 )112 2x 3 + 2x + x3 3x3 + 2x ( X2 + 1 )1/2 (X 2 + 1 )1/2 X ( 3 X2 + 2 ) ( X2 + 1 )112

83 .

.

--

�4x + 3 ·

,x > 5 � + �x - 5 . 5v4x � +3 2vx - 5 �x - 5 = �4x + 3 + ---.= 2�x - 5 5 �4x + 3 J4;+3 · 5 · J4;+3 + � · 2 · � 10�x - 5 �4x + 3 3 5 - 1 ( 4x x+ -)5+ 24x( x+- 5 ) O�( ) ( 3 ) 20x + 15 + 2x -10 10�( x - 5 )( 4x + 3 ) 22x + 5 10�( x - 5 )( 4x + 3 )

=

85.

_

81.

( �l + x - x · 2�:+ X ) (Ji+; - 2k l+x) l+x l+x -x ) ( 2Ji+;..,fl+; 2 �I + x l+x 2 ( 1 + x) - x 2( I + x) I /2 l + x 2+x 2 (1 + x) 3/ 2

( x+ 4)112 - 2x ( x + 4 rI l 2

[ [ [

x+4 ( X + 4 Y' 2 - X 2xY' 2 ( +4 x+4 2 ( X + 4 Y' 2 . ( x + 4 Y' 2 X 2x 2 ( x + 4 Y' ( + 4 Y' x+4 X + 4 - 2X ( x + 4 Y' 2 x+4 -x + 4 (X + 4( 2 x + 4 -x + 4 ( X + 4 ) 3/ 2 4-x ( x + 4 ) 3/ 2

J

J

J

2 - ( X2 -1 )112 x ( x2 -1 )1/2 ,x -1 x2 X 2 - ( x2 - 1 )112 . ( x 2 - 1 )112 ( x2 - 1 )112 2 x X2 - ( x2 -1 )1/2 . ( x 2 - 1 )1/2 ( x2 -1 )1/2 X2 x2 - ( x2 -1 ) ( x2 - 1 )1/2 X2 x( 2 _x2)1/2+ 1 2 x2 - 1 x

[

1

x 2 ( X2 -lt 2

22

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Section R. B: nth Roots; Rational Exponents

l + x2 - 2x"x -2Fx - -'x > 0 '----' ( 1 + X2 -)2 1 + X 2 - ( 2Fx) ( 2XFx) 2 Fx ( l + x2 t 1 + x2 - ( 2Fx) ( 2xFx) 2Fx r

87.

[

97.

) 99.

3X- 1 / 2 + % XII 2 ,X > 0 = l3/ 2 + 23 x 1 / 2 X ' = 3 · 2 +2X3xII //22 X 1 / 2 = 62X+l3x1 2 = 3(x2Xl+/ 22 ) J2 "'" 1 . 41

�(2)

1 . 4 1 42 1 3 5 6 2

_

101.

if4 "'" 1.59

' � � 4 ) 1 . 58740 1 0 52

89.

(x + l )3 / 2 + x .l2 (x + l) 1 / 2 1 03 .

2 + 1 "", 4 . 89 3 -"S

)

91.

93.

6Xl / 2 ( x2 + x) -8x3 / 2 - 8x l / 2 = 2X I / 2 ( 3(X2 + x) - 4x - 4) = 2X l / 2 ( 3x 2 - X -4 ) = 2X I l 2 (3x -4)(x + l) 3 ( X2 + 4f / 3 + x . 4 ( x 2 + 4t 3 · 2x

2 . 1 45268638

= ( X 2 + 4t 3 [ 3 ( x 2 + 4) + 8x 2 ] = ( X2 + 4t 3 [3x 2 + 12 + 8x 2 ] = ( x 2 + 4t 3 ( l lx2 + 12 )

95.

1 07. a.

b.

4(3x + St3 (2x + 3)3 / 2 + 3 (3x + St 3 ( 2x + 3t 2 = (3x + S) 1 / 3 ( 2x + 3 y / 2 [ 4(2x + 3) + 3(3x + S) ] = ( 3x + St3 (2x + 3t 2 (8x + 12 + 9x + lS ) = (3x + st3 (2x + 3) 1 1 2 (17x + 27) 3 . 2 where

x

�

4 . 8853 1 793 1

1 09.

T

V = 40(12/

�� - 0. 608

"'" IS, 660 . 4 gallons 9 V = 40 ( 1 ) 2 � 6 -0 . 608 "", 390.7 gallons 1

= 21r

m. = 21rJ2 "'" 8.89 seconds

--

23

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Chapter R: Review

111.

8 inches 8112 2/3 feet T = 2ffN) 2ff� 2ff( 41] "Jj6- 0. 9 1 seconds = 2Jj" = =

=

0

13.

0

1 5.

�

1 13 .

1 7.

Answers may vary. One possibility follows: If a = -5 , then .j;;i = �(_5)2 = J25 = 5 a Since we use the principal square root, which is always non-negative, .j;;i {a if a 0 = -a if a 0 which is the definition of l a l so .j;;i = l a l · *

1 9.

3 21.

- 43

I

71.-5 :0::; 4 -3x :0::; 2 -9:0::; -3x �-2 3 � x �-23

{ x l x :o:}:; } or ( -OO,}].

The solutionset is 1

67.

I-

I

o

I I I [I

I ..

I "I

I I

The solutionset is

The solutionset is

65.

I.

:0::;

The solutionset is

{xlx > -7} or (-7, 00) . I

0:0::; 2x -6 4 6:0::; 2x:o::; 10 3:O::; x :O::; 5

1

1

( 1 I 1

-6

I) I I .. 0

77.(x+2)( x-3»(x-l)(x+l) x2 -x -6 > x2 -1 -x-6> -1 -x > 5 x < -5 The solution set is {xl x < -5} or (-00, -5) .

"I"

I

1

"'1

) I I I I I-

-5

6

0

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 1: Equations and Inequalities

79.

x(4x + 3) � (2x + l)2 4x2 + 3x � 4x2 + 4x + l 3x � 4x + l -x � 1 x:::: - 1

87.

The solution set is

{xlx::::-l} , , , , [ , -I 0

81.

x + 1 < -3 -21 -< -3 4 6 � 4x + 4 < 9 2 � 4x < 5 -21 � x < -45

{l

or

Since

,

-

[

83.

I

1

,-

,

4

"2

89.

(4x + 2 t < 0 1 -4x + 2 < 0 4x + 2 < 0 x < --21

{l

91.

The solution set is x x < -�} or ( -00, -�) . I_I 85.

, )1

o

2

I"

_\

93.

0 < -x2 < 35 O < �x an d �x < l5 S ince �x > 0 , t his mean s tha t x > o . Therefor e, -X2 < 35 5X � < 5X 10 < 3x 10 -3 < x

95.

( ) (�)

{l }

The solution set is x x > �I or ('3° , 00 ) . 1

1

,

,

3

( 1

10

T

,

4

1

,

>

{} )

The solution set is x � � x < %} or [�, %). , 0

1 __

2x - 4 > 0 , this means that 2x - 4 0 . Therefore, 1 1 -2x - 4 < -2 1 1 2(x - 2) < -2 2(X - 2 2( - 2» < 2(x - 2> l < x-2 3<x The solution set is {x l x > 3} or (3, 00) .

[-1,00) .

, .

0 < ( 2x - 4 )-1 < '21 1 1 0 < -2x - 4 < -2 1 1 o < __ a nd __ < 1. 2x - 4 2 2x - 4

,

,

o

,

(�)

, ( 3

1.1-

If -1 < x < 1, then -1 + 4 < x + 4 < 1 + 4 3 < x+4 < 5 So, a = 3 an d b = 5. If 2 < x < 3, then -4(2) < -4(x) < -4(3) -12 < -4x < -8 S o,a = -12 a nd b = -8. If 0 < x < 4, then 2(0) < 2(x) < 2(4) 0 < 2x < 8 0 + 3 < 2x + 3 < 8 + 3 3 < 2x + 3 < 1 1 S o, a = 3 a nd b = 11. If -3 < x < 0, th en -3 + 4 < x + 4 < 0 + 4 l < x+4 -x + 4 > -4 1 -41 < -x+4 < 1 S o, a = "41 an d b = 1 .

'.1-

62

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Section 1.5: Solving Inequalities

97.

If 6 < 3x < 12, then 12 3x < -36 < 3 3 2<x- _1_ >- _1_ 1 . 12 1 + M 1 . 1 8 18, 000 > 1 8, 000 > 18, 000 1 . 12 - I + M - 1 . 1 8 1 6, 07 1 .43 � C � 1 5, 254.24 The dealer's cost varies from $ 15,254.24 to $ 1 6,071 .43, inclusive. =

+

1 -3 1 1 2x l > 3 2x < -3 or 2x > 3 x < - 23 or x > 23 x x < - or x >

-1

o

,-

o

[ ,

,

, , , , , , , , , , ,-

1 -4xl + I -5 1 � 1 1 -4xl + 5 � 1 1 -4xl � -4 This is impossible since absolute value always yields a non-negative number. The inequality has no solution. , ,

5 . 1

55.

, ..

1 1 - 2x I > 3 1 - 2x < -3 or 1 - 2x > 3 -2x < -4 or - 2x > 2 x > 2 or x < -1 {x l x < -l or x > 2} or ( -00, -I) u (2, 00) ,

- 1 2x- l l � -3 1 2x- 11 � 3 -3 � 2x- l � 3 - 2 � 2x� 4 -1 � x � 2 { x 1 -1 � x ::; 2} or [-1, 2] ,

2

, .. , , � , - I 0

4 . 9

53 .

1 1 - 4x l - 7 < -2 1 1 - 4x l < 5 -5 < 1 - 4x < 5 -6 < -4x < 4 -6 > x > 4 -4 -4 -32 > x > -1 or - 1 < x < -32 { x 1 - 1 < x < 1} or -1, 1 )

61.

, �, -

o

,) , 3

,-

4

5 + l x - l l > "21

9 I x - 1 1 > - "2 Absolute value yields a non-negative number, so this inequality is true for all real numbers, (-00, 00). , ... ,

o

67

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in wri ting from the publisher.


63.

65.

67.

69.

71.

73.

A temperature x that differs from 98.6° F by at least 1 .5°F . I x - 98.6° I � 1 .5° x - 98.6° S; -1 .5° or x - 98.6° � 1 .5° x S; 97. l° or x � 100. 1° The temperatures that are considered unhealthy are those that are less than 97. 1 of or greater than 100 . 1 "F, inclusive. The true average number of books read x should differ from 13.4 by less than 1 .35 books. I x - 13.4 1 < 1 .35 -1 .35 < x - 13.4 < 1 .35 12.05 < x < 14.75 Gallup is 99% confident that the actual average number of books read per year is between 12.05 and 14.75 books. x differs from 3 by less than I X -3 1 < 21 --21 < x - 3 < -21 -25 < x < -72

75.

I x - 21 S; 7 -7 S; x - 2 S; 7 -5 S; x S; 9 -15 S; x - l0 S; -1 1 1 -1 5 � -x - 10 � -1 1 -1 S; x -1 10 S; -15 --

a = -1 , b = -� 15 77.

�.

79.

x differs from -3 by more than 2. I x - (-3) 1 > 2 I x+3 1 > 2 x + 3 < - 2 or x + 3 > 2 x < -5 or x > -1 {x l x < -5 0r x > -I} I x - 11 < 3 -3 < x - 1 < 3 -3 + 5 « x - l) + 5 < 3 + 5 2 < x+4 < 8 a = 2, b = 8 I x + 41 S; 2 -2 S; x + 4 S; 2 -6 S; x S; -2 -12 S; 2x S; -4 -15 S; 2x - 3 S; -7 a = -15, b = -7

Given that a > 0, b > 0, and -.Ia 0 , we have b - a = ( .Jb + -.Ia) ( .Jb - -.Ia) > ° . Therefore, b - a > ° which means a < b. Prove l a + b l S; l a l + I b l · Note that l a + bl 2 = l a + b l · l a + bl · Case 1 : If a + b � 0, then l a + b l = a + b, so l a + b l · l a + b l = (a + b)(a + b) = a2 + 2ab + b2 s; l a l 2 + 2 I a l · l bl + l bI 2 = ( I a l + I b l) 2 by problem 78 Thus, (l a + b l) 2 S; (I a l + I bl )2 l a + bl S; l a l + I b l · Case 2: If a + b < O, then l a + b l = -(a + b), so l a + b l · l a + b l = ( -( a + b ) )( -(a + b) ) = (a + b) (a + b) = a2 + 2ab + b2 S; l a l 2 + 2 l a l ·l bl + I bl 2 = (I a l + I bl )2 by problem 78 Thus, ( I a + b\) 2 S; ( I a l + I bl f l a + b l S; l a l + I bl

68



Section 1. 7: Problem Solving: Interest, Mixture, Uniform Motion, and Constant Rate Job Applications

81.

Given that a > 0, x2 < a x2 - a < 0

91.

If x < -Fa , then x + Fa < 0 and x - Fa < -2Fa < o . Therefore, ( x + Fa ) ( x - Fa ) > 0 , which is a contradiction. If -Fa < x < Fa , then 0 < x + Fa < 2Fa and -2Fa < x - Fa < 0 . Therefore, ( x + Fa )( x - Fa ) < 0 . If x > Fa , then x + Fa > 2 Fa > 0 and x - Fa > 0 . Therefore, ( x + Fa )( x - Fa » O , which is a contradiction. So the solution set for x2 < a is { xl - Fa < x < Fa } . 83.

85.

x2 < 1 -Ji < x < Ji -1 < x < 1 The solution set is {x l - l < x < I} .

93 - 95.

89.

Answers will vary.

Section 1.7

x2 � 9 x � -.J9 or x � .J9 x � 3 or x � 3 The solution set is {xi x � -3 or x � 3} .

1. 3.

-

87.

1 3x - 12x + 11 1 = 4 3x - 12x + ll = 4 or 3x -12x + l l = -4 3x -12x + l l = 4 3x - 4 = 12x + 11 2x + 1 = 3x - 4 or 2x + 1 = -(3x -4) -x = -5 or 2x + 1 = -3x + 4 or 5x = 3 x=5 x=5 or x = -35 or 3x -12x + l l = -4 3x + 4 = 12x + ll 2x + 1 = 3x + 4 or 2x + 1 = -(3x + 4) -x = 3 or 2x + 1 = -3x - 4 x = -3 or 5x = -5 x = -3 or x = -1 The only values that check in the original equation are x = 5 and x = -1 . The solution set is {-I, 5} .

5.

x2 � 1 6 -.J16 � x � .J16 -4 � x � 4 The solution set is {xl - 4 � x � 4} .

7.

mathematical modeling uniform motion True; this is the uniform motion formula. Let A represent the area of the circle and the radius. The area of a circle is the product of times the square of the radius: A = r

1[

1[ r 2

9.

x2 > 4 x < -14 or x > 14 x < -2 or x > 2 The solution set is {xi x < -2 or x > 2} .

11.

13.

Let A represent the area of the square and the length of a side. The area of the square is the square of the length of a side: A = S2 Let F represent the force, the mass, and a the acceleration. Force equals the product of the mass times the acceleration: F = rna Let W represent the work, F the force, and d the distance. Work equals force times distance: s

rn

W = Fd

69




1 5.

1 7.

C = total variable cost in dollars, x = number of dishwashers manufactured: C = 1 50x Let x represent the amount of money invested in bonds. Then 50, 000 - x represents the amount of money invested in CD's. Since the total interest is to be $6,000, we have: 0. 15x + 0.07(50, 000 - x) 6, 000 (100)(0. I5x + 0.07(50, 000 - x)) = (6, 000)(100) 15x + 7(50, 000 - x) = 600, 000 1 5x + 350, 000 - 7x = 600, 000 8x + 350, 000 = 600, 000 8x = 250,000 x = 3 1, 250 $3 1,250 should be invested in bonds at 15% and $ 1 8,750 should be invested in CD's at 7%. Let x represent the amount of money loaned at 8%. Then 12, 000 - x represents the amount of money loaned at 1 8%. Since the total interest is to be $1,000, we have: 0.08x + 0. 1 8(12, 000 - x) = 1, 000 (1 00) (0.08x + 0. 18(12,000 - x)) = (1, 000)(100) 8x + 1 8(12, 000 - x) 100, 000 8x + 216, 000 - I 8x = 100, 000 -IOx + 216, 000 = 100, 000 -lOx = -1 16, 000 x = 1 1, 600 $ 1 1,600 is loaned at 8% and $400 is at 1 8%. Let x represent the number of pounds of Earl Gray tea. Then 100 - x represents the number of pounds of Orange Pekoe tea. 5x + 3(100 - x) = 4.50(100) 5x + 300 - 3x = 450 2x + 300 = 450 2x = 1 50 x = 75 75 pounds of Earl Gray tea must be blended with 25 pounds of Orange Pekoe.

23.

=

19.

25.

_

-

-

=

21.

Let x represent the number of pounds of cashews. Then x + 60 represents the number of pounds in the mixture. 9x + 3.50(60) = 7 .50(x + 60) 9x + 210 = 7 .50x + 450 1.5x = 240 x = 160 160 pounds of cashews must be added to the 60 pounds of almonds. Let r represent the speed of the current. Rate Time Distance 1 6-r I Upstream 1 6 - r 20 3 60 - 3 Downstream 16 + r �� = t 1 6+4-r Since the distance is the same in each direction: 1 6 - r = -16 + r -4 3 4(16 - r) = 3(16 + r) 64 - 4r = 48 + 3r 16 = 7r 16 � 2 .286 r=7 The speed of the current is approximately 2.286 miles per hour. Let r represent the speed of the current. Rate Time Distance Upstream l 5 - r 1 510- r 10 Downstream l 5 + r 1510+ r 10 Since the total time is 1 .5 hours, we have: � 15-r +� 15 + r = 1 .5 10(15 + r) + 1 0(15 - r) = 1 .5(15 - r)(15 + r) 150 + l Or + 1 50 - l Or = 1 .5(225 - r2 ) 300 = 1 .5(225 - r2 ) 200 = 225 - r2 r2 - 25 = 0 (r - 5)(r + 5) = 0 r = 5 or r = -5 Speed must be positive, so disregard r = -5 . The speed of the current is 5 miles per hour.

27.

--

--

70



Section 1. 7: Problem Solving: Interest, Mixture, Uniform Motion, and Constant Rate Job Applications

29.

Let r represent Karen's normal walking speed. Rate Time Distance With walkway r + 2.5 r +502.5 50 Against walkway r - 2.5 r -502.5 50 Since the total time is 40 seconds: � r + 2.5 + � r - 2.5 = 40 50(r - 2.5) + 50(r + 2.5) = 40(r - 2.5)(r + 2.5) 50r - 125 + 50r + 125 = 40(r2 - 6.25) 100r = 40r2 - 250 0 = 40r2 - 1 00r - 250 0 = 4r2 - l Or - 25 - 4(4)(-25) r = -(-10) ± �(-10)2 2(4) 10 ± v'5OO 10 ± lO.J5 5 ± 5.J5 8 8 4 r "" 4.05 or r "" -1.55 Speed must be positive, so disregard r "" -1 .55 . Karen' normal walking speed is approximately 4.05 feet per second. Let w represent the width of a regulation doubles tennis court. Then 2w+ 6 represents the length. The area is 2808 square feet: w(2w + 6) = 2808 2w2 + 6w = 2808 2W2 + 6w- 2808 = 0 w2 + 3w-1404 = 0 (w+ 39)(w - 36) = 0 w+ 39 = 0 or w - 36 = 0 w = 36 w = -39 or The width must be positive, so disregard w = -39 . The width of a regulation doubles tennis court is 36 feet and the length is 2(36) + 6 78 feet.

33.

I I tI

--

--

-,-_-'...-!"":",--=,,--,--,-,_,-,:,,, .-

31.

Let t represent the time it takes to do the job together. of job done Time to do job Part in one minute Trent 30 30 20 Lois 20 Together t 1 1 1 30 + 20 = ( 2t + 3t = 60 5t = 60 t = 12 Working together, the job can be done in 12 minutes. 1 = length of the garden w = width of the garden The length of the garden is to be twice its width. Thus, 1 = 2w . The dimensions of the fence are 1 + 4 and w+4 . The perimeter is 46 feet, so: 2(/ + 4) + 2( w + 4) = 46 2(2w+ 4) + 2(w + 4) = 46 4w + 8 + 2w + 8 = 46 6w + 16 = 46 6w = 30 w=5 The dimensions of the garden are 5 feet by 10 feet. b. Area = 1 . w = 5 . 10 = 50 square feet If the dimensions of the garden are the same, then the length and width of the fence are also the same (I + 4) . The perimeter is 46 feet, so: 2(/ + 4) + 2(/ + 4) = 46 21 + 8 + 21 + 8 = 46 4/ + 16 = 46 41 = 30 1 = 7.5 The dimensions of the garden are 7.5 feet by 7.5 feet. d. Area = 1 · w = 7.5(7.5) = 56.25 square feet.

35.

a.

c.

=

71

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37.

Let t represent the time it takes for the defensive back to catch the tight end. Time to Time Rate Distance 1 00 yards Tight 12 sec 1 00 25 25 t t 12 3 Ena -3 Def. t 11000 = 10 lOt 10 sec Back Since the defensive back has to 5 yards farther, we have: 25 3 t + 5 = 10t 25t + 1 5 = 30t 15 = 5 t t = 3 -)0 10t = 30 The defensive back will catch the tight end at the 45 yard line (15 + 30 45). Let x represent the number of gallons of pure water. Then x + 1 represents the number of gallons in the 60% solution. (%)(gallons) + (%)(gallons) = (%)(gallons) O(x) + 1(1) = 0.60(x + 1) 1 = 0.6x + 0.6 0.4 = 0.6x x = -46 = -23 2 "3 gallon of pure water should be added.

45.

run

_

run

47.

=

39.

41.

43.

_

_

Let t represent the time the auxiliary pump needs to run. Since the two pumps are emptying one tanker, we have: i+i = 1 4 9 27 + 4t = 36 4t = 9 t = = 2.25 The auxiliary pump must run for 2.25 hours. It must be started at 9:45 a.m. Let t represent the time for the tub to fill with the faucets on and the stopper removed. Since one tub is being filled, we have: /5 + - ;0 = 1 4t - 3t = 60 t = 60 60 minutes is required to fill the tub. Let t represent the time spent ing. Then 5 - t represents the time spent biking. Rate Time Distance Run 6 t 6t Bike 25 5 - t 25(5 - t) The total distance is 87 miles: 6t + 25(5 - t) = 87 6t + 125 - 25t = 87 -19t + 125 = 87 -19t = -38 t=2 The time spent ing is 2 hours, so the distance of the run is 6(2) = 12 miles. The distance of the bicycle race is 25(5 - 2) = 75 miles.

�

49.

( )

Let x represent the number of ounces of water to be evaporated; the amount of salt remains the same. Therefore, we get 0.04(32) = 0.06(32 - x) 1 .28 = 1 .92 - 0.06x 0.06x = 0.64 64 32 x - 0.64 0.06 - 6 - 3 - 101.3 10 1 "" 1 0.67 ounces of water need to be evaporated. Let x represent the number of grams of pure gold. Then 60 - x represents the number of grams of 12 karat gold to be used. x + � (60 - x) = � (60) x + 30 - 0.5x = 40 0.5x = 10 x = 20 20 grams of pure gold should be mixed with 40 grams of 12 karat gold. _

Let t represent the time it takes for Mike to catch up with Dan. Since the distances are the same, we have: 1 1 "6 t = "9 (t + l) 3t = 2t + 2 t=2 Mike will pass Dan after 2 minutes, which is a distance of � mile.

51.

-

runn

runn

72

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Chapter 1 Review Exercises

53.

55.

\020 meters/sec. In 9.99 seconds, Burke will run \020 (9.99) = 83.25 meters.

3.

Burke's rate is

Lewis would win by 1 6.75 meters. Let x be the original selling price of the shirt. Profit = Revenue - Cost

5.

4 = x - 0.40x - 20 � 24 = 0.60x � x = 40

The original price should be $40 to ensure a profit of $4 after the sale. If the sale is 50% off, the profit is:

40 - 0.50(40) - 20 = 40 - 20 - 20 = 0 At 50% off there will be no profit. 57.

7.

=

9.

(not possible) The time traveled with the tail wind

= 9 1 9 � 1 .6709 1 hours . 550 Since they were 20 minutes

(t hOur ) early, the

11.

time in still air would have been: 1 .6709 1 hrs + 20 min = (1 .6709 1 + 0.33333) hrs 2.00424 hrs Thus, with no wind, the ground speed is ��9 458.53 . Therefore, the tail wind is 2 . 424 550 - 458.53 = 9 1 .47 knots . �

�(x - ±) = % - � (12) ( �) ( x - ± ) = ( % -� } 12)

The solution set is {V} . 13.


�

2- =8

The solution set is

x(l - x) = 6 x - x2 = 6 o = x2 - x + 6 b2 - 4ac = (_1) 2 - 4 (1) (6) = 1 - 24 = -23

6x - 2 = 9 - 2x 8x = 1 1 11 X=8

�

6 - x = 24 x = -18

x = -6 -x-I 5 5x = 6x - 6 6=x Since x 6 does not cause a denominator to equal zero, the solution set is {6} .

Therefore, there are no real solutions.

t

1.

H}.

=

0.25x + 9.6 = 10.6 + 0.58x -0.33x = 1 x � -3 .03 liters

59.

3x - -x = 1 4 3 12 9x - 4x = 1 5x = 1 X = -1 5

The solution set is

It is impossible to mix two solutions with a lower concentration and end up with a new solution with a higher concentration. Algebraic Solution: Let x the number of liters of 25% solution. (% ) ( liters ) + (% ) ( liters ) = (% ) ( liters ) 0.25x + 0.48 ( 20) = 0.58( 20 + x)

was:

- 2(5 - 3x) + 8 = 4 + 5x - 1 0 + 6x + 8 = 4 + 5x 6x - 2 = 4 + 5x x=6 The solution set is {6} .

(x - 1)(2x + 3) = 3 2X2 + x - 3 = 3 2x2 + x - 6 = 0 (2x - 3)(x + 2) = 0 X = -3 2

or x = - 2

The solution set is

{- 1 8} . 73

{-2,%} .

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1 5.

2x+ 3 = 4x2 0 = 4x2 -2x-3 x -(-2) ± �(-2)2 = 2 ± ,J52 = 2 ± 2.Jlj = 1 ± .Jlj

25.

- 4(4)(-3)

2(4)

8

. The solutIOn set .

IS

17.

�X2 - 1 = 2 (�x2 _ d = (2)3 x2 -1 = 8 x2 = 9 x = ±3 Check x = -3 : �(_3)2 -1 = 2 :vB = 2 2=2

The solution set is 19.

{

8

4

1 - .Jl3 1 + .Jl3

}

-- ' -- . 4 4

27.

Check x = 3 :

�(3) 2 -1 = 2 :vB = 2

{-3,3} .

23 .

.Jx+l +.Jx-l = .J2x + l (.JX+l +.JX -I/ = ( .J2X + l t x + 1 + 2.Jx+l.Jx-I + x -1 2x + 1 2x + 2.Jx+1.Jx-l = 2x + 1 2.Jx+l.Jx- l = 1 (2.Jx+l.Jx- l) 2 = (1)2 4(x+ l)(x- l) = 1 4x2 -4 = 1 4x2 = 5 x2 = -45 x = -+ .J52 .J5 : Check x = 2 =

2=2

x(x+l)+2 = 0 x2 +x+2 = 0 x -1 ±�(1)2(1)2 - 4(1)(2) = -l ±2H

No real solutions. 21.

�2x+3 = 2 (�2x +3r = 24 2x +3 = 1 6 2x = 1 3 X = -132 13 : Check x = 2 4 2 ( 1� ) + 3 = �13 + 3 = � = 2 The solution set is { 1�} .

X4 -5x2 + 4 = 0 ( x2 -4 )( x 2 -1 ) = 0 x2 -4 = 0 or x2 - 1 = 0 x = ±2 or x = ±1 The solution set is {-2, -1, 1, 2} . .J2x-3 +x = 3 .J2x-3 = 3 - x 2x -3 = 9 - 6x + x2 x2 - 8x + 12 = 0 (x-2)(x-6) = 0 x = 2 or x = 6 Check x = 2: .j2(2) -3 + 2 = Jl + 2 = 3 Check x = 6: .j2(6) - 3 + 6 = .J9 +6 = 9 ,* 3 The solution set is {2} .

J1 +1 + J1 - 1 "N�) + I =

1 .79890743995 1 .79890743995

.J5 .' Check x = -2

�-� +I + �- � - I "N- �) + I

'

The second solution is not possible because it makes the radicand negative. The solution set is

{�} .

74

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29.


m or X = -m x = -l+n 1-n The solution set is {�, 1 + n n "# 1, n "# -1. 1 - n �},

2 X l/2 - 3 = 0 2Xl/2 = 3 ( 2X I /2 ) 2 = 32 4x = 9 x = '49 Check x = '49 : ( )1 () 2 '49 1 2 - 3 = 2 23 - 3 = 3 - 3 = 0

35.

{%} .

x-6 - 7x-3 - 8 = 0 Let u = x-3 so that u2 = x-6• u2 - 7u - 8 = 0 (u -8)(u + 1) = 0 or u = -1 u=8 or x-3 = -1 x-3 = 8 ( x-3 t 1 3 = (8rl1 3 or ( x -3 t l 3 = ( -1 r l/3 or x = -1 X = -21

3 7.

) () Check 21 : ( 2l � - 7 21 -3 - 8 = 64 - 56 - 8 = 0 Check - 1 : (-1 t - 7 ( -1 r3 - 8 = 1 + 7 - 8 = 0 The solution set is { -1, �} . 33 .

lOa2x2 - 2abx - 36b2 = 0 5a2x2 - abx - 1 8b2 = 0 (5ax + 9b}(ax - 2b) = 0 or ax - 2b = 0 5ax + 9b = 0 5ax = -9b ax = 2b 2b 9b X = -X= 5a a 2b } The solution set is { - 9b 5a , a , a "# O. �x2 + 3x + 7 - �x2 - 3x + 9 + 2 = 0 �x2 + 3x + 7 = �x2 - 3x + 9 - 2 ( �X 2 + 3x + 7 ) 2 = (�x2 - 3x + 9 _ 2) 2 x2 + 3x + 7 = x2 - 3x + 9 - 4 �x2 -3x + 9 + 4 6x - 6 = _4�X2 - 3x + 9 (6(x _ l) ) 2 = (-4�x2 _ 3x + 9) 2 36 ( x2 - 2x + 1 ) = 16 ( x2 - 3x + 9 ) 36x2 - 72x + 36 = 16x2 - 48x + 144 20x2 - 24x - 108 = 0 5x2 - 6x - 27 = 0 (5x + 9)( x -3) = 0 x = - 95 or x = 3 Check x = --95 ' ' -

x2 + m2 = 2mx + (nx) 2 x2 + m2 = 2mx + n2 x2 _ x2 n2x2 - 2mx + m2 = 0 ( 1 - n2 ) x2 - 2mx + m2 = 0 -m-i---4-(�1---n2--:-)-m-2 �r(--2_( -2m) -:-------±--'----, x=2 ( 1 - n2 ) 2m ± �4m2 --4m2 + 4m2n2 �-= --�� ) ( 2 1 - n2 2m ± � 2m ± 2mn 2 ( 1 - n2 ) 2 ( l - n2 ) _- 2m (l ± n) _- m (1 ± n) _- m (1 ± n) 2 ( 1 - n2 ) I - n2 (1 + n)(1 - n)

!!. _ 27 + 7 - !!. + 27 + 9 + 2 25 5 25 5 8 1 - 135 + 175 _ 81 + 135 + 225 + 2 25 25 �441 + 2 = .!.!. _ � + 2 = 0 _ = �121 25 25 5 5

75




x=3: �(3)2 +3(3) + 7 _ �(3)2 -3(3 ) +9 +2 = ../9 + 9 + 7 -../9 -9 + 9 + 2 = J25-J9+2 = 2+2 =4#0

Check


41 .

47.

{-�}.

1 2x+31 = 7 2x + 3 = 7 or 2x + 3 = -7 2x = 4 or 2x = -10 x = 2 or x = -5 The solution set is {-5, 2}. 1 2-3x l +2 = 9 1 2-3x l = 7 2-3x = 7 or 2-3x = -7 -3x = 5 or -3x = -9 x = 53 or x = 3 The solution set is {-�, 3 } 2x3 = 3x2 2x3 -3x2 = 0 x2 (2x-3 ) = 0 2x-3 = 0 x = 0 or x=-23 The solution set is

45.

•

t

14 49.

--

43 .

2x-3 2 �-x --+ 5 2 2(2x -3) + 10(2) � 5x 4x- 6 +20 � 5x 14 � x x � 14 {x\ x �14} or [14,(0 )

-9 � 2x+3 -4 � 7 3 6 � 2x+3 � -28 33 � 2x � -31 33 -> x -> _ � 2 2 33 31 - -2 -< x -< -2 {x \ - 321 � x � 3;} or [_ 321 , 3;] [

33

3 .1

2

2

51 .

2 < 3 -3x

)

-7

- 23

2x3 + 5x2 -8x -20 = 0 x2 (2x+ 5)-4(2x + 5 ) = 0 (2x+5)(x2 -4) = 0 2x + 5 = 0 or x2 -4 = 0 2x = -5 or x2 = 4 x = --25 or x = ±2 The solution set is { -%,-2,2 } .

53 .

\ 3x+4 1 < "2 -..!.. < 3x+4 b=0 or m1 - m2 =0 => m1 = m2 Since we are assuming that m1 *' m2, the only way that the two lines can have the same x-intercept is if b= O.

-s

13.

=

(x_h)2 + (y_k)2 =r2 (x -0)2 + (y 2 )2 = 2 2 x2 +( y_ 2 )2 4 General form: x 2 +/ - 4y + 4 = 4 x 2 + / - 4y =0 _

=

101

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Chapter 2: Graphs y

19.

5

x

( ) () (X-�J + / = ±

General form: x 2 - x + -1 + Y 2 -1 4 4 x2 + / - x = 0

-5

1 5.

(X _ h )2 + (y _ k)2 = r2 2 2 X - "21 + (y - O) 2 = "21

y 2

(x - h/ + (y _ k)2 = r2 (x _ 4)2 + (y _ ( _ 3»2 = 52 (X _ 4)2 + (y + 3)2 = 25 General form: x2 - 8x + 16 + / + 6y + 9 = 25 x2 + / - 8x + 6y = °

x

y

-2

x

21 .

x2 + y2 = 4 x2 + y2 = 22 a. Center: (0, 0) ; Radius = 2

b. 17.

( _ h) 2 + (y _ k) 2 = r2 (x _ ( _2) 2 + (y _ 1)2 = 42 (X + 2)2 + (y _ 1)2 = 1 6 General form: x2 + 4x + 4 + y2 - 2y + 1 = 16 x2 + / + 4x - 2y - 1 1 = ° X

y

5

x

y 8

-4 c.

x

-4

x-intercepts: x 2 + (0/ = 4 x2 = 4 x = ±J4 = ±2 y-intercepts: (0) 2 + y 2 = 4 y2 = 4 y = ± J4 = ±2 The intercepts are (-2, 0) , (2, 0) , (0, -2) , and (0, 2).

1 02

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Section 2.4: Circles

23.

c.

2 (x - 3f + 2/ = 8 (X_3) 2 + / = 4 Center: (3, 0); Radius = 2 a.

b.

y

5

x

y-intercepts: (0 _ 1)2 + (y - 2)2 = 32 ( _ 1)2 + (y _ 2)2 = 32 1 + (y_ 2) 2 = 9 (Y_ 2) 2 = 8 y - 2 = ±../8 y - 2 = ±2.fi y = 2 ± 2.fi The intercepts are ( 1 -.J5, 0 ) , ( 1 0) ,

-4 c.

25.

x-intercepts: (x - 3) 2 + (0) 2 = 4 (X_ 3) 2 = 4 x - 3 = ±.J4 x - 3 = ±2 x =3±2 x = 5 or x = 1 y-intercepts: (0 - 3/ + / = 4 (-3) 2 + / = 4 9 + y2 = 4 y2 = -5 No real solution. The intercepts are (1, 0) and (5, 0) .

( 0, 2 - 2.J2 ) , and ( 0, 2 + 2.J2 ) .

27.

+.J5,

x2 + / + 4x - 4y - I = 0 x2 + 4x + / - 4y = 1 (x2 + 4x + 4) + (/ - 4y + 4) = 1 + 4 + 4 (X + 2)2 + (y_ 2)2 = 32 Center: (-2, 2); Radius = 3 a.

b.

x2 + / - 2x - 4 y - 4 = ° x2 - 2x + /- 4 y = 4 (x2 - 2x + 1) + (/ - 4 y + 4) = 4 + 1 + 4 (X_ 1)2 + (y _ 2)2 = 32 Center: ( 1 , 2); Radius = 3 a.

b.

x-intercepts: (x _ 1)2 + (0 - 2)2 = 32 (X_ 1)2 + ( _2)2 = 32 (x - I) 2 + 4 = 9 (x - ll = 5 x - I = ±.J5 x = I ±.J5

y

-5

y

c.

x-intercepts: (x + 2)2 + (0 - 2)2 = 32 (X + 2)2 + 4 = 9 (X + 2)2 = 5 x+2 = x = -2±.J5

±.J5

x

-5

103

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Chapter 2: Graphs

(0+2)2+ (y -2)2=32 4+(y-2)2=9 (y _ 2)2 =5 y-2=±../5 y=2±../5 The intercepts are (-2-../5,0), (-2+../5,0), (0,2-../5), and (0,2+../5). x2 + i -x+2y+1=0 x2 - X +i + 2 y -1 (X2 -x+�) + (i +2y+1)=-1+�+ 1 (x- lJ + (y + 1)2 =(lJ Center: (.!. -1 ) ' Radius .!. 2 2" y-intercepts:

29.

31 .

2x2 +2i - 12x + 8y-24= x2 + i - 6x+4y = 1 2 x2 6x+ y2 + 4 1 2 (x2 -6 x+ 9) + (i +4 y+ 4)=12+ 9+ 4 (X-3)2+ (y+2)2 52 °

Y

_

a.

b.

Center: (3,-2); Radius Y 5

=

=

=

5

=

a.

=

b.

+ (0+2)2=52 (x-3)2+4=25 (X-3)2=21 x-3=±Jii. x=3±Jii. y-intercepts: (0-3i + (y +2)2 52 9 + (y + 2) 2=25 (y + 2)2 = 16 y+2=±4 -2±4 2 or y = -6 y The intercepts are (3 - Jii., 0) , (3+ Jii., 0) , (0,-6), and (0,2). 2X2+8x+2i = x2+4x+y2=0 x2 + 4 x+4+ i = 0+4 (x+ 2)2 y2 =2 2 Center: (-2,0); Radius: = 2

c.

y

2

x-intercepts: (x - 3)2

=

x

c.

=

( -lJ (0+ 1)2 =(lJ ( x-lJ + 1=� (x - �J 43

x-intercepts: x

+

33.

No real solutions

y-intercepts:

(0 -lJ +(y+ 1)2 (lJ 1 1 -+(y+ ) 2 =l 4 2 4 (y+ 1 ) = y + 1=0 y =- 1

=

°

a.

b.

=

Y

+

y

r

5

°

x

The only intercept is (0, -1) .

-5

104

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Section 2.4: Circles

(x+2)2+(0)2=22 (X+2)2=4 (X+2)2 =±J4 x+2=±2 x= 0 x=-2±2 or x=-4 y-intercepts: (0 + 2)2+ /= 22 4+/ =4 /=0 y=O The intercepts are (-4,0) and (0,0). Center at (0,0); containing point (- 2,3). r=�(-2-0)2+(3_ 0)2 =.J4+9 =J13 Equation: (x-0)2+ (y- oi =( J13t x2+ y2=13 Center at (2, 3); tangent to the x-axis. r=3 Equation: (X_2)2+(y_3)2=32 (X_2)2+(y_ 3)2=9 Endpoints of a diameter are (1, 4) and (-3,2). The center is at the midpoint of that diameter: Center: C+;-3)' T)=(-1,3) Radius: r= �(1-(_1))2+ (4 -3)2 =..J4+i=.J5 Equation: (x-(-1))2+(y_ 3)2=(.J5t (x+1)2+(y_3)2=5 Center at (-1,3); tangent to the line y 2. This means that the circle contains the point (-1,2), so the radius is r 1. Equation : (x+1)2+ ( y-3)2= (1)2 (x+1)2+(y_3)2=1 (c); Center: (1,-2 ) ; Radius 2 (b) ; Center: (-1,2); Radius 2

c.

35.

37.

39.

41.

47.

x-intercepts:

45.

(x,y)

x= y

x2+ y2=9 x2+X2=9 2x2=9 x2=-29 x= f2. = 3.J2 Vi 2

2x. Thus,

=,' = [2 3� J = (3J2)' =18 'qu", uniG. The diameter of the Ferris wheel was 250 feet, so the radius was 125 feet. The maximum height was 264 feet, so the center was at a height of 264-125=139 feet above the ground. Since the center of the wheel is on the y-axis, it is the point (0, 139). Thus, an equation for the wheel is: (x_O)2+(y_139)2=1252 x2+(y_139)2=15,625 x2+ y2+ 2x+4y-4091= 0 x2+2x+ /+4y-4091= 0 x2+2x+1+ / + 4y+4= 4091+5 (x+1)2+(y+2)2=4096 The circle representing Earth has center (-1,-2) and radius .J4096 = 64 . So the radius of the satellite's orbit is 64+0.6= 64.6 units. The equation of the orbit is (x+1)2+(y+2)2=(64.6)2 x2+ y2+2x+4y-4168.16= 0 The length of one side of the square is the area is A

49.

51.

=

=

=

43.

Let the upper-right comer of the square be the point . The circle and the square are both centered about the origin. Because of symmetry, we have that at the upper-right comer of the square. Therefore, we get

=

=

105

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Chapter 2: Graphs

53.

x2+/ = 9 Center: (0, 0) Slope from center to (1, 2.J2 ) is 2..Ji-0 = 2..Ji =2.J2 . 1-0 1 =--.J24 . SIope f the tangent I·me · 2,,2 Equation of the tangent line is: .J2 (x-l) y-2,,2 =- 4 y-2..Ji=- ..Ji4 x+ .J24 4y- 8..Ji= -..Ji x+..Ji ..Ji x+ 4y= 9..Ji .J2 x+4y- 9..Ji= 0 Let (h, k) be the center of the circle. x-2y+4=0 2y=x+4 1 y=-x+2 2 The slope of the tangent line is .!. . The slope 2 from (h, k) to (0,2) is -2. 2-k =_2 O-h 2-k=2h The other tangent line is y= 2x- , and it has slope 2. The slope from (h, k) to (3, -I) is _.!.. 2 -1-k -=--21 3-h 2+2k = 3 - h 2k= I -h h =1-2k Solve the two equations in h and k : 2-k=2(1-2k) 2-k=2-4k 3k=0 k=O h =1-2(0)=1 The center of the circle is (1, 0). 0

IS

57.

r::: -1

y=2 .

Therefore, the path of the center of the circle has the equation

r:::

55.

Consider the following diagram:

59.

(b), (c), (e) and (g) We need h, k > 0 and

61.

Answers will vary.

(0,0) on the graph.

Section 2.5 1. 3.

5.

7

7.

y=kx y=kx 2=10k k=-102 =-51 y=-x51 A =kx2 47r=k(2)2 47r= 4k 7r=k A = 7rX2 F=�d2 10=�52 10=�25 k=250 F = 250d2

106

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Section 2.5: Variation

9.

z

=k(x2+y2)

21.

5=k(32+42) 5=k(25) k=2.=� 25 5 I Z=S(x2 +y2 ) 11.

23.

kd2

M= J-; 24= 24=

k (42)

J9

16k 3

=

16=k(I)2 k=16 Therefore,we have equation s = 16t2. 1f t = 3 seconds,then s = 16(3)2 = 144 feet.

If s = 64 feet,then 64=1612 t2 = 4 t= ±2 Time must be positive,so we disregard t = -2. It takes 2 seconds to fall 64 feet.

( )

k=24 2 =� 16 2

13.

kB 6A9=k(1000) 0.00649=k Therefore we have the linear equation p = 0.00649B . If B=145000 ,then p = 0.00649(145000)= $941.05 . P

r2 = ka3 d2 k(23) 22 = 42 k(8) 4= 16

25.

E

3

=kW = k(20)

k=� 20 Therefore,we have the linear equation IfW=

4=!: 2 k=8 r2 = 8a3 d2

27.

E

=

�w. 20

15,then E=�(15)=2.25 . 20

R

=kg 47AO =k(12) 3.95= k Therefore,we have the linear equation R 3.95g . If g =10.5 ,then R =(3.95)(10.5) '" $41.48. =

17.

I

A=-bh 2

107


Chapter 2: Graphs

29.

D=� D=156, p=2.75; 156=_2.k7-5 k=429429 S D= . p 429 143 bags of candy D=-= 3 =� =600, =150; 600=� 150 k=90,000 . = 90,000 So, we have the equatIOn P 90,000 If = 200 , then = 200 450 cm3. W=�d2 If W=125, d= 3960 then k and k=1,960,200,000 125=-39602 . 1,960,200,000 So, we have the equation W= -'---'--'-d At the top of Mt. McKinley, we have d=3960+3.8=3963.8 , so W= 1,960,200,000 (3963.8)2 124.76 pounds.

= ksd3 36=k(75)(2)3 36 = 600k 0.06=k h

39.

P

a.

= 45 s =125, =0.06sd3. 45= (0.06)(125)d3 45=7.5d3 6=d3 d= :if6 1.82 inches =kmv2 1250=k(25)(10f 1250=2500k k=0.5 So, we have the equation = 0.5mv2. If m = 25 and v=15, then =0.5(25)(15)2=2812.5 Joules = kpd 100= k(25)(5) 75=125k0.75 0.6=k S we have the equatIOn = 0.6pd . If p = 40, d=8, and =0.50,then = 0.60.(40)(8) 50 =384 psi.

So, we have the equation h If h and then

0,

b. 31.

V

P

V

33.

V

K

K

--

-- =

S

43.

0,

t

.

S

t

--

t

S

45

�

37.

K

41.

P

V

P

�

-

47.

Answers will vary.


W=�d2 k_ 55=_ 39602 k=862,488,000

1.

fl a.

. W= 862,488,000 d2

b.

So, we have the equatIOn

d=3965,then W - 862,488,000 39652 54.86 pounds.

If

_ -

=(0,0) andP2=(4,2) d( fl ,Pz )=�(4_0)2+(2_0)2 = ../16 + 4 = Eo 2/5 The coordinates of the midpoint are: (X,Y)= (Xl +2 X2'Yl +2Y2 ) =(0+42 '�)=(� 2 2'�)=(2 2 '1) =

108

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1 = -= = 4-0 4 2 F or each run of 2, there is a rise of 1. F; = (1,-1) and P2 = (-2 , 3) d(F;,P2 ) = )(-2_1)2 +(3-(-1)/ = -h+16 =j2s = 5

c.

d.

3.

�y = 2-0 2 slope -

9.

LU

( )_( + 2 + ) = C+(- ) -1+3 ) = ( �1 , %) = (- �,1 ) �Y = 3-(-1) =4 = --4 slope =-2 -1 -3 3 For each run of 3, there is a rise of -4. F; = ( 4,-4) and 12 = ( 4,8) d(F;,P2 ) = )(4-4)2 +(8_(_4))2 = .JO+144 = .J144 = 12 LU

d.

5.

y = -y

The coordinates of the midpoint are: x Xl Y2 Yl - X,Y - 2 ' 2 2 2 ' 2

c.

11.

a.

b.

c.

d.

7.

Y

( ) = ( 4+42 '-4+8 ) = (�2'�2 ) = (4'2) 2

y = ±2 The intercepts are (-4, 0), (4, 0), (0, -2), and (0, 2). Test x-axis symmetry: Let Y

x2+4(-y)2 =16

= - = 8-(4-4-4) = -120 , undefined

slope �Y

y =-

x2 + 4/=16 same Test y-axis symmetry: Let x -x

LU

=

(_X)2 +4/=16 x2 +4/=16 same Test origin symmetry: Let x = -x and y = -y . (_x)2 +4(_y)2 =16 x2+4/=16 same

An undefined slope means the points lie on a vertical line. There is no change in x.

+4

2x = 3(-y)2 2x = 3y2 same Test y-axis symmetry: Let x = -x 2(-x) = 3/ -2x = 3/ different Test origin symm etry: Let x = -x and y = -y . 2(-x) = 3(_y)2 -2x = 3y2 different Therefore, the graph will have x-axis symmetry. x2+4y2=16 x-intercepts: y-intercepts: 2 x2 + 4 (0) =16 (Ol +4y2=16 x2 = 16 4/ =16 = ±4 y2 = 4 x

The coordinates of the midpoint are: + (x ,Y ) = xl + x2 ' Yl Y2 2 2

= x2

y-intercepts: 2(0) = 3/ y0 == Oy2

x =O The only intercept is (0, 0). Test x-axis symmetry: Let

a.

b.

= 3y2 x-intercepts: 2x = 3(0)2 2x = 0 2x

y

Therefore, the graph will have x-axis, y-axis, and origin symmetry.

109

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Chapter 2: Graphs

13.

= X4+2X2+1 y-intercepts: x-intercepts: y = (0)4+2(0)2+1 0 = x4+2x2+1 0 = (X2+1)(x2+1) = 1 x2+1 = 0 x2 = -1 no real solutions The only intercept is (0, 1). Test x-axis symmetry: Let y -y -y = X4+2X2+1 y = _x4-2X2-1 different Test y-axis symmetry: Let x = -x y = (-xt +2(_x)2+1 y = X4+2X2+1 same Test origin symmetry: Let x = -x and y = -y . -y = (_X)4+2(_X)2+1 -y = X4+2X2+1 y = -x4-2x2-1 different Y

17.

19.

=

21.

Y

=

x

Therefore, the graph will have y-axis symmetry.

1 5.

(x_h)2+(y_k)2 =r2 (x_(_2))2+(y-3/ = 42 (X+2)2+(y_3)2 = 16 (X_h)2+(y_k)2 = r2 (x-(_1))2+( y_(_2))2 = 12 (x+l)2+(y+2)2 = 1 x2+(y_1)2 = 4 x2+(y_1)2 = 22 Center: (0,1); Radius 2

x2+x+i +2y = 0 x-intercepts: x 2 +x+ (0)2 +2(0) = 0 x2+x == O0 x(x+l) x = 0, x = -1 y-intercepts: (0)2+ 0 +i + 2y = 0 iy(y+2) +2y == 00 y = O,y = -2 The intercepts are (-1,0), (0,0), and (0,-2). Test x-axis symmetry: Let y = -Y x2+X+(_y)2+2(-y) = 0 x2+x+i -2y = 0 different Test y-axis symmetry: Let x = -x (_X)2+(-x)+i +2y = 0 x2 X+ y2+2y = 0 different Test origin symmetry: Let x = -x and y = -y . (_x)2+(_x)+(_y)2+2(-y) = 0 x2-X+i -2y = 0 different

x2+(0-1/ = 4 x2+1 = 4 x2 = 3 x = ±.J3 y-intercepts: 02+ (Y -1/ = 4 (y_ l)2 = 4 y- l = ±2 = I±2 y y = 3 or y = -1 The intercepts are (-.J3, 0), (.J3, 0), (0, - 1 ) , and (0,3). x2+ y2_ 2x+ 4y-4 = 0 x2-2x+i +4y = 4 (X2-2x+l)+(i +4y+4) = 4+1+4 (x_ l)2+(y+2/ = 32 Center: (1, -2) Radius 3 x-intercepts:

23.

_

=

The graph has none of the indicated symmetries. 110

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Y 5

(

2 x-intercepts: (x _ 1) 2 + (0 + 2) 2 = (.J5 ) (x -l) 2 + 4 = 5 (x _ 1) 2 = 1 x -I = ± 1 x =1±1 x = 2 or x = ° 2 y-intercepts: (0 _ 1) 2 + (y + 2) 2 = J"S) 1 + (y + 2)2 = 5 (y + 2) 2 = 4 y + 2 = ±2 Y = -2 ± 2 y = O or y = -4 The intercepts are 0, 0 ) , 2, 0 ) , and 0,- 4 ) .

(

( ( (

x-intercepts: x _ l ) 2 + 0 + 2 ) 2 = 3 2 x _ l)2 + 4 = 9 x _l) 2 = 5 x -I = ± J"S x = 1 ± J"S 2 y-intercepts: 0 _ 1 ) + y + 2 ) 2 = 3 2 1 + y + 2)2 = 9 y+2 =8 y + 2 = ± .J8 y + 2 = ±2 J2 y = -2 ± 2J2 The intercepts are 1 - J"S , 0 ) , 1 + J"S , 0 ) ,

(

( (

( ( ( /

( 0, -2 - 2J2 ) , and ( 0, -2 + 2J2 ) . (

(

25.

(

(

(

)

Center: ( 1 , -2) Radius = J"S

2 7.

Slope =-2 ; containing (3,- 1 ) y - Y I = m x - XI ) y - (-I) = -2 x - 3 ) y +l = -2x + 6 y = -2x + 5 or 2x + y = 5

29.

vertical; containing (- 3,4) Vertical lines have equations of the form x = where is the x-intercept. Now, a vertical line containing the point (-3, 4) must have an x-intercept of -3, so the equation of the line is x = -3. The equation does not have a slope intercept form.

( (

a,

a

3x2 + 3y 2 - 6x + 1 2y = ° x 2 + i - 2x + 4 Y = ° x 2 - 2x + y 2 + 4y = ° x 2 - 2x + l + i +4y +4 = 1 + 4 x_l ) 2 + y + 2 ) 2 = J"S )2

)(

(

31.

y-intercept = -2; containing (5,- 3) Points are (5,-3) and (0,-2) - 2 - (-3) = 1 = --1 m= 5 0-5 -5 Y = mx +b 1 y = --x - 2 or x + 5y = - 1 0 5

33.

Parallel to 2x - 3y = -4 2x - 3y = - 4 -3y = -2x - 4 -3y -2x - 4 -3 -3 4 2 y = -x +3 3

(

y 5

-5

Slope = � ; containing (-5,3) 3 111

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Chapter 2: Graphs

35.

y-Yl=m (x-xl ) 2 Y -3=-(x-( 32 -5)) y-3=-(x+5) 32 1 0 y-3=-3 x+-3 2 1 9 or 2x-3y= - 1 9 y=-x+3 3 Perpendicular to x+Y = 2 x+y=2 y=-x+2 The slope of this line is -1 , so the slope of a line perpendicular to it is 1 . Slope 1 ; containing (4,-3) y-Yl=m(x-xl) y-(-3)=l(x-4) y+3=x-4 y = x- or x-y= 4x-5y=-20 -5y=-4x-20 4 y=-x+4 5 slope �; y-intercept= 4 x-intercept: Let y O. 4x-5(0)=-20 4x=-20 x=-5

39.

1 1 1 -x--y=-6 2 3 1 1 1 --y=--x-6 3 32 I y=-x+2 2 3 . ="21 sIope "2; y-mtercept x-intercept: Let y O. 1 1 O)=--1 -x--( 2 3 1 61 -x=-61 2 x=--3 =

=

=

7

37.

x

7

41.

2x-3y= 1 2 x-intercept: 2x - 3(0) =12 2x = 1 2 x=6

y-intercept: 2(0) - 3y =12 12

-3y= y=-4 The intercepts are (6,0) and (0, -4) .

=

=

Y 5

x

-5

11 2

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43.

1 1 -x+-y=2 3 x-i21 ntercept: y-intercept:1 1 1 2"x+3" 1(O)=2 2" (O)+3"1 y=2 -x=2 -y=2 2 x=4 3 y=6 The intercepts are (4,0) and (0,6)

49.

3.

.

45.

Y

Given the points A = (- 2,0),B = (-4,4),and e = (8,5). Find the distance between each pair of points. d(A,B)= �( -4_( _2))2 +( 4-0)2 =�4 +16 =Ea=215 d(B,e)=�( 8-( -4))2 +( 5-4)2 = �144+1 =Ms d(A,e)= �( 8-(-2))2 +( 5-0f =�100+25 =JW=515 [ d(A,B)J +[ d(A,e)J =[ d(B,e)J (Eat + (JW/ = (Ft45t 20+125=145 145=145 The isPythagorean Theorem is satisfied,so this a right triangle. Find the slopes: 4 =-4-(4-0- 2) =-=-2 -2 1 5-4 = 8-(-4)=12 5-0 = 5 = 1 =8-(-2) 10 "2 Since =-2 .�2 = -1, the sides and are perpendicular and the triangle is a rightAetriangle. 1-5 slope of -AB =--=-1 6-2 slope of A-e =-1-5 --= -1 8-2 slope of Be =-1-1 --=-1 8-6 lie on a line. Therefore,the points

=x3

b.

x

mAB

m BC

47.

m

slope �,3 containing the point (1,2) y

=

m ·m AC AB

5 1.

2

4

AC

x

113

AB

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Chapter 2: Graphs

53.

=kg 46.67=k(13) k=46.1367 =3.59 Therefore, we have the equation R=3.59g . If g=11.2 , then = 3.59(11.2) "" $40.2 1. R

Chapter 2 Test 1.

p

2.

3652=(k)(93)3 3652 k=933 Therefore, we have the equation T2= 3659332 a3 If 88 days, then 882=(396;2}a)3 a3 =(882)(�J a= 3 (882)(:::2 ) "" 36 million miles The graph of X= 0 is a vertical line passing through the origin. That is, X= 0 is the

d(�,P2)=�(5_(_1))2+(_1_3)2 =�62+(_4)2 =�36+16 =.J52=2Jl3 The coordinates of the midpoint are: x2 Yl +2 ) (x,Y)= (Xl +2' 3+(-1)) =(-1+5 2 ' 2 =(�'3) (2, 1) -1-3 -4-=--2 --m =--= 5-( -x X2 I -I) 6 3 If X increases by 3 units, will decrease by 2 units. =x2-9 Y2

=

T=

Y2- Yl

3. a.

b.

4.

Y

Y

x

57. a.

b.

c.

equation of the y-axis. The grapy of Y is a horizontal line passing through the origin. That is, Y the equation of the x-axis.

=0

x+ =0 y=-x Y

=

0 is 5.

x+ = 0

-1.

xy=0 y = 0 or x= 0 The graph of xy= 0 consists of the coordinate axes. x 2 + y 2= 0 =0 and x=0 The graph of x2 + y2=0 is consists of the d.

e.

y 5

The graph of y is line passing through the origin with slope =

=x

/

(0, -9)

-5

(1,-1)(4_2) ,

(9,-3)

Y

origin.

114

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exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing from the publi sher.

Chapter 2 Test

6.

x2+Y=9 x-2intercepts: x +0=9 x2=9 x= ±3

�

y-intercept: (0)2+Y=9 y=9

The intercepts are (-3,0), (3,0), and (0,9). Test x-axis symmetry: Let

7.

y=-y

5

x2+(-y)=9 x2- Y=9 different Test y-axis symmetry: Let x=-x (_x) 2+Y = 9 x2+ y=9 same Test origin symmetry: Let x=-x and y=-y (_x) 2+ (_Y ) =9 x2-y=9 different Therefore, the graph will have y-axis symmetry. Slope -2; containing (3,-4) y-y,= m(x-x,) y-(-4)=-2(x-3) y+4=-2x+6 y =-2x+2

-5

10.

=

2x+3y=6 3y=-2x+6 2 y=--x+2 3 Parallel line Any line parallel to

2x+3y=6 has slope

m= -�.3 The line contains (1,- 1) : y-y, = m(x-x,) 2 y-(-l)=--(x-I) 32 2 y+1=--x+32 31 y =--x-3 3 Perpendicular line Any line perpendicular to 2x+3y =6 has slope m= �.2 The line contains (0,3) : y-y, = m(x-x,) 3 y-3=-(x-O) 23 y-3=-x 23 y=-x+3 2

y

x

8.

�+y2 2+4x-2y-4=0 x +4x2+ y2-2y=4 2 (x +4x+4)+(y2 -2y+l)2=4+4+1 ( x+2) + (y_1) = 32 Center: (-2, I); Radius = 3 Y

(x_h)2+(y_k)2=r2 (X_4 ) 2 + ( y_(_3) ) 2=5 2 ( X_4) 2 + ( y+3 ) 2= 25 (x_4) 2 + ( y+3 ) 2= 25 General form: x2-8x+I6+ y2+6y+9=25 x2+ /-8x+6y=0 115

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Chapter 2: Graphs

11.

Let R = the resistance, length,and r = radius. Then R= k·-. r2 Now, R 10 ohms,when 50 feet and r= 6 x10-3 inch, so 50 10= k. (6x1O-3r (6xIO-3)2 = 7.2xI 0-6 = 10· Therefore,we50have the equation R= (7.2x1 0-6 )�r . If =100 feet and r = 7 10-3 inch,then R= (7.2x1 0-6 ) ( 7 x11000-3)2 ",,14.69 ohms. I

I=

7.

=

1=

9.

k

I

11.

x

3.

5.

2x - 3 � 7 2x � 1 0 x�5 {x x � 5}

!

or (-00,5]

5

13.

Chapter 2 Cumulative Review 1.

! x - 2!=1 x - 2 =1 or x - 2 =-1 x=3 The solution set is x=1 { I, 3 } . x2=-9 x=±H x= ±3i The solution set is {-3i, 3i} .

3x - 5 =0 3x = x=3

55 The solution set is {%}. 2X2-5x-3 = 0 ( 2x + I )( x - 3 )= 0 x = - -I or x= 3 2 The solution set is {-�, 3} . x 2 + 2x +5 = 0 r-2 -2 ± �'2 --;-:_-4--2( 1 )--'-( 1'-'-)-(5--'-) -----' x= -2±� 2 -2±..J-16 No real2solutions

!

x-2! � 1

-1 �x-2 � 1 l�x�3 { x l l � x:::;3 }

17.

] 3

= �(_1_4)2 + ( 3 - (_2))2 = �(_5)2 +(5)2 =.J25+25 = .J50 = 5.fi 3+(-2) = (�!) MidPoint= (-1+4 2 ' 2 J 2' 2 = x3 o

15.

[ 1

or [1, 3]

2

4

d ( P, Q)

y

x

116

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exist. No portion oflhis material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 2 Cumulative Review

19.

Perpendicular to Y=2x+ 1 ; Contains (3,5) Slope of perpendicular -.!.2 Y-YI =m (x-xl ) y-5=--2 (x-3) 1 -3 y-5=--x+ 21 132 y=--x+ 2 -2 =

1

y

-10

-6

-£.

2

-4 --6

2

4

6

8 10 12

x

-8

-10

117

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Chapter

3

Functions and Their Graphs Section 3.1

1. 3.

5. 7.

27.

We must not allow the denominator to be 0. x + 4 *- ° => x *- -4 ; Domain: { xl x *- -4} . independent; dependent

29.

[0, 5] We need the intersection of the intervals [0, 7] and [-2, 5] . That is, domain of/ n domain of g . "I

[

-2

0

-2

o

[

1

1

1

5

]

-2

0

7

1

5

7

5

7

�

g

31.

11.

True

13.

False; if the domain is not specified, we assume it is the largest set of real numbers for which the value of/is a real number.

19.

Not a function

21.

Function Domain: {I , 2, 3, 4} Range: { 3}

23.

Not a function

25.

Function Domain: {-2, - 1 , 0, I } Range: { O, 1 , 4}

/ = 4 - x2 =

33.

=

x =/ Solve for y: y ±.,J; For x 1, Y = ±1. Thus, (1, 1) and (1, -1) are on the graph. This is not a function, since a distinct x -value corresponds to two differenty-values. =

=

Function Domain: {Elvis, Colleen, Kaleigh, Marissa} Range: {Jan. 8, Mar. 15, Sept. 1 7} Not a function

= -

Solve for y: y = ±�4 _ X2 For x 0, y ±2 . Thus, (0, 2) and (0, -2) are on the graph. This is not a function, since a distinct x value corresponds to two differenty-values.

(g- /) (x) or g (x) - /(x)

17.

1 . The graph passes the vertical line x test Thus, the equation represents a function. Graph y

]�f

�"41-4-4-4[�--��]+-+-+1·· f+g

15.

=

(-1,3)

..III

9.

Graph y x 2 . The graph passes the vertical line

35.

Graph y 2 X 2 - 3x + 4 . The graph passes the vertical line test. Thus, the equation represents a function. =

118

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Section 3.1: Functions

37.

39.

2X2+3y2=1 Solve fory: 2X2+3i =1 3i =1-2x2 1-2x2 Y2=-3 1- X2 Y=±� � For x= 0, y= ±Jf . Thus, (o,Jf) and (0,-Jf) are on the graph. This is not a function, since a distinct x-value corresponds to two differenty-values. f(x)=3x2+2x-4 f(0)=3(0)2+2(0)-4=-4 f(I)=3(1)2+2(1)-4=3 +2-4=1 f(-I)=3(-1)2+2(-I)-4=3-2-4=-3 f(-x)=3(_X)2+2(-x)-4=3x2-2x-4 -f(x)=-(3x2+2x-4)=-3x2-2x+4 f(x+l)=3(x+l)2+2(x+l)-4 =3 (X2+2x+1)+2x+ 2-4 =3x2+6x+3 +2x+2-4 =3x2+8x+l f(2x)=3(2x)2+2(2x)-4=12x2+4x-4 f(x+h)=3(x+h)2+2(x+h)-4 =3(x2+2xh+h2)+2x+2h-4 =3x2+6xh+3h2+2x+2h-4 f(x)=+ x +1 o2 0 f(O)=-=-=O 0 +1 1 f(l)_ 121+1 _ 21

c.

d. e. f.

f(2x)=(2x)2x2+1 4x2x2+1 x+h x+h = f( x+h)=(x+h) 2+1 x2+2xh+h2+1 f(x)=lxl+4 f(0)=1 01+ 4=0+4=4 f(1)=111+4=1+4=5 f(-1)=I- l l+4=1+4=5 f(-x)=I-x 1+4=1 xl+4 -f(x)=-(Ixl+4)= -I x1-4 f(x+l)=lx+lj+4 f(2x)=1 2x1+4=21 x 1+4 f(x+h)=lx+hl+4

g.

a.

h.

b.

c.

43.

d.

a.

b.

e.

c.

f.

d.

e. f.

g.

g.

h.

41.

-1 =--1 f(-1)_- (_1)-12+1 1+1 2 f(-x)= (-x):+1 x -x -f(x)=- (x2+)1 =x2+1 f(x+l)=(x+l)x+l2+1 x+l 2 x +2x+1+1 x+l

h.

a.

b.

119

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Chapter 3: Functions and Their Graphs

45.

f (x)== 2x + 1 3x - 5 a. b. c.

d. e. f.

g. h.

47.

49.

51.

53.

55.

2 (0) + 1 0 + 1 _.!. = = 3 (0) - 5 0 - 5 5 2 (1) + 1 f (1)== 3 (1) == 2 + 1 ==2==_ � -5 3-5 -2 2 2 (-1) + 1 - 2 + 1 2 .!. == = = f (-I)== 3 (-1) - 5 -3 - 5 - 8 8

f (0) =

f (-x) =

57.

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

( )

_f (x) = _ 2X + l == - 2x - l 3x - 5 3x - 5 2 ( x + 1) + 1 2x + 2 + 1 2x + 3 == == f (x + 1) = 3(x + l) - 5 3x + 3 - 5 3x - 2 2 (2x) + 1 4x + l = f (2x)== 3 ( 2x ) - 5 6x - 5 f ( x + h ) ==

59.

61.

----

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

f(x) = -5x + 4 Domain: {x I x is any real number }

x f(x) ==-x2 + 1 Domain: {x I x is any real number }

--

g(x) == x x2 - 1 6 x 2 - 1 6:;t 0 x 2 :;t 1 6 => x:;t ±4 Domain: {xl x:;t -4, x:;t 4}

h e x) == -J3x - 1 2 3x - 12 � 0 3x � 1 2 x�4 Domain: {xl x �4} 4 -Jx - 9 x-9 > 0 x>9 Domain: {xl x > 9} f(x) =

g .f2

p(x) == -== ,---; x - I vx - l x-I > 0 x>1 Domain: {xl x > I} f(x)==3x + 4 g (x) == 2x - 3 a. (f + g)(x) == 3x + 4 + 2x - 3 == 5x + 1 Domain: {xl x is any real number } . b. (f - g)(x) = (3x + 4) - (2x - 3) = 3x + 4 - 2x + 3 = x+7 Domain: {xl x is any real number } . c. (f · g)(x) = (3x + 4)(2x - 3) = 6x 2 - 9x + 8x - 1 2 == 6x 2 - x - 1 2 Domain: {xl x is any real number } . d.

( fg ) (x)== 3x2x +- 43

2x - 3:;t 0 => 2x :;t 3 => x :;t -3 2

-2 F(x) = -x3 X +x x3 + x:;t 0 x(x 2 + 1):;t 0 x:;t 0, x 2 :;t -l Domain: {x l x:;t O}

e.

f.

g.

h.

(f + g )(3) = 5(3) + 1 = 1 5 + 1 == 1 6 (f - g )( 4) ==4 + 7 == 1 1 (f . g )(2) = 6(2) 2 - 2 - 12 = 24 - 2 - 12 == 10 f (1) = 3(1) + 4 = 3 + 4 ==� = -7 2(1) - 3 2 - 3 -1 g

()

120

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63.

f(x)=x-I g(x)= 2X2 (f+g)(x)=x-1+2x2= 2x2+x-1 Domain: { x I x is any real number } . (f-g)(x)=(x-1)2 - (2X2 ) = _2X +x-1 Domain: {xI x is any real number } . (f ·g)(x)= (x-1)(2x2)=2x3 _2x2 Domain: {xI x is any real number } . (gf}x)=� 2X2 Domain: {xlx;toO}. (f+g)(3)=2(3)2+3 -1=2(9)+3-1 =18+3-1= 20 (f-g)(4)=-2(4)2+4-1 =-2(16)+4-1 =-32+4-1=-29 (f.g)(2)=2(2)3 -2(2)2=2(8)-2(4) =16-8=8 (;)Cl)= ;(�)� = 2�1) = %=0 f(x)=../x g(x)=3x-5 (f+g)(x)=../x+3x-5 Domain: {xlx�O}. (f-g)(x)=../x-(3x-5)=../x-3x+5 Domain: {xlx�O}. (f.g)(x)= ../x(3x-5)=3x../x-5../x Domain: {xlx�O}. ../x (gf}x)= 3x-5 x� 0 and 3x-5;to 0 3x;to5 x;to-53 Domain: {xl x� 0 and x;to %} .

a.

f.

b.

g.

c.

h.

d.

67.

a.

e. f.

b.

g.

c.

h.

65.

(f+g)(3)=J3+3(3)-5 =J3+9-5=J3+4 (f -g)(4)= 14-3(4)+5 =2-12+5=-5 (f .g)(2)=3(2)-fi-5./2 = 6-fi-5-fi= ./2 _l_ =-2_l=_.!.2 (gf}l)= � 3(1)-5 = 3-5 f(x)=I+-x1 g(x)=-x1 1 1+-2 (f+g)(x)=1+-x1 +-= x x Domain: {xlx;toO}. 1 1 1 (f-g)(x)=1+---= xx Domain: {xlx;toO}. 1 1 (f .g)(x)= (1+-x1 r-=x X x2 Domain: {xlx;toO}. x+1 x+1 f( }x)= 1+.!.x = _x_ g .!.x .!.x = x . .:.1 =x+1 Domain: {xlx;toO}. 2 5 (f+g)(3)=1+-=3 3 (f-g)(4)=1 1 -1 =-+-=1 1 3 (f ·g)(2)=-+ 2 (2)2 2 4 4 ( ; )cl)=1 1= 2

e.

d.

a.

b.

+-

__

e.

f.

c.

g.

d.

h.

+

=>

121


exist. No portion of this material may be reproduced, in any form or by any means, without permi ssion in writing from the publisher.


69.

f(x) = 2x + 3 3x - 2 a.

4x -3x - 2

g.

--

4x (f + g)(x) = 2x + 3 + -3x - 2 3x - 2 2x + 3 + 4x 6x + 3 3x - 2 3x - 2 3x - 2 * 0 3x * 2 => x * Domain:

b.

g(x ) =

h.

i { x l x * i} .

71.

3 -� (f - g)(x) = 2x + 3x - 2 3x - 2 2x + 3 - 4x - 2x + 3 3x - 2 3x - 2 3x - 2 * ° 73.

c.

( ) (�)

2 = 8x + 12x (f . g)(x) = 2X + 3 3x - 2 3x - 2 (3x - 2) 2 3x - 2 * 0 75.

d.

(]

2x + 3 f (x) = 3x - 2 = 2x + 3 . 3x - 2 = 2x + 3 g � 3x - 2 4x 4x 3x - 2 3x - 2 * ° and x * ° 3x * 2 x * -2 3

{l %

(f . g)( 2) = 8(2i + 12(22 ) (3(2) - 2) 8(4) + 24 = 32 + 24 = 56 = 7 (6 _ 2) 2 (4)2 16 "2

( fg ) = 2(14(1)) + 3 = 2 4+ 3 = �4 ( ) I

1 (f + g)(x) = 6 - -x 2 1 6 - - x = 3x + l + g(x) 2 7 5 - - x = g(x) 2 g(x) = 5 - -7 x 2 f(x) = 4x + 3 f(x + h ) - f(x) 4(x + h) + 3 - (4x + 3) h h 4x + 4h + 3 - 4x - 3 h 4h =-=4 h

f(x) = 3x + 1

f(x) = x 2 - x + 4 f(x + h ) - f(x) h (X + h) 2 - (x + h ) + 4 - (x 2 - x + 4) h x 2 + 2xh + h 2 - x - h + 4 - x 2 + x - 4 h 2 2xh + h - h h = 2x + h - l

}

Domain: x x * and x * o . e.

f.

(f + g )(3) = 6(3) + 3 = 1 8 + 3 = E = 3 3(3 ) - 2 9 - 2 7 (f - g )( 4) = - 2(4 ) + 3 = - 8 + 3 = -=2 = _ .!. 3(4) - 2 12 - 2 1 0 2

122

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77.

f { x ) = 3x 2 - 2x + 6 f {x + h) - f {x) h 3 { X + h ) 2 - 2 { X + h ) + 6 - [3x 2 - 2x + 6 ] h 3 ( X 2 + 2xh + h 2 ) - 2x - 2h + 6 - 3x 2 + 2x - 6 h 3x 2 + 6xh + 3h 2 - 2h - 3x 2 6xh + 3h 2 - 2h h h = 6x + 3h - 2

[

79.

8 1.

83.


85.

J

87.

f(x) = 2x - A and f(4) = 0 x-3 2(4) -A f(4) = 4-3 8 0= A 1 0 = 8-A A=8 f is undefined when x = 3 . --

Let x represent the length of the rectangle. Then, � represents the width of the rectangle 2 since the length is twice the width. The function

f(x) = x3 - 2 f(x + h ) - f(x) h ( x + h ) 3 - 2 - ( x3 - 2 ) h x3 + 3x2 h + 3xh 2 + h 3 - 2 - x3 + 2 h 3 2 2 3x h + 3xh + h --- = 3 2 + 3 h + h 2 ---X X h

for the area .

IS:

89.

x 2 = -x 1 2 A(x) = x · -x = 2 2 2

Let x represent the number of hours worked. The function for the gross salary is:

G(x) = l Ox

91. a.

b.

f(x) = 2x3 + Ax 2 + 4x - 5 and f(2) = 5 f(2) = 2(2) 3 + A(2) 2 + 4(2) - 5 5 = 16 + 4A + 8 - 5 5 = 4A + 19 -14 = 4A A = -14 = _ 7... 4 2

c.

93. a.

f(x) = 3x + 8 and f(O) = 2 2x - A f(O) = 3(0) + 8 2(0) - A 2=� -A -2A = 8 A = -4

P is the dependent variable; independent variable

a

is the

P(20) = 0.01 5(20) 2 - 4.962(20) + 290.580 = 6 - 99.24 + 290.580 = 197.34 In 2005 there are 1 97.34 million people who are 20 years of age or older. P(O) = 0.0 1 5(0) 2 - 4.962(0) + 290.580 = 290.580 In 2005 there are 290.580 million people. H (l) = 20 - 4.9 ( 1 /

= 20 - 4.9 = 1 5 . 1 meters H ( 1 . 1 ) = 20 - 4.9 ( 1 . 1 ) 2 = 20 - 4.9 ( 1 .2 1 ) = 20 - 5.929 = 14.071 meters H (1.2) = 20 - 4.9 ( 1 .2 ) 2 = 20 - 4.9 ( 1 .44 ) = 20 - 7.056 = 12 .944 meters H ( 1 .3 ) = 20 - 4.9 ( 1 .3 / = 20 - 4.9 ( 1 .69 ) = 20 - 8.28 1 = 1 1 .719 meters

123

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b.

H (x) = 1 5 : 1 5 = 20 - 4.9x 2 -5 = - 4.9x 2 x 2 "" 1 . 0204 1 .0 1 seconds H (x) = lO : 1 0 = 20 - 4 . 9x 2 -10 = - 4 . 9x 2 x 2 "" 2 . 0408 x "" 1 . 43 seconds H (x) = 5 : 5 = 20 - 4.9x 2 -15 = -4 . 9x 2 x 2 "" 3 . 061 2 x "" 1 . 7 5 seconds H (x) = 0 0 = 20 - 4 . 9x 2 - 20 = -4 . 9x 2 x 2 "" 4 . 08 1 6 x "" 2 . 02 seconds

97.

99.

X""

c.

95.

( �}X) = ��:� H (x) = (p . I)(x) = P (x) . I (x) R (x) =

101. a.

P(x) =

=

b.

=

=

R (x) - C (x)

( - 1 .2x2 + 220x) - (0.OSx3 - 2X2

103. a.

b.

6Sx + SOO)

- 1 .2x2 + 220x - 0. OSx3 + 2x2 - 6Sx - SOO -0.OSX3 + 0. 8X2 + I S Sx - SOO

P(1 S)

=

-0.OS(1 S)3 + 0. 8(I S)2 + I S S(1 S) - SOO

= - 1 68.7S + 1 80 + 232S = $ 1 836.2S

c.

+

- s ao

When 1 5 hundred cell phones are sold, the profit is $ 1 836 . 25 .

h (x) = 2x h ( a + b) = 2 ( a + b) = 2a + 2b = h (a) + h (b) h (x) = 2x has the property. g (x) = x 2 g ( a + b) = ( a + b) 2 = a 2 + 2ab + b 2 Since

C (x) = 100 + � + 36, 000 x 10 C(500) = 100 + 500 + 36, 000 a. 10 500 = 100 + 50 + 72 = $222 450 + 36, 000 b. C (450) = 1 00 + 10 450 = 100 + 45 + 80 = $225 C ( 600) = 1 00 + 600 + 36, 000 c. 10 600 = 100 + 60 + 60 = $220 400 + 36, 000 d. C (400) = 100 + 10 400 = 1 00 + 40 + 90 = $230

c.

a 2 + 2ab + b 2 '" a 2 + b 2 = g ( a ) + g (b) , g (x) = x 2 does not have the property. F (x) = 5x - 2 F ( a + b) = 5 ( a + b) - 2 = 5a + 5b - 2

Since

5a + 5b - 2 '" 5a - 2 + 5b - 2 = F ( a ) + F (b) , F (x) = 5x - 2 does not have the property.

d.

G (x) = -1 x 1 "' -1 + -1 G (a) + G (b) G(a + b) = a+b a b G (x) = -1 does not have the property. x =

105.

Answers will vary.

124

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Section 3.2: The Graph of a Function

S ection 3 . 2 1.

n.

x 2 + 4/ = 1 6 x-intercepts: x2 + 4 (0) 2 16 x2 = 16 x = ±4 => (-4, 0) , ( 4, 0) y-intercepts: (0) 2 + 4y2 = 1 6 4y2 = 1 6 y2 = 4 y = ±2 => (0, -2) , (0, 2)

11.

13.

=

3. 5.

9.

a.

b. c.

d.

e. f.

g. h.

i.

j. k. I. m.

Not a function since vertical lines will intersect the graph in more than one point. Function Domain: {xl - 7t � X � 7t} ; Range: { Y I - 1 � y � l}

( % ) (% , 0) . (0, 1)

b.

Intercepts: - , 0 .

c.

Symmetry about y-axis.

15.

Not a function since vertical lines will intersect the graph in more than one point.

17.

Function Domain: {x l x > O} ; Range: { Y I y is any real number} b. Intercepts: ( 1 , 0) c. None

f (x) = ax 2 + 4 a (-1) 2 + 4 = 2 => a = -2 False; e.g. y = -1 . x

=

a.

vertical

7.

f(x) = - 2 when x = -5 and x 8.

a.

19.

f (O) = 3 since (0, 3) is on the graph. f(- 6) = -3 since (- 6, -3) is on the graph. f(6) = 0 since (6, 0) is on the graph. f(l l) = 1 since (1 1, 1) is on the graph. f(3) is positive sincef(3) ,., 3 .7. f( -4) is negative since f (-4) ,., - 1 . f (x) = 0 when x = -3, x = 6, and x = 10. f(x» O when - 3 < x < 6, and l 0 < x � 1 1 .

Function Domain: {x I x is any real number} ; Range: { Y I y � 2} b. Intercepts: (-3, 0), (3, 0), (0,2) c. Symmetry about y-axis. a.

21.

Function Domain: {x I x is any real number} ; Range: { Y I y � -3} b. Interc epts: ( 1 , 0), (3,0), (0,9) c. None a.

The domain offis {xl - 6 � x � l l} or [ - 6, 1 1] . 23.

The range offis { Y I - 3 � Y � 4} or [-3, 4] . The x-intercepts are -3 , 6, and 10. The y-intercept is 3. The line y = ! intersects the graph 3 times. 2 The line x = 5 intersects the graph 1 time. f (x) = 3 when x = 0 and x = 4.

f(x) = 2x 2 - x - l a.

b.

f(-I) = 2 ( _ 1) 2 - (-1) - 1 = 2 The point (-1, 2) is on the graph off f(-2) = 2( _2) 2 - ( -2) - 1 = 9 The point (-2, 9) is on the graph off

125

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publ isher.


c.

Solve for x : -1 = 2x 2 - x - 1 0 = 2X 2 _ X o = x ( 2x - 1 ) => = 0, x = 1

27.

(1 , -1 ) are on the graph of I · The domain oflis { x l x is any real number } .

a.

x

(0, -1) and

d. e.

I(x) = � X4 + 1

b.

x-intercepts: 1 ( x ) =0 => 2X 2 - x - 1 = 0

( 2x + 1 )( x - 1 ) = 0 => x = - "2'1 x = 1

25.

(-± 0) and ( 1, 0)

c.

y-intercept: 1 ( 0 ) =2 ( 0 )2 - 0 - 1 = -1 => ( 0, -1 ) I(x) = x + 2 x-6 a. 1(3) = 3 + 2 = -� * 14 3-6 3 The point ( 3, 14 ) is not on the graph off

b.

c.

d. e.

d.

1(4) = 4 + 2 = � = -3 4-6 -2 The point ( 4, -3 ) is on the graph off Solve for x : 2= x+2 x-6 2x - I 2 = x + 2 x = 14 ( 14, 2) is a point on the graph of 1 .

e.

