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PRELIMINARIES OVERVIEW This chapter reviews the basic...
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4100 AWL/Thomas_ch01p001072 8/19/04 10:49 AM Page 1
Chapter
1
PRELIMINARIES OVERVIEW This chapter reviews the basic ideas you need to start calculus. The topics include the real number system, Cartesian coordinates in the plane, straight lines, parabolas, circles, functions, and trigonometry. We also discuss the use of graphing calculators and computer graphing software.
1.1
Real Numbers and the Real Line This section reviews real numbers, inequalities, intervals, and absolute values.
Real Numbers Much of calculus is based on properties of the real number system. Real numbers are numbers that can be expressed as decimals, such as 
3 = 0.75000 Á 4 1 = 0.33333 Á 3
22 = 1.4142 Á The dots Á in each case indicate that the sequence of decimal digits goes on forever. Every conceivable decimal expansion represents a real number, although some numbers have two representations. For instance, the infinite decimals .999 Á and 1.000 Á represent the same real number 1. A similar statement holds for any number with an infinite tail of 9’s. The real numbers can be represented geometrically as points on a number line called the real line. –2
–1 – 3 4
0
1 3
1 兹2
2
3
4
The symbol denotes either the real number system or, equivalently, the real line. The properties of the real number system fall into three categories: algebraic properties, order properties, and completeness. The algebraic properties say that the real numbers can be added, subtracted, multiplied, and divided (except by 0) to produce more real numbers under the usual rules of arithmetic. You can never divide by 0.
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Chapter 1: Preliminaries
The order properties of real numbers are given in Appendix 4. The following useful rules can be derived from them, where the symbol Q means “implies.”
Rules for Inequalities If a, b, and c are real numbers, then:
5.
a 6 b Q a + c 6 b + c a 6 b Q a  c 6 b  c a 6 b and c 7 0 Q ac 6 bc a 6 b and c 6 0 Q bc 6 ac Special case: a 6 b Q b 6 a 1 a 7 0 Q a 7 0
6.
If a and b are both positive or both negative, then a 6 b Q
1. 2. 3. 4.
1 1 6 a b
Notice the rules for multiplying an inequality by a number. Multiplying by a positive number preserves the inequality; multiplying by a negative number reverses the inequality. Also, reciprocation reverses the inequality for numbers of the same sign. For example, 2 6 5 but 2 7 5 and 1>2 7 1>5. The completeness property of the real number system is deeper and harder to define precisely. However, the property is essential to the idea of a limit (Chapter 2). Roughly speaking, it says that there are enough real numbers to “complete” the real number line, in the sense that there are no “holes” or “gaps” in it. Many theorems of calculus would fail if the real number system were not complete. The topic is best saved for a more advanced course, but Appendix 4 hints about what is involved and how the real numbers are constructed. We distinguish three special subsets of real numbers. 1. 2. 3.
The natural numbers, namely 1, 2, 3, 4, Á The integers, namely 0, ;1, ;2, ;3, Á The rational numbers, namely the numbers that can be expressed in the form of a fraction m>n, where m and n are integers and n Z 0. Examples are 1 , 3

4 4 4 = = , 9 9 9
200 , 13
and
57 =
57 . 1
The rational numbers are precisely the real numbers with decimal expansions that are either (a) terminating (ending in an infinite string of zeros), for example, 3 = 0.75000 Á = 0.75 4
or
(b) eventually repeating (ending with a block of digits that repeats over and over), for example 23 = 2.090909 Á = 2.09 11
The bar indicates the block of repeating digits.
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Real Numbers and the Real Line
3
A terminating decimal expansion is a special type of repeating decimal since the ending zeros repeat. The set of rational numbers has all the algebraic and order properties of the real numbers but lacks the completeness property. For example, there is no rational number whose square is 2; there is a “hole” in the rational line where 22 should be. Real numbers that are not rational are called irrational numbers. They are characterized by having nonterminating and nonrepeating decimal expansions. Examples are 3 p, 22, 2 5, and log10 3. Since every decimal expansion represents a real number, it should be clear that there are infinitely many irrational numbers. Both rational and irrational numbers are found arbitrarily close to any point on the real line. Set notation is very useful for specifying a particular subset of real numbers. A set is a collection of objects, and these objects are the elements of the set. If S is a set, the notation a H S means that a is an element of S, and a x S means that a is not an element of S. If S and T are sets, then S ´ T is their union and consists of all elements belonging either to S or T (or to both S and T). The intersection S ¨ T consists of all elements belonging to both S and T. The empty set ¤ is the set that contains no elements. For example, the intersection of the rational numbers and the irrational numbers is the empty set. Some sets can be described by listing their elements in braces. For instance, the set A consisting of the natural numbers (or positive integers) less than 6 can be expressed as A = 51, 2, 3, 4, 56.
The entire set of integers is written as
50, ;1, ;2, ;3, Á 6.
Another way to describe a set is to enclose in braces a rule that generates all the elements of the set. For instance, the set A = 5x ƒ x is an integer and 0 6 x 6 66 is the set of positive integers less than 6.
Intervals A subset of the real line is called an interval if it contains at least two numbers and contains all the real numbers lying between any two of its elements. For example, the set of all real numbers x such that x 7 6 is an interval, as is the set of all x such that 2 … x … 5. The set of all nonzero real numbers is not an interval; since 0 is absent, the set fails to contain every real number between 1 and 1 (for example). Geometrically, intervals correspond to rays and line segments on the real line, along with the real line itself. Intervals of numbers corresponding to line segments are finite intervals; intervals corresponding to rays and the real line are infinite intervals. A finite interval is said to be closed if it contains both of its endpoints, halfopen if it contains one endpoint but not the other, and open if it contains neither endpoint. The endpoints are also called boundary points; they make up the interval’s boundary. The remaining points of the interval are interior points and together comprise the interval’s interior. Infinite intervals are closed if they contain a finite endpoint, and open otherwise. The entire real line is an infinite interval that is both open and closed.
Solving Inequalities The process of finding the interval or intervals of numbers that satisfy an inequality in x is called solving the inequality.
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Chapter 1: Preliminaries
TABLE 1.1 Types of intervals
Finite:
Notation
Set description
Type
(a, b)
5x ƒ a 6 x 6 b6
Open
5x ƒ a … x … b6
Closed
5x ƒ a … x 6 b6
Halfopen
5x ƒ a 6 x … b6
Halfopen
5x ƒ x 7 a6
Open
[a, q d
5x ƒ x Ú a6
Closed
s  q , bd
5x ƒ x 6 b6
Open
s  q , b]
5x ƒ x … b6
Closed
[a, b] [a, b) (a, b] sa, q d
Infinite:
(set of all real numbers)
s  q, q d
EXAMPLE 1
Picture
a
b
a
b
a
b
a
b
a
a
b
b
Both open and closed
Solve the following inequalities and show their solution sets on the real
line. (a) 2x  1 6 x + 3
0
1
x
4
–3 7
0
x
1 (b)
0
(c)
6 Ú 5 x  1
2x  1 6 x + 3 2x 6 x + 4 x 6 4
Add 1 to both sides. Subtract x from both sides.
The solution set is the open interval s  q , 4d (Figure 1.1a). x
11 5
1
x 6 2x + 1 3
Solution
(a)
(a)
(b) 
(c)
FIGURE 1.1 Solution sets for the inequalities in Example 1.
(b)

x 6 2x + 1 3
x 6 6x + 3 0 6 7x + 3 3 6 7x 
3 6 x 7
Multiply both sides by 3. Add x to both sides. Subtract 3 from both sides. Divide by 7.
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Real Numbers and the Real Line
5
The solution set is the open interval s 3>7, q d (Figure 1.1b). (c) The inequality 6>sx  1d Ú 5 can hold only if x 7 1, because otherwise 6>sx  1d is undefined or negative. Therefore, sx  1d is positive and the inequality will be preserved if we multiply both sides by sx  1d, and we have 6 Ú 5 x  1 6 Ú 5x  5 11 Ú 5x
Multiply both sides by sx  1d . Add 5 to both sides.
11 Ú x. 5
Or x …
11 . 5
The solution set is the halfopen interval (1, 11>5] (Figure 1.1c).
Absolute Value The absolute value of a number x, denoted by ƒ x ƒ , is defined by the formula ƒxƒ = e
EXAMPLE 2
x, x,
x Ú 0 x 6 0.
Finding Absolute Values ƒ 3 ƒ = 3,
ƒ 0 ƒ = 0,
ƒ 5 ƒ = s 5d = 5,
ƒ  ƒaƒƒ = ƒaƒ
Geometrically, the absolute value of x is the distance from x to 0 on the real number line. Since distances are always positive or 0, we see that ƒ x ƒ Ú 0 for every real number x, and ƒ x ƒ = 0 if and only if x = 0. Also, – 5 5 –5
3
ƒ x  y ƒ = the distance between x and y
0
3
4 1 1 4 3 1
4
FIGURE 1.2 Absolute values give distances between points on the number line.
on the real line (Figure 1.2). Since the symbol 2a always denotes the nonnegative square root of a, an alternate definition of ƒ x ƒ is ƒ x ƒ = 2x 2 . It is important to remember that 2a 2 = ƒ a ƒ . Do not write 2a 2 = a unless you already know that a Ú 0. The absolute value has the following properties. (You are asked to prove these properties in the exercises.)
Absolute Value Properties 1.
ƒ a ƒ = ƒ a ƒ
2.
ƒ ab ƒ = ƒ a ƒ ƒ b ƒ
3. 4.
ƒaƒ a ` ` = b ƒbƒ ƒa + bƒ … ƒaƒ + ƒbƒ
A number and its additive inverse or negative have the same absolute value. The absolute value of a product is the product of the absolute values. The absolute value of a quotient is the quotient of the absolute values. The triangle inequality. The absolute value of the sum of two numbers is less than or equal to the sum of their absolute values.
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Chapter 1: Preliminaries
Note that ƒ a ƒ Z  ƒ a ƒ . For example, ƒ 3 ƒ = 3, whereas  ƒ 3 ƒ = 3. If a and b differ in sign, then ƒ a + b ƒ is less than ƒ a ƒ + ƒ b ƒ . In all other cases, ƒ a + b ƒ equals ƒ a ƒ + ƒ b ƒ . Absolute value bars in expressions like ƒ 3 + 5 ƒ work like parentheses: We do the arithmetic inside before taking the absolute value. a
a –a
x
EXAMPLE 3
0
Illustrating the Triangle Inequality
a
ƒ 3 + 5 ƒ = ƒ 2 ƒ = 2 6 ƒ 3 ƒ + ƒ 5 ƒ = 8 ƒ3 + 5ƒ = ƒ8ƒ = ƒ3ƒ + ƒ5ƒ 3  5 ƒ = ƒ 8 ƒ = 8 = ƒ 3 ƒ + ƒ 5 ƒ ƒ
x
FIGURE 1.3 ƒ x ƒ 6 a means x lies between a and a.
The inequality ƒ x ƒ 6 a says that the distance from x to 0 is less than the positive number a. This means that x must lie between a and a, as we can see from Figure 1.3. The following statements are all consequences of the definition of absolute value and are often helpful when solving equations or inequalities involving absolute values.
Absolute Values and Intervals If a is any positive number, then 5. 6. 7. 8. 9.
ƒxƒ ƒxƒ ƒxƒ ƒxƒ ƒxƒ
= 6 7 … Ú
a a a a a
if and only if x = ;a if and only if a 6 x 6 a if and only if x 7 a or x 6 a if and only if a … x … a if and only if x Ú a or x … a
The symbol 3 is often used by mathematicians to denote the “if and only if ” logical relationship. It also means “implies and is implied by.”
EXAMPLE 4
Solving an Equation with Absolute Values
Solve the equation ƒ 2x  3 ƒ = 7. Solution
By Property 5, 2x  3 = ;7, so there are two possibilities: 2x  3 = 7 2x = 10 x = 5
2x  3 = 7 2x = 4 x = 2
Equivalent equations without absolute values Solve as usual.
The solutions of ƒ 2x  3 ƒ = 7 are x = 5 and x = 2.
EXAMPLE 5
Solving an Inequality Involving Absolute Values
2 Solve the inequality ` 5  x ` 6 1.
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1.1 Real Numbers and the Real Line Solution
7
We have 2 2 ` 5  x ` 6 1 3 1 6 5  x 6 1
Property 6
2 3 6 6  x 6 4
Subtract 5.
1 33 7 x 7 2
1 Multiply by  . 2
3
1 1 6 x 6 . 3 2
Take reciprocals.
Notice how the various rules for inequalities were used here. Multiplying by a negative number reverses the inequality. So does taking reciprocals in an inequality in which both sides are positive. The original inequality holds if and only if s1>3d 6 x 6 s1>2d. The solution set is the open interval (1>3, 1>2).
EXAMPLE 6 Solve the inequality and show the solution set on the real line: (a) ƒ 2x  3 ƒ … 1 x 1
2
(b) ƒ 2x  3 ƒ Ú 1
Solution
ƒ 2x  3 ƒ 1 … 2x  3 2 … 2x 1 … x
(a)
(a) x 1
2 (b)
FIGURE 1.4 The solution sets (a) [1, 2] and (b) s  q , 1] ´ [2, q d in Example 6.
… … … …
1 1 4 2
Property 8 Add 3. Divide by 2.
The solution set is the closed interval [1, 2] (Figure 1.4a). (b)
ƒ 2x  3 ƒ Ú 1 2x  3 Ú 1 or 2x  3 … 1 x 
3 1 Ú 2 2
or
x Ú 2
or
x 
Property 9
3 1 … 2 2
Divide by 2.
x … 1
Add
The solution set is s  q , 1] ´ [2, q d (Figure 1.4b).
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3 . 2
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1.1 Real Numbers and the Real Line
7
EXERCISES 1.1 Decimal Representations 1. Express 1>9 as a repeating decimal, using a bar to indicate the repeating digits. What are the decimal representations of 2>9? 3>9? 8>9? 9>9? 2. Express 1>11 as a repeating decimal, using a bar to indicate the repeating digits. What are the decimal representations of 2>11? 3>11? 9>11? 11>11?
Inequalities 3. If 2 6 x 6 6 , which of the following statements about x are necessarily true, and which are not necessarily true? a. 0 6 x 6 4 x c. 1 6 6 3 2 6 e. 1 6 x 6 3
b. 0 6 x  2 6 4 1 1 1 d. 6 x 6 6 2
g. 6 6 x 6 2
h. 6 6 x 6 2
f. ƒ x  4 ƒ 6 2
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Chapter 1: Preliminaries
4. If 1 6 y  5 6 1 , which of the following statements about y are necessarily true, and which are not necessarily true? a. 4 6 y 6 6
b. 6 6 y 6 4
c. y 7 4
d. y 6 6 y f. 2 6 6 3 2
e. 0 6 y  4 6 2 1 1 1 g. 6 y 6 6 4
Solve the inequalities in Exercises 35–42. Express the solution sets as intervals or unions of intervals and show them on the real line. Use the result 2a 2 = ƒ a ƒ as appropriate. 35. x 2 6 2 38.
h. ƒ y  5 ƒ 6 1
1 1 6 x2 6 9 4
41. x 2  x 6 0
In Exercises 5–12, solve the inequalities and show the solution sets on the real line.
36. 4 … x 2
37. 4 6 x 2 6 9
39. sx  1d2 6 4
40. sx + 3d2 6 2
42. x 2  x  2 Ú 0
Theory and Examples 43. Do not fall into the trap ƒ a ƒ = a . For what real numbers a is this equation true? For what real numbers is it false?
5. 2x 7 4
6. 8  3x Ú 5
7. 5x  3 … 7  3x
8. 3s2  xd 7 2s3 + xd
7 1 9. 2x  Ú 7x + 2 6 4 1 11. sx  2d 6 sx  6d 5 3
Quadratic Inequalities
6  x 3x  4 10. 6 4 2 x + 5 12 + 3x 12. … 2 4
Absolute Value
44. Solve the equation ƒ x  1 ƒ = 1  x . 45. A proof of the triangle inequality Give the reason justifying each of the numbered steps in the following proof of the triangle inequality. ƒ a + b ƒ 2 = sa + bd2 = a 2 + 2ab + b 2 … a2 + 2 ƒ a ƒ ƒ b ƒ + b2
Solve the equations in Exercises 13–18.
2
13. ƒ y ƒ = 3
14. ƒ y  3 ƒ = 7
15. ƒ 2t + 5 ƒ = 4
16. ƒ 1  t ƒ = 1
9 17. ƒ 8  3s ƒ = 2
18. `
s  1` = 1 2
Solve the inequalities in Exercises 19–34, expressing the solution sets as intervals or unions of intervals. Also, show each solution set on the real line. 19. ƒ x ƒ 6 2 22. ƒ t + 2 ƒ 6 1
20. ƒ x ƒ … 2 23. ƒ 3y  7 ƒ 6 4
21. ƒ t  1 ƒ … 3 24. ƒ 2y + 5 ƒ 6 1
25. `
26. `
1 1 27. ` 3  x ` 6 2
z  1` … 1 5
3 z  1` … 2 2
2 28. ` x  4 ` 6 3
29. ƒ 2s ƒ Ú 4
1 30. ƒ s + 3 ƒ Ú 2
31. ƒ 1  x ƒ 7 1
32. ƒ 2  3x ƒ 7 5
33. `
34. `
3r 2  1` 7 5 5
(1)
r + 1 ` Ú 1 2
= ƒaƒ + 2ƒaƒ ƒbƒ + ƒbƒ = s ƒ a ƒ + ƒ b ƒ d2 ƒa + bƒ … ƒaƒ + ƒbƒ
(2) 2
(3)
(4)
46. Prove that ƒ ab ƒ = ƒ a ƒ ƒ b ƒ for any numbers a and b. 47. If ƒ x ƒ … 3 and x 7 1>2 , what can you say about x? 48. Graph the inequality ƒ x ƒ + ƒ y ƒ … 1 . 49. Let ƒsxd = 2x + 1 and let d 7 0 be any positive number. Prove that ƒ x  1 ƒ 6 d implies ƒ ƒsxd  ƒs1d ƒ 6 2d . Here the notation ƒ(a) means the value of the expression 2x + 1 when x = a. This function notation is explained in Section 1.3. 50. Let ƒsxd = 2x + 3 and let P 7 0 be any positive number. Prove P that ƒ ƒsxd  ƒs0d ƒ 6 P whenever ƒ x  0 ƒ 6 . Here the no2 tation ƒ(a) means the value of the expression 2x + 3 when x = a. (See Section 1.3.) 51. For any number a, prove that ƒ a ƒ = ƒ a ƒ . 52. Let a be any positive number. Prove that ƒ x ƒ 7 a if and only if x 7 a or x 6 a . 53. a. If b is any nonzero real number, prove that ƒ 1>b ƒ = 1> ƒ b ƒ . ƒaƒ a for any numbers a and b Z 0 . b. Prove that ` ` = b ƒbƒ 54. Using mathematical induction (see Appendix 1), prove that n n ƒ a ƒ = ƒ a ƒ for any number a and positive integer n.
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1.2
This section reviews coordinates, lines, distance, circles, and parabolas in the plane. The notion of increment is also discussed.
y P(a, b)
b 3 2 Negative xaxis –3
–2
–1
1
Origin
0
1
–1 Negative yaxis
2
a3
x
Positive xaxis
–2 –3
FIGURE 1.5 Cartesian coordinates in the plane are based on two perpendicular axes intersecting at the origin. y (1, 3) 3 Second quadrant (, )
First quadrant (, )
2
1 (–2, 1)
(2, 1)
(0, 0) (1, 0)
–2
–1
0
1
2
x
(–2, –1) Third quadrant (, )
9
Lines, Circles, and Parabolas
1.2
Positive yaxis
Lines, Circles, and Parabolas
–1
Fourth quadrant (, )
–2 (1, –2)
FIGURE 1.6 Points labeled in the xycoordinate or Cartesian plane. The points on the axes all have coordinate pairs but are usually labeled with single real numbers, (so (1, 0) on the xaxis is labeled as 1). Notice the coordinate sign patterns of the quadrants.
Cartesian Coordinates in the Plane In the previous section we identified the points on the line with real numbers by assigning them coordinates. Points in the plane can be identified with ordered pairs of real numbers. To begin, we draw two perpendicular coordinate lines that intersect at the 0point of each line. These lines are called coordinate axes in the plane. On the horizontal xaxis, numbers are denoted by x and increase to the right. On the vertical yaxis, numbers are denoted by y and increase upward (Figure 1.5). Thus “upward” and “to the right” are positive directions, whereas “downward” and “to the left” are considered as negative. The origin O, also labeled 0, of the coordinate system is the point in the plane where x and y are both zero. If P is any point in the plane, it can be located by exactly one ordered pair of real numbers in the following way. Draw lines through P perpendicular to the two coordinate axes. These lines intersect the axes at points with coordinates a and b (Figure 1.5). The ordered pair (a, b) is assigned to the point P and is called its coordinate pair. The first number a is the xcoordinate (or abscissa) of P; the second number b is the ycoordinate (or ordinate) of P. The xcoordinate of every point on the yaxis is 0. The ycoordinate of every point on the xaxis is 0. The origin is the point (0, 0). Starting with an ordered pair (a, b), we can reverse the process and arrive at a corresponding point P in the plane. Often we identify P with the ordered pair and write P(a, b). We sometimes also refer to “the point (a, b)” and it will be clear from the context when (a, b) refers to a point in the plane and not to an open interval on the real line. Several points labeled by their coordinates are shown in Figure 1.6. This coordinate system is called the rectangular coordinate system or Cartesian coordinate system (after the sixteenth century French mathematician René Descartes). The coordinate axes of this coordinate or Cartesian plane divide the plane into four regions called quadrants, numbered counterclockwise as shown in Figure 1.6. The graph of an equation or inequality in the variables x and y is the set of all points P(x, y) in the plane whose coordinates satisfy the equation or inequality. When we plot data in the coordinate plane or graph formulas whose variables have different units of measure, we do not need to use the same scale on the two axes. If we plot time vs. thrust for a rocket motor, for example, there is no reason to place the mark that shows 1 sec on the time axis the same distance from the origin as the mark that shows 1 lb on the thrust axis. Usually when we graph functions whose variables do not represent physical measurements and when we draw figures in the coordinate plane to study their geometry and trigonometry, we try to make the scales on the axes identical. A vertical unit of distance then looks the same as a horizontal unit. As on a surveyor’s map or a scale drawing, line segments that are supposed to have the same length will look as if they do and angles that are supposed to be congruent will look congruent. Computer displays and calculator displays are another matter. The vertical and horizontal scales on machinegenerated graphs usually differ, and there are corresponding distortions in distances, slopes, and angles. Circles may look like ellipses, rectangles may look like squares, right angles may appear to be acute or obtuse, and so on. We discuss these displays and distortions in greater detail in Section 1.7.
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Chapter 1: Preliminaries
Increments and Straight Lines
y C(5, 6) 6
When a particle moves from one point in the plane to another, the net changes in its coordinates are called increments. They are calculated by subtracting the coordinates of the starting point from the coordinates of the ending point. If x changes from x1 to x2 , the increment in x is
B (2, 5)
5 4
y –5, x 0
3
¢x = x2  x1 .
2
EXAMPLE 1 In going from the point As4, 3d to the point B(2, 5) the increments in the x and ycoordinates are
y 8
1
D(5, 1)
0
1
2
3
4
x
5
¢x = 2  4 = 2,
–1
¢y = 5  s 3d = 8.
From C(5, 6) to D(5, 1) the coordinate increments are
–2
¢x = 5  5 = 0,
–3
(2, –3)
A(4, –3) x –2
FIGURE 1.7 Coordinate increments may be positive, negative, or zero (Example 1).
¢y = 1  6 = 5.
See Figure 1.7. Given two points P1sx1, y1 d and P2sx2, y2 d in the plane, we call the increments ¢x = x2  x1 and ¢y = y2  y1 the run and the rise, respectively, between P1 and P2 . Two such points always determine a unique straight line (usually called simply a line) passing through them both. We call the line P1 P2 . Any nonvertical line in the plane has the property that the ratio
HISTORICAL BIOGRAPHY*
¢y y2  y1 rise m = run = = x2  x1 ¢x
René Descartes (1596–1650)
has the same value for every choice of the two points P1sx1, y1 d and P2sx2, y2 d on the line (Figure 1.8). This is because the ratios of corresponding sides for similar triangles are equal. y P2
L
DEFINITION P2 (x2, y2) y (rise)
¢y y2  y1 rise m = run = = x2  x1 ¢x
y
is the slope of the nonvertical line P1 P2 .
P1 (x1, y1) x (run) P1 x 0
Slope
The constant
Q(x2, y1) Q x
FIGURE 1.8 Triangles P1 QP2 and P1 ¿Q¿P2 ¿ are similar, so the ratio of their sides has the same value for any two points on the line. This common value is the line’s slope.
The slope tells us the direction (uphill, downhill) and steepness of a line. A line with positive slope rises uphill to the right; one with negative slope falls downhill to the right (Figure 1.9). The greater the absolute value of the slope, the more rapid the rise or fall. The slope of a vertical line is undefined. Since the run ¢x is zero for a vertical line, we cannot evaluate the slope ratio m. The direction and steepness of a line can also be measured with an angle. The angle of inclination of a line that crosses the xaxis is the smallest counterclockwise angle from the xaxis to the line (Figure 1.10). The inclination of a horizontal line is 0°. The inclination of a vertical line is 90°. If f (the Greek letter phi) is the inclination of a line, then 0 … f 6 180°.
To learn more about the historical figures and the development of the major elements and topics of calculus, visit www.awbc.com/thomas.
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4100 AWL/Thomas_ch01p001072 8/19/04 10:50 AM Page 11
1.2 y
L2
m = tan f.
P4(3, 6)
P1(0, 5)
Straight lines have relatively simple equations. All points on the vertical line through the point a on the xaxis have xcoordinates equal to a. Thus, x = a is an equation for the vertical line. Similarly, y = b is an equation for the horizontal line meeting the yaxis at b. (See Figure 1.12.) We can write an equation for a nonvertical straight line L if we know its slope m and the coordinates of one point P1sx1, y1 d on it. If P(x, y) is any other point on L, then we can use the two points P1 and P to compute the slope,
4 3
P2(4, 2)
2 1 0 –1
1
2
3
4
5
x
6
y  y1 m = x  x1
P3(0, –2)
FIGURE 1.9
The slope of L1 is ¢y 6  s 2d 8 m = = = . 3  0 3 ¢x That is, y increases 8 units every time x increases 3 units. The slope of L2 is ¢y 3 2  5 m = = . = 4  0 4 ¢x That is, y decreases 3 units every time x increases 4 units.
this
so that y  y1 = msx  x1 d
y = y1 + msx  x1 d is the pointslope equation of the line that passes through the point sx1, y1 d and has slope m.
EXAMPLE 2 x
not this
y = y1 + msx  x1 d.
or
The equation
this x
not this
FIGURE 1.10 Angles of inclination are measured counterclockwise from the xaxis.
Write an equation for the line through the point (2, 3) with slope 3>2.
We substitute x1 = 2, y1 = 3, and m = 3>2 into the pointslope equation
Solution
and obtain y = 3 
3 Ax  2B, 2
or
y = 
3 x + 6. 2
When x = 0, y = 6 so the line intersects the yaxis at y = 6.
EXAMPLE 3 y
A Line Through Two Points
Write an equation for the line through s 2, 1d and (3, 4). P2
L
Solution
The line’s slope is
y P1
x m
11
The relationship between the slope m of a nonvertical line and the line’s angle of inclination f is shown in Figure 1.11:
L1 6
Lines, Circles, and Parabolas
m =
5 1  4 = = 1. 5 2  3
We can use this slope with either of the two given points in the pointslope equation:
y tan x
With sx1 , y1 d s2, 1d x
FIGURE 1.11 The slope of a nonvertical line is the tangent of its angle of inclination.
With sx1 , y1 d s3, 4d
+ 1 # sx  s 2dd
y = 4 + 1 # sx  3d y = 4 + x  3 y = x + 1
y = 1 y = 1 + x + 2 y = x + 1
Same result
Either way, y = x + 1 is an equation for the line (Figure 1.13).
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Chapter 1: Preliminaries
The ycoordinate of the point where a nonvertical line intersects the yaxis is called the yintercept of the line. Similarly, the xintercept of a nonhorizontal line is the xcoordinate of the point where it crosses the xaxis (Figure 1.14). A line with slope m and yintercept b passes through the point (0, b), so it has equation
y Along this line, x2
6 5
y = b + msx  0d,
Along this line, y3
4 3
or, more simply,
y = mx + b.
(2, 3)
2 1 1
0
2
3
The equation
x
4
y = mx + b
FIGURE 1.12 The standard equations for the vertical and horizontal lines through (2, 3) are x = 2 and y = 3 .
is called the slopeintercept equation of the line with slope m and yintercept b.
Lines with equations of the form y = mx have yintercept 0 and so pass through the origin. Equations of lines are called linear equations. The equation sA and B not both 0d
Ax + By = C
y 4
is called the general linear equation in x and y because its graph always represents a line and every line has an equation in this form (including lines with undefined slope).
(3, 4) yx1
EXAMPLE 4
Finding the Slope and yIntercept
Find the slope and yintercept of the line 8x + 5y = 20. –2
0 –1 (–2, –1)
1
2
3
x
Solution
FIGURE 1.13 The line in Example 3.
Solve the equation for y to put it in slopeintercept form: 8x + 5y = 20 5y = 8x + 20 y = 
8 x + 4. 5
The slope is m = 8>5. The yintercept is b = 4.
y
Parallel and Perpendicular Lines Lines that are parallel have equal angles of inclination, so they have the same slope (if they are not vertical). Conversely, lines with equal slopes have equal angles of inclination and so are parallel. If two nonvertical lines L1 and L2 are perpendicular, their slopes m1 and m2 satisfy m1 m2 = 1, so each slope is the negative reciprocal of the other:
b L
0
a
x
FIGURE 1.14 Line L has xintercept a and yintercept b.
1 m1 =  m2 ,
1 m2 =  m1 .
To see this, notice by inspecting similar triangles in Figure 1.15 that m1 = a>h, and m2 = h>a. Hence, m1 m2 = sa>hds h>ad = 1.
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1.2
Distance and Circles in the Plane
y L1
The distance between points in the plane is calculated with a formula that comes from the Pythagorean theorem (Figure 1.16).
C
1 A
y
Slope m 2 2
D
a
This distance is
兹 x2 – x12 y2 – y12 兹 (x2 – x1)2 (y2 – y1)2
d x
B
y2
FIGURE 1.15 ¢ADC is similar to ¢CDB . Hence f1 is also the upper angle in ¢CDB . From the sides of ¢CDB , we read tan f1 = a>h .
y1
Q(x2 , y2)
P(x1, y1)
1 h
Slope m1
L2
0
13
Lines, Circles, and Parabolas
y2 y1
C(x2 , y1)
x2 x1 0
x1
x2
x
FIGURE 1.16 To calculate the distance between Psx1 , y1 d and Qsx2 , y2 d , apply the Pythagorean theorem to triangle PCQ.
Distance Formula for Points in the Plane The distance between Psx1 , y1 d and Qsx2 , y2 d is d = 2s¢xd2 + s¢yd2 = 2sx2  x1 d2 + s y2  y1 d2 .
EXAMPLE 5
Calculating Distance
(a) The distance between Ps 1, 2d and Q(3, 4) is 2s3  s 1dd2 + s4  2d2 = 2s4d2 + s2d2 = 220 = 24 # 5 = 225 . (b) The distance from the origin to P(x, y) is
y
2sx  0d2 + s y  0d2 = 2x 2 + y 2 . P(x, y)
By definition, a circle of radius a is the set of all points P(x, y) whose distance from some center C(h, k) equals a (Figure 1.17). From the distance formula, P lies on the circle if and only if
a
2sx  hd2 + s y  kd2 = a,
C(h, k)
so (x h) 2 ( y k) 2 a 2 0
x
(x  h) 2 + (y  k) 2 = a 2.
(1)
FIGURE 1.17 A circle of radius a in the xyplane, with center at (h, k).
Equation (1) is the standard equation of a circle with center (h, k) and radius a. The circle of radius a = 1 and centered at the origin is the unit circle with equation x 2 + y 2 = 1.
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Chapter 1: Preliminaries
EXAMPLE 6 (a) The standard equation for the circle of radius 2 centered at (3, 4) is sx  3d2 + s y  4d2 = 22 = 4 . (b) The circle sx  1d2 + s y + 5d2 = 3 has h = 1, k = 5, and a = 23. The center is the point sh, kd = s1, 5d and the radius is a = 23. If an equation for a circle is not in standard form, we can find the circle’s center and radius by first converting the equation to standard form. The algebraic technique for doing so is completing the square (see Appendix 9).
EXAMPLE 7
Finding a Circle’s Center and Radius
Find the center and radius of the circle x 2 + y 2 + 4x  6y  3 = 0. We convert the equation to standard form by completing the squares in x and y:
Solution 2
y Exterior: (x h) 2 (y k) 2 a 2
2
x + y + 4x  6y  3 = 0
Start with the given equation.
sx 2 + 4x
Gather terms. Move the constant to the righthand side.
d + s y 2  6y
d = 3
2
2
6 4 ax 2 + 4x + a b b + ay 2  6y + a b b = 2 2
On: (x h)2 (y k)2 a2
2
6 4 3 + a b + a b 2 2
a
k
(h, k)
2
Add the square of half the coefficient of x to each side of the equation. Do the same for y. The parenthetical expressions on the lefthand side are now perfect squares.
sx 2 + 4x + 4d + s y 2  6y + 9d = 3 + 4 + 9 Write each quadratic as a squared linear expression.
sx + 2d2 + s y  3d2 = 16 Interior: (x h) 2 ( y k) 2 a 2
0
h
The center is s 2, 3d and the radius is a = 4. x
The points (x, y) satisfying the inequality sx  hd2 + s y  kd2 6 a 2
FIGURE 1.18 The interior and exterior of the circle sx  hd2 + s y  kd2 = a 2 .
make up the interior region of the circle with center (h, k) and radius a (Figure 1.18). The circle’s exterior consists of the points (x, y) satisfying sx  hd2 + s y  kd2 7 a 2 .
Parabolas The geometric definition and properties of general parabolas are reviewed in Section 10.1. Here we look at parabolas arising as the graphs of equations of the form y = ax 2 + bx + c.
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1.2
EXAMPLE 8
y
15
Lines, Circles, and Parabolas
The Parabola y = x 2
2
(–2, 4)
yx (2, 4)
4
Consider the equation y = x 2 . Some points whose coordinates satisfy this equation are 3 9 s0, 0d, s1, 1d, a , b, s 1, 1d, s2, 4d, and s 2, 4d. These points (and all others satisfying 2 4 the equation) make up a smooth curve called a parabola (Figure 1.19).
3 , 9 2 4 (–1, 1)
–2
–1
The graph of an equation of the form
(1, 1)
1 0
1
y = ax 2
x
2
is a parabola whose axis (axis of symmetry) is the yaxis. The parabola’s vertex (point where the parabola and axis cross) lies at the origin. The parabola opens upward if a 7 0 and downward if a 6 0. The larger the value of ƒ a ƒ , the narrower the parabola (Figure 1.20). Generally, the graph of y = ax 2 + bx + c is a shifted and scaled version of the parabola y = x 2 . We discuss shifting and scaling of graphs in more detail in Section 1.5.
FIGURE 1.19 The parabola y = x 2 (Example 8).
The Graph of y ax 2 bx c, a 0 The graph of the equation y = ax 2 + bx + c, a Z 0, is a parabola. The parabola opens upward if a 7 0 and downward if a 6 0. The axis is the line
y y 2x 2 symmetry
x = y
(2)
2
x 2
y
b . 2a
The vertex of the parabola is the point where the axis and parabola intersect. Its xcoordinate is x = b>2a; its ycoordinate is found by substituting x = b>2a in the parabola’s equation.
x2 10
1 –4 –3 –2
2
3
x
4
Vertex at origin
Axis of
–1 2
y –x 6 y –x 2
Notice that if a = 0, then we have y = bx + c which is an equation for a line. The axis, given by Equation (2), can be found by completing the square or by using a technique we study in Section 4.1.
EXAMPLE 9
Graphing a Parabola
Graph the equation y = FIGURE 1.20 Besides determining the direction in which the parabola y = ax 2 opens, the number a is a scaling factor. The parabola widens as a approaches zero and narrows as ƒ a ƒ becomes large.
Solution
1 2 x  x + 4. 2
Comparing the equation with y = ax 2 + bx + c we see that 1 a =  , 2
b = 1,
c = 4.
Since a 6 0, the parabola opens downward. From Equation (2) the axis is the vertical line x = 
s 1d b = = 1. 2a 2s 1>2d
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Chapter 1: Preliminaries
Vertex is –1, 9 2 Point symmetric with yintercept
y = 
Intercept at y 4 (0, 4) Axis: x –1
(–2, 4)
–3 –2
When x = 1, we have
y
3
1 y – x2 x 4 2
2
The vertex is s 1, 9>2d. The xintercepts are where y = 0: 
1 0
9 1 s 1d2  s 1d + 4 = . 2 2
1
x
Intercepts at x –4 and x 2
FIGURE 1.21 The parabola in Example 9.
1 2 x  x + 4 = 0 2
x 2 + 2x  8 = 0 sx  2dsx + 4d = 0 x = 2, x = 4 We plot some points, sketch the axis, and use the direction of opening to complete the graph in Figure 1.21.
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Chapter 1: Preliminaries
EXERCISES 1.2 Increments and Distance
18. Passes through s2, 3d with slope 1>2
In Exercises 1–4, a particle moves from A to B in the coordinate plane. Find the increments ¢x and ¢y in the particle’s coordinates. Also find the distance from A to B.
19. Passes through (3, 4) and s 2, 5d
1. As 3, 2d,
Bs 1, 2d
3. As 3.2, 2d, Bs 8.1, 2d
2. As 1, 2d,
Bs 3, 2d
4. As 22, 4d, Bs0, 1.5d
Describe the graphs of the equations in Exercises 5–8.
20. Passes through s 8, 0d and s 1, 3d 21. Has slope 5>4 and yintercept 6 22. Has slope 1>2 and yintercept 3 23. Passes through s 12, 9d and has slope 0 24. Passes through (1>3, 4), and has no slope 25. Has yintercept 4 and xintercept 1
5. x 2 + y 2 = 1
6. x 2 + y 2 = 2
26. Has yintercept 6 and xintercept 2
7. x 2 + y 2 … 3
8. x 2 + y 2 = 0
27. Passes through s5, 1d and is parallel to the line 2x + 5y = 15
Slopes, Lines, and Intercepts
28. Passes through A  22, 2 B parallel to the line 22x + 5y = 23
Plot the points in Exercises 9–12 and find the slope (if any) of the line they determine. Also find the common slope (if any) of the lines perpendicular to line AB.
30. Passes through (0, 1) and is perpendicular to the line 8x  13y = 13
9. As 1, 2d, 11. As2, 3d,
Bs 2, 1d Bs 1, 3d
10. As 2, 1d,
Bs2, 2d
12. As 2, 0d,
Bs 2, 2d
In Exercises 13–16, find an equation for (a) the vertical line and (b) the horizontal line through the given point. 13. s 1, 4>3d
15. A 0,  22 B
14.
A 22, 1.3 B
16. s p, 0d
In Exercises 17–30, write an equation for each line described. 17. Passes through s 1, 1d with slope 1
29. Passes through (4, 10) and is perpendicular to the line 6x  3y = 5
In Exercises 31–34, find the line’s x and yintercepts and use this information to graph the line. 31. 3x + 4y = 12
32. x + 2y = 4
33. 22x  23y = 26
34. 1.5x  y = 3
35. Is there anything special about the relationship between the lines Ax + By = C1 and Bx  Ay = C2 sA Z 0, B Z 0d ? Give reasons for your answer. 36. Is there anything special about the relationship between the lines Ax + By = C1 and Ax + By = C2 sA Z 0, B Z 0d ? Give reasons for your answer.
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1.2 Lines, Circles, and Parabolas
Increments and Motion
68. x 2 + y 2  4x + 2y 7 4,
37. A particle starts at As 2, 3d and its coordinates change by increments ¢x = 5, ¢y = 6 . Find its new position.
69. Write an inequality that describes the points that lie inside the circle with center s 2, 1d and radius 26 .
38. A particle starts at A(6, 0) and its coordinates change by increments ¢x = 6, ¢y = 0 . Find its new position.
70. Write an inequality that describes the points that lie outside the circle with center s 4, 2d and radius 4.
39. The coordinates of a particle change by ¢x = 5 and ¢y = 6 as it moves from A(x, y) to Bs3, 3d . Find x and y.
71. Write a pair of inequalities that describe the points that lie inside or on the circle with center (0, 0) and radius 22 , and on or to the right of the vertical line through (1, 0).
40. A particle started at A(1, 0), circled the origin once counterclockwise, and returned to A(1, 0). What were the net changes in its coordinates?
Circles In Exercises 41–46, find an equation for the circle with the given center C (h, k) and radius a. Then sketch the circle in the xyplane. Include the circle’s center in your sketch. Also, label the circle’s x and yintercepts, if any, with their coordinate pairs. 41. Cs0, 2d, 43. Cs 1, 5d,
a = 2
42. Cs 3, 0d,
a = 210
45. C A  23, 2 B ,
44. Cs1, 1d,
a = 2
46. Cs3, 1>2d,
a = 3
72. Write a pair of inequalities that describe the points that lie outside the circle with center (0, 0) and radius 2, and inside the circle that has center (1, 3) and passes through the origin.
Intersecting Lines, Circles, and Parabolas In Exercises 73–80, graph the two equations and find the points in which the graphs intersect. x2 + y2 = 1
73. y = 2x,
a = 22
74. x + y = 1,
sx  1d2 + y 2 = 1
a = 5
75. y  x = 1,
y = x2
76. x + y = 0,
y = sx  1d2
Graph the circles whose equations are given in Exercises 47–52. Label each circle’s center and intercepts (if any) with their coordinate pairs. 47. x 2 + y 2 + 4x  4y + 4 = 0 2
x 7 2
2
77. y = x ,
y = 2x 2  1
1 2 x , 4
y = sx  1d2
78. y =
2
48. x + y  8x + 4y + 16 = 0
79. x 2 + y 2 = 1,
sx  1d2 + y 2 = 1
50. x 2 + y 2  4x  s9>4d = 0
80. x 2 + y 2 = 1,
x2 + y = 1
51. x 2 + y 2  4x + 4y = 0
Applications
49. x 2 + y 2  3y  4 = 0
52. x 2 + y 2 + 2x = 3
81. Insulation By measuring slopes in the accompanying figure, estimate the temperature change in degrees per inch for (a) the gypsum wallboard; (b) the fiberglass insulation; (c) the wood sheathing.
Parabolas Graph the parabolas in Exercises 53–60. Label the vertex, axis, and intercepts in each case.
80° Sheathing
53. y = x 2  2x  3
54. y = x 2 + 4x + 3
55. y = x 2 + 4x
56. y = x 2 + 4x  5
57. y = x 2  6x  5
58. y = 2x 2  x + 3
60°
1 59. y = x 2 + x + 4 2
1 60. y =  x 2 + 2x + 4 4
50° Air
Describe the regions defined by the inequalities and pairs of inequalities in Exercises 61–68. 61. x 2 + y 2 7 7
70°
Temperature (°F)
Inequalities
Gypsum wallboard
Fiberglass between studs
inside room 40° at 72°F
Siding
Air outside at 0°F
30° 20°
62. x 2 + y 2 6 5 63. sx  1d2 + y 2 … 4
10°
64. x 2 + sy  2d2 Ú 4 65. x 2 + y 2 7 1, 2
2
66. x + y … 4,
0°
x2 + y2 6 4 2
2
sx + 2d + y … 4
67. x 2 + y 2 + 6y 6 0,
y 7 3
0
1
2 3 4 5 Distance through wall (inches)
6
The temperature changes in the wall in Exercises 81 and 82.
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Chapter 1: Preliminaries
82. Insulation According to the figure in Exercise 81, which of the materials is the best insulator? the poorest? Explain. 83. Pressure under water The pressure p experienced by a diver under water is related to the diver’s depth d by an equation of the form p = kd + 1 (k a constant). At the surface, the pressure is 1 atmosphere. The pressure at 100 meters is about 10.94 atmospheres. Find the pressure at 50 meters. 84. Reflected light A ray of light comes in along the line x + y = 1 from the second quadrant and reflects off the xaxis (see the accompanying figure). The angle of incidence is equal to the angle of reflection. Write an equation for the line along which the departing light travels.
88. Show that the triangle with vertices A(0, 0), B A 1, 23 B , and C (2, 0) is equilateral. 89. Show that the points As2, 1d , B(1, 3), and Cs 3, 2d are vertices of a square, and find the fourth vertex. 90. The rectangle shown here has sides parallel to the axes. It is three times as long as it is wide, and its perimeter is 56 units. Find the coordinates of the vertices A, B, and C. y
A
D(9, 2)
xy1 1
x
0
y B
Angle of Angle of incidence reflection
0
1
C
91. Three different parallelograms have vertices at s 1, 1d , (2, 0), and (2, 3). Sketch them and find the coordinates of the fourth vertex of each. x
The path of the light ray in Exercise 84. Angles of incidence and reflection are measured from the perpendicular.
92. A 90° rotation counterclockwise about the origin takes (2, 0) to (0, 2), and (0, 3) to s 3, 0d , as shown in the accompanying figure. Where does it take each of the following points? a. (4, 1)
b. s 2, 3d
c. s2, 5d
d. (x, 0)
e. (0, y)
f. (x, y)
g. What point is taken to (10, 3)? In the FCplane, sketch the graph of the
85. Fahrenheit vs. Celsius equation
y
C =
5 sF  32d 9
linking Fahrenheit and Celsius temperatures. On the same graph sketch the line C = F . Is there a temperature at which a Celsius thermometer gives the same numerical reading as a Fahrenheit thermometer? If so, find it. 86. The Mt. Washington Cog Railway Civil engineers calculate the slope of roadbed as the ratio of the distance it rises or falls to the distance it runs horizontally. They call this ratio the grade of the roadbed, usually written as a percentage. Along the coast, commercial railroad grades are usually less than 2%. In the mountains, they may go as high as 4%. Highway grades are usually less than 5%. The steepest part of the Mt. Washington Cog Railway in New Hampshire has an exceptional 37.1% grade. Along this part of the track, the seats in the front of the car are 14 ft above those in the rear. About how far apart are the front and rear rows of seats?
(0, 3) (0, 2)
(4, 1) x
(–3, 0)
(2, 0)
(–2, –3)
(2, –5)
93. For what value of k is the line 2x + ky = 3 perpendicular to the line 4x + y = 1 ? For what value of k are the lines parallel? 94. Find the line that passes through the point (1, 2) and through the point of intersection of the two lines x + 2y = 3 and 2x  3y = 1 . 95. Midpoint of a line segment Show that the point with coordinates
Theory and Examples 87. By calculating the lengths of its sides, show that the triangle with vertices at the points A(1, 2), B(5, 5), and Cs4, 2d is isosceles but not equilateral.
a
x1 + x2 y1 + y2 , b 2 2
is the midpoint of the line segment joining Psx1 , y1 d to Qsx2 , y2 d .
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1.2 Lines, Circles, and Parabolas 96. The distance from a point to a line We can find the distance from a point Psx0 , y0 d to a line L: Ax + By = C by taking the following steps (there is a somewhat faster method in Section 12.5):
19
Use these steps to find the distance from P to L in each of the following cases. a. Ps2, 1d,
L: y = x + 2
1. Find an equation for the line M through P perpendicular to L.
b. Ps4, 6d,
L : 4x + 3y = 12
2. Find the coordinates of the point Q in which M and L intersect.
c. Psa, bd,
L : x = 1
3. Find the distance from P to Q.
d. Psx0 , y0 d,
L : Ax + By = C
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1.3 Functions and Their Graphs
19
Functions and Their Graphs
1.3
Functions are the major objects we deal with in calculus because they are key to describing the real world in mathematical terms. This section reviews the ideas of functions, their graphs, and ways of representing them.
Functions; Domain and Range The temperature at which water boils depends on the elevation above sea level (the boiling point drops as you ascend). The interest paid on a cash investment depends on the length of time the investment is held. The area of a circle depends on the radius of the circle. The distance an object travels from an initial location along a straight line path depends on its speed. In each case, the value of one variable quantity, which we might call y, depends on the value of another variable quantity, which we might call x. Since the value of y is completely determined by the value of x, we say that y is a function of x. Often the value of y is given by a rule or formula that says how to calculate it from the variable x. For instance, the equation A = pr 2 is a rule that calculates the area A of a circle from its radius r. In calculus we may want to refer to an unspecified function without having any particular formula in mind. A symbolic way to say “y is a function of x” is by writing y = ƒsxd
s“y equals ƒ of x”d
In this notation, the symbol ƒ represents the function. The letter x, called the independent variable, represents the input value of ƒ, and y, the dependent variable, represents the corresponding output value of ƒ at x.
DEFINITION Function A function from a set D to a set Y is a rule that assigns a unique (single) element ƒsxd H Y to each element x H D.
x
Input (domain)
f
Output (range)
FIGURE 1.22 A diagram showing a function as a kind of machine.
f (x)
The set D of all possible input values is called the domain of the function. The set of all values of ƒ(x) as x varies throughout D is called the range of the function. The range may not include every element in the set Y. The domain and range of a function can be any sets of objects, but often in calculus they are sets of real numbers. (In Chapters 13–16 many variables may be involved.) Think of a function ƒ as a kind of machine that produces an output value ƒ(x) in its range whenever we feed it an input value x from its domain (Figure 1.22). The function
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Chapter 1: Preliminaries
x a D domain set
f (a)
f (x)
Y set containing the range
FIGURE 1.23 A function from a set D to a set Y assigns a unique element of Y to each element in D.
keys on a calculator give an example of a function as a machine. For instance, the 2x key on a calculator gives an output value (the square root) whenever you enter a nonnegative number x and press the 2x key. The output value appearing in the display is usually a decimal approximation to the square root of x. If you input a number x 6 0, then the calculator will indicate an error because x 6 0 is not in the domain of the function and cannot be accepted as an input. The 2x key on a calculator is not the same as the exact mathematical function ƒ defined by ƒsxd = 2x because it is limited to decimal outputs and has only finitely many inputs. A function can also be pictured as an arrow diagram (Figure 1.23). Each arrow associates an element of the domain D to a unique or single element in the set Y. In Figure 1.23, the arrows indicate that ƒ(a) is associated with a, ƒ(x) is associated with x, and so on. The domain of a function may be restricted by context. For example, the domain of the area function given by A = pr 2 only allows the radius r to be positive. When we define a function y = ƒsxd with a formula and the domain is not stated explicitly or restricted by context, the domain is assumed to be the largest set of real xvalues for which the formula gives real yvalues, the socalled natural domain. If we want to restrict the domain in some way, we must say so. The domain of y = x 2 is the entire set of real numbers. To restrict the function to, say, positive values of x, we would write “y = x 2, x 7 0.” Changing the domain to which we apply a formula usually changes the range as well. The range of y = x 2 is [0, q d. The range of y = x 2, x Ú 2, is the set of all numbers obtained by squaring numbers greater than or equal to 2. In set notation, the range is 5x 2 ƒ x Ú 26 or 5y ƒ y Ú 46 or [4, q d. When the range of a function is a set of real numbers, the function is said to be realvalued. The domains and ranges of many realvalued functions of a real variable are intervals or combinations of intervals. The intervals may be open, closed, or half open, and may be finite or infinite.
EXAMPLE 1
Identifying Domain and Range
Verify the domains and ranges of these functions. Function y y y y y
= = = = =
x2 1/x 2x 24  x 21  x 2
Domain (x)
Range (y)
s  q, q d s  q , 0d ´ s0, q d [0, q d s  q , 4] [1, 1]
[0, q d s  q , 0d ´ s0, q d [0, q d [0, q d [0, 1]
The formula y = x 2 gives a real yvalue for any real number x, so the domain is s  q , q d. The range of y = x 2 is [0, q d because the square of any real number is nonnegative and every nonnegative number y is the square of its own square root, y = A 2y B 2 for y Ú 0. The formula y = 1>x gives a real yvalue for every x except x = 0. We cannot divide any number by zero. The range of y = 1>x, the set of reciprocals of all nonzero real numbers, is the set of all nonzero real numbers, since y = 1>(1>y). The formula y = 2x gives a real yvalue only if x Ú 0. The range of y = 2x is [0, q d because every nonnegative number is some number’s square root (namely, it is the square root of its own square).
Solution
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21
Functions and Their Graphs
In y = 24  x, the quantity 4  x cannot be negative. That is, 4  x Ú 0, or x … 4. The formula gives real yvalues for all x … 4. The range of 24  x is [0, q d, the set of all nonnegative numbers. The formula y = 21  x 2 gives a real yvalue for every x in the closed interval from 1 to 1. Outside this domain, 1  x 2 is negative and its square root is not a real number. The values of 1  x 2 vary from 0 to 1 on the given domain, and the square roots of these values do the same. The range of 21  x 2 is [0, 1].
Graphs of Functions Another way to visualize a function is its graph. If ƒ is a function with domain D, its graph consists of the points in the Cartesian plane whose coordinates are the inputoutput pairs for ƒ. In set notation, the graph is 5sx, ƒsxdd ƒ x H D6. The graph of the function ƒsxd = x + 2 is the set of points with coordinates (x, y) for which y = x + 2. Its graph is sketched in Figure 1.24. The graph of a function ƒ is a useful picture of its behavior. If (x, y) is a point on the graph, then y = ƒsxd is the height of the graph above the point x. The height may be positive or negative, depending on the sign of ƒsxd (Figure 1.25).
y
f (1)
y
f (2) x
yx2
0
1
x
2 f(x)
2 (x, y) –2
0
x
FIGURE 1.24 The graph of ƒsxd = x + 2 is the set of points (x, y) for which y has the value x + 2 .
FIGURE 1.25 If (x, y) lies on the graph of f, then the value y = ƒsxd is the height of the graph above the point x (or below x if ƒ(x) is negative).
x
y x2
2 1 0 1
4 1 0 1
3 2
9 4
Graph the function y = x 2 over the interval [2, 2].
2
4
Solution
EXAMPLE 2
1.
Sketching a Graph
Make a table of xypairs that satisfy the function rule, in this case the equation y = x 2 .
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Chapter 1: Preliminaries
2.
Plot the points (x, y) whose coordinates appear in the table. Use fractions when they are convenient computationally.
3.
Draw a smooth curve through the plotted points. Label the curve with its equation.
y
y (–2, 4)
(2, 4)
4
4
3 , 9 2 4
2 (–1, 1)
1
–2
0
–1
y x2
3
3
2
(1, 1)
1
2
1 x
–2
–1
0
1
2
x
How do we know that the graph of y = x 2 doesn’t look like one of these curves?
Computers and graphing calculators graph functions in much this way—by stringing together plotted points—and the same question arises.
y
y
y x 2?
y x 2?
x
x
To find out, we could plot more points. But how would we then connect them? The basic question still remains: How do we know for sure what the graph looks like between the points we plot? The answer lies in calculus, as we will see in Chapter 4. There we will use the derivative to find a curve’s shape between plotted points. Meanwhile we will have to settle for plotting points and connecting them as best we can.
EXAMPLE 3 p 350 300 250 200 150 100 50 0
Evaluating a Function from Its Graph
The graph of a fruit fly population p is shown in Figure 1.26. (a) Find the populations after 20 and 45 days. (b) What is the (approximate) range of the population function over the time interval 0 … t … 50? Solution 10
20 30 Time (days)
40
50
FIGURE 1.26 Graph of a fruit fly population versus time (Example 3).
t
(a) We see from Figure 1.26 that the point (20, 100) lies on the graph, so the value of the population p at 20 is ps20d = 100. Likewise, p(45) is about 340. (b) The range of the population function over 0 … t … 50 is approximately [0, 345]. We also observe that the population appears to get closer and closer to the value p = 350 as time advances.
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23
Functions and Their Graphs
Representing a Function Numerically We have seen how a function may be represented algebraically by a formula (the area function) and visually by a graph (Examples 2 and 3). Another way to represent a function is numerically, through a table of values. Numerical representations are often used by engineers and applied scientists. From an appropriate table of values, a graph of the function can be obtained using the method illustrated in Example 2, possibly with the aid of a computer. The graph of only the tabled points is called a scatterplot.
EXAMPLE 4
A Function Defined by a Table of Values
Musical notes are pressure waves in the air that can be recorded. The data in Table 1.2 give recorded pressure displacement versus time in seconds of a musical note produced by a tuning fork. The table provides a representation of the pressure function over time. If we first make a scatterplot and then connect the data points (t, p) from the table, we obtain the graph shown in Figure 1.27.
p (pressure)
TABLE 1.2 Tuning fork data
Time
Pressure
Time
Pressure
0.00091 0.00108 0.00125 0.00144 0.00162 0.00180 0.00198 0.00216 0.00234 0.00253 0.00271 0.00289 0.00307 0.00325 0.00344
0.080 0.200 0.480 0.693 0.816 0.844 0.771 0.603 0.368 0.099 0.141 0.309 0.348 0.248 0.041
0.00362 0.00379 0.00398 0.00416 0.00435 0.00453 0.00471 0.00489 0.00507 0.00525 0.00543 0.00562 0.00579 0.00598
0.217 0.480 0.681 0.810 0.827 0.749 0.581 0.346 0.077 0.164 0.320 0.354 0.248 0.035
1.0 0.8 0.6 0.4 0.2 –0.2 –0.4 –0.6
Data
0.001 0.002 0.003 0.004 0.005 0.006 0.007
t (sec)
FIGURE 1.27 A smooth curve through the plotted points gives a graph of the pressure function represented by Table 1.2.
The Vertical Line Test Not every curve you draw is the graph of a function. A function ƒ can have only one value ƒ(x) for each x in its domain, so no vertical line can intersect the graph of a function more than once. Thus, a circle cannot be the graph of a function since some vertical lines intersect the circle twice (Figure 1.28a). If a is in the domain of a function ƒ, then the vertical line x = a will intersect the graph of ƒ in the single point (a, ƒ(a)). The circle in Figure 1.28a, however, does contain the graphs of two functions of x; the upper semicircle defined by the function ƒsxd = 21  x 2 and the lower semicircle defined by the function gsxd =  21  x 2 (Figures 1.28b and 1.28c). Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
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Chapter 1: Preliminaries
y
y
y
–1 –1
0
1
x
–1
0
x
0
(b) y 兹1 x 2
(a) x 2 y 2 1
1
x
1
(c) y –兹1 x 2
FIGURE 1.28 (a) The circle is not the graph of a function; it fails the vertical line test. (b) The upper semicircle is the graph of a function ƒsxd = 21  x 2 . (c) The lower semicircle is the graph of a function gsxd =  21  x 2 .
PiecewiseDefined Functions
y 3
y –x
2
y x
Sometimes a function is described by using different formulas on different parts of its domain. One example is the absolute value function
yx
ƒxƒ = e
1 –3 –2
–1
0
1
2
x
3
x, x,
x Ú 0 x 6 0,
whose graph is given in Figure 1.29. Here are some other examples.
FIGURE 1.29 The absolute value function has domain s  q , q d and range [0, q d .
EXAMPLE 5
Graphing PiecewiseDefined Functions
The function x, ƒsxd = • x 2, 1,
x 6 0 0 … x … 1 x 7 1
y y –x
2 1
–2
–1
is defined on the entire real line but has values given by different formulas depending on the position of x. The values of ƒ are given by: y = x when x 6 0, y = x 2 when 0 … x … 1, and y = 1 when x 7 1. The function, however, is just one function whose domain is the entire set of real numbers (Figure 1.30).
y f (x)
0
y1 y x2 1
x
2
FIGURE 1.30 To graph the function y = ƒsxd shown here, we apply different formulas to different parts of its domain (Example 5).
EXAMPLE 6
The Greatest Integer Function
The function whose value at any number x is the greatest integer less than or equal to x is called the greatest integer function or the integer floor function. It is denoted :x; , or, in some books, [x] or [[x]] or int x. Figure 1.31 shows the graph. Observe that :2.4; = 2, :2; = 2,
:1.9; = 1, :0.2; = 0,
:0; = 0, : 0.3; = 1
: 1.2; = 2, : 2; = 2.
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1.3
yx
2 y x
1 –2 –1
1
2
3
25
FIGURE 1.31 The graph of the greatest integer function y = :x; lies on or below the line y = x , so it provides an integer floor for x (Example 6).
y 3
Functions and Their Graphs
x
–2
EXAMPLE 7
y yx
3 2
y x
1 –2 –1
1
2
3
x
The Least Integer Function
The function whose value at any number x is the smallest integer greater than or equal to x is called the least integer function or the integer ceiling function. It is denoted < x = . Figure 1.32 shows the graph. For positive values of x, this function might represent, for example, the cost of parking x hours in a parking lot which charges $1 for each hour or part of an hour.
–1
EXAMPLE 8
–2
Write a formula for the function y = ƒsxd whose graph consists of the two line segments in Figure 1.33.
FIGURE 1.32 The graph of the least integer function y = < x = lies on or above the line y = x , so it provides an integer ceiling for x (Example 7).
Writing Formulas for PiecewiseDefined Functions
We find formulas for the segments from (0, 0) to (1, 1), and from (1, 0) to (2, 1) and piece them together in the manner of Example 5.
Solution
Segment from (0, 0) to (1, 1) The line through (0, 0) and (1, 1) has slope m = s1  0d>s1  0d = 1 and yintercept b = 0. Its slopeintercept equation is y = x. The segment from (0, 0) to (1, 1) that includes the point (0, 0) but not the point (1, 1) is the graph of the function y = x restricted to the halfopen interval 0 … x 6 1, namely,
y = x, y
The line through (1, 0) and (2, 1) has slope m = s1  0d>s2  1d = 1 and passes through the point (1, 0). The corresponding pointslope equation for the line is Segment from (1, 0) to (2, 1)
y f (x) (1, 1)
(2, 1)
1
0
0 … x 6 1.
y = 0 + 1(x  1),
1
2
x
y = x  1.
The segment from (1, 0) to (2, 1) that includes both endpoints is the graph of y = x  1 restricted to the closed interval 1 … x … 2, namely, y = x  1,
FIGURE 1.33 The segment on the left contains (0, 0) but not (1, 1). The segment on the right contains both of its endpoints (Example 8).
or
Piecewise formula
1 … x … 2.
Combining the formulas for the two pieces of the graph, we obtain ƒsxd = e
x, x  1,
0 … x 6 1 1 … x … 2.
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Chapter 1: Preliminaries
EXERCISES 1.3 Functions
12. Express the side length of a square as a function of the length d of the square’s diagonal. Then express the area as a function of the diagonal length.
In Exercises 1–6, find the domain and range of each function. 2. ƒsxd = 1  2x
1. ƒsxd = 1 + x 2 3. F std =
1
4. F std =
2t
5. g szd = 24  z
2
6. g szd =
13. Express the edge length of a cube as a function of the cube’s diagonal length d. Then express the surface area and volume of the cube as a function of the diagonal length.
1 1 + 2t 1
14. A point P in the first quadrant lies on the graph of the function ƒsxd = 2x . Express the coordinates of P as functions of the slope of the line joining P to the origin.
24  z 2
In Exercises 7 and 8, which of the graphs are graphs of functions of x, and which are not? Give reasons for your answers. 7. a.
b.
y
Functions and Graphs
y
Find the domain and graph the functions in Exercises 15–20.
x
0
8. a.
b.
y
x
0
15. ƒsxd = 5  2x
16. ƒsxd = 1  2x  x 2
17. g sxd = 2ƒ x ƒ 19. F std = t> ƒ t ƒ
18. g sxd = 2x
20. G std = 1> ƒ t ƒ 21. Graph the following equations and explain why they are not graphs of functions of x. a. ƒ y ƒ = x b. y 2 = x 2 22. Graph the following equations and explain why they are not graphs of functions of x.
y
a. ƒ x ƒ + ƒ y ƒ = 1
b. ƒ x + y ƒ = 1
PiecewiseDefined Functions Graph the functions in Exercises 23–26. x
0
0
x
9. Consider the function y = 2s1>xd  1 . a. Can x be negative? b. Can x = 0 ? c. Can x be greater than 1? d. What is the domain of the function? 10. Consider the function y = 22  1x . a. Can x be negative? b. Can 2x be greater than 2? c. What is the domain of the function?
23. ƒsxd = e
x, 2  x,
0 … x … 1 1 6 x … 2
24. g sxd = e
1  x, 2  x,
0 … x … 1 1 6 x … 2
25. F sxd = e
3  x, 2x ,
x … 1 x 7 1
26. G sxd = e
1>x , x,
x 6 0 0 … x
27. Find a formula for each function graphed. a.
b.
y 1
(1, 1) 2
Finding Formulas for Functions 11. Express the area and perimeter of an equilateral triangle as a function of the triangle’s side length x.
0
y
2
x 0
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1
2
3
4
t
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1.3 Functions and Their Graphs 28. a.
b.
y 2
Theory and Examples
y
37. A box with an open top is to be constructed from a rectangular piece of cardboard with dimensions 14 in. by 22 in. by cutting out equal squares of side x at each corner and then folding up the sides as in the figure. Express the volume V of the box as a function of x.
3
(2, 1) 2
2
x
5
1 –1
1
–1
x 2 (2, –1)
22 x
–2
x
x
x
14
–3
x
x x
29. a.
b.
y
y
(–1, 1) (1, 1) 1
38. The figure shown here shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 2 units long.
2 x
3
x
(–2, –1)
x
1 (1, –1) (3, –1)
a. Express the ycoordinate of P in terms of x. (You might start by writing an equation for the line AB.) b. Express the area of the rectangle in terms of x. y
30. a.
b.
y
y B
(T, 1)
1
A 0 0
T 2
T
–A
x
T 2
T 3T 2T 2
P(x, ?)
t
A –1
T 31. a. Graph the functions ƒsxd = x>2 and g sxd = 1 + s4>xd together to identify the values of x for which x 4 7 1 + x. 2 b. Confirm your findings in part (a) algebraically. T 32. a. Graph the functions ƒsxd = 3>sx  1d and g sxd = 2>sx + 1d together to identify the values of x for which
0
x
1
x
39. A cone problem Begin with a circular piece of paper with a 4 in. radius as shown in part (a). Cut out a sector with an arc length of x. Join the two edges of the remaining portion to form a cone with radius r and height h, as shown in part (b).
3 2 6 . x  1 x + 1 b. Confirm your findings in part (a) algebraically. 4 in.
The Greatest and Least Integer Functions 33. For what values of x is
b. < x = = 0 ?
a. :x; = 0 ?
34. What real numbers x satisfy the equation :x; = < x = ?
35. Does < x = =  :x; for all real x? Give reasons for your answer. 36. Graph the function ƒsxd = e
:x;, <x=,
Why is ƒ(x) called the integer part of x?
x Ú 0 x 6 0
4 in.
h r
x (a)
(b)
a. Explain why the circumference of the base of the cone is 8p  x . b. Express the radius r as a function of x . c. Express the height h as a function of x . d. Express the volume V of the cone as a function of x.
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Chapter 1: Preliminaries
40. Industrial costs Dayton Power and Light, Inc., has a power plant on the Miami River where the river is 800 ft wide. To lay a new cable from the plant to a location in the city 2 mi downstream on the opposite side costs $180 per foot across the river and $100 per foot along the land. 2 mi P
x
Q
Dayton
800 ft
Power plant (Not to scale)
a. Suppose that the cable goes from the plant to a point Q on the opposite side that is x ft from the point P directly opposite the
plant. Write a function C (x) that gives the cost of laying the cable in terms of the distance x. b. Generate a table of values to determine if the least expensive location for point Q is less than 2000 ft or greater than 2000 ft from point P. 41. For a curve to be symmetric about the xaxis, the point (x, y) must lie on the curve if and only if the point sx, yd lies on the curve. Explain why a curve that is symmetric about the xaxis is not the graph of a function, unless the function is y = 0 . 42. A magic trick You may have heard of a magic trick that goes like this: Take any number. Add 5. Double the result. Subtract 6. Divide by 2. Subtract 2. Now tell me your answer, and I’ll tell you what you started with. Pick a number and try it. You can see what is going on if you let x be your original number and follow the steps to make a formula ƒ(x) for the number you end up with.
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Chapter 1: Preliminaries
1.4
Identifying Functions; Mathematical Models There are a number of important types of functions frequently encountered in calculus. We identify and briefly summarize them here. Linear Functions A function of the form ƒsxd = mx + b, for constants m and b, is called a linear function. Figure 1.34 shows an array of lines ƒsxd = mx where b = 0, so these lines pass through the origin. Constant functions result when the slope m = 0 (Figure 1.35). y m –3
m2
y –3x
y 2x
m –1 m1 yx
y –x 0
1 y x 2
m
y
1 2 x
2
y3 2
1 0
FIGURE 1.34 The collection of lines y = mx has slope m and all lines pass through the origin.
1
2
x
FIGURE 1.35 A constant function has slope m = 0 .
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1.4
Identifying Functions; Mathematical Models
29
Power Functions A function ƒsxd = x a , where a is a constant, is called a power function. There are several important cases to consider. (a) a = n, a positive integer. The graphs of ƒsxd = x n , for n = 1, 2, 3, 4, 5, are displayed in Figure 1.36. These functions are defined for all real values of x. Notice that as the power n gets larger, the curves tend to flatten toward the xaxis on the interval s 1, 1d, and also rise more steeply for ƒ x ƒ 7 1. Each curve passes through the point (1, 1) and through the origin. y
y
yx
1 –1
y
y x2
1 0
–1
FIGURE 1.36
1
x
–1
0 –1
y
y x3
1 1
x
–1
0
y y x5
y x4
1 x
1
–1
–1
0
1 x
1
–1
–1
0
1
x
–1
Graphs of ƒsxd = x n, n = 1, 2, 3, 4, 5 defined for  q 6 x 6 q .
(b) a = 1
or
a = 2 .
The graphs of the functions ƒsxd = x 1 = 1>x and gsxd = x 2 = 1>x 2 are shown in Figure 1.37. Both functions are defined for all x Z 0 (you can never divide by zero). The graph of y = 1>x is the hyperbola xy = 1 which approaches the coordinate axes far from the origin. The graph of y = 1>x 2 also approaches the coordinate axes. y y y 12 x
y 1x 1 0
1 Domain: x 0 Range: y 0
(a)
x
1 x 1 Domain: x 0 Range: y 0
0
(b)
FIGURE 1.37 Graphs of the power functions ƒsxd = x a for part (a) a = 1 and for part (b) a = 2 .
(c) a =
1 1 3 2 , , , and . 2 3 2 3
3 The functions ƒsxd = x 1>2 = 2x and gsxd = x 1>3 = 2 x are the square root and cube root functions, respectively. The domain of the square root function is [0, q d, but the cube root function is defined for all real x. Their graphs are displayed in Figure 1.38 along with the graphs of y = x 3>2 and y = x 2>3 . (Recall that x 3>2 = sx 1>2 d3 and x 2>3 = sx 1>3 d2 .)
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Chapter 1: Preliminaries
y
y y x 3
y x
1 1 0
x
1
0
x
1
Domain: 0 x
Range: 0 y
Domain: – x
Range: – y
y y y x 32 y x 23 1
1
0
x
1
0 1 Domain: – x
Range: 0 y
Domain: 0 x
Range: 0 y
FIGURE 1.38 Graphs of the power functions ƒsxd = x a for a =
x
2 1 1 3 , , , and . 2 3 2 3
Polynomials A function p is a polynomial if psxd = an x n + an  1x n  1 + Á + a1 x + a0 where n is a nonnegative integer and the numbers a0 , a1 , a2 , Á , an are real constants (called the coefficients of the polynomial). All polynomials have domain s  q , q d. If the leading coefficient an Z 0 and n 7 0, then n is called the degree of the polynomial. Linear functions with m Z 0 are polynomials of degree 1. Polynomials of degree 2, usually written as psxd = ax 2 + bx + c, are called quadratic functions. Likewise, cubic functions are polynomials psxd = ax 3 + bx 2 + cx + d of degree 3. Figure 1.39 shows the graphs of three polynomials. You will learn how to graph polynomials in Chapter 4. 3 2 y x x 2x 1 3 3 2 y
4
y y
2
–2
14x 3
9x 2
y (x 2)4(x 1)3(x 1)
11x 1
16
2 –1
–4
y
8x 4
0
2
4
x
–2
1
–4
x
–1
–6 –2
2
0
1
2
–8 –10
–4
–16
–12 (a)
(b)
(c)
FIGURE 1.39 Graphs of three polynomial functions.
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31
Identifying Functions; Mathematical Models
Rational Functions A rational function is a quotient or ratio of two polynomials: psxd qsxd
ƒsxd =
where p and q are polynomials. The domain of a rational function is the set of all real x for which qsxd Z 0. For example, the function ƒsxd =
2x 2  3 7x + 4
is a rational function with domain 5x ƒ x Z 4>76. Its graph is shown in Figure 1.40a with the graphs of two other rational functions in Figures 1.40b and 1.40c.
y y
8 y
4
2 y 5x 2 8x 3 3x 2
6
2 2
–4
–2
2 y 2x 3 7x 4
2
4
x
–5
4
1
Line y 5 3
0
5
2
10
x
–4
–2
0
2
4
6
x
–2
–1
–2
y 11x3 2 2x 1
–2
–4 NOT TO SCALE
–4
–6 –8
(a)
FIGURE 1.40
(b)
(c)
Graphs of three rational functions.
Algebraic Functions An algebraic function is a function constructed from polynomials using algebraic operations (addition, subtraction, multiplication, division, and taking roots). Rational functions are special cases of algebraic functions. Figure 1.41 displays the graphs of three algebraic functions. Trigonometric Functions We review trigonometric functions in Section 1.6. The graphs of the sine and cosine functions are shown in Figure 1.42. Exponential Functions Functions of the form ƒsxd = a x , where the base a 7 0 is a positive constant and a Z 1, are called exponential functions. All exponential functions have domain s  q , q d and range s0, q d. So an exponential function never assumes the value 0. The graphs of some exponential functions are shown in Figure 1.43. The calculus of exponential functions is studied in Chapter 7. Logarithmic Functions These are the functions ƒsxd = loga x, where the base a Z 1 is a positive constant. They are the inverse functions of the exponential functions, and the Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
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y 4
y x(1 x)2/5
y
y x 1/3(x 4) y 3 (x 2 1) 2/3 4 y
3
1
2 1 –1 0 –1
4
x
x
0
0
–2
5 7
x
1
–1
–3
(b)
(a)
(c)
FIGURE 1.41 Graphs of three algebraic functions.
y
y
1 0
–
1
– 2
3
x
2
0 –1
–1
3 2
5 2
x
2
(b) f (x) ⫽ cos x
(a) f (x) ⫽ sin x
FIGURE 1.42 Graphs of the sine and cosine functions.
y
y y 10 x
y 10 –x
12
12
10
10
8
8 y 3 –x
6 4
y 3x
2 –1 –0.5 0 0.5 1 (a) y 2 x, y 3 x, y 10 x
6 4
y 2x
y 2 –x x
2
–1 –0.5 0 0.5 1 (b) y 2 –x, y 3 –x, y 10 –x
x
FIGURE 1.43 Graphs of exponential functions.
calculus of these functions is studied in Chapter 7. Figure 1.44 shows the graphs of four logarithmic functions with various bases. In each case the domain is s0, q d and the range is s  q , q d. Transcendental Functions These are functions that are not algebraic. They include the trigonometric, inverse trigonometric, exponential, and logarithmic functions, and many
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y
33
other functions as well (such as the hyperbolic functions studied in Chapter 7). An example of a transcendental function is a catenary. Its graph takes the shape of a cable, like a telephone line or TV cable, strung from one support to another and hanging freely under its own weight (Figure 1.45).
y log 2 x y log 3 x
1 x
1
0
Identifying Functions; Mathematical Models
y log5 x
–1
EXAMPLE 1
y log10 x
Recognizing Functions
Identify each function given here as one of the types of functions we have discussed. Keep in mind that some functions can fall into more than one category. For example, ƒsxd = x 2 is both a power function and a polynomial of second degree.
FIGURE 1.44 Graphs of four logarithmic functions.
(a) ƒsxd = 1 + x (d) ystd = sin Qt 
y
1 5 x 2
(b) gsxd = 7x
(c) hszd = z 7
p 4R
Solution
(a) ƒsxd = 1 + x 
(b) gsxd = 7x is an exponential function with base 7. Notice that the variable x is the exponent. (c) hszd = z 7 is a power function. (The variable z is the base.)
1
–1
1 5 x is a polynomial of degree 5. 2
0
1
FIGURE 1.45 Graph of a catenary or hanging cable. (The Latin word catena means “chain.”)
x
p (d) ystd = sin Qt  R is a trigonometric function. 4
Increasing Versus Decreasing Functions If the graph of a function climbs or rises as you move from left to right, we say that the function is increasing. If the graph descends or falls as you move from left to right, the function is decreasing. We give formal definitions of increasing functions and decreasing functions in Section 4.3. In that section, you will learn how to find the intervals over which a function is increasing and the intervals where it is decreasing. Here are examples from Figures 1.36, 1.37, and 1.38. Function y y y y y y
= = = = = =
x2 x3 1>x 1>x 2 2x x 2>3
Where increasing
Where decreasing
0 … x 6 q q 6 x 6 q Nowhere q 6 x 6 0 0 … x 6 q 0 … x 6 q
q 6 x … 0 Nowhere  q 6 x 6 0 and 0 6 x 6 q 0 6 x 6 q Nowhere q 6 x … 0
Even Functions and Odd Functions: Symmetry The graphs of even and odd functions have characteristic symmetry properties.
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Chapter 1: Preliminaries
DEFINITIONS Even Function, Odd Function A function y = ƒsxd is an even function of x if ƒs xd = ƒsxd, odd function of x if ƒs xd = ƒsxd, for every x in the function’s domain.
y y x2 (–x, y)
(x, y)
x
0 (a) y y x3 (x, y)
The names even and odd come from powers of x. If y is an even power of x, as in y = x 2 or y = x 4 , it is an even function of x (because s xd2 = x 2 and s xd4 = x 4 d. If y is an odd power of x, as in y = x or y = x 3 , it is an odd function of x (because s xd1 = x and s xd3 = x 3 d. The graph of an even function is symmetric about the yaxis. Since ƒs xd = ƒsxd, a point (x, y) lies on the graph if and only if the point s x, yd lies on the graph (Figure 1.46a). A reflection across the yaxis leaves the graph unchanged. The graph of an odd function is symmetric about the origin. Since ƒs xd = ƒsxd, a point (x, y) lies on the graph if and only if the point s x, yd lies on the graph (Figure 1.46b). Equivalently, a graph is symmetric about the origin if a rotation of 180° about the origin leaves the graph unchanged. Notice that the definitions imply both x and x must be in the domain of ƒ.
EXAMPLE 2 x
0
Recognizing Even and Odd Functions Even function: s xd2 = x 2 for all x; symmetry about yaxis. Even function: s xd2 + 1 = x 2 + 1 for all x; symmetry about yaxis (Figure 1.47a).
ƒsxd = x 2 ƒsxd = x 2 + 1
(–x, –y)
(b)
y
y y x2 1
FIGURE 1.46 In part (a) the graph of y = x 2 (an even function) is symmetric about the yaxis. The graph of y = x 3 (an odd function) in part (b) is symmetric about the origin.
yx1
y x2
yx 1 0
(a)
1 x
–1
0
x
(b)
FIGURE 1.47 (a) When we add the constant term 1 to the function y = x 2 , the resulting function y = x 2 + 1 is still even and its graph is still symmetric about the yaxis. (b) When we add the constant term 1 to the function y = x , the resulting function y = x + 1 is no longer odd. The symmetry about the origin is lost (Example 2).
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ƒsxd = x ƒsxd = x + 1
Identifying Functions; Mathematical Models
35
Odd function: s xd = x for all x; symmetry about the origin. Not odd: ƒs xd = x + 1 , but ƒsxd = x  1 . The two are not equal. Not even: s xd + 1 Z x + 1 for all x Z 0 (Figure 1.47b).
Mathematical Models To help us better understand our world, we often describe a particular phenomenon mathematically (by means of a function or an equation, for instance). Such a mathematical model is an idealization of the realworld phenomenon and is seldom a completely accurate representation. Although any model has its limitations, a good one can provide valuable results and conclusions. A model allows us to reach conclusions, as illustrated in Figure 1.48. Realworld data
Simplification
Analysis
Verification
Predictions/ explanations
Model
Interpretation
Mathematical conclusions
FIGURE 1.48 A flow of the modeling process beginning with an examination of realworld data.
Most models simplify reality and can only approximate realworld behavior. One simplifying relationship is proportionality.
DEFINITION Proportionality Two variables y and x are proportional (to one another) if one is always a constant multiple of the other; that is, if y = kx for some nonzero constant k.
The definition means that the graph of y versus x lies along a straight line through the origin. This graphical observation is useful in testing whether a given data collection reasonably assumes a proportionality relationship. If a proportionality is reasonable, a plot of one variable against the other should approximate a straight line through the origin.
EXAMPLE 3
Kepler’s Third Law
A famous proportionality, postulated by the German astronomer Johannes Kepler in the early seventeenth century, is his third law. If T is the period in days for a planet to complete one full orbit around the sun, and R is the mean distance of the planet to the sun, then Kepler postulated that T is proportional to R raised to the 3>2 power. That is, for some constant k, T = kR 3>2 .
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Let’s compare his law to the data in Table 1.3 taken from the 1993 World Almanac.
TABLE 1.3 Orbital periods and mean distances of planets
from the sun Planet Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto
T Period (days)
R Mean distance (millions of miles)
88.0 224.7 365.3 687.0 4,331.8 10,760.0 30,684.0 60,188.3 90,466.8
36 67.25 93 141.75 483.80 887.97 1,764.50 2,791.05 3,653.90
The graphing principle in this example may be new to you. To plot T versus R 3>2 we first calculate the value of R 3>2 for each value in Table 1.3. For example, 3653.90 3>2 L 220,869.1 and 363>2 = 216. The horizontal axis represents R 3>2 (not R values) and we plot the ordered pairs sR 3>2, Td in the coordinate system in Figure 1.49. This plot of ordered pairs or scatterplot gives a graph of the period versus the mean distance to the 3>2 power. We observe that the scatterplot in the figure does lie approximately along a straight line that projects through the origin. By picking two points that lie on that line we can easily estimate the slope, which is the constant of proportionality (in days per miles *10 4). k = slope =
90, 466.8  88 L 0.410 220,869.1  216
We estimate the model of Kepler’s third law to be T = 0.410R 3>2 (which depends on our choice of units). We need to be careful to point out that this is not a proof of Kepler’s third T
Period (Days)
36
90,000 60,000 30,000 0
0
80,000
160,000
240,000
R 3/2
(Miles 10 –4 )
FIGURE 1.49 Graph of Kepler’s third law as a proportionality: T = 0.410R 3>2 (Example 3).
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1.4 Identifying Functions; Mathematical Models
37
law. We cannot prove or verify a theorem by just looking at some examples. Nevertheless, Figure 1.49 suggests that Kepler’s third law is reasonable. The concept of proportionality is one way to test the reasonableness of a conjectured relationship between two variables, as in Example 3. It can also provide the basis for an empirical model which comes entirely from a table of collected data.
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37
EXERCISES 1.4 Recognizing Functions
b. y = 5x
6. a. y = 5x
In Exercises 1–4, identify each function as a constant function, linear function, power function, polynomial (state its degree), rational function, algebraic function, trigonometric function, exponential function, or logarithmic function. Remember that some functions can fall into more than one category.
y g
5 b. gsxd = 2 x
1. a. ƒsxd = 7  3x x2  1 c. hsxd = 2 x + 1 2. a. Fstd = t 4  t
h
x
d. rsxd = 8
x
0
b. Gstd = 5t 3 7 d. Rszd = 2 z
c. Hszd = 2z 3 + 1 3. a. y =
c. y = x 5
3 + 2x x  1
f
b. y = x 5>2  2x + 1
c. y = tan px
d. y = log7 x
1 4. a. y = log5 a t b
b. ƒszd =
z5 2z + 1
d. w = 5 cos a
c. gsxd = 21>x
t p + b 2 6
In Exercises 5 and 6, match each equation with its graph. Do not use a graphing device, and give reasons for your answer. 5. a. y = x 4
b. y = x 7
c. y = x 10
Increasing and Decreasing Functions Graph the functions in Exercises 7–18. What symmetries, if any, do the graphs have? Specify the intervals over which the function is increasing and the intervals where it is decreasing. 7. y = x 3
8. y = 
1 9. y =  x 13. y = x 3>8
g h
15. y = x
14. y = 42x
3>2 2>3
17. y = s xd
0 f
x
1 ƒxƒ 12. y = 2 x 10. y =
11. y = 2ƒ x ƒ
y
1 x2
16. y = s xd3>2 18. y = x 2>3
Even and Odd Functions In Exercises 19–30, say whether the function is even, odd, or neither. Give reasons for your answer. 20. ƒsxd = x 5
19. ƒsxd = 3 2
21. ƒsxd = x + 1
22. ƒsxd = x 2 + x
23. gsxd = x 3 + x
24. gsxd = x 4 + 3x 2  1
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Chapter 1: Preliminaries x x2  1
25. gsxd =
1 x2  1
26. gsxd =
27. hstd =
1 t  1
28. hstd = ƒ t 3 ƒ
29. hstd = 2t + 1
34. In October 2002, astronomers discovered a rocky, icy miniplanet tentatively named “Quaoar” circling the sun far beyond Neptune. The new planet is about 4 billion miles from Earth in an outer fringe of the solar system known as the Kuiper Belt. Using Kepler’s third law, estimate the time T it takes Quaoar to complete one full orbit around the sun.
30. hstd = 2 ƒ t ƒ + 1
Proportionality In Exercises 31 and 32, assess whether the given data sets reasonably support the stated proportionality assumption. Graph an appropriate scatterplot for your investigation and, if the proportionality assumption seems reasonable, estimate the constant of proportionality.
T 35. Spring elongation The response of a spring to various loads must be modeled to design a vehicle such as a dump truck, utility vehicle, or a luxury car that responds to road conditions in a desired way. We conducted an experiment to measure the stretch y of a spring in inches as a function of the number x of units of mass placed on the spring.
31. a. y is proportional to x y
1
2
3
x
5.9
12.1
17.9
b. y is proportional to x
4
5
6
7
8
23.9
29.9
36.2
41.8
48.2
1>2
y
3.5
5
6
7
8
x
3
6
9
12
15
32. a. y is proportional to 3x y
5
15
45
135
405
1215
3645
10,935
x
0
1
2
3
4
5
6
7
x (number of units of mass)
0
1
2
3
4
5
y (elongation in inches)
0
0.875
1.721
2.641
3.531
4.391
x (number of units of mass)
6
7
8
9
10
y (elongation in inches)
5.241
6.120
6.992
7.869
8.741
a. Make a scatterplot of the data to test the reasonableness of the hypothesis that stretch y is proportional to the mass x. b. Estimate the constant of proportionality from your graph obtained in part (a).
b. y is proportional to ln x y
2
4.8
5.3
6.5
8.0
10.5
14.4
15.0
x
2.0
5.0
6.0
9.0
14.0
35.0
120.0
150.0
c. Predict the elongation of the spring for 13 units of mass. 36. Ponderosa pines In the table, x represents the girth (distance around) of a pine tree measured in inches (in.) at shoulder height; y represents the board feet (bf) of lumber finally obtained.
T 33. The accompanying table shows the distance a car travels during the time the driver is reacting before applying the brakes, and the distance the car travels after the brakes are applied. The distances (in feet) depend on the speed of the car (in miles per hour). Test the reasonableness of the following proportionality assumptions and estimate the constants of proportionality.
x (in.)
17
19
20
23
25
28
32
38
39
41
y (bf)
19
25
32
57
71
113
123
252
259
294
Formulate and test the following two models: that usable board feet is proportional to (a) the square of the girth and (b) the cube of the girth. Does one model provide a better “explanation” than the other?
a. reaction distance is proportional to speed. b. braking distance is proportional to the square of the speed. Speed (mph)
20
25
30
35
40
45
50
55
60
65
70
75
80
Reaction distance (ft)
22
28
33
39
44
50
55
61
66
72
77
83
88
Braking distance (ft)
20
28
41
53
72
93
118
149
182
221
266
318
376
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1.5
Combining Functions; Shifting and Scaling Graphs In this section we look at the main ways functions are combined or transformed to form new functions.
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Combining Functions; Shifting and Scaling Graphs
39
Sums, Differences, Products, and Quotients Like numbers, functions can be added, subtracted, multiplied, and divided (except where the denominator is zero) to produce new functions. If ƒ and g are functions, then for every x that belongs to the domains of both ƒ and g (that is, for x H Dsƒd ¨ Dsgd), we define functions ƒ + g, ƒ  g, and ƒg by the formulas sƒ + gdsxd = ƒsxd + gsxd. sƒ  gdsxd = ƒsxd  gsxd. sƒgdsxd = ƒsxdgsxd. Notice that the + sign on the lefthand side of the first equation represents the operation of addition of functions, whereas the + on the righthand side of the equation means addition of the real numbers ƒ(x) and g(x). At any point of Dsƒd ¨ Dsgd at which gsxd Z 0, we can also define the function ƒ>g by the formula ƒ ƒsxd a g bsxd = gsxd
swhere gsxd Z 0d.
Functions can also be multiplied by constants: If c is a real number, then the function cƒ is defined for all x in the domain of ƒ by scƒdsxd = cƒsxd.
EXAMPLE 1
Combining Functions Algebraically
The functions defined by the formulas ƒsxd = 2x
and
g sxd = 21  x,
have domains Dsƒd = [0, q d and Dsgd = s  q , 1]. The points common to these domains are the points [0, q d ¨ s  q , 1] = [0, 1]. The following table summarizes the formulas and domains for the various algebraic combinations of the two functions. We also write ƒ # g for the product function ƒg. Function
Formula
Domain
ƒ + g ƒ  g g  ƒ ƒ#g
sƒ + gdsxd = 2x + 21  x sƒ  gdsxd = 2x  21  x sg  ƒdsxd = 21  x  2x sƒ # gdsxd = ƒsxdgsxd = 2xs1  xd ƒ ƒsxd x g sxd = gsxd = A 1  x
[0, 1] = Dsƒd ¨ Dsgd [0, 1] [0, 1] [0, 1]
ƒ>g g>ƒ
g gsxd 1  x sxd = = ƒ ƒsxd A x
[0, 1) sx = 1 excludedd (0, 1] sx = 0 excludedd
The graph of the function ƒ + g is obtained from the graphs of ƒ and g by adding the corresponding ycoordinates ƒ(x) and g(x) at each point x H Dsƒd ¨ Dsgd, as in Figure 1.50. The graphs of ƒ + g and ƒ # g from Example 1 are shown in Figure 1.51.
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Chapter 1: Preliminaries y
y
6
yfg
g(x) 兹1 x
8 y ( f g)(x) y g(x)
4 2
g(a)
y f (x)
1 2
f (a) g(a)
yf•g
f (a) 0
x
a
0
f(x) 兹x
1
FIGURE 1.50 Graphical addition of two functions.
1 5
2 5
3 5
4 5
1
x
FIGURE 1.51 The domain of the function ƒ + g is the intersection of the domains of ƒ and g, the interval [0, 1] on the xaxis where these domains overlap. This interval is also the domain of the function ƒ # g (Example 1).
Composite Functions Composition is another method for combining functions.
DEFINITION Composition of Functions If ƒ and g are functions, the composite function ƒ g (“ƒ composed with g”) is defined by sƒ gdsxd = ƒsgsxdd. The domain of ƒ g consists of the numbers x in the domain of g for which g(x) lies in the domain of ƒ.
The definition says that ƒ g can be formed when the range of g lies in the domain of ƒ. To find sƒ gdsxd, first find g(x) and second find ƒ(g(x)). Figure 1.52 pictures ƒ g as a machine diagram and Figure 1.53 shows the composite as an arrow diagram. f g f(g(x)) x g x
g
g(x)
f
f
f (g(x))
FIGURE 1.52 Two functions can be composed at x whenever the value of one function at x lies in the domain of the other. The composite is denoted by ƒ g.
g(x)
FIGURE 1.53
Arrow diagram for ƒ g .
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1.5
EXAMPLE 2
Combining Functions; Shifting and Scaling Graphs
41
Viewing a Function as a Composite
The function y = 21  x 2 can be thought of as first calculating 1  x 2 and then taking the square root of the result. The function y is the composite of the function gsxd = 1  x 2 and the function ƒsxd = 2x. Notice that 1  x 2 cannot be negative. The domain of the composite is [1, 1]. To evaluate the composite function g ƒ (when defined), we reverse the order, finding ƒ(x) first and then g(ƒ(x)). The domain of g ƒ is the set of numbers x in the domain of ƒ such that ƒ(x) lies in the domain of g. The functions ƒ g and g ƒ are usually quite different.
EXAMPLE 3
Finding Formulas for Composites
If ƒsxd = 2x and gsxd = x + 1, find (a) sƒ gdsxd
(b) sg ƒdsxd
(c) sƒ ƒdsxd
(d) sg gdsxd.
Solution Composite
Domain
(a) sƒ gdsxd = ƒsg sxdd = 2g sxd = 2x + 1 (b) sg ƒdsxd = g sƒsxdd = ƒsxd + 1 = 2x + 1
[1, q d [0, q d
(c) sƒ ƒdsxd = ƒsƒsxdd = 2ƒsxd = 21x = x 1>4 (d) sg gdsxd = g sg sxdd = g sxd + 1 = sx + 1d + 1 = x + 2
[0, q d s  q, q d
To see why the domain of ƒ g is [1, q d, notice that gsxd = x + 1 is defined for all real x but belongs to the domain of ƒ only if x + 1 Ú 0, that is to say, when x Ú 1. Notice that if ƒsxd = x 2 and gsxd = 2x, then sƒ gdsxd = the domain of ƒ g is [0, q d, not s  q , q d.
A 2x B 2 = x. However,
Shifting a Graph of a Function To shift the graph of a function y = ƒsxd straight up, add a positive constant to the righthand side of the formula y = ƒsxd. To shift the graph of a function y = ƒsxd straight down, add a negative constant to the righthand side of the formula y = ƒsxd. To shift the graph of y = ƒsxd to the left, add a positive constant to x. To shift the graph of y = ƒsxd to the right, add a negative constant to x.
Shift Formulas Vertical Shifts
y = ƒsxd + k
Shifts the graph of f up k units if k 7 0 Shifts it down ƒ k ƒ units if k 6 0
Horizontal Shifts
y = ƒsx + hd
Shifts the graph of f left h units if h 7 0 Shifts it right ƒ h ƒ units if h 6 0
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42
Chapter 1: Preliminaries
y
y x2 2 y x2 1 y x2
y x2 2 2 1 unit
1 –2
0 –1
2
EXAMPLE 4
Shifting a Graph
(a) Adding 1 to the righthand side of the formula y = x 2 to get y = x 2 + 1 shifts the graph up 1 unit (Figure 1.54). (b) Adding 2 to the righthand side of the formula y = x 2 to get y = x 2  2 shifts the graph down 2 units (Figure 1.54). (c) Adding 3 to x in y = x 2 to get y = sx + 3d2 shifts the graph 3 units to the left (Figure 1.55). (d) Adding 2 to x in y = ƒ x ƒ , and then adding 1 to the result, gives y = ƒ x  2 ƒ  1 and shifts the graph 2 units to the right and 1 unit down (Figure 1.56).
x
2 units Add a positive constant to x.
–2
FIGURE 1.54 To shift the graph of ƒsxd = x 2 up (or down), we add positive (or negative) constants to the formula for ƒ (Example 4a and b).
y
Add a negative constant to x.
y 4
y (x 3) 2
y x2
y x – 2 – 1
y (x 2) 2 1
1 –3
0
–4 1
2
–2
–1
2
4
6
x
x
FIGURE 1.55 To shift the graph of y = x 2 to the left, we add a positive constant to x. To shift the graph to the right, we add a negative constant to x (Example 4c).
FIGURE 1.56 Shifting the graph of y = ƒ x ƒ 2 units to the right and 1 unit down (Example 4d).
Scaling and Reflecting a Graph of a Function To scale the graph of a function y = ƒsxd is to stretch or compress it, vertically or horizontally. This is accomplished by multiplying the function ƒ, or the independent variable x, by an appropriate constant c. Reflections across the coordinate axes are special cases where c = 1.
Vertical and Horizontal Scaling and Reflecting Formulas For c 7 1, y = cƒsxd Stretches the graph of ƒ vertically by a factor of c. 1 y = c ƒsxd
Compresses the graph of ƒ vertically by a factor of c.
y = ƒscxd y = ƒsx>cd
Compresses the graph of ƒ horizontally by a factor of c. Stretches the graph of ƒ horizontally by a factor of c.
For c = 1, y = ƒsxd y = ƒs xd
Reflects the graph of ƒ across the xaxis. Reflects the graph of ƒ across the yaxis.
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1.5
EXAMPLE 5
43
Combining Functions; Shifting and Scaling Graphs
Scaling and Reflecting a Graph
(a) Vertical: Multiplying the righthand side of y = 2x by 3 to get y = 32x stretches the graph vertically by a factor of 3, whereas multiplying by 1>3 compresses the graph by a factor of 3 (Figure 1.57). (b) Horizontal: The graph of y = 23x is a horizontal compression of the graph of y = 2x by a factor of 3, and y = 2x>3 is a horizontal stretching by a factor of 3 (Figure 1.58). Note that y = 23x = 232x so a horizontal compression may correspond to a vertical stretching by a different scaling factor. Likewise, a horizontal stretching may correspond to a vertical compression by a different scaling factor. (c) Reflection: The graph of y =  2x is a reflection of y = 2x across the xaxis, and y = 2x is a reflection across the yaxis (Figure 1.59). y
y y 3兹x
5
compress
y 兹x
stretch
2
2
1
compress
1 0
y 兹3 x
3
3
–1
y 兹x
4
4
1
2
3
y 3 兹x 4
stretch
1
x
–1
FIGURE 1.57 Vertically stretching and compressing the graph y = 1x by a factor of 3 (Example 5a).
y
y 兹–x
0
1
2
3
4
1
y 兹x
–3
y 兹x兾3
–1
1
2
3
x
–1
x
y –兹x
FIGURE 1.58 Horizontally stretching and compressing the graph y = 1x by a factor of 3 (Example 5b).
EXAMPLE 6
–2
FIGURE 1.59 Reflections of the graph y = 1x across the coordinate axes (Example 5c).
Combining Scalings and Reflections
Given the function ƒsxd = x 4  4x 3 + 10 (Figure 1.60a), find formulas to (a) compress the graph horizontally by a factor of 2 followed by a reflection across the yaxis (Figure 1.60b). (b) compress the graph vertically by a factor of 2 followed by a reflection across the xaxis (Figure 1.60c). y 20
y
y 16x 4 32x 3 10 y f (x)
x4
4x 3
10 20 10
10 –1
0 –10
y – 12 x 4 2x 3 5
10 1
2
3
4
x
–2
–1
0 –10
1
x
–1
0
1
2
3
4
x
–10 –20
–20 (a)
(b)
(c)
FIGURE 1.60 (a) The original graph of f. (b) The horizontal compression of y = ƒsxd in part (a) by a factor of 2, followed by a reflection across the yaxis. (c) The vertical compression of y = ƒsxd in part (a) by a factor of 2, followed by a reflection across the xaxis (Example 6).
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44
Chapter 1: Preliminaries Solution
(a) The formula is obtained by substituting 2x for x in the righthand side of the equation for ƒ y = ƒs 2xd = s 2xd4  4s 2xd3 + 10 = 16x 4 + 32x 3 + 10 . (b) The formula is y = 
1 1 ƒsxd =  x 4 + 2x 3  5. 2 2
Ellipses Substituting cx for x in the standard equation for a circle of radius r centered at the origin gives c 2x 2 + y 2 = r 2 .
(1)
If 0 6 c 6 1, the graph of Equation (1) horizontally stretches the circle; if c 7 1 the circle is compressed horizontally. In either case, the graph of Equation (1) is an ellipse (Figure 1.61). Notice in Figure 1.61 that the yintercepts of all three graphs are always r and r. In Figure 1.61b, the line segment joining the points s ;r>c, 0d is called the major axis of the ellipse; the minor axis is the line segment joining s0, ;rd. The axes of the ellipse are reversed in Figure 1.61c: the major axis is the line segment joining the points s0, ;rd and the minor axis is the line segment joining the points s ;r>c, 0d. In both cases, the major axis is the line segment having the longer length.
y
y x2 y2 r2
r
–r
y
0
r
r
x
– cr
r
c 2x 2 y 2 r 2
0
r c
–r
x
– cr
–r
r c
x
–r
(a) circle
FIGURE 1.61
0
c 2x 2 y 2 r 2
(b) ellipse, 0 c 1
(c) ellipse, c 1
Horizontal stretchings or compressions of a circle produce graphs of ellipses.
If we divide both sides of Equation (1) by r 2 , we obtain y2 x2 + 2 = 1. 2 a b
(2)
where a = r>c and b = r. If a 7 b, the major axis is horizontal; if a 6 b, the major axis is vertical. The center of the ellipse given by Equation (2) is the origin (Figure 1.62).
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1.5 Combining Functions; Shifting and Scaling Graphs
45
FIGURE 1.62 Graph of the ellipse y2 x2 + = 1, a 7 b , where the major a2 b2 axis is horizontal.
y
b
–a
Major axis Center
a
x
–b
Substituting x  h for x, and y  k for y, in Equation (2) results in s y  kd2 sx  hd2 + = 1. a2 b2
(3)
Equation (3) is the standard equation of an ellipse with center at (h, k). The geometric definition and properties of ellipses are reviewed in Section 10.1.
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1.5 Combining Functions; Shifting and Scaling Graphs
45
EXERCISES 1.5 Sums, Differences, Products, and Quotients In Exercises 1 and 2, find the domains and ranges of ƒ, g, ƒ + g , and ƒ # g. 1. ƒsxd = x,
g sxd = 2x  1
2. ƒsxd = 2x + 1,
g sxd = 2x  1
In Exercises 3 and 4, find the domains and ranges of ƒ, g, ƒ>g, and g >ƒ. 3. ƒsxd = 2,
g sxd = x 2 + 1
4. ƒsxd = 1,
g sxd = 1 + 2x
Composites of Functions 2
5. If ƒsxd = x + 5 and g sxd = x  3 , find the following. a. ƒ( g (0))
b. g (ƒ(0))
c. ƒ( g (x))
d. g (ƒ(x))
e. ƒ(ƒ( 5))
f. g (g (2))
g. ƒ(ƒ(x))
h. g (g (x))
6. If ƒsxd = x  1 and gsxd = 1>sx + 1d , find the following.
7. If usxd = 4x  5, ysxd = x 2 , and ƒsxd = 1>x , find formulas for the following. a. u(y (ƒ(x)))
b. u (ƒ(y (x)))
c. y (u (ƒ(x)))
d. y (ƒ (u (x)))
e. ƒ (u (y (x)))
f. ƒ(y (u (x)))
8. If ƒsxd = 2x, g sxd = x>4 , and hsxd = 4x  8 , find formulas for the following. a. h(g (ƒ(x)))
b. h(ƒ(g (x)))
c. g (h (ƒ(x)))
d. g (ƒ(h(x)))
e. ƒ(g (h(x)))
f. ƒ(h(g(x)))
Let ƒsxd = x  3, g sxd = 2x , hsxd = x 3 , and jsxd = 2x . Express each of the functions in Exercises 9 and 10 as a composite involving one or more of ƒ, g, h, and j. 9. a. y = 2x  3
b. y = 22x
c. y = x 1>4
d. y = 4x
e. y = 2sx  3d
3
10. a. y = 2x  3
f. y = s2x  6d3 b. y = x 3>2
a. ƒ(g (1>2))
b. g (ƒ(1>2))
c. ƒ(g (x))
d. g (ƒ(x))
c. y = x 9
d. y = x  6
e. ƒ(ƒ(2))
f. g (g (2))
e. y = 22x  3
f. y = 2x 3  3
g. ƒ(ƒ(x))
h. g (g (x))
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46
Chapter 1: Preliminaries 16. The accompanying figure shows the graph of y = x 2 shifted to two new positions. Write equations for the new graphs.
11. Copy and complete the following table. g(x)
ƒ(x)
a. x  7
2x
y Position (a)
3x
b. x + 2
2x  5 x x  1 1 1 + x
c. d.
(ƒ g)(x)
x x  1
e.
y x2
2x 2  5 3
x
x
0
1 f. x
Position (b)
x
12. Copy and complete the following table. g(x)
ƒ(x)
(ƒ g)(x)
a.
1 x  1
ƒxƒ
?
b.
?
x  1 x
x x + 1
c.
?
2x
ƒxƒ
d.
2x
?
ƒxƒ
–5
17. Match the equations listed in parts (a)–(d) to the graphs in the accompanying figure. a. y = sx  1d2  4
b. y = sx  2d2 + 2
c. y = sx + 2d2 + 2
d. y = sx + 3d2  2 y
Position 2
Position 1
In Exercises 13 and 14, (a) write a formula for ƒ g and g ƒ and find the (b) domain and (c) range of each. 13. ƒ(x) = 2x + 1, 14. ƒ(x) = x 2,
1 g (x) = x
3 (–2, 2) Position 3
g (x) = 1  2x
Shifting Graphs
2
–4 –3 –2 –1 0
15. The accompanying figure shows the graph of y = x 2 shifted to two new positions. Write equations for the new graphs.
(2, 2)
1
x
1 2 3 Position 4
(–3, –2)
y (1, –4) –7
0
4
x
18. The accompanying figure shows the graph of y = x 2 shifted to four new positions. Write an equation for each new graph. y (1, 4)
Position (a)
y –x 2
(–2, 3)
Position (b) (b)
(a) (2, 0)
x
(–4, –1)
(c)
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(d)
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1.5 Combining Functions; Shifting and Scaling Graphs Exercises 19–28 tell how many units and in what directions the graphs of the given equations are to be shifted. Give an equation for the shifted graph. Then sketch the original and shifted graphs together, labeling each graph with its equation.
47
50. The accompanying figure shows the graph of a function g(t) with domain [4, 0] and range [3, 0] . Find the domains and ranges of the following functions, and sketch their graphs. y
19. x 2 + y 2 = 49
Down 3, left 2
20. x 2 + y 2 = 25
Up 3, left 4
21. y = x 3 22. y = x 2>3
Right 1, down 1
23. y = 2x
Left 0.81
24. y =  2x
Right 3
25. y = 2x  7
Up 7
26. y =
1 sx + 1d + 5 2
27. y = 1>x 28. y = 1>x
–4
Left 1, down 1
Down 5, right 1
Left 2, down 1
Graph the functions in Exercises 29–48.
x
0
y g(t)
Up 1, right 1 2
–2
–3
a. g s td
b. g std
c. g std + 3
d. 1  g std
e. g s t + 2d
f. g st  2d
g. g s1  td
h. g st  4d
Vertical and Horizontal Scaling
29. y = 2x + 4
30. y = 29  x
31. y = ƒ x  2 ƒ 33. y = 1 + 2x  1
32. y = ƒ 1  x ƒ  1 34. y = 1  2x
Exercises 51–60 tell by what factor and direction the graphs of the given functions are to be stretched or compressed. Give an equation for the stretched or compressed graph.
35. y = sx + 1d2>3
36. y = sx  8d2>3
51. y = x 2  1,
37. y = 1  x
2>3
38. y + 4 = x
3 39. y = 2 x  1  1
2>3
40. y = sx + 2d3>2 + 1
1 x  2
1 42. y = x  2
1 43. y = x + 2
1 44. y = x + 2
41. y =
45. y =
1 sx  1d2
46. y =
compressed horizontally by a factor of 2
1 53. y = 1 + 2 , x
compressed vertically by a factor of 2
1 , stretched horizontally by a factor of 3 x2 55. y = 2x + 1, compressed horizontally by a factor of 4
1  1 x2
56. y = 2x + 1,
stretched vertically by a factor of 3
57. y = 24  x 2,
stretched horizontally by a factor of 2
58. y = 24  x ,
compressed vertically by a factor of 3
2
59. y = 1  x 3,
compressed horizontally by a factor of 3
60. y = 1  x 3,
stretched horizontally by a factor of 2
Graphing
y
0
52. y = x  1,
54. y = 1 +
1 1 47. y = 2 + 1 48. y = x sx + 1d2 49. The accompanying figure shows the graph of a function ƒ(x) with domain [0, 2] and range [0, 1]. Find the domains and ranges of the following functions, and sketch their graphs.
1
stretched vertically by a factor of 3
2
In Exercises 61–68, graph each function, not by plotting points, but by starting with the graph of one of the standard functions presented in Figures 1.36–1.38, and applying an appropriate transformation.
y f (x)
2
x
a. ƒsxd + 2
b. ƒsxd  1
c. 2ƒ(x)
d. ƒsxd
e. ƒsx + 2d
f. ƒsx  1d
g. ƒs xd
h. ƒsx + 1d + 1
x 2
61. y =  22x + 1
62. y =
63. y = sx  1d3 + 2
64. y = s1  xd3 + 2
65. y =
1  1 2x
3 x 67. y =  2
A
1 
2 + 1 x2 68. y = s 2xd2>3 66. y =
69. Graph the function y = ƒ x 2  1 ƒ . 70. Graph the function y = 2ƒ x ƒ .
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Chapter 1: Preliminaries
Ellipses
Even and Odd Functions
Exercises 71–76 give equations of ellipses. Put each equation in standard form and sketch the ellipse.
79. Assume that ƒ is an even function, g is an odd function, and both ƒ and g are defined on the entire real line . Which of the following (where defined) are even? odd?
71. 9x 2 + 25y 2 = 225 2
72. 16x 2 + 7y 2 = 112
2
74. sx + 1d2 + 2y 2 = 4
73. 3x + s y  2d = 3 75. 3sx  1d2 + 2s y + 2d2 = 6 76. 6 ax +
2
2
3 1 b + 9 ay  b = 54 2 2
77. Write an equation for the ellipse sx 2>16d + sy 2>9d = 1 shifted 4 units to the left and 3 units up. Sketch the ellipse and identify its center and major axis.
78. Write an equation for the ellipse sx 2>4d + sy 2>25d = 1 shifted 3 units to the right and 2 units down. Sketch the ellipse and identify its center and major axis.
a. ƒg
b. ƒ>g
d. ƒ 2 = ƒƒ
e. g 2 = g g
g. g ƒ
h. ƒ ƒ
c. g >ƒ
f. ƒ g i. g g
80. Can a function be both even and odd? Give reasons for your answer. T 81. (Continuation of Example 1.) Graph the functions ƒsxd = 2x and g sxd = 21  x together with their (a) sum, (b) product, (c) two differences, (d) two quotients. T 82. Let ƒsxd = x  7 and g sxd = x 2 . Graph ƒ and g together with ƒ g and g ƒ .
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48
Chapter 1: Preliminaries
Trigonometric Functions
1.6
This section reviews the basic trigonometric functions. The trigonometric functions are important because they are periodic, or repeating, and therefore model many naturally occurring periodic processes.
B' s B 1 C
C ir
A' A r
Radian Measure
it cir cl
e
Un
θ
cle of ra diu
sr
FIGURE 1.63 The radian measure of angle ACB is the length u of arc AB on the unit circle centered at C. The value of u can be found from any other circle, however, as the ratio s>r. Thus s = r u is the length of arc on a circle of radius r when u is measured in radians.
In navigation and astronomy, angles are measured in degrees, but in calculus it is best to use units called radians because of the way they simplify later calculations. The radian measure of the angle ACB at the center of the unit circle (Figure 1.63) equals the length of the arc that ACB cuts from the unit circle. Figure 1.63 shows that s = ru is the length of arc cut from a circle of radius r when the subtending angle u producing the arc is measured in radians. Since the circumference of the circle is 2p and one complete revolution of a circle is 360°, the relation between radians and degrees is given by p radians = 180°. For example, 45° in radian measure is 45 #
Conversion Formulas
p p = rad, 180 4
and p>6 radians is
p 1 degree = s L0.02d radians 180 p Degrees to radians: multiply by 180 180 1 radian = p s L57d degrees 180 Radians to degrees: multiply by p
p # 180 = 30°. 6 p Figure 1.64 shows the angles of two common triangles in both measures. An angle in the xyplane is said to be in standard position if its vertex lies at the origin and its initial ray lies along the positive xaxis (Figure 1.65). Angles measured counterclockwise from the positive xaxis are assigned positive measures; angles measured clockwise are assigned negative measures.
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1.6
Degrees
Radians 4
45 兹2
1
45
49
Trigonometric Functions
兹2
1
4
90 1
2
1 y
y Terminal ray
6
30 兹3
2
60
90 1
Initial ray 兹3
2 3
2
x Positive measure
Initial ray
Negative measure
Terminal ray
x
1
FIGURE 1.64 The angles of two common triangles, in degrees and radians.
FIGURE 1.65 Angles in standard position in the xyplane.
When angles are used to describe counterclockwise rotations, our measurements can go arbitrarily far beyond 2p radians or 360°. Similarly, angles describing clockwise rotations can have negative measures of all sizes (Figure 1.66).
y
y 3 x
x
9 4
y
y – 5 2 x – 3 4
FIGURE 1.66 Nonzero radian measures can be positive or negative and can go beyond 2p .
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x
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50
Chapter 1: Preliminaries
hypotenuse
Angle Convention: Use Radians From now on in this book it is assumed that all angles are measured in radians unless degrees or some other unit is stated explicitly. When we talk about the angle p>3, we mean p>3 radians (which is 60°), not p>3 degrees. When you do calculus, keep your calculator in radian mode.
opposite
adjacent opp hyp adj cos hyp opp tan adj
hyp opp hyp sec adj adj cot opp
sin
csc
The Six Basic Trigonometric Functions You are probably familiar with defining the trigonometric functions of an acute angle in terms of the sides of a right triangle (Figure 1.67). We extend this definition to obtuse and negative angles by first placing the angle in standard position in a circle of radius r. We then define the trigonometric functions in terms of the coordinates of the point P(x, y) where the angle’s terminal ray intersects the circle (Figure 1.68). y r sine: sin u = r cosecant: csc u = y x r cosine: cos u = r secant: sec u = x y x tangent: tan u = x cotangent: cot u = y
FIGURE 1.67 Trigonometric ratios of an acute angle.
y
P(x, y) r
These extended definitions agree with the righttriangle definitions when the angle is acute (Figure 1.69). Notice also the following definitions, whenever the quotients are defined.
0
r
x
sin u cos u 1 sec u = cos u
1 tan u 1 csc u = sin u
tan u =
FIGURE 1.68 The trigonometric functions of a general angle u are defined in terms of x, y, and r.
As you can see, tan u and sec u are not defined if x = 0. This means they are not defined if u is ;p>2, ;3p>2, Á . Similarly, cot u and csc u are not defined for values of u for which y = 0, namely u = 0, ;p, ;2p, Á . The exact values of these trigonometric ratios for some angles can be read from the triangles in Figure 1.64. For instance,
y
0
p 1 = 4 22 p 1 cos = 4 22 p tan = 1 4 sin
hypotenuse r x adjacent
P(x, y) y opposite x
sin
p 1 = 6 2
sin
23 p = 3 2
cos
23 p = 6 2
cos
p 1 = 3 2
p p 1 tan = 23 = 3 6 23 The CAST rule (Figure 1.70) is useful for remembering when the basic trigonometric functions are positive or negative. For instance, from the triangle in Figure 1.71, we see that sin
FIGURE 1.69 The new and old definitions agree for acute angles.
cot u =
23 2p = , 3 2
tan
cos
2p 1 =  , 3 2
tan
2p =  23. 3
Using a similar method we determined the values of sin u, cos u, and tan u shown in Table 1.4.
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1.6
51
Trigonometric Functions
cos 2 , sin 2 – 1 , 兹3 3 3 2 2 y y P S sin pos
A all pos x
T tan pos
1
兹3 2
2 3 x
1 2
C cos pos
FIGURE 1.71 The triangle for calculating the sine and cosine of 2p>3 radians. The side lengths come from the geometry of right triangles.
FIGURE 1.70 The CAST rule, remembered by the statement “All Students Take Calculus,” tells which trigonometric functions are positive in each quadrant.
Most calculators and computers readily provide values of the trigonometric functions for angles given in either radians or degrees.
TABLE 1.4 Values of sin u, cos u, and tan u for selected values of u
Degrees
180
u (radians)
p
135 3p 4
90 p 2
45 p 4
0 0
30 p 6
45 p 4
60 p 3
90 p 2
120 2p 3
135 3p 4
150 5p 6
22 2
1 2
 22 2 1
sin u
0
 22 2
1
 22 2
0
1 2
22 2
23 2
1
23 2
cos u
1
 22 2
0
22 2
1
23 2
22 2
1 2
0

tan u
0
1
0
23 3
1
23
1
EXAMPLE 1
1 2
 23
180
270 3p 2
360
0
1
0
 23 2
1
0
1
 23 3
0
p
2p
0
Finding Trigonometric Function Values
If tan u = 3>2 and 0 6 u 6 p>2, find the five other trigonometric functions of u . From tan u = 3>2, we construct the right triangle of height 3 (opposite) and base 2 (adjacent) in Figure 1.72. The Pythagorean theorem gives the length of the hypotenuse, 24 + 9 = 213. From the triangle we write the values of the other five trigonometric functions:
Solution
cos u =
2 , 213
sin u =
3 213
,
sec u =
213 , 2
csc u =
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213 , 3
cot u =
2 3
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Chapter 1: Preliminaries
Periodicity and Graphs of the Trigonometric Functions
y
When an angle of measure u and an angle of measure u + 2p are in standard position, their terminal rays coincide. The two angles therefore have the same trigonometric function values:
兹13 3 0
2
cossu + 2pd = cos u secsu + 2pd = sec u
x
sinsu + 2pd = sin u cscsu + 2pd = csc u
tansu + 2pd = tan u cotsu + 2pd = cot u
Similarly, cos su  2pd = cos u, sin su  2pd = sin u, and so on. We describe this repeating behavior by saying that the six basic trigonometric functions are periodic.
FIGURE 1.72 The triangle for calculating the trigonometric functions in Example 1.
DEFINITION Periodic Function A function ƒ(x) is periodic if there is a positive number p such that ƒsx + pd = ƒsxd for every value of x. The smallest such value of p is the period of ƒ.
When we graph trigonometric functions in the coordinate plane, we usually denote the independent variable by x instead of u. See Figure 1.73.
y y
y y cos x
– – 2
2
0
y sin x
3 2 2
– – 2
3 2 2
Domain: x , 3 , . . . 2 2 Range: y ⱕ –1 and y ⱖ 1 Period: 2 (d)
2
0
y
y sec x
1 – 3 – – 0 2 2
x
3 2 2
y sinx Domain: – x
Range: –1 ⱕ y ⱕ 1 Period: 2 (b)
Domain: – x
Range: –1 ⱕ y ⱕ 1 Period: 2 (a) y
y tan x
x
x
– 3 – – 2 2
x
Domain: x , 3 , . . . 2 2 Range: – y
Period: (c) y
y csc x
1 – – 0 2
0 3 2 2
y cot x
1 2
3 2 2
Domain: x 0, , 2, . . . Range: y ⱕ –1 and y ⱖ 1 Period: 2
x
– – 0 2
2
3 2 2
Domain: x 0, , 2, . . . Range: – y
Period:
(e)
(f)
FIGURE 1.73 Graphs of the (a) cosine, (b) sine, (c) tangent, (d) secant, (e) cosecant, and (f) cotangent functions using radian measure. The shading for each trigonometric function indicates its periodicity.
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x
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1.6
Periods of Trigonometric Functions Period P :
tan sx + pd = tan x cot sx + pd = cot x
Period 2P :
sin sx + 2pd = sin x cos sx + 2pd = cos x sec sx + 2pd = sec x csc sx + 2pd = csc x
53
Trigonometric Functions
As we can see in Figure 1.73, the tangent and cotangent functions have period p = p. The other four functions have period 2p. Periodic functions are important because many behaviors studied in science are approximately periodic. A theorem from advanced calculus says that every periodic function we want to use in mathematical modeling can be written as an algebraic combination of sines and cosines. We show how to do this in Section 11.11. The symmetries in the graphs in Figure 1.73 reveal that the cosine and secant functions are even and the other four functions are odd: Even
Odd
coss xd = cos x
sin s xd = sin x
secs xd = sec x
tan s xd = tan x cscs xd = csc x cots xd = cot x
y
Identities
P(cos , sin )
The coordinates of any point P(x, y) in the plane can be expressed in terms of the point’s distance from the origin and the angle that ray OP makes with the positive xaxis (Figure 1.69). Since x>r = cos u and y>r = sin u, we have
x 2 y2 1
sin
x = r cos u,
x
cos
1
y = r sin u.
When r = 1 we can apply the Pythagorean theorem to the reference right triangle in Figure 1.74 and obtain the equation
cos2 u + sin2 u = 1.
FIGURE 1.74 The reference triangle for a general angle u .
(1)
This equation, true for all values of u, is the most frequently used identity in trigonometry. Dividing this identity in turn by cos2 u and sin2 u gives
1 + tan2 u = sec2 u. 1 + cot2 u = csc2 u.
The following formulas hold for all angles A and B (Exercises 53 and 54).
Addition Formulas cossA + Bd = cos A cos B  sin A sin B sinsA + Bd = sin A cos B + cos A sin B
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(2)
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Chapter 1: Preliminaries
There are similar formulas for cossA  Bd and sin sA  Bd (Exercises 35 and 36). All the trigonometric identities needed in this book derive from Equations (1) and (2). For example, substituting u for both A and B in the addition formulas gives
DoubleAngle Formulas cos 2u = cos2 u  sin2 u sin 2u = 2 sin u cos u
(3)
Additional formulas come from combining the equations cos2 u + sin2 u = 1,
cos2 u  sin2 u = cos 2u.
We add the two equations to get 2 cos2 u = 1 + cos 2u and subtract the second from the first to get 2 sin2 u = 1  cos 2u. This results in the following identities, which are useful in integral calculus.
HalfAngle Formulas cos2 u =
1 + cos 2u 2
(4)
sin2 u =
1  cos 2u 2
(5)
The Law of Cosines If a, b, and c are sides of a triangle ABC and if u is the angle opposite c, then
c 2 = a 2 + b 2  2ab cos u.
This equation is called the law of cosines. We can see why the law holds if we introduce coordinate axes with the origin at C and the positive xaxis along one side of the triangle, as in Figure 1.75. The coordinates of A are (b, 0); the coordinates of B are sa cos u, a sin ud. The square of the distance between A and B is therefore
y B(a cos , a sin )
c 2 = sa cos u  bd2 + sa sin ud2
c
a
C
(6)
b
A(b, 0)
x
FIGURE 1.75 The square of the distance between A and B gives the law of cosines.
= a 2scos2 u + sin2 ud + b 2  2ab cos u ('')''* 1 = a 2 + b 2  2ab cos u. The law of cosines generalizes the Pythagorean theorem. If u = p>2, then cos u = 0 and c 2 = a 2 + b 2 .
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1.6
Trigonometric Functions
55
Transformations of Trigonometric Graphs The rules for shifting, stretching, compressing, and reflecting the graph of a function apply to the trigonometric functions. The following diagram will remind you of the controlling parameters. Vertical stretch or compression; reflection about xaxis if negative
Vertical shift
y = aƒ(bsx + cdd + d Horizontal shift
Horizontal stretch or compression; reflection about yaxis if negative
EXAMPLE 2
Modeling Temperature in Alaska
The builders of the TransAlaska Pipeline used insulated pads to keep the pipeline heat from melting the permanently frozen soil beneath. To design the pads, it was necessary to take into account the variation in air temperature throughout the year. The variation was represented in the calculations by a general sine function or sinusoid of the form ƒsxd = A sin c
2p sx  Cd d + D, B
where ƒ A ƒ is the amplitude, ƒ B ƒ is the period, C is the horizontal shift, and D is the vertical shift (Figure 1.76).
y
(
Horizontal shift (C)
Amplitude (A)
D
DA
)
y A sin 2 (x C) D B
DA
This axis is the line y D.
Vertical shift (D) This distance is the period (B).
0
x
FIGURE 1.76 The general sine curve y = A sin [s2p>Bdsx  Cd] + D , shown for A, B, C, and D positive (Example 2).
Figure 1.77 shows how to use such a function to represent temperature data. The data points in the figure are plots of the mean daily air temperatures for Fairbanks, Alaska, based on records of the National Weather Service from 1941 to 1970. The sine function used to fit the data is ƒsxd = 37 sin c
2p sx  101d d + 25, 365
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Chapter 1: Preliminaries
where ƒ is temperature in degrees Fahrenheit and x is the number of the day counting from the beginning of the year. The fit, obtained by using the sinusoidal regression feature on a calculator or computer, as we discuss in the next section, is very good at capturing the trend of the data.
60 Temperature (°F)
56
40
20
0 20
Jan
Feb Mar Apr May Jun
Jul
Aug Sep Oct Nov Dec Jan
Feb Mar
FIGURE 1.77 Normal mean air temperatures for Fairbanks, Alaska, plotted as data points (red). The approximating sine function (blue) is ƒsxd
37 sin [s2p 365dsx
101d]
25 .
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Chapter 1: Preliminaries
EXERCISES 1.6 Radians, Degrees, and Circular Arcs 1. On a circle of radius 10 m, how long is an arc that subtends a central angle of (a) 4p>5 radians? (b) 110°? 2. A central angle in a circle of radius 8 is subtended by an arc of length 10p . Find the angle’s radian and degree measures. 3. You want to make an 80° angle by marking an arc on the perimeter of a 12in.diameter disk and drawing lines from the ends of the arc to the disk’s center. To the nearest tenth of an inch, how long should the arc be? 4. If you roll a 1mdiameter wheel forward 30 cm over level ground, through what angle will the wheel turn? Answer in radians (to the nearest tenth) and degrees (to the nearest degree).
5. Copy and complete the following table of function values. If the function is undefined at a given angle, enter “UND.” Do not use a calculator or tables.
sin u cos u tan u cot u sec u csc u
P
2P>3
0
U
3P>2
P>3
P>6
P>4
5P>6
sin u cos u tan u cot u sec u csc u In Exercises 7–12, one of sin x, cos x, and tan x is given. Find the other two if x lies in the specified interval.
Evaluating Trigonometric Functions
U
6. Copy and complete the following table of function values. If the function is undefined at a given angle, enter “UND.” Do not use a calculator or tables.
P>2
7. sin x =
3 , 5
p x c , pd 2
9. cos x =
1 , 3
x c
5 p , 0 d 10. cos x =  , 2 13
11. tan x =
1 , 2
x cp,
3p d 2
3P>4
8. tan x = 2,
1 12. sin x =  , 2
x c0,
p d 2
p x c , pd 2 x cp,
3p d 2
Graphing Trigonometric Functions Graph the functions in Exercises 13–22. What is the period of each function? 13. sin 2x
14. sin ( x>2)
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1.6 Trigonometric Functions
16. cos
15. cos px 17. sin
px 3
px 2
41. sin a
18. cos 2px
19. cos ax 
p b 2
20. sin ax +
p b 2
21. sin ax 
p b + 1 4
22. cos ax +
p b  1 4
Graph the functions in Exercises 23–26 in the tsplane (taxis horizontal, saxis vertical). What is the period of each function? What symmetries do the graphs have? 23. s = cot 2t
24. s = tan pt
25. s = sec a
t 26. s = csc a b 2
pt b 2
T 27. a. Graph y = cos x and y = sec x together for 3p>2 … x … 3p>2 . Comment on the behavior of sec x in relation to the signs and values of cos x. b. Graph y = sin x and y = csc x together for p … x … 2p . Comment on the behavior of csc x in relation to the signs and values of sin x. T 28. Graph y = tan x and y = cot x together for 7 … x … 7 . Comment on the behavior of cot x in relation to the signs and values of tan x. 29. Graph y = sin x and y = :sin x; together. What are the domain and range of :sin x; ? 30. Graph y = sin x and y = < sin x = together. What are the domain and range of < sin x = ?
Additional Trigonometric Identities
42. cos a
3p  xb 2
43. Evaluate sin
7p p p as sin a + b . 12 4 3
44. Evaluate cos
11p p 2p b. as cos a + 12 4 3
45. Evaluate cos
p . 12
46. Evaluate sin
5p . 12
p b = sin x 2
32. cos ax +
p b = sin x 2
33. sin ax +
p b = cos x 2
34. sin ax 
p b = cos x 2
3p + xb 2
Using the DoubleAngle Formulas Find the function values in Exercises 47–50. 47. cos2
p 8
48. cos2
p 12
49. sin2
p 12
50. sin2
p 8
Theory and Examples 51. The tangent sum formula The standard formula for the tangent of the sum of two angles is tan A + tan B . 1  tan A tan B
tansA + Bd = Derive the formula.
52. (Continuation of Exercise 51.) Derive a formula for tan sA  Bd . 53. Apply the law of cosines to the triangle in the accompanying figure to derive the formula for cos sA  Bd .
Use the addition formulas to derive the identities in Exercises 31–36. 31. cos ax 
57
y 1
A
35. cos sA  Bd = cos A cos B + sin A sin B (Exercise 53 provides a different derivation.)
1
B
0
1
x
36. sin sA  Bd = sin A cos B  cos A sin B 37. What happens if you take B = A in the identity cos sA  Bd = cos A cos B + sin A sin B ? Does the result agree with something you already know? 38. What happens if you take B = 2p in the addition formulas? Do the results agree with something you already know?
Using the Addition Formulas In Exercises 39–42, express the given quantity in terms of sin x and cos x. 39. cos sp + xd
40. sin s2p  xd
54. a. Apply the formula for cos sA  Bd to the identity sin u = cos a
p  u b to obtain the addition formula for sin sA + Bd . 2
b. Derive the formula for cos sA + Bd by substituting B for B in the formula for cos sA  Bd from Exercise 35. 55. A triangle has sides a = 2 and b = 3 and angle C = 60° . Find the length of side c.
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56. A triangle has sides a = 2 and b = 3 and angle C = 40° . Find the length of side c. 57. The law of sines The law of sines says that if a, b, and c are the sides opposite the angles A, B, and C in a triangle, then sin C sin A sin B a = b = c . Use the accompanying figures and the identity sin sp  ud = sin u , if required, to derive the law. A
c
B
A
c
b
h
b C
a
B
a
h
65. Temperature in Fairbanks, Alaska Find the (a) amplitude, (b) period, (c) horizontal shift, and (d) vertical shift of the general sine function ƒsxd = 37 sin a
2p sx  101db + 25 . 365
66. Temperature in Fairbanks, Alaska Use the equation in Exercise 65 to approximate the answers to the following questions about the temperature in Fairbanks, Alaska, shown in Figure 1.77. Assume that the year has 365 days. a. What are the highest and lowest mean daily temperatures shown? b. What is the average of the highest and lowest mean daily temperatures shown? Why is this average the vertical shift of the function?
C
COMPUTER EXPLORATIONS
58. A triangle has sides a = 2 and b = 3 and angle C = 60° (as in Exercise 55). Find the sine of angle B using the law of sines. 59. A triangle has side c = 2 and angles A = p>4 and B = p>3 . Find the length a of the side opposite A. T 60. The approximation sin x x It is often useful to know that, when x is measured in radians, sin x L x for numerically small values of x. In Section 3.8, we will see why the approximation holds. The approximation error is less than 1 in 5000 if ƒ x ƒ 6 0.1 .
In Exercises 67–70, you will explore graphically the general sine function ƒsxd = A sin a
2p sx  Cdb + D B
as you change the values of the constants A, B, C, and D. Use a CAS or computer grapher to perform the steps in the exercises. 67. The period B
Set the constants A = 3, C = D = 0 .
a. With your grapher in radian mode, graph y = sin x and y = x together in a viewing window about the origin. What do you see happening as x nears the origin?
a. Plot ƒ(x) for the values B = 1, 3, 2p, 5p over the interval 4p … x … 4p . Describe what happens to the graph of the general sine function as the period increases.
b. With your grapher in degree mode, graph y = sin x and y = x together about the origin again. How is the picture different from the one obtained with radian mode?
b. What happens to the graph for negative values of B? Try it with B = 3 and B = 2p .
c. A quick radian mode check Is your calculator in radian mode? Evaluate sin x at a value of x near the origin, say x = 0.1 . If sin x L x , the calculator is in radian mode; if not, it isn’t. Try it.
For
p 2 1 63. y =  p sin a tb + p 2
b. What happens to the graph for negative values of C?
69. The vertical shift D Set the constants A = 3, B = 6, C = 0 .
2p sx  Cdb + D, B
identify A, B, C, and D for the sine functions in Exercises 61–64 and sketch their graphs (see Figure 1.76). 61. y = 2 sin sx + pd  1
a. Plot ƒ(x) for the values C = 0, 1 , and 2 over the interval 4p … x … 4p . Describe what happens to the graph of the general sine function as C increases through positive values. c. What smallest positive value should be assigned to C so the graph exhibits no horizontal shift? Confirm your answer with a plot.
General Sine Curves ƒsxd = A sin a
68. The horizontal shift C Set the constants A = 3, B = 6, D = 0 .
62. y = 64. y =
1 1 sin spx  pd + 2 2 2pt L sin , L 2p
L 7 0
a. Plot ƒ(x) for the values D = 0, 1 , and 3 over the interval 4p … x … 4p . Describe what happens to the graph of the general sine function as D increases through positive values. b. What happens to the graph for negative values of D? 70. The amplitude A Set the constants B = 6, C = D = 0 . a. Describe what happens to the graph of the general sine function as A increases through positive values. Confirm your answer by plotting ƒ(x) for the values A = 1, 5 , and 9. b. What happens to the graph for negative values of A?
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1.7
1.7
Graphing with Calculators and Computers
59
Graphing with Calculators and Computers A graphing calculator or a computer with graphing software enables us to graph very complicated functions with high precision. Many of these functions could not otherwise be easily graphed. However, care must be taken when using such devices for graphing purposes and we address those issues in this section. In Chapter 4 we will see how calculus helps us to be certain we are viewing accurately all the important features of a function’s graph.
Graphing Windows When using a graphing calculator or computer as a graphing tool, a portion of the graph is displayed in a rectangular display or viewing window. Often the default window gives an incomplete or misleading picture of the graph. We use the term square window when the units or scales on both axis are the same. This term does not mean that the display window itself is square in shape (usually it is rectangular), but means instead that the xunit is the same as the yunit. When a graph is displayed in the default window, the xunit may differ from the yunit of scaling in order to fit the graph in the display. The viewing window in the display is set by specifying the minimum and maximum values of the independent and dependent variables. That is, an interval a … x … b is specified as well as a range c … y … d. The machine selects a certain number of equally spaced values of x between a and b. Starting with a first value for x, if it lies within the domain of the function ƒ being graphed, and if ƒ(x) lies inside the range [c, d], then the point (x, ƒ(x)) is plotted. If x lies outside the domain of ƒ, or ƒ(x) lies outside the specified range [c, d], the machine just moves on to the next xvalue since it cannot plot (x, ƒ(x)) in that case. The machine plots a large number of points (x, ƒ(x)) in this way and approximates the curve representing the graph by drawing a short line segment between each plotted point and its next neighboring point, as we might do by hand. Usually, adjacent points are so close together that the graphical representation has the appearance of a smooth curve. Things can go wrong with this procedure and we illustrate the most common problems through the following examples.
EXAMPLE 1
Choosing a Viewing Window
Graph the function ƒsxd = x 3  7x 2 + 28 in each of the following display or viewing windows: (a) [10, 10] by [10, 10]
(b) [4, 4] by [50, 10]
(c) [4, 10] by [60, 60]
Solution
(a) We select a = 10, b = 10, c = 10, and d = 10 to specify the interval of xvalues and the range of yvalues for the window. The resulting graph is shown in Figure 1.78a. It appears that the window is cutting off the bottom part of the graph and that the interval of xvalues is too large. Let’s try the next window. (b) Now we see more features of the graph (Figure 1.78b), but the top is missing and we need to view more to the right of x = 4 as well. The next window should help. (c) Figure 1.78c shows the graph in this new viewing window. Observe that we get a more complete picture of the graph in this window and it is a reasonable graph of a thirddegree polynomial. Choosing a good viewing window is a trialanderror process which may require some troubleshooting as well.
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10
10
60
–4 –10
4 –4
10
–10
–50
(a)
(b)
10
–60 (c)
The graph of ƒsxd = x 3  7x 2 + 28 in different viewing windows (Example 1).
FIGURE 1.78
EXAMPLE 2
Square Windows
When a graph is displayed, the xunit may differ from the yunit, as in the graphs shown in Figures 1.78b and 1.78c. The result is distortion in the picture, which may be misleading. The display window can be made square by compressing or stretching the units on one axis to match the scale on the other, giving the true graph. Many systems have builtin functions to make the window “square.” If yours does not, you will have to do some calculations and set the window size manually to get a square window, or bring to your viewing some foreknowledge of the true picture. Figure 1.79a shows the graphs of the perpendicular lines y = x and y x + 322, together with the semicircle y = 29  x 2 , in a nonsquare [6, 6] by [6, 8] display window. Notice the distortion. The lines do not appear to be perpendicular, and the semicircle appears to be elliptical in shape. Figure 1.79b shows the graphs of the same functions in a square window in which the xunits are scaled to be the same as the yunits. Notice that the [6, 6] by [4, 4] viewing window has the same xaxis in both figures, but the scaling on the xaxis has been compressed in Figure 1.79b to make the window square. Figure 1.79c gives an enlarged view with a square [3, 3] by [0, 4] window.
8
–6
4
4
6
–6
6
–3 –6 (a)
–4 (b)
3 0 (c)
FIGURE 1.79 Graphs of the perpendicular lines y = x and y = x + 322 , and the semicircle y = 29  x 2 , in (a) a nonsquare window, and (b) and (c) square windows (Example 2).
If the denominator of a rational function is zero at some xvalue within the viewing window, a calculator or graphing computer software may produce a steep nearvertical line segment from the top to the bottom of the window. Here is an example. Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
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1.7
EXAMPLE 3
Graphing with Calculators and Computers
61
Graph of a Rational Function
Graph the function y =
1 . 2  x
Figure 1.80a shows the graph in the [10, 10] by [10, 10] default square window with our computer graphing software. Notice the nearvertical line segment at x = 2. It is not truly a part of the graph and x = 2 does not belong to the domain of the function. By trial and error we can eliminate the line by changing the viewing window to the smaller [6, 6] by [4, 4] view, revealing a better graph (Figure 1.80b).
Solution
4
10
–10
10
–6
6
–10 (a)
–4 (b)
FIGURE 1.80 Graphs of the function y =
1 (Example 3). 2  x
Sometimes the graph of a trigonometric function oscillates very rapidly. When a calculator or computer software plots the points of the graph and connects them, many of the maximum and minimum points are actually missed. The resulting graph is then very misleading.
EXAMPLE 4
Graph of a Rapidly Oscillating Function
Graph the function ƒsxd = sin 100x. Figure 1.81a shows the graph of ƒ in the viewing window [12, 12] by [1, 1]. We see that the graph looks very strange because the sine curve should oscillate periodically between 1 and 1. This behavior is not exhibited in Figure 1.81a. We might experiment with a smaller viewing window, say [6, 6] by [1, 1], but the graph is not better (Figure 1.81b). The difficulty is that the period of the trigonometric function y = sin 100x is very small s2p>100 L 0.063d. If we choose the much smaller viewing window [0.1, 0.1] by [1, 1] we get the graph shown in Figure 1.81c. This graph reveals the expected oscillations of a sine curve. Solution
1
–12
1
12
–1 (a)
–6
1
6
–0.1
–1 (b)
0.1
–1 (c)
FIGURE 1.81 Graphs of the function y = sin 100x in three viewing windows. Because the period is 2p>100 L 0.063 , the smaller window in (c) best displays the true aspects of this rapidly oscillating function (Example 4).
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EXAMPLE 5
Another Rapidly Oscillating Function
Graph the function y = cos x +
1 sin 50x. 50
In the viewing window [6, 6] by [1, 1] the graph appears much like the cosine function with some small sharp wiggles on it (Figure 1.82a). We get a better look when we significantly reduce the window to [0.6, 0.6] by [0.8, 1.02], obtaining the graph in Figure 1.82b. We now see the small but rapid oscillations of the second term, 1>50 sin 50x, added to the comparatively larger values of the cosine curve.
Solution
1.02
1
–6
6
–0.6
0.6 0.8 (b)
–1 (a)
FIGURE 1.82 In (b) we see a closeup view of the function 1 sin 50x graphed in (a). The term cos x clearly dominates the y = cos x + 50 1 second term, sin 50x , which produces the rapid oscillations along the 50 cosine curve (Example 5).
EXAMPLE 6
Graphing an Odd Fractional Power
Graph the function y = x 1>3 . Many graphing devices display the graph shown in Figure 1.83a. When we 3 compare it with the graph of y = x 1>3 = 2 x in Figure 1.38, we see that the left branch for x 6 0 is missing. The reason the graphs differ is that many calculators and computer soft
Solution
2
–3
2
3
–2 (a)
–3
3
–2 (b)
FIGURE 1.83 The graph of y = x 1>3 is missing the left branch in (a). In x # 1>3 (b) we graph the function ƒsxd = ƒ x ƒ obtaining both branches. (See ƒxƒ Example 6.)
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ware programs calculate x 1>3 as e s1>3dln x . (The exponential and logarithmic functions are studied in Chapter 7.) Since the logarithmic function is not defined for negative values of x, the computing device can only produce the right branch where x 7 0. To obtain the full picture showing both branches, we can graph the function ƒsxd =
x # 1>3 ƒxƒ . ƒxƒ
This function equals x 1>3 except at x = 0 (where ƒ is undefined, although 0 1>3 = 0). The graph of ƒ is shown in Figure 1.83b.
Empirical Modeling: Capturing the Trend of Collected Data In Example 3 of Section 1.4, we verified the reasonableness of Kepler’s hypothesis that the period of a planet’s orbit is proportional to its mean distance from the sun raised to the 3>2 power. If we cannot hypothesize a relationship between a dependent variable and an independent variable, we might collect data points and try to find a curve that “fits” the data and captures the trend of the scatterplot. The process of finding a curve to fit data is called regression analysis and the curve is called a regression curve. A computer or graphing calculator finds the regression curve by finding the particular curve which minimizes the sum of the squares of the vertical distances between the data points and the curve. This method of least squares is discussed in the Section 14.7 exercises. There are many useful types of regression curves, such as straight lines, power, polynomial, exponential, logarithmic, and sinusoidal curves. Many computers or graphing calculators have a regression analysis feature to fit a variety of regression curve types. The next example illustrates using a graphing calculator’s linear regression feature to fit data from Table 1.5 with a linear equation.
EXAMPLE 7 TABLE 1.5 Price of a
U.S. postage stamp Year x
Cost y
1968 1971 1974 1975 1977 1981 1981 1985 1987 1991 1995 1998 2002
0.06 0.08 0.10 0.13 0.15 0.18 0.20 0.22 0.25 0.29 0.32 0.33 0.37
Fitting a Regression Line
Starting with the data in Table 1.5, build a model for the price of a postage stamp as a function of time. After verifying that the model is “reasonable,” use it to predict the price in 2010. We are building a model for the price of a stamp since 1968. There were two increases in 1981, one of three cents followed by another of two cents. To make 1981 comparable with the other listed years, we lump them together as a single fivecent increase, giving the data in Table 1.6. Figure 1.84a gives the scatterplot for Table 1.6.
Solution
TABLE 1.6 Price of a U.S postage stamp since 1968
x y
0 6
3 8
6 10
7 13
9 15
13 20
17 22
19 25
23 29
27 32
30 33
34 37
Since the scatterplot is fairly linear, we investigate a linear model. Upon entering the data into a graphing calculator (or computer software) and selecting the linear regression option, we find the regression line to be y = 0.94x + 6.10.
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Chapter 1: Preliminaries
Price of stamps (cents)
y 60 50 40 30 20 10 0
10 20 30 40 50 60 Year after 1968 (b)
(a)
x
FIGURE 1.84 (a) Scatterplot of (x, y) data in Table 1.6. (b) Using the regression line to estimate the price of a stamp in 2010. (Example 7).
Figure 1.84b shows the line and scatterplot together. The fit is remarkably good, so the model seems reasonable. Evaluating the regression line, we conclude that in 2010 sx = 42d, the price of a stamp will be y = 0.94s42d + 6.10 L 46 cents. The prediction is shown as the red point on the regression line in Figure 1.84b.
EXAMPLE 8
Finding a Curve to Predict Population Levels
We may want to predict the future size of a population, such as the number of trout or catfish living in a fish farm. Figure 1.85 shows a scatterplot of the data collected by R. Pearl for a collection of yeast cells (measured as biomass) growing over time (measured in hours) in a nutrient. y
Biomass
64
300 250 200 150 100 50 0
1
2
3 4 Time
5
6
7
x
FIGURE 1.85 Biomass of a yeast culture versus elapsed time (Example 8). (Data from R. Pearl, “The Growth of Population,” Quart. Rev. Biol., Vol. 2 (1927), pp. 532–548.)
The plot of points appears to be reasonably smooth with an upward curving trend. We might attempt to capture this trend by fitting a polynomial (for example, a quadratic y = ax 2 + bx + c), a power curve s y = ax b d, or an exponential curve s y = ae bx d. Figure 1.86 shows the result of using a calculator to fit a quadratic model.
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1.7
250 Biomass
65
The quadratic model y = 6.10x 2  9.28x + 16.43 appears to fit the collected data reasonably well (Figure 1.86). Using this model, we predict the population after 17 hours as ys17d = 1622.65. Let us examine more of Pearl’s data to see if our quadratic model continues to be a good one. In Figure 1.87, we display all of Pearl’s data. Now you see that the prediction of ys17d = 1622.65 grossly overestimates the observed population of 659.6. Why did the quadratic model fail to predict a more accurate value?
y
200 150 100 50 0
Graphing with Calculators and Computers
1
2
3
4 5 Time
6
7
x y 2000
FIGURE 1.86 Fitting a quadratic to Pearl’s data gives the equation y = 6.10x 2  9.28x + 16.43 and the prediction y s17d = 1622.65 (Example 8).
1800 1600
Yeast Population
1400 1200
Observed Predicted
1000 800 600 400 200
x 0
4.5
9 Time (hours)
13.5
18
FIGURE 1.87 The rest of Pearl’s data (Example 8).
The problem lies in the danger of predicting beyond the range of data used to build the empirical model. (The range of data creating our model was 0 … x … 7.) Such extrapolation is especially dangerous when the model selected is not supported by some underlying rationale suggesting the form of the model. In our yeast example, why would we expect a quadratic function as underlying population growth? Why not an exponential function? In the face of this, how then do we predict future values? Often, calculus can help, and in Chapter 9 we use it to model population growth.
Regression Analysis Regression analysis has four steps: 1. Plot the data (scatterplot). 2. Find a regression equation. For a line, it has the form y = mx + b, and for a quadratic, the form y = ax 2 + bx + c. 3. Superimpose the graph of the regression equation on the scatterplot to see the fit. 4. If the fit is satisfactory, use the regression equation to predict yvalues for values of x not in the table.
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EXERCISES 1.7 Choosing a Viewing Window In Exercises 1–4, use a graphing calculator or computer to determine which of the given viewing windows displays the most appropriate graph of the specified function. 1. ƒsxd = x 4  7x 2 + 6x a. [1, 1] by [1, 1]
b. [2, 2] by [5, 5]
c. [10, 10] by [10, 10]
d. [5, 5] by [25, 15]
3
2. ƒsxd = x  4x  4x + 16 a. [1, 1] by [5, 5]
b. [3, 3] by [10, 10]
c. [5, 5] by [10, 20]
d. [20, 20] by [100, 100]
3. ƒsxd = 5 + 12x  x 3 b. [5, 5] by [10, 10]
c. [4, 4] by [20, 20] 4. ƒsxd = 25 + 4x  x
32. Graph the upper branch of the hyperbola y 2  16x 2 = 1 . 33. Graph four periods of the function ƒsxd =  tan 2x . 34. Graph two periods of the function ƒsxd = 3 cot
x + 1. 2
35. Graph the function ƒsxd = sin 2x + cos 3x . 36. Graph the function ƒsxd = sin3 x .
2
a. [1, 1] by [1, 1]
31. Graph the lower half of the circle defined by the equation x 2 + 2x = 4 + 4y  y 2 .
d. [4, 5] by [15, 25]
Graphing in Dot Mode Another way to avoid incorrect connections when using a graphing device is through the use of a “dot mode,” which plots only the points. If your graphing utility allows that mode, use it to plot the functions in Exercises 37–40.
2
37. y =
a. [2, 2] by [2, 2]
b. [2, 6] by [1, 4]
c. [3, 7] by [0, 10]
d. [10, 10] by [10, 10]
1 38. y = sin x x3  1 40. y = 2 x  1
1 x  3
39. y = x:x;
Regression Analysis Determining a Viewing Window In Exercises 5–30, determine an appropriate viewing window for the given function and use it to display its graph. x3 x2  2x + 1 3 2
5. ƒsxd = x 4  4x 3 + 15
6. ƒsxd =
7. ƒsxd = x 5  5x 4 + 10
8. ƒsxd = 4x 3  x 4
9. ƒsxd = x29  x 2 11. y = 2x  3x 13. y = 5x
2>5
 2x
x2 + 2 x2 + 1 x  1 21. ƒsxd = 2 x  x  6
1 sin 30x 29. y = x + 10
2
sx  8d s5  xd 1 x + 3
x2  1 x2 + 1 8 22. ƒsxd = 2 x  9 20. ƒsxd =
6x 2  15x + 6 4x 2  10x 25. y = sin 250x x b 50
14. y = x
2>3
18. y = 1 
19. ƒsxd =
27. y = cos a
12. y = x
1>3
16. y = ƒ x 2  x ƒ
x + 3 x + 2
23. ƒsxd =
TABLE 1.7 Construction workers’ average
annual compensation Year
Annual compensation (dollars)
1980 1985 1988 1990 1992 1995 1999 2002
22,033 27,581 30,466 32,836 34,815 37,996 42,236 45,413
10. ƒsxd = x 2s6  x 3 d
2>3
15. y = ƒ x 2  1 ƒ 17. y =
T 41. Table 1.7 shows the mean annual compensation of construction workers.
24. ƒsxd =
x2  3 x  2
26. y = 3 cos 60x 28. y =
x 1 sin a b 10 10
1 cos 100x 30. y = x + 50 2
Source: U.S. Bureau of Economic Analysis.
a. Find a linear regression equation for the data. b. Find the slope of the regression line. What does the slope represent?
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1.7 Graphing with Calculators and Computers c. Superimpose the graph of the linear regression equation on a scatterplot of the data. d. Use the regression equation to predict the construction workers’ average annual compensation in 2010. T 42. The median price of existing singlefamily homes has increased consistently since 1970. The data in Table 1.8, however, show that there have been differences in various parts of the country. a. Find a linear regression equation for home cost in the Northeast. b. What does the slope of the regression line represent? c. Find a linear regression equation for home cost in the Midwest. d. Where is the median price increasing more rapidly, in the Northeast or the Midwest?
TABLE 1.8 Median price of singlefamily homes
Year
Northeast (dollars)
Midwest (dollars)
1970 1975 1980 1985 1990 1995 2000
25,200 39,300 60,800 88,900 141,200 197,100 264,700
20,100 30,100 51,900 58,900 74,000 88,300 97,000
Source: National Association of Realtors®
T 43. Vehicular stopping distance Table 1.9 shows the total stopping distance of a car as a function of its speed. a. Find the quadratic regression equation for the data in Table 1.9. b. Superimpose the graph of the quadratic regression equation on a scatterplot of the data. c. Use the graph of the quadratic regression equation to predict the average total stopping distance for speeds of 72 and 85 mph. Confirm algebraically. d. Now use linear regression to predict the average total stopping distance for speeds of 72 and 85 mph. Superimpose the regression line on a scatterplot of the data. Which gives the better fit, the line here or the graph in part (b)?
67
TABLE 1.9 Vehicular stopping distance
Speed (mph)
Average total stopping distance (ft)
20 25 30 35 40 45 50 55 60 65 70 75 80
42 56 73.5 91.5 116 142.5 173 209.5 248 292.5 343 401 464
Source: U.S. Bureau of Public Roads.
T 44. Stern waves Observations of the stern waves that follow a boat at right angles to its course have disclosed that the distance between the crests of these waves (their wave length) increases with the speed of the boat. Table 1.10 shows the relationship between wave length and the speed of the boat.
TABLE 1.10 Wave lengths
Wave length (m)
Speed (km/h)
0.20 0.65 1.13 2.55 4.00 5.75 7.80 10.20 12.90 16.00 18.40
1.8 3.6 5.4 7.2 9.0 10.8 12.6 14.4 16.2 18.0 19.8
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Chapter 1: Preliminaries a. Find a power regression equation y = ax b for the data in Table 1.10, where x is the wave length, and y the speed of the boat.
c. Use the graph of the power regression equation to predict the speed of the boat when the wave length is 11 m. Confirm algebraically.
b. Superimpose the graph of the power regression equation on a scatterplot of the data.
d. Now use linear regression to predict the speed when the wave length is 11 m. Superimpose the regression line on a scatterplot of the data. Which gives the better fit, the line here or the curve in part (b)?
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Chapter 1
Questions to Guide Your Review
1. How are the real numbers represented? What are the main categories characterizing the properties of the real number system? What are the primary subsets of the real numbers? 2. How are the rational numbers described in terms of decimal expansions? What are the irrational numbers? Give examples. 3. What are the order properties of the real numbers? How are they used in solving equations? 4. What is a number’s absolute value? Give examples? How are ƒ a ƒ , ƒ ab ƒ , ƒ a>b ƒ , and ƒ a + b ƒ related to ƒ a ƒ and ƒ b ƒ ? 5. How are absolute values used to describe intervals or unions of intervals? Give examples. 6. How do we identify points in the plane using the Cartesian coordinate system? What is the graph of an equation in the variables x and y? 7. How can you write an equation for a line if you know the coordinates of two points on the line? The line’s slope and the coordinates of one point on the line? The line’s slope and yintercept? Give examples. 8. What are the standard equations for lines perpendicular to the coordinate axes? 9. How are the slopes of mutually perpendicular lines related? What about parallel lines? Give examples. 10. When a line is not vertical, what is the relation between its slope and its angle of inclination? 11. How do you find the distance between two points in the coordinate plane? 12. What is the standard equation of a circle with center (h, k) and radius a? What is the unit circle and what is its equation? 13. Describe the steps you would take to graph the circle x 2 + y 2 + 4x  6y + 12 = 0 . 14. What inequality describes the points in the coordinate plane that lie inside the circle of radius a centered at the point (h, k)? That lie inside or on the circle? That lie outside the circle? That lie outside or on the circle? 15. If a, b, and c are constants and a Z 0 , what can you say about the graph of the equation y = ax 2 + bx + c ? In particular, how would you go about sketching the curve y = 2x 2 + 4x ?
16. What is a function? What is its domain? Its range? What is an arrow diagram for a function? Give examples. 17. What is the graph of a realvalued function of a real variable? What is the vertical line test? 18. What is a piecewisedefined function? Give examples. 19. What are the important types of functions frequently encountered in calculus? Give an example of each type. 20. In terms of its graph, what is meant by an increasing function? A decreasing function? Give an example of each. 21. What is an even function? An odd function? What symmetry properties do the graphs of such functions have? What advantage can we take of this? Given an example of a function that is neither even nor odd. 22. What does it mean to say that y is proportional to x? To x 3>2 ? What is the geometric interpretation of proportionality? How can this interpretation be used to test a proposed proportionality? 23. If ƒ and g are realvalued functions, how are the domains of ƒ + g, ƒ  g, ƒg , and ƒ>g related to the domains of ƒ and g? Give examples. 24. When is it possible to compose one function with another? Give examples of composites and their values at various points. Does the order in which functions are composed ever matter? 25. How do you change the equation y = ƒsxd to shift its graph vertically up or down by a factor k 7 0 ? Horizontally to the left or right? Give examples. 26. How do you change the equation y = ƒsxd to compress or stretch the graph by c 7 1 ? Reflect the graph across a coordinate axis? Give examples. 27. What is the standard equation of an ellipse with center (h, k)? What is its major axis? Its minor axis? Give examples. 28. What is radian measure? How do you convert from radians to degrees? Degrees to radians? 29. Graph the six basic trigonometric functions. What symmetries do the graphs have? 30. What is a periodic function? Give examples. What are the periods of the six basic trigonometric functions?
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Questions to Guide Your Review
69
31. Starting with the identity sin2 u + cos2 u = 1 and the formulas for cos sA + Bd and sin sA + Bd , show how a variety of other trigonometric identities may be derived.
stretching, compressing, and reflection of its graph? Give examples. Graph the general sine curve and identify the constants A, B, C, and D.
32. How does the formula for the general sine function ƒsxd = A sin ss2p>Bdsx  Cdd + D relate to the shifting,
33. Name three issues that arise when functions are graphed using a calculator or computer with graphing software. Give examples.
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Chapter 1 Practice Exercises
Chapter 1
69
Practice Exercises
Inequalities
16. through s 3, 6d and s1, 2d
In Exercises 1–4, solve the inequalities and show the solution sets on the real line.
17. the horizontal line through (0, 2) 18. through (3, 3) and s 2, 5d
1. 7 + 2x Ú 3
2. 3x 6 10
19. with slope 3 and yintercept 3
1 1 3. sx  1d 6 sx  2d 5 4
x  3 4 + x 4. Ú 2 3
20. through (3, 1) and parallel to 2x  y = 2 21. through s4, 12d and parallel to 4x + 3y = 12 22. through s 2, 3d and perpendicular to 3x  5y = 1
Absolute Value
23. through s 1, 2d and perpendicular to s1>2dx + s1>3dy = 1
Solve the equations or inequalities in Exercises 5–8. 5. ƒ x + 1 ƒ = 7
6. ƒ y  3 ƒ 6 4
7. ` 1 
8. `
3 x ` 7 2 2
2x + 7 ` … 5 3
Coordinates 9. A particle in the plane moved from As 2, 5d to the yaxis in such a way that ¢y equaled 3¢x . What were the particle’s new coordinates? 10. a. Plot the points As8, 1d, Bs2, 10d, Cs 4, 6d, Ds2, 3d , and E(14>3, 6). b. Find the slopes of the lines AB, BC, CD, DA, CE, and BD. c. Do any four of the five points A, B, C, D, and E form a parallelogram? d. Are any three of the five points collinear? How do you know? e. Which of the lines determined by the five points pass through the origin? 11. Do the points As6, 4d, Bs4, 3d , and Cs 2, 3d form an isosceles triangle? A right triangle? How do you know?
24. with xintercept 3 and yintercept 5
Functions and Graphs 25. Express the area and circumference of a circle as functions of the circle’s radius. Then express the area as a function of the circumference. 26. Express the radius of a sphere as a function of the sphere’s surface area. Then express the surface area as a function of the volume. 27. A point P in the first quadrant lies on the parabola y = x 2 . Express the coordinates of P as functions of the angle of inclination of the line joining P to the origin. 28. A hotair balloon rising straight up from a level field is tracked by a range finder located 500 ft from the point of liftoff. Express the balloon’s height as a function of the angle the line from the range finder to the balloon makes with the ground. In Exercises 29–32, determine whether the graph of the function is symmetric about the yaxis, the origin, or neither. 29. y = x 1>5
30. y = x 2>5
12. Find the coordinates of the point on the line y = 3x + 1 that is equidistant from (0, 0) and s 3, 4d .
31. y = x 2  2x  1
32. y = e x
Lines
33. y = x 2 + 1
34. y = x 5  x 3  x
In Exercises 13–24, write an equation for the specified line.
35. y = 1  cos x
36. y = sec x tan x
13. through s1, 6d with slope 3
x4 + 1 37. y = 3 x  2x 39. y = x + cos x
14. through s 1, 2d with slope 1>2 15. the vertical line through s0, 3d
2
In Exercises 33–40, determine whether the function is even, odd, or neither.
38. y = 1  sin x 40. y = 2x 4  1
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In Exercises 41–50, find the (a) domain and (b) range. 42. y = 2 + 21  x
41. y = ƒ x ƒ  2 43. y = 216  x 2 45. y = 2e
x
Composition with absolute values In Exercises 65–68, graph g1 and g2 together. Then describe how taking absolute values after applying g1 affects the graph.
44. y = 32  x + 1 46. y = tan s2x  pd
 3
48. y = x 2>5
65. x
49. y = ln sx  3d + 1
3 2  x 50. y = 1 + 2
66. 2x
4 … x … 0 0 6 x … 4
x  2, 52. y = • x, x + 2,
54.
69. y = cos 2x
0
y 5
1
1
4
x
Composition of Functions a. sƒ gds 1d .
b. sg ƒds2d .
c. sƒ ƒdsxd .
d. sg gdsxd .
56. ƒsxd = 2  x,
58. ƒsxd = 2x,
1
ƒ1sxd 60. x
3
61. x
2
ƒ2sxd ƒ1s ƒ x ƒ d
ƒxƒ
b. Express a in terms of A and b. 77. a. Express a in terms of B and b. b. Express c in terms of A and a.
g sxd = 2x + 2
2
76. a. Express a in terms of A and c.
78. a. Express sin A in terms of a and c.
g sxd = 21  x
ƒxƒ ƒ x ƒ3
75. a. Find a and b if c = 2, B = p>3 .
3 g sxd = 2 x + 1
Composition with absolute values In Exercises 59–64, graph ƒ1 and ƒ2 together. Then describe how applying the absolute value function before applying ƒ1 affects the graph.
59. x
In Exercises 75–78, ABC is a right triangle with the right angle at C. The sides opposite angles A, B, and C are a, b, and c, respectively.
2x + 2
In Exercises 57 and 58, (a) write a formula for ƒ g and g ƒ and find the (b) domain and (c) range of each. 57. ƒsxd = 2  x 2,
p b. 4
b. Find a and c if b = 2, B = p>3 .
In Exercises 55 and 56, find
g sxd =
p b. 3
74. Sketch the graph y = 1 + sin ax +
0
1 55. ƒsxd = x ,
71. y = sin px
x 2 px 72. y = cos 2 70. y = sin
73. Sketch the graph y = 2 cos ax 
(2, 5)
x
2
2 ƒx + xƒ
In Exercises 69–72, sketch the graph of the given function. What is the period of the function?
2 … x … 1 1 6 x … 1 1 6 x … 2
y
ƒ 2x ƒ ƒ 4  x2 ƒ
Trigonometry
In Exercises 53 and 54, write a piecewise formula for the function. 53.
2
68. x 2 + x
In Exercises 51 and 52, find the (a) domain and (b) range. 2 x, 2x,
3
67. 4  x
PiecewiseDefined Functions 51. y = e
g2sxd ƒ g1sxd ƒ ƒ x3 ƒ
g1sxd
47. y = 2 sin s3x + pd  1
b. Express sin A in terms of b and c. 79. Height of a pole Two wires stretch from the top T of a vertical pole to points B and C on the ground, where C is 10 m closer to the base of the pole than is B. If wire BT makes an angle of 35° with the horizontal and wire CT makes an angle of 50° with the horizontal, how high is the pole? 80. Height of a weather balloon Observers at positions A and B 2 km apart simultaneously measure the angle of elevation of a weather balloon to be 40° and 70°, respectively. If the balloon is directly above a point on the line segment between A and B, find the height of the balloon. T 81. a. Graph the function ƒsxd = sin x + cossx>2d . b. What appears to be the period of this function? c. Confirm your finding in part (b) algebraically.
1 62. x
1 ƒxƒ
63. 2x
2ƒ x ƒ
b. What are the domain and range of ƒ?
64. sin x
sin ƒ x ƒ
c. Is ƒ periodic? Give reasons for your answer.
T 82. a. Graph ƒsxd = sin s1>xd .
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Additional and Advanced Exercises
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Additional and Advanced Exercises
Functions and Graphs
13. Find sin B if a = 2, b = 3, c = 4 .
1. The graph of ƒ is shown. Draw the graph of each function. a. y = ƒs xd
b. y = ƒsxd
c. y = 2ƒsx + 1d + 1
d. y = 3ƒsx  2d  2 y
1
Derivations and Proofs 15. Prove the following identities. a.
1  cos x sin x = 1 + cos x sin x
b.
1  cos x x = tan2 1 + cos x 2
16. Explain the following “proof without words” of the law of cosines. (Source: “Proof without Words: The Law of Cosines,” Sidney H. Kung, Mathematics Magazine, Vol. 63, No. 5, Dec. 1990, p. 342.)
x
–1
14. Find sin C if a = 2, b = 4, c = 5 .
2. A portion of the graph of a function defined on [3, 3] is shown. Complete the graph assuming that the function is a. even.
2a cos b ac
b. odd.
c
y a
a
(3, 2)
2
b
a
1 0 –1
1
2
3
x
(1, –1)
17. Show that the area of triangle ABC is given by s1>2dab sin C = s1>2dbc sin A = s1>2dca sin B . C
3. Are there two functions ƒ and g such that ƒ g = g ƒ ? Give reasons for your answer. b
4. Are there two functions ƒ and g with the following property? The graphs of ƒ and g are not straight lines but the graph of ƒ g is a straight line. Give reasons for your answer. 5. If ƒ(x) is odd, can anything be said of g sxd = ƒsxd  2? What if ƒ is even instead? Give reasons for your answer. 6. If g (x) is an odd function defined for all values of x, can anything be said about g (0)? Give reasons for your answer. 7. Graph the equation ƒ x ƒ + ƒ y ƒ = 1 + x . 8. Graph the equation y + ƒ y ƒ = x + ƒ x ƒ .
Trigonometry In Exercises 9–14, ABC is an arbitrary triangle with sides a, b, and c opposite angles A, B, and C, respectively. 9. Find b if a = 23, A = p>3, B = p>4 .
A
a
c
18. Show that the area of triangle ABC is given by 2sss  adss  bdss  cd where s = sa + b + cd>2 is the semiperimeter of the triangle. 19. Properties of inequalities If a and b are real numbers, we say that a is less than b and write a 6 b if (and only if ) b  a is positive. Use this definition to prove the following properties of inequalities. If a, b, and c are real numbers, then: 1. a 6 b Q a + c 6 b + c 2. a 6 b Q a  c 6 b  c
10. Find sin B if a = 4, b = 3, A = p>4 .
3. a 6 b and c 7 0 Q ac 6 bc
11. Find cos A if a = 2, b = 2, c = 3 .
4. a 6 b and c 6 0 Q bc 6 ac (Special case: a 6 b Q b 6 a)
12. Find c if a = 2, b = 3, C = p>4 .
B
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Chapter 1: Preliminaries 1 5. a 7 0 Q a 7 0
b. b changes (a and c fixed, a Z 0)? c. c changes (a and b fixed, a Z 0)?
1 1 6. 0 6 a 6 b Q 6 a b 1 1 7. a 6 b 6 0 Q 6 a b 20. Prove that the following inequalities hold for any real numbers a and b. a. ƒ a ƒ 6 ƒ b ƒ if and only if a 2 6 b 2 b. ƒ a  b ƒ Ú ƒ ƒ a ƒ  ƒ b ƒ ƒ Generalizing the triangle inequality Prove by mathematical induction that the inequalities in Exercises 21 and 22 hold for any n real numbers a1, a2 , Á , an . (Mathematical induction is reviewed in Appendix 1.) 21. ƒ a1 + a2 + Á + an ƒ … ƒ a1 ƒ + ƒ a2 ƒ + Á + ƒ an ƒ 22. ƒ a1 + a2 + Á + an ƒ Ú ƒ a1 ƒ  ƒ a2 ƒ  Á  ƒ an ƒ 23. Show that if ƒ is both even and odd, then ƒsxd = 0 for every x in the domain of ƒ. 24. a. Evenodd decompositions Let ƒ be a function whose domain is symmetric about the origin, that is, x belongs to the domain whenever x does. Show that ƒ is the sum of an even function and an odd function:
27. Find all values of the slope of the line y = mx + 2 for which the xintercept exceeds 1>2 .
Geometry 28. An object’s center of mass moves at a constant velocity y along a straight line past the origin. The accompanying figure shows the coordinate system and the line of motion. The dots show positions that are 1 sec apart. Why are the areas A1, A2 , Á , A5 in the figure all equal? As in Kepler’s equal area law (see Section 13.6), the line that joins the object’s center of mass to the origin sweeps out equal areas in equal times. y t6
10
t5 Kilometers
72
A5
yt A4
5
yt
A3
t2
A2
t1
A1
ƒsxd = Esxd + O sxd , where E is an even function and O is an odd function. (Hint: Let Esxd = sƒsxd + ƒs xdd>2 . Show that Es xd = Esxd , so that E is even. Then show that O sxd = ƒsxd  Esxd is odd.) b. Uniqueness Show that there is only one way to write ƒ as the sum of an even and an odd function. (Hint: One way is given in part (a). If also ƒsxd = E1sxd + O1sxd where E1 is even and O1 is odd, show that E  E1 = O1  O . Then use Exercise 23 to show that E = E1 and O = O1 .)
0
5
10 Kilometers
15
29. a. Find the slope of the line from the origin to the midpoint P, of side AB in the triangle in the accompanying figure sa, b 7 0d . y
B(0, b)
Grapher Explorations—Effects of Parameters 25. What happens to the graph of y = ax 2 + bx + c as
P
a. a changes while b and c remain fixed? b. b changes (a and c fixed, a Z 0)?
O
c. c changes (a and b fixed, a Z 0)? 26. What happens to the graph of y = asx + bd3 + c as
x
b. When is OP perpendicular to AB?
a. a changes while b and c remain fixed?
Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
A(a, 0)
x
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Chapter 1: Preliminaries
Chapter 1
Technology Application Projects
An Overview of Mathematica An overview of Mathematica sufficient to complete the Mathematica modules appearing on the Web site.
Mathematica/Maple Module Modeling Change: Springs, Driving Safety, Radioactivity, Trees, Fish, and Mammals. Construct and interpret mathematical models, analyze and improve them, and make predictions using them.
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Chapter
2
LIMITS AND CONTINUITY OVERVIEW The concept of a limit is a central idea that distinguishes calculus from algebra and trigonometry. It is fundamental to finding the tangent to a curve or the velocity of an object. In this chapter we develop the limit, first intuitively and then formally. We use limits to describe the way a function ƒ varies. Some functions vary continuously; small changes in x produce only small changes in ƒ(x). Other functions can have values that jump or vary erratically. The notion of limit gives a precise way to distinguish between these behaviors. The geometric application of using limits to define the tangent to a curve leads at once to the important concept of the derivative of a function. The derivative, which we investigate thoroughly in Chapter 3, quantifies the way a function’s values change.
2.1
Rates of Change and Limits In this section, we introduce average and instantaneous rates of change. These lead to the main idea of the section, the idea of limit.
Average and Instantaneous Speed A moving body’s average speed during an interval of time is found by dividing the distance covered by the time elapsed. The unit of measure is length per unit time: kilometers per hour, feet per second, or whatever is appropriate to the problem at hand.
EXAMPLE 1
Finding an Average Speed
A rock breaks loose from the top of a tall cliff. What is its average speed (a) during the first 2 sec of fall? (b) during the 1sec interval between second 1 and second 2? In solving this problem we use the fact, discovered by Galileo in the late sixteenth century, that a solid object dropped from rest (not moving) to fall freely near the surface of the earth will fall a distance proportional to the square of the time it has been falling. (This assumes negligible air resistance to slow the object down and that gravity is
Solution
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Chapter 2: Limits and Continuity
HISTORICAL BIOGRAPHY* Galileo Galilei (1564–1642)
the only force acting on the falling body. We call this type of motion free fall.) If y denotes the distance fallen in feet after t seconds, then Galileo’s law is y = 16t 2 , where 16 is the constant of proportionality. The average speed of the rock during a given time interval is the change in distance, ¢y, divided by the length of the time interval, ¢t. (a) For the first 2 sec:
¢y 16s2d2  16s0d2 ft = = 32 sec 2  0 ¢t
(b) From sec 1 to sec 2:
¢y 16s2d2  16s1d2 ft = = 48 sec 2  1 ¢t
The next example examines what happens when we look at the average speed of a falling object over shorter and shorter time intervals.
EXAMPLE 2
Finding an Instantaneous Speed
Find the speed of the falling rock at t = 1 and t = 2 sec. We can calculate the average speed of the rock over a time interval [t0 , t0 + h], having length ¢t = h, as
Solution
¢y 16st0 + hd2  16t0 2 . = h ¢t
(1)
We cannot use this formula to calculate the “instantaneous” speed at t0 by substituting h = 0, because we cannot divide by zero. But we can use it to calculate average speeds over increasingly short time intervals starting at t0 = 1 and t0 = 2. When we do so, we see a pattern (Table 2.1). TABLE 2.1 Average speeds over short time intervals
Average speed:
¢y 16st0 + hd2  16t0 2 = h ¢t
Length of time interval h
Average speed over interval of length h starting at t0 1
Average speed over interval of length h starting at t0 2
1 0.1 0.01 0.001 0.0001
48 33.6 32.16 32.016 32.0016
80 65.6 64.16 64.016 64.0016
The average speed on intervals starting at t0 = 1 seems to approach a limiting value of 32 as the length of the interval decreases. This suggests that the rock is falling at a speed of 32 ft> sec at t0 = 1 sec. Let’s confirm this algebraically. To learn more about the historical figures and the development of the major elements and topics of calculus, visit www.awbc.com/thomas.
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If we set t0 = 1 and then expand the numerator in Equation (1) and simplify, we find that ¢y 16s1 + 2h + h 2 d  16 16s1 + hd2  16s1d2 = = h h ¢t 2 32h + 16h = 32 + 16h. = h For values of h different from 0, the expressions on the right and left are equivalent and the average speed is 32 + 16h ft>sec. We can now see why the average speed has the limiting value 32 + 16s0d = 32 ft>sec as h approaches 0. Similarly, setting t0 = 2 in Equation (1), the procedure yields ¢y = 64 + 16h ¢t for values of h different from 0. As h gets closer and closer to 0, the average speed at t0 = 2 sec has the limiting value 64 ft> sec.
Average Rates of Change and Secant Lines Given an arbitrary function y = ƒsxd, we calculate the average rate of change of y with respect to x over the interval [x1 , x2] by dividing the change in the value of y, ¢y = ƒsx2 d  ƒsx1 d, by the length ¢x = x2  x1 = h of the interval over which the change occurs.
DEFINITION Average Rate of Change over an Interval The average rate of change of y = ƒsxd with respect to x over the interval [x1 , x2] is ¢y ƒsx2 d  ƒsx1 d ƒsx1 + hd  ƒsx1 d = , = x2  x1 h ¢x
Geometrically, the rate of change of ƒ over [x1, x2] is the slope of the line through the points Psx1, ƒsx1 dd and Qsx2 , ƒsx2 dd (Figure 2.1). In geometry, a line joining two points of a curve is a secant to the curve. Thus, the average rate of change of ƒ from x1 to x2 is identical with the slope of secant PQ. Experimental biologists often want to know the rates at which populations grow under controlled laboratory conditions.
y y f (x) Q(x 2, f (x 2 ))
EXAMPLE 3
Secant
x h 0
x1
x2
FIGURE 2.1 A secant to the graph y = ƒsxd . Its slope is ¢y>¢x , the average rate of change of ƒ over the interval [x1 , x2] .
The Average Growth Rate of a Laboratory Population
Figure 2.2 shows how a population of fruit flies (Drosophila) grew in a 50day experiment. The number of flies was counted at regular intervals, the counted values plotted with respect to time, and the points joined by a smooth curve (colored blue in Figure 2.2). Find the average growth rate from day 23 to day 45.
y
P(x1, f(x1))
h Z 0.
x
There were 150 flies on day 23 and 340 flies on day 45. Thus the number of flies increased by 340  150 = 190 in 45  23 = 22 days. The average rate of change of the population from day 23 to day 45 was
Solution
Average rate of change:
¢p 340  150 190 = = L 8.6 flies>day. 45  23 22 ¢t
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Chapter 2: Limits and Continuity
p 350 Q(45, 340) Number of flies
300 p 190
250
p 8.6 flies/day t t 22
200 P(23, 150)
150 100 50 0
10
20 30 Time (days)
40
t
50
FIGURE 2.2 Growth of a fruit fly population in a controlled experiment. The average rate of change over 22 days is the slope ¢p>¢t of the secant line.
This average is the slope of the secant through the points P and Q on the graph in Figure 2.2. The average rate of change from day 23 to day 45 calculated in Example 3 does not tell us how fast the population was changing on day 23 itself. For that we need to examine time intervals closer to the day in question.
EXAMPLE 4
The Growth Rate on Day 23
How fast was the number of flies in the population of Example 3 growing on day 23? To answer this question, we examine the average rates of change over increasingly short time intervals starting at day 23. In geometric terms, we find these rates by calculating the slopes of secants from P to Q, for a sequence of points Q approaching P along the curve (Figure 2.3). Solution
p
Q (45, 340) (40, 330) (35, 310) (30, 265)
Slope of PQ ≤p/≤t (flies / day) 340 45 330 40 310 35 265 30