Solve for x :

2 1 = 2x 4 X +1 X 4 + 1 = 2x 2 X 4 _ 2x 2 + 1 = 0 (x 2 _ 1) 2 = 0 Z x - 1 = 0 => x = ±I ( 1 , 1 ) and (-1 , 1 ) are on the graph of 1 .

The domain of1 is { x I x is any real number } . x-intercept: z l (x ) =O => 2x = O X4 + 1 z 2x = 0 => x = 0 => ( 0, 0 ) -

f.

y-intercept: 2 (0 )z 0 = __ = 0 => ( 0, 0 ) 1(0) = 04 + 1 0 + 1

44x-z + x + 6 h ( x ) = -z -

The domain oflis { x l x * 6 } . x-intercepts: l (x ) =O => x + 2 = O x-6 x + 2 = O => x = -2 => ( -2, 0 ) y-intercept: 1 ( 0 ) = 0 + 2 = - .!. => 0, - .!. 0-6 3 3

29.

44 ( 8 ) 2 - + (8) + 6 28 2 = _ 28 1 6 + 14 784 "" lOA feet

a.

h (8) = -

b.

h (1 2) =

v

-

f.

( I� ) is on the graph off

The point 2,

'

f.

2 1(-1) = 2(-1) = � = 1 (_1) 4 + 1 2 The point (-1 , 1 ) is on the graph off 2 _� I( 2 ) _- 2(2) 4 (2) + 1 1 7

( )

-

44 ( 1 2 ) 2 + (1 2) + 6 28 2 = _ 6336 + 1 8 784 "" 9.9 feet

126

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Section 3.2: The Graph of a Function

c.

From part (a) we know the point ( 8, 1 0.4 ) is on the graph and from part (b) we know the point ( 1 2, 9.9 ) is on the graph. We could evaluate the function at several more values ofx (e.g. x = O , x = 1 5 , and x = 20 ) to obtain additional points. 44 ( 0 ) 2 h (O) = - + (0) + 6 = 6 28 2 44 ( 1 5 ) 2 h (1 5) = + ( 1 5 ) + 6 � 8.4 28 2 44 ( 20 ) 2 + ( 20 ) + 6 � 3.6 h ( 20 ) = 28 2 Some additional points are ( 0, 6 ) , ( 1 5, 8 .4 )

31.

b.

c.

and ( 20, 3.6 ) . The complete graph is given below.

d.

h

10

(8, 10.4)

5 x

d.

=

a.

-

15

-32x-2 hex) = - +X 1 30 2 -32(1 00) 2 + 1 00 h(IOO) 1 3 02 = -320, 000 + 1 00 � 8 1 . 07 fieet 1 6, 900

44 ( 1 5 ) 2 + ( 1 5 ) + 6 � 8.4 feet 28 2 No; when the ball is 1 5 feet in front of the foul line, it will be below the hoop. Therefore it cannot go through the hoop. h (1 5 ) =

In order for the ball to pass through the hoop, we need to have h ( 1 5 ) = 1 0 . 44 ( 1 5 ) 2 + ( 1 5) + 6 10 = v2 44 ( 1 5 ) 2 -1 1 =

e.

f + 300 h(300) = -32(300 1 30 2 - 2, 880, 000 + 300 � 129.59 feet = 1 6, 900 -32(500) 2 h(500) = + 500 1 30 2 - 8, 000, 000 + 500 � 26.63 feet = 1 6, 900 -32x 2 Solving hex) = --2- + x = 0 130 2 -32x +x=O 1 30 2 x -322X + 1 = 0 1 30 x = O or -32x2 + 1 = 0 130 1 = 32x2 130 1 30 2 = 32x 130 2 = 528. 13 feet x = -32 Therefore, the golf ball travels 528.13 feet. -32x 2 YI = --- + X 1 30 2

-( )

1 50

1::::::=:::=::1 600

v 2 = 4 ( 225 ) v 2 = 900 v = 30 ft/sec The ball must be shot with an initial velocity of 30 feet per second in order to go through the hoop. 127

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Chapter 3: Functions and Their Graphs f.

=

Use INTERSECT on the graphs of -32x 2 + x and =90 . Y2 Yl = 130

33. C(x) 1 00 + � + 36000 x 10

-

1 50

o

2

I..-______....J

-5

a.

600

I SO

b.

-5

200 225

� 300 325 350

h.

c.

'1 5 0 500

� 650 700 750

K

35. a.

1 2 '1 . 2 6 1 2 9 . 1 '1 B 1. 6 6 131.B 129 . 59 125 118.05

b.

264

Vl B2 B2.01

� 132.0l 132.03 132.02

X

260 261 262 2n 2 6 '1 265 266

c. d. e.

Vl 132 132. 01 132.02

f.

� B2.03 B2.02

Y l - 1 32 . 029 1 1 2426 Y l = 1 32 . 03 1 242604 x

"0 261 262 263 2 6 '1 265 266

ERROR B25 '1 7 0 l55 lOO 269 250

Y1

rru

225 222 2 2 0 . '1 5 220 220.3B 2 2 1 . '1 3 22l

X:... 6 00

The ball travels approximately feet before it reaches its maximum height of approximately 1 32.03 feet. 260 261 262 263 2 6 '1 265 266

1m

The cost per passenger is minimized to about $220 when the ground speed is roughiy 600 ·1 es per hour. X

Vl

�:... 2 75

x

1'/ 1 S 1 00+X/ 1 0+360 ...

The ball reaches a height of 90 feet twice. The first time is when the ball has traveled approximately 1 1 5.07 feet, and the second time is when the ball has traveled about 4 1 3 .05 feet. The ball travels approximately 275 feet before it reaches its maximum height of approxunate I IJ 1 3 1 8 feet. X

=

TblStart =0; �Tbl 50 0 50 100 150 200 250 300

O�S=60

g.

11======:::=1 1 000

Yl

=

-

37.

The graph of a function can have any number of x-intercepts. The graph of a function can have at most one y-intercept (otherwise the graph would fail the vertical line test).

39.

(a) III; (b) IV; (c) I; (d) V; (e) II

132 132.01 B2.02 B2.03

� 132.02

Yl = 1 32 . 029585799

(f +g)(2)= f(2) + g(2)=2 + 1=3 (f +g)(4)= f(4) +g(4) 1 + (-3)=-2 (f-g)(6)=f(6)-g(6) = 0 - 1=-1 (g /)(6)=g(6)-f(6)= 1 - 0=1 (f .g)(2)=f(2) ·g(2)=2(1)=2 (gf }4)=g(f(44)) = _-31 = _ .!.3

128

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Section 3.3: Properties of Functions

y

41.

origin: x � and y � -y (-Y ) = S (-x) 2 - 1 -x

-y = Sx2 - 1 y = -Sx2 + 1 different The equation has symmetry with respect to the y-axis only. (6, 0)

43.

a.

b. c. d. e.

f.

g. h. i.

45.

10

5.

Time (in minutes)

2 hours elapsed; Kevin was between 0 and 3 miles from home. O.S hours elapsed; Kevin was 3 miles from home. 0.3 hours elapsed; Kevin was between 0 and 3 miles from home. 0.2 hours elapsed; Kevin was at home. 0.9 hours elapsed; Kevin was between 0 and 2.8 miles from home. 0.3 hours elapsed; Kevin was 2.8 miles from home. 1 . 1 hours elapsed; Kevin was between 0 and 2.8 miles from home. The farthest distance Kevin is from home is 3 miles. Kevin returned home 2 times.

The intercepts are ( -3, 0 ) , ( 3, 0 ) , and ( 0, -9 ) .

Answers (graphs) will vary. Points of the form (S, y) and of the form (x, 0) cannot be on the graph of the function.

7.

even; odd

9.

True

11.

Yes

13.

No, it only increases on (S, 1 0).

15.

f is increasing on the intervals ( -8, -2 ) , ( 0, 2 ) , (S, oo ) .

17.

Yes. The local maximum at x = 2 is 10.

19.

f has local maxima at x = - 2 and x = 2 . The

21.

Section 3 . 3 1.

2<x<S

3.

x-axis: y � -y

y = x2 - 9 X-intercepts: 0 = x2 - 9 x2 = 9 � x = ±3 y-intercept: y = ( 0 ) 2 _ 9 = -9

local maxima are 6 and 1 0, respectively. Intercepts: (-2, 0), (2, 0), and (0, 3). b. Domain: {xl - 4 � x � 4} or [-4, 4 ) ; Range: { Y I 0 � y � 3} or [ 0, 3 ] . a.

c. d.

(-Y ) = Sx2 - 1 -y = Sx2 - 1 y = -Sx2 + 1 different y-axis: x � -x y = S (-x) 2 - 1 Y = Sx2 - 1 same

23.

a.

b. c.

d.

Increasing: (-2, 0) and (2, 4); Decreasing: (-4, -2) and (0, 2). Since the graph is symmetric with respect to the y-axis, the function is even. Intercepts: (0, 1). Domain: {x I x is any real number } ; Range: { y l y > O } or ( 0, 00 ) . Increasing: ( -00 , 00 ) ; Decreasing: never. Since the graph is not symmetric with respect to the y-axis or the origin, the function is neither even nor odd.

129

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25.

a.

b.

c.

d.

27.

a.

b. c.

d.

29.

31.

a.

b.

a.

b. 33.

35.

37.

Intercepts: (-1t, 0), (1t, 0), and (0, 0) .

39.

Domain: {xl - 1t � X � 1t} or [-1t, 1t] ; Range: { Y I - I � y � l} or [-l, l] . Increasing:

1 41. g(x) = 2

( -%' %}

( %) and (%' 1t)

Decreasing: -1t, -

x g (-x) = _1 _ = � = g (x) (-xi x2 Therefore, g is even.

.

Since the graph is symmetric with respect to the origin, the function is odd. Intercepts:

I(x) = x + 1 x l I(-x) = -x + l - x l = -x + 1 x l I is neither even nor odd.

3

-x43. h(x) = -9

G, 0) . (%' 0) . and ( 0, �) .

3x 2 3 3 h e -x) = _( _X)- 9 = x - 9 = -h (x) 3x2 3( _X)2 Therefore, h is odd.

Domain: {x l - 3 � x � 3 } or [-3, 3] ; Range: { y 1 - 1 � Y � 2} or [-1, 2] .

45. I (x) = x 3 - 3x + 2 on the interval ( -2, 2) Use MAXIMUM and MINIMUM on the graph of Yl = x3 - 3x + 2 .

Increasing: (2, 3) ; Decreasing: ( -1, 1 ) ; Constant: ( -3, - 1 ) and ( 1, 2) Since the graph is not symmetric with respect to the y-axis or the origin, the function is neither even nor odd.

5

2

I has a local maximum of 3 at x = O. I has a local minimum of 0 at both x = - 2 and x = 2.

I has a local maximum of 1 at x = ;

-

5

5 Y

I has a local minimum of -1 at x = - ; .

.J----....

/ -2 11

.

.

-

. " ..

" i r,i r,",ur,', X=.��77�

I

/

-'-...-. . �

'1 = 1 . !; !; [ - 1 1

2

x

-5

local maximum at: ( -1, 4) ; local minimum at: ( 1 , 0) I is increasing on: ( -2, -1 ) and ( 1 , 2) ; I is decreasing on: ( -1, 1 )

I(x) = 4x3 1(-X) = 4(-X)3 = - 4x3 = -/ (x) Therefore, I is odd. g (x) = -3x2 - 5 g (-x) = -3(-x)2 - 5 = - 3x2 - 5 = g (x) Therefore, g is even. F(x) = O , the shift is to the right k units; if y = (x + k) 2 , k > 0 , the shift is to the left k units. The graph of y = (x + 4) 2 is the same as the graph of y = x 2 , but shifted to the left 4 units. The graph of y = (x - 5) 2 is the graph of y = x 2 , but shifted to the right 5 units. X

I

l ./ /

,

""-

..._-,i---·"

, / . J

J

."..-'1'

x

8

-J

The graphs of y = xn , a positive odd integer, all have the same general shape. All go through the points (-1, -1) , (0, 0) , and (1, 1) . As increases, the graph of the function increases at a greater rate for I x I > 1 and is flatter around 0 for Ixl

r = S-2x Total Area = areasquare + areacircle = x 2 + r 2 S-2x) 2 = X 2 + 2S- 20x+4x2 A( X) = x 2 + 1t (-C = 21tr = 1 0 - 4x

7r

1t

7r

b.

9.

a.

In Quadrant I, X2 + y2 = 4� y = .J4-x2

A(x) = (2x)(2y) = 4x.J4 - x2 b.

------

1t

Since the lengths must be positive, we have: 1 O - 4x>O and x>O - 4x>-1 0 and x>O and x>0 x< 2.S Domain: {xl 0< x< 2.S}

p(x) = 2(2x) + 2(2y) = 4x+4.J4 - x2

151

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Chapter 3: Functions and Their Graphs c.

The total area is smallest when x � 1 .40 meters.

1 9.

8

a.

d2 =d\2 +d/ d2 = (2 - 30t) 2 + (3 - 40t) 2 d (t) = � (2 - 30t) 2 + (3 - 40t) 2 = � 4 - 1 20t + 900t2 + 9 - 240t + 1 600t 2 = �2500t2 - 360t + 13

0 ..._______ 2.5 o

d2 =3-401 d,=2-30t

13.

a.

b.

Since the wire of length x is bent into a circle, the circumference is x . Therefore, C (x) = x . Since C = x = 2n r, r =

b.

Xl'lin=0 Xl'lax=.15 Xscl=.05 Yl'lin=-l Yl'lax=4 Yscl=l Xres=l

-2n . x

21. 1 5.

a.

A = area, r = radius; diameter = 2r A(r) = (2r)(r ) = 2r2

b.

P

= perimeter per) = 2(2r ) + 2r 6 r

Area of the equilateral triangle 1 -J3 x=-J3 x 2 A=-x '-

2

2

4

x2 From problem 1 6, we have r 2 = 3

x=.onOOB5

V=.�0000001

r = radius of cylinder, h = height of cylinder, V = volume of cylinder H -h . 1es: H = -B Y SInU " 1ar tnang R r Hr = R ( H - h) Hr = RH - Rh Rh = R H -Hr RH -Hr H (R - r) h= = R R H ( - r) nH (R - r) r 2 V = nr 2 h = nr 2 = R

( :

.

Area inside the circle, but outside the triangle: A(x) = nr 2 -J3 x 2

__4

"ini ..... u .....

-

=

1 7.

The distance is smallest at t � 0.07 hours.

J

[�_ )

-J3 X2 = n�- -J3 x2 = 3 4 3 4

1 52

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23.

a.

The time on the boat is given by �. The

.

. on 1and'IS gIven by hme Island

12 - x -5

e.

3

.

_

f. Town

5.

12-x

d1

= �x2 + 22 = �x2 + 4

The total time for the trip is:

b. c.

d.

T(x) = 12 5- x +� = 12 - x + � 3 5 3 Domain: { xl 0 S; x $12 }

d.

T(4) = 12 5- 4 + f,i2;4 3 .J2O = -58 +--,:" 3 3.09 hours

e.

T(8) = 125- 8 + J'i,2;4 3 = -4 + J68 ,:" 3. 5 5 hours 3 5

f.

7.


,

This relation represents a function. Domain = {-I, 2, 4}; Range = { O 3}.

a.

b. c.

d.

f(x) = �x2_4 a. f(2) = � = -./ 4 - 4 = .JO = 0 b. f(-2) = �(_2) 2 - 4 = -./ 4 -4 = .JO = 0 c.

--

1.

f(x - 2) = 3(x - 2) (x _ 2)2 _1 3(x-2) 3x - 6 x2 - 4x + 4 -1 x2 -4x +3 f(2x) = 3(2x) = � (2X)2 -1 4x2 -1

= 4 6-1 _ = �3 = 2 1(2) = � (2)2 - 1 _ f(-2) = 3(-2) = 4--61 = -63 = -2 (_2)2 - 1 fe-x) = 3(-x) = -3x (-x)2_ 1 x2 - 1 3X = -3x -I(x) = - x2 - 1 x2- 1

f(-x) = �(-x)2_ 4 = �X2_ 4 -f(x) = _�X2 - 4 f(x - 2) = �(x _ 2)2 - 4 = �x2- 4x + 4 -4 = �x2 - 4x f(2x) = �(2x)2 _ 4 = �4x2 _4 = � 4 (x2 - 1) = 2�x2 -I

x2 - 4 f(x) = -x2 a. f(2) = 22 - 4 = 4 -4 4 = Q4 = o 22 (_2)2 - 4 = 4 - 4 = Q = O b. f(-2) = 4 4 (_2)2 x2- 4 c. fe -x) = (_X)2 - 4 = ( -x )2 X2d.

e.

( )

---------,

f(x - 2) - (x _ 2)2 - 4 - x2 -4x + 4 -4 (X _ 2)2 (x _ 2)2 x2 -4x x(x -4) (x - 2)2 (x - 2)2

1 53

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exist. No portion of this material may b e reproduced, in any form or by any means, without permission in wri ting from the publi sher.


f.

9.

f(2x) = (2X)2 - 4 = 4x2 - 4 4x2 (2X)2 4 (x2 -1) x2 _ 1

(f . g)(x) = f(x) ' g(x) = (2 -x) (3x+ 1) = 6x + 2 - 3x2 - X = -3x2 +5x+2 Domain: {xl x is any real number } f (x) = f (x) = 2 - x g g (x) 3x + l 3x+l * 0 3x*-I�x*--31

()

x f(x ) = ----Zx -9

The denominator cannot be zero:

x2 - 9 * 0 (x+3)(x - 3)*0 x * -3 or 3 Domain: {xl x* -3, x * 3} 11.

{ l -�}

Domain: x x *

f(x) = "'/2 - x

1 9.

The radicand must be non-negative:

2 -x�O x:s:2

Domain: {xlx:S:2} or (-00,2] 1 3.

f(x) = Fx r;r

(f -g)(x) = f (x) -g(x) = 3x2 + X + 1- 3x = 3x2 - 2x+l Domain: {x I x is any real number }

The radicand must be non-negative and the denominator cannot be zero: x> 0 Domain: {xlx>O} or (0,00 ) 15.

f(x) =

x x2 + 2x - 3

(f·g)(x) = f(x)·g(x } = (3x2 + X+I) (3x)

The denominator cannot be zero:

x2 + 2x - 3 * 0 (x + 3) (x - l) * 0 x* -3 or 1 Domain: {xlx * -3, x * I} 1 7.

f(x) = 3x2 +x+l g(x) = 3x (f+g)(x) = f (x)+g(x) = 3x2 + X + 1+ 3x = 3x2 + 4x + l Domain: {x I x is any real number }

= 9x3 + 3x2 +3x Domain: {x I x is any real number } f (x} 3x2 +x+l ( gf ) (x) = g(x) = 3x

f(x) = 2 -x g(x) = 3x+l (f+g)(x) = f (x}+ g (x) = 2 -x+ 3x + 1 = 2x + 3 Domain: {xl x is any real number }

3x*O� x * 0 Domain: {xl x * O}

(f -g)(x) = f (x) - g (x) = 2 -x - (3x + 1) = 2 -x - 3x - l = -4x + l Domain: {x I x is any real number } 1 54

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21.

I(x)= xx+l -I

g (x) = x..!.. (f + g)(x) = I (x)+ g (x) (x - l) x+l +-1 = x(x+l)+I --;-----7= -x-I x x (x - I ) x2 +x+x - l x2 + 2x - l x(x -l) x(x - l) Domain: {xl x"# O,X"# I }

�

---'-

Domain: {xl x"# O,X"# I }

()

d.

I (x ) = -3 when x

f.

-4

I(x) ° when ° x � 3 To graph y=/ (x - 3 ) ,shiftthe graphofl horizontally 3 units to the right. >

( )( ) ( 2 ( x2 + 2xh+ h 2 )+ 1 ) - ( 2X 2 + 1 )

= 2X2 + 4xh + 2h2 + 1 - 2X2 - 1 = 4xh + 2h2 The basic function is y = x3 so we start with the graph of this function. y

y=xJ

x

y�2x+l, x 6 -3x > 4 x < --43 Solution set: {x I x < --j} Interval notation: (-00, --j)

..

1 1.

]

3

-2

'2

x

3x-2y = 12 x-intercept: 3x -2( 0) = 12 3x = 12 x=4 The point (4,0) is on the graph. y-intercept: 3(0)-2y = 12 -2y = 12 y = -6 The point (0, -6) is on the graph. y

5

1 63

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in wri ting from the publi sher.


13.

x2 +(Y _ 3)2 = 16 This is the equation of a circle with radius = ..[I6 = 4 and center at (0,3). Starting at the center we can obtain some points on the graph by moving 4 units up, down, left, and right. The corresponding points are (0,7), (0, -1) , ( -4,3) , and (4,3), respectively. Yl
17.

f(x) = (x+2)2 -3

= x2 , shift the graph 2 units to the left [y = (x + 2 f ] and down 3 units [ Y =(X+2)2 -3 ] Starting with the graph of y

r

Y 5

x x

-3

15.

3x2 -4y = 12 3x2 -4(0) = 12 3x2 = 12 x2 = 4 x = ±2 y-intercept: 3(0)2 -4y = 12 -4y = 12 y = -3

(-2,' -3)

(0, -1)

-

19.

x-intercepts:

The intercepts are

if x�2 f(x) = fll2-x xl if x > 2 Graph the line y = 2 -x for x�2 . Two points on the graph are (0,2) and (2, 0) . Graph the line y = x for x > 2 . There is a hole in the graph at x = 2 . �t / / (4,4)

�5

(-2, 0), (2, 0) , and (0, -3) .

Check x-axis symmetry:

3x2 -4(-Y) = 12 3x2 +4y = 12 different Check y-axis symmetry: 3(-x)2 -4Y = 12 3x2 -4y = 12 same Check origin symmetry: 3( _X)2 -4(-y)= 12 3x2 + 4y = 12 different

5

'

, , , � (;, �)' -5

'

;,

X

•

The graph of the equation has y-axis symmetry. 164

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Chapter 4 Linear and Quadratic Functions c.

S ection 4.1 1.

d.

From the equation y = 2x - 3 , we see that the y intercept is -3 . Thus, the point (0, -3) is on the graph. To obtain a second point, we choose a value for x and find the corresponding value for y. Let x = 1 , then y = 2(1) - 3 = -1. Thus, the point (1, -1) is also on the graph. Plotting the two points and connecting with a line yields the graph below.

1 5.

average rate of change increasing

=

2

h(x) = -3x + 4 a. Slope -3 ; y-intercept 4 b. Plot the point (0, 4). Use the slope to find an additional point by moving 1 unit to the right and 3 units down. =

=

y

x

c.

3.

5. 7. 9. 11. 13.

d.

f(2) = 3(2)2 - 2 = 10 f(4) = 3(4)2 _ 2 = 46 i1y = f(4) - f(2) = 46 - 10 = 36 = 18 L1x 4-2 4-2 2

1 7.

f(-2) = (_2)2 - 4 = 4 - 4 = °

-3

=

f (x) = '41 x - 3

±; y-intercept

a.

Slope

b.

Plot the point (0, -3) . Use the slope to find an additional point by moving 4 units to the right and 1 unit up.

slope; y-intercept positive False. If x increases by 3, then y increases by 2.

=

=

-3

y

5

f(x) = 2x + 3 a. Slope 2; y-intercept 3 b. Plot the point (0,3). Use the slope to fmd an additional point by moving 1 unit to the right and 2 units up. =

average rate of change decreasing

=

-5 c.

d.

average rate of change increasing

=

1

'4

x

165

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AIl rights reserved. This material i s protected under all copyright laws as they currently

exist. No portion of this material may b e reproduced , in any form or by any means, without permi ssion in wri ting from the publi sher.

Chapter 4: Linear and Quadratic Functions

19.

F (x) = 4

a.

Slope 0; y-intercept = 4 Plot the point (0, 4) and draw a horizontal line through it.

25.

=

b.

.\'

5

(0.4)

0 1 2

x

-5

c.

d.

21

.

x -2 -1

average rate of change = 0 constant

27.

y = j (x) A vg. rate of change

�y &

=-

4 1

0 1 2

0 1 0

-2 -10

&

-4 - (-26) = -= 22 -1 - (-2) 1 22 2 - ( -4) 6 0 - ( -1) = -1 = 6

A vg. rate of change

=

x y = j (x) 8 -2 8 -1

�y

A vg. rate of change = &

8 - 8 = -0 = 0 -1 -(-2) 1 8 - 8 = -0 = 0 0 8 0 - ( -1) 1 8 - 8 -0 = 0 -= 1 8 1-0 1 8 - 8 -0 = 0 2 -= 8 2-1 1 This is a Imear functIOn wIth slope = 0, smce the

average rate of change is constant at O.

=

x y = j (x) -8 -2 -3 -1

2

�y

Avg. rate of change =-

This is not a linear functIOn, SInce the average rate of change is not constant.

-3 -3 1 - 4 = -= -1 - (-2) 1 -2 - 1 -3 -3 0 -2 1 0 - ( -1) = -= -5 - (-2) = -= -3 -3 -5 1 1-0 1 -8 - ( -5) = -= -3 -3 -8 2 1 2-1 ThIS IS a lInear functIOn wIth slope -3, SInce the average rate of change is constant at -3. 23.

x y = j (x) -26 -2 -4 -1

29.

�y &

-3 - ( -8 ) - -5 - 5 -1 -(-2) 1 0 - ( -3) = -3 = 3 0 - ( -1) 1

j (x) = 4x - l; g (x) = -2x +5 j (x) = 0 a. 4x - l = 0 X = -41

b.

j (x) > 0 4x - l > 0 x > -41

The solution set is

This IS not a linear functIon, SInce the average rate of change is not constant.

{x /x > ±} or (±, 00) .

166

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exist. No portion of thi s material may be reprod uced , in any form or by any means, without permi ssion in wri ting from the publi sher.

Section 4.1: L inear Functions and Their Properties

c.

d.

e.

= g(x) 4x-l = -2x+5 6x = 6 x=1 J ( x ):-::; g ( x ) 4x-l:-::; -2x+5 6x:-::; 6 x:-::; 1 J (x)

33.

a.

(x) =

g ( x ) when their graphs intersect. J Thus, x --4 . J ( x ):-::; g x when the graph ofJis above the graph of g. Thus, the solution is { x l x --4} or ( -00,-4) . =

b.

35.

The solution set is { x Ix:-::;

I} or (-00, 1].

a. b.

x

37.

()

< J ( x ) = g ( x ) when their graphs intersect. Thus, x = -6 . g(x) :-::; J ( x ) < h ( x ) when the graph ofJis

above or intersects the graph of g and below the graph of h. Thus, the solution is or x l :-::; x

{ -6 < 5} [-6, 5) . C(x) = 0. 2 5x+35 C(40) = 0. 2 5(40)+35 = $45 . Solve C(x) = 0. 2 5x+ 35 = 80 0. 2 5x + 35 = 80 0.25x = 45 x =� = 180 miles 0. 2 5 Solve C(x) = 0. 2 5x+35:-::; 100 0. 2 5x + 35:-::; 100 0. 2 5x:-::; 65 x:-::; � = 260 miles 0. 2 5 B(t) = 19. 2 5t+585.72 B(10) = 19. 2 5(10)+585.72 = $778. 22 Solve B(t) = 19. 2 5t+585. 7 2 =893. 7 2 19. 25t + 585.72 = 893. 72 19.25t = 308 308 = 16 years t = -19. 2 5 Therefore the average monthly benefit will be $893. 72 in the year 2006. Solve B(t) = 19. 2 5t+585. 7 2 > 1000 19. 25t + 585. 7 2 > 1000 19. 2 5t > 414. 2 8 414. 2 8 "" 21. 5 2 years t > --19.25 Therefore the average monthly benefit will exceed $1000 in the year 2012.

a.

b.

3 1 . a.

b.

c.

d.

e.

f.

(40, 50) 50 is = x = 40. The point (88, 80) is on the graph of y = J(x) , so the solution to J(x) = 80 is x = 88 . The point (-40, 0) is on the graph of y = J(x) , so the solution to J(x) = 0 is x = -40 . The y-coordinates of the graph of y = J(x) are above 50 when the x-coordinates are larger than 40. Thus, the solution to J(x) > 50 is {xix > 40} or (40, ) The y-coordinates of the graph of y = J(x) are below 80 when the x-coordinates are smaller than 88. Thus, the solution to J(x) :-::; 80 is { x l x:-::; 88} or (-00, 88]. The y-coordinates of the graph of y = J(x) are between 0 and 80 when the x-coordinates are between -40 and 88. Thus, the solution to 0 < J(x) < 80 is {xl-40 < x < 88} or (-40, 88) . is on the graph of The point y J(x) , so the solution to J(x) =

(0

c.

39.

a.

b.

.

c.

167

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41.

S ( p ) = - 200+50p; D( p ) = 1000 - 25p a. SOlveS ( p ) =D ( p ) , - 200+50p = 1 000 - 25p 75p = 1200 p = 1 200 = 16 75 S ( 1 6) = - 200+50 ( 1 6) = 600 Thus, the equilibrium price is $ 1 6, and the equilibrium quantity is 600 T-shirts. b. Solve D ( p » S ( p ) . 1 000 - 25p> - 200+ 50p 1 200>75p 1200>p 75 1 6>p The demand will exceed supply when the price is less than $ 1 6 (but still greater than $0). c. The price will eventually be increased.

T

€

� �

b.

c.

d.

4.nO()

3,O()()

2.(1)0 .1.000

(7300,730)

e.

--

43. a.

5,000

45.

We are told that the tax function T is for adjusted gross incomes x between $7,300 and $29,700, inclusive. Thus, the domain is { x i 7, 300�x�29, 700 } or [ 7300, 29700] .

T ( 1 8000 ) = 0. 1 5 ( 1 8000 - 7300 )+ 730 = 2335 If a single filer's adjusted gross income is $ 1 8,000, then his or her tax bill will be $2335. The independent variable is adjusted gross income, x. The dependent variable is the tax bill, T. Evaluate T at x = 7300, 1 8000, and 29700 . T ( 7300) = 0. 1 5 ( 7300 - 7300 )+ 730 = 730 T ( 1 8000) = 0. 1 5 ( 1 8000 - 7300)+ 730 = 2335 T ( 29700) = 0. 1 5 ( 29700 - 7300)+ 730 = 4090 Thus, the points ( 7300, 730 ) , ( 1 8000, 2335 ) , and ( 29700, 4090 ) are on the graph.

�.��

x

AdjusleJ Gn,ss Income ($)

We must solve T ( x ) = 2860 . 0. 1 5 ( x - 7300 )+ 730 = 2860 O. l 5x - 1095+ 730 = 2860 0. 1 5x - 365 = 2860 0. 1 5x = 3225 x = 2 1, 500 A single filer with an adjusted gross income of $2 1 ,500 will have a tax bill of $2860.

R ( x ) = 8x; C( x ) = 4.5x+17500 a. Solve R ( x ) = C ( x) . 8x = 4.5x + 1 7500 3.5x = 1 7500 x = 1 7500 � 5000 3.5 The break-even point occurs when the company sells 5000 units. Solve R ( x )> C ( x) b. 8x > 4.5x+ 1 7500 3.5x> 1 7500 x> 1 7500 = 5000 3.5 The company makes a profit if it sells more than 5000 units.

4 7 . a.

Consider the data points ( x, y) , where x the age in years of the computer and y the value in dollars of the computer. So we have the points ( 0,3000) and (3, 0) . The slope formula yields: m = �y = 0 - 3000 = -3000 = - 1 000 3-0 � 3 The y-intercept is ( 0,3000 ) , so b 3000 . Therefore, the linear function is V ( x ) = mx+b = -1000x+3000 . =

=

=

168

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exist. No portion of this material may be reproduced, in any form or by any mean s, without permi ssion in writing from the publi sher.

Section 4.1: L inear Functions and Their Properties

b.

The graph of V ( x ) = -1 000x+3000 j§ til

'0 .:3

b.

V(x)

CJ ::>

ca >

B

�

c::l

d.

b.

� QJ

..:l �

""

is

:0

c.

d.

49.

a.

2

Age

c.

d.

5 1 . a.

The graph of V ( x ) = - 1 2, 000x + 120, 000 �

1 000

X

V ( 2 ) = - 1 000 ( 2 )+3000 = 1 000 The computer's value after 2 years will be $ 1 000. Solve V ( x ) = 2000 -1 OOOx+ 3000 = 2000 -1 000x = - 1 000 x =1 The computer will be worth $2000 after 1 year.

19

( (x)

u

.::.:. 0 0

c.

The graph of C ( x ) = 90x+ 1 800

V(x)

b.

1 20,000 1 00,000 80,000

60,000

53.

40,000

20,000

2

4

Age

x

55.

V ( 4) = -12000 ( 4)+ 1 20000 = 72000 The machine's value after 4 years is given by $72,000. Solve V ( x ) = 72000 . -1 2000x+ 120000 = 72000 - 1 2000x = -48000 x = 4 The machine will be worth $72,000 after 4 years. Let x the number of bicycles manufactured. We can use the cost function C ( x ) = mx+b , with m 90 and b 1 800. Therefore C ( x ) = 90x+1 800 =

=

=

4

6

8 10 1 2 1 4

Number o f Bicycles

The cost of manufacturing 14 bicycles is given by C(14) = 90(14)+ 1 800 = $3060 . Solve C(x) = 90x+ 1 800 = 3780 90x+ 1 800 = 3780 90x = 1 980 x = 22 So 22 bicycles can be manufactured for $3780. Let x number of miles driven, and let C cost in dollars. Total cost (cost per mile)(number of miles) + fixed cost C(x) = 0.07x + 29 C(1 1 0) = (0.07)(1 1 0)+ 29 = $36.70 C(230) = (0.07)(230)+ 29 = $45.10 =

=

=

The graph shown has a positive slope and a positive y-intercept. Therefore, the function from (d) and (e) might have the graph shown. A linear function f(x) = mx+b will be odd provided fe-x) = -f(x) . That is, provided m ( -x)+b = -(mx+b) . -mx+b = -mx -b b = -b 2b = 0 b=O So a linear function f (x) = mx+b will be odd provided b = 0 . A linear function f (x) = mx+b will be even provided fe-x) = f(x) . That is, provided m (-x)+b = mx+b . -mx+b = mx+b -mxb = mx 0 = 2mx m =O So a linear function f(x) = mx+b will be even provided m = 0 .

1 69

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b.

Section 4.2 1.

scatter diagram

3.

Linear relation, m > 0

5.

Linear relation, m < 0

7.

Nonlinear relation

9.

a.

2 - (- 2)

The equation of the line is: Y - Y1

a

a

a

o

3

Answers will vary. We select (3, 4) and (9, 1 6). The slope of the line containing 12 = 2 . these pomts IS: m = 196_-34 = ""6 The equation of the line is: .

.

d.

Y - Yl

c.

= m (x - x1 ) y - 4 = 2(x - 3) y - 4 = 2x - 6 y = 2x - 2 20

e.

-6

Using the LINear REGression program, the line of best fit is: y = 2.2x + 1 .2

e.

3

13.

d.

1 y = -9 x + 4 2 6

c.

0 1.1:::::====:=..1 10 b.

a.

1 50

o ��=======� 1 0 o

a

Using the LINear REGression program, the line of best fit is: y = 2.0357x - 2.357 1

a

b.

a.

· ----'--' ...:.-'..---'_.=..l 0 -25 L 90

Answers will vary. We select (-20, 1 00) and (-1 0, 140). The slope of the line containing these points is: 140 -1 00 40 = 4 = - 1 0 - (-20) 1 0

The equation o f the line is:

6 a

Y - Yl

= m (x - x1 ) y - l 00 = 4 (x - (-20») y - l 00 = 4x + 80 y = 4x + 1 80

a

3

-3

a

a a

m= 11.

�

y + 4 = 4"9 x + "29

--,

_ _ _ __

a

= m (x - xl )

y - (- 4) = ( x - (- 2»)

2r.:0'--

a

Answers will vary. We select (-2,--4) and (2, 5). The slope of the line containing . IS. : = 5 - (- 4) = 9 . these pomts m 4"

a

-6 1 70

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exist. No portion of this material may b e reproduced, in any form or by any means , without permission in writing from the publisher.

Section 4.2: Building Linear Functions from Data c.

-25

d. �

a.

/ /a

- -�

oj Q ::::: o 0 '.0 "'Ci

0.. ......

§ ;,

�

.

o

U

a.

;g

0

t3 ' '-

'

.

10

10

20

y

280 ,3 '" 2(,0 v '0 240

.�

... ,

�

S

60 58 -

•

25.

23 is

� v 50 ;::: c 48

c.

1.

•

= x2 - 9

To fmd the y-intercept, let x 2 y

•

46

y

=

-.

x

Price (dollnrs/pair)

U s i n g the LIN ear REGression program, t h e

0 - 9 -9 . =

2

-9 = 0 2 x =9 x = ±.J9 = ±3

The intercepts are

D = - 1 .3355 P + 86. 1 974 . The corre l ation coeffi c i ent i s : "" -0.949 1 . As the price of the j ean s increases by $ 1 , the l in e of b e s t fit i s :

f.

5.

1 .34 pairs per day. D ( p ) = - 1 .3355p + 86. 1 974 Domain :

7.

{ p i 0 < p :-:; 64}

N ote th at the p-intercept is rough ly

=0:

(0, -9), (-3, 0),

and

(3, 0) .

25 4

deman d for the j eans d ecreases by about e.

=0:

To find the x- intercept(s), let y

r

d.

0 implies that the

Section 4.3

• v ;;: 4 4 o iJ 42 40 o '-.}---1--I-----L---1.---''--.L---'- P o 1 8 20 22 24 26 28 �O '

A corre l ation coe ffi c i ent of

data do n ot have a l i near re lation s h i p .

•

•

45

would not make sense to find the line of best fit.

•

�6 �/l � ::: :... )4 2.. t. 52 § �

56 and 53.

40

x

The data d o not fo l l ow a l i n e ar pattern s o it

inches. a.

•

Age 0[' M o t h e r

1 7.4 inches w o u l d h ave a he ight of about 26.98

A c h i l d w ith a h ead c ircumference of

21.

•

•

64.54

9.

and that the number of pairs of j e ans i n demand cannot be negative .

parabola

b

2a Tru e ;

-

b -

2a

=- 4 =2 2(-1) --

g. D(28) = - 1 .3355(28) + 86. 1 974 "" 48.8034 Deman d i s about 49 pairs.

1 72

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Section 4.3: Quadra tic Functions and Their Properties

11.

C

13.

F

1 5.

G

25.

by a factor of

1 7.

H

1 9.

j(x) = 4"1 x 2

Using the graph of y = x 2 , compress vertically by a factor of 4"1 .

i, then shift up 1 unit.

-s

y

-

27. x

--5

21.

f(x) = 4"1 x 2 + 1 Using the graph of y = x 2 , compress vertically

�

(0, 1 )

2

x

:)

f(x) = x 2 + 4x+ 2 = (x 2 +4x+4)+2 - 4 = (x+2) 2 _ 2 Using the graph of y = x 2 , shift left 2 units, then shift down 2 units.

j(X) = x 2 - 2 Using the graph of y = x 2 , compress vertically by a factor of 2, then shift down 2 units.

x

y

( - 2 . -2) -5

x

29. -s

23.

�

j(X) = X2 +2 Using the graph of y = x 2 , compress vertically by a factor of

� then shift up 2 units.

j(x) = 2x 2 - 4x+l = 2 ( X 2 - 2x )+ 1 = 2(x 2 - 2x+l)+1 - 2 = 2(x - l) 2 - 1 Using the graph of y = x 2 , shift right 1 unit, stretch vertically by a factor of 2, then shift down 1 unit. y '

,

Y

x

-5

-2

5

(1, - 1 )

x

-5

1 73

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31.

Thus, the vertex is (-1, -1) . The axis of symmetry is the line x = -1 . The discriminant is b2 -4ac = (2i -4(1)(0) = 4 0 , so the graph has x-intercepts. The two x-intercepts are found by solving: x2 + 0 x(x+2) = 0 x = 0 or x = -2 The x-intercepts are -2 and O . The y-intercept is f(O) = 0 .

f(x) = _x2 -2x = _ (x2 +2x) = -(x2 +2x+l)+1 = -(x+l)2 +1 Using the graph of = x2 , shift left 1 unit, reflect across the x-axis, then shift up 1 unit.

>

y

2x

y 5

x

33.

=

x

f(x) = 2"1 x2 +x-I = � ( x2 + 2x) -1 =�(X2 + 2x+I)-I-� =�(X+I) 2 _ % Using the graph of = x2 , shift left 1 unit, compress vertically by a factor of �, then shift down 2"3 unItS..

b. c.

y

37. a.

x

= -1

The domain is ( ) . The range is [-1, ) Decreasing on ( -1) . Increasing on (-1, ) . For f(x) = -x2 -6x, a = -l , b =-6 , c = O . Since a = -1 0, the graph opens down. The x-coordinate of the vertex is � x = -b2a = -(-6) 2(-1) = -2 = -3. The y-coordinate of the vertex is f ( ;: ) = f(-3) = -( _3)2 -6(-3) -9+ 18 = 9. Thus, the vertex is (-3, 9) . The axis of symmetry is the line x -3 . The discriminant is: b2 -4ac = ( _6)2 -4(-1)(0) = 36 0, soThethe graph has two x-intercepts. x-intercepts are found by _x2 -6x = 0 -x (x + 6) = 0 x = 0 or x = -6. The x-intercepts are -6 and The y-intercepts are f(O) = 0 . -00,

(0

(0

-00 ,

.

00

>

x =

-b

2a

=

-(2) 2(\)

=

-2 2

=

-

I

solving:

.

The y-coordinate of the vertex is f ( ;! ) = fe-I) = ( _ 1)2 + 2(-1) = 1-2 = -1.

o.

1 74

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Section 4.3: Quadratic Functions and Their Properties

41.

a.

For f(x) = x 2 + 2x - 8 , a 1 , b = 2 , c = -8 . Since a 1 > 0 , the graph opens up. The x-coordinate of the vertex is -b = -2 = -2 = -l . x=2a 2(1) 2 The y-coordinate of the vertex is = f( -1) = ( _ 1)2 + 2(-1) - 8 f =

=

(;!)

b. c.

The domain is ( -00, 00 ) . The range is ( -00, 9 ] . Increasing on ( -00 , - 3) . Decreasing on ( 3 00 ) . -

39. a.

= 1 - 2 - 8 = -9. Thus, the vertex is ( 1 - 9) . The axis of symmetry is the line x = -1 . The discriminant is: b 2 - 4ac = 2 2 - 4(1)(-8) = 4 + 32 = 36 0 , so the graph has two x-intercepts. The x-intercepts are found by solving: x 2 + 2x - 8 = 0 (x + 4)(x - 2) = 0 x = -4 or X = 2. The x-intercepts are -4 and 2 . The y-intercept is f(O) = -8 . = -1 -

>

,

For f(x) = 2X 2 - 8x, a = 2 , b = -8 , c = 0 . Since a 2 > 0 , the graph opens up. The x-coordinate of the vertex is x = -b = -(-8) = 8 = 2 . 2a 2(2) '4 The y-coordinate of the vertex is = f(2) 2(2) 2 - 8(2) = 8 - 1 6 -8. f =

( ;!)

,

=

x

=

Thus, the vertex is (2, - 8) . The axis of symmetry is the line x 2 . The discriminant is: b 2 - 4ac = ( _ 8)2 - 4(2)(0) = 64 > 0, so the graph has two x-intercepts. The x-intercepts are found by solving: 2x 2 - 8x = 0 2x(x - 4) = 0 x = 0 or x = 4. The x-intercepts are 0 and 4. The y-intercepts is f(O) = 0 .

�

x

=

b. c.

43. x

a.

The domain is ( -00, 00 ) . The range is [-9, 00 ) . Decreasing on ( -00, - 1) . Increasing on (-1, 00 ) . For f(x) = x2 + 2x + 1 , = 1 , b = 2 , c = 1 . Since a = 1 > 0 , the graph opens up. The x-coordinate of the vertex is -2 -b = -2 = - l . x=2a 2(1) = 2 The y-coordinate of the vertex is = f(-l) f a

( ;! )

b. c.

The domain is Decreasing on

= ( _ 1)2 + 2( -1) + 1 = 1 - 2 + 1 = O. Thus, the vertex is (-1, 0) . The axis of symmetry is the line x = -1 .

( -00, 00 ) . The range is [-8, 00 ) . ( -00 , 2) . Increasing on (2, 00 ) . 1 75

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Chapter 4: Linear and Qua dratic Functions

The discriminant is: b 2 - 4ac = 2 2 - 4(1)(1) = 4 - 4 = 0 , so the graph has one x-intercept. The x-intercept is found by solving: x 2 + 2x + l 0 =

(x + l) 2 = 0 x = -1.

-2

The x-intercept i s -1 . The y-intercept is 1(0) = 1 . b.

- 5 ( - 1 , 0) 1

b. c.

4 5 . a.

x =

1 1 1 1

-11

c.

x

47. a.

-5

The domain is ( -00, (0) . The range is [0, (0) . Decreasing on ( -00, - 1) . Increasing on (-1, (0) .

Thus, the vertex is

. =

�

=

= .!. - .!. + 2 = .!2 . 8

(�, - %)

The axis of symmetry is the line x . The discriminant is: b 2 - 4ac = 2 2 - 4(-2)(-3) 4 - 24 = -20 , so the graph has no x-intercepts. The y-intercept is 1(0) = -3 .

( ;� ) = I (�) = 2 (�y - � + 2

4 Thus, the vertex is

( ;�) = I (�) = -2 (�y + 2 (�) - 3 = - .!. + 1 - 3 = - � 2· 2

2(2)

8

00

For I (x) = -2x 2 + 2x - 3 , a = -2 , b = 2 , c = -3 . Since a = -2 < 0 , the graph opens down. The x-coordinate of the vertex is -b = --(2) -2 = -1 . x=2a 2(-2) -4 2 The y-coordinate of the vertex is I

The y-coordinate of the vertex is I

[l� , ) . Decreasing on (-oo,±) . Increasing on (±,oo) . ( -00 , (0 ) . The range is

The domain is

=

For l(x) = 2x 2 - x + 2 , a = 2 , b = - 1 , c = 2 . Since a = 2 > 0 , the graph opens up. The x-coordinate of the vertex is x = -b = -(- 1) = "41 . 2a

3

I 1 x= 4

x

(�, : ) .

I

The axis of symmetry is the line x = t . The discriminant is: b 2 - 4ac = (_ 1) 2 - 4(2)(2 ) = 1 - 1 6 = - 1 5 , so the graph has no x-intercepts. The y-intercept is 1(0) = 2 . 1

x =

I

2: x

b.

The domain is The range is

( -00 , (0 ) .

( - %] . -00,

1 76


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Section 4.3: Quadratic Functions and Their Properties

c.

( �) . Decreasing on (�, ) . For l( x) = 3 x 2 +6 x + 2 , a = 3 , b = 6 , c = 2 . Since a = 3 0 , the graph opens up. The x-coordinate of the vertex is x = -b = -6 = -6 = -1 . 2a 2(3) 6

Increasing on

51.

-�,

a.

1( ;� ) = / (- %) = -4 ( - %J -6 ( -%) + 2 = _ 2.+2.+ 4 2 2 = .!2.4 Thus, the vertex is ( _ %, 1; ) . The axis of symmetry is the line x = -% The discriminant is: b 2 -4ac = (_6)2 -4(-4)(2) = 36+32 68 ,

The y-coordinate of the vertex is

I( ;�) = 1(-1) = 3(-1)2 + 6(-1) + 2 = 3-6+2 = -l. (-1, -1)

.

Thus, the vertex is . The axis of symmetry is the line x -1 . The discriminant is:

= b2 -4ac = 62 -4(3)(2) = 36-24 = 12 ,

=

so the graph has two x-intercepts. The x-intercepts are found by solving: 0

so the graph has two x-intercepts. The x-intercepts are found by solving: 0

-4x2 -6x + 2 = x = - b ±�b 2 -4ac -(-6)±J(;8 2(-4) 2a 6±J68 6±2.JU 3±.JU -4 . The x-mtercepts are -3+.JU and -3-.JU . 4 4 The y-intercept is 1( 0 ) = 2 .

3x2 +6x+2 = ±�b 2--4ac x = --b-":":"' 2a �� -6±J!i. -6±2J3 -3±J3 3 6 6 J3 and -1 + 3"" . J3 . The x-mtercepts are - 1 - 3"" The y-intercept is 1( 0 ) = 2 .

-8

-8

IY

(0, 2)

(-1 + 1, 0) 4

b. c.

I

= -1 domain is ( -�, � ) .

The The range is

x

x

b.

[-1, ) . Decreasing on ( - 1) . Increasing on (-1, ) . �

The domain is The range is

-�,

c.

�

( -�, � ) .

( 1;] . _� ,

( -%, ) Increasing on ( -% ) .

Decreasing on

�

.

-�,

1 77

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exist. No portion of thi s material may be reproduced, in any form or by any means , without permi ssion in wri ting from the publi sher.

Chapter 4: Linear and Quadra tic Functions

53.

55.

Consider the form y = a ( x -h)2 + k . From the graph we know that the vertex is (-1, -2) so we have h = -1 and k = -2 . The graph also passes through the point (x, y) = (0, -1) . Substituting these values for x, y, h, and k, we can solve for a:

59.

-1 = a(0-(-I)) 2 + (-2) -1 = a(1)2 -2 -1 = a-2 l=a The quadratic function is f ( x) = (x + 1)2 -2 = x2 + 2x -1 . Consider the form y = a (x -h) 2 + . From the graph we know that the vertex is (-3,5) so we have h = -3 and = 5 . The graph also passes through the point (x, y) = (0, -4) . Substituting these values for x, h, and we can solve for a: -4 = a(0-(-3»)2 + 5 -4 = a(3 l + 5 -4 =9a+S -9 = 9a -1 = a The quadratic function is f(x) = _(X+3)2 +5 = _x2 -6x-4 . Consider the form y = a (x -h l + . From the graph we know that the vertex is (1, -3) so we have h = 1 and = -3 . The graph also passes through the point (x,y) = (3,5) . Substituting these values for x, y, h, and we can solve for a: 5 = a(3-1)2 +(-3) S = a(2)2 -3 S = 4a-3 8 =4a 2=a The quadratic function is f (x) = 2 (x -1 / -3 = 2x2 -4x -1 .

=

61.

>

.

63 .

0, the graph opens up, so the vertex is a minimum point. The minimum -b = --12 -12 occurs at x =2a 2(2) = -4 = -3. The minimum value is

k,

65.

=

1 80

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Section 4.3: Qua dratic Functions and Their Properties

81.

R(p) = _4p2 + 4000p , a = -4, = 4000, = O. Since a = -4 0 the graph is a parabola that opens down, so the vertex is a maximum point. The -4000 = 500 . maXlmum occurs at p = -2a = --2(-4) Thus, the unit price should be $500 revenue. The maximum revenue is for maximum R(500) = -4(500)2 + 4000(500) = -1000000 + 2000000 = $1,000,000 C(x) = x2 -140x+ 7400 , = 1, = -140, = 7400. Since = 1 > 0, the graph opens the vertexmarginal is a cost minimum point. up,Thesominimum 140 = 70 mp3 occurs at x = -2a = -(-140) = 2(1) 2 players produced. The minimum marginal cost is I(;!) = 1(70) = (70/ -140(70)+ 7400 = 4900 -9800+ 7400 = $2500 = -3. 24, = 242.1, = -738. 4 The maximum the income levelnumber is of hunters occurs when -242.1 ;:,: 37. 4 years old x = -2a = 2-242.1 = -( -3. 24 ) -6. 4 8 The number of hunters this old is: H(37.4) = -3. 24(37.4)2 + 242.1(37. 4 ) -738.4 ;:,: 3784 hunters The maximum occurs when x = 37.4 , so the function increases on the interval (0, 37.4) and decreases on the interval (37.4, ) the number ofbetween huntersages is decreasing for45Therefore, individuals who are 40 and years of age. b

-3 The solution set is { x l x > -3 } or ( -3,00 ) . f(x) > 0 when the graph offis above the x axis. Thus, { xi x < -2 or x > 2 } or, using interval notation, ( -00, -2 ) ( 2,00 ) . f(x)�0 when the graph offis below or intersects the x-axis. Thus, { xl -2 x 2 } or, using interval notation, [ -2, 2] . a.

b. •

200

u

�

�

x

From the graph, the data appear to be quadratic with a < 0 . 1 85

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Chapter 4: Linear and Qua dra tic Functions

5.

b.

g ( x ) when the graph offis above the graph of g. Thus, { xl x -2 or x > 1} or, using interval notation, ( -2 ) ( 1, ) x2 -3x-10 0 We graph the function f(x) = x2 -3x-10 . The intercepts are y-intercept: f(O) = -10 x-intercepts: x2 -3x -10 = 0 (x-5)(x+2) = 0 x = 5, x = - 2 The vertex is at x = -2ab = -(-3) 2(1) = i.2 Since f (i2 ) = - 494 ' the vertex is (i2 ' 494 ) . 10 10 a.

u

(0

10
0 We graph the function f(x) = x2 -1 . y-intercept: f(O) = -1 x-intercepts: x2 -1 = 0 (x+I)(x-I) = O x = -I, x = I The vertex is at x = -2ab = -2(1)(0) = O . Since f(O) - 1 , the vertex is (0,-1). d.

=

1 88

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Section 4.5: Inequalities In volving Quadratic Functions

g.

10

10

f(x) :2: 1 x2 - 1 :2: 1 x2 - 2 :2: 0 We graph the function p(x) x2 - 2 . The intercepts of p are y-intercept: p(O) = -2 x-intercepts: x2 - 2 = 0 x2 = 2 x = ±.fi = O . Since The vertex is at x = -2ab = -(0) 2(1) p(O) = -2 , the vertex is (0, -2). 10 =

-10 The graph is above the x-axis when x < -1 or x > 1 . Since the inequality is strict, the solution set is { xl x < -l or x > I} or, using interval notation, ( -1) u (1 , ) . g(x) � 0 3x + 3 � 0 3x � -3 x � -1 The solution set is { x I x � -I} or, using interval notation, ( - 1] . f(x) > g(x) x2 - 1 > 3x + 3 x2 - 3x - 4 > 0 We graph the function p( x) = x2 - 3x - 4 . The intercepts of p are y-intercept: p(O) = -4 x-intercepts: x2 - 3x - 4 = 0 (x - 4)(x + l) = 0 x = 4, x = -1 = � . Since The vertex is at x = -2ab = -(-3) 2(1) 2 CXl

-CXl,

e.

10

-CXl ,

f.

p

27.

(�2 ) = _ 254 ' the vertex is (�2 ' 25 ) . _

10

4

10 -10 The graph of p is above the x-axis when x < -l or x > 4 . Since the inequality is strict, the solution set is { xl x < -l or x > 4 } or, using interval notation, (-CXl, -1) u (4, CXl) .

-10 The graph ofp is above the x-axis when x < -.fi or x > .fi . Since the inequality is not strict, the solution set is { xl x � -.fi or X :2: .fi } or, using interval notation, (-CXl, -.fi J u [.fi, CXl) . f(x) = _x2 + 1; g(x) = 4x + 1 a. f(x) = 0 _x2 + 1 = 0 l - x2 = 0 (l - x) (l + x) = O x = 1; x = -1 Solution set: {-I, I} . g (x) = 0 b. 4x + l = 0 4x = -1 X = --41 Solution set:

{-�} .

1 89

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c.

d.

x-intercepts: _x2 -4x = 0 -x(x + 4) = 0 x = O; x = -4 The vertex is at x = -2ab = -(-4) 2(-1) = � -2 = -2 . Since p( -2) = 4 , the vertex is (-2, 4).

f(x) = g (x) _X2 + 1 = 4x + 1 0 = X2 + 4x 0 = x(x + 4) x = 0; x - 4 Solution set: {-4, O} . f(x) > 0 We graph the function f(x) = _x2 + I . y-intercept: f(O) = 1 _x2 + 1 = 0 x-intercepts: x2 - I = 0 (x + I)(x - I) = O x = -I;x = 1 -(0) = O . Since The vertex is at x = -2ab = 2(-1) f(O) = I , the vertex is (0, 1).

10

10

-10 The graph ofp is above the x-axis when -4 < x < O . Since the inequality is strict, the solution set is { x 1 - 4 < x < O} or, using interval notation, (-4, 0) . f(x) ;::: 1 g. _x2 + 1 ;::: 1 _x2 ;::: 0 We graph the function p(x) = _x2 . The -(0) = O . Since vertex is at x = -2ab = 2(-1) p(O) = 0 , the vertex is (0, 0). Since a = -1 < 0 , the parabola opens downward. 10

10

10 -10

c.

The graph is above the x-axis when -I < x < I . Since the inequality is strict, the solution set is { xl - I < x < I } or, using interval notation, (-I, I) . g(x) :o; O 4x + I :O; 0 4x :O; -I x :o; --41

10

-1 0

{ l -±} or, using interval notation, ( - ±] .

The solution set is x x :O;

29.

-00 ,

f.

f(x) > g(x) _x2 + 1 > 4x + 1 -x2 - 4x > 0 We graph the function p(x) = _x2 - 4x . The intercepts of p are y-intercept: p(O) = 0

The graph of p is never above the x-axis, but it does touch the x-axis at x = O. Since the inequality is not strict, the solution set is {O}. f(X) = X2 _ 4; g(x) = -x2 + 4 a. f(x) = 0 x2 - 4 = 0 (x - 2)(x + 2) = 0 x = 2;x = -2 Solution set: {-2, 2} .

1 90

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exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in wri tin g from the publi sher.

Section 4. 5: Inequalities Involving Quadra tic Functions

b.

c.

d.

g (x) = 0 _x2 + 4 = 0 x2 - 4 = 0 (x + 2)(x - 2) = 0 x = -2;x = 2 Solution set: {-2, 2} . f(x) = g(x) x2 - 4 = -x2 + 4 2X2 - 8 = 0 2(x - 2)(x + 2) = 0 x = 2;x = -2 Solution set: {-2, 2} . f(x) > 0 x2 - 4 > 0 We graph the function f(x) = x2 - 4 . y-intercept: f(O) = -4 x2 - 4 0 x-intercepts: (x + 2)(x - 2) = 0 x = -2; x = 2 -b -(0) The vertex is at x = 2a = -2(-1) = O . Since /(0) = -4 , the vertex is (0, -4) .

-b -(0) The vertex is at x = 2a = -2(-1) = O . Since g(O) = 4 , the vertex is (0, 4). 10

-101

-10

e.

The grapl;1 is below the x-axis when x < -2 or x > 2 . Since the inequality is not strict, the solution set is { xl x :"0: -2 or x � 2 } or, using interval notation, ( 00 -2] u [2, ) f(x) > g(x) x2 - 4 > _x2 + 4 2X2 - 8 > 0 We graph the function p( x) = 2X2 - 8 . y-intercept: p(O) = -8 2X2 - 8 = 0 x-intercepts: 2(x + 2)(x - 2) = 0 x -2; x 2 -b -(0) The vertex is at x = = - = o . Since 2a 2(2) p(O) = - 8 , the vertex is (0, -8). -

f.

=

10

�

10

-10

=

-101

I t '.

,

(0

.

=

10

10

-101

The graph is above the x-axis when x -2 or x > 2 . Since the inequality is strict, the solution set is { xl x < -2 or x > 2 } or, using interval notation, (-00,-2) u (2, (0 ) . g(x) :"O: O _x2 + 4 :"0: 0 We graph the function g(x) = _x2 + 4 . y-intercept: g(O) = 4 _x2 + 4 = 0 x-intercepts: x2 - 4 = 0 (x + 2)(x - 2) = 0 x = -2; x = 2

\ l ,

1 10

-10

2 . Since the inequality is strict, the solution set is { xl x < -2 or x > 2 } or, using interval notation, ( 00 -2) u (2, (0) . g. f(x) � 1 x2 - 4 � 1 2 x -5 � 0 We graph the function p(x) = x2 - 5 . y-intercept: p(O) = -5 x-intercepts: x2 - 5 = 0 x2 = 5 x = ±J5 -

,

191

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Chapter 4: Linear and Qua dratic Functions 10

The vertex is at x -b -(0) O . Since 2a 2(1) p(O) = -5 , the vertex is (0, -5). =

=

=

10

10

-10

10

The graph is above the x-axis when x < -1 or x > 2 . Since the inequality is strict, the solution set is {xl x < -1 or x > 2} or, using interval notation, ( -00, -1) u (2, )

-10

The graph ofp is above the x-axis when x < -.J5 or x > .J5 . Since the inequality is not strict, the solution set is { x i x � -.J5 or x � .J5 } or, using interval notation, (-00, -.J5] u [ .J5, ) (0

31.

(0

e.

.

f(X) = X2 _ X-2; g(x) = x2 +x-2 f(x) = 0 x2 -x-2 = 0 (x-2)(x+ 1) = 0 x = 2,x = -1

a.

Solution set:

b.

c.

d.

10 -10

.

'

g(x) � O x2 +x-2 � 0 We graph the function g(x) = x2 + X -2. y-intercept: g(O) = -2 x-intercepts: x2 +x-2 = 0 (x+2)(x-l) = 0 x = -2;x = 1 The vertex is at x = -2ab = -(1) 2(1) = _ .!.2 . Since f ( -�) -�, the vertex is (-�,-�} 10 =

{-I, 2} .

g(x) = O x2 +x-2 = 0 (x+2)(x-l) = 0 x = -2;x = 1 Solution set: {- 2, I} f(x) = g(x) x2 -X -2 = x2 + X -2 -2x = 0 x=o Solution set: { O } . f(x) > 0 x2 -x-2 > 0 We graph the function f(x) = x2 -X -2 . y-intercept: f(O) = -2 x-intercepts: x2 -x-2 = 0 (x-2)(x+l) = 0 x = 2;x = -1 The vertex is at x = -2ab = -(-1) 2(1) = .!.2 . Since f (.!.2 ) = _2. the vertex is (.!.2 ' _2. ) . 4

.

The graph is below the x-axis when -2 < x < 1 . Since the inequality is not strict, the solution set is { x 1 - 2 � x � I } or, using interval notation, [-2, 1] .

f.

g.

f(x) > g(x) x2 -X -2 > x2 + X -2 -2x > 0 x 96 -16t2 80t - 96> 0 We graph the function I(t)=-16t2+ 80t- 96 . The intercepts are y-intercept: 1(0) =-96 t-intercepts: -16t2 80t- 96=0 -16(t2 -5t+6)=0 16(t-2)(t- 3)=0 t=2, t =3

+

+

193

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250,000

Ol�--+-----��

1000 h.

-50,000

The graph offis above the p-axis when 276.39 < p < 723.61 . Since the inequality is strict, the solution set is {p!276.39 < p < 723.61} or, using interval notation, (276.39,723.61) . The revenue is more than $800,000 for prices between $276.39 and $723.61.

The graph offis above the c-axis when 0.112 < c < 81.907 . Since the inequality is strict, the solution set is {c! 0.112 < c < 81.907} or, using interval notation, (0.112,81.907) . Since the round is to be on the ground Note, 75 km = 75,000 m. So, x=75,000, v=897, and g=9.81.

y=0

.

( 2 )( 89 7 )

2 c(75 ,000)_(I+C2) 9.81 7S,000 =0

75,000c -34,290.724(1 + c2)=0 75,000c -34,290.724 -34,290.724c2=0 -34,290.724c2+ 75,000c -34,290.724=0 We graph

a.

f C c) = -34,290.724c2 + 75,000c - 34,290.724 .

Since the round must clear a hill 200 meters high, this mean y> 200 . Now x=2000, v=897, and g=9.81 .

c(2000) -(1+ c2)

The intercepts are y-intercept: f(O) =-34,290.724 c-intercepts:

( 9.�1)( 28�007 J > 200

-34,290.724c2 + 75,000c - 34,290.724 0 2 c = -(75,000)± �(75,000) -4(-34,290.724)(-34,290.724) =

2000c -24.3845(1 + c2)> 200 2000c -24.3845 -24.3845c2> 200 -24.3845c2+ 2000c -224.3845> 0

2(-34,290.724)

- 75,000± �921,584,990.2 -68,581.448 c"" 0.651 or c"" 1.536 =

We graph

f(c)=-24.3845c2+ 2000c -224.3845 .

It is possible to hit the target 75 kilometers away so long as c "" 0.651 or C"" 1.536 .

The intercepts are y-intercept: f(O)=-224.3845 c-intercepts:

39.

-24.3845c2 + 2000c -224.3845=0 -2000± �(200W - 4(-24.3825)(-224.3845)

We graph the function f(x)= (x -4)2. y-intercept: f(O)=16 x-intercepts: (x - 4)2=0

c= ------�--�--�----��----�

2(-24.3825) -2000± �3,9 78,113.985 = -48. 769 C"" 0.112 or c "" 81.90 7

x-4=0 x=4

The vertex is the vertex is (4, 0).

The vertex is at -

c= b= 2a

-(2000) =41.010 . 2(-24.3845)

(X_4)2 � 0

10

Since

f(41.010) "" 40,785.273, the vertex is

_IOI

(41.010, 40785.273).

�!\...a.....J

10

50,000

-10

The graph is never below the x-axis. Since the inequality is not strict, the only solution comes from the x-intercept. Therefore, the given inequality has exactly one real solution, namely

100

x=4 .

-5000 194

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41.

Solving x2 + x + 1> ° We graph the function f(x)=x2 + X + 1. y-intercept: f(O)=1 x-intercepts: b 2 - 4ac = 12 - 4(1) (1)=-3, so f has no x-intercepts. The vertex is at x= -b =

-(1)= _.!.. Since 2a 2(1) 2

( )

f -.!. =� the vertex is 2 4' 10

(-.!.2' �4 ) .

increasing

c.

5.

10

G ( x)=4 Slope = 0; y-intercept = 4 Plot the point (0, 4) and draw a horizontal line through it.

a.

b.

y

-10

The graph is always above the x-axis. Thus, the solution is the set of all real numbers or, using interval notation, (-00, (0 ) .

5

(-3.4)

f( x)=2x-5

-5

Slope = 2; y-intercept = -5

a.

b.

7.

y

3.

-54

b.

Slope =

�; y-intercept

Y

= f( x)

Avg. rate of change

-2

L'ly =

&

3-(-2) --5 - -5 0-(-1) 1 8-3 5 -=-=5 1 8 1-0 1 13-8 2 13 =�=5 2-1 1 18-13=�=5 3 18 3-2 1 This is a linear function with slope 5, since the average rate of change is constant at 5.

°

x

3

=

h(x)= x- 6 a.

x

-1

increasing

c.

constant

c.

Plot the point (0,-5). Use the slope to find an additional point by moving 1 unit to the right and 2 units up. 2

(4,4) x


1.

(0.4)

=

-6

Plot the point (0, -6 ) . Use the slope to find an additional point by moving 5 units to the right and 4 units up. 195

© 2008




9.

f(x) = (x - 2)2 + 2 Using the graph of y = x2 , shift right 2 units, then shift up 2 units.

b -4 -=2 4 = x = - -= . 2a 2(1) 2 The y-coordinate of the vertex is = f(2) = (2)2 - 4 ( 2 )+ 6 = 2 . f -

( :J

y

Thus, the vertex is (2, 2). The axis of symmetry is the line x = 2 The discriminant is: b2 - 4ac=(-4)2 - 4 (1) (6)=-8 0 , so the graph has two x-intercepts. The x-intercepts are found by solving: - 4x2+4x= 0 - 4x(x-1)=0 x= 0 or x =1 The x-intercepts are 0 and 1. The y-intercept is 1(0) =- 4(0)2 + 4(0)=0 .

the graph has no x-intercepts. The y intercept is 1(0)=2.(0 ) 2 + 3 (0) + 1= 1. 2

-2

I I I I 1 -12 I

2

x

197

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All rights reserved. This material is protected under all copyright laws as they currently



b.

c.

(-00, 00) . The range is [� ,00 J. Decreasing on ( -oo,-t J . Increasing on ( t 00) . -

23.

a.

b.

The domain is

c.

,

f(x) = 3x2+4x-1 a = 3, b = 4, c = -1. Since a = 3 > 0, the

25.

graph opens up. The x-coordinate of the

(-00,00) . The range is [-�, 00J. Decreasing on ( -00, -1J Increasing on (-1,00).

The domain is

f(x)=3x2-6x+4 a = 3, b = -6, c = 4. Since a = 3 > 0, the graph

. b 4 4 2 vertex x = - - = - -- = - - = - - . 2a 2(3) 6 3

opens up, so the vertex is a minimum point. . . The rrurumum occurs at x = - b = - -6 = -6 = 1 .

The y-coordinate of the vertex is

The minimum value is

IS

-

2a

( :J

f=

3

3

Thus, the vertex is

_

3

(-%,-�) .

27.

f(l) = 3(1) 2 -6(1)+ 4

f(x)=-x2+8x-4 a = -1, b = 8, c = -4. Since a = -1 < 0, the

3

The discriminant is:

x =- b =- 8 =- 8 = 4 2a 2(-1) -2

graph has two x-intercepts. The x-intercepts are found by solving:

The maximum value is

-

b2-4ac = (4)2-4(3)(-1) = 28 > 0 , so the

2a

( :J

f-

- 4±..fi8 2(3)

29.

-4±2.J7 6

-2±.J7 3 The x-intercepts are -2-.J7 3 -2+.J7 ",,0.22 . 3

""

x

I

--

= =

-

.

2

f( 4 ) = -( 4 ) +8(4)-4

-16+32-4 = 12

f(x) = -3x2+ 12x + 4 a = -3, b = 12, c = 4. Since a = -3 < 0, the

graph opens down, so the vertex is a maximum point. The maximum occurs at

-1.55 and

x=-

� = -� = -E = 2 . 2a 2(-3) -6

( :a )

The maximum value is

The y-intercept is f(O) = 3(0)2+ 4(0)-1 = -1 . =-

6

graph opens down, so the vertex is a maximum point. The maximum occurs at

The axis of symmetry is the line x = -� .

3x2+4x-1=0. x -b±�b2- 4ac

2(3)

=3-6+4=1

� �_1 = 2 _

=

-

32

f-

Y

=

2

f(2) = 3(2) +12(2)+4

= -1

_

2 + 24 + 4 = 16

-5 I

198





31.

{l

x2+ 6x -160, the graph opens up. The x-coordinate of the vertex is -12 b -12 x=--=---=--=2 . 2a 2 (3) 6 The y-coordinate of the vertex is = f (2)=3 (2)2 -12 (2)+ 4 f -

( :J

=12 - 24 + 4 =-8 Thus, the vertex is (2, -8) .

201

© 2008




c.

d.

10

The axis of symmetry is the line x= 2 . The discriminant is: b2 - 4ac =( _12)2 - 4( 3)( 4)=96 > 0 , so the graph has two x-intercepts. The x-intercepts are found by solving: 3x2 -12x+ 4 =0 . ±.J 2 - 4ac -(-12)± .J% x= -b b 2(3) 2a 12 ± 4 ..[6 6 ± 2 ..[6 3 6

-10 The graph is above the x-axis when x < 4 or x > 6 . Since the inequality is not strict, the solution set is {xl x ::; 4 or x � 6 } or, using interval notation, ( -ex), 4 ] u [6, ex)) .

-----

. The x-mtercepts are 6-Z.J6 � 0.37 and ---

6±Z.J6

---

3

3

9.

.

b.

' � 3.63 . The y-mtercept IS

f(O) = 3 (0) 2 -12(0)+ 4 = 4 . e.

)' 8

(4,4)

10.

7.

f(x) =_2X2+12x+ 3 a =-2, b =12, c = 3. Since a =-2 < 0, the graph opens down, so the vertex is a maximum point. The maximum occurs at b 12 12 x=--==--=3 . 2a -4 2(-2) The maximum value is f( 3)= _2( 3)2+12 ( 3)+3 =-18+ 36+ 3 = 21. --

8.

a.

x2-lOx+ 24�0 We graph the function f(x) = x2 -lOx+ 24 . The intercepts are y-intercept: f(O) = 24 x-intercepts: x2 -lOx+ 24 = 0 (x- 4)(x -6) = 0 x= 4, x= 6 . 10 -b -(-10) The vertex IS at x=-= --- = -=5 . 2 2a 2(1) Since f(5) =-1 , the vertex is (5, -1).

C(m) = 0.15m+129.50

C(860) = 0.15(860) +129.50 =129+129.50 = 258.50 If 860 miles are driven, the rental cost is $258.50. c. C(m) = 213.80 0.15m+129.50=213.80 0.15m = 84.30 m =562 The rental cost is $213.80 if 562 miles were driven. rex) = -0.115x2+1.183x+ 5.623; a =-0.115, b =1.183, c = 5.623 Since a =-0.115 < 0, the graph opens down, so the vertex is a maximum point. a. The maximum interest rate occurs at b -1.183 -1.183 5 = X=--= � .14 . 2a 2(-0.115) -0.23 The maximum interest rate was about r(5.14) =-0.115(5.14)2+1.183(5.14)+ 5.623 �8.67 Thus, the interest rate was highest in 1997, and the highest rate at this time was about 8.67%. b. The year 2010 corresponds to x=18 . r (18 ) =-0.115(18 )2+1.183( 18 )+5.623 �-10.34 The model estimates the rate in 2010 to be -10.34%. This rate does not make sense since an interest rate cannot be negative.

202

© 2008





1.

9.

P=(-1,3);Q=( 4,-2) Distance between P and Q:

d(P,Q) = �(4-( _1))2 + (_2 _ 3)2 = �(5)2 + ( _5)2 = ../25 + 25 = J50 = 5.fi

(

{ I �}

( ) ) 2 2' 2

Domain: z z*

Midpoint between P and Q: -1 + 4 3 - 2 = � � =( 1 . 5 0.5) ' ' 3.

2

5x + 3�0 5x�-3 3 x�- 5

11.

.

[

3 5

{xl x�-�} or [-�, (0).

b.

)

Slope of perpendicular = = m (x - xJ

�

( 4)

x+4

I f(I)=_ _= � * �, so 1) is not on

1+4 5 4

the graph off

c.

Perpendicular to Y = 2x + 1 ; Containing (3,5) Y-YI

x f(x) = a.

The solution set is

5.

3z - 1 6z-7 The denominator cannot be zero: 6z - 7 *0 6z*7 7 z*6 h(z) =

-�2 13.

y-5 = - ( X - 3)

y - 5 = --1 x + -3 2 2 13 y = --1 x + 2 2

-2 =-I, so (-2, - 1) is a f ( -2) = -2 =2 -2 + 4

point on the graph off Solve for x:

2 = _x_ x+4 2x +8= x x= -8 So, (-8, 2) is a point on the graph off

f ( x ) = x3 -5x + 4 on the interval (-4, 4) Use MAXIMUM and MINIMUM on the graph of Y I = x3 - 5x + 4 .

10

4

y

Haxi ..... ul'1'l H= '1. 5 -4x > 2 x < --1 2 1 {x l x < - -}

x

2

x5 - 1 y = - x-I

) 1

3. x

x 1 is not a vertical asymptote because of the =

following behavior: When x itd :

1,

"hole" with coordinates

-64

50

8

Negative

Positive

Positive

{ x l x < -2 } , or, using (-00, -2) .

x3 - 4x 2 > 0 x 2 (x - 4) > 0 f(x) = x3 - 4x 2 = x 2 (x - 4) x = 0, x = 4 are the zeros of f . Interval (-00, 0) (0, 4) Number

5

Value off

-5

-3

25

Conclusion

Negative

Negative

Positive

notation,

5 9 . Answers will vary.

)

00

1

The solution set is

(1, n ) .

(4,

-1

Chosen

will have a

6

Value off

interval notation, 5.

0

(5, 00)

Conclusion

The solution set is

In general, the graph of

. an rnteger,

-3

Chosen

x 3 - 1 = (x - 1) ( x 2 + x + I ) = X 2 + x + I y = -x-I x-I ( x 2 + 1 )( x 2 - 1 ) x 4 - 1 = -'---'-'-----'y = -x-I x-I ( X2 + I ) (x - I)(x + I) x-I 3 2 =x +X +x+1 ( x 4 + X3 + X 2 + x + I ) (x - I) x 5 - 1 = -'--' ----------'-y = -x-I x-I = x 4 + x3 + x 2 + X + 1 n ;;::

(x - 5f (x + 2) < 0 f(x) = (x - 5) 2 (X + 2) x = 5, x = - 2 are the zeros of f . Interval (-00, - 2) (-2, 5) Number

x 2 - 1 = (x + I) (x - I) = x + I y = -x-I x-I

xn - 1 y = -- , x-I

2

{ xl x > 4 } , or, using interval

(4, 00 ) .

61. Answers will vary, one example is

R (x) =

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

2 42

© 2008 Pearson Education, Inc., Upper Saddle River, NJ.


exist. No portion of this material may be reproduced, in any fonn or by any means, without pennission in writing from the publisher.

Section 5.4: Polynomial and Rational Inequalities

7.

13.

x3 -9x S 0 x(x-3)(x+3)sO f(x) x3 -9x x -3, x O,x 3 are the zeros of f .

=

=

=

=

=

Interval Number

(-3,0)

(0, 3)

-1

I

4

Value off

-28

8

-8

28

Conclusion

Negative

Positive

Negative

Positive

15.

(3, 00)

°

1.5

2.5

4

Value off

-6

0.375

-0.375

6

Conclusion

Negative

Positive

Negative

Positive

x3 _2X2 -3x>0 x ( x2 -2x-3 ) >0 x(x+1)(x-3)>0 f(x) x3 -2x2 -3x x -1, x 0, x 3 are the zeros of f . =

=

=

(-«>, - 1)

(-1,0)

(0,3)

(3, 00)

-2

-0.5

I

4

Value off

-1O

0.875

-4

20

Conclusion

Negative

Positive

Negative

Positive

Interval

=

Number Chosen

The solution set is { x l -1 < x < 0 or x>3 }, or, using interval notation, ( -1,0) or (3,00) .

17.

(x-1) ( x2+x+ 4) �0 f ( x) ( x -1) ( X2 + X + 4) x 1 is the zero of f. x2 + X + 4 0 has no real solution. Interval (-00, 1) (1, 00) Number 0 2 Chosen Value off -4 10 Conclusion Negative Positive The solution set is { x l x �1 } , or, using interval notation, [1, (0 ) .

X2(X2 -1) > 0 x2(x-1)(x+ 1)>0 f(x) x2 (x -1)(x + 1) x -1, x 0, x 1 are the zeros of f

=

=

(2, 3)

=

=

11.

(1,2)

The solution set is { x l xs1 or 2s xs3 }, or, using interval notation, ( -00,1] or [2,3] .

2x3> -8x2 2x3 +8x2>0 2X2 ( x+ 4) >0 f( x) 2x3 +8x2 X O,x -4 are the zeros off Interval (-00, -4) (-4,0) (0, 00) Number -5 1 -1 Chosen Value off -50 6 10 Conclusion Negative Positive Positive The solution set is { x 1-4 < x < 0 or x>O } , or, using interval notation, ( -4,0) or (0,00) . =

(-00, 1)

Chosen

The solution set is { x 1 xs -3 or 0s xs3 } , or, using interval notation, ( -00, -3] or [0,3 ] . 9.

=

Number

(3, 00)

-4

Chosen

=

Interval

=

(-«>,-3)

(x-1)(x-2)(x-3)s 0 f(x) (x -1)(x -2)(x -3) x 1, x 2, x 3 are the zeros of f .

=

=

=

Interval Number

=

=

(-00, -1)

(-1,0)

-2

-0.5

(0,

I)

0.5

(I, 00) 2

Chosen Value off

12

-0.1875

-0.1875

12

Conclusion

Positive

Negative

Negative

Positive

The solution set is { x I x < -1 or x>1 } , or, using interval notation, ( -00, -1) or (1,00) .

243

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Chapter 5: Polynomial and Rational Functions

1 9.

X 4>1 X4 -1>0 (x2 -1)(x2 + 1)>0 (x -l)(x + 1)(x2 + 1)>0 f(x) (x -l)(x + 1)(x2 + 1) x 1, x -1 are the zeros of f ; x2 + 1 has no real solution

23.

=

=

=

=

(

-

Number

<Xl,

-1)

-2

(-1,1) 0

(I,

<Xl

2

Chosen 15

-I

15

Conclusion

Positive

Negative

Positive

25.

x+1 >0 x-I f(x) x+l x-I The zeros and values where f is undefmed are x -1 and x 1 . Interval (-00, -1) (-1,1) (1, ) Number 0 2 -2 Chosen 1 Value off -1 3 3 Conclusion Positive Negative Positive The solution set is { x I x < -l or x>1 } , or, using interval notation, (-00, -1) or (1, 00) .

00 )

(0, I)

-2

-0.5

0.5

Chosen Value off

-1.5

1.5

-1.5

1.5

Conclusion

Negative

Positive

Negative

Positive

2

(x-2)2 � 0 x2 -1 (X_2)2 � 0 (x + l)(x -1) f(x) (x-2i x2 - 1 The zeros and values where f is undefmed are x -1, x 1 and x 2 .

=

=

(I,

(-1,0)

The solution set is { xlx:-::; -1 or O < x:-::;l } , or, using interval notation, (-00, -1] or (0,1] .

The solution set is { x I x < -l or x>1 } , or, using interval notation, (-00, -1) or (1, 00) . 21.

=

( -<Xl, - I)

Number

)

Value ofl

=

Interval

=

Interval

(x-1)(x+1) :-::; 0 x + 1) f(x) (x -1)(x x The zeros and values where f is undefmed are x -1, x 0 and x 1 .

=

=

=

(0

=

=

(-00, -I)

(- \, I)

(\, 2)

(2, 00)

-2

0

1.5

3

Valueofl

\6 -

-4

0.2

0.\25

Conclusion

Positive

Negative

Positive

Positive

Interval Number Chosen

3

The solution set is { x I x < -l or x>1 } , or, using interval notation, (-00, -1) or (1, 00) .

244 © 2008 Pearson Education, Inc., Upper Saddle River, NJ . All rights reserved. This material is protected under all copyright laws as they currently


Section 5.4: Polynomial and Rational Inequalities

6x-5 ° 2 0.25 [0,0.25] 1 ( 0.25 ) = 0. 1 5625 > ° [0. 1 25,0.25] 0. 1 25 3 1 ( 0. 125 ) 0:; -0.4043 < ° [0. 1 875,0.25] 4 0. 1 875 1 (0. 1 875 ) 0:; -0. 1 229 < ° [0. 1 875,0.2 1 875] 0.2 1 875 5 1 ( 0.2 1 875 ) 0:; 0.0 1 64 > ° [0.203 1 25,0.2 1 875] 0.203 1 2 5 6 1 ( 0.203 125 ) 0:; -0.0533 < ° [0.2 1 09375,0.2 1 875] 0.2 1 09375 7 1 ( 0.2 109375 ) 0:; -0.0 1 85 < ° Smce the endpomts of the new mterval at Step 7 agree to two decnnal places, r = 0.2 1 , correct to two decimal places. 2008 Pearson Education, Inc., Upper Saddle River, NJ.