150 23 150 23 150 23 150 23
L 8.6 L 10.6
B(35, 350)
350
Q(45, 340)
300 Number of flies
76
250 200 150
P(23, 150)
100
L 13.3
50
L 16.4
0
10 20 30 A(14, 0) Time (days)
40
50
FIGURE 2.3 The positions and slopes of four secants through the point P on the fruit fly graph (Example 4).
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t
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Rates of Change and Limits
77
The values in the table show that the secant slopes rise from 8.6 to 16.4 as the tcoordinate of Q decreases from 45 to 30, and we would expect the slopes to rise slightly higher as t continued on toward 23. Geometrically, the secants rotate about P and seem to approach the red line in the figure, a line that goes through P in the same direction that the curve goes through P. We will see that this line is called the tangent to the curve at P. Since the line appears to pass through the points (14, 0) and (35, 350), it has slope 350  0 = 16.7 flies>day (approximately). 35  14 On day 23 the population was increasing at a rate of about 16.7 flies> day. The rates at which the rock in Example 2 was falling at the instants t = 1 and t = 2 and the rate at which the population in Example 4 was changing on day t = 23 are called instantaneous rates of change. As the examples suggest, we find instantaneous rates as limiting values of average rates. In Example 4, we also pictured the tangent line to the population curve on day 23 as a limiting position of secant lines. Instantaneous rates and tangent lines, intimately connected, appear in many other contexts. To talk about the two constructively, and to understand the connection further, we need to investigate the process by which we determine limiting values, or limits, as we will soon call them.
Limits of Function Values Our examples have suggested the limit idea. Let’s begin with an informal definition of limit, postponing the precise definition until we’ve gained more insight. Let ƒ(x) be defined on an open interval about x0 , except possibly at x0 itself. If ƒ(x) gets arbitrarily close to L (as close to L as we like) for all x sufficiently close to x0 , we say that ƒ approaches the limit L as x approaches x0 , and we write lim ƒsxd = L,
x:x0
which is read “the limit of ƒ(x) as x approaches x0 is L”. Essentially, the definition says that the values of ƒ(x) are close to the number L whenever x is close to x0 (on either side of x0). This definition is “informal” because phrases like arbitrarily close and sufficiently close are imprecise; their meaning depends on the context. To a machinist manufacturing a piston, close may mean within a few thousandths of an inch. To an astronomer studying distant galaxies, close may mean within a few thousand lightyears. The definition is clear enough, however, to enable us to recognize and evaluate limits of specific functions. We will need the precise definition of Section 2.3, however, when we set out to prove theorems about limits.
EXAMPLE 5
Behavior of a Function Near a Point
How does the function ƒsxd =
x2  1 x  1
behave near x = 1? The given formula defines ƒ for all real numbers x except x = 1 (we cannot divide by zero). For any x Z 1, we can simplify the formula by factoring the numerator and canceling common factors:
Solution
ƒsxd =
sx  1dsx + 1d = x + 1 x  1
for
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x Z 1.
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Chapter 2: Limits and Continuity
The graph of ƒ is thus the line y = x + 1 with the point (1, 2) removed. This removed point is shown as a “hole” in Figure 2.4. Even though ƒ(1) is not defined, it is clear that we can make the value of ƒ(x) as close as we want to 2 by choosing x close enough to 1 (Table 2.2).
y
2 2 y f (x) x 1 x 1
1
TABLE 2.2 The closer x gets to 1, the closer ƒ(x) (x 2 1)/(x 1)
seems to get to 2 –1
0
x
1
y
2 yx1 1
–1
0
x
1
FIGURE 2.4 The graph of ƒ is identical with the line y = x + 1 except at x = 1 , where ƒ is not defined (Example 5).
x2 1 x 1, x1
Values of x below and above 1
ƒ(x)
0.9 1.1 0.99 1.01 0.999 1.001 0.999999 1.000001
1.9 2.1 1.99 2.01 1.999 2.001 1.999999 2.000001
x1
We say that ƒ(x) approaches the limit 2 as x approaches 1, and write lim ƒsxd = 2,
x:1
EXAMPLE 6
or
x2  1 = 2. x:1 x  1 lim
The Limit Value Does Not Depend on How the Function Is Defined at x0
The function ƒ in Figure 2.5 has limit 2 as x : 1 even though ƒ is not defined at x = 1. The function g has limit 2 as x : 1 even though 2 Z gs1d. The function h is the only one y
–1
y
y
2
2
2
1
1
1
0
2 (a) f (x) x 1 x 1
1
x
–1
0
1
x
x2 1 , x 1 (b) g(x) x 1 1, x1
–1
0
1
x
(c) h(x) x 1
FIGURE 2.5 The limits of ƒ(x), g(x), and h(x) all equal 2 as x approaches 1. However, only h (x) has the same function value as its limit at x = 1 (Example 6).
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79
whose limit as x : 1 equals its value at x = 1. For h, we have limx:1 hsxd = hs1d. This equality of limit and function value is special, and we return to it in Section 2.6.
y yx
Sometimes limx:x0 ƒsxd can be evaluated by calculating ƒsx0 d. This holds, for example, whenever ƒ(x) is an algebraic combination of polynomials and trigonometric functions for which ƒsx0 d is defined. (We will say more about this in Sections 2.2 and 2.6.)
x0
EXAMPLE 7
x
x0
Finding Limits by Calculating ƒ(x0)
(a) lim s4d = 4 x:2
(a) Identity function
(b)
y
lim s4d = 4
x: 13
(c) lim x = 3 x:3
(d) lim s5x  3d = 10  3 = 7 x:2
yk
k
(e)
x0
0
x
lim
x: 2
3x + 4 6 + 4 2 = = x + 5 2 + 5 3
EXAMPLE 8
The Identity and Constant Functions Have Limits at Every Point
(a) If ƒ is the identity function ƒsxd = x, then for any value of x0 (Figure 2.6a),
(b) Constant function
lim ƒsxd = lim x = x0 .
FIGURE 2.6
x:x0
The functions in Example 8.
x:x0
(b) If ƒ is the constant function ƒsxd = k (function with the constant value k), then for any value of x0 (Figure 2.6b), lim ƒsxd = lim k = k .
x:x0
x:x0
For instance, lim x = 3
x:3
and
lim s4d = lim s4d = 4.
x: 7
x:2
We prove these results in Example 3 in Section 2.3. Some ways that limits can fail to exist are illustrated in Figure 2.7 and described in the next example. y
1
y
x
0
y 1 , x0 y x 0, x 0
0, x 0 y 1, x ⱖ 0
0
1
x
x
0
xⱕ0 0, y 1 sin x , x 0
–1 (a) Unit step function U(x)
FIGURE 2.7
(b) g(x)
None of these functions has a limit as x approaches 0 (Example 9).
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(c) f(x)
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Chapter 2: Limits and Continuity
EXAMPLE 9
A Function May Fail to Have a Limit at a Point in Its Domain
Discuss the behavior of the following functions as x : 0. (a) Usxd = e
0, 1,
x 6 0 x Ú 0
(b) gsxd = L
1 x,
x Z 0
0,
x = 0
(c) ƒsxd = •
0,
x … 0
1 sin x ,
x 7 0
Solution
(a) It jumps: The unit step function U(x) has no limit as x : 0 because its values jump at x = 0. For negative values of x arbitrarily close to zero, Usxd = 0. For positive values of x arbitrarily close to zero, Usxd = 1. There is no single value L approached by U(x) as x : 0 (Figure 2.7a). (b) It grows too large to have a limit: g(x) has no limit as x : 0 because the values of g grow arbitrarily large in absolute value as x : 0 and do not stay close to any real number (Figure 2.7b). (c) It oscillates too much to have a limit: ƒ(x) has no limit as x : 0 because the function’s values oscillate between +1 and 1 in every open interval containing 0. The values do not stay close to any one number as x : 0 (Figure 2.7c).
Using Calculators and Computers to Estimate Limits Tables 2.1 and 2.2 illustrate using a calculator or computer to guess a limit numerically as x gets closer and closer to x0 . That procedure would also be successful for the limits of functions like those in Example 7 (these are continuous functions and we study them in Section 2.6). However, calculators and computers can give false values and misleading impressions for functions that are undefined at a point or fail to have a limit there. The differential calculus will help us know when a calculator or computer is providing strange or ambiguous information about a function’s behavior near some point (see Sections 4.4 and 4.6). For now, we simply need to be attentive to the fact that pitfalls may occur when using computing devices to guess the value of a limit. Here’s one example.
EXAMPLE 10
Guessing a Limit
Guess the value of lim
x:0
2x 2 + 100  10 . x2
Solution Table 2.3 lists values of the function for several values near x = 0. As x approaches 0 through the values ;1, ;0.5, ;0.10, and ;0.01, the function seems to approach the number 0.05. As we take even smaller values of x, ;0.0005, ;0.0001, ;0.00001, and ;0.000001, the function appears to approach the value 0. So what is the answer? Is it 0.05 or 0, or some other value? The calculator/computer values are ambiguous, but the theorems on limits presented in the next section will confirm the correct limit value to be 0.05 A = 120 B . Problems such as these demonstrate the
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2.1 Rates of Change and Limits
TABLE 2.3 Computer values of ƒ(x)
81
2x 2 + 100  10 Near x 0 x2
x
ƒ(x)
;1 ;0.5 ;0.1 ;0.01
0.049876 0.049969 t approaches 0.05? 0.049999 0.050000
;0.0005 ;0.0001 ;0.00001 ;0.000001
0.080000 0.000000 t approaches 0? 0.000000 0.000000
power of mathematical reasoning, once it is developed, over the conclusions we might draw from making a few observations. Both approaches have advantages and disadvantages in revealing nature’s realities.
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EXERCISES 2.1 Limits from Graphs 1. For the function g (x) graphed here, find the following limits or explain why they do not exist. a. lim g sxd
b. lim g sxd
x:1
c. lim g sxd
x: 2
x:3
3. Which of the following statements about the function y = ƒsxd graphed here are true, and which are false? a. lim ƒsxd exists. x:0
b. lim ƒsxd = 0 . x:0
c. lim ƒsxd = 1 .
y
x:0
d. lim ƒsxd = 1 . x:1
y g(x)
e. lim ƒsxd = 0 .
1
x:1
f. lim ƒsxd exists at every point x0 in s 1, 1d . 1
x
3
2
x:x0
y
2. For the function ƒ(t) graphed here, find the following limits or explain why they do not exist. a. lim ƒstd t: 2
b. lim ƒstd
1
y f (x)
c. lim ƒstd
t: 1
t: 0
–1
1
2
x
s –1 s f (t)
1
–1
0
–2
1 –1
t
4. Which of the following statements about the function y = ƒsxd graphed here are true, and which are false? a. lim ƒsxd does not exist. x:2
b. lim ƒsxd = 2 . x:2
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82
Chapter 2: Limits and Continuity 13. Let Gsxd = sx + 6d>sx 2 + 4x  12d .
c. lim ƒsxd does not exist. x:1
d. lim ƒsxd exists at every point x0 in s 1, 1d . x:x0
e. lim ƒsxd exists at every point x0 in (1, 3). x:x0
y
y f (x)
1 –1
1
2
3
x
–1 –2
Existence of Limits In Exercises 5 and 6, explain why the limits do not exist. x 1 6. lim x: 1 x  1 ƒxƒ 7. Suppose that a function ƒ(x) is defined for all real values of x except x = x0 . Can anything be said about the existence of limx:x0 ƒsxd ? Give reasons for your answer. 5. lim
x: 0
8. Suppose that a function ƒ(x) is defined for all x in [1, 1] . Can anything be said about the existence of limx:0 ƒsxd ? Give reasons for your answer. 9. If limx:1 ƒsxd = 5 , must ƒ be defined at x = 1 ? If it is, must ƒs1d = 5 ? Can we conclude anything about the values of ƒ at x = 1 ? Explain. 10. If ƒs1d = 5 , must limx:1 ƒsxd exist? If it does, then must limx:1 ƒsxd = 5 ? Can we conclude anything about limx:1 ƒsxd ? Explain.
Estimating Limits T You will find a graphing calculator useful for Exercises 11–20. 11. Let ƒsxd = sx 2  9d>sx + 3d . a. Make a table of the values of ƒ at the points x = 3.1, 3.01, 3.001 , and so on as far as your calculator can go. Then estimate limx:3 ƒsxd . What estimate do you arrive at if you evaluate ƒ at x = 2.9, 2.99, 2.999, Á instead? b. Support your conclusions in part (a) by graphing ƒ near x0 = 3 and using Zoom and Trace to estimate yvalues on the graph as x : 3 . c. Find limx:3 ƒsxd algebraically, as in Example 5.
12. Let g sxd = sx 2  2d> A x  22 B .
a. Make a table of the values of g at the points x = 1.4, 1.41, 1.414 , and so on through successive decimal approximations of 22 . Estimate limx:22 g sxd .
b. Support your conclusion in part (a) by graphing g near x0 = 22 and using Zoom and Trace to estimate yvalues on the graph as x : 22 . c. Find limx:22 g sxd algebraically.
a. Make a table of the values of G at x = 5.9, 5.99, 5.999, and so on. Then estimate limx:6 Gsxd . What estimate do you arrive at if you evaluate G at x = 6.1, 6.01, 6.001, Á instead? b. Support your conclusions in part (a) by graphing G and using Zoom and Trace to estimate yvalues on the graph as x : 6 . c. Find limx:6 Gsxd algebraically. 14. Let hsxd = sx 2  2x  3d>sx 2  4x + 3d . a. Make a table of the values of h at x = 2.9, 2.99, 2.999, and so on. Then estimate limx:3 hsxd . What estimate do you arrive at if you evaluate h at x = 3.1, 3.01, 3.001, Á instead? b. Support your conclusions in part (a) by graphing h near x0 = 3 and using Zoom and Trace to estimate yvalues on the graph as x : 3 . c. Find limx:3 hsxd algebraically. 15. Let ƒsxd = sx 2  1d>s ƒ x ƒ  1d . a. Make tables of the values of ƒ at values of x that approach x0 = 1 from above and below. Then estimate limx:1 ƒsxd . b. Support your conclusion in part (a) by graphing ƒ near x0 = 1 and using Zoom and Trace to estimate yvalues on the graph as x : 1 . c. Find limx:1 ƒsxd algebraically. 16. Let Fsxd = sx 2 + 3x + 2d>s2  ƒ x ƒ d . a. Make tables of values of F at values of x that approach x0 = 2 from above and below. Then estimate limx:2 Fsxd . b. Support your conclusion in part (a) by graphing F near x0 = 2 and using Zoom and Trace to estimate yvalues on the graph as x : 2 . c. Find limx:2 Fsxd algebraically. 17. Let g sud = ssin ud>u . a. Make a table of the values of g at values of u that approach u0 = 0 from above and below. Then estimate limu:0 g sud . b. Support your conclusion in part (a) by graphing g near u0 = 0 . 18. Let Gstd = s1  cos td>t 2 . a. Make tables of values of G at values of t that approach t0 = 0 from above and below. Then estimate limt:0 Gstd . b. Support your conclusion in part (a) by graphing G near t0 = 0 . 19. Let ƒsxd = x 1>s1  xd . a. Make tables of values of ƒ at values of x that approach x0 = 1 from above and below. Does ƒ appear to have a limit as x : 1 ? If so, what is it? If not, why not? b. Support your conclusions in part (a) by graphing ƒ near x0 = 1 .
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2.1 Rates of Change and Limits 20. Let ƒsxd = s3x  1d>x . a. Make tables of values of ƒ at values of x that approach x0 = 0 from above and below. Does ƒ appear to have a limit as x : 0 ? If so, what is it? If not, why not? b. Support your conclusions in part (a) by graphing ƒ near x0 = 0 .
Limits by Substitution In Exercises 21–28, find the limits by substitution. Support your answers with a computer or calculator if available.
36. The accompanying figure shows the plot of distance fallen versus time for an object that fell from the lunar landing module a distance 80 m to the surface of the moon. a. Estimate the slopes of the secants PQ1 , PQ2 , PQ3 , and PQ4 , arranging them in a table like the one in Figure 2.3. b. About how fast was the object going when it hit the surface? y
x: 0
lim s3x  1d
24. lim
x:1>3
x: 1
1 s3x  1d
80
3x 2 2x  1 cos x 28. lim x: p 1  p
25. lim 3xs2x  1d
26. lim
x: 1
27.
b. Then estimate the Cobra’s speed at time t = 20 sec .
Distance fallen (m)
x:2
23.
a. Estimate the slopes of secants PQ1 , PQ2 , PQ3 , and PQ4 , arranging them in order in a table like the one in Figure 2.3. What are the appropriate units for these slopes?
22. lim 2x
21. lim 2x
x: 1
lim x sin x
x:p>2
Average Rates of Change 29. ƒsxd = x 3 + 1 ;
Q3 Q2
40 Q1
20 0
P
Q4
60
In Exercises 29–34, find the average rate of change of the function over the given interval or intervals.
5 Elapsed time (sec)
10
t
T 37. The profits of a small company for each of the first five years of its operation are given in the following table:
b. [1, 1]
a. [2, 3]
83
2
30. g sxd = x ;
Year
Profit in $1000s
1990 1991 1992 1993 1994
6 27 62 111 174
a. [1, 1] b. [2, 0] 31. hstd = cot t ; a. [p>4, 3p>4] b. [p>6, p>2] 32. g std = 2 + cos t ; a. [0, p]
b. [p, p]
33. Rsud = 24u + 1;
[0, 2]
34. Psud = u3  4 u2 + 5u;
a. Plot points representing the profit as a function of year, and join them by as smooth a curve as you can.
[1, 2]
35. A Ford Mustang Cobra’s speed The accompanying figure shows the timetodistance graph for a 1994 Ford Mustang Cobra accelerating from a standstill. s P
650 600
Q3
Distance (m)
400
a. Find the average rate of change of F(x) over the intervals [1, x] for each x Z 1 in your table.
Q2
b. Extending the table if necessary, try to determine the rate of change of F(x) at x = 1 .
300 200
Q1
T 39. Let g sxd = 2x for x Ú 0 .
100 0
c. Use your graph to estimate the rate at which the profits were changing in 1992. T 38. Make a table of values for the function Fsxd = sx + 2d>sx  2d at the points x = 1.2, x = 11>10, x = 101>100, x = 1001>1000, x = 10001>10000 , and x = 1 .
Q4
500
b. What is the average rate of increase of the profits between 1992 and 1994?
5 10 15 20 Elapsed time (sec)
t
a. Find the average rate of change of g (x) with respect to x over the intervals [1, 2], [1, 1.5] and [1, 1 + h] . b. Make a table of values of the average rate of change of g with respect to x over the interval [1, 1 + h] for some values of h
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Chapter 2: Limits and Continuity approaching zero, say h = 0.1, 0.01, 0.001, 0.0001, 0.00001 , and 0.000001.
d. Calculate the limit as T approaches 2 of the average rate of change of ƒ with respect to t over the interval from 2 to T. You will have to do some algebra before you can substitute T = 2 .
c. What does your table indicate is the rate of change of g(x) with respect to x at x = 1 ?
COMPUTER EXPLORATIONS
d. Calculate the limit as h approaches zero of the average rate of change of g (x) with respect to x over the interval [1, 1 + h] .
Graphical Estimates of Limits
T 40. Let ƒstd = 1>t for t Z 0 . a. Find the average rate of change of ƒ with respect to t over the intervals (i) from t = 2 to t = 3 , and (ii) from t = 2 to t = T. b. Make a table of values of the average rate of change of ƒ with respect to t over the interval [2, T], for some values of T approaching 2, say T = 2.1, 2.01, 2.001, 2.0001, 2.00001 , and 2.000001. c. What does your table indicate is the rate of change of ƒ with respect to t at t = 2 ?
In Exercises 41–46, use a CAS to perform the following steps: a. Plot the function near the point x0 being approached. b. From your plot guess the value of the limit. 41. lim
x 4  16 x  2
42. lim
43. lim
21 + x  1 x
44. lim
45. lim
1  cos x x sin x
x:2
x: 1
3
x:0
x:0
x 3  x 2  5x  3 sx + 1d2 x2  9
2x 2 + 7  4 2x 2 46. lim x:0 3  3 cos x x:3
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Chapter 2: Limits and Continuity
2.2 HISTORICAL ESSAY* Limits
Calculating Limits Using the Limit Laws In Section 2.1 we used graphs and calculators to guess the values of limits. This section presents theorems for calculating limits. The first three let us build on the results of Example 8 in the preceding section to find limits of polynomials, rational functions, and powers. The fourth and fifth prepare for calculations later in the text.
The Limit Laws The next theorem tells how to calculate limits of functions that are arithmetic combinations of functions whose limits we already know.
THEOREM 1 Limit Laws If L, M, c and k are real numbers and lim ƒsxd = L
x:c
1. Sum Rule:
and
lim gsxd = M,
x:c
then
lim sƒsxd + gsxdd = L + M
x:c
The limit of the sum of two functions is the sum of their limits. 2. Difference Rule:
lim sƒsxd  gsxdd = L  M
x:c
The limit of the difference of two functions is the difference of their limits. 3. Product Rule:
lim sƒsxd # gsxdd = L # M
x:c
The limit of a product of two functions is the product of their limits.
To learn more about the historical figures and the development of the major elements and topics of calculus, visit www.awbc.com/thomas.
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2.2
Calculating Limits Using the Limit Laws
85
lim sk # ƒsxdd = k # L
4. Constant Multiple Rule:
x:c
The limit of a constant times a function is the constant times the limit of the function. ƒsxd L 5. Quotient Rule: lim = , M Z 0 M x:c gsxd The limit of a quotient of two functions is the quotient of their limits, provided the limit of the denominator is not zero. 6. Power Rule: If r and s are integers with no common factor and s Z 0, then lim sƒsxddr>s = L r>s
x:c
provided that L r>s is a real number. (If s is even, we assume that L 7 0.) The limit of a rational power of a function is that power of the limit of the function, provided the latter is a real number.
It is easy to convince ourselves that the properties in Theorem 1 are true (although these intuitive arguments do not constitute proofs). If x is sufficiently close to c, then ƒ(x) is close to L and g(x) is close to M, from our informal definition of a limit. It is then reasonable that ƒsxd + gsxd is close to L + M; ƒsxd  gsxd is close to L  M; ƒ(x)g(x) is close to LM; kƒ(x) is close to kL; and that ƒ(x)>g(x) is close to L>M if M is not zero. We prove the Sum Rule in Section 2.3, based on a precise definition of limit. Rules 2–5 are proved in Appendix 2. Rule 6 is proved in more advanced texts. Here are some examples of how Theorem 1 can be used to find limits of polynomial and rational functions.
EXAMPLE 1
Using the Limit Laws
Use the observations limx:c k = k and limx:c x = c (Example 8 in Section 2.1) and the properties of limits to find the following limits. (a) lim sx 3 + 4x 2  3d x:c
x4 + x2  1 x:c x2 + 5
(b) lim
(c) lim 24x 2  3 x: 2
Solution
(a) lim sx 3 + 4x 2  3d = lim x 3 + lim 4x 2  lim 3 x:c
x:c
x:c
x:c
= c 3 + 4c 2  3 4
x4 + x2  1 = x:c x2 + 5
(b) lim
Sum and Difference Rules Product and Multiple Rules
2
lim sx + x  1d
x:c
Quotient Rule
lim sx 2 + 5d
x:c
lim x 4 + lim x 2  lim 1
=
x:c
x:c
x:c
lim x 2 + lim 5
x:c
c4 + c2  1 = c2 + 5
Sum and Difference Rules
x:c
Power or Product Rule
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Chapter 2: Limits and Continuity
(c)
lim 24x 2  3 = 2 lim s4x 2  3d
x: 2
x: 2
Power Rule with r>s = 12
= 2 lim 4x 2  lim 3
Difference Rule
= 24s 2d2  3 = 216  3 = 213
Product and Multiple Rules
x: 2
x: 2
Two consequences of Theorem 1 further simplify the task of calculating limits of polynomials and rational functions. To evaluate the limit of a polynomial function as x approaches c, merely substitute c for x in the formula for the function. To evaluate the limit of a rational function as x approaches a point c at which the denominator is not zero, substitute c for x in the formula for the function. (See Examples 1a and 1b.)
THEOREM 2 Limits of Polynomials Can Be Found by Substitution If Psxd = an x n + an  1 x n  1 + Á + a0 , then lim Psxd = Pscd = an c n + an  1 c n  1 + Á + a 0 .
x:c
THEOREM 3
Limits of Rational Functions Can Be Found by Substitution If the Limit of the Denominator Is Not Zero If P(x) and Q(x) are polynomials and Qscd Z 0, then lim
x:c
EXAMPLE 2
Psxd Pscd = . Qsxd Qscd
Limit of a Rational Function s 1d3 + 4s 1d2  3 x 3 + 4x 2  3 0 = = = 0 6 x: 1 x2 + 5 s 1d2 + 5 lim
This result is similar to the second limit in Example 1 with c = 1, now done in one step.
Identifying Common Factors It can be shown that if Q(x) is a polynomial and Qscd = 0 , then sx  cd is a factor of Q(x). Thus, if the numerator and denominator of a rational function of x are both zero at x = c , they have sx  cd as a common factor.
Eliminating Zero Denominators Algebraically Theorem 3 applies only if the denominator of the rational function is not zero at the limit point c. If the denominator is zero, canceling common factors in the numerator and denominator may reduce the fraction to one whose denominator is no longer zero at c. If this happens, we can find the limit by substitution in the simplified fraction.
EXAMPLE 3
Canceling a Common Factor
Evaluate x2 + x  2 . x:1 x2  x lim
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2.2
We cannot substitute x = 1 because it makes the denominator zero. We test the numerator to see if it, too, is zero at x = 1. It is, so it has a factor of sx  1d in common with the denominator. Canceling the sx  1d’s gives a simpler fraction with the same values as the original for x Z 1:
2 x2 y x x2 x (1, 3)
sx  1dsx + 2d x + 2 x2 + x  2 = = x , 2 xsx 1d x  x –2
0
87
Solution
y
3
Calculating Limits Using the Limit Laws
x
1
if x Z 1.
Using the simpler fraction, we find the limit of these values as x : 1 by substitution: x2 + x  2 x + 2 1 + 2 = lim x = = 3. 1 x:1 x:1 x2  x lim
(a) y
See Figure 2.8.
yx2 x (1, 3)
3
EXAMPLE 4
Creating and Canceling a Common Factor
Evaluate
–2
0
1
lim
x
x:0
2x 2 + 100  10 . x2
This is the limit we considered in Example 10 of the preceding section. We cannot substitute x = 0, and the numerator and denominator have no obvious common factors. We can create a common factor by multiplying both numerator and denominator by the expression 2x 2 + 100 + 10 (obtained by changing the sign after the square root). The preliminary algebra rationalizes the numerator:
Solution (b)
FIGURE 2.8 The graph of ƒsxd = sx 2 + x  2d>sx 2  xd in part (a) is the same as the graph of g sxd = sx + 2d>x in part (b) except at x = 1 , where ƒ is undefined. The functions have the same limit as x : 1 (Example 3).
2x 2 + 100  10 2x 2 + 100  10 # 2x 2 + 100 + 10 = 2 x x2 2x 2 + 100 + 10 2 x + 100  100 = 2 x A 2x 2 + 100 + 10 B x2
x 2 A 2x 2 + 100 + 10 B 1 = . 2 2x + 100 + 10 =
Common factor x2 Cancel x2 for x 0
Therefore, lim
x:0
2x 2 + 100  10 1 = lim x:0 2x 2 + 100 + 10 x2 1 = 2 20 + 100 + 10 1 = = 0.05. 20
Denominator not 0 at x 0; substitute
This calculation provides the correct answer to the ambiguous computer results in Example 10 of the preceding section.
The Sandwich Theorem The following theorem will enable us to calculate a variety of limits in subsequent chapters. It is called the Sandwich Theorem because it refers to a function ƒ whose values are Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
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Chapter 2: Limits and Continuity
sandwiched between the values of two other functions g and h that have the same limit L at a point c. Being trapped between the values of two functions that approach L, the values of ƒ must also approach L (Figure 2.9). You will find a proof in Appendix 2.
y h f L
g x
c
0
THEOREM 4 The Sandwich Theorem Suppose that gsxd … ƒsxd … hsxd for all x in some open interval containing c, except possibly at x = c itself. Suppose also that lim gsxd = lim hsxd = L.
x:c
FIGURE 2.9 The graph of ƒ is sandwiched between the graphs of g and h.
y
0
EXAMPLE 5
y u(x)
1
–1
The Sandwich Theorem is sometimes called the Squeeze Theorem or the Pinching Theorem.
2 y1 x 2
2
Applying the Sandwich Theorem
Given that
2 y1 x 4
1
x:c
Then limx:c ƒsxd = L.
x
1 
for all x Z 0,
find limx:0 usxd, no matter how complicated u is. Solution
FIGURE 2.10 Any function u(x) whose graph lies in the region between y = 1 + sx 2>2d and y = 1  sx 2>4d has limit 1 as x : 0 (Example 5).
x2 x2 … usxd … 1 + 4 2
Since lim s1  sx 2>4dd = 1
and
x:0
lim s1 + sx 2>2dd = 1,
x:0
the Sandwich Theorem implies that limx:0 usxd = 1 (Figure 2.10).
EXAMPLE 6
More Applications of the Sandwich Theorem
(a) (Figure 2.11a). It follows from the definition of sin u that  ƒ u ƒ … sin u … ƒ u ƒ for all u, and since limu:0 s  ƒ u ƒ d = limu:0 ƒ u ƒ = 0, we have lim sin u = 0 .
u:0
y y 1
y
2 y sin
– –1
y –
y
1
–2
–1
0
y 1 cos 1
2
(b)
(a)
FIGURE 2.11 The Sandwich Theorem confirms that (a) limu:0 sin u = 0 and (b) limu:0 s1  cos ud = 0 (Example 6).
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2.2 Calculating Limits Using the Limit Laws
89
(b) (Figure 2.11b). From the definition of cos u, 0 … 1  cos u … ƒ u ƒ for all u, and we have limu:0 s1  cos ud = 0 or lim cos u = 1 .
u:0
(c) For any function ƒ(x), if limx:c ƒ ƒsxd ƒ = 0, then limx:c ƒsxd = 0. The argument:  ƒ ƒsxd ƒ … ƒsxd … ƒ ƒsxd ƒ and  ƒ ƒsxd ƒ and ƒ ƒsxd ƒ have limit 0 as x : c. Another important property of limits is given by the next theorem. A proof is given in the next section.
THEOREM 5 If ƒsxd … gsxd for all x in some open interval containing c, except possibly at x = c itself, and the limits of ƒ and g both exist as x approaches c, then lim ƒsxd … lim gsxd.
x:c
x:c
The assertion resulting from replacing the less than or equal to … inequality by the strict 6 inequality in Theorem 5 is false. Figure 2.11a shows that for u Z 0,  ƒ u ƒ 6 sin u 6 ƒ u ƒ , but in the limit as u : 0, equality holds.
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2.2 Calculating Limits Using the Limit Laws
EXERCISES 2.2 Limit Calculations
Find the limits in Exercises 19–36.
Find the limits in Exercises 1–18.
19. lim
1. lim s2x + 5d
2. lim s10  3xd
3. lim s x 2 + 5x  2d
4. lim sx 3  2x 2 + 4x + 8d
x  5 x 2  25 x 2 + 3x  10 21. lim x + 5 x: 5
5. lim 8st  5dst  7d
6. lim 3ss2s  1d
23. lim
x + 3 7. lim x:2 x + 6
4 8. lim x: 5 x  7
x: 7 x:2 t:6
y2 9. lim y: 5 5  y
4>3
13. lim s5  yd y: 3
17. lim
h: 0
s: 2>3
y: 2
x: 1
h: 0
x: 2
10. lim
11. lim 3s2x  1d2
15. lim
x: 12
3 23h + 1 + 1 23h + 1  1 h
x:5
y 2 + 5y + 6
12. lim sx + 3d1984 x: 4
1>3
14. lim s2z  8d
x: 3
t2 + t  2 t: 1 t2  1
24. lim
2x  4 x 3 + 2x 2
26. lim
25. lim
y + 2
x + 3 x 2 + 4x + 3 x 2  7x + 10 22. lim x  2 x:2 20. lim
x: 2
27. lim
u:1
u4  1 u3  1
2x  3 29. lim x:9 x  9
t: 1
y:0
h :0
18. lim
h :0
5 25h + 4 + 2 25h + 4  2 h
31. lim
x:1
x  1 2x + 3  2 2x + 12  4 x  2
3y 4  16y 2 y3  8 y4  16
28. lim
y: 2
30. lim
x:4
z: 0
16. lim
t 2 + 3t + 2 t2  t  2 5y 3 + 8y 2
4x  x 2 2  2x
32. lim
x: 1
2
33. lim
x:2
34. lim
x: 2
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2x 2 + 8  3 x + 1 x + 2 2x + 5  3 2
89
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Chapter 2: Limits and Continuity 2  2x 2  5 x + 3 x: 3
35. lim
42. Suppose that limx:2 psxd = 4, limx:2 r sxd = 0 , and limx:2 ssxd = 3 . Find
4  x
36. lim
5  2x + 9 2
x: 4
a. lim spsxd + r sxd + ssxdd x: 2
b.
Using Limit Rules 37. Suppose limx:0 ƒsxd = 1 and limx:0 g sxd = 5 . Name the rules in Theorem 1 that are used to accomplish steps (a), (b), and (c) of the following calculation. lim
x: 0
lim s2ƒsxd  g sxdd
2ƒsxd  g sxd sƒsxd + 7d2>3
x: 0
=
(a)
lim sƒsxd + 7d2>3
lim psxd # r sxd # ssxd
x: 2
c. lim s 4psxd + 5r sxdd>ssxd x: 2
Limits of Average Rates of Change Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form
x: 0
lim
lim 2ƒsxd  lim g sxd
x: 0
x:0
A lim A ƒsxd + 7 B B 2>3
=
h:0
(b)
x: 0
2 lim ƒsxd  lim g sxd x: 0
=
x:0
A lim ƒ(x) + lim 7 B x: 0
s1 + 7d2>3
lim 25hsxd
x:1
25hsxd x: 1 = psxds4  rsxdd lim spsxds4  rsxddd
(a)
A lim p(x) B A lim A 4  r(x) B B
(b)
x: 1
x: 1
A lim p(x) B A lim 4  lim r (x) B x: 1
=
x: 1
x:1
2s5ds5d 5 = 2 s1ds4  2d
39. Suppose limx:c ƒsxd = 5 and limx:c g sxd = 2 . Find a. lim ƒsxdg sxd
b. lim 2ƒsxdg sxd
c. lim sƒsxd + 3g sxdd
ƒsxd d. lim x: c ƒsxd  g sxd
x:c
x:c
x: c
40. Suppose limx:4 ƒsxd = 0 and limx:4 g sxd = 3 . Find a. lim sg sxd + 3d
b. lim xƒsxd
c. lim sg sxdd2
d. lim
x:4
x:4
x: 4
x: 4
g sxd ƒsxd  1
41. Suppose limx:b ƒsxd = 7 and limx:b g sxd = 3 . Find a. lim sƒsxd + g sxdd
b. lim ƒsxd # g sxd
c. lim 4g sxd
d. lim ƒsxd>g sxd
x:b x:b
x: b x: b
x = 0
x sin x x2 6 6 1 6 2  2 cos x
hold for all values of x close to zero. What, if anything, does this tell you about
25 lim hsxd =
x = 2
51. a. It can be shown that the inequalities 1 
x: 1
46. ƒsxd = 1>x,
48. ƒsxd = 23x + 1,
x = 7
50. If 2  x 2 … g sxd … 2 cos x for all x, find limx:0 g sxd .
2 lim 5hsxd x: 1
x = 2
x = 2
49. If 25  2x 2 … ƒsxd … 25  x 2 for 1 … x … 1 , find limx:0 ƒsxd .
x: 1
=
44. ƒsxd = x 2,
x = 1
Using the Sandwich Theorem
38. Let limx:1 hsxd = 5, limx:1 psxd = 1 , and limx:1 r sxd = 2 . Name the rules in Theorem 1 that are used to accomplish steps (a), (b), and (c) of the following calculation. lim
43. ƒsxd = x 2, 47. ƒsxd = 2x,
7 4
=
occur frequently in calculus. In Exercises 43–48, evaluate this limit for the given value of x and function ƒ. 45. ƒsxd = 3x  4,
x:0
s2ds1d  s 5d =
2>3
(c)
ƒsx + hd  ƒsxd h
lim
x:0
(c)
x sin x ? 2  2 cos x
Give reasons for your answer. T b. Graph
y = 1  sx 2>6d, y = sx sin xd>s2  2 cos xd, and y = 1
together for 2 … x … 2 . Comment on the behavior of the graphs as x : 0 . 52. a. Suppose that the inequalities 1  cos x x2 1 1 6 6 2 24 2 x2 hold for values of x close to zero. (They do, as you will see in Section 11.9.) What, if anything, does this tell you about lim
x:0
1  cos x ? x2
Give reasons for your answer.
b. Graph the equations y = s1>2d  sx 2>24d, y = s1  cos xd>x 2 , and y = 1>2 together for 2 … x … 2 . Comment on the behavior of the graphs as x : 0 .
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2.2 Calculating Limits Using the Limit Laws
Theory and Examples 4
2
2
4
53. If x … ƒsxd … x for x in [1, 1] and x … ƒsxd … x for x 6 1 and x 7 1 , at what points c do you automatically know limx:c ƒsxd ? What can you say about the value of the limit at these points? 54. Suppose that g sxd … ƒsxd … hsxd for all x Z 2 and suppose that lim g sxd = lim hsxd = 5 .
x: 2
x: 2
Can we conclude anything about the values of ƒ, g, and h at x = 2 ? Could ƒs2d = 0 ? Could limx:2 ƒsxd = 0 ? Give reasons for your answers. ƒsxd  5 55. If lim = 1 , find lim ƒsxd . x:4 x  2 x: 4 56. If lim
x: 2
ƒsxd x2
= 1 , find
a. lim ƒsxd x: 2
b.
lim
x: 2
91
ƒsxd x
ƒsxd  5 = 3 , find lim ƒsxd . x:2 x  2 x:2 ƒsxd  5 b. If lim = 4 , find lim ƒsxd . x:2 x  2 x:2 ƒsxd 58. If lim 2 = 1 , find x:0 x ƒsxd a. lim ƒsxd b. lim x x:0 x:0 57. a. If lim
T 59. a. Graph g sxd = x sin s1>xd to estimate limx:0 g sxd , zooming in on the origin as necessary. b. Confirm your estimate in part (a) with a proof. 2 3 T 60. a. Graph hsxd = x cos s1>x d to estimate limx:0 hsxd , zooming in on the origin as necessary.
b. Confirm your estimate in part (a) with a proof.
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2.3 The Precise Definition of a Limit
2.3
91
The Precise Definition of a Limit Now that we have gained some insight into the limit concept, working intuitively with the informal definition, we turn our attention to its precise definition. We replace vague phrases like “gets arbitrarily close to” in the informal definition with specific conditions that can be applied to any particular example. With a precise definition we will be able to prove conclusively the limit properties given in the preceding section, and we can establish other particular limits important to the study of calculus. To show that the limit of ƒ(x) as x : x0 equals the number L, we need to show that the gap between ƒ(x) and L can be made “as small as we choose” if x is kept “close enough” to x0 . Let us see what this would require if we specified the size of the gap between ƒ(x) and L.
EXAMPLE 1
Consider the function y = 2x  1 near x0 = 4. Intuitively it is clear that y is close to 7 when x is close to 4, so limx:4 s2x  1d = 7. However, how close to x0 = 4 does x have to be so that y = 2x  1 differs from 7 by, say, less than 2 units?
y y 2x 1 Upper bound: y9
9 To satisfy 7 this 5
We are asked: For what values of x is ƒ y  7 ƒ 6 2? To find the answer we first express ƒ y  7 ƒ in terms of x:
Solution
ƒ y  7 ƒ = ƒ s2x  1d  7 ƒ = ƒ 2x  8 ƒ . Lower bound: y5
3 4 5
0
A Linear Function
Restrict to this
FIGURE 2.12 Keeping x within 1 unit of x0 = 4 will keep y within 2 units of y0 = 7 (Example 1).
x
The question then becomes: what values of x satisfy the inequality ƒ 2x  8 ƒ 6 2? To find out, we solve the inequality: ƒ 2x  8 ƒ 2 6 3 1
6 6 6 6 6
2 2x  8 6 2 2x 6 10 x 6 5 x  4 6 1.
Keeping x within 1 unit of x0 = 4 will keep y within 2 units of y0 = 7 (Figure 2.12).
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Chapter 2: Limits and Continuity
In the previous example we determined how close x must be to a particular value x0 to ensure that the outputs ƒ(x) of some function lie within a prescribed interval about a limit value L. To show that the limit of ƒ(x) as x : x0 actually equals L, we must be able to show that the gap between ƒ(x) and L can be made less than any prescribed error, no matter how small, by holding x close enough to x0 .
y
L
1 10
f(x) f(x) lies in here
L L
Definition of Limit
1 10
for all x x0 in here
x
x0
0
x0
x0
Suppose we are watching the values of a function ƒ(x) as x approaches x0 (without taking on the value of x0 itself). Certainly we want to be able to say that ƒ(x) stays within onetenth of a unit of L as soon as x stays within some distance d of x0 (Figure 2.13). But that in itself is not enough, because as x continues on its course toward x0 , what is to prevent ƒ(x) from jittering about within the interval from L  (1>10) to L + (1>10) without tending toward L? We can be told that the error can be no more than 1>100 or 1>1000 or 1>100,000. Each time, we find a new dinterval about x0 so that keeping x within that interval satisfies the new error tolerance. And each time the possibility exists that ƒ(x) jitters away from L at some stage. The figures on the next page illustrate the problem. You can think of this as a quarrel between a skeptic and a scholar. The skeptic presents Pchallenges to prove that the limit does not exist or, more precisely, that there is room for doubt, and the scholar answers every challenge with a dinterval around x0 . How do we stop this seemingly endless series of challenges and responses? By proving that for every error tolerance P that the challenger can produce, we can find, calculate, or conjure a matching distance d that keeps x “close enough” to x0 to keep ƒ(x) within that tolerance of L (Figure 2.14). This leads us to the precise definition of a limit.
x
FIGURE 2.13 How should we define d 7 0 so that keeping x within the interval sx0  d, x0 + dd will keep ƒ(x) 1 1 within the interval aL ,L + b? 10 10
y
L L
f(x)
f(x) lies in here
DEFINITION Limit of a Function Let ƒ(x) be defined on an open interval about x0 , except possibly at x0 itself. We say that the limit of ƒ(x) as x approaches x0 is the number L, and write
L
lim ƒsxd = L,
x:x0
for all x x 0 in here x 0
x0
if, for every number P 7 0, there exists a corresponding number d 7 0 such that for all x,
x x0
x0
FIGURE 2.14 The relation of d and P in the definition of limit.
0 6 ƒ x  x0 ƒ 6 d
Q
ƒ ƒsxd  L ƒ 6 P.
One way to think about the definition is to suppose we are machining a generator shaft to a close tolerance. We may try for diameter L, but since nothing is perfect, we must be satisfied with a diameter ƒ(x) somewhere between L  P and L + P. The d is the measure of how accurate our control setting for x must be to guarantee this degree of accuracy in the diameter of the shaft. Notice that as the tolerance for error becomes stricter, we may have to adjust d. That is, the value of d, how tight our control setting must be, depends on the value of P, the error tolerance.
Examples: Testing the Definition The formal definition of limit does not tell how to find the limit of a function, but it enables us to verify that a suspected limit is correct. The following examples show how the
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2.3
y
L
y f (x) L
1 100
L
L 1 100
L
1 10
L x
x0
0
1 10 x x0 x 0 1/10 x 0 1/10 Response: x x 0 1/10 (a number)
1 100 L 1 L 100 L
y y f (x)
y f (x)
1 L 1000
1 L 1000
L
L
1 1000
1 L 1000
x0
0
x
Response: x x 0 1/1000
y
y y f (x) 1 L 100,000
L
L
1 L 100,000
1 L 100,000
x0 New challenge: 1 100,000
y y f (x)
1 100,000
0
x
x0
0
New challenge: 1 1000
L
0
New challenge: Make f (x) – L 1 100
y
L
x x0 x 0 1/100 x 0 1/100 Response: x x 0 1/100
x
x0
0
0
The challenge: Make f (x) – L 1 10
y f (x)
y f (x) L
93
y
y f (x)
1 10
L L
y
y
1 10
The Precise Definition of a Limit
x 0
y f (x)
L
x0
Response: x x 0 1/100,000
x
0
x0
x
New challenge: ...
definition can be used to verify limit statements for specific functions. (The first two examples correspond to parts of Examples 7 and 8 in Section 2.1.) However, the real purpose of the definition is not to do calculations like this, but rather to prove general theorems so that the calculation of specific limits can be simplified.
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Chapter 2: Limits and Continuity
y
EXAMPLE 2
y 5x 3
Testing the Definition
Show that
2
lim s5x  3d = 2.
x:1
2
Solution Set x0 = 1, ƒsxd = 5x  3, and L = 2 in the definition of limit. For any given P 7 0, we have to find a suitable d 7 0 so that if x Z 1 and x is within distance d of x0 = 1, that is, whenever
2 1 1 1 5 5
0
x
0 6 ƒ x  1 ƒ 6 d, it is true that ƒ(x) is within distance P of L = 2, so ƒ ƒsxd  2 ƒ 6 P. We find d by working backward from the Pinequality:
–3
ƒ s5x  3d  2 ƒ = ƒ 5x  5 ƒ 6 P 5ƒx  1ƒ 6 P ƒ x  1 ƒ 6 P>5.
NOT TO SCALE
FIGURE 2.15 If ƒsxd = 5x  3 , then 0 6 ƒ x  1 ƒ 6 P>5 guarantees that ƒ ƒsxd  2 ƒ 6 P (Example 2).
Thus, we can take d = P>5 (Figure 2.15). If 0 6 ƒ x  1 ƒ 6 d = P>5, then ƒ s5x  3d  2 ƒ = ƒ 5x  5 ƒ = 5 ƒ x  1 ƒ 6 5sP>5d = P, which proves that limx:1s5x  3d = 2. The value of d = P>5 is not the only value that will make 0 6 ƒ x  1 ƒ 6 d imply ƒ 5x  5 ƒ 6 P. Any smaller positive d will do as well. The definition does not ask for a “best” positive d, just one that will work.
EXAMPLE 3
Limits of the Identity and Constant Functions
y
Prove:
yx x0
(a) lim x = x0 x:x0
x0 x0 x0
x:x0
(k constant).
Solution
(a) Let P 7 0 be given. We must find d 7 0 such that for all x
x0
0
(b) lim k = k
0 6 ƒ x  x0 ƒ 6 d x0 x0 x0
x
FIGURE 2.16 For the function ƒsxd = x , we find that 0 6 ƒ x  x0 ƒ 6 d will guarantee ƒ ƒsxd  x0 ƒ 6 P whenever d … P (Example 3a).
implies
ƒ x  x0 ƒ 6 P.
The implication will hold if d equals P or any smaller positive number (Figure 2.16). This proves that limx:x0 x = x0 . (b) Let P 7 0 be given. We must find d 7 0 such that for all x 0 6 ƒ x  x0 ƒ 6 d
implies
ƒ k  k ƒ 6 P.
Since k  k = 0, we can use any positive number for d and the implication will hold (Figure 2.17). This proves that limx:x0 k = k.
Finding Deltas Algebraically for Given Epsilons In Examples 2 and 3, the interval of values about x0 for which ƒ ƒsxd  L ƒ was less than P was symmetric about x0 and we could take d to be half the length of that interval. When
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2.3
such symmetry is absent, as it usually is, we can take d to be the distance from x0 to the interval’s nearer endpoint.
y k k k
0
95
The Precise Definition of a Limit
yk
EXAMPLE 4
Finding Delta Algebraically
For the limit limx:5 2x  1 = 2, find a d 7 0 that works for P = 1. That is, find a d 7 0 such that for all x x0 x0 x0
FIGURE 2.17 For the function ƒsxd = k , we find that ƒ ƒsxd  k ƒ 6 P for any positive d (Example 3b).
Solution 1.
ƒ 2x  1  2 ƒ 6 1.
Q
0 6 ƒx  5ƒ 6 d
x
We organize the search into two steps, as discussed below.
Solve the inequality ƒ 2x  1  2 ƒ 6 1 to find an interval containing x0 = 5 on which the inequality holds for all x Z x0 . ƒ 2x  1  2 ƒ 6 1 1 6 2x  1  2 6 1 1 6 2x  1 6 3 1 6 x  1 6 9 2 6 x 6 10
2.
The inequality holds for all x in the open interval (2, 10), so it holds for all x Z 5 in this interval as well (see Figure 2.19). Find a value of d 7 0 to place the centered interval 5  d 6 x 6 5 + d (centered at x0 = 5) inside the interval (2, 10). The distance from 5 to the nearer endpoint of (2, 10) is 3 (Figure 2.18). If we take d = 3 or any smaller positive number, then the inequality 0 6 ƒ x  5 ƒ 6 d will automatically place x between 2 and 10 to make ƒ 2x  1  2 ƒ 6 1 (Figure 2.19) ƒ 2x  1  2 ƒ 6 1.
Q
0 6 ƒx  5ƒ 6 3
y y 兹x 1 3
2
1 3 2
3
3 5
8
10
3
x 0
1 2
5
8
10
NOT TO SCALE
FIGURE 2.18 An open interval of radius 3 about x0 = 5 will lie inside the open interval (2, 10).
FIGURE 2.19 The function and intervals in Example 4.
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x
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Chapter 2: Limits and Continuity
How to Find Algebraically a D for a Given f, L, x0 , and P>0 The process of finding a d 7 0 such that for all x 0 6 ƒ x  x0 ƒ 6 d
Q
ƒ ƒsxd  L ƒ 6 P
can be accomplished in two steps. 1. Solve the inequality ƒ ƒsxd  L ƒ 6 P to find an open interval (a, b) containing x0 on which the inequality holds for all x Z x0 . 2. Find a value of d 7 0 that places the open interval sx0  d, x0 + dd centered at x0 inside the interval (a, b). The inequality ƒ ƒsxd  L ƒ 6 P will hold for all x Z x0 in this dinterval.
EXAMPLE 5
Finding Delta Algebraically
Prove that limx:2 ƒsxd = 4 if ƒsxd = e
x 2, 1,
x Z 2 x = 2.
y
Solution
y x2
Our task is to show that given P 7 0 there exists a d 7 0 such that for all x 0 6 ƒx  2ƒ 6 d
4
1. (2, 4)
4
For x Z x0 = 2, we have ƒsxd = x 2 , and the inequality to solve is ƒ x 2  4 ƒ 6 P: ƒ x2  4 ƒ 6 P P 6 x 2  4 6 P 4  P 6 x2 6 4 + P 24  P 6 ƒ x ƒ 6 24 + P 24  P 6 x 6 24 + P.
(2, 1)
兹4
2
ƒ ƒsxd  4 ƒ 6 P.
Solve the inequality ƒ ƒsxd  4 ƒ 6 P to find an open interval containing x0 = 2 on which the inequality holds for all x Z x0 .
4
0
Q
x 兹4
FIGURE 2.20 An interval containing x = 2 so that the function in Example 5 satisfies ƒ ƒsxd  4 ƒ 6 P .
Assumes P 6 4 ; see below. An open interval about x0 = 2 that solves the inequality
The inequality ƒ ƒsxd  4 ƒ 6 P holds for all x Z 2 in the open interval A 24  P, 24 + P B (Figure 2.20). 2. Find a value of d 7 0 that places the centered interval s2  d, 2 + dd inside the interval A 24  P, 24 + P B . Take d to be the distance from x0 = 2 to the nearer endpoint of A 24  P, 24 + P B . In other words, take d = min E 2  24  P, 24 + P  2 F , the minimum (the smaller) of the two numbers 2  24  P and 24 + P  2. If d has this or any smaller positive value, the inequality 0 6 ƒ x  2 ƒ 6 d will automatically place x between 24  P and 24 + P to make ƒ ƒsxd  4 ƒ 6 P. For all x, 0 6 ƒx  2ƒ 6 d
Q
ƒ ƒsxd  4 ƒ 6 P.
This completes the proof.
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The Precise Definition of a Limit
97
Why was it all right to assume P 6 4? Because, in finding a d such that for all x, 0 6 ƒ x  2 ƒ 6 d implied ƒ ƒsxd  4 ƒ 6 P 6 4, we found a d that would work for any larger P as well. Finally, notice the freedom we gained in letting d = min E 2  24  P, 24 + P  2 F . We did not have to spend time deciding which, if either, number was the smaller of the two. We just let d represent the smaller and went on to finish the argument.
Using the Definition to Prove Theorems We do not usually rely on the formal definition of limit to verify specific limits such as those in the preceding examples. Rather we appeal to general theorems about limits, in particular the theorems of Section 2.2. The definition is used to prove these theorems (Appendix 2). As an example, we prove part 1 of Theorem 1, the Sum Rule.
EXAMPLE 6
Proving the Rule for the Limit of a Sum
Given that limx:c ƒsxd = L and limx:c gsxd = M, prove that lim sƒsxd + gsxdd = L + M.
x:c
Solution
Let P 7 0 be given. We want to find a positive number d such that for all x 0 6 ƒx  cƒ 6 d
ƒ ƒsxd + gsxd  sL + Md ƒ 6 P.
Q
Regrouping terms, we get ƒ ƒsxd + gsxd  sL + Md ƒ = ƒ sƒsxd  Ld + sgsxd  Md ƒ … ƒ ƒsxd  L ƒ + ƒ gsxd  M ƒ .
Triangle Inequality: ƒa + bƒ … ƒaƒ + ƒbƒ
Since limx:c ƒsxd = L, there exists a number d1 7 0 such that for all x 0 6 ƒ x  c ƒ 6 d1
Q
ƒ ƒsxd  L ƒ 6 P>2.
Similarly, since limx:c gsxd = M, there exists a number d2 7 0 such that for all x 0 6 ƒ x  c ƒ 6 d2
Q
ƒ gsxd  M ƒ 6 P>2.
Let d = min 5d1, d26, the smaller of d1 and d2 . If 0 6 ƒ x  c ƒ 6 d then ƒ x  c ƒ 6 d1 , so ƒ ƒsxd  L ƒ 6 P>2, and ƒ x  c ƒ 6 d2 , so ƒ gsxd  M ƒ 6 P>2. Therefore P P ƒ ƒsxd + gsxd  sL + Md ƒ 6 2 + 2 = P. This shows that limx:c sƒsxd + gsxdd = L + M. Let’s also prove Theorem 5 of Section 2.2.
EXAMPLE 7
Given that limx:c ƒsxd = L and limx:c gsxd = M, and that ƒsxd … g sxd for all x in an open interval containing c (except possibly c itself), prove that L … M. We use the method of proof by contradiction. Suppose, on the contrary, that L 7 M. Then by the limit of a difference property in Theorem 1,
Solution
lim s gsxd  ƒsxdd = M  L.
x:c
Therefore, for any P 7 0, there exists d 7 0 such that ƒ sgsxd  ƒsxdd  sM  Ld ƒ 6 P
whenever
0 6 ƒ x  c ƒ 6 d.
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Chapter 2: Limits and Continuity
Since L  M 7 0 by hypothesis, we take P = L  M in particular and we have a number d 7 0 such that ƒ sg sxd  ƒsxdd  sM  Ld ƒ 6 L  M
whenever
0 6 ƒ x  c ƒ 6 d.
whenever
0 6 ƒx  cƒ 6 d
Since a … ƒ a ƒ for any number a, we have sgsxd  ƒsxdd  sM  Ld 6 L  M which simplifies to gsxd 6 ƒsxd
whenever
0 6 ƒ x  c ƒ 6 d.
But this contradicts ƒsxd … gsxd. Thus the inequality L 7 M must be false. Therefore L … M.
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Chapter 2: Limits and Continuity
EXERCISES 2.3 Centering Intervals About a Point
9.
In Exercises 1–6, sketch the interval (a, b) on the xaxis with the point x0 inside. Then find a value of d 7 0 such that for all x, 0 6 ƒ x  x0 ƒ 6 d Q a 6 x 6 b . 1. a = 1, b = 7, x0 = 5 2. a = 1,
b = 7,
x0 = 2
3. a = 7>2,
b = 1>2,
x0 = 3
4. a = 7>2,
b = 1>2,
x0 = 3>2
b = 4>7,
5. a = 4>9,
6. a = 2.7591,
x0 = 1>2
b = 3.2391,
y 5 4 1 3 4
0
x0 = 3
10. y
f (x) 兹x x0 1 L1 1 y 兹x 4
f(x) 2兹x 1 x0 3 L4 0.2 y 2兹x 1
4.2 4 3.8
1
9 16
2
x
25 16
Finding Deltas Graphically
–1 0
2.61 3 3.41 NOT TO SCALE
In Exercises 7–14, use the graphs to find a d 7 0 such that for all x Q
0 6 ƒ x  x0 ƒ 6 d 7.
11.
ƒ ƒsxd  L ƒ 6 P .
12.
y 2x 4 f (x) 2x 4 x0 5 L6 0.2
6.2 6 5.8
0
y x2
y –3 x 3 2
y 4 x2
5 4
7.65 7.5 7.35
x
5 4.9
f (x) 4 x 2 x0 –1 L3 0.25
f (x) x 2 x0 2 L4 1
y
f (x) – 3 x 3 2 x0 –3 L 7.5 0.15
y
y
8.
y
x
3.25 3 2.75
3
5.1
0
NOT TO SCALE
兹3
2
x 兹5
NOT TO SCALE
–3.1
–3
–2.9
0
NOT TO SCALE
x –
兹5 –1 兹3 – 2 2 NOT TO SCALE
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0
x
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2.3 The Precise Definition of a Limit 13.
More on Formal Limits
14.
Each of Exercises 31–36 gives a function ƒ(x), a point x0, and a positive number P. Find L = lim ƒsxd. Then find a number d 7 0 such
y
y 2 兹–x x0 ⫽ –1 L⫽2 ⑀ ⫽ 0.5
f(x) ⫽
f (x) ⫽ 1x x0 ⫽ 1 2 L⫽2 ⑀ ⫽ 0.01
2.01
y⫽ 2 兹–x
x:x0
that for all x
Q
0 6 ƒ x  x0 ƒ 6 d x0 = 3,
31. ƒsxd = 3  2 x, 32. ƒsxd = 3x  2,
2
ƒ ƒsxd  L ƒ 6 P . P = 0.02
x0 = 1,
P = 0.03
2
2.5
1.99
2
y ⫽ 1x
1.5
33. ƒsxd =
x  4 , x  2
34. ƒsxd =
x 2 + 6x + 5 , x + 5
x0 = 2,
35. ƒsxd = 21  5x, 36. ƒsxd = 4>x,
P = 0.05
x0 = 5,
P = 0.05
x0 = 3,
x0 = 2,
P = 0.5
P = 0.4
Prove the limit statements in Exercises 37–50. –16 9
–1 – 16 25
x
0
x
1 1 1 2 2.01 1.99
0
37. lim s9  xd = 5
38. lim s3x  7d = 2
39. lim 2x  5 = 2
40. lim 24  x = 2
x:4
41. lim ƒsxd = 1 x:1
Finding Deltas Algebraically Each of Exercises 15–30 gives a function ƒ(x) and numbers L, x0 and P 7 0 . In each case, find an open interval about x0 on which the inequality ƒ ƒsxd  L ƒ 6 P holds. Then give a value for d 7 0 such that for all x satisfying 0 6 ƒ x  x0 ƒ 6 d the inequality ƒ ƒsxd  L ƒ 6 P holds. L = 5,
16. ƒsxd = 2x  2,
x0 = 4,
L = 6,
17. ƒsxd = 2x + 1, 18. ƒsxd = 2x,
L = 1, L = 1>2,
19. ƒsxd = 219  x, 20. ƒsxd = 2x  7, 21. ƒsxd = 1>x,
L = 3, L = 4,
L = 1>4,
P = 0.1
x0 = 1>4,
P = 0.1
x0 = 10, x0 = 4,
x0 = 23,
23. ƒsxd = x 2,
L = 4,
x0 = 2,
2
25. ƒsxd = x  5, 26. ƒsxd = 120>x,
L = 1, L = 11, L = 5,
x0 = 1, x0 = 4, x0 = 24,
44.
lim
x: 23
ƒsxd = e
if
x Z 1 x = 1
x 2, 1,
x Z 2 x = 2
45. lim
1 1 = 3 x2 x2  9 = 6 x + 3
46. lim
x:1
47. lim ƒsxd = 2
if
4  2x, ƒsxd = e 6x  4,
48. lim ƒsxd = 0
if
ƒsxd = e
x:0
2x, x>2,
x2  1 = 2 x  1 x 6 1 x Ú 1 x 6 0 x Ú 0
1 49. lim x sin x = 0
P = 0.1
x:0
y
P = 0.5 P = 0.1 P = 1 P = 1
L = 2m,
x0 = 2,
28. ƒsxd = mx, P = c 7 0
m 7 0,
L = 3m,
x0 = 3,
m 7 0,
x:0
x 2, ƒsxd = e 2,
1 43. lim x = 1 x:1
x:1
P = 0.05
m 7 0,
30. ƒsxd = mx + b, P = 0.05
x: 2
x: 3
P = 1
27. ƒsxd = mx,
m 7 0, 29. ƒsxd = mx + b, x0 = 1>2, P = c 7 0
42. lim ƒsxd = 4
if
P = 1
x0 = 23,
L = 3,
24. ƒsxd = 1>x,
P = 0.02
x0 = 0,
22. ƒsxd = x ,
2
P = 0.01
x0 = 2,
x:3
x:9
NOT TO SCALE
15. ƒsxd = x + 1,
99
P = 0.03
– 1 2 1 –
L = sm>2d + b, L = m + b,
y ⫽ x sin 1x
1 2
x0 = 1,
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1
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100
Chapter 2: Limits and Continuity
1 50. lim x 2 sin x = 0
volts and I is to be 5 ; 0.1 amp . In what interval does R have to lie for I to be within 0.1 amp of the value I0 = 5 ?
x: 0
y
V
y x2
1
0
2 –
2
1
R
When Is a Number L Not the Limit of ƒ(x) as x : x0 ?
y x 2 sin 1x –1
I
x
We can prove that limx:x0 ƒsxd Z L by providing an P 7 0 such that no possible d 7 0 satisfies the condition For all x, 0 6 ƒ x  x0 ƒ 6 d Q ƒ ƒsxd  L ƒ 6 P . We accomplish this for our candidate P by showing that for each d 7 0 there exists a value of x such that
–1
y –x 2
0 6 ƒ x  x0 ƒ 6 d
and
ƒ ƒsxd  L ƒ Ú P .
y y f(x)
Theory and Examples L
51. Define what it means to say that lim g sxd = k . x: 0
52. Prove that lim ƒsxd = L if and only if lim ƒsh + cd = L . x: c
L
h :0
53. A wrong statement about limits Show by example that the following statement is wrong.
L
The number L is the limit of ƒ(x) as x approaches x0 if ƒ(x) gets closer to L as x approaches x0 .
f (x)
Explain why the function in your example does not have the given value of L as a limit as x : x0 .
0 x0
Explain why the function in your example does not have the given value of L as a limit as x : x0 . T 55. Grinding engine cylinders Before contracting to grind engine cylinders to a crosssectional area of 9 in2 , you need to know how much deviation from the ideal cylinder diameter of x0 = 3.385 in. you can allow and still have the area come within 0.01 in2 of the required 9 in2 . To find out, you let A = psx>2d2 and look for the interval in which you must hold x to make ƒ A  9 ƒ … 0.01 . What interval do you find? 56. Manufacturing electrical resistors Ohm’s law for electrical circuits like the one shown in the accompanying figure states that V = RI . In this equation, V is a constant voltage, I is the current in amperes, and R is the resistance in ohms. Your firm has been asked to supply the resistors for a circuit in which V will be 120
x
x0
a value of x for which 0 x x 0 and f(x) L ⱖ
54. Another wrong statement about limits Show by example that the following statement is wrong. The number L is the limit of ƒ(x) as x approaches x0 if, given any P 7 0 , there exists a value of x for which ƒ ƒsxd  L ƒ 6 P.
x0
57. Let ƒsxd = e
x, x 6 1 x + 1, x 7 1. y yx1
2 y f(x) 1
1 yx
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4100 AWL/Thomas_ch02p073146 8/19/04 11:01 AM Page 101
2.3 The Precise Definition of a Limit a. Let P = 1>2 . Show that no possible d 7 0 satisfies the following condition: 0 6 ƒx  1ƒ 6 d Q ƒ ƒsxd  2 ƒ 6 1>2. That is, for each d 7 0 show that there is a value of x such that
For all x,
101
60. a. For the function graphed here, show that limx : 1 g sxd Z 2 . b. Does limx : 1 g sxd appear to exist? If so, what is the value of the limit? If not, why not? y
0 6 ƒx  1ƒ 6 d and ƒ ƒsxd  2 ƒ Ú 1>2. This will show that limx:1 ƒsxd Z 2 .
2
b. Show that limx:1 ƒsxd Z 1 . c. Show that limx:1 ƒsxd Z 1.5 . x 2, 58. Let hsxd = • 3, 2,
y g(x)
x 6 2 x = 2 x 7 2.
1
y –1
0
x
y h(x)
4
COMPUTER EXPLORATIONS 3
In Exercises 61–66, you will further explore finding deltas graphically. Use a CAS to perform the following steps:
y2
2
a. Plot the function y = ƒsxd near the point x0 being approached.
1
y x2
0
2
b. Guess the value of the limit L and then evaluate the limit symbolically to see if you guessed correctly. x
c. Using the value P = 0.2 , graph the banding lines y1 = L  P and y2 = L + P together with the function ƒ near x0 . d. From your graph in part (c), estimate a d 7 0 such that for all x
Show that a. lim hsxd Z 4
0 6 ƒ x  x0 ƒ 6 d
x:2
b. lim hsxd Z 3
ƒ ƒsxd  L ƒ 6 P .
Test your estimate by plotting ƒ, y1 , and y2 over the interval 0 6 ƒ x  x0 ƒ 6 d . For your viewing window use x0  2d … x … x0 + 2d and L  2P … y … L + 2P . If any function values lie outside the interval [L  P, L + P] , your choice of d was too large. Try again with a smaller estimate.
x:2
c. lim hsxd Z 2 x:2
59. For the function graphed here, explain why a. lim ƒsxd Z 4 x:3
e. Repeat parts (c) and (d) successively for P = 0.1, 0.05 , and 0.001.
b. lim ƒsxd Z 4.8 x:3
c. lim ƒsxd Z 3 x:3
y
61. ƒsxd =
x 4  81 , x  3
62. ƒsxd =
5x 3 + 9x 2 , 2x 5 + 3x 2
63. ƒsxd =
sin 2x , 3x
64. ƒsxd =
xs1  cos xd , x  sin x
65. ƒsxd =
3 2 x  1 , x  1
66. ƒsxd =
3x 2  s7x + 1d2x + 5 , x  1
4.8 4
y f (x)
3
0
Q
3
x
x0 = 3 x0 = 0
x0 = 0 x0 = 0
x0 = 1 x0 = 1
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102
Chapter 2: Limits and Continuity
OneSided Limits and Limits at Infinity
2.4
In this section we extend the limit concept to onesided limits, which are limits as x approaches the number x0 from the lefthand side (where x 6 x0) or the righthand side sx 7 x0 d only. We also analyze the graphs of certain rational functions as well as other functions with limit behavior as x : ; q .
y y x x 1
x
0
–1
FIGURE 2.21 Different righthand and lefthand limits at the origin.
OneSided Limits To have a limit L as x approaches c, a function ƒ must be defined on both sides of c and its values ƒ(x) must approach L as x approaches c from either side. Because of this, ordinary limits are called twosided. If ƒ fails to have a twosided limit at c, it may still have a onesided limit, that is, a limit if the approach is only from one side. If the approach is from the right, the limit is a righthand limit. From the left, it is a lefthand limit. The function ƒsxd = x> ƒ x ƒ (Figure 2.21) has limit 1 as x approaches 0 from the right, and limit 1 as x approaches 0 from the left. Since these onesided limit values are not the same, there is no single number that ƒ(x) approaches as x approaches 0. So ƒ(x) does not have a (twosided) limit at 0. Intuitively, if ƒ(x) is defined on an interval (c, b), where c 6 b, and approaches arbitrarily close to L as x approaches c from within that interval, then ƒ has righthand limit L at c. We write lim ƒsxd = L.
x:c +
The symbol “x : c + ” means that we consider only values of x greater than c. Similarly, if ƒ(x) is defined on an interval (a, c), where a 6 c and approaches arbitrarily close to M as x approaches c from within that interval, then ƒ has lefthand limit M at c. We write lim ƒsxd = M.
x:c 
The symbol “x : c  ” means that we consider only x values less than c. These informal definitions are illustrated in Figure 2.22. For the function ƒsxd = x> ƒ x ƒ in Figure 2.21 we have lim ƒsxd = 1
and
x:0 +
lim ƒsxd = 1.
x:0 
y
y
f (x)
L 0
c
x
M
f (x) x
0
(a) lim f (x) L x→c
FIGURE 2.22 (a) Righthand limit as x approaches c. approaches c.
x
c
(b) lim f (x) M x→c
(b) Lefthand limit as x
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2.4
EXAMPLE 1
y
OneSided Limits for a Semicircle
lim 24  x 2 = 0
x: 2 +
0
FIGURE 2.23
2
103
The domain of ƒsxd = 24  x 2 is [2, 2]; its graph is the semicircle in Figure 2.23. We have
y 兹4 x 2
–2
OneSided Limits and Limits at Infinity
and
lim 24  x 2 = 0.
x:2 
The function does not have a lefthand limit at x = 2 or a righthand limit at x = 2. It does not have ordinary twosided limits at either 2 or 2.
x
lim 24  x 2 = 0 and
x: 2 
lim 24  x 2 = 0 (Example 1).
x:  2 +
Onesided limits have all the properties listed in Theorem 1 in Section 2.2. The righthand limit of the sum of two functions is the sum of their righthand limits, and so on. The theorems for limits of polynomials and rational functions hold with onesided limits, as does the Sandwich Theorem and Theorem 5. Onesided limits are related to limits in the following way.
THEOREM 6 A function ƒ(x) has a limit as x approaches c if and only if it has lefthand and righthand limits there and these onesided limits are equal: lim ƒsxd = L
x:c
EXAMPLE 2
y
At x = 0:
y f (x)
2 1
0
At x = 1: 1
2
3
4
lim ƒsxd = L
x:c 
and
lim ƒsxd = L.
x:c +
Limits of the Function Graphed in Figure 2.24 limx:0+ ƒsxd = 1, limx:0 ƒsxd and limx:0 ƒsxd do not exist. The function is not defined to the left of x = 0. limx:1 ƒsxd = 0 even though ƒs1d = 1, limx:1+ ƒsxd = 1, limx:1 ƒsxd does not exist. The right and lefthand limits are not equal.
x
FIGURE 2.24 Graph of the function in Example 2.
3
At x = 2:
At x = 3: At x = 4:
limx:2 ƒsxd = 1, limx:2+ ƒsxd = 1, limx:2 ƒsxd = 1 even though ƒs2d = 2. limx:3 ƒsxd = limx:3+ ƒsxd = limx:3 ƒsxd = ƒs3d = 2. limx:4 ƒsxd = 1 even though ƒs4d Z 1, limx:4+ ƒsxd and limx:4 ƒsxd do not exist. The function is not defined to the right of x = 4.
At every other point c in [0, 4], ƒ(x) has limit ƒ(c).
Precise Definitions of OneSided Limits The formal definition of the limit in Section 2.3 is readily modified for onesided limits.
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104
Chapter 2: Limits and Continuity y
DEFINITIONS RightHand, LeftHand Limits We say that ƒ(x) has righthand limit L at x0 , and write lim ƒsxd = L
(See Figure 2.25)
x:x0 +
L
if for every number P 7 0 there exists a corresponding number d 7 0 such that for all x
f(x) f(x) lies in here
L
Q
x0 6 x 6 x0 + d
L
ƒ ƒsxd  L ƒ 6 P.
We say that ƒ has lefthand limit L at x0 , and write lim ƒsxd = L
for all x x 0 in here
if for every number P 7 0 there exists a corresponding number d 7 0 such that for all x
x 0
x
x0
x0
(See Figure 2.26)
x:x0 
Q
x0  d 6 x 6 x0
ƒ ƒsxd  L ƒ 6 P.
FIGURE 2.25 Intervals associated with the definition of righthand limit.
EXAMPLE 3 y
Applying the Definition to Find Delta
Prove that lim 2x = 0.
x:0 +
Let P 7 0 be given. Here x0 = 0 and L = 0, so we want to find a d 7 0 such that for all x
Solution L
f(x)
0 6 x 6 d
f(x) lies in here
L
or
L for all x x 0 in here
Squaring both sides of this last inequality gives x 6 P2
x
0 6 x 6 d.
If we choose d = P we have
x0
0 6 x 6 d = P2
FIGURE 2.26 Intervals associated with the definition of lefthand limit.
2x 6 P,
Q
or 0 6 x 6 P2
Q
ƒ 2x  0 ƒ 6 P.
According to the definition, this shows that limx:0+ 2x = 0 (Figure 2.27).
y f (x) 兹x
The functions examined so far have had some kind of limit at each point of interest. In general, that need not be the case.
f(x)
L0
if
2
x x0
2x 6 P.
Q
0 6 x 6 d
0
ƒ 2x  0 ƒ 6 P,
Q
EXAMPLE 4 x
FIGURE 2.27
2
x
lim 1x = 0 in Example 3.
x: 0 +
A Function Oscillating Too Much
Show that y = sin s1>xd has no limit as x approaches zero from either side (Figure 2.28). As x approaches zero, its reciprocal, 1>x, grows without bound and the values of sin (1>x) cycle repeatedly from 1 to 1. There is no single number L that the function’s
Solution
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OneSided Limits and Limits at Infinity
105
y 1
x
0
y sin 1x –1
FIGURE 2.28 The function y = sin s1>xd has neither a righthand nor a lefthand limit as x approaches zero (Example 4).
values stay increasingly close to as x approaches zero. This is true even if we restrict x to positive values or to negative values. The function has neither a righthand limit nor a lefthand limit at x = 0.
Limits Involving (sin U)/U A central fact about ssin ud>u is that in radian measure its limit as u : 0 is 1. We can see this in Figure 2.29 and confirm it algebraically using the Sandwich Theorem. y 1
–3
–2
y sin (radians)
–
2
3
y NOT TO SCALE
T
FIGURE 2.29 The graph of ƒsud = ssin ud>u .
1 P
THEOREM 7
tan
sin u = 1 u:0 u
1
lim
sin
cos Q
A(1, 0)
(1)
x
O
su in radiansd
1
FIGURE 2.30 The figure for the proof of Theorem 7. TA>OA = tan u , but OA = 1 , so TA = tan u .
Proof The plan is to show that the righthand and lefthand limits are both 1. Then we will know that the twosided limit is 1 as well. To show that the righthand limit is 1, we begin with positive values of u less than p>2 (Figure 2.30). Notice that Area ¢OAP 6 area sector OAP 6 area ¢OAT.
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106
Chapter 2: Limits and Continuity
Equation (2) is where radian measure comes in: The area of sector OAP is u>2 only if u is measured in radians.
We can express these areas in terms of u as follows: 1 1 1 base * height = s1dssin ud = sin u 2 2 2 u 1 1 Area sector OAP = r 2u = s1d2u = 2 2 2 1 1 1 Area ¢OAT = base * height = s1dstan ud = tan u. 2 2 2 Area ¢OAP =
(2)
Thus, 1 1 1 sin u 6 u 6 tan u. 2 2 2 This last inequality goes the same way if we divide all three terms by the number (1>2) sin u, which is positive since 0 6 u 6 p>2: 1 6
u 1 6 . cos u sin u
Taking reciprocals reverses the inequalities: 1 7
sin u 7 cos u. u
Since limu:0+ cos u = 1 (Example 6b, Section 2.2), the Sandwich Theorem gives lim+
u:0
sin u = 1. u
Recall that sin u and u are both odd functions (Section 1.4). Therefore, ƒsud = ssin ud>u is an even function, with a graph symmetric about the yaxis (see Figure 2.29). This symmetry implies that the lefthand limit at 0 exists and has the same value as the righthand limit: lim
u:0 
sin u sin u = 1 = lim+ , u u u:0
so limu:0 ssin ud>u = 1 by Theorem 6.
EXAMPLE 5
Using lim
u:0
Show that (a) lim
h:0
sin u = 1 u
cos h  1 sin 2x 2 = 0 and (b) lim = . 5 h x:0 5x
Solution
(a) Using the halfangle formula cos h = 1  2 sin2 sh>2d, we calculate 2 sin2 sh>2d cos h  1 = lim h h h:0 h:0 sin u =  lim sin u u:0 u lim
= s1ds0d = 0.
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Let u = h>2 .
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2.4
OneSided Limits and Limits at Infinity
107
(b) Equation (1) does not apply to the original fraction. We need a 2x in the denominator, not a 5x. We produce it by multiplying numerator and denominator by 2>5: s2>5d # sin 2x sin 2x = lim x:0 5x x:0 s2>5d # 5x lim
y 4
=
sin 2x 2 lim 5 x:0 2x
=
2 2 s1d = 5 5
3 y 1x
2
Now, Eq. (1) applies with u 2x.
1 –1 0 –1
FIGURE 2.31
1
2
3
4
The graph of y = 1>x .
x
Finite Limits as x : — ˆ The symbol for infinity s q d does not represent a real number. We use q to describe the behavior of a function when the values in its domain or range outgrow all finite bounds. For example, the function ƒsxd = 1>x is defined for all x Z 0 (Figure 2.31). When x is positive and becomes increasingly large, 1>x becomes increasingly small. When x is negative and its magnitude becomes increasingly large, 1>x again becomes small. We summarize these observations by saying that ƒsxd = 1>x has limit 0 as x : ; q or that 0 is a limit of ƒsxd = 1>x at infinity and negative infinity. Here is a precise definition.
DEFINITIONS Limit as x approaches ˆ or ˆ 1. We say that ƒ(x) has the limit L as x approaches infinity and write lim ƒsxd = L
x: q
if, for every number P 7 0, there exists a corresponding number M such that for all x x 7 M
Q
ƒ ƒsxd  L ƒ 6 P.
2. We say that ƒ(x) has the limit L as x approaches minus infinity and write lim ƒsxd = L
x:  q
if, for every number P 7 0, there exists a corresponding number N such that for all x x 6 N
Q
ƒ ƒsxd  L ƒ 6 P.
Intuitively, limx: q ƒsxd = L if, as x moves increasingly far from the origin in the positive direction, ƒ(x) gets arbitrarily close to L. Similarly, limx: q ƒsxd = L if, as x moves increasingly far from the origin in the negative direction, ƒ(x) gets arbitrarily close to L. The strategy for calculating limits of functions as x : ; q is similar to the one for finite limits in Section 2.2. There we first found the limits of the constant and identity functions y = k and y = x. We then extended these results to other functions by applying a theorem about limits of algebraic combinations. Here we do the same thing, except that the starting functions are y = k and y = 1>x instead of y = k and y = x.
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Chapter 2: Limits and Continuity
y y 1x
0
lim k = k
lim
x: ; q
1 x = 0.
(3)
We prove the latter and leave the former to Exercises 71 and 72.
x
M 1
and
x: ; q
y
N – 1 y –
The basic facts to be verified by applying the formal definition are
No matter what positive number is, the graph enters this band at x 1 and stays.
EXAMPLE 6
1 Limits at Infinity for ƒsxd = x
Show that
–
1 (a) lim x = 0 x: q
No matter what positive number is, the graph enters this band at x – 1 and stays.
(b)
lim
x:  q
1 x = 0.
Solution
(a) Let P 7 0 be given. We must find a number M such that for all x
FIGURE 2.32 The geometry behind the argument in Example 6.
x 7 M
1 1 ` x  0 ` = ` x ` 6 P.
Q
The implication will hold if M = 1>P or any larger positive number (Figure 2.32). This proves limx: q s1>xd = 0. (b) Let P 7 0 be given. We must find a number N such that for all x x 6 N
1 1 ` x  0 ` = ` x ` 6 P.
Q
The implication will hold if N = 1>P or any number less than 1>P (Figure 2.32). This proves limx: q s1>xd = 0. Limits at infinity have properties similar to those of finite limits.
THEOREM 8 Limit Laws as x : — ˆ If L, M, and k, are real numbers and lim ƒsxd = L
x: ; q
1. Sum Rule: 2. Difference Rule: 3. Product Rule: 4. Constant Multiple Rule: 5. Quotient Rule:
and
lim gsxd = M,
x: ; q
then
lim sƒsxd + gsxdd = L + M
x: ; q
lim sƒsxd  gsxdd = L  M
x: ; q
lim sƒsxd # gsxdd = L # M
x: ; q
lim sk # ƒsxdd = k # L
x: ; q
lim
x: ; q
ƒsxd L = , M gsxd
M Z 0
6. Power Rule: If r and s are integers with no common factors, s Z 0, then lim sƒsxddr>s = L r>s
x: ; q
provided that L r>s is a real number. (If s is even, we assume that L 7 0.)
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2.4
OneSided Limits and Limits at Infinity
109
These properties are just like the properties in Theorem 1, Section 2.2, and we use them the same way.
EXAMPLE 7
Using Theorem 8
1 1 (a) lim a5 + x b = lim 5 + lim x q q q x: x: x:
Sum Rule
= 5 + 0 = 5 (b)
lim
x:  q
p23 = x2 =
Known limits
1 1 lim p23 # x # x q x: 1 lim p23 # lim x x:  q
#
x:  q
2
–5
= p23 # 0 # 0 = 0
2 y 5x 2 8x 3 3x 2
y
1 lim x
x:  q
Product rule Known limits
Limits at Infinity of Rational Functions
1
Line y 5 3
0
5
10
x
To determine the limit of a rational function as x : ; q , we can divide the numerator and denominator by the highest power of x in the denominator. What happens then depends on the degrees of the polynomials involved.
EXAMPLE 8
–1 –2
Numerator and Denominator of Same Degree 5 + s8>xd  s3>x 2 d 5x 2 + 8x  3 = lim x: q x: q 3x 2 + 2 3 + s2>x 2 d lim
NOT TO SCALE
FIGURE 2.33 The graph of the function in Example 8. The graph approaches the line y = 5>3 as ƒ x ƒ increases.
=
EXAMPLE 9
11x + 2 = x:  q 2x 3  1 lim
y 6
See Fig. 2.33.
Degree of Numerator Less Than Degree of Denominator
y 8
5 + 0  0 5 = 3 + 0 3
Divide numerator and denominator by x 2.
11x 2 2x 3 1
=
lim
s11>x 2 d + s2>x 3 d
x:  q
2  s1>x 3 d
0 + 0 = 0 2  0
Divide numerator and denominator by x 3. See Fig. 2.34.
We give an example of the case when the degree of the numerator is greater than the degree of the denominator in the next section (Example 8, Section 2.5).
4 2
–4
–2
0
2
4
6
–2 –4 –6
x
Horizontal Asymptotes If the distance between the graph of a function and some fixed line approaches zero as a point on the graph moves increasingly far from the origin, we say that the graph approaches the line asymptotically and that the line is an asymptote of the graph. Looking at ƒsxd = 1>x (See Figure 2.31), we observe that the xaxis is an asymptote of the curve on the right because 1 lim x = 0 q x:
–8
FIGURE 2.34 The graph of the function in Example 9. The graph approaches the xaxis as ƒ x ƒ increases.
and on the left because 1 lim x = 0.
x:  q
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We say that the xaxis is a horizontal asymptote of the graph of ƒsxd = 1>x.
DEFINITION Horizontal Asymptote A line y = b is a horizontal asymptote of the graph of a function y = ƒsxd if either lim ƒsxd = b
or
x: q
lim ƒsxd = b.
x:  q
The curve ƒsxd =
5x 2 + 8x  3 3x 2 + 2
sketched in Figure 2.33 (Example 8) has the line y = 5>3 as a horizontal asymptote on both the right and the left because 5 3
lim ƒsxd =
x: q
EXAMPLE 10
and
lim ƒsxd =
x:  q
5 . 3
Substituting a New Variable
Find lim sin s1>xd. x: q
We introduce the new variable t = 1>x. From Example 6, we know that t : 0 + as x : q (see Figure 2.31). Therefore,
Solution
1 lim sin x = lim+ sin t = 0. t:0
x: q
The Sandwich Theorem Revisited The Sandwich Theorem also holds for limits as x : ; q .
EXAMPLE 11
A Curve May Cross Its Horizontal Asymptote
Using the Sandwich Theorem, find the horizontal asymptote of the curve y
y = 2 +
sin x x .
y 2 sinx x
Solution
2
We are interested in the behavior as x : ; q . Since sin x 1 0 … ` x ` … `x`
1 –3 –2 –
0
2
3
FIGURE 2.35 A curve may cross one of its asymptotes infinitely often (Example 11).
x
and limx:; q ƒ 1>x ƒ = 0, we have limx:; q ssin xd>x = 0 by the Sandwich Theorem. Hence, lim a2 +
x: ; q
sin x x b = 2 + 0 = 2,
and the line y = 2 is a horizontal asymptote of the curve on both left and right (Figure 2.35).
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2.4 OneSided Limits and Limits at Infinity
111
This example illustrates that a curve may cross one of its horizontal asymptotes, perhaps many times.
Oblique Asymptotes If the degree of the numerator of a rational function is one greater than the degree of the denominator, the graph has an oblique (slanted) asymptote. We find an equation for the asymptote by dividing numerator by denominator to express ƒ as a linear function plus a remainder that goes to zero as x : ; q . Here’s an example.
EXAMPLE 12
Finding an Oblique Asymptote
Find the oblique asymptote for the graph of ƒsxd =
2x 2  3 7x + 4
in Figure 2.36. Solution
y
ƒsxd =
4
2
By long division, we find
y
2x 2  3 7x + 4
8 115 2 = a x b + 7 49 49s7x + 4d ('')''* ('')''*
2x 2 3 7x 4
linear function gsxd
–4
–2
2
4
remainder
x
As x : ; q , the remainder, whose magnitude gives the vertical distance between the graphs of ƒ and g, goes to zero, making the (slanted) line
–2
gsxd =
–4
FIGURE 2.36 The function in Example 12 has an oblique asymptote.
8 2 x 7 49
an asymptote of the graph of ƒ (Figure 2.36). The line y = gsxd is an asymptote both to the right and to the left. In the next section you will see that the function ƒ(x) grows arbitrarily large in absolute value as x approaches 4>7, where the denominator becomes zero (Figure 2.36).
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2.4 OneSided Limits and Limits at Infinity
EXERCISES 2.4 Finding Limits Graphically
a.
1. Which of the following statements about the function y = ƒsxd graphed here are true, and which are false?
lim ƒsxd = 1
x: 1 +
y f (x)
d. lim ƒsxd = lim+ ƒsxd
e. lim ƒsxd exists
f. lim ƒsxd = 0
g. lim ƒsxd = 1
h. lim ƒsxd = 1
i. lim ƒsxd = 0
j. lim ƒsxd = 2
x:0
x:0
1
x:1
k. –1
0
1
2
x
x:0
c. lim ƒsxd = 1 x:0
y
b. lim ƒsxd = 0
lim  ƒsxd does not exist .
x: 1
x:0
x:0
x:0 x:1 x:2
l. lim+ ƒsxd = 0 x:2
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2. Which of the following statements about the function y = ƒsxd graphed here are true, and which are false?
a. Find limx:2+ ƒsxd, limx:2 ƒsxd , and ƒ(2). b. Does limx:2 ƒsxd exist? If so, what is it? If not, why not? c. Find limx:1 ƒsxd and limx:1+ ƒsxd .
y y f (x)
d. Does limx:1 ƒsxd exist? If so, what is it? If not, why not? 5. Let ƒsxd = •
2 1
–1
0
1
2
0,
x … 0
1 sin x ,
x 7 0.
y
x
3
1
a.
b. lim ƒsxd does not exist.
lim + ƒsxd = 1
x: 1
x: 2
c. lim ƒsxd = 2
d. lim ƒsxd = 2
e. lim+ ƒsxd = 1
f. lim ƒsxd does not exist.
x:2
x: 1
x:1
x: 1
g. lim+ ƒsxd = lim ƒsxd x:0
x
0
x: 0
xⱕ0 0, y 1 sin x , x 0
h. lim ƒsxd exists at every c in the open interval s 1, 1d . x:c
i. lim ƒsxd exists at every c in the open interval (1, 3). x:c
j.
lim ƒsxd = 0
–1
k. lim+ ƒsxd does not exist.
x: 1 
x: 3
3  x, 3. Let ƒsxd = • x + 1, 2
x 6 2 a. Does limx:0+ ƒsxd exist? If so, what is it? If not, why not?
x 7 2.
b. Does limx:0 ƒsxd exist? If so, what is it? If not, why not? y
c. Does limx:0 ƒsxd exist? If so, what is it? If not, why not? 6. Let g sxd = 2x sins1>xd .
y3x 3
y x1 2
y y 兹x
1 0
x
4
2
a. Find limx:2+ ƒsxd and limx:2 ƒsxd .
y 兹x sin 1x
b. Does limx:2 ƒsxd exist? If so, what is it? If not, why not? c. Find limx:4 ƒsxd and limx:4+ ƒsxd .
1 2
d. Does limx:4 ƒsxd exist? If so, what is it? If not, why not?
4. Let ƒsxd = d
0
3  x, x 6 2 2, x = 2 x , 2
1
2
1
x
x 7 2. y –1
y3x
y –兹x
3
a. Does limx:0+ g sxd exist? If so, what is it? If not, why not?
y x 2 –2
0
2
x
b. Does limx:0 g sxd exist? If so, what is it? If not, why not? c. Does limx:0 g sxd exist? If so, what is it? If not, why not?
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2.4 OneSided Limits and Limits at Infinity 7. a. Graph ƒsxd = e
x 3, 0,
sin U 1 U:0 U
x Z 1 x = 1.
Using lim
b. Find limx:1 ƒsxd and limx:1+ ƒsxd . c. Does limx:1 ƒsxd exist? If so, what is it? If not, why not? 1  x 2, 8. a. Graph ƒsxd = e 2,
Find the limits in Exercises 21–36. sin 22u
21. lim
sin kt t
sk constantd
b. Find limx:1+ ƒsxd and limx:1 ƒsxd . c. Does limx:1 ƒsxd exist? If so, what is it? If not, why not?
25. lim
tan 2x x
26. lim
27. lim
x csc 2x cos 5x
28. lim 6x 2scot xdscsc 2xd
x Z 1 x = 1.
u :0
x:0
Graph the functions in Exercises 9 and 10. Then answer these questions.
x:0
a. What are the domain and range of ƒ? b. At what points c, if any, does limx:c ƒsxd exist?
x, 10. ƒsxd = • 1, 0,
0 … x 6 1 1 … x 6 2 x = 2
Finding OneSided Limits Algebraically Find the limits in Exercises 11–18. x + 2 11. lim x: 0.5 A x + 1 lim + a
x: 2
x  1 12. lim+ x: 1 A x + 2
t: 0
15. lim+
2h 2 + 4h + 5  25 h
16. lim
26  25h + 11h + 6 h
ƒx + 2ƒ 17. a. lim +sx + 3d x + 2 x: 2 18. a. lim+ x:1
22x sx  1d ƒx  1ƒ
t:4
x 2  x + sin x 2x x:0
30. lim
32. lim
sin ssin hd sin h sin 5x sin 4x
h:0
34. lim
35. lim
tan 3x sin 8x
36. lim
x:0
2t tan t
x:0
sin u sin 2u
x:0
sin 3y cot 5y y cot 4y y:0
Calculating Limits as x : — ˆ In Exercises 37–42, find the limit of each function (a) as x : q and (b) as x :  q . (You may wish to visualize your answer with a graphing calculator or computer.) 2 37. ƒsxd = x  3
41. hsxd =
38. ƒsxd = p 
1 2 + s1>xd 5 + s7>xd 2
3  s1>x d
40. g sxd =
43. lim
x: q
45.
ƒx + 2ƒ b. lim sx + 3d x + 2 x: 2 b. limx: 1
22x sx  1d ƒx  1ƒ
b. limst  : t; d t: 4
lim
sin 2x x
3  s2>xd
42. hsxd =
4 +
t:  q
2  t + sin t t + cos t
44.
lim
u: q
46. lim
r: q
2 x2
1 8  s5>x 2 d
Find the limits in Exercises 43–46.
Use the graph of the greatest integer function y = :x; (sometimes written y = int x), Figure 1.31 in Section 1.3, to help you find the limits in Exercises 19 and 20. :u; :u; 19. a. lim+ b. limu u u :3 u: 3 20. a. lim+st  : t; d
t: 0
33. lim
2
h: 0
h:0
sin s1  cos td 1  cos t
39. g sxd =
2x + 5 x b ba 2 x + 1 x + x
x + 6 3  x 1 ba x ba b 14. lim a 7 x:1 x + 1 h: 0
31. lim
h sin 3h
24. lim
x + x cos x x:0 sin x cos x
u :0
1 … x 6 0, or 0 6 x … 1 x = 0 x 6 1, or x 7 1
t: 0
29. lim
c. At what points does only the lefthand limit exist? d. At what points does only the righthand limit exist?
13.
22. lim
22u sin 3y 23. lim y:0 4y
21  x 2, 9. ƒsxd = • 1, 2,
113
A 22>x 2 B
cos u 3u
r + sin r 2r + 7  5 sin r
Limits of Rational Functions In Exercises 47–56, find the limit of each rational function (a) as x : q and (b) as x :  q . 47. ƒsxd =
2x + 3 5x + 7
48. ƒsxd =
2x 3 + 7 x  x2 + x + 7
49. ƒsxd =
x + 1 x2 + 3
50. ƒsxd =
3x + 7 x2  2
51. hsxd =
7x 3 x  3x 2 + 6x
52. g sxd =
1 x 3  4x + 1
3
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Chapter 2: Limits and Continuity
53. g sxd =
10x 5 + x 4 + 31 x6
Formal Definitions of OneSided Limits 73. Given P 7 0 , find an interval I = s5, 5 + dd, d 7 0 , such that if x lies in I, then 2x  5 6 P . What limit is being verified and what is its value?
9x 4 + x 54. hsxd = 4 2x + 5x 2  x + 6 55. hsxd =
2x 3  2x + 3 3x 3 + 3x 2  5x
56. hsxd =
x 4 x  7x + 7x 2 + 9 4
74. Given P 7 0 , find an interval I = s4  d, 4d, d 7 0 , such that if x lies in I, then 24  x 6 P . What limit is being verified and what is its value? Use the definitions of righthand and lefthand limits to prove the limit statements in Exercises 75 and 76.
3
Limits with Noninteger or Negative Powers The process by which we determine limits of rational functions applies equally well to ratios containing noninteger or negative powers of x: divide numerator and denominator by the highest power of x in the denominator and proceed from there. Find the limits in Exercises 57–62. 2 2x + x 1 3x  7 x: q
57. lim
lim
61. lim
x: q
2  2x x 1 + x 4 60. lim 2 x: q x  x 3
5
3 2 x + 2x 2x 5>3  x 1>3 + 7
x:  q
2 + 2x
x: q
5
2x  2x 3
59.
58. lim
62.
x 8>5 + 3x + 2x
lim
x:  q
3 2 x  5x + 3 2x + x 2>3  4
Theory and Examples 63. Once you know limx:a ƒsxd and limx:a ƒsxd at an interior point of the domain of ƒ, do you then know limx:a ƒsxd ? Give reasons for your answer. +

x x  2 = 1 = 1 76. lim+ x:2 ƒ x  2 ƒ ƒxƒ 77. Greatest integer function Find (a) limx:400+ :x; and (b) limx:400 :x; ; then use limit definitions to verify your findings. (c) Based on your conclusions in parts (a) and (b), can anything be said about limx:400 :x; ? Give reasons for your answers. 75. limx:0
78. Onesided limits Let ƒsxd = e
Find (a) limx:0+ ƒsxd and (b) limx:0 ƒsxd ; then use limit definitions to verify your findings. (c) Based on your conclusions in parts (a) and (b), can anything be said about limx:0 ƒsxd ? Give reasons for your answer.
Grapher Explorations—“Seeing” Limits at Infinity Sometimes a change of variable can change an unfamiliar expression into one whose limit we know how to find. For example,
64. If you know that limx:c ƒsxd exists, can you find its value by calculating limx:c+ ƒsxd ? Give reasons for your answer. 65. Suppose that ƒ is an odd function of x. Does knowing that limx:0+ ƒsxd = 3 tell you anything about limx:0 ƒsxd ? Give reasons for your answer. 66. Suppose that ƒ is an even function of x. Does knowing that limx:2 ƒsxd = 7 tell you anything about either limx:2 ƒsxd or limx:2+ ƒsxd ? Give reasons for your answer. 67. Suppose that ƒ(x) and g (x) are polynomials in x and that limx: q sƒsxd>g sxdd = 2 . Can you conclude anything about limx: q sƒsxd>g sxdd ? Give reasons for your answer. 68. Suppose that ƒ(x) and g (x) are polynomials in x. Can the graph of ƒ(x)>g (x) have an asymptote if g (x) is never zero? Give reasons for your answer. 69. How many horizontal asymptotes can the graph of a given rational function have? Give reasons for your answer. 70. Find lim A 2x 2 + x  2x 2  x B .
1 lim sin x = lim+ sin u
x: q
Use the formal definitions of limits as x : ; q to establish the limits in Exercises 71 and 72.
u :0
Substitute u = 1>x
= 0. This suggests a creative way to “see” limits at infinity. Describe the procedure and use it to picture and determine limits in Exercises 79–84. 79. 80. 81.
1 lim x sin x
x: ; q
lim
cos s1>xd 1 + s1>xd
lim
3x + 4 2x  5
x:  q
x: ; q
1 82. lim a x b x: q 83.
x: q
x 2 sin s1>xd, x 6 0 2x, x 7 0.
1>x
1 2 lim a3 + x b acos x b
x: ; q
84. lim a x: q
3 1 1  cos x b a1 + sin x b x2
71. If ƒ has the constant value ƒsxd = k , then lim ƒsxd = k . x: q
72. If ƒ has the constant value ƒsxd = k , then lim ƒsxd = k . x:  q
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Infinite Limits and Vertical Asymptotes
115
Infinite Limits and Vertical Asymptotes
2.5
In this section we extend the concept of limit to infinite limits, which are not limits as before, but rather an entirely new use of the term limit. Infinite limits provide useful symbols and language for describing the behavior of functions whose values become arbitrarily large, positive or negative. We continue our analysis of graphs of rational functions from the last section, using vertical asymptotes and dominant terms for numerically large values of x.
Infinite Limits
y
B
You can get as high as you want by taking x close enough to 0. No matter how high B is, the graph goes higher.
Let us look again at the function ƒsxd = 1>x. As x : 0 + , the values of ƒ grow without bound, eventually reaching and surpassing every positive real number. That is, given any positive real number B, however large, the values of ƒ become larger still (Figure 2.37). Thus, ƒ has no limit as x : 0 + . It is nevertheless convenient to describe the behavior of ƒ by saying that ƒ(x) approaches q as x : 0 + . We write
y 1x x0
x
No matter how low –B is, the graph goes lower. You can get as low as you want by taking x close enough to 0.
FIGURE 2.37 1 lim = q x: 0 + x
1 lim ƒsxd = lim+ x = q . x:0
x
–B
Onesided infinite limits: 1 and lim = q x: 0  x
x:0 +
In writing this, we are not saying that the limit exists. Nor are we saying that there is a real number q , for there is no such number. Rather, we are saying that limx:0+ s1>xd does not exist because 1>x becomes arbitrarily large and positive as x : 0 + . As x : 0  , the values of ƒsxd = 1>x become arbitrarily large and negative. Given any negative real number B, the values of ƒ eventually lie below B. (See Figure 2.37.) We write 1 lim ƒsxd = lim x =  q . x:0
x:0 
Again, we are not saying that the limit exists and equals the number  q . There is no real number  q . We are describing the behavior of a function whose limit as x : 0  does not exist because its values become arbitrarily large and negative.
y
y
1 x1
EXAMPLE 1
1 –1
0
Find lim+ 1
2
3
x
x:1
OneSided Infinite Limits
1 x  1
and
lim
x:1 
1 . x  1
The graph of y = 1>sx  1d is the graph of y = 1>x shifted 1 unit to the right (Figure 2.38). Therefore, y = 1>sx  1d behaves near 1 exactly the way y = 1>x behaves near 0:
Geometric Solution
FIGURE 2.38 Near x = 1 , the function y = 1>sx  1d behaves the way the function y = 1>x behaves near x = 0 . Its graph is the graph of y = 1>x shifted 1 unit to the right (Example 1).
lim
x:1 +
1 = q x  1
and
lim
x:1 
1 = q. x  1
Think about the number x  1 and its reciprocal. As x : 1+ , we have sx  1d : 0 and 1>sx  1d : q . As x : 1 , we have sx  1d : 0  and 1>sx  1d :  q . Analytic Solution +
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Chapter 2: Limits and Continuity
EXAMPLE 2
y
Discuss the behavior of
No matter how high B is, the graph goes higher.
B
f(x) 12 x
x 0
x
x
(a) ƒsxd =
1 near x = 0, x2
(b) gsxd =
1 near x = 3. sx + 3d2
Solution
(a) As x approaches zero from either side, the values of 1>x 2 are positive and become arbitrarily large (Figure 2.39a):
(a)
g(x)
TwoSided Infinite Limits
1 (x 3)2
lim ƒsxd = lim
y
x:0
x:0
1 = q. x2
(b) The graph of gsxd = 1>sx + 3d2 is the graph of ƒsxd = 1>x 2 shifted 3 units to the left (Figure 2.39b). Therefore, g behaves near 3 exactly the way ƒ behaves near 0.
5 4
x: 3
2 1 –5
–4
–3
–2
–1
1 = q. x: 3 sx + 3d2
lim gsxd = lim
3
x
0
(b)
The function y = 1>x shows no consistent behavior as x : 0. We have 1>x : q if x : 0 + , but 1>x :  q if x : 0  . All we can say about limx:0 s1>xd is that it does not exist. The function y = 1>x 2 is different. Its values approach infinity as x approaches zero from either side, so we can say that limx:0 s1>x 2 d = q .
EXAMPLE 3
FIGURE 2.39 The graphs of the functions in Example 2. (a) ƒ(x) approaches infinity as x : 0 . (b) g (x) approaches infinity as x : 3 .
Rational Functions Can Behave in Various Ways Near Zeros of Their Denominators
(a) lim
sx  2d2 sx  2d2 x  2 = lim = lim = 0 x:2 sx  2dsx + 2d x:2 x + 2 x2  4
(b) lim
x  2 x  2 1 1 = lim = lim = 4 x:2 sx  2dsx + 2d x:2 x + 2 x2  4
x:2
x:2
(c)
lim
x:2 +
(d) limx:2
x  3 x  3 = q = lim+ 2 sx 2dsx + 2d x:2 x  4
The values are negative for x 7 2, x near 2.
x  3 x  3 = lim= q 2 sx 2dsx + 2d x:2 x  4
The values are positive for x 6 2, x near 2.
x  3 x  3 = lim does not exist. x:2 sx  2dsx + 2d x2  4 sx  2d 2  x 1 (f) lim = lim = lim = q x:2 sx  2d3 x:2 sx  2d3 x:2 sx  2d2 (e) lim
x:2
See parts (c) and (d).
In parts (a) and (b) the effect of the zero in the denominator at x = 2 is canceled because the numerator is zero there also. Thus a finite limit exists. This is not true in part (f), where cancellation still leaves a zero in the denominator.
Precise Definitions of Infinite Limits Instead of requiring ƒ(x) to lie arbitrarily close to a finite number L for all x sufficiently close to x0 , the definitions of infinite limits require ƒ(x) to lie arbitrarily far from the oriCopyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
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Infinite Limits and Vertical Asymptotes
117
gin. Except for this change, the language is identical with what we have seen before. Figures 2.40 and 2.41 accompany these definitions.
y y f (x)
DEFINITIONS
Infinity, Negative Infinity as Limits
1. We say that ƒ(x) approaches infinity as x approaches x0 , and write
B
lim ƒsxd = q ,
x:x0
0
x0
if for every positive real number B there exists a corresponding d 7 0 such that for all x
x
x0
x0
Q
0 6 ƒ x  x0 ƒ 6 d
ƒsxd 7 B.
2. We say that ƒ(x) approaches negative infinity as x approaches x0 , and write FIGURE 2.40 x : x0 .
ƒ(x) approaches infinity as
lim ƒsxd =  q ,
x:x0
if for every negative real number B there exists a corresponding d 7 0 such that for all x y
Q
0 6 ƒ x  x0 ƒ 6 d x0
x0
ƒsxd 6 B.
x0 x
0
The precise definitions of onesided infinite limits at x0 are similar and are stated in the exercises.
EXAMPLE 4
–B
Prove that lim
x:0
y f (x)
Solution
FIGURE 2.41 ƒ(x) approaches negative infinity as x : x0 .
Using the Definition of Infinite Limits 1 = q. x2
Given B 7 0, we want to find d 7 0 such that 0 6 ƒx  0ƒ 6 d
implies
1 7 B. x2
Now, 1 7 B x2
if and only if x 2 6
1 B
or, equivalently, ƒxƒ 6
1 . 2B
Thus, choosing d = 1> 2B (or any smaller positive number), we see that ƒxƒ 6 d
implies
1 1 7 2 Ú B. x2 d
Therefore, by definition, lim
x:0
1 = q. x2
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Chapter 2: Limits and Continuity
Vertical Asymptotes
y
Notice that the distance between a point on the graph of y = 1>x and the yaxis approaches zero as the point moves vertically along the graph and away from the origin (Figure 2.42). This behavior occurs because
Vertical asymptote
Horizontal asymptote
y 1x
1 lim x = q
1 0
Horizontal asymptote, y0
We say that the line x = 0 (the yaxis) is a vertical asymptote of the graph of y = 1>x. Observe that the denominator is zero at x = 0 and the function is undefined there.
DEFINITION Vertical Asymptote A line x = a is a vertical asymptote of the graph of a function y = ƒsxd if either lim ƒsxd = ; q
FIGURE 2.42 The coordinate axes are asymptotes of both branches of the hyperbola y = 1>x .
EXAMPLE 5
Horizontal asymptote, y1
3
Find the horizontal and vertical asymptotes of the curve y
y =
x3 x2
1
1 x2
1
3
We are interested in the behavior as x : ; q and as x : 2, where the denominator is zero. The asymptotes are quickly revealed if we recast the rational function as a polynomial with a remainder, by dividing sx + 2d into sx + 3d.
1 2
x + 3 . x + 2
Solution
2
–5 –4 –3 –2 –1 0 –1
lim ƒsxd = ; q .
x:a 
Looking for Asymptotes
6 4
or
x:a +
y
5
1 lim x =  q .
x:0 
x
1
Vertical asymptote, x0
Vertical asymptote, x –2
and
x:0 +
x
1 x + 2 x + 3 x + 2 1
–2 –3 –4
This result enables us to rewrite y: FIGURE 2.43 The lines y = 1 and x = 2 are asymptotes of the curve y = sx + 3d>sx + 2d (Example 5).
y = 1 +
1 . x + 2
We now see that the curve in question is the graph of y = 1>x shifted 1 unit up and 2 units left (Figure 2.43). The asymptotes, instead of being the coordinate axes, are now the lines y = 1 and x = 2.
EXAMPLE 6
Asymptotes Need Not Be TwoSided
Find the horizontal and vertical asymptotes of the graph of ƒsxd = 
8 . x2  4
We are interested in the behavior as x : ; q and as x : ;2, where the denominator is zero. Notice that ƒ is an even function of x, so its graph is symmetric with respect to the yaxis.
Solution
(a) The behavior as x : ; q . Since limx: q ƒsxd = 0, the line y = 0 is a horizontal asymptote of the graph to the right. By symmetry it is an asymptote to the left as well Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
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2.5
(Figure 2.44). Notice that the curve approaches the xaxis from only the negative side (or from below). (b) The behavior as x : ;2. Since
y
Vertical asymptote, x –2
8 7 6 5 4 3 2 1
–4 –3 –2 –1 0
y – 28 x 4
lim ƒsxd =  q
Vertical asymptote, x 2
and
x:2 +
lim ƒsxd = q ,
x:2 
the line x = 2 is a vertical asymptote both from the right and from the left. By symmetry, the same holds for the line x = 2.
Horizontal asymptote, y 0 1 2 3 4
119
Infinite Limits and Vertical Asymptotes
x
There are no other asymptotes because ƒ has a finite limit at every other point.
EXAMPLE 7
Curves with Infinitely Many Asymptotes
The curves 1 y = sec x = cos x FIGURE 2.44 Graph of y = 8>sx 2  4d . Notice that the curve approaches the xaxis from only one side. Asymptotes do not have to be twosided (Example 6).
and
sin x y = tan x = cos x
both have vertical asymptotes at oddinteger multiples of p>2, where cos x = 0 (Figure 2.45). y
y
y sec x
1 – 3 – – 2 2
y tan x
1 2
0
x
3 2
0 – 3 – – –1 2 2 2
3 2
x
FIGURE 2.45 The graphs of sec x and tan x have infinitely many vertical asymptotes (Example 7).
The graphs of y = csc x =
1 sin x
and
y = cot x =
cos x sin x
have vertical asymptotes at integer multiples of p, where sin x = 0 (Figure 2.46). y
y
y csc x
1
1 – – 2
0
y cot x
2
3 2 2
x
– – 2
0
FIGURE 2.46 The graphs of csc x and cot x (Example 7).
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2
3 2 2
x
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Chapter 2: Limits and Continuity
EXAMPLE 8
A Rational Function with Degree of Numerator Greater than Degree of Denominator
Find the asymptotes of the graph of y
3x1 1 2x 4 2 2x 4 The vertical distance between curve and line goes to zero as x →
6 5
3
We are interested in the behavior as x : ; q and also as x : 2, where the denominator is zero. We divide s2x  4d into sx 2  3d: x + 1 2 2x  4 x 2  3 x 2  2x 2x  3 2x  4 1
Oblique asymptote
x2 y x 1 2
2 1 –1 0 –1 –2 –3
x2  3 . 2x  4
Solution
y
4
ƒsxd =
x2
1
2
3
4
x
x
This tells us that
Vertical asymptote, x2
ƒsxd =
FIGURE 2.47 The graph of ƒsxd = sx 2  3d>s2x  4d has a vertical asymptote and an oblique asymptote (Example 8).
x2  3 x 1 . = + 1 + 2x  4 2 2x  4 123 linear
14243 remainder
Since limx:2+ ƒsxd = q and limx:2 ƒsxd =  q , the line x = 2 is a twosided vertical asymptote. As x : ; q , the remainder approaches 0 and ƒsxd : sx>2d + 1. The line y = sx>2d + 1 is an oblique asymptote both to the right and to the left (Figure 2.47). Notice in Example 8, that if the degree of the numerator in a rational function is greater than the degree of the denominator, then the limit is + q or  q , depending on the signs assumed by the numerator and denominator as ƒ x ƒ becomes large.
Dominant Terms Of all the observations we can make quickly about the function ƒsxd =
x2  3 2x  4
in Example 8, probably the most useful is that ƒsxd =
x 1 + 1 + . 2 2x  4
This tells us immediately that ƒsxd L
x + 1 2
For x numerically large
ƒsxd L
1 2x  4
For x near 2
If we want to know how ƒ behaves, this is the way to find out. It behaves like y = sx>2d + 1 when x is numerically large and the contribution of 1>s2x  4d to the total
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2.5
121
Infinite Limits and Vertical Asymptotes
value of ƒ is insignificant. It behaves like 1>s2x  4d when x is so close to 2 that 1>s2x  4d makes the dominant contribution. We say that sx>2d + 1 dominates when x is numerically large, and we say that 1>s2x  4d dominates when x is near 2. Dominant terms like these are the key to predicting a function’s behavior. Here’s another example.
EXAMPLE 9
Two Graphs Appearing Identical on a Large Scale
Let ƒsxd = 3x 4  2x 3 + 3x 2  5x + 6 and gsxd = 3x 4 . Show that although ƒ and g are quite different for numerically small values of x, they are virtually identical for ƒ x ƒ very large. The graphs of ƒ and g behave quite differently near the origin (Figure 2.48a), but appear as virtually identical on a larger scale (Figure 2.48b). Solution
y
y
20
500,000
15 300,000 10 f (x) 5 –2
–1
0
100,000
g(x) 1
2
x
–5
–20
–10
0
10
20
x
–100,000
(a)
(b)
FIGURE 2.48 The graphs of ƒ and g, (a) are distinct for ƒ x ƒ small, and (b) nearly identical for ƒ x ƒ large (Example 9).
We can test that the term 3x 4 in ƒ, represented graphically by g, dominates the polynomial ƒ for numerically large values of x by examining the ratio of the two functions as x : ; q . We find that ƒsxd = x: ; q gsxd lim
=
3x 4  2x 3 + 3x 2  5x + 6 x: ; q 3x 4 lim
lim a1 
x: ; q
5 2 1 2 + 2 + 4b 3 3x x 3x x
= 1, so that ƒ and g are nearly identical for ƒ x ƒ large.
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Chapter 2: Limits and Continuity
EXERCISES 2.5 x 2  3x + 2 as x 3  2x 2 a. x : 0 + c. x : 2
Infinite Limits
21. lim
Find the limits in Exercises 1–12. 1. lim+
1 3x
2. lim
5 2x
3. lim
3 x  2
4. lim+
1 x  3
x: 0
x: 2
5.
lim
x: 8 +
7. lim
x: 7
x: 0
x: 3
2x x + 8
6.
4 sx  7d2
9. a. lim+ x:0
10. a. lim+ x:0
8. lim
x: 0
2 3x 1>3
x: 0
x
3x 2x + 10
1 x 2sx + 1d
b. limx: 0
2
b. lim
x 1>5
x: 0
4
11. lim
lim
x: 5 
12. lim
2>5
x: 0
2 3x 1>3 2 x 1>5
15.
lim
x: sp>2d

tan x
14.
lim s1 + csc ud
x: sp>2d
+
sec x
16. lim s2  cot ud
u :0
u: 0
1 as x2  4 a. x : 2+
b. x : 2
c. x : 2+
d. x : 2
17. lim
x as x2  1 a. x : 1+
b. x : 1
c. x : 1+
d. x : 1
18. lim
x2 1  x b as 2
a. x : 0 +
b. x : 0 
c. x : 22
d. x : 1
20. lim
x2  1 as 2x + 4
a. x : 2+
b. x : 2
c. x : 1
d. x : 0
+
a. x : 2+ c. x : 0 
b. x : 2+ d. x : 1+
e. What, if anything, can be said about the limit as x : 0 ? Find the limits in Exercises 23–26. 23. lim a2 
3 b as t 1>3
a. t : 0 +
b. t : 0 
1 2 + b as x 2>3 sx  1d2>3 a. x : 0 + c. x : 1+

b. t : 0 
25. lim a
b. x : 0 d. x : 1
1 1 b as x 1>3 sx  1d4>3
a. x : 0 + c. x : 1+
Find the limits in Exercises 17–22.
3
x 2  3x + 2 as x 3  4x
26. lim a
Additional Calculations
19. lim a
22. lim
1 + 7b as t 3>5 a. t : 0 +
x 2>3
lim
e. What, if anything, can be said about the limit as x : 0 ?
24. lim a
1
Find the limits in Exercises 13–16. 13.
b. x : 2+ d. x : 2
b. x : 0 d. x : 1
Graphing Rational Functions Graph the rational functions in Exercises 27–38. Include the graphs and equations of the asymptotes and dominant terms. 27. y =
1 x  1
28. y =
1 x + 1
29. y =
1 2x + 4
30. y =
3 x  3
31. y =
x + 3 x + 2
32. y =
2x x + 1
33. y =
x2 x  1
34. y =
x2 + 1 x  1
35. y =
x2  4 x  1
36. y =
x2  1 2x + 4
37. y =
x2  1 x
38. y =
x3 + 1 x2
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2.5 Infinite Limits and Vertical Asymptotes
Inventing Graphs from Values and Limits In Exercises 39–42, sketch the graph of a function y = ƒsxd that satisfies the given conditions. No formulas are required—just label the coordinate axes and sketch an appropriate graph. (The answers are not unique, so your graphs may not be exactly like those in the answer section.) 39. ƒs0d = 0, ƒs1d = 2, ƒs 1d = 2, lim ƒsxd = 1, and x:  q
lim ƒsxd = 1
x: q
40. ƒs0d = 0, lim ƒsxd = 0, lim+ ƒsxd = 2, and x: ; q
x: 0
lim ƒsxd = 2
x:0 
41. ƒs0d = 0, lim ƒsxd = 0, lim ƒsxd = lim + ƒsxd = q , x: ; q
x: 1
x: 1
Modify the definition to cover the following cases. a. b.
lim ƒsxd = q
x:x0 
lim ƒsxd =  q
x:x0 +
c. lim  ƒsxd =  q x:x0
Use the formal definitions from Exercise 51 to prove the limit statements in Exercises 52–56. 1 52. lim+ x = q x:0 1 53. lim x =  q x:0
lim ƒsxd =  q , and lim  ƒsxd =  q
x:1 +
x: 1
42. ƒs2d = 1, ƒs 1d = 0, lim ƒsxd = 0, lim+ ƒsxd = q , x: q
x: 0
x:  q
In Exercises 43–46, find a function that satisfies the given conditions and sketch its graph. (The answers here are not unique. Any function that satisfies the conditions is acceptable. Feel free to use formulas defined in pieces if that will help.) 43. lim ƒsxd = 0, lim ƒsxd = q , and lim+ ƒsxd = q 44. 45.
x: 2
x: 2
lim g sxd = 0, lim g sxd =  q , and lim+ g sxd = q
x: ; q
x: 3
x: 3
lim hsxd = 1, lim hsxd = 1, lim hsxd = 1, and
x:  q
x: q
1 = q x  2
55. lim+
1 = q x  2
56. lim
1 = q 1  x2
x:2
x:2
Inventing Functions
x: ; q
54. lim
x: 0
lim ƒsxd =  q , and lim ƒsxd = 1
x: 0
x:1
Graphing Terms Each of the functions in Exercises 57–60 is given as the sum or difference of two terms. First graph the terms (with the same set of axes). Then, using these graphs as guides, sketch in the graph of the function. 1 57. y = sec x + x ,
lim hsxd = 1

p p 6 x 6 2 2
58. y = sec x 
1 , x2

p p 6 x 6 2 2
The Formal Definition of Infinite Limit
59. y = tan x +
1 , x2

p p 6 x 6 2 2
Use formal definitions to prove the limit statements in Exercises 47–50.
1 60. y = x  tan x,
x: 0 +
46.
lim k sxd = 1, lim k sxd = q , and lim+ k sxd =  q
x: ; q
x: 1
x: 1
1 = q 48. lim x: 0 ƒ x ƒ
1 47. lim 2 =  q x:0 x 49. lim
x:3
2 = q sx  3d2
50. lim
x: 5
1 = q sx + 5d2
Formal Definitions of Infinite OneSided Limits
lim
ƒsxd = q ,
if, for every positive real number B, there exists a corresponding number d 7 0 such that for all x x0 6 x 6 x0 + d
Q
p p 6 x 6 2 2
Graph the curves in Exercises 61–64. Explain the relation between the curve’s formula and what you see. 61. y =
x: x0 +