265

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©


(b)

1 (x) = X4 + 8x3 - x 2 + 2; - 1 � r � 0 We begin with the interval [-1 ,0). I( -1) = -6; 1 (0) = 2 Let m j = the midpoint of the interval being considered. So m1 = -0.5 n New interval mn I ( mn ) 1 -0.5 [-1 , -0.5] 1 (-0.5) 0.8 125 > 0 2 -0.75 [-0.75, -0.5] I( -0.75) ", - 1 .62 1 1 < 0 -0.625 3 [-0.625, -0.5] I( -0.625) ", -0. 1 9 1 2 < 0 4 -0.5625 [-0.625, -0.5625] 1 (-0.5625) ", 0.3599 > 0 -0.59375 5 [-0.625, -0.59375] 1 (-0.59375) ", 0.0972 > 0 6 -0.609375 [-0.609375, -0.59375] 1 (-0.609375) ", -0.0437 < 0 7 -0.60 1 5 625 [-0.609375, -0.601 5 625] I( -0.60 1 5625) ", 0.0275 > 0 Smce the endpomts of the new mterva1 at Step 7 agree to two decnna1 places, r = -0.60, correct to two decimal places. =

(c)

1 (x) = 2x3 + 6x2 - 8x + 2; - 5 � r � -4 We begin with the interval [-5,-4). 1 (-5) = -58; 1 (-4) = 2 Let mj = the midpoint of the interval being considered. So m1 = -4.5 n New interval mn I ( mn ) 1 -4.5 [-4.5,-4] I( -4.5) = -22.75 < 0 2 -4.25 [-4.25,-4] 1 (-4.25) ", -9. 1 56 < 0 3 -4. 1 25 [-4. 125,-4] I ( -4. 1 25) ", -3.2852 < 0 4 -4.0625 [-4.0625,-4] 1 ( -4.0625) ", -0.5708 < 0 5 -4.03 125 [-4.0625, -4.03 1 25] I( -4.03 125) ", 0.7324 > 0 6 -4.046875 [-4.0625, -4.046875] 1 (-4.046875) ", 0.0852 > 0 7 -4.0546875 [-4.0546875, -4.046875] 1 (-4.0546875) ", -0.24 1 7 < 0 8 -4.0507 8 1 25 [-4.05078 1 25, -4.046875] 1 (-4.05078 125) ", -0.0779 < 0 9 -4.048828 1 25 [-4.05078 1 25, -4.048828 1 25] 1 (-4.048828125) ", 0.0037 > 0 10 -4.0498046875 [-4.0498046875, -4.048828 125] 1 (-4.0498045875) ", -0.0371 < 0 Smce the endpomts of the new mterval at Step 1 0 agree to two decImal places, r = -4.05, correct to two decimal places.


266


Section 5.5: The Real Zeros of a Polynomial Function

(d)

l (x) = 3x3 - 1 0x + 9; - 3 : 0 6 -2. 1 7 1 875 [-2. 1 7 1 875, -2. 1 5 625] I( -2. 1 7 1 875) ,., -0.0157 < 0 7 -2. 1 640625 [-2. 1 7 1 875, -2. 1 640625] I( -2. 1 640625) ,., 0.2366 > 0 8 -2. 1 6796875 [-2. 1 7 1 875, -2. 1 6796875] I ( -2. 1 6796875) ,., 0. 1 108 > 0 9 -2. 1 6992 1 875 [-2. 1 7 1 875, -2. 1 6992 1 875] 1 (-2. 1 6992 1 875) ,., 0.0476 > 0 1 0 -2. 1 708984375 [-2. 1 7 1 875, -2. 1 708984375] I( -2. 1 708984375) ,., 0.0 1 60 > 0 Smce the endpomts of the new mterval at Step 10 agree to two deCImal places, r = -2. 1 7, correct to two decimal places.

(e)

1 : 0 [ 1 . 1 25 , 1 .25] 1 . 1 25 3 1 (1 . 125) ,., -0. 1855 < 0 4 1 . 1 875 [ 1 . 1 25, 1 . 1 875] 1 (1 . 1 875) ,., 0.2722 > 0 [ 1 . 1 25, 1 . 1 5625] 5 1 . 1 5 625 1 (1 . 1 5625) ,., 0.0390 > 0 6 1 . 140625 [ 1 . 140625, 1 . 1 5625] 1 ( 1 . 1 40625) ,., -0.0744 < 0 7 1 . 1484375 [ 1 . 1 484375, 1 . 1 5625] 1 (1 . 1484375) ,., -0.01 80 < 0 [ 1 . 1484375, 1 . 1 5234375] 1 . 1 5234375 8 1 (1 . 1 5234375) ,., 0.0 140 > 0 [ 1 . 1 50390625, 1 . 1 5234375] 9 1 . 1 50390625 1 (1 . 1 50390625) ,., -0.0038 < 0 Smce the endpomts of the new mterval at Step 9 agree to two deCImal places, r = 1 . 1 5 , correct to two decimal places.


267

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(f)

1 ( x)

2X4

=

+ x2

°�

- 1;

r

�1

We begin with the interval [0, 1 ] . 1 ( 0 ) - 1; 1 ( 1 ) = 2 =

mi

Let

m1

So

=

=

the midpoint of the interval being considered.

0.5

n

mn

1 ( 0.5 )

1

0.5

2

0.75

3

0 . 625

4

0. 6875

5

0.7 1 875

6

0.703 1 2 5

7

0.7 1 09375

8

0 .70703 1 2 5

9

0.708984375

New interval

I (mn )

=

-0.625

°

[0.5,0.75]

1 ( 0.625 ) "" -0.3042 < ° 1 ( 0.6875 ) "" -0.0805

°

[0.6875,0.7 1 875]

1 ( 0.7 1 09375 ) "" 0.0 1 64 > °

[0.703 1 25 , 0.7 1 09375]

1 ( 0.703 1 25 ) "" -0.0 1 68 < ° 1 ( 0.70703 1 25 ) "" -0.0003

°

[0.70703 1 25 , 0 .7 1 09375] [0.70703 1 2 5 , 0.7089843 75]

Smce the endpomts of the new mterval at Step 9 agree to two decImal places, 0.70, correct to two decimal places.

r =

(g)

I (x)

2X4 - 3x 3 - 4x 2 - 8;

=

We begin with the interval [2,3] 1 ( 2 ) = - 1 6; 1 ( 3 ) 37 =

Let So

mi

m1

=

=


2.5

n

mn

1

2.5

2

2.75

3

2 . 625

4

2 . 5 625

5

2.53 125

6

2. 546875

7

2 . 5 3 90625

1 (2.5)

=

- 1 .75

I (mn )

°

[2 . 5 ,2.75]

1 ( 2 . 5 625 ) "" 1 .4905 > °

[2 . 5 , 2 . 5 625]

1 ( 2 . 625 ) "" 5 . 1 3 5 3 > °

[2 .5,2.625]

1 ( 2 . 5 3 1 25 ) "" -0. 1 787 < °

[2. 5 3 1 25,2.5625]

1 ( 2.5390625 ) "" 0.2293 > °

[2. 5 3 1 25 , 2. 5 3 90625]

1 ( 2. 546875 ) "" 0 . 643 5 > °

[2. 5 3 1 25, 2 . 546875]

Smce the endpomts of the new mterval at Step 7 agree to two decnnal places, 2 . 5 3 , correct to two decimal places.

r =


268


Section 5.5: The Real Zeros of a Polynomial Function

1 ( x) = 3x 3 - 2X 2 - 20;

(h)

2�r�3

We begin with the interval [2,3] . 1 ( 2 ) = -4; 1 ( 3 ) = 43 Let

mi

=


So � = 2. 5 n

mn

New interval

I (mn )

1 ( 2 . 5 ) = 1 4.375 > 0

1

2.5

2

2.25

3

2 . 1 25

4

2 . 1 875

5

2 . 1 5 62 5

6

2 . 1 40625

7

2 . 1 3 28 1 25

8

2 . 1 2890625

9

2 . 1 30859375

1 ( 2 .25 ) "" 4.0469 > 0 1 ( 2 . 1 2 5 ) "" -0.244 1

0

[2 . 1 25,2. 1 875]

1 ( 2 . 1 40625 ) "" 0.2622 > 0

[2 . 1 25 , 2 . 1 40625]

1 ( 2 . 1 5 625 ) "" 0.777 1 > 0

[2 . 1 25,2. 1 5 625]

1 ( 2 . 1 328 1 25 ) "" 0.0080 > 0 1 ( 2 . 1 2 890625 ) "" -0. 1 1 83

3 }, or, using interval notation, [1,2] (3,00). -2

4 3.

1

E]

o

2

(

u

4

I�I I. 6

x2 -8x+12 > 0 x2 -16 f(x) = x2 x-8x+12 2 -16 (x-2)(x-6) >0 (x + 4)(x -4) The zeros and values where the expression is undefined are x = -4, x = 2, x = 4, and x = 6 . Interval

Number

Chosen

Value off

49.

-

-5

Negative

Positive

<Xl

0 The valueandof thenegatfunctive iatontheis posit iveSiatnceonethe endpoint ot h er. function isguarantees continuous,at least the Intermedi aiten Val ue Theorem one zero t h e given interval. f(x) =8x4 -4x3-2x-l; [0, 1] f(O) = -1 0 andf(1) = 1 > 0 The valueandof thenegatfunctiveiatonthiseposit iveSince at onethe endpoint ot h er. function isguarantees continuous,at ltehaste Intermediat e tValue Theorem one zero i n he given interval. f ( x ) = x3 -X 2 f ( 1 ) =-2; f( 2 ) =4 So by the Intermediate Value Theorem, f hasSubdivide a zero onthetheinterval interval[I,2[1,J i2n].to 10 equal subi ntervals: [1.1,1. [[1.I,1.1J; 5( 1,1.) 6];-2;1 [1. 6(,1.1.12];7)];[1.=[1.-1.2 ,1.7,1.7369];8];[1.[31.,1.8,1.4];9];[1.[41.,1.9,25];] f f ( 1.1 ) = -1.769;1 ( 1.2 ) = -1.472 f( 1. 2 ) = -1. 4 72;1 ( 1. 3) = -1.103 f( 1. 3) -1.103;1 ( 1. 4 ) = -0. 6 56 f (1.4) = -0.656;1 ( 1. 5 ) = -0.125 f( 1. 5 ) = -0.125;1 ( 1. 6) = 0.496 So f has a real zero on the interval [1. 5 ,1.6]. Subdivide the interval [1. 5 ,1. 6] into 10 equal subi n tervals: [ 1. 5 ,1. 5 1]; [1. 5 1,1. 5 2]; [1. 5 2,1. 5 3J; [1. 53,1.54J;

71.

0 and 1(4} =-36 2008 Pearson Education , Inc . , Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright l aws as they currently exist. No portion of thi s material may be reproduced, i n any form or by any means , without permission in writing from the publi sher.

Section 6.2: One-to-One Functions; Inverse Functions

69.

P = - -41 x+100' 0 �x �400 -41 x = 100x = 4(100-p)

Section 6.2 1.

P

3.

= �4(100-p) 25 +600 = 2.JI00-p 25 + 600, 0 � � 100 Thus, C(p) = 2.JI0025 + 600, 0 � p � 100. h = 2r p

5.

P

73 .

7.

9.

/(x) the number of Euros bought for x dollars; g ( x) the number of yen bought for x Euros / (x) = 0. 8 382x g (x) = 140.9687x (go /)(x) = g ( J(x)) = g (0. 8382x) = 140. 9 687(0. 8382x) = 118.15996x (go /)(1000) = 118.15996(1000) = 118,159.96 yen Githatven that/and g are odd functions, we know /(-x) =-/(x) and g(-x)=-g(x) for all x in the domain of/and g, respectively. The composite function (fog)(x) =/ ( g(x)) has the following property: (f 0 g)(-x) = / ( g ( -x) ) = / ( ( x )) since g is odd = -/ ( g ( x ) ) since / is odd = -(f 0 g)(x) Thus, / 0 g is an odd function. =

11.

=

a.

b.

13.

c.

1 5.

d.

75.

1 7.

1 9.

21.

-g

The setareofnoordered pairspairsiswia functi osame n because there ordered t h the element and different second elements. fIrst The function is not defined when x2 + 3x -18 = 0 . Solve: x2 +3x-18 = 0 (x + 6)(x -3) = 0 x = -6domaior nxis= 3{x I x "* -6, x"* 3} . The y=x False. If/andthegdomai are inverse functions, thennthof/e range of/is n of and t h e domai g is the range of g. The functioninisputsone-to-one because totherethe aresameno two distinct t h at correspond output. The function iinputs, s not one-to-one because there arethat two different 20 Hours and 50 Hours, correspond to the same output, $200. The functioninputs, is not one-to-one because there areto two distinct 2 and -3, t h at correspond the same output. The functioninputs is one-to-one becausetotherethe aresameno two distinct that correspond output. The zfunction/is one-to-onethe because every hori ontal l i n e i n tersects graph at exactly one point. The are horifunction/is zontal linesnot(forone-to-one example,because y = 1) there that intersect the graph at more than one point. The function/is one-to-onethe because every hori z ontal li n e i n tersects graph at exactly one point.

303 © 2008 Pearson Education , Inc . , Upper Saddle River, NJ. All rights reserved. Thi s material i s protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 6: Exponential and Logarithmic Functions

23.

Tothe find nterchangein thethe elements domaithen iwinverse, th the ielements range: in Annual Rainfall

460.00 202.01 1 96.46 1 91 .02 1 82.87

25.

33.

Location

Mt Waialeale , Hawaii Monrovia, Liberia Pago Pago, American Samoa Moulme i n , Burm a Lae , Papua N ew G u inea

Domain: {460. 00, 202. 0 1, 196.46, 191.02, 182. 8 7} Range: Waialeale, Monrovia, Pago Pago, Moulmei{Mtn, Lea} Tothe find nterchangein thethe elrange: ements in domaithen iwinverse, th the ielements Monthly Cost

of Life Insurance

$7.09 $8.40 $1 1 .29

27.

2 9.

31.

Age

30 40 45

Domain: {$7. 09, $8. 40, $11.2 9} Range: {30, 40, 45} Interchange the entri(2,-1), es in (11,0), each ordered { (5,-3), (9,-2), (-5,l)}pair: Domain: {5, 9, 2, 11, -5} Range: {-3,-2,-1, 0, 1} Interchange the entries { (1,-2), ( 2,-3), (4,2)} pair: (0,-10in), each (9,1),ordered Domain: {1, 2, 0, 9, 4} Range: {-2,-3,-1 O, 1, 2} f(x) =3x+4; g(x) = -(31 x-4) f ( g(X) ) = f G (X-4) ) = 3 G (X-4) ) +4 = (x-4) +4 = x g ( !(x) ) = g(3x+4) = '31 ( (3x + 4) -4) = '31 (3x) = x Thus, f and g are inverses of each other.

37.

x f(x) = 4x-8; g(x) =-+2 4 f ( g(x» ) = f(�+ 2 ) =4 (� +2) -8 = x+8-8 =x g ( !(x» ) = g(4x-8) 4x-8 +2 = -4 =x-2+2 =x Thus,!and g are inverses of each other. g(x) = 4x+8 f ( g(x» ) = f ( 4x+8 ) = (4x+8 t -8 = x+8-8 =x g ( !(x» ) = g(X3 -8) = 4(x3 -8)+8 = (;3 =x Thus,!and g are inverses of each other. f(x) =-;x1 g(x) =-x1 11 x f ( g(x) ) = f (-x1 ) =-=l·-=x 1 g ( !(x» ) = g (-x1 ) =-=1·-=x 11 x1 x Thus,!and g are inverses of each other. x

3 04 © 2008 Pearson Education , Inc . , Upper Saddle River, NJ. All rights reserved. This material i s protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


39.

4x-3 -x) = I(x) = 2x+3 g( ; x+4 2-x 4X-3 2 ( ) +3 2-x = l(g(X))=/ ( 4x-3 ) 4x-3 +4 2-x -2-x (= 2 (�) +3) c2-X) ( 4X-3 2-x +4) ( 2-X) 2(4x-3) +3(2 -x) 4x-3+4(2- x) 8x-6+6-3x 4x-3+8-4x 5x =x5 2X+3 ) - 3 4 (� 2X+3 g(J(x)) = g ( x+4 ) = 2- 2x+3 x+4

43.

41 .

2

/

/

' / y =

y 2

x

/

45.

/

/

/

-2

Graphing the inverse: )!

3

/

47.

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

4(2x + 3) -3(x + 4) 2(x+4)-(2x+3) 8x+12-3x-12 5x2x+8-2x-3 = x5 Thus, Iand g are inverses of each other. Graphing the inverse:y

Graphing the inverse:( 1 , 2)

/

/

/

/

/

/

/

/

/

/

/

= X

I- I

x

-3

I(x) = 3x y =3x x = 3y Inverse y = -x3 r l (x) = �3 x Verifying: 1(1-1 (x) ) = I (± X) = 3 (± X) = x r l (I(x)) = rl (3x) = �3 (3x) = x Domain ofl Range of I-I All real numbers Range ofl Domain of I-I All real numbers =

=

=

X

/

/

/ /v / ..

=

/ (2 , 1) f- 1 / /

/

( - 2, -2)

/

X

-2

-5

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49.

4x + 2 f(x) == 4x+2 x = 4y + 2 Inverse 4y = x-2 x-2 y= -41 x y=--4 -2 rl(X) =::._4 2-2 Verifying: f { !-I (X» ) = f (� -�) = 4 (� -�) + 2 =x-2+2=x 4x+2 rl(J(x» ) = r l (4x+2) = 4 _ 2-2 =x+ -21 --21 =x Domain off Range of f-I All real numbers Range off Domain of f- I All real numbers

Domain off Range of f-I All real numbers Range off Domain of f-I All real numbers =

Y

=

Y+S f- t /"

/

/

/

/

/

/

/ /y

=

�

Y 5

/

/

/

/

/

/ y = x

x

-5 53 .

= x

f - l (x )

=

/

=

f(x ) = 4x + 2

51.

=

=

=

=

-�

f(x) = x3 -1 =X3 -1 x = i -1 Inverse i =x+l y=rx+i rl(x) =rx+i Verifying: f { !-I (x) ) = f( rx+!) = (rx+!t -1 =x+l-l =x r l (J(x» ) = rl ( x3 -1) = 4(x3 -1 ) + 1 =.(1 =X

f(x) = x2 + 4, x � 0 y = x2 +4, x � O x = i + 4, y � 0 Inverse y 2 = x-4, x � 4 y =.)x-4, x � 4 rl(x) =.)x-4, x�4 Verifying: f(rl(x» ) = f (.)x-4 ) = (.)x-4 r +4 =x-4+4 =x rl (J(x» ) = rl ( X2 +4 ) = �( X2 +4 ) -4 =H =l xl =x, x � 0 Domain off Range of r l {x I x � O} or [0, (0) Range off Domain of rl {x I x � 4} or [4, ) =

Y

=

=

(0

y 8

=

f(x) =

x2 +

X �

//

0

/

/

/

4,

/

/ f- l (x) - 2 /7 /�2 /

,Y

= X

--Ix

-4

///

=

x

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55.

Verifying: x +1 f V-I (X) ) = f e x ) 2x+1 _ 2 x l·x

f(x) = -x4 y=x4 x = y4 Inverse xy=4 y = -4x rl(x) =�x Verifying: f(rl(x) ) =f (�) = � = 4 {�) =X x -

x 2x+1-2x x = -1 = X l) rl (J(x)) = f -I (_ x-2 1 2 (_ x-2 )+1 x-2 ( 2(�)+1 }X- 2) 1 )(X-2) (_ x-2 2+ (x-2) = 1 = -x1 =x Domain off Range of f- I All real numbers except 2. Range off Domain of f- I All real numbers except O.

x Domain off Range of f -I All real numbers except O . Range off Domain of f-I All real numbers except =

=

=

o.

=

=

=

X

/

57.

5

=

=

x=o Y

-5

1 f(x) = x-21 y= - x-21 x = y-2 Inverse xy-2x = 1 xy = 2x+1 2x+1 y=- x 2x+1 rl(x) = x

x=2 'Y=x I / I / / I / I / /

----Y= 2 �====tr��X y = O /

--

/

/

/

/

/

/

/

/

307 © 2008 Pearson Education , Inc . , Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright laws as they currently exist. No portion of thi s material may be reproduced, in any form or by any means , without permi ssion in wri ting from the publisher.


59.

f(x) = 3+x2 2 y = -3+x2 x = -3+ y Inverse x(3+ y) = 2 3x+xy = 2 xy = 2-3x 2-3x y=- x 2-3x r l(x) = x Verifying: 2-3X 2 f(rl(x) ) = f (-x ) = 3+ 2-3x x 2x 2·x (3+ 2 �3X )x 3x+2-3x = -2x2 = X 2) 2 3( 2 rl (I(x)) = rl (_3+x_) = 2� 3+x ( 2-3(�)}3+X) C�J(3+X) 2(3+x)-3(2) _ 6+2x-6 2 2 2x =-2 =X Domain off Range of f-I All real numbers except -3 . Range off Domain of f-I All real numbers except

61.

_

=

=

=

=

=

=

=

f(x) = � x+23x y = - x+2 x=� y+2 Inverse x(y+2) = 3y xy+2x =3y xy-3y = -2x y(x-3) = -2x -2x y=x-3 -2x rl (x)= x-3 Verifying: f(rl(x) ) = f( ::� ) 3 (�) _ (3 ( �)}X-3) -2x +2 ( -2x +2 )(X-3) x-3 x-3 -6x -6x - = --=x ----2x+2x-6 -6 rl (I(x)) = rl (� x+2 ) -2 (� ) ( -2 (� ) }X+2) 3x - (--3 3X --3 x+2 x+2 ) (x+2) -6x -6x --3x-3x-6 =-=x -6 Domain off Range of f-I All real numbers except -2 . Range off = Domain of f-I All real numbers except 3.

O.

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63.

f(x) = � 3x-l 2x y = 3x-l 2y Inverse x = -3y-l 3xy-x = 2y 3xy-2y =x y(3x-2) = x x y = -3x-2 x rl(x) = _ 3x-2_ Verifying: x ) = 2 (6) f ( rl(x) ) = f (x ) -1 3x-2 3 (_ 3x-2 ( 2 (6) ) (3X-2) ( 3 ( 3x2x� 2 ) -1)2x(3x -2) ---- =-=x 3x-(3x-2) 2 2x M rl (J(x)) = f (� 3x-l ) = 3 (� 3x-l ) -2 (� 3x- l ) (3X-l) (3 ( 3��2x1 ) -2) (3X-l) 3(2x) -2(3x -1) 2x = -2x =x ---6x-6x+2 2 Domain off Range of f- I All real numbers except -31 . Range off Domain of f-I All real numbers except -23 .

65.

=

=

f(x) = 3x+4 2x-3 3x+4 y = -2x-3 3y+4 x = -2y-3 Inverse x(2y-3) = 3y+4 2xy -3x = 3y + 4 2xy -3y = 3x+4 y(2x-3) = 3x+4 y = 3x+4 2x-3 3x+4 rl(x) = 2x-3 Verifying: 3X+4 ) + 4 3 x 2x-3 ( ) = f ( r\x) ) = f (x+2 2 ( 3X+4 ) _ 3 2x-3 ( 3 (�) +4}2X-3) (3(3x2 ( �:+ 4):�+)4(2x-3}-3)2X-3) 2(3x + 4) -3(2x -3) 9x+12+8x-12 -- = 17x -6x+8-6x+9 17 =x 3X+4 ) +4 3 ( 3X+4 rl (J(x)) =f-I ( 2x-3 ) = 2 2x-3 ( 3X+4 2x-3 ) _ 3 (3 (�) +4}2X-3) (3(23x( �:+ 4):�+)4(2x-3}-3)2X-3) 2(3x + 4) -3(2x -3) 9x+12+8x-12 17x =x --= 6x+8-6x+9 17 Domain off Range of f-I All real numbers except -23 . Range off Domain of f-I All real numbers except -23 . =

=

=

=

=

=

309

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67.

l(x) = 2x+3 x+2 2x+3 y= - x+2+ 3 2y x = y+2 Inverse xy + 2x = 2y + 3 xy -2y = -2x+3+3 y(x-2) = -2x y= -2x+3 x-2 -2x+3 rl(x) = x-2 Verifying: -2X+3 ) +3 ( 2 +3 x-2 1(1 -1 (x) ) = 1 ( -2X x-2 ) = -2x+3 x-2 +2 ( 2 ( -:�� 3 ) +3}X-2) ( -2X+3 +2 ) (X-2) x-2 +3(x-2) 2(-2x+3) -2) -2x + 3 + 2(x -4x+6+3x-6 -x =x ----- = -2x+3+2x-4 -1 2X+3 ) + 3 ( 2 � rl (l(x))= rl ( 2X+3 x+2 ) = 2x+3 x+2 _ 2 (�) +3}X+2) ( -22X+3 ( x+2 -2) (X+2) -2(2x + 3) + 3(x + 2) 2x + 3 -2(x + 2) -4x-6+3x+6--- = --x =x -2x+3-2x-4 -1 Domain ofl Range of I- I All real numbers except -2. Range ofl Domain of I-I All real numbers except 2.

69.

x2 -4- ' x > O I(x) = 2x2 2 x -4- ' x > O y =-2x2 X = -/2y--42- ' y > O Inverse 2x/ = / -4, x < -21 2xy 2 - / =-4, x< -21 / ( 2x-l) = -4, x< -21 / ( 1-2x) = 4, x < -21 4 ' x < -1 y2 = 1-2x 2 1 4 y = �1_ 2X ' x O} or (0, Range ofl Domain of I- I =

=

00

83.

)

=

Because the ordered pair (-1, 0) is on the graph, /(-I) = O . Because the ordered pair (1,2) is on the graph, 1(1) = 2 . Because the ordered pair (0,1) is on the graph, r l(1) = O . Because the ordered pair (1,2) is on the graph, r l(2) = 1. Since 1(7) 13 , we have r l (13) = 7; the iwhen nput ofthetheoutput function is function the outputis ofthethine putinverse of the of the inverse. nce theanddomai of a function is the range the n iSinverse, the nrange of the function is the ofdomai of the inverse, we get the following for I- I : Domain: [-2,00) Range: [5,00) nce theanddomain of a function is the range the n iSinverse, the range of the function is the ofdomai of the inverse, we get the following for g-I : Domain: [0,00) Range: all real numbers Since I (x) is increasing on the interval (0,5), it is one-to-one on the interval and has an inverse, rl (x) . In addition, we can say that r l (x) is increasing on the interval ( I ( 0), I (5)) . I(x) = mx+b, m "" 0 y= mx+b x = my + b Inverse x-b= my y = -m1 (x-b) rl(x) = �m (x-b), m "" O

7 1 . a.

85.

b.

77.

79.

81 .

�

�

(x)

87. a.

d.

75.

�

�

c.

73.

If { a ,b) is on the graph off, then (b, a ) is on the graph of I-I . Since the graph of I-I lies in quadrant I, both coordinates of (a,b) are positive, which means that both coordinates of (b, a) are positive. Thus, the graph of I-I must lie in quadrant I. Answers may vary. One possibility follows: I(x) = / x / ' x 0 is one-to-one. Thus, I(x) = x, x 0 y = x, x 0 rl = x, x 0 d = 6. 9 7r -90. 3 9 d + 90. 3 9 = 6. 9 7r d---+90. 3 9 = r 6. 9 7 we would write Therefore, 39 r( d ) = d +90. 6. 9 7 r (d(r)) = (6. 9 7r-90. 3 9)+90. 3 9 6.---97r + 90.6.3997 -90.39 -- = 6. 97r r 6.97 6.97 d d ( r (d )) = 6. 9 7 ( +90. 3 9 ) _90. 3 9 6.97 39 = dd +90. 3 9-90. = 39 "" 56. 0 1 r(300) = 300+90. 6.requi 97 red to stop was 300 feet, Ifthe distance the speed of the car was roughly 56 mph. 6 feet 72 inches W (72) = 50+ 2. 3 (72-60) = 50 + 2. 3 (12) = 50 + 27.6 = 77.6 The ideal weight ofa 6-foot male is 77.6 kg. = 50+ 2. 3(h -60) -50 = 2. 3h -138 +88 = 2. 3 h +88 2. 3 we would write Therefore, h(W) = 2.+883

=

b.

--

=

c.

89.

a.

b.

=

W

W

W

W

=h W

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c.

» )+88 h( W(h» )= (50+2. 3(2.h-60 3 +88 2.3h 50+ 2. 3 h -138 = 2. 3 = 2. 3 = h W (h(W») =50 + 2. 3 ( W+88 -60) 2.3

= 5 0 + W + 88 - 138 = W

d.

91.

a.

b.

c.

0

h(80) = 8 +88 = � 73.04 The heiofght80ofkga malis roughly e who is73atinhisches.ideal weight From the resttheriction given in the problem statement, domai n i s { g I 30,650 ::; g ::;74,200} or [30650, 74200]. T(15, 1 00) = 4220 + 0. 2 5(30, 650 -30, 650) =4220 T(74,200) = 4220 + 0. 2 5(74, 200 -30, 650) = 15,107. 5 Since T is linear and increasing, we have that the range is {T 1 4220::; T ::; 15,107. 5 } or [4220, 15107. 5 ] . T = 4220 + 0. 2 5 ( g -30, 650) T -4220 = 0. 2 5(g-30,650) T -4220 = g -30, 650 5 T -4220 +300.2' 650 =g 0. 2 5 Therefore, we would write g(T) = + 30 650 Domain: {T 1 4220::; T ::; 15,107. 5 } Range: { g 1 30,650::; g ::; 74, 200} The graph of is symmetri c about the y-axis. Sithencerock t represents the number of seconds after begins to fall, we know that t ;;: : 0 . The graph i s stri c tl y decreasing over i t s domain, so it is one-to-one. H = 100 -4. 9t2 H + 4.9t2 = 100 4.9t2 = 100 - H t2 = t = )'OO-H 2.3

2.3

4220 T0.25

93.

a.

b.

�

c.

95.

Therefore, we would write t ( H) = lOO4.-9 H . need tthe;;:: 0pri) ncipal square (rootNote:sincewe weonlyknow H (t ( H)) = 100 -4 . 9 ( l O��H J ) = 100_4. 9 ( 100-H 4.9 = lOO - l OO + H =H 9t 2 ) t(H(t)) = 100- ( 100-4. 4. 9 = )4. 9t 2 = J? = t ( since t ;;:: 0) 4. 9 2.02 t(80) = lOO4.-80 9 Itfalwil 80l take meters.the rock about 2. 02 seconds to �

ax + b I (x) = ex +d ax + b y = - ex + d ay + b x ey +d x(ey + d) ay + b exy + dx ay + b exy - ay = b - dx y( ex - a) b - dx b - dx y = - ex - a +b F I (X) -dx ex - a I I-I

= -- Inverse = = =

'

H

97. 99.

100 - H 4.9

4.9

= ax + b = ---dx + b . Now, = prOVl' ded that -ex + d ex - a This is only true if a = - d . Answers will vary. No, not every odd functi on is one-to-one. For 3 example, I(x) = x - x is an odd function, but it is not one-to-one.

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Section 6.3: Exponential Functions

Section 6.3 1.

3.

43 = 64 ; 82/3 = (.�8r = 22 = 4 ; T2 = �3 =..!.9 . False. To obtain the graph of = (x -2)3, we would shift the graph of = x3 to the 2 units. True False 32.2 "" 11. 2 12 32.23 "" 1 1 . 5 87 32.236 "" 11. 664 3.Js "" 11.665 23. 14 "" 8. 8 15 23. 1 4 1 "" 8. 821 23. 1 41 5 "" 8. 824 211 "" 8. 825 3.f·7 "" 21. 2 17 3.142.7 1 ",, 22.2 17 3.1412.7 1 8 "" 22.440 22.459 "" 3.320 e-O·85 "" 0.427 y

y

5. 7. 9. 11.

23.

right

a.

b.

d.

25.

a.

-

b. C.

d.

1 5.

a.

b. C.

d.

1 7. 1 9.

21.

ne ""

y

= f (x)

27.

f (x + l) f (x)

�3 = 2 -1 3 12 = 2 0 6 6 18 = 3 1 12 2 12 23 3018 since the ratio of an exponenti Not not constant. ve termsalisfunction consecuti -

H (x + H (x )

--

a

el . 2

x

Y = H (x)

a

C.

13.

1) 1 =4 -1 -41 -( 1 1 4) i1 =4 0 1 �4 = 4 1 4 64 =4 2 16 16 3 64 siwintceh the=rati4 . oSoofthe ails function an exponenti Yes, constant terms consecutive base is 4. f (x + l) x Y = f (x) f (x) -1 23 ( 3 3/ 2) =3 · -23 =2 �3 =2 0 3 .!3. = 2 1 6 6 24 =2 2 12 12 3 24 on siwintceh the=rati2 . oSoofthe aisl functi an exponenti Yes, constant terms consecutive base is 2. = H (x) HH(x(x)+ l) x i2 =2 -1 2 -46 = -32 0 4 231 1068 on since the ratio of an exponenti Not not constant. termsalisfuncti consecutive x

-

29. 31.

y

B

D

313

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Horizontal Asymptote: y

33. A 35. E 37.

1 Using the graph of y 2x , shift the graph up 1 uniDomai t. n: All real numbers Range: {y \ y > O} or (1, 00) Horizontal Asymptote: y 1 f(x)

==

==

°

2X +

==

==

43.

f(x)

==

TX

-2

Using the graph of y 3x , reflect the graph about theDomai y-axis,n: Alandl realshiftnumbers down 2 units. Range: {y \ y > -2} or (-2, 00) Horizontal Asymptote: y -2 ==

==

39.

f (x) 3x-1 ==

Using the graph of y 3x , shift the graph right 1 uniDomain: t. All real numbers Range: { y\y> O} or (0, 00) Horizontal Asymptote: y 0 ==

x

y = -2

==

45.

-5

2 + 4x-1 Using the graph of y 4x , shift the graph to the riDomain: ght one uniAllt real and numbers up 2 units. Range: {y \ y > 2} or (2, 00) Horizontal Asymptote: y 2 f(x)

==

==

-3

41.

-1

f (x) 3 ==

(0, i)

x

3

( 1, ��)

{�J

-

Using the graph of y (�J vertically stretch the graph bymula tfactor ofy-coordi 3. That is, for each point on the graph, i p l y the Domain: All real numbers nate by 3. Range: { y \ y > O} or (0, 00) ==

y ' 8

==

\

'

-4

-2

6

x

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47.

2 Using the graph of y = 3x , stretch the graph hori zontaln: Ally lbyreala factor of 2, and shift up 2 units. Domai numbers Range: > 2} or (2, Horizontal Asymptote: y = 2 f(x) =

+ }' 1 2

{y I y

00

)

y

x

8

-

53.

x

-2

2

f(x) = S - e-x

Using the graph of y = eX , reflect the graph about the y-axis, reflect about the x-axis, and shift up 5 units. Domain: {yAlI lyrealS}numbers Range: or ( 5) Horizontal Asymptote: y = 5
O} or (0, ) Horizontal Asymptote: y 00

=

y

y

8

°

x

55. x

-2

51.

-00 ,

2 Using the graph of y = eX , reflect the graph about thereflecty-axis,aboutstretch horizontal ly byupa factor of2, the x-axis, and shift 2 uni t s. Domai n : All real numbers Range: {y I y 2} or ( 2) Horizontal Asymptote: y = 2 f(x) = - e-x /2

O} or (0, ) Horizontal Asymptote: y 00

=

-00 ,

y

5

°

- - - - - - -

- - - - - - -

x

y

=2

-5

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57.

7x = 7

3

67.

Wesideshaveof thea single termTherefore, with the same basesetontheboth equation. we can exponents equal to each other: The solution set is { } x=3.

3 .

59.

-3x + 42 = 4x

The solution set is { } 42 = 7x

rx = 1 6 4 =2

6=x

rX

6 .

The solution set is { } -x = 4

x = -4

69.

-4 .

61. (.!.)X GJ (�J GJ

3x

2 . 71.

= 22

x = -1

( 2 )X

2

( )

4X . 2 X = 1 6 2 2 2 · 2x = 24 2 2

22 x . 2X = 2 2 2 x+x =

3 x=2

28

28

( )( )

x 2 + 2x = 8

The solution set is {%} .

x 2 + 2x - 8 = 0 x+4

x-2 =0

or 2 The solution set is { } x+4 = 0

3 J" = 9 x 3 3x = 3

x = -4

( 2 )X

x-

=0

x=2

-4, 2 .

3 3 X = 3 2x 3 x = 2x

73.

3 eX = e x + 8

x = 3x + 8

The solution set is { } -2x = 8

3 x - 2x = 0

x = -4

x x2 - 2 = 0

or

x+1 =0

-1, 7 .

2x = 3

x=0

= 3 6.
0, I . The graph contains the points ( -l, �) . (0,1) , (1,3) , and (2,9) . In other words, 1(-1)=-31 , /(0)= 1, 1(1)=3, and 1(2)=9. Therefore, 1 (0) = k· 1 = k· 1 = k·l 1=k and 1 (1) = 3 Let' s use = 3, = 1. Then I (x) = 3x• Now we need to verify that this function yields the other a

a

aP· x

*"

a dO) aD

8 1 . a.

a P-( I )

= aP a

P

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,


known points on the graph. 1 (-1) =

TI

= .!.3 ;

9 3.

1(2) = 32 = 9

89.

So we have the function 1 ( x) = 3x . need a function of the form 1 ( x) = k . aP'x , with a > 0, a 1 . The graph contains the points (-1' - �} (0,-1), (1, - 6), and (2, -36) . Inother words, 1(-1) = - "61 ' 1 (0) = -1 , 1 ( 1)= - 6 , and

We

{

I

y

0.5

f

"*

1(2) = -36 . 1 (0) = k . a P '( O) -1 = k · ao -1 = k · l -1 = k 1(1) = -a P - ( l) -6 = -aP 6 = aP a = 6, p = 1 . I (x) = _6x

Domain: ( ) Range: {Y I - l :O:; y < O} or [-1, 0) Intercept: (0, -1) -00 , 00

Therefore,

95.

and

97.

•

1(2) = -62 = -36

99.

So we have the function 1 (x) = _6x . 1 ( x ) = { e x if x < 0 eX if x 0 y 4

101.

x

p(x) = 1 6, 630(0.90 )X a. p(3) = 16, 630(0.90) 3 � $12, 123 b. p(9) = 16, 630(0.90)9 � $6, 443

D (h) = 5e-O.4 h D (I) = 5e-0.4 ( I ) 5e-O .4 � 3.35 1 3.35 D ( 6) = 5e-O.4 ( 6) = 5e-2 .4 � 0.45 6 0.45 F (t) l _ e-o . 1t a. F(10) = I - e-O·1 (1 0) = 1 - e-1 � 0.632

After hours, mil igrams wil be present. milligrams After hours, milligrams wil be present. =

b.

-2

c.

- 00 , 00

�

00

oflight oflight

=

�

Domain: ( ) Range: {Y I Y I} or [1, Intercept: (0, 1)

p(n) = 100(0.97) n p(10) ::: 100(0.97io � 74% a. b. p(25) 100(0.97)25 � 47% =

Let's use Then Now we need to verify this function yields the other known points on thethatgraph. I ( -1) = _6-1 = -.!.6 91.

-e: �fx < 0 I (x) = -e x x � 0

)

The probabi lity thatPMa iscar0.632. will arrive within 10 minutes of 12:00 F (40) = I - e-o. I ( 40) = 1 - e-4 � 0.982 The probabi lity thatPMa iscar0.982. will arrive within 40 minutes of 12:00 As t � oo, F (t) = I - e-O·1t � 1 - 0 = 1

31 8 © 2008 Pearson Education , Inc. , Upper Saddle River, NJ. All rights reserved. Thi s material i s protected under all copyright l aws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publi sher.


d.

Graphing the function: 1

�

V'

o

F

r =![,- e{H] r, = \2: [I_ e-(',")o ' ] =12 [I- e"'] = S .4 1 4 amperes after 0. 3 second

3.

40 (7) "" 0. 5 0 , so about minutes are needed for the probability to reach 50%. o

e.

1 07 .

7

1

V = . 5 0 3 � 1�7

1 03 .

�

1

40

20X e-20 P(x) =-- x! 1 0 P(15) = 20 15!5 e-2 "" 0. 0516 or 5.16% The probabi5: 00litPMy thatand156:cars00 PMwillisarri5.16%. ve between P( 20) = 20220!0 e-20 "" 0. 0 888 or The probabi5: 00litPMy thatand206:cars00 PMwillisarrive between

b.

,

t

- I

a.

b.

amperes after 0. 5 second = \�O[I- e{',")'] = 12[I- e-'] = 103 76 amperes after 1 second As � co, e-(�} � O . Therefore, as, 120 [ 1- e-( 150 ) 1 � 12[1-0]=12, t � co, JI =10 which amperes.means the maximum current is 12 See the graph at the bottom of the page. = I �O [I- e-(�)o' 1 = 24 [I- e" "] "" 3. 343 amperes after 0. 3 second = I�O [ I_ e-(�)" ' 1 = 24 [I- e" " ] "" 5. 309 amperes after 0. 5 second = I�O [I_ e{M] = 24[I- e" ' ] "" 9.443 amperes after 1 second As � co, e-(�} � O . Therefore, as, t �co, 11 = -1 205- [l- e-( 150) 1 � 24[1-0]=24 , which amperes.means the maximum current is 24 See the graph on the next page.

c.

8 . 88%

d.

8 . 88%.

1

,

1

a.

b.

1

c.

e.

,

,

t

- I

f.

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y 24

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

22 20 18 16

12(/ )

14 12 10

- - - - - - - - - - - - -

(0.5. 7.585)

(0.3, 5.4 1 4)

�

4 2

o

109.

(1

.

0 . 10 . 376 ) - - - -

( 1.0, 9.443)

1 1 (/) .

=

12(1

-

e

-

3.0

2.0

1 1 5.

S

inh x= -l ( x - ) f(-x) = sinh(-x) =-21 ( e

a.

e-

=_

n

b.

e

x

�( x _ e

n

-x

-e

e

x

)

e-

x)

= -sinh x =-f(x) Therefore, f(x) = sinh x is an odd function. Let >-; = � ( ex _e x ) -

.

3 .5

x

1 13.

O .5

- - - - - - - - - - - - - - - - - - - - - - - - -

0. 1 0.2 0 . 3 0.4 0.5 0 . 6 0.7 0.8 0.9 l.O

1 1 1 2+ -2!1 +-+-+ 3! 4! . . +-n! 1 = 4; 2+ -2!1 + -3!1 + -"" 4! 2.7083 = 6; 2+ -2!1 + -3!1 + -4!1 + -5!1 + -6!1 "" 2. 7 181 = 8; 2+ -2!1 + -3!1 + -4!1 + -5!1 + -6!1 + -7!1 + -8!1 "" 2.7 182788 1 1 1 1 + -1 + -1 + -1 + -1 + -1 n =10; 2+-+-+-+2! 3! 4! 5! 6! 7! 8! 9! 10! "" 2. 7 182818 "" 2. 7 18281828 f(x) = aX f(x+ h)- f(x) ax+h _ a h hh aX ( a -1 ) hh = ax [ a h-I ) f(x) = aX f(-x) = a-x =_a1X =_f(x)1_

24( \

- e - 2f )

n

111.

=

- 3.5

1 1 7.

f(x) = 2(21X ) +1 f(1) = 2(2 ) + 1 = 2 2 + 1 = 4 + 1 = 5 f(2) = 2(22 ) + 1 = 24 + 1 = 16 + 1 = 17 f(3) = 2(23 ) +1 = 28 +1 = 256+1 =257 f(4) = 2(254 ) +1 = i6 +1 = 65,536+ = 65,537 2(2 ) + 1 = 232 + 1 = 4,294,967,296+ 1 f(5) == 4,294,967,297 = 641x6,700,417 Answers will vary. 1

1 1 9.

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I)

Section 6. 4: Logarithmic Functions

121.

()

Given the function J x =

a

X

with

,

>1,

a

If x > 0 , the graph becomes steeper as a increases. If x < 0 , the graph becomes less steep as increases.

Section 6.4 1.

3x - 7 � 8

-

{ l

3} .

The solution set is x x �

3.

-x-I

x+4

>0

()

J x =

x-I

x+4 Jis zero or undefined when x = l or x = -4 . Interval

(-00, -4)

(-4, 1)

(1, 00)

Test Value

-5

0

Value ofJ

6

Conclusion

positive

-

{ l

The solution set is x x

(

0 2

= loga x , then x = aY

5 2 = 25 .

The domain of J is

a

9 = 3 2 is equivalent to 2 = log3 9 .

© 2008 Pearson

log5 25 = 2

) ( , l)

1, 0 ,

9.

e

27.

x>3

u

False. If

1 5.

log 1 = 0 since 2° = 1 . 2

e

x-3 > 0

1

-4 or x > I

7.

X

25.

2

1 -4 negative

=4 .

in 4 = x is equivalent to

a

2x

5x � 1 5 x�3

X

23.

=3.

1 requires __ > O . x+l

1 _ is undefined when (x) = _ x+l

x

Interval

(-00, - 1)

(-1, 00)

Test Value

-2

0

Value ofp

-1

1

Conclusion

negative

positive

{ l

= -1 .

} (

)

The domain ofJis x x > - I or - 1, 00 .

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exist. No portion of this material may be reprod uced, i n any form or by any means, without permission in writing from the publisher.


45.

g(X) =IOgs ( X ; I )

X

requires

;

I

59.

>O.

= x x+ 1 is zero or undefmed when - l or x = O .

p (x)

x

=

Interval

(-00, - 1)

( - 1, 0)

(0, 00)

Test Value

-2

1 -2

1

Value ofp

-1

-1

2

negative

positive

2 positive

Conclusion

61.

}

The domain of g is x x < - l or x > O ;

47.

{ l

(-oo,-I)u( O,oo) . f(x) = .J

.

{(x) .

=

(l)X 2

in x requires In x � 0 and x > 0

eO

In x � O x�

x�1

The domain of h is x x �

{ l

I}

or [ 1, (0

).

-3 6 3.

_ ---In

51.

53 .

55.

65.

10

3 "" 30.099 0.04 1n 4 + ln 2

log 4 + log 2

"" 2 .303

2 ln 5 + log 50 log 4 - In 2

57. If the graph of f(x) (2, 2) , then f(2)

= a2 =

log" 2

2

=

loga 2

=

) og l X 2

D

67.

A

69.

E

71.

f(x) b.

= loga x

=

B

a.

",, -5 3 . 99 1

(2. - 1 ) f- I (x)

=

In(x + 4)

Domain: (- 4,

(0

Using the graph of 4 units to the left. X = -4 I I I I I I I

contains the point

2 . Thus,

2

) y

=

In x , shift the graph

y 4 X

a = ±J2 Since the base a must be positive by definition, we have that a = J2 .

-4 c.

Range: (-00,

(0

)

Vertical Asymptote : x

=-4

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f(x)= ln(x+4) y = ln(x+4) x = In(y + 4) y+4= ex -4 rl(x) = -4

d.

y

=

e

e

f.

Inverse

Range off (-00, 00)

f.

Shift the graph of y

=

y 5

e

X

down

4

y = O

units .

75 .

-

a.

...,

- �)

Using the graph of y y 5

= In x,

shift up

7

-2

Domain: (0, 00)

Using the graph of y

shift down y 2

Domain: (0, 00)

2

3

f(x) = 2+ln x y = 2+ln x x = 2+ln y x-2 = ln y = r l(x) = e

x=

© 2008

!. , and

2

3

8

x = ()

c.

Range : ( - 00 , 00)

d.

f(x) = In(2x)-3 = In(2x)-3 x = In(2y)-3 x+3 = In(2y) 2y = y=-21 e r 1 ( x) =!.2

°

Vertical Asymptote :

x=

°

y

e

Inverse

x +3

x+ 3

Inverse

e

x-2

x 2 e -

e.

compress the

'0 . )

x=o

Vertical Asymptote :

X

( 1 , - 2.3)

-5

y

2

X

units.

-

Range: (-00, 00)

= ln x,

units.

-

d.

, shift the graph

graph horizontally by a factor of

y = -4

5

f(x) = 2+ln x

c.

X

f(x) = In(2x) -3

b.

b.

e

X

e.

a.

units to the right. y 8

=

X

x

73.


Range off (-00, 00)

X +3

e.

Range off (-00, 00 )

f.

Using the graph of

y=

e

X

, reflect the graph

about the y-axis, and reflect about the x-axis.

323 Pearson Education , Inc . , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyri ght laws as they currently

exist. No portion of this material may be reproduced, in any form or by any means, without permi ssion in writing from the publi sher.


79.

1

f(x) = - log ( 2x ) 2 Domain: a.

b.

(0, 00)

Using the graph of

y

= log x , compress the

graph horizontally by a factor of -

77.

3

b.

Domain: (4,

00

)

Using the graph of

y

(5, n

= log x , shift the graph

�

4 units to the right and 2 units up. x = 4 y 2

i� I

10

-2 c.

x

d.

-2 c.

d.

(

-00, 00

)

f(x) = log(x - 4) + 2

y

= log( x - 4) + 2 x = log(y - 4) + 2 x - 2 = log(y - 4)

y_4=10x-2 y l Ox-2 r l 10x-2 =

+4

(x) =

+4

e.

Range off

f.

Using the graph of

(

-00, 00

)

Vertical Asymptote: x =

0

1 y '21 ( ) 1 ( y) ( ) 102x ,!,, 10 2x r l ,!,, 102x ( ) y lOX , f(x) = '2 log ( 2x ) log 2x

Inverse

Y=

2

(x) =

)

y lOX , =

e.

2 Range off

f.

Using the graph of

- 00 , 00

=

compress the

graph horizontally by a factor of .!. , and 2

shift the graph

compress vertically by a factor of y 7

4

4

x

y = O

-4 2008 Pearson

-00 , 00

x

2y =

Inverse

2 units to the right and 4 units up. y 6

©

(

10

x = - log 2 2 2x = log 2Y

Vertical Asymptote: x = 4

-

x = o

Range :

0. 0)

=

Range :

, and

2

compress vertically by a factor of .!. . 2 Y 2

f(x) = log ( x - 4 ) + 2

a.

.!.

-5

-3

0 , 5) (O, n

5

.!. . 2

x

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exist. No portion of this material may be reproduced, in any form or by any means, without pennission i n wri ting from the publisher.


81.

f(x) = 3 + log 3 ( x + 2 )

a.

b.

Domain: (-2,

00

)

Using the graph of y = log 3 X , shift 2 units to the left, and shift up 3 units. x = -2 ( 1 Y 1 , 4) 1 .) _

..,

1

( - 1 , .J) 1

-

-5

c.

d.

5

c.

x

d.

1-5

Range:

(

)

Horizontal Asymptote: y = -3 f(x) = eX +2 - 3 y = eX +2 - 3

y = 3 + log 3 ( x + 2 )

r l (x) = In(x + 3) - 2 e.

Inverse

f.

x - 3 = log 3 ( y + 2 ) 3 y + 2 = 3 x3 y = 3 x- _ 2 3 rl (x) = 3 x - - 2 Range off

f.

Using the graph of y = 3 x , shift 3 units to

)

I I

X

x

------

f(x) = eX +2 - 3 a.

Domain:

b.

Using the graph of y = eX , shift the graph

-00 , 00

the left, and shift down 2 units . I y 5

I

I

( - 2, -2)

8 5.

-4

Using the graph of y = ln x , shift 3 units to

-5

the right, and shift down 2 units. Y 6

y = - 2 �- - - - -

Range off (-3, 00)

II

e.

-00, 00

Inverse

y = In(x + 3) - 2

f(x) = 3 + log 3 ( x + 2 )

(

)

x + 3 = ey +2 y + 2 = ln(x + 3)

Vertical Asymptote : x = -2

x = 3 + log 3 ( y + 2 )

83 .

00

x = ey +2 - 3

-00, 00

(

Range: (-3,

5

I I I

= -3

3 f(x) = 2 x / + 4 a.

Domain:

b.

Using the graph of y = 2 x , stretch the graph

(

- 00 , 00

)

horizontally by a factor of 3 , and shift units up. y 10

4

)

two units to the left, and shift 3 units down. -3


7

325 All rights reserved. Thi s material is protected under all copyright laws as they currently

exist. No portion of this material may be reproduced, in any form or by any means, without permi ssion in writing from the publisher.


(4, 00) =4 f(x) = 2xI3 + 4 = 2xl3 + 4 x = yl3 + 4 x-4 = 2yI3 (x-4) 2::. = = (x-4) rl(x) = (x-4) (4, 00) = x, 4

c.

Range :

Horizontal Asymptote :

d.

93.

y

eX

Inverse

y

log

95.

2

3 log

3 10g

2

Range off

f.

Using the graph of

log 4

The solution set is

2

e.

e5

The solution set is

Y

3

=5 = x=5 {5} . 64 = x 4x = 64 4x = 43 x=3 {3} . 243 = 2x + I = 243 = 2x+ 1 = 5 2x =4 x=2 {2} . = 10 3x = Inl0 X=--10 { In; O } .

In e x

97.

log 3

3 2 X+ l 32X+ l

shift units log y 2 to the right, and stretch vertically by a factor of 3 . y 8

35

The solution set is 99. x

In

3

-2

87.

log 3

The solution set is

x=2 X =32 x=9 {9} . ( 2x + 1) = 2x + 1 = 23 2x+l =8 2x =7 X =-27

101.


log

2

3

The solution set is 91 .

1 03 .

log

.

IS

2

.

d x 2 + 1 ) =2

x2 + 1 = x2 + 1 = 9 x2 =8 x = ±J8 = ±2.J2 { -2.J2, 2.J2 } . 32

{-i} .

=

*'

=8 2x+5 = 1n 8 2x = -5+ln 8 -5+ln 8 x= --2 {-S+ In 8} e2 x + 5

· set The s oIutJon

4 2 x2 =4 x = 2 (x -2, {2} .

log x

3 e x

The solution set is

base is positive)

The solution set is

© 2008

326 Pearson Education , Inc . , Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright laws as they currently

exist. No portion of this material may be reproduced, in any form or by any means , without permi ssion in writing from the publisher.

Section 6.4: Logarithmic Functions

105.log 28X=-3 8x = 2-3

c.

(23f = r3 23x 2 -3 3x=-3 x=-1 The solution set is {-I} . =

In(-x ) if x 0

107. SeO.2x = 7

7

S

G(x )= 2 log (2x+1)= 2 3 2x+l = 32 2x+l = 9 2 x=8 x= 4 The point (4,2) is on the graph of G.

y

)

,.,

7

.

0.2x= In S

s(

S(0.2x) = In�)

x

7

x=SIn S

The solution set is {SIn

�} .

Domain: { x I x -:;; O} Range: ( -00, 00) Intercepts: (-1,0), (1,0)

109.2·102 - X=S

102-x =� 2

S 2-x= log2

S -x=-2+1og2 5 x= 2-log2

The solution set is {2 111.a.

-IOg%} .

{-InX if 0 < x 0 2x>-1 1 x>-2 Domain:

b.

115.f(x) =

-3

Domain: {xlx>O}; ( 0, 00 ) Range: {Y I Y � O}; [0,(0) Intercept: (1,0)

{XIX>-�} or (-�,oo)

117.pH = -loglO [H+ ]

G( 40)= log (2·40+1) 3 = log 81 3 =4 The point (40,4) is on the graph of G.

a.

b. c.

pH =-loglO [0.1] =-(-1) 1 =

pH =-loglo [0.01] = -(-2) = 2

pH = -loglO [0.001] = -(-3) 3 =

327

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d.

b.

As the H+ decreases, the pH increases.

[

e.

3 . 5 = - logl o H+

[ ]

-3 . 5 = 10g l O H+

]

In(0.2) = - 0. 1 t

[ H+ ] = 1 0 -35.

t=

'" 3.16xlO-4

= 0.0003 1 6

f.

[

J

P a.

=

loglo [ H+ ]

123.

'" 3 . 98 1 x l 0-8 0.0000000398 1

760e-01. 45h 320 = 760e-{)1· 45h 320 -01. 45h -=e 760 In

( ) 320

760

h=

( )

760e-{)1· 45h -{).145h -=e 760 667 In ( ) = -O.l45h 760 In ( 667 ) 760 h = -0. 1 45 '" 0.90

a.

1-

�� [

1-

e-(1015)r ]

L

= 5 , and

12

e-2r = -7

12 -2t = In(7 / 1 2) In(7 / 1 2)

'" 0.2695 -2 It takes approximately 0.2695 second to obtain a current of 0.5 ampere. t=

Substituting E = 1 2 , R = 1 0 , [ = 1 .0 , we obtain:

e-o.lr

I e-0lr

0.5 = -0 5 = .

![l-e-(RIL)rJ

� = 1_e-2r

Approximately 0.90 kilometers. =

5e-{).4h 2 = 5e-0.4h 0.4 = e-O.4h In(O.4) = - O.4h h = In(O.4) '" 2 .29 - 0.4 =

0.5 =

667 = 667

121. F(t)

e-{)J. r will never equal zero; thus, F(t) = 1 - e-Olr will never equal 1 00% because

Substituting E = 1 2 , R = 1 0 , [ = 0.5 , we obtain:

'" 5 .97

-0. 1 4 5 Approximately 5 . 97 kilometers. b.

It is impossible for the probability to reach

D

125. [ =

320 760

'" 1 6 .09 - 0. 1 Approximately 1 6.09 minutes.

Approximately 2.29 hours, or 2 hours and 1 7 minutes.

= -0. 1 45h In

In(0.2)

I.

H+ = 1 0 - 7.4

=

119.

c.

[ ]

7 .4 = - logl O H+ -7.4 =

l-e-{)I· r - 0.2 = _e-O.lr 0.2 = e-{)·lr 0.8 =

1 .0 =

.

_e-Ol. r 0.5 = e-Ol. r In (0. 5 ) = -O.lt In(0.5) ", 6.93 t=

�� [

.!..Q 1 12

=

e-2r -2t

--

-0.1

e-(1015)r ]

=

5, and

e -2 r

1 =

t=

Approximately 6.93 minutes.

1-

L

6 In(1 / 6) In(1 I6) -2

'" 0.8959

328

© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright laws as they currently exist. No portion of thi s material may be reproduced, in any form or by any mean s, without permission in wri ting from the publi sher.

Section 6.5: Properties of Logarithms

It takes approximately 0.8959 second to obtain a current of 0.5 ampere. Graphing:

2.0 t

a..

1.6 1.2

� 0.8

(0.8959,1) 135.

(0.2695,0.5 ) Seconds

127.

129.

( ) 3

1.

10=10 log (109 ) =10 ·9 =90 decibels

3. 5. 7. 9.

133. R a.

= e""

1.4= i (O .03) 1.4= eO. 03k

In(I.4)=0.03k k = In(I.4) 0.03 "" 11.216 c.

sum rloga M False log3 371 =71 In e

-4 =

4

-

ios,7 = 7

13.

log8 2 + log8 4= 10g8 ( 4 . 2)= log8 8=1

17.

=

= eI 1.2 1 6 (O. 17 ) e1 .9 0672 "" 6.73 1 00 = el 1 .216 x 100= el1.2 16 x In(100)=11.216x ) x= In(100 11.216 "" 0.41

R

x.

n.

--

b.

=

Section 6.5

L (IO-3 )=10 10g ( 1 0 12 ) -

a

x

7 L (IO-7 ) 10 10g 10-1 2 10=10 log (105 ) =10·5 =50 decibels =

=

=

f

r/)

5= el 1 .216 x In 5 11.216x x � 11.216 "" 0. 1 43 At a percent concentration of 0.143 or higher, the driver should be charged with a DUI. e. Answers will vary. If the base of logarithmic function equals 1, we would have the following: f(x)= log I (x) I- I ( )=IX 1 for every real number In other words, I -I would be a constant function and, therefore, I-I would not be one to-one.

d.

log26 ·log6 4= log6 4 108,6 6 = log6 (2 2yag, = log6 2 2108,6 = log6 ios, 6' = log6 6 2 =2

329

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19. 21.

310g,S-log,4 =310g'4� =-4S

43.

e log,I6 Let a=loge' 16, then (e2r =16. e 2a =16 e 20 =4 2 (e 20),,2=(4 2t2 eO =4 a=ln4 log,I6 = eIn4 = 4 . Th e' •

45.

47.

us,

23.

ln6=In(2·3)= ln2+ln3= a+b

25.

ln1.S=ln�=ln3ln2=b- a 2 1n8=ln23 =3·ln2=3a 1nif6=In6\15 1 =-In6 S

27. 29.

In( )= In e+Inx=1+Inx

37.

In(xex)=lnx+lnex =lnx+x

39.

41.

=3log2x-log2(x-3)

[

]

(X+21 =log[x(x+2)] - log(x+3)2 log X(x+3) = logx + log(x +2) -2log(x+3)

[

]

2 3 In X(X+4)22 =.!.3 In (X-2)(x+2 1) _X_

[

1I

(X+ 4)

]

=l[In(x -2)(x+1)-In(x+4)2] 1 - 2) +In(x+1)- 2ln(x+4)] =-[In(x 3 1 2 +4) I =-In(x 3 +1)--In(x 3 3 - 2)+-In(x 49.

In Sx.Jl+3x

(X- 4)3

=In( Sx.Jl+3x) -In(x- 4)3 = lnS+ lnx+In.Jl+3x -3ln(x-4) =lnS+lnx+ln(1+3xY'2-3ln(x-4) = lnS+lnx+'21 ln(1+3x)-3ln(x-4)

=� ln(2 . 3) =S1 (ln2+ln3) =.!.(a+b) S

35.

(�)

log2 x -3 = log2 x3 -log2 (x-3)

ex

loga ( U2V3 ) =loga u 2+loga v3 =2loga u +3logo v

51.

3logs u +4logs v= logs u 3 +logs v4 =logs ( U3V4)

53.

log3 Fx -log3 x3 = log3

(J) = log3 ( X��2 ) =log3 X-S/2 S =--log 2 3X

In (X 2�)=lnx 2+ln� =lnx 2+ 1n(I - xt2 =2lnx+-21 In(l-x) 330

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Section 6.5: Properties of Logarithms

55.

log 4 (x 2 -1)- 51og 4 (x+l) =log 4 (X 2 -1)- log 4 (x+1r =log 4 -15 (x+l) =lOg 4 (X+l)(X;I) (x+l)

63.

[ X2 ] [ ] =IOg' [ ::I�' ] (

57.

21og2 (x+l)-log2(x+3)-log2 (x-I) =log2(x+1) 2-log2 +3)-log2 (x-1) +1) 2 log2 (x-1) =log2 (x(x+3 ) 2 (x+l) =lOg2 (x+3)(x-l)

(X

lnC�J+ln( X;I)_ln(X2 -1) X X+lx ] _ln(X 2-1) =In [ x-I = In[�-;x-I (X2 -1) ] = h{ (x- ;)(:' -I) 1 In [ (x-1)(:: �)(X+1)] =lnCX�I)2) 2 __.

65.

log3 21= log21 log3 '" 2.771

67.

log71 = log71 '" -3.880 log l/3 71= log(I/3) -log3

69.

log7 5.615 logvrz 7= � log,,2

71.

lne '" 0.874 log e=-

73.

logx lnx or y= -y=log 4 =log4 ln4

'"

81og2 .J3x- 2 -log2

2

75.

+2) y=log2(x+2) = In(xln2+2) or y= log(x log2 ----=-'--= ---'-

3

=log2 (.J3x-2 r -(log24-log2x) +log2 4 = log2 (3x-2)4 -log24+log2 +log2 4 =log2 (3x -2)4 +log2 =log2[x(3x- 2t ] 61.

X

-2

(�)+log2 4 X

lnn

1[

=In(x-lr =-2ln(x-l)

59.

]

[

X

5 -2

-�

2loga(5x 3 ) logo (2x +3) =logo (5x 3 r -logo (2x+3)1/2 =logo (25x6 )-log a.J2x+3

77.

log(x+l) y=logx- I(x+l)= In(x+l) In(x -1) or y= log(x-1) 4

OII---; = 2 10g " x is { xl x > O } . These two domains are different because the logarithm property loga xn = ·loga x holds only when log" x exists. Answers may vary. One possibility follows: Let x = 4 and y = 4 . Then log2(x + y) = log2(4+4) = log28 = 3 . But log2X + log2y = log24 + log24 = 2+ 2 = 4 . Thus, log2(4 + 4) :t log24 + log24 and, in general, log2(x + y) :t log2X +log2y .

9.

log4 (x + 2) = log4 8 x +2 = 8

x=6

n

The solution set is {6} . 11.

-1 log3 X = 2 log 3 2 2 log3 X II2 = log3 22 x 1 l2 = 4 x = 16

13. Section 6.6

1.

=

X = 16

loga M -loga N

� f(

107.

log4 X

-10

-1 . 43 , so the solution set is { - 1 .43 } .

x = 42

= A-B =

x :::;

The solution set is {1 6} . 3log2 X = -log2 27 log2 x3 = log2 27 -1 x 3 = 2 TI

1

x2 - 7 x - 30 = 0

x3 = 27

(x + 3)(x - 1 O) = 0

1

or x - l 0 = 0 x = -3 or x = 10 The solution set is { -3, 1 O } . x+3 = 0

X=3

The solution set is

{�}.

333

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23.

15. 3 log (x - 1) + log 4 = 5

2 2 log (x - l) 3 + log 4 = 5 2 2 log 4(x - l) 3 = 5 2 4(x - l) 3 = 2 5

(

log8(x + 6) = I - logg(x + 4) log8(x + 6) + logg(x + 4) = 1

[

(X - I) 3 =

]

logg (x + 6)(x + 4) = I

)

(x + 6)(x + 4) = 8 1 x 2 + 4x + 6x + 24 = 8

32 4

x2 + l Ox + 1 6 = 0

(x - l) 3 = 8

(x + 8)(x + 2) = 0 x = -8 or x = -2

x-I = 2

(

) { -2 } .

{}

solution set is

17. log x + log(x + 1 5) = 2

25. ln x + In(x + 2) = 4

log x(x + 1 5) = 2

In x(x + 2) = 4

(

)

(

x2 + 2x - e4 = 0 r- - - - - - - 2 ± 2 2 - 4 (I )( - e 4 ) ----''-----.:... ---'---.:. .:.. .. x=

x 2 + 1 5x - 1 00 = 0 (x + 20)(x - 5) = 0

2(1)

)

Since log -20 is undefined, the solution set is

19.

=

log(2x + 1) - log(x - 2) =

( 21 ) 2x + x-

1

(

) ( ) log 3 [( x+l )( x + 4 ) ] = 2 ( x + l )( x + 4 ) = 32

-2 1 2 1 X= -=8 -8

x 2 + 4x + x + 4 = 9

{�)}.

x2 + 5x - 5 = 0

�

- 5 ± 5 2 - 4(1)(- 5) ...:. x = ----''---....:.---'...:.2(1)

21. log (x + 7) + log (x + 8) = 1 2

[ (x + 7)(x + 8) ] = 1 2

-5±

.J45

2

(x + 7)(x + 8) = i X=

x 2 + 8x + 7x + 5 6 = 2

-5 -3

J5

2 "" -5.854

x 2 + 1 5x + 54 = 0

-5±3 2 or X =

J5

- 5 + 3 J5 --2

"" 0 . 854 Since log 3 -8 .854 + 1 = log 3 -7 .854 is undefined, the solution set is

(x + 9)(x + 6) = 0 x = -9 or x = -6

( - 9 + 7 ) = log 2( - 2 ) 2 the solution set is { -6 } .

�

27. log 3 x + 1 + log 3 x + 4 = 2

-8x = -2 1

Since log

2

( ) is undefined, the solution set is { - I + �I + e4 } "" { 6 . 456 } .

2x + l = 1 0x - 20

log

2

Since In -8 .456

=1

= 101 x-2 2x + l = 1 O(x - 2)

2

� = - 2 ±2 .,Jl;l = - I ± �I+e4

�

2x + l

The solution set is

-2±

x = - 1 - I + e 4 or x = - 1 + I + e 4 "" -8 . 456 "" 6 . 456

log(2x + 1) = 1 + log(x - 2)

log

�

_

x = - 20 or x = 5

{ 5} .

)

x(x + 2) = e 4

x( x + 1 5) = 1 02

(

( )

Since log8 - 8 + 6 = log8 - 2 is undefined, the

x=3 The solution set is 3 .

{

is undefined,

-

S+3 2

.J5

(

)

}

"" 0.854 .

{

(

)

}

334

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Section 6.6: Logarithmic and Exponential Equations

37.

( xx:-x+ J=-1 ::2 ::=(�r X

logl13

The solution set is

{-logg1.2}=

x +x --=3 x 2 -x x2 +X=3 ( X2 -x ) x2 + X=3x2 -3x _2X2 +4x=0 -2x(x-2)=0 -2x=0 or x-2=0 x=O or x=2

x=0 , {2}.

39.

x=2 ,

The solution set is IOg2

In ( 3 1 - 2 X

{

43.

y-5 =8 2 x-5 =23

2" =10 "" 3.322 x= log2 1 0= lnl0 ln2

{

}

(%J =i- X In(%J = In ( i- x )

x ln(3/S)= (l-x)ln7 xln(3/5)=ln7-xln7 xln(3/5) +x ln 7= In 7 x(ln(3/5)+ In 7 )= In 7 x= In(3/5ln7+ln7 "" 1.356 ) In7 The solution set is In(3/5) +ln7 "" {1.356}.

{8}.

The solution set is Iog21

)= In (4x )

(1-2x)ln3=xln4 In3-2x In 3=x In 4 In3= 2x In3+x In 4 ln3=x(2 ln3+ln4) x= ln3 "" 0.307 2ln3+ln4 In3 "" {0.307}. The solution set is 2ln3+ln4

x= �, the solution set is {�}.

35.

5 (23.:c )=8 23x =�S

log2

Since each of the original logarithms is defined for

The solution set is

}

(1.2) ",, {-0.088}. - log 8

log

(%) ) "" 0.226 x= .!. (�)= In(8/5 5 3ln2 3 {� (%)}={ ln3�;) } "" {0.226} .

x-I x-2 x+6 x+3 (x-l) (x+3)=(x-2) (x+6) x2 +2x-3=x2 +4x-12 2x-3= 4x-12 9= 2x X=-29

x-5=3 x=8

{

3x= log2

Since each of the original logarithrns are not defined for but are defined for the solution set is

33.

8- x =1.2 -x=logg1.2 ) "" -0.088 x=-logg1.2=- loglog(1.2 8

O}=f�l�} "" {3 .322}.

{

}

335

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exist. No portion of this material may be reproduced, in any form or by any mean s , without permi ssion in writing from the publi sher.


45.

47.

1.2x= (0.5)-X In 1.2x = In (0.5fx xln(1.2)=-xln(0.5) xln(1.2)+xln(0.5)=0 x(ln(1.2)+10(0.5))=0 x=o The solution set is {O} .

53.

a

---

nl-x= eX 10 nl-x=10 eX (l-x)lnn=x Inn -xln n=x Inn=x+xlnn 10 n=x(1+ 10 n) Ion ",,0.534 x= l+lnn

Therefore, we get

�

4x=-2+17 = 10g 4 (-2+17 ) (we ignore the first solution since 4x is never

--

The solution set is

49.

51.

55.

25x -8·5"" =-16 ( 52 r -8·5'< =-16 ( 5x )2 -8 . 5'< =-16 Let u = 5'< u2 -8u=-16 u2 -8u+16=0 (U-4)2 =0 u=4 5'2 } or (-00,1) u(2,00) . Interval (-00,1) Test Value 0

FI (X)= 273 x Domain of I Range of I- I All real numbers except 0 Range of I Domain of I- I All real numbers except 0

35.

In e.f2 = J2

39.

log3

=

=

=

=

23.

a.

b.

25. 27. 29.

1(4)=34 = 8 1 g(9)= log3 (9)= log3 (32 )=2

C.

1(-2)=T2 =.!9

d.

g

( 2\ )=log3 ( 2\ )

log3 ( T3 )=-3

52 = is equivalent to 2= logs logs =13 is equivalent to 5 13= z

z

u

43.

W

log3 + log3 v2 -log3 =log3 + 2 log3 -log3 log (x 2 .Jx 3 +1 )= logx 2 +log (x 3+1(2 =2 l0gx + log (x 3+1) U

V

U

W

W

�

In

[ x� )=In(Ah l )-In(X-3) j

I (x )= log(3x-2) requires: 3x-2> 0 x>-2 3 Domain: x x >% or

45.

(x ) log2 (X2 - 3x + 2 ) requires p(x)=x 2 -3x +2> 0 (x-2)(x-l)> 0 x =2 and x = 1 are the zeros of p . H

2

3 =lnX +ln (X2 +1t -In(x-3) = lnx + ln (x2 +1)- In(x - 3)

U

{ l } (%,00)

31.

u

=

41. =

[ : )=log3 uv2 -log3

3 3log4 X2 + "21 log4 "rX = log4 (X2 ) +log4 (X 1 /2 )1/2 = 1og4 X6 + 1og4 X 1 /4 = log4 (x6 . XI /4 ) log4 X2S /4 25 log x =4 4 =

=

354

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47.

In ( X� I +lnC: J-ln(X 2 -1) JI x ( ( x I 2 =In --:;- . x + 1 J- n x -1) X-I =In X2x +_1I 1 =In X - I ' x + I (x - l)(x + l) I =In-(x + 1)2 =In(x+lr 2 =-2ln(x+1) I Iog(x+3)+ log(x 2) ] 210g2 +310gx --[ 2 = log2 2 + logx3 -.!.. log [ (x + 3 )(x-2) ] 2 = log (4x3)- log( x+3)(x -2) ( 2 4x3 = log [(x + 3)(x 2)]112

[ ] (

49.

d.

)

e. f.

f(x)=2 x-3 y = 2x -3 =2y- 3 Inverse y-3= log2 x Y =3+log2 x l F (X) = 3+ log2 x

Range off (0, 00 ) Using the graph of y= log2 X , shift the graph vertically 3 units up.

- J

-5 -5

57.

ln x

= log3 X = ln3 1 :/

b.

18

I'

x = o

±

f(X)= ( 3-X ) 3.

3

x

5

�

-1 1

y=0

y 5

ln19 2.124 log4 19=-ln4

53. r;

-3

55.

Range: (0, 00 ) Horizontal Asymptote: X

-

(

51.

c.

Domain: (-00, 00 ) Using the graph of y = 3x reflect the graph about the y-axis, and compress vertically by a factor of -2I .

,

f(x) =2x-3 3.

b.

Domain: (-00, 00 ) Using the graph of y=2x shift the graph horizontally 3 units to the right.

,

y 7 t-

-1

c.

7

-3

Range: (0, 00 ) Horizontal Asymptote:

y =0

x

355

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d.

� (TX ) = � ( TX ) = � ( TY )

d.

f(x) = y x

Inverse

e.

f.

=

log 3 (2x)

-

= - log 3 (2x) rl (x) = - log 3 (2x) 2x > 0 y

x>O Range off

= l - e-x x = l - e-Y Inverse x - l = -e- Y l - x = e- Y -Y = ln ( l - x ) y = ln ( l x ) r l ( x) = - In ( l - x ) y

2x = TY -y

f (x) = l - e-x

e.

(0, 00)

0 -1 x

-x >

Range off


= log 3 x , compress the

graph horizontally by a factor of reflect about the x-axis. y 5

f.

.! , and 2

-

(-00, 1)


= ln x , reflect the

graph about the y-axis, shift to the right unit, and reflect about the x-axis. y� I

1

5 x

-5

I

x = O

7

( � , - 2)

(- 1 .72,- 1 )

-5

1

f(x) = l - e- x

59. a.

Domain:

b.


x :: l

61. f(x) = -In ( x + 3 ) 2

(-00 , 00) = eX , reflect about the

y-axis, reflect about the x-axis, and shift up unit.

1

a.

Domain:

b.


Y

(-3, 00)

to the left

= ln x , shift the graph

3 units and compress vertically by

a factor of

t. y

5

5

(0, 0.55) c.

Range:

(-00, 1)

Horizontal Asymptote : y

-5 =

c.

1

Range:

(-00 , 00)

Vertical Asymptote : x =

-3

356

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6 7.

.!.2 In (x+3) y=.!.ln(x+3) 2 1 x = -In(y +3) 2 2x= In (y+3) y+3=e2 x y=e2x -3 F I (x) = e2 x _3

d.

I(x) =

(0

e.

Range off ( - 00,

f.

Using the graph of

3 units.

( r

Inverse

6 1 1 = -1 x= --

�

4

The solution set is

y = eX , compress

1- ' and shift

x ln 5 = x ln 3 + 2 ln 3

x In 5 - x ln 3 = 2 ln 3

x(ln5 -ln3)=2 ln3

y

2 ln 3 ln5-ln3 4.301 21 3 The solution set is { 4 . 3 0 1} . 5 3 x=

41-2x = 2

12 X=-

5

The solution set is

4

The solution set is

}

""

32

../x - 2 = 9

x - 2 = 92

=81 x=83 log 3 ../83 2 = log 3 J8i

x-2

1

x +X=-

2

+ 2x - l=0

Check:

2 _-4-(2-)(-_-1) -2 ± �r-2x= 2(2) -2 ± ..f0. -2 ± 2.J3 -1 ± .J3 2 4 4

{ �.J3 , �.J3 }

{I: } .

73 . log 3 "/x - 2 = 2 ../x - 2 =

{�} .

y2+X = .J3 3x2+x = 3 1 / 2 2x 2

- In

-5x = - 1 2

22 -4x = i 2 - 4x = 1 - 4x=-1 1 X=-

2

n

In

4x = 9x - 1 2

( 22 y- 2 X = 2

65 .

{

""

4 (32 t = (33 t34x = 39x-12

-5

63 .

{±} .

In (5x ) = In (3 H2 ) x ln 5 = ( x+ 2) ln 3

)

horizontally by a factor of down

=-3

log 64 x x-3 = 64 l/3 = 4- 1 3 x-3

-

= log 3 9

=2

The solution set is

{83} .

The solution is -I

-I

""

{-1.366, 0.366}. 3 57

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83.

a.

2

2 3 = 2 2 x +5.1 3 = 2X 2 + 5x 0 = 2X 2 + 5x - 3 o = (2x - 1)(x + 3) x = -1 or x = -3 2 The solution set is 77.

79 .

81.

Y x= 2

10

{- 3, �} .

I (x)= \og2 (x - 2)+ 1

( 6, 3) -10

log 6 (x + 3) + log 6 (x + 4) = 1 log 6 ( x + 3)(x + 4») = I (x + 3)(x + 4) = 61 x 2 + 7x + 12 = 6 x 2 + 7x + 6 = 0 (x + 6)(x + I) = 0 x = - 6 or x = -1 Since log 6 (-6 + 3) = log 6 (-3) is undefined, the solution set is {-I} . e1 - x = 5 l - x = 1n 5 -x = -I + 1n 5 x = 1 - 1n 5 -0.609 The solution set is { 1 - ln 5 } {-0.609} . ""

f ( x) = log 2 ( x - 2) + 1 Using the graph of y = log 2 X , shift the graph right 2 units and up 1 unit.

""

9x + 4 · 3 x - 3 = 0 ( 32 y + 4 · 3x - 3 = 0 ( 3x ) 2 + 4 · 3 x - 3 = 0 Let u = 3.1 . u 2 + 4u - 3 = 0 a = l, b = 4, c = -3 -----4) ± �r( ( 4 ) 2 - 4(1)( -3) ��--�� u= 2(1) 17 .fi8 -4 ± = = -4 ± 2 = -2 ± 17 2 2 � or Y = -2 + 17 J' can't be negative x = log 3 ( -2 + 17 ) _

b.

f (6 ) = log 2 ( 6 - 2 ) + 1 = log 2 (4) + 1 = 2 + 1 = 3 The solution set is {3 } . The point (6, 3) is on the graph off

c.

f ( x) = 4 log 2 (x - 2) + 1 = 4 log 2 (x - 2) = 3 x - 2 = 23 x-2 = 8 x = 10 The solution set is { I O} . The point (10, 4) is on the graph off

d.

f (x ) = 0 log 2 (x - 2) + 1 = 0 log 2 ( x - 2) = -1 x - 2 = T1 x - 2 = -1 2 x = -5 2 Based on the graph drawn in part (a), f ( x ) > 0 when x > 2. . The solution set is 2 x I x > or

{ %} (%,00).

The solution set is { log 3 ( -2 + 17) } "" {-0.398} . 358

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e.

) y = log 2 ( X - 2 ) + 1 x = log2 ( Y - 2 ) + 1 x - I = log2 ( )

f ( x ) = log2 (x - 2 + 1

y _ 2 = 2x-

1

y-2

10

;

;

;

;

;

;

I-I (x) =

2x-1 2 +

1n 0.05 =

;

t = 20 1 5 - 2005 = 1 0

)

o 5 P = Pa il = 6, 45 1, 05 8, 790e . O I 1 ( I O)

( ) 760

"'" 7, 237, 27 1, 5 0 1 people

300

97. A =

87. P = 25 e o. 1 d

P = 25 eo. 1 ( 4) = 25eo.4 "'" 3 7 . 3 watts

a.

2 = e O. 1 d

b.

n=

99.

0. 1

1 70

a.

a

log 1 0, 000 - log 90, 000 10g(l - 0.20)

( )

o

()

n = _--'-�'--_...o....!. 10g(1 - 0. 1 5)

( �)

10g 0 . 5

log 0 . 8 5

log 0 . 8 5

nt

"'" $4 1, 668.97

b.

. .

1 50

a a

a

a

a

a

a

a

a

. .. .. . . .

.

.

a

.

a

a

a

. . .

16

Using EXPonential REGression on the data

(

){ 0.995 1 Y Yj = ( 1 65 . 73 )( 0.995 1 Y

yields : y = 1 65 .73

0 i ' I

( ;r

a

""' 9 . 85 years

log 0.5i - log i

91. P = A I +

)1 0 "", 4 8 3 . 67

The government would have to pay back approximately $483 .67 billion in 20 1 5 .

In 2 "'" 6.9 decibels

10g

11

= 3 1 9 ( 1 .042 5

1n 2 = O . ld

89.

( r ( 0.0�25 ) ( 0)

p I+�

A = 3 19 1 +

50 = 25eO I d

d=

"'" 24 203

95. Pa = 6, 45 1, 058, 790 , k = 0. 0 1 1 5 , and

"'" 3229.5 meters

b.

=

(In O. 5 ) t 5600 InO. 05 5600 , (�)

The man died approximately 24,203 years ago.

-1 0

(

I

0.05 = e

t

85 . h(300) = 30(0) + 8000 IOg

a.

In5600 1 ( In5600o. 5 )

0.05 Ao = Aoe

y x=2

;

In O. 5 5600( 0. 5 )

k=

=

r 1 (x)

y = x/ ;

0.5 = e 5600 k In 0 . 5 = 5 600 k

Inverse

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

Y

o 0.5 Ao = Ao ek( 5600 ) A = A e kl

93.

( �r

= 85, 000 1 +

c.

""' 4 . 27 years

0. 4

2

1 70

(J 8)

359

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e.

d. Find x when y= 1 10 .

(165.73)(0.9951)' = 110 (0.9951Y = � 165.73 xlnO.9951 = In � 165.73 110 In 165.73 "" 83 x= lnO.9951 Therefore, it will take approximately 83

Find

46.9292 = 10 1 21.2733e-0. 73061 46.9292 = 10 (1 + 21.2733e- 0 7306 1 ) 46.9292 + 06 = 1 21.2733e-O · 73 1 10 46.9292 -1 = 2 1.2733e 10 3.69292 = 21.2733e- 0 73061 +

( ) ( )

-0.73061

3.69292 e-0.7 3 061 = 21.2733 3.69292 -0.7306t = In 21.2733 3.69292 In 2 1.2733 -'---� = I -0.7306 1 ",,2.4

( (

seconds for the probe to reach a temperature of 110°F.

101.

--, _ ...::;.; 5 0::... ,... _ __

a.

t when C = 10 .

---

-

- 1 �=======.l 9 o

b.

c.

y.

f.

46.93 1 + 2 1.273e- . 061 46 .93 _ + 1 21.273e-0.73061

I -

-I

d.

Therefore, after approximately 2.4 days (during the 10th hour on the 3rd day), 10 people had caught the cold.

The data appear to have a logistic relation Using LOGISTIC REGression on the data yields : C=

0 73

50

/,

o

1� 21.2733e-0.73 061 � 0 , which . means 1 + 21.2733e- 0 73 061 � 1 , so 46.9292 C= � 46.9292 + 1 2 1.2733e- 0.7 3 061

As

) )

oc) ,

1 when C= 46 . 46.9292 = 46 1 + 2 1.2733e-0 7306 1 46.9292 = 46 (1 + 21.2733e-0 73 06 1 ) 46.9292 = 1 + 21.2733e- 0. 73 061 46 46.9292 1 = 21.2733e-O·73061 46 0.0202 = 21.2733e-0 73 061 0.0202 = e-O · 73061 21.2733 _0_.0_2_0_2_= e-0.73061 21.2733 0.0202 = In 21.2733 -0.73061 0.0202 In 21.2733 -- = 1 -0.7306 I "" 9.5 Therefore, after approximately 9.5 days (during the 12th hour on the 10th day), 46

Find

( (

Therefore, according to the function, a maximum of about 47 people can catch the cold.

--'-

In reality, all 50 people living in the town might catch the cold.

) )

"';:"

people had caught the cold.

360

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Chapter 6 Test

Chapter 6 Test

1.

I( ) = x+2 g(x) = 2x + 5 x-2 The domain of I is {x I x * 2} . The domain of g is all real numbers. (jog)(x) = I(g(x») = 1 (2x + 5) X

3.

a.

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

b.

c.

2.

a.

(go/ )(x) = g(j(-2») = g C�: �) = g( O) = 2(0) + 5 =5 (jog)(x) = I{g( -2)) = 1(2(-2) + 5) = / (1) = 11 +- 22 = 2 -1 = -3 Graph y = 4x2 + 3 y

-5

=

=

=

=

4.

•

:

3x = 243 3 x = 35 x=5 6. 10gb 16 = 2 b2 = 16 b = ±M = ±4 Since the base of a logarithm must be positive, the only viable solution is b = 4 . 7. log5 =4 x = 54 x = 625 8. + 2 22.086 9. log 20 1 .301 5.

5

x

The function is not one-to-one because it fails the horizontal line test. A horizontal line (for example, y = 4 ) intersects the graph twice. b. Graph y = .Jx + 3 - 5

(-3,- 5)

X

:

y

-5

The function is one-to-one because it passes the horizontal line test. Every horizontal line intersects the graph at most once. 2 j(x) = -3x - 5 2 y = -3x - 5 2 Inverse x = -3y - 5 x(3y - 5) = 2 3.xy- 5x = 2 3.xy = 5x + 2 5x + 2 y = -3x 5x l r (x) = 3x+ 2 Domain of I Range of I-I All real numbers except -53 . Range of I Domain of I-I All real numbers except O. If the point (3, -5) is on the graph off, then the point (-5, 3) must be on the graph of rl

5

e3

x

�

�

10.

-8

In 21 2.771 log 3 21 = -In3 �

361

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exist. No portion of this material may be reproduced, in any fonn or by any means, without pennission in writing /Tom the publisher.


11. 12.

In

133 "" 4.89 0 f(x) = 4x+1 -2 Domain: (-00, 00 ) " b. Using the graph of y = 4 , shift the graph 1 unit to the left, and shift 2 units down.

13.

!(x) = I - log5 ( x - 2 ) Domain: (2, 00 ) b. Using the graph of y = log5 , shift the graph to the right 2 units, reflect vertically about the y-axis, and shift up I unit. a.

X

a.

y

y

6

(3,1) -

5

5

x

y

-----

=-

I I I I

2

-5 c.

d.

e.

f.

c.

Range: (-2, 00 ) Horizontal Asymptote: y = -2 f( x) = 4"+1 - 2 y = 4 x+1 _ 2 x = 4y+I - 2 Inverse x + 2 = 4 y+ 1 y + 1 = log4 ( x + 2) y = log4 ( x + 2) - 1 l r (x) = log4 (x + 2) - 1 Range off (-2, 00) Using the graph of y = log4 x , shift the graph 2 units to the left, and shift down I unit.

d.

e. f.

x=2

Range: ( - 00, 00) Vertical Asymptote: x = 2 !(x) = I - log5 ( x - 2 ) y = I - 10gs ( x - 2 ) x = 1 -10g5 ( y - 2 ) Inverse x - I = -log5 ( y - 2 ) l - x = log5 ( y-2 ) y - 2 = 51-x y = 51-x + 2 rl(x) = 51-x + 2 Range off (-00, 00 ) Using the graph of y = 5x, reflect the graph horizontally about the y-axis, shift to the right I unit, and shift up 2 units. y

7

y 5

(2,0)

x

x

-3

I

1-5 I x =-2

-3

(0, 7)

7

362

x

© 2008 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under an copyright laws as they currently


Chapter 6 Test

5 H2=::125 5 H2=::53 x + 2=::3 x=:: 1 The solution set is {I}. 15 . log(x + 9)=::2 x + 9 =:: 102 x + 9 =:: 100 x=:: 91 The solution set is { 91 }.

14.

19 . log2 (x - 4) + log2 (x + 4)=::3 log2 [( - 4)( x + 4) ]=::3 log2 ( x2 - 16 )=::3 x2 - 1 6 =:: 23 x2 -16 =:: 8 x2 =:: 24 x=::±J24=::±216 Because x -216 results in a negative arguments for the original logarithms, the only viable solution is x=::216. That is, the solution set is {216} "" {4.899}. 3 20. log2 2 4x x - 3x - 18 23 log2 (x +23)(xx - 6) == log2 ( 22 x3 ) - log2 [(x - 6)(x + 3)] =::log2 22 + log2 x3 - [log2 (x - 6) + log2 (x + 3)] =::2 + 31og2 - log2 (x - 6) - log2 (x +3) 21 . A == Ao ela 34 =:: 50/ (30) 0 . 68=::e30k In O .68 30k k== In O .68 30 Cno.68)1 Thus, the decay model is A == 50e 30 . We need to find t when A== 2 : (Ino.68)1 2 == 50e 30 (Ino.68)1 0.04 e 30 In��6 8 In 0. 04 == In 0.04 t == In��68 "" 250.39 X

=::

16 . 8-2e-x=::4 -2e-x=::-4 e-x=::2 -x=::In2 x=:: - ln2 "" -0.693 The solution set is { -In 2 } "" { -0.693} . 17 . 1og ( x2 + 3 )=:: log (x + 6} x2 + 3=::x + 6 x2 - x -3 == 0 - - - - - -=- 2 - 4(1 )(-3) l± .Ji3 ± �r( 1 ) -(-1) =:: -x=:: 2(1) 2 The solutIOn set 1S -"" { -1 . 303, 2.303} .

[

=::

7x+3== eX In7x+3=::In eX (x + 3)In7=:: x xIn7 + 3In7=::x xIn7 - x=::-3In7 x(In7 - 1)== -3In7 x== In-31n-I7 7

==

31n7

I-In 7

[

)

X

. . {1--JI32- ' I+JI3} 2

18 .

)

=::

==

",, -6.172

31n7 } ",, { -6.172}. The solution set is { 1-l n7

( } ( )

There will be 2 mg of the substance remaining after about 250.39 days.

363

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22.

a.

b.

Note that 8 months -23 year. Thus, 2 P = 1000 , r = 0.05 , n = 12 , and t = 3 =

Let n represent the number of people who must shout. Then the intensity will be nx 10-4. If D = 125 , then 125 = 1010g nx l 125 = 10 log ( nx 108 ) 12.5 = log ( nx108 ) nx 108 = 1012 .5 n = 104.5 "" 3 1, 623 About 3 1,623 people would have to shout at the same time in order for the resulting sound level to meet the pain threshold .

( O��:)

( (3 So, A = 1000 ( 1 +-1'-2 ) 12) 2/ ) = 1000 ( 1 + O��5r 005

b.

",,$ 1033.82 Note that 9 months -43 year. Thus, A = 1000 , r = 0.05 , n = 4 , and t = -43 So, 0.05 1000 = Ao 1 +4 1000 = Ao ( 1.0125 ) 3 Ao = 1000 3 ""$ 963.42 ( 1.0125 ) r = 0. 0 6 and n = 1 . So, 2Ao = Ao l + 0. 6 ( 2Ao = Ao(1.06Y 2 = (1 .06Y ln 2 t = log . 06 2 = l ln1 .06 "" 1 1 .9 It will take about 1 1.9 years to double your money under these conditions. =

(

c.

)(4)(3/4)

( � ) 1)


3.

1

a.

C ) C ) 10-12

80 = 10l0g o�12 8 = IOg :-12 8 = logI - log 8 = logI - (-12) 8 = logI + 12 -4 = log I I = 10-4 = 0.0001 If one person shouts, the intensity is 10-4 watts per square meter. Thus, if two people shout at the same time, the intensity will be 2 x 10-4 watts per square meter. Thus, the loudness will be = l O log ( 2 x 108 ) "" 83 D = 10l0g decibels

2

a.

--

23.

The graph represents a function since it passes the Vertical Line Test. The function is not a one-to-one function since the graph fails the Horizontal Line Test. x +l =1 2 ! ! "'" ' 2 + 2 = 4 4 = 2 l ' 2 ' 2 is not on the graph. ! + J3 = ! +� = l ' J3 is on b. 4 4 ' 2' 2 2 2 the graph. 2x -4y = 16 x-intercept: y-intercept: 2 ( 0 ) -4y = 16 2x -4 ( 0 ) = 16 -4y = 16 2x = 16 x=8 y = -4

5.

(!J2 ( ) +! ! (!!J ( ) 2 ( )2 (! )

y 10

x �;4 J ( 21O�

-10

364




Given that the graph of I(x) = ax2 + bx + c has vertex (4, -8) and passes through the point b =4, (0,24) , we can conclude -2a I( 4) = -8 , and 1(0) = 24 . Noti ce that 1 (0) = 24 a (0) 2 + b ( O ) + c = 24 c = 24 Therefore, I (x) = ax2 + bx + c = + bx + 24 . b = 4 , so that b = -8a , Furthermore, -2a and 1 (4) = -8 a ( 4) 2 + b ( 4) + 24 = -8 l6a + 4b + 24 = -8 l 6a + 4b = -32 4a + b = -8 Replacing b with -8a in this equation yields 4a -8a = -8 -4a = -8 a=2 So b = -8a = -8(2) = -16 . Therefore, we have the function I (x) = 2X2 - 16x + 24 . 2 g(x) = 9. I (x) = x2 + 2 x- 3 I(g(x)) = I � 3 = _2 x-3 + 2 4 ---=-2 + 2 (x-3) The domain ofIis {xI x is any real number} . The domain of g is {xl x:;to 3} . So, the domain of I(g(x)) is {xl x:;to 3} . 4 + 2 = -4 + 2 = 3 l (g(S)) = (5 _43) 2 + 2 = 4 22

7.

11.

a.

g(x) = 3 x + 2 Using the graph of y = 3 x , shift up 2 units . y

10

y

=

2

-

- - �-=-:::� x

ax2

-5

Domain of g: (- 00, 00) Range of g: (2, 00 ) Horizontal Asymptote for g: y 2 g(x) = 3X +2 Y = 3x + 2 x = Y + 2 Inverse x-2 = Y y = log 3 (x-2) g-I (x) log 3 (x-2) Domain of g-I : (2, 00 ) Range of g -I : (-00, 00 ) Vertical Asymptote for g-I : x = 2 =

b.

=

c.

e ) ( )

v=x / /

.'

2

/

/

/

/

/

/

/

/

x g -l(x)

-5

365

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13.

log3 (x + 1) + log3 (2x - 3) = log9 9 log3 ( x + l)(2x - 3)) = 1 (x + l)(2x -3) = 31 2X2 - x - 3 = 3 2X2 - x - 6 = 0 (2x+ 3)(x-2) = 0 x = 23 or x = 2 --

( ) (-�) is undefined

Since log3 -% + 1 = log3 the solution set is {2} . 15.

a.

2;,;:o:...-_____-.

a

0�=====l80 b.

c.

o

Answers will vary. Answers will vary.

366


exist. No portion of this material may be reproduced , in any form or by any means, without permission in writing from the publisher.

Chapter 7 Trigonometric Functions Section 7.1 1. 3.

5.

25.

C

= 27rr standard position �,!!.-

27.

7. True 9. True

29.

11.

13.

15. 31.

17.

19. 33.

21.

23.

(

)

1 .� .� 0 40010'25"= 40 + 10 ._ 60 + 25 60 60 "" (40 + 0 . 1667 + 0.00694)° "" 40.17°

(

)

(

)o

.�. � 0 1°2'3" = 1 + 2 . � + 3 60 60 60 "" (1 + 0.0 333 + 0.00083t "" 1 .03° .� . � 9°9'9" = 9 + 9· � 60 + 9 60 60 = (9 + 0.15 +0.0025t "" 9. 15° 40.32°= 40°+ 0.32° = 40°+ 0.32(60') = 40°+19.2' = 40°+19'+0.2' = 40°+19'+0.2(60") = 40°+19'+ 12" = 40°19'12" 18.255°= 18°+ 0.255° = 18°+ 0.255(60') = 1 8°+15.3' = 1 8°+15'+0.3' = 18°+15'+ 0.3(60") = 1 8°+15'+ 18" = 18°15'18" 19.99°= 19°+ 0.99° = 19°+ 0.99(60') = 19°+59.4' = 19°+59'+0.4' = 19°+59'+0.4(60") = 19°+59'+ 24" = 19°59'24"

35.

300 = 30 . � 180 radian = 2:.6 radian

37.

41t rad'lans 1t d' 2400 = 240·ra Ian 3 1 80 = -

367



Chapter 7: Trigonometric Functions

180 degrees",362.1 1° . 6. 32 rad·lans = 6.32·7t

39.

- 600 = -60 · � 180 radian = _2:3 radian

69

41.

1800 = 180 . � 180 radian = 7t radians

71.

43.

-135 0 = - 135 · � 180 radian

45.

-900 = -90 . � 1 80 radian = - 2:2 radians

47.

49.

51.

53.

55.

57.

59.

61.

63.

= _

s

37t radians 4 73.

--

75.

-_.-

67.

r = 5 miles; = 3 miles; = r() () = !..r = �5 = 0.6 radian s

s

77.

_.-

· r = 2 inches; () = 300= 30·� 1 80 = 2:6 radian ' s

r()

= = 2 · 2:6 = 2:3'" 1.047 inches

-

81.

-400 = - 40·� 1 80 radian 27t rad Ian . = -9 '" -0.70 radian

() = 3.!. radian·' A = 2 fe A

= '!'r2() 2

=�r2G) 2 = .!.6 r2 12 = r2 r = = 213'" 3 .464 feet 2

1250 = 125 ·� 1 80 radian 257t radlans . 36 '" 2.1 8 radians 3.14 radians = 3.14·180 7t degrees'" 179.910 2 radians = 2 · 180 7t degrees'" 1 14.59°

.Jl2

= --

65.

s

s

-

-

() = 3.!. radian; = 2 feet; = r() 2 = 6 feet r = -() = (1/3)

s

180 degrees = 600 -7t3 = 7t_3 . 7t 57t = 57t 180 degrees = - 225 0 4 4 7t 7t = 7t_. 180 degrees = 900 2 2 7t 7t 7t 180 12 = 12 7t degrees = 150 180 degrees = -90 0 --7t2 = --7t2 . 7t - 7t6 = - -7t6 . 180 7t degrees = -30 0 1 77t radian'" 0.30 radian 170 = 17·� 180 radian = 180 --

r = 10 meters; () = .!.2 radian; = r() = 10·-21 = 5 meters

368


exist. No portion of thi s material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Section 7.1: Angles and Their Measure

83.

85.

87.

r =5 miles; A =3 mi2 A=.!.2r20 3=.!.2 (5)2 0 3= 252 0 6 0.24 radian 0=-= 25 1t 1t ' r = 2 inches; 0=30°=30·-= 180 -6 radIan rc3 1.047 m. 2 A=-r12 20=-21 (2)2 (-rc6 ) =-:::! r = 2 feet; 0= rc3 radians 1t 21t 2.094 feet =rO = 2·-= 3 -:::! 3 2rc :::!2.094 fe = A =.!.2 r20 .!.2 (2 )2 (rc)= 3 3 1t -71tradIans ' r 12 yards; 0=70°= 70·-= 80 18 1

93.

97.

t

99.

t

d

rm

=

=

101. 91.

t

r =5 cm; = 20 seconds; 0=3.!. radian O)=!!...= (120/3) =3.!.._201 =�60 radian/sec v='::=rO = 5·(201/3) =�3 .�20 =�12 em/sec = 26 inches; r =13 inches; v=35 12 in. 1 35 mi ' 5280 ft '--'--v=-hr ft 60 min =36,960 in.lmin v 36,960 in.lmin 0)=-= r 13 in. :::! 2843.08 radians/min 2843.08 rad 1 rev min 21trad :::!452.5 rev/min r0==3960 miles 35°9'- 29°57' =5°12' =5.2° =5.2·� 180 :::!0.09076 radian =rO =3960·0.09076 :::!359 miles r =3429.5 miles 0)=1 rev/day = 27t radians/day =�12 radianslhr v=rO)=3429.5·�:::! 12 898 mileslhr t

S

89.

1t 1tradIan' r=4m, 0=45°=45·-=180 4 A =.!.2 r20=.!.2 (4)2 (rc)4 = 2rc:::! 6.28 m2

rIn=156 minutes, inches 1 90°=-1t. 0=-6015 rev =-·360°= 2 radIans 4 =rO =6·�2 =31t:::! 9.42 inches In 25 minutes, 51t. 5 0=-6025 rev=-·3600=1500=12 6 radIans 57t 57t :::! 15.71.mches =rO =6 . -= 6 s

hr

milhr

s

103.

s

369

© 2008 Pearson Education, Inc . , Upper Saddle River, Nl. All rights reserved . Thi s material i s protected under all cop yri ght laws as they currently

exist. No portion of this material may be reproduced, in any form or by any means , without permission in writing from the publisher.


105.

r= UJ= = 2n

x105 miles 2.39 3 days 1 rev/27. radiansl27.3 days 12·27.3 radianslhr =rUJ= (2.39x 105)._n_ "" 2292 miles/hr 327.6 = 2 inches; r2 = 8 inches; UJI =3 rev/min =67r radians/min Find UJ2:

115.

s

7t

_ __ _

v

107.

s

'i

C

= 1.5n

radians/min = l.5n revmm 2n ./ =-3 rev/nun. 4 r = 4 feet; UJ= 10 rev/min = 20n radians/min v=rUJ = 4 · 20n ft = 80n min 80re ft 1 mi 60 min ft hr "" 2.86 milhr5280 = 8.5 feet; r = 4.25 feet; V= 9.55 V 9.55 UJ=-= r 4.25 ft 9.55 mi 5280 ft 1 rev hr 4.25 ft nu 60 min 2n "" 31.47 rev/min inis the24 hours. one fullin 24rotation makestraveled earth The hours distance The the At the equator earth. miles. circumference Therefore, circumference ofis the2n(3960 ) must travel to keep person a linearthevelocity theup with sun is: 2n(3960 ) v=!.- = 24 "" 1037 mileslhr

117.

--

119.

I

113.

hr

7r

'i

of the is r andangletheislength radius ofbya circle also central the subtended arctheIf themeasure r, then of the angle is radian. Also, 180 rad·lan = degrees. per traveled the distance speedandmeasures Linear the measures speed angular time, unit In other time.traveled per unit angle in a central change distance describes speed linear words, andthe of a circle, on thetheedgeturning a pointspeedlocated byangular of rate describes circle itself. Answers will vary. I

----

.--_.

--- =

VI =v2 'iUJI =r2UJ2 'iUJI = r2UJ2 r2UJI r2UJI 'i UJ2 =-

rni!hr

rni!hr

7r'

•

min = -_._--.---

d

25 . 500= 12,500 "" 3979 miles . = 2 trr = 2 12,500 25,000 nules. miles, 397925,000 of Earth is approximately The theradiuscircumference approximately is and miles. rotates at UJI rev/min , so VI = 'iUJI' r2 rotates at UJ2 rev/min, so v2 =r2UJ2 the belt speedweofhave Since theis linear that:connecting the the same, pulleys r=

12n 8

111.

=rB

500 =r . .!!. 25...

VI = v2 'iUJI =r2UJ2 2(6n) = 8UJ2 UJ2 =-

109.

betweenSinceAlexandria thatto bethe =distance We know theis 7.20, miles. Syene and 500 Alexandria in rays Sun's the of measure of Earth the center atSyene angle formed the centralAlexandria also must and between 7.20. Converting to radians, we have be 7.20= 7.20. � 1800=.!!.25... radian . Therefore,

I

121.

._--.--

I

-

7r

123 -125.

t

370



Section 7.2: Right Triangle Trigonometry

15.

Section 7.2 1.

3.

7. 9. 11.

13.

c2 =a2+b2 =62+102 =36+100=136 c =56 = 2134 complementary True True opposite 5; adjacent 12; hypotenuse=? (hypotenuse)2 =52 + 122 =169 hypotenuse =.J169 =13 hyp sinB = opp hyp =2-13 cscB= opp =�5 adj =.!2 secB= hyp =� cosB= hyp adj 12 13 adj .!2 tanB = opp adj =2-12 cotB= opp = 5 opposite 2; adjacent 3; hypotenuse ? (hypotenuse)2 = 22 +32 =13 hypotenuse =Jl3 _2_ _2_ . Jl3 2J13 sinB = opp hyp = Jl3 = Jl3 Jl3 = 13 adj = _3_ =_3_. Jl3 =3J13 cosB = hyp Jl3 Jl3 Jl3 13 tanB= opp adj =�3 hyp = .J13 cscB = opp 2 hyp Jl3 secB=-=-adj 3 adj =l cotB= opp 2 =

=

=

=

17.

=

=?

adjacent 2; hypotenuse 4; opposite (opposite)2 + 22 =42 (opposite)2 = 16-4 =12 opposite=.J12= 2J3 2J3 = J3 sinB = opp = hyp 4 2 adj =� =.!.. cosB = hyp 4 2 2J3 tanB = opp adj = 2 =J3 hyp =_4_ =_4_ . J3 = 2J3 cscB = opp 2J3 2J3 J3 3 secB = hyp adj =.i2 = 2 adj =_2_ =_2_ . J3 = J3 cotB = opp 2J3 2J3 J3 3 opposite J2; adjacent 1; hypotenuse=? (hypotenuse)2 =(J2f + 12 =3 hypotenuse =J3 J3= 16 B= = J3J2 =J3J2 . J3 cosB= adj =_J31 =_J31 . J3J3 = J3 = oadjpp= J21 = J2 cscB= opp= J2J3 = J3J2. J2J2= 162 secB= adj = J31 =J3 cotB= oadjpp=_J21_=_J21_. J2J2= J22 sin

=

=

=

=

°PP hyp

3

hyp

3

tan e

hyp

hyp

371

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19.

23.

opposite 1 ; hypotenuse = J5 ; adjacent 12 + (adjacent)2 = (J5 (adjacent)2 = 5 - 1 = 4 adjacent = J4 = 2 _1_ = _1_ .J5 = J5 sinB = hopp = J5 J5 J5 5 yp adj = � = � .J5 = 2 J5 cos () = hyp J5 J5 J5 5 ! tan B = opp adj = 2 J5 = J5 csc () = hyp = opp 1 J5 sec B = hyp = adj -2 adj = � = 2 cot B = opp 1 =

t

=?

1 _ _1_ 2.J5 3J5 secB = _ cosB = J5 = J5 J5 = 5 3 1_ = _1_ = _5_J5 = 5J5 = J5 cot () = _ tan () 2J5 2J5 J5 10 2 5 25.

sin () = .fi2 corresponds to the right triangle: b=v'2

� � =2

a

(.fit

Using the Pythagorean Theorem: a2 + = 22 a2 = 4 - 2 = 2 a = .fi So the triangle is:

cos () = J3 2 1 B in �=__1 =__1 . J3=J3 s "2 = = tanB co B J3 =�2 . J3 J3 J3 J3 3 s 2 1 1 =1·2 = 2 csc () = -= sin () 1 2 1 =1 =2 = -2 . J3 2J3 -sec ()=-= cos () J3 J3 J3 J3 3 2 1_ = _1_ = � = �. J3 = 3J3 = J3 cot () = _ tan B J3 J3 J3 J3 3 3 J5 sin () = � '' cos () = 3 3 2 sinB =.l=?:. . �=�=� . .Js=2.Js tanB=cosB .Js 3.Js .Js .Js.Js 5 3 1 = -1 = -3 csc B = -sin () 2 2 3

b=�

� a

=.J2

.fi

adj = cos B = hyp 2 .fi opp tan B = -= adj .fi = 1 yp = � = � . sec B = hadj hyp = � = � . csc B = opp .fi adj - = 1 cot () = -= opp

.fi=.fi .fi .fi.fi .fi=.fi .fi.fi .fi .fi

372

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27.

cosf) = -31 Using the Pythagorean Identities: sin2f) + cos2 f) = 1 =1

.J5 = .J5 = csc f) = hyp opp 1 .J5 secf) = hyp adj = 2 adj = � = 2 cotf) = opp 1

Sin2B+GJ

sin 2 f) + .!. 9

=1 sin2f)=�

31.

9

= fI 2.J2 '19 3 (Note: sin f) must be positive since f) is acute . ) sinf) = 2f = 2.J2 . � = 2.J2 tanf) = cos f) t 3 1 sin f)

=

1 =1 = -3 = 3 ..Ji= -3..Ji cscf) = -sinf) 2f 2..Ji 2..Ji..Ji 4 1 = -1 = 1 ·3 = 3 sec f) = -cos f) t 1 ..Ji ..Ji 1 =-1 = -.- = cot f) = -tanf) 2..Ji 2..Ji..Ji 4 tanf) = .!.2 corresponds to the right triangle: -_.-

29.

b= l

=

�

-_.-

33.

a=2

Using the Pythagorean Theorem: c2 = f + 22 = 5 c = .J5 So, the triangle is: b= l

t:s:.f5= a=2

secf) = 3 Using the Pythagorean Identities: tan2 f) + 1 = sec2 f) tan2 f) + 1 = 32 tan2 f) = 32 - 1 = 8 tanf) = .J8 = 2..Ji (Note: tanf) must be positive since f) is acute.) 1 1 cos f) = -sec f) = 3 sin f) tan f) = -cos f) , so 2J2 sm. f) = (tanf) )( cos f))=2"11�2·-31 = 3 1 = -1 = -3 -3 .J2 . - = 3J2 cscf) = -sinf) 2f 2.J2 2..Ji.J2 -4 1 = -1 = 1 ..Ji= ..Ji cot f) = -tanf) 2..Ji 2..Ji..Ji -4 tanf)=..Ji Using the Pythagorean Identities: sec2 f) = tan2 f) + 1 sec2 f) = (..Jir + 1 = 3 sec f) = J3 (Note: secf) must be positive since f) is acute . ) I -I = I J3 =J3 cos f) = -secf) = J3 J3 J3 3 sinf) so tan f) = --, cosf) J3 = J6 . = (tan f)) ( cosf)) = "11�2·smf) 3 -3 1_ = � = 2- = 2-. J6 = 3J6 = J6 cscf) = _ sinf) .Jf J6 J6 J6 6 2 1 1 = 1 .J2 =.J2 cot f) = --2 tanf) = .J2 ..Ji .J2 _.-

e

opp = _1 = _I_ . .J5 = .J5 sin e = hyp .J5 .J5 .J5 5 adj = l:...- = l:...- . .J5 = 2 .J5 cos e = hyp .J5 .J5.J5 5

_.-

373

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35.

�=2

csce= 2 corresponds to the right triangle: b= l

a

43.

8

Using the Pythagorean Theorem: a2 +e = 22 a2 + 1 = 4 a2 = 4 - 1 = 3 a = .J3 So the triangle is: b= l

39.

41.

cos 1 0° = sin(9 0° - 10°) = sin 80° = 1 sin 8 0° sin 80° sin 8 0° using the identity cos e = sin ( 90° - e)

1 - cos2 20° - cos2 70° = 1-cos2 200 -sin2(9 00 -700 ) = 1 - cos2 20°- sin2(200 ) = 1 - ( cos2 20°+ sin2(200» ) =1- 1 =0 using the identities cos e = sin ( 90° - e) and sin 2 e + cos2 e = 1 . sin(90° cos 70° - 70°) � -'--cos 20° 49 . tan 20 0 - --- = tan 200 cos 20° sin 20° = tan 200 - --cos 20° = tan 20° - tan 20° =0 using the identities cos e = sin ( 900 - e) and sin e tan e = -cos e .

47.

�=2 =.J3 a

37.

45.

sin38° - cos52° = sin38°- sin(900 -52°) = sin 38° - sin38° =0 using the identity cos e = sin ( 90° - e)

8

.!. sin e = °PP hyp = 2 adj .J3 cose= hyp -= 2 _1 = _1 . .J3 = .J33 tane= °PP = adj .J3 .J3.J3 l:...- = l:...- . .J3= 2.J3 sec e = hyp = adj .J3 .J3.J3 3 cot e = 0adj = .J3 = .J3 PP 1 sin2 20° + cos2 20°= 1 , using the identity sin2 e + cos2 e = 1

--

51.

_ = 1, using the sin80°csc800= sin 80°. sin180° 1 · csc e = -1· d entity sine sin 50° = tan500 - tan 500= 0 , usmg . the tan 50° ---cos 50° sine 1·d entity · tan e = -cos e __

--

sin 35° ) sec55o . cOS 350 tan 35° . sec55° . cos35° = ( cos 35° = sin 35°·sec55° = sin 35° ·csc (900-55° ) = sin35°·csc35° 1 = sin 350 . _ sin_ 35° =1 sin e , . the l·dentitles .. tan e = -usmg cose 1 sec e = csc (9 0° - e) , and csce = -sm.e.

374




53.

55.

cos35° ·sin 55°+ cos 55°·sin35° = cos35°·cos(900-55°) + sin(900-55°) . sin35° = cos 35°· cos 35° + sin 35° ·sin35° = cos2 35°+ sin2 35° =1 using the identities sin8 = cos ( 90° -8 ) , cos8 = sin ( 90° -8 ) , and sin2 8 + cos2 8 = 1 . Given: sin 30° = .!.2 a.

b.

c.

61.

a.

i

cos 60° = sin ( 90° - 60° ) = sin 30° = cos2 30° = 1 - sin2 30° = 1 - 2 43 _ = .!.1 = 2 csc 2:6 = csc30° = sin130° 2

(i)

__

---

d. 57.

f.

g.

Given: tan8 = 4 sec28 = I + tan28 = 1 + 42 = 1 + 16 = 17 1 =1 b. cot8 = tan8 4

h.

a.

--

c.

d.

63.

( )

cot � -8 = tan8 = 4

csc2 8 = 1 + coe 8 1 17 1 = 1 +-1 = 1 + -== 1 + tan28 16 16 42

65.

--

59.

Given: esc8 = 4 1 sin8 = _ csc8_ = 4.!. a.

b. c.

d.

Given: sin 38° "" 0.62 cos 38° "" ? sin2 38° + cos2 38° = 1 cos2 38° = 1 - sin2 38° cos38° = .J1 - sin2 38° "" �1 - ( 0 . 62 ) 2 "" 0.78 sin38° "" 0.62 "" 0 . 79 tan380 = cos38° b. 0.78 5 cos 38° "" 0.78 5 "" 1 .2 7 cot380 = sin c. 38° 0 . 62 1 1 d. sec380 = __ ",, __ "" 1.27 cos 38° 0 . 78 5 1 1 . 61 csc38° = sin 138° ""--,,,, e. 0 . 62

sin 52° =cos ( 90° - 52° ) = cos38° ",, 0 .7 8 cos 52° = sin ( 90° - 52° ) sin 38° "" 0.62 tan 52° cot ( 90° - 52° ) = cot38° "" 1 .27 =

=

Given: sin8 = 0 . 3 sin8 + cos � -8 = sin8 + sin8 = 0 . 3 + 0.3 = 0 . 6

( )

The equation sin8 = cos ( 28 + 30° ) will be true when 8 = 90° - ( 28 + 30° ) 8 = 60° - 28 38= 60° 8 = 20°

coe 8 = esc 2 8 - 1 42 - 1 = 16 - 1 = 15 sec(90° -8) = esc8= 4 1 =16 sec 2 8 = 1 + tan 2 8 = 1 + coe1 8 1 + 15 15 =

-- =

375

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67.

a.

b. c.

l S00 + -= SOO S + S = 10 mmutes . T = -300 100

f.

SOO l-S00 = S+ l S = 20 mmutes . T = -+ 100 100 SOO SOO so X = -tan O = -, tan 0 x SOO �n. 0 = distance in sand , so SOO . d·Istance In. sand = -sin O -x distance in sand T ( 0) = l S00 100 300 + S O SOO l S00 ---=:ta:=.On:..: O'- sin 0 300 + 100 S_ + _S_ = S__ 3tan O sin O 1_ + _ 1 _ = S 1__ 3 tan 0 sin 0 S OO 1 . tan 0 = -lS00 = -3 , so we can consIder the triangle:

20

00 �======l90 ° o

Use the MINIMUM feature: 20

00

_ _

)

(

d.

g.

I�

69.

a.

3

S_ + _S_ T= S__ 3 tan O sinO = S - Sl + -lS ;-

() 3"

e.

S_ + _S_ with the Let 1'; = S _ _ 3 tanx sinx calculator in DEGREE mode.

b.

"iniMU,""

X=70.S2B77S _ V=9. 71�O�SZ o

•

900

The time is least when the angle is approximately 70.5". The value ofx for this angIe IS. x = tan SOO 70.S3° "" 177 feet . The least time is approximately 9.7 minutes. Answers will vary. Z2 =X2 + R2 Z = �X2+ R2 = � 4002+ 6002 = � S20,000 = 200.J1j "" 721.1 ohms The impedance is about 721 . 1 ohms. sinq$ = XZ = 400 2.J1j 200.J1j 13 cosq$ = -R = 600 3.J1j Z 200.J1j 13 X 400 = -2 tanq$ = -R = 600 3 cscq$ = X � = 200.J1j = .J1j 2 400 .J1j 200.J1j secq$ = ZR = 600 = 3 600 = -3 cotq$ = X -R = 400 2 ----;:=

M

= S - S + SM "" I S.8 minutes 1000 feet along the paved path leaves an additional SOO feet in the direction of the path, so the angle of the path across the sand is 45". S + _S_ T = S - 3 tan4So sin4So S S S+10 = S-+ = S3 J2 3 · 1 J2 2 "" 1 0.4 minutes 376




71.

Since IOAI = IocI =1 , �OAC is isosceles. Thus,LOAC LOAC = LOCA.. Now LOCA LAOC = 1 800 LOAC LOCA (1800- B) = 1800 LOAC+LOCA = 2( LOAC ) = B

a.

+

+

+

+

B

+

LOAC=�2

C DI J CD I = CDI sinB = II OC I 1 1 cosB = IOlODCIi = IO1DI = O1 DI sinO tan�= IIACDIDI IAOIIC+DIloDI=l+ICloDIDI = l+cosO h = x·-xh = xtanB -h = (I-x) tan(nB) h (1 - x) I-x xtanB = (1 - x)tan(nO) xtanB = tan(nB)-xtan (nB) xtanB xtan(nB)= tan(nB) x( tanB tan(nO))= (nB) tan(nB)x = ------'--',tanB tan--(nO) Area �OAC =-21 OC 1 I ·I AC I = LI2 OCI .I ACI 1 . =-cosasma 2 1 . =-smacosa 2 Area �OCB = 211 oc1·1 BC I = .!.2 'IOB 12 . o1Ol cBI 1 .Ol1 BCBII = .!.2 IOB 1 2 cosfJsinfJ = .!.2 IOB 12 sinfJcosfJ

b.

c.

73.

d.

2

2

·

+

+

2

,8)

tan

+

75.

+

e.

2

=

Area �OAB = 21 1 BD1 ·O1 AI = �I BDI·l = L2 IOBI·IOl BDBII = .!.2 IOBI sin(a fJ) Ol C i cosa = 01 AI Jocl . ol BI =IOBI cosfJ Ol C i 1 Ol Ci Ol BI Area �OAB = Area �OA C Area �OCB .!.I DB Isin(a+,8) =.!.sinacosa +.!.I DB 12 sin,8cos,8 cosa . --sm( cos,8 a +,8) . . +--cos2a- sm,8cos,8 =smacosa 2 cos ,8 cos,8 . cosa . sin(a+ ,8)=--smacosa +--sm,8cos,8 cosa cos,8 sinea+ =sinacos,8 + cosasin,8 sina . = --·cosa sma = tancosaa cosa = cosfJcosa = cosfJtan,8 sinfJ = cosfJ·_cos,8 = sinfJ sin2a+ cos2a=1 sin2a + tan2,8= sin2,8 sin2a+--=1 cos2,8 . sm2a =1 sm. 2a+ I-sin 2a (1-sin2a) (sin2a+ I�i�i��a)=(1)(1-sin2a)

c.

77.

a.

I

b.

377

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a

a

sin2 -sin4a+sin2 =1-sin2 a sin4a-3sin2a+l=0 Using the quadratic formula: 3±J5 sm. 2 -2 J5 ± . �3-SIna= 2 . �3-J5 But �3-+J5 2->1 . So, sma= -2- ' Consider the right triangle: a=

79.

7.

f(60°) =sin 60° =

9.

f

11.

13.

15. a

If () is an acute angle in this triangle, then: a a>0, b 0 and c>O. So cos () = ->0 . c 2 2 Also, since2 a +2 b2 =c , we know that:

17.

>

0 ° and cot e > 0. If a is the reference angle for e , then cos a = i5 . Now draw the appropriate triangle and use the Pythagorean Theorem to find the values of the other trigonometric functions of a .

12 cos a = 13

5 tana = 12

13 sec e = -12

12 cot e = -5

13 csca = 5

13 12 seca = cot a = 12 5 Finally, assign the appropriate signs to the values of the other trigonometric functions of e . 13 5 csc e = 12 tan e = -cos e = -5 12 13 387




95.

cos 0=--,13 1800 < 0< 2700 (quadrant III) Since 0 is in quadrant III, cos 0 and secO > 0. IfNowa draw is the reference angle for 0 , then seca = 2 . theTheorem appropriate triangl evalues and useof thethe Pythagorean to find the other trigonometric functions of a . °

°

y 4

°

y

sma=-3 cosa =-4 cota =-4 csca =-3 seca =-4 Finally, the appropriate signs ofto O.the values of the otherassigntrigonometric functions smO = --3 cosO = --4 cotO 4 cscO= - -3 secO = --4 tan 0 = -'3'1 sinO> (quadrant II) Since 0 is in quadrant II, cos 0 < 0, sec 0 < 0, tan 0 < and cotO < 0, while sinO> and cscO> 0. IfNowa draw is thethereference angletriangl for 0 e, then tanthe= 1 / 3 . appropriate and use Pythagorean Theorem trigonometric functionstooffinda .the values of the other ·

4

101.

III ,

103.

3

5

5

= -

5

5

.J3 = .J3 .J3 cosa =-1 . =tan a = sma 1 2 2 2 .J3 2.J3 1 .J3 -J3 cota=-'-= csca=-·-=-J3 .J3 3 .J3 .J3 3 Final l y , assign the appropriate signs to of the other trigonometric functions of O.the values .!. tan 0 = -.J3 sin 0 = _ .J3 2 cos 0 = 2 2.J3 cotO= --.J3 cscO= --3 3 tanO=�,4 sinO and cot 0 > 0. If a is the reference angle for 0 , then tana = �4 . Now draw theTheorem appropriate triangl evalues and useof thethe Pythagorean to fmd the other trigonometric functions of a. °

5

5

·

3

5

5

°

°

°

a

y

°

1 J10 J10 sma=--·--=-J10 J10 10 csc = .JlO1 =.JlO J10 = 3J1O seca =-3 ._.JlO cosa =-_ J10 J10 10 3 cota = -=1 3 Finally, the appropriate signs ofto O.the values of the otherassigntrigonometric functions cscO=vlO cos f} = - 3Fa sm f} = Fa sec f} = FlO cote = -3 3 a

·

--

3

·

10

-

r.;;

10

--

- --

389




105.

csc 0 = -2, tan 0 > 0 => 0 in quadrant III Since 0 is in quadrant III, cos 0 < 0, sec 0< 0, sin 0< 0 and csc 0 < 0, while tan 0 > 0 and cot O > O. If a is the reference angle for 0 , then csc a = 2 . Now draw the appropriate triangle and use the Pythagorean Theorem to find the values of the other trigonometric functions of a .

113.

y

115. x

J3 . = -1 sma cos a = 2 2 J3 = -J3 = J3 seca = 2 ._ 2J3 tana = -1 · 3 J3 J3 J3 J3 3 cot a = J31 = J3 Finally, assign the appropriate signs to the values of the other trigonometric functions of 0 . sinO = .!..2 cos O = _ J32 J3 2J3 tan B = secO = 3 3 cot B = J3 sin 40° +sinI300+sin 2200+sin310° = sin 40° +sin( 40° +90°) + sin( 40°+ 1 80°) + sin( 40° +270°) = sin 40° + sin 40° - sin 40° -sin 40° =0 Since f (B) = sin B = 0.2 is positive, 0 must lie either in quadrant I or II. Therefore, B + tr must lie either in quadrant III or IV. Thus, f ( B + 7t) = sin( 0 + 7t) = -0.2

117.

1 = -1 = 5 G·Iven sm. B = -51 , then csc B = -sin O 1 5 Since csc 0 > 0 , B must lie in quadrant I or II. This means that csc ( 0 + tr) must lie in quadrant III or IV with the same reference angle as B . Since cosecant is negative in quadrants III and IV, we have csc(B + tr) = -5 . sin 1° + sin 2° + sin 3° +... + sin 357° + sin 358° + sin 359° = sin 1° +sin 2° + sin3° + . . . + sin(360° _3°) + sin(360° - 2°) + sin(360° _1°) = sin 1° +sin 2° +sin3° + . . . +sin( _3°) + sin( -2°) + sin( -10) . = sin 1° +sin 2° +sin3° + . . -sin 3° -sin 2° -sin 1°

a.

�

b.

--

109.

111.

.

�

-

107.

322.J2 [sm( 2( 60°)) - cos( 2( 600)) -1 = --:n] 32.J2 (0.866 - (-0.5) - 1) 16.6 ft . Let 1'; = 322.J2 32 [sm(2x) - cos {2x) - I] R

c.

Since F ( 0) = tan 0 = 3 is positive, B must lie either in quadrant I or III. Therefore, B + 7t must also lie either in quadrant I or III. Thus, F (O+ 7t) = tan (B + tr) = 3 .

119.

--

Using the MAXIMUM feature, we find: 20

Answers will vary.

390



Section 7.5: Unit Circle Approach: Properties of the Trigonometric Functions

Section 7.5

3.

5.

7.

. = -2 smt 3

even

1 = (1)(Js) = � Js = 2; 3 csc t = l = 1 (% ) = % 3 sec t = Js = ( Js ) = Js Js = 3;

All real number, except odd multiples of 7r2 -0.2, 0.2

. = --21 smt

cost

tant =

.J3 2

= -

3

1 tant = ] = ( - �) (�) = - � � = - � 2 csct = I.!. = { - f = -2 _ 2 ) sect = � = ( �) = � � = 2� 2 .J3 cot! = �.!. = ( � ) ( -f) =-.J3 2

11.

J5 cott = 15.

17.

fi cos t = -2

tant = _2_ = 1

cott =_fi2_ = 1

fi

fi

2

csc t =

fi

2

19.

� = ( -1) = -1 � = -fi

2

sect =

For the point (3, -4) , x = 3, y = -4, r = �x2 + y2 = �9 + 16 = J2s = 5 SIn. () = --45 cos () = -35 tan () = --43 sec () = -35

cot () = --43

For the point (-2, 3), x=-2 , y=3, r = �x2 + y2 =�4 + 9 = 03 sin () = _3_ 03 = 3 03 csc () = .j13 3 13 03 03 .j13 cos () = __2_ 03 = _ 203 13 sec () = - 2 03 03 tan () = --23 cot () =--23

-

_ fi = b p = _ fi2 ' _ fi2 '. a = _ fi 2' 2

. = -fi smt 2

I3 = (�)(%) = �

csc () = --45

)

(

2

J5 cos t = 3

� = { - 1) - � � = -fi =

For the point (- I, - I), x = - I, y = - I, r = �x2 + y2 = .J1+1 = fi sin () = -=!. fi = _ fi2 csc () = fi -1 =- fi fi fi cos () = -=!. fi = _ fi2 sec () = fi -1 = -fi fi fi -1 = 1 -=1 cot () = -1 tan () = -1 1

-

2

-

391

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49.

25.

27.

29.

31.

33.

csc4500= csc(3600+ 90°) = csc900= 1 cot 390° = cot(180° + 1 80° + 30° ) = cot 30° = J3 337t = cos 7t + 87t cos 4 4" = COs � + 4 ' 27t = cos -7t4 .fi 2 tan(217t) = tan( O + 217t) tan ( 0) = 0

( ) ( )

55.

57.

59.

=

( ) ( )

177t = sec 7t + 47t sec 4 4" sec � + 2 · 27t = sec-47t = .fi J3 tan 197t 6 = tan 2:6 + 37t = tan 2:6 = 3 =

35.

53.

( )

61.

65.

67.

tan

( )

sec -2:6 = sec2:6 = 2 J3 3 sin ( -7t) + cos (57t) = - sin ( 7t) + cos ( 7t + 47t) = 0 + COS7r = -1 sec( -7t) + csc

(-%) = sec7t - csc %

= -1 - 1 = -2 sin _ 9; _ tan _ 9; 97t . 97t tan= -Sln-+ 4 4 = -sin � + 27t + tan � + 27t . 7t + tan -7t = -Slll4 4 2 - .fi .fi + 1 or -= -2 2 '

( ) ( )

( ) ( )

63.

45.

tan(-7t) = - tan7t = 0

(-�) = -tan � = -1 69.

The domain of the sine function is the set of all real numbers. I ((}) = tan () is not defined for numbers that are odd multiples of 2:2 . I( (}) = sec ()

is not defined for numbers that are odd multiples of 2:2 . The range of the sine function is the set of all real numbers between -1 and 1 , inclusive. That is, the interval [ -1, 1 ] . The range of the tangent function is the set of all real numbers. That is, (-00 , 00 ) .

392



Section 7.5: Unit Circle Approach: Properties of the Trigonometric Functions

71.

73.

75.

77.

79.

81.

83.

85.

87.

The range of the secant function is the set of all real number greater than or equal to 1 and all real numbers less than or equal to -1 . That is, the interval (-ex), -1 ] or [ 1 ,ex) ) . The sine function is odd because sine-(}) = -sin () . Its graph is symmetric with respect to the origin. The tangent function is odd because tan(-(}) = -tan () . Its graph is symmetric with respect to the origin. The secant function is even because sec( -(}) = sec () . Its graph is symmetric with respect to the y-axis. If sin () = 0.3 , then sin () + sin ( () + 21t) + sin ( () + 41t) = 0.3 + 0.3 + 0.3 = 0.9 If tan () = 3 , then tan () + tan ( () + 1t) + tan ( () + 21t) =3+3+3 =9 fe -a) = -f (a) = --31 b. f (a) + f (a + 21t) + f (a + 41t) = f (a) + f (a) + f (a) = -31 + -3I + -3I =1 fe -a) = -f (a) = -2 b. f (a) + f (a + 1t) + f (a + 21t) = f (a) + f (a) + f (a) = 2+2+2 =6 f e -a) = f (a) = - 4 b. f (a) + f (a + 21t) + f (a + 41t) = f (a) + f (a) + f (a) = -4 + (-4) + (-4) = -12

89.

When t = 1 , the coordinate on the unit circle is approximately (0 . 5, 0.8) . Thus, 1 "" 1.3 sinl "" 0.8 cscl "" 0.8 1 = 2.0 sec l ",, cosl "" 0.5 0.5 0.5 ",, O . 6 0.8 tan I "" 0.5 = 1.6 cot l "" 0.8 on RADIAN mode: Set the calculator S u /s � a.

-

ln � 8414709848 os ( l ) . 5403023059 �,- an ( l ) 1 . 557407725 b.

ln� �88395 106 l/cos ( l ) 1 . 8508 1 5718 l/t an ( l ) . 6420926159

When t = 5.1 , the coordinate on the unit circle is approximately (0.4, -0.9) . Thus, 1 "" -1.1 sin 5.1 "" -0.9 csc5.l "" -0.9 I = 2.5 cos 5 . 1 "" 0.4 sec5.l "" 0.4 -0 9 tan 5. 1 "" -' 0.4- "" -2.3 cotS.l "" � -0.9 "" -0.4 Set the calculator on RADIAN mode: --

JSlm � : §158146823 Il /s ln�r : ��012977 pos(5. 1) l/cos (5. 1) 2. 645658426 . 3779777427 (5. 1) 1/t an -2. 449389416 an- (5.1) . 4082650123

a.

91.

Let P = (x, y) be the point on the unit circle that corresponds to an angle t. Consider the equation tan t = 2:.x = a . Then y = ax . Now x2 + l 1 , 1 and so x2 + a2 x2 = 1 . Thus, x = ± � l +a2 y = ± � ; that is, for any real number a , l + a2 there is a point P = (x, y) on the unit circle for which tan t = a . In other words, -ex) < tan t < ex) , and the range of the tangent function is the set of all real numbers. =

a.

a.

393




93.

Suppose there is a number p, 0 < p < 21t, for which sine 0 + p) sin 0 for all O . If 0 0 , then sin (0 + p) sin p sin 0 0 ; so that p n . If 0 = � then sin % + p sin . 3n - 1 sm. "2n 1, But p n . Thus, sm. 2 =

( ) ( %) ( ) ()

=

=

=

5.

=

=

-1

3.

=

=

=

=

7.

=

9.

or = 1 . This is impossible. The smallest positive number p for which sine 0 + p) = sin 0 for all 0 is therefore p 2n .

1cos_O since cos 0 has period =

95.

f ( 0) = sec 0 =

_

1 1.

:

3 ,. 27r6 = 7r3 False The graph of y = sin x crosses the y-axis at the point (0, 0), so the y-intercept is O . The graph of y = sin x is increasing for n <X< n 2 2 The largest value of y = sin x is 1 . sin x = 0 when x = 0, n, 2 n . - -

2n , so does f (O) = sec O. 97. If P = (a, b ) is the point on the unit circle corresponding to 0 , then Q = (-a, -b) is the point on the unit circle corresponding to 0 + n . -b b Thus, tan( 0 + n) = = - = tan 0 . Ifthere -a a exists a number p, 0 < p < Jr, for which tan(0 + p) = tan 0 for all 0 , then if 0 = 0 tan(p) = tan( O ) = O . But this means thatp is a multiple of n . Since no multiple of n exists in the interval (0, n) , this is impossible. Therefore, the fundamental period of f ( 0) tan 0 is Jr. sin O - O sin O = tan 0 . Since L 99. Slope ofM = cosO - O = -cosO is parallel to M, the slope of L = tan 0 . 1 01 - 103. Answers will vary.

13.

15.

1 7.

,

19.

- .

3n -n ' . = l when x = -smx 2 ' 2' 3n . = - 1 when x = --n smx ' 2 2 y = 2 sin x This is in the form y = A sin«(i}x) where A = 2 and = Thus, the amplitude is I A I = 1 2 1 = 2 2n = 2n = 2 n . and the peno. d ' T = 1

1.

(i)

IS

=

21.

(i)

y = -4cos(2x) This is in the form y A cos( (i}x) where A -4 and = 2 . Thus, the amplitude is I A I = 1 -4 1 = 4 and the period is 2n = 2n n . T= 2 =

(i)

=

Section 7.6 1.

I ; !!:..2 + 2Jrk , k is any integer

=

(i)

y = 3x2 Using the graph of y = x2 , vertically stretch the graph by a factor of 3.

23.

y = 6sin(nx) This is in the form y = A sine (i}x) where A = 6 and Jr . Thus, the amplitude is I A I = 1 6 1 6 2n = 2n = 2 . and the peno. d ' T = n (i)

=

=

IS

(i)

x

-5

394

© 2008 Pearson Education, Inc . , Upper Saddle River, NJ. AlJ rights reserved. This material is protected under alJ copyright laws as they currently


Section 7. 6: Graphs o f the Sine and Cosine Functions

-

3 " , and 2" 0, "2 , " , 2 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 for y 4 cos x , we multiply the y-coordinates of the five key points for y = cos x by A 4 . The five key points are (2", 4) (", -4) , We plot these five points and fill in the graph of the curve . We then extend the graph in either direction to obtain the graph shown below .

This is in the form y = A cos( OJx) where A = - -2I and OJ = �2 . Thus, the amplitude is 1A1= T

=

1- i 1 i and the period is =

=

(0,4), (;,0).

27t 27t 4 7t - -;- - 1 - 3 ' _

_

_

2

This is in the form y = A sin(OJx) where A and OJ 1A1= T

29. 31. 33. 35.

=

=

2" . Thus, the amplitude is 3

1 -% 1

=

27t = 27t

OJ

2 Jf

=

(11T. 4)

5

3

- -

x

% and the period is =

3

.

3

45.

F A H C

41. 43.

Comparing y -4 sin x to y = A sin ( ) , we find A = -4 and OJ = I . Therefore, the amplitude is 1 -41 = 4 and the period is 2"I 2" . Because the amplitude is 4, the graph of y = -4sinx will lie between -4 and 4 on the y-axis. Because the period is 2" , one cycle will begin at x = ° and end at x = 2" . We divide the interval into four subintervals, each of length 2"4 = !!... by OJX

=

=

37. J 39.

e;,o).

A B Comparing y = 4 cos x to Y = A cos ( OJx) , we find A = 4 and OJ = I Therefore, the amplitude is 1 41 = 4 and the period is 2"I = 2" . Because the amplitude is 4, the graph of y = 4 cos x will lie between -4 and 4 on the y-axis. Because the period is 2" , one cycle will begin at x = ° and end at x = 2" . We divide the interval 27r] into four subintervals, each of length 2"4 = "2 by finding the following values:

[0,2,,] 2 3"

0,

finding the following values: "2 , " , 2 , 2" 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 for y = -4 sin x , we multiply the y-coordinates of the five key points for y sin x by A -4 . The five key points are e; ,4 . (2", 0) ,-4 , We plot these five points and fill in the graph of the curve. We then extend the graph in either

.

=

(0,0), (; ) (",0),

[0,

=

)

395




direction to obtain the graph shown below. , y 5 C�, 4)

49.

Since sine is an odd function, we can plot the equivalent form y = - sin ( 2x ) . Comparing y = -sin(2x) to y = A Sin(wx) , we find A = -1 and w = 2 . Therefore, the amplitude is 1 - 11 = 1 and the period is 22Jr = Because the amplitude is 1 , the graph of y = - sin (2x) will lie between -1 and 1 on the y-axis. Because the period is Jr , one cycle will begin at x = ° and end at x = Jr . We divide the interval [0, ] into four subintervals, each of length 4 by finding the following values: Jr .

47.

(_3;. -4) -5 (� , -4) Comparing y cos( 4x) to A Cos( wx) , we find A 1 and w 4 . Therefore, the amplitude is III 1 and the period is 24Jr Jr2 . Because the amplitude is 1, the graph of y = cos ( 4x) will lie between -1 and 1 on the y-axis. Because the period is 2 , one cycle will begin at x ° and =

=

Jr

Y =

=

=

Jr

=

Jr

=

0, -Jr4 , -Jr2 , -34Jr , and

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 for y sin (2x) , we multiply the y-coordinates of the five key points for y = sin x by A = -1 . The five key points are (0, 0) , (: ' -1), ( ; , 0), e: ,1), ( Jr, 0) We plot these five points and fill in the graph of the curve . We then extend the graph in either direction to obtain the graph shown below. y 2

]

=

end at x = ; . We divide the interval [0, ; into four subintervals, each of length 4/ 2 = Jr8 by finding the following values: 3 Jr ' °, 8 Jr4 ' -8Jr and -Jr2 These values of x determine the x-coordinates of the five key points on the graph. The five key points are Jr

'

-

C;r, l) e:, l)

(0,1), ( ; ,0), (:,-1), e;,o) , (;,1) (-� . l ) (_3; . 0) 11"

(0, 1 )

(i · O)

(11", 0) 211"

We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below. y 2

Jr

X

-2

(� , 1 )

x

-2/

(-i , 0) (-� . - 1) -2

396



Section 7. 6: Graphs o f the Sine and Cosine Functions

51.

( )

Comparing y = 2 sin � x to y = A Sin(mx ) , we find A = 2 and m = -21 . Therefore, the amplitude is 121 = 2 and the period is 127r1 2 = 47r . Because the amplitude is 2, the graph of y = 2 sin ( x will lie between -2 and 2 on the y-axis. Because the period is 47r , one cycle will begin at x = ° and end at x = 47r . We divide the interval [0, 47r] into four subintervals, each 47r = 7r by finding the following of length 4 values: 0, 7r , 27r , 37r , and 47r 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 for y = 2 sin x ) , we multiply the y-coordinates of the five key points for y = sin x by A = 2 . The five key points are (0,0) , (7r, 2) , (27r,0) , (37r,-2) , (47r,0) We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.

53.

I 1

�)

Because the amplitude is !2 , the graph of

y = - !2 cos (2x) will lie between -!2 and !2 on the y-axis . Because the period is 7r , one cycle will begin at x = ° and end at x = 7r . We divide the interval [0, 7r] into four subintervals, each of length 7r4 by finding the following values: 37r , and 7r 0, -7r4 , -7r2 , 4 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 for y = -!2 cos (2x) , we multiply the y-coordinates of the five key points for y = cos x by A = - !2 .The five key points are

(�

( - 3'IT , 2)

y

3

Comparing y = -!2 cos (2x ) to Y = A cos (mx ) , we find A = - -21 and m = 2 . Therefore, the 27r = 7r . . 1 1 amp11tu · de IS - - = - and the peno . d I' S 2 2 2

(o,-�) , ( : ,0) , (;,�) , e:,o) , (7r'-�)

We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.

('IT. 2)

y 2

-3

(311", -2)

- 21T -2

397

© 2008 Pearson Education , Inc. , Upper Saddle River, N l . A l l rights reserved. This material i s protected under a l l copyright laws a s they currently



55.

We begin by considering y = 2 sin x . Comparing y = 2sinx to y = A Sin(mx) , we find A = 2 and m = 1 . Therefore, the amplitude is 121 = 2 and the period is 27(1 = 27( . Because the amplitude is 2, the graph of y = 2 sin x will lie between -2 and 2 on the y-axis. Because the period is 27( , one cycle will begin at x = ° and end at x = 27( . We divide the interval [0, 27( ] 27( = 7( by into four subintervals, each of length ""4 "2 finding the following values: 37( , and 27( 0, -7(2 , 7( , 2 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 for y = 2 sin x + 3 , we multiply the y-coordinates of the five key points for y = sin x by A = 2 and then add 3 units. Thus, the graph of y = 2 sin x + 3 will lie between 1 and 5 on the y axis. The five key points are (0,3) , (7(,3) , e;,l), (27(,3) We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.

57.

We begin by considering y = 5 cos ( 7(x) . Comparing y = 5 cos(7(x) to Y = A cos(mx) , we find A = 5 and m = 7( . Therefore, the amplitude is 1 51 = 5 and the period is 27(7( = 2 . Because the amplitude is 5, the graph of y = 5 cos ( 7( x) will lie between and 5 on the y-axis. Because the period is 2 , one cycle will begin at x = ° and end at x = 2 . We divide the interval [0,2] into four subintervals, each of length -42 = 21 by finding the following values: 0, 21 , 1, -23 , and 2 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 for y = 5 cos (7(x ) -3 , we multiply the y-coordinates of the five key points for y = cos x by A = 5 and then subtract 3 units. Thus, the graph of y = cos ( 7(x ) - 3 will lie between -8 and 2 on the y-axis. The five key points are (0, 2) , -3 , (1, - 8 ) , -3 , (2, 2) We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.

-5

-

-

5

(;,5),

(% ' )

(� , )

y

-4

398

© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is prote cted under all copyright laws as they currently

. exist. No portion of this material may be reproduced, in any form or by any means, without permIssIon m wntmg from the pubhsher.

Section 7. 6: Graphs of the Sine and Cosine Functions

59.

( )

61.

We begin by considering y = -6sin � x .

(� )

Comparing y = -6 Sin x to y = A sin(wx) , 7r

we find A = -6 and w = -3 . Therefore, the amplitude is 1 -61 = 6 and the period is 7r27r/3 6 . Because the amplitude is 6, the graph of y = 6sin � x will lie between -6 and 6 on the y-axis. Because the period is 6, one cycle will begin at x = ° and end at x = 6 . We divide the interval [0,6] into four subintervals, each of length i4 l2 by finding the following values: 0, -23 , 3, -92 , and 6 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 for y = -6sin � x + 4 , we multiply the y coordinates of the five key points for y = sin x by A = -6 and then add 4 units. Thus, the graph of y = -6sin � x + 4 will lie between -2 and lO on the y-axis. The five key points are (0,4) , % ,-2 (3,4) , (6,4) We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.

y = 5 - 3 sin ( 2x ) = -3 sin ( 2x ) + 5 We begin by considering y = -3 sin (2x ) . Comparing y = -3sin( 2x) to Y = A sin( wx) , we find A = -3 and w = 2 . Therefore, the amplitude is 1 -31 = 3 and the period is 227! 7! . Because the amplitude is 3, the graph of y = -3 sin ( 2x) will lie between -3 and 3 on the y-axis. Because the period is one cycle will begin at x = ° and end at x = We divide the interval [0, ] into four subintervals, each of length 4 by finding the following values: 37r , and 0, -4 , -2 , 4 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 for y = -3sin (2x) + 5 , we multiply the ycoordinates of the five key points for y = sin by A = -3 and then add 5 units. Thus, the graph of y = -3sin(2x) + 5 will lie between 2 and 8 on the y-axis . The five key points are 3� ,8 , (7!,5) (0,5) , : , 2 , We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below. =

=

( )

7r ,

7! .

7r

7r

7r

=

7r

7r

x

( )

( ) (;,5), ( )

( ) ( } (�,1O),

y

10

(7t ,5) (6, 4) 8

-7t

7t X

x

399




63.

Since sine is an odd function, we can plot the 21i ) . .5 (3x eqUlva ent orm y = -3sm 21i ) to 5 . (3x Companng. y =-3sm y = sin(OJx) , we find = - 23 and OJ = 21i3 . Therefore, the amplitude is \ - %\ =% and the period is � 21i / 3 =3 . Because the amplitude is 21i x) will lie ( 2.,3 the graph of y = _2.sin 3 3 5 5 . Because the d -3 on the y-aXIS. between --an 3 period is 3 , one cycle will begin at x = and end at x =3 . We divide the interval [0, 3 ] into four subintervals, each of length � by fmding the following values: 9 and 3 3 -, 3 -, 0, -, 4 2 4 Thesefivevalues of x determine the x-coordinates ofy thecoordinates key ofpoints on the graph. To obtain the the five key points for 2 y =-%sin ( ; x) , we multiply the y coordinates of the five key points for y sin x by =_2.3 .The five key points are .

1

65.

fi

A

A

A

A

We begin by considering y =-%cos ( : x) . Comparing y = -% cos ( : x) to y = cos(OJx) , we find = --23 and OJ =-1i4 . Therefore, the amplitude is \ - %\ =% and the period is 21i =8 . Because the amplitude is � , the graph 2 1i /4 of y = -�cos 2 (1i4 x) will lie between -�2 and �2 on the y-axis. Because the period is 8, one cycle will begin at x = and end at x =8 . We divide the interval [0, 8] into four subintervals, 8 2 by finding the following each of length 4-= values: 0 , 2, 4, 6, and 8 These of x determine the x-coordinates ofy thecoordinates fivevalues key ofpoints on the graph. To obtain the the five key points for y=-�cos 2 (1i4 x) +.!.,2 we multiply theycoordinates of the five key points for y cos x by -�2 and then add .!.2 unit. Thus, the graph of y = -�cos 2 (1i4 x) +.!.2 will lie between 1 and 2 on the y-axis. The five key points are ) ) ( 0 , - 1 ) , (2,� . (4,2), (6,� . ( 8, -1 ) We plot these five points andthefillgraphin thein either graph of thedirection curve. We then extend to obtain the graph shown below. (2, +)(4, 2) (-4, 2) °

°

4

=

A

=

A

=

-

,0 ,0 ) . (%' % ). ( 3 ) (�,-%). (% We plot these five points and fill in the graph of ( 0, 0 ) ,

the curve.toWeobtain thentheextend graphbelow. in either direction graphtheshown y 2

(- 8, - 1)

-2

-2 (0, - 1 )

(8, - 1 )

400



Section 7.6: Graphs of the Sine and Cosine Functions

6 7.

I A I = 3,' T = n ,' m = 2Tn = 2nn = 2 y = ±3sin(2x)

69.

2n = 2 2n = n I A I = 3; T = 2; m = T y = ±3sin(nx) The graph is a cosine graph with amplitude 5 and period 8. Find m : 8 = 2nm 8m = 2n 2n = -n m=8 4

71.

79.

(� )

81.

( )

The equation is: y = 5 cos � x .

73.

83.

(� )

77.

The graph is a sine graph with amplitude 3 and period 4 . Find m : 4 = 2n m 4m = 2n 2n = -n m=4 2 The equation is: y = 3 sin �

( x) .

The graph is a reflected cosine graph with amplitude 3 and period 4n . Find m : 4n = 2nm 4nm = 2n 2n = -1 m=4n 2 The equation is: y = -3COS X .

75.

The graph is a reflected cosine graph, shifted up 1 unit, with amplitude 1 and period � . 2 3 2n Find m : 2 m 3m = 4n 4n m=3 The equation is: y = - cos n x + 1 .

The graph is a sine graph with amplitude �4 and period 1 . Find m : 1 = 2nm m = 2n The equation is: y = �sin 4 ( 2nx ) .

85.

The graph is a reflected sine graph with amplitude 1 and period 41l"3 . Find m : 4n3 2n m 4nm = 6n 6n = -3 m=4n 2

The graph is a reflected cosine graph with amplitude 4 and period 21l"3 . 2n = 2n Find m : 3 m 2nm = 6n 6n = 3 m=2n The equation is: y = -4 cos ( 3x ) . sin ( 1l" I 2 ) - sin ( 0) 1l" 1 2 - 0 1l" I 2 1-0 2 1l" 1 2 1l" The average rate of change is 3.1l" . f ( 1l" I 2 ) - f ( 0)

_

(% )

The equation is: y = - sin x . 401

© 2008 Pearson Education , Inc . , Upper Saddle River, NJ. All rights reserved . Thi s material is protected under all copyright laws as they currently

exist. No portion of this material may be reproduced , in any form or by any means, without permission in writing from the publisher.


87.

(� �) (� )

91.

. - Sin 0 f (" / 2) - f ( O ) Sin ,, / 2 ,, / 2 - 0 sin ( / 4) - sin ( 0 ) .

_

(J

0

g)(x) -2( cosx) -2cosx =

=

"

_

x

J2

2 J2 2 J2 ,, / 2 2 " " The average rate of change is J2 " .

= __ = _ 0 _ = -

89.

(g f) ( x) cos( -2x) =

0

(J g)(x) = sin (4x) 0

x

x

(g

0

f )(x) = 4(sinx) 4sinx =

93.

I (t) = 220 sin(60n t), t � 0

2n � Period: T 2nOJ 60n 30 Amplitude: 1 A 1 1 220 1 = 220 =

=

=

=

x

I

402



Section 7.6: Graphs of the Sine and Cosine Functions

95.

V(t) = 220sin(1207tt) Amplitude: I A I = 1 220 I = 220 Period: T = 27tcv = � 1207t = � 60

99.

a.

a.

b, e.

b.

c.

d.

97.

a.

27t . Physical potential: cv = 23 ' 2 7t = -7t ; EmotlOna . 1 potentia. 1 : cv = 28 14

27t Intellectual potential: cv = 33 Graphing: 1 10

0

V = IR 220sin(1207tt) = 101 22 sin(1207tt) = I � l(t) = 22sin(120m) Amplitude: I A I = 1 22 1 = 22 _1 Period: T = 27tcv = � = 1207t 60

0

33

# 1 , #2, #3

( �; t) + 50 p(t) = 50sin e : t ) + 50 p(t) 50sin (�; t ) + 50

#1 : p(t) = 50sin #2 : c.

J p(t) = [V (t) R (Vo sin (27t/ )t t V0 2 sin2 (27t/ )t R R = _VR0_2 sin2 (27tft ) The graph is the reflected cosine graph translated up a distance equivalent to the amplitude. The period is _2/1_ , so cv = 47t / . Va2 . . IS. 1 V0 2 = The amphtude 2 R 2R The equation is: V02 p(t) = V2R0 2 cos( 47t/ )t + _ 2R V 2 0 =_ 2R (l - cos( 47t/ )t) Comparing the formulas: sin2 (27tft ) = .!.2 ( 1 - cos (47tft ))

d.

#3 : No.

=

1 10

_

b.

7 3 0 5 �=====�=�� o

_ . -

_

c.

101.

_

Physical potential peaks at 15 days after the 20th birthday, with minimums at the 3rd and 26th days. Emotional potential is 50% at the 17th day, with a maximum at the 10th day and a minimum at the 24th day. Intellectual potential starts fairly high, drops to a minimum at the 13th day, and rises to a maximum at the 29th day. Y = lsin xl , - 2,, '5, x '5, 2!r x

-I

1 03 - 105. Answers will

vary.

403

© 2008 Pearson Education, Inc. , Upper Saddle River, Nl. All rights reserved. Thi s material is protected under all copyright laws as they currently

exist. No portion of this material may be reproduced, i n any form or by any means, without permission in writing from the publisher.


S ection 7.7 1.

21.

x = 4 souveniers

3.

origin; x = odd multiples of 2:2

5.

Y

7.

9.

11.

13.

1 5.

1 7.

( )

y = tan � x ; The graph of y = tan x is . honzontally compressed by a factor of 2

7r

= cosx The y-intercept of y = tan x is 0 . The y-intercept of y = sec x is 1 . secx = 1 when x = -27t, 0, 27t; secx = -1 when x = -7t, 7t y = sec x has vertical asymptotes when 37t - 7t 7t 37t x = -T ' 2' 2' T ·

23.

.

(±x) ;

y = cot The graph of y = cot x is horizontally stretched by a factor of 4. y

Y

= tanx has vertical asymptotes when 37t - -7t -7t 37t . x = -' ' ' 2 2 2 2

5

x

y = 3 tan x ; The graph of y = tan x is stretched vertically by a factor of 3.

111 6 1

IY

x

25.

Y

= 2 sec x ; The graph of y = sec x is stretched vertically by a factor of2. (0 , 2) y

1 9.

y = 4 cot x ; The graph of y = cot x is stretched vertically by a factor of 4.

I 'IT i1 2 (- 'IT, - )� it ¥, 4

y

16

1 'IT 'IT -x2 i, l\t, ( , ) I

I

I

x

404

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Section 7. 7: Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions

27.

y = -3 cscx ; The graph of y = cscx is vertically stretched by a factor of 3 and reflected about the x-axis.

2

, ,

,

,"

,

, x

,

,

Ao:""-r----+--

,

,

(T '

, I

,

29.

'

:U' e�, 3)

, ,

, , ,

,

,

3) ' (--, - n 3 1T

,

- 1T

'

(� )

y = 4sec x ; The graph of y = sec x is horizontally stretched by a factor of 2 and vertically stretched by a factor of 4. y

,'

-

33.

3)

di �

( - 37T, 2 ) -

4

:(0, 1

47T

7T

-8

35.

( )

y = sec 2; X + 2 ; The graph of y = secx is horizontally. compressed by a factor of � 2n and shifted up 2 units .

l ,

, , , , , (0, 4) : , ,

( )

y = tan � x + 1 ; The graph of y = tan x is horizontally stretched by a factor of 4 and shifted up 1 unit.

y

, 31T (2 1T, - 4) , X

, ,

, ,

I

31.

-4

y = -2 csc ( nx ) ; The graph of y = cscx is horizontally compressed by a factor of � , vertically stretched by a factor of 2, and reflected about the x-axis.

Ji 7\

y 6

(-t, 2) :

-2.0 , - 1 .0 ,

(_1.2 , - 2) '

,

, ,

I

:

, , ,

! -6

n

l\l'

, , ,

,

37.

, , , ,

y

, (+ , 2) 2.0 , x ( ) _+--r-- T, -2 :

f

� (� )

y = tan x - 2 ; The graph of y = tanx is horizontally stretched by a factor of 4, vertically compressed by a factor of .!2 , and shifted down 2 units. 8

1

(0, -8

( 1T , -t) 37T

-2)r--

1T

I

X

405




39.

Y = 2 CSC

(� )

(g f)( x) = 4( tan x ) = 4tanx

X -1;

The graph of y = cscx is horizontally stretched by a factor of 3, vertically stretched by a factor of 2 , and shifted down 1 unit.

0

y

J

y

6

x

{ 47.

(J o g)(x) = -2 (cot x) = -2 cotx y

f 41.

-O (�)- f ( O) = tan(Jr I 6) - tan(0) = --� Jr 1 6 Jr 1 6 �6 - O

43.

I I

x

-2

J3 0 _6 = _2J3 =_ 3 Jr Jr The average rate of change is 2J3 Jr . f

i\

2

(g o f )(x) = cot(-2x)

(�) - f ( O) _ tan(2 . JrI 6) - tan(2 . 0) �6 - O

Jr l 6

Jr l 6 Jr The average rate of change is 6J3 Jr . 45.

( J o g)(x) = tan(4x)

49.

x

a.

Consider the length of the line segment in two sections, x, the portion across the hall that is 3 feet wide and y, the portion across that hall that is 4 feet wide . Then, and SIn. f) = -y4 cos f) = -x3 3 4 x = -y = - cos f) sin f) Thus, L x + y = cos3 f) + sin4 f) = 3 sec f) + 4 csc f) =

--

--

406



.

Section 7. 8: Phase Shift; Sinusoidal Curve Fitting

h.

c.

Let

25

1'; =

-cos3 x + -.smx-4 . -

u

O lk=== ===;;!1 1 o 25

u

[!, ! J [ J 4 4 (�,O} ( } e;,-4} e;,o)

2!. 2

4

y

=

5.

tan x

-cot(x

y

= 2COs

(3X + %)

Amplitude: I A I = 1 2 1 = 2 Period: T = 21tOJ = 21t3

(-%) =-�

x

y=

1 1 141 4 OJ

is least when e "" 0.83 . 3 + L "" cos ( 0.83 ) sin ( 0 . 83 ) ",, 9.86 feet . Note that rounding up will result in a ladder that won ' t fit around the comer. Answers will vary. L

phase shift y = 4sin(2x - 1t) Amplitude: A = = 21t 21t Period: T = = - = 1t 2 OJ Phase Shift: .t = �2 Interval defining one cycle: + T = � , 3; Subinterval width: T 7r Key points: 3; 4 ( 7r, 0 ) , '

Use MINIMUM to find the least value:

" i n i l''Iurl')

51.

1.

3.

o X=.Bn2 729 �V=9.B6S6626 o

d.

Section 7.8

Phase Shift: .tOJ = 3 6 Interval defining one cycle:

[!,! +TJ=[-�,�] Subinterval width: 27r 3 4 -4 6

+ �)

Yes, the two functions are equivalent.

-= T

/

=-

7r

407

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publi sher.


Key points: (-� ' 2), (0, 0) , (�'-2) (�,o), (;,2)

9.

y = 4sin(7U+2)-5 Amplitude: 1 A 1 = 1 4 1 = 4 Period: 1t 2 Phase Shift: t = -1t = _ 21t Interval defining one cycle: [�,� +T] = [-�,2-�] Subinterval width: T 2 -=-= 4 poi4 nts:2 Key (_2 ) , (!_2 2 ) ( 1 _ 2 -5 ) (%-�,-9), ( 2 - �,-5) I1J

I1J

1

7.

y = -3sin (2X+�) Amplitude: 1 A 1 = 1 -3 1 = 3 Period: T = 21t = 21t2 = 1t 1t Phase Shift: t = -22 = _ .::4 Interval defining one cycle: [ � , � + T ] = [ � , 3: ] Subinterval width: T 4 points: 4 Key (- : ,0) , (0, -3) , ( : , 0) , (;,3 ) , e: , 0)

Jr

I1J

'

-

5

Jr

'

.2

'IT

,

Jr

'

,

y

(1 _ 1. -1)'

I1J

-1

2

,

_

Jr

11.

y =3COS(7U-2)+5 Amplitude: 1 A 1 = 1 3 1 = 3 Period: 1t Phase Shift: t = 21t Interval defining one cycle: [�, � +T] = [�,2+ �] Subinterval width: T 2 Key4 poi4 nts:2 I1J

x

I1J

1

408

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Section 7. 8: Phase Shift; Sinusoidal Curve Fitting

( � ,8J , (� + � , 5 J . ( 1 + � , 2J . (% + � ,5 J . ( 2 + � , 8J

15.

¢

13.

x

(

=

1

Assuming A is positive, we have that Y = A sin(wx - ¢) = 2 sin(2x - 1) 17.

-1

�=�

[w = t2 = _2

( ; , 8)

y

IA 1 = 2; T = 1[;

( )J = 3 sin ( 2X - % )

)

Y = -3sin - 2X + % = -3sin ( - 2X - % Amplitude: 1 A 1 = 1 3 1 = 3 Period: T = 27t = 27t2 = 7t 7t Phase Shift: [w = 2"2 = �4 Interval defining one cycle: + T = : ' 5: Subinterval width: T Jr 4 4 Key points: (:,0J. ; ,3 , (Jr, -3) ,

(

19.

W

y

( _ � , 3)

5

)

y = 2 tan(4x - Jr) Begin with the graph of y = tan x and apply the following transformations: 1) Shift right units [Y = tan (x )] Jr

[ �, � ] [ ]

( ] e:,o] ,

I A I = 3; T = 31[; [w = -3 27t = 27t = -2 [ = [ = -w=T 37t 3 w 2 3 3 ¢ = _ .!.3 . 3.3 = _ 3.9 Assuming A is positive, we have that Y = A sin(wx - ¢) = 3sin % x+ %

- Jr

2) Horizontally compress by a factor of .!.4 [ Y = tan (4x - Jr) ] 3) Vertically stretch by a factor of2 [ Y = 2 tan ( 4x ) ] - Jr

e:,oJ

( T ,OJ

( � , 3)

-5 409

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21.

( )

Y = 3CSC 2x - : Begin with the graph of = csc x and apply the following transformations: 1)

Shift right

:

Y

units

[Y ( x-:)1 =

csc

2) Horizontally compress by a factor of ..!.2

25.

3) Vertically stretch by a factor of 3 y = 3CSC 2X - :

[

( )]

JJ I lUI ,I I I

5 (- 0; , 3)

1

I Y 1 : 1 1

1

1

1 (-*1 ' -3)

1

3

1 1

I 1 (3: , 3)

1

1

'IT

"2

I

1

1

= -sec( 27l"x + ) Begin with the graph of y = sec x and apply the following transformations: 1) Shift left 7l" units [ Y = sec( x + 7l" )] 2) Horizontally compress by a factor of _27l"1_ [ Y = sec( 2Jrx + 7l" ) ] 3) Reflect about the x-axis [Y = -sec( 27l"x + 7l" ) ] y

1 1 1

1

1

( - 1,

I

23.

{ )

y = -CO 2X + � Begin with the graph of y = cot x and apply the following transformations: 1) Shift left units = cot x +

� [Y

(

7l"

�)]

27.

2) Horizontally compress by a factor of ..!.2 3) Reflect about the x-axis

1 1 1

y

1) )1

i\ TnT, ] , (

1

3

I, 1

1 1

1

}

I

(1 , I )

x

b-, - 1 )

1

I (t) = 120sin 307t t - � t � O 27t = � Period: T = 27tOJ = 307t 15 Amplitude: 1 A I = 1 120 1 = 120 7t Phase Shift: tOJ = 307t "3 = � 90

410

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exist. No portion of this material may be reproduced, in any fonn or by any means, without pennission in writing from the publisher.

Section 7.8: Phase Shift; Sinusoidal Curve Fitting

29.

a.

y 60

50

•

40

30

•

•

•

•

••

•

•

•

•

•

20 0 I 2 3 4 5 6 7 8 9101112

b.

o

x

31.

56.0 - 24.2 = 3 1 .8 = 1 5.9 Amplitude: A 2 2 + 24.2 80.2 56.0 = 40. 1 Vertical Shift: 2 2 = 21t 1t 12 6 Phase shift (use y 24 .2, x = 1):

a.

=

-

.

.

y 80 •

50

40

30

=-

•

•

•

•

.

.

. .

.

.

•

•• •

•

.

•

.

•

0 2 01 2 3 456 7 8 9 lOll 12

(� . 1 - ¢) + 40. 1 1 5 .9 sin (� - ¢ )

b.

24.2 = 1 5 .9 sin =

.

60

=

- 1 5 .9

..

20

70

=

0)

60

c.

13

x

.4 - .5 49.9 = 24.95 Amplitude: A = 75 25 = 2 2 + 100.9 25.5 75.4 = = 50.45 Vertical Shift: 2 2 21t 1t 12 6 Phase shift (use y 25.5, x = 1): 0)=-=-

=

- � = �- ¢ 2 6 21t ¢= 3 Thus, y 1 5 .9 sin

(� x- 231t ) +40. 1 y = 1 5 .9 Sin [� ( x - 4 ) ] + 40. 1 . =

c.

- 24.95

=

-

or

-� = �- ¢ 2 6 ¢ = 21t 3 Thus, y = 24.95 Sin

y 60 50

40 x 20 �� 0123456789101112

d.

(� ·1 - ¢) + 50.45 24.95 sin ( � ¢ )

25.5 = 24.95 sin

c.

y = 1 5 .62 sin ( 0.5 17x- 2 .096 ) + 40.377

(� x_ 231t ) + 50.45 y = 24.95 sin [ � (x - 4 ) ] + 50.45 .

or

y 80

70

plnl(e9

60

'!:I=a>l<sin(bx+c)+d a=15.61996209 b=. 517364549 c= -2.095883506 d=40.37675696

50

40

30 20

0123456789101112

x

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d.

y = 25 .693 sin(0.476x - 1 .8 1 4) +49.854

3 5 . a.

5��:�in(bx+c)+d a=25.6934405 b=.4764311009 c=-1.813776523 d=49.85374426 80

e.

33.

a.

b.

1/\13 20

3.6333 + 12.5 = 1 6 . 1 333 hours which is at 4:08 PM. Amplitude: A = 8 .2 - (- 0.6) = � = 4.4 2 2 7.6 .2 8 0.6) +( = - = 3.8 VertlcaISh'fit: 2 2 2 rc = -47Z' rc = (1) = -12.5 6.25 25 Phase shift (use y = 8.2, x = 3 .6333):

(

1

.