Grapher Explorations—Comparing Graphs with Formulas
51. Here is the definition of infinite righthand limit.
We say that ƒ(x) approaches infinity as x approaches x0 from the right, and write
123
62. y =
x 24  x 2 1
24  x 2 1 63. y = x 2>3 + 1>3 x 64. y = sin a
p b x2 + 1
ƒsxd 7 B.
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Chapter 2: Limits and Continuity
Continuity
2.6 y
Distance fallen (m)
80
P
Q4 Q3
60 Q2
40 Q1
20 0
5 Elapsed time (sec)
10
t
FIGURE 2.49 Connecting plotted points by an unbroken curve from experimental data Q1 , Q2 , Q3 , Á for a falling object.
When we plot function values generated in a laboratory or collected in the field, we often connect the plotted points with an unbroken curve to show what the function’s values are likely to have been at the times we did not measure (Figure 2.49). In doing so, we are assuming that we are working with a continuous function, so its outputs vary continuously with the inputs and do not jump from one value to another without taking on the values in between. The limit of a continuous function as x approaches c can be found simply by calculating the value of the function at c. (We found this to be true for polynomials in Section 2.2.) Any function y = ƒsxd whose graph can be sketched over its domain in one continuous motion without lifting the pencil is an example of a continuous function. In this section we investigate more precisely what it means for a function to be continuous. We also study the properties of continuous functions, and see that many of the function types presented in Section 1.4 are continuous.
Continuity at a Point To understand continuity, we need to consider a function like the one in Figure 2.50 whose limits we investigated in Example 2, Section 2.4.
EXAMPLE 1
Investigating Continuity
y
Find the points at which the function ƒ in Figure 2.50 is continuous and the points at which ƒ is discontinuous.
y f (x)
2
The function ƒ is continuous at every point in its domain [0, 4] except at x = 1, x = 2, and x = 4. At these points, there are breaks in the graph. Note the relationship between the limit of ƒ and the value of ƒ at each point of the function’s domain.
Solution 1
0
1
2
3
x
4
Points at which ƒ is continuous:
FIGURE 2.50 The function is continuous on [0, 4] except at x = 1, x = 2 , and x = 4 (Example 1).
At x = 0, At x = 3, At 0 6 c 6 4, c Z 1, 2,
lim ƒsxd = ƒs0d.
x:0 +
lim ƒsxd = ƒs3d.
x:3
lim ƒsxd = ƒscd.
x:c
Points at which ƒ is discontinuous:
At x = 1, Continuity from the right
Twosided continuity
Continuity from the left
At x = 2, At x = 4,
y f (x)
At c 6 0, c 7 4, a
c
b
lim ƒsxd does not exist.
x:1
lim ƒsxd = 1, but 1 Z ƒs2d.
x:2
lim ƒsxd = 1, but 1 Z ƒs4d.
x:4 
these points are not in the domain of ƒ.
x
FIGURE 2.51 Continuity at points a, b, and c.
To define continuity at a point in a function’s domain, we need to define continuity at an interior point (which involves a twosided limit) and continuity at an endpoint (which involves a onesided limit) (Figure 2.51).
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2.6
Continuity
125
DEFINITION Continuous at a Point Interior point: A function y = ƒsxd is continuous at an interior point c of its domain if lim ƒsxd = ƒscd.
x:c
Endpoint: A function y = ƒsxd is continuous at a left endpoint a or is continuous at a right endpoint b of its domain if lim ƒsxd = ƒsad
x:a +
y 2
–2
0
y 兹4 x 2
2
x
FIGURE 2.52 A function that is continuous at every domain point (Example 2).
or
lim ƒsxd = ƒsbd,
x:b 
respectively.
If a function ƒ is not continuous at a point c, we say that ƒ is discontinuous at c and c is a point of discontinuity of ƒ. Note that c need not be in the domain of ƒ. A function ƒ is rightcontinuous (continuous from the right) at a point x = c in its domain if limx:c+ ƒsxd = ƒscd. It is leftcontinuous (continuous from the left) at c if limx:c ƒsxd = ƒscd. Thus, a function is continuous at a left endpoint a of its domain if it is rightcontinuous at a and continuous at a right endpoint b of its domain if it is leftcontinuous at b. A function is continuous at an interior point c of its domain if and only if it is both rightcontinuous and leftcontinuous at c (Figure 2.51).
EXAMPLE 2
A Function Continuous Throughout Its Domain
The function ƒsxd = 24  x 2 is continuous at every point of its domain, [2, 2] (Figure 2.52), including x = 2, where ƒ is rightcontinuous, and x = 2, where ƒ is leftcontinuous.
y 1
0
EXAMPLE 3
y U(x)
The Unit Step Function Has a Jump Discontinuity
The unit step function U(x), graphed in Figure 2.53, is rightcontinuous at x = 0, but is neither leftcontinuous nor continuous there. It has a jump discontinuity at x = 0. x
We summarize continuity at a point in the form of a test.
FIGURE 2.53 A function that is rightcontinuous, but not leftcontinuous, at the origin. It has a jump discontinuity there (Example 3).
Continuity Test A function ƒ(x) is continuous at x = c if and only if it meets the following three conditions. 1. 2. 3.
ƒ(c) exists limx:c ƒsxd exists limx:c ƒsxd = ƒscd
(c lies in the domain of ƒ) (ƒ has a limit as x : c) (the limit equals the function value)
For onesided continuity and continuity at an endpoint, the limits in parts 2 and 3 of the test should be replaced by the appropriate onesided limits.
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Chapter 2: Limits and Continuity
EXAMPLE 4
y 4
The function y = :x; or y = int x, introduced in Chapter 1, is graphed in Figure 2.54. It is discontinuous at every integer because the limit does not exist at any integer n:
y int x or y x
3
lim int x = n  1
1 1
2
and
x:n 
2
–1
The Greatest Integer Function
3
4
x
lim int x = n
x:n +
so the lefthand and righthand limits are not equal as x : n. Since int n = n, the greatest integer function is rightcontinuous at every integer n (but not leftcontinuous). The greatest integer function is continuous at every real number other than the integers. For example, lim int x = 1 = int 1.5.
x:1.5
In general, if n  1 6 c 6 n, n an integer, then
–2
lim int x = n  1 = int c.
FIGURE 2.54 The greatest integer function is continuous at every noninteger point. It is rightcontinuous, but not leftcontinuous, at every integer point (Example 4).
x:c
Figure 2.55 is a catalog of discontinuity types. The function in Figure 2.55a is continuous at x = 0. The function in Figure 2.55b would be continuous if it had ƒs0d = 1. The function in Figure 2.55c would be continuous if ƒ(0) were 1 instead of 2. The discontinuities in Figure 2.55b and c are removable. Each function has a limit as x : 0, and we can remove the discontinuity by setting ƒ(0) equal to this limit. The discontinuities in Figure 2.55d through f are more serious: limx:0 ƒsxd does not exist, and there is no way to improve the situation by changing ƒ at 0. The step function in Figure 2.55d has a jump discontinuity: The onesided limits exist but have different values. The function ƒsxd = 1>x 2 in Figure 2.55e has an infinite discontinuity. The function in Figure 2.55f has an oscillating discontinuity: It oscillates too much to have a limit as x : 0.
y
y
y y f (x) 1
0
2
y f (x) 1
x
x
0
(b)
(c) y
y y f (x) 12 x
1
1 0
y f (x)
1
0
(a)
y
1 x
y f(x)
0
(d) y sin 2 x
0
x
x –1
(e)
(f)
FIGURE 2.55 The function in (a) is continuous at x = 0 ; the functions in (b) through (f ) are not.
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x
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2.6
127
Continuous Functions
y y 1x
0
Continuity
x
A function is continuous on an interval if and only if it is continuous at every point of the interval. For example, the semicircle function graphed in Figure 2.52 is continuous on the interval [2, 2], which is its domain. A continuous function is one that is continuous at every point of its domain. A continuous function need not be continuous on every interval. For example, y = 1>x is not continuous on [1, 1] (Figure 2.56), but it is continuous over its domain s  q , 0d ´ s0, q d.
EXAMPLE 5
FIGURE 2.56 The function y = 1>x is continuous at every value of x except x = 0 . It has a point of discontinuity at x = 0 (Example 5).
Identifying Continuous Functions
(a) The function y = 1>x (Figure 2.56) is a continuous function because it is continuous at every point of its domain. It has a point of discontinuity at x = 0, however, because it is not defined there. (b) The identity function ƒsxd = x and constant functions are continuous everywhere by Example 3, Section 2.3. Algebraic combinations of continuous functions are continuous wherever they are defined.
THEOREM 9 Properties of Continuous Functions If the functions ƒ and g are continuous at x = c, then the following combinations are continuous at x = c. 1. 2. 3. 4. 5. 6.
Sums: Differences: Products: Constant multiples: Quotients: Powers:
ƒ + g ƒ  g ƒ#g k # ƒ, for any number k ƒ>g provided gscd Z 0 f r>s , provided it is defined on an open interval containing c, where r and s are integers
Most of the results in Theorem 9 are easily proved from the limit rules in Theorem 1, Section 2.2. For instance, to prove the sum property we have lim sƒ + gdsxd = lim sƒsxd + gsxdd
x:c
x:c
= lim ƒsxd + lim gsxd,
Sum Rule, Theorem 1
= ƒscd + gscd = sƒ + gdscd.
Continuity of ƒ, g at c
x:c
x:c
This shows that ƒ + g is continuous.
EXAMPLE 6
Polynomial and Rational Functions Are Continuous
(a) Every polynomial Psxd = an x n + an  1x n  1 + Á + a0 is continuous because lim Psxd = Pscd by Theorem 2, Section 2.2. x:c
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(b) If P(x) and Q(x) are polynomials, then the rational function Psxd>Qsxd is continuous wherever it is defined sQscd Z 0d by the Quotient Rule in Theorem 9.
EXAMPLE 7
Continuity of the Absolute Value Function
The function ƒsxd = ƒ x ƒ is continuous at every value of x. If x 7 0, we have ƒsxd = x, a polynomial. If x 6 0, we have ƒsxd = x, another polynomial. Finally, at the origin, limx:0 ƒ x ƒ = 0 = ƒ 0 ƒ . The functions y = sin x and y = cos x are continuous at x = 0 by Example 6 of Section 2.2. Both functions are, in fact, continuous everywhere (see Exercise 62). It follows from Theorem 9 that all six trigonometric functions are then continuous wherever they are defined. For example, y = tan x is continuous on Á ´ s p>2, p>2d ´ sp>2, 3p>2d ´ Á .
Composites All composites of continuous functions are continuous. The idea is that if ƒ(x) is continuous at x = c and g(x) is continuous at x = ƒscd, then g f is continuous at x = c (Figure 2.57). In this case, the limit as x : c is g(ƒ(c)). g f ˚ Continuous at c f
g
Continuous at c
Continuous at f (c) f (c)
c
g( f(c))
FIGURE 2.57 Composites of continuous functions are continuous.
THEOREM 10 Composite of Continuous Functions If ƒ is continuous at c and g is continuous at ƒ(c), then the composite g f is continuous at c.
Intuitively, Theorem 10 is reasonable because if x is close to c, then ƒ(x) is close to ƒ(c), and since g is continuous at ƒ(c), it follows that g(ƒ(x)) is close to g(ƒ(c)). The continuity of composites holds for any finite number of functions. The only requirement is that each function be continuous where it is applied. For an outline of the proof of Theorem 10, see Exercise 6 in Appendix 2.
EXAMPLE 8
Applying Theorems 9 and 10
Show that the following functions are continuous everywhere on their respective domains. x 2>3 1 + x4
(a) y = 2x 2  2x  5
(b) y =
(c) y = `
(d) y = `
x  2 ` x2  2
x sin x ` x2 + 2
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2.6 Solution
y 0.4 0.3 0.2 0.1 –2
–
0
129
Continuity
2
FIGURE 2.58 The graph suggests that y = ƒ sx sin xd>sx 2 + 2d ƒ is continuous (Example 8d).
x
(a) The square root function is continuous on [0, q d because it is a rational power of the continuous identity function ƒsxd = x (Part 6, Theorem 9). The given function is then the composite of the polynomial ƒsxd = x 2  2x  5 with the square root function gstd = 2t. (b) The numerator is a rational power of the identity function; the denominator is an everywherepositive polynomial. Therefore, the quotient is continuous. (c) The quotient sx  2d>sx 2  2d is continuous for all x Z ; 22, and the function is the composition of this quotient with the continuous absolute value function (Example 7). (d) Because the sine function is everywherecontinuous (Exercise 62), the numerator term x sin x is the product of continuous functions, and the denominator term x 2 + 2 is an everywherepositive polynomial. The given function is the composite of a quotient of continuous functions with the continuous absolute value function (Figure 2.58).
Continuous Extension to a Point The function y = ssin xd>x is continuous at every point except x = 0. In this it is like the function y = 1>x. But y = ssin xd>x is different from y = 1>x in that it has a finite limit as x : 0 (Theorem 7). It is therefore possible to extend the function’s domain to include the point x = 0 in such a way that the extended function is continuous at x = 0. We define Fsxd = L
sin x x ,
x Z 0
1,
x = 0.
The function F(x) is continuous at x = 0 because lim
x:0
sin x x = Fs0d
(Figure 2.59).
y
y
(0, 1) – , 2 2 – 2
(0, 1)
f (x) , 2 2
0
– , 2 2 2
x
– 2
(a)
F(x) , 2 2
0
2
(b)
FIGURE 2.59 The graph (a) of ƒsxd = ssin xd>x for p>2 … x … p>2 does not include the point (0, 1) because the function is not defined at x = 0 . (b) We can remove the discontinuity from the graph by defining the new function F(x) with Fs0d = 1 and Fsxd = ƒsxd everywhere else. Note that Fs0d = lim ƒsxd . x:0
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More generally, a function (such as a rational function) may have a limit even at a point where it is not defined. If ƒ(c) is not defined, but limx:c ƒsxd = L exists, we can define a new function F(x) by the rule Fsxd = e
ƒsxd, L,
if x is in the domain of f if x = c.
The function F is continuous at x = c. It is called the continuous extension of ƒ to x = c. For rational functions ƒ, continuous extensions are usually found by canceling common factors.
EXAMPLE 9
A Continuous Extension
Show that ƒsxd =
x2 + x  6 x2  4
y 2
y
has a continuous extension to x = 2, and find that extension.
x2 x 6 x2 4
Solution
1 –1
0
1
2
3
4
–1
x 3 y x2
Fsxd =
1 0
sx  2dsx + 3d x2 + x  6 x + 3 . = = x + 2 sx  2dsx + 2d x2  4
The new function
y 2
ƒsxd =
x
(a)
5 4
Although ƒ(2) is not defined, if x Z 2 we have
1
2
3
4
x
(b)
x + 3 x + 2
is equal to ƒ(x) for x Z 2, but is continuous at x = 2, having there the value of 5>4. Thus F is the continuous extension of ƒ to x = 2, and x2 + x  6 5 = lim ƒsxd = . 4 x:2 x:2 x2  4 lim
FIGURE 2.60 (a) The graph of ƒ(x) and (b) the graph of its continuous extension F(x) (Example 9).
The graph of ƒ is shown in Figure 2.60. The continuous extension F has the same graph except with no hole at (2, 5>4). Effectively, F is the function ƒ with its point of discontinuity at x = 2 removed.
Intermediate Value Theorem for Continuous Functions Functions that are continuous on intervals have properties that make them particularly useful in mathematics and its applications. One of these is the Intermediate Value Property. A function is said to have the Intermediate Value Property if whenever it takes on two values, it also takes on all the values in between.
THEOREM 11 The Intermediate Value Theorem for Continuous Functions A function y = ƒsxd that is continuous on a closed interval [a, b] takes on every value between ƒ(a) and ƒ(b). In other words, if y0 is any value between ƒ(a) and ƒ(b), then y0 = ƒscd for some c in [a, b].
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131
y y f (x) f (b)
y0 f (a)
0
a
c
b
x
Geometrically, the Intermediate Value Theorem says that any horizontal line y = y0 crossing the yaxis between the numbers ƒ(a) and ƒ(b) will cross the curve y = ƒsxd at least once over the interval [a, b]. The proof of the Intermediate Value Theorem depends on the completeness property of the real number system and can be found in more advanced texts. The continuity of ƒ on the interval is essential to Theorem 11. If ƒ is discontinuous at even one point of the interval, the theorem’s conclusion may fail, as it does for the function graphed in Figure 2.61.
y 3
x
A Consequence for Graphing: Connectivity Theorem 11 is the reason the graph of a function continuous on an interval cannot have any breaks over the interval. It will be connected, a single, unbroken curve, like the graph of sin x. It will not have jumps like the graph of the greatest integer function (Figure 2.54) or separate branches like the graph of 1>x (Figure 2.56).
FIGURE 2.61 The function 2x  2, 1 … x 6 2 ƒsxd = e 3, 2 … x … 4 does not take on all values between ƒs1d = 0 and ƒs4d = 3 ; it misses all the values between 2 and 3.
A Consequence for Root Finding We call a solution of the equation ƒsxd = 0 a root of the equation or zero of the function ƒ. The Intermediate Value Theorem tells us that if ƒ is continuous, then any interval on which ƒ changes sign contains a zero of the function. In practical terms, when we see the graph of a continuous function cross the horizontal axis on a computer screen, we know it is not stepping across. There really is a point where the function’s value is zero. This consequence leads to a procedure for estimating the zeros of any continuous function we can graph:
2 1
0
1
2
3
4
1. 2.
Graph the function over a large interval to see roughly where the zeros are. Zoom in on each zero to estimate its xcoordinate value.
You can practice this procedure on your graphing calculator or computer in some of the exercises. Figure 2.62 shows a typical sequence of steps in a graphical solution of the equation x 3  x  1 = 0.
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5
1
1 –1
1.6
2
–2
–1 (a)
(b)
0.02
0.003
1.320
1.330
–0.02
1.3240
1.3248
–0.003 (c)
(d)
FIGURE 2.62 Zooming in on a zero of the function ƒsxd = x 3  x  1 . The zero is near x = 1.3247 .
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EXERCISES 2.6 Exercises 5–10 are about the function
Continuity from Graphs In Exercises 1–4, say whether the function graphed is continuous on [1, 3] . If not, where does it fail to be continuous and why? 1.
2. y
y y f (x)
–1
y g(x)
2
2
1
1
0
1
2
3
x
3.
–1
1
2
3
x
y h(x)
1
1
0
1
2
3
x
–1
0
x x 1 x x
6 0 6 1 6 2 6 3
y f (x)
(1, 2) 2 y 2x y –2x 4 1 (1, 1)
y
2
… 6 = 6 6
y
4.
2
1 0 x 1 2
graphed in the accompanying figure.
0
y
–1
x 2  1, 2x, ƒsxd = e 1, 2x + 4, 0,
y k(x)
–1 y x2 1
1
2
3
x
0
1
2
–1
The graph for Exercises 5–10.
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3
x
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133
5. a. b. c. d.
Does ƒs 1d exist? Does limx: 1+ ƒsxd exist? Does limx:1+ ƒsxd = ƒs 1d ? Is ƒ continuous at x = 1 ?
33. lim cos a
6. a. b. c. d.
Does ƒ(1) exist? Does limx:1 ƒsxd exist? Does limx:1 ƒsxd = ƒs1d ?
Continuous Extensions
Is ƒ continuous at x = 1 ?
36. Define h(2) in a way that extends hstd = st 2 + 3t  10d>st  2d to be continuous at t = 2 .
t: 0
34.
p 219  3 sec 2t
b
lim 2csc2 x + 513 tan x
x:p>6
35. Define g (3) in a way that extends g sxd = sx 2  9d>sx  3d to be continuous at x = 3 .
7. a. Is ƒ defined at x = 2 ? (Look at the definition of ƒ.) b. Is ƒ continuous at x = 2 ? 8. At what values of x is ƒ continuous?
37. Define ƒ(1) in a way that extends ƒssd = ss 3  1d>ss 2  1d to be continuous at s = 1 .
9. What value should be assigned to ƒ(2) to make the extended function continuous at x = 2 ?
38. Define g(4) in a way that extends g sxd = sx 2  16d> sx 2  3x  4d to be continuous at x = 4 .
10. To what new value should ƒ(1) be changed to remove the discontinuity?
39. For what value of a is ƒsxd = e
Applying the Continuity Test At which points do the functions in Exercises 11 and 12 fail to be continuous? At which points, if any, are the discontinuities removable? Not removable? Give reasons for your answers. 11. Exercise 1, Section 2.4
1  3x 13. y = x  2 15. y =
1 + 4 14. y = sx + 2d2
x + 1 x 2  4x + 3
17. y = ƒ x  1 ƒ + sin x 19. y =
cos x x
16. y =
x + 3 x 2  3x  10
18. y =
x2 1 2 ƒxƒ + 1
x + 2 20. y = cos x px 2
x 6 3 x Ú 3
continuous at every x? 40. For what value of b is g sxd = e
12. Exercise 2, Section 2.4
At what points are the functions in Exercises 13–28 continuous?
x 2  1, 2ax,
x, bx 2,
x 6 2 x Ú 2
continuous at every x? T In Exercises 41–44, graph the function ƒ to see whether it appears to have a continuous extension to the origin. If it does, use Trace and Zoom to find a good candidate for the extended function’s value at x = 0 . If the function does not appear to have a continuous extension, can it be extended to be continuous at the origin from the right or from the left? If so, what do you think the extended function’s value(s) should be? 41. ƒsxd =
10 x  1 x
42. ƒsxd =
10 ƒ x ƒ  1 x
sin x ƒxƒ
44. ƒsxd = s1 + 2xd1>x
21. y = csc 2x
22. y = tan
x tan x 23. y = 2 x + 1 25. y = 22x + 3
2x 4 + 1 24. y = 1 + sin2 x 4 26. y = 23x  1
43. ƒsxd =
27. y = s2x  1d1>3
28. y = s2  xd1>5
45. A continuous function y = ƒsxd is known to be negative at x = 0 and positive at x = 1 . Why does the equation ƒsxd = 0 have at least one solution between x = 0 and x = 1 ? Illustrate with a sketch.
Composite Functions Find the limits in Exercises 29–34. Are the functions continuous at the point being approached? 29. lim sin sx  sin xd x:p
30. lim sin a t:0
p cos stan tdb 2
31. lim sec sy sec2 y  tan2 y  1d y:1
32. lim tan a x:0
p cos ssin x 1>3 db 4
Theory and Examples
46. Explain why the equation cos x = x has at least one solution. 47. Roots of a cubic Show that the equation x 3  15x + 1 = 0 has three solutions in the interval [4, 4] . 48. A function value Show that the function Fsxd = sx  ad2 # sx  bd2 + x takes on the value sa + bd>2 for some value of x. 49. Solving an equation If ƒsxd = x 3  8x + 10 , show that there are values c for which ƒ(c) equals (a) p ; (b)  23 ; (c) 5,000,000.
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50. Explain why the following five statements ask for the same information. a. Find the roots of ƒsxd = x 3  3x  1 . b. Find the xcoordinates of the points where the curve y = x 3 crosses the line y = 3x + 1 . c. Find all the values of x for which x 3  3x = 1 . d. Find the xcoordinates of the points where the cubic curve y = x 3  3x crosses the line y = 1 . e. Solve the equation x 3  3x  1 = 0 . 51. Removable discontinuity Give an example of a function ƒ(x) that is continuous for all values of x except x = 2 , where it has a removable discontinuity. Explain how you know that ƒ is discontinuous at x = 2 , and how you know the discontinuity is removable. 52. Nonremovable discontinuity Give an example of a function g(x) that is continuous for all values of x except x = 1 , where it has a nonremovable discontinuity. Explain how you know that g is discontinuous there and why the discontinuity is not removable.
58. Stretching a rubber band Is it true that if you stretch a rubber band by moving one end to the right and the other to the left, some point of the band will end up in its original position? Give reasons for your answer. 59. A fixed point theorem Suppose that a function ƒ is continuous on the closed interval [0, 1] and that 0 … ƒsxd … 1 for every x in [0, 1]. Show that there must exist a number c in [0, 1] such that ƒscd = c (c is called a fixed point of ƒ). 60. The signpreserving property of continuous functions Let ƒ be defined on an interval (a, b) and suppose that ƒscd Z 0 at some c where ƒ is continuous. Show that there is an interval sc  d, c + dd about c where ƒ has the same sign as ƒ(c). Notice how remarkable this conclusion is. Although ƒ is defined throughout (a, b), it is not required to be continuous at any point except c. That and the condition ƒscd Z 0 are enough to make ƒ different from zero (positive or negative) throughout an entire interval. 61. Prove that ƒ is continuous at c if and only if lim ƒsc + hd = ƒscd .
h:0
53. A function discontinuous at every point a. Use the fact that every nonempty interval of real numbers contains both rational and irrational numbers to show that the function 1, if x is rational ƒsxd = e 0, if x is irrational is discontinuous at every point.
62. Use Exercise 61 together with the identities
b. Is ƒ rightcontinuous or leftcontinuous at any point?
Solving Equations Graphically
54. If functions ƒ(x) and g(x) are continuous for 0 … x … 1 , could ƒ(x)>g (x) possibly be discontinuous at a point of [0, 1]? Give reasons for your answer. 55. If the product function hsxd = ƒsxd # g sxd is continuous at x = 0 , must ƒ(x) and g(x) be continuous at x = 0 ? Give reasons for your answer. 56. Discontinuous composite of continuous functions Give an example of functions ƒ and g, both continuous at x = 0 , for which the composite ƒ g is discontinuous at x = 0 . Does this contradict Theorem 10? Give reasons for your answer. 57. Neverzero continuous functions Is it true that a continuous function that is never zero on an interval never changes sign on that interval? Give reasons for your answer.
sin sh + cd = sin h cos c + cos h sin c , cos sh + cd = cos h cos c  sin h sin c to prove that ƒsxd = sin x and g sxd = cos x are continuous at every point x = c .
T Use a graphing calculator or computer grapher to solve the equations in Exercises 63–70. 63. x 3  3x  1 = 0 64. 2x 3  2x 2  2x + 1 = 0 65. xsx  1d2 = 1
sone rootd
66. x x = 2 67. 2x + 21 + x = 4 68. x 3  15x + 1 = 0 69. cos x = x 70. 2 sin x = x
sthree rootsd
sone rootd . Make sure you are using radian mode. sthree rootsd . Make sure you are using radian mode.
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2.7
Tangents and Derivatives This section continues the discussion of secants and tangents begun in Section 2.1. We calculate limits of secant slopes to find tangents to curves.
What Is a Tangent to a Curve? For circles, tangency is straightforward. A line L is tangent to a circle at a point P if L passes through P perpendicular to the radius at P (Figure 2.63). Such a line just touches
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the circle. But what does it mean to say that a line L is tangent to some other curve C at a point P? Generalizing from the geometry of the circle, we might say that it means one of the following:
P L
1. 2. 3.
O
FIGURE 2.63 L is tangent to the circle at P if it passes through P perpendicular to radius OP.
L passes through P perpendicular to the line from P to the center of C. L passes through only one point of C, namely P. L passes through P and lies on one side of C only.
Although these statements are valid if C is a circle, none of them works consistently for more general curves. Most curves do not have centers, and a line we may want to call tangent may intersect C at other points or cross C at the point of tangency (Figure 2.64).
L
L
C L
C
C P
P P
L meets C only at P but is not tangent to C.
L is tangent to C at P but meets C at several points.
L is tangent to C at P but lies on two sides of C, crossing C at P.
FIGURE 2.64 Exploding myths about tangent lines.
HISTORICAL BIOGRAPHY Pierre de Fermat (1601–1665)
To define tangency for general curves, we need a dynamic approach that takes into account the behavior of the secants through P and nearby points Q as Q moves toward P along the curve (Figure 2.65). It goes like this: 1. 2. 3.
We start with what we can calculate, namely the slope of the secant PQ. Investigate the limit of the secant slope as Q approaches P along the curve. If the limit exists, take it to be the slope of the curve at P and define the tangent to the curve at P to be the line through P with this slope.
This approach is what we were doing in the fallingrock and fruit fly examples in Section 2.1.
Secants
Tangent
P
P Q
Tangent
Secants
Q
FIGURE 2.65 The dynamic approach to tangency. The tangent to the curve at P is the line through P whose slope is the limit of the secant slopes as Q : P from either side.
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EXAMPLE 1
Tangent Line to a Parabola
Find the slope of the parabola y = x 2 at the point P(2, 4). Write an equation for the tangent to the parabola at this point. We begin with a secant line through P(2, 4) and Qs2 + h, s2 + hd2 d nearby. We then write an expression for the slope of the secant PQ and investigate what happens to the slope as Q approaches P along the curve: Solution
Secant slope =
¢y s2 + hd2  22 h 2 + 4h + 4  4 = = h h ¢x =
h 2 + 4h = h + 4. h
If h 7 0, then Q lies above and to the right of P, as in Figure 2.66. If h 6 0, then Q lies to the left of P (not shown). In either case, as Q approaches P along the curve, h approaches zero and the secant slope approaches 4: lim sh + 4d = 4.
h:0
We take 4 to be the parabola’s slope at P. The tangent to the parabola at P is the line through P with slope 4: y = 4 + 4sx  2d y = 4x  4.
Pointslope equation
y y x2
Secant slope is
(2 h) 2 4 h 4. h
Q(2 h, (2 h) 2) Tangent slope 4 ∆y (2 h)2 4 P(2, 4) ∆x h 0
2
2h
x
NOT TO SCALE
FIGURE 2.66 Finding the slope of the parabola y = x 2 at the point P(2, 4) (Example 1).
Finding a Tangent to the Graph of a Function The problem of finding a tangent to a curve was the dominant mathematical problem of the early seventeenth century. In optics, the tangent determined the angle at which a ray of light entered a curved lens. In mechanics, the tangent determined the direction of a body’s motion at every point along its path. In geometry, the tangents to two curves at a point of intersection determined the angles at which they intersected. To find a tangent to an arbitrary curve y = ƒsxd at a point Psx0 , ƒsx0 dd, we use the same dynamic procedure. We calculate the slope of the secant through P and a point Qsx0 + h, ƒsx0 + hdd. We then investigate the limit of the slope as h : 0 (Figure 2.67). If the limit exists, we call it the slope of the curve at P and define the tangent at P to be the line through P having this slope. Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
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Tangents and Derivatives
137
y
DEFINITIONS Slope, Tangent Line The slope of the curve y = ƒsxd at the point Psx0 , ƒsx0 dd is the number
y f (x) Q(x 0 h, f (x 0 h))
ƒsx0 + hd  ƒsx0 d h h:0
f (x 0 h) f (x 0)
m = lim
(provided the limit exists).
The tangent line to the curve at P is the line through P with this slope.
P(x 0, f(x 0)) h 0
x0
FIGURE 2.67
x0 h
The slope of the tangent ƒsx0 + hd  ƒsx0 d line at P is lim . h h :0
x
Whenever we make a new definition, we try it on familiar objects to be sure it is consistent with results we expect in familiar cases. Example 2 shows that the new definition of slope agrees with the old definition from Section 1.2 when we apply it to nonvertical lines.
EXAMPLE 2
Testing the Definition
Show that the line y = mx + b is its own tangent at any point sx0, mx0 + bd. Solution
1.
We let ƒsxd = mx + b and organize the work into three steps.
Find ƒsx0 d and ƒsx0 + hd. ƒsx0 d = mx0 + b ƒsx0 + hd = msx0 + hd + b = mx0 + mh + b
2.
Find the slope lim sƒsx0 + hd  ƒsx0 dd>h. h:0
ƒsx0 + hd  ƒsx0 d smx0 + mh + bd  smx0 + bd lim = lim h h h:0 h:0 = lim
h:0
3.
mh = m h
Find the tangent line using the pointslope equation. The tangent line at the point sx0, mx0 + bd is y = smx0 + bd + msx  x0 d y = mx0 + b + mx  mx0 y = mx + b.
Let’s summarize the steps in Example 2.
Finding the Tangent to the Curve y ƒsxd at sx0 , y0 d 1. Calculate ƒsx0 d and ƒsx0 + hd. 2. Calculate the slope ƒsx0 + hd  ƒsx0 d . m = lim h h:0 3. If the limit exists, find the tangent line as y = y0 + msx  x0 d.
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EXAMPLE 3
Slope and Tangent to y = 1>x, x Z 0
(a) Find the slope of the curve y = 1>x at x = a Z 0. (b) Where does the slope equal 1>4? (c) What happens to the tangent to the curve at the point (a, 1>a) as a changes? Solution
(a) Here ƒsxd = 1>x. The slope at (a, 1>a) is 1 1  a a + h ƒsa + hd  ƒsad = lim lim h h:0 h:0 h 1 a  sa + hd = lim h:0 h asa + hd = lim
h hasa + hd
= lim
1 1 =  2. asa + hd a
h:0
h:0
Notice how we had to keep writing “limh:0” before each fraction until the stage where we could evaluate the limit by substituting h = 0. The number a may be positive or negative, but not 0. (b) The slope of y = 1>x at the point where x = a is 1>a 2 . It will be 1>4 provided that 
1 1 =  . 4 a2
This equation is equivalent to a 2 = 4, so a = 2 or a = 2. The curve has slope 1>4 at the two points (2, 1>2) and s 2, 1>2d (Figure 2.68). (c) Notice that the slope 1>a 2 is always negative if a Z 0. As a : 0 +, the slope approaches  q and the tangent becomes increasingly steep (Figure 2.69). We see this situation again as a : 0  . As a moves away from the origin in either direction, the slope approaches 0  and the tangent levels off. y y
y 1x y 1x
slope is – 1 4
slope is – 12 a 2, 1 2 x
–2, – 1 2
0
a
x
slope is – 1 4
FIGURE 2.68 The two tangent lines to y = 1>x having slope 1>4 (Example 3).
FIGURE 2.69 The tangent slopes, steep near the origin, become more gradual as the point of tangency moves away.
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Tangents and Derivatives
139
Rates of Change: Derivative at a Point The expression ƒsx0 + hd  ƒsx0 d h is called the difference quotient of ƒ at x0 with increment h. If the difference quotient has a limit as h approaches zero, that limit is called the derivative of ƒ at x0. If we interpret the difference quotient as a secant slope, the derivative gives the slope of the curve and tangent at the point where x = x0 . If we interpret the difference quotient as an average rate of change, as we did in Section 2.1, the derivative gives the function’s rate of change with respect to x at the point x = x0 . The derivative is one of the two most important mathematical objects considered in calculus. We begin a thorough study of it in Chapter 3. The other important object is the integral, and we initiate its study in Chapter 5.
EXAMPLE 4
Instantaneous Speed (Continuation of Section 2.1, Examples 1 and 2)
In Examples 1 and 2 in Section 2.1, we studied the speed of a rock falling freely from rest near the surface of the earth. We knew that the rock fell y = 16t 2 feet during the first t sec, and we used a sequence of average rates over increasingly short intervals to estimate the rock’s speed at the instant t = 1. Exactly what was the rock’s speed at this time? We let ƒstd = 16t 2 . The average speed of the rock over the interval between t = 1 and t = 1 + h seconds was Solution
ƒs1 + hd  ƒs1d 16s1 + hd2  16s1d2 16sh 2 + 2hd = = = 16sh + 2d. h h h The rock’s speed at the instant t = 1 was lim 16sh + 2d = 16s0 + 2d = 32 ft>sec.
h:0
Our original estimate of 32 ft > sec was right.
Summary We have been discussing slopes of curves, lines tangent to a curve, the rate of change of a function, the limit of the difference quotient, and the derivative of a function at a point. All of these ideas refer to the same thing, summarized here:
1. 2. 3. 4. 5.
The slope of y = ƒsxd at x = x0 The slope of the tangent to the curve y = ƒsxd at x = x0 The rate of change of ƒ(x) with respect to x at x = x0 The derivative of ƒ at x = x0 ƒsx0 + hd  ƒsx0 d The limit of the difference quotient, lim h h:0
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EXERCISES 2.7 Slopes and Tangent Lines
In Exercises 19–22, find the slope of the curve at the point indicated.
In Exercises 1–4, use the grid and a straight edge to make a rough estimate of the slope of the curve (in yunits per xunit) at the points P1 and P2 . Graphs can shift during a press run, so your estimates may be somewhat different from those in the back of the book.
19. y = 5x 2,
y
1.
2.
2
y
P2
1
0
2
x
P1 –1
P1 0
x  1 , x + 1
x = 0
22. y =
Tangent Lines with Specified Slopes 24. g sxd = x 3  3x
26. Find an equation of the straight line having slope 1>4 that is tangent to the curve y = 2x .
Rates of Change
x
1
x = 2
25. Find equations of all lines having slope 1 that are tangent to the curve y = 1>sx  1d .
1 –1
x = 3
23. ƒsxd = x 2 + 4x  1
1 –2
1 , x  1
20. y = 1  x 2,
At what points do the graphs of the functions in Exercises 23 and 24 have horizontal tangents?
P2
2
21. y =
x = 1
27. Object dropped from a tower An object is dropped from the top of a 100mhigh tower. Its height above ground after t sec is 100  4.9t 2 m. How fast is it falling 2 sec after it is dropped?
–2
–1
28. Speed of a rocket At t sec after liftoff, the height of a rocket is 3t 2 ft. How fast is the rocket climbing 10 sec after liftoff? y
3.
4.
y
29. Circle’s changing area What is the rate of change of the area of a circle sA = pr 2 d with respect to the radius when the radius is r = 3 ?
4
30. Ball’s changing volume What is the rate of change of the volume of a ball sV = s4>3dpr 3 d with respect to the radius when the radius is r = 2 ?
2 3 2
P2
P1
1
P2
P1
Testing for Tangents
1 0
1
2
x
–2
–1
0
31. Does the graph of 1
2
In Exercises 5–10, find an equation for the tangent to the curve at the given point. Then sketch the curve and tangent together. 5. y = 4  x 2,
s1, 2d
7. y = 22x, 3
6. y = sx  1d2 + 1,
s 1, 3d
8. y =
1 10. y = 3 , x
s 2, 8d
9. y = x ,
1 , x2
ƒsxd = e
x
have a tangent at the origin? Give reasons for your answer. 32. Does the graph of
s1, 1d
g sxd = e
s 1, 1d 1 a2,  b 8
x 2 sin s1>xd, x Z 0 0, x = 0
x sin s1>xd, x Z 0 0, x = 0
have a tangent at the origin? Give reasons for your answer.
Vertical Tangents
In Exercises 11–18, find the slope of the function’s graph at the given point. Then find an equation for the line tangent to the graph there.
We say that the curve y = ƒsxd has a vertical tangent at the point where x = x0 if lim h:0 sƒsx0 + hd  ƒsx0 dd>h = q or  q .
11. ƒsxd = x 2 + 1,
s2, 5d
12. ƒsxd = x  2x 2,
Vertical tangent at x = 0 (see accompanying figure):
x , x  2
s3, 3d
14. g sxd =
13. g sxd =
15. hstd = t 3,
s2, 8d
17. ƒsxd = 2x,
s4, 2d
s1, 1d
8 , s2, 2d x2 16. hstd = t 3 + 3t, s1, 4d 18. ƒsxd = 2x + 1,
s8, 3d
ƒs0 + hd  ƒs0d h 1>3  0 = lim h h h:0 h:0 lim
= lim
h:0
1 = q h 2>3
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141
34. Does the graph of
y
Usxd = e
y f (x) x 1兾3
x 6 0 x Ú 0
have a vertical tangent at the point (0, 1)? Give reasons for your answer.
x
0
0, 1,
T
a. Graph the curves in Exercises 35–44. Where do the graphs appear to have vertical tangents? b. Confirm your findings in part (a) with limit calculations. But before you do, read the introduction to Exercises 33 and 34.
VERTICAL TANGENT AT ORIGIN
35. y = x 2>5
36. y = x 4>5
1>5
38. y = x 3>5
37. y = x No vertical tangent at x = 0 (see next figure):
39. y = 4x 2>5  2x
g s0 + hd  g s0d h 2>3  0 lim = lim h h h: 0 h :0 = lim
h :0
41. y = x
43. y = e
1 h
1>3
does not exist, because the limit is q from the right and  q from the left.
40. y = x 5>3  5x 2>3 1>3
42. y = x 1>3 + sx  1d1>3
 sx  1d
 2ƒ x ƒ , x … 0 2x, x 7 0
44. y = 2 ƒ 4  x ƒ
COMPUTER EXPLORATIONS
Graphing Secant and Tangent Lines Use a CAS to perform the following steps for the functions in Exercises 45–48.
y y g(x)
2>3
x 2兾3
a. Plot y = ƒsxd over the interval sx0  1>2d … x … sx0 + 3d . b. Holding x0 fixed, the difference quotient x
0 NO VERTICAL TANGENT AT ORIGIN
33. Does the graph of 1, ƒsxd = • 0, 1,
x 6 0 x = 0 x 7 0
have a vertical tangent at the origin? Give reasons for your answer.
qshd =
ƒsx0 + hd  ƒsx0 d h
at x0 becomes a function of the step size h. Enter this function into your CAS workspace. c. Find the limit of q as h : 0 . d. Define the secant lines y = ƒsx0 d + q # sx  x0 d for h = 3, 2 , and 1. Graph them together with ƒ and the tangent line over the interval in part (a). 5 45. ƒsxd = x 3 + 2x, x0 = 0 46. ƒsxd = x + x , x0 = 1 47. ƒsxd = x + sin s2xd, x0 = p>2 48. ƒsxd = cos x + 4 sin s2xd,
x0 = p
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Chapter 2 Questions to Guide Your Review
Chapter 2
141
Questions to Guide Your Review
1. What is the average rate of change of the function g(t) over the interval from t = a to t = b ? How is it related to a secant line? 2. What limit must be calculated to find the rate of change of a function g(t) at t = t0 ? 3. What is an informal or intuitive definition of the limit lim ƒsxd = L ?
x: x0
4. Does the existence and value of the limit of a function ƒ(x) as x approaches x0 ever depend on what happens at x = x0 ? Explain and give examples. 5. What function behaviors might occur for which the limit may fail to exist? Give examples. 6. What theorems are available for calculating limits? Give examples of how the theorems are used.
Why is the definition “informal”? Give examples.
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7. How are onesided limits related to limits? How can this relationship sometimes be used to calculate a limit or prove it does not exist? Give examples. 8. What is the value of lim u:0 sssin ud>ud ? Does it matter whether u is measured in degrees or radians? Explain. 9. What exactly does limx:x0 ƒsxd = L mean? Give an example in which you find a d 7 0 for a given f, L, x0 , and P 7 0 in the precise definition of limit. 10. Give precise definitions of the following statements. a. limx:2 ƒsxd = 5 c. limx:2 ƒsxd = q
b. limx:2+ ƒsxd = 5 d. limx:2 ƒsxd =  q
11. What exactly do limx: q ƒsxd = L and limx: q ƒsxd = L mean? Give examples. 12. What are limx:; q k (k a constant) and limx:; q s1>xd ? How do you extend these results to other functions? Give examples. 13. How do you find the limit of a rational function as x : ; q ? Give examples. 14. What are horizontal, vertical, and oblique asymptotes? Give examples. 15. What conditions must be satisfied by a function if it is to be continuous at an interior point of its domain? At an endpoint? 16. How can looking at the graph of a function help you tell where the function is continuous? 17. What does it mean for a function to be rightcontinuous at a point? Leftcontinuous? How are continuity and onesided continuity related? 18. What can be said about the continuity of polynomials? Of rational functions? Of trigonometric functions? Of rational powers and al
gebraic combinations of functions? Of composites of functions? Of absolute values of functions? 19. Under what circumstances can you extend a function ƒ(x) to be continuous at a point x = c ? Give an example. 20. What does it mean for a function to be continuous on an interval? 21. What does it mean for a function to be continuous? Give examples to illustrate the fact that a function that is not continuous on its entire domain may still be continuous on selected intervals within the domain. 22. What are the basic types of discontinuity? Give an example of each. What is a removable discontinuity? Give an example. 23. What does it mean for a function to have the Intermediate Value Property? What conditions guarantee that a function has this property over an interval? What are the consequences for graphing and solving the equation ƒsxd = 0 ? 24. It is often said that a function is continuous if you can draw its graph without having to lift your pen from the paper. Why is that? 25. What does it mean for a line to be tangent to a curve C at a point P? 26. What is the significance of the formula lim
h:0
ƒsx + hd  ƒsxd ? h
Interpret the formula geometrically and physically. 27. How do you find the tangent to the curve y = ƒsxd at a point sx0, y0 d on the curve? 28. How does the slope of the curve y = ƒsxd at x = x0 relate to the function’s rate of change with respect to x at x = x0 ? To the derivative of ƒ at x0 ?
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Chapter 2
Practice Exercises
Limits and Continuity
2. Repeat the instructions of Exercise 1 for 0, 1>x, ƒsxd = d 0, 1,
1. Graph the function 1, x, ƒsxd = e 1, x, 1,
x … 1 1 6 x 6 0 x = 0 0 6 x 6 1 x Ú 1.
Then discuss, in detail, limits, onesided limits, continuity, and onesided continuity of ƒ at x = 1, 0 , and 1. Are any of the discontinuities removable? Explain.
x 0 x x
… 6 = 7
1 ƒxƒ 6 1 1 1.
3. Suppose that ƒ(t) and g(t) are defined for all t and that limt:t0 ƒstd = 7 and limt:t0 g std = 0 . Find the limit as t : t0 of the following functions. a. 3ƒ(t) c. ƒstd # g std e. cos (g(t)) g. ƒstd + g std
b. sƒstdd2 ƒstd d. g std  7 f. ƒ ƒstd ƒ h. 1>ƒ(t)
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Chapter 2 4. Suppose that ƒ(x) and g (x) are defined for all x and that limx:0 ƒsxd = 1>2 and limx:0 g sxd = 22 . Find the limits as x : 0 of the following functions. a. g sxd
b. g sxd # ƒsxd
c. ƒsxd + g sxd
d. 1>ƒ(x) ƒsxd # cos x f. x  1
e. x + ƒsxd
In Exercises 5 and 6, find the value that limx:0 g sxd must have if the given limit statements hold. 4  g sxd b = 1 5. lim a 6. lim ax lim g sxdb = 2 x x: 0
x: 4
x:0
7. On what intervals are the following functions continuous? a. ƒsxd = x 1>3
b. g sxd = x 3>4
c. hsxd = x 2>3
d. ksxd = x 1>6
8. On what intervals are the following functions continuous? a. ƒsxd = tan x
b. g sxd = csc x
cos x c. hsxd = x  p
d. ksxd =
sin x x
Finding Limits In Exercises 9–16, find the limit or explain why it does not exist. x 2  4x + 4 x + 5x 2  14x a. as x : 0
9. lim
b. as x : 1
1  2x x: 1 1  x
x2  a2 x: a x 4  a 4 sx + hd2  x 2 14. lim h x: 0 12. lim
sx + hd2  x 2 h h:0
13. lim
1 1 2 + x 2 15. lim x x: 0
s2 + xd3  8 x x: 0
16. lim
In Exercises 17–20, find the limit of g (x) as x approaches the indicated value. 17. lim+ s4g sxdd1>3 = 2 x: 0
x: 1
3x 2 + 1 = q g sxd
18.
lim
x: 25
20. lim
x: 2
1 = 2 x + g sxd 5  x2 2g sxd
= 0
x: q
x + sin x + 22x x + sin x
2x + 3 5x + 7
30. lim
x: q
x 2>3 + x 1 x 2>3 + cos2 x
Continuous Extension 31. Can ƒsxd = xsx 2  1d> ƒ x 2  1 ƒ be extended to be continuous at x = 1 or 1 ? Give reasons for your answers. (Graph the function—you will find the graph interesting.) 32. Explain why the function ƒsxd = sin s1>xd has no continuous extension to x = 0 . T In Exercises 33–36, graph the function to see whether it appears to have a continuous extension to the given point a. If it does, use Trace and Zoom to find a good candidate for the extended function’s value at a. If the function does not appear to have a continuous extension, can it be extended to be continuous from the right or left? If so, what do you think the extended function’s value should be? x  1 4
x  2x
,
a = 1
34. g sud =
a = 0 36. k sxd =
5 cos u , 4u  2p x , 1  2ƒ x ƒ
a = p>2 a = 0
Roots T 37. Let ƒsxd = x 3  x  1 . a. Show that ƒ has a zero between 1 and 2. b. Solve the equation ƒsxd = 0 graphically with an error of magnitude at most 10 8 . c. It can be shown that the exact value of the solution in part (b) is 269 1>3 269 1>3 1 1 + b b + a 2 18 2 18 Evaluate this exact answer and compare it with the value you found in part (b). a
T 38. Let ƒsud = u3  2u + 2 . a. Show that ƒ has a zero between 2 and 0. b. Solve the equation ƒsud = 0 graphically with an error of magnitude at most 10 4 . a
Find the limits in Exercises 21–30. x: q
29. lim
x: q
c. It can be shown that the exact value of the solution in part (b) is
Limits at Infinity 21. lim
24. lim
x:  q
35. hstd = s1 + ƒ t ƒ d1>t,
5
11. lim
19. lim
lim
33. ƒsxd =
b. as x : 2
143
1 x 2  7x + 1 x4 + x3 26. lim x: q 12x 3 + 128 sIf you have a grapher, try graphing the function sin x 27. lim x: q :x; for 5 … x … 5.d sIf you have a grapher, try graphing cos u  1 28. lim ƒsxd = xscos s1>xd  1d near the origin to u u: q “see” the limit at infinity.d 23.
3
x2 + x x + 2x 4 + x 3 a. as x : 0
10. lim
x 2  4x + 8 3x 3 2 x  7x 25. lim x:  q x + 1
Practice Exercises
22.
lim
x:  q
2x 2 + 3 5x 2 + 7
1>3
1>3
19 19  1b  a + 1b A 27 A 27 Evaluate this exact answer and compare it with the value you found in part (b).
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Chapter 2 T
Additional and Advanced Exercises
1. Assigning a value to 00 The rules of exponents (see Appendix 9) tell us that a 0 = 1 if a is any number different from zero. They also tell us that 0 n = 0 if n is any positive number. If we tried to extend these rules to include the case 0 0 , we would get conflicting results. The first rule would say 0 0 = 1 , whereas the second would say 0 0 = 0 . We are not dealing with a question of right or wrong here. Neither rule applies as it stands, so there is no contradiction. We could, in fact, define 0 0 to have any value we wanted as long as we could persuade others to agree. What value would you like 0 0 to have? Here is an example that might help you to decide. (See Exercise 2 below for another example.) a. Calculate x x for x = 0.1 , 0.01, 0.001, and so on as far as your calculator can go. Record the values you get. What pattern do you see? x
b. Graph the function y = x for 0 6 x … 1 . Even though the function is not defined for x … 0 , the graph will approach the yaxis from the right. Toward what yvalue does it seem to be headed? Zoom in to further support your idea. T
2. A reason you might want 00 to be something other than 0 or 1 As the number x increases through positive values, the numbers 1>x and 1> (ln x) both approach zero. What happens to the number 1 ƒsxd = a x b
1>sln xd
as x increases? Here are two ways to find out. a. Evaluate ƒ for x = 10 , 100, 1000, and so on as far as your calculator can reasonably go. What pattern do you see? b. Graph ƒ in a variety of graphing windows, including windows that contain the origin. What do you see? Trace the yvalues along the graph. What do you find? 3. Lorentz contraction In relativity theory, the length of an object, say a rocket, appears to an observer to depend on the speed at which the object is traveling with respect to the observer. If the observer measures the rocket’s length as L0 at rest, then at speed y the length will appear to be L = L0 1 
B
x Exit rate y ft 3兾min
Suppose that y = 2x>2 for a certain tank. You are trying to maintain a fairly constant exit rate by adding water to the tank with a hose from time to time. How deep must you keep the water if you want to maintain the exit rate a. within 0.2 ft3>min of the rate y0 = 1 ft3>min ?
b. within 0.1 ft3>min of the rate y0 = 1 ft3>min ?
5. Thermal expansion in precise equipment As you may know, most metals expand when heated and contract when cooled. The dimensions of a piece of laboratory equipment are sometimes so critical that the shop where the equipment is made must be held at the same temperature as the laboratory where the equipment is to be used. A typical aluminum bar that is 10 cm wide at 70°F will be y = 10 + st  70d * 10 4 centimeters wide at a nearby temperature t. Suppose that you are using a bar like this in a gravity wave detector, where its width must stay within 0.0005 cm of the ideal 10 cm. How close to t0 = 70°F must you maintain the temperature to ensure that this tolerance is not exceeded? 6. Stripes on a measuring cup The interior of a typical 1L measuring cup is a right circular cylinder of radius 6 cm (see accompanying figure). The volume of water we put in the cup is therefore a function of the level h to which the cup is filled, the formula being V = p62h = 36ph . How closely must we measure h to measure out 1 L of water s1000 cm3 d with an error of no more than 1% s10 cm3 d ?
y2 . c2
This equation is the Lorentz contraction formula. Here, c is the speed of light in a vacuum, about 3 * 10 8 m>sec . What happens to L as y increases? Find limy:c L . Why was the lefthand limit needed? 4. Controlling the flow from a draining tank Torricelli’s law says that if you drain a tank like the one in the figure shown, the rate y at which water runs out is a constant times the square root of the water’s depth x. The constant depends on the size and shape of the exit valve.
Stripes about 1 mm wide (a)
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18. The Dirichlet ruler function If x is a rational number, then x can be written in a unique way as a quotient of integers m>n where n 7 0 and m and n have no common factors greater than 1. (We say that such a fraction is in lowest terms. For example, 6>4 written in lowest terms is 3>2.) Let ƒ(x) be defined for all x in the interval [0, 1] by
r 6 cm
Liquid volume V 36h
h
Additional and Advanced Exercises
ƒsxd = e
1>n, 0,
if x = m>n is a rational number in lowest terms if x is irrational.
For instance, ƒs0d = ƒs1d = 1, ƒs1>2d = 1>2, ƒs1>3d = ƒs2>3d = 1>3, ƒs1>4d = ƒs3>4d = 1>4 , and so on.
(b)
A 1L measuring cup (a), modeled as a right circular cylinder (b) of radius r = 6 cm
a. Show that ƒ is discontinuous at every rational number in [0, 1].
Precise Definition of Limit
b. Show that ƒ is continuous at every irrational number in [0, 1]. (Hint: If P is a given positive number, show that there are only finitely many rational numbers r in [0, 1] such that ƒsrd Ú P .)
In Exercises 7–10, use the formal definition of limit to prove that the function is continuous at x0 .
c. Sketch the graph of ƒ. Why do you think ƒ is called the “ruler function”?
7. ƒsxd = x 2  7,
8. g sxd = 1>s2xd,
x0 = 1
9. hsxd = 22x  3,
x0 = 1>4
x0 = 2 10. Fsxd = 29  x,
x0 = 5
19. Antipodal points Is there any reason to believe that there is always a pair of antipodal (diametrically opposite) points on Earth’s equator where the temperatures are the same? Explain.
11. Uniqueness of limits Show that a function cannot have two different limits at the same point. That is, if limx:x0 ƒsxd = L1 and limx:x0 ƒsxd = L2 , then L1 = L2 .
20. If lim sƒsxd + g sxdd = 3 and lim sƒsxd  g sxdd = 1 , find
12. Prove the limit Constant Multiple Rule: lim kƒsxd = k lim ƒsxd for any constant k .
21. Roots of a quadratic equation that is almost linear The equation ax 2 + 2x  1 = 0 , where a is a constant, has two roots if a 7 1 and a Z 0 , one positive and one negative:
x: c
x: c
13. Onesided limits If limx:0+ ƒsxd = A and limx:0 ƒsxd = B , find 3
3
a. limx:0+ ƒsx  xd
b. limx:0 ƒsx  xd
c. limx:0+ ƒsx 2  x 4 d
d. limx:0 ƒsx 2  x 4 d
x:c
lim ƒsxdg sxd .
x:c
x:c
r+sad =
1 + 21 + a , a
rsad =
1  21 + a . a
a. What happens to r+sad as a : 0 ? As a : 1+ ?
14. Limits and continuity Which of the following statements are true, and which are false? If true, say why; if false, give a counterexample (that is, an example confirming the falsehood).
b. What happens to rsad as a : 0 ? As a : 1+ ?
a. If limx:a ƒsxd exists but limx:a g sxd does not exist, then limx:asƒsxd + g sxdd does not exist.
d. For added support, graph ƒsxd = ax 2 + 2x  1 simultaneously for a = 1, 0.5, 0.2, 0.1, and 0.05.
c. Support your conclusions by graphing r+sad and rsad as functions of a. Describe what you see.
b. If neither limx:a ƒsxd nor limx:a g sxd exists, then limx:a sƒsxd + g sxdd does not exist.
22. Root of an equation Show that the equation x + 2 cos x = 0 has at least one solution.
c. If ƒ is continuous at x, then so is ƒ ƒ ƒ .
23. Bounded functions A realvalued function ƒ is bounded from above on a set D if there exists a number N such that ƒsxd … N for all x in D. We call N, when it exists, an upper bound for ƒ on D and say that ƒ is bounded from above by N. In a similar manner, we say that ƒ is bounded from below on D if there exists a number M such that ƒsxd Ú M for all x in D. We call M, when it exists, a lower bound for ƒ on D and say that ƒ is bounded from below by M. We say that ƒ is bounded on D if it is bounded from both above and below.
d. If ƒ ƒ ƒ is continuous at a, then so is ƒ. In Exercises 15 and 16, use the formal definition of limit to prove that the function has a continuous extension to the given value of x. 15. ƒsxd =
x2  1 , x + 1
x = 1
16. g sxd =
x 2  2x  3 , 2x  6
x = 3
17. A function continuous at only one point Let ƒsxd = e
x, 0,
if x is rational if x is irrational.
a. Show that ƒ is continuous at x = 0 . b. Use the fact that every nonempty open interval of real numbers contains both rational and irrational numbers to show that ƒ is not continuous at any nonzero value of x.
a. Show that ƒ is bounded on D if and only if there exists a number B such that ƒ ƒsxd ƒ … B for all x in D. b. Suppose that ƒ is bounded from above by N. Show that if limx:x0 ƒsxd = L , then L … N . c. Suppose that ƒ is bounded from below by M. Show that if limx:x0 ƒsxd = L , then L Ú M .
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24. Max 5a, b6 and min 5a, b6
sin x 2 sin x 2 x2 lim x = 1 # 0 = 0 . = lim x 2 x:0 x:0 x x:0
b. lim
a. Show that the expression ƒa  bƒ a + b max 5a, b6 = + 2 2 equals a if a Ú b and equals b if b Ú a . In other words, max {a, b} gives the larger of the two numbers a and b.
b. Find a similar expression for min 5a, b6 , the smaller of a and b.
Generalized Limits Involving
sin U U
The formula limu:0 ssin ud>u = 1 can be generalized. If limx:c ƒsxd = 0 and ƒ(x) is never zero in an open interval containing the point x = c, except possibly c itself, then sin ƒsxd lim = 1. x: c ƒsxd Here are several examples. sin x 2 = 1. x: 0 x 2
a. lim
sin sx 2  x  2d sin sx 2  x  2d = lim x + 1 x: 1 x: 1 sx 2  x  2d
c. lim
#
sx 2  x  2d sx + 1dsx  2d = 1 # lim = 3 . x + 1 x + 1 x: 1 x: 1 lim
d. lim
sin A 1  2x B x  1
x:1
1 # lim
x:1
sin A 1  2x B 1  2x = x  1 x:1 1  2x
= lim
A 1  2x B A 1 + 2x B sx  1d A 1 + 2x B
= lim
x:1
Find the limits in Exercises 25–30. 25. lim
sin s1  cos xd x
27. lim
sin ssin xd x
x:0
x:0
sin sx 2  4d x  2 x:2
29. lim
1  x
sx  1d A 1 + 2x B
26. lim+
sin x
sin 2x sin sx 2 + xd 28. lim x x:0 x:0
30. lim
sin A 2x  3 B
x:9
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x  9
1 = . 2
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Chapter 2: Limits and Continuity
Chapter 2
Technology Application Projects
MathematicaMaple Module Take It to the Limit Part I Part II (Zero Raised to the Power Zero: What Does it Mean?) Part III (OneSided Limits) Visualize and interpret the limit concept through graphical and numerical explorations. Part IV (What a Difference a Power Makes) See how sensitive limits can be with various powers of x.
MathematicaMaple Module Going to Infinity Part I (Exploring Function Behavior as x : q or x :  q ) This module provides four examples to explore the behavior of a function as x : q or x :  q . Part II (Rates of Growth) Observe graphs that appear to be continuous, yet the function is not continuous. Several issues of continuity are explored to obtain results that you may find surprising.
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Chapter
3
DIFFERENTIATION OVERVIEW In Chapter 2, we defined the slope of a curve at a point as the limit of secant slopes. This limit, called a derivative, measures the rate at which a function changes, and it is one of the most important ideas in calculus. Derivatives are used to calculate velocity and acceleration, to estimate the rate of spread of a disease, to set levels of production so as to maximize efficiency, to find the best dimensions of a cylindrical can, to find the age of a prehistoric artifact, and for many other applications. In this chapter, we develop techniques to calculate derivatives easily and learn how to use derivatives to approximate complicated functions.
3.1
The Derivative as a Function At the end of Chapter 2, we defined the slope of a curve y = ƒsxd at the point where x = x0 to be
HISTORICAL ESSAY lim
The Derivative
h:0
ƒsx0 + hd  ƒsx0 d . h
We called this limit, when it existed, the derivative of ƒ at x0 . We now investigate the derivative as a function derived from ƒ by considering the limit at each point of the domain of ƒ.
DEFINITION Derivative Function The derivative of the function ƒ(x) with respect to the variable x is the function ƒ¿ whose value at x is ƒsx + hd  ƒsxd , h h:0
ƒ¿sxd = lim provided the limit exists.
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148
Chapter 3: Differentiation y f (x) Secant slope is f (z) f (x) zx
Q(z, f (z))
f (z) f (x)
P(x, f(x))
We use the notation ƒ(x) rather than simply ƒ in the definition to emphasize the independent variable x, which we are differentiating with respect to. The domain of ƒ¿ is the set of points in the domain of ƒ for which the limit exists, and the domain may be the same or smaller than the domain of ƒ. If ƒ¿ exists at a particular x, we say that ƒ is differentiable (has a derivative) at x. If ƒ¿ exists at every point in the domain of ƒ, we call ƒ differentiable. If we write z = x + h, then h = z  x and h approaches 0 if and only if z approaches x. Therefore, an equivalent definition of the derivative is as follows (see Figure 3.1).
hzx x
z
Alternative Formula for the Derivative
Derivative of f at x is f(x h) f (x) f '(x) lim h h→0 lim
z→x
ƒ¿sxd = lim
f (z) f (x) zx
z:x
FIGURE 3.1 The way we write the difference quotient for the derivative of a function ƒ depends on how we label the points involved.
ƒszd  ƒsxd z  x .
Calculating Derivatives from the Definition The process of calculating a derivative is called differentiation. To emphasize the idea that differentiation is an operation performed on a function y = ƒsxd, we use the notation d ƒsxd dx as another way to denote the derivative ƒ¿sxd. Examples 2 and 3 of Section 2.7 illustrate the differentiation process for the functions y = mx + b and y = 1>x. Example 2 shows that d smx + bd = m. dx For instance, 3 d 3 a x  4b = . 2 dx 2 In Example 3, we see that d 1 1 a b =  2. dx x x Here are two more examples.
EXAMPLE 1
Applying the Definition
Differentiate ƒsxd =
Solution
x . x  1
Here we have ƒsxd =
x x  1
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The Derivative as a Function
149
and sx + hd , so sx + hd  1 ƒsx + hd  ƒsxd ƒ¿sxd = lim h h:0 ƒsx + hd =
= lim
x x + h x  1 x + h  1
h sx + hdsx  1d  xsx + h  1d 1 = lim # sx + h  1dsx  1d h:0 h h:0
= lim
h 1 # h sx + h  1dsx  1d
= lim
1 1 . = sx + h  1dsx  1d sx  1d2
h:0
h:0
EXAMPLE 2
a c ad  cb  = b d bd
Derivative of the Square Root Function
(a) Find the derivative of y = 1x for x 7 0. (b) Find the tangent line to the curve y = 1x at x = 4.
Solution
You will often need to know the derivative of 1x for x 7 0 :
(a) We use the equivalent form to calculate ƒ¿ : ƒszd  ƒsxd z  x z:x
d 1 1x = . dx 2 1x
ƒ¿sxd = lim
= lim
z:x
= lim
z:x
= lim
y
z:x
y 1x1 4
(4, 2)
1z  1x
A 1z  1x B A 1z + 1x B 1 1 = . 1z + 1x 21x
(b) The slope of the curve at x = 4 is ƒ¿s4d =
y 兹x
1 0
1z  1x z  x
1 1 = . 4 224
The tangent is the line through the point (4, 2) with slope 1>4 (Figure 3.2): 4
x
FIGURE 3.2 The curve y = 1x and its tangent at (4, 2). The tangent’s slope is found by evaluating the derivative at x = 4 (Example 2).
y = 2 + y =
1 sx  4d 4
1 x + 1. 4
We consider the derivative of y = 1x when x = 0 in Example 6.
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Chapter 3: Differentiation
Notations There are many ways to denote the derivative of a function y = ƒsxd, where the independent variable is x and the dependent variable is y. Some common alternative notations for the derivative are ƒ¿sxd = y¿ =
dƒ dy d = = ƒsxd = Dsƒdsxd = Dx ƒsxd. dx dx dx
The symbols d>dx and D indicate the operation of differentiation and are called differentiation operators. We read dy>dx as “the derivative of y with respect to x,” and dƒ>dx and (d>dx)ƒ(x) as “the derivative of ƒ with respect to x.” The “prime” notations y¿ and ƒ¿ come from notations that Newton used for derivatives. The d>dx notations are similar to those used by Leibniz. The symbol dy>dx should not be regarded as a ratio (until we introduce the idea of “differentials” in Section 3.8). Be careful not to confuse the notation D(ƒ) as meaning the domain of the function ƒ instead of the derivative function ƒ¿ . The distinction should be clear from the context. To indicate the value of a derivative at a specified number x = a, we use the notation ƒ¿sad =
df dy d ` = ` = ƒsxd ` . dx x = a dx x = a dx x=a
For instance, in Example 2b we could write ƒ¿s4d =
d 1 1 1 1x ` = ` = = . 4 dx 21x x=4 x=4 224
To evaluate an expression, we sometimes use the right bracket ] in place of the vertical bar ƒ .
Graphing the Derivative We can often make a reasonable plot of the derivative of y = ƒsxd by estimating the slopes on the graph of ƒ. That is, we plot the points sx, ƒ¿sxdd in the xyplane and connect them with a smooth curve, which represents y = ƒ¿sxd.
EXAMPLE 3
Graphing a Derivative
Graph the derivative of the function y = ƒsxd in Figure 3.3a. We sketch the tangents to the graph of ƒ at frequent intervals and use their slopes to estimate the values of ƒ¿sxd at these points. We plot the corresponding sx, ƒ¿sxdd pairs and connect them with a smooth curve as sketched in Figure 3.3b.
Solution
What can we learn from the graph of y = ƒ¿sxd? At a glance we can see 1. 2. 3.
where the rate of change of ƒ is positive, negative, or zero; the rough size of the growth rate at any x and its size in relation to the size of ƒ(x); where the rate of change itself is increasing or decreasing.
Here’s another example.
EXAMPLE 4
Concentration of Blood Sugar
On April 23, 1988, the humanpowered airplane Daedalus flew a recordbreaking 119 km from Crete to the island of Santorini in the Aegean Sea, southeast of mainland Greece. Dur
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151
The Derivative as a Function
y
Slope 0
y f (x)
A Slope –1 10
B
Slope – 4 3
C
E
D
5
Slope 0
Slope 8 2 yunits/xunit 4 8 yunits
0
4 xunits 10 15 (a)
5
x
Slope
4 y f '(x)
3
E'
2 1 A'
D' –1 –2
5 C'
10
15
x
B' Vertical coordinate –1 (b)
FIGURE 3.3 We made the graph of y = ƒ¿sxd in (b) by plotting slopes from the graph of y = ƒsxd in (a). The vertical coordinate of B¿ is the slope at B and so on. The graph of ƒ¿ is a visual record of how the slope of ƒ changes with x.
ing the 6hour endurance tests before the flight, researchers monitored the prospective pilots’ bloodsugar concentrations. The concentration graph for one of the athletepilots is shown in Figure 3.4a, where the concentration in milligrams> deciliter is plotted against time in hours. The graph consists of line segments connecting data points. The constant slope of each segment gives an estimate of the derivative of the concentration between measurements. We calculated the slope of each segment from the coordinate grid and plotted the derivative as a step function in Figure 3.4b. To make the plot for the first hour, for instance, we observed that the concentration increased from about 79 mg> dL to 93 mg> dL. The net increase was ¢y = 93  79 = 14 mg>dL. Dividing this by ¢t = 1 hour gave the rate of change as ¢y 14 = 14 mg>dL per hour. = 1 ¢t Notice that we can make no estimate of the concentration’s rate of change at times t = 1, 2, Á , 5, where the graph we have drawn for the concentration has a corner and no slope. The derivative step function is not defined at these times.
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Chapter 3: Differentiation
Athens
110
a ge Ae
GREECE
TURKEY
nS ea
Concentration, mg/dL
y
100 90 SANTORINI
80
RHODES
0
1
2
3 4 Time (h)
5
Sea of Crete
t
6
(a)
Heraklion
Mediterranean Sea
0
50
CRETE
100
150 km
Daedalus's flight path on April 23, 1988 15 10
▼
Rate of change of concentration,
mg/dL h
y'
5 0
1
2
3
4
5
6
t
FIGURE 3.4 (a) Graph of the sugar concentration in the blood of a Daedalus pilot during a 6hour preflight endurance test. (b) The derivative of the pilot’s bloodsugar concentration shows how rapidly the concentration rose and fell during various portions of the test.
–5
Differentiable on an Interval; OneSided Derivatives
–10 Time (h)
A function y = ƒsxd is differentiable on an open interval (finite or infinite) if it has a derivative at each point of the interval. It is differentiable on a closed interval [a, b] if it is differentiable on the interior (a, b) and if the limits
(b)
lim
ƒsa + hd  ƒsad h
Righthand derivative at a
lim
ƒsb + hd  ƒsbd h
Lefthand derivative at b
h:0 +
h:0 
Slope f(a h) f(a) lim h h→0
exist at the endpoints (Figure 3.5). Righthand and lefthand derivatives may be defined at any point of a function’s domain. The usual relation between onesided and twosided limits holds for these derivatives. Because of Theorem 6, Section 2.4, a function has a derivative at a point if and only if it has lefthand and righthand derivatives there, and these onesided derivatives are equal.
Slope f (b h) f (b) lim h h→0
EXAMPLE 5
Show that the function y = ƒ x ƒ is differentiable on s  q , 0d and s0, q d but has no derivative at x = 0.
y f (x)
Solution
a
ah h0
y = ƒ x ƒ Is Not Differentiable at the Origin
bh h0
b
To the right of the origin, d d d sxd = sxd = s1 # xd = 1. dx ƒ ƒ dx dx
x
d smx + bd = m, ƒ x ƒ = x dx
To the left,
FIGURE 3.5 Derivatives at endpoints are onesided limits.
d d d sxd = s xd = s 1 # xd = 1 dx ƒ ƒ dx dx
ƒ x ƒ = x
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3.1
y x
ƒ0 + hƒ  ƒ0ƒ ƒhƒ = lim+ h h:0 h:0 h h = lim+ ƒ h ƒ = h when h 7 0. h:0 h = lim+1 = 1
Righthand derivative of ƒ x ƒ at zero = lim+
y' 1
0
153
(Figure 3.6). There can be no derivative at the origin because the onesided derivatives differ there:
y
y' –1
The Derivative as a Function
x y' not defined at x 0: righthand derivative lefthand derivative
h:0
ƒ0 + hƒ  ƒ0ƒ ƒhƒ = limh h:0 h:0 h h = limƒ h ƒ = h when h 6 0. h:0 h = lim  1 = 1.
Lefthand derivative of ƒ x ƒ at zero = lim
FIGURE 3.6 The function y = ƒ x ƒ is not differentiable at the origin where the graph has a “corner.”
h:0
EXAMPLE 6
y = 1x Is Not Differentiable at x = 0
In Example 2 we found that for x 7 0, d 1 1x = . dx 21x We apply the definition to examine if the derivative exists at x = 0: lim
h:0 +
20 + h  20 1 = q. = lim+ h h:0 1h
Since the (righthand) limit is not finite, there is no derivative at x = 0. Since the slopes of the secant lines joining the origin to the points A h, 1h B on a graph of y = 1x approach q , the graph has a vertical tangent at the origin.
When Does a Function Not Have a Derivative at a Point? A function has a derivative at a point x0 if the slopes of the secant lines through Psx0, ƒsx0 dd and a nearby point Q on the graph approach a limit as Q approaches P. Whenever the secants fail to take up a limiting position or become vertical as Q approaches P, the derivative does not exist. Thus differentiability is a “smoothness” condition on the graph of ƒ. A function whose graph is otherwise smooth will fail to have a derivative at a point for several reasons, such as at points where the graph has 1.
a corner, where the onesided derivatives differ.
2. a cusp, where the slope of PQ approaches q from one side and  q from the other. P
P
Q Q
Q
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Q
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Chapter 3: Differentiation
3.
a vertical tangent, where the slope of PQ approaches q from both sides or approaches  q from both sides (here,  q ).
Q
P
Q
4.
a discontinuity.
P
P
Q
Q
Q
Q
Differentiable Functions Are Continuous A function is continuous at every point where it has a derivative.
THEOREM 1 Differentiability Implies Continuity If ƒ has a derivative at x = c, then ƒ is continuous at x = c.
Proof Given that ƒ¿scd exists, we must show that limx:c ƒsxd = ƒscd, or equivalently, that limh:0 ƒsc + hd = ƒscd. If h Z 0, then ƒsc + hd = ƒscd + sƒsc + hd  ƒscdd ƒsc + hd  ƒscd # h. = ƒscd + h
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155
Now take limits as h : 0. By Theorem 1 of Section 2.2, ƒsc + hd  ƒscd h h:0
lim ƒsc + hd = lim ƒscd + lim
h:0
h:0
# lim h h:0
= ƒscd + ƒ¿scd # 0 = ƒscd + 0 = ƒscd. Similar arguments with onesided limits show that if ƒ has a derivative from one side (right or left) at x = c then ƒ is continuous from that side at x = c. Theorem 1 on page 154 says that if a function has a discontinuity at a point (for instance, a jump discontinuity), then it cannot be differentiable there. The greatest integer function y = :x; = int x fails to be differentiable at every integer x = n (Example 4, Section 2.6).
CAUTION The converse of Theorem 1 is false. A function need not have a derivative at a point where it is continuous, as we saw in Example 5.
y
1
The Intermediate Value Property of Derivatives
y U(x)
Not every function can be some function’s derivative, as we see from the following theorem. x
0
FIGURE 3.7 The unit step function does not have the Intermediate Value Property and cannot be the derivative of a function on the real line.
THEOREM 2 If a and b are any two points in an interval on which ƒ is differentiable, then ƒ¿ takes on every value between ƒ¿sad and ƒ¿sbd.
Theorem 2 (which we will not prove) says that a function cannot be a derivative on an interval unless it has the Intermediate Value Property there. For example, the unit step function in Figure 3.7 cannot be the derivative of any realvalued function on the real line. In Chapter 5 we will see that every continuous function is a derivative of some function. In Section 4.4, we invoke Theorem 2 to analyze what happens at a point on the graph of a twicedifferentiable function where it changes its “bending” behavior.
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3.1 The Derivative as a Function
EXERCISES 3.1 1  z ; 2z
Finding Derivative Functions and Values
4. k szd =
Using the definition, calculate the derivatives of the functions in Exercises 1–6. Then find the values of the derivatives as specified.
5. psud = 23u ;
1. ƒsxd = 4  x 2;
ƒ¿s 3d, ƒ¿s0d, ƒ¿s1d 2
2. Fsxd = sx  1d + 1; 3. g std =
1 ; t2
F¿s 1d, F¿s0d, F¿s2d
g¿s 1d, g¿s2d, g¿ A 23 B
k¿s 1d, k¿s1d, k¿ A 22 B p¿s1d, p¿s3d, p¿s2>3d
6. r ssd = 22s + 1 ;
r¿s0d, r¿s1d, r¿s1>2d
In Exercises 7–12, find the indicated derivatives. dy s3 dr if y = 2x 3 if r = + 1 7. 8. 2 dx ds
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155
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Chapter 3: Differentiation
9.
ds dt
if
s =
10.
dy dt
if
1 y = t  t
11.
dp dq
if
p =
dz 12. dw
if
z =
Graphs
t 2t + 1
Match the functions graphed in Exercises 27–30 with the derivatives graphed in the accompanying figures (a)–(d). y'
2q + 1 1 23w  2
2
15. s = t  t ,
(b)
y'
y'
x = 2
x
0
x
0
t = 1
16. y = sx + 1d3,
x = 2
In Exercises 17–18, differentiate the functions. Then find an equation of the tangent line at the indicated point on the graph of the function. 17. y = ƒsxd =
(a)
x = 3
1 , 2 + x
3
x
0
In Exercises 13–16, differentiate the functions and find the slope of the tangent line at the given value of the independent variable. 9 13. ƒsxd = x + x ,
x
0
Slopes and Tangent Lines
14. k sxd =
y'
1
8 2x  2
,
(c)
27.
(d) y
28.
y
sx, yd = s6, 4d
18. w = g szd = 1 + 24  z,
In Exercises 19–22, find the values of the derivatives. 19.
ds ` dt t = 1
20.
dy ` dx x = 23
21.
dr ` du u = 0
if
r =
22.
dw ` dz z = 4
if
w = z + 1z
y f 2 (x)
y f1(x)
sz, wd = s3, 2d
x
0
x
0
s = 1  3t 2
if
1 y = 1  x
if
29.
30.
y
y
y f3(x)
2
y f4(x)
24  u
Using the Alternative Formula for Derivatives Use the formula ƒ¿sxd = lim
z: x
x
0
ƒszd  ƒsxd z  x
31. a. The graph in the accompanying figure is made of line segments joined end to end. At which points of the interval [4, 6] is ƒ¿ not defined? Give reasons for your answer. y
to find the derivative of the functions in Exercises 23–26. 23. ƒsxd =
1 x + 2
24. ƒsxd =
1 sx  1d2
25. g sxd =
x x  1
x
0
(6, 2)
(0, 2) y f(x)
(– 4, 0)
0
1
(1, –2)
26. g sxd = 1 + 1x
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6
(4, –2)
x
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3.1 The Derivative as a Function b. Graph the derivative of ƒ. The graph should show a step function.
157
p 350
32. Recovering a function from its derivative
300
a. Use the following information to graph the function ƒ over the closed interval [2, 5] .
250
i) The graph of ƒ is made of closed line segments joined end to end. ii) The graph starts at the point s 2, 3d . iii) The derivative of ƒ is the step function in the figure shown here.
200 150 100 50
y'
0
10
20
y' f '(x)
30
40
50
t
Time (days)
1 –2
0
1
3
b. During what days does the population seem to be increasing fastest? Slowest?
x
5
OneSided Derivatives
–2
Compare the righthand and lefthand derivatives to show that the functions in Exercises 35–38 are not differentiable at the point P. b. Repeat part (a) assuming that the graph starts at s 2, 0d instead of s 2, 3d . 33. Growth in the economy The graph in the accompanying figure shows the average annual percentage change y = ƒstd in the U.S. gross national product (GNP) for the years 1983–1988. Graph dy>dt (where defined). (Source: Statistical Abstracts of the United States, 110th Edition, U.S. Department of Commerce, p. 427.)
35.
y f (x) yx
y2 2
37.
0
38.
y
3
1
1985
1986
1987
1988
34. Fruit flies (Continuation of Example 3, Section 2.1.) Populations starting out in closed environments grow slowly at first, when there are relatively few members, then more rapidly as the number of reproducing individuals increases and resources are still abundant, then slowly again as the population reaches the carrying capacity of the environment. a. Use the graphical technique of Example 3 to graph the derivative of the fruit fly population introduced in Section 2.1. The graph of the population is reproduced here.
1 0
x
y f (x) P(1, 1)
y 2x 1 1984
2
y
y f (x)
2 1 0 1983
P(1, 2) 1 x
P(0, 0)
6 4
y 2x
y f (x)
2
yx
7% 5
y
36.
y
P(1, 1) y 兹x 1
1
x
yx
1
y 1x x
Differentiability and Continuity on an Interval Each figure in Exercises 39–44 shows the graph of a function over a closed interval D. At what domain points does the function appear to be a. differentiable? b. continuous but not differentiable? c. neither continuous nor differentiable?
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Give reasons for your answers. 39.
40. y
y
y f (x) D: –3 ⱕ x ⱕ 2 2
2
1
1
–3
–2
–1
0 1
2
x
–2
–1
0
–1
–1
–2
–2
41.
y f (x) D: –2 ⱕ x ⱕ 3
1
2
3
x
42. y
y y f (x) D: –3 ⱕ x ⱕ 3
y f (x) D: –2 ⱕ x ⱕ 3 3
1 –3 –2 –1 0 –1
1
2
3
2
x
1
–2
–2
43.
–1
1
0
3
x
44.
b. Show that y
y y f (x) D: –1 ⱕ x ⱕ 2
4
0
y f (x) D: –3 ⱕ x ⱕ 3
2
1
–1
2
1
2
x
51. Tangent to a parabola Does the parabola y = 2x 2  13x + 5 have a tangent whose slope is 1 ? If so, find an equation for the line and the point of tangency. If not, why not? 52. Tangent to y 1x Does any tangent to the curve y = 1x cross the xaxis at x = 1 ? If so, find an equation for the line and the point of tangency. If not, why not? 53. Greatest integer in x Does any function differentiable on s  q , q d have y = int x , the greatest integer in x (see Figure 2.55), as its derivative? Give reasons for your answer. 54. Derivative of y ƒ x ƒ Graph the derivative of ƒsxd = ƒ x ƒ . Then graph y = s ƒ x ƒ  0d>sx  0d = ƒ x ƒ >x . What can you conclude? 55. Derivative of ƒ Does knowing that a function ƒ(x) is differentiable at x = x0 tell you anything about the differentiability of the function ƒ at x = x0 ? Give reasons for your answer. 56. Derivative of multiples Does knowing that a function g (t) is differentiable at t = 7 tell you anything about the differentiability of the function 3g at t = 7 ? Give reasons for your answer. 57. Limit of a quotient Suppose that functions g(t) and h(t) are defined for all values of t and g s0d = hs0d = 0 . Can limt:0 sg stdd>shstdd exist? If it does exist, must it equal zero? Give reasons for your answers. 58. a. Let ƒ(x) be a function satisfying ƒ ƒsxd ƒ … x 2 for 1 … x … 1 . Show that ƒ is differentiable at x = 0 and find ƒ¿s0d .
–3 –2 –1 0
1 2
3
x
ƒsxd = L
1 x 2 sin x ,
x Z 0
0, x = 0 is differentiable at x = 0 and find ƒ¿s0d . T 59. Graph y = 1> A 21x B in a window that has 0 … x … 2 . Then, on the same screen, graph y =
for h = 1, 0.5, 0.1 . Then try h = 1, 0.5, 0.1 . Explain what is going on.
Theory and Examples In Exercises 45–48, a. Find the derivative ƒ¿sxd of the given function y = ƒsxd . b. Graph y = ƒsxd and y = ƒ¿sxd side by side using separate sets of coordinate axes, and answer the following questions. c. For what values of x, if any, is ƒ¿ positive? Zero? Negative? d. Over what intervals of xvalues, if any, does the function y = ƒsxd increase as x increases? Decrease as x increases? How is this related to what you found in part (c)? (We will say more about this relationship in Chapter 4.) 45. y = x 2
47. y = x 3>3
1x + h  1x h
46. y = 1>x 48. y = x 4>4
49. Does the curve y = x 3 ever have a negative slope? If so, where? Give reasons for your answer. 50. Does the curve y = 2 1x have any horizontal tangents? If so, where? Give reasons for your answer.
T 60. Graph y = 3x 2 in a window that has 2 … x … 2, 0 … y … 3 . Then, on the same screen, graph y =
sx + hd3  x 3 h
for h = 2, 1, 0.2 . Then try h = 2, 1, 0.2 . Explain what is going on. T 61. Weierstrass’s nowhere differentiable continuous function The sum of the first eight terms of the Weierstrass function ƒ(x) = g nq= 0 s2>3dn cos s9npxd is g sxd = cos spxd + s2>3d1 cos s9pxd + s2>3d2 cos s92pxd + s2>3d3 cos s93pxd + Á + s2>3d7 cos s97pxd . Graph this sum. Zoom in several times. How wiggly and bumpy is this graph? Specify a viewing window in which the displayed portion of the graph is smooth.
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3.1 The Derivative as a Function COMPUTER EXPLORATIONS Use a CAS to perform the following steps for the functions in Exercises 62–67. a. Plot y = ƒsxd to see that function’s global behavior. b. Define the difference quotient q at a general point x, with general step size h. c. Take the limit as h : 0 . What formula does this give? d. Substitute the value x = x0 and plot the function y = ƒsxd together with its tangent line at that point.
159
f. Graph the formula obtained in part (c). What does it mean when its values are negative? Zero? Positive? Does this make sense with your plot from part (a)? Give reasons for your answer. 62. ƒsxd = x 3 + x 2  x, 63. ƒsxd = x
1>3
+ x
2>3
,
x0 = 1 x0 = 1
4x , x0 = 2 x2 + 1 66. ƒsxd = sin 2x, x0 = p>2 64. ƒsxd =
x  1 , x0 = 1 3x 2 + 1 67. ƒsxd = x 2 cos x, x0 = p>4 65. ƒsxd =
e. Substitute various values for x larger and smaller than x0 into the formula obtained in part (c). Do the numbers make sense with your picture?
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Differentiation Rules
3.2
This section introduces a few rules that allow us to differentiate a great variety of functions. By proving these rules here, we can differentiate functions without having to apply the definition of the derivative each time.
Powers, Multiples, Sums, and Differences The first rule of differentiation is that the derivative of every constant function is zero.
RULE 1 Derivative of a Constant Function If ƒ has the constant value ƒsxd = c, then dƒ d = scd = 0. dx dx
EXAMPLE 1 If ƒ has the constant value ƒsxd = 8, then
y c
(x h, c)
(x, c)
df d s8d = 0. = dx dx
yc
Similarly, h 0
x
xh
FIGURE 3.8 The rule sd>dxdscd = 0 is another way to say that the values of constant functions never change and that the slope of a horizontal line is zero at every point.
x
d p a b = 0 2 dx
and
d a23b = 0. dx
Proof of Rule 1 We apply the definition of derivative to ƒsxd = c, the function whose outputs have the constant value c (Figure 3.8). At every value of x, we find that ƒsx + hd  ƒsxd c  c = lim = lim 0 = 0. h h h:0 h:0 h:0
ƒ¿sxd = lim
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The second rule tells how to differentiate x n if n is a positive integer.
RULE 2 Power Rule for Positive Integers If n is a positive integer, then d n x = nx n  1 . dx
To apply the Power Rule, we subtract 1 from the original exponent (n) and multiply the result by n.
EXAMPLE 2
HISTORICAL BIOGRAPHY Richard Courant (1888–1972)
Interpreting Rule 2
ƒ
x
x2
x3
x4
Á
ƒ¿
1
2x
3x 2
4x 3
Á
First Proof of Rule 2 The formula z n  x n = sz  xdsz n  1 + z n  2 x + Á + zx n  2 + x n  1 d can be verified by multiplying out the righthand side. Then from the alternative form for the definition of the derivative, ƒszd  ƒsxd zn  xn = lim z  x z x z:x z:x
ƒ¿sxd = lim
= lim sz n  1 + z n  2x + Á + zx n  2 + x n  1 d z:x
= nx n  1 Second Proof of Rule 2 If ƒsxd = x n , then ƒsx + hd = sx + hdn . Since n is a positive integer, we can expand sx + hdn by the Binomial Theorem to get ƒsx + hd  ƒsxd sx + hdn  x n = lim h h h:0 h:0
ƒ¿sxd = lim
= lim
cx n + nx n  1h +
h:0
nx n  1h + = lim
h:0
= lim cnx n  1 + h:0
nsn  1d n  2 2 Á x h + + nxh n  1 + h n d  x n 2 h
nsn  1d n  2 2 Á x h + + nxh n  1 + h n 2 h nsn  1d n  2 x h + Á + nxh n  2 + h n  1 d 2
= nx n  1 The third rule says that when a differentiable function is multiplied by a constant, its derivative is multiplied by the same constant.
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161
RULE 3 Constant Multiple Rule If u is a differentiable function of x, and c is a constant, then d du scud = c . dx dx
In particular, if n is a positive integer, then
y y 3x 2
3
d scx n d = cnx n  1 . dx Slope 3(2x) 6x Slope 6(1) 6
(1, 3)
EXAMPLE 3 (a) The derivative formula d s3x 2 d = 3 # 2x = 6x dx
y x2 2
1
0
Slope 2x Slope 2(1) 2 (1, 1)
1
2
says that if we rescale the graph of y = x 2 by multiplying each ycoordinate by 3, then we multiply the slope at each point by 3 (Figure 3.9). (b) A useful special case The derivative of the negative of a differentiable function u is the negative of the function’s derivative. Rule 3 with c = 1 gives d d d du s ud = s 1 # ud = 1 # sud =  . dx dx dx dx
x
FIGURE 3.9 The graphs of y = x 2 and y = 3x 2 . Tripling the ycoordinates triples the slope (Example 3).
Proof of Rule 3 cusx + hd  cusxd d cu = lim dx h h:0 usx + hd  usxd h h:0
= c lim = c
du dx
Derivative definition with ƒsxd = cusxd Limit property
u is differentiable.
The next rule says that the derivative of the sum of two differentiable functions is the sum of their derivatives. Denoting Functions by u and Y The functions we are working with when we need a differentiation formula are likely to be denoted by letters like ƒ and g. When we apply the formula, we do not want to find it using these same letters in some other way. To guard against this problem, we denote the functions in differentiation rules by letters like u and y that are not likely to be already in use.
RULE 4 Derivative Sum Rule If u and y are differentiable functions of x, then their sum u + y is differentiable at every point where u and y are both differentiable. At such points, d du dy su + yd = + . dx dx dx
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Chapter 3: Differentiation
EXAMPLE 4
Derivative of a Sum y = x 4 + 12x dy d 4 d = sx d + s12xd dx dx dx = 4x 3 + 12
Proof of Rule 4 We apply the definition of derivative to ƒsxd = usxd + ysxd: [usx + hd + ysx + hd]  [usxd + ysxd] d [usxd + ysxd] = lim dx h h:0 = lim c h:0
ysx + hd  ysxd usx + hd  usxd + d h h
usx + hd  usxd ysx + hd  ysxd du dy + lim = + . h h dx dx h:0 h:0
= lim
Combining the Sum Rule with the Constant Multiple Rule gives the Difference Rule, which says that the derivative of a difference of differentiable functions is the difference of their derivatives. d dy d du du dy su  yd = [u + s 1dy] = + s 1d = dx dx dx dx dx dx The Sum Rule also extends to sums of more than two functions, as long as there are only finitely many functions in the sum. If u1 , u2 , Á , un are differentiable at x, then so is u1 + u2 + Á + un , and dun du1 du2 d su + u2 + Á + un d = + + Á + . dx 1 dx dx dx
EXAMPLE 5
Derivative of a Polynomial y = x3 +
4 2 x  5x + 1 3
dy d 3 d 4 2 d d = x + a x b s5xd + s1d dx dx dx 3 dx dx = 3x 2 +
4# 2x  5 + 0 3
= 3x 2 +
8 x  5 3
Notice that we can differentiate any polynomial term by term, the way we differentiated the polynomial in Example 5. All polynomials are differentiable everywhere. Proof of the Sum Rule for Sums of More Than Two Functions
We prove the statement
dun du1 du2 d su + u2 + Á + un d = + + Á + dx 1 dx dx dx by mathematical induction (see Appendix 1). The statement is true for n = 2, as was just proved. This is Step 1 of the induction proof. Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
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163
Step 2 is to show that if the statement is true for any positive integer n = k, where k Ú n0 = 2, then it is also true for n = k + 1. So suppose that duk du2 du1 d + + Á + . su + u2 + Á + uk d = dx 1 dx dx dx
(1)
Then d (u1 + u2 + Á + uk + uk + 1) dx (++++)++++* ()* Call the function defined by this sum u.
Call this function y.
=
duk + 1 d su + u2 + Á + uk d + dx 1 dx
Rule 4 for
=
duk duk + 1 du1 du2 + + Á + + . dx dx dx dx
Eq. (1)
d su + yd dx
With these steps verified, the mathematical induction principle now guarantees the Sum Rule for every integer n Ú 2.
EXAMPLE 6 y
Finding Horizontal Tangents
Does the curve y = x 4  2x 2 + 2 have any horizontal tangents? If so, where?
y ⫽ x 4 ⫺ 2x 2 ⫹ 2
Solution
The horizontal tangents, if any, occur where the slope dy>dx is zero. We have, dy d 4 sx  2x 2 + 2d = 4x 3  4x. = dx dx
(0, 2)
(–1, 1)
–1
1
0
Now solve the equation
(1, 1)
1
dy = 0 for x: dx 4x 3  4x = 0 4xsx 2  1d = 0 x = 0, 1, 1.
x
FIGURE 3.10 The curve y = x 4  2x 2 + 2 and its horizontal tangents (Example 6).
The curve y = x 4  2x 2 + 2 has horizontal tangents at x = 0, 1, and 1. The corresponding points on the curve are (0, 2), (1, 1) and s 1, 1d. See Figure 3.10.
Products and Quotients While the derivative of the sum of two functions is the sum of their derivatives, the derivative of the product of two functions is not the product of their derivatives. For instance, d 2 d sx # xd = sx d = 2x, dx dx
while
d d sxd # sxd = 1 # 1 = 1. dx dx
The derivative of a product of two functions is the sum of two products, as we now explain.
RULE 5 Derivative Product Rule If u and y are differentiable at x, then so is their product uy, and d dy du suyd = u + y . dx dx dx
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Picturing the Product Rule If u(x) and y(x) are positive and increase when x increases, and if h 7 0 ,
The derivative of the product uy is u times the derivative of y plus y times the derivative of u. In prime notation, suyd¿ = uy¿ + yu¿ . In function notation, d [ƒsxdgsxd] = ƒsxdg¿sxd + gsxdƒ¿sxd. dx
y(x h) y
u(x) y
u y
u(x)y(x)
y(x) u
y(x)
EXAMPLE 7
Using the Product Rule
Find the derivative of
0
u(x)
u u(x h) Solution
then the total shaded area in the picture is usx + hdysx + hd  usxdysxd = usx + hd ¢y + ysx + hd ¢u  ¢u¢y . Dividing both sides of this equation by h gives usx + hdysx + hd  usxdysxd h ¢u ¢y + ysx + hd = usx + hd h h ¢y  ¢u . h As h : 0 + , ¢u #
dy ¢y :0# = 0, h dx
leaving
1 1 y = x ax 2 + x b . We apply the Product Rule with u = 1>x and y = x 2 + s1>xd: d dy du suyd = u + y , and dx dx dx d 1 1 a x b =  2 by dx x Example 3, Section 2.7.
d 1 2 1 1 1 1 1 c ax + x b d = x a2x  2 b + ax 2 + x b a 2 b dx x x x = 2 
1 1  1  3 x3 x
= 1 
2 . x3
Proof of Rule 5 usx + hdysx + hd  usxdysxd d suyd = lim dx h h:0 To change this fraction into an equivalent one that contains difference quotients for the derivatives of u and y, we subtract and add usx + hdysxd in the numerator: usx + hdysx + hd  usx + hdysxd + usx + hdysxd  usxdysxd d suyd = lim dx h h:0 = lim cusx + hd
dy du d suyd = u + y . dx dx dx
h:0
ysx + hd  ysxd usx + hd  usxd + ysxd d h h
= lim usx + hd # lim h:0
h:0
ysx + hd  ysxd usx + hd  usxd + ysxd # lim . h h h:0
As h approaches zero, usx + hd approaches u(x) because u, being differentiable at x, is continuous at x. The two fractions approach the values of dy>dx at x and du>dx at x. In short, d dy du suyd = u + y . dx dx dx In the following example, we have only numerical values with which to work.
EXAMPLE 8
Derivative from Numerical Values
Let y = uy be the product of the functions u and y. Find y¿s2d if us2d = 3, Solution
u¿s2d = 4,
ys2d = 1,
and
y¿s2d = 2.
From the Product Rule, in the form y¿ = suyd¿ = uy¿ + yu¿ ,
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165
we have y¿s2d = us2dy¿s2d + ys2du¿s2d = s3ds2d + s1ds 4d = 6  4 = 2.
EXAMPLE 9
Differentiating a Product in Two Ways
Find the derivative of y = sx 2 + 1dsx 3 + 3d. Solution
(a) From the Product Rule with u = x 2 + 1 and y = x 3 + 3, we find d dx
C A x 2 + 1 B A x 3 + 3 B D = sx 2 + 1ds3x 2 d + sx 3 + 3ds2xd = 3x 4 + 3x 2 + 2x 4 + 6x = 5x 4 + 3x 2 + 6x.
(b) This particular product can be differentiated as well (perhaps better) by multiplying out the original expression for y and differentiating the resulting polynomial: y = sx 2 + 1dsx 3 + 3d = x 5 + x 3 + 3x 2 + 3 dy = 5x 4 + 3x 2 + 6x. dx This is in agreement with our first calculation. Just as the derivative of the product of two differentiable functions is not the product of their derivatives, the derivative of the quotient of two functions is not the quotient of their derivatives. What happens instead is the Quotient Rule.
RULE 6 Derivative Quotient Rule If u and y are differentiable at x and if ysxd Z 0, then the quotient u>y is differentiable at x, and d u a b = dx y
y
du dy  u dx dx . 2 y
In function notation, gsxdƒ¿sxd  ƒsxdg¿sxd d ƒsxd c d = . dx gsxd g 2sxd
EXAMPLE 10
Using the Quotient Rule
Find the derivative of y =
t2  1 . t2 + 1
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Chapter 3: Differentiation Solution
We apply the Quotient Rule with u = t 2  1 and y = t 2 + 1: dy st 2 + 1d # 2t  st 2  1d # 2t = dt st 2 + 1d2 =
2t 3 + 2t  2t 3 + 2t st 2 + 1d2
=
4t . st 2 + 1d2
ysdu>dtd  usdy>dtd d u a b = dt y y2
Proof of Rule 6 usx + hd usxd ysx + hd ysxd d u a b = lim dx y h h:0 ysxdusx + hd  usxdysx + hd hysx + hdysxd h:0
= lim
To change the last fraction into an equivalent one that contains the difference quotients for the derivatives of u and y, we subtract and add y(x)u(x) in the numerator. We then get ysxdusx + hd  ysxdusxd + ysxdusxd  usxdysx + hd d u a b = lim dx y hysx + hdysxd h:0 ysxd = lim
h:0
usx + hd  usxd ysx + hd  ysxd  usxd h h . ysx + hdysxd
Taking the limit in the numerator and denominator now gives the Quotient Rule.
Negative Integer Powers of x The Power Rule for negative integers is the same as the rule for positive integers.
RULE 7 Power Rule for Negative Integers If n is a negative integer and x Z 0, then d n sx d = nx n  1 . dx
EXAMPLE 11 (a)
d 1 d 1 1 a b = sx d = s 1dx 2 =  2 dx x dx x
(b)
d 4 d 12 a b = 4 sx 3 d = 4s 3dx 4 =  4 dx x 3 dx x
Agrees with Example 3, Section 2.7
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Proof of Rule 7 The proof uses the Quotient Rule. If n is a negative integer, then n = m, where m is a positive integer. Hence, x n = x m = 1>x m , and d n d 1 sx d = a b dx dx x m xm #
y
=
0  mx m  1 x 2m = mx m  1 = nx n  1 .
y x 2x
=
4 3
Quotient Rule with u = 1 and y = x m
Since m 7 0,
d m sx d = mx m  1 dx
Since m = n
(1, 3)
EXAMPLE 12
2
1
2
3
Tangent to a Curve
Find an equation for the tangent to the curve
y –x 4
1
0
d d A 1 B  1 # dx A x m B dx sx m d2
2 y = x + x
x
at the point (1, 3) (Figure 3.11). FIGURE 3.11 The tangent to the curve y = x + s2>xd at (1, 3) in Example 12. The curve has a thirdquadrant portion not shown here. We see how to graph functions like this one in Chapter 4.
Solution
The slope of the curve is dy d d 1 1 2 = sxd + 2 a b = 1 + 2 a 2 b = 1  2 . dx dx dx x x x
The slope at x = 1 is dy 2 ` = c1  2 d = 1  2 = 1. dx x = 1 x x=1 The line through (1, 3) with slope m = 1 is y  3 = s 1dsx  1d y = x + 1 + 3 y = x + 4.
Pointslope equation
The choice of which rules to use in solving a differentiation problem can make a difference in how much work you have to do. Here is an example.
EXAMPLE 13
Choosing Which Rule to Use
Rather than using the Quotient Rule to find the derivative of y =
sx  1dsx 2  2xd , x4
expand the numerator and divide by x 4 : y =
sx  1dsx 2  2xd x 3  3x 2 + 2x = = x 1  3x 2 + 2x 3 . 4 x x4
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Then use the Sum and Power Rules: dy = x 2  3s 2dx 3 + 2s 3dx 4 dx 6 6 1 =  2 + 3  4. x x x