Amplitude: A = 13.75 - 1 0.53 = 1.6 1 2 +10.53 3.75 1 = 1 2. 14 Vertical Shift: 2 2 rc (1) = 365 Phase shift (use y = 1 3 .75, x = 1 7 2): 1 3.75 = 1 .6 1 sin 2 rc . 1 72 - tP +12. 14 365 1 .6 1 = 1 .6 1 Sin 2 rc . 1 72 - tP 365 1 = sin 344 rc - tP 365 344 rc -tP �= 2 365 tP 1 .3900 Thus, y = 1 .6 1 sin 2 rc X- 1 .39 +12 . 14 or 365 Y= 1 .6 1 Sin 2 rc ( X- 80.75 ) +12. 14 . 365 �

( �� · 3.6333 - tP) +3.8 4.4 = 4.4 Sin (�� . 3.6333 - tP ) . ( 14.5332 rc ) 1 = sm 25 8.2 = 4.4 Sin

b.

c.

rc 40.5332 rc tP 2 25 tP 0 . 2555

(�� x - 0.2555 ) +3.8 or y = 4.4 Sin [ �� ( X- O.5083 ) ] +3.8 .

10

y 9

O

7 5

d.

3

1 2.42 hours

� �-L��-L-L��__ 140 2 80 420 x

The actual hours of sunlight on April 1 , 2005 was 12.43 hours. This is very close to the predicted amount of 12.42 hours.

x 2

Y = 4.4 Sin �

)

y

Thus, Y= 4.4 Sin

d.

(

20

�

1

)

) )

[ ] Y = 1 .6 1 Sin [ 21t ( 9 1 - 80.75 ) ] +12. 14 365 �

"' 'I'

c.

( (

( �� (1 6 . 1 333 ) - 0.2555 ) + 3.8

8 .2 feet

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37.

a.

0)=

-

( (

) )

(- do)

b.

c.


Amplitude: A = 1 9.42 - 5 .47 = 6.975 2 19.42 + 5 .47 = 1 2.445 Vertical Shift: 2 2 7t 365 Phase shift (use y = 1 9.42, x = 1 72): 19.42 = 6.975 sin 2 7t . 1 72 - rp + 12.445 365 6.975 = 6.975 sin 2 7t ·172 - ¢ 365 344 7t 1 - sm . --'I' 365 2: = 344 7t _¢ 2 365 ¢ :d.39 Thus, y = 6.975 sin 2 7t x - 1 .39 + 12.445 365 or y = 6.975 sin 2 7t ( X- 80.75 ) + 1 2.445 . 365 Y

(

[

(

)

) ]

= 6.975 sin 2 7t ( 9 1 ) - 1 .39 + 1 2.445 365 "'" 13.67 hours y

1.

135° = 1 3 5 · � radian = 3 7t radians 4 1 80

3.

1 8° 1 8 . � radian = .2:.. radian 10 1 80

5.

3 7t = _ 3 7t . 1 80 - degrees = 1 35 ° 4 4 7t

7.

5 7t = - 5_ 1 80 degrees = - 450 ° 7t . -2 2 7t

9.

. -7t = I - -1 = -1 tan-7t - sm 4 2 2 6

=

11.

3.J2 - 4-J2 - 4 · . 45 ° - 4 tan -7t = 3·./3 = -J3 3 sm 6 2 3 2 3

13.

6 cos 3 + 2 tan

;

(-%) = 6 ( - �) + 2 ( -J3) = -3J2 - 2./3

( -%) - cot ( - 547t ) = sec % + cot 5;

=

=

2+1 3

1 5.

sec

1 7.

tan 7t + sin 7t = 0 + 0 = 0

1 9.

cos 540° - tan( - 405° ) = -1 - (-1) = -1 + 1 0

25.

sin 50° sin 50° = sin 50° = 1 cos 40° sin ( 90° - 40° ) sin 50°

=

d.

39.

The actual hours of sunlight on April 1 , 2005 was 1 3 .43 hours. This i s close to the predicted amount of 1 3 .67 hours. Answers will vary.

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29.

sin 400° · sec( -50°) = sin( 40° +360°) . sec 50° 1= sin400.cos 50° = sin 400, sin (90°1 -50°) 1 . = sm40°· sin 40° =1 () is acute,so () lies in quadrant and sin () = i5 corresponds to the right triangle: --

31.

I

b=4 �='5 a

35.

(3

Using the Pythagorean Theorem: a2 +42 = 52 a2 = 25 -16 = 9 a =..[9 = 3 So the triangle is:

b=4�'=5 aadj=33 cos() =-hy p =-5

33.

()
0 y 2 = 4 - 1 = 3, Y > 0 Y = .j3 .j3 tan fJ = = .j3 1

O x2 = 25 - 9 = 1 6, x > O x=4 4 tan a = -3 cos a = - , 4 5

( 2$ f + y 2 = 5 2 ,

( )

� - ( - �) = % = 2 1 + (� () �) !

. a = -3 , 0 < a < -1t sm 2 5 $ 2 - -1t < fJ < 0 cos fJ = -5 ' 2 y

sine a + fJ) = sin a cos fJ + cos a sin fJ $ $ = � . 2 +i . _ 5 5 5 5 6$ - 4 $ 2 $ 25 25 cos(a + fJ) = cos a cos fJ - sin a sin fJ 2$ _ � . _ $ =i. 5 5 5 5 8$ + 3 $ 1 1 $ 25 25 sine a - fJ) = sin a cos fJ - cos a sin fJ 2$ _ i . _ $ =L 5 5 5 5 2$ 1 0$ = -6$ + 4$ = -5 25 25 tan a - tan fJ tan( a - fJ) = ----'--1 + tan a · tan fJ

0 , 0 a �2 and so (�-a) 2 and fJ both lie in the interval (0, � ) . Then cot (� -a ) = cot fJ implies % - a = fJ or a = --fJ 2 . Thus, tan_ 1 (�1 ) = "2 -tan-I ' f v > 0 . 7:

a

93.

7:

a

'; = 16sin x(cos x + 1) and use the MAXIMUM feature:
0 Find the intersection of >'; = 6 sin x - eX and Y2 = 2 : 6

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493 Education , Inc . , Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright laws as they curren tly

exist. No portion of this material may be reproduced, in any form or by any mean s, without permi ssion in writing from the publi sher.

Chapter 8: Analytic Trigonometry

Ch apter 8 Test 1.

Let () = sec - I

4.

(�) . We seek the angle (), such

1(1 -1 (x)) = tan ( tan- I x) = x . Since the domain of the inverse tangent is all real numbers, we can directly apply this equation to get tan tan- I � = � .

that 0 � () � re and () "* 2're ' whose secant equals � . The only value in the restricted range with a secant of � is � . Thus, sec - I 2.

Let () = sin - I

( )

(�) = � .

5.

(-,,?) . We seek the angle () , such

3.

(

.

t l

)

sin - I sin 1 �re follows the form of the equation rl (I( x)) = sin- I ( sin (x)) = but because re re , we cannot l Ire . . ' I --,-- IS not m the mterva 2 2 5 directly use the equation. We need to fmd an angle () in the interval re , re for which sin l Ire = sin () . The angle S 2 2 l Ire is in quadrant I. The reference angle of S l Ire = sm. -re . S mce l Ire . re . -re IS. m. ' - IS - and sm S S S

[ ] x ,

6.

[_ ]

"lV ,

_

( %) sec [ COS- I ( -%)] = sec ()

Let () = COS- I

-

.

- 4' = --43 7. sin-I ( 0.382 ) :::; 0.392 radian 3

[-; , ; ] , we can apply the equation . l lre-) = 5re ' above and get sm. ( smS

the interval

-I

© 2008 Pearson Educati on , Inc . , Upper Saddle

y

y

cos ()

S

S

cot ( csc- I .JiO ) Since csc- I () = !... = 10 re2 -< () -< re2 ' let = J10 and = 1 . Solve for x: x2 + e = (J1O f x 2 + I = 10 x2 = 9 x=3 () is in quadrant I. Thus, cot ( csc -I J1O ) = cot () = � = T = 3 . r

. equals -� The that --re2 0 .

S ection 9.4

False a = 2, c =4, B = 45° = �acsinB = �(2)(4)sin45°"" 2. 8 3 7. a = 2, b = 3, C = 95° = �abSinC = �(2)(3)sin95°"" 2.99 9. a = 6, b = 5, c = 8 s = 21 (a+b+c) =21 (6+5+8) = 219 = .Js(s -a)(s -b)(s -c) = C;)(i)(�)(%) = �3��1 "" 14. 9 8 11. a = 9, b = 6, c = 4 s = 21 (a+b+c) = 21 (9+6+4) = 219 = .Js(s -a)(s -b)(s -c) = C;)( �)(i)C21 ) = �1 ��3 "" 9. 5 6 13. a = 3, b = 4, C = 40° = � absin C = �(3)(4)Sin 40°"" 3. 8 6 15. b = 1, c = 3, A = 80° = �bcsin A = �(I)(3)sin800"" 1.48 17. a = 3, c= 2, B = 110° = �acsin B = �(3)(2)sin 1 10°"" 2. 82 19. a= 12, b = 13, c = 5 s = 21 (a+b+c) = 21 (12 +13+5) = 15 = .Js(s -a)(s -b)(s -c) = �(15)(3)(2)(10) = .)900 = 30 3. 5.

K

K

.

�

47.

�

K

. C l -cosc sm-= 2 2

K

_(a2 -2ab+b2 _c2) 4ab

49

c)( a -b -c) = "/-(a -b+4ab (a -b+c)(b+c-a) 4ab (2s -2b)(2s -2a) 4ab 4(s-b)(s-a) 4ab (s-a)(s-b) ab - 51. Answers will vary.

K

K

K

K

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Chapter 9: Applications of Trigonometric Functions

Find the area of thec =lo75t using Heron's Formula: a= 100, b= 50, 225 s = "21 (a+b+c) = 21 (100+50+75) = 2 = �s(s-a)(s-b)(s-c) ( 2�5 )( 2� )( 1 �5 )( 7� ) 52,734,375 16 ",, 1 815.4 6 Cost = ($3)(1815.4 6) = $5446. 3 8 37. Divide home plate into a rectangle and a triangle.

35.

a = 2, b = 2, c = 2 1 =3 s = 21 (a+b+c) = 2(2+2+2) = �s(s-a)(s-b)(s-c) = �(3)(1)(1)(1) = .fj "" 1. 73 23. a = 5, b = 8, c = 9 s = 21 (a+b+c) = 21 (5 +8+9) = 11 = �s(s-a)(s-b)(s-c) = �(11)( 6)(3)(2) = .J396 "" 19.90 sin -A = sinB . we know 25. From the Law of Smes a b- . asinB Solvmg for b, so we have that b = sm-A- ' Thus, sin B )sm. C = 21 ab sm. C = 21 a ( asinA a2 sin BsinC 2sin A 27. A= 40°, B = 20°, a = 2 C = 180° A-B = 180° - 40° - 20° = 120° 22 sin 20° . sin 120° "" 0.92 = a2 sinBsinC 2sin40° 2sinA 29. B= 70°, C= 10°, b=5 A= 180°- B -C = 180°- 70°- 10°= 100° n AsinC 52 sin 100°· sin 1 0° "" 2. 2 7 = b2 si2sinB 2sin 70° 31. = 110°, C = 30°, c = 3 B = 180°- A-C = 180°- 110°- 30°= 40° AsinB 32 sin1 �Oo. sin40° ",,5 .44 = c2 sin2sinC 2sm30° 33. Area of a sector = 21 r2 (} where () is in radians. () = 70 ' 1801t = 1871t A = L2 82 . 71t18 = 1121t9 A = � . 8·8 sin 70° = 32 sin 70° = 11:1t -32sin700 "" 9. 03 As

21.

K

K

K

.

.

17"

= = (17)(8. 5) = 144. 5 in2 U sing Heron's formula we get = �s(s-a)(s - b)(s -c) 1 12 +17) = 20.5 1 +b+c) = -(12+ s = -(a 2 Thus,2 =� = �(20. 5 )(8. 5 )(8. 5)(3. 5 ) = .J5183. 9 375 "" 72. 0 sq. in. So, + = = 144. 5 +72. 0 = 216. 5 in2 The area of home plate is about 216. 5 in2 . ARectangle

------

ATriangle

K A

ATotal

K

Sector

Triangle

egment

lw

Ani,ogle

-

K

8.5"

8.5"

K

(20.5)(20.5 -12)(20.5 -12)(20.5 17) -

ARectangle

ATriangle

ft2

fe

fe

518 © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Section 9.4: Area of a Triangle

The sector.area is the sum of the area of a triangle and a = �r . r sine1t - B) = �r2 sine1t - B) = 2"1 r2 B As = � r2 sin ( 1t - B) +�r2 B As = �r2 (sin(1t - B) +B) = � r2 ( sin 1tCOSB -cos 1tsin B+B) = �r2 ( 0 . cos B -(-1) sin B + B) = � r2 (B sin B) 41. Use Heron's formula: a = 87 , b = 190 , = 173 =-21 ( a + b+ ) =-21 ( 87 +190+173) = 225 = � ( )( - b )( ) = �225 ( 225 -87 )( 225 -190)( 225 -173) = �225 ( 138 )( 35 ) ( 52 ) = �56,511,000 "" 7517.4 The feet ofbuilding groundcovers area. approximately 7 517. 4 square Area /).OAC = 2"1 1oc1·1ACI 43. = L2 Io1c1.1A1CI . = 2'1 cos asma . a = 2'1 smacos Area /).OCB = 2'1 1 OCI·IBCI I 2 .1oOBcI1.1BC = 1.2 ,1OB1l lOBI = �IOB12 cos fJsin fJ = �IOBI2 sin fJcos fJ

39.

c.

ArriangJe ec!or

haded region

d.

e.

+

I OB Isin(a+ fJ)

cs o a . sm( cs o

c

c

s

K

s

s -a

s

Area /).OAB = 2"11 BD1·1 OAI = 2"1IBDI'1 = 21 " 1 OBI·II BOBDII = �IOB I sin(a + fJ) lIOC i cos A = OA I = ' oc1.1 OBI =IOBI cos BlOCI 1 IocI lOBI Area /).OAB = Area /).OAC+Area /).OCB � � � 12 -fJ a + =

sn i a cs o a+ I OB

sn i fJ c o s fJ

. cs o 2a . fJ) =sma c osa+--2-smfJ c osfJ cs o fJ

cs o a . . c osfJ . sm( a+ fJ) =--sma c s o a+- sm fJ c osfJ cs o a cs o fJ

s -c

a = a 45. The grazing area must be considered in sections. Region AJ represents three-fourth of a circle with radius 1 00 feet. Thus, AJ = � 1t ( I00l = 75001t "" 23, 5 61.94 ft2 Angles are needed to find regions � and A3 : (see the figure) sine

a.

+ fJ)

sin

cs o fJ +c s o asin fJ

�c B

10 ft

A

In MBC, LCBA = 45°, AB = 10, AC =90. Find LBCA: sin LCBA sinLBCA 90 10 sin 45° sinLBCA 90 10 sinLBCA = IOsin45° 90 "" 0. 0786 LBCA = sin -J COS�450 ) "" 4.51° LBAC = 180°-45°- 4. 5 1°= 130.49°

b.



LDAC = 130.49°- 90°= 40.49° A3 = i (l O)(90)sin40.49 ", 292.19 fe The angle for the sector � is 90°- 40.49°= 49. 5 1° . A2 = i ( 90 ) 2 ( 49. 5 1 . 1 ;0 ) '" 3499. 66 fe Since the cow can go in either direction around the bam, both � and � must be doubled. Thus, the total grazing area is: 23,561.94+2(3499. 66) +2(292.19) ", 31,146 47. Perimeter: = a+b+c = 9+10+17 = 36 Area: s =-21 (a+b+c) =-21 ( 9+10+17) = 18 = �s( s -a)( s-b)( s-c) = �18( 18 -9 )( 18 -10)( 18 -17) = �18( 9 )(8 )( 1 ) ..11 296 = 36 Siequal, nce thetheperimeter and area are numerically given triangle is a perfect triangle. Perimeter: = a+b+c = 6+25 +29 = 60 Area: s =-21 (a+b+c) =-21 ( 6+25 +29) = 30 = �s( s-a)( s-b)( s-c) = �30 (30 -6)( 30 -25 )( 30 -29 ) = �30( 24 )( 5 )( 1 ) = ..13 600 = 60 Since thetheperimeter and area are numerically equal, given triangle is a perfect triangle.

49.

0

K

.

_

51.

P

=

53.

P

fi

_

__ .

K

b.

K

--

fe

a.

We know = i ah and = i abSin C , whic h means h = bsinC . From the Law of Sines, we asinB sinB know sinA a = -b- ' so b = SIn-A- . There ore, asisinn AB )Slll. C - a sin B sin C . h - (-sin A . -A ·Slll. -B C·Slll C 2 2 cosC 2_ cot 2 = = ----;===r=== Slll-2 c· ac C

C ab r ac (s-a)(s-b) c ab(s-a)(s-b)(s-c)2 r abc2 (s-a)(s-b) 2 = �. s-c = �r �(S-C) c2 r c s-c r = Area MOQ+ Area MOR + Area !1QOR 1 +-rb 1 +-ra 1 = -rc 21 2 2 =-r2 ( a+b+c) = rs Now, = �s(s-a)(s -b)(s-c) , so rs �s(s-a)(s-b)(s-c) -b)(s-c) r = --'-�s(s--a)(s -----s K

K

K

=

520 © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyri ght laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Section 9.S: Simple Harmonic Motion; Damped Motion; Combining Wa ves

S ection 9.5 1. 3. 5. 7. 9. 11. 13.

21.

d(t) = cos ( 2t) e

-tlTC

y

= 2:r4 = :r2 simple harmonic ; damped d = - Scos ( ttt) d = -6cos ( 2t) d = - Ssin (7tt) d = -6sin ( 2t) d = Ssin(3t) Simple harmonic S meters 27t seconds 3 � oscillatio n/second 27t d = 6cos(7tt) Simple harmonic 6 meters 2 seconds .!. oscillatio n/second 2 d= -3Sin (k t) Simple harmonic 3 meters 47t seconds 1 47t oscillation/second d = 6+2cos(27tt) Simple harmonic 2 meters 1 second oscillatio n/second lsi s· '

2'1l'

23.

d( t)

-tI21t =e

X

cost

y

a.

b. c.

d.

15.

25.

j ( x ) = x+cosx y 2'7T

a.

b. c.

d.

17.

-'IT

a.

27.

b.

y

c.

d.

19.

j ( x ) = x-sin x

_ _

a.

b. c.

d.

-21T'

I



29.

j ( x) = sinx+cosx

37. a.

y 2

d

d

(0. 8)2 = - e- 0.St/2 (10) cos [ (-23n)2 - --4(10)2 J = _5 e-0.St/2 0 cos [ 49n2 - 0.40064 ] t

5

t

b. -2

31.

j ( x ) = sinx+sin ( 2x)

39. a.

y

b.

c.

y

=

d

d

(0. 7) 2 = _10 e-0.7t/2 (2 5) cos [ ( 2 n)2 - 4(25) 2 0.49 ] = _10 e-0.7t/50 cos [ 425n2 - 2500

d. t

5

t

J e.

b. 41. a.

o IE+-t"+-+-I-+'-t-'I-++

35. a.

d

d

20

sin(2x)

-2

33. a.

Damped motionfactor with aofbob0. 7 ofkg/sec. mass 20 kg and a damping 20 meters leftward

b.

The displacement of theis bob of the second oscillation aboutat 18.the3start 3 meters. The displacementt/ of0 the bob approaches zero, since e-{J·7 4 0 as Damped motionfactor with ofa bob0 . 6 ofkg/sec. mass 40 kg and a damping 30 meters leftward �

t � 00 .

c.

(0.6)2 - - 18 e-0.6 t/2 (3 0) cos [(-2n)2 - --4(30/ J 0. 3 6 ] = _18 e- 0.6 t/60 cos [ �4 3600

o 11+-+-+++4+..Ir--l...... 1

t

_

-20

t

d.

b.

e.

The displacement of theis bob of the second oscillation aboutat28the.4start 7 meters. The displacement6t/Sof the bob approaches zero, since e-{J· O 0 as �

t

� 00 .


Section 9.5: Simple Harmonic Motion; Da mped Motion; Combining Wa ves

43. a.

b.

Damped motion with a bob of mass 1 5 kg and a damping factor of 0.9 kg/sec. 1 5 meters leftward

-15

d.

The displacement of the bob at the start of the second oscillation is about 1 2.53 meters.

c.

The displacement of the bob approaches zero, since e-O·9t130 � 0 as t � 00 .

O�320 f\ 1.25

The maximum displacement is the amplitude so we have a = 0.80 . The frequency is given by (1)

f = - = 520 . Therefore,

(1)

Irlt�rstctio� X=.�SlS216S

= 1 0401Z" and the

I I

f = - = 440 . Therefore,

OJ = 8801Z" and the 21Z" movement of the tuning fork is described by the equation d = 0.0 1 sin ( 880m ) .

rs�o

r\

[\

�

51.

L"'-

1. � ."---. .L

I�t. � �=2.1BBSSQ6

-1.25

;c(

.

25

"-

I / V=.�

V=.�

3.2 0

��

.....

�

I !

"il'lirvlu X=2.966�S23

-1.25

�

"'-

V= -3699522

Let r; = sin ( 21Z" ( 852 ) x ) + sin ( 21Z" ( 1209 ) x ) .

-1

On the interval O:=;t :=; 3 , the graph of V touches the graph of y = e-t1 3 when t = 0, 2 . The graph of V touches the graph of y = _ e-t1 3 when t = 1, 3 .

53.

3.2

'"

2.5

b.

3.2

'--.

1 25

rs�

V

1

'I \

Irlt'-':rs ol'l X=1.75�9n� -

1 . 25

o -"-

"-

-1.25

-1.25

(1)

�

IMt � X=.666�B6�2 v= -.�

V=.�

-1.25

The maximum displacement is the amplitude so we have a = 0.01. The frequency is given by

49. a.

1 25

\ \

21Z" motion of the diaphragm is described by the equation d = 0.80 cos ( 1 040m ) .

47.

We need to solve the inequality -0.4< e-tl 3 cos ( m )< 0.4 on the interval o:=; t :=; 3 . To do so, we consider the graphs of y = -0.4, y=e-tI 3 cos ( 1Z" t ) , andy = O.4 . On the interval O:=;t :=;3 , we can use the INTERSECT feature on a calculator to determine that y = e -t1 3 cos ( m ) intersects y = 0.4 when t '" 0.35, t '" 1 .75 , and t '" 2. 1 9, y = e -t1 3 cos ( 1Z"t ) intersects y = -0.4 when t '" 0.67 and t '" 1 .29 and the graph shows that -0.4< e-tl 3 cos ( 1Z"t )< 0.4 when t = 3 . Therefore, the voltage V is between -0.4 and 0.4 on the intervals 0.35 < t< 0.67 , 1 .29
z

3

r = � x2 + l = � = .fi tan 0 = 2:'.x = 1 0= 45° The polar form of z = 1 + i is z = r (cosO + i sinO) = .fi (cos 45° + isin 45°)

. Imaginary axis

--_ -3'----t-1r-'---�3

---

Real

axis

-3

17.

Imaginal)' axis 1

Real -----+---LJ'---axis 1

r = � x2 +l = � 4 2 + ( _ 4) 2 = F32 = 4.fi tan 0 = 2:'.x = -4 4 = -1 0= 3 1 5° The polar form of = 4 - 4i is = r ( cos 0 + i sin 0) = 4.fi ( cos 3 15° + i sin 3 1 5°) z

z

Imaginary axis 4

13.

2

) f

r = �x 2 + y 2 = ( .J3 + ( _ 1) 2 = -/4 = 2 -1 = -.J3 tanO = -yx = 3 .J3 0= 330° The polar form of = .J3 - i is z = r ( cos 0 + i sin 0) = 2 ( cos 330° + i sin 3300)

-+-""*--1.l...-----L___+_

___

Real axis

z

550


Section 10.3: The Complex Plane; De Moivre's Theorem

19.

r = � x 2 + y 2 = � 3 2 + ( _ 4) 2 = 55 = 5 tanB= yx = -34 B::::: 306.9° The polar fonn of z = 3 - 4i is z = r ( cosB+ i sinB) = 5 ( cos 306 . 9° + i sin 306 . 9°) -

-

--�--�--r--u----��

3

2 9.

0 . 2 ( cos 1 00° + i sin 100°) ::::: 0.2 ( -0.1736 + 0 . 9848i) ::::: -0.035 + 0. 197i

31.

Imaginary axis Real

33.

aXIS

( cos 3; + i sin 3; ) = 3 (0 - l i) = -3 i

27.

(

z

i[cos (40° - 20°) + i sin(40° - 20° )] = � (cos 20° + i sin 20°)

35.

B::::: 123 . 7° The polar fonn of z - 2 + 3i is z = r (cosB+ isinB) = .J13 (cos 123 . 7° + i sin 123 . r) =

zw

Imagi.nary aXIS

z

=

3 (cos 130° + i sin 130°) 4( cos 2700+ i s in 270°)

%[ cos (130° - 270°) + i s i n ( l 300- 270°)] %( cos (-140°) + isin (-140°») %[ cos (360°-140°) + i s in (3600-1400)] = %(cos 2200+ isin 220°) =

-1

[ i � i)

=

2( cos 120° + i sin 1200) = 2 - +

(

3· 4[ cos (1300+ 270°) + isin (1300+ 270°)] = 12( cos 4000+ isi n 400°) = 12[cos (400°- 360°) + i s in (400°- 360°)] = 12( cos 40° + isin 40°)

=

Real --�-----+--�--�� axis -2 2

25.

= 3( cos 130° + i s in 130°). 4( cos 2700+ isin 270°) =

w

23.

= 2 ( cos 40° + i sin 400) . 4( cos 20° + isin 200) = 2 . 4[cos(400 + 200 ) + i sin(400 + 20°)] = 8 ( cos 60° + i sin 60° ) 2 ( cos 40° + i sin 40° ) = 4(cos 200 + i sin 200)

=

r = � X 2 + y 2 = � (_ 2) 2 + 3 2 = .J13 _l tanB= zx = � -2 = 2

3

::::: 1 . 970 + 0 . 347 i

zw

w

21.

)

2 cos l� + i sin t8 ::::: 2 (0 . 9848 + 0. 1736i)

= -I + .J3 i

) [ � -� )

4 cos 741r + i sin 741r = 4 i = 2J2 - 2J2 i


551 All rights reserved. This material is protected under all copyright laws as they currently


Chapter 10: Polar Coordinates; Vectors

41.

37.

[ 4 ( cos 400 + i sin 400)r := 43 [cos (3 · 40°) + i Sin(3 . 400)] = 64 ( cos 120° + i sin 120°)

( � � i)

= 64 - +

:=-32 + 32.J3 i 43.

1t

= cos 40 + 39.

. I

' sm

1t

40

z = 2 + 2i r = ../2 2 + 2 2 = .J8 = 2 .fi tanB= �2 = l B = 45° z = 2.fi (cos 45° + i sin 45°)

= 32 (0 + l i) = 32 i 45.

f

:=27

-1 = -.J3 tanB= 3 .J3 B = 330° w = 2 ( cos 3300 + isin3300) w =2-/2 (cos 45°+ isin 45°)· 2( cos3300+ isin 3300) = 2-/2· 2[ cos ( 45°+ 330°) + i sin (45°+ 330°)] :=

:=

(� + � ) i

27 +--1 27.J3 . =2 2 47.

z

:=

[.J3 ( cos 1 0° + i sin 10°) r :=( .J3 f [ cos( 6 . 10° ) + i sine 6 . 10° ) ] :=27 (cos 60° + i sin 60°)

w=.J3 - i r = .J3 + ( _ 1) 2 = -14 = 2

�(

[2 ( cos 1� + i sin 1�)r := 25 [ cos ( 5 . 1� ) + i sin ( 5 . 1� ) ] :=32 ( cos % + i Sin % )

[� (cos �� + i sin ��)r = ( �r [ cos ( 4 . �� ) + i sin ( 4 �� ) ] 31t + 1. SIll , 3n = 2S( cos4 4) = 25 ( - V; + V; i J .

4-/2 ( cos375°+ i sin 375°) 4-/2[ cos (375°- 360°) + i sin (375°- 360°) ] 4-/2 (cos 15° + i sin 15°)

z := 2.fi ( cos 45° + i sin 45°) w 2(cos 3300+ i sin 3300) :=2 [cos ( 45° - 330°) + i sin( 45° - 330°)]

f

2S.fi + --1 25 .fi . = --2 2

:=.fi[ cos ( - 285°) + i sin( - 285°)] :=.fi[ cos( 360° - 285°) + i sin (3600 - 285°)] = .fi(cos 75° + i sin 75°) 552 © 2008 Pearson Education,



Section 10.3: The Complex Plane; De Moivre's Theorem

49.

l-i r = �12 + (_ 1)2 = .fi tanO = -1-= 17 -1 0= 4 1t 1 .=.fi( COS 47 1t + 1S..lli 47 1t ) (1-i) 5 = [.fi ( cos 74 1t + i sin 74 1t )] 5 = (.fit [ cos( 5· 74 1t ) + ism( 5· 74 1t )] 35 1t + 1S..lli 4 35 1t ) = 4.fi2 ( COS 4 = 4 .fi( - � + � i) = -4+4 i .fi-i r = �(.fit + (_ 1) 2 = ./3 -1 -.fi2 tanO =-= .fi 0"",324 .736° .fi -i "",./3(cos324.736°+ isin 324 .736°) ( .fi_ i)6 ""'[ ./3(COS324 . 7360+ ism324 . 7360)T (./3)6 [cos( 6 ·324 .736°) + isin( 6 . 324 .736°)] = 27(cos1 948.4 16°+ i sin 1 948.4 16°) "'" 27( -0.8519 + 0.5237i) ""'-23+14 . 142i

zk

Zo

51.

Z2

55.

) (

+

4-4./3i r �4 2 + ( - 4 ./3f = J64 = 8 tan0 = -44./3 ./3 0= 300° 4 -4./3i = 8( cos 300°+ ism3000) The four complex fourth roots of 4 -4./3i 8( cos 300°+ ism3000) are: =

=-

=

zk

Zo

zl

Z2

=

Z3

57.

53. l+i

[ (

) (

3600k 3000 3600k =� cos 3000 -4- +4- +isin -4-+- 4=�[cos(7So+ 90°k)+i sin (7So +90°k)J =�[cos(7So+ 90°.0)+isin (7So + 90°· O)J =� (cos7So+ isin7S0) =�[cos(7So+ 90°.1)+isin(7So+ 90°.1)J = � (cos 16S0+ isin 165°) =�[cos(7So+90°· 2)+isin(75°+90°· 2)J =� (cos2S5°+ isin 2SS0) =�[cos(7So+ 90°·3)+isin(7SO+90°· 3)J =� (cos345°+ isin34S0)

-16i r = �02 +( _ 16)2 = ../2 56 1 6 tanO = -16 -0 0= 2 70° -16 i = 16( cos 270°+ i sin2 70°) The four complex fourth roots of -16i = 16( cos 270°+ ism270°) are: =

r = � = .fi tanO = 11-= 1 0= 45° 1 +i = .fi ( cos 45°+i sin 45°) The three complex cube roots of + i = .fi ( cos 45° +i sin 45°) are: 1

553 «:I 2008 Pearson Education, Inc., Upper

)]

+

ZI

-I

[ (

4S0 360°k . 4S0 360°k . = ff2 J2 cos 3 +- - +/sm 3 +-33 = �[cos(ISO+120°k) isin(ISO+120°k)J = �[cos(lso+120°. 0)+isin(ISo+120°· O)J = � (cosIS0+ isinISO) = �[cos(ISo+120°·1) isin(ISO+120°.1)J = � (cos 13So+ isinl3S0) = �[cos(ISO+120°.2)+isin(ISO+120°.2)J = � (cos2SSo+ isin2SSo)

Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently


)]


3600-k ) +i sin (4 00 + 3600-k )] = � [ cos (400 + 4 4 1 [ cos ( 90° k ) + isin ( 90° k )J Zo = cos ( 90° . 0 ) + i sin ( 90° · 0 ) = cos 0° + i sin 0° = 1 + 0 i = 1 = cos( 90°. 1) + sin ( 90° . 1) = cos90° + i sin 90°= 0 + Ii = i Z2 = cos (90° . 2 ) + i sin ( 90°· 2 ) cos 180°+ i sin 180°= -1 + Oi = -1 z3 = cos ( 90° . 3) + i sin ( 90° . 3) = cos 270°+ isin2700= 0-li = -i The complex fourth roots of unity are: l, i, - I, -i .

[ (2700 3600 ) (2700 3600 )]

k 4r;;: . . 6 cos -+ -- +/sm zk = ,,10

(

4

4

-

4

k

zk

+ -4

= 2[cos 67.50 + 900 k) isin 67.5 0 + 900 k)]

+ ( Zo = 2 [ cos ( 67. so + 90° · 0) + i sin( 67. so + 90°· 0) J = 2 ( cos 67. so + i sin67. so ) z, = 2[ cos(67S+ 90 °· 1 ) + isin(67S+ 90° . 1 )J = 2 ( cos 157. so + i sinI 57. so ) Z2 = 2 [ cos ( 67. s o + 90°· 2 ) + isin( 67. s o + 90°· 2 )J = 2 ( cos 247. so + i sin247. so ) z3 = 2 [ cos ( 67. s o + 90°· 3 ) + isin( 67. so + 90°· 3)J = 2 ( cos337. s o + i sin337. so )

-

=

zl

i

=

59.

r = �02 + 12 = Ji = 1 tan 0 = -I o 0= 90° i = 1 ( cos 90° + i sin 90°) The five complex fifth roots of i = 1 (cos 90° + i sin 9 0° ) are: [ ( 900+-3600k-) +isin (5900 +-3600k5-)] zk = 1f1 cos 5 5 = I [cos(ISO + 72° k ) + i sin (ISO + 72° k )] Zo = 1 [cos(ISo+ 72° · O)+isin(ISo+ 72° · O)J = cos l S o + i sin l S o = 1 [cos ( 1 8° + 72° . 1 ) + i sin ( 1 8° + 72° · 1)J = cos 90°+ i sin 90° Z2 = 1 [ cos ( 1 8° + 72° · 2 ) + i sin ( 18° + 72° · 2 )J = cos I 62° + i sin l62° Z3 = 1 [ cos ( 1 8° + 72°· 3 ) + i sin ( 1 8° + 72°· 3)J = cos 234° + i sin 234° Z4 = I [ cos ( 1 8° + 72° . 4 ) + i sin ( 1 8° + 72° · 4 )J = cos 306° + i sin 306° 61. 1 = I+Oi r = �12 +02 = Ji= 1

Imagi.nary

I 1

f

-II \ \

63.

zl

/

�

-

-- ....

....

'-

"

"

,

\ \

\

\

"

"

'-

....

....

..--

-i

/

w = r ( 0+ i 0) w:f= n = * [ (� : )

/

/

/

I

I \ I

Real aXIs

Let cos sin be a complex number. If 0 , there are distinct nth roots of given by the formula: zk cos + 2 n + i sin + 2 n IZk

65.

/

/

/

:Ins

1 = * for all k

w,

(� : )}

= 1, ... , n -I

where k 0, 2,

Examining the formula for the distinct complex nth roots of the complex number

w= r(cosO+isinO), . . (-;;-0 + ---;;-)] , = \Inr[r cos(-;;-0 + -2kn- ) + zsm where k = 0, 1, 2, . .. , n -I , we see that the zk are 2n . spaced apart by an angle of n zk

tan O = -o =O 1

0= 0° I + 0 i = 1 ( cos 0° + i sin 0° )

n

2kn

The four complex fourth roots of unity are: 554 © 2008 Pearson Education, Inc., Upper



Section 10.4: Vectors

Iv I = 4 , then 1 3vl = 1 3 1 . 1 v I = 3· 4 = 12 . = (0, 0), Q = (3, 4) 25. v = (3 -O)i+ (4 - 0)j = 3i+ 4 j = (3, 2), Q = (S, 6) 27. v= (S-3)i+(6-2)j= 2i+4j 29. = (-2, -1), Q = (6,-2) v = [6-(-2)]i+[-2-(-1)]j = 8i-j = (1, 0), Q = (0, 1) 31. v = (0 -l)i+ (1 - O)j= -i+ j 33. For v= 3i-4j, I Ivl = �32 +(-4) 2 =55=s. 35. For v = i -j, I v I = �1 2 + ( _ 1) 2 = Ji. . 37. For v= -2i+3j, I vll = �( _ 2) 2 +32 =J13. 39. 2v+3w = 2(3i-Sj)+3(-2i +3j) = 6i-1 OJ -6i+ 9j = -j 41. I v-wl = I (3i-Sj)-(-2i+3j) 1 = I S i-8jl = �S 2 +(-8) 2 = ../89 43. I v I - Iw I = 11 3i -Sj 1 - 1I -2i+3jl = �32 + (_ 5)2 �r-(_-2)--:-2-+ --:-32 = ..f34- J13 v Si Si = -Si5 = .1 45. U = - = - = I Iv I I Si I -12S + 0 v 3i-4j 3i-4j 47. = U R = P i- 4 jl = �32 +(_ 4)2 3i-4j 55 3i -4j 5 4. 3. = -I--J S S

Section 10.4

23. If

5.

unit horizontal, vertical True

7.

v+w

1. 3.

P

P

P

P

v

9.

3v

3v

• • • _---_---__ ---l.. ....

11.

13.

v-w

.

v

3v+u-2w

3v

_

15. 17. 19. 21.

True False False True

= -F +E-D D-E = H+G

C

555 © 2008 Pearson Education,




I vI = 5, a = 60° v = I v I(cosai + sin aj) = 5 ( cos(600)i+ sin ( 60°) j) 1 . J3. ) = 5 (-I+-j 2 2 5 . 5 ../3. = -I+-j 2 2 57. Ilvll = 14, a = 120° v = Iv I(cos ai+ sinaj ) = 14 ( cos(1200)i + sin(120°) j) = 14 ( - � i+ � j) = -7i+ 7../3j 59. I Iv I = 25, a = 330° v = Iv I(cosai+ sinaj ) = 25 [cos(3300)i+ sin(330°) j] = 25 (� i- � j) 25../3. 25 . =--I--j 2 2 61. Let =(3,-1). Then = +v = (3,-1)+(-4,5)= (-1,4). The new coordinate will be (-1, 4).

55.

v= ai+ bj . We want Il vll = 4 and a= 2b. Il vll = �a2 + b2 = �(2b)2 + b2 = �5b2 �5b2 = 4 5b2 = 16 b2 = 165 4J5 + [i6 = -+ �= + b= -Vs J5 5 a= 2b = 2 ( ± 4�) = ± 8� 8J5. 4J5 . or V= --I--j 8J5. 4J5. v=-I+-j 5 5 5 5 53. v= 2i-j, w= xi+3j, Ilv+wll = 5 I v+wI = 1 2i-j+ xi+ 3j I = I (2+x)i+2j l = �(2 + X)2 + 22 = �X2 +4x+4+4 = �X2 +4x+8 Solve for x: �x2 +4x+8 = 5 x2 +4x+8 = 25 x2 +4x-1 7= 0 x = -4 ± �162(1)-4(1)(-1 7) -4±J84 2 -4±2m 2 = -2± m The solution set is { -2+m, -2-m}.

51. Let

u

a.

u'

u

b.

_----!._----'. �-c-

-5

63.

F

= 40 [ cos(300)i+Sin(300)j] ../3. 1 . ) - 40 (-I+-j 2 2 = 20../3 i+ 20j _

556 © 2008 Pearson Education, Inc., Upper Saddle River, NJ.



Section 10.4: Vectors

F] F2 F3 F. I F.I [ (175.8 ) + sin(175.8°) jJ I F.I ( -0.9973li+ 0. 0 7324j) F2 = I F2 1 [cos{3.70)i+ sin{3.70)j] ""I F2 1 1( 0 .99792i+ 0 . 0 6453j) F3 = -150j

Fj = 40[ cos (30° ) i+ sin (30°) j] = 40 [� i+ -i j) = 20v'3i+20j F2 = 60[cos(-45°)i+ sin( 45° )j] = 60 [ � i- � j) = 30.fi i- 30.fij F = Fj + F2 = 20v'3 i+ 20j+ 30.fi i-30v'2j = ( 20v'3+ 30.fi) i+ ( 20 -30v'2) j 67. Let FI be the tension on the left cable and F2 be the tension on the right cable. Let F3 represent

69. Let be the tension on the left end of the rope and be the tension on the right end of the rope. Let represent the force of the weight of the tightrope walker. = cos ° i

65.

-

""

For equilibrium, the sum of the force vectors must be zero.

F.+ F2 + F3 = -0.99731 I1F.l i+0. 0 7324 I1F.ll j +0.99792 I1F2 I1i+0 . 06453I1 F2 I1 j-150 j = (-0.99731 I1 FIII +0.99792 I1 F2 1I) i + (0. 0 7324 I1 F.I +0.06453I1 F2 1 -150) j =0 Set the i and j components equal to zero and solve: {-0.99731 I1 F)I +0.99792 I1 F2 1 = 0 0. 0 7324 1 F)1 +0 . 064531 F2 1 -150 = 0 Solve the fIrst equation for I F2 I and substitute the result into the second equatic.n to solve the system: I F2 I = �:::�� I F)I ""0.999391 F)I 0. 0 7324 1 1 F) I + 0 . 06453(0.999391 F.I ) -150 = 0 0.1377311F) 1 = 150 I F.I = 1089. 1 I F2 1 = 0 .99939(1089.1) ""1088.4 The tension in the left end of the rope is about 1 089. 1 pounds and the tension in the right end of the rope is about 1 088. 4 pounds.

the force of the weight of the box. cos ( 155° ) i +sin ( 155° )j]

F. = I FI I [ ""I F.I ( -0 .9063i+ 0.4 226j) F2 = I F2 I [cos ( 40°) i+ sin ( 40°) j] ""I F2 1 (0.7660i+0.6428j) F3 = -1000j

For equilibrium, the sum of the force vectors must be zero.

F.+ F2 + F3 = -0 .90631I F.Ii i+0 .4226 1 FI Il j +0.7660 1I F2 1I i+0 . 64281IF2 1I j-1000j = (-0.9063 1IF. I +0. 7660 1I F2 1I) i + (0.4226 1 F.I + 0 . 64281 F2 1 -1000 ) j =0 Set the i and j components equal to zero and solve: {0.-04.9063 1IFII + 0.7660 1F2 1 = 0 226 1IF.I +0. 64281IF2 1 -1000 = 0 Solve the fIrst equation for I F2 1 and substitute

the result into the second equation to solve the system:

IF211 = �:��:� I FIII "" 1 . 1832 1IF.11 0.4226 1IF.11 + 0. 6428(1.1832 1IF.ID -1000 = 0 1 .1832 1 F.11 = 1000 I F.I "" 845. 2 I F2 11 ""1 . 1832{845 . 2 ) ""1000 The tension in the left cable is about 845 . 2 pounds and the tension in the right cable is about 1000 pounds. 557 © 2008 Pearson Education, Inc., Upper Saddle River, NJ.




71.

=3000i = 2000[cos (45°) i+ sin (45°)j] = 2000 ( '7 i+ '7 j) =1000J2 i+ 1000J2j = + =3000i+ 1000J2 i+ 1000J2j =( 3000 + 1 000J2 ) i+ 1000J2j I I = ( 3000 + lOOOJ2 t + (lOOOJ2 t ""4635. 2 The monster truck must pull with a force of approximately 4635.2 pounds in order to remain unmoved. Fl

Section 10.5

F2

F

Fl

1.

5. True 7.

c.

9.

FI

F2

F3

F4

V·W

The vectors are orthogonal.

v=2i+j, w=i- 2j w= 2(1) + 1(- 2) = 2 - 2 =0 0 v·w = cos () = I villiwI � 22 +12 �12 +(_ 2)2 O_ = � =O =_ FsFs 5 90° () "" a.

F4

F3

W

b.

=-3i; =-i+4j; =4i- 2j; =-4j A vector v=a i + b j needs to be added for equilibrium. Find vector v=a i + b j : + + + + =0 -3i+ (-i+4j) + (4i- 2j) + (-4j) + (ai+ bj) =0 Oi+ - 2j+ (ai+ bj) =0 ai+ (-2+b)j=0 a = 0; -2+ b =0 b= 2 Therefore, v= 2j . F2

v=i -j, =i+ j =1(1)+ (-1)(1) =1-1 =0 0 v·w = cos () = lI vl l wl �1 2 + (_1)2 � =0 () =90° a.

F

FI

C

3. parallel

F2

73. The given forces are:

c2 =a2 + b2 - 2ab cos

V·

b.

V

c.

11.

The vectors are orthogonal.

v=.j3i-j, w=i+ j =v'3(1)+ (-1)(1) = v'3-1 v'3-1 :--= cos () = Il vlll wl �(v'3)2 + (_ 1)2 � v'3-1 =v'3-1 F6-J2 4 .J4J2 2J2 () =75° a.

b.

c.

13.

-;===-r====

V·W

V·W

-;=====

--

The vectors are neither parallel nor orthogonal.

v=3i+ 4 j, w=4i+ 3j v,w=3(4)+4(3)=1 2+1 2= 24 24 cos () = = I vil il wI �32 +42 �4 2 +32 24 24 J25J25 25 () ""16.3° a.

b.

75. Answers will vary.

c.

V·W

-;:===--===

The vectors are neither parallel nor orthogonal.




Section 10.5: The Dot Product

v= 4i, w= j a. V· w = 4(0) + 0(1) = 0 + 0 = 0 0 cos () = I vVli ·liwwI = -===--=== ../4 2+ 02../0 2+ 12 =� = �4 = O 4·1 () = 900 The vectors are orthogonal. 17. v = i- a j, w = 2i + 3j Two vectors are orthogonal if the dot product is : zero. Solve for a v·w = O 1(2) + (-a)(3) = 0 2-3a = 0 3a = 2 a= -23 19. v= 2i-3j, w= i-j v, = v·w)2w= 2(1)+ (-3)(-1) (i_ j) (I 1 (�12 +(-1)2 r 5, 5. =-52 (.I-J.) =-I--J 2 2 5. 5 ') = --1--) 1. 1. v2 =V-VI = (21-' 3)') - (-1--) 2 2 2 2 21. v= i-j, w= -i-2j VI = vw·w2 = 1(-1)+(-1)(-2) (-i-2j) (I 1 1) ( �( _ 1)2 +( _2) 2J 1, 2. = --51 (1+' 2)') = --1--) 5 5 . v2 =V-VI = ( ) - ( - 51,1- 52.) ) = 56 1- 53,) 23. v= 3i+ j, w= -2i-j v, = vw·w) 2w= 3(-2)+1(-1) (-2i-j) (lI l ( �(-2)2+(-1)2 r 14. 7. ' ') =-1+-) = --57( - 21-) 5 5 1, 2. 14. 7 ) =-1--) ' ' v2 =v-vl = (31+) ) - (-1+-) 5 5 5 5

15.

b.

c.

b.

W

c.

29,

I

=

W

1-).

lI Ii I =�(-0. 02)2+(-0. 0 1f = ../0.0005 "'" 0.022 The the sun'scentimeter. rays is about 0.022intensity watts perof square I A I = �(300) 2 + (400)2 = ../250,000 = 500 The area centimeter.of the solar panel is 500 square W= II . AI = I ( -0.02i-O. Olj)· (300i+ 400j)1 = 1(-0.02)(300) + (-0. 0 1)(400)1 = 1 -6+ (-4)1 = 1- 101 = 10 This means 10 watts of energy is collected. Toandcollect the maximum number ofsolar watts, A should be parallel with the panels facing the sun. Let va = the velocity of the plane in still air, vw = the velocity of the wind, and vg the velocity of the plane relative to the ground. Vg=Va+Vw va = 550 [ cos(225°) i+sin(225°)j] = 550 [ - � i- � j) = -2 75.fi i-275.fi j Vw = 80i Vg= Va+Vw = -275.fi i-275.fi j+ 80i = (80 -275.fi ) i-275.fi j

27. a.

.

'





Find the angle between vwand j: cos = I vvww il'lji jI mt(1) = mt 0.9887 = -3(0)+ 20";02 + 12 20 �8.6 0 The heading of the boat needs to be about 8.60 upstream. The velocity of1 .the8 kilometers boat directlyper across the river is about 9 hour. The time to cross the river is: 0.5 = --� 1 9.8 0.025 hours or = 19.0.58 . 60 �1.5 2 minutes. 33. Split the force into the components going down the hill and perpendicular to the hill.

The speed of the plane relative to the ground is: Ilvg 1 = �(80-275./2t + ( _ 275./2)2 = �6400 -44,000./2+ 151, 250+ 151,250 .�";246,674.6 �4 96 .7 miles per hour To find the direction, find the angle between vg and a convenient vector such as due south, -j . cos = vv .(1 -_j)j I I ;1 (80 - 275./2 )( 0) + ( - 275./2) (-1) 496.7�02 + (_1) 2 275./2�0.7830 496.7 �38. 5 0 The plane is miles traveling with ainground speed of about 4 . per hour an approximate 7 96 direction of 38.50 degrees west of south. 31. Let the positive x-axis point downstream, so that the velocity of the current is v = 3i. Let vw = the velocity of the boat in the water, and Vg = the velocity of the boat relative to the land. Then vg = vw + V and vg = since the boat is going directly across the river. The speed of the boat is I vw I = 20 ; we need to find the direction. Let vw = a i+ b j , so I VW I = ";a2 +b2 20 a2 +b2 = 400 Since vg = vw + , k j= a i+ b j+ 3i = (a + 3) i+ bj a+3 = 0 a = -3 k= b a2 +b2 = 400 9+b2 = 400 b2 = 3 91 k= b= mt�1 9.8 Vw = -3i+mtj and Vg = .J3 91j

e

=

e

e

t

t

=

e

c

c

F d = F sin8°= 5300sin8°� 73 7.6 F p = F cos8°= 5300cos8°�5248.4 The force requiisrabout ed to keep thepounds. SiennaThe fromforce rolling down the hill 3 6 7. 7 perpendicular to the hill is approximately 5248.4 pounds. 35. Let va = the velocity of the plane in still air, vw = the velocity of the wind, and Vg = the velocity of the plane relative to the ground. Vg =va +vw va = 500 [ cos( 45°) i + sin( 45°) j] ./2 ./2· -_ 500 [-1+ 2 2 'J J = 250./2i+ 250./2j vw = 60 [ cos ( 120o ) i + sin( 1200) j] = 60 [ - � i+ � jJ = -30i+ 30.J3 j

k

=

Vc

©

560 2008 Pearson Education, Inc., Upper Saddle River, NJ.



Section 10.5: The Dot Product

Vg =va+vw = 2S0..fi i+ 2S0..fi j -30i+ 30J3 j = (2S0..fi - 30) + (2S0..fi + 30J3) j The speed of the plane relative to the ground is: Ivg II = �( -30 + 2S0..fit + ( 2S0..fi + 30J3t "".J26 9,12 9. 1 "" S18.8 kmlhr To find the direction, find the angle between v and a convenient vector such as due north, j. cos () = ..,.--Vv-"g,,--. jj-I II g II 1 (0) + ( 2S0..fi + 30J3) (1) - ( -30 + 2S0..fi)SI8.8.J0 2 + 12 +30J3 "" 0.7816 = 2S0..fiS18.8 () "" 38.6 ° The travelingin awith a ground of of aboutplane S18.8is kmlhr direction of 38.speed 6 ° east north (N3 8.6°E) . 37. We must determine the component force going down the ramp . Fd = F sin200= 2S0sin 20°"" 8S . S Timmy about 8S . S pounds of force to hold themust pianoexert in position . 39. W = F · AB W= 2, AB = 4i F = cos a i-sin a j 2 = ( cosa i-sina j) · 4i 2 = 4cosa -21 = cosa a = 60° 41. Since = 0 i + 0 j and v=a i+ b j, we have that O·v= O· a+O·b = O . 43. If v= ,i +bd = c o sai+sina j and w = a2 i+b2 j = cosfJi+sinfJ j, then cos(a - fJ) = V· = a,a2 + b,b2 = cosa cos fJ + sina sinfJ

Let u = a, i + bd and v= a2 i+ b2j. Then, (u+v) · (u -v) = [(a,+ a2 )i+ (b,+ b2)j] [ (a,-a2 )i+ (b, -b2 ) j] = (a,+ a2 )( a,- a2 ) + (b, + b2 )( b, -b2 ) = a, 2 - a2 2 + b, 2 - b22 = ( a,2 + b/ ) -( a/ + b/ ) = I ul12 - 1vI 2 Since the vectors have the same magnitudes and (u +v)( u -v) = 0 , the vectors are orthogonal. Because the vectors u and v are radii of the circle, we know they have the same magnitude. Since u +v and u -v are sides ofsincethe weangleknowinscribed in arethe semi circle and that they orthogonal part (a), this angle must be a right angle.by 47. (IIw I v + IlvIw) '( 1 w Iv - l lvIw) = (IIw10 2 V· v- Iwil i vI V·w + 1 wi l i vIlvw , - (II V 102 = (11 1 ) 2 V·v- (II V 1 ) 2w·w = (II 1 )2 (II V 102 - (I V1 )2 (I I 1 )2 =0 Therefore, the vectors are orthogonal. 49. (II u +v Ilf - (II u -v Ilf = (u +v) · (u +v)-(u -v) · (u -v) = (u·u + u · v+ V· u + V·v) -(u·u -u · v-v· u + V· v) = 2(u ·v)+2(u ·v) = 4(u ·v)

45.

i

g

a.

b.

w · w

W

W

W

0

a

W

©

561 2008 Pearson Education,




Polar coordinates of the point (-3, 3) are (-3.[i, -%) or ( 3.[i, 3;) . 9. The point (0, - 2) lies on the negative y-axis. r = �x2 +y2 = �02 + (_ 2)2 = 2

Ch apter 10 Review Exercises

3J3 ' y = 3S .lli -1t =-3 x = 3cos-61t = -2' 6 2 Rectangular coordinates of ( 3, i) are ( 3J32 ' l). 2

- 2 is undefined, so B = - 21t ; - 21t +7r =27r ' Polar coordinates of the point (0, - 2) are o

The point (3, 4) lies in quadrant I . r = �x2 + / = .J32 +42 = 5 B = tan-I (�) = tan-I (�)� 0.93 tan- I (�) + 7r � 4 . 0 7 Polar coordinates of the point (3, 4) are (5, 0.93) or (-5, 4 . 0 7) . r = 2sinB 13. r 2 = 2rsinB x2 + y2 = 2y x2 + y2 -2y = 0 x2 + / -2y+l= 1 x2 +(Y _ l)2 = e The graph is a circle with radius 1 and center (0,1) in rectangular coordinates.

11.

x = -2cos 431t = l'' Y = - 2sin 431t = J3 Rectangular coordinates of ( -2, � 1t ) are (1, J3) . 5.

a.

(-3, - �) ---..

0 .....,... ..

b.

x = -3 cos( -�) 0; y = -3sin ( -�) 3 Rectangular coordinates of (-3,-%) are (0, 3) . 7. The point (-3, 3) lies in quadrant II. r = �x2 +y 2 = �(_ 3)2 + 32 = 3.[i B = tan-I (�) = tan-I ( �3) = tan-I (-I) = - % =

=

Ytn

9 = '2


All rights reserved. This material i s protected under all copyright laws a s they currently



15.

a.

b.

r=5 r 2 = 25 x2 + y 2 = 5 2

19.

The a circle with radius and centegraph r at theis pole. 5

r = 4cosO The graph will be a circle with radius 2 and center (2, 0) . Check for symmetry: B by -B . r 4 cos( -0) = 4 cos 0 . The aXIS. graph is symmetric with respect to the polar The line 0 = � : Replace 0 by 7t 0 . r = 4 cos( 7t 0) = 4( cos cos 0 + sirl sin0) = 4(-cosB+0) = -4cosO The test fails. The pole: Replace r by - r . -r 4 cos O . The test fails. Due to symmetry with respect to the polar axis, assign values to 0 from 0 to 7t . 0 r = 4 cosO 0 4 Polar axis : Re place

The result is

=

Yt o=I

-

-

7r

17.

=

cos 0 + 3r sin0 = 6

a.

r

b.

y = - 3 x+ 2 The graph is a line withy-intercept (0, 2) and slope 3I in rectangular coordinates.

=

x + 3y 6 3y = -x + 6 1

7t 2.J3 "" 3.5 6 7t 2 3 7t 2 27t -2 3 57t - 2.J3 "" -3.5 6 -

-

Ye t=

7r

-

�

-

0

-

-

7r

-4

e

=

31T 2





21.

r = 3 - 3sin O The graph will be a cardioid. Check for symmetry: Polar axis: Replace () by - O . The result is r = 3 - 3sin( - 0) = 3 + 3 sin 0 . The test fails.

23.

The line () = � : Replace () by () . r = 3 - 3sin(7t - (}) = 3 - 3( sin cos () - cos sin (}) = 3 - 3(0 + sin 0) = 3 - 3 sin O The graph is symmetric with respect to the line O = !!:..2 ' 7t -

Jr

-r =

assign values to - -7t2 - -7t3 7t -6 0 7t 6 7t 3 7t 2

Jr

3 - 3 sin 0 .

Due to symmetry with respect to the line 0 = 0

i:

The line 0 = Replace 0 by 7t - 0 . r = 4 - cos( 7t - 0) = 4 - (cos cos 0 + sin sin 0) = 4 - (- cos O + 0) = 4 + cos O The test fails. The pole: Replace r by - r . -r = 4 -cos 0 The test fails. Due to symmetry with respect to the polar axis, assign values to 0 from 0 to 7t . 0 r = 4 - cos O 3 0 7t 4 -13 � 3 . 1 6 2 7t7 3 2 7t4 2 9 27t 3 2 57t 4 + 13 � 4.9 2 6 5 Jr

Jr

The pole: Replace r by - r . The test fails.

i

r = 4 - cos O The graph will be a limayon without inner loop. Check for symmetry: Polar axis: Replace 0 by - O . The result is r = 4 - cos( - 0) = 4 - cos 0 . The graph is symmetric with respect to the polar axis.

Jr

.

'

0

from - !!:..2 to !!:..2 . r = 3 - 3 sin O 6 3 + 313 2 � 5.6 9 2 3 3 2 3 - 313 2 � 0.4 0





r = �x2 + y 2 = �(_1)2 +(_1)2 = .J2 tanB =Lx = -1-1 = 1 B = 225° The polar form of = -1-i is = r( cosB+isinB) = .J2 (cos225°+ isin225°) . 27. r = �x 2 + y 2 = �4 2 +( _ 3)2 = .J25 = 5 tanB =Lx = _ i4 B ", 323. 1° The polar form of = 4 -3 i is = r(cosB+isinB) = 5( cos 323.1°+ isin323. 1°) . 29. 2(COS1500+isin1500) = 2 ( -.J} + l i ) = --13 +i

33.

25.

0.1(cos3500+isin3500) '" O. l(0.9848-0.l736i) '" 0.10-0.02i 0.06

z

z

0.02

- 0.06

z

35.

z

2

- ..

-1

-' 2-

= (cos 80°+ i sin 80°)( cos 50°+ i sin 50°) = 1 · 1 [cos(80°+ 50°) + i sin(80°+ 50°)] = cos 130°+ isin130° cos 80°+ i sin 80° cos 50°+ i sin 50° = i [cos(80°- 50°) + i sin(80°- 50°)] = cos 30°+ i sin 30°

zw

z

w

Imaginary axis

-;! �----'t--' --"---'--------o-----,

Imagi n a ry '1 xis

Real axis

6(cos 21t + isin 21t ) 6 (cosO + isin 0) =6 91t . ' 91t ) . ( COS-+1Sl\l= .( 1t . ' 1t ) 2 COS S +1Sl\l S =

-2

=

31.

z

w

3

5

5

Imaginary axis 3

Real -+----:---t-...l-.Jl� axis -1

�

565 2008 Pearson Education,




39.

z W = 5 ( cos l00 + i sin l00) . ( cos 355° + i sin355°) = 5 · 1 [ cos(1 0° + 355° ) + i sin(10° + 355°) ] = 5 ( cos365° + i sin 365° ) = 5 ( cos 5° +isin 5°) 5 ( cos 1 0° + i sin 1 0°) W cos 355° + i sin 355° = -i[ cos(10° - 355° ) + i sin(10° - 355°) ] = 5 [ cos(-345°) + i sin(-345° ) ] = 5 ( cos 1 5° +i sin 1 5° )

47.

r

i 3 e "" 5 3 . 1 3 °

tan e

[3 ( cos 200 + i sin 200 )T 33 [ cos(3 . 20° ) + i sin(3 . 20° ) ] = 27 ( cos 60° + i sin 60° )

49.

(� + � i ) ../3

27 + 27 1. =2 2 --

43.

[ (

.J2 cos 5 1t + i sin 5 1t 8 8

[ (4 ·

5 1t + i sin 8 . 2 51t + I. SlD 51t cos2

= (.J2f cos

=4

.

(

= 4(0 + l i) = 4i 45.

)

)r

)

Zk

(4 . 5;)]

--

) ( ]

Zl

-%

�e + ( -../3t = 2 tan 8 = -../3 = -../3

Z2

=

1 8 = 300° 1 - i = 2 ( cos 300° + isin 300° ) i = [2( cos 300° + isin 3000 ) r = 26 [cos ( 6 . 300° ) + i sin ( 6 . 300° ) J = 64 · ( cos 1 800° + i sin 1 800° ) = 64 · ( cos 0° +i sin 0° ) = 64 + 0 i = 64

--

�

= 3 [ cos ( 120° . 2 ) + i sin ( 120° . 2 )J 3 i = 3 ( cos 240° + i sin 2400 ) =

-%- �

51.

../3 (1-../3 r


[ (

Zo

1 - ../3 i r

27 + O i = .J27 2 + 0 2 = 27 tan 8 = ..Q.. 27 = 0 8 = 0° 27 + Oi = 27 ( cos Oo +isin OO) The three complex cube roots of 27 + Oi = 27 ( cos oo +isin Oo ) are: 0 0 360 0 . k 00 36 0 0 .k + 3 + i sin "3 + 3 = :iffi . cos "3 = 3 [ cos ( 120° k ) + i sin ( 1 20° ·k )J = 3 [ cos ( 120° · 0) + i sin ( 120° . O)J = 3 ( cos 0° + i sin 0°) = 3 = 3 [ cos ( 120° . 1 ) + isin ( 120° . 1 ) = 3 ( cos 120° + i sin l200) = + 3 i r

=

= 27

=

3 + 4i = 5 [ cos ( 53. 13° ) + i sin ( 53. 13°)J (3 + 4it = ( 54 [cos ( 4 . 53. 1 3° ) + i sin ( 4 . 53. 13°) Jf = 625 [ cos ( 212 .52° ) + i sin ( 212.52° )J "" 625 ( -0.8432 - 0.5376i) = - 527 - 336i

Z

41.

3 + 4i = .J3 2 + 4 2 = 5

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)]


The angle between v and i equals 60° . Thus, . . y x x cos60° = I vvl · li ili = ( xi+3· 1j) i 3·1 = -3 We also conclude that v lies in Quadrant I. cos 60° = "3 -1 = -x 2 3 x =-23 x2 + y 2 = 9 (%J + y2 = 9 y 2 = 9 - (%J = 9 - � = 364-9 = 247 (27 = +- 3/3 y = -V4 2 Since v lies in Quadrant I, y = 3f . So, v = X·I+. y . .J = "23.I + -3/32- .J . 71. v = -2i + j, w = 4i -3j v.w = -2(4) + 1(-3) = -8 -3 = -11 cos () = I vv·1 w1wI -11 2 _ 2 �( ) +12 �4 2 + (_ 3)2 -11 = -11 J5·5 5../5 16 () 9.7° 73. v= i- 3j, w= -i+ j v·w = 1(-1)+ (-3)(1) = -1-3 = -4 cos () = I vv·1 w1wI -4 �e + (_3)2 �(_1)2 +e -4 -2 = --2J5MF2 J5 5 () 153.4 °

53.

x

= (1, -2), Q = (3, -6) v = (3-1)i+( -6- (-2))j = 2i-4j Ilvl = �2 2 + (_4)2 = hO = 2J5 57. P = (0, -2), Q = (-1, 1) v = (-1-0)i+(I- (-2))j = -i+ 3j I vll = �(_1)2 + 32 = M 59. v+w = (-2i + j )+(4i-3j) = 2i-2 j 61. 4v-3w = 4( -2i + j)-3(4i - 3j) = -8i+4j-12i + 9j = -20 i+1 3j 63. I vl = I -2i + jl = �(-2) 2 +e = J5 65. I v � + I w I = 11 - 2i + j I + 1 4 i - 3j I = �(_2)2 +12 + �42 + (_ 3)2 = J5 5 7.24 v -2i+ j = -2i + j 67. U =- = I v� 1 -2i+ jl �(_2)2 +12 -2i+ j = -../5 . 2J5.I + -J = -5 5 J5 69. Let v = x . i + y . j, I v I = 3 , with the angle between v and i equal to 60° . I vl = 3 �X2 + y2 = 3 x2 + y 2 = 9

55. P

+

+

r:::;

r:::;

r:::;





75.

v = 2i + 3j , w = -4i - 6j v · w = (2)(-4) + (3) (-6) = - 8 - 1 8 = -26 cos = vvil ·liww II () II -26 .J2 2 + 3 2 �( _4) 2 ( _ 6) 2 -26 = -26 = -26 = -1 J13J52 .J676 26 () = cos- I ( -1 ) = 1 80° Thus, the vectors are parallel. v = 3i - 4j , w = -3i + 4j V · w = (3)(-3) + (-4)(4) = -9 - 1 6 = -25 cos = I vil ·liw () I v w II -25 �3 2 + (_4) 2 �( _3) 2 + 4 2 -25 = -25 = - 1 J25J25 25 () = cos- I ( -1) = 180° Thus, the vectors are parallel. v = 3i - 2j , w = 4i + 6j v . w = (3) ( 4) + ( -2) (6) = 12 - 12 = 0 Thus, the vectors are orthogonal. v = 2i + j , w = -4i + 3j The decomposition of v into 2 vectors V I and v 2 so that v I is parallel to w and v 2 is perpendicular to w is given by: v I = v · w2 W w1 and v 2 = V - V I v . w -_ (2i + j) . (-4i + 3j) ( -41. + 3J. ) V I _- -Il w ll 2 �( _4)2 + 3 2 2 = (2)( -4)25+ (1)(3) (-4i + 3j) = --51 ( -4 1' + 3 J' ) = -45 I. --53 J.

83.

c

79.

81.

(

©

g

w

c '

w

g

c

w

Since the river is 1 mile wide, it takes the swimmer about 0.2 hour to cross the river . The swimmer will end up (0.2)(2) 0.4 miles downstream. Let F, the tension on the left cable, F2 the tension on the right cable, and FJ the force of the weight of the box. F, = I F, II [ cos(1400) i + sin (140 ) j ] ,., II F, -0.766044i + 0.6427 88 j) F2 = II F2 11[ cos (300) i + sin ( 300 ) j] ,., II F2 11 ( 0. 8 66025i + 0.5 j) F3 = - 2000j For equilibrium, the sum of the force vectors must be zero. Fl + F2 + F3 = - 0.766044 1 FJ Ii i + 0.6427 88 11 FJ I l j + 0. 8 66025 1I F2 1I i + 0.5 1I F2 1I j - 2000j = (-0.766044 11 Fl II + 0. 8 66025 11 F2 11 ) i + (0.6427 88 1I FJ I + 0.5 1I F2 11 - 2000 ) j =0 Set the i and j components equal to zero and solve: - 0.766044 1I FJ II + 0.866025 1I F2 11 = 0 0.6427 8811 FJ II + 0.5 11 F2 11 - 2000 = 0 Solve the first equation for IIF2 11 and substitute the result into the second equation to solve the system: F2 1 II FJ I ,., 0. 88 4552 1 FJ 0.6427 8811 FJ I + 0.5 (0.8 84552 11 Fl II ) - 2000 = 0 =

85.

=

=

I(

)

V 2 = V - V I = 2 1. + J. - 54 1. - 53 J. 3. = 2I. + J. - -45 1. + -J 5 6 . . 8 = -I 5 +-J 5

=

w

-lI

(

w

g

=

+

77.

Let the positive x-axis point downstream, so that the velocity of the current is v = 2i . Let v the velocity of the swimmer in the water and v the velocity of the swimmer relative to the land. Then v = v + V The speed of the swimmer is II v II = 5 , and the heading is directly across the river, so v = 5 j . Then v = v + V = 5 j + 2i = 2i + 5 j

°

=

{

)

I ,., �:�:��;

I

568 2008 Pearson Education, Inc., Upper



Chapter 1 0

= =

1 .085064 1 F\ II 2000 II F\ II 1 843.2 1 1b I F2 1 0.884552(1 843.2 1) 1630.41 1b The tension in the left cable is about 1 843.21 pounds and the tension in the right cable is about 1630.41 pounds. ""

AB W

5.

=

=7 �X2 + y 2 = 7 ( �x 2 + y 2 t = 72 x2 y 2 = Thus the equation is a circle with center (0, 0) and radius 7. r

+

49

=

=

20i F . AB

= = [% i 5� j). 20i = (%}20) + [ 5�} 0 ) = 50 ft-lb +

e

6. Ch apter 10 Test 7t

1 - 3.

=

3 7t 2

= =

tan g 3 sin g 3 cos g r sin g 3 r cos g l. 3 or Thus the equation is a straight line with and b O.

=

2

x=

7t

6

y = 3x

=

4.

est

In =

3

x = 2 and y = 2"-J3--- --=r = �x 2 + y 2 = �(2) 2 + ( 2 J33 ) 2 = J16 = 4 and tan g =l. = 2J3 = J3 x 2 Since (x, y) is in quadrant !, we have g = 3" .

The point (2, 2J3) in rectangular coordinates is the point ( 4, 1-) in polar coordinates.


Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all cop

yright laws as they


c urr

ntly


7.

r sin 2 8 + 8 sin 8 = r r 2 sin 2 8 + 8r sin 8 = r 2 y 2 + 8y = x 2 + y 2 8y = x 2 or 4 ( 2 ) y = x2 The graph is a parabola with vertex (0, 0) and focus (0, 2) in re c tangular coordinat s Yt

e

9.

.

e =� 2

x

i+li+l-H�

e = 1t

e =0

e=

8.

31t 2

r 2 cos 8 = 5 Polar axis: Replace 8 with -8 : r 2 cos( -8) = 5 r 2 cos 8 = 5 Since the resulting equation is the same as the original, the graph is symmetrical with respect to the polar axis. The line 8 = 1 : Replace 8 with :r - 8 : r 2 cos(:r - 8) = 5 r 2 (cos :r cos 8 + sin :r sin 8) = 5 r 2 (-cos 8) = 5 _ r 2 cos 8 = 5 Since the resulting equation is not the same as the original, the graph may or may not be symmetrical with respect to the line 8 = :r . 2 The pole: Replacing r with -r : ( _r) 2 cos 8 = 5 r 2 cos 8 = 5 Since the resulting equation is the same as the original, the graph is symmetrical with respect to the pole. Note: Since we have now established symmetry about the pole and the polar axis, it can be shown that the graph must also be symmetric about the line 8 = 1 .

1 0.

11.

r = 5 sin 8 cos 2 8 The polar axis: Replace 8 with -8 : r = 5 sine -8) cos 2 (-8) r = 5(-sin 8) cos 2 8 r = -5 sin 8 cos 2 8 Since the resulting equation is the not same as the original, the graph may or may not be symmetrical with respect to the polar axis . The line 8 = � : Replace 8 with :r - 8 : r = 5 sin(:r - 8) cos 2 (:r - 8) r = 5( sin :r cos 8 - cos :r sin 8)( - cos 8) 2 r = 5(0 · cos 8 - (-1) . sin 8)( cos 2 8) r = 5 sin 8 cos 2 8 Since the resulting equation is the same as the original, the graph is symmetrical with respect to the line 8 = � . The pole: Replacing r with -r : -r = 5 sin 8cos 2 8 r = -5 sin 8cos 2 8 Since the resulting equation is not the same as the original, the graph may or may not be symmetrical with respect to the pole. = (2( cos 85° i sin 850)][ 3 ( cos 22° i sin 220)] = 2 · 3[cos(85° + 22°) + i sin(85° + 22°)] = 6(cos 1 07° + i sin 107°) zw

+

+

= 3 ( cos 22° + isin 22° ) 2 ( cos 85° + i sin 85° ) = � [cos(22° - 85°) + i sin(22° - 85°)] 2 = � [cos( -63°) + i sine -63°)] 2 Since -63 ° has the same terminal side as 297° , we can write = � (cos 297° + i sin 297°) 2 5 = [3(cos 22° + isin 22o)] = 3 5 [cos(5 · 22°) + i sin(5 · 22°)] = 243(cos 1 10° + i sin 1 10°) w z

w z

12.

w

5

570


All rights reserved. This material i s protected under all copyright laws a s they currently


Chapter 10 Test

13.

1 8.

Let w = -8 + 8.J3 i ; then \ wi = �( 8)2 + (8.J3)2 = ..)64 + 192 = ..)256 = 16 so we can write w in polar form as W = 16 - � + � i = 1 6(cos I 200 + i sin I 200) Using De Moivre ' s Theorem, the three distinct 3 rd roots of w are = .ift6[ cos ( 1 2r r ) + i sin ( 1 2r + r )] = 2ifi [cos( 40° + 1200k) + i sin(40° + 1200k)] where k = 0, 1, and 2 . So we have = 2ifi[cos( 40° + 120° . 0) + i sin( 40° + 120° . 0)] = 2ifi ( cos 40° + i sin 40°) "" 1.93 + 1.62i = 2ifi [cos( 40° + 120° · 1) + i sin(40° + 120° . 1)] = 2ifi (cos 160° + sin 160°) "" -2.37 + 0 . 86i Z2 = 2ifi [cos(40° + 120° . 2) + i sin( 40° + 120° . 2)] = 2ifi ( cos 280° + i sin 280°) "" 0.44 - 2.4 8i

From Exercise 17, we can write v = 10(cos315°i + sin315°j)

_

[

i

�{; ,+ �})�svthe origin 10 as=>b=l the center ofpartithecle path hyperbola, the equation of the would be x2 / x�o ---=1 100 100 Assume -ax22 --b2/=1. If",thae eccentri co.ity When is closeb tois close 1, thento 0, the and b", hyperbola is veryarenarrow, the asymptotes close tobecause O . the slopes of If tishmuch e eccentrilargercitythanis verya andlarge,b isthen very large. The resul t i s a hyperbol a that i s very wide,large. because the slopes of the asymptotes arevery For ':;'a2 -.;..b2=1 , the opposite is true. When the ricity isthecloseslopes to 1, ofthethehyperbola isvery wicleccent dosee because asymptotes are to O. When the eccentricity i s very large, narrowlarge.because the slopes ofthethehyperbola asymptotesisveryarevery --

69.

73 .

a.

b.

71.

x2 y2=1 (a=2, b=1) This s a hyperbola with horizontal transverse axis, icentered at (0, 0) and has asymptotes: y=±21 x x42 (a=1, b=2) y2 --=1 This is a hyperbola withvertical transverse axis,1 centered at (0, 0) and has asymptotes: y=±2x . Siasymptotes, nce the twotheyhyperbolas have the same areyconjugates. 4-

3

2

x 2 -y =1

4

75.

C

C

-3

Put the equation in standard hyperbola form: AX2+c/ + F = 0 A· C 0, F 0 Ax2+C /=-F A-FX2 + Cy-F2=1 2 +( y�)2 =l ( x�) _ _ Sithinsceis -Fa hyperbola / A and -Fwith/ Ccenter have(0,opposite signs, 0). The ransvverse x-axis ifif --FF / A> The ttrans erse axis axis isis thethe y-axis A o0 ..
0 andA= C , the equation defmes a circle.

11.

13.

X

-

23.

7t

,

,

.

7t

7t

.

,

7t

,

7t

,

7t

,

,

--

,

,

Y

25.

cos(2(})=-25 1 -1+ - =�16 =4=>(}�36.90 -f-fsin(}=� =&9 =s3 ; cos (}=� S 3 ,=-1 (4x ' -3Y )' x= x, cos() -y, sm. () =-x45 , --y 5 5 y=x' sin(}+ Y 'cos (}=�5 x'+ i5 y' =.!.5(. 3x'+ 4y') 600


Section 11.5:

Rotation of Axes; General Form of a Conic

16(� ( 4x' -3y')J +24(� ( 4x'- 3Y'))(� ( 3x'+4y'))+9G ( 3x'+4y')J -130(� ( 4x'- 3y'))+90G ( 3X '+4Y ') )= 0 �� (16x,2- 24x )/ + 9y,2)+�: ( 12x,2 +7 x'y '-12y,2) +:5( 9X,2+24x'y '+16y'2)-104x'+78y'+ 54x'+ny'= 0 288 ,2 288 168" --y 384" 144 ,2+-x 256 ,2--xy -x 25 25 +-y 25 8125 ,2+-xy 25 25 216" 144 ,2- 50x '+150y=' 0 +-x 25 25 ,2 25 ,2+-xy+-y 25x -50x'+2150y'= 0 x, - 2x2'=-6 Y ' (x'- 1) =-6y' +1 (X ' -1)2=-6( Y ' -�) Parabola; vertex (1, �) , focus (1, -�) ; axis of symmetry parallel to the y' axis. y 5

x

43. 45. 47. 49. 51.

A=l, B=3, C=-2 B2-4AC=32- 4(1)(- 2)=17>0 ; hyperbola A=1, B -7, C 3 B2- 4AC=(_7)2-4(1)(3) 37> 0 ; hyperbola A=9, B 12, C=4 B2-4AC=122-4(9)(4)= 0 ; parabola A=lO , B=-12, C=4 B2- 4AC=(-12)2_4(10)(4)=-16 0 , the equation defines a

24

=

7

hyperbola. 8.

We have cot ( 28 ) = - 14 so it follows that cos ( 28 ) = 15

3x 2 - .xy + 2 / + 3 y + 1 = 0 is in the form Ax 2 + Bxy + C/ + Dx + Ey + F = 0 where A = 3 , B = -1 , and C = 2 . B 2 - 4A C = ( _1 ) 2 - 4 ( 3 )( 2 ) = 1 - 24 = -23 Since B 2 - 4A C < 0 , the equation defines an

-

,in 0

()

� ( ;) 1- -

�

2

5 = fl6 � V 25 5 �

( ;5 ) = {9 = �

1+ -

V25

2

5

8 = cos- 1 % ", 53. 130 With these values, the rotation formulas are 3 4 3 4 x = - x ,- - y , and y = - x '+ - y ' 5 5 5 5 Substituting these values into the original equation and simplifying, we obtain

x2 - 6xy + 9 / + 2x - 3y - 2 = 0 is in the form AX 2 + Bxy + C/ + Dx + £y + F = 0 where A = 1 , B = -6 , and C = 9 . B 2 - 4A C = ( _6 ) 2 - 4 ( 1 ) ( 9 ) = 36 - 36 =0 Since B 2 - 4A C = 0 , the equation defmes a

4 1x 2 - 24xy + 34/ - 25 = 0 4 3 4 2 3 4 3 4 1 S" x ,- S" y , - 24 S" x , - S" y , S" x , + S" y ,

(

)(

)2 ( +34 S" x '+ S" ) = 25 (4 3

parabola. 10.

1 - co, ( 20 ) = 2

�

l + co, ( 20 ) = ,", 0 = 2

ellipse. 9.

.

4 lx 2 - 24.xy + 34y 2 - 25 = 0 Substituting x = x ' cos 8 - y ' sin 8 and y = x ' sin 8 + y ' cos 8 transforms the equation

Y

'

)

Multiply both sides by 25 and expand to obtain

into one that represents a rotation through an angle 8 . To eliminate the x ' y ' term in the new equation, we need cot ( 28) = A B- C . That is, we need to solve 4 1 - 34 cot ( 2 ll ) = ---24

(

)

(

2 2 2 2 41 9X , - 24x ' y ' + 16y , - 24 1 2x , - 7x ' Y '- 1 2 y ,

+34 ( l 6x ,2 + 24x ' y '+ 9y ,2 ) = 625

,

62 5 x 2 + 1 2 5 0y , 2 = 625 X ,2 + 2y ,2 = 1 x,2 y ,2 -+- = 1

u

cot ( 28 ) = - � 24 Since cot ( 28 ) < 0 , we may choose 90° < 28 < 1 80° , or 45° < 8 < 90° .

1

1

2

.J2

Thus: Ji. = 1 and b = II = 2 V"2 This is the equation of an ellipse with center at ( 0, 0 ) in the x ' y '- plane . The vertices are at a =

624


)

Chapter 11 Test

�

c

-

center

c=-

plane

-2rcosB = 3 = 2rcosB+3 = (2r cos B + 3)2 x2 + y2 =(2x+3)2 x2 +l =4x2 +12x+9 3x2 + 12x -l =-9 3(X2 +4x)-l =-9 3(X2 +4x+4)-y2 = -9+12 3(x+2)2 _ y2 =3 (X+2)2 --= y23 1 (1) {X=3t-2 y = l - Ji (2) t x Y 0 x=3(0)-2=-2 y=I-JQ=1 1 x=3(1)-2=1 y=I-J! =0 4 x=3(4)-2=10 y=I-v'4=-1 9 x = 3 (9) -2 = 25 y = I-J9 =-2 r

(-1,0) and (1,0) in the x'y'- plane. J22 12 21 2 =a2 - b2 =1--=The foci are located at (± � ,0] in the x' y '- plane. In summary: x'y' xy(0,0) (0,0) (±1,0) (�,�}( -�,-�) ( 2J25 ' 3J210 ] ' (O,± '7) _2J2 3J2 ( 5 ' _ 10 ) ( 3J210 ' 2J25 ) ' (± '7,0) _ 3J2 2J2 ( 10 ' _ 5 ) The graph is given below. plane

vertices

minor axis intercepts

r r2

�----':.....

12.

foci

(x,y) (-2,1) (1,0) (10,-1) (25,-2)

y :)

-5

11.

Towe find thtoe elrectangul athr eequati oblneforfrtomhe curve, need i m i n ate vari a the equations. Wex =can3t-2start by solving equation ( 1) for 3t = x+2 x+2 t =-3 ng this result for into equation Substi t uti gives T-3- ' -2 :S;x:S;25 y=I- V/x+2

3 = l-ecosB 1-2cosB Therefore, = 2 and = �2 Since I , this is the equation of a hyperbola. 3 ( I -2 cos B) = 3I-2cosB r-2rcosB = 3 Since x = rcosB and x2 + y2 = we get r

ep

e

r=

p

t

t.

e>

.

. ( 10 , - 1 )

---

r

t

r2 ,

625

(2)


Chapter 11: 13.

Analytic Geometry

We can draw the parabola used to form the reflector 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. y

-2

1.

4 ft

2

(-2, 1 . 5)


(2, 1 .5 )

o

x

2

-2

The form of the equation of the parabola is x 2 = 4ay and its focus is at (0, a) . Since the point (2, 1 .5) is on the graph, we have

3.

2 2 = 4a (1 .5) 4 = 6a a - 13

The microphone should be located i feet (or 8 inches) from the base of the reflector, along its axis of symmetry. y

(-2, 1 . 5)

=

(2, 1 .5 )

x

F

- (0 �

6 - x � x2 0 � X2 + x - 6 x2 + x - 6 � O (x + 3)(x - 2) � 0 f(x) x 2 + x - 6 x = - 3, x 2 are the zeros of f . Interval (-00 , -3) ( -3, 2) Test Value -4 0 Value off 6 -6 =

4 ft

2

f (x + h) - f (x) h -3 ( X +h) 2 + 5 ( X + h) - 2 - ( -3x 2 + 5x - 2 ) h -3 ( X2 + 2xh + h2 ) + 5x + 5h - 2 + 3x2 - 5x + 2 h -3x 2 - 6xh - 3h 2 + 5x + 5h - 2 + 3x 2 - 5x + 2 h -6xh - 3h 2 + 5h = -6x - 3h + 5 h

5.

3

, 1. )

(2 , 00 ) 3 6

Conclusion Positive Negative Positive The solution set is { x 1 - 3 � x � 2 } , or [-3, 2] . f (x) log4 (x - 2) a. f (x) = log4 (x - 2) = 2 =

x - 2 = 42 x-2 = 16 x = 18

b.

The solution set is { 1 8} . f (x) = log4 (x - 2) � 2 x - 2 � 4 2 and x - 2 > 0 x - 2 � 1 6 and x > 2 � 18 and x > 2 x

2 < x � 18 (2, 1 8]

626

© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permi ssion in writing from the publisher.

Chapter 11 Review Exercises 7.

sin ( 28 ) = 0.5

28 = !!:. + 2krr. => 6 511" + 2krr. => or 28 = 6 where k is any integer.

9.

8 = .!!:.... + k7t 12 5rr. 8 = - + krr. 12

Using rectangular coordinates, the circle with center point (0, 4) and radius 4 has the equation: (X _ h ) 2 + ( y _ k ) 2 = r 2 (X _ O)2 + ( y _ 4 )2 = 4 2 X2 + / - 8y + 1 6 = 1 6 X 2 + / - 8y = 0

Converting to polar coordinates : r 2 - 8r sin 8 = 0 r 2 = 8r sin 8 r = 8 sin 8 y

9

.x

-5

11.

8 < 90° rr. rr. krr. 28 = - + krr. => 8 = - + - where k is any 4 8 2 ' integer. On the interval 0° < 8 < 90° , the solution is 8 = � = 22.5° .

cot ( 28) = 1, where

0°

[!4 =� ! �] {:: �1 x= y=-1,

b.

27.

31.

'i

[� =� � =�l [-2(2)+5 -5 � 2 �5 ->[: T �l -4

25.

Section 12.2: Systems of Linear Equations: Matrices

zi s any

or usi ng ordered

639



Chapter 12: Systems of Equations and Inequalities

39.

{2X-4Y=-2

3x+2y= 3

45.

I-�]�[� -� I-�] (Rl==-3t�) + �[� -� I-�] (R2 � r2) �[� -� 1-�] (R2= r2) �[ 1 01 !] (Rl= 2r2+�)

-�

o

t

The so Iutl· On IS·

(�'i). 41. { X+ 2Y=4 2x+4y=8

x=-,12 y=-43 4"

o

o

.

[� �I:]�[� �I�] (R2=-2�+r2)

o

This is a dependent system.

is a ny rea l number is a ny rea l number}

o

I

_ 1.

I -I

The solution is

8

o

-1-

or

t

o

W rite the a ugmented ma trix:

1.

or -, 1

W rite the a ugmented ma trix: 1 -1


43.

I

1

The so Iuti· on IS·

47.

The solution is I or

-'3

'3

. or usm g ordered pairs

x+2y=4 x=4-2y x=4 -2y,y {(x,y) x= 4-2y,y 12X+3Y= 6 x- y=-2 2[ 3 1 6]�[1 t 3] (Rl=t�) t 1 -1 t �[10 -f 3] (R2=-�+r2) �[� 11�] (R2=-ir2) �[� � 11] (Rl=-tr2+�) x=%, y=1 (%, )

3[15 -55 1 213]�[151 -15 1 211] (Rl=trl) � [ 1 -301 1 61] (R2=-15r1 +r2) 5 [ 1 � 15]1 (R2=30r2) �[01 01154] (Rl= 1 r2+�) x=-,43 y=-51 ( 43 -5 ). {2xx- y-3z==166 2y+ z= 4 [2 0 -30 166] 2 14 [� 01 -12 -30 46] (R2=-2�+r2) 2 14 r� 01 -11 -t0 261 (R2= r2) 2 14 [� 01 01 -t 21 (Rl= r2+r1 0 4 0 R3 =-2r2 +r3) [� 01 01 _-t1. 21 (R3=tr3) 0 0 1 [� 01 01 00 28] [Rl==trr33++r21 0 1 0x= 8,yR2= 2, = 0 (8,2,0).



[�

3x-5y= 3 {15x+5y=21

8

o

o

1 .

The solution is

640

t

Z

1j

or



49.

1

l


x-2y+ 3z = 7 2x+ y+ z = 4 -3x+2y-2z = -IO

53.

-x+ y+ z = -I -x+2y-3z = -4 3x-2y-7 z = 0 W rit e t he a ugment ed mat rix:

[=: � -� ��] [ : -� =� ] [ : �� -�] [� � �! ��] l l 3 -2 -7

0

1 -4 3 -2 -7 0

� -

1 � 0

0

�

1

2x-2y-2z = 2 51. 2x+ 3y+ z = 2 3x+2y = 0 W rite the a ugment ed ma trix: 3 2

�

�

o

o

55.

5 3 -3 -

1 -4 -3 0

�

3 -7

�

0 0 -3 There is no solut ion. The system is inconsist ent .

z

I

-2 3 -7 1

1 -1 t 3 0 -4 -1 2 _4 7 0 0

(R3 = -r2 + r3)

z

2x-2Y+ 3 z = 6 4x- 3 y+ 2z = 0 -2x+ 3y-7 z =1 W rite the a ugmented mat rix:

�

(R, �t,,)

0

1 [ ! �� ] [ � =� ! �1 [ 1( [� =1 ��1 ( -2

0 0

�

-

0 The ma trix i n the la st step rep resents the system X - 5z = -2 X = 5Z-2 y-4z = -30 r, equiva lent ly, y=4z -3 0 =0 0 =0 The solution is x= 5z -2, y= 4z -3 , is a ny rea l number or {(x,y,) z I x= 5z -2,y= 4z - 3, is a ny rea l number} .

The solution is x= 2, y= -1, z = 1 or (2,-1,1) .

[� -� -� �] [i -� -1 �] [� -� -� �] [� � � �]

o

(Rl = -Ij )

o

(Rl=trl) R2 =-41j+r2 R3 21j + r3 =

Rl r2 + Ij R3 -r2 + r3 =

]

]

0 0 -9 1 There is no solut ion. The syst em is inconsist ent . =

641



57.

l

X+Y- Z= 6 3x-2y+ z=-5 x+3y-2z= 14

[� -� -: -�] ->[i -� �� 1] ->[i 0 �� l

W rite the augmented matri x: 3 -2

1

14

-2

[�o 0 =! �j [I0 0 ¥1.] o0 0 [�o 0 � �l I, 2

�

�

�

1

t-s -t -1 1 -2 -2

The solution is x== 59.

!

The solution is X=-3,y=�,Z=1 or (R, --h)

6 1.

(-3,�,1).

3X+ y- z=j

2x- y+ z=1 4x+2y

8

=3

= (R3 b)

y== 3, Z= -2,

or

(1, 3,-2) .

x+2y- z=-3 2x-4y+ z==-7 -2x+2y- 3z==4

[ � -� -: =�l [[�oI -� -I-� -�=�]l 0 o [0I 0 j

W rite the augmented matrix: -2

�

�

�

2 -3

4

6 -5 -2

2 1 -t "8 6 -5 -2

o0

(R2 =-tr2)

--;\- -¥ -t "8I II II -4 -4

642



63.

1

{


x+ y+ z+ W= 4 2x- y+ z = 0 3x+ 2Y+ z - W= 6 x-2y-2z +2w = -1 W rite the a ugmented ma trix: 1

65.

[� -� � �1 � [ -� -� -:1 and [� -:1 (Interchange) [� : 1 [f � -� � ��1 [f ; -j �� 1 [f ; ! -� �1 [ � ; ! -1 ;1 [f ; P �1

X + 2Y + z = l 2x - y + 2z = 2 3x + y + 3z = 3

3 2 1 -I 6 1 -2 -2 2 -1

�

o

o

�

o o

�

o

o

-I

-2 -3 -3

-I

-4

-6

-5

- I -2 - 4 -3 - I -2 - 8 -3 -3 -5

1 2

4

-2 - 8 -3 -3 1 -5 -3

-I

r2

( R2

{

r3

=

The ma trix in the la st step rep resents t he system X + 2Y + Z = 1 - 5y = 0 0=0 Substitute a nd solve: -5y = 0 x + 2(0) +z = 1 y=o z = l-x The solution is y 0, Z 1 - x, x is a ny rea l number or { (x, y,z) I y = 0, z = I - x, x is a ny rea l number} .

-r2 )

=

4

( R, �h)

67.

4

=

{

x-y+ z= 5 3x + 2y - 2z = ° W rite the a ugmented ma trix: l � 2 �1

[� � 1 �] �[�

4

( R, �h)

4

1 ]

0 2 1 -1 -3 The ma trix in the la st step rep resents the syst em X=2 or, equiva lently, x = 2 y =z-3 y -z = -3 Thus, the solution is x = 2 , y z - 3 , z is a ny rea l number or { (x, y,z)lx = 2, y = z -3, is a ny rea l number} . °

4

The solution is x= I , Y= 2, z= 0 , W =1 or ( 1 , 2, 0, 1 ).

{

{

=

643



1 r � �1 =� �01


2X + 3 Y - z = 3 x - y -z = O 69. -x + y +z = O x + y + 3z = 5 W rit e t he a ugment ed mat rix:

Y-7

=��jl

� [� I

n -1

----'- [� 1 � o 10 [� o1� -8 11 � oo 01.!2.0 rj � ----,

o 2 o 0 =�0 �0 -S -7

7 4 5

C:) = -7

13 x =9 1 3 7 1 9 or Th us, t h e so 1utlOll · IS· = -, Y = - , z = 9 3 7 19 9'18'18 .

18 18

(1 ) { 4x + y + z - w = 4 x- y + 2z + 3w = 3 Writ e t he a ugment ed mat rix: 1 [; -1 2 � 1;] (int ercha nge) -1 2 �1 1:] a nd r2 X

�1 ( int erCha nge ) -11 11 31 05 a nd r2 � �1 �1 o 0 00 o 2 45 (int ercha nge)

� rr� �

X-S

Y = 18

1 3 5

-1 1 1 1

C:) = 7 7

71.

1j

1j

{

The mat rix in t he la st st ep rep resent s t he syst em x - y + 2z + 3w = 3 5y - 7z - 1 3w = -S The second equat ion yields 5y - 7z - l 3w = -S 5y = 7z + 1 3w - S 13 S 7 + -w-y = -z 5 5 5 The fl rst equat ion yields x - y + 2z + 3w = 3 x = 3 + y - 2z - 3w Subst itut ing for y: x = 3 + - S + 7 z + 1 3 w - 2z - 3w S S S 3 x = --z - -2 w+ -7 5 5 5 3 - -2 w+ -7 , · · x = --z Thus,t h e so lutlOnlS 5 5 5 1 3 --S , z a nd wa re a ny rea l numbers or y = -7 + 5 5 5 , x, y , Z) x = - � z - � w + .?' y = .? z + � w- !' 5 5 5 5 5 5

r) a nd r4

( R2 = -2r) + r2 )

-7 -7 IS 9

(

-7 -7 -7 18

The mat rix in t he la st st ep rep resent s t he syst em X - SZ = -7 y - 7z = -7 19 z =IS Subst it ut e a nd solve:

{(W

z

)

i

W

}

z a nd w a re a ny rea l numbers .

644

© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without pennission in writing from the publisher.


(I nt ercha nge)

73. Ea ch oft he point s must sat isfy t he equat ion 2 y = ax + bx + c . (1, 2) : 2 = a + b +c (-2, -7): - 7 = 4a - 2b + c (2 3) - 3 = 4a + 2b + c Set up a mat rix a nd solve:

r3 a nd fj

:

,-

[� -� -�l 4

2

-3

o o o

The solut ion is a = - 2, b = 1, c = 3 ; so t he equat ion is y = _2x 2 + + 3 .

o o

X

75. Ea ch oft he point s must sat isfy t he equat ion f(x) = ax 3 + bx 2 + cx + d. f(-3) = -12 : -27a + 9b - 3c + d = - 1 1 2 f(-I) = -2 : -a + b- c + d = -2 f(l) = 4 : a + b+c+d= 4 8a + 4b + 2c + d = 1 3 f(2) = 1 3 : Set up a mat rix a nd solve: -1 2

I

The solut ion is a = 3, b = -4, c = 0, d = 5 ; so t he equat ion is f(x) = 3x3 - 4x2 + 5 .

[-�� : =� ��1 1 1 8 4

o 0 "3I .!±3 0 -"35 - "3 0 0 0 -5 - 25 o 0 "3I .!±3 0 -1. _1. 0 3 3

4 13

645

© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws

as

they currently



== =

Condition on investment equa tion: z = O.5x x - 2z = 0 Set up a ma trix a nd solve:

77. Let x the number of servings of sa lmon stea k. Let y the number of servings of ba ked eggs. Let z the number of servings ofa corn squa sh. P rotein equa tion: 30x + 15 Y + 3z = 78 Ca rbohydra te equa tion: 20x + 2y + 25z = 59 V ita min A equa tion: 2x + 20y + 32z = 75

0.07

o

1

10,000 1 0.01 0.02 80 -3 -10,000 -1

(I ntercha nge) r3

1 10,000 0.08 680 0 -2

1 0,000 2

1 -1

a nd r 1

-

3

-� �� -� ��

o

o

]

-1 - 2000

o

]

8000 -10,000

]

1 2000

o

0

4000

1 0

o

4000

1 2000

]

]

Ca rletta should invest $4000 in Trea sury bills, $4000 in Trea sury bonds,a nd $2000 in corpora te bonds.

== =

81. Let x the number of D elta s produced. Let y the number of Beta s produced. Let z the number of Sigma s produced. Pa inting equa tion: l Ox + 1 6 y + 8z = 240 D rying equa tion: 3x + 5 y + 2z = 69 P olishing equa tion: 2x + 3 y + z = 41 Set up a ma trix a nd solve:

Substitute z = 1 a nd solve: -1 98y - 295(l) = - 69 1 -198y = -396 y=2 x + 10(2) + 16(1) = 37.5 x + 36 = 37.5 x = 1 .5 The dietitia n should serve 1 .5 servings of sa lmon stea k, 2 servings of ba ked eggs, a nd 1 serving of a corn squa sh.

] [ [� � � ��l [� � � �:O] ->[i 10 16 3 5 3

2

-t

2

== =

79. Let x the a mount invested in Trea sury bills. Let y the a mount invested in Trea sury bonds. Let z the a mount invested in corpora te bonds. Tota l investment equa tion: x + y + z = 10, 000 Annua l income equa tion: 0.06x + O.07y + 0.08z = 680

-t

646

o

3

8 240 2 69 1

41

1 41

-3

-25

© 2008 Pearson Education, Inc., Upper Saddle River, Nl. All rights reserved. This material is protected under all copyright laws as they currently


1


--)

1 0 0 0 * o I 0 0 2 '6 o 0 I 0 23 o 0 0 1 �

16 44 -, -, 12 =2, I) =23 The so Iuti· on IS· I, =23

= =

{-

3.

83. Rewrite the system to set up the ma trix a nd solve: 212 = 4 4 + 8 - 212 = 0 8 = 514 + 1, � I, + 514 = 8 I, + 313 = 4 4 = 313 + I, � -� -� = O �+� = �

z

{ [�I � � �:] (Interchange) [� j � -� -�� ]�] and --) [� � -I� �] � 0 0

3

-I

o 0 o 0

3 -5 -4 -6 -8

--)[�� :] ] �r� ; � - 6 -8 3 -5 -4

-I

0 -2

5 8 o 2 6 8 I

]

1

I 20,000 I I 0.07 0.09 0.11 2000 1 20,000 --) 7 9 11 200,000 (R2 =100'2 ) 20,000 ----, o 2 1460,000 (R2 ='2 -7/i)

[I I 1 ] -'-[I 1 ] --)[� �I��:��] (R2=t'2 ) [ 1 - 11 - 1 0, 000] (R[ = - '2) ---,

0 4 0

[--) � � � o 0 o 0

[

-I

'2

28 23

85. Let x = the a mount invested in Trea sury bills. Let y the a mount invested in corp ora te bonds. Let z the a mount invested in j unk bonds. Tota l investment equa tion: x+y+ =20,000 Annua l income equa tion: 0.07x+ 0.09y+ O.llz=2000 Set up a ma trix a nd solve:

The compa ny should p roduce 8 D elta s,S Beta s, a nd 10 Sigma s.

I

14

Ii

----'0.

a

'i o 1 2 30, 000 The ma trix in the la st step rep resents the -z = -10' 000 system y + 2z = 30, 000 Therefore the solution is = -10, 000 + z , y = 30, 000 -2z , z is a ny rea l nu mber. P ossible investment stra tegies:

{X

x

(InterChange) ') and'4

Amount Invested At

7% 0 1 000 2000 3000 4000 5000

-2

;�

9% 1 0,000 8000 6000 4000 2000 0

1 1% 1 0,000 1 1 ,000 12,000 1 3,000 14,000 1 5,00 0

647


Chapter 12: Systems of Equations and Inequalities b.

1

Tota l invest ment equa tion: x + y + z = 25, 000 Annua l income equa tion: 0.07x + 0.09y + 0. l lz = 2000 Set u p a ma trix a nd solve: 1 1 1 25, 000 0.07 0.09 0.11 2000 1 1 I 25, 000 = 1 00 7 9 1 1 200, 000 1 2 5, 000 --"- 1 = - 71j ) 0 2 4 25, 000 1 2 5, 000 --"- 1 = 2 12, 500 0 - 1 1 2, 500 = 1j 2 1 2, 500 The ma trix in the la st step rep resents the x - z = 12, 500 system y + 2z = 1 2, 500 T hu s, the solu tion i s x = z + 1 2, 500 , y = -2z + 12, 500 , z is a ny rea l nu mber. P ossible invest ment stra tegies:

[ �[ [ [

] 1 ] (R 2 r2) 1 ] (R 2 r2 1 ] (R 2 tr2) 1 ] (R( -r2) 1

----r

----r

{

{

=

Amount Invested At

7% 30,000

c.

9% 12,500 8500 4500 0

9% 0

1 1% 0

This will yield ($30,000)(0.07) $2100, which is more tha n the requ ired income. =

d.

87.

Amount Invested At

7% 12,500 14,500 1 6,500 1 8,750

] (R2 = tr2) 1 ] (R( -r2)

1 3 0, 000 2 -5000 o -1 3 5, 000 = 1j I 2 -5000 T he ma trix in the la st step represents the x - z = 35, 000 system y + 2z = -5000 Thu s, the solu tion is x = z + 35, 000 , y = -2z - 5000 , z is a ny rea l nu mber. H owever, y a nd z ca nn ot be nega tive. From y = -2z - 5000 , we mu st ha ve y z = O. O ne p ossible invest ment stra tegy

=

Let x the a mou nt of su pplement 1 . Let y the a mou nt of su pp lement 2. Let z the a mou nt of su pplement 3. 0.20X + 0.40 y + 0.30z = 40 V ita min C 0.30x + 0.20y + 0.50z = 30 V ita min D Mu ltiplying ea ch equa tion by 1 0 yields 2X + 4 Y + 3z = 400 3x + 2y + 5z = 300 Set up a mat rix a nd solve:

{ {

1 1% 0 2000 4000 6250

Answers will va ry. =

=

Tota l investment equa tion: x + y + z = 30, 000 Annua l income equa tion: 0.07x + 0.09y + 0. llz = 2000 Set up a ma trix a nd solve: I 1 I 3 0, 000 0.07 0.09 0. 1 1 2000 1 1 1 30, 000 = I 00r2 7 9 1 1 200, 000 1 30, 000 --"- 1 = r2 - 71j ) o 2 4 - 10, 000

[

�[ [ -r

1

1

1

] ] (R 2 ] (R(

)

648


Section 12.3: Systems of Linear Equations: Determinants

{

{

The ma trix in the la st step rep resents the system X + i Z = 50 y - i z = 75

15. x + y = 8 x-y = 4 D = I 1 =-1 - 1 =-2 1 -I D = 8 1 =-8 -4=-12

Therefore the solution is x = 50 - 2. 4 Y = 75 + � Z , z is a ny rea l number. 8 P ossible combina tions: Supp lement 1 Supp lement 2 Supp lement 3 Omg 50mg 75mg 8mg 36mg 76mg 16mg 77mg 22mg 78mg 8mg 24mg Z ,

x

y

17.

Section 12.3

{2x5X +- 3yy == 11 32 D = I � -� 1 = 15+2 = 17 D = 1 13 -1 1 = 39+12 = 51 12 3 D = 1 5 13 1 = 60 - 26 = 34 y 2 12 x

1. determina nts 3. Fa lse

I! � 1 = 3(2) - 4(1) = 6 - 4 = 2 7. I_� � 1 = 6(3) - (-1)(4) = 18+4 = 22 9. I-! -� 1 = -3(2) - 4(-1) = -6+4 = -2 11. J � �l �JI�� j�41: � H : ��I

Find the solutions by Cra mer's Rule: Dy 34 Dx = 5 1 = 3 y =x ==-= 2 D 17 D 17 The solution i s (3, 2).

5.

19.

�

�

= 3[( -1) ( -2) -2 (5)]-4[1 (-2) - 1 (5)]

=3 ( -8) -4 ( -7)+2 (3) =-24 +28+6 =10 4 6

1 1

D = 1 8 =4 -8=-4 1 4 Find the solutions by Cra mer's Rule: Dy -4 Dx = -= -12 6 y = =-= 2 x=_ D -2 D -2 The solution is (6, 2).

89 - 91. Answers will var y.

13.

I 1 1 4 -1 1

+2[1 (2) - 1 ( -1)]

=

Find the solutions by Cra mer's Rule: Dy -24 Dx = 48 = 8 y = x == = -4 D 6 D 6 The solution is (8, -4) . -

1 1 1 1 1 1 -3 1

-I 2 -I ° =4 -1 -3 -3 4

{3Xx + 2y == 240 D =I � � 1 = 6-0 = 6 Dx = 1 2� � 1 = 48 - 0 = 48 Dy 1 3 24 1 = 0 - 24 = - 24 1 0

° ( 1) 6 ° +2 6 -1 4 __ 1 4

=4[ -1 (4) -0 (-3)] + 1[6 (4) - 1 (0) ] +2[6 (-3) - 1 ( -1)] = 4( -4) + 1 (24) +2 (-1 7) =-16 +24 - 34 =-26

649




{

3X -6Y = 24 21. 5x+4y = 12 D = - = 12 - (-30) = 42

29.

I� �1 Dx = 1 24 -6 1 = 96 - (-72) = 168 12 4 D = 1 3 24 1 = 36 - 120 = -84 y 5 12

1 1 1 D, = I � : 1 = 1 -6= -5

Find the solutions by Cra mer's Rule: 15 D y -5 D 3 y=-=-=l -"2 =X=-x =-D -5 2 D -5 The solution is 1 .

3X - 2Y = 4 23. 6x -4y = 0 D = 3 -2 = -12- (-12) = 0 6 -4 Since D = 0, Cra mer's Rule does not app ly. -4Y = -2 25. 2X 3x+2y = 3 D = - = 4+12 = 16

(%, )

1 1

{

r::1 :� �

D= 2 3 =- 2 - 3=-5 1 -1 6 3 =_6_�= _� Dx = .1 -1 2 2 2

Find the solutions by Cra mer's Rule: D -84 Dx =168 =4 y=--..L x=_ =-- = -2 D 42 D 42 The solution is (4,-2) .

{

3

31 .

I � �1 Dx = 1 -2 -4 1 = -4+12 = 8 3 2 2 D = 1 -2 1 = 6+6 = 12 y 3 3

{ 3x - 5y= 3 1 5x + 5y= 2 1 D= \ 3 -5 \ = 1 5- (-75)=90 15 5 Dx = 1 3 -5 \ = 1 5 - (-105)=120 21 5 y = \ 3 3 \ =63 - 45= 1 8 1 5 21 D

Find the solutions by C ra mer's Rule: D x= Dx = 120 = 4 y= y = 1 8 ="51 Ii 90 "3 1) 90 The solution is

Find the solutions by Cra mer's Rule: 8 = 1 Y = Dy = 12 = 3 x = D; = 16 "2 1) 16 4 The solution is .

(�, %)

2x - 3y = -1 27. { lOx+ lOy = 5 D = 1 2 -3 1 = 20- (-30) = 50 10 10 D = 1 -1 -3 1 = -10 - (-15) = 5 5 10 D = 1 2 -1 1 =10- (-10) = 20 y 10 5

33.

x

(�, �)

{

x + y - z= 6 3x - 2y + z =-5 x + 3y - 2z = 14 -1 D= 3 - 2 3 -2 =1 - 2 1 _ 1 3 1 + ( _ 1) 3 3 -2 1 -2 1 = 1(4 - 3) - 1(-6 - 1) - 1(9+2) =1+7 - 1 1 =-3

1

Find the solutions by Cra mer's Rule: D 20 2 Dx = 5 = -1 x=D 50 10 y =Dy = 50 = -5 The solution is 10 5

(J..., 3.).

.

1 1

1 1

650




Dx

Dy

6 -1 = -5 - 2 1 14 3 - 2 =6 - 2 1 _ 1 -5 1 + ( _ 1) -5 -2 3 -2 14 -2 14 3 =6(4 - 3) - 1(10 - 1 4) - 1(-1 5 + 28) =6+ 4 - 1 3 =-3 6 -1 = 3 -5 1 1 14 -2 1 + (_1) 3 = 1 -5 14 - 2 1 -2 1 14 = 1(1 0 - 14) -6(-6 - 1) - 1(42 + 5) -4 + 42 - 47 =-9 1 6 = 3 - 2 -5 1 3 14

=

Dz

1

1 1

/

1 / _ 6/ 3 /

1 1

Dx = -7-34 -422 -3-11 =-3 1 - � _�1 -2 1 -� _�I + (- I)I -� - �I ==-30-34-2 -3(12 -2) -2(21-4) -1(-14 + 16) =-66 -3 -1 Dy = 2 -7 1 -2 4 -3 =1 1 -74 -31 1 _(_3)1 -22 -31 1+(_1) 1 - 22 -47 1 = 1(21-4)+ 3(-6 + 2) -1(8 -14) = 1 7-12+6 =11 2 -3 Dz = -22 -42 -74 = I I - � -: 1 - 21 _ � -:1 +(-3) 1_� -�I ==-2+12+12 1(-16+14) -2(8 -14) -3(4 -8) =22 Find the solutions by Cra mer's Rule: Dy 1 1 1 Dx = --66 = -3 y -=-=x=_ = D 22 2 D 2222 D z = z = =1 D 22

1

/ -5 /

/ � �� / - 1 / � �� / +6/ � -� /

=I -

= 1(-28 + 1 5) - 1(42 + 5) +6(9 + 2) = -13 - 47 + 66

=6 Find the solutions by Cra mer's Rule: -3 -9 =-=1 y= -L=-=3 -3 -3 6 -2 =-= -3 The solution is (1, 3, -2) .

x=_DDx z = _DDz

35.

{

D D

{

( �)

The solution is -3, , 1 .

37.

x + 2y - z=-3 2x - 4y z = -7 -2x + 2y - 3z= 4 -1 -4

+ D= -22 22 -31 =1 1 - � _�1 -2 1 _ � _�I +(- I) I _ � ==10+8+4 1(12 -2) -2(-6 + 2 ) -1 ( 4 -8) =22

x - 2y + 3z= 1 3x + y - 2z = 0 2x - 4y +6z = 2

1 2 3 D = 3 - 1 -2 2 -4 6 = 1 1� - �1-(- 2) 1� - �1 + 3 1� �l = 1(6 -8) + 2( 18 4) + 3(-12 -2 ) =-2 + 44-42 =0 Since D = 0 , Cra mer's Rule does not app ly. +

651


{


39.

x + 2y - z=O 2x -4y + z = 0 -2x + 2y - 3z = 0 1 2 -1 D= 2 -4 - 2 2 -3 =1 -4 1 _ 2 2 1 + (_ 1) 2 -4 2 -3 -2 -3 -2 2 = 1(12 - 2) - 2(-6+ 2) - 1(4- 8) = 10+ 8 + 4 = 22 o 2 -1 Dx = 0 -4 1 = 0 [By Theorem (12)] o 2 -3 0 -1 D = 2 0 1 =0 [By Theorem (12)] -2 0 -3 I 2 0 Dz = 2 -4 o =0 [By Theorem (12)] -2 2 0 Find the solutions by Cramer's Rule: Dy 0 Dx =-= 0 O -=O y=-= x= D 22 D 22 z = Dz = � = O D 22 The solution is (0, 0, 0).

\

\ \

\ \

43.

Solve for x: =5

I: � 1

3x -4x = 5 -x= 5 x=-5

\

45.

Solve for x: x 1 4 3 2 =2 -1 2 5 x

l � � 1 - l l _� � 1 + l l _� � 1=2

x ( 1 5 -4) - ( 20+ 2) + ( 8 + 3 ) = 2 l lx - 22 + 1 1 = 2 l lx = 1 3 13 x= II

y

41.

47.

x 2 3 1 x 0 =7 -2 6 x

{

1 1

1 1

l � x_( -�2x1 -)2-1 �2(-_2�) \++33(11�-6x�)1==77

- 2x2 + 4 + 3 - 1 8x = 7 _ 2X 2 - 1 8x = 0 - 2x ( x + 9 )= 0 x =0 or x=-9

x - 2y + 3z=0 3x + y - 2z= 0 2x -4y +6z = 0 1 -2 3 -2 D= 3 2 -4 6 = 1 1 - 2 _ (_ 2) 3 - 2 + 3 3 1 -4 6 2 6 2 -4 = 1(6 - 8) + 2(1 8 +4) + 3(-12 - 2) = -2+44-42 =0 Since D= 0 , Cramer's Rule does not apply.

1

Solve for x:

x y z 49.

u

v

w

=4

2 3 By Theorem ( 1 1), the value of a determinant changes sign if any two rows are interchanged. 1 2 3 Thus, =-4 . x y z

1

U

v

w

652



51.

xY z Let \ u v wi = 4 . 2 3 x -3 -6 -9 1 = -3 1 1 x

u

Y

v

55. Y

z 2 3 v

z

w

u =-3 - I l u

x

(

)

2 2

v

z w

2 3

i

[Theorem ( l l )]

= 3(4) =12

53.

z

�1

=4

3 2x 2y 2z u - I v - 2 w- 3 2 3 [Theorem (14)] z = 21 x Y u-I v-2 w-3 x Y z = 2(-1) \ 2 3 [Theorem (1 1)] u- I v - 2 w - 3 x Y z = 2(-I)( - I) \ U - I v - 2 w - 3 1 [Theorem (1 1)] 2 3 xYz [Theorem (1 5)] = 2(-1)(- 1) I u v w (R2 -r3 +r2 ) I 2 3

[Theorem ( 14)]

w

Y

xY Let I u v

xY z Let \ u v w \ = 4 2 3 2 3 x-3 y-6 z - 9 2u 2v 2 w 3 2 = 21 x -3 y- 6 z - 9 [Theorem (14)] u v w x-3 y-6 z-9 2 = 2(-1) 1 3 [Theorem (11)] u v w x-3 y-6 z-9 = 2(-1)(-1) \ u w [Theorem (1 1)] v 3 2 xYz [Theorem (1 5)] = 2(-1)(-1) u v w (Rl = -3r3 +rl ) I 23

=

= 2(-1)(-1)(4) =8

57.

x Y xl Yl I I = 0 X2 Y2 X I Y2Yl II I _ y I X2Xl II I + I I X2xl Y2YI I = 0 X(YI - Y 2 ) - Y (XI - X2 ) + (xI Y2 - X2Yl ) = 0 X(YI - Y 2 ) + Y (X2 - XI ) X2YI - XI Y 2 Y (X2 - Xl) = X2YI - X1 Y 2 + X(Y2 - Yi ) Y (X2 - Xl) - YI (X2 - Xl ) = X2YI - XI Y2 + X(Y2 - YI ) - YI (X2 - XI ) (X2 - XI ) (y - Yl ) = X (Y2 - Yl ) + X2YI - X1 Y 2 - Y1 X 2 + Y1 XI (X2 - XI )(Y-Yl ) = (Y2 - Yl )X - (Y2 - YI )XI (X2 - XI )(y - YI ) (Y2 -YI )(X - XI ) (Y - YI ) = (YX22 -Yl - Xl ) (X-Xl )

Expa nding t he det ermina nt :

=

= 2(-1)(-1)(4) =8

=

f

\

This is t he 2-point form oft he equat ion fora line.

653


as

they currently



59.

ExpandiXng D,x wexobtain D = -21 YII Y22 Y33 1 1 1

LetA =(xl, YI ),B = (x2 , Y2 ) , and C= (x3 , Y3 ) represent cesonofoura trtriiangle. Forn the first siquadrant. mplicity,theSeewevertipositi a ngle i be the poifromnt closest toaxithes, andy-axiB sbe, thefigure. be poithenpoiLett "between" nAt farthest theA andY poi n ts C. C

A (x,

= �1 ( XI (Y2 - Y3 )-x2 (YI - Y3 )+x3 (YI - Y2 ) ) = -(21 xIY2 -xIY3 -x2 YI x2 Y3 + x3� -X3 Y2 ) which is the same as the area of triangle ABC. Ifdiftheferentl vertiycthan es ofshown the triainnglethearefigure, posittheionedsigns may the absolute value of D will gibevereversed. the area ofThus,the triangle. ces of a triangle are (2, 3), (5, 2), and If(6,the5),verti then: 2 5 6 D= .!.2 3 2 5 1 = � ( 2 1 � � 1 - 5 1 � � 1 +6 1 � � I ) = � [2(2 -5) -5(3 - 5) + 6(3 - 2)] = � [2(-3) - 5(-2) + 6(1)] = �1 [-6+ 10+ 6] The=5area of the triangle is 151 = 5 square units. If a = 0, then b 0 and e 0 since { by = s . ad -be 0 , and the system is ex+dy=t The solution of the system is Y = :.-,b t -dy t - d(t) = tb - sd . . Cramer' s X = --= e e be Rule, we get D = I � � I = -be , D = I st db I = sd -tb ' Dy = I � ; 1 = 0 -se = -se , so +

D (x" O)

E (x" O)

Let D = (xl ' O), E = (x2 ' 0) , and F= (x3 ' 0). Wethe areas find theof tarea ngle ABCandbyBEFC subtracti rapezoiof tridsaADEB fromng the area of trapezoid ADFC. Note: AD = YI , BE = Y2 , CF = Y3 ' DF = x3 -XI ' DE = x2 -xI , and EF = X3 - x2 . Thus, the areas of the three trapezoids Iare as follows: = -(X -X + 2 3 I )( YI Y3 ) I -X = -(X + 2 2 I )( YI Y2 ) ' and I -X + = -(X 2 3 2 )( Y2 Y3 ) ' The area= of our triangle ABC is I -X )( + ) = -(2I X3 -XI )( YI + Y3 ) --(X 2 2 I YI Y2 - -(2I x3 -X2 )( Y2 + Y3 ) I I I 1 - �� I = �� + �� - �� - �� --x21 2 Y2 + -2I xIYI + -2I xI Y2 --x2I 3Y2 --x21 3Y3 + -2I x2 Y2 + -2I x2 Y3 I I 1 + ��Y 1 2 - ��Y 1 2 + �� I = �� - �� - �� = -(21 x3 YI -xIY3 - x2 YI + xIY2 - x3 Y2 + X2 Y3 ) KADFC KADEB

KBEFC KABC

KADFC - KADEB - KBEFC

61.

*

*

*

--

Usmg

x

654

© 2008 Pearson Education, Inc., Upper Saddle River, N J. All rights reserved. This material is protected under all copyright laws a s they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


x = DDx = ds--betb = tdbe- sd and Dy --= --bese -bs , wh' h ' the so iutl' On. Note Y = -= that these solutions agree if = O. If b = 0, then a;t 0 and d 0 since ad - be ;t 0 and the system is {axex+dy=t=s . The solution of the system is x = a y = t -dex atad- es . Usmg. Cramer' s Rule, we get D = I : � 1 = ad, Dx = I ; � 1 = sd , and DY = I ae st l =at _ es , so x= DDx = aSdd = !...a . Dy at - es h' h ' the soiutl' On. and y = -= Note that Dthese soladutions agree if e = O . If e = 0, then a;t 0 and d ;t 0 since ad - be ;t 0 , and the system is {ax+bY ==st . The solution of the system is y = d x = s -aby = sda-dtb Using Cramer' s Rule , we get D = 1 aO db 1 = ad ' Dx = 1 st db 1 = sd - tb ' and Dy = 1 a0 st 1 = at , so x = DDx = sdad- tb and Dy at t ' h' h 1S' the soiutl'On. Note Y =-=-=D ad d that these solutions agree if b = O. If d = 0, then b;t 0 and e ;t 0 since ad - be 0 , and the system is {exax+bY ==ts . The solution of the system is x = !-.e , s -bax esbe- at Usmg. Cramer ' s Rule, we y = --= IC

D

IS

x

d

;t

!.... ,

63.

-- --=

IC

IS

=

=

D x

IS

-- --

(

dy

!.... ,

-- --- .

W

=

D

,

--- , w

get D = I: �I = 0 - be = -be, D Ist bol 0 - tb -tb ' and DY = lae st l = at - es , so x = D = --betb = !-.e and y at - es = es - at , whi.ch . the Y = -= sola =0.ution.D Note-bethat thesebe solutions agree if Evaluating the determinant to show the relationship : all a12 a13 ka21 ka22 ka23 a31 a32 a33 _- all 1 kan ka23 I -a1 2 1 ka2 1 ka23 1 + a13 1 ka21 a32 a33 a31 a33 a31 = all (ka22a33 - ka23a32 ) -a1 2 (ka21 a33 -ka23a31 ) + a13 (ka21a32 -ka22a31 ) = kall (a22a33 -a23a32 ) -kaI 2 (a21 a33 -a23a31 ) + ka13 (a21 a32 - a22a31 ) = k all (a22a33 - a23a32 ) -a12 (a21 a33 -a23a31) +a13 (a21a32 -a22a31 ) ) = k ( all 1 a22a32 a23a33 1 -a12 I a21a31 a23a33 1 + a13 1 a21a31 a2a322 I) = k a21all a22a1 2 a23a13 a31 a32 a33 relEvalatiuoatinshingpthe: determinant to show the all + ka21 a12 + ka22 a13 ka23 a21 a22 a23 a31 a32 a33 = (all + ka21) 1 a22a32 a23a33 1 _ (a12 + ka22 ) 1 a31a21 + (aI3 + ka23 ) 1 a21a31 a22a32 1 = (all + ka21)(a22a33 -a23a32 ) - (aI2 + ka22 )(a21 a33 -a23a31 ) +(aI3 + ka23 )(a21 a32 -a22a31 )

65.

+

IC

;t

--- .

655



= all (a22a33 - a23a32 ) + ka2l (a22a33 -a23a32 ) -a12 (a2I a33 - a23a31 ) - ka22 (a21 a33 - a23a31 ) a21a32 - a22a31)+ ka23 (a21a32 - a22a31) = all +a13( (a22a33 - a23a32) + ka21a22a33 -ka2la23a32 - a1 2 (a21a33 - a23a31 ) -ka22a2l a33 + ka22a23a3l + a13 (a2Ia32 -a22a31) + ka23a2l a32 ka23a22a31 = all -(a22a33 - a23a32 ) - a12 (a21 a33 - a23a31 ) +a13(a2Ia32 -a22a31 ) = all l ::� ::: I -a1 2 I::: ::: I +a13 I :: : ::� I = a2lall a22al2 a23a13 a31 a32 a33

13.

1 5.

Section 12.4 1. 3. 5. 7.

9.

1 1.

1 7.

inverse identity False A+B = [� � -�]+[_ � � _�] 3 + 1 -S + 0] = [ 1 + 0+4 (- 2) 2 +3 6+ (-2) = [ -� : -:] 4A = 4 [0I 23 -S6] = [44.·01 44.3· 2 4(4·6-S)] = [40 128 -2024] 3A - 2B=3 [01 23 -6S] _ 2 [-24 31 -20]

=[� � = [-8 7

7 °

AC�[� � -�lU !] 0(1) + 3(2) + (-5)(3)] = [0(41(4) )++ 3(62(6) +)+(-65)(-2) (- 2) 1(1)+ 2(2)+ 6(3) = [284 -239] � -�] CA = -2 3 = 6(04(0)) ++ 2(11(1)) 6(34(3)) ++ 21((22)) 64(-(-55))+1(+ 2(6)6) - 2(0)+ 3(1) - 2(3)+ 3(2) - 2(-5) + 3(6) = [� �� 3 0 28

[

[ : �l ' [�

1

=�:l

U �l([� � -�H-� U[1 5 �}l-� : -!] = 22 :� =��l -11 22

c(A + B) �

�

1 9.

21.

AC - 312 CA - CB

=[� �

3

-:Jl

7

-9] _ [ 3 0] 23 0 3 -9] 20

- ��] - [-! � -�] - IS] 22

656


Section 12.4: Matrix Algebra

23 .

all = 2(2)+(-2)(3) = -2 al Z = 2(1)+(-2)(-1) =4 al 3 = 2(4)+(-2)(3) = 2 al 4 = 2(6)+(-2)(2)= 8 aZ I = 1(2) + 0(3) = 2 azz = 1(1)+0(- 1) 1 aZ3 =1(4)+0(3)=4 aZ4 = 1(6) + 0(2) = 6 [ � - �][ � _: ; �] = [- � 4 � :]

31.

=

25.

=

27.

°

� [� : I-�

!l[ �l

5

1

[�[ : m: _;] 1(1)

=

2 9.

/'j

[� �l �l 1(1)+2(- 1)+3(2) 1(2) + 2(0) + 3(4) ] [0(1)+(-1)(-1)+4(2) 0(2) + (-1)(0) + 4(4) [9 146]

=

+ 0(6) + $ 8)

\(3) + 0(2)

+

1(-1)

2 ($ + 4 (6) + \ (8)

2(3) + 4 (2) + $ - I )

3(1)

3(3)

+ 6(6) + \(8)

+ 6 (2) + \( - $

A = [� �] identity and use the matrito findx wittheh theinverse: Augment row operations [� � I � �] ) � [� � I � �] ( Interchange and rz � [ 2 -21 1 01 -31 ] (Rz =-3/'j +rz )

33.

]

A = [� �] wherea*O. identity and use Augment row operatithoensmatrito findx wittheh theinverse: [a2 a1 1 1 01] [a1 ta i t 01] ( Rl = t rl ) �1 (Rz = - a + rz ) � [� � [� �

[ � �] the identity and use row Augment find thex witinhverse: operatio1nstheto matri [� � 1 �1 0 �]1 ( Interchange) �U 1 1 1 0 0] 1 and rz � [� -1 1 1 - 2] (Rz = -2rl + rz ) �[� � 1 _� �]- ( Rz = -rz ) � [� � 1 _ � 1 �]- 1](Rl = -rz + ) Thus, [-1 2

t]

°

°

/'j

A=

--[;

�

�

A

-I

=

.

6 57




35.

[ �l

A = � �� - 2 -3 0 Augment theto matri h the identity and use row operations fmd thex wiintverse:

->

[-2� ��- 0� �0 0� �l1 (R, = 2 r, + r, J �l I � [� �� 2 l = R -h , ( �l i � ! [� �: - o1 0 0 2 2 l� ! =1 2 ] l� : -� -�

->

3

,

->

->

_

_

.1

2.

39.

1

->

41.

37.

Augment h the identity and use row operationstheto matri find thex wiintverse: � -� � � 3 1 2 0 0 1 - 4 -3 � - 2 -3 0 4 3 -� (R2 = - r2 ) - 2 - 1 -3 0

[� � [o� [o�

�

-1

�l

-1

�l �ll

�] - 11 ['{! -!1 5

3

-;;

2 -7

-7

7

9

7

7

1. 7

_

.£. 7

-;;

1. 7

7

->

-t

l l�

0 -3 - 2 1 1 4 3 - 1 ( R3 = t r3 ) � 00 0 I 0 I (RR2I :--3r34+r3 �+ r2 ) 0 I I Thu, 2 I {2X+ y = 8 x+y =5 Rewrite the system of equations in matrix form: A = [� �J X = [; J B = [!] Find the inverse of A and solve1 X1]= A-I B : From Problem 29, A- I = [- 1 -2 , so X = £ I B= [_: -�] [!] = [�l The solution is x = 3, y = 2 or (3, 2) . { 2X+ y = 0 x+y = 5 Rewrite the system of equations in matrix form: A = [ � � l X = [;l B = [ �] Find the inverse of A and solve X = A- I B : From Problem 29, A- I = [- 11 -21] , so X = £I B= [_ � -n[�] = [��l The solution is x = -5, y = 10 or (-5, 10) . 7 -7

7

.

7

658



Section 12. 4: Matrix Algebra

43.

{2x6X ++ 2y5Y == 2 Rewrite the system of equations in matrix form: A = [� �J. X [;J. B = GJ Find the inverse of A and solve X = A- I B : From Problem 31, A- I = [- 11 _2.23] , so X=A- I B=[_ � - t] [�] = [-�l The solution is x=2,y = - 1 or (2, - 1) . { 2x6X ++ 2y5Y == 135 Rewrite the system of equations in matrix form: A= [ � �l X = [;l B = [ I �] Find the inverse of A and solve_X = A- I B : From Problem 31, A- I = [- 11 2.]32 , so X=[I B= [_: -t][I�] = [!l The solution is x = �, y = 2 or (�, 2) . X + y== -3a a*"O { 2ax+ay Rewrite the system of equations in matrix form: A = [� �l X = [ ;l B = [=:] Find the inverse of A and solve X = A- I B : From Problem 33, A- I = [- 11 -1.-; ] , so X � A-'B � [_ : - �] [=!lf�l The solution is x = -2, y = 1 or (-2, 1) . 7

49.

=

45.

47.

X + y =2a a*"O { 2ax+ay =5 Rewrite the system of equations in matrix form: A =[� �J x=[;J B= [;] Find the inverse of A and solve X = [ I B : From Problem 33, A- I = [- 1 -1.-; ] , so _ 1 [ I X =A- B = - 1 1.�][�7 ] = [1.�] . The soIutl· On · x = -a2 , y = -a3 or (-,a2 a3) X - 2y+y+z=z = 01 y - 2x - 3 = -5 Rewrite the system of equations in matrix form: X� A� : Find the inverse of A and solve X = A- I B : 3 -3 From Problem 35, A- I = - 2 2 so -4 5 - 2 3 3 X =A- I B = - 2 -2 -4 5 The solution is x = - 2, y = 3, z = 5 or (-2, 3, 5) . X =2�:;:� - 2x - 3y 2 Rewrite the system of equations in matrix form: X� B� A� : Find the inverse of A and solve A-I B : 1

a

51.

53.

a

{

1

IS

-

.

U �� n [ l B�[=!l [ -�l ' [ =�Elf�l U �� il [ l m X

=

659




[ - �l ' [ �mH-;j

From Problem 35, A- I = - � � so -4 5 -2 X =A-I B= - � -� -4 5 12 21 z= 1 or The so1uhon· lS· x =-,y=--,

X'

59.

9

55.

3x+ y+2z = 1 Rewrite the system of equations in matrix form: X� B� A� � � Find the inverse of A and solve X = A-I B : From Problem 37, A-I = t 7 so 2

-;;

I

-;;

t

61.

-;;

7

57.

7

-;;

-;;

[; � I � �] � [� � I -; �] � [� ! I -t �]

[ -ll m m [-:3 _1] ' 3� 4] 1. ] [ 9 ] [ t -!7' � �7: ' 34 =-,85 z = -12 or The so1utl·On lS· x=--,y -;;

Fmm Pmblem 37, I X =A- B= L 7 7 The solution is x = .!.3 , y = 1, z = �3 or (.!.,3 1, �)3 . A= [; �] Augment identity and use row operatitheonsmatrito findx withthetheinverse:

[ i

{3x+2yx+ y+ z = z=8

-� [ : ; -;], so -

y matrix on theThereleft.is noThus,waythitos obtai matrinx thehasidnoentiintverse. A = [1015 23] Augment identity and use row operatitheonsmatrito findx witheth theinverse:

G� � I � �] � [1 � � I -i �] ( R2 =- t � + r2 ) � [ol t0 -I)� 01] ( RI = I� � ) There is no way to obtain the identity matrix on

7

x+ y+ z = 2 3x+2y- z = -73 3x+ y+2z = -103 Rewrite the system of equations in matrix form:

the left; thus, there is no inverse.

[ -l} [ l [,t]

A� � � X� : B� Find the inverse of A and solve X = A- I B :

660


63.

[A = -31 -� -�-1]

Section 12.4: Matrix A lgebra

67.

Augment identity and use row operatitheonsmatrito findx wittheh theinverse:

17� 02 [ - . 132 [ . 132 [ - . 132

r3

69.

6

6

o 3

A=

17� 01 L ei [ -. 132

Thus,

8 35

. 135 - . 13 . .. - . 132 . 131 ... 131 133 ... •

A- I :..

..13 1

.. 8 1

. 135

- . 13 1 1

.. •

•

•

l !:l ! -

132

[Ft�.. - . 13 1 . 13 1 - . 13 3 13 3

- . e ... . 133... - . e ... . 137...

J J - . 134 4. 88 E -4J :: . 137 . 86 J) ..

•

-0.02 0. 06 0.07 0.06 61 -12 1 0 - 12 7 ; B= -9 4 - 1 12 Enter the matrices into a graphing utility and use below:to solve the system. The result is shown

] []

[ [ 4 . 56661 7862 J [ -6 . 44363 1 134 J [ -24 . e7467857J J

71.

- . 132 . 81 J 13 1 83 J J

Thus, 5 7 , 07 . y,., - 6.the44,solutiz :.o. n- 24.to the07 system or (4. 5i7,s -6.44.4,-24. ) 25 61 - 12 21 18 - 12 7 7 3 - 1 -2 A=

[

4

t7�i'��7488492J [2. 4568138199 J [ 8 . 26513137321 J J

[ 0.0.0011 -0.0.0025 -0.0.0011] •

4

A- I B

"6

65.

- . 134 . 135 . 13 1 . 136

A - I :..

->

->

r�- l1 r 0. 02 - 0.04 - 0.0 1 0.0 1 1 Thus, - 0.0.0022 0.0.0051 - 0.0.0043 -0.0.0003 . A

-16

[-�[ :-�2-��5 �0� �0 0�1] �] ) � -3 --1 �1 0 0 (Interchange and � [i -� ;; ! -�] � - � �l [� � I! � r� : � � -; Jl There no waythereto obtai the left;is thus, is nonintheverse.identity matrix on [2185 -612 rl

2 1 18 = 21 1012 -1152 8

- 0.02 0. 0 1 0.03

] [] ;

x :..

B=

Thus, z :.. 8o.n27toorthe(-system 1.19, 2.is46,8.2-71.19) . , y:.. 2 .the46 ,soluti A = [: �1 B = [1:��00] AB= [: 1�] [1:1��] 1853.40 ] = [36((71.71.0000))+12+9((158.158.6600)) ] [2116. 20' Nitotalkki'tuis ttotal tion is2$1853. ion istui$2116. 0. 40, and Joe s x :..

73.

a.

b.

=

661

© 2008 Pearson Education, Inc., Upper Saddle River, N J . All rights reserved. This material i s protected under all copyright laws a s they currently



75.

a.

b.

2 3 [700500 350500 400850] 2 3 500 700 The 3 by 2 matrix is: 350 500 . 400 850 The 3 by 1 matrix representing the amount of The rows of the by matrix represent stainless steel and aluminum. The columns represent l O-gallon, 5-gallon, and I -gallon. . The by matrix is:

nmtm,l i c.

d. e.

{ :]

a.

[]

Thus, r'

11,500] [700500 350500 400850] . 1 � [17,050 3 Thus, 11, 5 00 pounds of stainless steel and 17,050 pounds of aluminum were used that day. The 1 by 2 matrix representing cost is: [ 0.10 0. 0 5 ] . The total cost of the day ' s production was: 11 500] [ 0 . 10 0. 0 5 ] . [ 17,050 [ 2002.50] . The total cost of the day ' s production was $2002. 5 0. 21 1 110 11 1 The days usage of materials is:

'

77.

[ ]

K=

b.

=

=

=

al2 =

=

al 3 =

=

=

=

a23

a3 1 =

=

a32 =

=

=

a33 =

c.

fun.

(interChange ) and 'i

al l =

a22 =

Augment the matrix with the identity and use row operations to find the inverse:

->

]

20 19 14 because 47(1) + 34(- 1) + 33(0) 13 47(0) + 34(1) + 33(-1) 1 47(-1) + 34(1) + 33(1) 20 44(1) + 36(-1) + 27(0) 8 44(0) +36(1)+27(- 1) 9 =44(-1)+ 36(1)+27(1)=19 47(1)+ 41(-1) + 20(0) 6 47(0) +41(1) +20(- 1) 2 1 47(-1)+41(1)+20(1) 14 13 � M;l � A; 20�T;8�H;9�I; 19� S;6�F; 21 �U; 14� N The message: Math is a2 1 =

[ ]

[�1 1� � 0� �0 �]1 0 [� : 0 : H] [� _� ! =�

=H

r2

->

662



Section 12.5: Partial Fraction Decomposition

79.

A=

�]

[:

7.

If D = ad - be � 0 , then a � 0 and d � 0 , or b � 0 and e � O . Assuming the former, then

[: � I � �] �[: ; !I � �] l O

d

-

;;

bc a

a

-;;;

-, --'-

1

0

0

0

0

ad-bc

c_ _ 1. + _b a a(ad-bc) -c ad-bc

d ad-bc -c ad-bc

-b ad-bc a ad-bc

Thus , A- l = ic

- /5

9 The proper rational expression is: x2 + 5 = 1 + _2 9_

0 1

_ .f.

->

/5

-b ad-be

x2 - 4

9

a ' (R2 -- ad-bc ,.. ) 2

/bc

a

=

.

=

x -4

11.

d

D -e

-b a

]

x _4

The rational expression x(x - l) x 2 - x is improper, so --:---(x + 4)(x - 3) x 2 + x - 12 perform the division: 1 2 2 x + X-12 X -x+ 0 x 2 + x - 12 - 2x + 1 2 The proper rational expression is: x( x - I) = 1 + -2 2x + 1 2 1 + - 2(x - -6)

)

Section 12.5

( x + 4)(x - 3)

1.

True

3.

3x + 6x3 + 3x 2 = 3x2 ( x 2 + 2x + 1 )

5.

x -4

. . 1 expreSSIOn 5x3 +2 2x - I IS. Improper, . The ratIOna x -4 so perform the division: 5x + 2x - l - 20x 22x - l The proper rational expression is: -I 5x3 + 2x - I 5x + 22x 2 2

where D = ad - be .

4

1

)

x 2 - 4 x2 + 5 x2 - 4

[ ] [� !",�" �1 �[� : ",�J [ 1 1 -+[; -+ [; _� -;l [ -fr] �[

�

. . x2 + 5 . . -- IS Improper, so The ratIOnal expreSSIOn x2 - 4 perform the division:

13.

= 3x 2 ( x + 1 ) 2

Find the partial fraction decomposition: 4 B --= -A + -x(x - I)

X (X -

The rational expression + is proper, since x -1 the degree of the numerator is less than the degree of the denominator.

(x + 4)(x 3)

x + x - 12

l)

(__) = 4

x(x - I )

x

x-I

X (X - l)

(

A x

+� x-I

4 = A(x - l) + Bx

)

Let x = 1 , then 4 = A(O) + B B=4 Let x = O , then 4 = A(-1) + B(O) A = -4 4 =-4 + 4 x(x - l) x x - I

---

--

663



15.

Find the partial fraction decomposition: A Bx+C 1 -+ 2 x(x +1) x x2 +1 A B +C X(X2 + 1) x(x;+I) = X (X2 + 1) x + x� +1 1 = A(x 2 + 1) + (Bx + C)x Let x = 0 , then 1 = A(0 2 +I) +(B(O)+C)(O) A=1 1= A(12 +I) +(B(l) +C)(I) Let x = I , then 1= 2A+B+C 1= 2(1) +B+C B+C = -I Let x = -I , then 1= A« 1) 2 +I) +(B(-I)+C)(-I) 1= A(I +1)+(-B +C)(-I) 1= 2A + B - C 1 = 2(1)+B - C B - C = -I Solve the system of equations: B + C = -l B - C = -1 2B = - 2 B = -1 -1 + C = -1 C=O -x 1 = -I + ---:-2 2 x(x + 1) X x + 1

J

(

-(

1 9.

)

B=

x

.!.2

Let x = -1 , then (_1) 2 = A(- 1 - 1)(-1 + 1) + B(-1 + 1) + C(- 1 - 1) 2 1 = A( -2)(0) + B(O) + C (-2 i 1 = 4C C = -1 4 Let x = 0 , then 0 2 = A(0 - 1)(0 + 1) + B(0 + 1) + C(0 - 1) 2 O = -A + B + C A = B+C A = -1 + -1 = -3 2 4 4 1. .1 1. x2 = _4_ + 2 2 + _4_ 2 (x - l) (x + l) x - I (x - l) x + l _____

Find the partial fraction decomposition: x A + B ---- = (x - l)(x - 2) x - I x - 2 Multiplying both sides by (x - 1)(x - 2) , we obtain: x = A(x - 2) + B(x - l) Let x = l, then I = A(l - 2) + B(I - I) 1 = -A A = -1 Let x = 2, then 2 = A(2 - 2) + B(2 - 1) 2=B x -1 + 2 = -(x - l)(x - 2) x - I x - 2

-- --

----

--

Let x = 1 , then 1 2 = A(l - 1)(l + I) + B(1 + 1) + C(1 - 1) 2 1 = A(0)(2) + B(2) + C(0) 2 1 = 2B

_

1 7.

Find the partial fraction decomposition: A + B + -C x2 --:;---- = 2 2 (x - I) (x + l) x - I (x - l) x + l Multiplying both sides by (x _ I)2 ( +I) , we obtain: x2 = A(x - l)(x + l) + B(x + l) + C(x - l) 2

21.

--

Find the partial fraction decomposition: 1 1 (X _ 2)(X2 +2x+4) Bx+C A +-=----2 2 x 2 (x- 2)(x +2x+4) x +2x+4 Multiplying both sides by (x - 2)(x2 +2x+4) , we obtain: 1 = A(x2 + 2x + 4) + (Bx + C)(x - 2) Let x = 2 , then 1 = A ( 22 + 2(2) + 4 ) + (B(2) + C) (2 - 2) 1 = 1 2A A =� 12

664


as

they currently



Let x = 0, then 1 = A ( 02 + 2(0) + 4 ) + (B(O) + C)(O-2) 1 = 4A -2C 1 = 4 ( 1112 )- 2C -2C=�3 C =-.!.3 Let x = 21 , then 1 = A ( 1 2(1) 4) (B (I) C)(I -2) 1=7A-B-C 1 = 7 (1 1l2 ) -B+-31 B=--121 __ 3x -8 = _x -_2 - -L2xx- 134 rr -2rr( x+4) = -x-2 x + 2x+4 23. Find the partial fraction decomposition: -L

12

+

+

+

+

+

Let x = 0 , then 02 = A(O - 1)(0 + 1)2 + B(O + 1)2 + C(O _1)2 (0 + 1) + D(0-1)2 O =-A+B+C+D A-C= B+D A - C = -41 + -41 = -21 Let x = 2 , then 2 2 = A(2 - 1)(2 + 1) 2 + B(2 + 1)2 + C(2 _1)2 (2+ 1) + D(2-1/ 4=9A+9B+3C+D 9A+3C = 4 - 9B - D 9A+3C = 4 - 9 (±) - ± = % 3A+C= -21 Solve the system of equations: A-C=.!.2 3A+ C = .!.2 4A = 1 A = .!.4 �+C=.!. 4 21 C= --4

12 2 x +

+

+ ---7-'''-'--'':'''

- -- - - Mul , we obtaitinp:lying both sides by x2

A

(x _ I) 2 (x + I ) 2

x-I

x2

=

A(x - I)(x + 1) 2 +

+

B

(x _ I ) 2

B(x

+

C

x+1

+

D

(x + I ) 2

(x _ 1 ) 2 ( x + 1) 2

+ 1 )2

+ C(x _ I ) 2 (x + 1) + D(x - I ) 2

Let x = 1 , then = A(1 -1)(1 + Ii + B(l + 1)2 + C(1 - 1)2 (l + 1)+ D(l-1)2 1 =4B B= .!.4 Let x = - 1 , then (_1)2 = A(-1-1)(-1+1) 2 +B(- 1+1)2 + C(- 1-1)2 (-1 + 1)+D(-1-1)2 1=4D D=.!.4 e

25.

Find x-3 the partial fractiA on decomposi tion: B C ------: x+2 -x+l -(x+2)(x+ I) 2 :- = -(x+ 1)2 2 Multiplying both sides2by 2)(x 1) , we obtain : - 3 = A (x 1) + B(x 2)(x 1) C(x Let x = - 2 , then 2 B(-2+ 2)(-2+ 1) + C(- 2+ 2) 1 -2 -3-5== A(-2 ) A=-5A +

x

+

+

+

(x +

+

+

+

+

+ 2)

+

665




Let x = - 1 then 3 = A(- 1 + 1)2 + B(- 1 + 2)(-1 + 1) + C(- 1 + 2) - 1 -- 4=C C= - 4 Let x = 0 then 0 - 3 = A(O+ 1)2 + B(O+ 2)(0+ 1) + C(O+ 2) -3 = A+2B+2C -3 = -5 + 2B + 2(-4) 2B =10 B=5 x-3 --+--+--5 5 - 4 7'" (x+2)(x+l)2 x+2 x+1 (x+l)2 Find the partiAal fractiB oCx+D n decomposition: ( ) x x2 x2 + 4 Multiplying both sides by x2 (x2 + 4) we obtain : x+4 = Ax(x2 +4)+ B(x2 +4)+ (Cx+D)x2 Let x = 0 then 0+ 4 = A(0)(02 4) B(02 + 4) ( C(O) + D) (0)2 4=4B B=1 Let x = I then A(l)(12 +4)+B(12 +4)+(C(I)+D)(l)2 1+4=5=5A+5B+C+D 5=5A+5+C+D 5A+C+D= 0 Let x = - 1 then -1+4 = A(-I)« - 1)2 +4)+ B«- 1)2 +4) + (C(- I)+ D)(- 1)2 33== -5A+5-C+D 5A+5B-C+D -5A - C+D= - 2 Let x = 2 then 2 + 46=16A+8B+8C+4D = A(2)(22 + 4) + B(22 + 4) + (C(2) + D)(2)2 6 = 16A+8+8C+4D 16A+8C+4D= 2 Sol5A+C+D= ve the system 0of equations: -5A- C+D = - 2 2D=-2 D= - 1 ,

27.

x+4 2 x x2 + 4

--

X

.L .i.

29.

,

,

,

+

---,4 -'-----'-

X

- + - + -

+

5A+C-lC = 01-5A + 4(-1)4 =-2 16A +16A+8 8(1-5A)40A - = -2 - - 24A=-6 A = .!.4 C=I-5 (±) =I- % = - ± x+4 = -t + -+ 1 --'-- t-x-l�--::x2(x2 +4) x2 x2 +4 ) = + _x21 + - .Lx2( x+4 +4 Find the partial fraction decomposition: A + --::-Bx+C x2 +2x+3 ----= ,---.., x+ +2x+4 (x + 1)(x2 +2x+4) 1 x2 Multiplying both sides by (x + 1)(x2 + 2x + 4) we obtain: x2 + 2x+ 3 = A(x2 + 2x+4) + (Bx+ C)(x+l) Let x = - 1 then (_1)2 +2(-1)+3 = A«-1)2 +2(-1)+4) +(B(-I)+C)(-I+I) 2=3A A = �3 Let x = 0 then 02 + 2(0) 3 = A(02 2(0) + 4) (B(O) C)(O I) 3=4A+C 3=4 ( 2/3 ) +C C= .!.3 Let x = 1 then 12 +2(1)+3 = A(12 +2(l)+4)+(B(I)+ C)(l+ I) 6=7A+2B+2C 6 = 7 ( 2 3 ) + 2B + 2 ( 1 3 ) 2B = 6 _H._1. = 1.3 3 3 B=.L3 x2 + 2x + 3 =_3_+ 3 x + 3 (x+l)(x2 +2x+4) x+l x2 +2x+4 3 +2x+ 4 =_x+l3 _+�x2l(x+l) =

,

+

,

,

,

+

,

+

+

+

+

,

,

/

/

1.

______

.L

1

1.

_ _ _

666

© 2008 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


31.

Find thexpartial fractiAon decomposi tion: B (3x-2)(2x+1) = 3x-2 + 2x+1 MUltiplying both sides by (3x -2)(2x + I) we obtain: = A(2x + I) + B(3x -2) Let x = - -21 , then -� = A ( 2 ( -1/2 ) + 1 ) + B ( 3 ( -1/2) -2 ) - �2 =- 22 B B = -71 Let x = -23 , then � = A ( 2 ( 2/3 ) + 1 ) + B ( 3 ( 2/3 ) -2 ) 3 �=2 3 3 A= �7 2 =�+� (3x-2)(2x+1) 3x-2 2x+1 Find thex partial fractixon decomposi tion:B A x2 +2x-3 (x+3)(x-l) =--+-x+3 x-I Multiplying both sides by (x + 3)(x -1) , we obtain : x= A(x-1)+B(x+3) Let x = l , then 1 = A(l-1)+B(l+3) 1 =4B B =-41 Let x = -3 , then -3 = A(-3 -1) + B(-3+ 3) -3 =-4A A = �4 x =_4_+_4_ x2 +2x-3 x+3 x-I -----

--

35.

--

--::---::c-

,

,

x

=

37.

A

33.

7(0)

----

----=-___

1

Find7x+3 the partial fracti7x+3 on decomposition: x3 _2x2 -3x x(x-3)(x+1) A B+C =-+ Multiplying bothxsidesx-3by x(xx+1-3)(x + 1) , we obtain: 7x + 3 = A(x-3)(x + 1) + Bx(x+ 1) + Cx(x-3) Let x+= =0 , then I) + 1) --

I

X

Find the partial fraction decomposition: x2 + 2x + 3 = --+ Ax + B --::-Cx +--:D :(x2 +4)2 x2 +4 (x2 +4)2 Multiplying both sides by (x2 + 4)2 we obtain : x2 +2x+3 = (Ax+B)(x2 +4)+Cx+D x2 + 2x + 3 Ax3 + Bx2 + 4Ax + 4B + Cx + D x2 +2x+3 = Ax3 +Bx2 +(4A+C)x+4B+D A=O; B=l; 4A+C= 2 4B+D=3 4(0)+C = 2 4(l)+D = 3 D=-l C=2

3

--

A(O - 3)(0 + + B(O)(O

+ C(O)(O - 3)

3 = -3 A

A = -I

Let x = 3 then + + + 24 12B B=2 Let7(-1)+3 x = -1=, A(-1-3)(-1+1)+B(-1)(-1+1) then +C(-1)(-1-3) -4=4C C=-l 7x+3 -1 -1 2 +---+ -= ----,x3 _ 2x2 -3x x x-3 x+1 7(3) + 3 = A (3 - 3)(3 ,

\)

B(3)(3

\) + C(3)(3 - 3)

=

1

--

667




39.

41.

Perform synthetic division to find a factor: 2)1 -4 5 -2 2 -4 2 -2 0 x3 -4x2 +5x-2 = (x-2)(x2 -2x+l) = (x-2)(x-l) 2 Find the partial fraction decomposition: x3 -4x2 +5x-2 (x-2)(x-l)2 A B C =--+--+--x-2 x-I (x-l)2 Multiplying both sides by (x -2)(x _1) 2 , we obtai2 n: 2 x = A(x-l) +B(x-2)(x-l)+C(x-2) Let x = 2 , then 22 = A(2-1) 2 +B(2-2)(2-1)+C(2-2) Let4=Ax = 1 , then 12 = A(1-1)2 +B(1-2)(1-1)+C(1-2) 1=-C C=-1 Let x = 0 , then 02 = A(0-1)2 +B(0-2)(0-1)+C(0-2) 0 = A+2B-2C 0=4+2B -2(-I) -2B=6 B=-3 -1 -3 --4 --+ --= --+ --:-x3 --4x--::-x2 2-+ 5x-2 x-2 x-I (x-l)2 Find the3 partial fraction decomposition: Ex+F 3 Ax+B + Cx+D + ---x2 3 = --(x +l6) x2 +16 (x2 +16)2 (x2 +16) Multiplying both sides by (x2 + 16)3 , we obtain: +16) x3 = (Ax+B)(x2 +16)2 + (Cx+D)(x2+Ex+F x3 = (Ax + B)(x4 + 32x2 + 256) + Cx3 + Dx2 + 16Cx+ 16D+ Ex+ F

x3 = Ax5 + BX4 + 32Ax3 + 32Bx2 +3 256Ax 2 + DxEx+ + 256B + CX16D+ + 16Cx+ F X3 = Ax5 + BX4 + (32A + C)X3 + (32B + D)x2 + (256A + 16C + E)x +(256B + 16D+ A=O; B = O ; 32A+C= 1 32(0)+ C = 1 C =1 256A+16C+E = 0 32B+D =0 256(0) + 16(1) + E = 0 32(0)+D = 0 E=-16 D=O 256B+16D+F = 0 256(0)+16(0)+F = 0 F = O -16x 3 ----:(x2:-x+3-16)- ::-=3 (x2 +x16)2 + ---.,(x2 + 16) Find the4 partial fractio4n decomposiAtion: B x-3 2x+l (x-3)(2x+l) - = --+-2X2 -5x-3 ---Multiplying both sides by (x-3)(2x+ 1), we obtain: 4 = A(2x + 1) + B(x -3) Let x = - -21 , then 4 = A ( 2 ( - �) + 1 ) +B (- � -3) 4=- :"2 B B = - �7 Let x = 3 , then 4 = A(2(3) + 1) + B(3 -3) 4 = 7A A ='±7 4 ____ 2x2 -5x-3 = _x-3 + 2x+l F)

43.

�----::-

7

±

_

7_

_l

_ _

668


Section 12. 6: Systems of Nonlinear Equations

45.

on decomposition: d the partial fracti Fin2x+3 2x+3 X4 -9x2 x2(x-3)(x+3) A B C D =-+-+--+-X x2 x-3 x+3 Multiplying both sides by x2 (x -3)(x + 3) , we obtain: 2x+ 3 = Ax(x-3)(x + 3) + B(x -3)(x+ 3) + Cx2(X+ 3) + DX2(X-3) then Let2·0+3x = =0 ,A·0(0-3)(0+3)+B(0-3)(0+3) + C . 02(0+ 3)+ D· 02(0-3) 3=-9B B =-.!.3 Let2·3 x+ =3 =3 ,Athen . 3(3 -3)(3 + 3) + B(3 -3)(3 + 3) + C . 32(3 +3) + D·32(3 -3) 9=54C C=.!.6 then x = -3=,A(-3)(-3-3)(-3+3) Let2(-3)+3 + B(-3-3)(-3 +3) +C(-3)2(-3 +3) +D(-3)2(-3-3) -3 = -54D D =�18 , then Let2·1+3x ==1 A·l(I-3)(1+3)+B(I-3)(1+3) + C . 12(1 +3) + D . f(I-3) 5 = -8A-8B+4C-2D 5 = -8A-8 ( -113 ) + 4 ( 116 ) -2(1118 ) 5 = -8A+-+83 -32 --91 -8A=�9 A=-�9 2x+3 x-3 + _x+318_ X4 -9x2 = _9X + _3x2 + _6_ _

___

.2.

_

1

1

Section 1 2.6 1.

y = 3x+ 2 graph is a line. The x-i0=ntercept: 3x+2 3x = -2 X =- -23 y-intercept: y = 3 (0 ) + 2 = 2 y

(0 , 2)

2

3.

x

y2 = x2 -1 x2 y2 = 1 x2 y2 1 ---= 12 graph is a hyperbola with center (0, 0), The at es c verti and x-axis, the along s axi transverse (-1,0) and (1,0). The asymptotes are y =-x and y=x . _

e

y

5.

{y = x2 +1 y=x+l

x

..L

y =x + l 669



7.

9.

(0, 1) and (I , 2) are the intersection points. Solve by substitution: x2 + 1 = x+ 1 x2 -x = 0 x(x-1) = 0 x = 0 or x = 1 y=l y = 2 Solutions: (0, 1) and ( 1 , 2) {yy=8-x = �36-x2 -""""'_( 4

-

"2, 4 + "1/2)

(4 + "2, 4

-

{y = J;

y=2-x

-5

"1/2)

-8

poi(2.5n9,ts.5.4 1) and (5.4 1, 2. 59) are the intersection Solve by substitution: �36-x2 = 8-x 36-x2 = 64-16x+x2 2x2 -16x+28= 0 x2 -8x+ 14 = 0 x = 8 ± .J64-56 2 8 ± 2J2 2 =4 ± J2 Ifx = 4+J2, y = 8- ( 4+J2) = 4-J2 Ifx = 4-J2, y = 8- ( 4-J2) = 4+J2 Solutions: (4 + J2, 4 - J2 ) and (4 - J2, 4 + J2)

11.

y ;:;:: 2 - x

(1, I ) is the intersection point. Solve by substitution: J; = 2-x x=4-4x+x2 x2 -5x+4=0 (x-4)(x-1) = 0 x = 4 or x = I y = -2 or y=l Eliminate (4, -2); we must have y � 0 . Solution: ( I , 1) {X = 2Y x= i -2y x = 2y

-----

x

-5

(0, 0) and (8, 4) are the intersection points. Solve by substitution: 2y =i -2y i -4y=0 y(y-4) = 0 y =0 or y=4 Solutions:x = 0(0,or0)xand=8 (8, 4) 670


Section 12.6: Systems of Nonlinear Equations

13.

X2 + y2 = 4 {X2 +2x+ l = 0

1 7.

y

y

i -X = 4

-5

-5

(-2, 0) is the intersection point. Substitute 4 for x2 + y2 in the second equation. 2x+4= 0 2x=-4 x=-2 y = �4-(-2)2 = 0 Solution: (-2, 0) { y = 3x-5 15. x2 + y2 = y

Y

5

5

5

{X2 2+-x=4 l =4

(-1, 1. 7 3),on(-1,poin-1.ts.7 3), (0, 2), and (0, -2) are the intersecti Substitute x + 4 for l in the first equation: x2 +x+4= 4 x2 +x= 0 x(x+1)x == 00 or x = - 1 l =3 l =4 y = ±2 Y = ±.J3 Solutions: (0, -2), (0, 2), ( .J3) , ( - J3) { xy =4 x2 + l =8 - 1,

y = 3x - 5

1 9.

- 1,

y

x

(1, -2) and (2, 1) are the intersection points. Solve by substitution: x2 +(3x-5)2 = 5 x2 +9x2 -30x+ 25 = 5 10x2 -30x+ 20 = 0 x2 -3x+2 = 0 (x - 1)(x-2) = 0 x = 1 or x = 2 y=-2 y = l Solutions: (1, -2) and 1) (2,

671

(-2, -2) and (2, 2) are the intersection points. Solve by substitution: x2 + (;J = 8 x2 +�= 8 x2 4 + 16 = 8x2 X x4 -8x2 +16= 0 (x2 _4)2 = 0 Xy== 22 oror x2yx ===4--22 Solutions: (-2, -2) and (2, 2)




21.

{X2 + y2 = 4

25.

y = x2 -9 y

{

secondon equati on for y, substitute into theSolvfirste theequati and solve: 2X2 + y2 = 18 xy=4 => y=4x 2X2 +(�J 18 2X2 +�= x2 18 2x4 + 16 = 18x2 2x4 -18x2 +16=0 x4 -9x2 +8=0 ( x2 -8 ) ( x2 -1 ) = 0 or x2 = 1 x2 = 8 x = ±.J8= ± 2.J2 or x = ± l If x = 2.J2 : y = _2.J24_ = .J2 4 _=_.J2 Ifx =-2.J2: y =_ -2.J2 Ifx = 1 : y = -41 =4 4 Ifx=-l : y=-=-4 -1 Solutions: ( 2.J2, .J2) , ( -2.J2, -.J2), (1, 4), (-1, -4) Substiotuten andthesolfirstve:equation into the second equati { y= 2x+l 2X2 + i = 1 2x2 + ( 2x + 1 )2 = 1 2x2 + 4x2 + 4x + 1 = 1 6x2 +4x 0 2x ( 3x+2) = 0 2x= 0 or 3x+2 = 0 x=O or X =- -23 Ifx = 0: y = 2(0)+ 1 = 1 Ifx=- � : y = 2 ( - �) +I=- � +1=- � Solutions: (0" 1) (-�3 ' -.!.3 ) =

-10

23.

No solution; Inconsistent. Solve by substitution: x2 +(x2 _9)2 = 4 x2 + X4 -18x2 + 81 = 4 X4 -17 x2 + 77 = 0 17 ± �289 -4(77) 2 l7 ± r-l9 2 real solutions to this expression. There are no Inconsistent. -4 {yy ==x26x-13 27.

(3, 5) is the intersection point. Solve by substitution: x2 -4 = 6x-13 x2 -6x+9= 0 (X-3)2 = 0 x-3= 0 x=3 Solution: y(3,=5(3)2) -4 = 5

=

672




29.

Solve theequati firstoequati osolve: n for y, substitute into the second n and x+y+l = 0 � y=-x-l +6y-x=-5 {x2x2++(-x/ _1)2 + -x -x = x2 +X2 + 2x+ 1-6x-6-x = -5 2x2 -5x=0= 0 x(2x-5) x= 0 or x = � Ifx = O:5 y = -(0)-1 = -1 Ifx = -2 : y = - -25 -I = - -27 Solutions: (0,-1), (f ,- f) secondon and equatisolve: on for y, substitute into tSolhe vfirste theequati 4X2 -3xy+9/ = 15 2x+3y = 5 � y = - -32 x+-35 4x2 -3X ( - � X+ �) +9 ( - � x+ �r = 15 4x2 +2x2 -5x+4x2 -20x+25 =15 IOx2 -25x + 10 = 0 2X2 -5x+2 = 0 (2x -1)(x -2) 0 X = -21 or x = 2 Ifx = -21 : y=- �(�) + � = � 5 1 Ifx=2: y=- -23 (2)+-=3 3 Solutions: (�, �} ( 2, �) Mul each sidetoofelithemisecond add ttheiplyequations nate y: equation by 4 and {3x2X2 +-4// =31=-7 � 12x2x2 +4/ _4y2 = -7 =124 13x2 = 117 x2x==93 ± If x = 3 : 3(3)2 + / 31 � y2 = 4 � = ±2 If x=-3 : 3(-3)2 +y2 = 31 � / =4� y= ±2 Solutions: (3, 2), (3, -2), (-3, 2), (-3, -2) 6(

31.

-1)

3 5.

3x2 +5y2 = 12 Mul tisipldye each sidsecond e of thequati e first oequati onanbyd a5dandd each of the n by 3 the equations to eliminate y: 35x2 -15/ = -25 9x2 + 15y2 = 36 44x2 = 11 x2 = -4 X=± -21 Ifx = ! : 3 (�J +5/ =12 � / = � � y= ± % Ifx= - ! : 2y 2 = 80 "9 => y 2 = 40 => y = ± 2 M 9 3 8 Ifx = -- : 3 2 8 -'3 + 2y 2 = 16 => 2y 2 = 80 "9 40 => y = ± -2M => y 2 = 9 3 Solutions: 2M 2M 2M 3' 3 ' 3' 3 ' 3' 3 ' _ � _ 2M 3' 3

()

45.

( )

[� ) [� _ ) [_� [ )

� -3 -2 y2 = x2

)

{

( )( )( )( ) -

I +� = 6 _ x4 l �-� = 19 X4 l Multiply each side of the first equation by -2 and add the equations to eliminate x: -2 - 1 2 = -12 X4 y 4 � - � = 19 x4 y 4 - �4 = 7 y l = -2 There are no real solutions. The system is inconsistent.

674


Section 12.6: Systems of Nonlinear Equations 47.

{X2 -3XY 2+2/ = 0 x +xy = 6

Subtract the second equation from the first to eliminate the x 2 term.

49.

-4xy+2y2 =-6 2xy-/ =3 Since y"# 0 , we can solve for x in this equation to get y2 +3 ' y "# O x= � Now substitute for x in the second equation and solve for y. x2 +xy = 6 ( Y�; 3 J + ( y�; 3 )y = 6 y4 +6y2 +9 + y22+3 = 6 4/ y4 +6/ +9+2y4 +6y2 = 24/ 3/ -12/ +9 = 0 y4 _4y2 +3 = 0 (/ - 3 )(/ -1)= 0 Thus, y = ± 13 or y = ± 1 . Ify=l: x = 2·1 = 2 Ify=-I : x= 2(-I) = -2 Ify=13: x= 13 Ify = -13 : x = -13 Solutions: (2, I), (-2,-1), (13 , 13 ), ( -13 ,-13) {/ + y + x2 -X -2 = 0 x-2 y+l+--=O y Multiply each side of the second equation by -y and add the equations to eliminate y: / + y + x2 - X -2 = 0 -/ -y -x+2 =O x2 -2x = 0 x ( x-2 ) = 0 x = 0 or x = 2

51.

Ifx =O: y2 + Y + 02 _ 0 -2 = 0 => / + Y -2 = 0 => (y+2)(y-l) = 0 => y=-2 or y=1 Ifx = 2: / + y + 22 -2 -2 = 0 => / + y = 0 => y(y + I) = 0 => y = 0 or y = -I Note: y "# 0 because of division by zero. Solutions: (0, -2), (0, I), (2, -I) Rewrite each equation in exponential form: {loglogx (4y)x y == 35 4yy == xx53 Substitute the first equation into the second and solve: 4x3 = x5 x5 _ 4x3 = 0 X3 (x2 -4) =20 x3 = 0 or x = 4 => x = 0 or x = ±2 The base of a logarithm must be positive, thus x "# 0 and x "# - 2 . Ifx = 2 : y = 23 = 8 Solution: (2, 8) Rewrite each equation in exponential form: lnx = 4lny => x = e4 1n y = e1n y' = y4 log 3 X = 2 + 210g 3 y x _- 32+2Iog, y -- 32 ' 32 1og, y -- 32 ' 3log, y' - 9y2 x = y4 2 So we have the system { x=9y Therefore we have : 9/ = / => 9/ - / = 0 => / (9 -/) = 0 /(3+ y)(3-y)- = 0 y = 0 or y = 3 or y = 3 Since In y is undefined when y ::; 0 , the only solution is y = 3 . Ify=3: X =y4 => x=34 =81 Solution: (81, 3 ) �

�

53.

675


{ +x+x+l+/ -3Y -+2Y =O= 0


55.

X2

59.

;

x 2 {(XX + t) + (Yy - tf == t ( +t)2 + ( -tf t

X2 +

x +

i

-

3y

+2

=

0

)'

--I-�-

(-t· t) 2

-2

(-i-. -!.) .

57.