Second and HigherOrder Derivatives If y = ƒsxd is a differentiable function, then its derivative ƒ¿sxd is also a function. If ƒ¿ is also differentiable, then we can differentiate ƒ¿ to get a new function of x denoted by ƒ– . So ƒ– = sƒ¿d¿ . The function ƒ– is called the second derivative of ƒ because it is the derivative of the first derivative. Notationally, ƒ–sxd =
d 2y dx
2
=
dy¿ d dy a b = = y– = D 2sƒdsxd = D x2 ƒsxd. dx dx dx
The symbol D 2 means the operation of differentiation is performed twice. If y = x 6 , then y¿ = 6x 5 and we have y– =
How to Read the Symbols for Derivatives “y prime” y¿ “y double prime” y– d 2y “d squared y dx squared” dx 2 y‡ “y triple prime” y snd “y super n” d ny “d to the n of y by dx to the n” dx n n D “D to the n”
dy¿ d = A 6x 5 B = 30x 4 . dx dx
Thus D 2 A x 6 B = 30x 4 . If y– is differentiable, its derivative, y‡ = dy–>dx = d 3y>dx 3 is the third derivative of y with respect to x. The names continue as you imagine, with y snd =
d ny d sn  1d y = = D ny dx dx n
denoting the nth derivative of y with respect to x for any positive integer n. We can interpret the second derivative as the rate of change of the slope of the tangent to the graph of y = ƒsxd at each point. You will see in the next chapter that the second derivative reveals whether the graph bends upward or downward from the tangent line as we move off the point of tangency. In the next section, we interpret both the second and third derivatives in terms of motion along a straight line.
EXAMPLE 14
Finding Higher Derivatives
The first four derivatives of y = x 3  3x 2 + 2 are First derivative: Second derivative: Third derivative: Fourth derivative:
y¿ = 3x 2  6x y– = 6x  6 y‡ = 6 y s4d = 0.
The function has derivatives of all orders, the fifth and later derivatives all being zero.
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EXERCISES 3.2 Derivative Calculations In Exercises 1–12, find the first and second derivatives. 1. y = x 2 + 3 3
3. s = 5t  3t
4. w = 3z 7  7z 3 + 21z 2 3
4x  x 3
6. y =
1 7. w = 3z 2  z
2
5 1 11. r = 2 2s 3s
x x x + + 3 2 4
4 t2 10. y = 4  2x  x 3 12 1 4  3 + 4 12. r = u u u
In Exercises 13–16, find y¿ (a) by applying the Product Rule and (b) by multiplying the factors to produce a sum of simpler terms to differentiate. 3
1 1 1 15. y = sx 2 + 1d ax + 5 + x b 16. y = ax + x b ax  x + 1b Find the derivatives of the functions in Exercises 17–28. 2x + 5 3x  2
19. g sxd =
18. z =
x2  4 x + 0.5 1s  1 1s + 1
t2  1 t + t  2 22. w = s2x  7d1sx + 5d 24. u =
1 + x  4 1x 25. y = x
2
5x + 1 2 1x
26. r = 2 a
1
+ 2ub 2u sx + 1dsx + 2d 1 27. y = 2 28. y = 2 sx  1dsx  2d sx  1dsx + x + 1d Find the derivatives of all orders of the functions in Exercises 29 and 30. 29. y =
3 x4  x2  x 2 2
30. y =
x5 120
33. r =
x3 + 7 x
32. s =
su  1dsu 2 + u + 1d
35. w = a
us0d = 5,
u3 1 + 3z bs3  zd 3z
34. u =
u¿s0d = 3,
ys0d = 1,
y¿s0d = 2 .
Find the values of the following derivatives at x = 0 . a.
d suyd dx
b.
d u a b dx y
c.
d y a b dx u
d.
d s7y  2ud dx
40. Suppose u and y are differentiable functions of x and that us1d = 2,
u¿s1d = 0,
ys1d = 5,
y¿s1d = 1 .
Find the values of the following derivatives at x = 1 . a.
d suyd dx
b.
d u a b dx y
c.
d y a b dx u
d.
d s7y  2ud dx
Slopes and Tangents
b. Smallest slope What is the smallest slope on the curve? At what point on the curve does the curve have this slope? c. Tangents having specified slope Find equations for the tangents to the curve at the points where the slope of the curve is 8. 42. a. Horizontal tangents Find equations for the horizontal tangents to the curve y = x 3  3x  2 . Also find equations for the lines that are perpendicular to these tangents at the points of tangency. b. Smallest slope What is the smallest slope on the curve? At what point on the curve does the curve have this slope? Find an equation for the line that is perpendicular to the curve’s tangent at this point. 43. Find the tangents to Newton’s serpentine (graphed here) at the origin and the point (1, 2).
Find the first and second derivatives of the functions in Exercises 31–38. 31. y =
sq  1d3 + sq + 1d3
41. a. Normal to a curve Find an equation for the line perpendicular to the tangent to the curve y = x 3  4x + 1 at the point (2, 1).
2x + 1 x2  1
20. ƒstd =
21. y = s1  tds1 + t 2 d1 23. ƒssd =
q2 + 3
39. Suppose u and y are functions of x that are differentiable at x = 0 and that
2
13. y = s3  x dsx  x + 1d 14. y = sx  1dsx + x + 1d
17. y =
38. p =
Using Numerical Values
8. s = 2t 1 +
9. y = 6x 2  10x  5x 2
2
q2 + 3 q4  1 b ba 12q q3
2. y = x 2 + x + 8
5
3
5. y =
37. p = a
t 2 + 5t  1 t2 sx 2 + xdsx 2  x + 1d x4
y y 24x x 1
(1, 2) 2 1 0
1 2 3 4
36. w = sz + 1dsz  1dsz 2 + 1d
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44. Find the tangent to the Witch of Agnesi (graphed here) at the point (2, 1).
51. Suppose that the function y in the Product Rule has a constant value c. What does the Product Rule then say? What does this say about the Constant Multiple Rule?
y y 2 1 0
8 x2 4
52. The Reciprocal Rule
(2, 1) 1 2 3
how to find the amount of medicine to which the body is most sensitive.
x
45. Quadratic tangent to identity function The curve y = ax 2 + bx + c passes through the point (1, 2) and is tangent to the line y = x at the origin. Find a, b, and c. 46. Quadratics having a common tangent The curves y = x 2 + ax + b and y = cx  x 2 have a common tangent line at the point (1, 0). Find a, b, and c. 47. a. Find an equation for the line that is tangent to the curve y = x 3  x at the point s 1, 0d .
a. The Reciprocal Rule says that at any point where the function y(x) is differentiable and different from zero, d 1 1 dy a b =  2 . dx y y dx Show that the Reciprocal Rule is a special case of the Quotient Rule. b. Show that the Reciprocal Rule and the Product Rule together imply the Quotient Rule. 53. Generalizing the Product Rule The Product Rule gives the formula
T b. Graph the curve and tangent line together. The tangent intersects the curve at another point. Use Zoom and Trace to estimate the point’s coordinates. T
c. Confirm your estimates of the coordinates of the second intersection point by solving the equations for the curve and tangent simultaneously (Solver key).
48. a. Find an equation for the line that is tangent to the curve y = x 3  6x 2 + 5x at the origin. T b. Graph the curve and tangent together. The tangent intersects the curve at another point. Use Zoom and Trace to estimate the point’s coordinates. T
c. Confirm your estimates of the coordinates of the second intersection point by solving the equations for the curve and tangent simultaneously (Solver key).
d dy du suyd = u + y dx dx dx for the derivative of the product uy of two differentiable functions of x. a. What is the analogous formula for the derivative of the product uyw of three differentiable functions of x ? b. What is the formula for the derivative of the product u1 u2 u3 u4 of four differentiable functions of x? c. What is the formula for the derivative of a product u1 u2 u3 Á un of a finite number n of differentiable functions of x? 54. Rational Powers d 3>2 A x B by writing x 3>2 as x # x 1>2 and using the Product dx Rule. Express your answer as a rational number times a rational power of x. Work parts (b) and (c) by a similar method.
a. Find
Theory and Examples 49. The general polynomial of degree n has the form Psxd = an x n + an  1 x n  1 + Á + a2 x 2 + a1 x + a0 where an Z 0 . Find P¿sxd . 50. The body’s reaction to medicine The reaction of the body to a dose of medicine can sometimes be represented by an equation of the form R = M2 a
C M  b, 2 3
where C is a positive constant and M is the amount of medicine absorbed in the blood. If the reaction is a change in blood pressure, R is measured in millimeters of mercury. If the reaction is a change in temperature, R is measured in degrees, and so on. Find dR>dM . This derivative, as a function of M, is called the sensitivity of the body to the medicine. In Section 4.5, we will see
b. Find
d 5>2 sx d . dx
c. Find
d 7>2 sx d . dx
d. What patterns do you see in your answers to parts (a), (b), and (c)? Rational powers are one of the topics in Section 3.6. 55. Cylinder pressure If gas in a cylinder is maintained at a constant temperature T, the pressure P is related to the volume V by a formula of the form P =
an 2 nRT  2, V  nb V
in which a, b, n, and R are constants. Find dP>dV . (See accompanying figure.)
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56. The best quantity to order One of the formulas for inventory management says that the average weekly cost of ordering, paying for, and holding merchandise is hq km Asqd = q + cm + , 2 where q is the quantity you order when things run low (shoes, radios, brooms, or whatever the item might be); k is the cost of placing an order (the same, no matter how often you order); c is the cost of one item (a constant); m is the number of items sold each week (a constant); and h is the weekly holding cost per item (a constant that takes into account things such as space, utilities, insurance, and security). Find dA>dq and d 2A>dq 2 .
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3.3
171
The Derivative as a Rate of Change In Section 2.1, we initiated the study of average and instantaneous rates of change. In this section, we continue our investigations of applications in which derivatives are used to model the rates at which things change in the world around us. We revisit the study of motion along a line and examine other applications. It is natural to think of change as change with respect to time, but other variables can be treated in the same way. For example, a physician may want to know how change in dosage affects the body’s response to a drug. An economist may want to study how the cost of producing steel varies with the number of tons produced.
Instantaneous Rates of Change If we interpret the difference quotient sƒsx + hd  ƒsxdd>h as the average rate of change in ƒ over the interval from x to x + h, we can interpret its limit as h : 0 as the rate at which ƒ is changing at the point x.
DEFINITION Instantaneous Rate of Change The instantaneous rate of change of ƒ with respect to x at x0 is the derivative ƒ¿sx0 d = lim
h:0
ƒsx0 + hd  ƒsx0 d , h
provided the limit exists.
Thus, instantaneous rates are limits of average rates. It is conventional to use the word instantaneous even when x does not represent time. The word is, however, frequently omitted. When we say rate of change, we mean instantaneous rate of change.
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EXAMPLE 1
How a Circle’s Area Changes with Its Diameter
The area A of a circle is related to its diameter by the equation A =
p 2 D . 4
How fast does the area change with respect to the diameter when the diameter is 10 m? Solution
The rate of change of the area with respect to the diameter is dA p pD = # 2D = . 4 2 dD
When D = 10 m, the area is changing at rate sp>2d10 = 5p m2>m. Position at time t … s f(t)
∆s
Motion Along a Line: Displacement, Velocity, Speed, Acceleration, and Jerk
and at time t ∆ t s ∆s f (t ∆t)
FIGURE 3.12 The positions of a body moving along a coordinate line at time t and shortly later at time t + ¢t .
s
Suppose that an object is moving along a coordinate line (say an saxis) so that we know its position s on that line as a function of time t: s = ƒstd. The displacement of the object over the time interval from t to t + ¢t (Figure 3.12) is ¢s = ƒst + ¢td  ƒstd, and the average velocity of the object over that time interval is yay =
ƒst + ¢td  ƒstd displacement ¢s = = . travel time ¢t ¢t
To find the body’s velocity at the exact instant t, we take the limit of the average velocity over the interval from t to t + ¢t as ¢t shrinks to zero. This limit is the derivative of ƒ with respect to t.
DEFINITION Velocity Velocity (instantaneous velocity) is the derivative of position with respect to time. If a body’s position at time t is s = ƒstd, then the body’s velocity at time t is ystd =
EXAMPLE 2
ƒst + ¢td  ƒstd ds . = lim dt ¢t ¢t:0
Finding the Velocity of a Race Car
Figure 3.13 shows the timetodistance graph of a 1996 Riley & Scott Mk IIIOlds WSC race car. The slope of the secant PQ is the average velocity for the 3sec interval from t = 2 to t = 5 sec; in this case, it is about 100 ft> sec or 68 mph. The slope of the tangent at P is the speedometer reading at t = 2 sec, about 57 ft> sec or 39 mph. The acceleration for the period shown is a nearly constant 28.5 ft>sec2 during
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s 800 700
Distance (ft)
600 500
Secant slope is average velocity for interval from t 2 to t 5.
400 300
Q Tangent slope is speedometer reading at t 2 (instantaneous velocity).
200 100
P
0
1
2
3
4
5
6
7
8
t
Elapsed time (sec)
FIGURE 3.13 The timetodistance graph for Example 2. The slope of the tangent line at P is the instantaneous velocity at t = 2 sec.
each second, which is about 0.89g, where g is the acceleration due to gravity. The race car’s top speed is an estimated 190 mph. (Source: Road and Track, March 1997.) Besides telling how fast an object is moving, its velocity tells the direction of motion. When the object is moving forward (s increasing), the velocity is positive; when the body is moving backward (s decreasing), the velocity is negative (Figure 3.14).
s
s s f (t)
s f (t) ds 0 dt
ds 0 dt
t
0 s increasing: positive slope so moving forward
t
0 s decreasing: negative slope so moving backward
FIGURE 3.14 For motion s = ƒstd along a straight line, y = ds/dt is positive when s increases and negative when s decreases.
If we drive to a friend’s house and back at 30 mph, say, the speedometer will show 30 on the way over but it will not show 30 on the way back, even though our distance from home is decreasing. The speedometer always shows speed, which is the absolute value of velocity. Speed measures the rate of progress regardless of direction.
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DEFINITION Speed Speed is the absolute value of velocity. ds Speed = ƒ ystd ƒ = ` ` dt
EXAMPLE 3
Horizontal Motion
Figure 3.15 shows the velocity y = ƒ¿std of a particle moving on a coordinate line. The particle moves forward for the first 3 sec, moves backward for the next 2 sec, stands still for a second, and moves forward again. The particle achieves its greatest speed at time t = 4, while moving backward.
y MOVES FORWARD
FORWARD AGAIN
(y 0)
(y 0)
y f '(t) Speeds up
Steady (y const)
Slows down
Speeds up Stands still (y 0)
0
1
2
3
4
5
6
7
t (sec)
Greatest speed
Speeds up
Slows down
MOVES BACKWARD
(y 0)
FIGURE 3.15 The velocity graph for Example 3.
HISTORICAL BIOGRAPHY Bernard Bolzano (1781–1848)
The rate at which a body’s velocity changes is the body’s acceleration. The acceleration measures how quickly the body picks up or loses speed. A sudden change in acceleration is called a jerk. When a ride in a car or a bus is jerky, it is not that the accelerations involved are necessarily large but that the changes in acceleration are abrupt.
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DEFINITIONS Acceleration, Jerk Acceleration is the derivative of velocity with respect to time. If a body’s position at time t is s = ƒstd, then the body’s acceleration at time t is dy d 2s = 2. dt dt Jerk is the derivative of acceleration with respect to time: astd =
jstd =
da d 3s = 3. dt dt
Near the surface of the Earth all bodies fall with the same constant acceleration. Galileo’s experiments with free fall (Example 1, Section 2.1) lead to the equation s =
1 2 gt , 2
where s is distance and g is the acceleration due to Earth’s gravity. This equation holds in a vacuum, where there is no air resistance, and closely models the fall of dense, heavy objects, such as rocks or steel tools, for the first few seconds of their fall, before air resistance starts to slow them down. The value of g in the equation s = s1>2dgt 2 depends on the units used to measure t and s. With t in seconds (the usual unit), the value of g determined by measurement at sea level is approximately 32 ft>sec2 (feet per second squared) in English units, and g = 9.8 m>sec2 (meters per second squared) in metric units. (These gravitational constants depend on the distance from Earth’s center of mass, and are slightly lower on top of Mt. Everest, for example.) The jerk of the constant acceleration of gravity sg = 32 ft>sec2 d is zero: j = t (seconds) t0
s (meters)
t1
5
d sgd = 0. dt
An object does not exhibit jerkiness during free fall.
0
EXAMPLE 4
Modeling Free Fall
10
Figure 3.16 shows the free fall of a heavy ball bearing released from rest at time t = 0 sec.
15 t2
(a) How many meters does the ball fall in the first 2 sec? (b) What is its velocity, speed, and acceleration then?
20 25 30
Solution
35
(a) The metric freefall equation is s = 4.9t 2 . During the first 2 sec, the ball falls
40 t3
ss2d = 4.9s2d2 = 19.6 m.
45
(b) At any time t, velocity is the derivative of position: FIGURE 3.16 A ball bearing falling from rest (Example 4).
ystd = s¿std =
d s4.9t 2 d = 9.8t. dt
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At t = 2, the velocity is
s
ys2d = 19.6 m>sec
y0
smax
in the downward (increasing s) direction. The speed at t = 2 is
Height (ft)
Speed = ƒ ys2d ƒ = 19.6 m>sec. t?
256
The acceleration at any time t is astd = y¿std = s–std = 9.8 m>sec2 . At t = 2, the acceleration is 9.8 m>sec2 .
EXAMPLE 5
A dynamite blast blows a heavy rock straight up with a launch velocity of 160 ft> sec (about 109 mph) (Figure 3.17a). It reaches a height of s = 160t  16t 2 ft after t sec.
s0
(a) How high does the rock go? (b) What are the velocity and speed of the rock when it is 256 ft above the ground on the way up? On the way down? (c) What is the acceleration of the rock at any time t during its flight (after the blast)? (d) When does the rock hit the ground again?
(a) s, y 400
s 160t 16t 2
160
0 –160
Modeling Vertical Motion
Solution 5
10
t
y ds 160 32t dt (b)
FIGURE 3.17 (a) The rock in Example 5. (b) The graphs of s and y as functions of time; s is largest when y = ds/dt = 0 . The graph of s is not the path of the rock: It is a plot of height versus time. The slope of the plot is the rock’s velocity, graphed here as a straight line.
(a) In the coordinate system we have chosen, s measures height from the ground up, so the velocity is positive on the way up and negative on the way down. The instant the rock is at its highest point is the one instant during the flight when the velocity is 0. To find the maximum height, all we need to do is to find when y = 0 and evaluate s at this time. At any time t, the velocity is y =
ds d s160t  16t 2 d = 160  32t ft>sec. = dt dt
The velocity is zero when 160  32t = 0
or
t = 5 sec.
The rock’s height at t = 5 sec is smax = ss5d = 160s5d  16s5d2 = 800  400 = 400 ft. See Figure 3.17b. (b) To find the rock’s velocity at 256 ft on the way up and again on the way down, we first find the two values of t for which sstd = 160t  16t 2 = 256. To solve this equation, we write 16t 2  160t + 256 = 0 16st 2  10t + 16d = 0 st  2dst  8d = 0 t = 2 sec, t = 8 sec.
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The rock is 256 ft above the ground 2 sec after the explosion and again 8 sec after the explosion. The rock’s velocities at these times are
y (dollars) Slope marginal cost
0
The Derivative as a Rate of Change
y c(x)
xh
x (tons/week)
ys2d = 160  32s2d = 160  64 = 96 ft>sec. ys8d = 160  32s8d = 160  256 = 96 ft>sec. At both instants, the rock’s speed is 96 ft> sec. Since ys2d 7 0, the rock is moving upward (s is increasing) at t = 2 sec; it is moving downward (s is decreasing) at t = 8 because ys8d 6 0.
x
FIGURE 3.18 Weekly steel production: c(x) is the cost of producing x tons per week. The cost of producing an additional h tons is csx + hd  csxd .
(c) At any time during its flight following the explosion, the rock’s acceleration is a constant a =
d dy = s160  32td = 32 ft>sec2 . dt dt
The acceleration is always downward. As the rock rises, it slows down; as it falls, it speeds up. (d) The rock hits the ground at the positive time t for which s = 0. The equation 160t  16t 2 = 0 factors to give 16ts10  td = 0, so it has solutions t = 0 and t = 10. At t = 0, the blast occurred and the rock was thrown upward. It returned to the ground 10 sec later.
Derivatives in Economics Engineers use the terms velocity and acceleration to refer to the derivatives of functions describing motion. Economists, too, have a specialized vocabulary for rates of change and derivatives. They call them marginals. In a manufacturing operation, the cost of production c(x) is a function of x, the number of units produced. The marginal cost of production is the rate of change of cost with respect to level of production, so it is dc>dx. Suppose that c(x) represents the dollars needed to produce x tons of steel in one week. It costs more to produce x + h units per week, and the cost difference, divided by h, is the average cost of producing each additional ton:
y y c(x)
csx + hd  csxd average cost of each of the additional = h tons of steel produced. h
x 1
c
dc dx
The limit of this ratio as h : 0 is the marginal cost of producing more steel per week when the current weekly production is x tons (Figure 3.18). csx + hd  csxd dc = lim = marginal cost of production. dx h h:0 Sometimes the marginal cost of production is loosely defined to be the extra cost of producing one unit:
0
x
x1
x
FIGURE 3.19 The marginal cost dc>dx is approximately the extra cost ¢c of producing ¢ x = 1 more unit.
csx + 1d  csxd ¢c , = 1 ¢x which is approximated by the value of dc>dx at x. This approximation is acceptable if the slope of the graph of c does not change quickly near x. Then the difference quotient will be close to its limit dc>dx, which is the rise in the tangent line if ¢x = 1 (Figure 3.19). The approximation works best for large values of x.
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Economists often represent a total cost function by a cubic polynomial csxd = ax 3 + bx 2 + gx + d where d represents fixed costs such as rent, heat, equipment capitalization, and management costs. The other terms represent variable costs such as the costs of raw materials, taxes, and labor. Fixed costs are independent of the number of units produced, whereas variable costs depend on the quantity produced. A cubic polynomial is usually complicated enough to capture the cost behavior on a relevant quantity interval.
EXAMPLE 6
Marginal Cost and Marginal Revenue
Suppose that it costs csxd = x 3  6x 2 + 15x dollars to produce x radiators when 8 to 30 radiators are produced and that rsxd = x 3  3x 2 + 12x gives the dollar revenue from selling x radiators. Your shop currently produces 10 radiators a day. About how much extra will it cost to produce one more radiator a day, and what is your estimated increase in revenue for selling 11 radiators a day? Solution
The cost of producing one more radiator a day when 10 are produced is about
c¿s10d: c¿sxd =
d 3 A x  6x 2 + 15x B = 3x 2  12x + 15 dx
c¿s10d = 3s100d  12s10d + 15 = 195. The additional cost will be about $195. The marginal revenue is r¿sxd =
d 3 A x  3x 2 + 12x B = 3x 2  6x + 12. dx
The marginal revenue function estimates the increase in revenue that will result from selling one additional unit. If you currently sell 10 radiators a day, you can expect your revenue to increase by about r¿s10d = 3s100d  6s10d + 12 = $252 if you increase sales to 11 radiators a day.
EXAMPLE 7
Marginal Tax Rate
To get some feel for the language of marginal rates, consider marginal tax rates. If your marginal income tax rate is 28% and your income increases by $1000, you can expect to pay an extra $280 in taxes. This does not mean that you pay 28% of your entire income in taxes. It just means that at your current income level I, the rate of increase of taxes T with respect to income is dT>dI = 0.28. You will pay $0.28 out of every extra dollar you earn in taxes. Of course, if you earn a lot more, you may land in a higher tax bracket and your marginal rate will increase.
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Sensitivity to Change When a small change in x produces a large change in the value of a function ƒ(x), we say that the function is relatively sensitive to changes in x. The derivative ƒ¿sxd is a measure of this sensitivity.
EXAMPLE 8
Genetic Data and Sensitivity to Change
The Austrian monk Gregor Johann Mendel (1822–1884), working with garden peas and other plants, provided the first scientific explanation of hybridization. His careful records showed that if p (a number between 0 and 1) is the frequency of the gene for smooth skin in peas (dominant) and s1  pd is the frequency of the gene for wrinkled skin in peas, then the proportion of smoothskinned peas in the next generation will be y = 2ps1  pd + p 2 = 2p  p 2 . The graph of y versus p in Figure 3.20a suggests that the value of y is more sensitive to a change in p when p is small than when p is large. Indeed, this fact is borne out by the derivative graph in Figure 3.20b, which shows that dy>dp is close to 2 when p is near 0 and close to 0 when p is near 1. dy /dp 2 y dy 2 2p dp
1 y 2p p 2 1
0 (a)
p
0
1
p
(b)
FIGURE 3.20 (a) The graph of y = 2p  p 2 , describing the proportion of smoothskinned peas. (b) The graph of dy>dp (Example 8).
The implication for genetics is that introducing a few more dominant genes into a highly recessive population (where the frequency of wrinkled skin peas is small) will have a more dramatic effect on later generations than will a similar increase in a highly dominant population.
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EXERCISES 3.3 Motion Along a Coordinate Line Exercises 1–6 give the positions s = ƒstd of a body moving on a coordinate line, with s in meters and t in seconds. a. Find the body’s displacement and average velocity for the given time interval.
b. Find the body’s speed and acceleration at the endpoints of the interval. c. When, if ever, during the interval does the body change direction? 1. s = t 2  3t + 2, 2
2. s = 6t  t ,
0 … t … 2
0 … t … 6
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3. s = t 3 + 3t 2  3t,
4. s = st >4d  t + t , 4
5. s =
25 5  t, t2
6. s =
25 , t + 5
3
2
0 … t … 3 0 … t … 3
1 … t … 5 4 … t … 0
7. Particle motion At time t, the position of a body moving along the saxis is s = t 3  6t 2 + 9t m . a. Find the body’s acceleration each time the velocity is zero. b. Find the body’s speed each time the acceleration is zero. c. Find the total distance traveled by the body from t = 0 to t = 2. 8. Particle motion At time t Ú 0 , the velocity of a body moving along the saxis is y = t 2  4t + 3 . a. Find the body’s acceleration each time the velocity is zero. b. When is the body moving forward? Backward? c. When is the body’s velocity increasing? Decreasing?
FreeFall Applications
height above ground t sec into the fall would have been s = 179  16t 2 . a. What would have been the ball’s velocity, speed, and acceleration at time t? b. About how long would it have taken the ball to hit the ground? c. What would have been the ball’s velocity at the moment of impact? 14. Galileo’s freefall formula Galileo developed a formula for a body’s velocity during free fall by rolling balls from rest down increasingly steep inclined planks and looking for a limiting formula that would predict a ball’s behavior when the plank was vertical and the ball fell freely; see part (a) of the accompanying figure. He found that, for any given angle of the plank, the ball’s velocity t sec into motion was a constant multiple of t. That is, the velocity was given by a formula of the form y = kt . The value of the constant k depended on the inclination of the plank. In modern notation—part (b) of the figure—with distance in meters and time in seconds, what Galileo determined by experiment was that, for any given angle u , the ball’s velocity t sec into the roll was y = 9.8ssin udt m>sec .
9. Free fall on Mars and Jupiter The equations for free fall at the surfaces of Mars and Jupiter (s in meters, t in seconds) are s = 1.86t 2 on Mars and s = 11.44t 2 on Jupiter. How long does it take a rock falling from rest to reach a velocity of 27.8 m> sec (about 100 km> h) on each planet?
Freefall position
10. Lunar projectile motion A rock thrown vertically upward from the surface of the moon at a velocity of 24 m> sec (about 86 km> h) reaches a height of s = 24t  0.8t 2 meters in t sec. a. Find the rock’s velocity and acceleration at time t. (The acceleration in this case is the acceleration of gravity on the moon.)
?
θ (a)
(b)
b. How long does it take the rock to reach its highest point?
a. What is the equation for the ball’s velocity during free fall?
c. How high does the rock go?
b. Building on your work in part (a), what constant acceleration does a freely falling body experience near the surface of Earth?
d. How long does it take the rock to reach half its maximum height? e. How long is the rock aloft? 11. Finding g on a small airless planet Explorers on a small airless planet used a spring gun to launch a ball bearing vertically upward from the surface at a launch velocity of 15 m> sec. Because the acceleration of gravity at the planet’s surface was gs m>sec2 , the explorers expected the ball bearing to reach a height of s = 15t  s1>2dgs t 2 meters t sec later. The ball bearing reached its maximum height 20 sec after being launched. What was the value of gs ? 12. Speeding bullet A 45caliber bullet fired straight up from the surface of the moon would reach a height of s = 832t  2.6t 2 feet after t sec. On Earth, in the absence of air, its height would be s = 832t  16t 2 ft after t sec. How long will the bullet be aloft in each case? How high will the bullet go? 13. Free fall from the Tower of Pisa Had Galileo dropped a cannonball from the Tower of Pisa, 179 ft above the ground, the ball’s
Conclusions About Motion from Graphs 15. The accompanying figure shows the velocity y = ds>dt = ƒstd (m> sec) of a body moving along a coordinate line. y(m/sec) y f (t)
3 0
2
4
6
8 10
t (sec)
–3
a. When does the body reverse direction? b. When (approximately) is the body moving at a constant speed? c. Graph the body’s speed for 0 … t … 10 . d. Graph the acceleration, where defined.
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3.3 The Derivative as a Rate of Change 16. A particle P moves on the number line shown in part (a) of the accompanying figure. Part (b) shows the position of P as a function of time t. P
s (cm)
0
f. When was the rocket’s acceleration greatest? g. When was the acceleration constant? What was its value then (to the nearest integer)? 18. The accompanying figure shows the velocity y = ƒstd of a particle moving on a coordinate line. y
(a) s (cm)
y f(t) s f (t)
2 0
181
1
2
3
4
0 5
6
1 2 3 4 5 6 7 8 9
t (sec)
t (sec)
–2 (6, 4)
–4
a. When does the particle move forward? Move backward? Speed up? Slow down?
(b)
b. When is the particle’s acceleration positive? Negative? Zero? c. When does the particle move at its greatest speed?
a. When is P moving to the left? Moving to the right? Standing still? b. Graph the particle’s velocity and speed (where defined). 17. Launching a rocket When a model rocket is launched, the propellant burns for a few seconds, accelerating the rocket upward. After burnout, the rocket coasts upward for a while and then begins to fall. A small explosive charge pops out a parachute shortly after the rocket starts down. The parachute slows the rocket to keep it from breaking when it lands. The figure here shows velocity data from the flight of the model rocket. Use the data to answer the following.
d. When does the particle stand still for more than an instant? 19. Two falling balls The multiflash photograph in the accompanying figure shows two balls falling from rest. The vertical rulers are marked in centimeters. Use the equation s = 490t 2 (the freefall equation for s in centimeters and t in seconds) to answer the following questions.
a. How fast was the rocket climbing when the engine stopped? b. For how many seconds did the engine burn? 200
Velocity (ft/sec)
150 100 50 0 –50 –100 0
2
4 6 8 10 Time after launch (sec)
12
c. When did the rocket reach its highest point? What was its velocity then? d. When did the parachute pop out? How fast was the rocket falling then? e. How long did the rocket fall before the parachute opened?
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a. How long did it take the balls to fall the first 160 cm? What was their average velocity for the period?
y A
b. How fast were the balls falling when they reached the 160cm mark? What was their acceleration then? c. About how fast was the light flashing (flashes per second)? 20. A traveling truck The accompanying graph shows the position s of a truck traveling on a highway. The truck starts at t = 0 and returns 15 h later at t = 15 .
t
0
B
a. Use the technique described in Section 3.1, Example 3, to graph the truck’s velocity y = ds>dt for 0 … t … 15 . Then repeat the process, with the velocity curve, to graph the truck’s acceleration dy>dt. C
b. Suppose that s = 15t 2  t 3 . Graph ds>dt and d 2s>dt 2 and compare your graphs with those in part (a). FIGURE 3.22
The graphs for Exercise 22.
Position, s (km)
500
Economics
400
23. Marginal cost Suppose that the dollar cost of producing x washing machines is csxd = 2000 + 100x  0.1x 2 .
300
a. Find the average cost per machine of producing the first 100 washing machines.
200
b. Find the marginal cost when 100 washing machines are produced.
100 0
5 10 Elapsed time, t (hr)
15
21. The graphs in Figure 3.21 show the position s, velocity y = ds>dt , and acceleration a = d 2s>dt 2 of a body moving along a coordinate line as functions of time t. Which graph is which? Give reasons for your answers.
c. Show that the marginal cost when 100 washing machines are produced is approximately the cost of producing one more washing machine after the first 100 have been made, by calculating the latter cost directly. 24. Marginal revenue Suppose that the revenue from selling x washing machines is 1 rsxd = 20,000 a1  x b
y
dollars. A
a. Find the marginal revenue when 100 machines are produced.
B C
t
0
b. Use the function r¿sxd to estimate the increase in revenue that will result from increasing production from 100 machines a week to 101 machines a week. c. Find the limit of r¿sxd as x : q . How would you interpret this number?
Additional Applications
FIGURE 3.21
The graphs for Exercise 21.
22. The graphs in Figure 3.22 show the position s, the velocity y = ds>dt , and the acceleration a = d 2s>dt 2 of a body moving along the coordinate line as functions of time t. Which graph is which? Give reasons for your answers.
25. Bacterium population When a bactericide was added to a nutrient broth in which bacteria were growing, the bacterium population continued to grow for a while, but then stopped growing and began to decline. The size of the population at time t (hours) was b = 10 6 + 10 4t  10 3t 2 . Find the growth rates at a. t = 0 hours . b. t = 5 hours . c. t = 10 hours .
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3.3 The Derivative as a Rate of Change 26. Draining a tank The number of gallons of water in a tank t minutes after the tank has started to drain is Qstd = 200s30  td2 . How fast is the water running out at the end of 10 min? What is the average rate at which the water flows out during the first 10 min? T 27. Draining a tank It takes 12 hours to drain a storage tank by opening the valve at the bottom. The depth y of fluid in the tank t hours after the valve is opened is given by the formula 2
t y = 6 a1 b m. 12 a. Find the rate dy>dt (m> h) at which the tank is draining at time t.
b. When is the fluid level in the tank falling fastest? Slowest? What are the values of dy>dt at these times?
c. Graph y and dy>dt together and discuss the behavior of y in relation to the signs and values of dy>dt. 28. Inflating a balloon The volume V = s4>3dpr 3 of a spherical balloon changes with the radius.
T Exercises 31–34 give the position function s = ƒstd of a body moving along the saxis as a function of time t. Graph ƒ together with the velocity function ystd = ds>dt = ƒ¿std and the acceleration function astd = d 2s>dt 2 = ƒ–std . Comment on the body’s behavior in relation to the signs and values of y and a. Include in your commentary such topics as the following: a. When is the body momentarily at rest? b. When does it move to the left (down) or to the right (up)? c. When does it change direction? d. When does it speed up and slow down? e. When is it moving fastest (highest speed)? Slowest? f. When is it farthest from the axis origin? 31. s = 200t  16t 2, 0 … t … 12.5 (a heavy object fired straight up from Earth’s surface at 200 ft> sec)
32. s = t 2  3t + 2, 3
0 … t … 5
2
a. At what rate sft3>ftd does the volume change with respect to the radius when r = 2 ft ?
33. s = t  6t + 7t,
b. By approximately how much does the volume increase when the radius changes from 2 to 2.2 ft?
35. Thoroughbred racing A racehorse is running a 10furlong race. (A furlong is 220 yards, although we will use furlongs and seconds as our units in this exercise.) As the horse passes each furlong marker (F ), a steward records the time elapsed (t) since the beginning of the race, as shown in the table:
29. Airplane takeoff Suppose that the distance an aircraft travels along a runway before takeoff is given by D = s10>9dt 2 , where D is measured in meters from the starting point and t is measured in seconds from the time the brakes are released. The aircraft will become airborne when its speed reaches 200 km> h. How long will it take to become airborne, and what distance will it travel in that time? 30. Volcanic lava fountains Although the November 1959 Kilauea Iki eruption on the island of Hawaii began with a line of fountains along the wall of the crater, activity was later confined to a single vent in the crater’s floor, which at one point shot lava 1900 ft straight into the air (a world record). What was the lava’s exit velocity in feet per second? In miles per hour? (Hint: If y0 is the exit velocity of a particle of lava, its height t sec later will be s = y0 t  16t 2 ft . Begin by finding the time at which ds>dt = 0 . Neglect air resistance.)
0 … t … 4
34. s = 4  7t + 6t 2  t 3,
0 … t … 4
F
0
1
2
3
4
5
6
7
8
9
10
t
0
20
33
46
59
73
86
100
112
124
135
a. How long does it take the horse to finish the race? b. What is the average speed of the horse over the first 5 furlongs? c. What is the approximate speed of the horse as it passes the 3furlong marker? d. During which portion of the race is the horse running the fastest? e. During which portion of the race is the horse accelerating the fastest?
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3.4
183
Derivatives of Trigonometric Functions Many of the phenomena we want information about are approximately periodic (electromagnetic fields, heart rhythms, tides, weather). The derivatives of sines and cosines play a key role in describing periodic changes. This section shows how to differentiate the six basic trigonometric functions.
Derivative of the Sine Function To calculate the derivative of ƒsxd = sin x, for x measured in radians, we combine the limits in Example 5a and Theorem 7 in Section 2.4 with the angle sum identity for the sine: sin sx + hd = sin x cos h + cos x sin h. Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
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If ƒsxd = sin x, then ƒsx + hd  ƒsxd h h:0 sin sx + hd  sin x = lim h h:0 ssin x cos h + cos x sin hd  sin x = lim h h:0 sin x scos h  1d + cos x sin h = lim h h:0
ƒ¿sxd = lim
= lim asin x # h:0
Derivative definition
Sine angle sum identity
cos h  1 sin h b + lim acos x # b h h h:0
cos h  1 sin h + cos x # lim h h:0 h = sin x # 0 + cos x # 1 = cos x. = sin x # lim
h:0
Example 5(a) and Theorem 7, Section 2.4
The derivative of the sine function is the cosine function: d ssin xd = cos x. dx
EXAMPLE 1
Derivatives Involving the Sine
(a) y = x 2  sin x: dy d = 2x A sin x B dx dx
Difference Rule
= 2x  cos x. (b) y = x 2 sin x: dy d = x2 A sin x B + 2x sin x dx dx
Product Rule
= x 2 cos x + 2x sin x. (c) y =
sin x x : d x# A sin x B  sin x # 1 dy dx = dx x2 x cos x  sin x = . x2
Quotient Rule
Derivative of the Cosine Function With the help of the angle sum formula for the cosine, cos sx + hd = cos x cos h  sin x sin h, Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
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Derivatives of Trigonometric Functions
185
we have cossx + hd  cos x d scos xd = lim dx h h:0
y y cos x
1 –
0 –1 y'
scos x cos h  sin x sin hd  cos x h h:0
= lim
x
= lim
y' –sin x
h:0
1 –
0 –1
Cosine angle sum identity
cos xscos h  1d  sin x sin h h
= lim cos x #
x
Derivative definition
h:0
cos h  1 sin h  lim sin x # h h h:0
cos h  1 sin h  sin x # lim h h:0 h:0 h
= cos x # lim FIGURE 3.23 The curve y¿ = sin x as the graph of the slopes of the tangents to the curve y = cos x .
= cos x # 0  sin x # 1 = sin x.
Example 5(a) and Theorem 7, Section 2.4
The derivative of the cosine function is the negative of the sine function: d scos xd = sin x dx
Figure 3.23 shows a way to visualize this result.
EXAMPLE 2
Derivatives Involving the Cosine
(a) y = 5x + cos x: dy d d = s5xd + A cos x B dx dx dx
Sum Rule
= 5  sin x. (b) y = sin x cos x: dy d d = sin x A cos x B + cos x dx A sin x B dx dx
Product Rule
= sin xs sin xd + cos xscos xd = cos2 x  sin2 x. (c) y =
cos x : 1  sin x d d A 1  sin x B dx A cos x B  cos x dx A 1  sin x B dy = dx s1  sin xd2 s1  sin xds sin xd  cos xs0  cos xd s1  sin xd2 1  sin x = s1  sin xd2 1 . = 1  sin x
Quotient Rule
=
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sin2 x + cos2 x = 1
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Simple Harmonic Motion The motion of a body bobbing freely up and down on the end of a spring or bungee cord is an example of simple harmonic motion. The next example describes a case in which there are no opposing forces such as friction or buoyancy to slow the motion down.
–5
0
Rest position
5
Position at t0
EXAMPLE 3
A body hanging from a spring (Figure 3.24) is stretched 5 units beyond its rest position and released at time t = 0 to bob up and down. Its position at any later time t is s = 5 cos t. What are its velocity and acceleration at time t ?
s
FIGURE 3.24 A body hanging from a vertical spring and then displaced oscillates above and below its rest position. Its motion is described by trigonometric functions (Example 3).
We have Position:
Solution
s = 5 cos t ds d Velocity: y = = s5 cos td = 5 sin t dt dt dy d Acceleration: a = = s 5 sin td = 5 cos t. dt dt Notice how much we can learn from these equations:
s, y
1.
y –5 sin t
s 5 cos t
2.
0
2
3 2
Motion on a Spring
2 5 2
t
3.
FIGURE 3.25 The graphs of the position and velocity of the body in Example 3.
4.
As time passes, the weight moves down and up between s = 5 and s = 5 on the saxis. The amplitude of the motion is 5. The period of the motion is 2p. The velocity y = 5 sin t attains its greatest magnitude, 5, when cos t = 0, as the graphs show in Figure 3.25. Hence, the speed of the weight, ƒ y ƒ = 5 ƒ sin t ƒ , is greatest when cos t = 0, that is, when s = 0 (the rest position). The speed of the weight is zero when sin t = 0. This occurs when s = 5 cos t = ;5, at the endpoints of the interval of motion. The acceleration value is always the exact opposite of the position value. When the weight is above the rest position, gravity is pulling it back down; when the weight is below the rest position, the spring is pulling it back up. The acceleration, a = 5 cos t, is zero only at the rest position, where cos t = 0 and the force of gravity and the force from the spring offset each other. When the weight is anywhere else, the two forces are unequal and acceleration is nonzero. The acceleration is greatest in magnitude at the points farthest from the rest position, where cos t = ;1.
EXAMPLE 4
Jerk
The jerk of the simple harmonic motion in Example 3 is j =
da d = s 5 cos td = 5 sin t. dt dt
It has its greatest magnitude when sin t = ;1, not at the extremes of the displacement but at the rest position, where the acceleration changes direction and sign.
Derivatives of the Other Basic Trigonometric Functions Because sin x and cos x are differentiable functions of x, the related functions sin x tan x = cos x ,
cot x =
cos x , sin x
1 sec x = cos x ,
and
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csc x =
1 sin x
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Derivatives of Trigonometric Functions
187
are differentiable at every value of x at which they are defined. Their derivatives, calculated from the Quotient Rule, are given by the following formulas. Notice the negative signs in the derivative formulas for the cofunctions.
Derivatives of the Other Trigonometric Functions d stan xd = sec2 x dx d ssec xd = sec x tan x dx d scot xd = csc2 x dx d scsc xd = csc x cot x dx
To show a typical calculation, we derive the derivative of the tangent function. The other derivations are left to Exercise 50.
EXAMPLE 5 Find d(tan x)> dx. Solution
d sin x d A tan x B = dx a cos x b = dx
cos x
d d A sin x B  sin x dx A cos x B dx cos2 x
Quotient Rule
cos x cos x  sin x s sin xd cos2 x cos2 x + sin2 x = cos2 x 1 = sec2 x = cos2 x
=
EXAMPLE 6 Find y– if y = sec x. Solution
y = sec x y¿ = sec x tan x y– =
d ssec x tan xd dx
= sec x
d d A tan x B + tan x dx A sec x B dx
= sec xssec2 xd + tan xssec x tan xd = sec3 x + sec x tan2 x
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Product Rule
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The differentiability of the trigonometric functions throughout their domains gives another proof of their continuity at every point in their domains (Theorem 1, Section 3.1). So we can calculate limits of algebraic combinations and composites of trigonometric functions by direct substitution.
EXAMPLE 7 lim
x:0
Finding a Trigonometric Limit
22 + sec x 22 + sec 0 22 + 1 23 = = = =  23 1 cossp  tan xd cossp  tan 0d cossp  0d
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EXERCISES 3.4 Derivatives
26. Find y s4d = d 4 y>dx 4 if
In Exercises 1–12, find dy>dx.
a. y = 2 sin x .
1. y = 10x + 3 cos x
3 2. y = x + 5 sin x
3. y = csc x  4 1x + 7
4. y = x 2 cot x 
1 x2
5. y = ssec x + tan xdssec x  tan xd
7. y =
cot x 1 + cot x
8. y =
cos x 1 + sin x
cos x x 10. y = x + cos x
4 1 9. y = cos x + tan x
11. y = x 2 sin x + 2x cos x  2 sin x
28. y = tan x,
p>2 6 x 6 p>2
x = p>3, 0, p>3 29. y = sec x,
p>2 6 x 6 p>2 3p>2 … x … 2p
x = p>3, 3p>2 14. s = t 2  sec t + 1
1 + csc t 1  csc t
3p>2 … x … 2p
x = p, 0, 3p>2
30. y = 1 + cos x,
In Exercises 13–16, find ds>dt.
15. s =
In Exercises 27–30, graph the curves over the given intervals, together with their tangents at the given values of x. Label each curve and tangent with its equation.
x = p>3, p>4
12. y = x 2 cos x  2x sin x  2 cos x
13. s = tan t  t
Tangent Lines
27. y = sin x,
6. y = ssin x + cos xd sec x
b. y = 9 cos x .
16. s =
sin t 1  cos t
T Do the graphs of the functions in Exercises 31–34 have any horizontal tangents in the interval 0 … x … 2p ? If so, where? If not, why not? Visualize your findings by graphing the functions with a grapher. 31. y = x + sin x
In Exercises 17–20, find dr>du . 17. r = 4  u 2 sin u
18. r = u sin u + cos u
19. r = sec u csc u
20. r = s1 + sec ud sin u
32. y = 2x + sin x 33. y = x  cot x 34. y = x + 2 cos x
In Exercises 21–24, find dp>dq. 1 21. p = 5 + cot q 23. p =
sin q + cos q cos q
22. p = s1 + csc qd cos q 24. p =
tan q 1 + tan q
25. Find y– if a. y = csc x .
35. Find all points on the curve y = tan x, p>2 6 x 6 p>2 , where the tangent line is parallel to the line y = 2x . Sketch the curve and tangent(s) together, labeling each with its equation. 36. Find all points on the curve y = cot x, 0 6 x 6 p , where the tangent line is parallel to the line y = x . Sketch the curve and tangent(s) together, labeling each with its equation.
b. y = sec x .
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3.4 Derivatives of Trigonometric Functions In Exercises 37 and 38, find an equation for (a) the tangent to the curve at P and (b) the horizontal tangent to the curve at Q. 37.
48. Is there a value of b that will make g sxd = e
38. y
y
x + b, x 6 0 cos x, x Ú 0
continuous at x = 0 ? Differentiable at x = 0 ? Give reasons for your answers.
Q P , 2 2
2
189
49. Find d 999>dx 999 scos xd .
50. Derive the formula for the derivative with respect to x of 1
0
2 2 y 4 cot x 2csc x 1
a. sec x.
P , 4 4
4
b. csc x.
T 51. Graph y = cos x for p … x … 2p . On the same screen, graph y =
x Q
1 4
0
2
3
T 52. Graph y = sin x for p … x … 2p . On the same screen, graph y =
Trigonometric Limits 1 1 39. lim sin a x  b 2 x:2 x:  p>6
T 53. Centered difference quotients The centered difference quotient
21 + cos sp csc xd
41. lim sec ccos x + p tan a x:0
ƒsx + hd  ƒsx  hd 2h
p b  1d 4 sec x
is used to approximate ƒ¿sxd in numerical work because (1) its limit as h : 0 equals ƒ¿sxd when ƒ¿sxd exists, and (2) it usually gives a better approximation of ƒ¿sxd for a given value of h than Fermat’s difference quotient
p + tan x b 42. lim sin a tan x  2 sec x x:0 43. lim tan a1 t:0
44. lim cos a u :0
cos sx + hd  cos x h
for h = 1, 0.5, 0.3 , and 0.1. Then, in a new window, try h = 1, 0.5 , and 0.3 . What happens as h : 0 + ? As h : 0  ? What phenomenon is being illustrated here?
Find the limits in Exercises 39–44.
lim
sin sx + hd  sin x h
for h = 1, 0.5, 0.3 , and 0.1. Then, in a new window, try h = 1, 0.5 , and 0.3 . What happens as h : 0 + ? As h : 0  ? What phenomenon is being illustrated here?
x
y 1 兹2 csc x cot x
40.
c. cot x.
sin t t b
ƒsx + hd  ƒsxd . h
pu b sin u
See the accompanying figure. y
Simple Harmonic Motion
Slope f '(x)
The equations in Exercises 45 and 46 give the position s = ƒstd of a body moving on a coordinate line (s in meters, t in seconds). Find the body’s velocity, speed, acceleration, and jerk at time t = p>4 sec . 46. s = sin t + cos t
45. s = 2  2 sin t
B
f(x h) f(x) h
A Slope
Theory and Examples
f(x h) f(x h) 2h
y f (x)
47. Is there a value of c that will make sin2 3x , x2 ƒsxd = L c,
Slope
C
x Z 0 x = 0
continuous at x = 0 ? Give reasons for your answer.
h 0
xh
h x
xh
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a. To see how rapidly the centered difference quotient for ƒsxd = sin x converges to ƒ¿sxd = cos x , graph y = cos x together with y =
sin sx + hd  sin sx  hd 2h
over the interval [p, 2p] for h = 1, 0.5 , and 0.3. Compare the results with those obtained in Exercise 51 for the same values of h. b. To see how rapidly the centered difference quotient for ƒsxd = cos x converges to ƒ¿sxd = sin x , graph y = sin x together with y =
T 57. Exploring (sin kx) / x Graph y = ssin xd>x , y = ssin 2xd>x , and y = ssin 4xd>x together over the interval 2 … x … 2 . Where does each graph appear to cross the yaxis? Do the graphs really intersect the axis? What would you expect the graphs of y = ssin 5xd>x and y = ssin s 3xdd>x to do as x : 0 ? Why? What about the graph of y = ssin kxd>x for other values of k? Give reasons for your answers. T 58. Radians versus degrees: degree mode derivatives What happens to the derivatives of sin x and cos x if x is measured in degrees instead of radians? To find out, take the following steps.
cos sx + hd  cos sx  hd 2h
over the interval [p, 2p] for h = 1, 0.5 , and 0.3. Compare the results with those obtained in Exercise 52 for the same values of h. 54. A caution about centered difference quotients of Exercise 53.) The quotient
T 56. Slopes on the graph of the cotangent function Graph y = cot x and its derivative together for 0 6 x 6 p . Does the graph of the cotangent function appear to have a smallest slope? A largest slope? Is the slope ever positive? Give reasons for your answers.
(Continuation
ƒsx + hd  ƒsx  hd 2h may have a limit as h : 0 when ƒ has no derivative at x. As a case in point, take ƒsxd = ƒ x ƒ and calculate ƒ0 + hƒ  ƒ0  hƒ lim . 2h h :0 As you will see, the limit exists even though ƒsxd = ƒ x ƒ has no derivative at x = 0 . Moral: Before using a centered difference quotient, be sure the derivative exists. T 55. Slopes on the graph of the tangent function Graph y = tan x and its derivative together on s p>2, p>2d . Does the graph of the tangent function appear to have a smallest slope? a largest slope? Is the slope ever negative? Give reasons for your answers.
a. With your graphing calculator or computer grapher in degree mode, graph sin h ƒshd = h and estimate limh:0 ƒshd . Compare your estimate with p>180 . Is there any reason to believe the limit should be p>180 ? b. With your grapher still in degree mode, estimate cos h  1 lim . h h:0 c. Now go back to the derivation of the formula for the derivative of sin x in the text and carry out the steps of the derivation using degreemode limits. What formula do you obtain for the derivative? d. Work through the derivation of the formula for the derivative of cos x using degreemode limits. What formula do you obtain for the derivative? e. The disadvantages of the degreemode formulas become apparent as you start taking derivatives of higher order. Try it. What are the second and third degreemode derivatives of sin x and cos x?
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The Chain Rule and Parametric Equations We know how to differentiate y = ƒsud = sin u and u = gsxd = x 2  4, but how do we differentiate a composite like Fsxd = ƒsgsxdd = sin sx 2  4d? The differentiation formulas we have studied so far do not tell us how to calculate F¿sxd. So how do we find the derivative of F = ƒ g? The answer is, with the Chain Rule, which says that the derivative of the composite of two differentiable functions is the product of their derivatives evaluated at appropriate points. The Chain Rule is one of the most important and widely used rules of differentiation. This section describes the rule and how to use it. We then apply the rule to describe curves in the plane and their tangent lines in another way.
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Derivative of a Composite Function We begin with examples.
EXAMPLE 1
Relating Derivatives
3 1 1 x = s3xd is the composite of the functions y = u and u = 3x. 2 2 2 How are the derivatives of these functions related? The function y =
Solution 2
dy 3 = , 2 dx
3
Since
1
C: y turns
We have
B: u turns
and
du = 3. dx
3 1 = # 3, we see that 2 2 dy dy du # . = dx du dx
A: x turns
FIGURE 3.26 When gear A makes x turns, gear B makes u turns and gear C makes y turns. By comparing circumferences or counting teeth, we see that y = u>2 (C turns onehalf turn for each B turn) and u = 3x (B turns three times for A’s one), so y = 3x>2 . Thus, dy>dx = 3>2 = s1>2ds3d = sdy>dudsdu>dxd .
dy 1 = , 2 du
Is it an accident that dy dy du # ? = dx du dx If we think of the derivative as a rate of change, our intuition allows us to see that this relationship is reasonable. If y = ƒsud changes half as fast as u and u = gsxd changes three times as fast as x, then we expect y to change 3>2 times as fast as x. This effect is much like that of a multiple gear train (Figure 3.26).
EXAMPLE 2 The function y = 9x 4 + 6x 2 + 1 = s3x 2 + 1d2 is the composite of y = u 2 and u = 3x 2 + 1. Calculating derivatives, we see that dy du # = 2u # 6x du dx = 2s3x 2 + 1d # 6x = 36x 3 + 12x. Calculating the derivative from the expanded formula, we get dy d = A 9x 4 + 6x 2 + 1 B dx dx = 36x 3 + 12x. Once again, dy du # = dy . du dx dx The derivative of the composite function ƒ(g(x)) at x is the derivative of ƒ at g(x) times the derivative of g at x. This is known as the Chain Rule (Figure 3.27).
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Composite f ˚ g Rate of change at x is f '(g(x)) • g'(x).
x
g
f
Rate of change at x is g'(x).
Rate of change at g(x) is f '( g(x)).
u g(x)
y f (u) f(g(x))
FIGURE 3.27 Rates of change multiply: The derivative of ƒ g at x is the derivative of ƒ at g(x) times the derivative of g at x.
THEOREM 3 The Chain Rule If ƒ(u) is differentiable at the point u = gsxd and g(x) is differentiable at x, then the composite function sƒ gdsxd = ƒsgsxdd is differentiable at x, and sƒ gd¿sxd = ƒ¿sgsxdd # g¿sxd. In Leibniz’s notation, if y = ƒsud and u = gsxd, then dy dy du # , = dx du dx where dy>du is evaluated at u = gsxd.
Intuitive “Proof” of the Chain Rule: Let ¢u be the change in u corresponding to a change of ¢x in x, that is ¢u = gsx + ¢xd  gsxd Then the corresponding change in y is ¢y = ƒsu + ¢ud  ƒsud. It would be tempting to write ¢y ¢y ¢u # = ¢x ¢u ¢x
(1)
and take the limit as ¢x : 0: dy ¢y = lim dx ¢x:0 ¢x ¢y ¢u # = lim ¢x:0 ¢u ¢x ¢y # lim ¢u = lim ¢x:0 ¢u ¢x:0 ¢x ¢y # lim ¢u = lim ¢u:0 ¢u ¢x:0 ¢x dy du = . du dx
(Note that ¢u : 0 as ¢x : 0 since g is continuous.)
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The only flaw in this reasoning is that in Equation (1) it might happen that ¢u = 0 (even when ¢x Z 0) and, of course, we can’t divide by 0. The proof requires a different approach to overcome this flaw, and we give a precise proof in Section 3.8.
EXAMPLE 3
Applying the Chain Rule
An object moves along the xaxis so that its position at any time t Ú 0 is given by xstd = cosst 2 + 1d. Find the velocity of the object as a function of t. We know that the velocity is dx>dt. In this instance, x is a composite function: x = cossud and u = t 2 + 1. We have
Solution
dx = sin sud du
x = cossud
du = 2t. dt
u = t2 + 1
By the Chain Rule, dx dx # du = dt du dt = sinsud # 2t = sinst 2 + 1d # 2t = 2t sin st 2 + 1d.
dx evaluated at u du
As we see from Example 3, a difficulty with the Leibniz notation is that it doesn’t state specifically where the derivatives are supposed to be evaluated.
“OutsideInside” Rule It sometimes helps to think about the Chain Rule this way: If y = ƒsgsxdd, then dy = ƒ¿sgsxdd # g¿sxd. dx In words, differentiate the “outside” function ƒ and evaluate it at the “inside” function g(x) left alone; then multiply by the derivative of the “inside function.”
EXAMPLE 4
Differentiating from the Outside In
Differentiate sin sx 2 + xd with respect to x. Solution
d sin (x 2 + x) = cos (x 2 + x) # (2x + 1) dx (+)+* (+)+* (+)+* inside
inside derivative of left alone the inside
Repeated Use of the Chain Rule We sometimes have to use the Chain Rule two or more times to find a derivative. Here is an example.
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HISTORICAL BIOGRAPHY
EXAMPLE 5
Johann Bernoulli (1667–1748)
Find the derivative of gstd = tan s5  sin 2td.
A ThreeLink “Chain”
Notice here that the tangent is a function of 5  sin 2t, whereas the sine is a function of 2t, which is itself a function of t. Therefore, by the Chain Rule,
Solution
g¿std =
d A tan A 5  sin 2t B B dt
d A 5  sin 2t B dt d = sec2 s5  sin 2td # a0  cos 2t # A 2t B b dt = sec2 s5  sin 2td #
Derivative of tan u with u = 5  sin 2t Derivative of 5  sin u with u = 2t
= sec2 s5  sin 2td # s cos 2td # 2 = 2scos 2td sec2 s5  sin 2td.
The Chain Rule with Powers of a Function If ƒ is a differentiable function of u and if u is a differentiable function of x, then substituting y = ƒsud into the Chain Rule formula dy dy du # = dx du dx leads to the formula du d ƒsud = ƒ¿sud . dx dx Here’s an example of how it works: If n is a positive or negative integer and ƒsud = u n , the Power Rules (Rules 2 and 7) tell us that ƒ¿sud = nu n  1 . If u is a differentiable function of x, then we can use the Chain Rule to extend this to the Power Chain Rule: d n du u = nu n  1 . dx dx
EXAMPLE 6 (a)
d A u n B = nu n  1 du
Applying the Power Chain Rule
d d s5x 3  x 4 d7 = 7s5x 3  x 4 d6 A 5x 3  x 4 B dx dx = 7s5x 3  x 4 d6s5 # 3x 2  4x 3 d = 7s5x 3  x 4 d6s15x 2  4x 3 d
(b)
Power Chain Rule with u = 5x 3  x 4, n = 7
d d 1 b = s3x  2d1 a dx 3x  2 dx = 1s3x  2d2
d s3x  2d dx
Power Chain Rule with u = 3x  2, n = 1
= 1s3x  2d2s3d 3 = s3x  2d2 In part (b) we could also have found the derivative with the Quotient Rule.
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sinn x means ssin xdn, n Z 1 .
EXAMPLE 7
The Chain Rule and Parametric Equations
195
Finding Tangent Slopes
(a) Find the slope of the line tangent to the curve y = sin5 x at the point where x = p>3. (b) Show that the slope of every line tangent to the curve y = 1>s1  2xd3 is positive. Solution
(a)
dy d = 5 sin4 x # sin x dx dx
Power Chain Rule with u = sin x, n = 5
= 5 sin4 x cos x The tangent line has slope dy 23 4 1 45 ` = 5a b a b = . 2 2 32 dx x = p>3 (b)
dy d = s1  2xd3 dx dx = 3s1  2xd4 #
d s1  2xd dx
Power Chain Rule with u = s1  2xd, n = 3
= 3s1  2xd4 # s 2d =
6 s1  2xd4
At any point (x, y) on the curve, x Z 1>2 and the slope of the tangent line is dy 6 = , dx s1  2xd4 the quotient of two positive numbers.
EXAMPLE 8
Radians Versus Degrees
It is important to remember that the formulas for the derivatives of both sin x and cos x were obtained under the assumption that x is measured in radians, not degrees. The Chain Rule gives us new insight into the difference between the two. Since 180° = p radians, x° = px>180 radians where x° means the angle x measured in degrees. By the Chain Rule, d d px px p p b = b = sin sx°d = sin a cos a cossx°d. 180 180 180 180 dx dx See Figure 3.28. Similarly, the derivative of cossx°d is sp>180d sin sx°d. The factor p>180, annoying in the first derivative, would compound with repeated differentiation. We see at a glance the compelling reason for the use of radian measure.
Parametric Equations Instead of describing a curve by expressing the ycoordinate of a point P(x, y) on the curve as a function of x, it is sometimes more convenient to describe the curve by expressing both coordinates as functions of a third variable t. Figure 3.29 shows the path of a moving particle described by a pair of equations, x = ƒstd and y = gstd. For studying motion,
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y sin(x°) sin x 180
y 1
x
y sin x
180
FIGURE 3.28 Sin sx°d oscillates only p/180 times as often as sin x oscillates. Its maximum slope is p/180 at x = 0 (Example 8).
t usually denotes time. Equations like these are better than a Cartesian formula because they tell us the particle’s position sx, yd = sƒstd, gstdd at any time t. Position of particle at time t
( f (t), g(t))
DEFINITION Parametric Curve If x and y are given as functions x = ƒstd, y = gstd over an interval of tvalues, then the set of points sx, yd = sƒstd, gstdd defined by these equations is a parametric curve. The equations are parametric equations for the curve.
FIGURE 3.29 The path traced by a particle moving in the xyplane is not always the graph of a function of x or a function of y.
The variable t is a parameter for the curve, and its domain I is the parameter interval. If I is a closed interval, a … t … b, the point (ƒ(a), g(a)) is the initial point of the curve. The point (ƒ(b), g(b)) is the terminal point. When we give parametric equations and a parameter interval for a curve, we say that we have parametrized the curve. The equations and interval together constitute a parametrization of the curve.
y t 2
x2 y2 1 P(cos t, sin t)
EXAMPLE 9 t
t 0
Moving Counterclockwise on a Circle
Graph the parametric curves t0 (1, 0)
x
(a) x = cos t, (b) x = a cos t,
y = sin t, y = a sin t,
0 … t … 2p. 0 … t … 2p.
Solution t 3 2
FIGURE 3.30 The equations x = cos t and y = sin t describe motion on the circle x 2 + y 2 = 1 . The arrow shows the direction of increasing t (Example 9).
(a) Since x 2 + y 2 = cos2 t + sin2 t = 1, the parametric curve lies along the unit circle x 2 + y 2 = 1. As t increases from 0 to 2p, the point sx, yd = scos t, sin td starts at (1, 0) and traces the entire circle once counterclockwise (Figure 3.30). (b) For x = a cos t, y = a sin t, 0 … t … 2p, we have x 2 + y 2 = a 2 cos2 t + a 2 sin2 t = a 2 . The parametrization describes a motion that begins at the point (a, 0) and traverses the circle x 2 + y 2 = a 2 once counterclockwise, returning to (a, 0) at t = 2p.
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y
EXAMPLE 10
y x 2, x ⱖ 0
The Chain Rule and Parametric Equations
197
Moving Along a Parabola
The position P(x, y) of a particle moving in the xyplane is given by the equations and parameter interval P(兹t, t)
x = 1t,
t1
t Ú 0.
Identify the path traced by the particle and describe the motion.
(1, 1) 0 Starts at t0
y = t,
x
FIGURE 3.31 The equations x = 1t and y = t and the interval t Ú 0 describe the motion of a particle that traces the righthand half of the parabola y = x 2 (Example 10).
We try to identify the path by eliminating t between the equations x = 1t and y = t. With any luck, this will produce a recognizable algebraic relation between x and y. We find that Solution
y = t =
A 1t B 2 = x 2 .
Thus, the particle’s position coordinates satisfy the equation y = x 2 , so the particle moves along the parabola y = x 2 . It would be a mistake, however, to conclude that the particle’s path is the entire parabola y = x 2 ; it is only half the parabola. The particle’s xcoordinate is never negative. The particle starts at (0, 0) when t = 0 and rises into the first quadrant as t increases (Figure 3.31). The parameter interval is [0, q d and there is no terminal point.
EXAMPLE 11
Parametrizing a Line Segment
Find a parametrization for the line segment with endpoints s 2, 1d and (3, 5). Solution
Using s 2, 1d we create the parametric equations x = 2 + at,
y = 1 + bt.
These represent a line, as we can see by solving each equation for t and equating to obtain y  1 x + 2 a = b . This line goes through the point s 2, 1d when t = 0. We determine a and b so that the line goes through (3, 5) when t = 1. 3 = 2 + a 5 = 1 + b
Q Q
a = 5 b = 4
x = 3 when t = 1 . y = 5 when t = 1 .
Therefore, x = 2 + 5t,
y = 1 + 4t,
0 … t … 1
is a parametrization of the line segment with initial point s 2, 1d and terminal point (3, 5).
Slopes of Parametrized Curves A parametrized curve x = ƒstd and y = gstd is differentiable at t if ƒ and g are differentiable at t. At a point on a differentiable parametrized curve where y is also a differentiable function of x, the derivatives dy>dt, dx>dt, and dy>dx are related by the Chain Rule: dy dy dx # . = dt dx dt If dx>dt Z 0, we may divide both sides of this equation by dx>dt to solve for dy>dx.
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Parametric Formula for dy/dx If all three derivatives exist and dx>dt Z 0, dy>dt dy = . dx dx>dt
EXAMPLE 12
(2)
Differentiating with a Parameter
If x = 2t + 3 and y = t 2  1, find the value of dy>dx at t = 6. Solution
Equation (2) gives dy>dx as a function of t: dy>dt dy x  3 2t = = . = t = 2 2 dx dx>dt
When t = 6, dy>dx = 6. Notice that we are also able to find the derivative dy>dx as a function of x.
EXAMPLE 13
Moving Along the Ellipse x2>a2 + y2>b2 = 1
Describe the motion of a particle whose position P(x, y) at time t is given by x = a cos t,
y = b sin t,
0 … t … 2p.
Find the line tangent to the curve at the point A a> 22, b> 22 B , where t = p>4. (The constants a and b are both positive.) Solution We find a Cartesian equation for the particle’s coordinates by eliminating t between the equations
x cos t = a ,
sin t =
y . b
The identity cos2 t + sin2 t = 1, yields 2
y2 x2 + = 1. a2 b2
2
y x a a b + a b = 1, b
or
The particle’s coordinates (x, y) satisfy the equation sx 2>a 2 d + sy 2>b 2 d = 1, so the particle moves along this ellipse. When t = 0, the particle’s coordinates are x = a coss0d = a,
y = b sin s0d = 0,
so the motion starts at (a, 0). As t increases, the particle rises and moves toward the left, moving counterclockwise. It traverses the ellipse once, returning to its starting position (a, 0) at t = 2p. The slope of the tangent line to the ellipse when t = p>4 is dy>dt dy ` = ` dx t = p>4 dx>dt t = p>4 =
b cos t ` a sin t t = p>4 b> 22
=
a> 22
b = a.
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The tangent line is y 
b 22
b =  a ax 
y =
b 22
a 22
b
b  a ax 
a 22
b
or b y =  a x + 22b. If parametric equations define y as a twicedifferentiable function of x, we can apply Equation (2) to the function dy>dx = y¿ to calculate d 2y>dx 2 as a function of t: d 2y dx
2
=
dy¿>dt d s y¿d = . dx dx>dt
Eq. (2) with y¿ in place of y
Parametric Formula for d 2y/dx2 If the equations x = ƒstd, y = gstd define y as a twicedifferentiable function of x, then at any point where dx>dt Z 0, d 2y dx
2
=
dy¿>dt . dx>dt
(3)
Finding d 2y>dx2 for a Parametrized Curve
EXAMPLE 14
Find d 2y>dx 2 as a function of t if x = t  t 2, y = t  t 3 . Finding d 2y>dx 2 in Terms of t 1. Express y¿ = dy>dx in terms of t. 2. Find dy¿>dt . 3. Divide dy¿>dt by dx>dt.
Solution
1.
Express y¿ = dy>dx in terms of t. y¿ =
dy>dt dy 1  3t 2 = = 1  2t dx dx>dt
2.
Differentiate y¿ with respect to t.
3.
dy¿ 2  6t + 6t 2 d 1  3t 2 = a b = dt dt 1  2t s1  2td2 Divide dy¿>dt by dx>dt. d 2y dx 2
EXAMPLE 15
=
Quotient Rule
s2  6t + 6t 2 d>s1  2td2 dy¿>dt 2  6t + 6t 2 = = 1  2t dx>dt s1  2td3
Eq. (3)
Dropping Emergency Supplies
A Red Cross aircraft is dropping emergency food and medical supplies into a disaster area. If the aircraft releases the supplies immediately above the edge of an open field 700 ft long and if the cargo moves along the path x = 120t
and
y = 16t 2 + 500,
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t Ú 0
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Position of aircraft at release
500 Path of dropped cargo
does the cargo land in the field? The coordinates x and y are measured in feet, and the parameter t (time since release) in seconds. Find a Cartesian equation for the path of the falling cargo (Figure 3.32) and the cargo’s rate of descent relative to its forward motion when it hits the ground. Solution
0
Open field
? 700
x
The cargo hits the ground when y = 0, which occurs at time t when 16t 2 + 500 = 0 525 500 = sec. 2 A 16
t =
FIGURE 3.32 The path of the dropped cargo of supplies in Example 15.
Set y = 0 . t Ú 0
The xcoordinate at the time of the release is x = 0. At the time the cargo hits the ground, the xcoordinate is x = 120t = 120 a
525 b = 30025 ft. 2
Since 30025 L 670.8 6 700, the cargo does land in the field. We find a Cartesian equation for the cargo’s coordinates by eliminating t between the parametric equations: y = 16t 2 + 500 = 16 a = 
2
x b + 500 120
1 2 x + 500. 900
Parametric equation for y Substitute for t from the equation x = 120t . A parabola
The rate of descent relative to its forward motion when the cargo hits the ground is dy>dt dy ` = ` dx t = 525>2 dx>dt t = 525>2 =
32t ` 120 t = 525>2
= 
225 L 1.49. 3
Thus, it is falling about 1.5 feet for every foot of forward motion when it hits the ground.
USING TECHNOLOGY
Simulation of Motion on a Vertical Line
The parametric equations xstd = c, x (t ) = 2 y (t ) = 160t –16t 2 and
x (t ) = t y (t ) = 160t –16t 2 in dot mode
ystd = ƒstd
will illuminate pixels along the vertical line x = c. If ƒ(t) denotes the height of a moving body at time t, graphing sxstd, ystdd = sc, ƒstdd will simulate the actual motion. Try it for the rock in Example 5, Section 3.3 with xstd = 2, say, and ystd = 160t  16t 2 , in dot mode with t Step = 0.1. Why does the spacing of the dots vary? Why does the grapher seem to stop after it reaches the top? (Try the plots for 0 … t … 5 and 5 … t … 10 separately.) For a second experiment, plot the parametric equations xstd = t,
ystd = 160t  16t 2
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together with the vertical line simulation of the motion, again in dot mode. Use what you know about the behavior of the rock from the calculations of Example 5 to select a window size that will display all the interesting behavior.
Standard Parametrizations and Derivative Rules CIRCLE
x2 + y2 = a2 :
ELLIPSE
x = a cos t y = a sin t 0 … t … 2p FUNCTION
y = ƒsxd :
x = t y = ƒstd
y2 x2 + 2 = 1: 2 a b
x = a cos t y = b sin t 0 … t … 2p DERIVATIVES
y¿ =
dy>dt dy , = dx dx>dt
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d 2y dx
2
=
dy¿>dt dx>dt
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EXERCISES 3.5 Derivative Calculations In Exercises 1–8, given y = ƒsud and u = gsxd , find dy>dx = ƒ¿sgsxddg¿sxd . u = s1>2dx
1. y = 6u  9,
4
3
2. y = 2u ,
u = 8x  1
3. y = sin u,
u = 3x + 1
4. y = cos u,
u = x>3
5. y = cos u,
u = sin x
6. y = sin u,
u = x  cos x
7. y = tan u,
u = 10x  5
8. y = sec u,
u = x 2 + 7x
In Exercises 9–18, write the function in the form y = ƒsud and u = gsxd . Then find dy>dx as a function of x. 9. y = s2x + 1d5 11. y = a1 
x b 7
10. y = s4  3xd9 12. y = a
7
x2 1 13. y = a + x  x b 8
4
x  1b 2
10
x 1 b 14. y = a + 5 5x
5
22. s = sin a
3pt 3pt b + cos a b 2 2
23. r = scsc u + cot ud1
24. r = ssec u + tan ud1
25. y = x 2 sin4 x + x cos2 x
x 1 26. y = x sin5 x  cos3 x 3
27. y =
1 1 s3x  2d7 + a4 b 21 2x 2 1 2 a + 1b 8 x
28. y = s5  2xd3 +
1
4
29. y = s4x + 3d4sx + 1d3
30. y = s2x  5d1sx 2  5xd6
31. hsxd = x tan A 21x B + 7
1 32. k sxd = x 2 sec a x b
33. ƒsud = a
sin u b 1 + cos u
2
34. g std = a
1 + cos t b sin t
15. y = sec stan xd
1 16. y = cot ap  x b
35. r = sin su2 d cos s2ud
1 36. r = sec2u tan a b u
17. y = sin3 x
18. y = 5 cos4 x
37. q = sin a
sin t 38. q = cot a t b
Find the derivatives of the functions in Exercises 19–38. 20. q = 22r  r 2
19. p = 23  t 21. s =
4 4 sin 3t + cos 5t 5p 3p
t 2t + 1
b
1
In Exercises 39–48, find dy>dt. 39. y = sin2 spt  2d
40. y = sec2 pt
41. y = s1 + cos 2td
42. y = s1 + cot st>2dd2
4
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202
Chapter 3: Differentiation t 44. y = cos a5 sin a b b 3
43. y = sin scos s2t  5dd 45. y = a1 + tan4 a
t bb 12
47. y = 21 + cos st 2 d
Find the derivatives with respect to x of the following combinations at the given value of x,
1 A 1 + cos2 A 7t B B 3 6
3
46. y =
48. y = 4 sin A 21 + 1t B
50. y = A 1  1x B
3
Choices in Composition What happens if you can write a function as a composite in different ways? Do you get the same derivative each time? The Chain Rule says you should. Try it with the functions in Exercises 63 and 64.
In Exercises 53–58, find the value of sƒ gd¿ at the given value of x. 53. ƒsud = u 5 + 1,
u = g sxd = 1x,
1 54. ƒsud = 1  u ,
u = g sxd =
2u , 57. ƒsud = 2 u + 1 58. ƒsud = a
x = 1
1 , 1  x
u = g sxd = 5 1x,
x = 1
u = g sxd = px,
u = g sxd = 10x + x + 1,
g (x)
ƒ(x)
g(x)
2 3
8 3
2 4
1>3 2p
3 5
u = 1>sx  1d .
and and
u = 1x u = x3 .
b. ƒsxd + g sxd,
x = 2 x = 3
d. ƒsxd>g sxd,
x = 3
x = 2
x = 2
f. 2ƒsxd,
x = 3
h. 2f sxd + g 2sxd, 2
x = 2 x = 2
60. Suppose that the functions ƒ and g and their derivatives with respect to x have the following values at x = 0 and x = 1 . x
ƒ(x)
g (x)
ƒ(x)
g(x)
0 1
1 3
1 4
5 1>3
1>3 8>3
b. Slopes on a tangent curve What is the smallest value the slope of the curve can ever have on the interval 2 6 x 6 2 ? Give reasons for your answer. 66. Slopes on sine curves a. Find equations for the tangents to the curves y = sin 2x and y = sin sx>2d at the origin. Is there anything special about how the tangents are related? Give reasons for your answer. b. Can anything be said about the tangents to the curves y = sin mx and y = sin sx>md at the origin sm a constant Z 0d ? Give reasons for your answer.
Find the derivatives with respect to x of the following combinations at the given value of x.
g. 1>g sxd,
and
65. a. Find the tangent to the curve y = 2 tan spx>4d at x = 1 .
x = 0
u = g sxd =
ƒ(x)
2
b. y = 1 + s1>ud
Tangents and Slopes
x
e. ƒsg sxdd,
u = 5x  35
x = 1>4
2
c. ƒsxd # g sxd,
and
a. y = u 3
1  1, x = 1 x2 59. Suppose that functions ƒ and g and their derivatives with respect to x have the following values at x = 2 and x = 3 .
a. 2ƒsxd,
a. y = su>5d + 7
64. Find dy>dx if y = x 3>2 by using the Chain Rule with y as a composite of
x = 1
2
u  1 b , u + 1
63. Find dy>dx if y = x by using the Chain Rule with y as a composite of
b. y = 1u
1 , cos2 u
x = 1
x = 0
62. Find dy>dt when x = 1 if y = x 2 + 7x  5 and dx>dt = 1>3 .
Finding Numerical Values of Derivatives
56. ƒsud = u +
f. sx 11 + ƒsxdd2,
x = 0
x = 0
1
x 52. y = 9 tan a b 3
1 51. y = cot s3x  1d 9
pu , 10
e. g sƒsxdd,
d. ƒsg sxdd,
x = 0
61. Find ds>dt when u = 3p>2 if s = cos u and d u>dt = 5 .
Find y– in Exercises 49–52.
55. ƒsud = cot
b. ƒsxdg 3sxd,
g. ƒsx + g sxdd,
Second Derivatives 1 49. y = a1 + x b
a. 5ƒsxd  g sxd, x = 1 ƒsxd c. , x = 1 g sxd + 1
c. For a given m, what are the largest values the slopes of the curves y = sin mx and y = sin sx>md can ever have? Give reasons for your answer. d. The function y = sin x completes one period on the interval [0, 2p] , the function y = sin 2x completes two periods, the function y = sin sx>2d completes half a period, and so on. Is there any relation between the number of periods y = sin mx completes on [0, 2p] and the slope of the curve y = sin mx at the origin? Give reasons for your answer.
Finding Cartesian Equations from Parametric Equations Exercises 67–78 give parametric equations and parameter intervals for the motion of a particle in the xyplane. Identify the particle’s path by
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3.5 The Chain Rule and Parametric Equations finding a Cartesian equation for it. Graph the Cartesian equation. (The graphs will vary with the equation used.) Indicate the portion of the graph traced by the particle and the direction of motion. 68. x = cos sp  td,
0 … t … p
y = sin sp  td,
0 … t … p
69. x = 4 cos t,
y = 2 sin t,
70. x = 4 sin t,
y = 5 cos t, 0 … t … 2p q 6 t 6 q
71. x = 3t,
0 … t … 2p
y = 9t 2,
72. x =  1t,
y = t,
t Ú 0
73. x = 2t  5,
y = 4t  7,
74. x = 3  3t,
y = 2t,
75. x = t,
76. x = 2t + 1, 78. x = sec t,
1 … t … 0
y = 1t,
77. x = sec2 t  1,
t Ú 0
y = tan t,
y = tan t,
p>2 6 t 6 p>2
p>2 6 t 6 p>2
Determining Parametric Equations 79. Find parametric equations and a parameter interval for the motion of a particle that starts at (a, 0) and traces the circle x 2 + y 2 = a 2 a. once clockwise.
b. once counterclockwise.
c. twice clockwise.
d. twice counterclockwise.
(There are many ways to do these, so your answers may not be the same as the ones in the back of the book.) 80. Find parametric equations and a parameter interval for the motion of a particle that starts at (a, 0) and traces the ellipse sx 2>a 2 d + sy 2>b 2 d = 1 a. once clockwise.
b. once counterclockwise.
c. twice clockwise.
d. twice counterclockwise.
(As in Exercise 79, there are many correct answers.) In Exercises 81–86, find a parametrization for the curve. 81. the line segment with endpoints s 1, 3d and (4, 1) 82. the line segment with endpoints s 1, 3d and s3, 2d 83. the lower half of the parabola x  1 = y 2 84. the left half of the parabola y = x 2 + 2x 85. the ray (half line) with initial point (2, 3) that passes through the point s 1, 1d 86. the ray (half line) with initial point s 1, 2d that passes through the point (0, 0)
Tangents to Parametrized Curves In Exercises 87–94, find an equation for the line tangent to the curve at the point defined by the given value of t. Also, find the value of d 2y>dx 2 at this point. 87. x = 2 cos t, 88. x = cos t, 89. x = t,
y = 2 sin t, y = 23 cos t,
y = 1t,
4
t = 1>4
t = p>4 t = 2p>3
t = 3
91. x = 2t + 3,
y = t ,
92. x = t  sin t,
y = 1  cos t,
93. x = cos t,
t = 1
y = 1 + sin t,
94. x = sec2 t  1,
y = tan t,
t = p>3
t = p>2 t = p>4
Theory, Examples, and Applications 95. Running machinery too fast Suppose that a piston is moving straight up and down and that its position at time t sec is
q 6 t 6 q
0 … t … 1
y = 21  t 2,
y = 23t,
s = A cos s2pbtd , with A and b positive. The value of A is the amplitude of the motion, and b is the frequency (number of times the piston moves up and down each second). What effect does doubling the frequency have on the piston’s velocity, acceleration, and jerk? (Once you find out, you will know why machinery breaks when you run it too fast.) 96. Temperatures in Fairbanks, Alaska The graph in Figure 3.33 shows the average Fahrenheit temperature in Fairbanks, Alaska, during a typical 365day year. The equation that approximates the temperature on day x is y = 37 sin c
2p sx  101d d + 25 . 365
a. On what day is the temperature increasing the fastest? b. About how many degrees per day is the temperature increasing when it is increasing at its fastest?
y 60 40 20 0 ...... ..
. .. .. .. ... .... ....
....... .... . ........ .. . .... .... ... . .... . ... ... ... ... ... ... ..
. ... .... ............ ..... .
. .... ....
x
–20
Ja n Fe b M ar A pr M ay Ju n Ju l A ug Se p O ct N ov D ec Ja n Fe b M ar
y = sin 2t,
2
Temperature (˚F)
67. x = cos 2t,
90. x =  2t + 1,
FIGURE 3.33 Normal mean air temperatures at Fairbanks, Alaska, plotted as data points, and the approximating sine function (Exercise 96).
97. Particle motion The position of a particle moving along a coordinate line is s = 21 + 4t , with s in meters and t in seconds. Find the particle’s velocity and acceleration at t = 6 sec. 98. Constant acceleration Suppose that the velocity of a falling body is y = k1s m>sec (k a constant) at the instant the body has fallen s m from its starting point. Show that the body’s acceleration is constant.
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99. Falling meteorite The velocity of a heavy meteorite entering Earth’s atmosphere is inversely proportional to 1s when it is s km from Earth’s center. Show that the meteorite’s acceleration is inversely proportional to s 2 . 100. Particle acceleration A particle moves along the xaxis with velocity dx>dt = ƒsxd . Show that the particle’s acceleration is ƒsxdƒ¿sxd . 101. Temperature and the period of a pendulum For oscillations of small amplitude (short swings), we may safely model the relationship between the period T and the length L of a simple pendulum with the equation L T = 2p g , A
T The curves in Exercises 107 and 108 are called Bowditch curves or Lissajous figures. In each case, find the point in the interior of the first quadrant where the tangent to the curve is horizontal, and find the equations of the two tangents at the origin. 107.
y
dL = kL . du
108.
x sin t y sin 2t
–1
where g is the constant acceleration of gravity at the pendulum’s location. If we measure g in centimeters per second squared, we measure L in centimeters and T in seconds. If the pendulum is made of metal, its length will vary with temperature, either increasing or decreasing at a rate that is roughly proportional to L. In symbols, with u being temperature and k the proportionality constant,
1
y 1
x
x sin 2t y sin 3t
–1
x
1 –1
Using the Chain Rule, show that the power rule sd>dxdx n = nx n  1 holds for the functions x n in Exercises 109 and 110. 109. x 1>4 = 2 1x
110. x 3>4 = 2x1x
COMPUTER EXPLORATIONS
Assuming this to be the case, show that the rate at which the period changes with respect to temperature is kT>2. 102. Chain Rule composites
for h = 1.0, 0.7, and 0.3 . Experiment with other values of h. What do you see happening as h : 0 ? Explain this behavior.
Suppose that ƒsxd = x 2 and g sxd = ƒ x ƒ . Then the
Trigonometric Polynomials 111. As Figure 3.34 shows, the trigonometric “polynomial” s = ƒstd = 0.78540  0.63662 cos 2t  0.07074 cos 6t
sƒ gdsxd = ƒ x ƒ 2 = x 2
and
sg ƒdsxd = ƒ x 2 ƒ = x 2
are both differentiable at x = 0 even though g itself is not differentiable at x = 0 . Does this contradict the Chain Rule? Explain. 103. Tangents Suppose that u = g sxd is differentiable at x = 1 and that y = ƒsud is differentiable at u = g s1d . If the graph of y = ƒsg sxdd has a horizontal tangent at x = 1 , can we conclude anything about the tangent to the graph of g at x = 1 or the tangent to the graph of f at u = g s1d ? Give reasons for your answer. 104. Suppose that u = g sxd is differentiable at x = 5, y = ƒsud is differentiable at u = g s 5d , and sƒ gd¿s 5d is negative. What, if anything, can be said about the values of g¿s 5d and ƒ¿sg s 5dd ?
0.02546 cos 10t  0.01299 cos 14t gives a good approximation of the sawtooth function s = g std on the interval [p, p] . How well does the derivative of ƒ approximate the derivative of g at the points where dg>dt is defined? To find out, carry out the following steps. a. Graph dg>dt (where defined) over [p, p] . b. Find dƒ>dt. c. Graph dƒ>dt. Where does the approximation of dg>dt by dƒ>dt seem to be best? Least good? Approximations by trigonometric polynomials are important in the theories of heat and oscillation, but we must not expect too much of them, as we see in the next exercise.
T 105. The derivative of sin 2x Graph the function y = 2 cos 2x for 2 … x … 3.5 . Then, on the same screen, graph y =
s
sin 2sx + hd  sin 2x h
for h = 1.0, 0.5 , and 0.2. Experiment with other values of h, including negative values. What do you see happening as h : 0 ? Explain this behavior. 2
2
T 106. The derivative of cos sx d Graph y = 2x sin sx d for 2 … x … 3 . Then, on the same screen, graph cos ssx + hd2 d  cos sx 2 d y = h
2
–
0
s g(t) s f(t)
FIGURE 3.34 The approximation of a sawtooth function by a trigonometric “polynomial” (Exercise 111).
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t
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3.5 The Chain Rule and Parametric Equations 112. (Continuation of Exercise 111.) In Exercise 111, the trigonometric polynomial ƒ(t) that approximated the sawtooth function g (t) on [p, p] had a derivative that approximated the derivative of the sawtooth function. It is possible, however, for a trigonometric polynomial to approximate a function in a reasonable way without its derivative approximating the function’s derivative at all well. As a case in point, the “polynomial” s = hstd = 1.2732 sin 2t + 0.4244 sin 6t + 0.25465 sin 10t + 0.18189 sin 14t + 0.14147 sin 18t
s s k(t)
–
– 2
2
–1
b. Find dh>dt. c. Graph dh>dt to see how badly the graph fits the graph of dk>dt. Comment on what you see.
Parametrized Curves Use a CAS to perform the following steps on the parametrized curves in Exercises 113–116. b. Find dy>dx and d 2y>dx 2 at the point t0 . c. Find an equation for the tangent line to the curve at the point defined by the given value t0 . Plot the curve together with the tangent line on a single graph. 113. x =
s h(t)
0
a. Graph dk>dt (where defined) over [p, p] .
a. Plot the curve for the given interval of t values.
graphed in Figure 3.35 approximates the step function s = kstd shown there. Yet the derivative of h is nothing like the derivative of k.
1
205
t
1 3 t , 3
y =
1 2 t , 2
0 … t … 1,
114. x = 2t 3  16t 2 + 25t + 5, t0 = 3>2 115. x = t  cos t, t
116. x = e cos t,
y = t 2 + t  3,
y = 1 + sin t, t
y = e sin t,
t0 = 1>2
p … t … p,
0 … t … p,
FIGURE 3.35 The approximation of a step function by a trigonometric “polynomial” (Exercise 112).
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0 … t … 6, t0 = p>4
t0 = p>2
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3.6 Implicit Differentiation
Implicit Differentiation
3.6 y
y1 兹25 x 2
Most of the functions we have dealt with so far have been described by an equation of the form y = ƒsxd that expresses y explicitly in terms of the variable x. We have learned rules for differentiating functions defined in this way. In Section 3.5 we also learned how to find the derivative dy>dx when a curve is defined parametrically by equations x = xstd and y = ystd. A third situation occurs when we encounter equations like x 2 + y 2  25 = 0,
–5
0
5
x
(3, – 4) y2 –兹25 x 2
205
Slope – xy 3 4
FIGURE 3.36 The circle combines the graphs of two functions. The graph of y2 is the lower semicircle and passes through s3, 4d .
y 2  x = 0,
or
x 3 + y 3  9xy = 0.
(See Figures 3.36, 3.37, and 3.38.) These equations define an implicit relation between the variables x and y. In some cases we may be able to solve such an equation for y as an explicit function (or even several functions) of x. When we cannot put an equation Fsx, yd = 0 in the form y = ƒsxd to differentiate it in the usual way, we may still be able to find dy>dx by implicit differentiation. This consists of differentiating both sides of the equation with respect to x and then solving the resulting equation for y¿ . This section describes the technique and uses it to extend the Power Rule for differentiation to include rational exponents. In the examples and exercises of this section it is always assumed that the given equation determines y implicitly as a differentiable function of x.
Implicitly Defined Functions We begin with an example.
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206
y
Chapter 3: Differentiation
EXAMPLE 1
y2 x
Differentiating Implicitly 2
Slope 1 1 2y1 2兹x
Find dy>dx if y = x. y1 兹x
The equation y 2 = x defines two differentiable functions of x that we can actually find, namely y1 = 1x and y2 =  1x (Figure 3.37). We know how to calculate the derivative of each of these for x 7 0:
Solution
P(x, 兹x ) x
0
dy1 1 = dx 21x
Q(x, 兹x ) y 2 兹x Slope 1 1 2y 2 2兹x
FIGURE 3.37 The equation y 2  x = 0 , or y 2 = x as it is usually written, defines two differentiable functions of x on the interval x Ú 0 . Example 1 shows how to find the derivatives of these functions without solving the equation y 2 = x for y.
and
dy2 1 = . dx 21x
But suppose that we knew only that the equation y 2 = x defined y as one or more differentiable functions of x for x 7 0 without knowing exactly what these functions were. Could we still find dy>dx? The answer is yes. To find dy>dx, we simply differentiate both sides of the equation y 2 = x with respect to x, treating y = ƒsxd as a differentiable function of x: y2 = x dy 2y = 1 dx dy 1 = . 2y dx
d 2 Ay B = dx dy d C ƒ A x B D 2 = 2ƒsxdƒ¿sxd = 2y dx . dx
The Chain Rule gives
y
5
This one formula gives the derivatives we calculated for both explicit solutions y1 = 1x and y2 =  1x:
y f1(x) (x 0, y 1)
dy1 1 1 = = 2y1 dx 21x
A
and
x 3 y 3 9xy 0 y f2(x)
(x 0, y 2) x0
0
5
EXAMPLE 2 x
dy2 1 1 1 = = = . 2y2 dx 21x 2 A  1x B
Slope of a Circle at a Point
Find the slope of circle x 2 + y 2 = 25 at the point s3, 4d. The circle is not the graph of a single function of x. Rather it is the combined graphs of two differentiable functions, y1 = 225  x 2 and y2 =  225  x 2 (Figure 3.36). The point s3, 4d lies on the graph of y2 , so we can find the slope by calculating explicitly:
Solution
(x 0, y3)
y f3 (x)
FIGURE 3.38 The curve x 3 + y 3  9xy = 0 is not the graph of any one function of x. The curve can, however, be divided into separate arcs that are the graphs of functions of x. This particular curve, called a folium, dates to Descartes in 1638.
dy2 6 3 2x ` = ` = = . 2 x=3 4 dx x = 3 2225  x 2225  9 But we can also solve the problem more easily by differentiating the given equation of the circle implicitly with respect to x: d 2 d d A x B + dx A y 2 B = dx A 25 B dx dy 2x + 2y = 0 dx dy x = y. dx x The slope at s3, 4d is  y `
= s3, 4d
3 3 = . 4 4
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Implicit Differentiation
207
Notice that unlike the slope formula for dy2>dx, which applies only to points below the xaxis, the formula dy>dx = x>y applies everywhere the circle has a slope. Notice also that the derivative involves both variables x and y, not just the independent variable x. To calculate the derivatives of other implicitly defined functions, we proceed as in Examples 1 and 2: We treat y as a differentiable implicit function of x and apply the usual rules to differentiate both sides of the defining equation. y
EXAMPLE 3 y2 x2 sin xy
4
Differentiating Implicitly 2
Find dy>dx if y = x 2 + sin xy (Figure 3.39). Solution
2
y 2 = x 2 + sin xy –4
–2
0
2
4
x
d 2 d d A y B = dx A x 2 B + dx A sin xy B dx
–2
2y
dy d = 2x + scos xyd A xy B dx dx
Á treating y as a function of x and using the Chain Rule.
2y
dy dy = 2x + scos xyd ay + x b dx dx
Treat xy as a product.
–4
FIGURE 3.39 The graph of y 2 = x 2 + sin xy in Example 3. The example shows how to find slopes on this implicitly defined curve.
2y
Differentiate both sides with respect to x Á
dy dy  scos xyd ax b = 2x + scos xydy dx dx s2y  x cos xyd
Collect terms with dy>dx Á
dy = 2x + y cos xy dx
Á and factor out dy>dx .
dy 2x + y cos xy = 2y  x cos xy dx
Solve for dy>dx by dividing.
Notice that the formula for dy>dx applies everywhere that the implicitly defined curve has a slope. Notice again that the derivative involves both variables x and y, not just the independent variable x.
Light ray
Tangent Curve of lens surface
Normal line A
Point of entry P
Implicit Differentiation 1. Differentiate both sides of the equation with respect to x, treating y as a differentiable function of x. 2. Collect the terms with dy>dx on one side of the equation. 3. Solve for dy>dx.
B
Lenses, Tangents, and Normal Lines FIGURE 3.40 The profile of a lens, showing the bending (refraction) of a ray of light as it passes through the lens surface.
In the law that describes how light changes direction as it enters a lens, the important angles are the angles the light makes with the line perpendicular to the surface of the lens at the point of entry (angles A and B in Figure 3.40). This line is called the normal to the surface at the point of entry. In a profile view of a lens like the one in Figure 3.40, the normal is the line perpendicular to the tangent to the profile curve at the point of entry.
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Chapter 3: Differentiation y
EXAMPLE 4
t
en
g an
Tangent and Normal to the Folium of Descartes
Show that the point (2, 4) lies on the curve x 3 + y 3  9xy = 0. Then find the tangent and normal to the curve there (Figure 3.41).
T
4 No rm
x 3 y 3 9xy 0
al
0
2
The point (2, 4) lies on the curve because its coordinates satisfy the equation given for the curve: 23 + 43  9s2ds4d = 8 + 64  72 = 0. To find the slope of the curve at (2, 4), we first use implicit differentiation to find a formula for dy>dx:
Solution
x
FIGURE 3.41 Example 4 shows how to find equations for the tangent and normal to the folium of Descartes at (2, 4).
x 3 + y 3  9xy d 3 d d A x B + dx A y 3 B  dx A 9xy B dx dy dy dx 3x 2 + 3y 2  9 ax + y b dx dx dx dy s3y 2  9xd + 3x 2  9y dx dy 3sy 2  3xd dx dy dx
= 0 d = A0B dx
Differentiate both sides with respect to x. Treat xy as a product and y as a function of x.
= 0 = 0 = 9y  3x 2 3y  x 2 =
y 2  3x
.
Solve for dy>dx.
We then evaluate the derivative at sx, yd = s2, 4d: dy 3y  x 2 3s4d  22 8 4 ` = 2 ` = 2 = = . 5 10 dx s2, 4d y  3x s2, 4d 4  3s2d The tangent at (2, 4) is the line through (2, 4) with slope 4>5: y = 4 + y =
4 x  2B 5A
4 12 x + . 5 5
The normal to the curve at (2, 4) is the line perpendicular to the tangent there, the line through (2, 4) with slope 5>4: 5 sx  2d 4 5 13 . y =  x + 4 2
y = 4 
The quadratic formula enables us to solve a seconddegree equation like y 2  2xy + 3x 2 = 0 for y in terms of x. There is a formula for the three roots of a cubic equation that is like the quadratic formula but much more complicated. If this formula is used to solve the equation x 3 + y 3 = 9xy for y in terms of x, then three functions determined by the equation are y = ƒsxd =
x3 x6 x3 x6 +  27x 3 + 3  27x 3 C 2 C 2 B4 B4 3