2

_

,.\'

/ +4y+4=x-l+4 (y+2)2 = x+3 Substitute this result into the first equation. (x-l) 2 +x+3 =4 x2 -2x + 1 + x + 3 = 4 x2 -x = 0 x(x-l) = 0 x = 0 or x = 1 Ifx = 0 : (y+2) 2 = 0+3 y + 2 = ±.J3 => y -2 ± .J3 If x = 1 : (y + 2) 2 = 1 + 3 y + 2 ±2 => y = -2 ± 2 The points of intersection are: ( 0, -2 - J-j), ( 0, -2 + J-j ) , (1, -4 ) , ( 1, 0) . =

v)' x + l + 'x- = O

-2

Complete the square on the second equation.

=

x, => x =-2y { 2 x+2y=0 (x-l) +(y_l)2 5 (-2y-l) 2 +(y-l/ =5 4/ +4y+l+ / -2y+l = 5 => 5/ +2y-3 = 0 (5y-3)(y+l) = 0 y = -35 =0 . 6 or y=-l x = 65 = -1.2 or x = 2 The points of intersection are ( -%, %), (2, -1) .

y

Solve the first equation for substitute into the second equation and solve:

s

(O, .J3"- 2)

=

i + 4y - x +

5

(x -

Ii

+

(y

- Ii

=

61.

{y=2 _x-34

x,

Solve the first equation for substitute into the second equation and solve:

x -6x + / + 1 = 0 4 y=- x-3 x-3=-y4 4 x=-+3 y (; +3J -6 (; +3) +/ +1 =0 1 6 24 -+-+9---18+ 2y Y 24Y / +1=0 1 6 2 -8 = 0 -+y y2

5

x

x +

1 =0

(O, -.J3"- 2)

- -

Y

x

2y = 0

-5

676



y4

=

1 6 + _ 8y 2 0 / - 8/ + 1 6 = 0 (/ =0 y2 - 4 = 0 y2 = 4 y = ±2 x = -4 + 3 = 5

_4) 2

Ify = 2: Ify = - 2:

67.

YI = �12-x4; Y2 = -�12-x4; Y3 = ..J2j"-;; Y4 =-·J2lx Use INTERSECT to solve: Graph:

nT��-----' XM i n= -4 . 1 XMax=4 . 7 Xsc l = l YM i n= -3 . 1 YMax=3 . 1 Ysc l = l Xres= l

24

x = -+3 = 1 -2 The points of intersection are: ( 1 , -2), (5, 2). Y

I I I I

5

x

X=.5B009379

x=

Graph: YI = (2 / 3); Use INTERSECT to solve:

X

1\

3. 1

x =

I n t � r s t

-w-w 2 2=A w2 --w+A 2 =O � ± -4A ± v� 1f 2' V4 w= 2 = 2 4 E2 +- �p2 2-l6A p ± �p2 -16A 4 2 2 If W= p+�p -16A then 4 1 = 2 _ p+�p42 -16A = p_�p42 -l6A 2 If w= p_�p -l6A then 4 1=-2 p_�p42 -16A p+�p42 -16A If it is required that length be greater than width, then the solution is: -l6A --.:..-42 -w p_�p42 -l6A and l = -p+�p Solve the equation: m 2 -4(2m -4) = 0 m2 -8m+16 = 0 ( m_4)2 = 0 m=4 Use the point-slope equation with slope 4 and the point (2, 4) to obtain the equation of the tangent line: y-4 = 4(x-2) y-4 = 4x-8 y = 4x-4 P

P

95.

Y

P

mx

P

=>

=>

=>

P

91.

{y = x2 + 2 y=mx+b Solve the system by substitution: x2 + 2 = + b x2 - + 2 -b = 0 Note that the tangent line passes through ( 1, 3). Find the relation 3 = m(l)+b b = 3-m m a d Substitute into the quadratic to eliminate b: x2 -mx+ 2-(3-m) = 0 x2 -mx+ (m-I) = 0 Find when the discriminant equals 0: ( _m )2 -4 (1 )( m-1) = 0 m2 -4m+4= 0 ( m_2 )2 =0 m-2 = 0 m1 = 2 b=3-m =3-2= The equation of the tangent line is y = 2x + 1 . Solve the system: {2x2 +3i = 14 = mx+b Solve the system by substitution: 2X2 + 3 ( + b )2 = 14 2X2 + 3m2x2 + 6mbx + 3b2 = 14 (3m2 + 2 ) x2 + 6mbx+3b2 -14 = 0 Note that the tangent line passes through (1, 2). Find the relation between m and b: 2 = m(l) + b b = 2 -m Substitute into the quadratic to eliminate b: (3m2 + 2)x2 + 6m(2 -m)x + 3(2 -m)2 -14 = 0 (3m2 + 2)x2 + (l2m -6m 2 )x + (3m2 -12m -2) = 0 Find when the discriminant equals 0: (l2m -6m 2 )2 -4(3m2 + 2)(3m2 -12m -2) = 0 144m2 +96m+ 16 = 0 9m 2 +6m+1= 0 (3m + 1)2 = 0 3m+1 = 0 m=--31 b = 2 -m = 2 - ( -�) = f The equation of the tangent line is y = -.!. x + 2. . 3 3 mx

P

/ 2

Solve the system:

=>

=>

680




97.

{x2- /y==3mx+b

Solve the system:

1j

x2_(mx+b ) 2 =3 x2 _ m2 x2 -2mbx -b2 = (l _ m2) x2-2mbx-b2 -3 = 0 = m(2) + b b = 1-2m m b:

Solve the system by substitution:

3

(2, b: (1-m2)x2-2m(l -2m)x -(1-2m)2- = 0 (l_ m2)x2 +(-2m+4m2)x- l+4m-4m2-3 = 0 (l_ m2)x2+(-2m+4m2)x+( -4m2 +4m-4) = 0 0: ( -2m+4m2t -4(1- m2)( -4m2+ 4m -4) = 0 4m2-16m3 + 16m4 -16m4 + 16m3 -16m+ 16 = 0 4m2-16m+ 16 = 0 m2 -4m+4=0 (m-2f =0 = 2x -3m=. 2 r2 : + r, �� r1r2=-a =-r2- ab (-r2- �}2 = � c -r22 --rab 2 --= a 0 ar/ +br2+c = 0

Note that the tangent line passes through Find the relation between and

I).

1j

Substitute into the quadratic to eliminate

3

101.

Find when the discriminant equals

The equation of the tangent line is 99.

{"

-

The solutions are:

=>

I

= -r2 ab = _[ -b±.Jb22a -4ac l - �a b"+.Jb2 -4ac 2b 2a 2a -b"+.Jb2-4ac 2a = -b+.Jb22a -4ac r2 = -b-.Jb22a -4ac . and

10010

10

Since the area of the square piece of sheet metal is square feet, the sheet's dimensions are feet by feet. Let the length of the cut.

x=

I I

10

x

:x

'

--I

10

10 -2x;

= 10 -2x; = x . x 0 10 -2x 0 x x =( ) ( ) ( ) = (10 -2x)(10 -2x)( x) = (10-2x) 2(x) 9 (1O-2x) 2(x)=9. (1O-2x) 2(x) = 9 ( 100-40x+4x2 ) x = 9 100x -40x2+ 4x3 = 9 4x3 -40x2 + 100x -9 = 0 . = 4x3 -40x2+ 1OOx -9

width

height

--

r-

,

The dimensions of the box are: length

y

I

=

--

Note that each of

these expressions must be positive. So we must have > and > => < 5, that is, o < < 5. So the volume of the box is given by V length . width . height

Solve for 1j and �

Substitute and solve:

a.

1j

In order to get a volume equal to

cubic feet,

we solve

So we need to solve the equation

Graphing

YI

on a

calculator yields the graph

681

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Chapter 12: Systems of Equations and Inequalities 80

-,

4 z.r�· H=.09��6ns

-40

10

4 ZoQyo· H=�.Z7��796

-40

True

7.

satisfied

9.

False

v=o

1 1.

80

-,

5.

10

x �0 Graph the line x = O . Use a solid line since the inequality uses 2:. Choose a test point not on the line, such as (2, 0). Since 2 2: 0 is true, shade the side of the line containing (2, 0). Y 5

v=o

80

-,UIO z.r�· H=S.6nOS91

-40

v= -ZE -11

The graph indicates that there are three real zeros on the interval [0, 6]. Using the ZERO feature of a graphing calculator, we find that the three roots shown occur at x::::: 0.093 , x::::: 4.274 and x::::: 5 . 632 . But we've already noted that we must have 0<x 0,> SFl20C96C. > 96C, SF + 120C + F > 0, 10F > SF, 10F + 120C> SF + 120C. 10 120 x -1 y S SSO��xy��16120 Sx �y

The comer points are and Evaluate the objective function: Vertex

c.

Answers will vary.

Chapter 12 Review Exercises .

1

so

Since

{ 5x2X-+ 2yy =S= 5

y = 2x-5 .

Solve the first equation for y:

so

5x 2(2x -5) = S 5x + 4x-1O = S 9x = IS y = 2(2) -5 =x 4= -52 = -1 x = 2, y=-1 (2,-1). =� {3X-4Y x-3y =-2 x: x =3y +-21 3(3Y+±)-4Y = 4 3 =4 9y + --4y 2 5 5y =-2 y =-12 X=3(�)+�= 2 x = 2, y = ± (2, �) .

Substitute and solve: +

Thus, the maximum revenue occurs when the aircraft is configured with first class seats and coach seats. - 0, 120C>(15,96C,120). 15

Value of

(15,120)

or

Solve the second equation for

Substitute into the first equation and solve:

(15, 120), (10, SO).(S, SO), (S, 120),

The comer points are and Evaluate the objective function:

The solution is

or

695

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5.

{3x+2y-4=0 x-2y-4 = 0

Solve the first equation for

x: x = 2Y + 4

3(2y+4)+ 2y-4 == 00 6y+12+2y-4 8yy == -8 -l x = 2(-1) + 4 = 2 x = 2, y = (2, -1) . {yx == 3y+4 2x-5 x = 3(2x -5) + 4 x=6x-15+4 -5x =-11 X=-115 Y = 2 C51) -5 = - � 11 3 ) . x = 5'11 y = -53 (5'-5 {�x-x- �3y+4=0 2 2 y+ �3 = 0 3 {-3x+ 3X-9Y+12 = 0 9y-8 = 0 4=0

Substitute into the second equation and solve:

The solution is

7.

The solution is =

13.

9.

1 5.

3x-2y+3z = -16

x; -2

{ -2X-4Y+2Z = -12 2x- y+3z =-13

or

-5y+5z=-25 y-z =5 -3 x: -3x-6y+3z = -18 3x-2y+3z = -16 -8y+6z = -34 8 y: 8y-8z = 40 -8y+ 6z =-34 -2z = 6 z=-3 y-(-3) = 5 x+2(2)-(-3) = 6 y = 2 x+4+3 = 6 x = -1,y = X2,=z-1= -3 (-1, 2,-3) .

Multiply each side of the first equation by add to the third equation to eliminate

Multiply each side of the first result by to the second result to eliminate

3, 2

-3

and

and

and add

Substituting and solving for the other variables:

MUltiply each side of the first equation by and each side of the second equation by and add to eliminate

{4X+9x-6y6y-26y: == 00

{2x-X+ 2Y-y+3zZ==-136

Multiply each side of the first equation by add to the second equation to eliminate

There is no solution to the system. The system is inconsistent.

{2X3x-2y + 3y -13 = 0 =0

8

There is no solution to the system. The system of equations is inconsistent.

Multiply each side of the first equation by and each side of the second equation by -6 and add:

1 1.

or

Multiply each side of the second equation by and add to eliminate

- l or

Substitute the first equation into the second equation and solve:

· · The so1 utlOn IS

y: 3(2)-2y = 0 -2yy =-6 =3 x = 2, y = 3 (2, 3). {3X-22Y x--y3 =12 x: {-3x+3x-2y2y == -368 0 =-28

Substitute and solve for

13x -26 = 0 13xx == 226

The solution i s

696

or

© 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright l aws as they currently exist. No portion of this material may b e reproduced, in any form or by any means, without permission in writing from the publisher.

1 7.


{ x+2y-4z 2x -4Y +Z == 27-15

5x- 6y-2z = -3 Multiply bythe2,firandst equation byto -1eliminate and thex:second equation then add -2X+4Y-Z = 15 { 2x+ 4y -8z = 54 8y -9z = 69 Mul t ipl y the secondto equation byx: -5 and add to the-5x-l0y+20z third equation el i minate = -135 5x- 6y -2z =-3 -16y+18z =-138 both sidesresulof thet to fieliminate rst resulty:by 2 and addMultiply to the second 16y-18z = 138 -16y+18z = -138 The-16y+18z system0=0is= -138 dependent. 18z+138=16y y=-z+forSubstituting x: ( into) the second equation and solving x+2 8z+8 -4z= 27 x+-z+--4z= 27 4 4 x=-z+4 4 7 39 The so utlOn x = -z4 +-4 = -z98 +-698 , z . 7 39 ' any real number or {(x,y,z) 1 x="4z+4 9 69 .zls anyrea numb} y=-z+-, er . 8 8 = 8 3X+2Y {x+4y= -1 + [ 42]+[35 -42] [ 2+11+3 +4+(-4)5] = [43 -49] -1+5 4 4 9

9

2 5.

2 7.

29.

39

·

IS

,

y

IS

21.

A

C

1

=

2

0

1

-1

-4

=[� �] Augment the matrix identity and use row operations to finwith d thetheinverse: 6 11 [� 3 3 ° �] Interchange [� 613 °1 �] ( and r2 ) [� -61°1 -41] (R =-4 + r2 ) [� �I-� ;]-1 (R2 =- ir2) [� °11 -i t-1] (RI = -3r2 + ) Thus A-I = [ t 2 ] . A

Ij

�

2

�

1

1 9.

_

CB

22 -13

69

·

-8

=

8

7

-3

-I

=

69

9

1

0

23 .

69

8

.0] [ 6 ] [ 1 0 ] [ 6·1 6 6A= 6· -12 42 = 6(-1) 6·2 6·2 6·4 = -612 2412 AB�[ � 2�H41 1 _20] [ 2(1(44))++4(0(11)) 2(1(--33))++0(4(11)) 2(1(00))++0(4(--22))1 [-1124(4)-2+-32(1) 0]-1(-3)+2(1) -1(0)+2(-2) -2 5-4 = [31 -45] -[41 -31 02 ] 5 2 3(0) -4(-2)] [= 3(4)1(4)+5(1) -4(1) 3(-3) -4(1) 1(-3)+5(1) 1(0)+5(-2) 5(4)+2(1) 5(-3)+2(1) 5(0)+2(-2) [= 98 -132 -108]

�

�

Ij

�

,

5

1j

-� 6

3"

0

=

2+2

697



31.

A=

[� � �l 1 -1

3 5.

2

x witheth theinveridenti rAugment ow opera3thetionsmatri to find se: ty and use [�1 -1� 2 �0 �0 �l1 1 3 3 1 0 O� 0 -1-4 -2-1 -1-1 0 �

l [ [� -! -� --: -! �l [� ! -� : �! �l [01 0 -32 -2 3 0 1 2. .=-1 ->l� : � -t J -11 (�� :�;;,: ,J Thus , [ 1 (�

=

-

"

o

£1

7

=

-

=

�

)

� �

7

37.

[� �� I �] (interchange ) [� �16� \ �]2 and r2 [ 1 16-50 \ -5] (R2 = -3fj + 2 ) [ 1 10�l [0l O1 / ] Ii

1

o

r

o

1

t

The so utlOn x -,25 y -101 or (-,25 101 ) 5X-6Y-3Z 6 4x-7 y-2z=-3 3x+ y-7z 1 Write the augmented matrix:

{

1 ·

0

·

IS

=

=

-

.

=

=

[:3 =� ��-7 -�11

7

2. 7

-7

->[:1 �� -�1 [00 -11-2 -4-� -26-3:1

Augment identity and use row operathetionsmatri to fixnwid theth theinverse:

[-� - -� I � �] ( interchange ) � [-� -� I � �] and r2 �[ � � I �0 �]-1 1 -2 .. [0 0 1 1 4] (RI ) There is no inverse because there is no way to

�

I'j

----r

Write the augmented matrix: [1� �� \ �] �

->

�

=

�

o

->

{10x+lOy 3x- 2y 1 5

->

= - I'j

obtai singularn th.e identity on the left side. The matrix is

�

l�1

[00 � =� it] o

21 41 104 II

698

© 2008 Pearson Education, Inc . , Upper Saddle River, NJ. AlJ rights reserved. This material is protected under alJ copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


60 1

9 2 39 1 3

11

- IT

3 9.

IT

41.

11

The so utlOn x = , y = -,133 z = -133 or (9 , �3 ' �3 ) . -2z= 1 2x+3y =-3 4x-3y-4z = 3 · IS

I ·

{ x-x- y+y- 5zz== 6 2x-2y+ Z = 1 Write the augmented matrix: [2� -2=� -�I �lI �[i -i �� ] =-hj �[i -i : -I:]] I [i -i The system is dependent. 0

11

9

{X

(R2

_

o

�

{zx== -1y+1

43.

o

-1

0

The solution is x = y + z = -I , y is any real number or {(x,y,z)l x = y+l, z =-I, yisany real number} . x- y- z- t = 1 2x + - z + 2t = 3 x-2y-2z-3t = 3x-4y+ z+ 5t = -3 Write -the augmented matrix: 1,

{ [�1 -2 -2 -3-� �] 3 -4 [i �-1� �:4 ��8 -6-:] ( Interchange ] 1 [� �i =j �� -6 and Y

0

: =:

1

T he so utlOn x = --2 ' y = --23 ' z = --3 or ( !2 ' 3.3 ' �4 ) . IS ·

I ·

_

_

1

�

4

_

�

0

5

-3

o

-:

1

r2

699

r3

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51.

[R)Rl == r-3r2 +2'1+r) R4 = r2 +r4

)

D

[RRl2=-r= -r44 +rl+r2)

4 5.

47.

4 9.

R) = -r4 +r) The solution is x = 4, Y = 2, = 3, = -2 or (4, 2, 3,-2) . . 5 3 4 =3(3)-4(1) = 9-4= 11 3/ t

D

8

Dx D

- 13

D --.-1.. D

'

-8 8

+

x

y

1

-� ; : = I I � :H-� :H-� �I = 1(6-6) -4(-3 -24) + 0(-1-8) = 1(0) -4(-27) + 0(-9) = 0+ 108+ 0 = 108 2 -3 0 5 0 1 = 21 6 6 = 2(0 -6) -1(0 -2) -3(30 -0) = 2(-6) -1(-2) -3(30) = -12+2-90 =-100 2

Set up ands Rule: evaluate the determinants to use Cramer' D= I � �2 1 =1(2)-3(-2)=2+6=8 Dx = 1 : �2 1 =4(2)-4(-2)=8+8=16 Dy = 1 � : 1 =1(4)-3(4)=4-12=-8 16 2 y = = -= -1 The solution is x = = -= or (2,-1) . 3y -13 = 0 {2X 3x-2y= 0 Writ e the system is standard form: 2X+3Y= 13 {3x-2y =0 Set up ands Rule: evaluate the determinants to use Cramer' D= 1 23 -23 1 =-4-9=-13 D = 1 130 -23 1 =-26-0= -26 D = 1 23 130 1 = 0-39=-39 - 26 2 Th e so utlOn x=-=-= y= [i = _ 13 =3 or (2, 3). x+2y- z= 6 2x- y+3z= - 13 3x-2y+3z =-16 evaluate the determinants to use Set up ands Rule: Cramer' 1 2 -1 D= 2 -1 3 3 -2 3 = 1 1 =� � 1 - 21 =� � 1 +(-1) 1 � =� I = 1( -3 6) -2(-3 +6) +(-1)(-4+3) =3+6+1=10 ---L

53 .

Z

{3x+2y=4 x-2y=4

5 5.

o

1

D y

·

IS ·

-3 9

'

+

700

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Letx = 0, then 6 = A(0-4)+B(0) -4A = 6

6 2 -1 -13 = -1 3 -16 -2 3 = 6 1 =� � 1 -2 1 =� ! � 1 + (- 1) 1 =�! =� 1 = 6( -3 +6) - 2(-39 + 48) +(-1)(26 -16) = 18-18-10 6 -1= -10 = 2 -13 3 3 -16 3 =11 =� ! � 1 -6 1 � � 1 +(-1)1 � =�! 1 = 1( -39 + 48) -6(6 -9) +(- 1) (-32 + 39) = 9+1 18-7 2 =620 = 2 -1 -13 3 -2 -16 = 1 1 =� =:! 1 -2 1 � = : ! 1 +6 1 � =� 1 = 1 (16-26) -2(-32 + 39) + 6(-4+ 3) = -10-14-6 = -30 -10 The so utJOn = - = 10 = -1 , -30 = -3 or = � = 1020 = 2 , z = ; = 10 Dx

A =--23

61.

D y

----=---

Dz

y

D

1 ·

· IS

X

Dx D

(-1, 2,-3) .

57.

D

8.

+-

1-4 = A(l)(l-I)+B(I-1)+C(1) -3 = C C=-3 Let x = 0 , then 0-4 = A(O)(O-I) + B(O-I)+C(0)2 -4= -B Let Bx==42 then 2 -4 = A(2)(2-1) +B(2 -1) +C(2)2 -2 = 2A+B+4C 2A = -2-4-4(-3) 2A = 6 A = 3 4 -3 x-4 2x (x-l) = -+X3 -+- x 2 x-I Find the partial fraction decomposition: A Bx+C x --+ ---= ---:(x2-+9)(x+l) x+l -x2 +9 Multiply both sides by (x + 1)(x2 + 9) . x = A(x2 +9)+(Bx+ C)(x+1) Let x = -1 , then -1 = A ( ( _1 ) 2 +9 ) +( B ( -I ) +C )( -1+1 ) -1 = A(lO)+(-B+C)(O) -1 = 10A A = -J.10 .. Let x = 0 then ----=---

8

63 .

k

k.

59.

-

,

-

Let I ; � 1 = Then 1 2X2a Y 1 = 2 ( ) = 16 by Theorem (14). The of the determinant by by whenvalue the elements of a columnis multiplied are multiplied Find the partial fraction decomposition: b

6 = ---.2.3 + _32_ x(x-4) x x-4 Find the partial fraction decomposition: A B2 +-C = 2x x-4 (x-l) X x x-I2 Multiply both sides by x (x-l) x-4 = Ax(x-l)+B(x-l)+ Cx2 Let x = 1 , then 2

6 ) = X(X_4) ( A + �) X(X-4) ( x(x-4) x x-4 6= A(x-4)+Bx Letx=4, then 6= A(4-4)+B(4) 4B = 6 B = 'i2

,

701

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o=A

( 02 + 9 ) + ( B ( 0 ) + C) ( 0 + 1 )

67.

( I�) + C

0 = 9A + C

Let

,

Multiply both sides by

then 1 = A ( 12 + 9 ) + ( B ( 1 ) + C) (1 + 1)

Let

1 = A(l O) + (B + C)(2) 1 = 1 0A + 2B + 2C

( 1�) + 2B + 2 ( :0 )

1 = 10 -

9 1 = - 1 + 2B + S

then

,

A = .!. 4 x = -1 ( _ 1)2 = A(- 1 + 1)« - 1)2 + 1) + B(- I - I )« - 1)2 + 1) + ( C (- I ) + D)( -1 - 1)(- 1 + 1 ) 1 = - 4B

Let

2B = .!. S B =� 10

65.

--

(x2 + 1)(x2 - 1) (x2 + l )(x - l )(x + 1) B + Cx + D A + -= -x - I x + 1 x2 + 1 (x - l )(x + 1)(x2 + 1) . x2 = A(x + 1)(x2 + 1) + B(x _ 1)(x2 + 1) + ( Cx + D)(x - l )(x + 1) x=1 12 = A( l + I )(12 + 1) + B(l - I ) (l 2 + 1) + (C(l) + D )(1 - 1) ( 1 + 1) 1 = 4A

0=9 -

C=� 10 x=1

Find the partial fraction decomposition:

,

then

B = -.!. 4 x=0 02 = A(0 + 1)(02 + 1) + B(0 - 1)(02 + 1) + (ceO) + D)(O - l) (O + 1) O= A-B-D

Find the partial fraction decomposition:

Let

--

Cx+ ----: D :x3 ---:- = Ax + B + ----: ---,(x2 + 4)2 x2 + 4 (x2 + 4)2 (x2 + 4)2 . x3 = (Ax + B)(x2 + 4) + Cx + D x3 = Ax3 + Bx2 + 4 Ax + 4 B + Cx + D x3 = Ax3 + Bx2 + (4A + C)x + 4B + D A = 1; B = 0 4A + C = O 4(1) + C = 0 C = -4 4B + D = 0 4(0) + D = 0 D=O x + ---:-:4x-x3 -- --::----::- --::- = -(x2 + 4)2 x2 + 4 (x2 + 4)2

Multiply both sides by

,

then

± ( ±) - D

O= - -

D = .!. 2 x=2 22 = A(2 + 1)(22 + 1) + B(2 - 1)(22 + 1) + (ce2) + D)(2 - 1)(2 + 1) 4 = I SA + S B + 6C + 3D

Let

,

then

6C = 0 C=O

702

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as they currently

exist. No portion of this material may b e reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 12 Review Exercises .

69.

Solve theequation first equation second and solve:for y, substitute into the

{

75.

2X + y + 3 = 0 � y = - 2x - 3 x2 + y2 = S x2 + (- 2x - 3)2 = S � x2 + 4x2 + 12x + 9 = S Sx2 + 1 2x + 4 = 0 � ( Sx + 2)(x + 2) = 0 2 => x = x= -2

Substitute

S

71.

y=1

Solutions: (-�, _ lSI ) ' (-2, 1) . Multiply side of theto eliminate second equati and add theeachequations xy: on by 2 S

f 2xy +

y2 = 1 0 l-xy + 3l = 2

�

�

( �) � ( ) �

J2 J2 : X = -2 2 =3 Ify = - 2J2 : X = - 2 - 2 J2 = 4 3 3 Ify = J2 : x = - J2 Ify = - J2 : x = J2 Ify = 2

2x(-J2) + (- J2t = 1 0 � - 2J2x = 8 � x = - 2J2 (2J2, J2), (-2J2 , -J2)

Solutions: Substitouten andintosolve: the second equation into the first equati

(

Solutions:

{

)(

)

, ,

- 4J2 2J2 4J2 - 2J2 -J2 J2 ) ( 3 ' 3 ' 3 ' 3 ' ( J2 , -J2)

X2 + l = 6y x2 = 3y 3y + l = 6y l - 3y = 0 y(y - 3) = 0 � y = 0 y = 3 y = 0 : x2 = 3(0) � x2 = 0 � x = 0 x2 = 3(3) � x2 = 9 � x = ± 3 Y=3: (0, 0) , (-3, 3), (3, 3).

If If Solutions:

and solve:

3x2 + 4xy + sl = 8 3( _y)2 + 4(-y)y + sl = 8 3l - 4l + sl = 8 4l = 8 l = 2 => y = ± J2

2x (J2) + (J2t = 1 0 � 2J2x = 8 � x = 2J2 Ify = - J2 :

73.

or

Substitute x = -y and solve:

2xy + y2 = 1 0 - 2xy + 6l = 4

7l = 14 y2 = 2 y = ± J2

Ify = J2 :

{

x,

3x2 + 4.xy + sl = 8 x2 + 3.xy + 2l = 0 x2 + 3.xy + 2 y2 = 0 (x + 2y)( x + y) = 0 � x = - 2y x = -y x = - 2y 3x2 + 4xy + sl = 8 3(- 2y)2 + 4(- 2y)y + Sy2 = 8 1 2y2 _ 8y2 + sl = 8 9l = 8 2J2 8 Y 2 = - => y = ± -3 9

- or

11 y = --

Factor the into second substitute the equation first equati, solve on andforsolve:

or

703

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77.

{

X2 - 3x + / + y = - 2 X2 - X + y + l = 0 y

83.

--

Multiply side of theto eliminate second equation and add theeachequations y: by -y _

79.

Graph the line - 2x + y 2 . Use a solid line since the inequality uses ::;. Choose a test point not on the line, such as (0, 0). Since - 2(0) + 0 � 2 is true, line acontaining Graphshade the linethe sixd+eyof=the2 . Use solid line (0,since0). the inequality not on the line, such asuses(0, 0).Choose Since 0a +test0 ;:::point 2 is false, shade theoverlapping opposite regi sideoofn isthethelinsolution. e from (0, 0). The =

x2 3x + y2 + Y = - 2 _ x2 + y2 - Y = 0 - 2x = -2 x=1 Ifx = l : 12 - 3(1) + / + y = - 2 / +y = o y(y + l) = 0 y = 0 y = -1 y "* 0 X

{- 2Xx ++ yy �;::: 22

2:.

_

y

or Note that because that would cause division in the original system. Solution:by( 1zero , -1) 3x + 4y � 1 2 Graph the line 3x + 4y = 1 2 . Use a solid line sinceonthetheinequality not line, suchuses as (0,� 0).. Choose Since a test point 3( 0) + 4( 0) � 1 2 is true, shade the side of the line containing (0, 0).

The unbounded.ofFind the vertices: To findgraphtheisintersection x + y 2 and -2x + y = 2 , solve the system:

{- 2xx ++ y == 22

=

Y

Solve the first equation for x: x Substitute and solve:

y

=

2

-

y

.

- 2(2 - y) + y = 2 - 4 + 2y + y = 2 3y = 6 y=2 x = 2-2=0

The The point comerofpoiinntersection t is (0, 2).is (0, 2). 81.

y � x2

Graph the parabola y = x2 • Use a solid curve since the inequality uses � . Choose a test point not on the parabola, such as (0, 1). Since 0 � 12 is false,(0, shade the opposite side of the parabola from 1).

85.

{

�;�

x+ 2x + 3y � 6 x ;::: 0; y ;::: O .

Graph Shaded region is the first quadrant. Graph the line x + y = 4 . Use a solid line since thetheinequality usesas ::;(0,. Choose a test point not on line, such 0). Since 0 + 0 � 4 is true, shade the side of the line containing (0, 0). Graph the line 2x + 3y = Use uses as::;. (0, 0). Choosea solida testlinepointsincenottheoninequality the line, such Since 2(0) + 3(0) � is true, shade the side of the line containing (0, 0).

y

5

6

x

.

6

704

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Chapter 12 Review Exercises y

89.

8

02 0

x

The overlapping region is the solution. Theandgraph isaxisbounded. Find the vertices: The x-axis y i n tersect at The intersection of 2x + y and the y-axis is 2). The intersection of 2x + y and the x-axis is 2), and The three comer points are x ;::: 2x+ ; : � x+2y ;::: 2 Graph x ;::: y ;::: Shaded region is the first quadrant. Graph the line 2x + y Use a solid line since thetheinequality usesas Choose a test point not on line, such Since + ng is true,Graph shadethethelinesidexof+ the2y line2 . Use containi atestsolidpointlinenotsinceon thetheline, inequality uses . Choose a such as Since + line from 2 is false, shade the opposite side of the =

6

(0,

3

{

=

87.

16 :S . (0, 0). =

16 (0, 0). x

�

=

° ° ;:::

(0, 0), (0,

°

0;

0.

=

8.

91.

:S. (0, 0).

2(0)

°

y2

(0, 0).

6

(3, 0). (3, 0).

16

(0, 0).

(0, 0).

3

{X2 y2

Graph the system of inequalities: + � x+y ;::: 2 Graph the circle x2 + .Use a solid line notsinceonthe2theinequality circle, suchusesas ChooseSincea test point + � is true, shade the side of the circle containing Graph the line + y 2 . Use a solid line since inequality thethe line, such asuses . Choose Since a +test poi2 nits notfalse,on shade the oppositregion e side ofis thethe solution. line from The overlapping

° :S 8 (0, 0).

�

4

=

=

x2

2

:S

(0, 0).

2(0) � (0, 0).

x2

Graph the system of inequalities: { y� xy � Graph the parabola y . Use a solid line since the inequality uses . Choose a test point not on theshade parabola, such asside(1, 2).of theSinparabol ce 2 � a ifrom s false, the opposite Graph(1,the2).hyperbola xy Use a solid line sinceonthetheinequali ty usessuch .asChoose aSitestnce point not hyperbola, (1, 2). 2 � 4 is astrue,(1, shade theoverl sameappisidengofregitheon is hyperbola 2). The the solution.

y

=

1

4.

:S

1·

x

9

overlapping region is the solution. The graphof isThe bounded. Find the vertices: The intersection x + 2y 2 and the y-axis is 1). The intersection of x + 2y 2 and the x-axis is (2, The intersection of 2x and the y-axis is The intersection of 2x + y and the x axis is (2, andThe four comer points are 1), =

=

+

(0, 8).

(4, 0). (0, 8), 0),

y

(0,

0).

=

8

=

8

(0,

(4, 0).

7 05

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93.

Maximize z = 3x + 4y subject to x:2: 0, y:2: 0 , 3x + 2Y :2: 6 , x y � Graph the constraints. +

8.

x = 12_3( 247 ) = 12- 772 = g7 The pomt. f'mtersectton , ( 12 24 ) . The comer points are (0, 0), (0,4), (4,0), (g7 ' 247 ) . Evaluate the objective function: Vertex Value of z = 3x + 5y = 3(0) + 5(1) = 5 (0,(0, 4)1) zzz==3(0)3(1)++5(4)5(0)= 203 (1, 0) (4, 0) z = 3(4)+ 5(0) = 12 ( ' 24 Z =3 + 5 4 = 1 6 7 7 ) C;) e7 ) � The minimum value is 3 at (1, 0). 5Y= 5 { 2X+ 4x+l0y= A Multiply side of the first equation by -2 and eliminate each x: = -10 {-4X-I0Y 4x+1Oy = A 0= A-I0 If thereofareelimination to be infinitely many solutions, the result should be 0 O Therefore, . A -10 = 0 or = . 10 y = ax2 +bx+c At (0, 2the equation becomes: 1 = a(0) + b(O) + c c=1 At 0) the equation becomes: 2 0= a(1) +b(I)+c 0= a+b+c a+b+c= O At (-2, 1) the equation2 becomes: 1 = a(-2) +b(-2)+c 1 = 4a-2b+c 4a-2b+c The system =of1 equations is: {4a-2b+c a+ b+c== O1 c= 1 Substitute c = 1 into the first and second equations and simplify: .

0

IS

7 ' -:;-

1), ( 1 ,

The comerthepoints are (0,function: 3), (2, 0), (0, 8), 0) . Evaluate objective Vertex Value ofz = 3x+4y (0, 3) z = 3(0) + 4(3) = 12 (0, z = 3(0) + 4(8) = 32 (2, 0) z = 3(2)+4(0) = 6 0) = 3(8) + 4(0) = 24 The maximum value is 32 at (0,.8) Minimize z = 3x+ 5y subject to x:2: 0, y:2: 0 , x+y :2: 1 , 3x+2y�12 , x+3y � 12. Graph the constraints.

=

(8,

g

8)

(8

95.

97.

z

A

99.

=

1)

(1,

To find the intersection of 3x+ 2y = 12 and x + 3y = 12 , solve the system:

= 12 {3X+2Y x+3y=12

Solve the second equation for x: x = 12 -3Y Substitute and solve:

3(12-3y)+2y = 12 36-9y+2y=12 = -24 y= 24-7y

7

706

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a+b+ l = O

Multiaddplytoeachthe side the second and first ofequation to eliequati minateon by

{

4a - 2b + l = 1

a+ b = -1

4a - 2b = 0

Solve first equation secondtheequation and solve:for substitute into the a = -b - l

a,

y =5

5 + 3z = 1 1

x + 9 = 10

Thus, medium boxes, and large boxes ofsmall cookiesbox,should be purchased. Let the speed of the boat in still water, and let the speed of the river current. The distance from Chiritza to the Flotel Orellana is kilometers. Rate Time Distance trip downstream trip downstream The system of equations is: z=2 I

y=

3

x2 -

3

1 05 .

x+1 .

=

$6.00

y

x=I

5

$9.00

x = y =

{%

6x + 9 y = 6 .90(1 00) .

5/2

1 00

x-y

3

1 00

(x + y) = 1 00

3(x - y) = 1 00

Solve the first equation for Solve by substitution: 6x + 9y = 690

6,

y: y = 1 00 - x .

6x + 9(1 00 - x) = 690

1 5x + 1 5y = 600

6x + 900 - 9x = 690

1 5x - 1 5y = 5 00

-3x = - 2 1 0 x = 70 y = 1 00 - 70 = 3 0

The blend iscoffee madeandup of pounds poundsof theof the per-pound per-pound coffee. Let the number of small boxes, let the number ofof large mediuboxes. m boxes, and let the number Oatmeal raisin equation: Chocolate chip equation: Shortbread equation: 70

$6 .00$9.00-

30

y

=

z

3

5,

30x = 1 1 00 1 1 00 1 1 0 X= =3 30 0 _ 3Y = 1 00

C� )

--

1 1 0 - 3y = 1 00

=

1 0 = 3y

=

10 y=3

The speed of the boat is the speed of the current is

x + 2 y + 2z = 1 5

{

x+y

Multiply both sides sides ofofthethe second first equati on obyn by multiply both equati and add the results.

x + y = 1 00

x

2

1 00

=

x + y = 1 00

1 03 .

x + 5 + 2(2) = 1 0

3z = 6

The quadratic function is -.!. '!: Let the number of pounds of coffee that costs per pound, and let the number ofThenpounds of coffeerepresents that coststhe totalperamount pound.of in the blend. value of the blend will becoffee represented by theTheequation: Solve the system of equations:

{

-x - y - 2z = - 1 0

Substituting and solving for the other variables:

b = - '!: 3 2 1 a = - - I = -3 3

x

x + 2y + 2z = 1 5 y +3z = 1 1

4(-b - I) - 2b = 0 - 4b - 4 - 2b = 0 - 6b = 4

101.

-1

x:

x + y + 2z = 1 0

y + 3z = 1 1

1 1 0 / 3 � 36.67 km/hr ; 1 0 / 3 '" 3 . 3 3 kmIhr .

x + 2y + 2z = 1 5 x + y + 2z = 1 0 y + 3z = 1 1

707

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1 07.

Let x the number of hours for Bruce to do the job alone, let y = the number of hours for Bryce to do theforjobMartyaloneto ,doandtheletjobz alone. the number of hours Then lIx represents the fraction of the job that Bruce does in one hour. lIy represents the fraction of the job that Bryce does in one hour. liz represents the fraction of the job that Marty doesequation in one hour. The representing working together is: Bruce and Bryce

1 09.

=

Let x theeachnumber engines produced week,ofandgasoline let y the number of diesel produced each week. The total cost is:engines C = 450x + 550y . Cost is to be minimized; thusare: , this is the objective function. The constraints 20 ::; x ::; 60 number of gasoline engines needed and capacity each week. 1 5 ::; y ::; 40 number of diesel engines needed and capacity each week. x + y 50 number of engines produced to prevent layoffs. Graph the constraints. =

=

=

: : . : . ; . : . : -: - : - : . ; ' ; ' ; ' ; ' ; ' - : ' ; ' ; ' : ' - : ' ; ' ; , " , ' , . �

1 = -3 = 0.75 -1 + -1 = -x y (4 /3) 4

The equation representing working together is: Bryce and Marty

_ The equation representing Bruce and Marty

. .

.!. + .!. = 1 _ = � = 0.625 y z (8 / 5) 6

.

working together is:

.!. + .!. = _1_ = � = 0.375 x z ( 8 / 3) 8

{y-X-II y-I

Solve the system of equations:

!

+ = 0.75 + z-I = 0.625 x-I + z-I = 0.375 u = x - I , v = y -I , u + v = 0.75 V + W = 0.625 u + w = 0.375

: : : : : : : : : : JO : : :

The comer points are (20, 30), (20, 40), (35, 1 5), (60, 1 5), (60, 40) Evaluate the objective function: Vertex Value ofC = 450x + 550y (20, 30) = 450(20) + 550(30) = 25, 500 (20, 40) C = 450(35) + 550(40) = 3 1, 000 (35, 1 5) C = 450(35) + 550(1 5) = 24, 000 (60, 1 5) C = 450(60) + 550(1 5) = 35, 250 i60, 401 C = 450i60}+ 550140) = 49, 000 The minimum cost is $24,000, when 35 gasoline engines and 1 5 diesel engines are produced. The excess capacity is 15 gasoline engines, since only 20 gasoline engines had to be delivered.

Let

C

Solve the first equation for u: u = 0.75 - v . Solve the second equation for w: w = 0.625 - v . Substitute into the third equation and solve: (0.75 - v) + (0.625 - v) = 0.375 -2v = -1 v = 0.5 u = 0.75 - 0.5 = 0.25 w = 0.625 - 0.5 = 0. 1 25

Solve for x, y, and z : x = 4, y = 2, z = 8 (reciprocals) Bruce the jobin 8inhours. 4 hours , Bryce in 2 hours , can and doMarty

708

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Chapter 12 Test 1.

{-2X + y = -7

2.

{

Chapter 12 Test

.!. X - 2Y = 1 3 5 x _ 30y = 1 8

We choosethetofirstuseequation the methodby of- 1 eli5 tomiobtai nationn theand multiply equivalent system

4x + 3y = 9

Substitution: We solve the first equation for y, obtaining y = 2x - 7 Next we substitute equation and solve this for x.result for y in the second

{-5X5x +- 30y 30Y = -1 5 = 18

replace the second equatitheonequi by thevalentsumsystem of theWe two equations to obtain

{-5X + 30Y0 == 3-1 5

4x + 3y = 9 4x + 3 (2x - 7) = 9 4x + 6x - 2 1 = 9 l Ox = 30 30 = 3 x=10

second equation is athatcontradiction anditselfhashas nonoThesolution solution. This means the system and is therefore inconsistent.

Wex = can now obtain the value for y by letting 3 in our substitution for y.

3.

y = 2x - 7 y = 2 (3) - 7 = 6 - 7 = -1

{

x - y + 2z = 5 (I) 3x + 4y - z = -2 (2) 5x + 2y + 3z = 8 (3)

We use the method of elimination and begin by eliminating the variable y from equation (2). Multiply each side of(2). equation (1) by 4 and add theour result to equation This result becomes new equation (2).

The solution of the system is x = 3 , y = - 1 or (3, -1) . Elimination: Multiply each side ofofthex infirsttheequation by 2 soare that the coefficients two equations negatives each other. The result is the equivalentofsystem

x - y + 2z = 5 3x + 4y - z = -2

{-4X + 2Y = -14

4x - 4y + 8z = 20 3x + 4y - z = -2 7x + 7z = 1 8 (2)

We now eliminate the variable y from equation (3) by multiplying each side of equation by 2 and addingourthenewresultequatito equati on (3). The result becomes o n (3). x - y + 2z = 5 2x - 2y + 4z = 10

4x + 3y = 9

We canbyreplace the ofsecond equation of thisThe system the sum the two equations. result is the equivalent system

(1)

{-4X + 2y = -14

5x + 2y + 3z = 8

5y = -5

5x + 2y + 3z = 8 + 7z = 1 8 (3) 7x

Now we solve the second equation for y. 5y = -5

Our (equivalent) system now looks like

We thisandvaluesolveforfory ix.nto the orig-2x iback-substitute nal+firsty = equation -7

Treat equations andtwo variables, as a systemandof two equations containing eliminate x variable by multiplying each side ofequation equation(3).the(2)The byresult -I and adding the result to becomes our new equation (3). 7x + 7z = 1 8 - 7x - 7z = - 1 8

{

-5 = -l y=5

-2x + ( -I) = -7 -2x = -6 -6 = 3 x=-2

The solution of the system is x = 3 , y = -l or

x - y + 2z = 5 (I) 7x + 7z = 1 8 (2) 7x + 7z = 1 8 (3) (2) (3)

7x + 7z = 1 8

7x + 7z = 1 8 0 = 0 (3)

We now have the equivalent system

(3, -1) .

709

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{

x - y + 2z = 5 7x + 7z = 1 8 0=0

adding our the result to equation resultandbecomes new equati on The -2

(1) (2) (3)

3x + 2y - 8z = -3 6x - 3y + 1 5z = 8

This is equivalent to aSisystem ofoftwothe equati oonsns with three variables. n ce one equati contains three variables andwilonel becontains only two vari a bles, the system dependent. There infinitely manyfor solutions. We solveare equation and determine that . . SubstI. tute th'1S expreSSlOn. mto equation ( 1 ) to obtain in terms of (2)

x = -z +

(2)

7

1

z.

-5z = 0 z=O

Back-substitute solve for

_ y + 2z = 5

z=0 y. -7y + 3 lz = 1 4 -7y + 3 l (0) = 1 4 -7y = 1 4 y = -2

18 -z + - - y + 2z = 5 7 17 -y + z = 7 17 y = z-7 18 17 · x = -z + - , Y = z - - , 7 7

The so utlOn 1S is any real number or { \ ; is any real number} . 1 ·

z

{

4.

1

\

8

{

,

x=

(3)

5.

(2) 3.

We of elimination and begin byThe elicoefficients museinatitheng method theon variable from equation inother equations ( 1 ) and arethe two negatives of each so we si m ply add equations together. This result becomes our new equation x

z=0

into

, y = -2 , z = o

{

4X - 5Y + Z = 0 -2x - y + 6 = - 1 9 x + 5y - 5z = 1 0

{

O. 4x - 5y + z = 0 -2x - y + Oz -25 x + 5y - 5 z = 1 0

, -2, 0

We first terms check arethe onequations tosidmake sureequati that oalln variable the left e of the and the constants areweonputtheitriinghtwithside.a coeffici If a ent variable i s missing, of Our system can be rewritten as

3x + 2y - 8z = -3 ( 1 ) -3x - 2y + 3z = 3 (2) 6x - 3y + 1 5z = 8 (3)

(2).

x

and

x.

The solution of the original system is � or (� }

(2)

start by clearing the fraction in equationby byWemultiplying both sides of the equation 6x - 3y + 1 5z = 8

(3)

3x + 2y - 8z = -3 3x + 2(-2) - 8(0) = -3 3x - 4 = -3 3x = 1 1 X=3

3x + 2y - 8z = -3 ( 1 )

t

into equation and

Finally, back-substitute equation ( 1 ) and solve for

,z

-x - y + z = 1

z -5 .

y = -2

(X, y, Z) x = -z +

y = z-

3X + 2Y - 8Z = -3 ( 1 ) -5z = 0 (2) -7y + 3 1z = 14 (3)

We sidessolve of theequation equation byfor by dividing both

18

-

( ;) -z +

{

- 7y + 3 1z = 1 4 (3)

Our (equivalent) system now looks like

x

y x - y + 2z = 5

(3). (3). - 6x - 4y + 1 6z = 6 6x - 3y + 1 5z = 8

(2)

=

The augmented matrix is

(2) . 3x + 2y - 8z = -3 -3x - 2y + 3z = 3

[-� =� � � 1

- 5z = 0 (2)

Webynowmultiplying eliminate theeachvariable fromoequati side of equati n ( 1 ) byon

1

x

(3)

5

-5

- 5 10

710

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Chapter 12 Test

6.

The matriwithx hasthreethreeequations. rows andTherepresents aumns system three col tosystem the lehasft ofthree the vertical barWeindicate thatx, y,theand Z variables. can let these variables. The column to the rionghtthe ofridenote the vertical bar represents the constants ght side of the equations. The system is 3x + 2y + 4 z = -6 3X + 2 Y + 4Z = -6 Ix + 0Y + 8z = 2 or x + 8z = 2

{

7.

-2x + ly + 3z = -1 l

3

2

3

-1

8

3

8

4

=

- 1

5

3

2

-3

24

=

12

6

1 2.

-22

[ ] -+ [� : ;] [ 1 ] Therefore, [ =

2 _�

[ A I 12 ]

=

( R2 = -5 lj + r2 )

(R, +, )

�l] .

(R, = -tr2 + 'i )

We first form the matrix [B I 13 ] =

cannot be computed because the dimensions are mismatched. To multiply two matrices, wex to need the number of columns in the first matri bematrix. the same asx Athehasnumber of rows in the second Matri 2 columns, but matrix has 3 rows. Therefore, the operation cannot be performed. Here we are taking the product of a 2 3 matrix and a 3 x 2 matrix. Since the number of columns inrowstheinfirstthematri x ismatri the same as the number of second x (3 in both cases), the operati on can be performed and will result in a 2 x 2 matrix. AC

[AI12]=[� ! I � �] Next we use row operations to transform into reduced row echelon form. [� 1� I -� �] ° � -----" 5 4 [ I -oj 1 ] (R, 1 lj )

£'

11

�

We first form the matrix

1 -� 11, 0 -----" 0 _ _ 1 � t _% 1 0 2 -1 -----" 0 1 _ _� 1

A - 3C "

=

9.

-2x + y + 3z = - 1 1

[� =�] + [; � ] = + l �] [� =�l [-;I � [� [� �]-t �1 [� =�l -[I: �!l [ 3 -�9l

2A + C = 2

6

8.

{

1 1.

[� �l �l � � �l 2 3

0 0 ° 1

N ext we use row operations to transform [B I 13 ]

C

1 0.

x

BA

�][� �j [ ]

"[� [

�2

= 1 ' 1 + (-2) . 0 + 5 ' 3 1 ' (- 1 ) + (-2 ) ' (-4) + 5 ' 2 0·1+3·0+1·3 0 · (-1) + 3 (--4) + 1 . 2

=

©

16 17 3 -10

] 71 1

2008 Pearson Education, I nc . , Upper Saddle River, NJ.




�[�

0

1

3

0

3

71

-2

-2

0

[Rl :��r3 1

1

-5

-4

3

-2

B-' =

0'

� - 7 r3 + r2

[� Tl {6X+3Y

Thus,

0

ofequatis,othen insystem actually The consists of one two variables. system isnumber dependent and therefore has an infinite of solutions. Any ordered pair satisfying the equation or is a solution to the system.

+�

-5

1 3.

1 5.

= 12

2x - Y = -2

We start by writing the augmented matrix for the system.

2

12

-1

-2

3

12

-1

-2

-? ----'-

-r

� -,-

-r

-r

----'-

2

1

3

12

-

-!-

3

1

--!-

o

- -!-

-1

0

1

3

1 2 3

0 1

=

=

x = -!- '

4

+

1 13

7

1 13

o

3

3 _ 2 -10

-?

r +�

=3

7

2

4

-2 0

1

row represents the equation . Using back-substitute into the equation (from the second row) and obtain

8X + Y = 56

z=0

] [� ; I :6

y + 3z = -2 y + 3(0) = -2 y = -2

Next we usematri rowxoperations to transform augmented into row echelon form.the

Using and the equation and obtain y = -2

G ; I : ] R2 = RI r2 � I �] matrix is now in row echelon [The� augmented -8

7

[� � � � ] �[� !I i3 �l [� � ; -6=�] ->[� � } -3��] 3 ->[�matri� x is now]in row echelon form. The last The


6

3 12 -10


7

�

x + 2y + 4z = -3 2x + 7y + 1 5z = - 1 2 4x + 7y + 1 3z = - 1 0

4

12

18

= -4x + 28 ,

[� � � � ]

-1

1

----'- 1 0

x + .!.

-2

6

-r

14.

-1

{

4


1 ] [� we use row operations to transform the Next augmented matrix into row echelon form. [6 1 ] [6 1 ] (RR2I == r2fj ) [ 1 ] (RI = t fj ) [ 6 1-1] (R2 -6fj rJ [[ 1 ] (R2 = i r2 ) 1 ] (R2 t 2 ) The solution of the system is Y or (-!- , 3) { Y= 3

Y Y

x + .!. = 7 ,

z=0,

z 0 y + 3z = -2 =

we

we back-substitute into (from the first row)

x + 2y + 4z = -3

+

x + 2y + 4z = -3

x + 2(-2) + 4(0) = -3

The solution is

form. Because the bottom row consists entirely

x=1 x = l , y = -2 ,

z = O or

712

(1, -2, 0) .

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1 6.

Chapter 12 Test

{2X+2Y-3 =5 Z x-y+2z = 8

1 9.

3x+5y-8z = -2 We start by writing the augmented matrix for the system. �1 -; 3 5 -8 -2

[� : 1 [�3 �15 -8�3 -2�] 1 = [� � �3 :] 3 5 -8 -2 1 =[ 74 � : � =u = [� : �:� =��j 1 = [� � -:t j'l


20.

(R, =*" )

The last row represents the equation 0 = -4 whichno isolution s a contradiction. has and is be Therefore, inconsistent.the system 1 7.

1-32 75 1 =(-2)( 7 )-(5)(3)=-14-15 =-29 2 6 4 0 -1 2 -4 - 2 1: �41 -(-4) 1� 1 �41 +6 1� 1 :1 = 2[4(-4)-2(0)]+4[1(-4)-(-1)(0)] + 6[1(2) -(-1)4] = 2(-16)+4(-4)+6(6) =-32-16+36 =-12 -4

1 8.

4X+3Y = -23 { 3x-5y = 19 The determinant variables is D of the coefficients of the D = 1 43 -53 1 = (4)( -5)-(3)(3) = -20-9 = -29 Since D;t; 0 , Cramer's Rule can be applied. Dx = 1��3 �51 = (-23)( -5)-(3)(19) = 58 Dy =1 ; ��31 =(4)(19)-(-23)(3)=145 x = DxD =�= -2 -29 145 =-5 y = DDy = -29 The solution of the system is x = -2 , y = -5 or (-2,-5) . 4X-3Y+2Z = 15 -2x+ y-3z = -15 5x -5y + 2z = 18 The determinant variables is D of the coefficients of the 4 -3 2 D = -2 -3 5 -5 2 = 4 1�5 -;1 - ( _3) 1 �2 -;1 +2 1-52 �51 = 4(2-15) + 3( -4+ 15) + 2(10 -5) = 4( -13)+3(11)+2(5) =-52+33+10 =-9 Since D ;t; 0 , Cramer's Rule can be applied. 15 -3 2 Dx = -15 -3 18 -5 2 -_ 15 1 -51 -32 1 -(-3) 1-1518 -32 1 + 2 1-1518 -51 1 = 15(2 -15) + 3( -30 + 54) + 2 ( 75 -18) = 15( -13)+ 3(24)+ 2( 5 7 ) =-9

{

713

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4 15 2 Dy = -2 - 1 5 -3 5 18 2 -1 5 -3 _ - 2 -3 -2 - 1 5 +2 15 =4 18 2 5 2 5 18 = 4 ( -30 + 54) - 1 5 (-4 + 1 5) + 2 ( -36 + 75) = 4(24) - 1 5 (1 1) + 2 (39) = -9 4 -3 1 5 Dz = -2 -15 5 -5 1 8 2 5 5 _ 2 =4 +15 _ ( 3) 5 5 = 4 (1 8 - 75) + 3 ( -36 + 75) + 1 5 (10 - 5) = 4( -57) + 3 (39) + 1 5 (5) = -36 D 9 Dx = -9 = 1 , y = _ x=_ Y = - = -1 , D -9 D -9 z = Dz = -36 = 4 D -9 x = 1 , Y = -1 , z = 4 (1, -1, 4) .

1

1 1

1 1

22.

1

�

y = x+l

solSubsti ve fortutex:x + for y into the first equation and 1

2 (x + l) 2 - 3x 2 = 5 2 (x2 + 2x + l) - 3x2 = 5 2X 2 + 4x + 2 - 3x 2 = 5 _x 2 + 4x - 3 = 0 x 2 - 4x + 3 = 0 (x - l)(x - 3) = 0 x=1 x=3

orthese values into the second Back substi t u t e equati on to determine y: x=l : y = I+I = 2

1� �� 1 1� �� 1 1� � 1

x = 3 : y = 3+1 = 4

The solutions of the system are (1, 2) and (3, 4) .

23.

The solution of the system is or

21.

{2y2 -y3X2- x == 51

{

{4xX2 +- l3y ::;� 1000

Graph the circle x2 + l = 100 . Use a solid ity usesas (0,::; .0).Choose poicurvent notsin ceonththee inciequal rcle , such Sincea test ci02r cle02as::;(0,1000);isthtrue, at is ,shade insidethethesame cir clesi.de of the Graph the line 4x - 3y = Use a solid line since thethe ilninequali a test- 3(1) poin t not0 ison e, suchtyasuses(0, 1).Choose Sin ce 4(0) fal(0,se,I).shade the opposi side ofisthethe lsolineufrom tion. The overl appingteregion +

3X 2 + l = 12 l = 9x 9x l x: 3x2 + (9x) = 1 2 3x2 + 9x - 1 2 = 0 x 2 + 3x - 4 = 0 (x - l )(x + 4) = 0 x=1 x = -4

Substitute for into the first equation and solve for

2:

o.

.

�

y 12

or Back substi ute thesene values into the second equati on to tdetermi y: x = 1 : l = 9(1) 9 y = ±3 x = -4 : l = 9(-4) = -36 y = ±.J-36 =

(n ot real) The solutions of the system are (1, -3) and (1, 3) .

4x - 3y = 0

714

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Appendix: Graphing Utilities

11.

9.

XMin=-1(a XMax=20 Xsc.l=2 YMin=-1500 YMax=500 Ysc.l=100 Xres:;:l

Zt�O �=.31n09B9

Y=O

The positive x-intercepts are 0.32 and 1 2.30. 13.

Answers may vary. A possible answer follows: Xmax -Xmin = 8 - ( -4) = 1 2 We want a ratio of 3:2, so the difference between Ymax and YmiD should be 8. In order to see the point (4,8) , the Ymax value must be greater than 8. We might choose Ymax = 1 0 , which means 1 0 - Ymin = 8 , or YmiD = 2 . Since we are on the order of 1 0, we would use a scale of 1. Thus, YmiD = 2 , Ymax = 1 0 , and Y.c1 = 1 will make the point (4,8) visible and have a square screen.

XMin=-50 XfVlax=50 Xsc.l=10 YMin=-10000 YfVlax=25000 Ysc.l=5000 Xres=l

The positive X-intercepts are 1 .00 and 23.00. Section 5

Problems 1-8 assume that a ratio of 3:2 is required for a square screen, as with a TI-84 Plus. 3 -( -3) 6 3 2 -(-2) 4 2 A ratio of 3:2 results in a square screen.

3.

5.

7.

9- 0 9 3 4- (-2) 6 2 Ymax - YmiD A ratio of 3:2 results in a square screen. Xmax -Xmin

Xmax -Xmin = Ymax-Ymin

6- (-6) g

= 3 2 -(-2) 4 A ratio of 3: 1 results in a screen that is not square. =

Xmax -Xmin

9-0 9 3 4 - (-2) 6 2 A ratio of 3:2 results in a square screen. 760

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Section 3: Using a Graphing Utility to Locate Intercepts and Check for Symmetry

3.

27.

XMin=-4 XMax=l Xscl=l YMin=-4 YMax=4 Yscl=l Xres=l

Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are (-3,-7), (-2,-2), and (- 1, 1). Ztr9 x= -1.707107

29.

y=o

The smaller x-intercept is roughly -1.7 1 . 5.

Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are (-3,7.5), (-2,6) , and (- 1,4.5).

XMin=-5 XMax=5 Xscl=l YMin=-5 YMax=5 Yscl=l Xres=l

31.

The smaller x-intercept is roughly -0.28.

Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are (-3,- 1.5), (-2,0), and (- 1, 1.5).

7.

XMin=-8 XMax=8 Xscl=l YMin=-20 YMax=50 Yscl=5 Xres=l

Section 3

1.

1JJ��-6

\

X�lax=4 Xscl=l YMin=-5 YMax=5 Yscl=l Xres=l

\

V

"--./

The positive x-intercept is 3.00.

j

9.

"--./

Ztr9 X=-3.�1�t;1�

y=o

The smaller x-intercept is roughly -3.4 1.

XMin=-8 XMax=8 Xscl=l YMin=-50 YMax=350 Yscl=50 Xres=l

The positive X-intercept is 4.50.

759

© 2008 Pearson Education, me., Upper S addle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of thi s material may be reproduced , in any form or by any means, without permi ssion in writing from the publisher.

Appendix: Graphing Utilities c.

8

17.

10

Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are (-3,-1), (-2,0) , and (-1,1).

-8 20

d.

19. 5

-5

15.

- 20

a.

Each ordered pair from the table corresponds to a point on the graph . Three points on the graph are (-3,5), (-2,4) , and (-1,3). 21.

5 -4

Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are (-3,-4), (-2,-2) , and (-1,0).

8

b.

10

- 10

23.

-8

c.

8

-10

Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are (-3,8), (-2,6) , and (-1,4) .

10 25.

d.

-8

20

Each ordered pair from the table corresponds to a point on the graph. Three points on the graph are (-3,11), (-2,6) , and (-1,3).

5

758

© 2008 Pearson Educati on, Inc . , Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright laws as they currently exist. No portion of thi s material may be reproduced, in any form or by any mean s , without permi ssion in writing from the publisher.

Section 2: Using a Graphing Utility to Graph Equations c.

11.

•

"',"

"

J\.

P:' v' :,

a.

"110

-

5.

-4

-8 d.

b.

- 5. ,. ';

9.

"

. ...,.... 'uc'

"'

: 15

I' ' ( .,

-5 I

� ,

""'� . n: .

,L

:lJ r.,.'.-

,

I 10

'c 'I 5

10 -8

-4 d.

b.

-5. "in _ ;....q--.. ..

10 -8

c.

-10 V"

.

'. j'c,

13.

.. ..

5

- 20

a.

-5t�;

110

,

t '<J )' '\:'.,

15

-4

-8

d.

'

-8

c.

a.

5

l '1, �'m '

'

2U

b. - 10

5

' 10 -8

757

© 2008 Pearson Education , Inc . , Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright laws as they currently

exist. No portion of thi s material may be reproduced, in any form or by any means , without permi ssion in writing from the publisher.

Appendix: Graphing Utilities c.

5.

-

10

a.

10

4

-8

-5

3.

-4

20

d.

b. 5

5

5

10 -8

c.

4

-

8

-10

- 20

a.

5

-5

8

, j

-10

10

-4

8

b.

d.

-)()

-8 8

c.

7.

-10

d.

-5

10

��

-20

a.

to

4

-5

20

5

8

b. -1 0

5

10 -8

756

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Appendix Graphing Utilities 1 5.

Section 1 1. 3. 5.

7.

9.

11.

13.

( -1, 4) ; Quadrant II

(3, 1) ; Quadrant I Xmin = -6 Xmax = 6 X sci = 2 Ymin = -4 Ymax = 4 Y sci = 2

17.

Xmin = - 1 0 X max = 1 1 0 X sci = 10 Ymin = - 1 0 Ymax = 160 Y sci = 1 0

fi =(1,3);P =( 5,1 5) 2 d (F; ,12 ) = �,-(5---1)-2 +- (-1 5- --3)-2 = � (4) 2 + (12) 2 = .J16 + 144 = .J160 = 4.jlo

Xmin = -6 Xmax = 6 X sci = 2 Ymin = -1 Ymax = 3 Y sci = 1

1 9.

fi =(-4,6);12 =( 4,-8)

d (fi, 12 ) = �,-(4---(--4-)) 2- +- (-_ 8- _-6)--:- 2 = �(8) 2 + (-14) 2 = .J64+ 196 = .J 260 = 2.J6s

Xmin = 3 Xmax = 9 X sci = 1 Ymin = 2 Ymax = 1 0 Y sci = 2

Section 2

Xmin = -1 1 Xmax = 5 X sci = 1 Ymin = -3 Ymax = 6 Y sci = 1

1.

4

a.

-4

Xmin = -30 X max = 50 X sci = 1 0 Ymin = -90 Ymax = 50 Y sci = 10

8

b. -\0

755

10

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Chapter 14: Counting and Probability

9.

x; 2-3

Multiply each side of the first equation by and add to the second equation to eliminate multiply each side of the first equation by and add to the third equation to eliminate

{3x+x- y-2y+3zz == -815

x:

-2x+4y- z = -27 -3x+6y-3z = -45 3x+ y - 3z = -8 7y-6z = -53 x- 2y+ z = 15 � 2x-4y+2z = 30 -2x+4y- z = -27 -2x+4y-z = -27 z=3 Substituting and solving for the other variables: z = 3:::> 7y-6 ( 3 ) = -53 7y = -35 y = -5 z = 3,y = -5:::> x-2( -5)+3 = 15 x+1 O +3 = 15:::> x = 2 The solution is x = 2, y = -5, z = 3 or (2, -5, 3) . �

11.

= 3 sin ( 2x + ll' ) = 3 sin ( 2 ( x + ; )) Amplitude: 1 A 1 = 131 = 3 21t T =-=ll' Period: 2 ll' ¢ -ll' . Phase Shift: (() = -2 = --2 Y

y

x

754

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3x2 - 2x = -1 3x2 - 2x+l = 0 -b±Jb2 - 4ac -x = ---2a -(-2) ± �(_ 2) 2 - 4(3)(1) 2 ± v'4=12 2(3) 6 2±.,/-8 2± 2.[ii l± .[ii 3 6 6

. . {l-

The solutIOn set IS 3.

Xl"lin=-l Xl"lax=4 Xscl=l Yl"lin=-H30 Yl"lax=20 Yscl=20 Xres=l

hi 1 + hi 3 3

-- ' --

}

Step 4: From the graph we see that there are x-intercepts at -0. 2 and 3 . Using synthetic division with 3: 3)5 -9 - 7 - 3 1 - 6 1 5 1 8 33 6 2 5 6 11 o Since the remainder is 0, x- 3 is a factor. The other factor is the quotient: 5x3 +6x2 + 1 1x+ 2 . Using synthetic division with 2 on the quotient: -0.2) 5 6 1 1 2 -1 -1 -2 5 5 10 0 Since the remainder is 0, x- ( -0.2) = x+ 0.2 is a factor. The other factor is the quotient: 5x2 + 5x + 1 0 = 5 (x2 + + 2 ) .

.

y = 2 (x+1) 2 - 4 Using the graph of y = x2 , horizontally shift to the left 1 unit, vertically stretch by a factor of 2, and vertically shift down 4 units. y

x

-5

(-2, -2)

(-1. -4)

-5

5.

X

Factoring, f(x) = 5(x2 + + 2) (x - 3)(x + 0.2) The real zeros are 3 and - 0.2. The complex zeros come from solving x2 +x+2 = o. - 1± �1 2 - 4 ( 1)(2) 2 x = -b±Jb - 4ac 2 ( 1) 2a -1±v1-8 -1±H 2 2 - 1 ±.J7i 2 Therefore, over the set of complex numbers, f(x) = 5x4 -9x3 - 7x2 - 3 1x-6 has zeros

f(x) = 5x4 -9x3 - 7x2 - 3 1x -6 Step 1: f(x) has at most 4 real zeros. Step 2: Possible rational zeros: p = ± 1,± 2,±3,±6; q = ±1,± 5; 2 3 6 1 2,± -,±3,±-,±6,± P = ± 1,±-,± q 5 5 5 5 Step 3: Using the Bounds on Zeros Theorem: f(x) = 5 (x4 - 1. 8x3 -1.4x2 -6. 2x - 1 .2 ) a = -1 .8, a = - 1 .4, at = -6.2, ao = - 1 .2 2 3 Max { 1,1- 1 . 21+1- 6.21 + 1 - 1.4 1+1 - 1 .8 1} = Max {I, 1 0.6} 1 0.6

X

------

=

{_1..2 +.J72 i' _1..2 _ .J72'i _1..5' 3}.

1 + Max {I - 1 . 2 1 ,I -6.21 ,1 - 1.41 ,1 - 1 .81} = 1+6. 2 = 7.2 The smaller of the two numbers is 7.2. Thus, every zero of f lies between -7.2 and 7.2. Graphing using the bounds: (Second graph has a better window.)

7.

log (9) = log (3 2 ) = 2 3 3

753

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11.

12.

310

4

26

We are choosing letters from distinct letters and digits from distinct digits. The letters and numbers are placed in order following the format LLL DDDD with repetitions being allowed. Using the Multiplication Principle, we get Note that there are only possibilities for the third letter. There are possible license plates using the new format. Let A Kiersten accepted at USC, and B Kiersten accepted at FSU. Then, we get P (A) P(B) and

=

1 5.

26· 26·23 ·10 ·10·10·10 = 155,480,000 155,23 4 80,000

6 . =C (53,5 ) ·C (42,1 )

53! 42! 5!·48! 1!·4 1! 53.52.5 1.50.49 42.4 1 ! 4 1! 5·4·3·2·1 53·52·5 1 ·50·49·42 5·4·3·2

=(

a.

=

1 3 . a.

p( win on $1 play )

1-0. 70 = 0. 3 0

Kiersten has a chance of not being admitted to FSU.

16.

Since the bottle is chosen at random, all bottles are equally likely to be selected. Thus,

5 =-5 =-1 = 0.25 8+5+4+3 20 4 There is a 25% chance that the selected bottle contains Coke. 11 0. 5 5 P( Pepsl. U IBC ) = 8+3 = -= 8+5+4+3 20 There is a 55% chance that the selected bottle contains either Pepsi or IBC

$1

1 120,526,770 "" 0. 0000000083 =

The number of elements in the sample space can be obtained by using the Multiplication Principle:

6·6·6·6·6 = 7,776 C 5,2

2

55

5·5·5 = 125 2 . 125 = 5·4·125 C ( 5, 2 ) . 125 = � = 1250 2! ·3! 2 The probability of getting exactly 2 fours on 5 rolls of a die is given by 1250 p(exactly 2 fOurs ) = -7776 "" 0.1608 .

.

1 4.

)

Consider the rolls as a sequence of slots. The number of ways to position fours in slots is ( ) . The remaining three slots can be filled with any of the five remaining numbers from the die. Repetitions are allowed so this can be done in different ways. Therefore, the total number of ways to get exactly fours is

P( Coke ) =

b.

120,526,770

)(

Since each possible combination is equally likely, the probability of winning on a play is

to at least one of the universities. b. Here we need the Complement of an event.

30%

n

53! 42! 5!·( 53-5) ! ]!. ( 42-1 ) !

= 0. 60 , = 0. 7 0 , P ( AnB ) = 0.35 . Here we need to use the Addition Rule. P(A UB) = P(A)+P(B)-P(AnB) = 0. 60+0. 7 0-0. 3 5 = 0. 9 5 Kiersten has a 95% chance of being admitted =

1

n

=

p('B) = I-P(B)

65

The number of different selections of numbers is the number of ways we can choose white balls and red ball, where the order of the white balls is not important. This requires the use of the Multiplication Principle and the combination formula. Thus, the total number of distinct ways to pick the numbers is given by (white balls ) (red ball )

Since the ages cover all possibilities and the age groups are mutually exclusive, the sum of all the probabilities must equal

0.1-0.03 8 1=0.19 0. 23+0.29 +0. 25 +1. 0. 0 1 = 0. 8 1 The given probabilities sum to 0. 8 1. This means the missing probability (for 18-20) must be 0.19. +

752

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Chapter 14 Test

33. a.

b.

c.

35. a.

b.

37.

39.

365· 364· 363····348 = 8 .6 3462 8387 x1045

P(10,6 ) =

7.

C (11, 5 ) =

P(no one has same birthday) 348 = 365· 364· 363··· 0.6 531 365 1 8 �

P(at least 2 have same birthday) = 1 - P(no one has same birthday) = 1 - 0.6 531 = 0.3469 P(unemployed) = 0.058

2.

P($1 bill) n($1 bill) = .i 9 n(S)

8.

Let S be all possible selections, so n(S) = 100 . Let D be a card that is divisible by 5, so neD) = 20. Let PN be a card that is 1 or a prime number, so n(PN) = 26 . 20 = .!. = 0.2 P(D) = neD) = n(S) 100 5 n(P P(PN) = N) = 26 = � = 0.26 n(S) 100 50

9.

From the figure: n( biology chemistry physics ) = 22 + 8 + 2 + 4 +9+ 7 +15 = 67 Therefore, n( none of the three ) = 70 -67 = 3 or

n ( biol.

and

8·7·6·5·4! 4!· 2·1

8·7 . 6 . 5 = 4 . 7 .6 . 5 2

= 840 There are 840 distinct arrangements of the letters in the word REDEEMED.

chem. ) -n ( biol. and chern. and phys. )

10.

From the figure: n( physics chemistry ) = 4 +2 + 7 +9+15 +8 = 45 or

5.

8! 4!2!l!l!

--= ----

=

= ( 8+2 ) -2 = 8

4.

Because the letters are not distinct and order matters, we use the permutation formula for non distinct objects. We have four different letters, two of which are repeated (E four times and D two times) . n!

From the figure: n( only biology and chemistry ) =

Since the order in which the colors are selected doesn't matter, this is a combination problem. We have n = 21 colors and we wish to select = 6 of them. 21! 21! C ( 21,6 ) = 6!( 21 -6 )! 6!15! 21· 20·19·1 8·17 ·16·15! 6!1 5! 21·20·19·1 8·17 ·16 6· 5· 4· 3·2·1 = 54, 264 There are 54,264 ways to choose 6 colors from the 21 available colors. r

From the figure: n( physics ) = 4 +2 + 7 +9 = 22 or

3.

11! = J..!.!.. 5!(1 1 - 5)! 5!6! 11·10·9· 8· 7 ·6! 5· 4· 3·2·1·6! 11·10·9· 8· 7 5· 4· 3·2·1 = 462

P(not unemployed) = 1 - P(unemployed) = 1 - 0.058 = 0.942

Chapter 1 4 Test 1.

10! = 10! (1 0 -6)! 4! 10·9· 8· 7·6· 5· 4! 4! =10·9· 8· 7·6· 5 = 1 51,200

6.

Since the order of the horses matters and all the horses are distinct, we use the permutation formula for distinct objects. 87 ! P( 8,2 ) = _8_ _ = 8! = · ·6! = 8 . 7 = 56 ( 8 - 2 ) ! 6! 6! There are 56 different e a t bets for 8-horse race. x c a

7! = 7 ·6· 5· 4· 3 .2 ·1 = 5040 751

an

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69. a.

P(freshman or female)

=

7.

P(freshman) + P(female) - P(freshman and female) n(freshman) + n(female) - n(freshman and female)

9.

n(S)

1 8+1 5 - 8 25 33 33 P(sophomore or male)

b.

=

P(sophomore) + P( male) - P(sophomore and male) n(sophomore) + n(male) - n(sophomore and male)

71.

73 .

25 33

C(8, 3) =

1 7.

1 9.

=

.

-=

There are 2 choices of material, 3 choices of color, and 10 choices of size. The complete assortment would have: 2 · 3 · 1 0 = 60 suits. There are two possible outcomes for each game or 2 · 2 · 2 · 2·2 · 2 · 2 = 27 = 1 28 outcom for 7 games.

Since order is significant, this is a permutation. 7 P(9, 4) = _9_! _ 9! = 9 · 8 · · 6 · 5! =3024 (9 - 4)! 5! 5! ways to seat 4 people in 9 seats. =

0 .167

21.

23.

25.

Chapter 14 Review E xercises

5.

8! = � 8· 7 6 . 5! 56 ( 8 - 3)! 3! 5! 3! 5!·3·2 · 1

s

27.

3.

= n(A u C) = 20+ 5 = 25

13.

-

1.

n(neither in A nor in C)

8! 8 . 7 . 6 . 5! = 3 36 8! P(8, 3) = _ _ _ = = (8 - 3)! 5! 5!

1 5.

peat least 2 with same birthday) = 1 - P(none with same birthday) = 1 - n(different birthdays) n(S) · 363·362·361 · 360 .. · 354 = 1 - 365·364 36512 "" 1 - 0.833 = The sample space for picking 5 out of 10 numbers in a particular order contains 0 10! = 10 ! = Pl( ,5) = 30, 240 POSSI'ble (10 - 5)! 5! outcomes. One of these is the desired outcome. Thus, the probability of winning is: neE) = n(winning) peE) = n(S) n(total possible outcomes) 1 = __ "" 0.000033069 30, 240

From the figure:

11.

n(S)

15+ 1 8 - 8 33

From the figure: n(A and C) = n(An C) = 1+6 = 7

Choose 4

ers --order is significant: 8 7 · · 5·4! = 8! P(8 , 4) = __ = � = · 6 1680 (8 - 4)! 4! 4! ways a squad can be chosen. Choose 2 teams from 14-order is not significant: 14! = � = 14 . 13 . 12! =9 1 C (1 4 , 2) = ( 1 4-2)!2! 1 2! 2! 12! · 2 · [ ways to choose 2 teams. runn

_

There are 8·10·1 0·10·10 ·10· 2 1,600, 000 possible phone numbers. =

There are 24 · 9 · 1 0 · 1 0 · 10 = 2 16, 000 possible license plates.

29. Since there are repeated letters: 7''- = 7 ·6 ·5 · 4 · 3·2 · 1 = 1 260 2! · 2! 2·1·2 · 1 can be formed .

{ Dave } , {Joanne} , {Erica} , { Dave, Joanne} , {Dave, Erica} , {Joanne, Erica } , {Dave, Joanne,Erica } 0,

3 1 . a.

n(A) = 8, nCB) 12, n(An B) = 3 n(A u B) = n(A)+ nCB) - n(An B) = 8+1 2 - 3 = 1 7 =

different words

C(9, 4) · C(9, 3) · C(9, 2) = 126·84·36 = 38 1 024 committees can be formed. C(9, 4) · C(5, 3) · C(2, 2) = 1 26·1 0·1 = 1260 committees can be formed. ,

b.

6

From the figure: n(A) = 2 0+ 2 + + 1 = 29 750


Section 14.3: Probability

neE) = n {2, 4,6, 8, 1 0} = 2- = ! n(S) 10 10 2

55.

33.

P(E) =

35.

5 5 = -1 P(white) = n(white) = n(S)· 5+1 0+8+7 3 0 6

37.

The sample space is: S {BBB,BBG, BGB, GBB,BGG,GBG,GGB,GGG} =

P(3 boys) =

39.

n(3 boys) = .!. n(S) 8

43.

45.

47. 49.

51.

53.

59.

The sample space is: S {BBBB, BBBG, BBGB, =

BGBB,GBBB, BBGG, BGBG,GBBG,BGGB, GBGB,GGBB,BGGG,GBGG, GGBG,GGGB, GGGG}

P(1 girl , 3 boys) = 41.

57.

n(l girl, 3 boys) = � = .!. 16 4 n(S)

61.

d

63 .

3)

n(sum of two ice is P(sum f two d·Ice IS. 3) = --'--------'n(S) n { 1 ,2 or 2, 1 } 2 � = n(S) 36 1 8 0

d

P(white or green) = P(white)+ P(green) n(white) + n(green) n(S) 9+8 17 9+8+3 20 P(not white) = 1 - P(white) = 1 n(white) n(S) � = 1 - = .!..!. 20 20 P(strike or one) = P(strike)+ P(one) n(strike) + (one) n(S) 3+1 -4 8 8 2 n

n(sum of two ice is7) P(sum f two d·Ice IS. 7) = ----'-------'n(S) n {1 ,6 or 2,5 or 3,4 or 4,3 or 5,2 or 6, 1 } 6 = -1 n(S) 36 6 0

P(never gambled online) = 1 - P(gambled online) = 1- 0.05 = 0.95

There are 30 households out of 1 00 with an income of $30,000 or more. peE) = neE) = n(30, 000 or more) = � � n(S) n(total households) 100 10 =

65.

peA U B) = peA)+ PC B ) - peAn B) = 0.25+ 0.45 - 0. 1 5 = 0.55

There are 40 households out of 1 00 with an income of less than $20,000. peE) = neE) = n(less than $20,000) = � = � n(S) n(total households) 100 5

67. a.

peA U B) = peA)+ PCB ) = 0.25+ 0.45 = 0.70

b.

P(AuB) = P(A)+P(B)-P(An B) 0.85 = 0.60 + PCB)-0.05 PCB) = 0.85 - 0.60+0.05 = 0.30

c.

d.

P(theft not cleared) = 1 - P( theft cleared) = 1-0. 1 3 = 0.87

e.

f.

P(does not own cat) = 1 - P(owns cat) = 1 - 0.34 = 0.66

g.

h.

P(l or 2) = P(1)+ P(2) = 0.24 + 0.33 = 0. 57 P(1 or more) = 1 - P(none) = 1- 0.05 = 0 .95 P(3 or fewer) = 1 - p( 4 or more) = 1- 0.17 = 0.83 P(3 or more) = P(3)+ P(4 or more) = 0.2 1 + 0.1 7 = 0.38 P(fewer than 2) = P(O) + = 0.05+ 0.24 = 0.29 P(fewer than 1) = P(O) = 0.05 + P(3) P(l, 2, or 3) = P(l) + = 0.24+0.33+0.2 1 = 0.78 P(2 or more) P(2) + P(3) + P(4 or more) = 0.33 + 0.2 1 + 0.1 7 = 0.7 1

P(I)

P(2)

=

749

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1 3 . The sample space of tossing two fair coins and a

2 1 . The sample space is:

fair die is: S = {HH1, HH2, HH3, HH4, HH5, HH6,

{I 1 Yellow, 1 1 Red, 1 1 Green, 1 2 Yellow, 1 2 Red, 1 2 Green, 1 3 Yellow, 1 3 Red, 1 3 Green, 1 4 Yellow, 1 4 Red, 1 4 Green, 2 1 Yellow, 2 1 Red, 2 1 Green, 2 2 Yellow, 2 2 Red, 2 2 Green, 2 3 Yellow, 2 3 Red, 2 3 Green, 2 4 Yellow, 2 4 Red, 2 4 Green, 3 1 Yellow, 3 1 Red, 3 1 Green, 3 2 Yellow, 3 2 Red, 3 2 Green, 3 3 Yellow, 3 3 Red, 3 3 Green, 3 4 Yellow, 3 4 Red, 3 4 Green, 4 1 Yellow, 4 1 Red, 4 1 Green, 4 2 Yellow, 4 2 Red, 4 2 Green, 4 3 Yellow, 4 3 Red, 4 3 Green, 4 4 Yellow, 4 4 Red, 4 4 Green} There are 48 equally likely events and the

S

HTl, HT2, HT3, HT4, HT5, HT6, TH1, TH2, TH3, TH4, TH5, TH6, TTl, TT2, TT3, TT4, TT5, TT6} There are 24 equally likely outcomes and the probability of each is J..-. 24

1 5. The sample space for tossing three fair coins is:

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} There are 8 equally likely outcomes and the

probability of each is

probability of each is 2.. 8 1 7. The sample space is: S

=

h.

I

1 1 1 1 1 = -+-+-+ - = 48 48 4 8 48 1 2

The probability of

23 . A , B,

getting a 2 or 4 followed by a Red is 1 1 1 P(2 Red) + P(4 Red) = -+- = - . 12 12 6

27. Let P(tails) = x, then P(heads) = 4x

x + 4x = 1

=

{I Yellow Forward, 1 Yellow Backward, 1 Red Forward, 1 Red Backward, 1 Green Forward, 1 Green Backward, 2 Yellow Forward, 2 Yellow Backward, 2 Red Forward, 2 Red Backward, 2 Green Forward, 2 Green B ackward, 3 Yellow Forward, 3 Yellow B ackward, 3 Red Forward, 3 Red Backward, 3 Green Forward, 3 Green B ackward, 4 Yellow Forward, 4 Yellow Backward, 4 Red Forward, 4 Red Backward, 4 Green Forward, 4 Green Backward} There are 24 equally likely events and the


5x = 1 1 X=5 P( tails) = .!. , 5 29.