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Implicit Differentiation
209
and y =
x6 x6 x3 x3 1 cƒsxd ; 2 3 a 3 +  27x 3  3  27x 3 b d. 2 C 2 C 2 B4 B4
Using implicit differentiation in Example 4 was much simpler than calculating dy>dx directly from any of the above formulas. Finding slopes on curves defined by higherdegree equations usually requires implicit differentiation.
Derivatives of Higher Order Implicit differentiation can also be used to find higher derivatives. Here is an example.
EXAMPLE 5 2
Finding a Second Derivative Implicitly
2
Find d y>dx if 2x 3  3y 2 = 8. To start, we differentiate both sides of the equation with respect to x in order to find y¿ = dy>dx.
Solution
d d A 2x 3  3y 2 B = dx s8d dx 6x 2  6yy¿ = 0 x 2  yy¿ = 0 x2 y¿ = y ,
Treat y as a function of x.
when y Z 0
Solve for y¿ .
We now apply the Quotient Rule to find y– . y– =
2xy  x 2y¿ d x2 2x x2 # ay b = = y¿ y dx y2 y2
Finally, we substitute y¿ = x 2>y to express y– in terms of x and y. 2x x2 x2 2x x4 y– = y  2 a y b = y  3 , y y
when y Z 0
Rational Powers of Differentiable Functions We know that the rule d n x = nx n  1 dx holds when n is an integer. Using implicit differentiation we can show that it holds when n is any rational number.
THEOREM 4 Power Rule for Rational Powers If p>q is a rational number, then x p>q is differentiable at every interior point of the domain of x sp>qd  1 , and p d p>q x = q x sp>qd  1 . dx
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EXAMPLE 6 (a) (b) (c)
Using the Rational Power Rule
d 1>2 A x B = 12 x 1>2 = 1 dx 21x d 2>3 A x B = 23 x 1>3 dx
d 4>3 A x B =  34 x 7>3 dx
for x 7 0 for x Z 0 for x Z 0 q
Proof of Theorem 4 Let p and q be integers with q 7 0 and suppose that y = 2x p = x p>q . Then yq = xp. Since p and q are integers (for which we already have the Power Rule), and assuming that y is a differentiable function of x, we can differentiate both sides of the equation with respect to x and get qy q  1
dy = px p  1 . dx
If y Z 0, we can divide both sides of the equation by qy q  1 to solve for dy>dx, obtaining dy px p  1 = dx qy q  1 p xp1 = q # p>q q  1 sx d p xp1 = q # p  p>q x p = q # x s p  1d  s p  p>qd
y = x p>q p p q sq  1d = p  q A law of exponents
p = q # x s p>qd  1, which proves the rule. We will drop the assumption of differentiability used in the proof of Theorem 4 in Chapter 7, where we prove the Power Rule for any nonzero real exponent. (See Section 7.3.) By combining the result of Theorem 4 with the Chain Rule, we get an extension of the Power Chain Rule to rational powers of u: If p>q is a rational number and u is a differentiable function of x, then u p>q is a differentiable function of x and p d p>q du u = q u s p>qd  1 , dx dx provided that u Z 0 if s p>qd 6 1. This restriction is necessary because 0 might be in the domain of u p>q but not in the domain of u s p>qd  1 , as we see in the next example.
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3.6 Implicit Differentiation
EXAMPLE 7
Using the Rational Power and Chain Rules
function defined on [1, 1] $++%++&
(a)
d A 1  x 2 B 1>4 = 41 A 1  x 2 B 3>4(2x) dx =
Power Chain Rule with u = 1  x 2
x 2 A 1  x 2 B 3>4
(+++)+++* derivative defined only on s 1, 1d
(b)
d d 1 scos xd1>5 =  scos xd6>5 scos xd 5 dx dx = =
1 scos xd6>5 s sin xd 5
1 ssin xdscos xd6>5 5
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EXERCISES 3.6 Derivatives of Rational Powers
Find dr>du in Exercises 3336.
Find dy>dx in Exercises 110.
33. u1>2 + r 1>2 = 1
1. y = x 9>4
2. y = x 3>5
3. y = 22x
4. y = 25x
5. y = 72x + 6
6. y = 22x  1
7. y = s2x + 5d1>2
8. y = s1  6xd2>3
9. y = xsx 2 + 1d1>2
10. y = xsx 2 + 1d1>2
4
3
35. sin sr ud =
Find the first derivatives of the functions in Exercises 1118. 7
1 2
12. r = 2u3
3 2>3 4 u + u3>4 2 3
36. cos r + cot u = r u
Second Derivatives In Exercises 3742, use implicit differentiation to find dy>dx and then d 2y>dx 2 . 37. x 2 + y 2 = 1
4
11. s = 2t 2
34. r  22u =
38. x 2>3 + y 2>3 = 1
13. y = sin [s2t + 5d2>3]
14. z = cos [s1  6td2>3]
39. y = x + 2x
40. y 2  2x = 1  2y
15. ƒsxd = 21  1x
16. gsxd = 2s2x 1>2 + 1d1>3
41. 2 1y = x  y
42. xy + y 2 = 1
3 1 + cos s2ud 17. hsud = 2
18. ksud = ssin su + 5dd5>4
2
2
3
3
43. If x + y = 16 , find the value of d 2y>dx 2 at the point (2, 2). 44. If xy + y 2 = 1 , find the value of d 2y>dx 2 at the point s0, 1d .
Differentiating Implicitly Use implicit differentiation to find dy>dx in Exercises 1932.
Slopes, Tangents, and Normals
19. x 2y + xy 2 = 6
In Exercises 45 and 46, find the slope of the curve at the given points.
20. x 3 + y 3 = 18xy
2
3
21. 2xy + y = x + y 2
2
2
23. x sx  yd = x  y x  1 x + 1
3
22. x  xy + y = 1 2
2
24. s3xy + 7d = 6y x  y 26. x 2 = x + y
45. y 2 + x 2 = y 4  2x
at
46. sx 2 + y 2 d2 = sx  yd2
s 2, 1d and s 2, 1d at
s1, 0d and s1, 1d
27. x = tan y
28. xy = cot sxyd
In Exercises 4756, verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point.
29. x + tan sxyd = 0
30. x + sin y = xy
47. x 2 + xy  y 2 = 1,
1 32. y 2 cos a y b = 2x + 2y
48. x 2 + y 2 = 25,
25. y 2 =
1 31. y sin a y b = 1  xy
49. x 2y 2 = 9,
s2, 3d
s3, 4d
s 1, 3d
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50. y 2  2x  4y  1 = 0, 2
51. 6x + 3xy + 2y + 17y  6 = 0,
A 23, 2 B
52. x  23xy + 2y = 5, 2
2
54. x sin 2y = y cos 2x,
sp>4, p>2d
55. y = 2 sin spx  yd,
s1, 0d
2
s 1, 0d
y
y 4 4y 2 x 4 9x 2
s1, p>2d
53. 2xy + p sin y = 2p,
2
61. The devil’s curve (Gabriel Cramer [the Cramer of Cramer’s rule], 1750) Find the slopes of the devil’s curve y 4  4y 2 = x 4  9x 2 at the four indicated points.
s 2, 1d
2
56. x cos y  sin y = 0,
57. Parallel tangents Find the two points where the curve x 2 + xy + y 2 = 7 crosses the xaxis, and show that the tangents to the curve at these points are parallel. What is the common slope of these tangents? 58. Tangents parallel to the coordinate axes Find points on the curve x 2 + xy + y 2 = 7 (a) where the tangent is parallel to the xaxis and (b) where the tangent is parallel to the yaxis. In the latter case, dy>dx is not defined, but dx>dy is. What value does dx>dy have at these points? 59. The eight curve Find the slopes of the curve y 4 = y 2  x 2 at the two points shown here. y
–3
x
3
(–3, –2)
(3, –2)
–2
62. The folium of Descartes (See Figure 3.38) a. Find the slope of the folium of Descartes, x 3 + y 3  9xy = 0 at the points (4, 2) and (2, 4). b. At what point other than the origin does the folium have a horizontal tangent?
Implicitly Defined Parametrizations
兹3 , 1 4 2 y4 y2 x2
(3, 2)
c. Find the coordinates of the point A in Figure 3.38, where the folium has a vertical tangent.
兹3 , 兹3 4 2
1
2
(–3, 2)
s0, pd
x
0
Assuming that the equations in Exercises 63–66 define x and y implicitly as differentiable functions x = ƒstd, y = g std , find the slope of the curve x = ƒstd, y = g std at the given value of t. 63. x 2  2tx + 2t 2 = 4, 64. x = 25  1t,
y st  1d = 1t,
65. x + 2x 3>2 = t 2 + t, 66. x sin t + 2x = t, –1
2y 3  3t 2 = 4,
t = 2
t = 4
y2t + 1 + 2t1y = 4,
t sin t  2t = y,
Theory and Examples
60. The cissoid of Diocles (from about 200 B.C.) Find equations for the tangent and normal to the cissoid of Diocles y 2s2  xd = x 3 at (1, 1). y y 2(2 x) x 3
67. Which of the following could be true if ƒ–sxd = x 1>3 ? 3 2>3 x  3 2 1 c. ƒ‡sxd =  x 4>3 3
a. ƒsxd =
9 5>3 x  7 10 3 d. ƒ¿sxd = x 2>3 + 6 2 b. ƒsxd =
68. Is there anything special about the tangents to the curves y 2 = x 3 and 2x 2 + 3y 2 = 5 at the points s1, ;1d ? Give reasons for your answer. y y2 x3
(1, 1)
1
2x 2 3y 2 5 0
t = 0
t = p
1
(1, 1)
x
x
0 (1, –1)
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70. Normals parallel to a line Find the normals to the curve xy + 2x  y = 0 that are parallel to the line 2x + y = 0 .
predicted the general behavior of the derivative graphs from looking at the graph of x 4 + 4y 2 = 1 ? Could you have predicted the general behavior of the graph of x 4 + 4y 2 = 1 by looking at the derivative graphs? Give reasons for your answers.
71. Normals to a parabola Show that if it is possible to draw three normals from the point (a, 0) to the parabola x = y 2 shown here, then a must be greater than 1>2. One of the normals is the xaxis. For what value of a are the other two normals perpendicular?
76. a. Given that sx  2d2 + y 2 = 4 find dy>dx two ways: (1) by solving for y and differentiating the resulting functions with respect to x and (2) by implicit differentiation. Do you get the same result each way?
69. Intersecting normal The line that is normal to the curve x 2 + 2xy  3y 2 = 0 at (1, 1) intersects the curve at what other point?
y x y2
(a, 0)
0
x
b. Solve the equation sx  2d2 + y 2 = 4 for y and graph the resulting functions together to produce a complete graph of the equation sx  2d2 + y 2 = 4 . Then add the graphs of the functions’ first derivatives to your picture. Could you have predicted the general behavior of the derivative graphs from looking at the graph of sx  2d2 + y 2 = 4 ? Could you have predicted the general behavior of the graph of sx  2d2 + y 2 = 4 by looking at the derivative graphs? Give reasons for your answers. Use a CAS to perform the following steps in Exercises 77–84. a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point P satisfies the equation.
72. What is the geometry behind the restrictions on the domains of the derivatives in Example 6(b) and Example 7(a)? T In Exercises 73 and 74, find both dy>dx (treating y as a differentiable function of x) and dx>dy (treating x as a differentiable function of y). How do dy>dx and dx>dy seem to be related? Explain the relationship geometrically in terms of the graphs. 73. xy 3 + x 2y = 6
74. x 3 + y 2 = sin2 y
b. Using implicit differentiation, find a formula for the derivative dy>dx and evaluate it at the given point P. c. Use the slope found in part (b) to find an equation for the tangent line to the curve at P. Then plot the implicit curve and tangent line together on a single graph. 77. x 3  xy + y 3 = 7, 5
3
2
Ps2, 1d 4
78. x + y x + yx + y = 4,
Ps1, 1d
COMPUTER EXPLORATIONS
2 + x 79. y 2 + y = , 1  x
75. a. Given that x 4 + 4y 2 = 1 , find dy>dx two ways: (1) by solving for y and differentiating the resulting functions in the usual way and (2) by implicit differentiation. Do you get the same result each way?
80. y 3 + cos xy = x 2, Ps1, 0d y p 81. x + tan a x b = 2, P a1, b 4
b. Solve the equation x 4 + 4y 2 = 1 for y and graph the resulting functions together to produce a complete graph of the equation x 4 + 4y 2 = 1 . Then add the graphs of the first derivatives of these functions to your display. Could you have
Ps0, 1d
82. xy 3 + tan (x + yd = 1, 83. 2y 2 + sxyd1>3 = x 2 + 2, 84. x21 + 2y + y = x 2,
p P a , 0b 4 Ps1, 1d P s1, 0d
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213
Related Rates In this section we look at problems that ask for the rate at which some variable changes. In each case the rate is a derivative that has to be computed from the rate at which some other variable (or perhaps several variables) is known to change. To find it, we write an equation that relates the variables involved and differentiate it to get an equation that relates the rate we seek to the rates we know. The problem of finding a rate you cannot measure easily from some other rates that you can is called a related rates problem. Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
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Related Rates Equations Suppose we are pumping air into a spherical balloon. Both the volume and radius of the balloon are increasing over time. If V is the volume and r is the radius of the balloon at an instant of time, then V =
4 3 pr . 3
Using the Chain Rule, we differentiate to find the related rates equation dV dr dV dr = = 4pr 2 . dt dr dt dt So if we know the radius r of the balloon and the rate dV>dt at which the volume is increasing at a given instant of time, then we can solve this last equation for dr>dt to find how fast the radius is increasing at that instant. Note that it is easier to measure directly the rate of increase of the volume than it is to measure the increase in the radius. The related rates equation allows us to calculate dr>dt from dV>dt. Very often the key to relating the variables in a related rates problem is drawing a picture that shows the geometric relations between them, as illustrated in the following example.
EXAMPLE 1
r
Pumping Out a Tank
How rapidly will the fluid level inside a vertical cylindrical tank drop if we pump the fluid out at the rate of 3000 L> min?
dh ? dt h
Solution We draw a picture of a partially filled vertical cylindrical tank, calling its radius r and the height of the fluid h (Figure 3.42). Call the volume of the fluid V. As time passes, the radius remains constant, but V and h change. We think of V and h as differentiable functions of time and use t to represent time. We are told that
dV –3000 L/min dt
FIGURE 3.42 The rate of change of fluid volume in a cylindrical tank is related to the rate of change of fluid level in the tank (Example 1).
dV = 3000. dt
We pump out at the rate of 3000 L> min. The rate is negative because the volume is decreasing.
dh . dt
How fast will the fluid level drop?
We are asked to find
To find dh>dt, we first write an equation that relates h to V. The equation depends on the units chosen for V, r, and h. With V in liters and r and h in meters, the appropriate equation for the cylinder’s volume is V = 1000pr 2h because a cubic meter contains 1000 L. Since V and h are differentiable functions of t, we can differentiate both sides of the equation V = 1000pr 2h with respect to t to get an equation that relates dh>dt to dV>dt: dV dh . = 1000pr 2 dt dt
r is a constant.
We substitute the known value dV>dt = 3000 and solve for dh>dt: 3 3000 dh =  2. = 2 dt 1000pr pr
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Related Rates
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The fluid level will drop at the rate of 3>spr 2 d m>min. The equation dh>dt = 3>pr 2 shows how the rate at which the fluid level drops depends on the tank’s radius. If r is small, dh>dt will be large; if r is large, dh>dt will be small. If r = 1 m: If r = 10 m:
3 dh =  p L 0.95 m>min = 95 cm>min. dt 3 dh = L 0.0095 m>min = 0.95 cm>min. 100p dt
Related Rates Problem Strategy 1. Draw a picture and name the variables and constants. Use t for time. Assume that all variables are differentiable functions of t. 2. Write down the numerical information (in terms of the symbols you have chosen). 3. Write down what you are asked to find (usually a rate, expressed as a derivative). 4. Write an equation that relates the variables. You may have to combine two or more equations to get a single equation that relates the variable whose rate you want to the variables whose rates you know. 5. Differentiate with respect to t. Then express the rate you want in terms of the rate and variables whose values you know. 6. Evaluate. Use known values to find the unknown rate.
EXAMPLE 2 Balloon
d 0.14 rad/min dt when /4
Range finder
A Rising Balloon
A hot air balloon rising straight up from a level field is tracked by a range finder 500 ft from the liftoff point. At the moment the range finder’s elevation angle is p>4, the angle is increasing at the rate of 0.14 rad> min. How fast is the balloon rising at that moment? dy ? y dt when /4
500 ft
FIGURE 3.43 The rate of change of the balloon’s height is related to the rate of change of the angle the range finder makes with the ground (Example 2).
We answer the question in six steps. 1. Draw a picture and name the variables and constants (Figure 3.43). The variables in the picture are u = the angle in radians the range finder makes with the ground. y = the height in feet of the balloon. We let t represent time in minutes and assume that u and y are differentiable functions of t. The one constant in the picture is the distance from the range finder to the liftoff point (500 ft). There is no need to give it a special symbol. 2. Write down the additional numerical information. Solution
du = 0.14 rad>min dt 3.
when
u =
p 4
Write down what we are to find. We want dy>dt when u = p>4.
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4.
Write an equation that relates the variables y and u. y = tan u 500
5.
or
y = 500 tan u
Differentiate with respect to t using the Chain Rule. The result tells how dy>dt (which we want) is related to du>dt (which we know). dy du = 500 ssec2 ud dt dt
6.
Evaluate with u = p>4 and du>dt = 0.14 to find dy>dt. dy = 500 A 22 B 2s0.14d = 140 dt
sec
p = 22 4
At the moment in question, the balloon is rising at the rate of 140 ft> min.
EXAMPLE 3
y
A police cruiser, approaching a rightangled intersection from the north, is chasing a speeding car that has turned the corner and is now moving straight east. When the cruiser is 0.6 mi north of the intersection and the car is 0.8 mi to the east, the police determine with radar that the distance between them and the car is increasing at 20 mph. If the cruiser is moving at 60 mph at the instant of measurement, what is the speed of the car?
Situation when x 0.8, y 0.6 y dy –60 dt 0
ds 20 dt
dx ? dt
A Highway Chase
x
x
We picture the car and cruiser in the coordinate plane, using the positive xaxis as the eastbound highway and the positive yaxis as the southbound highway (Figure 3.44). We let t represent time and set
Solution
FIGURE 3.44 The speed of the car is related to the speed of the police cruiser and the rate of change of the distance between them (Example 3).
x = position of car at time t y = position of cruiser at time t s = distance between car and cruiser at time t. We assume that x, y, and s are differentiable functions of t. We want to find dx>dt when x = 0.8 mi,
dy = 60 mph, dt
y = 0.6 mi,
ds = 20 mph. dt
Note that dy>dt is negative because y is decreasing. We differentiate the distance equation s2 = x2 + y2 (we could also use s = 2x 2 + y 2), and obtain 2s
dy ds dx = 2x + 2y dt dt dt dy ds dx 1 = s ax + y b dt dt dt =
1 2x + y 2
2
ax
dy dx + y b. dt dt
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Finally, use x = 0.8, y = 0.6, dy>dt = 60, ds>dt = 20, and solve for dx>dt. 20 =
dx 1 + A 0.6 B A 60 B b a0.8 dt 2s0.8d2 + s0.6d2
202s0.8d2 + s0.6d2 + s0.6ds60d dx = = 70 0.8 dt At the moment in question, the car’s speed is 70 mph.
EXAMPLE 4
dV 9 ft 3/min dt
Water runs into a conical tank at the rate of 9 ft3>min. The tank stands point down and has a height of 10 ft and a base radius of 5 ft. How fast is the water level rising when the water is 6 ft deep?
5 ft
x dy ? dt when y 6 ft
Filling a Conical Tank
Solution 10 ft
V = volume sft3 d of the water in the tank at time t smind x = radius sftd of the surface of the water at time t y = depth sftd of water in tank at time t.
y
FIGURE 3.45 The geometry of the conical tank and the rate at which water fills the tank determine how fast the water level rises (Example 4).
Figure 3.45 shows a partially filled conical tank. The variables in the problem are
We assume that V, x, and y are differentiable functions of t. The constants are the dimensions of the tank. We are asked for dy>dt when y = 6 ft
dV = 9 ft3>min. dt
and
The water forms a cone with volume V =
1 2 px y. 3
This equation involves x as well as V and y. Because no information is given about x and dx>dt at the time in question, we need to eliminate x. The similar triangles in Figure 3.45 give us a way to express x in terms of y: 5 x y = 10
or
x =
y . 2
Therefore, V =
y 2 p 3 1 pa b y = y 3 2 12
to give the derivative dV p # 2 dy p dy = 3y = y2 . 12 4 dt dt dt Finally, use y = 6 and dV>dt = 9 to solve for dy>dt. 9 =
dy p 6B2 A 4 dt
dy 1 = p L 0.32 dt At the moment in question, the water level is rising at about 0.32 ft> min. Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
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EXERCISES 3.7 1. Area Suppose that the radius r and area A = pr 2 of a circle are differentiable functions of t. Write an equation that relates dA>dt to dr>dt.
b. How is ds>dt related to dx>dt and dy>dt if neither x nor y is constant?
2. Surface area Suppose that the radius r and surface area S = 4pr 2 of a sphere are differentiable functions of t. Write an equation that relates dS>dt to dr>dt.
8. Diagonals If x, y, and z are lengths of the edges of a rectangular box, the common length of the box’s diagonals is s = 2x 2 + y 2 + z 2 .
3. Volume The radius r and height h of a right circular cylinder are related to the cylinder’s volume V by the formula V = pr 2h . a. How is dV>dt related to dh>dt if r is constant? b. How is dV>dt related to dr>dt if h is constant? c. How is dV>dt related to dr>dt and dh>dt if neither r nor h is constant? 4. Volume The radius r and height h of a right circular cone are related to the cone’s volume V by the equation V = s1>3dpr 2h .
c. How is dx>dt related to dy>dt if s is constant?
a. Assuming that x, y, and z are differentiable functions of t, how is ds>dt related to dx>dt, dy>dt, and dz>dt? b. How is ds>dt related to dy>dt and dz>dt if x is constant? c. How are dx>dt, dy>dt, and dz>dt related if s is constant? 9. Area The area A of a triangle with sides of lengths a and b enclosing an angle of measure u is A =
a. How is dV>dt related to dh>dt if r is constant?
1 ab sin u . 2
b. How is dV>dt related to dr>dt if h is constant?
a. How is dA>dt related to du>dt if a and b are constant?
c. How is dV>dt related to dr>dt and dh>dt if neither r nor h is constant?
b. How is dA>dt related to du>dt and da>dt if only b is constant?
5. Changing voltage The voltage V (volts), current I (amperes), and resistance R (ohms) of an electric circuit like the one shown here are related by the equation V = IR . Suppose that V is increasing at the rate of 1 volt> sec while I is decreasing at the rate of 1> 3 amp> sec. Let t denote time in seconds. V
I
R
a. What is the value of dV>dt?
c. How is dA>dt related to du>dt, da>dt , and db>dt if none of a, b, and u are constant? 10. Heating a plate When a circular plate of metal is heated in an oven, its radius increases at the rate of 0.01 cm> min. At what rate is the plate’s area increasing when the radius is 50 cm? 11. Changing dimensions in a rectangle The length l of a rectangle is decreasing at the rate of 2 cm> sec while the width w is increasing at the rate of 2 cm> sec. When l = 12 cm and w = 5 cm , find the rates of change of (a) the area, (b) the perimeter, and (c) the lengths of the diagonals of the rectangle. Which of these quantities are decreasing, and which are increasing? 12. Changing dimensions in a rectangular box Suppose that the edge lengths x, y, and z of a closed rectangular box are changing at the following rates:
b. What is the value of dI>dt? c. What equation relates dR>dt to dV>dt and dI>dt? d. Find the rate at which R is changing when V = 12 volts and I = 2 amp. Is R increasing, or decreasing? 6. Electrical power The power P (watts) of an electric circuit is related to the circuit’s resistance R (ohms) and current I (amperes) by the equation P = RI 2 . a. How are dP>dt, dR>dt, and dI>dt related if none of P, R, and I are constant? b. How is dR>dt related to dI>dt if P is constant? 7. Distance Let x and y be differentiable functions of t and let s = 2x 2 + y 2 be the distance between the points (x, 0) and (0, y) in the xyplane. a. How is ds>dt related to dx>dt if y is constant?
dx = 1 m>sec, dt
dy = 2 m>sec, dt
dz = 1 m>sec . dt
Find the rates at which the box’s (a) volume, (b) surface area, and (c) diagonal length s = 2x 2 + y 2 + z 2 are changing at the instant when x = 4, y = 3 , and z = 2 . 13. A sliding ladder A 13ft ladder is leaning against a house when its base starts to slide away. By the time the base is 12 ft from the house, the base is moving at the rate of 5 ft> sec. a. How fast is the top of the ladder sliding down the wall then?
b. At what rate is the area of the triangle formed by the ladder, wall, and ground changing then? c. At what rate is the angle u between the ladder and the ground changing then?
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a. At what rate is the water level changing when the water is 8 m deep?
y
b. What is the radius r of the water’s surface when the water is y m deep?
y(t)
c. At what rate is the radius r changing when the water is 8 m deep?
13ft ladder
0
x(t)
x
14. Commercial air traffic Two commercial airplanes are flying at 40,000 ft along straightline courses that intersect at right angles. Plane A is approaching the intersection point at a speed of 442 knots (nautical miles per hour; a nautical mile is 2000 yd). Plane B is approaching the intersection at 481 knots. At what rate is the distance between the planes changing when A is 5 nautical miles from the intersection point and B is 12 nautical miles from the intersection point? 15. Flying a kite A girl flies a kite at a height of 300 ft, the wind carrying the kite horizontally away from her at a rate of 25 ft> sec. How fast must she let out the string when the kite is 500 ft away from her? 16. Boring a cylinder The mechanics at Lincoln Automotive are reboring a 6in.deep cylinder to fit a new piston. The machine they are using increases the cylinder’s radius onethousandth of an inch every 3 min. How rapidly is the cylinder volume increasing when the bore (diameter) is 3.800 in.? 17. A growing sand pile Sand falls from a conveyor belt at the rate of 10 m3>min onto the top of a conical pile. The height of the pile is always threeeighths of the base diameter. How fast are the (a) height and (b) radius changing when the pile is 4 m high? Answer in centimeters per minute. 18. A draining conical reservoir Water is flowing at the rate of 50 m3>min from a shallow concrete conical reservoir (vertex down) of base radius 45 m and height 6 m. a. How fast (centimeters per minute) is the water level falling when the water is 5 m deep?
20. A growing raindrop Suppose that a drop of mist is a perfect sphere and that, through condensation, the drop picks up moisture at a rate proportional to its surface area. Show that under these circumstances the drop’s radius increases at a constant rate. 21. The radius of an inflating balloon A spherical balloon is inflated with helium at the rate of 100p ft3>min . How fast is the balloon’s radius increasing at the instant the radius is 5 ft? How fast is the surface area increasing? 22. Hauling in a dinghy A dinghy is pulled toward a dock by a rope from the bow through a ring on the dock 6 ft above the bow. The rope is hauled in at the rate of 2 ft> sec. a. How fast is the boat approaching the dock when 10 ft of rope are out?
b. At what rate is the angle u changing then (see the figure)? Ring at edge of dock
6'
23. A balloon and a bicycle A balloon is rising vertically above a level, straight road at a constant rate of 1 ft> sec. Just when the balloon is 65 ft above the ground, a bicycle moving at a constant rate of 17 ft> sec passes under it. How fast is the distance s(t) between the bicycle and balloon increasing 3 sec later? y
b. How fast is the radius of the water’s surface changing then? Answer in centimeters per minute. 19. A draining hemispherical reservoir Water is flowing at the rate of 6 m3>min from a reservoir shaped like a hemispherical bowl of radius 13 m, shown here in profile. Answer the following questions, given that the volume of water in a hemispherical bowl of radius R is V = sp>3dy 2s3R  yd when the water is y meters deep.
y(t)
Center of sphere s(t)
13 Water level r y 0
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x(t)
x
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24. Making coffee Coffee is draining from a conical filter into a cylindrical coffeepot at the rate of 10 in3>min . a. How fast is the level in the pot rising when the coffee in the cone is 5 in. deep?
b. How fast is the level in the cone falling then?
a. r sxd = 9x, c sxd = x 3  6x 2 + 15x, when x = 2 b. r sxd = 70x, c sxd = x 3  6x 2 + 45>x, when x = 1.5
6"
6" How fast is this level falling?
How fast is this level rising? 6"
25. Cardiac output In the late 1860s, Adolf Fick, a professor of physiology in the Faculty of Medicine in Würzberg, Germany, developed one of the methods we use today for measuring how much blood your heart pumps in a minute. Your cardiac output as you read this sentence is probably about 7 L> min. At rest it is likely to be a bit under 6 L> min. If you are a trained marathon runner running a marathon, your cardiac output can be as high as 30 L> min. Your cardiac output can be calculated with the formula Q y = , D where Q is the number of milliliters of CO2 you exhale in a minute and D is the difference between the CO2 concentration (ml> L) in the blood pumped to the lungs and the CO2 concentration in the blood returning from the lungs. With Q = 233 ml>min and D = 97  56 = 41 ml>L , y =
26. Cost, revenue, and profit A company can manufacture x items at a cost of c(x) thousand dollars, a sales revenue of r(x) thousand dollars, and a profit of psxd = r sxd  c sxd thousand dollars . Find dc>dt, dr>dt, and dp>dt for the following values of x and dx>dt.
233 ml>min L 5.68 L>min , 41 ml>L
fairly close to the 6 L> min that most people have at basal (resting) conditions. (Data courtesy of J. Kenneth Herd, M.D., Quillan College of Medicine, East Tennessee State University.) Suppose that when Q = 233 and D = 41 , we also know that D is decreasing at the rate of 2 units a minute but that Q remains unchanged. What is happening to the cardiac output?
and and
dx>dt = 0.1 dx>dt = 0.05
27. Moving along a parabola A particle moves along the parabola y = x 2 in the first quadrant in such a way that its xcoordinate (measured in meters) increases at a steady 10 m> sec. How fast is the angle of inclination u of the line joining the particle to the origin changing when x = 3 m ? 28. Moving along another parabola A particle moves from right to left along the parabolic curve y = 1x in such a way that its xcoordinate (measured in meters) decreases at the rate of 8 m> sec. How fast is the angle of inclination u of the line joining the particle to the origin changing when x = 4 ? 29. Motion in the plane The coordinates of a particle in the metric xyplane are differentiable functions of time t with dx>dt = 1 m>sec and dy>dt = 5 m>sec . How fast is the particle’s distance from the origin changing as it passes through the point (5, 12)?
30. A moving shadow A man 6 ft tall walks at the rate of 5 ft> sec toward a streetlight that is 16 ft above the ground. At what rate is the tip of his shadow moving? At what rate is the length of his shadow changing when he is 10 ft from the base of the light?
31. Another moving shadow A light shines from the top of a pole 50 ft high. A ball is dropped from the same height from a point 30 ft away from the light. (See accompanying figure.) How fast is the shadow of the ball moving along the ground 1>2 sec later? (Assume the ball falls a distance s = 16t 2 ft in t sec .) Light Ball at time t 0 1/2 sec later 50ft pole
Shadow 0
30
x(t)
x
NOT TO SCALE
32. Videotaping a moving car You are videotaping a race from a stand 132 ft from the track, following a car that is moving at 180 mi> h (264 ft> sec). How fast will your camera angle u be changing when the car is right in front of you? A half second later?
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221
36. Walkers A and B are walking on straight streets that meet at right angles. A approaches the intersection at 2 m> sec; B moves away from the intersection 1 m> sec. At what rate is the angle u changing when A is 10 m from the intersection and B is 20 m from the intersection? Express your answer in degrees per second to the nearest degree.
132'
A
Car
33. A melting ice layer A spherical iron ball 8 in. in diameter is coated with a layer of ice of uniform thickness. If the ice melts at the rate of 10 in3>min , how fast is the thickness of the ice decreasing when it is 2 in. thick? How fast is the outer surface area of ice decreasing? 34. Highway patrol A highway patrol plane flies 3 mi above a level, straight road at a steady 120 mi>h. The pilot sees an oncoming car and with radar determines that at the instant the lineofsight distance from plane to car is 5 mi, the lineofsight distance is decreasing at the rate of 160 mi>h. Find the car’s speed along the highway. 35. A building’s shadow On a morning of a day when the sun will pass directly overhead, the shadow of an 80ft building on level ground is 60 ft long. At the moment in question, the angle u the sun makes with the ground is increasing at the rate of 0.27° >min. At what rate is the shadow decreasing? (Remember to use radians. Express your answer in inches per minute, to the nearest tenth.)
O
B
37. Baseball players A baseball diamond is a square 90 ft on a side. A player runs from first base to second at a rate of 16 ft> sec. a. At what rate is the player’s distance from third base changing when the player is 30 ft from first base?
b. At what rates are angles u1 and u2 (see the figure) changing at that time? c. The player slides into second base at the rate of 15 ft> sec. At what rates are angles u1 and u2 changing as the player touches base? Second base 90' Third base
1
2
Player 30'
First base
Home
80'
38. Ships Two ships are steaming straight away from a point O along routes that make a 120° angle. Ship A moves at 14 knots (nautical miles per hour; a nautical mile is 2000 yd). Ship B moves at 21 knots. How fast are the ships moving apart when OA = 5 and OB = 3 nautical miles?
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221
Linearization and Differentials Sometimes we can approximate complicated functions with simpler ones that give the accuracy we want for specific applications and are easier to work with. The approximating functions discussed in this section are called linearizations, and they are based on tangent lines. Other approximating functions, such as polynomials, are discussed in Chapter 11.
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We introduce new variables dx and dy, called differentials, and define them in a way that makes Leibniz’s notation for the derivative dy>dx a true ratio. We use dy to estimate error in measurement and sensitivity of a function to change. Application of these ideas then provides for a precise proof of the Chain Rule (Section 3.5).
Linearization As you can see in Figure 3.46, the tangent to the curve y = x 2 lies close to the curve near the point of tangency. For a brief interval to either side, the yvalues along the tangent line give good approximations to the yvalues on the curve. We observe this phenomenon by zooming in on the two graphs at the point of tangency or by looking at tables of values for the difference between ƒ(x) and its tangent line near the xcoordinate of the point of tangency. Locally, every differentiable curve behaves like a straight line.
4
2 y x2
y x2
y 2x 1
y 2x 1
(1, 1)
(1, 1) –1
3 0
0
2 0
y x 2 and its tangent y 2x 1 at (1, 1). 1.2
Tangent and curve very close near (1, 1). 1.003
y x2
y x2 y 2x 1
(1, 1)
(1, 1)
y 2x 1 0.8
1.2 0.8
Tangent and curve very close throughout entire xinterval shown.
y
0.997 0.997
1.003
Tangent and curve closer still. Computer screen cannot distinguish tangent from curve on this xinterval.
FIGURE 3.46 The more we magnify the graph of a function near a point where the function is differentiable, the flatter the graph becomes and the more it resembles its tangent.
y f (x) Slope f '(a)
In general, the tangent to y = ƒsxd at a point x = a, where ƒ is differentiable (Figure 3.47), passes through the point (a, ƒ(a)), so its pointslope equation is
(a, f(a))
y = ƒsad + ƒ¿sadsx  ad. 0
x
a
FIGURE 3.47 The tangent to the curve y = ƒsxd at x = a is the line Lsxd = ƒsad + ƒ¿sadsx  ad .
Thus, this tangent line is the graph of the linear function Lsxd = ƒsad + ƒ¿sadsx  ad. For as long as this line remains close to the graph of ƒ, L(x) gives a good approximation to ƒ(x).
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DEFINITIONS Linearization, Standard Linear Approximation If ƒ is differentiable at x = a, then the approximating function Lsxd = ƒsad + ƒ¿sadsx  ad is the linearization of ƒ at a. The approximation ƒsxd L Lsxd of ƒ by L is the standard linear approximation of ƒ at a. The point x = a is the center of the approximation.
EXAMPLE 1
Finding a Linearization
Find the linearization of ƒsxd = 21 + x at x = 0 (Figure 3.48). y
y 5 x 4 4
y1 x 2 2 y 兹1 x 1
–1
0
1
2
3
4
x
FIGURE 3.48 The graph of y = 21 + x and its linearizations at x = 0 and x = 3 . Figure 3.49 shows a magnified view of the small window about 1 on the yaxis. Solution
Since
y1 x 2
1.1
ƒ¿sxd =
y 兹1 x 1.0
1 A 1 + x B 1>2 , 2
we have ƒs0d = 1 and ƒ¿s0d = 1>2, giving the linearization Lsxd = ƒsad + ƒ¿sadsx  ad = 1 +
x 1 Ax  0B = 1 + 2 . 2
See Figure 3.49. 0.9 –0.1
0
0.1
0.2
FIGURE 3.49 Magnified view of the window in Figure 3.48.
Look at how accurate the approximation 21 + x L 1 + sx>2d from Example 1 is for values of x near 0. As we move away from zero, we lose accuracy. For example, for x = 2, the linearization gives 2 as the approximation for 23 , which is not even accurate to one decimal place. Do not be misled by the preceding calculations into thinking that whatever we do with a linearization is better done with a calculator. In practice, we would never use a linearization to find a particular square root. The utility of a linearization is its ability to replace a complicated formula by a simpler one over an entire interval of values. If we have to work with 21 + x for x close to 0 and can tolerate the small amount of error involved, we can
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True value
ƒ True value approximation ƒ
0.2 = 1.10 2
1.095445
610 2
0.05 = 1.025 2
1.024695
610 3
0.005 = 1.00250 2
1.002497
610 5
Approximation 21.2 L 1 + 21.05 L 1 + 21.005 L 1 +
work with 1 + sx>2d instead. Of course, we then need to know how much error there is. We have more to say on the estimation of error in Chapter 11. A linear approximation normally loses accuracy away from its center. As Figure 3.48 suggests, the approximation 21 + x L 1 + sx>2d will probably be too crude to be useful near x = 3. There, we need the linearization at x = 3.
EXAMPLE 2
Finding a Linearization at Another Point
Find the linearization of ƒsxd = 21 + x at x = 3. Solution
We evaluate the equation defining Lsxd at a = 3. With ƒs3d = 2,
ƒ¿s3d =
1 1 = , A 1 + x B 1>2 ` 2 4 x=3
we have Lsxd = 2 +
5 x 1 Ax  3B = 4 + 4 . 4
At x = 3.2, the linearization in Example 2 gives 21 + x = 21 + 3.2 L
5 3.2 + = 1.250 + 0.800 = 2.050, 4 4
which differs from the true value 24.2 L 2.04939 by less than one onethousandth. The linearization in Example 1 gives 21 + x = 21 + 3.2 L 1 + y
3.2 = 1 + 1.6 = 2.6, 2
a result that is off by more than 25%.
EXAMPLE 3
Finding a Linearization for the Cosine Function
Find the linearization of ƒsxd = cos x at x = p>2 (Figure 3.50). 0
2
x y cos x
y –x 2
FIGURE 3.50 The graph of ƒsxd = cos x and its linearization at x = p>2 . Near x = p>2, cos x L x + sp>2d (Example 3).
Since ƒsp>2d = cossp>2d = 0, ƒ¿sxd = sin x, and ƒ¿sp>2d = sin sp>2d = 1, we have
Solution
Lsxd = ƒsad + ƒ¿sadsx  ad p = 0 + s 1d ax  b 2 p = x + . 2
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An important linear approximation for roots and powers is s1 + xdk L 1 + kx
sx near 0; any number kd
(Exercise 15). This approximation, good for values of x sufficiently close to zero, has broad application. For example, when x is small, 21 + x L 1 +
1 x 2
k = 1>2
1 = s1  xd1 L 1 + s 1ds xd = 1 + x 1  x 5 1 3 2 1 + 5x 4 = s1 + 5x 4 d1>3 L 1 + A 5x 4 B = 1 + x 4 3 3 1 1 1 = s1  x 2 d1>2 L 1 + a bs x 2 d = 1 + x 2 2 2 21  x 2
k = 1; replace x by x . k = 1>3; replace x by 5x 4 . k = 1>2; replace x by x 2 .
Differentials We sometimes use the Leibniz notation dy>dx to represent the derivative of y with respect to x. Contrary to its appearance, it is not a ratio. We now introduce two new variables dx and dy with the property that if their ratio exists, it will be equal to the derivative.
DEFINITION Differential Let y = ƒsxd be a differentiable function. The differential dx is an independent variable. The differential dy is dy = ƒ¿sxd dx.
Unlike the independent variable dx, the variable dy is always a dependent variable. It depends on both x and dx. If dx is given a specific value and x is a particular number in the domain of the function ƒ, then the numerical value of dy is determined.
EXAMPLE 4
Finding the Differential dy
(a) Find dy if y = x 5 + 37x. (b) Find the value of dy when x = 1 and dx = 0.2. Solution
(a) dy = s5x 4 + 37d dx (b) Substituting x = 1 and dx = 0.2 in the expression for dy, we have dy = s5 # 14 + 37d0.2 = 8.4. The geometric meaning of differentials is shown in Figure 3.51. Let x = a and set dx = ¢x. The corresponding change in y = ƒsxd is ¢y = ƒsa + dxd  ƒsad.
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y
y f (x)
y f (a dx) f (a) L f '(a)dx (a, f (a)) dx x When dx is a small change in x, the corresponding change in the linearization is precisely dy.
Tangent line 0
x
a dx
a
FIGURE 3.51 Geometrically, the differential dy is the change ¢L in the linearization of ƒ when x = a changes by an amount dx = ¢x .
The corresponding change in the tangent line L is ¢L = L(a + dx)  L(a) = ƒ(a) + ƒ¿(a)[(a + dx)  a]  ƒ(a) (++++++)++++++* L(a dx)
()* L(a)
= ƒ¿(a) dx. That is, the change in the linearization of ƒ is precisely the value of the differential dy when x = a and dx = ¢x. Therefore, dy represents the amount the tangent line rises or falls when x changes by an amount dx = ¢x. If dx Z 0, then the quotient of the differential dy by the differential dx is equal to the derivative ƒ¿sxd because ƒ¿sxd dx dy = ƒ¿sxd = . dx dx
dy , dx = We sometimes write
df = ƒ¿sxd dx in place of dy = ƒ¿sxd dx, calling dƒ the differential of ƒ. For instance, if ƒsxd = 3x 2  6, then df = ds3x 2  6d = 6x dx. Every differentiation formula like dsu + yd du dy = + dx dx dx
or
dssin ud du = cos u dx dx
or
dssin ud = cos u du.
has a corresponding differential form like dsu + yd = du + dy
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EXAMPLE 5
Linearization and Differentials
227
Finding Differentials of Functions
(a) dstan 2xd = sec2s2xd ds2xd = 2 sec2 2x dx sx + 1d dx  x dsx + 1d x x dx + dx  x dx dx (b) d a b = = = 2 2 x + 1 sx + 1d sx + 1d sx + 1d2
Estimating with Differentials Suppose we know the value of a differentiable function ƒ(x) at a point a and want to predict how much this value will change if we move to a nearby point a + dx. If dx is small, then we can see from Figure 3.51 that ¢y is approximately equal to the differential dy. Since ƒsa + dxd = ƒsad + ¢y, the differential approximation gives ƒsa + dxd L ƒsad + dy where dx = ¢x. Thus the approximation ¢y L dy can be used to calculate ƒsa + dxd when ƒ(a) is known and dx is small.
EXAMPLE 6
dr 0.1
Estimating with Differentials
The radius r of a circle increases from a = 10 m to 10.1 m (Figure 3.52). Use dA to estimate the increase in the circle’s area A. Estimate the area of the enlarged circle and compare your estimate to the true area.
a 10
Solution
dA = A¿sad dr = 2pa dr = 2ps10ds0.1d = 2p m2 .
A ≈ dA 2 a dr
FIGURE 3.52 When dr is small compared with a, as it is when dr = 0.1 and a = 10 , the differential dA = 2pa dr gives a way to estimate the area of the circle with radius r = a + dr (Example 6).
Since A = pr 2 , the estimated increase is
Thus, As10 + 0.1d L As10d + 2p = ps10d2 + 2p = 102p. The area of a circle of radius 10.1 m is approximately 102p m2 . The true area is As10.1d = ps10.1d2 = 102.01p m2 . The error in our estimate is 0.01p m2 , which is the difference ¢A  dA.
Error in Differential Approximation Let ƒ(x) be differentiable at x = a and suppose that dx = ¢x is an increment of x. We have two ways to describe the change in ƒ as x changes from a to a + ¢x: The true change: The differential estimate:
¢ƒ = ƒsa + ¢xd  ƒsad dƒ = ƒ¿sad ¢x.
How well does dƒ approximate ¢ƒ?
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We measure the approximation error by subtracting dƒ from ¢f: Approximation error = ¢ƒ  dƒ = ¢ƒ  ƒ¿sad¢x = ƒ(a + ¢x)  ƒ(a)  ƒ¿(a)¢x (++++)++++* ƒ
= a
ƒ(a + ¢x)  ƒ(a)  ƒ¿(a)b # ¢x ¢x
(+++++++)+++++++* Call this part P
= P # ¢x. As ¢x : 0, the difference quotient ƒsa + ¢xd  ƒsad ¢x approaches ƒ¿sad (remember the definition of ƒ¿sad), so the quantity in parentheses becomes a very small number (which is why we called it P). In fact, P : 0 as ¢x : 0. When ¢x is small, the approximation error P ¢x is smaller still. ¢ƒ = ƒ¿(a)¢x + P ¢x ()* true change
(+)+* estimated change
()* error
Although we do not know exactly how small the error is and will not be able to make much progress on this front until Chapter 11, there is something worth noting here, namely the form taken by the equation.
Change in y ƒsxd near x a If y = ƒsxd is differentiable at x = a and x changes from a to a + ¢x, the change ¢y in ƒ is given by an equation of the form ¢y = ƒ¿sad ¢x + P ¢x
(1)
in which P : 0 as ¢x : 0.
In Example 6 we found that ¢A = p(10.1) 2  p(10) 2 = (102.01  100)p = (2p + 0.01p) m2 ()* dA
()* error
so the approximation error is ¢A  dA = P ¢r = 0.01p and P = 0.01p>¢r = 0.01p>0.1 = 0.1p m. Equation (1) enables us to bring the proof of the Chain Rule to a successful conclusion. Proof of the Chain Rule Our goal is to show that if ƒ(u) is a differentiable function of u and u = gsxd is a differentiable function of x, then the composite y = ƒsg sxdd is a differentiable function of x.
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More precisely, if g is differentiable at x0 and ƒ is differentiable at gsx0 d, then the composite is differentiable at x0 and dy ` = ƒ¿s gsx0 dd # g¿sx0 d. dx x = x0 Let ¢x be an increment in x and let ¢u and ¢y be the corresponding increments in u and y. Applying Equation (1) we have, ¢u = g¿sx0 d¢x + P1 ¢x = sg¿sx0 d + P1 d¢x, where P1 : 0 as ¢x : 0. Similarly, ¢y = ƒ¿su0 d¢u + P2 ¢u = sƒ¿su0 d + P2 d¢u, where P2 : 0 as ¢u : 0. Notice also that ¢u : 0 as ¢x : 0. Combining the equations for ¢u and ¢y gives ¢y = sƒ¿su0 d + P2 dsg¿sx0 d + P1 d¢x, so ¢y = ƒ¿su0 dg¿sx0 d + P2 g¿sx0 d + ƒ¿su0 dP1 + P2P1 . ¢x Since P1 and P2 go to zero as ¢x goes to zero, three of the four terms on the right vanish in the limit, leaving dy ¢y ` = lim = ƒ¿su0 dg¿sx0 d = ƒ¿sgsx0 dd # g¿sx0 d. dx x = x0 ¢x:0 ¢x This concludes the proof.
Sensitivity to Change The equation df = ƒ¿sxd dx tells how sensitive the output of ƒ is to a change in input at different values of x. The larger the value of ƒ¿ at x, the greater the effect of a given change dx. As we move from a to a nearby point a + dx, we can describe the change in ƒ in three ways:
Absolute change Relative change Percentage change
EXAMPLE 7
True
Estimated
¢f = ƒsa + dxd  ƒsad ¢f ƒsad ¢f * 100 ƒsad
df = ƒ¿sad dx df ƒsad df * 100 ƒsad
Finding the Depth of a Well
You want to calculate the depth of a well from the equation s = 16t 2 by timing how long it takes a heavy stone you drop to splash into the water below. How sensitive will your calculations be to a 0.1sec error in measuring the time? Solution
The size of ds in the equation ds = 32t dt
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depends on how big t is. If t = 2 sec, the change caused by dt = 0.1 is about ds = 32s2ds0.1d = 6.4 ft. Three seconds later at t = 5 sec, the change caused by the same dt is ds = 32s5ds0.1d = 16 ft. The estimated depth of the well differs from its true depth by a greater distance the longer the time it takes the stone to splash into the water below, for a given error in measuring the time.
Opaque dye Blockage
EXAMPLE 8
Unclogging Arteries
In the late 1830s, French physiologist Jean Poiseuille (“pwaZOY”) discovered the formula we use today to predict how much the radius of a partially clogged artery has to be expanded to restore normal flow. His formula, Angiography
V = kr 4 ,
An opaque dye is injected into a partially blocked artery to make the inside visible under Xrays. This reveals the location and severity of the blockage.
says that the volume V of fluid flowing through a small pipe or tube in a unit of time at a fixed pressure is a constant times the fourth power of the tube’s radius r. How will a 10% increase in r affect V? Solution
Inflatable balloon on catheter
The differentials of r and V are related by the equation dV =
dV dr = 4kr 3 dr. dr
The relative change in V is dV 4kr 3 dr dr = = 4 r. V kr 4
Angioplasty A balloontipped catheter is inflated inside the artery to widen it at the blockage site.
The relative change in V is 4 times the relative change in r, so a 10% increase in r will produce a 40% increase in the flow.
EXAMPLE 9
Converting Mass to Energy
Newton’s second law, F =
d dy smyd = m = ma, dt dt
is stated with the assumption that mass is constant, but we know this is not strictly true because the mass of a body increases with velocity. In Einstein’s corrected formula, mass has the value m =
m0
21  y2>c 2
,
where the “rest mass” m0 represents the mass of a body that is not moving and c is the speed of light, which is about 300,000 km> sec. Use the approximation 1 1 L 1 + x2 2 2 21  x to estimate the increase ¢m in mass resulting from the added velocity y.
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231
When y is very small compared with c, y2>c 2 is close to zero and it is safe to use the approximation Solution
1 1 y2 L 1 + a 2b 2 c 21  y2>c 2
to obtain m = or
m0
21  y2>c 2
y Eq. (2) with x = c
L m0 c1 +
1 y2 1 1 a b d = m0 + m0 y2 a 2 b , 2 c2 2 c
m L m0 +
1 1 m y2 a 2 b . 2 0 c
(3)
Equation (3) expresses the increase in mass that results from the added velocity y. Energy Interpretation
In Newtonian physics, s1>2dm0 y2 is the kinetic energy (KE) of the body, and if we rewrite Equation (3) in the form sm  m0 dc 2 L
1 m y2 , 2 0
we see that sm  m0 dc 2 L
1 1 1 m y2 = m0 y2  m0s0d2 = ¢sKEd, 2 0 2 2
or s¢mdc 2 L ¢sKEd. So the change in kinetic energy ¢sKEd in going from velocity 0 to velocity y is approximately equal to s¢mdc 2 , the change in mass times the square of the speed of light. Using c L 3 * 10 8 m>sec, we see that a small change in mass can create a large change in energy.
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EXERCISES 3.8 Finding Linearizations In Exercises 1–4, find the linearization L(x) of ƒ(x) at x = a . 1. ƒsxd = x 3  2x + 3, 2. ƒsxd = 2x + 9, 2
1 3. ƒsxd = x + x , 3 x, 4. ƒsxd = 2
a = 2
a = 4
a = 1
subsequent work as simple as possible, you want to center each linearization not at x0 but at a nearby integer x = a at which the given function and its derivative are easy to evaluate. What linearization do you use in each case? 5. ƒsxd = x 2 + 2x, 6. ƒsxd = x 1,
x0 = 0.1
x0 = 0.9
7. ƒsxd = 2x 2 + 4x  3,
a = 8
Linearization for Approximation You want linearizations that will replace the functions in Exercises 5–10 over intervals that include the given points x0 . To make your
8. ƒsxd = 1 + x, 3 9. ƒsxd = 2 x,
10. ƒsxd =
x , x + 1
x0 = 0.9
x0 = 8.1 x0 = 8.5 x0 = 1.3
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Chapter 3: Differentiation
Linearizing Trigonometric Functions
y
y f (x)
In Exercises 11–14, find the linearization of ƒ at x = a . Then graph the linearization and ƒ together. 11. ƒsxd = sin x
at
sad x = 0,
sbd x = p
f f(x 0 dx) f(x 0)
12. ƒsxd = cos x
at
sad x = 0,
sbd x = p>2
13. ƒsxd = sec x
at
sad x = 0,
sbd x = p>3
14. ƒsxd = tan x
at
sad x = 0,
(x 0, f (x 0 )) dx Tangent
sbd x = p>4 0
The Approximation s1 xdk « 1 kx 15. Show that the linearization of ƒsxd = s1 + xdk at x = 0 is Lsxd = 1 + kx . 16. Use the linear approximation s1 + xdk L 1 + kx to find an approximation for the function ƒ(x) for values of x near zero. a. ƒsxd = s1  xd6 c. ƒsxd =
b. ƒsxd =
1
e. ƒsxd = s4 + 3xd1>3
f. ƒsxd =
3
a1 
1 b 2 + x
Derivatives in Differential Form In Exercises 19–30, find dy.
21. y =
20. y = x21  x
2x 1 + x2
22. y =
2
21x 3s1 + 1xd
23. 2y 3>2 + xy  x = 0
24. xy 2  4x 3>2  y = 0
25. y = sin s51xd
26. y = cos sx 2 d
27. y = 4 tan sx 3>3d 29. y = 3 csc s1  2 1xd
3
33. ƒsxd = x  x, 34. ƒsxd = x 4,
x0 = 1,
x0 = 1,
dx = 0.1
x0 = 1,
dx = 0.1
dx = 0.1
dx = 0.1
x0 = 0.5,
dx = 0.1 x0 = 2,
dx = 0.1
Differential Estimates of Change 2
18. Find the linearization of ƒsxd = 2x + 1 + sin x at x = 0 . How is it related to the individual linearizations of 2x + 1 and sin x at x = 0 ?
19. y = x  31x
x0 = 1,
32. ƒsxd = 2x 2 + 4x  3,
36. ƒsxd = x 3  2x + 3,
B 17. Faster than a calculator Use the approximation s1 + xdk L 1 + kx to estimate the following. 3 a. s1.0002d50 b. 2 1.009
3
31. ƒsxd = x 2 + 2x,
x
x 0 dx
x0
35. ƒsxd = x 1,
2 1  x
d. ƒsxd = 22 + x 2
21 + x
df f '(x 0 ) dx
In Exercises 37–42, write a differential formula that estimates the given change in volume or surface area. 37. The change in the volume V = s4>3dpr 3 of a sphere when the radius changes from r0 to r0 + dr 38. The change in the volume V = x 3 of a cube when the edge lengths change from x0 to x0 + dx 39. The change in the surface area S = 6x 2 of a cube when the edge lengths change from x0 to x0 + dx 40. The change in the lateral surface area S = pr2r 2 + h 2 of a right circular cone when the radius changes from r0 to r0 + dr and the height does not change 41. The change in the volume V = pr 2h of a right circular cylinder when the radius changes from r0 to r0 + dr and the height does not change 42. The change in the lateral surface area S = 2prh of a right circular cylinder when the height changes from h0 to h0 + dh and the radius does not change
28. y = sec sx 2  1d
Applications
30. y = 2 cot a
43. The radius of a circle is increased from 2.00 to 2.02 m.
1 b 1x
Approximation Error In Exercises 31–36, each function ƒ(x) changes value when x changes from x0 to x0 + dx . Find a. the change ¢ƒ = ƒsx0 + dxd  ƒsx0 d ; b. the value of the estimate df = ƒ¿sx0 d dx ; and c. the approximation error ƒ ¢ƒ  dƒ ƒ .
a. Estimate the resulting change in area. b. Express the estimate as a percentage of the circle’s original area. 44. The diameter of a tree was 10 in. During the following year, the circumference increased 2 in. About how much did the tree’s diameter increase? The tree’s crosssection area? 45. Estimating volume Estimate the volume of material in a cylindrical shell with height 30 in., radius 6 in., and shell thickness 0.5 in.
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233
d (“delta”) is the weight density of the blood, y is the average velocity of the exiting blood, and g is the acceleration of gravity. When P, V, d , and y remain constant, W becomes a function of g, and the equation takes the simplified form
0.5 in.
b W = a + g sa, b constantd .
30 in.
6 in.
46. Estimating height of a building A surveyor, standing 30 ft from the base of a building, measures the angle of elevation to the top of the building to be 75°. How accurately must the angle be measured for the percentage error in estimating the height of the building to be less than 4%? 47. Tolerance The height and radius of a right circular cylinder are equal, so the cylinder’s volume is V = ph 3 . The volume is to be calculated with an error of no more than 1% of the true value. Find approximately the greatest error that can be tolerated in the measurement of h, expressed as a percentage of h. 48. Tolerance a. About how accurately must the interior diameter of a 10mhigh cylindrical storage tank be measured to calculate the tank’s volume to within 1% of its true value? b. About how accurately must the tank’s exterior diameter be measured to calculate the amount of paint it will take to paint the side of the tank to within 5% of the true amount? 49. Minting coins A manufacturer contracts to mint coins for the federal government. How much variation dr in the radius of the coins can be tolerated if the coins are to weigh within 1>1000 of their ideal weight? Assume that the thickness does not vary. 50. Sketching the change in a cube’s volume The volume V = x 3 of a cube with edges of length x increases by an amount ¢V when x increases by an amount ¢x . Show with a sketch how to represent ¢V geometrically as the sum of the volumes of a. three slabs of dimensions x by x by ¢x b. three bars of dimensions x by ¢x by ¢x c. one cube of dimensions ¢ x by ¢ x by ¢ x . The differential formula dV = 3x 2 dx estimates the change in V with the three slabs. 51. The effect of flight maneuvers on the heart The amount of work done by the heart’s main pumping chamber, the left ventricle, is given by the equation 2
W = PV +
Vdy , 2g
where W is the work per unit time, P is the average blood pressure, V is the volume of blood pumped out during the unit of time,
As a member of NASA’s medical team, you want to know how sensitive W is to apparent changes in g caused by flight maneuvers, and this depends on the initial value of g. As part of your investigation, you decide to compare the effect on W of a given change dg on the moon, where g = 5.2 ft>sec2 , with the effect the same change dg would have on Earth, where g = 32 ft>sec2 . Use the simplified equation above to find the ratio of dWmoon to dWEarth . 52. Measuring acceleration of gravity When the length L of a clock pendulum is held constant by controlling its temperature, the pendulum’s period T depends on the acceleration of gravity g. The period will therefore vary slightly as the clock is moved from place to place on the earth’s surface, depending on the change in g. By keeping track of ¢T , we can estimate the variation in g from the equation T = 2psL>gd1>2 that relates T, g, and L. a. With L held constant and g as the independent variable, calculate dT and use it to answer parts (b) and (c). b. If g increases, will T increase or decrease? Will a pendulum clock speed up or slow down? Explain. c. A clock with a 100cm pendulum is moved from a location where g = 980 cm>sec2 to a new location. This increases the period by dT = 0.001 sec . Find dg and estimate the value of g at the new location. 53. The edge of a cube is measured as 10 cm with an error of 1%. The cube’s volume is to be calculated from this measurement. Estimate the percentage error in the volume calculation. 54. About how accurately should you measure the side of a square to be sure of calculating the area within 2% of its true value? 55. The diameter of a sphere is measured as 100 ; 1 cm and the volume is calculated from this measurement. Estimate the percentage error in the volume calculation. 56. Estimate the allowable percentage error in measuring the diameter D of a sphere if the volume is to be calculated correctly to within 3%. 57. (Continuation of Example 7.) Show that a 5% error in measuring t will cause about a 10% error in calculating s from the equation s = 16t 2 . 58. (Continuation of Example 8.) By what percentage should r be increased to increase V by 50%?
Theory and Examples 59. Show that the approximation of 21 + x by its linearization at the origin must improve as x : 0 by showing that lim
x:0
21 + x = 1. 1 + sx>2d
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Chapter 3: Differentiation
60. Show that the approximation of tan x by its linearization at the origin must improve as x : 0 by showing that lim
x: 0
tan x x = 1.
61. The linearization is the best linear approximation (This is why we use the linearization.) Suppose that y = ƒsxd is differentiable at x = a and that g sxd = msx  ad + c is a linear function in which m and c are constants. If the error Esxd = ƒsxd  g sxd were small enough near x = a , we might think of using g as a linear approximation of ƒ instead of the linearization Lsxd = ƒsad + ƒ¿sadsx  ad . Show that if we impose on g the conditions 1. Esad = 0
The approximation error is zero at x = a .
Esxd 2. lim x  a = 0 x:a
The error is negligible when compared with x  a .
then g sxd = ƒsad + ƒ¿sadsx  ad . Thus, the linearization L(x) gives the only linear approximation whose error is both zero at x = a and negligible in comparison with x  a . The linearization, L(x): y f (a) f '(a)(x a)
Some other linear approximation, g(x): y m(x a) c y f (x)
T 63. Reading derivatives from graphs The idea that differentiable curves flatten out when magnified can be used to estimate the values of the derivatives of functions at particular points. We magnify the curve until the portion we see looks like a straight line through the point in question, and then we use the screen’s coordinate grid to read the slope of the curve as the slope of the line it resembles. a. To see how the process works, try it first with the function y = x 2 at x = 1 . The slope you read should be 2. b. Then try it with the curve y = e x at x = 1, x = 0, and x = 1. In each case, compare your estimate of the derivative with the value of e x at the point. What pattern do you see? Test it with other values of x. Chapter 7 will explain what is going on. 64. Suppose that the graph of a differentiable function ƒ(x) has a horizontal tangent at x = a . Can anything be said about the linearization of ƒ at x = a ? Give reasons for your answer. 65. To what relative speed should a body at rest be accelerated to increase its mass by 1%? T 66. Repeated roottaking a. Enter 2 in your calculator and take successive square roots by pressing the square root key repeatedly (or raising the displayed number repeatedly to the 0.5 power). What pattern do you see emerging? Explain what is going on. What happens if you take successive tenth roots instead? b. Repeat the procedure with 0.5 in place of 2 as the original entry. What happens now? Can you use any positive number x in place of 2? Explain what is going on.
(a, f (a))
a
x
COMPUTER EXPLORATIONS 62. Quadratic approximations a. Let Qsxd = b0 + b1sx  ad + b2sx  ad2 be a quadratic approximation to ƒ(x) at x = a with the properties: i. Qsad = ƒsad
b. Find the linearization L of the function at the point a.
iii. Q–sad = ƒ–sad Determine the coefficients b0 , b1 , and b2 . b. Find the quadratic approximation to ƒsxd = 1>s1  xd at x = 0. c. Graph ƒsxd = 1>s1  xd and its quadratic approximation at x = 0 . Then zoom in on the two graphs at the point (0, 1). Comment on what you see.
T d. Find the quadratic approximation to gsxd = 1>x at x = 1 . Graph g and its quadratic approximation together. Comment on what you see. T
In Exercises 67–70, use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval I. Perform the following steps: a. Plot the function ƒ over I.
ii. Q¿sad = ƒ¿sad
T
Comparing Functions with Their Linearizations
e. Find the quadratic approximation to hsxd = 21 + x at x = 0 . Graph h and its quadratic approximation together. Comment on what you see. f. What are the linearizations of ƒ, g, and h at the respective points in parts (b), (d), and (e)?
c. Plot ƒ and L together on a single graph. d. Plot the absolute error ƒ ƒsxd  Lsxd ƒ over I and find its maximum value. e. From your graph in part (d), estimate as large a d 7 0 as you can, satisfying ƒx  aƒ 6 d
Q
ƒ ƒsxd  Lsxd ƒ 6 P
for P = 0.5, 0.1, and 0.01 . Then check graphically to see if your destimate holds true. 67. ƒsxd = x 3 + x 2  2x,
[1, 2],
a = 1
3 x  1 1 , c , 1 d, a = 4 2 4x 2 + 1 69. ƒsxd = x 2>3sx  2d, [2, 3], a = 2 68. ƒsxd =
70. ƒsxd = 1x  sin x,
[0, 2p],
a = 2
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Chapter 3
Chapter 3
Questions to Guide Your Review
235
Questions to Guide Your Review
1. What is the derivative of a function ƒ? How is its domain related to the domain of ƒ? Give examples.
13. What is the relationship between a function’s average and instantaneous rates of change? Give an example.
2. What role does the derivative play in defining slopes, tangents, and rates of change?
14. How do derivatives arise in the study of motion? What can you learn about a body’s motion along a line by examining the derivatives of the body’s position function? Give examples.
3. How can you sometimes graph the derivative of a function when all you have is a table of the function’s values? 4. What does it mean for a function to be differentiable on an open interval? On a closed interval? 5. How are derivatives and onesided derivatives related? 6. Describe geometrically when a function typically does not have a derivative at a point. 7. How is a function’s differentiability at a point related to its continuity there, if at all?
16. Give examples of still other applications of derivatives. 17. What do the limits limh:0 sssin hd>hd and limh:0 sscos h  1d>hd have to do with the derivatives of the sine and cosine functions? What are the derivatives of these functions? 18. Once you know the derivatives of sin x and cos x, how can you find the derivatives of tan x, cot x, sec x, and csc x? What are the derivatives of these functions? 19. At what points are the six basic trigonometric functions continuous? How do you know?
8. Could the unit step function Usxd = e
15. How can derivatives arise in economics?
0, 1,
x 6 0 x Ú 0
possibly be the derivative of some other function on [1, 1] ? Explain. 9. What rules do you know for calculating derivatives? Give some examples. 10. Explain how the three formulas
20. What is the rule for calculating the derivative of a composite of two differentiable functions? How is such a derivative evaluated? Give examples. 21. What is the formula for the slope dy>dx of a parametrized curve x = ƒstd, y = g std ? When does the formula apply? When can you expect to be able to find d 2y>dx 2 as well? Give examples. 22. If u is a differentiable function of x, how do you find sd>dxdsu n d if n is an integer? If n is a rational number? Give examples.
a.
d n sx d = nx n  1 dx
23. What is implicit differentiation? When do you need it? Give examples.
du d scud = c dx dx
24. How do related rates problems arise? Give examples.
b.
du1 du2 Á dun d su + u2 + Á + un d = + + + dx 1 dx dx dx enable us to differentiate any polynomial. c.
11. What formula do we need, in addition to the three listed in Question 10, to differentiate rational functions? 12. What is a second derivative? A third derivative? How many derivatives do the functions you know have? Give examples.
25. Outline a strategy for solving related rates problems. Illustrate with an example. 26. What is the linearization L(x) of a function ƒ(x) at a point x = a ? What is required of ƒ at a for the linearization to exist? How are linearizations used? Give examples. 27. If x moves from a to a nearby value a + dx , how do you estimate the corresponding change in the value of a differentiable function ƒ(x)? How do you estimate the relative change? The percentage change? Give an example.
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Chapter 3 Practice Exercises
Chapter 3
235
Practice Exercises
Derivatives of Functions
5. y = sx + 1d2sx 2 + 2xd
6. y = s2x  5ds4  xd1
Find the derivatives of the functions in Exercises 140.
7. y = su2 + sec u + 1d3
8. y = a1 
1. y = x 5  0.125x 2 + 0.25x
2. y = 3  0.7x 3 + 0.3x 7
3. y = x 3  3sx 2 + p2 d
4. y = x 7 + 27x 
1 p + 1
9. s =
1t 1 + 1t
10. s =
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1 1t  1
csc u u2  b 2 4
2
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236
Chapter 3: Differentiation 1 2 sin x sin2 x
11. y = 2 tan2 x  sec2 x
12. y =
13. s = cos4 s1  2td
2 14. s = cot3 a t b
15. s = ssec t + tan td5
16. s = csc5 s1  t + 3t 2 d
17. r = 22u sin u
18. r = 2u 2cos u
ƒ(x)
g (x)
ƒ(x)
g(x)
0 1
1 3
1 5
3 1>2
1>2 4
Find the first derivatives of the following combinations at the given value of x.
20. r = sin A u + 2u + 1 B
19. r = sin 22u
x
23. y = x 1>2 sec s2xd2
24. y = 1x csc sx + 1d3
a. 6ƒsxd  g sxd, x = 1 ƒsxd , x = 1 c. g sxd + 1
25. y = 5 cot x 2
26. y = x 2 cot 5x
e. g sƒsxdd,
2 1 2 x csc x 2
21. y =
22. y = 2 1x sin 1x
27. y = x 2 sin2 s2x 2 d 29. s = a
4t b t + 1
31. y = a 33. y =
1x 2 b 1 + x
x2 + x B x2
35. r = a
sin u b cos u  1
2 2 1x b 2 1x + 1
34. y = 4x2x + 1x 2
36. r = a
1 + sin u b 1  cos u
3 2 s5x + sin 2xd3>2
40. y = s3 + cos3 3xd1>3
x
ƒ(x)
ƒ(x)
0 1
9 3
2 1>5
Find the first derivatives of the following combinations at the given value of x. a. 1x ƒsxd,
b. 2ƒsxd,
x = 1
x = 0
d. ƒs1  5 tan xd,
x = 0
px f. 10 sin a b ƒ 2sxd, 2
x=1
57. Find the value of dy>dt at t = 0 if y = 3 sin 2x and x = t 2 + p .
In Exercises 41–48, find dy>dx. 41. xy + 2x + 3y = 1
42. x 2 + xy + y 2  5x = 2
43. x 3 + 4xy  3y 4>3 = 2x
44. 5x 4>5 + 10y 6>5 = 15
45. 1xy = 1
46. x 2y 2 = 1
x x + 1
x = 1
x = 0
c. ƒs 1xd, x = 1 ƒsxd , x = 0 e. 2 + cos x
Implicit Differentiation
47. y 2 =
x = 0
2
38. y = 20s3x  4d1>4s3x  4d1>5
37. y = s2x + 1d22x + 1 39. y =
1 15s15t  1d3
32. y = a
x = 0
56. Suppose that the function ƒ(x) and its first derivative have the following values at x = 0 and x = 1 .
2
30. s =
f. sx + ƒsxdd3>2,
d. ƒsg sxdd,
x = 0
g. ƒsx + g sxdd,
28. y = x 2 sin2 sx 3 d
b. ƒsxdg 2sxd,
58. Find the value of ds>du at u = 2 if s = t 2 + 5t and t = su 2 + 2ud1>3 .
59. Find the value of dw>ds at s = 0 if w = sin A 1r  2 B and r = 8 sin ss + p>6d . 60. Find the value of dr>dt at t = 0 if r = su2 + 7d1>3 and u2t + u = 1 .
1 + x A1  x
48. y 2 =
In Exercises 49 and 50, find dp>dq.
61. If y 3 + y = 2 cos x , find the value of d 2y>dx 2 at the point (0, 1).
49. p 3 + 4pq  3q 2 = 2
62. If x 1>3 + y 1>3 = 4 , find d 2y>dx 2 at the point (8, 8).
50. q = s5p 2 + 2pd3>2
In Exercises 51 and 52, find dr>ds. 51. r cos 2s + sin2 s = p 2
Derivative Definition
52. 2rs  r  s + s 2 = 3
In Exercises 63 and 64, find the derivative using the definition.
2
53. Find d y>dx by implicit differentiation: 3
3
a. x + y = 1 54. a. By differentiating dy>dx = x>y .
2 b. y = 1  x x 2  y 2 = 1 implicitly, 2
63. ƒstd = show
that
1 2t + 1
64. g sxd = 2x 2 + 1
65. a. Graph the function
b. Then show that d 2y>dx 2 = 1>y 3 .
ƒsxd = e
x 2, x 2,
Numerical Values of Derivatives
b. Is ƒ continuous at x = 0 ?
55. Suppose that functions ƒ(x) and g(x) and their first derivatives have the following values at x = 0 and x = 1 .
c. Is ƒ differentiable at x = 0 ?
1 … x 6 0 0 … x … 1.
Give reasons for your answers.
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Chapter 3 66. a. Graph the function ƒsxd = e
x, tan x,
1 … x 6 0 0 … x … p>4.
b. Is ƒ continuous at x = 0 ? c. Is ƒ differentiable at x = 0 ? 67. a. Graph the function x, 0 … x … 1 2  x, 1 6 x … 2.
78. Slope of tangent Show that the tangent to the curve y = x 3 at any point sa, a 3 d meets the curve again at a point where the slope is four times the slope at sa, a 3 d .
80. Normal to a circle Show that the normal line at any point of the circle x 2 + y 2 = a 2 passes through the origin.
Tangents and Normals to Implicitly Defined Curves
b. Is ƒ continuous at x = 1 ? c. Is ƒ differentiable at x = 1 ?
In Exercises 81–86, find equations for the lines that are tangent and normal to the curve at the given point.
Give reasons for your answers. 68. For what value or values of the constant m, if any, is sin 2x, ƒsxd = e mx,
77. Tangent parabola The parabola y = x 2 + C is to be tangent to the line y = x . Find C.
79. Tangent curve For what value of c is the curve y = c>sx + 1d tangent to the line through the points s0, 3d and s5, 2d ?
Give reasons for your answers. ƒsxd = e
237
Practice Exercises
x … 0 x 7 0
81. x 2 + 2y 2 = 9, 3
2
82. x + y = 2,
s1, 2d s1, 1d
83. xy + 2x  5y = 2, 2
a. continuous at x = 0 ?
84. s y  xd = 2x + 4,
b. differentiable at x = 0 ?
85. x + 1xy = 6,
Give reasons for your answers.
86. x 3>2 + 2y 3>2 = 17,
Slopes, Tangents, and Normals 69. Tangents with specified slope Are there any points on the curve y = sx>2d + 1>s2x  4d where the slope is 3>2 ? If so, find them.
s6, 2d
s4, 1d s1, 4d
87. Find the slope of the curve x 3y 3 + y 2 = x + y at the points (1, 1) and s1, 1d . 88. The graph shown suggests that the curve y = sin sx  sin xd might have horizontal tangents at the xaxis. Does it? Give reasons for your answer.
70. Tangents with specified slope Are there any points on the curve y = x  1>s2xd where the slope is 3? If so, find them. 71. Horizontal tangents Find the points on the curve y = 2x 3  3x 2  12x + 20 where the tangent is parallel to the xaxis.
s3, 2d
y
y sin (x sin x)
1
–2
0
–
72. Tangent intercepts Find the x and yintercepts of the line that is tangent to the curve y = x 3 at the point s 2, 8d .
2
x
–1
73. Tangents perpendicular or parallel to lines Find the points on the curve y = 2x 3  3x 2  12x + 20 where the tangent is a. perpendicular to the line y = 1  sx>24d .
Tangents to Parametrized Curves
b. parallel to the line y = 22  12x .
In Exercises 89 and 90, find an equation for the line in the xyplane that is tangent to the curve at the point corresponding to the given value of t. Also, find the value of d 2y>dx 2 at this point.
74. Intersecting tangents Show that the tangents to the curve y = sp sin xd>x at x = p and x = p intersect at right angles. 75. Normals parallel to a line Find the points on the curve y = tan x, p>2 6 x 6 p>2 , where the normal is parallel to the line y = x>2 . Sketch the curve and normals together, labeling each with its equation. 76. Tangent and normal lines Find equations for the tangent and normal to the curve y = 1 + cos x at the point sp>2, 1d . Sketch the curve, tangent, and normal together, labeling each with its equation.
89. x = s1>2d tan t, 90. x = 1 + 1>t 2,
y = s1>2d sec t, y = 1  3>t,
t = p>3
t = 2
Analyzing Graphs Each of the figures in Exercises 91 and 92 shows two graphs, the graph of a function y = ƒsxd together with the graph of its derivative ƒ¿sxd . Which graph is which? How do you know?
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Chapter 3: Differentiation
91.
92.
y A
y 4
2
A
Number of rabbits
2000
Initial no. rabbits 1000 Initial no. foxes 40
3 2
1
(20, 1700) B
1 –1
x
1
0
0
1
2
1000 x Number of foxes
–1 0
50
150
200
100 150 Time (days) Derivative of the rabbit population (b)
200
B –2
93. Use the following information to graph the function y = ƒsxd for 1 … x … 6 . i. The graph of ƒ is made of line segments joined end to end. ii. The graph starts at the point s 1, 2d .
100 Time (days) (a)
100 50 (20, 40)
iii. The derivative of ƒ, where defined, agrees with the step function shown here.
0 –50
y –100
y f '(x)
0
50
1 –1
1
2
3
4
5
6
x
–1
FIGURE 3.53 Rabbits and foxes in an arctic predatorprey food chain. –2
94. Repeat Exercise 93, supposing that the graph starts at s 1, 0d instead of s 1, 2d . Exercises 95 and 96 are about the graphs in Figure 3.53 (righthand column). The graphs in part (a) show the numbers of rabbits and foxes in a small arctic population. They are plotted as functions of time for 200 days. The number of rabbits increases at first, as the rabbits reproduce. But the foxes prey on rabbits and, as the number of foxes increases, the rabbit population levels off and then drops. Figure 3.53b shows the graph of the derivative of the rabbit population. We made it by plotting slopes. 95. a. What is the value of the derivative of the rabbit population in Figure 3.53 when the number of rabbits is largest? Smallest? b. What is the size of the rabbit population in Figure 3.53 when its derivative is largest? Smallest (negative value)? 96. In what units should the slopes of the rabbit and fox population curves be measured?
Trigonometric Limits 97. lim
x: 0
sin x 2x 2  x
3x  tan 7x 2x x: 0
98. lim
99. lim
r: 0
101.
sin r tan 2r
lim
u :sp>2d 
sin ssin ud u u :0
100. lim
4 tan2 u + tan u + 1 tan2 u + 5
102. lim+
1  2 cot2 u 5 cot u  7 cot u  8
103. lim
x sin x 2  2 cos x
u :0
x:0
2
104. lim
u :0
1  cos u u2
Show how to extend the functions in Exercises 105 and 106 to be continuous at the origin. 105. g sxd =
tan stan xd tan x
106. ƒsxd =
tan stan xd sin ssin xd
Related Rates 107. Right circular cylinder The total surface area S of a right circular cylinder is related to the base radius r and height h by the equation S = 2pr 2 + 2prh . a. How is dS>dt related to dr>dt if h is constant? b. How is dS>dt related to dh>dt if r is constant?
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Chapter 3
Practice Exercises
c. How is dS>dt related to dr>dt and dh>dt if neither r nor h is constant?
239
4'
d. How is dr>dt related to dh>dt if S is constant? 108. Right circular cone The lateral surface area S of a right circular cone is related to the base radius r and height h by the equation S = pr2r 2 + h 2 .
r
a. How is dS>dt related to dr>dt if h is constant?
10'
b. How is dS>dt related to dh>dt if r is constant? c. How is dS>dt related to dr>dt and dh>dt if neither r nor h is constant? 109. Circle’s changing area The radius of a circle is changing at the rate of 2>p m>sec. At what rate is the circle’s area changing when r = 10 m ? 110. Cube’s changing edges The volume of a cube is increasing at the rate of 1200 cm3>min at the instant its edges are 20 cm long. At what rate are the lengths of the edges changing at that instant? 111. Resistors connected in parallel If two resistors of R1 and R2 ohms are connected in parallel in an electric circuit to make an Rohm resistor, the value of R can be found from the equation
h Exit rate: 5 ft3 /min
116. Rotating spool As television cable is pulled from a large spool to be strung from the telephone poles along a street, it unwinds from the spool in layers of constant radius (see accompanying figure). If the truck pulling the cable moves at a steady 6 ft> sec (a touch over 4 mph), use the equation s = r u to find how fast (radians per second) the spool is turning when the layer of radius 1.2 ft is being unwound.
1 1 1 + . = R R1 R2
1.2' R1
R2 R
If R1 is decreasing at the rate of 1 ohm> sec and R2 is increasing at the rate of 0.5 ohm> sec, at what rate is R changing when R1 = 75 ohms and R2 = 50 ohms ? 112. Impedance in a series circuit The impedance Z (ohms) in a series circuit is related to the resistance R (ohms) and reactance X (ohms) by the equation Z = 2R 2 + X 2 . If R is increasing at 3 ohms> sec and X is decreasing at 2 ohms> sec, at what rate is Z changing when R = 10 ohms and X = 20 ohms ? 113. Speed of moving particle The coordinates of a particle moving in the metric xyplane are differentiable functions of time t with dx>dt = 10 m/sec and dy>dt = 5 m/sec . How fast is the particle moving away from the origin as it passes through the point s3, 4d ? 114. Motion of a particle A particle moves along the curve y = x 3>2 in the first quadrant in such a way that its distance from the origin increases at the rate of 11 units per second. Find dx>dt when x = 3 . 115. Draining a tank Water drains from the conical tank shown in the accompanying figure at the rate of 5 ft3>min . a. What is the relation between the variables h and r in the figure?
b. How fast is the water level dropping when h = 6 ft ?
117. Moving searchlight beam The figure shows a boat 1 km offshore, sweeping the shore with a searchlight. The light turns at a constant rate, du>dt = 0.6 rad/sec. a. How fast is the light moving along the shore when it reaches point A?
b. How many revolutions per minute is 0.6 rad> sec?
x
A 1 km
118. Points moving on coordinate axes Points A and B move along the x and yaxes, respectively, in such a way that the distance r (meters) along the perpendicular from the origin to the line AB remains constant. How fast is OA changing, and is it increasing, or decreasing, when OB = 2r and B is moving toward O at the rate of 0.3r m> sec?
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Chapter 3: Differentiation
Linearization
124. Controlling error
119. Find the linearizations of a. tan x at x = p>4
b. sec x at x = p>4 .
Graph the curves and linearizations together. 120. We can obtain a useful linear approximation of the function ƒsxd = 1>s1 + tan xd at x = 0 by combining the approximations 1 L 1  x 1 + x
and
tan x L x
to get 1 L 1  x. 1 + tan x Show that this result is the standard linear approximation of 1>s1 + tan xd at x = 0 . 121. Find the linearization of ƒsxd = 21 + x + sin x  0.5 at x = 0 . 122. Find the linearization of ƒsxd = 2>s1  xd + 21 + x  3.1 at x = 0 .
Differential Estimates of Change 123. Surface area of a cone Write a formula that estimates the change that occurs in the lateral surface area of a right circular cone when the height changes from h0 to h0 + dh and the radius does not change.
a. How accurately should you measure the edge of a cube to be reasonably sure of calculating the cube’s surface area with an error of no more than 2%? b. Suppose that the edge is measured with the accuracy required in part (a). About how accurately can the cube’s volume be calculated from the edge measurement? To find out, estimate the percentage error in the volume calculation that might result from using the edge measurement. 125. Compounding error The circumference of the equator of a sphere is measured as 10 cm with a possible error of 0.4 cm. This measurement is then used to calculate the radius. The radius is then used to calculate the surface area and volume of the sphere. Estimate the percentage errors in the calculated values of a. the radius. b. the surface area. c. the volume. 126. Finding height To find the height of a lamppost (see accompanying figure), you stand a 6 ft pole 20 ft from the lamp and measure the length a of its shadow, finding it to be 15 ft, give or take an inch. Calculate the height of the lamppost using the value a = 15 and estimate the possible error in the result.
h h 6 ft r
20 ft a
V 1 r 2h 3 S r 兹r 2 h 2 (Lateral surface area)
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Chapter 3: Differentiation
Chapter 3
Additional and Advanced Exercises
1. An equation like sin2 u + cos2 u = 1 is called an identity because it holds for all values of u . An equation like sin u = 0.5 is not an identity because it holds only for selected values of u , not all. If you differentiate both sides of a trigonometric identity in u with respect to u , the resulting new equation will also be an identity. Differentiate the following to show that the resulting equations hold for all u . a. sin 2u = 2 sin u cos u b. cos 2u = cos2 u  sin2 u
2. If the identity sin sx + ad = sin x cos a + cos x sin a is differentiated with respect to x, is the resulting equation also an identity? Does this principle apply to the equation x 2  2x  8 = 0 ? Explain. 3. a. Find values for the constants a, b, and c that will make ƒsxd = cos x
and
g sxd = a + bx + cx 2
satisfy the conditions ƒs0d = g s0d,
ƒ¿s0d = g¿s0d,
and
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ƒ–s0d = g–s0d .
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Chapter 3 b. Find values for b and c that will make ƒsxd = sin sx + ad
and
g sxd = b sin x + c cos x
satisfy the conditions ƒs0d = g s0d
and
ƒ¿s0d = g¿s0d .
Additional and Advanced Exercises
8. Designing a gondola The designer of a 30ftdiameter spherical hot air balloon wants to suspend the gondola 8 ft below the bottom of the balloon with cables tangent to the surface of the balloon, as shown. Two of the cables are shown running from the top edges of the gondola to their points of tangency, s 12, 9d and s12, 9d . How wide should the gondola be?
c. For the determined values of a, b, and c, what happens for the third and fourth derivatives of ƒ and g in each of parts (a) and (b)? 4. Solutions to differential equations
241
y x 2 y 2 225
a. Show that y = sin x, y = cos x , and y = a cos x + b sin x (a and b constants) all satisfy the equation y– + y = 0 . b. How would you modify the functions in part (a) to satisfy the equation y– + 4y = 0 ? Generalize this result. 5. An osculating circle Find the values of h, k, and a that make the circle sx  hd2 + s y  kd2 = a 2 tangent to the parabola y = x 2 + 1 at the point (1, 2) and that also make the second derivatives d 2y>dx 2 have the same value on both curves there. Circles like this one that are tangent to a curve and have the same second derivative as the curve at the point of tangency are called osculating circles (from the Latin osculari, meaning “to kiss”). We encounter them again in Chapter 13.
x
0 (–12, –9)
(12, –9)
Suspension cables Gondola
15 ft
8 ft Width
NOT TO SCALE
9. Pisa by parachute The photograph shows Mike McCarthy parachuting from the top of the Tower of Pisa on August 5, 1988. Make a rough sketch to show the shape of the graph of his speed during the jump.
6. Marginal revenue A bus will hold 60 people. The number x of people per trip who use the bus is related to the fare charged ( p dollars) by the law p = [3  sx>40d]2 . Write an expression for the total revenue r(x) per trip received by the bus company. What number of people per trip will make the marginal revenue dr>dx equal to zero? What is the corresponding fare? (This fare is the one that maximizes the revenue, so the bus company should probably rethink its fare policy.) 7. Industrial production a. Economists often use the expression “rate of growth” in relative rather than absolute terms. For example, let u = ƒstd be the number of people in the labor force at time t in a given industry. (We treat this function as though it were differentiable even though it is an integervalued step function.) Let y = g std be the average production per person in the labor force at time t. The total production is then y = uy . If the labor force is growing at the rate of 4% per year sdu>dt = 0.04ud and the production per worker is growing at the rate of 5% per year sdy>dt = 0.05yd , find the rate of growth of the total production, y. b. Suppose that the labor force in part (a) is decreasing at the rate of 2% per year while the production per person is increasing at the rate of 3% per year. Is the total production increasing, or is it decreasing, and at what rate?
Photograph is not available.
Mike McCarthy of London jumped from the Tower of Pisa and then opened his parachute in what he said was a world record lowlevel parachute jump of 179 ft. (Source: Boston Globe, Aug. 6, 1988.)
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Chapter 3: Differentiation
10. Motion of a particle The position at time t Ú 0 of a particle moving along a coordinate line is
have a derivative at x = 0 ? Explain. 17. a. For what values of a and b will
s = 10 cos st + p>4d .
ƒsxd = e
a. What is the particle’s starting position st = 0d ? b. What are the points farthest to the left and right of the origin reached by the particle? c. Find the particle’s velocity and acceleration at the points in part (b).
ax, ax 2  bx + 3,
be differentiable for all values of x? b. Discuss the geometry of the resulting graph of ƒ. 18. a. For what values of a and b will
d. When does the particle first reach the origin? What are its velocity, speed, and acceleration then?
g sxd = e
11. Shooting a paper clip On Earth, you can easily shoot a paper clip 64 ft straight up into the air with a rubber band. In t sec after firing, the paper clip is s = 64t  16t 2 ft above your hand.
be differentiable for all values of x?
a. How long does it take the paper clip to reach its maximum height? With what velocity does it leave your hand? b. On the moon, the same acceleration will send the paper clip to a height of s = 64t  2.6t 2 ft in t sec. About how long will it take the paper clip to reach its maximum height, and how high will it go? 12. Velocities of two particles At time t sec, the positions of two particles on a coordinate line are s1 = 3t 3  12t 2 + 18t + 5 m and s2 = t 3 + 9t 2  12t m . When do the particles have the same velocities? 13. Velocity of a particle A particle of constant mass m moves along the xaxis. Its velocity y and position x satisfy the equation 1 1 msy 2  y0 2 d = k sx0 2  x 2 d , 2 2 where k, y0 , and x0 are constants. Show that whenever y Z 0 , m
dy = kx . dt
x 6 2 x Ú 2
ax + b, ax 3 + x + 2b,
x … 1 x 7 1
b. Discuss the geometry of the resulting graph of g. 19. Odd differentiable functions Is there anything special about the derivative of an odd differentiable function of x? Give reasons for your answer. 20. Even differentiable functions Is there anything special about the derivative of an even differentiable function of x? Give reasons for your answer. 21. Suppose that the functions ƒ and g are defined throughout an open interval containing the point x0 , that ƒ is differentiable at x0 , that ƒsx0 d = 0 , and that g is continuous at x0 . Show that the product ƒg is differentiable at x0 . This process shows, for example, that although ƒ x ƒ is not differentiable at x = 0 , the product x ƒ x ƒ is differentiable at x = 0 . 22. (Continuation of Exercise 21.) Use the result of Exercise 21 to show that the following functions are differentiable at x = 0 . b. x 2>3 sin x
a. ƒ x ƒ sin x d. hsxd = e
3 x s1  cos xd c. 2
2
x sin s1>xd, x Z 0 0, x = 0
23. Is the derivative of
14. Average and instantaneous velocity a. Show that if the position x of a moving point is given by a quadratic function of t, x = At 2 + Bt + C , then the average velocity over any time interval [t1, t2] is equal to the instantaneous velocity at the midpoint of the time interval. b. What is the geometric significance of the result in part (a)? 15. Find all values of the constants m and b for which the function sin x, x 6 p y = e mx + b, x Ú p
hsxd = e
x 2 sin s1>xd, x Z 0 0, x = 0
continuous at x = 0 ? How about the derivative of k sxd = xhsxd ? Give reasons for your answers. 24. Suppose that a function ƒ satisfies the following conditions for all real values of x and y: i. ƒsx + yd = ƒsxd # ƒs yd . ii. ƒsxd = 1 + xg sxd , where limx:0 g sxd = 1 . Show that the derivative ƒ¿sxd exists at every value of x and that ƒ¿sxd = ƒsxd .
is a. continuous at x = p .
25. The generalized product rule Use mathematical induction to prove that if y = u1 u2 Á un is a finite product of differentiable functions, then y is differentiable on their common domain and
b. differentiable at x = p . 16. Does the function ƒsxd = L
1  cos x , x
x Z 0
0,
x = 0
dy dun du2 Á du1 Á u un + u1 = un + Á + u1 u2 Á un  1 . dx dx 2 dx dx
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Chapter 3 26. Leibniz’s rule for higherorder derivatives of products Leibniz’s rule for higherorder derivatives of products of differentiable functions says that a. b.
d 2suyd 2
=
du dy d 2y d 2u y + 2 + u 2 2 dx dx dx dx
=
d 2u dy du d 2y d 3y d 3u y + 3 + u + 3 dx dx 2 dx 3 dx 2 dx dx 3
dx d 3suyd
dx 3 n d suyd d n  1u dy Á d nu = + c. n n y + n dx dx dx n  1 dx nsn  1d Á sn  k + 1d d n  ku d ky + k! dx n  k dx k n
d y + Á + u n. dx The equations in parts (a) and (b) are special cases of the equation in part (c). Derive the equation in part (c) by mathematical induction, using m m m! m! a b + a b = + . k!sm  kd! sk + 1d!sm  k  1d! k k + 1 27. The period of a clock pendulum The period T of a clock pendulum (time for one full swing and back) is given by the formula T 2 = 4p2L>g , where T is measured in seconds, g = 32.2 ft>sec2 , and L, the length of the pendulum, is measured in feet. Find approximately
Additional and Advanced Exercises
243
a. the length of a clock pendulum whose period is T = 1 sec . b. the change dT in T if the pendulum in part (a) is lengthened 0.01 ft. c. the amount the clock gains or loses in a day as a result of the period’s changing by the amount dT found in part (b). 28. The melting ice cube Assume an ice cube retains its cubical shape as it melts. If we call its edge length s, its volume is V = s 3 and its surface area is 6s 2 . We assume that V and s are differentiable functions of time t. We assume also that the cube’s volume decreases at a rate that is proportional to its surface area. (This latter assumption seems reasonable enough when we think that the melting takes place at the surface: Changing the amount of surface changes the amount of ice exposed to melt.) In mathematical terms, dV = k s6s 2 d, dt
k 7 0.
The minus sign indicates that the volume is decreasing. We assume that the proportionality factor k is constant. (It probably depends on many things, such as the relative humidity of the surrounding air, the air temperature, and the incidence or absence of sunlight, to name only a few.) Assume a particular set of conditions in which the cube lost 1> 4 of its volume during the first hour, and that the volume is V0 when t = 0 . How long will it take the ice cube to melt?
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Chapter 3 Technology Application Projects
Chapter 3
243
Technology Application Projects
Mathematica/Maple Module Convergence of Secant Slopes to the Derivative Function You will visualize the secant line between successive points on a curve and observe what happens as the distance between them becomes small. The function, sample points, and secant lines are plotted on a single graph, while a second graph compares the slopes of the secant lines with the derivative function.
Mathematica/Maple Module Derivatives, Slopes, Tangent Lines, and Making Movies Parts I–III. You will visualize the derivative at a point, the linearization of a function, and the derivative of a function. You learn how to plot the function and selected tangents on the same graph. Part IV (Plotting Many Tangents) Part V (Making Movies). Parts IV and V of the module can be used to animate tangent lines as one moves along the graph of a function.
Mathematica/Maple Module Convergence of Secant Slopes to the Derivative Function You will visualize righthand and lefthand derivatives.
Mathematica/Maple Module Motion Along a Straight Line: Position : Velocity : Acceleration Observe dramatic animated visualizations of the derivative relations among the position, velocity, and acceleration functions. Figures in the text can be animated.
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Chapter
4
APPLICATIONS OF DERIVATIVES OVERVIEW This chapter studies some of the important applications of derivatives. We learn how derivatives are used to find extreme values of functions, to determine and analyze the shapes of graphs, to calculate limits of fractions whose numerators and denominators both approach zero or infinity, and to find numerically where a function equals zero. We also consider the process of recovering a function from its derivative. The key to many of these accomplishments is the Mean Value Theorem, a theorem whose corollaries provide the gateway to integral calculus in Chapter 5.
Extreme Values of Functions
4.1
This section shows how to locate and identify extreme values of a continuous function from its derivative. Once we can do this, we can solve a variety of optimization problems in which we find the optimal (best) way to do something in a given situation.
DEFINITIONS Absolute Maximum, Absolute Minimum Let ƒ be a function with domain D. Then ƒ has an absolute maximum value on D at a point c if ƒsxd … ƒscd y
and an absolute minimum value on D at c if
1
y sin x
y cos x
– 2
for all x in D
0
2
ƒsxd Ú ƒscd
for all x in D.
x
–1
FIGURE 4.1 Absolute extrema for the sine and cosine functions on [p>2, p>2] . These values can depend on the domain of a function.
Absolute maximum and minimum values are called absolute extrema (plural of the Latin extremum). Absolute extrema are also called global extrema, to distinguish them from local extrema defined below. For example, on the closed interval [p>2, p>2] the function ƒsxd = cos x takes on an absolute maximum value of 1 (once) and an absolute minimum value of 0 (twice). On the same interval, the function gsxd = sin x takes on a maximum value of 1 and a minimum value of 1 (Figure 4.1). Functions with the same defining rule can have different extrema, depending on the domain.
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4.1 Extreme Values of Functions
EXAMPLE 1
245
Exploring Absolute Extrema
The absolute extrema of the following functions on their domains can be seen in Figure 4.2.
y
y
y x2 D (–, )
y x2 D [0, 2]
2
x
2
(a) abs min only
(b) abs max and min
y
y
y x2 D (0, 2]
x
y x2 D (0, 2)
2
x
2
(c) abs max only
x
(d) no max or min
FIGURE 4.2 Graphs for Example 1.
Function rule
Domain D
Absolute extrema on D
(a) y = x 2
s  q, q d
No absolute maximum. Absolute minimum of 0 at x = 0.
(b) y = x 2
[0, 2]
Absolute maximum of 4 at x = 2. Absolute minimum of 0 at x = 0.
(c) y = x 2
(0, 2]
Absolute maximum of 4 at x = 2. No absolute minimum.
(d) y = x 2
(0, 2)
No absolute extrema.
HISTORICAL BIOGRAPHY Daniel Bernoulli (1700–1789)
The following theorem asserts that a function which is continuous at every point of a closed interval [a, b] has an absolute maximum and an absolute minimum value on the interval. We always look for these values when we graph a function.
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Chapter 4: Applications of Derivatives
THEOREM 1 The Extreme Value Theorem If ƒ is continuous on a closed interval [a, b], then ƒ attains both an absolute maximum value M and an absolute minimum value m in [a, b]. That is, there are numbers x1 and x2 in [a, b] with ƒsx1 d = m, ƒsx2 d = M, and m … ƒsxd … M for every other x in [a, b] (Figure 4.3).
(x2, M) y f (x)
y f (x)
M
M x1
a
x2
b
m
m
x
b
a
x
Maximum and minimum at endpoints
(x1, m) Maximum and minimum at interior points
y f (x) y f (x)
M
m
m a
M
x2
b
x
Maximum at interior point, minimum at endpoint
a
x1
b
x
Minimum at interior point, maximum at endpoint
FIGURE 4.3 Some possibilities for a continuous function’s maximum and minimum on a closed interval [a, b].
y No largest value 1 yx 0 x1 0
1 Smallest value
x
FIGURE 4.4 Even a single point of discontinuity can keep a function from having either a maximum or minimum value on a closed interval. The function y = e
x, 0 … x 6 1 0, x = 1
is continuous at every point of [0, 1] except x = 1 , yet its graph over [0, 1] does not have a highest point.
The proof of The Extreme Value Theorem requires a detailed knowledge of the real number system (see Appendix 4) and we will not give it here. Figure 4.3 illustrates possible locations for the absolute extrema of a continuous function on a closed interval [a, b]. As we observed for the function y = cos x, it is possible that an absolute minimum (or absolute maximum) may occur at two or more different points of the interval. The requirements in Theorem 1 that the interval be closed and finite, and that the function be continuous, are key ingredients. Without them, the conclusion of the theorem need not hold. Example 1 shows that an absolute extreme value may not exist if the interval fails to be both closed and finite. Figure 4.4 shows that the continuity requirement cannot be omitted.
Local (Relative) Extreme Values Figure 4.5 shows a graph with five points where a function has extreme values on its domain [a, b]. The function’s absolute minimum occurs at a even though at e the function’s value is
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Extreme Values of Functions
Absolute maximum No greater value of f anywhere. Also a local maximum.
Local maximum No greater value of f nearby.
Local minimum No smaller value of f nearby.
y f (x)
Absolute minimum No smaller value of f anywhere. Also a local minimum.
Local minimum No smaller value of f nearby. a
c
e
d
b
x
FIGURE 4.5 How to classify maxima and minima.
smaller than at any other point nearby. The curve rises to the left and falls to the right around c, making ƒ(c) a maximum locally. The function attains its absolute maximum at d.
DEFINITIONS Local Maximum, Local Minimum A function ƒ has a local maximum value at an interior point c of its domain if ƒsxd … ƒscd
for all x in some open interval containing c.
A function ƒ has a local minimum value at an interior point c of its domain if ƒsxd Ú ƒscd
for all x in some open interval containing c.
We can extend the definitions of local extrema to the endpoints of intervals by defining ƒ to have a local maximum or local minimum value at an endpoint c if the appropriate inequality holds for all x in some halfopen interval in its domain containing c. In Figure 4.5, the function ƒ has local maxima at c and d and local minima at a, e, and b. Local extrema are also called relative extrema. An absolute maximum is also a local maximum. Being the largest value overall, it is also the largest value in its immediate neighborhood. Hence, a list of all local maxima will automatically include the absolute maximum if there is one. Similarly, a list of all local minima will include the absolute minimum if there is one.
Finding Extrema The next theorem explains why we usually need to investigate only a few values to find a function’s extrema.
THEOREM 2 The First Derivative Theorem for Local Extreme Values If ƒ has a local maximum or minimum value at an interior point c of its domain, and if ƒ¿ is defined at c, then ƒ¿scd = 0.
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Proof To prove that ƒ¿scd is zero at a local extremum, we show first that ƒ¿scd cannot be positive and second that ƒ¿scd cannot be negative. The only number that is neither positive nor negative is zero, so that is what ƒ¿scd must be. To begin, suppose that ƒ has a local maximum value at x = c (Figure 4.6) so that ƒsxd  ƒscd … 0 for all values of x near enough to c. Since c is an interior point of ƒ’s domain, ƒ¿scd is defined by the twosided limit
Local maximum value
y f (x)
ƒsxd  ƒscd x  c . x:c lim
Secant slopes 0 (never negative)
x
Secant slopes 0 (never positive)
c
x
This means that the righthand and lefthand limits both exist at x = c and equal ƒ¿scd. When we examine these limits separately, we find that
x
FIGURE 4.6 A curve with a local maximum value. The slope at c, simultaneously the limit of nonpositive numbers and nonnegative numbers, is zero.
ƒ¿scd = lim+
ƒsxd  ƒscd … 0. x  c
ƒ¿scd = lim
ƒsxd  ƒscd Ú 0. x  c
x:c
Because sx  cd 7 0 and ƒsxd … ƒscd
(1)
Because sx  cd 6 0 and ƒsxd … ƒscd
(2)
Similarly, x:c
Together, Equations (1) and (2) imply ƒ¿scd = 0. This proves the theorem for local maximum values. To prove it for local minimum values, we simply use ƒsxd Ú ƒscd, which reverses the inequalities in Equations (1) and (2). Theorem 2 says that a function’s first derivative is always zero at an interior point where the function has a local extreme value and the derivative is defined. Hence the only places where a function ƒ can possibly have an extreme value (local or global) are 1. 2. 3.
interior points where ƒ¿ = 0, interior points where ƒ¿ is undefined, endpoints of the domain of ƒ.
The following definition helps us to summarize.
DEFINITION Critical Point An interior point of the domain of a function ƒ where ƒ¿ is zero or undefined is a critical point of ƒ.
Thus the only domain points where a function can assume extreme values are critical points and endpoints. Be careful not to misinterpret Theorem 2 because its converse is false. A differentiable function may have a critical point at x = c without having a local extreme value there. For instance, the function ƒsxd = x 3 has a critical point at the origin and zero value there, but is positive to the right of the origin and negative to the left. So it cannot have a local extreme value at the origin. Instead, it has a point of inflection there. This idea is defined and discussed further in Section 4.4. Most quests for extreme values call for finding the absolute extrema of a continuous function on a closed and finite interval. Theorem 1 assures us that such values exist; Theorem 2 tells us that they are taken on only at critical points and endpoints. Often we can Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
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249
simply list these points and calculate the corresponding function values to find what the largest and smallest values are, and where they are located.
How to Find the Absolute Extrema of a Continuous Function ƒ on a Finite Closed Interval 1. Evaluate ƒ at all critical points and endpoints. 2. Take the largest and smallest of these values.
EXAMPLE 2
Finding Absolute Extrema
Find the absolute maximum and minimum values of ƒsxd = x 2 on [2, 1]. The function is differentiable over its entire domain, so the only critical point is where ƒ¿sxd = 2x = 0, namely x = 0. We need to check the function’s values at x = 0 and at the endpoints x = 2 and x = 1:
Solution
Critical point value: ƒs0d = 0 Endpoint values: ƒs 2d = 4 ƒs1d = 1
y 7 –2
–1
(1, 7) t
1 y 8t t
4
The function has an absolute maximum value of 4 at x = 2 and an absolute minimum value of 0 at x = 0.
EXAMPLE 3
Absolute Extrema at Endpoints
Find the absolute extrema values of gstd = 8t  t 4 on [2, 1]. The function is differentiable on its entire domain, so the only critical points occur where g¿std = 0. Solving this equation gives
Solution (–2, –32)
– 32
8  4t 3 = 0 FIGURE 4.7 The extreme values of g std = 8t  t 4 on [2, 1] (Example 3).
or
3 t = 2 2 7 1,
a point not in the given domain. The function’s absolute extrema therefore occur at the endpoints, gs 2d = 32 (absolute minimum), and gs1d = 7 (absolute maximum). See Figure 4.7.
EXAMPLE 4
Finding Absolute Extrema on a Closed Interval
Find the absolute maximum and minimum values of ƒsxd = x 2>3 on the interval [2, 3]. We evaluate the function at the critical points and endpoints and take the largest and smallest of the resulting values. The first derivative
Solution
2 1>3 2 x = 3 3 32 x has no zeros but is undefined at the interior point x = 0. The values of ƒ at this one critical point and at the endpoints are ƒ¿sxd =
Critical point value: ƒs0d = 0 3 Endpoint values: ƒs 2d = s 2d2>3 = 2 4 3 2>3 ƒs3d = s3d = 29. Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
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y
3 9 L 2.08, and it We can see from this list that the function’s absolute maximum value is 2 occurs at the right endpoint x = 3. The absolute minimum value is 0, and it occurs at the interior point x = 0. (Figure 4.8).
y x 2/3, –2 ≤ x ≤ 3 Absolute maximum; also a local maximum
Local maximum 2 1 –2
–1
0
1 2 3 Absolute minimum; also a local minimum
x
FIGURE 4.8 The extreme values of ƒsxd = x 2>3 on [2, 3] occur at x = 0 and x = 3 (Example 4).
While a function’s extrema can occur only at critical points and endpoints, not every critical point or endpoint signals the presence of an extreme value. Figure 4.9 illustrates this for interior points. We complete this section with an example illustrating how the concepts we studied are used to solve a realworld optimization problem.
EXAMPLE 5
Piping Oil from a Drilling Rig to a Refinery
A drilling rig 12 mi offshore is to be connected by pipe to a refinery onshore, 20 mi straight down the coast from the rig. If underwater pipe costs $500,000 per mile and landbased pipe costs $300,000 per mile, what combination of the two will give the least expensive connection?
y
Solution
y x3 1
–1
We try a few possibilities to get a feel for the problem:
(a) Smallest amount of underwater pipe
0
1
x
–1
Rig
12
(a)
Refinery 20
y
Underwater pipe is more expensive, so we use as little as we can. We run straight to shore (12 mi) and use land pipe for 20 mi to the refinery.
1 y x1/3 –1
0
1
x
Dollar cost = 12s500,000d + 20s300,000d = 12,000,000 (b) All pipe underwater (most direct route)
–1 Rig (b)
FIGURE 4.9 Critical points without extreme values. (a) y¿ = 3x 2 is 0 at x = 0 , but y = x 3 has no extremum there. (b) y¿ = s1>3dx 2>3 is undefined at x = 0 , but y = x 1>3 has no extremum there.
兹144 + 400 12 Refinery 20
We go straight to the refinery underwater. Dollar cost = 2544 s500,000d L 11,661,900 This is less expensive than plan (a). Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
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(c) Something in between
Rig
x
12 mi
Refinery y
20 – y 20 mi
Now we introduce the length x of underwater pipe and the length y of landbased pipe as variables. The right angle opposite the rig is the key to expressing the relationship between x and y, for the Pythagorean theorem gives x 2 = 122 + s20  yd2 x = 2144 + s20  yd2 .
(3)
Only the positive root has meaning in this model. The dollar cost of the pipeline is c = 500,000x + 300,000y. To express c as a function of a single variable, we can substitute for x, using Equation (3): cs yd = 500,0002144 + s20  yd2 + 300,000y. Our goal now is to find the minimum value of c(y) on the interval 0 … y … 20. The first derivative of c(y) with respect to y according to the Chain Rule is 1 # 2s20  yds 1d + 300,000 2 2144 + s20  yd2 20  y = 500,000 + 300,000. 2144 + s20  yd2
c¿s yd = 500,000 #
Setting c¿ equal to zero gives 500,000 s20  yd = 300,0002144 + s20  yd2 5 A 20  y B = 2144 + s20  yd2 3 25 A 20  y B 2 = 144 + s20  yd2 9 16 A 20  y B 2 = 144 9 3 s20  yd = ; # 12 = ;9 4 y = 20 ; 9 y = 11
or
y = 29.
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Only y = 11 lies in the interval of interest. The values of c at this one critical point and at the endpoints are cs11d = 10,800,000 cs0d = 11,661,900 cs20d = 12,000,000 The least expensive connection costs $10,800,000, and we achieve it by running the line underwater to the point on shore 11 mi from the refinery.
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EXERCISES 4.1 Finding Extrema from Graphs
9.
In Exercises 1–6, determine from the graph whether the function has any absolute extreme values on [a, b]. Then explain how your answer is consistent with Theorem 1. y
1.
2.
10.
y
y
5
(1, 2)
2
–3
y
y h(x)
x
2 –1
y f (x) –2
x
2
0
In Exercises 1114, match the table with a graph. 0
a
c1
c2
b
x 0
y
3.
4.
a
c
x
b
11.
y
y f (x) y h(x)
13. 0
a
c
b
x
y
5.
a
0
c
y g(x)
0
a
c
b
ƒ (x)
a b c
0 0 5
x
ƒ (x)
a b c
does not exist 0 2
b
x
0
a
c
8.
y
ƒ(x)
a b c
0 0 5
x
ƒ(x)
a b c
does not exist does not exist 1.7
y g(x)
x
b
a
In Exercises 7–10, find the extreme values and where they occur. 7.
14.
x
x
y
6.
12.
x
b c
(a)
a
b
c
(b)
y 2
1 –1
1 –1
x –2
0
2
x
a
b
c
(c)
Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
a (d)
b
c
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Absolute Extrema on Finite Closed Intervals
Local Extrema and Critical Points
In Exercises 15–30, find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates.
In Exercises 45–52, find the derivative at each critical point and determine the local extreme values.
15. ƒsxd =
2 x  5, 3
2 … x … 3 4 … x … 1
16. ƒsxd = x  4, 17. ƒsxd = x 2  1,
1 … x … 2
18. ƒsxd = 4  x 2,
3 … x … 1
45. y = x 2>3sx + 2d 47. y = x24  x
46. y = x 2>3sx 2  4d 48. y = x 2 23  x
2
49. y = e
4  2x, x + 1,
51. y = e
x 2  2x + 4, x 2 + 6x  4,
52. y = •