I

I

24

P(heads) = � 5

P(2) = P(4) = P(6) = x P(1) = P(3) = P(5) = 2x P(l) + P(2) + P(3) + P(4) + P(5) + P(6) = 1 2x + x + 2x + x + 2x + x = 1 9x = 1

l4 ' The probability of

1 X=9

getting a I, followed by a Red or Green, followed by a Backward is: pel Red Backward)+pel Green Backward) 24

C, F

25. B

1 9. The sample space is:

S

�. The probability of

48 getting a 2, followed by a 2 or 4, followed by a Red or Green is P( 2 2 Red) + P(2 4 Red ) + P(2 2 Green ) + P( 2 4 Green )

{I Yellow, 1 Red, 1 Green, 2 Yellow, 2 Red, 2 Green, 3 Yellow, 3 Red, 3 Green, 4 Yellow, 4 Red, 4 Green} There are 12 equally likely events and the


=

P(2) = P(4) = P(6) = .!. 9

I

12

P(l) = P(3) = P(5) = � 9

=-+-=-

31.

peE) =

neE) n(S)

=

n{1, 2, 3} 10

=

� 10

748

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exist. No portion of this material may be reproduced, in any form or by any means , without permi ssion in writing from the publi sher.

Section 14.3: Probability

6 1 . Choose 2 players from a group of 6 players. Thus,

47. The 1 st person can have any of 365 days, the 2nd

person can have any of the remaining 364 days. Thus, there are (365)(364) 132,860 possible ways two people can have different birthdays.

there are C(6, 2) =15 different teams possible.

=

63. a .

49. Choosing 2 boys from the 4 boys can be done

C(4,2) ways. Choosing 3 girls from the 8 girls can be done in C(8,3) ways. Thus, there are a total of: C(4, 2)· C(8, 3)=

8!

4!

b.

(4- 2)! 2! (8- 3)!3! 4! 8! 2! 2! 5!3! 4· 3 · 2! 8·7·6·5 ! 2·1· 2!

5!3!

=6·56 =336 51. This is a permutation with repetition. There are

�=90, 720 2! 2!

53. a.

b.

c.

different words.

65.

If numbers can be repeated, there are (50)(50)(50) 125,000 different lock combinations. If no number can be repeated, then there are 50·49 · 48=117, 600 different lock combinations. =

Answers will vary. Typical combination locks require two full clockwise rotations to the first number, followed by a full counter-clockwise rotation past the first number to the second number, followed by a clockwise rotation to the third number (not past the second). This is not clear from the given directions. Perhaps a better name for a combination lock would be a permutation lock since the order in which the numbers are entered matters.

Answers will vary.

C(7, 2)·C(3,1)= 21·3=63 Section 14.3

C(7, 3)· C(3, 0) = 35·1= 35 C(3, 3)· C(7, 0)=1·1=1

1 . equally likely

55. There are C(100, 22) ways to form the first

3. False; probability may equal

O.

In such cases, the corresponding event will never happen.

committee . There are 78 senators left, so there are C(78, 13) ways to form the second committee . There are C(65, 10) ways to form the third committee. There are C(55, 5) ways to form the fourth committee . There are C(50, 16) ways to form the fifth committee. There are C(34, 17) ways to form the sixth committee. There are C(17, 17) ways to form the seventh committee . The total number of committees = C(100, 22)· C(78, 13)·C(65, 10)·C(55, 5) . C(50, 16)·C(34,17)· C(17, 17)

5. Probabilities must be between 0 and 1 , inclusive.

Thus, 0, 0.01, 0.35, and 1 could be probabilities. 7. All the probabilities are between 0 and 1.

+

+

The sum of the probabilities is 0.2 0.3 + 0.1 0.4 = 1. This is a probability model. 9. All the probabilities are between 0 and 1.

+

The sum of the probabilities is 0.3 + 0.2 + 0.1 0.3 0.9. This is not a probability model.

"" 1.157x10 76

=

1 1 . The sample space is: S =

57. There are 9 choices for the first position, 8

choices for the second position, 7 for the third position, etc. There are 9·8·7·6·5 · 4· 3·2·1=9!= 362, 880 possible batting orders.

{HH, HT, TH, TT}.

Each outcome is equally likely to occur; so

peE)= neE) n(S) 1

. The probabilities are: 1

I

I

P(HH)=-, P(HT)=-, P(TH)=-, P(TT)=-. 4

59. The team must have 1 pitcher and 8 position

4

4

4

players (non-pitchers). For pitcher, choose 1 player from a group of 4 players, i.e., C(4, 1). For position players, choose 8 players from a group of 11 players, i.e., C(11, 8). Thus, the number different teams possible is C( 4,1)· C(11, 8)= 4 ·165= 660. 747



27.

Section 14.2 1.

C(5,3) =

1; 1

3 . pennutation 29. 5. True 7.

9.

6! 6! 6 . 5 . 4! = 30 P(6, 2) = _ _ _ = = (6 - 2)! 4! 4!

11.

13.

· . 6 . 5 ·4! 8 = 1680 P(8, 4) = _ !_ _ = � = 8 7 (8 - 4)! 4! 4!

1 5.

C(8, 2) =

8! = � = 8 . 7 . 6! = 28 (8 - 2)! 2! 6! 2! 6! · 2 · 1

1 7.

C(7,4) =

7! = � = 7 . 6 .5 . 4! = 35 (7 - 4)! 4! 3! 4! 4! · 3 · 2 · 1

1 9.

C(15, 15) =

21.

26! = � = 10,400,600 C(26, 13) = (26 - 1 3)! 1 3! 1 3! 1 3!

25.

{123, 124, 1 34, 234} 4 · 3! = 4 C(4,3) = 4! = (4 - 3)! 3! I! 3! 4 choices for the first letter in the code and 4 choices for the second letter in the code; there are (4)(4) 16 possible two-letter codes. =

33. There are two choices for each of three

positions; there are digit numbers.

(2)(2)(2) 8 possible three=

4 choices for the first position, 3 choices for the second position, 2 choices for the third position, and 1 choice for the fourth position. Thus there are (4)(3)(2)( 1) 24 possible ways four people can be lined up.

35. To line up the four people, there are

=

37. Since no letter can be repeated, there are 5

choices for the first letter, 4 choices for the second letter, and 3 choices for the third letter. Thus, there are (5)(4)(3) 60 possible threeletter codes. =

15! = �= �=1 (15 - 15)! 15 ! O! 15! 15! · 1

39. There are 26 possible one-letter names. There

are (26)(26) 676 possible two-letter names. There are (26)(26)(26) 1 7,576 possible threeletter names. Thus, there are 26 + 676 + 1 7,576 1 8,278 possible companies that can be listed on the New York Stock Exchange. =

=

{abe,abd,abe,aeb,aed,aee,adb,ade, ade,aeb,aee,aed,bae,bad,bae,bea, bcd,bee,bda,bde,bde,bea, bee,bed, cab,cad,cae,eba,ebd, ebe,eda,edb, ede,eea,eeb,eed,dab, dae,dae,dba, dbe,dbe,dca,deb,dee,dea,deb,dec, eab,eae,ead,eba,ebe,ebd,eea,eeb, eed, eda,edb, ede}

P(5,3) =

5! = 5 . 4 . 3! = 10 (5 - 3)! 3! 2 · 1 · 3!

3 1 . There are

4! 4! 4 . 3 . 2 . 1 = 24 P(4,4) = _ _ _ = = (4 - 4)! O! 1 P(7, 0) = _7 !_ _ = 7! = 1 (7 - O)! 7!

23.

{abc,abd,abe,aed,ace,ade,bcd,bee,bde,ede}

=

4 1 . A committee of 4 from a total of 7 students is

given by:

C(7,4) =

7! = � = 7 ·6 · 5 · 4! = 35 (7 - 4)! 4! 3! 4! 3 · 2 · 1 · 4!

35 committees are possible. 43. There are

2 possible answers for each question.

Therefore, there are i 0 = 1 024 different possible arrangements of the answers .

5! = � = 5 · 4 · 3 · 2! = 60 (5 - 3)! 2! 2!

45. There are 5 choices for the first position,

{ 1 23, 124, 132, 1 34, 142, 143, 2 1 3, 2 1 4, 23 1, 234, 24 1, 243, 3 1 2, 3 1 4,32 1, 324, 34 1,342,4 1 2, 4 1 3,421, 423, 43 1, 432} 4! 4! 4 · 3 · 2 · 1 = 24 P(4,3) = _ _ _ = = 1 (4 - 3)! 1!

4

choices for the second position, 3 choices for the third position, 2 choices for the fourth position, and 1 choice for the fifth position. Thus, there are (5)(4)(3)(2)(1) 120 possible arrangements of the books. =

746


Chapter 14 Counting and Probability Section 14.1

A = { those who will purchase a major appliance }

25. Let

B = { those who will buy a car} n(U) = 500, n(A) = 200, n(B) = 150, n(A n B) = 25 n(A u B) = n(A) + nCB) - n(A n B)

1 . union 3. True; the union of two sets includes those elements

or

that are in one both of the sets. The intersection consists of the elements that are in sets. Thus, the intersection is a subset of the union.

both

= 200 + 150 - 25 = 325

n(purchase neither) = 5. 7.

9.

11.

n(A)+ n(B)- n(AnB)

u

B)

= 500 - 325 = 175

{a},{b},{c},{ d},{a,b},{a,c},{a, d}, {b,c},{b,d},{c,d},{a,b,c},{a,b,d}, {a,� d},{b,� d},{a,b,c, d}

= 150 - 25 = 125 27. Construct a Venn diagram: 15

n(A) = 1 5, nCB) = 20, n(A n B) = 10 n(A U B) = n(A) + nCB) - n(A n B) = 1 5 + 20 - 1 0 = 25

n(A u B) = 50, n(A nB) = 10, nCB) = 20 n(A u B) = n(A) + nCB) - n(A n B) 50 = n( A) + 20 -10 40 = n( )

1 3 . From the figure:

A

n(purchase only a car) = n(B) - n(A u B)

0,

A

n(U)- n(

and

n(A) = 15 + 3 + 5 + 2 = 25

(a) 15

(b) 15

(c) 15

(d) 25

(e) 40 29. a.

1 5. From the figure :

n(A orB) = n(AuB)

=

1 1, 597 There were 11,5971 thousand males 18 years old and older who were widowed or divorced.

1 7. From the figure:

n(A but not C) = n(A) - n(AnC) = 25 - 7 = 18 n(A and B and C) = n(

A

n

b.

B n C) = 5

2 1 . There are 5 choices of shirts and 3 choices of

ties; there are (5)(3)

=

+ n(divorced)

= 2, 643 + 8, 954

= 15 + 2 + 5 + 3 + 10 + 2 = 37

1 9. From the figure:

n(widowed or divorced) = n( widowed)

n(married, widowed or divorced) = n(married) + n( widowed ) + n (divorced) = 62, 486 + 2, 643 + 8,954 = 74,083

There were 74,083 thousand males 18 years old and older who were married, widowed, or divorced.

15 different arrangements.

23. There are 9 choices for the first digit, and 10

choices for each of the other three digits. Thus, there are (9)(10)(10)(10) 9000 possible four digit numbers.

3 1 . There are 8 choices for the DOW stocks, 1 5

=

choices for the NASDAQ stocks, and 4 choices for the global stocks. Thus, there are (8)(15)(4) 480 different portfolios. =

33. Answers will vary. 745



Chapter 13: Sequences; Induction; the Binomial Theorem

9.


3.

=

2ex ex

( )

1n eX = 1n

x = 1n

(%)

(%)

�

0. 9 16


=

Ix2 1 = 9 x2 = 9 or x2 = -9 x =±3 or x = ±3i =5 = -5 2

Center point (0, 4); passing through the pole (0,4) implies that the radius 4 using rectangular coordinates: (x _ h ) 2 + (y -k l r2 (X _ O ) 2 + (y _4 ) 2 = 4 2 x2 +y2 - 8y+16 = 16 x2 +/ - 8y = 0 converting to polar coordinates: r2 - 8rsinB = ° r2 = 8rsinB r = 8sinB

{1n(%)} .

11.

cos-I (-0.5) We are fmding the angle B, �B � cosine equals -0.5 . cosB = -0.5 �B � 27t 27t B = -=:>cos -I ( -0.5 ) = 3 3 -1[

Given a circle with center -( 1 , 2) and containing the point (3, 5), we first use the distance formula to determine the radius. r = ( 3 _ ( _I ) ) 2 + (5 _ 2) 2

-1[

�

1[,

whose

1[

= �4 2 +32 =.J16+9 = 55 =5 Therefore, the equation of the circle is given by (x _ ( _ 1) )2 +(y _ 2 )2 = 52 (x+ l) 2 + (y _ 2) 2 = 52 x2 + 2x+l+/ - 4y+ 4 = 25 x2 + / + 2x - 4y- 20 = °

7.

Center: (0, 0); Focus: (0 , 3); Vertex: (0, 4); Major axis is the y-axis; a = 4; c = 3 . Find b: b2 = a 2 _ c 2 = 16 - 9 = 7=:>b = .fi Write the equation using rectangular coordinates: x2 / -+- = 1 7 16 Parametric equations for the ellipse are: x = .fi cos ( t ) ; y = 4 sin ( t ) ; ° � t � 2 1[

1[

744

© 2008 Pearson Education, Inc., Upper Saddle River, NJ . All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 13 Test

13.

{3m+ 2 ) 5 =

(�)(3mt +(�)(3mt (2 )+G)(3m)3 (2)2 +C)(3m)2 (2 )3 +(!)(3m)(2t +G} 2)5

= 243m5 +5·8 1m4 ·2+10·27 m3 ·4+1O · 9 m 2 ·8+5 · 3m · 16+32 = 243m5 + 8 1 0m4 +1 080m3 + 720 m 2 +240 m+32

1 4 . First we show that the statement holds for n =

(1+1) = 1+1 = 2

1.

The equality is true for n = 1 so Condition I holds. Next we assume that true for some k, and we determine whether the formula then holds for

(1+iXl+�X 1+j) . . ( 1+;) = +I .

n

is

k+ 1. We assume that

(1+iX 1+�X l+j) ... ( 1+I) = k+ 1 . (1+iX l+�X l+±}.{ 1+-}X1+ k� 1 ) = (k+ 1)+ 1 = k+ 2 (1+iXI+�X 1+j}.. ( 1+-}X1+ k� 1 ) = [(1+iXI+�Xl+j) . . . ( 1+-})]( 1+ k� 1 )

Now we need to show that

.

We do this as follows:

(using the induction assumption)

= {k+ l ) . I+ {k+I ) · k-1- = k+l+1 +1 = k+2

Condition II also holds. Thus, formula holds true for all natural numbers.

1 6. The weights for each set form an arithmetic

1 5. The yearly values of the Durango form a

geometric sequence with first term and common ratio r 15% loss in value).

= 0.85

al = 31, 000

sequence with first term

d

a, = 100

and common

difference = 30 . If we imagine the weightlifter only performed one repetition per set, the total weight lifted in 5 sets would be the sum of the first five terms of the sequence.

(which represents a

an = 3 1, 000 . (0.85r -'

a n = a, +( n -1) d a = 1 00 +(5-1)(30) = 1 00 + 4 ( 30 ) = 220 5 Sn = i(a+an )

The nth term of the sequence represents the value of the Durango at the beginning of the nth year. Since we want to know the value after 1 0 years, we are looking for the 1 1 th term o f the sequence. That is, the value of the Durango at the beginning of the 1 1 th year. a l = a, . r"-l = 3 1, 000 · ( 0. 85)'0 = 6, 1 03. 1 1

S = �(1 00+220 ) = �(320) = 800 5 Since he performs 1 0 repetitions in each set, we multiply the sum by 1 0 to obtain the total weight

l

1 0 years, the Durango will be worth $6, 1 03. 1 1 .

After

lifted.

1 0 ( 800 ) = 8000

The weightlifter will have lifted a total of 8000 pounds after 5 sets .

743




8.

-2,- 1 0,-1 8,-26,... -10 - ( -2) = -8 , - 1 8- ( - 1 0) = -8 , -26 - (-18) = -8

1 0.

The ratio of consecutive terms is constant. Therefore, the sequence is geometric with

a l = -2 . an = al + ( n - 1) d =-2 + (n- 1) (-8) = -2- 8n+ 8 = 6-8n The sum of the fIrst n terms of the sequence is

common ratio

n

n-I

HH] % [ (�J ) � [ (�J )

11.

[ J[

]

al =- 1 + 7 = 1 3 2' 2

-�

_

2n- 3 2n+ 1 2 ( n- 1) - 3 2 (n- 1)+ 1

and fIrst term

2n - 3 . 2n - 1 (2n- 3) (2n-1) 2n+ 1 2n- 5 (2n+ 1)(2n-5)

The ratio of consecutive terms is not constant. Therefore, the sequence is not geometric.

The sum of the fIrst n terms of the sequence is given by

. . 12. F or this geometrIc senes we have

Sn = % ( al + an )

and

a l = 256 . Since I r I =

converges and we get

� e -�) = � ( 27- n)

=

n-I

The difference of consecutive terms is not constant. Therefore, the sequence is not arithmetic.

The difference between consecutive terms is constant. Therefore, the sequence is arithmetic with common difference d =

-3 5

2n- 3 a n = -2n+ 1 2n - 3 2n- 5 2n - 3 2 ( n - 1) - 3 = ----a - a = --2n+ 1 2 (n- 1)+ 1 2n+ 1 2n- 1 (2n - 3) ( 2n- 1) - ( 2n - 5) (2n+ 1 ) ( 2n+ 1)( 2n - 1 ) ( 4n2 - 8n+ 3 ) - ( 4n2 - 8n - 5 ) 4n2 - 1 8 4n2 - 1 n

= -!!..+ 7 - -�+ 7 2 2 n 7 + n- 1 7 = --+ --2 2 1 2

a l = 25 .

m

� = (-2+ 6- 8n) � = (4- 8n) � = n( 2 - 4n)

a -a

and fIrst term

" ' " ' - r "25 �25 Sn " a 1 1-r 1--2 5 = 1 5 1= 25 ' 1 -

Sn = (a + an )

an =-!!..+ 7 2

r =�

The sum of the fIrst n terms of the sequence is given by

given by

•

8

lQ = 2' � = 2 .I = � ..!. = 2 25 5 1 0 5 ' 4 5 4 5

The difference between consecutive terms is constant. Therefore, the sequence is arithmetic with common difference d = -8 and fIrst term

9

.

25,1 0,4'"58' . .

7 2

r = -64 = 1 256 -4

I-�I = � < 1 ,

the series

S = � = 256 = 256 = 1 024 5 1 - r l-(-t) � 00

742


Chapter 13 Test

69. This is an ordinary annuity with

( )( ) -[[ f ]

P= $500

and

4.

n = 4 30 = 1 20 payment periods. The . . d IS' 0 . 08 = 0 . 02 . Thus, mterest rate per peno 4 2 1+0.02 0 - 1 "" $244, 1 29.08 A = 500 0. 02

-al = -- = -0 = 0 1+8 9

2 l . an = n - 1 n+8

2 a2 = 2 - 1 = 3 2+8 1 0 2 a3 = 3 - 1 = 8 3+8 0 15 - -5 42 - 1 - a4 - -4+8 1 2 4 2 as = 5 -1 = 24 5+8 13

The first five terms of the sequence are 1 34, and 404.

Notice that the signs o f each term alternate, with the first term being negative. This implies that the general term will include a power of - 1 Also note that the numerator is always 1 more than the term number and the denominator is 4 more than the term number. Thus, each term is in the form

( t ( ::!) �) ) (k ) -1

. The last numerator is

11

1 0 terms. 2 3 --+ 4 .. +1 1 = -1 k + 1 --+5 6 7 . 14 k=1 +4

which indicates that there are 10

6.

3 The first five terms of the sequence are 0 , 10 ' 24 8 5 11 ' 4" ' and 13 '

) ))

- 2 + 3 - 4 + '+ 1 1 5 6 7 " 14

.

e -1

)

[(�r -l]+[(�)' -2]+[(�r -3]+[(�r -4]

1i � � � = 3 - 1+ 9 - 2+ 27 - 3+ 8 1 - 4 = 1 3 0 - 1 0 = _ 680 81 81 5.

a l = 4; an = 3an _ 1 + 2 a 2 = 3al + 2 = 3 ( 4 + 2 = 14 a3 = 3a2 + 2 = 3 ( 14 + 2 = 44 a4 = 3a3 + 2 = 3 ( 44 + 2 = 1 34 as = 3a4 + 2 = 3 ( 1 34 + 2 = 404

-k

=

Chapter 13 Test

2.

�[(�J ]

k

6, 12,36, 144, ... 1 2 -6 = 6 and 36 - 12 = 24

The difference between consecutive terms is not constant. Therefore, the sequence is not arithmetic . .ll =

6

2

and

36 = 3 12

The ratio of consecutive terms is not constant. Therefore, the sequence is not geometric. 7.

4, 14, 44,

a n = _ .!.. 4 n 2 I n � -2 ' 4 a n - I _' L 4 n-1 2

_

_ L 4n -I . 4 2 1 n- = 4 .4 2 1

Since the ratio of consecutive terms is constant, the sequence is geometric with common ratio

r=4

and first term

The sum of the first given by Sn

n = a I 1-r 1-r n = -2· 1 - 4 1-4 = 1 - 4n

a l = - 21 . 4 I

n

=

-2 .

terms of the sequence is

. -

�(

--

)

74 1




57.

(X+2)5

=

( �}5 + ( � JX4 ' 2+G}3 . 22 +GJX2 . 23 +( :JXl . 24 +G} 25

= X5 + 5 · 2 x 4 + 10 · 4x 3 + 10 · 8x 2 + 5 ·16x + 1 · 32 = x5 + lOx4 + 40x 3 + 80x 2 + 80x + 32 59.

(2x+3)5 =

( �J c2X)5 +( � J c2X)4 ' 3+GJ c2X)3 . 32 +GJ c2X)2 . 33 +(:J c2X)1 . 34 +G} 35

4 3 2 = 32x5 + 5 · 16x · 3 + 10 · 8x · 9 + 10 · 4x · 27 + 5 · 2x · 81 + 1 . 243 2 = 32x5 + 240x 4 + nOX3 +1080x + 81 Ox+ 243

61.

n

= 9, j = 2, x = x,

a

=

( 92JX7 ·22 = � . 4X7 = 2!7!

c.

2 9·8 . 2 ·1

(�r (�r () ) n � n ( ) ()

4x 7 = 144x 7

The coefficient of x 7 is 144.

63.

n

= 7, j = 5, x = 2 x,

( 7J (2X)2 . 15

a

=1

7 .6 . 2 = � ' 4x 2 (1) = 4x = 84x 2 5 5! 2! 2 ·1 The coefficient of x 2 is 84.

,

a.

= 80, d = -3,

n

a

=

25

2

= 25 d.

(80 + 8) = 25(44) = 1100 bricks

a.

r

3 = 4

b.

() ( 2 %r

3 20 � 4

=

13 5 16

height is

0

feet .

nth

10g

log 0.025

"" 12 .82 3 log 4 The height is less than 6 inches after the 13th strike . Since this is a geometric sequence with I r I < 1 , the distance is the sum of the two

�) �) ( (

"" 8 .44 feet .

After striking the ground the

�

Distance going up: 15 15 Sup = = = 60 feet. 1-

After striking the ground the third time, the height is

�

(l-�) m

67. This is a geometric sequence with

= 20,

0.025

20

infinite geometric series - the distances going down plus the distances going up. Distance going down: 20 20 Sd,w. � � � 80 roo'

1100 bricks are needed to build the steps.

Q,

�

�

2 5 = 80 + (25 -1)(-3) = 80 - 72 = 8 bricks

b. S2 5

0.5

log ( 0.02 5

65. This is an arithmetic sequence with a

If the height is less than 6 inches or 0.5 feet, then:

time, the

The total distance traveled is 140 feet.

740



= 1: 2 . 31-1 = 2 and 31 -1 = 2 ... . + 2 · -13k-1 = 3k+-1,I-1 then 2+6+18+ 2 + 6 + 18 + . +. 2. . 3k k+ 12 · 3k k = [ 2 + 6+ 18+ . + 2 3 - J + 2·3 = 3 k - 1 + 2 · 3k = 3 · 3k - 1 = 3 k + l - l Condi true. tions and are satisfied; the statement is = 1: (3 ·1-2)2 =land-·l21 (6·12 -3· 1 -1) =1 +42 + . . +(3k-2)2 = � ' k ( 6k2 -3k-l) , th12en+42 +72 + . . . +(3k-2)2 +(3(k+l) -2)2 = [12 +42 +72 + . . +(3k-2)2 J +(3k+l)2 = ± . k ( 6k2 -3k-l) +(3k+l)2 = �. [6k3 -3k2 -k + 18e + 12k + 2J = ± -[6e + 15e + 11k + 2J = � ' (k+l) [6k2 +9k+2J = ± - [6k3 +6e +9k 2 +9k+2k+2J = ± . [6k2 ( k + 1)+9k(k + 1) + 2(k +1)J = ± ' (k+l) [6k2 +12k+6-3k-3-1] = ± ' (k+l) [6(k2 +2k+l) -3(k+l) -IJ = ± ' (k+l) [6(k+l)2 -3(k+l)-IJ tCondi rue. tions and are satisfied; the statement is 5·42·1 l O ( 25 ) - -2!3!5! 5·2·41 ··3·2· 3 · 2·11 ---

51. I:

II:

n

If

·

53. I:

II:

I

II

I

II

n

If

e

55.

739




31.

33.

Arial =th3,metdic= 4, an = al +(n-l)d a9 =3+(9-1)4=3+8(4)=3+32=35 Geometric 1 al = 1, r = 10 ' n = 1 1; all = al rn-I all _- 1· (-101 ) -_ ( 101 ) 1 10, 000, 000, 000 Arial =thmJ2et,icd = J2, n = 9, a = al + (n -1)d n J2 J2 J2 a9 = J2+""(9 -1) = + 8J2 = 9 12.7279 a20 =ioans:l + 19d = 96 ; Solaa7lv=+ea6dtlh+e=6dsyst31=em31 of equat aSubtl +19dract t=he96second equation from the first equat -13ddio==n-655and solve for d. aal ==31-6( 5) = 31-30 = 1 al + (n -1) d n =1+(n-l)(5) 1+5n-5 =5n-4 General formula: {an } = { 5n -4} a =a +17d =8; alSolQv=ae thl e+9d=0 s l l aal l+17d +9d ==syst8 em of equations: Subt ractiontandhe second equatd. ion from the first equat sol v e for -8d ==-81 al = -9(1) = -9 1 1-1

35.

37.

-

10

41.

3

S

43.

45.

l-r

2

a, = 21 ' r 3 Since IrI 1 , the series diverges. al = 4, r =-1 Since I rI < , the series converges. n - al _ 4 _ 4 -- 8 - � - (I - i) - (i) 32·1 + 1) = n = 1: 3· 1 = 3 and -(1 f 3+6+9+ . . +3k =-(3k2 k +1) , then + 9 + . . +. .3k+3k]+3(k+1 3=[+36+6+9+ + 3(k + 1) ) 3k2 =-(k+1)+3(k+1) = (k + l) e: 3) = 3(\+ « k + 1) Condi true. tions and are satisfied; the statement is >

47.

=2

2

1

S

=

39.

an == -9+(n al +(n-l)d - 1)(1) = -9+n-l =n-lO General formula: { an } = { n-l0} al = 3, r =-1 Since I r I < 1, th3e series 3converges. n = � = _(1 -_�) =_(�)_= 2.2 al = 2, r = --1 Since IrI < 1 , the series converges.

49. 1 :

a

II:

d

3

I

+

I

1)

+

1)

II

738

© 2008 Pearson Education , Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws

as

they c urrently

exist. No portion of this material may be reproduced, in any form or by any means, without permission in wtiting froni the publisher.

Chapter 13 Re view Exercises

Chapter 13 Review Exercises 1.

7.

9.

11.

13.

= _�

5+3 7 4+3 = "6'7 a5 =(_ 1)5 5+2 3+3 =- 56 ' a4 = (-1)4 4+2 2+3 = 4"5 ' a3 =(-1)3 3+2 1+3 =- "34 ' a2 =(-1)2 2+2 al =(-1)1 1+2

=

al = 2, a2 = 2 -2 = 0, a3 = 2 - = 2, a4 = 2 -2 0, a5 = 2 - = 2 4 + 2) = (4 . 1 + 2)+( 4·2 + 2)+(4·3+ 2)+(4 . 4 +2) = ( 6)+(10) +(14)+(18) = 48 �)4k 12 31 41 -131 = L:\ 3 ( -1) (-k1 ) 3, %, � , %, 136 ' ' ' Geometric 1--+---+,,·+ {all}= (n= +{ 1n++5)5}-(Arin +t5)hm=etnic+ 6 -n -5 = 1 (%)3 2 -31 =-21 r=-=-· Sn = '::2 [6+n+5]='::2 (n+ll) S. =6 [1- (�J l {en } = { 2n3 } Examine the terms of the sequence:is no2,common 16,54, di128,250, . .there is no There f f e rence; Neicommon ther. ratThere common ratio; neither. io. is no common difference or 1) {sn } =(11{ +2)3n } Geomet r i c 50(50+ ) 3825 ( 3 = k I 3 = 3k I 2 n+ 1 3 3 23 2 r = --= 23n --= 23n 12311+3-3n = 23n= 8 Sn = 8 [\�8�' ) = 8 [ =�" ) = % (8 -1) 30(3� + 1) ) -30(9) = 3 ( 0,4, 8, 12, . . Arithmetic = 4 -0= 4 1395 -270 = 1125 Sn = '::2 ( 2(0) + (n -1)4 ) ='::2 (4(n -1)) 2n(n -1) °

k=1

°

hI

k=1

21.

3

d

1 5.

={I��y ] ={I Nl ]

23.

1 7.

25.

k=1

=

k=1

k=1

1 9.

d

k=1

k=1

k=1

k =1

=

=

= ± (1- 21187 ) 1093 = L2 2186 2187 2187 "" 0.49977

=

737



39.

The 1xO term in ) 1 f( �) ( X2 t-) (.!.x ) = f( �) x24-3) occurs when: 24-3j24=3j =0 j=8 The12 coeffici ent12·11· is 1 0· 9 � ( 8 ) = 8!4! = 4·3·2·1 =495 }=o

43.

J

}=o

41.

J

The X4 term in l �) ( -2)) xl O-%) ) f f ( l �) (X)I O-) (2) ( occurs when: Fx 10-�j=4 2 j =-6 2 j=4 The10 coefficient is 10 . 9 . 8 . 7 ( 4 ) ( _2)4 = � 6!4! . 16= 4·3·2·1 . 1 6=3360 )=0

J

=

)=0

J

-�

(1 .001)5 =(1+1O-3 t = (�}5 + (� }4 . 1 O-3+ G) e . ( 1 O-3 t + G ) e . (1 O-3 t + . . ... 1= 1 ++ 05(0.005.001)+ 0+.010(0. 0 00001) + 10(0 0 00000001) + . 00010+ 0 0 00000010+· · . = 1 .00501 (correct to 5 decimal places) n! = (n-1)n! !(1)! = n(n-1)! ( n-1n ) = ( n-1)! ( n-(n-1))! n-1)! = ( ( n ) = n!(nn!-n)! = n!n!O ! = n!n!. 1 = n!n! = 1 =

45.

n

n

49.

= (�} 11/ + ( � } ln-l . 1+ G} ln-2 . 12 + . . + ( :} In-n ·1/1 = ( �) + C) +· · ·+ (:) 5 +( ) + (�)(�) � (�J (�) G) (�)3 (�J + G) (�J (�J + (�) (�) (�J + G ) (�J = (� + �J = (1)5 = 1

736

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Section 13.5: The Binomial Theorem

1 9.

21.

2 5.

2 7.

29.

31.

(X-2)6 = ( �}6 + ( �}5(_ 2)+ ( �}4(_ 2)2 + (�) X\-2)3 + (!}2 (_ 2)4 + G) X(-2)5 + ( :}O (_ 2)6 . 4 +20x\-8)+15x2 · 1 6+6x· (-32)+64 == x6x6 -12x5 +6x5(-2)+15x4 + 60x4 -160x3 + 240x2 -192x + 64 (3x+1)4 = (�}3X)4 + (:}3X)3 + G}3X)2 + (;) (3X)+ (:) = 81x4 +4·27x3 +6· 9x2 +4·3x+1 = 81x4 +108x3 +54x2 +12x+1 (Fx

+ J2 t = (�)(Fx)6 + ( �) (Fxt (J2 y + G)Fxf (J2/ + ( �) (Fx t (J2)3 +(!)(Fxt(J2f +G}Fx)(J2t + (:)(J2)6

3/2 + 15· 4x + 16·2 4J2x1 / 2 + 8 = x33 + 6J2J2x5/ 22 + 15· 2x2 2 + 20·J2 2J2x = x + 6 x5/ + 30x + 40 x3/ 2 + 60x + 24J2x / + 8 (ax+by)5 = (�} ax)5 + (� }ax)4 . by + (�}ax)3 (bY/ + G}ax)2 (by)3 + (:) ax(by)4 + G}by)5 = a5x5 + 5a4x4by +10a3 x3b2/ +10a2x2b3/ +5axb4/ +b5 / n9= 9, j = 2, x = 2x, a = 3 n10= 10, j = 4, x = x, a = 3 ·8·7 9 10· . 128x\9) . 8 1x6 = ( 2) (2X)7 ·32 =� ( 4 ) X6 . 34 = � 2!7! 4!6! 4·3·2·1 . 8 1x6 9·2·18 . 128x7 . 9 Ox6 = 17, 0 1 = The coefficient ofx6 is 17,010. 7 = 41, 472x n=12, j =5, x=2x, a=-l 7 The coef fi c i e nt of x i s 41, 4 72 . 12( 5 ) (2x)7 · (-1)5 =-·128x 12!5!7! 7 (-1) n7= 7, j = 4, x = x, a = 37 ·6·5 9 · 8 � 3 . 34 x = ( ) . 7 ( -128)x = 12·11·10· 4 4!3! . 8 1x3 = 3·2· 1 . 8 1x3 = 2835x3 5· 4 · 3 ·2·1 -101 , 3 76x7 = n9= 9, j = 2, x = 3x, a = The coefficient ofx7 is -101, 376. . 2 187x7 . 4 (2 ) (3X)7 . (_ 2)2 =� 2!7! = 9·2 ·81 . 8748x7 = 314, 928x7 33.

35.

37.

-

2

735




. nsum=3:of (3-2) 180°=180° whi c hi s t h e t h e angl e s ofa t r i a ngl e . Assume atoffaorconvex any intpolegerygonk, thwie tsumh siofdes tihs e(kangl-2)etsh.180° A convex pol y gon wi t h . witkhe+thanglI ksisidedessesisconsi plus satstrofiangla convex e. Thus,polthyegonsum of (k -2) .180°+ 180° = «k + I) -2) · 1 80°. Condi true. tions and are satisfied; the statement is

2 . . If 2+4+6+ +2k =k +k+2, t h en 2+4+6+"·+2k+2(k+l ) = [22 + 4 + 6 + . . + 2k] + 2k + 2 = k + k + 2 + 2k + 2 = + 2k2 + I) + (k + I) + 2 =(k+I) +(k+I) +2 n=l: 2· 1 =2andI2 +1+2=4:;t2 n = 1: a + (1-1)d = a and 1· a + d 1(1-12 ) = a f a + (a + d)k(k+ (a-1)+ 2d) + " . + [a + (k -l)d] = ka +d 2 then a + (a + d) + (a + 2d) + + [a + - l)d] + (a + kd) [a + (a + d) + (a + 2d) + . . . + [a + - l)d]] + (a + kd) =ka+d k(k2-1) +(a+ kd) = (k + I)a + d [ k(k2-1) + k ] = (k + I)a + d [ k2 - � + 2k ] k2 2+k-] =(k +l)a+d [=(k+l)a+d [ (k �l)k ] ] ) = (k+1) a+d [ (k +I)((k+1)-1 2 tCondi rue. tions and are satisfied; the statement is

33. I:

29. II:

II:

k

(e

I:

I

31. I:

II:

Section 13.5

I

. . ·

=

I

(k

II

1.

3.

(k

5.

7. 9.

II.

13.

II

1 5.

Pascal Triangle n! False', (jn ) = j! ( n-j)! (35 ) 3!2!5! 5·4·3·2 3·2·1·2·1·1 _ 5.2·14 - 10 7·6 (75 ) - 5!2!7! 7·6·5·4·3·2·1 -5·4·3·2 ·1·2· 1 2· 1 21 50·4 ! 50 9 (4509) = � = = 49 !1! 49 !·1 1 =50 1000) = 1000! 1000!O! = I = I (1000 (5523) =� 23! 32! "" 1 . 8664 X I OI 5 (4725 ) =�"" 25! 22! 1 .4834 X I OI 3 _

_

- -

1

734

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Section 13.4: Mathematical Induction

= 1: 1(1 + 1) = 2 and -·31 1(1 + 1)(1 + 2) = 2 f 1·2+2· 3 +3·4+ . . +k(k+l) =-·31 k(k +1)(k +2), then 1· 2 + 2·3 +3·4 + . . . +k(k + 1) + (k + 1)(k + 1+1) = [1 . 2 + 2 · 3 +3· 4 + . . + k(k + 1)] + (k + 1)(k + 2) = � . k(k + 1)(k + 2) + (k + 1)(k + 2) = (k + 1)(k + 2) [� k + 1] = 31 (k + 1)(k + 2)(k + 3) = 31 (k + 1)((k + 1) + 1)((k + 1) + 2) Conditions and are satisfied; the statement is true. = 1: a -b is a factor of al -bl a-b. =l 12 +1=2is divisible by2 fa -ba -bis ias faafctactoroofr ofaakk+1 -bk_ bk+, 1show. that If k2 +2k is divisible by 2 , then (k + 1) + (k + 1) k 2 + 2k + 1 + k + 1 ak+1 _ bk+1 = aa .. akk -b-a .. bk a . -b· k =(e +k)+(2k+2) = a bk +k bk b Sidivnicesibkle2 by+ k2,isthdienvis(ikb+le1)by2 +2 (kand+ 12k) is+ 2 is =a(ak -bk)+b (a-b) di v i s i b l e by 2 . are satisfied; the statement is Siis nacefactao-br ofisaa-bfact, tohrenof aa-bk -bkis aandfactaor-bof Condi t i o ns and true. k+1 _ bk+1 . a Condi tions and are satisfied; the statement is =l 12 -1+2=2is divisible by2 t r ue . f k2 -k + 2 is divisible2by 2 , then (l+a y =1+a I +1 . a (k + V -(k + 1) + 2 = k 2+ 2k + 1-k -1 + 2 Assume thatitythholeredis.s anWeinneedtegertok show for whithcath if =(k -k+2)+(2k) t h e i n equal Sidivnicesibkle2 by-k2,+ t2henis di(kvi+si1)b2le-(kby 2+and1) + 22kisis (1 + a t 1 + ka then di v i s i b l e by 2. ( 1 + a )k + 1 1 + ( k + 1 ) a . (l+a)k+1 =(I +at(l+a) tCondi rue. tions and are satisl fied; the statement is (I +ka)(I +a) = 1: If x > 1 then X = x > 1 . = +a + ka l+ka2 Assume, for some nat u ral number k, t h at i f = 1+ (k +1) a +ka2 xThen> 1 ,xtk+hen1 >x1k, for> 1 x>. 1, 1+(k+l )a 1xk+ = xk . x> 1· x = x > 1 Condi true. tions and are satisfied, the statement is k > 1) ( x tCondi rue. tions and are satisfied; the statement is

1 7. I :

II:

n

I

_ .

_ .

I

1 9. I:

n

II

25. I :

:

II:

II:

n

=

I

=

I

21. I:

II:

n

II

I

:

II

I

27. I :

�

n=l :

II:

�

I

23. I :

II:

�

II

�

n

�

I

II

t

I

II

733

© 2008 Pearson Education, Inc ., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


1 1 . I:

II :

1 = -1 and-1 =-1 n = 1 : 1(1+1) 2 1+1 2 1 1 + -1 + ··· + 1 k , then If 1·2 + 2·3 3·4 k(k+1) k+1 1 + 1 + 1 + ... + 1 + 1 = [1.21 + 1 + 1 + .. . + 1 ] + 1 1.2 2·3 3 · 4 2·3 3·4 k(k + 1) (k + 1)(k + 2) k(k + 1) (k + 1)(k + 1 + 1) k 1 k k+2 1 =--+ k + 1 (k + 1)(k + 2) =-_.--+ k + 1 k + 2 (k + 1)(k + 2) k 2 +2k+1 (k+1)(k+1) k+1 k+1 (k+1)(k+2) (k+1)(k+ 2) k+2 (k+1)+1 --

--

----

Conditions and are satisfied; the statement is true. I

13. I:

II:

II

1 =1 n = 1 : 1 2 = 1 and-·1(1+1)(2·1+1) 6 1 12 +2 2 +3 2 + ... +e =-·k(k+1)(2k+1) , then 6 12 +22 +32 + ... +k 2 +(k+1) 2 = [12 +22 +32 + ... +k 2 J +(k+1)2 = � k(k+1)(2k+1)+(k+1) 2 = (k + 1) [� k(2k + 1) +k + 1 ] = (k + 1{� k 2 + � k +k +1] = (k + 1{� e + � k + 1 ] = � (k + 1) [ 2k 2 + 7k +6] = -61 . (k + 1)(k + 2)(2k + 3) = i · (k + 1) ((k + 1)+ 1) ( 2(k + 1)+ 1)

If

Conditions and are satisfied; the statement is true. 1 -1) = 4 n = 1 : 5 -1 = 4 and -·1(9 2 1 4+ 3 +2+ ... +(5-k) = 2 ·k(9-k) , then I

II

1 5. I :

II:

If

1 4 +3 + 2+ ··· + (5 -k) +( 5 - (k + 1)) = [4+3 + 2 + ... + (5 -k)]+ (4-k) = -k(9-k) 2 +(4 -k) = 29 k- 21 k 2 +4-k = - 21 k 2 + 27 k+4 = - 21 ' [ k 2 -7k-8] 1 1 = - -21 ·(k + 1)(k - 8) = -·(k 2 + 1)(8-k) = -2 . (k + 1)[9 -(k + 1)]

Conditions and are satisfied; the statement is true. I

II

732

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Section 13.4: Mathematical Induction

1 07. 1 09.

Answers wil vary. Answers iwially,varybut .thBote domai h increase (ogeomet r decrease) exponent n of a rliecthe sequence i s t h e set of nat u ral numbers whi of an exponent aldomai l realnnumbers . ial function is the set of

5.

I: n = 1: 3 . 1 -1 = 2 and -21 . 1 (3 . 1 + 1) = 2 II: If 2 + 5 + 8 + . . + (3k = 21 " k(3k + , th2en+ 5 + 8 + . . + (3k -1) + [3(k + 1) -1] = [2 + 5 + 8 + . . + (3k 1 ] + (3k + 2) =.!..2 k(3k + 1) +(3k + 2) = l..2 e +�k2 +3k + 2 = l..2 e + 2.2 k + 2 = .!..2 (3e + 7k + 4) = -·21 (k +1)(3k +4) = l1 ·(k +1) (3(k + 1)+ 1) tCondi rue. tions I and I are satisfied; the statement is I: n=l: 21-1 =landi-l=1 II: If + 2 +222 +. .. . +k-2k-1 1 =k2+Ik-1 , then + 2 + 2 + + 2 k-+1 2 k = [1 + 2 + 22 + ... + 2 ] + 2 = 2kk+-1_+ 2k = 2 ·2k -1 = 2 l l tCondi rue. tions I and I are satisfied; the statement is - 1)

-

Section 13.4 1.

3.

I: n = 1: 2·1 = 2 and 1(1 + 1) = 2 II: If2+4+6+···+2k=k(k+l ) ,t h en 2= +[2+4+6+ 4 + 6 + . . +. .2k+2k]+2(k+l + 2(k + 1) ) ==k(k(k +1+ 1))(k++2)2(k + 1) =(k+l) ((k+l)+I) tCondi rue. tions I and I are satisfied; the statement is I: n = 1 + 2 = 3 and -·21 1(1 + 5) = 3 1 then II: If 3+4+5+ . . . +(k+2) =-·k(k+5), 2 k + 2) + [(k + 1) + 2] 3= [34+4+5+ + 5 + . ....+ (+(k+2)]+(k+3) = -·21 k(k +5) +(k +3) =�k2 2 +�k+k+3 722 1=-k2 2 +-k+3 = � . ( k 2 + 7 k + 6) =-·(k+1 21 )(k+6) = � . (k+1) ((k+l)+5) tCondi rue. tions I and I are satisfied; the statement is

7.

1

1:

+

1

1)

)

-1

II: If 1+4+42 + . . . +4k-l = 31 · (4k -1) ,then 1+4+42 + . . +4k-1k-+41 k+I-k1 = [1+4+42 + . . +4 J +4 = 3l · ( 4k -1) +4k = 31 . 4k - 31 + 4 = � . 4k _ ± = ± (4 . 4k -1) = � . (4k +1 -1) Condi true. tions I and I are satisfied; the statement is k

73 1

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89.

This nis=an(12)(30) ordinary=annui y with peri= $100ods. Theand 360 tpayment mt. erest rate per peno. d · 012. 1 2 001 . Thus, [1+0.00.01tO1 -1 ] "" $349,496.4 1 A = 100 [ Thin =s(is4)an(20)ordi=narypayment annuity periwithods.=The$500intanderest rate per period is 4 = 0.02 . Thus, [ [1+00..002tO2 -1 ] "" $96, 5.9 A = 500 Thiands nis=an(12)(10) ordinary= annui t y wi t h A = $50,000 120 payment peri o ds. The mt. erest rate per perl.od · 012.06 = 000. 5 . Thus, 50,000 = [ 0.005 -1 ] = 50,000 [[1 + 00.0.05t005 20 -1 1 "" $305 . 1 0 Thial =s 1is, argeomet ric= sequence with = 2, n 64 . Find the 1-264 sum of the1-264 geometric series: S64 = 1 ( 1-2 ) = -1 = 264 _ 1 grains The common rat i o , r = 0 9 0 < 1 . The sum i s : . 1 .9 0.11 0 . S=--=-=lO 1-0 The multiplier is 10. This is an infinit1e.0geomet ric series with 3 a=4, and r=-1 . 09 4 Find th"um (1- 1 .03 f $72 .67 . 1 . 09 IS

91.

80

--

=

-

)

>

n

0.08

8

I S --

1 [1 + 0.005] 2 0

1 03 .

P �----'''----

99.

Pde

""

1 1 0. 2 1

=

B:

= i9

= 1 .845 x l 0 1 9

97.

--'----"-

d

P

95.

GiFinvden:n when al = 1000, an < 0r.0=1 0: .9 100 (0.9 < 0.0 1 < (n -1 log(0.9) < log (0.00001) n -1 loglo(0g(0.00001) 9) . loglo(0g .(0000 1 ) +1 0 9 ) . OnamolithnetIlltwil hbedaylessortDecember 20, 2007, the h an $0 0 1 . . Find the sum ofn the geometric series:l l-r1-r ) = 1000 [ 1-(01-0.9.t9 ) S111 = al (-.0. 1 l ) = $9999 .92 = 1000 ( 1-(09t thethsumetm iofc serieachessequence: A:Fia n=dAri$1000, wi t h : Filnd the1000sum of=th-1,e aritnhmet= 1000ic series: SIOOO = -(1000+ 1) = 500(1001) $500,500 2 r1i9c .sequence with n = aFin1d=Thith1,es sirusm=a 2,geomet of t h e geomet r i c seri e s: SI9 =1 ( I-1-2i 9 ) = 1--1i9 -1= $524,287 results in more money. Thesequence amountwitpaih adl each day f o rms a geomet r i c = 0.0 1 and_ r = 2 . 1 1-2222 =41, 943.03 1_r22 S2 =al · -=0.0 1 · -1-r Theworkedtotaall payment l 22 days.would be $41,943.03 if you =a ·r =0 0 1 2 =20, 9 71 5 2 . . l The paymentwil onvary.theWi220dth tdayhis payment is $20,971pl.a5n,2. the Answers bulonekdayof thcane payment isalatly treduce he end tshoemioveralssingl even dramat i c payment .seNottheicamount e that wipaith doneonsithcek 220dayd youday woul d l o which is about half the total payment for the 22 days. 0 ),, 1 1 ( 0.9 ) "- 0.0000 1

.

P

88

93.

101.

P

B

1 05.

a

"

22

22 - 1

( )2 1

730

© 2008 Pearson Educati on, Inc . , Upper Saddle River, NJ. All rights reserved. Thi s material is protected under all copyright laws as they c u rrently

exist. No portion of thi s material may be reproduced , in any form or by any means, without permi ssion in writing from the publi sher.

Section 13.3: Geometric Sequences; Geometric Series

69.

= 8, r = -21 Since�I r I < 1, th8e serie_8s converges . S = 1-r = _(1-_�) = (�_) =16 = 2, r =--41 Since�I r I < 1, the 2series converges. S = 1-r = 1- (- ) = _2_ = �5 ( �) (%) = 8 , r = 23 Since \rI > 1, the series diverges. = 5, r =41 Since I r 1 < th5e series converges . S - G; - (1- � - (�)5 - 320 ) = 21 ' r= 3 Since IrI > 1 , the series diverges. =6, r= -32 Since�I r I < 1, the 6series converges . S = 1-r = 1- ( j ) = _6(�)_ = �5 ( ) �)k;1 (-23 )k = �)k;1 . -23 . (-23 )k-1 =�)k;1 (-23 )k-1 = 2, r = 32 Since I r I < 1, t2he seri2es converges. S - l-r - 1- 2. - 6

83.

al

�

00

71 .

85.

al

75.

77.

79.

_

_

_

an

c.

_

_

00

n

+

al

00

81.

al

b.

al

00

4

al O

1,

al

-

al

al

_

=

87. a.

al

00

�

as

00

73.

Fisystndemtheofcommon ratio ofthe terms and solve the equat i o ns: x+2x --=r x+3 --=r; x+2 x + 3 x2 + 4x + 4 = x2 + 3x x x +x 2 = x+2 Thi=s $18,000, is a geometrri=c seri1.05,es win =th5. Find the 5th term:= 18000(1 .05)S-1 = 18000(1.05t = $21, 879. 1 1 Fisequence: nd the 10th term of the geometric = 2, r =100-1.9, n = 10 = 2(0.9) = 2(0.9)9 = 0.775 feet Find when -1 < 1 : 2(0.9)n I < 1 (0.9r- < 0.5 (n -1) log(0.9) < log(0.5 ) 5 ) n-l > lloog(0. g (0.9) n > 1l0og(0g(0..95 )) 1 "" 7.5 8 On the 8th swing the arc is less than 1 foot. Find the sum of t1he first 15 swings:5 SI S - 2 [ 1_(0.1-09.9) 5 ) -_ 2 ( 1-(0.0. 19Y ] = 20 (1-(0. 9Y S ) = 15. 8 8 feet Find the 2infinite2sum ofthe geometric series: S =--=-=20 1-0.9 0. 1 feet d.

00

00

a1

00

a1

_

_

3

-1 3

729

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exist. No portion of thi s material may be reproduced , in any form or by any means, without permi ssion in writing from the publisher.


43.

45.

47.

49.

a1 = 1, r = -1, n = 9 a9 = 1 . ( _ 1)9 - 1 = ( _ 1)8 = 1

8

a1 = 004, r = 0. 1, n = a8 = Oo4 . (O. lt l = Oo4 (O. lf = 0.00000004 14 = 2 a = a n- l a1 = 7 , r = ' n 1 r 7 a n = 7 · 2 n-1 •

1 a 1 = -3 , r = __ = -.!. ' an = a 1 r n- 1 3 -3 n-l n- 2 a n = -3 = -

( ) ( ) -�

51.

243 = a 1 ( -3 t· •

•

59.

�

8 1 9 1 . 75

Therefore,

4- 1 � � =� = = r2 a a1 r 2 - 1 r 2 1 575 = 225 r2 = 7 r = .J225 = 1 5 an = a1 · r n- l = a 1 . 1 52- 1 7 = 1 5a1 7 a1 = 15 , an = 2 · 1 5 n - 1 = 7 · 1 5 n - 2 15 •

[ ) [ )

61.

1

243 = a 1 ( -3 f 243 = -243a 1 -1 = a 1 nl an = - ( -3 ) -

53.

a1 = -1, r = 2 � � Sn = al = 1 - 2n = -I 1-2 l-r

63.

3

65.

-32767

7

Therefore 55.

•

67.

Since

1 a1 = "4 ' r = 2 � .!. � .!. = = - (1 - 2n ) S11 = a1 1-2 4 l-r =

1 a1 = 1, r = 3 I r I < 1,

the series converges.

� S = � = _1_ = _1_ = l-r 12

[ ) 4[ )

ro

± ( 211 - 1 )

( �) (�)

728


exist. No portion of this material may be reproduced, in any form or by any mean s, without permission in writing from the publi sher.

Section 13.3: Geometric Sequences; Geometric Series

1 5.

n+l ) n+l n ( 2 r = _ = 2(-3--3) = i/3 Thetherefore rat2(�)io tofheconsecut iveistgeomet erms isrconst a. nt, sequence i c - 21 13 ' 2 -- 2 2 /3 - 23/ 3 - 2' 4 - 24/3 ( 3n+1-1 ) 3n-n .-2nn+ r = ( 23�:n+11 ) = -3121 = 3n-(n -l) ·2"-(n+l) = 3.2-1 = �2 3

e1 -

1 7.

27.

__

3 1 -11 2 r = eJ (�J = eJ+ 3 3

e

, e3 -

e

Thetherefratoreio tofheconsecut erms isrconst sequenceiveistgeomet ic. ant, 29.

19.

1 = -i = -2 =-2 ' t2 = -22 =22 =-4 ' 344-1 334 27 33-1 32 t3 = -- = -3 = - t4 = -- =-=23 2 8 ' 2 2 16 { n+2} = (n + 1 + 2) -(n + 2) = n 3 -n -2 = 1 d

21.

23.

2 5.

-2, -8, . .. -2 = -8 = 2 -4 = r=-4 -1 -2

-

1

,

--4,

erms isrconst tTheherefratoreio tofheconsecut sequenceiviestgeomet ic. ant,

Thetherefratoreio tofheconsecut iveistgeomet erms isrconst ant, sequence i c . 3 1 -1 3° 1 3 2-1 3 1 3 t

{m}+1

9

Thetherefratoreio tofheconsecut erms isrconst sequenceiveistgeomet ic. ant,

+

Theconstdiantffe, rence consecutiisvearitethrmsmetiisc. thereforebetwtheene sequence { 4n 2 } Examine the terms of the sequence: 4, 16, 36,is64,no 100, . . difference; there is no There common common ratio; neither.

33.

35.

{3-�n } d = (3-�(n+1»)- (3-�n) =3--n23 - -23 -3+-n23 = - -23 The dif erence between consecutive terms is

37.

39.

constant, therefore the sequence is arithmetic. 1, 3, 6,is10,no .common . Neithdierfference or common ratio. There

41.

as = 2·3s -1 = 2 . 34 = 2·81 = 162 an = 2 . 3"-1 as = 5(_1)s-1 = 5(-1)4 = 5·1 = 5 an = 5 . (_1)"-1 as = 0 · ( 2"1 )S-1 = 0· (2"1 )4 = l an = O · ( � ) n - = 0 as = h . (htl =h.(hf = h·4=4.[i an =h . (h ),, -1 = (h )" a1 = 1, r = -. 21 ' n = 7 °

64

727

-.

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61 .

al = 35 , d = 37-35 = 2 , an = al +(n-1)d = 35+(27 -1)(2) = 35+26(2) = 87 27 27 = 1647 = 2(35+87) = 2(122) The amphitheater has 1647 seats. Thewithyearlal =y35,000, salaries dform= 1400,an arithmet= 280,000 ic sequence. d the $280, number000of. years for the aggregate salary tFionequal =!:[2a 2 l +(n-1)d] 280,000 = !:[2(35, 2 000) + (n -1)1400] 280,000 = n [35,000 + 700n -700] 280,000 = n(700n + 34,300) 280,000 = 700n 2 +34,300n 400 = n 2 +49n n 2 +49n-400 = 0 492-----4(1)(-400) -49± �n = ----'2(1) -49 ± .J4001 -49 63.2 5 2 nIt t::::a: kes7.l3about or2 n8::::years : - 56.13to have an aggregate salary ofyearsat lewiastl be$280,$319,0002.00The. aggregate salary after 8

7.

a 27

S27

63 .

9.

5.

=3

1 2

erms isrconst tTheherefratoreio tofhe1consecut sequenceiviestgeomet ic. a3nt, al = -3 (�) -%, a2 = -3 (�J a3 = -3 (�J -%, a4 = -3 (�J = 136

13.

erms isrconst tTheherefratoreio tofheconsecut sequenceiveistgeomet ic. ant, i- 1 2° -2 ci =-4 = -2 2 = 2 4' c2 = -224--1 = '22i = 2 21 ' 23-1 = '222 2 = 1, C3 = 4 24-1 23 = 2 c4 =--=4 22

Answers h increasen of (anor aridecrease) at asequence constantwiisratlthevary ,e butset. oftBothenatdomai t h met i c unralisnumbers whialllreale the domai n of a l i n ear funct i o t h e set of numbers.

_I

Section 1 3 .3

3.

n l n 3 +-

11.

±

A

3n+n 1 = = -3

l 3 2 24 S3 = 3 = 27, S4 = 3 =81

Sn

----- ::::: ----

1.

r

Thetherefore ratio tofheconsecut iviestgeomet erms isrconst ant, sequence i c . S = 31 = 3, S = 3 = 9,

Sn

65.

Falnegatse;ivthe e(ocommon rathisioresulcantsbeinposia sequence tive or of r 0, but t only Os).

= 1000 (1 + 0.�4 J'2 = $1082. 4 3

geometric annuity

726

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Section 13.2: A rithmetic Sequences

45.

a1 = 4 , d = 4. 5 -4 = 0. 5 , an = a1 +(n-l)d 100 = 4+(n-l)(0. 5 ) 96 =0. 5 (n-l) 192 = n-l 193 = n = �2 (al +an)= 1932 (4+100) = 1 �3 (104) = 1 O,036 a\ = 2(1)-5 = -3, agO = 2(80)-5 = 155 80 (-3+155) = 40(152) = 6080 2 1 1 11 a\ = 6--(1) 2 =-2 , alOo = 6- -2 (100) = -44 = 1�O C21 +(-44)) = 50 ( - 7; ) = -1925 a1 = 14, d = 16-14 = 2 , an = a\ +(n-1)d a120 = 14+(120-1)(2) = 14+ 119( 2) = 252 120 =-(14+252) = 60(266) = 15,960 2

55.

S

57.

SgO =

49.

S

SIOO

51.

=

S120

53 .

= 2 [2(25) + (30 -1)(1)] = 15(50 + 29) = 15(79) There= 1185 are 1 185 seats in the theater. Thetolmightrower colandored1 tilteileins have 20p rowtiles iThe n. the botnumber t h e t o up thwie th ta\ria=ngl20,e.decreases Thid =s-1,is anandbyarin1th=asmet20.weicmove sequence Fi n d t h e sum: = 202 [2(20)+(20-1)(-1) ] = 10(40 -19) = 10(21) There= 210are 210 lighter tiles. colandored1 titleileins have 19p row tiles. iThen the botThenumber todarker m rowdecreases t h e t o up thwie th ta\riangl19,e. dThi=s-1,is anandbyarin1t=hasmet19we.icFimove sequence nd the sum: =�[2(19)+(19-1)(-1) ] 2 =�(38-18) =�(20) = 190 2 2 There are 190 darker tiles. The Siairncecooln srepresent at the ratsethofousands 5. 5 ° F per 1000, wefeet. of feet have d = -5. 5 . The ground temperature is 67°F so we have I; = 67 -5. 5 = 61. 5 . Therefore, {Tn} = { 61. 5 + (n -1)(-5. 5 ) } = {-5. 5 n+67} or {67-5. 5 n} Afthaveer 1'sthe=parcel of ai r has ri s en 5000 feet, we 61.5 +(5-1)(-5. 5 )=39. 5 . The5000parcel feet. of air wil be 39.5°P after it has risen S30

Sn

47.

The= t25ota+l 26number of seat s i s : . . + 27 + . + (25 + 29 (1)) Thid =s1,is a\the=sum25, ofandan arin =th30met. ic sequence with Find th30e sum of the sequence:

S

Fisolnvdetthheecommon diequat fferenceions:of the terms and syst e m of (2x+1)-(x+3) = d x-2 = d �

(5x+2)-(2x+l) = d 3x+l = d 3x+l = x-2 2x = -3 X = - -23 �

59.

725


exist. No portion of this material may be reproduced , in any form or by any means , without permi ssion in writing from the publisher.

Chapter 13: Sequences; Induction; the Binomia l Theorem

25.

27.

29.

31.

al = 2, d = "25 -2 = "21 ' an = al +(n-l)d agO = 2+(80-1)-21 = -832 ag = al +7d = 8 a20 = al +19d = 44

33.

ions by subtracting the Solfirstveeqthuate systion efromm oftheequatsecond: 12d = 36 � d = 3 al =8-7(3) = 8-21 = -13 : a rmul o f e v Recursi al = -13 an = a, _1 + 3 nth term: an = al + ( n -1) d

alg = al +17d = -9 ions by subtracting the Solfialr4stv=eeqaltuathe+13dsystion efr=mom-1ofthequat second: e 4d = -8 � d = -2 al = -1-13(-2) = -1+26 = 25 al = 25 a" = an _I -2 : a rmul o f e v Recursi nth term: an = al +(n-l)d = 25+(n-l)(-2) = 25 -2n+2 = 27 -2n

= -13+(n-l)(3) = -13+3n -3 = 3n-16 a9 = al +8d = -5 aI 5 = al +14d =31 ions by subtracting the Solfirstveeqtuathe systion efromm oftheequatsecond: 6d =36 � d = 6 al = -5-8(6) = -5-48 = -53 al = -53 a" = an_I + 6 : a rmul o f e v Recursi nth term: an = al + ( n -1) d = -53+(n -l)(6) = -53+6n-6 = 6n-59 al 5 = al +14d = 0 a40 = al +39d = -50 ions by subtracting the e systion efrmomoftheequatsecond: fiSol25drstvee=qthuat-50� d = -2 al = -14(-2) = 28 : a rmul o f e v Recursi al = 28 an = an _I -2 nth term: an = al +(n -1)d = 28+(n-l)(-2) = 28-2n+ 2 = 30-2n

39.

al = 2, d = 4 -2 = 2, an = al + (n -1)d 70 = 2 +(n-l)2 70 = 2+2n-2 70 = 2n n = 35 =-n2 ( al + an ) = -352 (2+70) =�(72) 2 = 35(36) = 1260 al = 5, d = 9-5 = 4, an = al +(n- l)d 49 =5+(n-l)4 49 = 5+4n -4 48 = 4n n = 12 Sn = "2n (al +an ) = 212 (5+49) = 6(54) =324 al = 73 , d = 78 -73 = 5, an = a1 + ( n ) d 558 = 73+(n-l)(5) 485 = 5(n -1) 97 = n-l 98 = n =!:(2 al + an) = 982 (73 + 558) = 49( 631) = 30,919

S"

41.

43.

-1

S"

724

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Section 13.2: Arithmetic Sequences

Section 13.2 1. 3.

5.

7.

arithmetic

d = sn - sn _1 = (n + 4) -(n -1 + 4) = (n + 4) -(n + 3) = w1 een consecutive terms is Theconst= din+4-n-3 fanterence bet , therefore the sequence is arithmetic SI = 1+4 = 5, s2 = 2+4 = 6, S3 = 3+4 = 7, . s4 =4+4 = 8 d = an -an_I = (2n -5)-(2(n -1)-5) =(2n-5)-(2n-2-5) = 2n -5 -2n + 7 = 2 Theconstdiantf e, rence betorewtheene seconsecut isvearitethrmsmetiisc. t h eref uence i q al = 2 ·1-5 = -3, a2 = 2 . 2 -5 = -1, a3 = 2 . 3 -5 = 1, a4 = 2 ·4 -5 = 3 d =cn -cn_1 = (6 -2n) -(6-2(n-l» = ( 6 -2n ) -( 6 -2n + 2) = 6 -2n -6 + 2n -2 = -2

13.

1 5.

=

1 7.

Theconstdiafntfe, rence betorewtheene seconsecut isvearitethrmsmetiisc. t h eref uence i q cl = 6 -2 ·1 = 4, c = 6 -2·2 = 2,

9.

= In ( 3n ) - In ( 3n-l ) = nln(3 )-(n-l)ln(3) = (ln3)(n -( n -1) = (ln3)( n -n + 1) Theconst= ln3diantffe, rence betorewtheene seconsecut isvearitethrmsmetiisc. t h eref uence i q SI = In ( 31 ) = ln(3), S2 = In ( 32 ) = 2ln(3), S3 = In (33 ) = 3ln(3), S4 = In ( 34 ) = 4 ln (3) an = al +(n-l)d = 2+(n-l)3 = 2+3n-3 = 3n-l aS I = 3·51-1 = 152 an = al +(n-l)d 5 +(n -1)(-3) = 5-3n+3 = 8-3n aS I = 8-3·51 = -145 an = al +(n-l)d = O +(n-l)-21 1 1 =-n-2 2 =!(n-l) 2 1 aS I =-(51 2 -1) = 25 an = al +(n-l)d = ..fi +'(n -1)..fi = ..fi + ..fin - ..fi = ..fin aS I = 51..fi al = 2, d = 2, an = al + (n-l)d al Oo = 2 + (100 -1)2 = 2 + 99(2) 2 + 1 9 8 = 200 al = 1, d = -2-1 = -3, an =al +(n-l)d a90 = 1+(90-1)(-3) = 1+89(-3) = 1-267 = -266

c3 = 6-2·3 = 0, c24 = 6-2·4 = -2 d = tn - tn_I = (� - ± n ) - (� - ± (n -1» ) = (� - ± n ) - (� - ± n+ ±) 1 -1 n- -1 = - -1 =-21 - -31 n--+ 2 3ween3consecut 3 ive terms is Theconstdiantf e, rence bet therefore the se1 quence is ari1 thmetic. 1 1 1 tI =-2 - -3 ·1 =-6' t2 =-2 - -3 ·2 = - -6' 1 -1 ·3 = - -1 t4 =--1 -1 ·4 = - -5 t3 =-2 3 2' 2 3 6

1 9.

1

21.

=

23 .

723

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81.

BI = 1.0 IBo -100 = 1. 0 1(3000) -100 = $2930 $2930

John's balance is payment. 83.

Phil' s balance is first payment.

b. c.

d.

b.

After months there are rabbits . 87.

This is the Fibonacci sequence.

after making the

k=O

k =O

' '

�. '(, 1 . �� 669296668

It will take

n

= 12

to approximate

1 ( 1. 3 ) = el .3

correct to

8

decimal places.

al = 0.4 , a2 = 0.4 + 0. 3 . 22-2 = 0.4 + 0.3 = 0. 7 , a3 = 0.4 + 0. 3 . 23-2 = 0.4 + 0. 3 ( 2 ) = 1. 0 , a4 = 0.4 +0 .3 . 264-2 = 0.4 +0. 3 ( 4 ) = 1. 6 , as =0.4 +0. 3 . 25-27 = 0.4 +0.3 ( 8 ) = 2. 8 , a6 = 0.4 +0.3 . 2 -2 = 0.4 +0. 3 ( 16) = 5. 2 , a7 = 0.4 +0. 3 . 2 -2 = 0.4 +0. 3 ( 32 ) = 10.0 , as = 0.4 + 0.3 · 2S-2 = 0.4 + 0.3 ( 64) = 19. 6 0.4, 0.7 , 1.0, 1. 6, 2. 8 , 5.2 , 10. 0, 19. 6 . 5, a5 = 2. 8 , a9 = 0.4 + 0. 3 . 29-2 = 0.4 + 0. 3 ( 128) = 38. 8 alO = 0.4 + 0.3 · i O-2 = 0.4 + 0. 3 ( 256) = 77.2 a9 , all = 0.4 +0. 3 ·il-2 = 0.4 +0. 3 ( 512 ) = 154 of2003 313 154

The first eight terms of the sequence are

and

Except for term which has no match, Bode 's formula provides excellent approximations for the mean distances of the planets from the sun.

c.

The mean distance of Ceres from the Sun is approximated by

e.

Pluto 's distance is approximated by

d.

f.

mature pairs of

1 ( 1. 3 ) = el .3 "" ± It: = 1 O!.30 + .!2.l ! + ... + 1 4.34! ",, 3 . 630170833 7 1 ( 1. 3 ) = el .3 "" ± I ;: = 1.O3!0 + 1.l!3 1 + ... + 1.7!3 "" 3. 669060828 1 ( 1. 3 ) = el .3 "" 3.669296668

uM ( se� ( 1 . 3"n/n ! , n , \3 , 1 3 » 3 . 669296667 uM ( se"l ( 1 . 3"n/n ! , n , \3 , 1 2 » 3 . 669296662 9 1 . a.

=

after making the first

BI = 1. 005Bo -534.47 = 1. 005(18,500) -534.47 = $18,058.03 $18,058.03

89. a.

al = 1, a2 1, a3 = 2, a4 = 3, as = 5, a6 = 8, a7 = 13, as = 21, an = an_I + an-2 ag = a7 + a6 = 13 + 8 = 21 7 21 1, 1, 2, 3, 5, 8, 13

85.

and that of Uranus is

as = 19. 6 .

but no term approximates Neptune ' s mean distance from the sun.

According to Bode ' s Law, the mean orbital distance

UB

will be

AU from the sun.

93 . Answers will vary.

722

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Section 13. 1: Sequences

35.

3 7.

39.

a1 = 2, a2 = 3 + 2 = 5, a3 = 3 + 5 = 8, a4 =3+8=11, as = 3+11 = 14 a1 = -2, a2 = 2+(-2) = 0, a3 =3+0=3, a4 = 4+3 = 7, as = 5+7 = 12 a1 = 5, a2 = 2·5 = 10, a3 = 2·10 = 20, a4 = 2· 20 = 40, as = 2 . 40 = 80

61.

63.

67.

1 2 -3 + . .. +-13 = I13 --k -+-+ 2 3 4 13 + 1 k + 1 1 + "' +(_1) 6 (�) = ± (_1)k (�) 1_�+� 3 9 27 3 k=O 3 hI

_

a + (a + d ) + (a + 2d) + . . · + (a + nd) = or

43.

45.

47.

49. 51.

53.

55.

57.

59.

a1 = 1, a2 = 2, a3 = 2 ·1 = 2, a4 = 2 . 2 4, as =4·2 = 8 a1 = A, a2 = A+d, a3 = (A+d)+d = A+2d, a4 = (A + 2d) + d = A + 3d, as = (A +3d)+d = A+4d a1 = .fi, a2 = f2;J2, a3 = �2 + �2 + .fi , Q4 = �2+ �2+ f2;J2 , as = �2+ �2+ �2+ �2+.fi k=1I (k +2) = 3+4+5 +6+ 7 + . .. +(n+ 2} n2 Ik=1 -k22 = 2 1 . . · + -1 Ik=O -31k = I+ -31 + -91 +-+ 27 3" 1 "' + -1 Ik=O k3 1+1 = -31 + -91 +-+ 27 311 k k=I2 (-I) lnk = ln2 - ln 3+ ln 4- .. . +(- lt lnn 1+2+3+ . . · +20 = k=1I20 k

69.

=

71.

73.

75.

II

II

9 49 25 + 2 + - + 8 + - + 1 8 + - + 32 + . . . + 2 2 2 2

1

77.

II

11 -1

79.

_

= k=1I (a+(k-1}d) II

II

= 40(5} = 200 Ik=140 5 = � 04 I40 k = 40(40+1) 2 = 20(41} = 820 20 = 5Ik+ 20 I20 3 Ik�20 (5k +3) = k�I20 (5k )+ k�I3 k� k� = 5 ( 20(2� + I} ) + 3(20} = 1050 + 60 = 1110 Ik=116 (k 2 + 4 ) = k=1I16 k 2 + k=1I16 4 = 16(16+1}(26 . 16+1} +4(16} = 1496 + 64 = 1560 k�0 2k = 2k�/k = 2 [� k - � k ] = 2 [ 60( 6� + I} _ 9(92+ 1} ] = 2[1830-45] = 3570 Ik=S20 k3 =k=1I20 k3 - k=1I4 e = [ 20(2� +I} _ [ 4(:+I} = 2102 -102 = 44,000 times

hi

J

II

k=OI (a+kd)

J

72 1

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Chapter 13 Sequences ; Induction; the Binomial Theorem Section 1 3 . 1 1.

1(2) = 2-12 =.!.2 '. 1(3) = 3 3-1 = 3.3

23 .

7. True; a sequence is a function whose domain is

the set of positive integers.

11.

13.

1 5.

1 7.

19.

10! = 10·9·8·7 ·6·5 ·4·3 ·2·1 = 3,628,800 9! = 9 .8 . 7 . 6! = 9 . 8 . 7 = 504 6! 6! 3!· 7! 3·2·1·7·6·5·4! 4! . 4! . = 3 2 ·1· 7 ·6 5 = 1,260 SI = 1, S2 = 2, S3 = 3, S4 = 4, S5 = 5 2 2 1 1 1 a1 - -1+2 - -3' a2 - -2+2 - -4 - -2 ' 4 4 2 a3 - 3+23 - 5'3 a4 - -4+2 - -6 - -3 ' 5 5 a5 =--=5+2 7 cI = (_1)1 +1 (1 2 ) = 1, c2 = (_1)2+ 1 (22 ) = _4, c3 = (-1)3+1 (32 ) = 9,c4 = ( _ 1)4+ 1 (42 ) = _16, (_1)5+1 (5 2 ) 25 2 2 = -4 =-2 S -- 31i+-1 -- 24 - -21 ' S2 =-3 2 + 41 10 5 ' S3 = 3323+ 1 = 288 = 72 ' S4 = 3 42 + 1 = 8216 = 418 ' S5 = 352+5 1 = 24432 = 618 Cs

21.

=

-

t

3. sequence

9.

1 (_1)1 t1 - (1+1)(1+2) 2·3 - - "6 ' 2 - 1 -1 (_1) 2 (2+1)(2+2) 3·4 - 12 ' -1 = - 1 = t3 = (3 +(_1)3 4· 1)(3 + 2) 5 20' (_1)4 - 1 - 1 t4 - (4+1)(4+2) 5·6 30' (_1) 5 = -1 = - 1 t5 = (5+1)(5+2) 6 .7 42 1

-

27. Each telm is a fraction with the numerator equal

to the term number and the denominator equal to one more than the term number.

n an = - n+1 1

2.

29. Each term is a fraction with the numerator equal

to and the denominator equal to a power of The power is equal to one less than the term number.

a

Ii

=-n2 1-1

-1 n -1 1.

3 1 . The terms form an alternating sequence. Ignoring

the sign, each term always contains a The sign alternates by raising to a power. Since the first term is positive, we use as the power .

all = (_1)"-1

=

I

33. The terms ( ignoring the sign) are equal to the

. ( l)n+1 . an = ( -1 )11+1 ·n

term number. The alternating sign is obtained by

usmg -

720



j.

x 2 - 2x - 4y + I = O x 2 - 2x + 1 = 4y 4y = (x _ l)

13.

2

4

(2 cos x - I)(cos x + 2) =

I.

y = 3 sin(2x)

00

I cos x = - or cos x = -2 2 Since cos x = -2 is impossible, we are left with 1 cos x = 2 Jr x = ± + 2Jrk, where k is an integer

3

The solution set is

{i ;

-2 Y

}

2 cos 2 x + 3 cos x - 2 =

-1

k.

(

2 - 2 cos 2 x = 3 cos x

1 y = - (x - l) 2

y

2 sin 2 x = 3 cos x 2 I - cos 2 x = 3 cos x

xx=±

= sm x

+ 2Jrk, k is any integer

}

2Jr . Peno d : - = Jr 2 Amplitude: 3 y 3 x

71 9

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y = X3

c.

g.

Y

y = In x

X=O y

x x

y

d.

=

1

h.

-

x

Y�

-2 2 1�

�l (- 1 , - 1)

y = ..Jx

e.

Yt 2

�

-

x

n

=

=

1

v

1

�

/'

! 'Ii

71 -\

x

-2

y

5

+

( -'M-i' ,O.) (1

2[-

2

·

(O,-or f.

[� r [ H

The graph is an ellipse. 2 x2 � + =1

2

2

2X2 + 5y 2 = 1

i.

(o.vi-) . ' \

\

( 't ·O)

x

x 2 _ 3y 2 = 1 The graph is a hyperbola x2 y 2 =

eX

I

- - -

1

(fJ' x

y = o- -

-[ � )' = 3

1

1 y

-1

718

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Chapter 1 2 Cumulative Review

Chapter 1 2 Cumulative Review

1.

9.

x(x2X2=2x-10-xor=) =002x -1 = 0 x=The solution set is { �} . -8x-3 =2x30 -3x2 -8x -3 appears to Thehave2x3 -3x2 graph of = an x-intercept at x = 3 . Using synthetic division: -3 -8 6 9 3 3 Therefore, 2x3 -3x2 -8x -3 = 0 (x -3) (2X2 + 3x + 1) = 0 (x -3)(2x+ 1)(1 x+ 1) = 0 x = 3 or x = -- or x = -1 The solution set is {-1, -�, 3} . log3 (lxo-1g ) + log3 ++ 1) )== 3 ((x -1)(2x =32 2X2 -x-1= 9 2X25)(x + == 00 X=-5 orx=-2 Since x = makes the original logarithms undefined, the solution set is {%} . g(x) = -x4-x3+ 1 g(-x) = (-x + 1 =-x-4-+x 1 = -g(x) Thus, griics anwitoddh respect functitoonthande oriitsgigraph symmet n. i s

f(x) = 3x 2 + 1 Usihorinzgontthaelgraph of y = 3x , shi ft t h e graph ts to the riunightt. , then shift the graph vertiycallyuniupward 2

5

1

2

......---1'.:-.- - - - - x

O,

3.

5

-5

1';

-5

Domai n : Range: (1, Horizontal Asymptote: y = 6 is a line. Thex-y i=nt3xgraph e3x+6 rcept: y-y=3(0)+6 intercept: 0= 3x=-6 =6 x=-2 ( -00, (0 )

-3

11

o

2

1

y

.

a.

1

(0

)

+

2

5.

(2x

I)

2 2

(x - 1)(2x + l)

(2x -

- x - lO 2)

b.

2

)4

y

and

( 0, 2) .t

( - 2, 0)

2

2(-X)3

4

y2 = is a circle with center (0, 0) Theradix2 +ugraph s 2.

-2

7.

.Y

2

2 3

(0. -2)

717

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2x-3y � 2 2(0)-3(0)� 2 �2 (0, 0) °

false

The point

2x -3 � 2 y = �3 x - �3 .

is not a solution. Thus, the

graph of the inequality

Y

includes the

28. Let

=

=

unit price for flare j eans, unit price for camisoles, and unit price for t-shirts. The given information yields a system of equations with each of the three women yielding an equation. (Megan)

Because the inequality is non-strict, the line is also part of the graph of the solution. The overlapping shaded region (that is, the shaded region in the graph below) is the solution to the system of linear inequalities. 8

Value of obj . function,

From the table, we can see that the maximum value of is and it occurs at the point

half-plane below the line

.Y

(x, y) z (0,1) z = 5(0)+8(1) = 8 (3,2) z = 5(3)+8(2) = 31 (0,8) z = 5(0)+8(8) =64 z 64, (0,8) . i c t

Comer point,

?

=

{2ii + 2c+3t+ 4t==42.905

2 2 Y = -x 3 3 --

i + 3c 2t = 62 +

(Paige) (Kara)

We can solve this system by using matrices.

y=

-

5= 4 5 ( R) = til ) 1 3 2 ��62 1 3 2 ��62 = 1 -5 �2 � ��17 1 = 2 � ;'17� = 1 � � �2 12 = ( R, = h l z=6 y -z = 2. 5 y-z = 2. 5 y-6 = 2. 5 y = 8. 5 z = 6 x + 3z = 42. 5 x+3z = 42. 5 x+3( 6) = 42. 5 x = 24. 5 $24 5 0, $8. 50, $6.. 00.

[� � �

1 2X + 4

( 4,2) (8,0) . z = 5x + 8y. z x O �:+ y � 8 x-3y � -3 2x + y = 8 x -3y = -3 y = -2x+8 -3y = -x-3 y = -31 x+1

The graph is unbounded. The comer points are and 27. The objective function is

We seek

the largest value of that can occur if and y are solutions of the system of linear inequalities

{

l [� � � l [� l [� l [ ��:l [� ! �l �:l °

°

°

°

° °

The last row represents the equation Substituting this result into

The graph of this system (the feasible points) is shown as the shaded region in the figure below. The comer points of the feasible region are and

. (from the

second row) gives

(0,1) , (3,2), (0,8) . y

Substituting first row) gives x - 3y = -3

into

Thus, flare j eans cost and t-shirts cost

2x + y = 8

(from the

camisoles cost

716

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Chapter 12 Test

24.

A+B = O C=O 6A+3B+D = 4 3C+E = 0 9A = -3

3x+7 (x +3)2 x+3 3x+7 --+ A B ----;:-= (X+3)2 x+3 (X+3)2

The denominator contains the repeated linear factor . Thus, the partial fraction decomposition takes on the form -"'7'

-

From the last equation we get

(x + 3)2 3x + 7 = A ( x + 3) + B 3x+ 7 = Ax+(3A+ B)

Substituting this value into the first equation

Clear the fractions by multiplying both sides by .

gives

The result is the identity

x

A=3.

7 = 3A+B 7 = 3(3)+B -2 = B 3x+7 =--+ 3 -2 ' x+3 (x+3)2 (x+3)2 4x2 -3 X(X2 +3t

2 6.

and

{�2x-3y:L�8

2

The inequalities

x x2 + 3 .

The denominator contains the linear factor

A = _ .l3 6 ( - t) +3 (t) +D = 4 -2+1+D = 4 D=5 4x2 -3 -3 3 x +---. 5x ---.=-+ x 2 +3) (x 2 (x2 +3)2 x(x2 +3)

into the third

Therefore, the partial fraction decomposition is 1 1

Thus, the partial fraction decomposition is

x�0 y�0

x+2y � 8 1 y � --x+4 2 (0,0) . x+2y � 8 0+2(0) � 8 �8 (0,0)

and

require that

the graph be in quadrant I .

and

the repeated irreducible quadratic factor The partial fraction decomposition takes on the form

4x2 -3 = -+ A --Bx+ C + ----::Dx+E -----;:X x2 + 3 X ( x2 + 3 t ( x2 3 t x( x2 + 3 r 4x2 - 3 = A (X2 + 3) 2 + X(X2 + 3)(Bx + C) + x(Dx + E) 4x2 -3 = (A + B)X4 + Cx3 +( 6A+3B+D)x2 +(3C+E)x+(9A)

Test the point

+

o

We clear the fractions by multiplying both sides by

E=0. B =.l3

equation gives us

Therefore, we have Substituting this result into the second equation gives

25.

From the second equation, we

Substituting

W e equate coefficients of like powers of to obtain the system

7 = 3A+B

B =!3 . C=0.

know Substituting this value into the fourth equation yields

or

{3 = A

A = -!3 .

to obtain the identity

?

false

The point

graph of the inequality

Collecting like terms yields

x + 2y � 8 y = -!2 x 4 .

is not a solution. Thus, the

half-plane above the line

includes the +

Because

the inequality is non-strict, the line is also part of the graph of the solution.

Equating coefficients, we obtain the system

2x-3y � 2 2 2 y :'O: -x-3 3 (0,0) .

Test the point 715

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Trigonometry: A Unit Circle Approach Student Solutions Manual

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