50. y = e
x … 1 x 7 1
3  x, x 6 0 3 + 2x  x 2, x Ú 0
x … 1 x 7 1
15 1 2 1 x  x + , 4 2 4
x … 1
1 19. Fsxd =  2 , x
0.5 … x … 2
1 20. Fsxd =  x ,
2 … x … 1
In Exercises 53 and 54, give reasons for your answers.
21. hsxd = 2x,
1 … x … 8
53. Let ƒsxd = sx  2d2>3 .
3
22. hsxd = 3x 2>3,
23. g sxd = 24  x , 24. g sxd =  25  x 2,
 25 … x … 0

5p p … u … 2 6
26. ƒsud = tan u,

p p … u … 3 4
27. g sxd = csc x,
p 2p … x … 3 3
28. g sxd = sec x,
p p  … x … 3 6
29. ƒstd = 2  ƒ t ƒ , 30. ƒstd = ƒ t  5 ƒ ,
d. Repeat parts (a) and (b) for ƒsxd = sx  ad2>3 , replacing 2 by a. 54. Let ƒsxd = ƒ x 3  9x ƒ . a. Does ƒ¿s0d exist? b. Does ƒ¿s3d exist? c. Does ƒ¿s 3d exist? d. Determine all extrema of ƒ.
Optimization Applications
4 … t … 7
In Exercises 31–34, find the function’s absolute maximum and minimum values and say where they are assumed. 1 … x … 8
32. ƒsxd = x 5>3,
1 … x … 8
3>5
33. g sud = u , 34. hsud = 3u2>3,
32 … u … 1 27 … u … 8
Finding Extreme Values In Exercises 35–44, find the extreme values of the function and where they occur. 35. y = 2x 2  8x + 9 3
38. y = x 3  3x 2 + 3x  2
39. y = 2x 2  1
40. y =
1
3 2 1  x2 x 43. y = 2 x + 1
Whenever you are maximizing or minimizing a function of a single variable, we urge you to graph the function over the domain that is appropriate to the problem you are solving. The graph will provide insight before you begin to calculate and will furnish a visual context for understanding your answer. 55. Constructing a pipeline Supertankers offload oil at a docking facility 4 mi offshore. The nearest refinery is 9 mi east of the shore point nearest the docking facility. A pipeline must be constructed connecting the docking facility with the refinery. The pipeline costs $300,000 per mile if constructed underwater and $200,000 per mile if overland.
36. y = x 3  2x + 4
37. y = x + x  8x + 5
41. y =
2
x 7 1
c. Does the result in part (b) contradict the Extreme Value Theorem?
1 … t … 3
31. ƒsxd = x 4>3,
x  6x + 8x,
b. Show that the only local extreme value of ƒ occurs at x = 2 .
2 … x … 1
25. ƒsud = sin u,
2
a. Does ƒ¿s2d exist?
1 … x … 1 2
3
1
4 mi
21  x 2
42. y = 23 + 2x  x 2 44. y =
Docking Facility
x + 1 x 2 + 2x + 2
Shore A
B
Refinery
9 mi
a. Locate Point B to minimize the cost of the construction.
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b. The cost of underwater construction is expected to increase, whereas the cost of overland construction is expected to stay constant. At what cost does it become optimal to construct the pipeline directly to Point A? 56. Upgrading a highway A highway must be constructed to connect Village A with Village B. There is a rudimentary roadway that can be upgraded 50 mi south of the line connecting the two villages. The cost of upgrading the existing roadway is $300,000 per mile, whereas the cost of constructing a new highway is $500,000 per mile. Find the combination of upgrading and new construction that minimizes the cost of connecting the two villages. Clearly define the location of the proposed highway. 150 mi A
B 50 mi
50 mi
60. The function Psxd = 2x +
200 x ,
0 6 x 6 q,
models the perimeter of a rectangle of dimensions x by 100>x. a. Find any extreme values of P. b. Give an interpretation in terms of perimeter of the rectangle for any values found in part (a). 61. Area of a right triangle What is the largest possible area for a right triangle whose hypotenuse is 5 cm long? 62. Area of an athletic field An athletic field is to be built in the shape of a rectangle x units long capped by semicircular regions of radius r at the two ends. The field is to be bounded by a 400m racetrack. a. Express the area of the rectangular portion of the field as a function of x alone or r alone (your choice).
Old road
b. What values of x and r give the rectangular portion the largest possible area?
57. Locating a pumping station Two towns lie on the south side of a river. A pumping station is to be located to serve the two towns. A pipeline will be constructed from the pumping station to each of the towns along the line connecting the town and the pumping station. Locate the pumping station to minimize the amount of pipeline that must be constructed.
63. Maximum height of a vertically moving body body moving vertically is given by s = 
1 2 gt + y0 t + s0, 2
The height of a
g 7 0,
with s in meters and t in seconds. Find the body’s maximum height. 64. Peak alternating current Suppose that at any given time t (in seconds) the current i (in amperes) in an alternating current circuit is i = 2 cos t + 2 sin t . What is the peak current for this circuit (largest magnitude)?
10 mi P
2 mi
b. Interpret any values found in part (a) in terms of volume of the box.
5 mi
A
Theory and Examples B
58. Length of a guy wire One tower is 50 ft high and another tower is 30 ft high. The towers are 150 ft apart. A guy wire is to run from Point A to the top of each tower.
50' A
30'
150'
a. Locate Point A so that the total length of guy wire is minimal. b. Show in general that regardless of the height of the towers, the length of guy wire is minimized if the angles at A are equal. 59. The function
65. A minimum with no derivative The function ƒsxd = ƒ x ƒ has an absolute minimum value at x = 0 even though ƒ is not differentiable at x = 0 . Is this consistent with Theorem 2? Give reasons for your answer. 66. Even functions If an even function ƒ(x) has a local maximum value at x = c , can anything be said about the value of ƒ at x = c ? Give reasons for your answer. 67. Odd functions If an odd function g (x) has a local minimum value at x = c , can anything be said about the value of g at x = c ? Give reasons for your answer. 68. We know how to find the extreme values of a continuous function ƒ(x) by investigating its values at critical points and endpoints. But what if there are no critical points or endpoints? What happens then? Do such functions really exist? Give reasons for your answers. 69. Cubic functions Consider the cubic function ƒsxd = ax 3 + bx 2 + cx + d .
V sxd = xs10  2xds16  2xd,
0 6 x 6 5,
models the volume of a box.
a. Show that ƒ can have 0, 1, or 2 critical points. Give examples and graphs to support your argument.
a. Find the extreme values of V.
b. How many local extreme values can ƒ have?
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4.1 Extreme Values of Functions T 70. Functions with no extreme values at endpoints a. Graph the function ƒ(x) = •
1 sin x ,
x 7 0
0,
x = 0.
Explain why ƒs0d = 0 is not a local extreme value of ƒ. b. Construct a function of your own that fails to have an extreme value at a domain endpoint. T Graph the functions in Exercises 71–74. Then find the extreme values of the function on the interval and say where they occur. 71. ƒsxd = ƒ x  2 ƒ + ƒ x + 3 ƒ , 72. gsxd = ƒ x  1 ƒ  ƒ x  5 ƒ , 73. hsxd = ƒ x + 2 ƒ  ƒ x  3 ƒ , 74. ksxd = ƒ x + 1 ƒ + ƒ x  3 ƒ ,
5 … x … 5 2 … x … 7 q 6 x 6 q
a. Plot the function over the interval to see its general behavior there. b. Find the interior points where ƒ¿ = 0 . (In some exercises, you may have to use the numerical equation solver to approximate a solution.) You may want to plot ƒ¿ as well. c. Find the interior points where ƒ¿ does not exist. d. Evaluate the function at all points found in parts (b) and (c) and at the endpoints of the interval. e. Find the function’s absolute extreme values on the interval and identify where they occur. 75. ƒsxd = x 4  8x 2 + 4x + 2, 76. ƒsxd = x 4 + 4x 3  4x + 1, 77. ƒsxd = x
2>3
s3  xd,
[20>25, 64>25] [3>4, 3]
[2, 2]
78. ƒsxd = 2 + 2x  3x 2>3,
[1, 10>3]
79. ƒsxd = 2x + cos x,
[0, 2p]
80. ƒsxd = x 3>4  sin x +
1 , 2
q 6 x 6 q
COMPUTER EXPLORATIONS
255
[0, 2p]
In Exercises 75–80, you will use a CAS to help find the absolute extrema of the given function over the specified closed interval. Perform the following steps.
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4.2 The Mean Value Theorem
The Mean Value Theorem
4.2
We know that constant functions have zero derivatives, but could there be a complicated function, with many terms, the derivatives of which all cancel to give zero? What is the relationship between two functions that have identical derivatives over an interval? What we are really asking here is what functions can have a particular kind of derivative. These and many other questions we study in this chapter are answered by applying the Mean Value Theorem. To arrive at this theorem we first need Rolle’s Theorem.
y f '(c) 0 y f (x)
0
a
c
x
b
Rolle’s Theorem
(a)
Drawing the graph of a function gives strong geometric evidence that between any two points where a differentiable function crosses a horizontal line there is at least one point on the curve where the tangent is horizontal (Figure 4.10). More precisely, we have the following theorem.
y f '(c1 ) 0
0
a
c1
255
f '(c3 ) 0 y f (x)
f '(c2 ) 0
c2
c3
b
x
THEOREM 3 Rolle’s Theorem Suppose that y = ƒsxd is continuous at every point of the closed interval [a, b] and differentiable at every point of its interior (a, b). If ƒsad = ƒsbd,
(b)
FIGURE 4.10 Rolle’s Theorem says that a differentiable curve has at least one horizontal tangent between any two points where it crosses a horizontal line. It may have just one (a), or it may have more (b).
then there is at least one number c in (a, b) at which ƒ¿scd = 0.
Proof Being continuous, ƒ assumes absolute maximum and minimum values on [a, b]. These can occur only
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HISTORICAL BIOGRAPHY
1. 2. 3.
Michel Rolle (1652–1719)
at interior points where ƒ¿ is zero, at interior points where ƒ¿ does not exist, at the endpoints of the function’s domain, in this case a and b.
By hypothesis, ƒ has a derivative at every interior point. That rules out possibility (2), leaving us with interior points where ƒ¿ = 0 and with the two endpoints a and b. If either the maximum or the minimum occurs at a point c between a and b, then ƒ¿scd = 0 by Theorem 2 in Section 4.1, and we have found a point for Rolle’s theorem. If both the absolute maximum and the absolute minimum occur at the endpoints, then because ƒsad = ƒsbd it must be the case that ƒ is a constant function with ƒsxd = ƒsad = ƒsbd for every x H [a, b]. Therefore ƒ¿sxd = 0 and the point c can be taken anywhere in the interior (a, b). The hypotheses of Theorem 3 are essential. If they fail at even one point, the graph may not have a horizontal tangent (Figure 4.11). y
y
y
y f (x)
a
y f (x)
b
x
(a) Discontinuous at an endpoint of [a, b]
a
x0 b
y f(x)
x
(b) Discontinuous at an interior point of [a, b]
a
x0
b
x
(c) Continuous on [a, b] but not differentiable at an interior point
FIGURE 4.11 There may be no horizontal tangent if the hypotheses of Rolle’s Theorem do not hold. y
(– 兹3, 2兹3 )
EXAMPLE 1
3 y x 3x 3
Horizontal Tangents of a Cubic Polynomial
The polynomial function ƒsxd = –3
0
3
x
(兹3, –2兹3 )
graphed in Figure 4.12 is continuous at every point of [3, 3] and is differentiable at every point of s 3, 3d. Since ƒs 3d = ƒs3d = 0, Rolle’s Theorem says that ƒ¿ must be zero at least once in the open interval between a = 3 and b = 3. In fact, ƒ¿sxd = x 2  3 is zero twice in this interval, once at x =  23 and again at x = 23.
EXAMPLE 2
FIGURE 4.12 As predicted by Rolle’s Theorem, this curve has horizontal tangents between the points where it crosses the xaxis (Example 1).
x3  3x 3
Solution of an Equation ƒsxd = 0
Show that the equation x 3 + 3x + 1 = 0 has exactly one real solution. Solution
Let y = ƒsxd = x 3 + 3x + 1.
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4.2
is never zero (because it is always positive). Now, if there were even two points x = a and x = b where ƒ(x) was zero, Rolle’s Theorem would guarantee the existence of a point x = c in between them where ƒ¿ was zero. Therefore, ƒ has no more than one zero. It does in fact have one zero, because the Intermediate Value Theorem tells us that the graph of y = ƒsxd crosses the xaxis somewhere between x = 1 (where y = 3) and x = 0 (where y = 1). (See Figure 4.13.)
y (1, 5)
y x 3 3x 1
1 –1
0
1
257
The Mean Value Theorem
Our main use of Rolle’s Theorem is in proving the Mean Value Theorem.
x
The Mean Value Theorem The Mean Value Theorem, which was first stated by JosephLouis Lagrange, is a slanted version of Rolle’s Theorem (Figure 4.14). There is a point where the tangent is parallel to chord AB.
(–1, –3)
FIGURE 4.13 The only real zero of the polynomial y = x 3 + 3x + 1 is the one shown here where the curve crosses the xaxis between 1 and 0 (Example 2).
THEOREM 4 The Mean Value Theorem Suppose y = ƒsxd is continuous on a closed interval [a, b] and differentiable on the interval’s interior (a, b). Then there is at least one point c in (a, b) at which ƒsbd  ƒsad = ƒ¿scd. b  a
Tangent parallel to chord
y Slope f '(c)
B Slope
(2)
hsxd = ƒsxd  gsxd
a y f(x)
c
x
b
= ƒsxd  ƒsad 
y f (x)
B(b, f (b))
A(a, f(a))
b
h¿sxd = ƒ¿sxd 
ƒsbd  ƒsad b  a
Derivative of Eq. (3) p
h¿scd = ƒ¿scd 
ƒsbd  ƒsad b  a
p with x = c
0 = ƒ¿scd 
ƒsbd  ƒsad b  a
h¿scd = 0
x
ƒ¿scd = FIGURE 4.15 The graph of ƒ and the chord AB over the interval [a, b].
ƒsbd  ƒsad sx  ad. b  a
(3)
Figure 4.16 shows the graphs of ƒ, g, and h together. The function h satisfies the hypotheses of Rolle’s Theorem on [a, b]. It is continuous on [a, b] and differentiable on (a, b) because both ƒ and g are. Also, hsad = hsbd = 0 because the graphs of ƒ and g both pass through A and B. Therefore h¿scd = 0 at some point c H sa, bd. This is the point we want for Equation (1). To verify Equation (1), we differentiate both sides of Equation (3) with respect to x and then set x = c:
FIGURE 4.14 Geometrically, the Mean Value Theorem says that somewhere between A and B the curve has at least one tangent parallel to chord AB.
a
ƒsbd  ƒsad sx  ad b  a
(pointslope equation). The vertical difference between the graphs of ƒ and g at x is
A 0
Proof We picture the graph of ƒ as a curve in the plane and draw a line through the points A(a, ƒ(a)) and B(b, ƒ(b)) (see Figure 4.15). The line is the graph of the function gsxd = ƒsad +
f (b) f (a) ba
(1)
ƒsbd  ƒsad , b  a
which is what we set out to prove.
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HISTORICAL BIOGRAPHY
y f (x)
JosephLouis Lagrange (1736–1813)
B h(x)
y
y g(x)
A
y 兹1 x 2, –1 x 1 1
h(x) f (x) g(x)
a
x
b
x
FIGURE 4.16 The chord AB is the graph of the function g (x). The function hsxd = ƒsxd  g sxd gives the vertical distance between the graphs of ƒ and g at x.
y 4
–1
0
1
x
FIGURE 4.17 The function ƒsxd = 21  x 2 satisfies the hypotheses (and conclusion) of the Mean Value Theorem on [1, 1] even though ƒ is not differentiable at 1 and 1.
B(2, 4)
The hypotheses of the Mean Value Theorem do not require ƒ to be differentiable at either a or b. Continuity at a and b is enough (Figure 4.17).
y x2
The function ƒsxd = x 2 (Figure 4.18) is continuous for 0 … x … 2 and differentiable for 0 6 x 6 2. Since ƒs0d = 0 and ƒs2d = 4, the Mean Value Theorem says that at some point c in the interval, the derivative ƒ¿sxd = 2x must have the value s4  0d>s2  0d = 2. In this (exceptional) case we can identify c by solving the equation 2c = 2 to get c = 1.
EXAMPLE 3
3
2
1
(1, 1)
A Physical Interpretation 1
A(0, 0)
x
2
FIGURE 4.18 As we find in Example 3, c = 1 is where the tangent is parallel to the chord.
If we think of the number sƒsbd  ƒsadd>sb  ad as the average change in ƒ over [a, b] and ƒ¿scd as an instantaneous change, then the Mean Value Theorem says that at some interior point the instantaneous change must equal the average change over the entire interval.
EXAMPLE 4 If a car accelerating from zero takes 8 sec to go 352 ft, its average velocity for the 8sec interval is 352>8 = 44 ft>sec. At some point during the acceleration, the Mean Value Theorem says, the speedometer must read exactly 30 mph (44 ft>sec) (Figure 4.19).
s
Distance (ft)
400
s f (t) (8, 352)
320
At the beginning of the section, we asked what kind of function has a zero derivative over an interval. The first corollary of the Mean Value Theorem provides the answer.
240 160 80 0
Mathematical Consequences
At this point, the car’s speed was 30 mph. t
5 Time (sec)
FIGURE 4.19 Distance versus elapsed time for the car in Example 4.
COROLLARY 1 Functions with Zero Derivatives Are Constant If ƒ¿sxd = 0 at each point x of an open interval (a, b), then ƒsxd = C for all x H sa, bd, where C is a constant.
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The Mean Value Theorem
259
Proof We want to show that ƒ has a constant value on the interval (a, b). We do so by showing that if x1 and x2 are any two points in (a, b), then ƒsx1 d = ƒsx2 d. Numbering x1 and x2 from left to right, we have x1 6 x2 . Then ƒ satisfies the hypotheses of the Mean Value Theorem on [x1 , x2]: It is differentiable at every point of [x1, x2] and hence continuous at every point as well. Therefore, y
y x2 C
ƒsx2 d  ƒsx1 d = ƒ¿scd x2  x1
C2
at some point c between x1 and x2 . Since ƒ¿ = 0 throughout (a, b), this equation translates successively into
C1
ƒsx2 d  ƒsx1 d = 0, x2  x1
C0
C –2
1 0
and
ƒsx1 d = ƒsx2 d.
At the beginning of this section, we also asked about the relationship between two functions that have identical derivatives over an interval. The next corollary tells us that their values on the interval have a constant difference.
C –1
2
ƒsx2 d  ƒsx1 d = 0,
x
–1 –2
COROLLARY 2 Functions with the Same Derivative Differ by a Constant If ƒ¿sxd = g¿sxd at each point x in an open interval (a, b), then there exists a constant C such that ƒsxd = gsxd + C for all x H sa, bd. That is, ƒ  g is a constant on (a, b).
Proof At each point x H sa, bd the derivative of the difference function h = ƒ  g is
FIGURE 4.20 From a geometric point of view, Corollary 2 of the Mean Value Theorem says that the graphs of functions with identical derivatives on an interval can differ only by a vertical shift there. The graphs of the functions with derivative 2x are the parabolas y = x 2 + C , shown here for selected values of C.
h¿sxd = ƒ¿sxd  g¿sxd = 0. Thus, hsxd = C on (a, b) by Corollary 1. That is, ƒsxd  gsxd = C on (a, b), so ƒsxd = gsxd + C. Corollaries 1 and 2 are also true if the open interval (a, b) fails to be finite. That is, they remain true if the interval is sa, q d, s  q , bd, or s  q , q d. Corollary 2 plays an important role when we discuss antiderivatives in Section 4.8. It tells us, for instance, that since the derivative of ƒsxd = x 2 on s  q , q d is 2x, any other function with derivative 2x on s  q , q d must have the formula x 2 + C for some value of C (Figure 4.20).
EXAMPLE 5
Find the function ƒ(x) whose derivative is sin x and whose graph passes through the point (0, 2). Since ƒ(x) has the same derivative as gsxd = cos x, we know that ƒsxd = cos x + C for some constant C. The value of C can be determined from the condition that ƒs0d = 2 (the graph of ƒ passes through (0, 2)):
Solution
ƒs0d = cos s0d + C = 2,
so
C = 3.
The function is ƒsxd = cos x + 3.
Finding Velocity and Position from Acceleration Here is how to find the velocity and displacement functions of a body falling freely from rest with acceleration 9.8 m>sec2 .
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We know that y(t) is some function whose derivative is 9.8. We also know that the derivative of g std = 9.8t is 9.8. By Corollary 2, ystd = 9.8t + C for some constant C. Since the body falls from rest, ys0d = 0. Thus 9.8s0d + C = 0,
and
C = 0.
The velocity function must be ystd = 9.8t. How about the position function s(t)? We know that s(t) is some function whose derivative is 9.8t. We also know that the derivative of ƒstd = 4.9t 2 is 9.8t. By Corollary 2, sstd = 4.9t 2 + C for some constant C. If the initial height is ss0d = h, measured positive downward from the rest position, then 4.9s0d2 + C = h,
and
C = h.
The position function must be sstd = 4.9t 2 + h. The ability to find functions from their rates of change is one of the very powerful tools of calculus. As we will see, it lies at the heart of the mathematical developments in Chapter 5.
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EXERCISES 4.2 Finding c in the Mean Value Theorem
9. The function
Find the value or values of c that satisfy the equation ƒsbd  ƒsad = ƒ¿scd b  a in the conclusion of the Mean Value Theorem for the functions and intervals in Exercises 1–4. 1. ƒsxd = x 2 + 2x  1, 2. ƒsxd = x 2>3,
[0, 1]
4. ƒsxd = 2x  1,
1 c , 2d 2 [1, 3]
Checking and Using Hypotheses Which of the functions in Exercises 5–8 satisfy the hypotheses of the Mean Value Theorem on the given interval, and which do not? Give reasons for your answers. 5. ƒsxd = x 2>3,
[1, 8]
6. ƒsxd = x 4>5,
[0, 1]
7. ƒsxd = 2xs1  xd, 8. ƒsxd =
L
x, 0,
0 … x 6 1 x = 1
is zero at x = 0 and x = 1 and differentiable on (0, 1), but its derivative on (0, 1) is never zero. How can this be? Doesn’t Rolle’s Theorem say the derivative has to be zero somewhere in (0, 1)? Give reasons for your answer. 10. For what values of a, m and b does the function
[0, 1]
1 3. ƒsxd = x + x ,
ƒsxd = e
3, ƒsxd = • x 2 + 3x + a, mx + b,
x = 0 0 6 x 6 1 1 … x … 2
satisfy the hypotheses of the Mean Value Theorem on the interval [0, 2]?
Roots (Zeros) 11. a. Plot the zeros of each polynomial on a line together with the zeros of its first derivative.
[0, 1]
sin x x ,
p … x 6 0
0,
x = 0
i) y = x 2  4 ii) y = x 2 + 8x + 15 iii) y = x 3  3x 2 + 4 = sx + 1dsx  2d2 iv) y = x 3  33x 2 + 216x = xsx  9dsx  24d
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4.2 The Mean Value Theorem b. Use Rolle’s Theorem to prove that between every two zeros of x n + an  1x n  1 + Á + a1 x + a0 there lies a zero of
In Exercises 33–36, find the function with the given derivative whose graph passes through the point P. 33. ƒ¿sxd = 2x  1,
nx n  1 + sn  1dan  1x n  2 + Á + a1 . 12. Suppose that ƒ– is continuous on [a, b] and that ƒ has three zeros in the interval. Show that ƒ– has at least one zero in (a, b). Generalize this result. 13. Show that if ƒ– 7 0 throughout an interval [a, b], then ƒ¿ has at most one zero in [a, b]. What if ƒ– 6 0 throughout [a, b] instead?
261
Ps0, 0d
1 34. g¿sxd = 2 + 2x, x
Ps 1, 1d
35. r¿sud = 8  csc2 u,
p P a , 0b 4
36. r¿std = sec t tan t  1,
Ps0, 0d
14. Show that a cubic polynomial can have at most three real zeros. Show that the functions in Exercises 15–22 have exactly one zero in the given interval. 15. ƒsxd = x 4 + 3x + 1,
[2, 1]
4 + 7, x2
s  q , 0d
16. ƒsxd = x 3 +
Exercises 37–40 give the velocity y = ds>dt and initial position of a body moving along a coordinate line. Find the body’s position at time t. 37. y = 9.8t + 5, 39. y = sin pt,
17. g std = 2t + 21 + t  4, 18. g std =
Finding Position from Velocity
u 19. r sud = u + sin a b  8, 3 2
2t 2 40. y = p cos p ,
ssp2 d = 1
ss0d = 0
Finding Position from Acceleration
s 1, 1d
Exercises 41–44 give the acceleration a = d 2s>dt 2 , initial velocity, and initial position of a body moving on a coordinate line. Find the body’s position at time t.
s  q, q d
20. r sud = 2u  cos2 u + 22,
s  q, q d
s0, p>2d
22. r sud = tan u  cot u  u,
ss0.5d = 4
s0, q d
1 + 21 + t  3.1, 1  t
1 21. r sud = sec u  3 + 5, u
ss0d = 10 38. y = 32t  2,
41. a = 32,
ys0d = 20,
42. a = 9.8,
ys0d = 3,
ss0d = 5 ss0d = 0
43. a = 4 sin 2t,
ys0d = 2,
ss0d = 3
3t 9 cos p , p2
ys0d = 0,
ss0d = 1
s0, p>2d 44. a =
Finding Functions from Derivatives 23. Suppose that ƒs 1d = 3 and that ƒ¿sxd = 0 for all x. Must ƒsxd = 3 for all x? Give reasons for your answer. 24. Suppose that ƒs0d = 5 and that ƒ¿sxd = 2 for all x. Must ƒsxd = 2x + 5 for all x. Give reasons for your answer. 25. Suppose that ƒ¿sxd = 2x for all x. Find ƒ(2) if a. ƒs0d = 0
b. ƒs1d = 0
c. ƒs 2d = 3 .
26. What can be said about functions whose derivatives are constant? Give reasons for your answer. In Exercises 27–32, find all possible functions with the given derivative. 27. a. y¿ = x
b. y¿ = x 2
c. y¿ = x 3
28. a. y¿ = 2x
b. y¿ = 2x  1
c. y¿ = 3x 2 + 2x  1
1 29. a. y¿ =  2 x
1 b. y¿ = 1  2 x
1 c. y¿ = 5 + 2 x
30. a. y¿ =
1 2 2x
b. y¿ =
1 2x
c. y¿ = 4x 
Applications 45. Temperature change It took 14 sec for a mercury thermometer to rise from 19°C to 100°C when it was taken from a freezer and placed in boiling water. Show that somewhere along the way the mercury was rising at the rate of 8.5°C>sec. 46. A trucker handed in a ticket at a toll booth showing that in 2 hours she had covered 159 mi on a toll road with speed limit 65 mph. The trucker was cited for speeding. Why? 47. Classical accounts tell us that a 170oar trireme (ancient Greek or Roman warship) once covered 184 sea miles in 24 hours. Explain why at some point during this feat the trireme’s speed exceeded 7.5 knots (sea miles per hour). 48. A marathoner ran the 26.2mi New York City Marathon in 2.2 hours. Show that at least twice the marathoner was running at exactly 11 mph. 49. Show that at some instant during a 2hour automobile trip the car’s speedometer reading will equal the average speed for the trip.
1 2x
31. a. y¿ = sin 2t
t b. y¿ = cos 2
c. y¿ = sin 2t + cos
t 2
32. a. y¿ = sec2 u
b. y¿ = 2u
c. y¿ = 2u  sec2 u
50. Free fall on the moon On our moon, the acceleration of gravity is 1.6 m>sec2 . If a rock is dropped into a crevasse, how fast will it be going just before it hits bottom 30 sec later?
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Theory and Examples 51. The geometric mean of a and b The geometric mean of two positive numbers a and b is the number 2ab . Show that the value of c in the conclusion of the Mean Value Theorem for ƒsxd = 1>x on an interval of positive numbers [a, b] is c = 2ab . 52. The arithmetic mean of a and b The arithmetic mean of two numbers a and b is the number sa + bd>2 . Show that the value of c in the conclusion of the Mean Value Theorem for ƒsxd = x 2 on any interval [a, b] is c = sa + bd>2 . T 53. Graph the function ƒsxd = sin x sin sx + 2d  sin2 sx + 1d . What does the graph do? Why does the function behave this way? Give reasons for your answers. 54. Rolle’s Theorem a. Construct a polynomial ƒ(x) that has zeros at x = 2, 1, 0, 1, and 2 . b. Graph ƒ and its derivative ƒ¿ together. How is what you see related to Rolle’s Theorem? c. Do gsxd = sin x and its derivative g¿ illustrate the same phenomenon? 55. Unique solution Assume that ƒ is continuous on [a, b] and differentiable on (a, b). Also assume that ƒ(a) and ƒ(b) have opposite signs and that ƒ¿ Z 0 between a and b. Show that ƒsxd = 0 exactly once between a and b. 56. Parallel tangents Assume that ƒ and g are differentiable on [a, b] and that ƒsad = g sad and ƒsbd = g sbd . Show that there is at least one point between a and b where the tangents to the graphs of ƒ and g are parallel or the same line. Illustrate with a sketch.
57. If the graphs of two differentiable functions ƒ(x) and g(x) start at the same point in the plane and the functions have the same rate of change at every point, do the graphs have to be identical? Give reasons for your answer. 58. Show that for any numbers a and b, the inequality ƒ sin b  sin a ƒ … ƒ b  a ƒ is true. 59. Assume that ƒ is differentiable on a … x … b and that ƒsbd 6 ƒsad . Show that ƒ¿ is negative at some point between a and b. 60. Let ƒ be a function defined on an interval [a, b]. What conditions could you place on ƒ to guarantee that min ƒ¿ …
ƒsbd  ƒsad … max ƒ¿ , b  a
where min ƒ¿ and max ƒ¿ refer to the minimum and maximum values of ƒ¿ on [a, b]? Give reasons for your answers. T 61. Use the inequalities in Exercise 60 to estimate ƒs0.1d if ƒ¿sxd = 1>s1 + x 4 cos xd for 0 … x … 0.1 and ƒs0d = 1 . T 62. Use the inequalities in Exercise 60 to estimate ƒs0.1d if ƒ¿sxd = 1>s1  x 4 d for 0 … x … 0.1 and ƒs0d = 2 . 63. Let ƒ be differentiable at every value of x and suppose that ƒs1d = 1 , that ƒ¿ 6 0 on s  q , 1d , and that ƒ¿ 7 0 on s1, q d . a. Show that ƒsxd Ú 1 for all x. b. Must ƒ¿s1d = 0 ? Explain. 64. Let ƒsxd = px 2 + qx + r be a quadratic function defined on a closed interval [a, b]. Show that there is exactly one point c in (a, b) at which ƒ satisfies the conclusion of the Mean Value Theorem.
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4.3
Monotonic Functions and The First Derivative Test In sketching the graph of a differentiable function it is useful to know where it increases (rises from left to right) and where it decreases (falls from left to right) over an interval. This section defines precisely what it means for a function to be increasing or decreasing over an interval, and gives a test to determine where it increases and where it decreases. We also show how to test the critical points of a function for the presence of local extreme values.
Increasing Functions and Decreasing Functions What kinds of functions have positive derivatives or negative derivatives? The answer, provided by the Mean Value Theorem’s third corollary, is this: The only functions with positive derivatives are increasing functions; the only functions with negative derivatives are decreasing functions.
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Monotonic Functions and The First Derivative Test
263
DEFINITIONS Increasing, Decreasing Function Let ƒ be a function defined on an interval I and let x1 and x2 be any two points in I. 1. 2.
If ƒsx1 d 6 ƒsx2 d whenever x1 6 x2 , then ƒ is said to be increasing on I. If ƒsx2 d 6 ƒsx1 d whenever x1 6 x2 , then ƒ is said to be decreasing on I.
A function that is increasing or decreasing on I is called monotonic on I.
y 4
y x2
3
Function decreasing
Function increasing
2
y' 0
y' 0
1
–2
–1
0
1 y' 0
It is important to realize that the definitions of increasing and decreasing functions must be satisfied for every pair of points x1 and x2 in I with x1 6 x2 . Because of the inequality 6 comparing the function values, and not … , some books say that ƒ is strictly increasing or decreasing on I. The interval I may be finite or infinite. The function ƒsxd = x 2 decreases on s  q , 0] and increases on [0, q d as can be seen from its graph (Figure 4.21). The function ƒ is monotonic on s  q , 0] and [0, q d, but it is not monotonic on s  q , q d. Notice that on the interval s  q , 0d the tangents have negative slopes, so the first derivative is always negative there; for s0, q d the tangents have positive slopes and the first derivative is positive. The following result confirms these observations.
x
2 2
FIGURE 4.21 The function ƒsxd = x is monotonic on the intervals s  q , 0] and [0, q d , but it is not monotonic on s  q, q d.
COROLLARY 3 First Derivative Test for Monotonic Functions Suppose that ƒ is continuous on [a, b] and differentiable on (a, b). If ƒ¿sxd 7 0 at each point x H sa, bd, then ƒ is increasing on [a, b]. If ƒ¿sxd 6 0 at each point x H sa, bd, then ƒ is decreasing on [a, b].
Proof Let x1 and x2 be any two points in [a, b] with x1 6 x2 . The Mean Value Theorem applied to ƒ on [x1, x2] says that ƒsx2 d  ƒsx1 d = ƒ¿scdsx2  x1 d for some c between x1 and x2 . The sign of the righthand side of this equation is the same as the sign of ƒ¿scd because x2  x1 is positive. Therefore, ƒsx2 d 7 ƒsx1 d if ƒ¿ is positive on (a, b) and ƒsx2 d 6 ƒsx1 d if ƒ¿ is negative on (a, b). Here is how to apply the First Derivative Test to find where a function is increasing and decreasing. If a 6 b are two critical points for a function ƒ, and if ƒ¿ exists but is not zero on the interval (a, b), then ƒ¿ must be positive on (a, b) or negative there (Theorem 2, Section 3.1). One way we can determine the sign of ƒ¿ on the interval is simply by evaluating ƒ¿ for some point x in (a, b). Then we apply Corollary 3.
EXAMPLE 1
Using the First Derivative Test for Monotonic Functions
Find the critical points of ƒsxd = x 3  12x  5 and identify the intervals on which ƒ is increasing and decreasing. Solution
The function ƒ is everywhere continuous and differentiable. The first derivative ƒ¿sxd = 3x 2  12 = 3sx 2  4d = 3sx + 2dsx  2d
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is zero at x = 2 and x = 2. These critical points subdivide the domain of ƒ into intervals s  q , 2d, s 2, 2d, and s2, q d on which ƒ¿ is either positive or negative. We determine the sign of ƒ¿ by evaluating ƒ at a convenient point in each subinterval. The behavior of ƒ is determined by then applying Corollary 3 to each subinterval. The results are summarized in the following table, and the graph of ƒ is given in Figure 4.22.
y y x3 – 12x – 5 20 (–2, 11) 10
–4 –3 –2 –1 0
1
2
3
x
4
 q 6 x 6 2 ƒ¿s 3d = 15 + increasing
Intervals ƒ œ Evaluated Sign of ƒ œ Behavior of ƒ
–10
–20
2 6 x 6 2 ƒ¿s0d = 12 decreasing
2 6 x 6 q ƒ¿s3d = 15 + increasing
(2, –21)
FIGURE 4.22 The function ƒsxd = x 3  12x  5 is monotonic on three separate intervals (Example 1).
Corollary 3 is valid for infinite as well as finite intervals, and we used that fact in our analysis in Example 1. Knowing where a function increases and decreases also tells us how to test for the nature of local extreme values.
HISTORICAL BIOGRAPHY
First Derivative Test for Local Extrema
Edmund Halley (1656–1742)
In Figure 4.23, at the points where ƒ has a minimum value, ƒ¿ 6 0 immediately to the left and ƒ¿ 7 0 immediately to the right. (If the point is an endpoint, there is only one side to consider.) Thus, the function is decreasing on the left of the minimum value and it is increasing on its right. Similarly, at the points where ƒ has a maximum value, ƒ¿ 7 0 immediately to the left and ƒ¿ 6 0 immediately to the right. Thus, the function is increasing on the left of the maximum value and decreasing on its right. In summary, at a local extreme point, the sign of ƒ¿sxd changes.
Absolute max f ' undefined Local max f'0 No extreme f'0 f'0
y f(x)
No extreme f'0
f'0
f'0
f'0
f'0 Local min
Local min f'0
f'0 Absolute min a
FIGURE 4.23
c1
c2
c3
c4
c5
b
x
A function’s first derivative tells how the graph rises and falls.
These observations lead to a test for the presence and nature of local extreme values of differentiable functions.
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Monotonic Functions and The First Derivative Test
265
First Derivative Test for Local Extrema Suppose that c is a critical point of a continuous function ƒ, and that ƒ is differentiable at every point in some interval containing c except possibly at c itself. Moving across c from left to right, 1. 2. 3.
if ƒ¿ changes from negative to positive at c, then ƒ has a local minimum at c; if ƒ¿ changes from positive to negative at c, then ƒ has a local maximum at c; if ƒ¿ does not change sign at c (that is, ƒ¿ is positive on both sides of c or negative on both sides), then ƒ has no local extremum at c.
The test for local extrema at endpoints is similar, but there is only one side to consider. Proof Part (1). Since the sign of ƒ¿ changes from negative to positive at c, these are numbers a and b such that ƒ¿ 6 0 on (a, c) and ƒ¿ 7 0 on (c, b). If x H sa, cd, then ƒscd 6 ƒsxd because ƒ¿ 6 0 implies that ƒ is decreasing on [a, c]. If x H sc, bd, then ƒscd 6 ƒsxd because ƒ¿ 7 0 implies that ƒ is increasing on [c, b]. Therefore, ƒsxd Ú ƒscd for every x H sa, bd. By definition, ƒ has a local minimum at c. Parts (2) and (3) are proved similarly.
EXAMPLE 2
Using the First Derivative Test for Local Extrema
Find the critical points of ƒsxd = x 1>3sx  4d = x 4>3  4x 1>3 . Identify the intervals on which ƒ is increasing and decreasing. Find the function’s local and absolute extreme values. Solution The function ƒ is continuous at all x since it is the product of two continuous functions, x 1>3 and sx  4d. The first derivative
ƒ¿sxd = =
d 4>3 4 4 x  4x 1>3R = x 1>3  x 2>3 3 3 dx Q 4sx  1d 4 2>3 x Qx  1R = 3 3x 2>3
is zero at x = 1 and undefined at x = 0. There are no endpoints in the domain, so the critical points x = 0 and x = 1 are the only places where ƒ might have an extreme value. The critical points partition the xaxis into intervals on which ƒ¿ is either positive or negative. The sign pattern of ƒ¿ reveals the behavior of ƒ between and at the critical points. We can display the information in a table like the following:
Intervals Sign of ƒ¿ Behavior of ƒ
x 6 0 decreasing
0 6 x 6 1 decreasing
x 7 1 + increasing
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y 4 y x1/3(x 4) 2 1 –1 0 –1
1
2
3
4
x
Corollary 3 to the Mean Value Theorem tells us that ƒ decreases on s  q , 0d, decreases on (0, 1), and increases on s1, q d. The First Derivative Test for Local Extrema tells us that ƒ does not have an extreme value at x = 0 (ƒ¿ does not change sign) and that ƒ has a local minimum at x = 1 (ƒ¿ changes from negative to positive). The value of the local minimum is ƒs1d = 11>3s1  4d = 3. This is also an absolute minimum because the function’s values fall toward it from the left and rise away from it on the right. Figure 4.24 shows this value in relation to the function’s graph. Note that limx:0 ƒ¿sxd =  q , so the graph of ƒ has a vertical tangent at the origin.
–2 –3
(1, 3)
FIGURE 4.24 The function ƒsxd = x 1/3 sx  4d decreases when x 6 1 and increases when x 7 1 (Example 2).
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EXERCISES 4.3 Analyzing ƒ Given ƒ
17. ƒsxd = x 4  8x 2 + 16
Answer the following questions about the functions whose derivatives are given in Exercises 1–8:
19. Hstd =
b. On what intervals is ƒ increasing or decreasing?
2. ƒ¿sxd = sx  1dsx + 2d
3. ƒ¿sxd = sx  1d sx + 2d
2
x2  3 , x  2
25. ƒsxd = x 1>3sx + 8d 27. hsxd = x 1>3sx 2  4d
28. k sxd = x 2>3sx 2  4d
2
Extreme Values on HalfOpen Intervals In Exercises 29–36:
8. ƒ¿sxd = x 1>2sx  3d
a. Identify the function’s local extreme values in the given domain, and say where they are assumed.
Extremes of Given Functions
b. Which of the extreme values, if any, are absolute?
In Exercises 9–28: a. Find the intervals on which the function is increasing and decreasing. b. Then identify the function’s local extreme values, if any, saying where they are taken on. c. Which, if any, of the extreme values are absolute?
T
c. Support your findings with a graphing calculator or computer grapher. 29. ƒsxd = 2x  x 2,  q 6 x … 2 30. ƒsxd = sx + 1d2,
q 6 x … 0
2
31. g sxd = x  4x + 4, 2
d. Support your findings with a graphing calculator or computer grapher.
32. g sxd = x  6x  9,
9. g std = t 2  3t + 3
34. ƒstd = t 3  3t 2,
3
11. hsxd = x + 2x 2
24. ƒsxd =
x Z 2
4. ƒ¿sxd = sx  1d sx + 2d
6. ƒ¿sxd = sx  7dsx + 1dsx + 5d
T
22. g sxd = x 2 25  x
2
5. ƒ¿sxd = sx  1dsx + 2dsx  3d 7. ƒ¿sxd = x 1>3sx + 2d
20. Kstd = 15t 3  t 5
x3 3x + 1 26. g sxd = x 2>3sx + 5d
23. ƒsxd =
c. At what points, if any, does ƒ assume local maximum and minimum values? 2
3 4 t  t6 2
21. g sxd = x28  x 2
a. What are the critical points of ƒ?
1. ƒ¿sxd = xsx  1d
18. g sxd = x 4  4x 3 + 4x 2
2
3
10. g std = 3t 2 + 9t + 5
3
33. ƒstd = 12t  t ,
3
12. hsxd = 2x  18x 3
13. ƒsud = 3u  4u
14. ƒsud = 6u  u
15. ƒsrd = 3r 3 + 16r
16. hsrd = sr + 7d3
35. hsxd =
1 … x 6 q
4 … x 6 q 3 … t 6 q q 6 t … 3
x3  2x 2 + 4x, 3
0 … x 6 q
36. k sxd = x 3 + 3x 2 + 3x + 1,
q 6 x … 0
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4.3 Monotonic Functions and The First Derivative Test
Graphing Calculator or Computer Grapher
c. ƒ¿sxd 7 0 for x Z 1 ;
In Exercises 37–40:
d. ƒ¿sxd 6 0 for x Z 1 .
a. Find the local extrema of each function on the given interval, and say where they are assumed. T b. Graph the function and its derivative together. Comment on the behavior of ƒ in relation to the signs and values of ƒ¿ . x x 37. ƒsxd =  2 sin , 2 2
0 … x … 2p
38. ƒsxd = 2 cos x  cos2 x, 2
267
44. Sketch the graph of a differentiable function y = ƒsxd that has a. a local minimum at (1, 1) and a local maximum at (3, 3); b. a local maximum at (1, 1) and a local minimum at (3, 3); c. local maxima at (1, 1) and (3, 3); d. local minima at (1, 1) and (3, 3). 45. Sketch the graph of a continuous function y = g sxd such that
p … x … p
39. ƒsxd = csc x  2 cot x,
0 6 x 6 p
40. ƒsxd = sec2 x  2 tan x,
p p 6 x 6 2 2
Theory and Examples Show that the functions in Exercises 41 and 42 have local extreme values at the given values of u , and say which kind of local extreme the function has. 41. hsud = 3 cos
u , 2
0 … u … 2p,
42. hsud = 5 sin
u , 2
0 … u … p,
at u = 0 and u = 2p at u = 0 and u = p
43. Sketch the graph of a differentiable function y = ƒsxd through the point (1, 1) if ƒ¿s1d = 0 and a. ƒ¿sxd 7 0 for x 6 1 and ƒ¿sxd 6 0 for x 7 1 ;
a. g s2d = 2, 0 6 g¿ 6 1 for x 6 2, g¿sxd : 1 as x : 2, 1 6 g¿ 6 0 for x 7 2, and g¿sxd : 1+ as x : 2+ ; b. g s2d = 2, g¿ 6 0 for x 6 2, g¿sxd :  q as x : 2, g¿ 7 0 for x 7 2, and g¿sxd : q as x : 2+ . 46. Sketch the graph of a continuous function y = hsxd such that a. hs0d = 0, 2 … hsxd … 2 for all x, h¿sxd : q as x : 0 , and h¿sxd : q as x : 0 + ; b. hs0d = 0, 2 … hsxd … 0 for all x, h¿sxd : q as x : 0 , and h¿sxd :  q as x : 0 + . 47. As x moves from left to right through the point c = 2 , is the graph of ƒsxd = x 3  3x + 2 rising, or is it falling? Give reasons for your answer. 48. Find the intervals on which the function ƒsxd = ax 2 + bx + c, a Z 0 , is increasing and decreasing. Describe the reasoning behind your answer.
b. ƒ¿sxd 6 0 for x 6 1 and ƒ¿sxd 7 0 for x 7 1 ;
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4.4 Concavity and Curve Sketching
Concavity and Curve Sketching
4.4
In Section 4.3 we saw how the first derivative tells us where a function is increasing and where it is decreasing. At a critical point of a differentiable function, the First Derivative Test tells us whether there is a local maximum or a local minimum, or whether the graph just continues to rise or fall there. In this section we see how the second derivative gives information about the way the graph of a differentiable function bends or turns. This additional information enables us to capture key aspects of the behavior of a function and its graph, and then present these features in a sketch of the graph.
y
CA
VE
UP
y x3
CO
N W
0
N
f ' increases x
Concavity
CO NC AV E
DO
f ' decreases
267
FIGURE 4.25 The graph of ƒsxd = x 3 is concave down on s  q , 0d and concave up on s0, q d (Example 1a).
As you can see in Figure 4.25, the curve y = x 3 rises as x increases, but the portions defined on the intervals s  q , 0d and s0, q d turn in different ways. As we approach the origin from the left along the curve, the curve turns to our right and falls below its tangents. The slopes of the tangents are decreasing on the interval s  q , 0d. As we move away from the origin along the curve to the right, the curve turns to our left and rises above its tangents. The slopes of the tangents are increasing on the interval s0, q d. This turning or bending behavior defines the concavity of the curve.
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DEFINITION Concave Up, Concave Down The graph of a differentiable function y = ƒsxd is (a) concave up on an open interval I if ƒ¿ is increasing on I (b) concave down on an open interval I if ƒ¿ is decreasing on I.
If y = ƒsxd has a second derivative, we can apply Corollary 3 of the Mean Value Theorem to conclude that ƒ¿ increases if ƒ– 7 0 on I, and decreases if ƒ– 6 0.
The Second Derivative Test for Concavity Let y = ƒsxd be twicedifferentiable on an interval I. 1. 2.
If y = ƒsxd is twicedifferentiable, we will use the notations ƒ– and y– interchangeably when denoting the second derivative.
y 4
y x2
EXAMPLE 1
3
P
–2
–1
P
EU AV CO
EU
y'' 0
1
y'' 0
0
1
x
2
Applying the Concavity Test
(a) The curve y = x 3 (Figure 4.25) is concave down on s  q , 0d where y– = 6x 6 0 and concave up on s0, q d where y– = 6x 7 0. (b) The curve y = x 2 (Figure 4.26) is concave up on s  q , q d because its second derivative y– = 2 is always positive.
NC
CA V CON
2
If ƒ– 7 0 on I, the graph of ƒ over I is concave up. If ƒ– 6 0 on I, the graph of ƒ over I is concave down.
EXAMPLE 2
Determining Concavity
Determine the concavity of y = 3 + sin x on [0, 2p]. FIGURE 4.26 The graph of ƒsxd = x 2 is concave up on every interval (Example 1b).
The graph of y = 3 + sin x is concave down on s0, pd, where y– = sin x is negative. It is concave up on sp, 2pd, where y– = sin x is positive (Figure 4.27).
Solution
Points of Inflection The curve y = 3 + sin x in Example 2 changes concavity at the point sp, 3d. We call sp, 3d a point of inflection of the curve.
y 4
y 3 sin x
3 2 1 0 –1 –2
2
x
DEFINITION Point of Inflection A point where the graph of a function has a tangent line and where the concavity changes is a point of inflection.
y'' – sin x
FIGURE 4.27 Using the graph of y– to determine the concavity of y (Example 2).
A point on a curve where y– is positive on one side and negative on the other is a point of inflection. At such a point, y– is either zero (because derivatives have the Intermediate Value Property) or undefined. If y is a twicedifferentiable function, y– = 0 at a point of inflection and y¿ has a local maximum or minimum.
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4.4
EXAMPLE 3
y y x4
1
EXAMPLE 4
y'' 0 0
1
y
0
An Inflection Point May Not Exist Where y– = 0
An Inflection Point May Occur Where y– Does Not Exist
The curve y = x 1>3 has a point of inflection at x = 0 (Figure 4.29), but y– does not exist there.
x
FIGURE 4.28 The graph of y = x 4 has no inflection point at the origin, even though y– = 0 there (Example 3).
y'' does not exist.
269
The curve y = x 4 has no inflection point at x = 0 (Figure 4.28). Even though y– = 12x 2 is zero there, it does not change sign.
2
–1
Concavity and Curve Sketching
y x1/3 x
y– =
We see from Example 3 that a zero second derivative does not always produce a point of inflection. From Example 4, we see that inflection points can also occur where there is no second derivative. To study the motion of a body moving along a line as a function of time, we often are interested in knowing when the body’s acceleration, given by the second derivative, is positive or negative. The points of inflection on the graph of the body’s position function reveal where the acceleration changes sign.
EXAMPLE 5 FIGURE 4.29 A point where y– fails to exist can be a point of inflection (Example 4).
d2 d 1 2>3 2 ax 1>3 b = a x b =  x 5>3 . 9 dx 3 dx 2
Studying Motion Along a Line
A particle is moving along a horizontal line with position function sstd = 2t 3  14t 2 + 22t  5,
t Ú 0.
Find the velocity and acceleration, and describe the motion of the particle. Solution
The velocity is ystd = s¿std = 6t 2  28t + 22 = 2st  1ds3t  11d,
and the acceleration is astd = y¿std = s–std = 12t  28 = 4s3t  7d. When the function s(t) is increasing, the particle is moving to the right; when s(t) is decreasing, the particle is moving to the left. Notice that the first derivative sy = s¿d is zero when t = 1 and t = 11>3. Intervals Sign of Y s œ Behavior of s Particle motion
0 6 t 6 1 + increasing right
1 6 t 6 11>3 decreasing left
11>3 6 t + increasing right
The particle is moving to the right in the time intervals [0, 1) and s11>3, q d, and moving to the left in (1, 11>3). It is momentarily stationary (at rest), at t = 1 and t = 11>3. The acceleration astd = s–std = 4s3t  7d is zero when t = 7>3. Intervals Sign of a s ﬂ Graph of s
0 6 t 6 7>3 concave down
7>3 6 t + concave up
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The accelerating force is directed toward the left during the time interval [0, 7>3], is momentarily zero at t = 7>3, and is directed toward the right thereafter.
Second Derivative Test for Local Extrema Instead of looking for sign changes in ƒ¿ at critical points, we can sometimes use the following test to determine the presence and character of local extrema.
THEOREM 5 Second Derivative Test for Local Extrema Suppose ƒ– is continuous on an open interval that contains x = c. 1. 2. 3.
f ' 0, f '' 0 ⇒ local max
f ' 0, f '' 0 ⇒ local min
If ƒ¿scd = 0 and ƒ–scd 6 0, then ƒ has a local maximum at x = c. If ƒ¿scd = 0 and ƒ–scd 7 0, then ƒ has a local minimum at x = c. If ƒ¿scd = 0 and ƒ–scd = 0, then the test fails. The function ƒ may have a local maximum, a local minimum, or neither.
Proof Part (1). If ƒ–scd 6 0, then ƒ–sxd 6 0 on some open interval I containing the point c, since ƒ– is continuous. Therefore, ƒ¿ is decreasing on I. Since ƒ¿scd = 0, the sign of ƒ¿ changes from positive to negative at c so ƒ has a local maximum at c by the First Derivative Test. The proof of Part (2) is similar. For Part (3), consider the three functions y = x 4, y = x 4 , and y = x 3 . For each function, the first and second derivatives are zero at x = 0. Yet the function y = x 4 has a local minimum there, y = x 4 has a local maximum, and y = x 3 is increasing in any open interval containing x = 0 (having neither a maximum nor a minimum there). Thus the test fails. This test requires us to know ƒ– only at c itself and not in an interval about c. This makes the test easy to apply. That’s the good news. The bad news is that the test is inconclusive if ƒ– = 0 or if ƒ– does not exist at x = c. When this happens, use the First Derivative Test for local extreme values. Together ƒ¿ and ƒ– tell us the shape of the function’s graph, that is, where the critical points are located and what happens at a critical point, where the function is increasing and where it is decreasing, and how the curve is turning or bending as defined by its concavity. We use this information to sketch a graph of the function that captures its key features.
EXAMPLE 6
Using ƒ¿ and ƒ– to Graph ƒ
Sketch a graph of the function ƒsxd = x 4  4x 3 + 10 using the following steps. (a) (b) (c) (d)
Identify where the extrema of ƒ occur. Find the intervals on which ƒ is increasing and the intervals on which ƒ is decreasing. Find where the graph of ƒ is concave up and where it is concave down. Sketch the general shape of the graph for ƒ.
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Concavity and Curve Sketching
271
(e) Plot some specific points, such as local maximum and minimum points, points of inflection, and intercepts. Then sketch the curve. ƒ is continuous since ƒ¿sxd = 4x 3  12x 2 exists. The domain of ƒ is q q s  , d, and the domain of ƒ¿ is also s  q , q d. Thus, the critical points of ƒ occur only at the zeros of ƒ¿ . Since
Solution
ƒ¿sxd = 4x 3  12x 2 = 4x 2sx  3d the first derivative is zero at x = 0 and x = 3. Intervals Sign of ƒ œ Behavior of ƒ
x 6 0 decreasing
0 6 x 6 3 decreasing
3 6 x + increasing
(a) Using the First Derivative Test for local extrema and the table above, we see that there is no extremum at x = 0 and a local minimum at x = 3. (b) Using the table above, we see that ƒ is decreasing on s  q , 0] and [0, 3], and increasing on [3, q d. (c) ƒ–sxd = 12x 2  24x = 12xsx  2d is zero at x = 0 and x = 2. Intervals Sign of ƒ œ Behavior of ƒ
x 6 0 + concave up
0 6 x 6 2 concave down
2 6 x + concave up
We see that ƒ is concave up on the intervals s  q , 0d and s2, q d, and concave down on (0, 2). (d) Summarizing the information in the two tables above, we obtain xx 2 d + 1
.
Expanding numerator
Dividing by x 2
We see that ƒsxd : 1+ as x : q and that ƒsxd : 1 as x :  q . Thus, the line y = 1 is a horizontal asymptote. Since ƒ decreases on s  q , 1d and then increases on s 1, 1d, we know that ƒs 1d = 0 is a local minimum. Although ƒ decreases on s1, q d, it never crosses the horizontal asymptote y = 1 on that interval (it approaches the asymptote from above). So the graph never becomes negative, and ƒs 1d = 0 is an absolute minimum as well. Likewise, ƒs1d = 2 is an absolute maximum because the graph never crosses the asymptote y = 1 on the interval s  q , 1d, approaching it from below. Therefore, there are no vertical asymptotes (the range of ƒ is 0 … y … 2). The graph of ƒ is sketched in Figure 4.31. Notice how the graph is concave down as it approaches the horizontal asymptote y = 1 as x :  q , and concave up in its approach to y = 1 as x : q .
Learning About Functions from Derivatives As we saw in Examples 6 and 7, we can learn almost everything we need to know about a twicedifferentiable function y = ƒsxd by examining its first derivative. We can find where the function’s graph rises and falls and where any local extrema are assumed. We can differentiate y¿ to learn how the graph bends as it passes over the intervals of rise and fall. We can determine the shape of the function’s graph. Information we cannot get from the derivative is how to place the graph in the xyplane. But, as we discovered in Section 4.2, the only additional information we need to position the graph is the value of ƒ at one point. The derivative does not give us information about the asymptotes, which are found using limits (Sections 2.4 and 2.5).
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y f (x)
y f (x)
Differentiable ⇒ smooth, connected; graph may rise and fall
y' 0 ⇒ rises from left to right; may be wavy
or
y'' 0 ⇒ concave up throughout; no waves; graph may rise or fall
y f (x)
y' 0 ⇒ falls from left to right; may be wavy
or
y'' 0 ⇒ concave down throughout; no waves; graph may rise or fall
y'' changes sign Inflection point
or y' changes sign ⇒ graph has local maximum or local minimum
y' 0 and y'' 0 at a point; graph has local maximum
y' 0 and y'' 0 at a point; graph has local minimum
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EXERCISES 4.4 Analyzing Graphed Functions
3.
Identify the inflection points and local maxima and minima of the functions graphed in Exercises 1–8. Identify the intervals on which the functions are concave up and concave down. 1.
3 2 y x x 2x 1 3 2 3 y
0
2.
4 y x 2x2 4 4 y
4.
y 3 (x 2 1)2/3 4 y
0
y 9 x1/3(x 2 7) 14 y
x
x
0
x
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0
x
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4.4 Concavity and Curve Sketching 5.
6. y tan x 4x, – x
y x sin 2x, – 2 x 2 3 3 y
– 2 3
2 3
0
y
x
2
2
Sketching the General Shape Knowing y Each of Exercises 41–62 gives the first derivative of a continuous function y = ƒsxd . Find y– and then use steps 2–4 of the graphing procedure on page 272 to sketch the general shape of the graph of ƒ. 41. y¿ = 2 + x  x 2
x
0
42. y¿ = x 2  x  6
2
44. y¿ = x 2s2  xd
43. y¿ = xsx  3d
45. y¿ = xsx 2  12d
46. y¿ = sx  1d2s2x + 3d
2
7. y sin x , – 2 x 2
8.
y
x
0
47. y¿ = s8x  5x ds4  xd 3 y 2 cos x 兹2 x, – x 2 y
–
3 2
0
x
NOT TO SCALE
Graphing Equations Use the steps of the graphing procedure on page 272 to graph the equations in Exercises 9–40. Include the coordinates of any local extreme points and inflection points. 9. y = x 2  4x + 3
10. y = 6  2x  x 2 12. y = xs6  2xd2
13. y = 2x 3 + 6x 2  3
14. y = 1  9x  6x 2  x 3 3
16. y = 1  sx + 1d
15. y = sx  2d + 1 17. y = x 4  2x 2 = x 2sx 2  2d
18. y = x 4 + 6x 2  4 = x 2s6  x 2 d  4 19. y = 4x 3  x 4 = x 3s4  xd 4
x  5b 2
4
23. y = x + sin x,
0 … x … 2p
24. y = x  sin x,
0 … x … 2p
25. y = x 1>5
50. y¿ = tan x,

p p 6 x 6 2 2
u 51. y¿ = cot , 2
u 0 6 u 6 2p 52. y¿ = csc2 , 2 
54. y¿ = 1  cot2 u,
0 6 u 6 p
55. y¿ = cos t,
0 … t … 2p
56. y¿ = sin t,
0 … t … 2p 58. y¿ = sx  2d1>3
57. y¿ = sx + 1d2>3 sx  1d
60. y¿ = x 4>5sx + 1d
2>3
61. y¿ = 2 ƒ x ƒ = e 62. y¿ = e
0 6 u 6 2p
p p 6 u 6 2 2
53. y¿ = tan2 u  1,
2x, x … 0 2x, x 7 0
x 2, x … 0 x 2, x 7 0
63.
y
64.
y y f '(x)
y f'(x)
P
26. y = x 3>5
27. y = x 2>5
28. y = x 4>5 2>3
30. y = 5x 2>5  2x
5  xb 2
32. y = x 2>3sx  5d
29. y = 2x  3x 31. y = x 2>3 a
p p 6x6 2 2
Each of Exercises 63–66 shows the graphs of the first and second derivatives of a function y = ƒsxd . Copy the picture and add to it a sketch of the approximate graph of ƒ, given that the graph passes through the point P.
4
21. y = x  5x = x sx  5d 22. y = x a

Sketching y from Graphs of y and y
20. y = x 4 + 2x 3 = x 3sx + 2d 5
48. y¿ = sx 2  2xdsx  5d2
49. y¿ = sec2 x,
59. y¿ = x
11. y = x 3  3x + 3 3
2
33. y = x28  x 2
34. y = s2  x 2 d3>2
2
x  3 , 35. y = x  2
x Z 2
37. y = ƒ x 2  1 ƒ 39. y = 2 ƒ x ƒ = e
2 x, 2x,
x3 36. y = 2 3x + 1 38. y = ƒ x 2  2x ƒ x … 0 x 7 0
x
x P
y f ''(x)
65.
y P
y f '(x)
x
0 y f ''(x)
40. y = 2 ƒ x  4 ƒ
Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
y f''(x)
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276
Motion Along a Line The graphs in Exercises 71 and 72 show the position s = ƒstd of a body moving back and forth on a coordinate line. (a) When is the body moving away from the origin? Toward the origin? At approximately what times is the (b) velocity equal to zero? (c) Acceleration equal to zero? (d) When is the acceleration positive? Negative?
y y f '(x) x
0 y f ''(x)
71.
P
s Displacement
66.
Chapter 4: Applications of Derivatives
Theory and Examples 67. The accompanying figure shows a portion of the graph of a twicedifferentiable function y = ƒsxd . At each of the five labeled points, classify y¿ and y– as positive, negative, or zero.
s f (t)
0
5
10 Time (sec)
y S
y f (x)
Displacement
T
Q
x
0
0
68. Sketch a smooth connected curve y = ƒsxd with ƒs 2d = 8,
ƒ¿sxd 6 0
for
ƒs2d = 0,
ƒ–sxd 6 0
for
ƒ x ƒ 6 2, x 6 0,
ƒ–sxd 7 0
for
x 7 0.
for
ƒ x ƒ 7 2,
69. Sketch the graph of a twicedifferentiable function y = ƒsxd with the following properties. Label coordinates where possible. x
s f (t)
5
10 Time (sec)
ƒ¿s2d = ƒ¿s 2d = 0,
ƒs0d = 4, ƒ¿sxd 7 0
y
t
s
72. R
P
15
15
t
73. Marginal cost The accompanying graph shows the hypothetical cost c = ƒsxd of manufacturing x items. At approximately what production level does the marginal cost change from decreasing to increasing? c
Derivatives
2
1
2 6 x 6 4 4
4
4 6 x 6 6 6
7
x 7 6
y¿ 6 0,
y– 7 0
y¿ = 0,
y– 7 0
y¿ 7 0,
y– 7 0
y¿ 7 0,
y– = 0
y¿ 7 0,
y– 6 0
y¿ = 0,
y– 6 0
y¿ 6 0,
y– 6 0
Cost
c f (x)
x 6 2
20 40 60 80 100 120 Thousands of units produced
74. The accompanying graph shows the monthly revenue of the Widget Corporation for the last 12 years. During approximately what time intervals was the marginal revenue increasing? decreasing?
70. Sketch the graph of a twicedifferentiable function y = ƒsxd that passes through the points s 2, 2d, s 1, 1d, s0, 0d, s1, 1d and (2, 2) and whose first two derivatives have the following sign patterns: y¿:
+
y–:


+ 0
2 + 1
y y r(t)
2
1
x
0
5
10
Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
t
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4.4 Concavity and Curve Sketching 75. Suppose the derivative of the function y = ƒsxd is y¿ = sx  1d2sx  2d . At what points, if any, does the graph of ƒ have a local minimum, local maximum, or point of inflection? (Hint: Draw the sign pattern for y¿ .) 76. Suppose the derivative of the function y = ƒsxd is y¿ = sx  1d2sx  2dsx  4d .
277
xaxis related to the graph of the function? In what other ways are the graphs of the derivatives related to the graph of the function? 85. y = x 5  5x 4  240
86. y = x 3  12x 2
87. y =
4 5 x + 16x 2  25 5
88. y =
x4 x3  4x 2 + 12x + 20 4 3
At what points, if any, does the graph of ƒ have a local minimum, local maximum, or point of inflection?
89. Graph ƒsxd = 2x 4  4x 2 + 1 and its first two derivatives together. Comment on the behavior of ƒ in relation to the signs and values of ƒ¿ and ƒ– .
77. For x 7 0 , sketch a curve y = ƒsxd that has ƒs1d = 0 and ƒ¿sxd = 1>x . Can anything be said about the concavity of such a curve? Give reasons for your answer.
90. Graph ƒsxd = x cos x and its second derivative together for 0 … x … 2p . Comment on the behavior of the graph of ƒ in relation to the signs and values of ƒ– .
78. Can anything be said about the graph of a function y = ƒsxd that has a continuous second derivative that is never zero? Give reasons for your answer.
91. a. On a common screen, graph ƒsxd = x 3 + k x for k = 0 and nearby positive and negative values of k. How does the value of k seem to affect the shape of the graph?
79. If b, c, and d are constants, for what value of b will the curve y = x 3 + bx 2 + cx + d have a point of inflection at x = 1 ? Give reasons for your answer. True, or false? Explain.
80. Horizontal tangents
a. The graph of every polynomial of even degree (largest exponent even) has at least one horizontal tangent. b. The graph of every polynomial of odd degree (largest exponent odd) has at least one horizontal tangent. 81. Parabolas a. Find the coordinates of the vertex of the parabola y = ax 2 + bx + c, a Z 0 . b. When is the parabola concave up? Concave down? Give reasons for your answers. 82. Is it true that the concavity of the graph of a twicedifferentiable function y = ƒsxd changes every time ƒ–sxd = 0 ? Give reasons for your answer. 83. Quadratic curves What can you say about the inflection points of a quadratic curve y = ax 2 + bx + c, a Z 0 ? Give reasons for your answer. 84. Cubic curves What can you say about the inflection points of a cubic curve y = ax 3 + bx 2 + cx + d, a Z 0 ? Give reasons for your answer.
COMPUTER EXPLORATIONS In Exercises 85–88, find the inflection points (if any) on the graph of the function and the coordinates of the points on the graph where the function has a local maximum or local minimum value. Then graph the function in a region large enough to show all these points simultaneously. Add to your picture the graphs of the function’s first and second derivatives. How are the values at which these graphs intersect the
b. Find ƒ¿sxd . As you will see, ƒ¿sxd is a quadratic function of x. Find the discriminant of the quadratic (the discriminant of ax 2 + bx + c is b 2  4ac). For what values of k is the discriminant positive? Zero? Negative? For what values of k does ƒ¿ have two zeros? One or no zeros? Now explain what the value of k has to do with the shape of the graph of ƒ. c. Experiment with other values of k. What appears to happen as k :  q ? as k : q ? 92. a. On a common screen, graph ƒsxd = x 4 + kx 3 + 6x 2, 2 … x … 2 for k = 4 , and some nearby integer values of k. How does the value of k seem to affect the shape of the graph? b. Find ƒ–sxd . As you will see, ƒ–sxd is a quadratic function of x. What is the discriminant of this quadratic (see Exercise 91(b))? For what values of k is the discriminant positive? Zero? Negative? For what values of k does ƒ–sxd have two zeros? One or no zeros? Now explain what the value of k has to do with the shape of the graph of ƒ. 93. a. Graph y = x 2>3sx 2  2d for 3 … x … 3 . Then use calculus to confirm what the screen shows about concavity, rise, and fall. (Depending on your grapher, you may have to enter x 2>3 as sx 2 d1>3 to obtain a plot for negative values of x.) b. Does the curve have a cusp at x = 0 , or does it just have a corner with different righthand and lefthand derivatives? 94. a. Graph y = 9x 2>3sx  1d for 0.5 … x … 1.5 . Then use calculus to confirm what the screen shows about concavity, rise, and fall. What concavity does the curve have to the left of the origin? (Depending on your grapher, you may have to enter x 2>3 as sx 2 d1>3 to obtain a plot for negative values of x.) b. Does the curve have a cusp at x = 0 , or does it just have a corner with different righthand and lefthand derivatives? 95. Does the curve y = x 2 + 3 sin 2x have a horizontal tangent near x = 3 ? Give reasons for your answer.
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Chapter 4: Applications of Derivatives
Applied Optimization Problems
4.5
To optimize something means to maximize or minimize some aspect of it. What are the dimensions of a rectangle with fixed perimeter having maximum area? What is the least expensive shape for a cylindrical can? What is the size of the most profitable production run? The differential calculus is a powerful tool for solving problems that call for maximizing or minimizing a function. In this section we solve a variety of optimization problems from business, mathematics, physics, and economics.
x
12
Examples from Business and Industry EXAMPLE 1
x x
An opentop box is to be made by cutting small congruent squares from the corners of a 12in.by12in. sheet of tin and bending up the sides. How large should the squares cut from the corners be to make the box hold as much as possible?
x
12
Fabricating a Box
(a)
We start with a picture (Figure 4.32). In the figure, the corner squares are x in. on a side. The volume of the box is a function of this variable:
Solution
Vsxd = xs12  2xd2 = 144x  48x 2 + 4x 3 .
V = hlw
x
Since the sides of the sheet of tin are only 12 in. long, x … 6 and the domain of V is the interval 0 … x … 6. A graph of V (Figure 4.33) suggests a minimum value of 0 at x = 0 and x = 6 and a maximum near x = 2. To learn more, we examine the first derivative of V with respect to x:
12 2x 12 12 2x
x
x
dV = 144  96x + 12x 2 = 12s12  8x + x 2 d = 12s2  xds6  xd. dx
(b)
FIGURE 4.32 An open box made by cutting the corners from a square sheet of tin. What size corners maximize the box’s volume (Example 1)?
Of the two zeros, x = 2 and x = 6, only x = 2 lies in the interior of the function’s domain and makes the criticalpoint list. The values of V at this one critical point and two endpoints are Criticalpoint value: Endpoint values:
Maximum y
Vs6d = 0.
3
The maximum volume is 128 in. . The cutout squares should be 2 in. on a side.
Volume
y x(12 – 2x)2, 0x6
EXAMPLE 2
min
min 0
Vs2d = 128 Vs0d = 0,
2
6
NOT TO SCALE
Designing an Efficient Cylindrical Can
You have been asked to design a 1liter can shaped like a right circular cylinder (Figure 4.34). What dimensions will use the least material?
x
Solution Volume of can: If r and h are measured in centimeters, then the volume of the can in cubic centimeters is
FIGURE 4.33 The volume of the box in Figure 4.32 graphed as a function of x.
pr 2h = 1000. Surface area of can:
1 liter = 1000 cm3
2 + 2prh A = 2pr ()* ()*
circular ends
circular wall
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How can we interpret the phrase “least material”? First, it is customary to ignore the thickness of the material and the waste in manufacturing. Then we ask for dimensions r and h that make the total surface area as small as possible while satisfying the constraint pr 2h = 1000. To express the surface area as a function of one variable, we solve for one of the variables in pr 2h = 1000 and substitute that expression into the surface area formula. Solving for h is easier:
2r
h
h =
1000 . pr 2
Thus, A = 2pr 2 + 2prh
FIGURE 4.34 This 1L can uses the least material when h = 2r (Example 2).
= 2pr 2 + 2pr a = 2pr 2 +
1000 b pr 2
2000 r .
Our goal is to find a value of r 7 0 that minimizes the value of A. Figure 4.35 suggests that such a value exists. A Tall and thin can Short and wide can —— , r 0 A 2r 2 2000 r
Tall and thin
min r
0
3
Short and wide
500
FIGURE 4.35 The graph of A = 2pr 2 + 2000>r is concave up.
Notice from the graph that for small r (a tall thin container, like a piece of pipe), the term 2000>r dominates and A is large. For large r (a short wide container, like a pizza pan), the term 2pr 2 dominates and A again is large. Since A is differentiable on r 7 0, an interval with no endpoints, it can have a minimum value only where its first derivative is zero. 2000 dA = 4pr dr r2 0 = 4pr 
2000 r2
4pr 3 = 2000 r =
3 500 L 5.42 A p
3 500>p? What happens at r = 2
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Set dA>dr = 0 . Multiply by r 2 . Solve for r.
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The second derivative 4000 d 2A = 4p + 2 dr r3 is positive throughout the domain of A. The graph is therefore everywhere concave up and 3 500>p an absolute minimum. the value of A at r = 2 The corresponding value of h (after a little algebra) is h =
500 1000 = 2 3 p = 2r. A pr 2
The 1L can that uses the least material has height equal to the diameter, here with r L 5.42 cm and h L 10.84 cm.
Solving Applied Optimization Problems 1. Read the problem. Read the problem until you understand it. What is given? What is the unknown quantity to be optimized? 2. Draw a picture. Label any part that may be important to the problem. 3. Introduce variables. List every relation in the picture and in the problem as an equation or algebraic expression, and identify the unknown variable. 4. Write an equation for the unknown quantity. If you can, express the unknown as a function of a single variable or in two equations in two unknowns. This may require considerable manipulation. 5. Test the critical points and endpoints in the domain of the unknown. Use what you know about the shape of the function’s graph. Use the first and second derivatives to identify and classify the function’s critical points.
Examples from Mathematics and Physics EXAMPLE 3
Inscribing Rectangles
A rectangle is to be inscribed in a semicircle of radius 2. What is the largest area the rectangle can have, and what are its dimensions? y
Let sx, 24  x 2 d be the coordinates of the corner of the rectangle obtained by placing the circle and rectangle in the coordinate plane (Figure 4.36). The length, height, and area of the rectangle can then be expressed in terms of the position x of the lower righthand corner: Solution
x 2 y2 4 x, 兹4 x 2
Length: 2x,
2 –2 –x
0
x 2
FIGURE 4.36 The rectangle inscribed in the semicircle in Example 3.
x
Height: 24  x 2,
Area: 2x # 24  x 2 .
Notice that the values of x are to be found in the interval 0 … x … 2, where the selected corner of the rectangle lies. Our goal is to find the absolute maximum value of the function Asxd = 2x24  x 2 on the domain [0, 2].
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Applied Optimization Problems
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The derivative 2x 2 dA = + 224  x 2 dx 24  x 2 is not defined when x = 2 and is equal to zero when 2x 2
+ 224  x 2 24  x 2 2x 2 + 2s4  x 2 d 8  4x 2 x2
= 0 = 0 = 0 = 2 or x = ; 22.
Of the two zeros, x = 22 and x =  22, only x = 22 lies in the interior of A’s domain and makes the criticalpoint list. The values of A at the endpoints and at this one critical point are Criticalpoint value: Endpoint values:
A A 22 B = 22224  2 = 4 As0d = 0, As2d = 0.
The area has a maximum value of 4 when the rectangle is 24  x 2 = 22 units high and 2x = 222 unit long. HISTORICAL BIOGRAPHY Willebrord Snell van Royen (1580–1626)
EXAMPLE 4
Fermat’s Principle and Snell’s Law
The speed of light depends on the medium through which it travels, and is generally slower in denser media. Fermat’s principle in optics states that light travels from one point to another along a path for which the time of travel is a minimum. Find the path that a ray of light will follow in going from a point A in a medium where the speed of light is c1 to a point B in a second medium where its speed is c2 . Since light traveling from A to B follows the quickest route, we look for a path that will minimize the travel time. We assume that A and B lie in the xyplane and that the line separating the two media is the xaxis (Figure 4.37). In a uniform medium, where the speed of light remains constant, “shortest time” means “shortest path,” and the ray of light will follow a straight line. Thus the path from A to B will consist of a line segment from A to a boundary point P, followed by another line segment from P to B. Distance equals rate times time, so Solution
y A a
1
Angle of incidence 1
Medium 1 P
0
x Medium 2
b
2 dx
d Angle of refraction
Time =
x
The time required for light to travel from A to P is 2a 2 + x 2 AP t1 = c1 = . c1
B
FIGURE 4.37 A light ray refracted (deflected from its path) as it passes from one medium to a denser medium (Example 4).
distance rate .
From P to B, the time is 2b 2 + sd  xd2 PB . t2 = c2 = c2
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Chapter 4: Applications of Derivatives
The time from A to B is the sum of these: t = t1 + t2 =
2b 2 + sd  xd2 2a 2 + x 2 + . c1 c2
This equation expresses t as a differentiable function of x whose domain is [0, d]. We want to find the absolute minimum value of t on this closed interval. We find the derivative dt x d  x = . 2 2 2 dx c1 2a + x c2 2b + sd  xd2 In terms of the angles u1 and u2 in Figure 4.37, sin u1 sin u2 dt = c1  c2 . dx
dt/dx negative
0
dt/dx zero
If we restrict x to the interval 0 … x … d, then t has a negative derivative at x = 0 and a positive derivative at x = d. By the Intermediate Value Theorem for Derivatives (Section 3.1), there is a point x0 H [0, d] where dt>dx = 0 (Figure 4.38). There is only one such point because dt>dx is an increasing function of x (Exercise 54). At this point
dt/dx positive
x0
sin u1 sin u2 c1 = c2 .
x d
FIGURE 4.38 The sign pattern of dt>dx in Example 4.
This equation is Snell’s Law or the Law of Refraction, and is an important principle in the theory of optics. It describes the path the ray of light follows.
Examples from Economics In these examples we point out two ways that calculus makes a contribution to economics. The first has to do with maximizing profit. The second has to do with minimizing average cost. Suppose that rsxd = the revenue from selling x items csxd = the cost of producing the x items psxd = rsxd  csxd = the profit from producing and selling x items. The marginal revenue, marginal cost, and marginal profit when producing and selling x items are dr = marginal revenue, dx dc = marginal cost, dx dp = marginal profit. dx The first observation is about the relationship of p to these derivatives. If r(x) and c(x) are differentiable for all x 7 0, and if psxd = rsxd  csxd has a maximum value, it occurs at a production level at which p¿sxd = 0. Since p¿sxd = r¿sxd  c¿sxd, p¿sxd = 0 implies that r¿sxd  c¿sxd = 0
or
r¿sxd = c¿sxd.
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Therefore
At a production level yielding maximum profit, marginal revenue equals marginal cost (Figure 4.39).
y
Dollars
Cost c(x)
Revenue r(x) Breakeven point Maximum profit, c'(x) r'(x)
B
Local maximum for loss (minimum profit), c'(x) r'(x) x Items produced
0
FIGURE 4.39 The graph of a typical cost function starts concave down and later turns concave up. It crosses the revenue curve at the breakeven point B. To the left of B, the company operates at a loss. To the right, the company operates at a profit, with the maximum profit occurring where c¿sxd = r¿sxd . Farther to the right, cost exceeds revenue (perhaps because of a combination of rising labor and material costs and market saturation) and production levels become unprofitable again.
EXAMPLE 5
Maximizing Profit
Suppose that rsxd = 9x and csxd = x 3  6x 2 + 15x, where x represents thousands of units. Is there a production level that maximizes profit? If so, what is it? y
Solution
Notice that r¿sxd = 9 and c¿sxd = 3x 2  12x + 15. 3x 2  12x + 15 = 9 3x 2  12x + 6 = 0
c(x) x 3 6x2 15x
Set c¿sxd = r¿sxd.
The two solutions of the quadratic equation are r(x) 9x Maximum for profit
Local maximum for loss 0 2 兹2
2
2 兹2
NOT TO SCALE
FIGURE 4.40 The cost and revenue curves for Example 5.
x
x1 =
12  272 = 2  22 L 0.586 6
x2 =
12 + 272 = 2 + 22 L 3.414. 6
and
The possible production levels for maximum profit are x L 0.586 thousand units or x L 3.414 thousand units. The second derivative of psxd = rsxd  csxd is p–sxd = c–sxd since r–sxd is everywhere zero. Thus, p–sxd = 6s2  xd which is negative at x = 2 + 22 and positive at x = 2  22. By the Second Derivative Test, a maximum profit occurs at about x = 3.414 (where revenue exceeds costs) and maximum loss occurs at about x = 0.586. The graph of r(x) is shown in Figure 4.40.
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284 y
Chapter 4: Applications of Derivatives
EXAMPLE 6
c(x) 5000 x 25x
A cabinetmaker uses plantationfarmed mahogany to produce 5 furnishings each day. Each delivery of one container of wood is $5000, whereas the storage of that material is $10 per day per unit stored, where a unit is the amount of material needed by her to produce 1 furnishing. How much material should be ordered each time and how often should the material be delivered to minimize her average daily cost in the production cycle between deliveries?
Cost
y 25x
y 5000 x
x min Cycle length
Minimizing Costs
If she asks for a delivery every x days, then she must order 5x units to have enough material for that delivery cycle. The average amount in storage is approximately onehalf of the delivery amount, or 5 x>2. Thus, the cost of delivery and storage for each cycle is approximately
Solution x
Cost per cycle = delivery costs + storage costs
FIGURE 4.41 The average daily cost c(x) is the sum of a hyperbola and a linear function (Example 6).
Cost per cycle = 5000 ()*
+
a
5x b 2
#
()* average amount stored
delivery cost
x ()* number of days stored
#
10 ()* storage cost per day
We compute the average daily cost c(x) by dividing the cost per cycle by the number of days x in the cycle (see Figure 4.41). csxd =
5000 x + 25x,
x 7 0.
As x : 0 and as x : q , the average daily cost becomes large. So we expect a minimum to exist, but where? Our goal is to determine the number of days x between deliveries that provides the absolute minimum cost. We find the critical points by determining where the derivative is equal to zero: 5000 + 25 = 0 x2 x = ; 2200 L ;14.14.
c¿sxd = 
Of the two critical points, only 2200 lies in the domain of c(x). The criticalpoint value of the average daily cost is c A 2200 B =
5000 2200
+ 25 2200 = 50022 L $707.11.
We note that c(x) is defined over the open interval s0, q d with c–sxd = 10000>x 3 7 0. Thus, an absolute minimum exists at x = 2200 L 14.14 days. The cabinetmaker should schedule a delivery of 5s14d = 70 units of the exotic wood every 14 days. In Examples 5 and 6 we allowed the number of items x to be any positive real number. In reality it usually only makes sense for x to be a positive integer (or zero). If we must round our answers, should we round up or down?
EXAMPLE 7
Sensitivity of the Minimum Cost
Should we round the number of days between deliveries up or down for the best solution in Example 6?
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The average daily cost will increase by about $0.03 if we round down from 14.14 to 14 days:
Solution
cs14d =
5000 + 25s14d = $707.14 14
and cs14d  cs14.14d = $707.14  $707.11 = $0.03. On the other hand, cs15d = $708.33, and our cost would increase by $708.33 $707.11 = $1.22 if we round up. Thus, it is better that we round x down to 14 days.
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EXERCISES 4.5 Whenever you are maximizing or minimizing a function of a single variable, we urge you to graph it over the domain that is appropriate to the problem you are solving. The graph will provide insight before you calculate and will furnish a visual context for understanding your answer.
Applications in Geometry 1. Minimizing perimeter What is the smallest perimeter possible for a rectangle whose area is 16 in.2 , and what are its dimensions? 2. Show that among all rectangles with an 8m perimeter, the one with largest area is a square. 3. The figure shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 2 units long. a. Express the ycoordinate of P in terms of x. (Hint: Write an equation for the line AB.) b. Express the area of the rectangle in terms of x. c. What is the largest area the rectangle can have, and what are its dimensions? y B
the box of largest volume you can make this way, and what is its volume? 6. You are planning to close off a corner of the first quadrant with a line segment 20 units long running from (a, 0) to (0, b). Show that the area of the triangle enclosed by the segment is largest when a = b. 7. The best fencing plan A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a singlestrand electric fence. With 800m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions? 8. The shortest fence A 216 m2 rectangular pea patch is to be enclosed by a fence and divided into two equal parts by another fence parallel to one of the sides. What dimensions for the outer rectangle will require the smallest total length of fence? How much fence will be needed? 9. Designing a tank Your iron works has contracted to design and build a 500 ft3 , squarebased, opentop, rectangular steel holding tank for a paper company. The tank is to be made by welding thin stainless steel plates together along their edges. As the production engineer, your job is to find dimensions for the base and height that will make the tank weigh as little as possible. a. What dimensions do you tell the shop to use?
P(x, ?)
b. Briefly describe how you took weight into account. A
–1
0
x
1
x
4. A rectangle has its base on the xaxis and its upper two vertices on the parabola y = 12  x 2 . What is the largest area the rectangle can have, and what are its dimensions? 5. You are planning to make an open rectangular box from an 8in.by15in. piece of cardboard by cutting congruent squares from the corners and folding up the sides. What are the dimensions of
10. Catching rainwater A 1125 ft3 opentop rectangular tank with a square base x ft on a side and y ft deep is to be built with its top flush with the ground to catch runoff water. The costs associated with the tank involve not only the material from which the tank is made but also an excavation charge proportional to the product xy. a. If the total cost is c = 5sx 2 + 4xyd + 10xy , what values of x and y will minimize it? b. Give a possible scenario for the cost function in part (a).
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11. Designing a poster You are designing a rectangular poster to contain 50 in.2 of printing with a 4in. margin at the top and bottom and a 2in. margin at each side. What overall dimensions will minimize the amount of paper used? 12. Find the volume of the largest right circular cone that can be inscribed in a sphere of radius 3.
3
c. Use a graphical method to find the maximum volume and the value of x that gives it. d. Confirm your result in part (c) analytically. T 17. Designing a suitcase A 24in.by36in. sheet of cardboard is folded in half to form a 24in.by18in. rectangle as shown in the accompanying figure. Then four congruent squares of side length x are cut from the corners of the folded rectangle. The sheet is unfolded, and the six tabs are folded up to form a box with sides and a lid. a. Write a formula V(x) for the volume of the box. b. Find the domain of V for the problem situation and graph V over this domain.
3
y
c. Use a graphical method to find the maximum volume and the value of x that gives it.
x
d. Confirm your result in part (c) analytically. 13. Two sides of a triangle have lengths a and b, and the angle between them is u . What value of u will maximize the triangle’s area? (Hint: A = s1>2dab sin u .)
e. Find a value of x that yields a volume of 1120 in.3 . f. Write a paragraph describing the issues that arise in part (b).
14. Designing a can What are the dimensions of the lightest opentop right circular cylindrical can that will hold a volume of 1000 cm3 ? Compare the result here with the result in Example 2. 15. Designing a can You are designing a 1000 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in cutting the aluminum for the side, but the top and bottom of radius r will be cut from squares that measure 2r units on a side. The total amount of aluminum used up by the can will therefore be
x
x
x
x
24"
24" x
x x
36"
x 18"
The sheet is then unfolded.
A = 8r 2 + 2prh rather than the A = 2pr 2 + 2prh in Example 2. In Example 2, the ratio of h to r for the most economical can was 2 to 1. What is the ratio now? T 16. Designing a box with a lid A piece of cardboard measures 10 in. by 15 in. Two equal squares are removed from the corners of a 10in. side as shown in the figure. Two equal rectangles are removed from the other corners so that the tabs can be folded to form a rectangular box with lid.
NOT TO SCALE
x
Base
36"
x
x 10"
24"
Base
x
18. A rectangle is to be inscribed under the arch of the curve y = 4 cos s0.5xd from x = p to x = p . What are the dimensions of the rectangle with largest area, and what is the largest area?
x
19. Find the dimensions of a right circular cylinder of maximum volume that can be inscribed in a sphere of radius 10 cm. What is the maximum volume?
Lid
x x
x 15"
a. Write a formula V(x) for the volume of the box. b. Find the domain of V for the problem situation and graph V over this domain.
20. a. The U.S. Postal Service will accept a box for domestic shipment only if the sum of its length and girth (distance around) does not exceed 108 in. What dimensions will give a box with a square end the largest possible volume?
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4.5 Applied Optimization Problems
24. The trough in the figure is to be made to the dimensions shown. Only the angle u can be varied. What value of u will maximize the trough’s volume?
Girth distance around here
1'
1' 1'
Length
287
20'
Square end
T b. Graph the volume of a 108in. box (length plus girth equals 108 in.) as a function of its length and compare what you see with your answer in part (a). 21. (Continuation of Exercise 20.) a. Suppose that instead of having a box with square ends you have a box with square sides so that its dimensions are h by h by w and the girth is 2h + 2w . What dimensions will give the box its largest volume now? Girth
25. Paper folding A rectangular sheet of 8.5in.by11in. paper is placed on a flat surface. One of the corners is placed on the opposite longer edge, as shown in the figure, and held there as the paper is smoothed flat. The problem is to make the length of the crease as small as possible. Call the length L. Try it with paper. a. Show that L 2 = 2x 3>s2x  8.5d .
b. What value of x minimizes L 2 ?
c. What is the minimum value of L? D
C
R
h
兹L2 x 2
L
w
Crease
Q (originally at A)
h x
T b. Graph the volume as a function of h and compare what you see with your answer in part (a). 22. A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only half as much light per unit area as clear glass does. The total perimeter is fixed. Find the proportions of the window that will admit the most light. Neglect the thickness of the frame.
x A
P
B
26. Constructing cylinders Compare the answers to the following two construction problems. a. A rectangular sheet of perimeter 36 cm and dimensions x cm by y cm to be rolled into a cylinder as shown in part (a) of the figure. What values of x and y give the largest volume? b. The same sheet is to be revolved about one of the sides of length y to sweep out the cylinder as shown in part (b) of the figure. What values of x and y give the largest volume?
y x
23. A silo (base not included) is to be constructed in the form of a cylinder surmounted by a hemisphere. The cost of construction per square unit of surface area is twice as great for the hemisphere as it is for the cylindrical sidewall. Determine the dimensions to be used if the volume is fixed and the cost of construction is to be kept to a minimum. Neglect the thickness of the silo and waste in construction.
Circumference x
x y
y
(a)
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(b)
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27. Constructing cones A right triangle whose hypotenuse is 23 m long is revolved about one of its legs to generate a right circular cone. Find the radius, height, and volume of the cone of greatest volume that can be made this way.
h
a. Find the dimensions of the strongest beam that can be cut from a 12in.diameter cylindrical log. b. Graph S as a function of the beam’s width w, assuming the proportionality constant to be k = 1 . Reconcile what you see with your answer in part (a). c. On the same screen, graph S as a function of the beam’s depth d, again taking k = 1 . Compare the graphs with one another and with your answer in part (a). What would be the effect of changing to some other value of k? Try it.
兹3 r
28. What value of a makes ƒsxd = x 2 + sa>xd have a. a local minimum at x = 2 ?
12"
b. a point of inflection at x = 1 ?
d
29. Show that ƒsxd = x 2 + sa>xd cannot have a local maximum for any value of a. 30. What values of a and b make ƒsxd = x 3 + ax 2 + bx have
w
a. a local maximum at x = 1 and a local minimum at x = 3 ? b. a local minimum at x = 4 and a point of inflection at x = 1 ?
Physical Applications 31. Vertical motion The height of an object moving vertically is given by 2
s = 16t + 96t + 112 , with s in feet and t in seconds. Find
a. Find the dimensions of the stiffest beam that can be cut from a 12in.diameter cylindrical log. b. Graph S as a function of the beam’s width w, assuming the proportionality constant to be k = 1 . Reconcile what you see with your answer in part (a). c. On the same screen, graph S as a function of the beam’s depth d, again taking k = 1 . Compare the graphs with one another and with your answer in part (a). What would be the effect of changing to some other value of k? Try it.
a. the object’s velocity when t = 0 b. its maximum height and when it occurs c. its velocity when s = 0 . 32. Quickest route Jane is 2 mi offshore in a boat and wishes to reach a coastal village 6 mi down a straight shoreline from the point nearest the boat. She can row 2 mph and can walk 5 mph. Where should she land her boat to reach the village in the least amount of time? 33. Shortest beam The 8ft wall shown here stands 27 ft from the building. Find the length of the shortest straight beam that will reach to the side of the building from the ground outside the wall.
Beam
T 35. Stiffness of a beam The stiffness S of a rectangular beam is proportional to its width times the cube of its depth.
Building
36. Motion on a line The positions of two particles on the saxis are s1 = sin t and s2 = sin st + p>3d , with s1 and s2 in meters and t in seconds. a. At what time(s) in the interval 0 … t … 2p do the particles meet? b. What is the farthest apart that the particles ever get? c. When in the interval 0 … t … 2p is the distance between the particles changing the fastest? 37. Frictionless cart A small frictionless cart, attached to the wall by a spring, is pulled 10 cm from its rest position and released at time t = 0 to roll back and forth for 4 sec. Its position at time t is s = 10 cos pt . a. What is the cart’s maximum speed? When is the cart moving that fast? Where is it then? What is the magnitude of the acceleration then? b. Where is the cart when the magnitude of the acceleration is greatest? What is the cart’s speed then?
8' wall 27'
T 34. Strength of a beam The strength S of a rectangular wooden beam is proportional to its width times the square of its depth. (See accompanying figure.)
0
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10
s
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4.5 Applied Optimization Problems 38. Two masses hanging side by side from springs have positions s1 = 2 sin t and s2 = sin 2t , respectively.
289
Normal Light receiver
a. At what times in the interval 0 6 t do the masses pass each other? (Hint: sin 2t = 2 sin t cos t .)
Light source A
b. When in the interval 0 … t … 2p is the vertical distance between the masses the greatest? What is this distance? (Hint: cos 2t = 2 cos2 t  1 .)
B
Angle of reflection 2
Angle of incidence 1
Plane mirror
m1
s1 0 s2
m2
s
39. Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical miles per hour; a nautical mile is 2000 yd) and continued to do so all day. Ship B was sailing east at 8 knots and continued to do so all day.
41. Tin pest When metallic tin is kept below 13.2°C, it slowly becomes brittle and crumbles to a gray powder. Tin objects eventually crumble to this gray powder spontaneously if kept in a cold climate for years. The Europeans who saw tin organ pipes in their churches crumble away years ago called the change tin pest because it seemed to be contagious, and indeed it was, for the gray powder is a catalyst for its own formation. A catalyst for a chemical reaction is a substance that controls the rate of reaction without undergoing any permanent change in itself. An autocatalytic reaction is one whose product is a catalyst for its own formation. Such a reaction may proceed slowly at first if the amount of catalyst present is small and slowly again at the end, when most of the original substance is used up. But in between, when both the substance and its catalyst product are abundant, the reaction proceeds at a faster pace. In some cases, it is reasonable to assume that the rate y = dx>dt of the reaction is proportional both to the amount of the original substance present and to the amount of product. That is, y may be considered to be a function of x alone, and y = kxsa  xd = kax  kx 2 ,
a. Start counting time with t = 0 at noon and express the distance s between the ships as a function of t. b. How rapidly was the distance between the ships changing at noon? One hour later?
where
c. The visibility that day was 5 nautical miles. Did the ships ever sight each other?
a = the amount of substance at the beginning
T d. Graph s and ds>dt together as functions of t for 1 … t … 3 , using different colors if possible. Compare the graphs and reconcile what you see with your answers in parts (b) and (c). e. The graph of ds>dt looks as if it might have a horizontal asymptote in the first quadrant. This in turn suggests that ds>dt approaches a limiting value as t : q . What is this value? What is its relation to the ships’ individual speeds? 40. Fermat’s principle in optics Fermat’s principle in optics states that light always travels from one point to another along a path that minimizes the travel time. Light from a source A is reflected by a plane mirror to a receiver at point B, as shown in the figure. Show that for the light to obey Fermat’s principle, the angle of incidence must equal the angle of reflection, both measured from the line normal to the reflecting surface. (This result can also be derived without calculus. There is a purely geometric argument, which you may prefer.)
x = the amount of product k = a positive constant . At what value of x does the rate y have a maximum? What is the maximum value of y? 42. Airplane landing path An airplane is flying at altitude H when it begins its descent to an airport runway that is at horizontal ground distance L from the airplane, as shown in the figure. Assume that the landing path of the airplane is the graph of a cubic polynomial function y = ax 3 + bx 2 + cx + d, where y s Ld = H and y s0d = 0 . a. What is dy>dx at x = 0 ? b. What is dy>dx at x = L ? c. Use the values for dy>dx at x = 0 and x = L together with y s0d = 0 and y s Ld = H to show that 3
2
x x y sxd = H c2 a b + 3 a b d . L L
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Chapter 4: Applications of Derivatives 48. Production level Suppose that csxd = x 3  20x 2 + 20,000x is the cost of manufacturing x items. Find a production level that will minimize the average cost of making x items.
y
Landing path
49. Average daily cost In Example 6, assume for any material that a cost of d is incurred per delivery, the storage cost is s dollars per unit stored per day, and the production rate is p units per day.
H = Cruising altitude Airport x L
a. How much should be delivered every x days? b. Show that
Business and Economics 43. It costs you c dollars each to manufacture and distribute backpacks. If the backpacks sell at x dollars each, the number sold is given by a n = x  c + bs100  xd , where a and b are positive constants. What selling price will bring a maximum profit? 44. You operate a tour service that offers the following rates: $200 per person if 50 people (the minimum number to book the tour) go on the tour. For each additional person, up to a maximum of 80 people total, the rate per person is reduced by $2. It costs $6000 (a fixed cost) plus $32 per person to conduct the tour. How many people does it take to maximize your profit?
cost per cycle = d +
c. Find the time between deliveries x* and the amount to deliver that minimizes the average daily cost of delivery and storage. d. Show that x* occurs at the intersection of the hyperbola y = d>x and the line y = psx>2 . 50. Minimizing average cost Suppose that csxd = 2000 + 96x + 4x 3>2 , where x represents thousands of units. Is there a production level that minimizes average cost? If so, what is it?
Medicine 51. Sensitivity to medicine (Continuation of Exercise 50, Section 3.2.) Find the amount of medicine to which the body is most sensitive by finding the value of M that maximizes the derivative dR>dM , where
45. Wilson lot size formula One of the formulas for inventory management says that the average weekly cost of ordering, paying for, and holding merchandise is hq km , Asqd = q + cm + 2 where q is the quantity you order when things run low (shoes, radios, brooms, or whatever the item might be), k is the cost of placing an order (the same, no matter how often you order), c is the cost of one item (a constant), m is the number of items sold each week (a constant), and h is the weekly holding cost per item (a constant that takes into account things such as space, utilities, insurance, and security). a. Your job, as the inventory manager for your store, is to find the quantity that will minimize A(q). What is it? (The formula you get for the answer is called the Wilson lot size formula.) b. Shipping costs sometimes depend on order size. When they do, it is more realistic to replace k by k + bq , the sum of k and a constant multiple of q. What is the most economical quantity to order now?
px sx . 2
R = M2 a
C M  b 2 3
and C is a constant. 52. How we cough a. When we cough, the trachea (windpipe) contracts to increase the velocity of the air going out. This raises the questions of how much it should contract to maximize the velocity and whether it really contracts that much when we cough. Under reasonable assumptions about the elasticity of the tracheal wall and about how the air near the wall is slowed by friction, the average flow velocity y can be modeled by the equation y = csr0  rdr 2 cm>sec,
r0 … r … r0 , 2
46. Production level Prove that the production level (if any) at which average cost is smallest is a level at which the average cost equals marginal cost.
where r0 is the rest radius of the trachea in centimeters and c is a positive constant whose value depends in part on the length of the trachea. Show that y is greatest when r = s2>3dr0 , that is, when the trachea is about 33% contracted. The remarkable fact is that Xray photographs confirm that the trachea contracts about this much during a cough.
47. Show that if rsxd = 6x and csxd = x 3  6x 2 + 15x are your revenue and cost functions, then the best you can do is break even (have revenue equal cost).
T b. Take r0 to be 0.5 and c to be 1 and graph y over the interval 0 … r … 0.5 . Compare what you see with the claim that y is at a maximum when r = s2>3dr0 .
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Theory and Examples Show that if a, b, c, and d
53. An inequality for positive integers are positive integers, then
sa 2 + 1dsb 2 + 1dsc 2 + 1dsd 2 + 1d Ú 16 . abcd
291
59. a. How close does the curve y = 2x come to the point (3>2, 0)? (Hint: If you minimize the square of the distance, you can avoid square roots.) T b. Graph the distance function and y = 2x together and reconcile what you see with your answer in part (a). y
54. The derivative dt>dx in Example 4 a. Show that ƒsxd =
(x, 兹x)
x 2a + x 2
2
y 兹x
is an increasing function of x. b. Show that g sxd =
d  x
x
3 , 0 2
0
2b 2 + sd  xd2 60. a. How close does the semicircle y = 216  x 2 come to the point A 1, 23 B ?
is a decreasing function of x. c. Show that dt d  x x = dx c1 2a 2 + x 2 c2 2b 2 + sd  xd2
T b. Graph the distance function and y = 216  x 2 together and reconcile what you see with your answer in part (a).
COMPUTER EXPLORATIONS
is an increasing function of x. 55. Let ƒ(x) and g(x) be the differentiable functions graphed here. Point c is the point where the vertical distance between the curves is the greatest. Is there anything special about the tangents to the two curves at c? Give reasons for your answer.
In Exercises 61 and 62, you may find it helpful to use a CAS. 61. Generalized cone problem A cone of height h and radius r is constructed from a flat, circular disk of radius a in. by removing a sector AOC of arc length x in. and then connecting the edges OA and OC. a. Find a formula for the volume V of the cone in terms of x and a. b. Find r and h in the cone of maximum volume for a = 4, 5, 6, 8 .
y f (x)
c. Find a simple relationship between r and h that is independent of a for the cone of maximum volume. Explain how you arrived at your relationship.
y g(x)
A a
c
b
x
56. You have been asked to determine whether the function ƒsxd = 3 + 4 cos x + cos 2x is ever negative. a. Explain why you need to consider values of x only in the interval [0, 2p] .
x
a O
4" a
C A
O
r C NOT TO SCALE
b. Is ƒ ever negative? Explain. 57. a. The function y = cot x  22 csc x has an absolute maximum value on the interval 0 6 x 6 p . Find it. T b. Graph the function and compare what you see with your answer in part (a). 58. a. The function y = tan x + 3 cot x has an absolute minimum value on the interval 0 6 x 6 p>2 . Find it. T b. Graph the function and compare what you see with your answer in part (a).
h
62. Circumscribing an ellipse Let P(x, a) and Qs x, ad be two points on the upper half of the ellipse sy  5d2 x2 + = 1 100 25 centered at (0, 5). A triangle RST is formed by using the tangent lines to the ellipse at Q and P as shown in the figure.
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Chapter 4: Applications of Derivatives where y = ƒsxd is the function representing the upper half of the ellipse.
y R
b. What is the domain of A? Draw the graph of A. How are the asymptotes of the graph related to the problem situation? P(x, a)
Q(–x, a)
c. Determine the height of the triangle with minimum area. How is it related to the y coordinate of the center of the ellipse?
5 S
T
a. Show that the area of the triangle is Asxd = ƒ¿sxd cx 
ƒsxd 2 d , ƒ¿sxd
x
d. Repeat parts (a) through (c) for the ellipse sy  Bd2 x2 + = 1 C2 B2 centered at (0, B). Show that the triangle has minimum area when its height is 3B.
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4.6
Indeterminate Forms and L’Hôpital’s Rule
HISTORICAL BIOGRAPHY Guillaume François Antoine de l’Hôpital (1661–1704)
John Bernoulli discovered a rule for calculating limits of fractions whose numerators and denominators both approach zero or + q . The rule is known today as l’Hôpital’s Rule, after Guillaume de l’Hôpital. He was a French nobleman who wrote the first introductory differential calculus text, where the rule first appeared in print.
Indeterminate Form 0/0 If the continuous functions ƒ(x) and g(x) are both zero at x = a, then lim
x:a
ƒsxd gsxd
cannot be found by substituting x = a. The substitution produces 0>0, a meaningless expression, which we cannot evaluate. We use 0>0 as a notation for an expression known as an indeterminate form. Sometimes, but not always, limits that lead to indeterminate forms may be found by cancellation, rearrangement of terms, or other algebraic manipulations. This was our experience in Chapter 2. It took considerable analysis in Section 2.4 to find limx:0 ssin xd>x. But we have had success with the limit ƒsxd  ƒsad , x  a x:a
ƒ¿sad = lim
from which we calculate derivatives and which always produces the equivalent of 0>0 when we substitute x = a. L’Hôpital’s Rule enables us to draw on our success with derivatives to evaluate limits that otherwise lead to indeterminate forms.
THEOREM 6 L’Hôpital’s Rule (First Form) Suppose that ƒsad = gsad = 0, that ƒ¿sad and g¿sad exist, and that g¿sad Z 0. Then lim
x:a
ƒsxd ƒ¿sad = . gsxd g¿sad
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4.6
Caution To apply l’Hôpital’s Rule to ƒ>g, divide the derivative of ƒ by the derivative of g. Do not fall into the trap of taking the derivative of ƒ>g. The quotient to use is ƒ¿>g¿ , not sƒ>gd¿ .
Indeterminate Forms and L’Hôpital’s Rule
293
Proof Working backward from ƒ¿sad and g¿sad, which are themselves limits, we have ƒ¿sad = g¿sad
ƒsxd  ƒsad x  a x:a lim
gsxd  gsad x  a x:a lim
= lim
x:a
EXAMPLE 1 (a) lim
x:0
= lim
x:a
ƒsxd  ƒsad x  a gsxd  gsad x  a
ƒsxd  ƒsad ƒsxd  0 ƒsxd = lim = lim . gsxd  gsad x:a gsxd  0 x:a gsxd
Using L’Hôpital’s Rule
3x  sin x 3  cos x = ` = 2 x 1 x=0
1 21 + x  1 221 + x 3 1 (b) lim = = x 2 x:0 1 x=0 Sometimes after differentiation, the new numerator and denominator both equal zero at x = a, as we see in Example 2. In these cases, we apply a stronger form of l’Hôpital’s Rule.
THEOREM 7 L’Hôpital’s Rule (Stronger Form) Suppose that ƒsad = gsad = 0, that ƒ and g are differentiable on an open interval I containing a, and that g¿sxd Z 0 on I if x Z a. Then lim
x:a
ƒsxd ƒ¿sxd = lim , gsxd x:a g¿sxd
assuming that the limit on the right side exists. Before we give a proof of Theorem 7, let’s consider an example.
EXAMPLE 2 (a) lim
Applying the Stronger Form of L’Hôpital’s Rule
21 + x  1  x>2
0 0
x2
x:0
s1>2ds1 + xd1>2  1>2 2x x:0
= lim = lim
x:0
(b) lim
x:0
s1>4ds1 + xd3>2 1 = 2 8
x  sin x x3
Still
0 ; differentiate again. 0
Not
0 ; limit is found. 0
0 0
= lim
1  cos x 3x 2
Still
0 0
= lim
sin x 6x
Still
0 0
= lim
cos x 1 = 6 6
0 Not ; limit is found. 0
x:0
x:0
x:0
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The proof of the stronger form of l’Hôpital’s Rule is based on Cauchy’s Mean Value Theorem, a Mean Value Theorem that involves two functions instead of one. We prove Cauchy’s Theorem first and then show how it leads to l’Hôpital’s Rule. HISTORICAL BIOGRAPHY
THEOREM 8 Cauchy’s Mean Value Theorem Suppose functions ƒ and g are continuous on [a, b] and differentiable throughout (a, b) and also suppose g¿sxd Z 0 throughout (a, b). Then there exists a number c in (a, b) at which
AugustinLouis Cauchy (1789–1857)
ƒ¿scd ƒsbd  ƒsad = . g¿scd gsbd  gsad
Proof We apply the Mean Value Theorem of Section 4.2 twice. First we use it to show that gsad Z gsbd. For if g(b) did equal g (a), then the Mean Value Theorem would give g¿scd =
gsbd  gsad = 0 b  a
for some c between a and b, which cannot happen because g¿sxd Z 0 in (a, b). We next apply the Mean Value Theorem to the function Fsxd = ƒsxd  ƒsad 
ƒsbd  ƒsad [ gsxd  gsad]. gsbd  gsad
This function is continuous and differentiable where ƒ and g are, and Fsbd = Fsad = 0. Therefore, there is a number c between a and b for which F¿scd = 0. When expressed in terms of ƒ and g, this equation becomes F¿scd = ƒ¿scd 
ƒsbd  ƒsad [ g¿scd] = 0 gsbd  gsad
or ƒsbd  ƒsad ƒ¿scd = . g¿scd gsbd  gsad
y
(g(b), f (b))
(g(c), f (c))
slope
f (b) f (a) g(b) g(a)
(g(a), f (a)) 0
x
FIGURE 4.42 There is at least one value of the parameter t = c, a 6 c 6 b , for which the slope of the tangent to the curve at (g (c), ƒ(c)) is the same as the slope of the secant line joining the points (g (a), ƒ(a)) and (g (b), ƒ(b)).
Notice that the Mean Value Theorem in Section 4.2 is Theorem 8 with gsxd = x. Cauchy’s Mean Value Theorem has a geometric interpretation for a curve C defined by the parametric equations x = gstd and y = ƒstd. From Equation (2) in Section 3.5, the slope of the parametric curve at t is given by dy>dt ƒ¿std , = dx>dt g¿std so ƒ¿scd>g¿scd is the slope of the tangent to the curve when t = c. The secant line joining the two points (g(a), ƒ(a)) and (g(b), ƒ(b)) on C has slope ƒsbd  ƒsad . gsbd  gsad Theorem 8 says that there is a parameter value c in the interval (a, b) for which the slope of the tangent to the curve at the point (g(c), ƒ(c)) is the same as the slope of the secant line joining the points (g(a), ƒ(a)) and (g(b), ƒ(b)). This geometric result is shown in Figure 4.42. Note that more than one such value c of the parameter may exist. We now prove Theorem 7.
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Indeterminate Forms and L’Hôpital’s Rule
295
Proof of the Stronger Form of l’Hôpital’s Rule We first establish the limit equation for the case x : a + . The method needs almost no change to apply to x : a , and the combination of these two cases establishes the result. Suppose that x lies to the right of a. Then g¿sxd Z 0, and we can apply Cauchy’s Mean Value Theorem to the closed interval from a to x. This step produces a number c between a and x such that ƒ¿scd ƒsxd  ƒsad = . g¿scd gsxd  gsad But ƒsad = gsad = 0, so ƒ¿scd ƒsxd = . g¿scd gsxd As x approaches a, c approaches a because it always lies between a and x. Therefore, lim+
x:a
ƒsxd ƒ¿scd ƒ¿sxd = lim+ = lim+ , gsxd c:a g¿scd x:a g¿sxd
which establishes l’Hôpital’s Rule for the case where x approaches a from above. The case where x approaches a from below is proved by applying Cauchy’s Mean Value Theorem to the closed interval [x, a], x 6 a. Most functions encountered in the real world and most functions in this book satisfy the conditions of l’Hôpital’s Rule.
Using L’Hôpital’s Rule To find lim
x:a
ƒsxd gsxd
by l’Hôpital’s Rule, continue to differentiate ƒ and g, so long as we still get the form 0>0 at x = a. But as soon as one or the other of these derivatives is different from zero at x = a we stop differentiating. L’Hôpital’s Rule does not apply when either the numerator or denominator has a finite nonzero limit.
EXAMPLE 3
Incorrectly Applying the Stronger Form of L’Hôpital’s Rule lim
x:0
= lim
x:0
1  cos x x + x2
sin x 0 = = 0 1 + 2x 1
0 0 0 Not ; limit is found. 0
Up to now the calculation is correct, but if we continue to differentiate in an attempt to apply l’Hôpital’s Rule once more, we get lim
x:0
1  cos x sin x cos x 1 = lim = , = lim 2 x:0 1 + 2x x:0 2 x + x2
which is wrong. L’Hôpital’s Rule can only be applied to limits which give indeterminate forms, and 0>1 is not an indeterminate form. Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
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Chapter 4: Applications of Derivatives
L’Hôpital’s Rule applies to onesided limits as well, which is apparent from the proof of Theorem 7.
EXAMPLE 4 (a) lim+ x:0
Recall that q and + q mean the same thing.
Using L’Hôpital’s Rule with OneSided Limits
sin x x2
0 0
= lim+ x:0
(b) limx:0
cos x = q 2x
Positive for x 7 0 .
sin x x2
0 0
= limx:0
cos x = q 2x
Negative for x 6 0 .
Indeterminate Forms ˆ > ˆ , ˆ # 0, ˆ ˆ Sometimes when we try to evaluate a limit as x : a by substituting x = a we get an ambiguous expression like q > q , q # 0, or q  q , instead of 0>0. We first consider the form q > q . In more advanced books it is proved that l’Hôpital’s Rule applies to the indeterminate form q > q as well as to 0>0. If ƒsxd : ; q and gsxd : ; q as x : a, then lim
x:a
ƒsxd ƒ¿sxd = lim gsxd x:a g¿sxd
provided the limit on the right exists. In the notation x : a, a may be either finite or infinite. Moreover x : a may be replaced by the onesided limits x : a + or x : a  .
EXAMPLE 5
Working with the Indeterminate Form q > q
Find (a)
lim
x:p>2
(b) lim
x: q
sec x 1 + tan x
x  2x 2 3x 2 + 5x
Solution
(a) The numerator and denominator are discontinuous at x = p>2, so we investigate the onesided limits there. To apply l’Hôpital’s Rule, we can choose I to be any open interval with x = p>2 as an endpoint. lim
x:sp>2d 
sec x 1 + tan x =
q q from the left
lim
x:sp>2d 
sec x tan x = sec2 x
lim
x:sp>2d 
sin x = 1
The righthand limit is 1 also, with s  q d>s  q d as the indeterminate form. Therefore, the twosided limit is equal to 1. (b) lim
x: q
x  2x 2 1  4x 4 2 = lim =  . = lim 3 x: q 6x + 5 x: q 6 3x 2 + 5x
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Indeterminate Forms and L’Hôpital’s Rule
Next we turn our attention to the indeterminate forms q # 0 and q  q . Sometimes these forms can be handled by using algebra to convert them to a 0>0 or q > q form. Here again we do not mean to suggest that q # 0 or q  q is a number. They are only notations for functional behaviors when considering limits. Here are examples of how we might work with these indeterminate forms.
EXAMPLE 6
Working with the Indeterminate Form q # 0
Find 1 lim ax sin x b q x: Solution
1 lim ax sin x b
q #0
x: q
1 = lim+ a sin hb h:0 h
Let h = 1>x .
= 1
EXAMPLE 7
Working with the Indeterminate Form q  q
Find lim a
x:0
Solution
1 1  xb. sin x
If x : 0 + , then sin x : 0 + and 1 1  x : q  q. sin x
Similarly, if x : 0  , then sin x : 0  and 1 1  x :  q  s qd =  q + q . sin x Neither form reveals what happens in the limit. To find out, we first combine the fractions: x  sin x 1 1  x = sin x x sin x
Common denominator is x sin x
Then apply l’Hôpital’s Rule to the result: lim a
x:0
x  sin x 1 1  x b = lim sin x x:0 x sin x = lim
1  cos x sin x + x cos x
= lim
sin x 0 = = 0. 2 2 cos x  x sin x
x:0
x:0
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0 0 Still
0 0
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Chapter 4: Applications of Derivatives
EXERCISES 4.6 Finding Limits In Exercises 1–6, use l’Hôpital’s Rule to evaluate the limit. Then evaluate the limit using a method studied in Chapter 2. 1. lim
x: 2
x  2 x2  4
2. lim
x: 0
2
sin 5x x 3
5x  3x 7x 2 + 1
4. lim
x  1 4x 3  x  3
1  cos x x: 0 x2
6. lim
2x 2 + 3x x3 + x + 1
3. lim
x: q
x: 1
5. lim
x: q
Applying l’Hôpital’s Rule Use l’Hôpital’s Rule to find the limits in Exercises 7–26.
11. 13.
8.
sin u p  u
u :p
10.
sin x  cos x x  p>4 x: p>4 lim
lim
x: sp>2d
12.
p  ax  btan x 2
2asa + xd  a , x
a 7 0
xscos x  1d x: 0 sin x  x
19. lim
asr n  1d , r:1 r  1
21. lim
1 22. lim+ a x x: 0
1 2x
lim
1  sin x 1 + cos 2x
lim
cos x  0.5 x  p>3
x: p>2
x: p>3
2x
x + 72x 2x 2 + 5  3 16. lim x: 2 x2  4 10ssin t  td 18. lim t: 0 t3 sin sa + hd  sin a 20. lim h h :0
0 x  3 = = 0 6 x2  3 32. ˆ / ˆ Form Give an example of two differentiable functions ƒ and g with lim x: q ƒsxd = lim x: q g sxd = q that satisfy the following. ƒsxd ƒsxd = 3 = 0 a. lim b. lim x: q g sxd x: q g sxd b. lim
x:3
c. lim
23. lim sx  2x 2 + xd x: q
25.
lim
x: ; q
3x  5 2x 2  x + 2
Find a value of c that makes the function
9x  3 sin 3x , 5x 3 ƒsxd = • c,
x Z 0 x = 0
continuous at x = 0 . Explain why your value of c works. 34. Let ƒsxd = e
x + 2, 0,
x Z 0 x = 0
g sxd = e
and
x + 1, 0,
x Z 0 x = 0.
a. Show that lim
x:0
ƒ¿sxd = 1 g¿sxd
but
lim
x:0
ƒsxd = 2. g sxd
Estimate the value of 2x 2  s3x + 1d2x + 2 x  1 x:1
by graphing. Then confirm your estimate with l’Hôpital’s Rule. T 36. ˆ ˆ Form a. Estimate the value of lim A x  2x 2 + x B
x: q
L’Hôpital’s Rule does not help with the limits in Exercises 27–30. Try it; you just keep on cycling. Find the limits some other way. 2x + 1 sec x 29. lim x: sp>2d  tan x
33. Continuous extension
lim
Theory and Applications 29x + 1
ƒsxd = q g sxd
T 35. 0/0 Form
b
sin 7x 26. lim x: 0 tan 11x
x: q
x  3 1 1 = lim = 2 2x 6 x:3 x  3
b. Explain why this does not contradict l’Hôpital’s Rule.
n a positive integer
1 24. lim x tan x x: q
27. lim
x:3
x: 0
2x 2  s3x + 1d2x + 2 x  1 x: 1 x: 0
2x  p x: p>2 cos x lim
14. lim
15. lim 17. lim
a. lim
x: q
sin t 2 t t:0
7. lim
9. lim
31. Which one is correct, and which one is wrong? Give reasons for your answers.
28. lim+
2x
2sin x cot x 30. lim+ csc x x: 0 x: 0
by graphing ƒsxd = x  2x 2 + x over a suitably large interval of xvalues. b. Now confirm your estimate by finding the limit with l’Hôpital’s Rule. As the first step, multiply ƒ(x) by the fraction A x + 2x 2 + x B > A x + 2x 2 + x B and simplify the new numerator.
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4.6 Indeterminate Forms and L’Hôpital’s Rule T 37. Let
299
Interpret this geometrically. ƒsxd =
1  cos x 6 . x 12
y
C
Explain why some graphs of ƒ may give false information about limx:0 ƒsxd . (Hint: Try the window [1, 1] by [0.5, 1] .)
a. ƒsxd = x,
g sxd = x 2,
sa, bd = s 2, 0d
b. ƒsxd = x,
g sxd = x 2,
sa, bd arbitrary
c. ƒsxd = x >3  4x, 3
g sxd = x 2,
P(x, 0)
O
D
a. Show that the length of PA is
x
us1  cos ud . u  sin u
40. A right triangle has one leg of length 1, another of length y, and a hypotenuse of length r. The angle opposite y has radian measure u . Find the limits as u : p>2 of a. r  y. b. r 2  y 2 . c. r 3  y 3 . r
y
b. Find lim s1  xd . u :0
c. Show that lim [s1  xd  s1  cos ud] = 0. u: q
A(1, 0)
sa, bd = s0, 3d
39. In the accompanying figure, the circle has radius OA equal to 1, and AB is tangent to the circle at A. The arc AC has radian measure u and the segment AB also has length u . The line through B and C crosses the xaxis at P(x, 0).
1  x =
38. Find all values of c, that satisfy the conclusion of Cauchy’s Mean Value Theorem for the given functions and interval.
B(1, )
1
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4.7 Newton’s Method
4.7
Newton’s Method
HISTORICAL BIOGRAPHY Niels Henrik Abel (1802–1829)
299
One of the basic problems of mathematics is solving equations. Using the quadratic root formula, we know how to find a point (solution) where x 2  3x + 2 = 0. There are more complicated formulas to solve cubic or quartic equations (polynomials of degree 3 or 4), but the Norwegian mathematician Niels Abel showed that no simple formulas exist to solve polynomials of degree equal to five. There is also no simple formula for solving equations like sin x = x 2 , which involve transcendental functions as well as polynomials or other algebraic functions. In this section we study a numerical method, called Newton’s method or the Newton–Raphson method, which is a technique to approximate the solution to an equation ƒsxd = 0. Essentially it uses tangent lines in place of the graph of y = ƒsxd near the points where ƒ is zero. (A value of x where ƒ is zero is a root of the function ƒ and a solution of the equation ƒsxd = 0.)
Procedure for Newton’s Method The goal of Newton’s method for estimating a solution of an equation ƒsxd = 0 is to produce a sequence of approximations that approach the solution. We pick the first number x0 of the sequence. Then, under favorable circumstances, the method does the rest by moving step by step toward a point where the graph of ƒ crosses the xaxis (Figure 4.43). At each
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Chapter 4: Applications of Derivatives
step the method approximates a zero of ƒ with a zero of one of its linearizations. Here is how it works. The initial estimate, x0 , may be found by graphing or just plain guessing. The method then uses the tangent to the curve y = ƒsxd at sx0, ƒsx0 dd to approximate the curve, calling the point x1 where the tangent meets the xaxis (Figure 4.43). The number x1 is usually a better approximation to the solution than is x0 . The point x2 where the tangent to the curve at sx1, ƒsx1 dd crosses the xaxis is the next approximation in the sequence. We continue on, using each approximation to generate the next, until we are close enough to the root to stop. We can derive a formula for generating the successive approximations in the following way. Given the approximation xn , the pointslope equation for the tangent to the curve at sxn, ƒsxn dd is
y y f (x) (x0, f (x0))
(x1, f (x1)) (x2, f(x2 )) Root sought
y = ƒsxn d + ƒ¿sxn dsx  xn d. x
0
x3 Fourth
x2 Third
x1 Second
x0 First
We can find where it crosses the xaxis by setting y = 0 (Figure 4.44). 0 = ƒsxn d + ƒ¿sxn dsx  xn d ƒsxn d = x  xn ƒ¿sxn d ƒsxn d x = xn ƒ¿sxn d
APPROXIMATIONS
FIGURE 4.43 Newton’s method starts with an initial guess x0 and (under favorable circumstances) improves the guess one step at a time.
If ƒ¿sxn d Z 0
This value of x is the next approximation xn + 1 . Here is a summary of Newton’s method. y y f (x) Point: (xn, f (xn )) Slope: f'(xn ) Tangent line equation: y f (xn ) f '(xn )(x xn ) (xn, f (xn))
Tangent line (graph of linearization of f at xn )
Procedure for Newton’s Method 1. Guess a first approximation to a solution of the equation ƒsxd = 0. A graph of y = ƒsxd may help. 2. Use the first approximation to get a second, the second to get a third, and so on, using the formula xn + 1 = xn 
ƒsxn d , ƒ¿sxn d
if ƒ¿sxn d Z 0
(1)
Root sought 0
x
xn xn1 xn
f (xn ) f '(xn )
Applying Newton’s Method
FIGURE 4.44 The geometry of the successive steps of Newton’s method. From xn we go up to the curve and follow the tangent line down to find xn + 1 .
Applications of Newton’s method generally involve many numerical computations, making them well suited for computers or calculators. Nevertheless, even when the calculations are done by hand (which may be very tedious), they give a powerful way to find solutions of equations. In our first example, we find decimal approximations to 22 by estimating the positive root of the equation ƒsxd = x 2  2 = 0.
EXAMPLE 1
Finding the Square Root of 2
Find the positive root of the equation ƒsxd = x 2  2 = 0.
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301
With ƒsxd = x 2  2 and ƒ¿sxd = 2x, Equation (1) becomes
Solution
xn + 1 = xn 
xn 2  2 2xn
= xn 
xn 1 + xn 2
=
xn 1 + xn . 2
The equation xn + 1 =
xn 1 + xn 2
enables us to go from each approximation to the next with just a few keystrokes. With the starting value x0 = 1, we get the results in the first column of the following table. (To five decimal places, 22 = 1.41421.)
y 20
x0 x1 x2 x3
y x3 x 1
15 10
0
2
1
3
x
FIGURE 4.45 The graph of ƒsxd = x 3  x  1 crosses the xaxis once; this is the root we want to find (Example 2).
Root sought x1 x 1
1.5 1.3478 (1, –1)
FIGURE 4.46 The first three xvalues in Table 4.1 (four decimal places).
1 1 3 5
EXAMPLE 2
Using Newton’s Method
Find the xcoordinate of the point where the curve y = x 3  x crosses the horizontal line y = 1. The curve crosses the line when x 3  x = 1 or x 3  x  1 = 0. When does ƒsxd = x  x  1 equal zero? Since ƒs1d = 1 and ƒs2d = 5, we know by the Intermediate Value Theorem there is a root in the interval (1, 2) (Figure 4.45). We apply Newton’s method to ƒ with the starting value x0 = 1. The results are displayed in Table 4.1 and Figure 4.46. At n = 5, we come to the result x6 = x5 = 1.3247 17957. When xn + 1 = xn , Equation (1) shows that ƒsxn d = 0. We have found a solution ofƒsxd = 0 to nine decimals. 3
(1.5, 0.875)
x2
0.41421 0.08579 0.00246 0.00001
Solution
y x3 x 1
x0
1 1.5 1.41667 1.41422
Number of correct digits
Newton’s method is the method used by most calculators to calculate roots because it converges so fast (more about this later). If the arithmetic in the table in Example 1 had been carried to 13 decimal places instead of 5, then going one step further would have given 22 correctly to more than 10 decimal places.
5
–1
= = = =
Error
In Figure 4.47 we have indicated that the process in Example 2 might have started at the point B0s3, 23d on the curve, with x0 = 3. Point B0 is quite far from the xaxis, but the tangent at B0 crosses the xaxis at about (2.12, 0), so x1 is still an improvement over x0 . If we use Equation (1) repeatedly as before, with ƒsxd = x 3  x  1 and ƒ¿sxd = 3x 2  1, we confirm the nineplace solution x7 = x6 = 1.3247 17957 in seven steps.
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TABLE 4.1 The result of applying Newton’s method to ƒsxd = x3  x  1
with x0 = 1 n
xn
ƒ(xn)
ƒ(xn)
xn1 xn
0 1 2 3 4 5
1 1.5 1.3478 26087 1.3252 00399 1.3247 18174 1.3247 17957
1 0.875 0.1006 82173 0.0020 58362 0.0000 00924 1.8672E13
2 5.75 4.4499 05482 4.2684 68292 4.2646 34722 4.2646 32999
1.5 1.3478 26087 1.3252 00399 1.3247 18174 1.3247 17957 1.3247 17957
ƒsxn d ƒ¿sxn d
The curve in Figure 4.47 has a local maximum at x = 1> 23 and a local minimum at x = 1> 23. We would not expect good results from Newton’s method if we were to start with x0 between these points, but we can start any place to the right of x = 1> 23 and get the answer. It would not be very clever to do so, but we could even begin far to the right of B0 , for example with x0 = 10. It takes a bit longer, but the process still converges to the same answer as before.
y 25 B0(3, 23) 20 y x3 x 1 15
Convergence of Newton’s Method 10 B1(2.12, 6.35) 5 –1兾兹3
Root sought 1兾兹3
0
–1
x2 x 1 1
1.6 2.12
x0
x
3
FIGURE 4.47 Any starting value x0 to the right of x = 1> 23 will lead to the root.
y f (x)
x0
r
In practice, Newton’s method usually converges with impressive speed, but this is not guaranteed. One way to test convergence is to begin by graphing the function to estimate a good starting value for x0 . You can test that you are getting closer to a zero of the function by evaluating ƒ ƒsxn d ƒ and check that the method is converging by evaluating ƒ xn  xn + 1 ƒ . Theory does provide some help. A theorem from advanced calculus says that if
x'0
x
`
sƒ¿sxdd2
` 6 1
(2)
for all x in an interval about a root r, then the method will converge to r for any starting value x0 in that interval. Note that this condition is satisfied if the graph of ƒ is not too horizontal near where it crosses the xaxis. Newton’s method always converges if, between r and x0 , the graph of ƒ is concave up when ƒsx0 d 7 0 and concave down when ƒsx0 d 6 0. (See Figure 4.48.) In most cases, the speed of the convergence to the root r is expressed by the advanced calculus formula ƒ xn + 1  r ƒ … (++)++* error en1
FIGURE 4.48 Newton’s method will converge to r from either starting point.
ƒsxdƒ–sxd
max ƒ ƒ– ƒ ƒ x  r ƒ 2 = constant # ƒ xn  r ƒ 2, 2 min ƒ ƒ¿ ƒ n (+)+*
(3)
error en
where max and min refer to the maximum and minimum values in an interval surrounding r. The formula says that the error in step n + 1 is no greater than a constant times the square of the error in step n. This may not seem like much, but think of what it says. If the constant is less than or equal to 1 and ƒ xn  r ƒ 6 10 3 , then ƒ xn + 1  r ƒ 6 10 6 . In a single step, the method moves from three decimal places of accuracy to six, and the number of decimals of accuracy continues to double with each successive step.
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4.7 Newton’s Method
But Things Can Go Wrong
(xn, f (xn ))
x
xn
FIGURE 4.49 If ƒ¿sxn d = 0 , there is no intersection point to define xn + 1 .
y y f (x)
r 0
x0
x1
x
Newton’s method stops if ƒ¿sxn d = 0 (Figure 4.49). In that case, try a new starting point. Of course, ƒ and ƒ¿ may have the same root. To detect whether this is so, you could first find the solutions of ƒ¿sxd = 0 and check ƒ at those values, or you could graph ƒ and ƒ¿ together. Newton’s method does not always converge. For instance, if ƒsxd = e
 2r  x, 2x  r,
x 6 r x Ú r,
the graph will be like the one in Figure 4.50. If we begin with x0 = r  h, we get x1 = r + h, and successive approximations go back and forth between these two values. No amount of iteration brings us closer to the root than our first guess. If Newton’s method does converge, it converges to a root. Be careful, however. There are situations in which the method appears to converge but there is no root there. Fortunately, such situations are rare. When Newton’s method converges to a root, it may not be the root you have in mind. Figure 4.51 shows two ways this can happen. y f (x) Starting point
FIGURE 4.50 Newton’s method fails to converge. You go from x0 to x1 and back to x0 , never getting any closer to r.
Root sought
x0
Root found
y f (x) x1
x
x1 Root found
x2
x
x0 Root sought
Starting point
FIGURE 4.51 If you start too far away, Newton’s method may miss the root you want.
Fractal Basins and Newton’s Method The process of finding roots by Newton’s method can be uncertain in the sense that for some equations, the final outcome can be extremely sensitive to the starting value’s location. The equation 4x 4  4x 2 = 0 is a case in point (Figure 4.52a). Starting values in the blue zone on the xaxis lead to root A. Starting values in the black lead to root B, and starting values in the red zone lead to root C. The points ; 22>2 give horizontal tangents. The points ; 221>7 “cycle,” each leading to the other, and back (Figure 4.52b). The interval between 221>7 and 22>2 contains infinitely many open intervals of points leading to root A, alternating with intervals of points leading to root C (Figure 4.52c). The boundary points separating consecutive intervals (there are infinitely many) do not lead to roots, but cycle back and forth from one to another. Moreover, as we select points that approach 221>7 from the right, it becomes increasingly difficult to distinguish which lead to root A and which to root C. On the same side of 221>7, we find arbitrarily close together points whose ultimate destinations are far apart. If we think of the roots as “attractors” of other points, the coloring in Figure 4.52 shows the intervals of the points they attract (the “intervals of attraction”). You might think that points between roots A and B would be attracted to either A or B, but, as we see, that is not the case. Between A and B there are infinitely many intervals of points attracted to C. Similarly between B and C lie infinitely many intervals of points attracted to A. We encounter an even more dramatic example of such behavior when we apply Newton’s method to solve the complexnumber equation z 6  1 = 0. It has six solutions: 1, 1, and the four numbers ;s1>2d ; A 23>2 B i. As Figure 4.53 suggests, each of the Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
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y y 4x 4 – 4x 2
– 兹21 7
Root A –1
Root B 0
– 兹2 2
兹21 7 兹2 2
– 兹21 7
Root C 1
兹21 7 0
x
x 兹2 2
– 兹2 2
–1 (a)
(b)
兹21 7
兹2 2 (c)
FIGURE 4.52 (a) Starting values in A  q ,  22>2 B , A  221>7, 221>7 B , and A 22>2, q B lead respectively to roots A, B, and C. (b) The values x = ; 221>7 lead only to each other. (c) Between 221>7 and 22>2 , there are infinitely many open intervals of points attracted to A alternating with open intervals of points attracted to C. This behavior is mirrored in the interval A  22>2,  221>7 B .
FIGURE 4.53 This computergenerated initial value portrait uses color to show where different points in the complex plane end up when they are used as starting values in applying Newton’s method to solve the equation z 6  1 = 0 . Red points go to 1, green points to s1>2d + A 23>2 B i , dark blue points to s 1>2d + A 23>2 B i , and so on. Starting values that generate sequences that do not arrive within 0.1 unit of a root after 32 steps are colored black.
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six roots has infinitely many “basins” of attraction in the complex plane (Appendix 5). Starting points in red basins are attracted to the root 1, those in the green basin to the root s1>2d + A 23>2 B i, and so on. Each basin has a boundary whose complicated pattern repeats without end under successive magnifications. These basins are called fractal basins.
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EXERCISES 4.7 RootFinding 1. Use Newton’s method to estimate the solutions of the equation x 2 + x  1 = 0 . Start with x0 = 1 for the lefthand solution and with x0 = 1 for the solution on the right. Then, in each case, find x2 . 2. Use Newton’s method to estimate the one real solution of x 3 + 3x + 1 = 0 . Start with x0 = 0 and then find x2 . 3. Use Newton’s method to estimate the two zeros of the function ƒsxd = x 4 + x  3 . Start with x0 = 1 for the lefthand zero and with x0 = 1 for the zero on the right. Then, in each case, find x2 . 4. Use Newton’s method to estimate the two zeros of the function ƒsxd = 2x  x 2 + 1 . Start with x0 = 0 for the lefthand zero and with x0 = 2 for the zero on the right. Then, in each case, find x2 . 5. Use Newton’s method to find the positive fourth root of 2 by solving the equation x 4  2 = 0 . Start with x0 = 1 and find x2 . 6. Use Newton’s method to find the negative fourth root of 2 by solving the equation x 4  2 = 0 . Start with x0 = 1 and find x2 .
Theory, Examples, and Applications 7. Guessing a root Suppose that your first guess is lucky, in the sense that x0 is a root of ƒsxd = 0 . Assuming that ƒ¿sx0 d is defined and not 0, what happens to x1 and later approximations? 8. Estimating pi You plan to estimate p>2 to five decimal places by using Newton’s method to solve the equation cos x = 0 . Does it matter what your starting value is? Give reasons for your answer. 9. Oscillation Show that if h 7 0 , applying Newton’s method to ƒsxd = e
2x, 2x,
x Ú 0 x 6 0
leads to x1 = h if x0 = h and to x1 = h if x0 = h . Draw a picture that shows what is going on.
iii) Find the xcoordinates of the points where the curve y = x 3  3x crosses the horizontal line y = 1 . iv) Find the values of x where the derivative of g sxd = s1>4dx 4  s3>2dx 2  x + 5 equals zero. 12. Locating a planet To calculate a planet’s space coordinates, we have to solve equations like x = 1 + 0.5 sin x . Graphing the function ƒsxd = x  1  0.5 sin x suggests that the function has a root near x = 1.5 . Use one application of Newton’s method to improve this estimate. That is, start with x0 = 1.5 and find x1 . (The value of the root is 1.49870 to five decimal places.) Remember to use radians. T 13. A program for using Newton’s method on a grapher Let ƒsxd = x 3 + 3x + 1 . Here is a home screen program to perform the computations in Newton’s method. a. Let y0 = ƒsxd and y1 = NDER ƒsxd . b. Store x0 = 0.3 into x .
c. Then store x  sy0>y1 d into x and press the Enter key over and over. Watch as the numbers converge to the zero of ƒ.
d. Use different values for x0 and repeat steps (b) and (c). e. Write your own equation and use this approach to solve it using Newton’s method. Compare your answer with the answer given by the builtin feature of your calculator that gives zeros of functions. T 14. (Continuation of Exercise 11.) a. Use Newton’s method to find the two negative zeros of ƒsxd = x 3  3x  1 to five decimal places. b. Graph ƒsxd = x 3  3x  1 for 2 … x … 2.5 . Use the Zoom and Trace features to estimate the zeros of ƒ to five decimal places. c. Graph g sxd = 0.25x 4  1.5x 2  x + 5 . Use the Zoom and Trace features with appropriate rescaling to find, to five decimal places, the values of x where the graph has horizontal tangents.
10. Approximations that get worse and worse Apply Newton’s method to ƒsxd = x 1>3 with x0 = 1 and calculate x1 , x2 , x3 , and x4 . Find a formula for ƒ xn ƒ . What happens to ƒ xn ƒ as n : q ? Draw a picture that shows what is going on.
T 15. Intersecting curves The curve y = tan x crosses the line y = 2x between x = 0 and x = p>2 . Use Newton’s method to find where.
11. Explain why the following four statements ask for the same information:
T 16. Real solutions of a quartic Use Newton’s method to find the two real solutions of the equation x 4  2x 3  x 2  2x + 2 = 0 .
i) Find the roots of ƒsxd = x 3  3x  1 . ii) Find the xcoordinates of the intersections of the curve y = x 3 with the line y = 3x + 1 .
T 17. a. How many solutions does the equation sin 3x = 0.99  x 2 have? b. Use Newton’s method to find them.
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T 18. Intersection of curves a. Does cos 3x ever equal x? Give reasons for your answer. b. Use Newton’s method to find where. T 19. Find the four real zeros of the function ƒsxd = 2x 4  4x 2 + 1 . T 20. Estimating pi Estimate p to as many decimal places as your calculator will display by using Newton’s method to solve the equation tan x = 0 with x0 = 3 . 21. At what values(s) of x does cos x = 2x ?
a. Show that the value of x that minimizes the distance between the submarine and the buoy is a solution of the equation x = 1>sx 2 + 1d . b. Solve the equation x = 1>sx 2 + 1d with Newton’s method. 27. Curves that are nearly flat at the root Some curves are so flat that, in practice, Newton’s method stops too far from the root to give a useful estimate. Try Newton’s method on ƒsxd = sx  1d40 with a starting value of x0 = 2 to see how close your machine comes to the root x = 1 .
22. At what value(s) of x does cos x = x ? 23. Use the Intermediate Value Theorem from Section 2.6 to show that ƒsxd = x 3 + 2x  4 has a root between x = 1 and x = 2 . Then find the root to five decimal places.
y
24. Factoring a quartic Find the approximate values of r1 through r4 in the factorization 8x 4  14x 3  9x 2 + 11x  1 = 8sx  r1 dsx  r2 dsx  r3 dsx  r4 d . y
y 8x 4 14x 3 9x 2 11x 1
2 –1
1
–2 –4 –6 –8 –10 –12
Slope –40
x
2
y (x 1) 40 Slope 40
1
(2, 1)
Nearly flat 0
T 25. Converging to different zeros Use Newton’s method to find the zeros of ƒsxd = 4x 4  4x 2 using the given starting values (Figure 4.52). a. x0 = 2 and x0 = 0.8 , lying in A  q ,  22>2 B b. x0 = 0.5 and x0 = 0.25 , lying in A  221>7, 221>7 B c. x0 = 0.8 and x0 = 2 , lying in A 22>2, q B
d. x0 =  221>7 and x0 = 221>7
26. The sonobuoy problem In submarine location problems, it is often necessary to find a submarine’s closest point of approach (CPA) to a sonobuoy (sound detector) in the water. Suppose that the submarine travels on the parabolic path y = x 2 and that the buoy is located at the point s2, 1>2d .
x
2
28. Finding a root different from the one sought All three roots of ƒsxd = 4x 4  4x 2 can be found by starting Newton’s method near x = 221>7 . Try it. (See Figure 4.52.) 29. Finding an ion concentration While trying to find the acidity of a saturated solution of magnesium hydroxide in hydrochloric acid, you derive the equation 3.64 * 10 11 = [H3 O+] + 3.6 * 10 4 [H3 O+]2 for the hydronium ion concentration [H3 O+] . To find the value of [H3 O+] , you set x = 10 4[H3 O+] and convert the equation to x 3 + 3.6x 2  36.4 = 0 . You then solve this by Newton’s method. What do you get for x? (Make it good to two decimal places.) For [H3 O+] ?
y y x2 Submarine track in two dimensions
1
T 30. Complex roots If you have a computer or a calculator that can be programmed to do complexnumber arithmetic, experiment with Newton’s method to solve the equation z 6  1 = 0 . The recursion relation to use is
CPA
zn + 1 = zn 0
1
1
2
1 Sonobuoy 2, – 2
x
z n6  1 6z n5
or
zn + 1 =
5 1 z + . 6 n 6z n5
Try these starting values (among others): 2, i, 23 + i .
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Antiderivatives We have studied how to find the derivative of a function. However, many problems require that we recover a function from its known derivative (from its known rate of change). For instance, we may know the velocity function of an object falling from an initial height and need to know its height at any time over some period. More generally, we want to find a function F from its derivative ƒ. If such a function F exists, it is called an antiderivative of ƒ.
Finding Antiderivatives
DEFINITION Antiderivative A function F is an antiderivative of ƒ on an interval I if F¿sxd = ƒsxd for all x in I.
The process of recovering a function F(x) from its derivative ƒ(x) is called antidifferentiation. We use capital letters such as F to represent an antiderivative of a function ƒ, G to represent an antiderivative of g, and so forth.
EXAMPLE 1
Finding Antiderivatives
Find an antiderivative for each of the following functions. (a) ƒsxd = 2x (b) gsxd = cos x (c) hsxd = 2x + cos x Solution
(a) Fsxd = x 2 (b) Gsxd = sin x (c) Hsxd = x 2 + sin x Each answer can be checked by differentiating. The derivative of Fsxd = x 2 is 2x. The derivative of Gsxd = sin x is cos x and the derivative of Hsxd = x 2 + sin x is 2x + cos x.
The function Fsxd = x 2 is not the only function whose derivative is 2x. The function x + 1 has the same derivative. So does x 2 + C for any constant C. Are there others? Corollary 2 of the Mean Value Theorem in Section 4.2 gives the answer: Any two antiderivatives of a function differ by a constant. So the functions x 2 + C, where C is an arbitrary constant, form all the antiderivatives of ƒsxd = 2x. More generally, we have the following result. 2
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If F is an antiderivative of ƒ on an interval I, then the most general antiderivative of ƒ on I is Fsxd + C where C is an arbitrary constant.
Thus the most general antiderivative of ƒ on I is a family of functions Fsxd + C whose graphs are vertical translates of one another. We can select a particular antiderivative from this family by assigning a specific value to C. Here is an example showing how such an assignment might be made.
EXAMPLE 2
Finding a Particular Antiderivative
Find an antiderivative of ƒsxd = sin x that satisfies Fs0d = 3. Solution
Since the derivative of cos x is sin x, the general antiderivative Fsxd = cos x + C
gives all the antiderivatives of ƒ(x). The condition Fs0d = 3 determines a specific value for C. Substituting x = 0 into Fsxd = cos x + C gives Fs0d = cos 0 + C = 1 + C. Since Fs0d = 3, solving for C gives C = 4. So Fsxd = cos x + 4 is the antiderivative satisfying Fs0d = 3. By working backward from assorted differentiation rules, we can derive formulas and rules for antiderivatives. In each case there is an arbitrary constant C in the general expression representing all antiderivatives of a given function. Table 4.2 gives antiderivative formulas for a number of important functions. TABLE 4.2 Antiderivative formulas
Function
General antiderivative
1.
xn
xn+1 + C, n + 1
2.
sin kx

3.
cos kx
sin kx + C, k
4. 5. 6. 7.
sec2 x csc2 x sec x tan x csc x cot x
tan x + C cot x + C sec x + C csc x + C
cos kx + C, k
n Z 1, n rational k a constant, k Z 0 k a constant, k Z 0
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The rules in Table 4.2 are easily verified by differentiating the general antiderivative formula to obtain the function to its left. For example, the derivative of tan x + C is sec2 x, whatever the value of the constant C, and this establishes the formula for the most general antiderivative of sec2 x.
EXAMPLE 3
Finding Antiderivatives Using Table 4.2
Find the general antiderivative of each of the following functions. (a) ƒsxd = x 5 1 2x (c) hsxd = sin 2x (b) gsxd =
(d) isxd = cos
x 2
Solution
x6 + C 6
(a) Fsxd =
Formula 1 with n = 5
(b) gsxd = x 1>2 , so Gsxd =
x 1>2 + C = 22x + C 1>2
Formula 1 with n = 1>2
cos 2x + C 2
(c) Hsxd =
Formula 2 with k = 2
sin sx>2d Formula 3 x + C = 2 sin + C with k = 1>2 2 1>2 Other derivative rules also lead to corresponding antiderivative rules. We can add and subtract antiderivatives, and multiply them by constants. The formulas in Table 4.3 are easily proved by differentiating the antiderivatives and verifying that the result agrees with the original function. Formula 2 is the special case k = 1 in Formula 1. (d) Isxd =
TABLE 4.3 Antiderivative linearity rules
1. 2. 3.
Constant Multiple Rule: Negative Rule: Sum or Difference Rule:
EXAMPLE 4
Function
General antiderivative
kƒ(x) ƒsxd ƒsxd ; gsxd
kFsxd + C, k a constant Fsxd + C, Fsxd ; Gsxd + C
Using the Linearity Rules for Antiderivatives
Find the general antiderivative of ƒsxd =
3 2x
+ sin 2x.
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We have that ƒsxd = 3gsxd + hsxd for the functions g and h in Example 3. Since Gsxd = 22x is an antiderivative of g(x) from Example 3b, it follows from the Constant Multiple Rule for antiderivatives that 3Gsxd = 3 # 22x = 62x is an antiderivative of 3g sxd = 3> 2x. Likewise, from Example 3c we know that Hsxd = s 1>2d cos 2x is an antiderivative of hsxd = sin 2x. From the Sum Rule for antiderivatives, we then get that Solution
Fsxd = 3Gsxd + Hsxd + C = 62x 
1 cos 2x + C 2
is the general antiderivative formula for ƒ(x), where C is an arbitrary constant. Antiderivatives play several important roles, and methods and techniques for finding them are a major part of calculus. (This is the subject of Chapter 8.)
Initial Value Problems and Differential Equations Finding an antiderivative for a function ƒ(x) is the same problem as finding a function y(x) that satisfies the equation dy = ƒsxd. dx This is called a differential equation, since it is an equation involving an unknown function y that is being differentiated. To solve it, we need a function y(x) that satisfies the equation. This function is found by taking the antiderivative of ƒ(x). We fix the arbitrary constant arising in the antidifferentiation process by specifying an initial condition ysx0 d = y0 . This condition means the function y(x) has the value y0 when x = x0 . The combination of a differential equation and an initial condition is called an initial value problem. Such problems play important roles in all branches of science. Here’s an example of solving an initial value problem.
EXAMPLE 5
Finding a Curve from Its Slope Function and a Point
Find the curve whose slope at the point (x, y) is 3x 2 if the curve is required to pass through the point s1, 1d. In mathematical language, we are asked to solve the initial value problem that consists of the following.
Solution
dy = 3x 2 dx
The differential equation: The initial condition: 1.
The curve’s slope is 3x 2 .
ys1d = 1
Solve the differential equation: The function y is an antiderivative of ƒsxd = 3x 2 , so y = x 3 + C. This result tells us that y equals x 3 + C for some value of C. We find that value from the initial condition ys1d = 1.
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2.
y
y x3 C 2 1
C2 C1 C0 C –1 C –2 x
0 –1
(1, –1)
–2
Antiderivatives
311
Evaluate C: y = x3 + C 1 = s1d3 + C C = 2.
Initial condition ys1d = 1
The curve we want is y = x 3  2 (Figure 4.54). The most general antiderivative Fsxd + C (which is x 3 + C in Example 5) of the function ƒ(x) gives the general solution y = Fsxd + C of the differential equation dy>dx = ƒsxd. The general solution gives all the solutions of the equation (there are infinitely many, one for each value of C). We solve the differential equation by finding its general solution. We then solve the initial value problem by finding the particular solution that satisfies the initial condition ysx0 d = y0 .
Antiderivatives and Motion FIGURE 4.54 The curves y = x 3 + C fill the coordinate plane without overlapping. In Example 5, we identify the curve y = x 3  2 as the one that passes through the given point s1, 1d .
We have seen that the derivative of the position of an object gives its velocity, and the derivative of its velocity gives its acceleration. If we know an object’s acceleration, then by finding an antiderivative we can recover the velocity, and from an antiderivative of the velocity we can recover its position function. This procedure was used as an application of Corollary 2 in Section 4.2. Now that we have a terminology and conceptual framework in terms of antiderivatives, we revisit the problem from the point of view of differential equations.
EXAMPLE 6
Dropping a Package from an Ascending Balloon
A balloon ascending at the rate of 12 ft>sec is at a height 80 ft above the ground when a package is dropped. How long does it take the package to reach the ground? Solution Let y (t) denote the velocity of the package at time t, and let s(t) denote its height above the ground. The acceleration of gravity near the surface of the earth is 32 ft>sec2 . Assuming no other forces act on the dropped package, we have
dy = 32. dt
Negative because gravity acts in the direction of decreasing s.
This leads to the initial value problem. Differential equation:
dy = 32 dt
Initial condition:
ys0d = 12,
which is our mathematical model for the package’s motion. We solve the initial value problem to obtain the velocity of the package. 1.
Solve the differential equation: The general formula for an antiderivative of 32 is y = 32t + C.
2.
Having found the general solution of the differential equation, we use the initial condition to find the particular solution that solves our problem. Evaluate C: 12 = 32s0d + C C = 12.
Initial condition ys0d = 12
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The solution of the initial value problem is y = 32t + 12. Since velocity is the derivative of height and the height of the package is 80 ft at the time t = 0 when it is dropped, we now have a second initial value problem. Differential equation:
ds = 32t + 12 dt
Initial condition:
ss0d = 80
Set y = ds>dt in the last equation.
We solve this initial value problem to find the height as a function of t. 1.
Solve the differential equation: Finding the general antiderivative of 32t + 12 gives s = 16t 2 + 12t + C.
2.
Evaluate C: 80 = 16s0d2 + 12s0d + C C = 80.
Initial condition ss0d = 80
The package’s height above ground at time t is s = 16t 2 + 12t + 80. Use the solution: To find how long it takes the package to reach the ground, we set s equal to 0 and solve for t: 16t 2 + 12t + 80 = 0 4t 2 + 3t + 20 = 0 t =
3 ; 2329 8
t L 1.89,
Quadratic formula
t L 2.64.
The package hits the ground about 2.64 sec after it is dropped from the balloon. (The negative root has no physical meaning.)
Indefinite Integrals A special symbol is used to denote the collection of all antiderivatives of a function ƒ.
DEFINITION Indefinite Integral, Integrand The set of all antiderivatives of ƒ is the indefinite integral of ƒ with respect to x, denoted by L
ƒsxd dx.
The symbol 1 is an integral sign. The function ƒ is the integrand of the integral, and x is the variable of integration.
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Using this notation, we restate the solutions of Example 1, as follows: L L L
2x dx = x 2 + C,
cos x dx = sin x + C,
s2x + cos xd dx = x 2 + sin x + C.
This notation is related to the main application of antiderivatives, which will be explored in Chapter 5. Antiderivatives play a key role in computing limits of infinite sums, an unexpected and wonderfully useful role that is described in a central result of Chapter 5, called the Fundamental Theorem of Calculus.
EXAMPLE 7
Indefinite Integration Done TermbyTerm and Rewriting the Constant of Integration
Evaluate L
sx 2  2x + 5d dx.
If we recognize that sx 3>3d  x 2 + 5x is an antiderivative of x 2  2x + 5, we can evaluate the integral as
Solution
antiderivative $++%++&
L
(x 2  2x + 5) dx =
x3  x 2 + 5x + C. 3 ()* arbitrary constant
If we do not recognize the antiderivative right away, we can generate it termbyterm with the Sum, Difference, and Constant Multiple Rules: L
sx 2  2x + 5d dx =
L
x 2 dx 
L
2x dx +
L
5 dx
x 2 dx  2
x dx + 5 1 dx L L L x3 x2 = a + C1 b  2 a + C2 b + 5sx + C3 d 3 2 =
=
x3 + C1  x 2  2C2 + 5x + 5C3 . 3
This formula is more complicated than it needs to be. If we combine C1, 2C2 , and 5C3 into a single arbitrary constant C = C1  2C2 + 5C3 , the formula simplifies to x3  x 2 + 5x + C 3
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and still gives all the antiderivatives there are. For this reason, we recommend that you go right to the final form even if you elect to integrate termbyterm. Write L
sx 2  2x + 5d dx =
x 2 dx 2x dx + 5 dx L L L x3 =  x 2 + 5x + C. 3
Find the simplest antiderivative you can for each part and add the arbitrary constant of integration at the end.
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EXERCISES 4.8 Finding Antiderivatives
Finding Indefinite Integrals
In Exercises 1–16, find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation.
In Exercises 17–54, find the most general antiderivative or indefinite integral. Check your answers by differentiation.
1. a. 2x
b. x 2
c. x 2  2x + 1
2. a. 6x
b. x 7
c. x 7  6x + 8
3. a. 3x 4
b. x 4
c. x 4 + 2x + 3
4. a. 2x 3
b.
1 5. a. 2 x
5 b. 2 x
5 c. 2  2 x
2 6. a.  3 x 3 7. a. 2x 2
1 b. 2x 3 1 b. 2 2x 1 b. 3 32 x
1 c. x  3 x
4 3 x 8. a. 2 3
x 3 + x2 2
9. a.
2 1>3 x 3
b.
10. a.
1 1>2 x 2
b. 
12. a. p cos px
b.
13. a. sec2 x
c. 2x + 3
1 3>2 x 2
b. 3 sin x p cos 2 2 b. sec2 3
L
px 2 x 3
a3t 2 +
20.
t b dt 2
1 1 a 2  x 2  b dx 3 L x
24.
L
t2 + 4t 3 b dt L 2 L
L
L
3 x B dx A 2x + 2
28.
L
a8y 
30.
27.
1 4>3 x 3
29.
c. 
L
c. 
3 5>2 x 2
31.
L
2 b dy y 1>4
s1  x 2  3x 5 d dx
1 2 a  3 + 2xb dx x L 5
26.
2x 1
s5  6xd dx a
x 1>3 dx
L
x 5>4 dx a
2x 2 b dx + 2 2x
1 1 a  5>4 b dy y L 7
2xs1  x 3 d dx
32.
t2t + 2t dt t2 L
34.
L
s 2 cos td dt
36.
L
s 5 sin td dt
L
7 sin
u du 3
38.
L
3 cos 5u du
s 3 csc2 xd dx
40.
csc u cot u du 41. 2 L
42.
c. sin px  3 sin 3x
33.
px + p cos x 2 3x c. sec2 2
35.
L
x 3sx + 1d dx
4 + 2t dt t3 L
c. cos
c. 1  8 csc 2x
15. a. csc x cot x
b. csc 5x cot 5x
c. p csc
b. 4 sec 3x tan 3x
18.
22.
25.
3 2 x
sx + 1d dx
s2x 3  5x + 7d dx
1
14. a. csc x
16. a. sec x tan x
21. 23.
3x 3 b.  csc2 2 2
2
L
c. x 3 + x  1
c. 2x +
L
19.
3
1 2>3 x 3
11. a. p sin px
17.
2
c. sec
px px cot 2 2
px px tan 2 2
37. 39.
L
L
a
sec2 x b dx 3
2 sec u tan u du L5
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4.8 Antiderivatives
43. 45. 47.
L L
s4 sec x tan x  2 sec2 xd dx 44. ssin 2x  csc2 xd dx
46.
1 + cos 4t dt 2 L
48.
1 scsc2 x  csc x cot xd dx L2 L
1  cos 6t dt 2 L
49.
s1 + tan2 ud du L (Hint: 1 + tan2 u = sec2 u)
50.
L
51.
cot2 x dx L (Hint: 1 + cot2 x = csc2 x)
52.
L
cos u stan u + sec ud du
54.
53.
L
s2 cos 2x  3 sin 3xd dx
s2 + tan2 ud du
63. Right, or wrong? Say which for each formula and give a brief reason for each answer. a.
L
b.
L
s2x + 1d2 dx =
3s2x + 1d2 dx = s2x + 1d3 + C
6s2x + 1d2 dx = s2x + 1d3 + C L 64. Right, or wrong? Say which for each formula and give a brief reason for each answer. c.
s1  cot2 xd dx
csc u du csc u  sin u L
s2x + 1d3 + C 3
a.
L
b.
L
c.
Checking Antiderivative Formulas
22x + 1 dx = 2x 2 + x + C 22x + 1 dx = 2x 2 + x + C 22x + 1 dx =
L
1 A 22x + 1 B 3 + C 3
Verify the formulas in Exercises 55–60 by differentiation.
Initial Value Problems
s7x  2d4 s7x  2d3 dx = 55. + C 28 L s3x + 5d1 s3x + 5d2 dx = + C 56. 3 L
65. Which of the following graphs shows the solution of the initial value problem
57.
L
58.
L
sec2 s5x  1d dx = csc2 a
dy = 2x, dx
1 tan s5x  1d + C 5
y
y
4
1 1 dx = + C 2 x + 1 L sx + 1d x 1 dx = + C 60. 2 x + 1 L sx + 1d 61. Right, or wrong? Say which for each formula and give a brief reason for each answer. x2 sin x + C 2
a.
L
x sin x dx =
b.
L
x sin x dx = x cos x + C
3
b. c.
L
3
2 1
1
1
–1 0
L
x
1
1 2 tan u + C 2
tan u sec2 u du =
1 sec2 u + C 2
–1 0
(a)
(1, 4)
2
1
x
–1 0
(b)
1
x
(c)
Give reasons for your answer. 66. Which of the following graphs shows the solution of the initial value problem dy = x, dx
y = 1 when x = 1 ? y
y
sec3 u + C 3
tan u sec2 u du =
4
(1, 4)
3 2
x sin x dx = x cos x + sin x + C L 62. Right, or wrong? Say which for each formula and give a brief reason for each answer. tan u sec2 u du =
4
(1, 4)
c.
L
y
x  1 x  1 b dx = 3 cot a b + C 3 3
59.
a.
y = 4 when x = 1 ?
y
(–1, 1) (–1, 1) 0 (a)
x
(–1, 1) 0 (b)
Give reasons for your answer.
Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
x
0 (c)
x
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Chapter 4: Applications of Derivatives
Solve the initial value problems in Exercises 67–86. dy 67. = 2x  7, y s2d = 0 dx 68.
dy = 10  x, dx
y s0d = 1
88. a. Find a curve y = ƒsxd with the following properties: d 2y = 6x i) dx 2 ii) Its graph passes through the point (0, 1), and has a horizontal tangent there. b. How many curves like this are there? How do you know?
dy 1 = 2 + x, x 7 0; y s2d = 1 dx x dy = 9x 2  4x + 5, y s 1d = 0 70. dx 69.
dy = 3x 2>3, 71. dx
Solution (Integral) Curves Exercises 89–92 show solution curves of differential equations. In each exercise, find an equation for the curve through the labeled point.
y s 1d = 5
89.
dy 1 , y s4d = 0 = 72. dx 2 2x ds = 1 + cos t, s s0d = 4 73. dt 74.
ds = cos t + sin t, dt
2 1 (1, 0.5)
s spd = 1
0
1
2
91.
p y a b = 7 2
80. 81.
dx 2
y¿s0d = 4,
d 2r 2 = 3; dt 2 t
1
2
x
y¿s0d = 2,
1
y s0d = 1
0
2
y s0d = 0
dr = 1, ` dt t = 1
r s1d = 1
3t d 2s ds = ; = 3, s s4d = 4 ` 8 dt t = 4 dt 2 d 3y = 6; y–s0d = 8, y¿s0d = 0, 83. dx 3
x
4
2
0
(1, 2)
1
2
3
–2
82.
y s0d = 5
d3 u 1 = 0; u–s0d = 2, u¿s0d =  , us0d = 22 84. 2 dt 3 85. y s4d = sin t + cos t ; y‡s0d = 7, y–s0d = y¿s0d = 1, y s0d = 0 86. y s4d = cos x + 8 sin 2x ; y‡s0d = 0, y–s0d = y¿s0d = 1,
dy 1 sin x dx 2兹x
6
(–, –1)
= 0;
92. y
dy sin x cos x dx y
2
dx 2 d 2y
0
r s0d = 1
dy = 8t + csc2 t, 78. dt = 2  6x;
1
–1
y s0d = 1
d y
x
(–1, 1)
–1
r s0d = 0
dy 1 = sec t tan t, 77. 2 dt
79.
y dy x 1 dx
–1
dr = p sin pu, 75. du dr = cos pu, 76. du
90.
dy 1 4 x1/3 3 dx
y
y s0d = 3
Applications 93. Finding displacement from an antiderivative of velocity a. Suppose that the velocity of a body moving along the saxis is ds = y = 9.8t  3 . dt i) Find the body’s displacement over the time interval from t = 1 to t = 3 given that s = 5 when t = 0 .
Finding Curves
ii) Find the body’s displacement from t = 1 to t = 3 given that s = 2 when t = 0 .
87. Find the curve y = ƒsxd in the xyplane that passes through the point (9, 4) and whose slope at each point is 3 2x .
iii) Now find the body’s displacement from t = 1 to t = 3 given that s = s0 when t = 0 .
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4.8 Antiderivatives b. Suppose that the position s of a body moving along a coordinate line is a differentiable function of time t. Is it true that once you know an antiderivative of the velocity function ds>dt you can find the body’s displacement from t = a to t = b even if you do not know the body’s exact position at either of those times? Give reasons for your answer. 94. Liftoff from Earth A rocket lifts off the surface of Earth with a constant acceleration of 20 m>sec2 . How fast will the rocket be going 1 min later?
99. Motion with constant acceleration The standard equation for the position s of a body moving with a constant acceleration a along a coordinate line is s =
2
Differential equation: Initial conditions:
d s = k dt 2
sk constantd
2. Find the value of t that makes ds>dt = 0 . (The answer will involve k.) 3. Find the value of k that makes s = 242 for the value of t you found in Step 2. 96. Stopping a motorcycle The State of Illinois Cycle Rider Safety Program requires riders to be able to brake from 30 mph (44 ft>sec) to 0 in 45 ft. What constant deceleration does it take to do that? 97. Motion along a coordinate line A particle moves on a coordinate line with acceleration a = d 2s>dt 2 = 15 2t  A 3> 2t B , subject to the conditions that ds>dt = 4 and s = 0 when t = 1 . Find
Initial conditions:
T 98. The hammer and the feather When Apollo 15 astronaut David Scott dropped a hammer and a feather on the moon to demonstrate that in a vacuum all bodies fall with the same (constant) acceleration, he dropped them from about 4 ft above the ground. The television footage of the event shows the hammer and the feather falling more slowly than on Earth, where, in a vacuum, they would have taken only half a second to fall the 4 ft. How long did it take the hammer and feather to fall 4 ft on the moon? To find out, solve the following initial value problem for s as a function of t. Then find the value of t that makes s equal to 0. Differential equation: Initial conditions:
d 2s = 5.2 ft>sec2 dt 2 ds = 0 and s = 4 when t = 0 dt
ds = y0 and s = s0 when t = 0 . dt
100. Free fall near the surface of a planet For free fall near the surface of a planet where the acceleration due to gravity has a constant magnitude of g lengthunits>sec2 , Equation (1) in Exercise 99 takes the form s = 
1 2 gt + y0 t + s0 , 2
(2)
where s is the body’s height above the surface. The equation has a minus sign because the acceleration acts downward, in the direction of decreasing s. The velocity y0 is positive if the object is rising at time t = 0 and negative if the object is falling. Instead of using the result of Exercise 99, you can derive Equation (2) directly by solving an appropriate initial value problem. What initial value problem? Solve it to be sure you have the right one, explaining the solution steps as you go along.
Theory and Examples 101. Suppose that ƒsxd =
a. the velocity y = ds>dt in terms of t b. the position s in terms of t.
(1)
d 2s = a dt 2
Differential equation:
ds = 88 and s = 0 when t = 0 . dt Measuring time and distance from when the brakes are applied.
a 2 t + y0 t + s0 , 2
where y0 and s0 are the body’s velocity and position at time t = 0 . Derive this equation by solving the initial value problem
95. Stopping a car in time You are driving along a highway at a steady 60 mph (88 ft>sec) when you see an accident ahead and slam on the brakes. What constant deceleration is required to stop your car in 242 ft? To find out, carry out the following steps. 1. Solve the initial value problem
317
d A 1  2x B dx
and
g sxd =
d sx + 2d . dx
Find: a.
L
ƒsxd dx
b.
L
g sxd dx
c.
L
[ƒsxd] dx
d.
L
[g sxd] dx
e.
L
[ƒsxd + g sxd] dx
f.
L
[ƒsxd  g sxd] dx
102. Uniqueness of solutions If differentiable functions y = Fsxd and y = Gsxd both solve the initial value problem dy = ƒsxd, dx
y sx0 d = y0 ,
on an interval I, must Fsxd = Gsxd for every x in I ? Give reasons for your answer.
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Chapter 4: Applications of Derivatives
COMPUTER EXPLORATIONS
105. y¿ =
Use a CAS to solve the initial problems in Exercises 103–106. Plot the solution curves. 2
103. y¿ = cos x + sin x, 1 104. y¿ = x + x,
y spd = 1
1 24  x 2
2 106. y– = x + 2x,
,
y s0d = 2 y s1d = 0,
y¿s1d = 0
y s1d = 1
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Chapter 4: Applications of Derivatives
Chapter 4
Questions to Guide Your Review
1. What can be said about the extreme values of a function that is continuous on a closed interval?
13. List the steps you would take to graph a polynomial function. Illustrate with an example.
2. What does it mean for a function to have a local extreme value on its domain? An absolute extreme value? How are local and absolute extreme values related, if at all? Give examples.
14. What is a cusp? Give examples.
3. How do you find the absolute extrema of a continuous function on a closed interval? Give examples.
16. Outline a general strategy for solving maxmin problems. Give examples.
4. What are the hypotheses and conclusion of Rolle’s Theorem? Are the hypotheses really necessary? Explain.
17. Describe l’Hôpital’s Rule. How do you know when to use the rule and when to stop? Give an example.
5. What are the hypotheses and conclusion of the Mean Value Theorem? What physical interpretations might the theorem have?
18. How can you sometimes handle limits that lead to indeterminate forms q > q , q # 0 , and q  q . Give examples.
6. State the Mean Value Theorem’s three corollaries. 7. How can you sometimes identify a function ƒ(x) by knowing ƒ¿ and knowing the value of ƒ at a point x = x0 ? Give an example. 8. What is the First Derivative Test for Local Extreme Values? Give examples of how it is applied. 9. How do you test a twicedifferentiable function to determine where its graph is concave up or concave down? Give examples. 10. What is an inflection point? Give an example. What physical significance do inflection points sometimes have? 11. What is the Second Derivative Test for Local Extreme Values? Give examples of how it is applied. 12. What do the derivatives of a function tell you about the shape of its graph?
15. List the steps you would take to graph a rational function. Illustrate with an example.
19. Describe Newton’s method for solving equations. Give an example. What is the theory behind the method? What are some of the things to watch out for when you use the method? 20. Can a function have more than one antiderivative? If so, how are the antiderivatives related? Explain.
21. What is an indefinite integral? How do you evaluate one? What general formulas do you know for finding indefinite integrals? 22. How can you sometimes solve a differential equation of the form dy>dx = ƒsxd ? 23. What is an initial value problem? How do you solve one? Give an example. 24. If you know the acceleration of a body moving along a coordinate line as a function of time, what more do you need to know to find the body’s position function? Give an example.
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Chapter 4: Applications of Derivatives
Chapter 4
Practice Exercises
Existence of Extreme Values 1. Does ƒsxd = x 3 + 2x + tan x have any local maximum or minimum values? Give reasons for your answer.
3. Does ƒsxd = s7 + xds11  3xd1>3 have an absolute minimum value? An absolute maximum? If so, find them or give reasons why they fail to exist. List all critical points of ƒ.
2. Does g sxd = csc x + 2 cot x have any local maximum values? Give reasons for your answer.
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Chapter 4
ax + b x2  1
has a local extreme value of 1 at x = 3 . Is this extreme value a local maximum, or a local minimum? Give reasons for your answer.
5. The greatest integer function ƒsxd = :x; , defined for all values of x, assumes a local maximum value of 0 at each point of [0, 1). Could any of these local maximum values also be local minimum values of ƒ? Give reasons for your answer.
6. a. Give an example of a differentiable function ƒ whose first derivative is zero at some point c even though ƒ has neither a local maximum nor a local minimum at c. b. How is this consistent with Theorem 2 in Section 4.1? Give reasons for your answer. 7. The function y = 1>x does not take on either a maximum or a minimum on the interval 0 6 x 6 1 even though the function is continuous on this interval. Does this contradict the Extreme Value Theorem for continuous functions? Why? 8. What are the maximum and minimum values of the function y = ƒ x ƒ on the interval 1 … x 6 1 ? Notice that the interval is not closed. Is this consistent with the Extreme Value Theorem for continuous functions? Why? T
319
The Mean Value Theorem
4. Find values of a and b such that the function ƒsxd =
Practice Exercises
9. A graph that is large enough to show a function’s global behavior may fail to reveal important local features. The graph of ƒsxd = sx 8>8d  sx 6>2d  x 5 + 5x 3 is a case in point.
a. Graph ƒ over the interval 2.5 … x … 2.5 . Where does the graph appear to have local extreme values or points of inflection?
b. Now factor ƒ¿sxd and show that ƒ has a local maximum at x = 3 2 5 L 1.70998 and local minima at x = ; 23 L ;1.73205 .
11. a. Show that g std = sin2 t  3t decreases on every interval in its domain. b. How many solutions does the equation sin2 t  3t = 5 have? Give reasons for your answer. 12. a. Show that y = tan u increases on every interval in its domain. b. If the conclusion in part (a) is really correct, how do you explain the fact that tan p = 0 is less than tan sp>4d = 1 ? 13. a. Show that the equation x 4 + 2x 2  2 = 0 has exactly one solution on [0, 1]. T b. Find the solution to as many decimal places as you can. 14. a. Show that ƒsxd = x>sx + 1d increases on every interval in its domain. b. Show that ƒsxd = x 3 + 2x has no local maximum or minimum values. 15. Water in a reservoir As a result of a heavy rain, the volume of water in a reservoir increased by 1400 acreft in 24 hours. Show that at some instant during that period the reservoir’s volume was increasing at a rate in excess of 225,000 gal>min. (An acrefoot is 43,560 ft3 , the volume that would cover 1 acre to the depth of 1 ft. A cubic foot holds 7.48 gal.) 16. The formula Fsxd = 3x + C gives a different function for each value of C. All of these functions, however, have the same derivative with respect to x, namely F¿sxd = 3 . Are these the only differentiable functions whose derivative is 3? Could there be any others? Give reasons for your answers. 17. Show that x d d 1 a ab = b x + 1 dx x + 1 dx even though
c. Zoom in on the graph to find a viewing window that shows the 3 presence of the extreme values at x = 2 5 and x = 23 .
x 1 Z . x + 1 x + 1
The moral here is that without calculus the existence of two of the three extreme values would probably have gone unnoticed. On any normal graph of the function, the values would lie close enough together to fall within the dimensions of a single pixel on the screen. (Source: Uses of Technology in the Mathematics Curriculum, by Benny Evans and Jerry Johnson, Oklahoma State University, published in 1990 under National Science Foundation Grant USE8950044.)
Doesn’t this contradict Corollary 2 of the Mean Value Theorem? Give reasons for your answer. 18. Calculate the first derivatives of ƒsxd = x 2>sx 2 + 1d and g sxd = 1>sx 2 + 1d . What can you conclude about the graphs of these functions?
T 10. (Continuation of Exercise 9.)
a. Graph ƒsxd = sx 8>8d  s2>5dx 5  5x  s5>x 2 d + 11 over the interval 2 … x … 2 . Where does the graph appear to have local extreme values or points of inflection?
Conclusions from Graphs In Exercises 19 and 20, use the graph to answer the questions. 19. Identify any global extreme values of ƒ and the values of x at which they occur. y y f (x)
7
b. Show that ƒ has a local maximum value at x = 25 L 1.2585 3 and a local minimum value at x = 2 2 L 1.2599 . c. Zoom in to find a viewing window that shows the presence of 7 3 the extreme values at x = 25 and x = 2 2.
(1, 1)
0
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2, 1 2 x
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Chapter 4: Applications of Derivatives
20. Estimate the intervals on which the function y = ƒsxd is
35. y¿ = 6xsx + 1dsx  2d 4
2
36. y¿ = x 2s6  4xd 38. y¿ = 4x 2  x 4
a. increasing.
37. y¿ = x  2x
b. decreasing.
In Exercises 39–42, graph each function. Then use the function’s first derivative to explain what you see.
c. Use the given graph of ƒ¿ to indicate where any local extreme values of the function occur, and whether each extreme is a relative maximum or minimum. y
39. y = x 2>3 + sx  1d1>3 41. y = x
y f ' (x)
–1
44. y =
2x x + 5
45. y =
x2 + 1 x
46. y =
x2  x + 1 x
47. y =
x3 + 2 2x
48. y =
x4  1 x2
49. y =
x2  4 x2  3
50. y =
x2 x  4
–2
Each of the graphs in Exercises 21 and 22 is the graph of the position function s = ƒstd of a body moving on a coordinate line (t represents time). At approximately what times (if any) is each body’s (a) velocity equal to zero? (b) Acceleration equal to zero? During approximately what time intervals does the body move (c) forward? (d) Backward? s s f (t)
Use l’Hôpital’s Rule to find the limits in Exercises 51–62. 51. lim
x 2 + 3x  4 x  1
52. lim
xa  1 xb  1
53. lim
tan x x
54. lim
tan x x + sin x
56. lim
sin mx sin nx
x:1
sin2 x x:0 tan sx 2 d 57. lim  sec 7x cos 3x 55. lim
9
12 14
t
x:p>2
22.
s
59. lim scsc x  cot xd
s f (t)
2
Applying l’Hôpital’s Rule
x:p
6
42. y = x 2>3  sx  1d1>3
x + 1 x  3
x
3
+ sx  1d
40. y = x 2>3 + sx  1d2>3
43. y =
(–3, 1)
0
1>3
Sketch the graphs of the functions in Exercises 43–50.
(2, 3)
21.
1>3
x:0
x:1
x:0
x:0
58. lim+ 2x sec x x:0
60. lim a x:0
61. lim A 2x 2 + x + 1  2x 2  x B
1 1  2b x4 x
x: q
0
2
4
6
8
t
x: q
Graphs and Graphing Graph the curves in Exercises 23–32. 23. y = x 2  sx 3>6d
24. y = x 3  3x 2 + 3
25. y = x 3 + 6x 2  9x + 3 26. y = s1>8dsx 3 + 3x 2  9x  27d 27. y = x 3s8  xd
28. y = x 2s2x 2  9d
29. y = x  3x 2>3
30. y = x 1>3sx  4d
31. y = x23  x
32. y = x24  x 2
Each of Exercises 33–38 gives the first derivative of a function y = ƒsxd . (a) At what points, if any, does the graph of ƒ have a local maximum, local minimum, or inflection point? (b) Sketch the general shape of the graph. 33. y¿ = 16  x 2
62. lim a
34. y¿ = x 2  x  6
x3 x3  2 b x  1 x + 1 2
Optimization 63. The sum of two nonnegative numbers is 36. Find the numbers if a. the difference of their square roots is to be as large as possible. b. the sum of their square roots is to be as large as possible. 64. The sum of two nonnegative numbers is 20. Find the numbers a. if the product of one number and the square root of the other is to be as large as possible. b. if one number plus the square root of the other is to be as large as possible. 65. An isosceles triangle has its vertex at the origin and its base parallel to the xaxis with the vertices above the axis on the curve y = 27  x 2 . Find the largest area the triangle can have.
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Chapter 4 Practice Exercises 66. A customer has asked you to design an opentop rectangular stainless steel vat. It is to have a square base and a volume of 32 ft3 , to be welded from quarterinch plate, and to weigh no more than necessary. What dimensions do you recommend? 67. Find the height and radius of the largest right circular cylinder that can be put in a sphere of radius 23 . 68. The figure here shows two right circular cones, one upside down inside the other. The two bases are parallel, and the vertex of the smaller cone lies at the center of the larger cone’s base. What values of r and h will give the smaller cone the largest possible volume?
Newton’s Method 73. Let ƒsxd = 3x  x 3 . Show that the equation ƒsxd = 4 has a solution in the interval [2, 3] and use Newton’s method to find it. 74. Let ƒsxd = x 4  x 3 . Show that the equation ƒsxd = 75 has a solution in the interval [3, 4] and use Newton’s method to find it.
Finding Indefinite Integrals Find the indefinite integrals (most general antiderivatives) in Exercises 75–90. Check your answers by differentiation. 75. 76.
L
77. h 6'
69. Manufacturing tires Your company can manufacture x hundred grade A tires and y hundred grade B tires a day, where 0 … x … 4 and 40  10x y = . 5  x
L L
a32t +
70. Particle motion The positions of two particles on the saxis are s1 = cos t and s2 = cos st + p>4d . a. What is the farthest apart the particles ever get? b. When do the particles collide? T 71. Opentop box An opentop rectangular box is constructed from a 10in.by16in. piece of cardboard by cutting squares of equal side length from the corners and folding up the sides. Find analytically the dimensions of the box of largest volume and the maximum volume. Support your answers graphically. 72. The ladder problem What is the approximate length (in feet) of the longest ladder you can carry horizontally around the corner of the corridor shown here? Round your answer down to the nearest foot. y
(8, 6)
0
8
4 b dt t2
3 1 a  4 b dt t L 22t dr 79. 2 L sr + 5d 6 dr
80.
L A r  22 B 3
81.
L
3u2u2 + 1 du
u du L 27 + u2
83.
L
85.
L
87.
L
89.
L
90.
L
x 3s1 + x 4 d1>4 dx sec2
s ds 10
csc 22u cot 22u du
84.
L
86.
L
88.
L
csc2 ps ds sec
u u tan du 3 3
sin2
x dx 4
cos2
1 + cos 2u x dx QHint: cos2 u = R 2 2
Solve the initial value problems in Exercises 91–94. dy x2 + 1 91. , y s1d = 1 = dx x2 2 dy 1 = ax + x b , dx
y s1d = 1
3 d 2r = 152t + ; r¿s1d = 8, r s1d = 0 dt 2 2t d 3r 94. = cos t; r–s0d = r¿s0d = 0, r s0d = 1 dt 3 93.
x
s2  xd3>5 dx
Initial Value Problems
92.
6
t2 + tb dt 2
78.
82.
Your profit on a grade A tire is twice your profit on a grade B tire. What is the most profitable number of each kind to make?
sx 3 + 5x  7d dx a8t 3 
12' r
321
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Chapter 4
Additional and Advanced Exercises
1. What can you say about a function whose maximum and minimum values on an interval are equal? Give reasons for your answer.
a. If ƒ and g are positive, with local maxima at x = a , and if ƒ¿ and g¿ change sign at a, does h have a local maximum at a?
2. Is it true that a discontinuous function cannot have both an absolute maximum and an absolute minimum value on a closed interval? Give reasons for your answer.
b. If the graphs of ƒ and g have inflection points at x = a , does the graph of h have an inflection point at a?
3. Can you conclude anything about the extreme values of a continuous function on an open interval? On a halfopen interval? Give reasons for your answer. 4. Local extrema Use the sign pattern for the derivative df = 6sx  1dsx  2d2sx  3d3sx  4d4 dx
In either case, if the answer is yes, give a proof. If the answer is no, give a counterexample. 11. Finding a function Use the following information to find the values of a, b, and c in the formula ƒsxd = sx + ad> sbx 2 + cx + 2d . i) The values of a, b, and c are either 0 or 1. ii) The graph of ƒ passes through the point s 1, 0d . iii) The line y = 1 is an asymptote of the graph of ƒ.
to identify the points where ƒ has local maximum and minimum values. 5. Local extrema a. Suppose that the first derivative of y = ƒsxd is y¿ = 6sx + 1dsx  2d2 . At what points, if any, does the graph of ƒ have a local maximum, local minimum, or point of inflection? b. Suppose that the first derivative of y = ƒsxd is
12. Horizontal tangent For what value or values of the constant k will the curve y = x 3 + kx 2 + 3x  4 have exactly one horizontal tangent? 13. Largest inscribed triangle Points A and B lie at the ends of a diameter of a unit circle and point C lies on the circumference. Is it true that the area of triangle ABC is largest when the triangle is isosceles? How do you know? 14. Proving the second derivative test The Second Derivative Test for Local Maxima and Minima (Section 4.4) says:
y¿ = 6x sx + 1dsx  2d .
a. ƒ has a local maximum value at x = c if ƒ¿scd = 0 and ƒ–scd 6 0
At what points, if any, does the graph of ƒ have a local maximum, local minimum, or point of inflection?
b. ƒ has a local minimum value at x = c if ƒ¿scd = 0 and ƒ–scd 7 0 .
6. If ƒ¿sxd … 2 for all x, what is the most the values of ƒ can increase on [0, 6]? Give reasons for your answer. 7. Bounding a function Suppose that ƒ is continuous on [a, b] and that c is an interior point of the interval. Show that if ƒ¿sxd … 0 on [a, c) and ƒ¿sxd Ú 0 on (c, b], then ƒ(x) is never less than ƒ(c) on [a, b].
To prove statement (a), let P = s1>2d ƒ ƒ–scd ƒ . Then use the fact that ƒ–scd = lim
h:0
ƒ¿sc + hd  ƒ¿scd ƒ¿sc + hd = lim h h h:0
to conclude that for some d 7 0 ,
8. An inequality a. Show that 1>2 … x>s1 + x 2 d … 1>2 for every value of x. b. Suppose that ƒ is a function whose derivative is ƒ¿sxd = x>s1 + x 2 d . Use the result in part (a) to show that 1 ƒ ƒsbd  ƒsad ƒ … 2 ƒ b  a ƒ for any a and b. 9. The derivative of ƒsxd = x 2 is zero at x = 0 , but ƒ is not a constant function. Doesn’t this contradict the corollary of the Mean Value Theorem that says that functions with zero derivatives are constant? Give reasons for your answer. 10. Extrema and inflection points Let h = ƒg be the product of two differentiable functions of x.
0 6 ƒhƒ 6 d
Q
ƒ¿sc + hd 6 ƒ–scd + P 6 0 . h
Thus, ƒ¿sc + hd is positive for d 6 h 6 0 and negative for 0 6 h 6 d . Prove statement (b) in a similar way. 15. Hole in a water tank You want to bore a hole in the side of the tank shown here at a height that will make the stream of water coming out hit the ground as far from the tank as possible. If you drill the hole near the top, where the pressure is low, the water will exit slowly but spend a relatively long time in the air. If you drill the hole near the bottom, the water will exit at a higher velocity but have only a short time to fall. Where is the best place, if any, for the hole? (Hint: How long will it take an exiting particle of water to fall from height y to the ground?)
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Tank kept full, top open
323
18. Minimizing a parameter Find the smallest value of the positive constant m that will make mx  1 + s1>xd greater than or equal to zero for all positive values of x.
y
19. Evaluate the following limits. h Exit velocity ⫽ 兹64(h ⫺ y) y
a. lim
x:0
2 sin 5x 3x
b. lim sin 5x cot 3x x:0
c. lim x csc 22x 2
d.
x:0
e. lim
x:0
Ground
0
x
Range
16. Kicking a field goal An American football player wants to kick a field goal with the ball being on a right hash mark. Assume that the goal posts are b feet apart and that the hash mark line is a distance a 7 0 feet from the right goal post. (See the accompanying figure.) Find the distance h from the goal post line that gives the kicker his largest angle b . Assume that the football field is flat. Goal posts b
Goal post line
a
Right hash mark line h

x  sin x x  tan x
lim ssec x  tan xd
x:p>2
f. lim
x:0
sin x 2 x sin x
sec x  1 x3  8 g. lim h. lim 2 2 x:0 x:2 x x  4 20. L’Hôpital’s Rule does not help with the following limits. Find them some other way. a. lim
x: q
2x + 5
2x
b. lim
2x + 5
x: q
x + 72x
21. Suppose that it costs a company y = a + bx dollars to produce x units per week. It can sell x units per week at a price of P = c  ex dollars per unit. Each of a, b, c, and e represents a positive constant. (a) What production level maximizes the profit? (b) What is the corresponding price? (c) What is the weekly profit at this level of production? (d) At what price should each item be sold to maximize profits if the government imposes a tax of t dollars per item sold? Comment on the difference between this price and the price before the tax. 22. Estimating reciprocals without division You can estimate the value of the reciprocal of a number a without ever dividing by a if you apply Newton’s method to the function ƒsxd = s1>xd  a . For example, if a = 3 , the function involved is ƒsxd = s1>xd  3 . a. Graph y = s1>xd  3 . Where does the graph cross the xaxis?
Football
17. A maxmin problem with a variable answer Sometimes the solution of a maxmin problem depends on the proportions of the shapes involved. As a case in point, suppose that a right circular cylinder of radius r and height h is inscribed in a right circular cone of radius R and height H, as shown here. Find the value of r (in terms of R and H) that maximizes the total surface area of the cylinder (including top and bottom). As you will see, the solution depends on whether H … 2R or H 7 2R .
b. Show that the recursion formula in this case is xn + 1 = xns2  3xn d , so there is no need for division. q
23. To find x = 2a , we apply Newton’s method to ƒsxd = x q  a . Here we assume that a is a positive real number and q is a positive integer. Show that x1 is a “weighted average” of x0 and a>x 0q  1 , and find the coefficients m0, m1 such that x1 = m0 x0 + m1 a
q  1 b, x0
a
m0 7 0, m1 7 0, m0 + m1 = 1. q1
What conclusion would you reach if x0 and a>x 0 What would be the value of x1 in that case?
r H
h
were equal?
24. The family of straight lines y = ax + b (a, b arbitrary constants) can be characterized by the relation y– = 0 . Find a similar relation satisfied by the family of all circles sx  hd2 + sy  hd2 = r 2 ,
R
where h and r are arbitrary constants. (Hint: Eliminate h and r from the set of three equations including the given one and two obtained by successive differentiation.)
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25. Assume that the brakes of an automobile produce a constant deceleration of k ft>sec2 . (a) Determine what k must be to bring an automobile traveling 60 mi>hr (88 ft>sec) to rest in a distance of 100 ft from the point where the brakes are applied. (b) With the same k, how far would a car traveling 30 mi>hr travel before being brought to a stop? 26. Let ƒ(x), g (x) be two continuously differentiable functions satisfying the relationships ƒ¿sxd = g sxd and ƒ–sxd = ƒsxd . Let hsxd = f 2sxd + g 2sxd . If hs0d = 5 , find h(10). 27. Can there be a curve satisfying the following conditions? d 2y>dx 2 is everywhere equal to zero and, when x = 0, y = 0 and dy>dx = 1 . Give a reason for your answer. 28. Find the equation for the curve in the xyplane that passes through the point s1, 1d if its slope at x is always 3x 2 + 2 . 29. A particle moves along the xaxis. Its acceleration is a = t 2 . At t = 0 , the particle is at the origin. In the course of its motion, it reaches the point x = b , where b 7 0 , but no point beyond b. Determine its velocity at t = 0 .
a. the velocity y in terms of t. b. the position s in terms of t. 31. Given ƒsxd = ax 2 + 2bx + c with a 7 0 . By considering the minimum, prove that ƒsxd Ú 0 for all real x if, and only if, b 2  ac … 0 . 32. Schwarz’s inequality a. In Exercise 31, let ƒsxd = sa1 x + b1 d2 + sa2 x + b2 d2 + Á + san x + bn d2 , and deduce Schwarz’s inequality: sa1 b1 + a2 b2 + Á + an bn d2 … sa12 + a22 + Á + an2 dsb12 + b 22 + Á + bn2 d . b. Show that equality holds in Schwarz’s inequality only if there exists a real number x that makes ai x equal bi for every value of i from 1 to n.
30. A particle moves with acceleration a = 2t  A 1> 2t B . Assuming that the velocity y = 4>3 and the position s = 4>15 when t = 0 , find
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Chapter 4: Applications of Derivatives
Chapter 4
Technology Application Projects
Mathematica/Maple Module Motion Along a Straight Line: Position : Velocity : Acceleration You will observe the shape of a graph through dramatic animated visualizations of the derivative relations among the position, velocity, and acceleration. Figures in the text can be animated.
Mathematica/Maple Module Newton’s Method: Estimate p to How Many Places? Plot a function, observe a root, pick a starting point near the root, and use Newton’s Iteration Procedure to approximate the root to a desired accuracy. The numbers p, e , and 22 are approximated.
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Chapter
5
INTEGRATION OVERVIEW One of the great achievements of classical geometry was to obtain formulas for the areas and volumes of triangles, spheres, and cones. In this chapter we study a method to calculate the areas and volumes of these and other more general shapes. The method we develop, called integration, is a tool for calculating much more than areas and volumes. The integral has many applications in statistics, economics, the sciences, and engineering. It allows us to calculate quantities ranging from probabilities and averages to energy consumption and the forces against a dam’s floodgates. The idea behind integration is that we can effectively compute many quantities by breaking them into small pieces, and then summing the contributions from each small part. We develop the theory of the integral in the setting of area, where it most clearly reveals its nature. We begin with examples involving finite sums. These lead naturally to the question of what happens when more and more terms are summed. Passing to the limit, as the number of terms goes to infinity, then gives an integral. While integration and differentiation are closely connected, we will not see the roles of the derivative and antiderivative emerge until Section 5.4. The nature of their connection, contained in the Fundamental Theorem of Calculus, is one of the most important ideas in calculus.
Estimating with Finite Sums
5.1
This section shows how area, average values, and the distance traveled by an object over time can all be approximated by finite sums. Finite sums are the basis for defining the integral in Section 5.3.
y
1
Area
y 1 x2
The area of a region with a curved boundary can be approximated by summing the areas of a collection of rectangles. Using more rectangles can increase the accuracy of the approximation.
0.5 R
EXAMPLE 1 0
0.5
1
FIGURE 5.1 The area of the region R cannot be found by a simple geometry formula (Example 1).
x
Approximating Area
What is the area of the shaded region R that lies above the xaxis, below the graph of y = 1  x 2 , and between the vertical lines x = 0 and x = 1? (See Figure 5.1.) An architect might want to know this area to calculate the weight of a custom window with a shape described by R. Unfortunately, there is no simple geometric formula for calculating the areas of shapes having curved boundaries like the region R.
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Chapter 5: Integration
y
1
y y 1 x2
(0, 1)
1
(0, 1)
y 1 x2
1 , 15 4 16
1 , 3 2 4
1 , 3 2 4
0.5
3 , 7 4 16
0.5 R
0
R
0.5
x
1
0
0.25
(a)
0.5
0.75
1
x
(b)
FIGURE 5.2 (a) We get an upper estimate of the area of R by using two rectangles containing R. (b) Four rectangles give a better upper estimate. Both estimates overshoot the true value for the area.
While we do not yet have a method for determining the exact area of R, we can approximate it in a simple way. Figure 5.2a shows two rectangles that together contain the region R. Each rectangle has width 1>2 and they have heights 1 and 3>4, moving from left to right. The height of each rectangle is the maximum value of the function ƒ, obtained by evaluating ƒ at the left endpoint of the subinterval of [0, 1] forming the base of the rectangle. The total area of the two rectangles approximates the area A of the region R, A L 1#
3 1 + 2 4
#
7 1 = = 0.875. 2 8
This estimate is larger than the true area A, since the two rectangles contain R. We say that 0.875 is an upper sum because it is obtained by taking the height of each rectangle as the maximum (uppermost) value of ƒ(x) for x a point in the base interval of the rectangle. In Figure 5.2b, we improve our estimate by using four thinner rectangles, each of width 1>4, which taken together contain the region R. These four rectangles give the approximation A L 1#
15 1 + 4 16
3 1 + 4 4
#
#
7 1 + 4 16
#
25 1 = = 0.78125, 4 32
which is still greater than A since the four rectangles contain R. Suppose instead we use four rectangles contained inside the region R to estimate the area, as in Figure 5.3a. Each rectangle has width 1>4 as before, but the rectangles are shorter and lie entirely beneath the graph of ƒ. The function ƒsxd = 1  x 2 is decreasing on [0, 1], so the height of each of these rectangles is given by the value of ƒ at the right endpoint of the subinterval forming its base. The fourth rectangle has zero height and therefore contributes no area. Summing these rectangles with heights equal to the minimum value of ƒ(x) for x a point in each base subinterval, gives a lower sum approximation to the area, A L
15 16
#
3 1 + 4 4
#
7 1 + 4 16
#
17 1 1 + 0# = = 0.53125. 4 4 32
This estimate is smaller than the area A since the rectangles all lie inside of the region R. The true value of A lies somewhere between these lower and upper sums: 0.53125 6 A 6 0.78125.
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5.1
y
y 1 , 15 4 16
1
y 1 x2
327
Estimating with Finite Sums
1 , 63 8 64
1
3 , 55 8 64
y 1 x2
1 , 3 2 4 5 , 39 8 64 3 , 7 4 16
0.5
0.5 7 , 15 8 64
0
0.25
0.5 (a)
0.75
x
1
0.25 0.5 0.75 1 0.125 0.375 0.625 0.875
0
x
(b)
FIGURE 5.3 (a) Rectangles contained in R give an estimate for the area that undershoots the true value. (b) The midpoint rule uses rectangles whose height is the value of y = ƒsxd at the midpoints of their bases.
y 1 y 1 x2
1
0
x
(a)
By considering both lower and upper sum approximations we get not only estimates for the area, but also a bound on the size of the possible error in these estimates since the true value of the area lies somewhere between them. Here the error cannot be greater than the difference 0.78125  0.53125 = 0.25. Yet another estimate can be obtained by using rectangles whose heights are the values of ƒ at the midpoints of their bases (Figure 5.3b). This method of estimation is called the midpoint rule for approximating the area. The midpoint rule gives an estimate that is between a lower sum and an upper sum, but it is not clear whether it overestimates or underestimates the true area. With four rectangles of width 1>4 as before, the midpoint rule estimates the area of R to be A L
y
#
55 1 + 4 64
#
39 1 + 4 64
#
15 1 + 4 64
#
172 1 = 4 64
#
1 = 0.671875. 4
In each of our computed sums, the interval [a, b] over which the function ƒ is defined was subdivided into n subintervals of equal width (also called length) ¢x = sb  ad>n, and ƒ was evaluated at a point in each subinterval: c1 in the first subinterval, c2 in the second subinterval, and so on. The finite sums then all take the form
1 y1x
63 64
2
ƒsc1 d ¢x + ƒsc2 d ¢x + ƒsc3 d ¢x + Á + ƒscn d ¢x.
1
0
x
(b)
FIGURE 5.4 (a) A lower sum using 16 rectangles of equal width ¢x = 1>16. (b) An upper sum using 16 rectangles.
By taking more and more rectangles, with each rectangle thinner than before, it appears that these finite sums give better and better approximations to the true area of the region R. Figure 5.4a shows a lower sum approximation for the area of R using 16 rectangles of equal width. The sum of their areas is 0.634765625, which appears close to the true area, but is still smaller since the rectangles lie inside R. Figure 5.4b shows an upper sum approximation using 16 rectangles of equal width. The sum of their areas is 0.697265625, which is somewhat larger than the true area because the rectangles taken together contain R. The midpoint rule for 16 rectangles gives a total area approximation of 0.6669921875, but it is not immediately clear whether this estimate is larger or smaller than the true area.
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Chapter 5: Integration
TABLE 5.1 Finite approximations for the area of R
Number of subintervals
Lower sum
Midpoint rule
Upper sum
2 4 16 50 100 1000
.375 .53125 .634765625 .6566 .66165 .6661665
.6875 .671875 .6669921875 .6667 .666675 .66666675
.875 .78125 .697265625 .6766 .67165 .6671665
Table 5.1 shows the values of upper and lower sum approximations to the area of R using up to 1000 rectangles. In Section 5.2 we will see how to get an exact value of the areas of regions such as R by taking a limit as the base width of each rectangle goes to zero and the number of rectangles goes to infinity. With the techniques developed there, we will be able to show that the area of R is exactly 2>3 .
Distance Traveled Suppose we know the velocity function y(t) of a car moving down a highway, without changing direction, and want to know how far it traveled between times t = a and t = b. If we already know an antiderivative F(t) of y(t) we can find the car’s position function s(t) by setting sstd = Fstd + C. The distance traveled can then be found by calculating the change in position, ssbd  ssad (see Exercise 93, Section 4.8). If the velocity function is determined by recording a speedometer reading at various times on the car, then we have no formula from which to obtain an antiderivative function for velocity. So what do we do in this situation? When we don’t know an antiderivative for the velocity function y(t), we can approximate the distance traveled in the following way. Subdivide the interval [a, b] into short time intervals on each of which the velocity is considered to be fairly constant. Then approximate the distance traveled on each time subinterval with the usual distance formula distance = velocity * time and add the results across [a, b]. Suppose the subdivided interval looks like t a
t1
t
t
t2
t3
b
t (sec)
with the subintervals all of equal length ¢t. Pick a number t1 in the first interval. If ¢t is so small that the velocity barely changes over a short time interval of duration ¢t, then the distance traveled in the first time interval is about yst1 d ¢t. If t2 is a number in the second interval, the distance traveled in the second time interval is about yst2 d ¢t. The sum of the distances traveled over all the time intervals is D L yst1 d ¢t + yst2 d ¢t + Á + ystn d ¢t, where n is the total number of subintervals. Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
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5.1
EXAMPLE 2
Estimating with Finite Sums
329
Estimating the Height of a Projectile
The velocity function of a projectile fired straight into the air is ƒstd = 160  9.8t m>sec. Use the summation technique just described to estimate how far the projectile rises during the first 3 sec. How close do the sums come to the exact figure of 435.9 m? We explore the results for different numbers of intervals and different choices of evaluation points. Notice that ƒ(t) is decreasing, so choosing left endpoints gives an upper sum estimate; choosing right endpoints gives a lower sum estimate. Solution
(a) Three subintervals of length 1, with ƒ evaluated at left endpoints giving an upper sum: t1
t2
t3
0
1
2
t
3
t
With ƒ evaluated at t = 0, 1, and 2, we have D L ƒst1 d ¢t + ƒst2 d ¢t + ƒst3 d ¢t = [160  9.8s0d]s1d + [160  9.8s1d]s1d + [160  9.8s2d]s1d = 450.6. (b) Three subintervals of length 1, with ƒ evaluated at right endpoints giving a lower sum:
0
t1
t2
t3
1
2
3
t
t
With ƒ evaluated at t = 1, 2, and 3, we have D L ƒst1 d ¢t + ƒst2 d ¢t + ƒst3 d ¢t = [160  9.8s1d]s1d + [160  9.8s2d]s1d + [160  9.8s3d]s1d = 421.2. (c) With six subintervals of length 1>2, we get t1 t 2 t 3 t 4 t 5 t 6 0 t
1
2
t1 t 2 t 3 t 4 t 5 t 6 3
t
0
1
2
3
t
t
An upper sum using left endpoints: D L 443.25; a lower sum using right endpoints: D L 428.55. These sixinterval estimates are somewhat closer than the threeinterval estimates. The results improve as the subintervals get shorter. As we can see in Table 5.2, the leftendpoint upper sums approach the true value 435.9 from above, whereas the rightendpoint lower sums approach it from below. The true Copyright © 2005 Pearson Education, Inc., publishing as Pearson AddisonWesley
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Chapter 5: Integration
TABLE 5.2 Traveldistance estimates
Number of subintervals
Length of each subinterval
Upper sum
Lower sum
3 6 12 24 48 96 192
1 1>2 1>4 1>8 1>16 1>32 1>64
450.6 443.25 439.57 437.74 436.82 436.36 436.13
421.2 428.55 432.22 434.06 434.98 435.44 435.67
value lies between these upper and lower sums. The magnitude of the error in the closest entries is 0.23, a small percentage of the true value. Error magnitude = ƒ true value  calculated value ƒ = ƒ 435.9  435.67 ƒ = 0.23. Error percentage =
0.23 L 0.05%. 435.9
It would be reasonable to conclude from the table’s last entries that the projectile rose about 436 m during its first 3 sec of flight.
Displacement Versus Distance Traveled If a body with position function s(t) moves along a coordinate line without changing direction, we can calculate the total distance it travels from t = a to t = b by summing the distance traveled over small intervals, as in Example 2. If the body changes direction one or more times during the trip, then we need to use the body’s speed ƒ ystd ƒ , which is the absolute value of its velocity function, y(t), to find the total distance traveled. Using the velocity itself, as in Example 2, only gives an estimate to the body’s displacement, ssbd  ssad, the difference between its initial and final positions. To see why, partition the time interval [a, b] into small enough equal subintervals ¢t so that the body’s velocity does not change very much from time tk  1 to tk . Then ystk d gives a good approximation of the velocity throughout the interval. Accordingly, the change in the body’s position coordinate during the time interval is about ystk d ¢t. The change is positive if ystk d is positive and negative if ystk d is negative. In either case, the distance traveled during the subinterval is about ƒ ystk d ƒ ¢t. The total distance traveled is approximately the sum ƒ yst1 d ƒ ¢t + ƒ yst2 d ƒ ¢t + Á + ƒ ystn d ƒ ¢t.
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5.1
y
331
y y g(x)
yc
c
0
Estimating with Finite Sums
a
(a)
b
c
x
0
a
(b)
b
x
FIGURE 5.5 (a) The average value of ƒsxd = c on [a, b] is the area of the rectangle divided by b  a . (b) The average value of g (x) on [a, b] is the area beneath its graph divided by b  a .
Average Value of a Nonnegative Function The average value of a collection of n numbers x1, x2 , Á , xn is obtained by adding them together and dividing by n. But what is the average value of a continuous function ƒ on an interval [a, b]? Such a function can assume infinitely many values. For example, the temperature at a certain location in a town is a continuous function that goes up and down each day. What does it mean to say that the average temperature in the town over the course of a day is 73 degrees? When a function is constant, this question is easy to answer. A function with constant value c on an interval [a, b] has average value c. When c is positive, its graph over [a, b] gives a rectangle of height c. The average value of the function can then be interpreted geometrically as the area of this rectangle divided by its width b  a (Figure 5.5a). What if we want to find the average value of a nonconstant function, such as the function g in Figure 5.5b? We can think of this graph as a snapshot of the height of some water that is sloshing around in a tank, between enclosing walls at x = a and x = b. As the water moves, its height over each point changes, but its average height remains the same. To get the average height of the water, we let it settle down until it is level and its height is constant. The resulting height c equals the area under the graph of g divided by b  a. We are led to define the average value of a nonnegative function on an interval [a, b] to be the area under its graph divided by b  a. For this definition to be valid, we need a precise understanding of what is meant by the area under a graph. This will be obtained in Section 5.3, but for now we look at two simple examples.
y
6 f(x) 3x
EXAMPLE 3 4
The Average Value of a Linear Function
What is the average value of the function ƒsxd = 3x on the interval [0, 2]? The average equals the area under the graph divided by the width of the interval. In this case we do not need finite approximation to estimate the area of the region under the graph: a triangle of height 6 and base 2 has area 6 (Figure 5.6). The width of the interval is b  a = 2  0 = 2. The average value of the function is 6>2 = 3.
Solution 2
0
1
2
3
x
EXAMPLE 4
The Average Value of sin x
Estimate the average value of the function ƒsxd = sin x on the interval [0, p]. FIGURE 5.6 The average value of ƒsxd = 3x over [0, 2] is 3 (Example 3).
Looking at the graph of sin x between 0 and p in Figure 5.7, we can see that its average height is somewhere between 0 and 1. To find the average we need to
Solution
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Chapter 5: Integration y
y f (x) sin x
f (x) sin x
1
1
2 (a)
0
x
2 (b)
0
x
FIGURE 5.7 Approximating the area under ƒsxd = sin x between 0 and p to compute the average value of sin x over [0, p] , using (a) four rectangles; (b) eight rectangles (Example 4).
calculate the area A under the graph and then divide this area by the length of the interval, p  0 = p. We do not have a simple way to determine the area, so we approximate it with finite sums. To get an upper sum estimate, we add the areas of four rectangles of equal width p>4 that together contain the region beneath the graph of y = sin x and above the xaxis on [0, p]. We choose the heights of the rectangles to be the largest value of sin x on each subinterval. Over a particular subinterval, this largest value may occur at the left endpoint, the right endpoint, or somewhere between them. We evaluate sin x at this point to get the height of the rectangle for an upper sum. The sum of the rectangle areas then estimates the total area (Figure 5.7a): A L asin = a
p b 4
p p + asin b 4 2
#
p p + asin b 4 2
1 1 + 1 + 1 + b 22 22
#
p p L s3.42d # L 2.69. 4 4
#
#
3p p + asin b 4 4
#
p 4
To estimate the average value of sin x we divide the estimated area by p and obtain the approximation 2.69>p L 0.86. If we use eight rectangles of equal width p>8 all lying above the graph of y = sin x (Figure 5.7b), we get the area estimate A L asin
3p 5p 3p 7p # p p p p p + sin + sin + sin + sin + sin + sin + sin b 8 4 8 2 2 8 4 8 8 p p L s.38 + .71 + .92 + 1 + 1 + .92 + .71 + .38d # = s6.02d # L 2.365. 8 8
Dividing this result by the length p of the interval gives a more accurate estimate of 0.753 for the average. Since we used an upper sum to approximate the area, this estimate is still greater than the actual average value of sin x over [0, p]. If we use more and more rectangles, with each rectangle getting thinner and thinner, we get closer and closer to the true average value. Using the techniques of Section 5.3, we will show that the true average value is 2>p L 0.64. As before, we could just as well have used rectangles lying under the graph of y = sin x and calculated a lower sum approximation, or we could have used the midpoint rule. In Section 5.3, we will see that it doesn’t matter whether our approximating rectangles are chosen to give upper sums, lower sums, or a sum in between. In each case, the approximations are close to the true area if all the rectangles are sufficiently thin.
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Summary The area under the graph of a positive function, the distance traveled by a moving object that doesn’t change direction, and the average value of a nonnegative function over an interval can all be approximated by finite sums. First we subdivide the interval into subintervals, treating the appropriate function ƒ as if it were constant over each particular subinterval. Then we multiply the width of each subinterval by the value of ƒ at some point within it, and add these products together. If the interval [a, b] is subdivided into n subintervals of equal widths ¢x = sb  ad>n, and if ƒsck d is the value of ƒ at the chosen point ck in the kth subinterval, this process gives a finite sum of the form ƒsc1 d ¢x + ƒsc2 d ¢x + ƒsc3 d ¢x + Á + ƒscn d ¢x. The choices for the ck could maximize or minimize the value of ƒ in the kth subinterval, or give some value in between. The true value lies somewhere between the approximations given by upper sums and lower sums. The finite sum approximations we looked at improved as we took more subintervals of thinner width.
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5.1 Estimating with Finite Sums
EXERCISES 5.1 Area In Exercises 1–4 use finite approximations to estimate the area under the graph of the function using a. a lower sum with two rectangles of equal width. b. a lower sum with four rectangles of equal width. c. an upper sum with two rectangles of equal width. d. an upper sum with four rectangles of equal width. 1. ƒsxd = x 2 between x = 0 and x = 1 .
Time (sec)
Velocity (in. / sec)
Time (sec)
Velocity (in. / sec)
0 1 2 3 4 5
0 12 22 10 5 13
6 7 8 9 10
11 6 2 6 0
2. ƒsxd = x 3 between x = 0 and x = 1 . 3. ƒsxd = 1>x between x = 1 and x = 5 . 4. ƒsxd = 4  x 2 between x = 2 and x = 2 . Using rectangles whose height is given by the value of the function at the midpoint of the rectangle’s base (the midpoint rule) estimate the area under the graphs of the following functions, using first two and then four rectangles. 5. ƒsxd = x 2 between x = 0 and x = 1 . 6. ƒsxd = x 3 between x = 0 and x = 1 .
10. Distance traveled upstream You are sitting on the bank of a tidal river watching the incoming tide carry a bottle upstream. You record the velocity of the flow every 5 minutes for an hour, with the results shown in the accompanying table. About how far upstream did the bottle travel during that hour? Find an estimate using 12 subintervals of length 5 with a. leftendpoint values. b. rightendpoint values.
7. ƒsxd = 1>x between x = 1 and x = 5 . 8. ƒsxd = 4  x 2 between x = 2 and x = 2 .
Distance 9. Distance traveled The accompanying table shows the velocity of a model train engine moving along a track for 10 sec. Estimate the distance traveled by the engine using 10 subintervals of length 1 with a. leftendpoint values. b. rightendpoint values.
Time (min)
Velocity (m / sec)
Time (min)
Velocity (m / sec)
0 5 10 15 20 25 30
1 1.2 1.7 2.0 1.8 1.6 1.4
35 40 45 50 55 60
1.2 1.0 1.8 1.5 1.2 0
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Chapter 5: Integration
11. Length of a road You and a companion are about to drive a twisty stretch of dirt road in a car whose speedometer works but whose odometer (mileage counter) is broken. To find out how long this particular stretch of road is, you record the car’s velocity at 10sec intervals, with the results shown in the accompanying table. Estimate the length of the road using a. leftendpoint values. b. rightendpoint values.
Time (sec)
Velocity (converted to ft / sec) (30 mi / h 44 ft / sec)
0 10 20 30 40 50 60
0 44 15 35 30 44 35
Time (sec)
Velocity (converted to ft / sec) (30 mi / h 44 ft / sec)
70 80 90 100 110 120
15 22 35 44 30 35
a. Use rectangles to estimate how far the car traveled during the 36 sec it took to reach 142 mi> h.
b. Roughly how many seconds did it take the car to reach the halfway point? About how fast was the car going then?
Velocity and Distance 13. Free fall with air resistance An object is dropped straight down from a helicopter. The object falls faster and faster but its acceleration (rate of change of its velocity) decreases over time because of air resistance. The acceleration is measured in ft>sec2 and recorded every second after the drop for 5 sec, as shown: t
0
1
2
3
4
5
a
32.00
19.41
11.77
7.14
4.33
2.63
a. Find an upper estimate for the speed when t = 5 . b. Find a lower estimate for the speed when t = 5 . c. Find an upper estimate for the distance fallen when t = 3 . 14. Distance traveled by a projectile An object is shot straight upward from sea level with an initial velocity of 400 ft> sec.
12. Distance from velocity data The accompanying table gives data for the velocity of a vintage sports car accelerating from 0 to 142 mi> h in 36 sec (10 thousandths of an hour). Time (h)
Velocity (mi / h)
Time (h)
Velocity (mi / h)
0.0 0.001 0.002 0.003 0.004 0.005
0 40 62 82 96 108
0.006 0.007 0.008 0.009 0.010
116 125 132 137 142
a. Assuming that gravity is the only force acting on the object, give an upper estimate for its velocity after 5 sec have elapsed. Use g = 32 ft>sec2 for the gravitational acceleration.
b. Find a lower estimate for the height attained after 5 sec.
Average Value of a Function In Exercises 15–18, use a finite sum to estimate the average value of ƒ on the given interval by partitioning the interval into four subintervals of equal length and evaluating ƒ at the subinterval midpoints. 15. ƒsxd = x 3 on [0, 2] 17. ƒstd = s1>2d + sin2 pt
16. ƒsxd = 1>x on [1, 9] on [0, 2]
y y 1 sin 2 t 2
1.5 mi/hr
1
160
0.5
140
0
120
18. ƒstd = 1  acos
100
1
pt b 4
t
2
4
on [0, 4]
80 y
4 y 1 cos t 4
60 1
40 20 0
0.002 0.004 0.006 0.008 0.01
hours
0
1
2
3
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4
t
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5.1 Estimating with Finite Sums
Pollution Control 19. Water pollution Oil is leaking out of a tanker damaged at sea. The damage to the tanker is worsening as evidenced by the increased leakage each hour, recorded in the following table. Time (h)
0
1
2
3
4
Leakage (gal> h)
50
70
97
136
190
Time (h)
5
6
7
8
265
369
516
720
Leakage (gal> h)
c. The tanker continues to leak 720 gal> h after the first 8 hours. If the tanker originally contained 25,000 gal of oil, approximately how many more hours will elapse in the worst case before all the oil has spilled? In the best case?
20. Air pollution A power plant generates electricity by burning oil. Pollutants produced as a result of the burning process are removed by scrubbers in the smokestacks. Over time, the scrubbers become less efficient and eventually they must be replaced when the amount of pollution released exceeds government standards. Measurements are taken at the end of each month determining the rate at which pollutants are released into the atmosphere, recorded as follows. Month
Jan
Feb
Mar
Apr
May
Jun
Pollutant Release rate (tons> day)
0.20
0.25
0.27
0.34
0.45
0.52
Month
Jul
Aug
Sep
Oct
Nov
Dec
0.70
0.81
0.85
0.89
b. In the best case, approximately when will a total of 125 tons of pollutants have been released into the atmosphere?
Area of a Circle
a. 4 (square)
b. Repeat part (a) for the quantity of oil that has escaped after 8 hours.
0.63
a. Assuming a 30day month and that new scrubbers allow only 0.05 ton> day released, give an upper estimate of the total tonnage of pollutants released by the end of June. What is a lower estimate?
21. Inscribe a regular nsided polygon inside a circle of radius 1 and compute the area of the polygon for the following values of n:
a. Give an upper and a lower estimate of the total quantity of oil that has escaped after 5 hours.
Pollutant Release rate (tons> day)
335
0.95
b. 8 (octagon)
c. 16
d. Compare the areas in parts (a), (b), and (c) with the area of the circle. 22. (Continuation of Exercise 21) a. Inscribe a regular nsided polygon inside a circle of radius 1 and compute the area of one of the n congruent triangles formed by drawing radii to the vertices of the polygon. b. Compute the limit of the area of the inscribed polygon as n: q. c. Repeat the computations in parts (a) and (b) for a circle of radius r. COMPUTER EXPLORATIONS In Exercises 23–26, use a CAS to perform the following steps. a. Plot the functions over the given interval. b. Subdivide the interval into n = 100 , 200, and 1000 subintervals of equal length and evaluate the function at the midpoint of each subinterval. c. Compute the average value of the function values generated in part (b). d. Solve the equation ƒsxd = saverage valued for x using the average value calculated in part (c) for the n = 1000 partitioning. 23. ƒsxd = sin x on [0, p] 1 25. ƒsxd = x sin x 1 26. ƒsxd = x sin2 x
on on
24. ƒsxd = sin2 x on [0, p]
p c , pd 4 p c , pd 4
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5.2 Sigma Notation and Limits of Finite Sums
5.2
335
Sigma Notation and Limits of Finite Sums In estimating with finite sums in Section 5.1, we often encountered sums with many terms (up to 1000 in Table 5.1, for instance). In this section we introduce a notation to write sums with a large number of terms. After describing the notation and stating several of its properties, we look at what happens to a finite sum approximation as the number of terms approaches infinity.
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Finite Sums and Sigma Notation Sigma notation enables us to write a sum with many terms in the compact form n
Á + an  1 + an . a ak = a1 + a2 + a3 +
k=1
The Greek letter © (capital sigma, corresponding to our letter S), stands for “sum.” The index of summation k tells us where the sum begins (at the number below the © symbol) and where it ends (at the number above © ). Any letter can be used to denote the index, but the letters i, j, and k are customary. The index k ends at k n.
5 n
The summation symbol (Greek letter sigma)
ak
a k is a formula for the kth term.
k1 The index k starts at k 1.
Thus we can write 11
12 + 22 + 32 + 42 + 52 + 62 + 72 + 82 + 92 + 10 2 + 112 = a k 2 , k=1
and 100
ƒs1d + ƒs2d + ƒs3d + Á + ƒs100d = a ƒsid. i=1
The sigma notation used on the right side of these equations is much more compact than the summation expressions on the left side.
EXAMPLE 1
Using Sigma Notation
The sum in sigma notation
The sum written out, one term for each value of k
The value of the sum
ak
1 + 2 + 3 + 4 + 5
15
k a s 1d k
s 1d1s1d + s 1d2s2d + s 1d3s3d
1 + 2  3 = 2
k a k + 1 k=1
1 2 + 1 + 1 2 + 1
7 1 2 + = 2 3 6
52 42 + 4  1 5  1
16 25 139 + = 3 4 12
5 k=1 3 k=1 2
5
k2 a k=4 k  1
The lower limit of summation does not have to be 1; it can be any integer.
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5.2
EXAMPLE 2
Sigma Notation and Limits of Finite Sums
337
Using Different Index Starting Values
Express the sum 1 + 3 + 5 + 7 + 9 in sigma notation. The formula generating the terms changes with the lower limit of summation, but the terms generated remain the same. It is often simplest to start with k = 0 or k = 1.
Solution
4
1 + 3 + 5 + 7 + 9 = a s2k + 1d
Starting with k = 0:
k=0 5
1 + 3 + 5 + 7 + 9 = a s2k  1d
Starting with k = 1:
k=1 6
1 + 3 + 5 + 7 + 9 = a s2k  3d
Starting with k = 2:
k=2
1
1 + 3 + 5 + 7 + 9 = a s2k + 7d
Starting with k = 3:
k = 3
When we have a sum such as 3 2 a sk + k d
k=1
we can rearrange its terms, 3 2 2 2 2 a sk + k d = s1 + 1 d + s2 + 2 d + s3 + 3 d
k=1
= s1 + 2 + 3d + s12 + 22 + 32 d 3
Regroup terms.
3
= a k + a k2 k=1
k=1
This illustrates a general rule for finite sums: n
n
n
a sak + bk d = a ak + a bk
k=1
k=1
k=1
Four such rules are given below. A proof that they are valid can be obtained using mathematical induction (see Appendix 1).
Algebra Rules for Finite Su