Chapter 5 Applications of Derivatives PDF

7535_Ch05_pp190-229.qxd 1/10/11 10:08 AM Page 190 CHAPTER Applications 5 of Derivatives An automobile’s gas mileage is...

7535_Ch05_pp190-229.qxd 1/10/11 10:08 AM Page 190 CHAPTER Applications 5 of Derivatives An automobile’s gas mileage is a function of many variables, including road 5.1 Extreme Values surface, tire type, velocity, fuel octane rating, road angle, and the speed of Functions and direction of the wind. If we look only at velocity’s effect on gas mileage, the mileage of a certain car can be approximated by 5.2 Mean Value Theorem m(v ) = 0.00015v 3 - 0.032v 2 + 1.8v + 1.7 5.3 Connecting f  and f  with the Graph of f (where v is velocity). 5.4 Modeling and At what speed should you drive this car to obtain the best gas mileage? The Optimization ideas in Section 5.1 will help you find the answer. 5.5 Linearization and Differentials 5.6 Related Rates 190 7535_Ch05_pp190-229.qxd 1/10/11 10:08 AM Page 191 Section 5.1 Extreme Values of Functions 191 CHAPTER 5 Overview In the past, when virtually all graphing was done by hand—often laboriously—derivatives were the key tool used to sketch the graph of a function. Now we can graph a function quickly, and usually correctly, using a grapher. However, confirmation of much of what we see and conclude true from a grapher view must still come from calculus. This chapter shows how to draw conclusions from derivatives about the extreme val- ues of a function and about the general shape of a function’s graph. We will also see how a tangent line captures the shape of a curve near the point of tangency, how to deduce rates of change we cannot measure from rates of change we already know, and how to find a function when we know only its first derivative and its value at a single point. The key to recovering functions from derivatives is the Mean Value Theorem, a theorem whose corollaries provide the gateway to integral calculus, which we begin in Chapter 6. 5.1 Extreme Values of Functions What you will learn about... Absolute (Global) Extreme Values Absolute (Global) Extreme Values One of the most useful things we can learn from a function’s derivative is whether the Local (Relative) Extreme Values function assumes any maximum or minimum values on a given interval and where these values are located if it does. Once we know how to find a function’s extreme val- Finding Extreme Values ues, we will be able to answer such questions as “What is the most effective size for a and why... dose of medicine?” and “What is the least expensive way to pipe oil from an offshore Finding maximum and minimum values well to a refinery down the coast?” We will see how to answer questions like these in of functions, called optimization, is Section 5.4. an important issue in real-world problems. DEFINITION Absolute Extreme Values Let f be a function with domain D. Then f (c) is the (a) absolute maximum value on D if and only if f(x) … f(c) for all x in D. (b) absolute minimum value on D if and only if f(x) Ú f(c) for all x in D. Absolute (or global) maximum and minimum values are also called absolute extrema (plural of the Latin extremum). We often omit the term “absolute” or “global” and just say maximum and minimum. y Example 1 shows that extreme values can occur at interior points or endpoints of intervals. 1 y  sin x y  cos x EXAMPLE 1 Exploring Extreme Values x On [- p/2, p/2], f(x) = cos x takes on a maximum value of 1 (once) and a mini-  – –– 0  –– 2 2 mum value of 0 (twice). The function g(x) = sin x takes on a maximum value of 1 and a minimum value of -1 (Figure 5.1). Now Try Exercise 1. –1 Functions with the same defining rule can have different extrema, depending on the Figure 5.1 (Example 1) domain. 7535_Ch05_pp190-229.qxd 1/10/11 10:08 AM Page 192 192 Chapter 5 Applications of Derivatives y EXAMPLE 2 Exploring Absolute Extrema The absolute extrema of the following functions on their domains can be seen in Figure 5.2. y  x2 D  (–  ,  ) Function Rule Domain D Absolute Extrema on D (a) y = x2 (- q, q) No absolute maximum x Absolute minimum of 0 at x = 0 2 (b) y = x2 [0, 2] Absolute maximum of 4 at x = 2 (a) abs min only Absolute minimum of 0 at x = 0 (c) y = x2 (0, 2] Absolute maximum of 4 at x = 2 y No absolute minimum (d) y = x2 (0, 2) No absolute extrema y  x2 Now Try Exercise 3. D  [0, 2] Example 2 shows that a function may fail to have a maximum or minimum value. This x cannot happen with a continuous function on a finite closed interval. 2 (b) abs max and min y THEOREM 1 The Extreme Value Theorem 2 yx If f is continuous on a closed interval [a, b], then f has both a maximum value and a D  (0, 2] minimum value on the interval. (Figure 5.3) x 2 (c) abs max only (x 2, M) y y  f (x) M y  f (x) M y  x2 x1 m D  (0, 2) x x a x2 b a b m Maximum and minimum at endpoints (x1, m) x Maximum and minimum 2 at interior points (d) no abs max or min Figure 5.2 (Example 2) y  f (x) y  f(x) M M m m x x a x2 b a x1 b Maximum at interior point, Minimum at interior point, minimum at endpoint maximum at endpoint Figure 5.3 Some possibilities for a continuous function’s maximum (M) and minimum (m) on a closed interval [a, b]. 7535_Ch05_pp190-229.qxd 1/10/11 10:08 AM Page 193 Section 5.1 Extreme Values of Functions 193 Absolute maximum. No greater value of f anywhere. Also a local maximum. Local maximum. No greater value of f nearby. Local minimum. y  f (x) No smaller value of f nearby. Absolute minimum. No smaller value Local minimum. of f anywhere. Also a No smaller value of local minimum. f nearby. x a c e d b Figure 5.4 Classifying extreme values. Local (Relative) Extreme Values Figure 5.4 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 smaller than at any other point nearby. The curve rises to the left and falls to the right around c, making f(c) a maximum locally. The function attains its absolute maxi- mum at d. DEFINITION Local Extreme Values Let c be an interior point of the domain of the function f. Then f(c) is a (a) local maximum value at c if and only if f(x) … f(c) for all x in some open interval containing c. (b) local minimum value at c if and only if f(x) Ú f(c) for all x in some open interval containing c. A function f has a local maximum or local minimum at an endpoint c if the appro- priate inequality holds for all x in some half-open domain interval containing c. Local extrema are also called relative extrema. An absolute extremum is also a local extremum, because being an extreme value over- all makes it an extreme value in its immediate neighborhood. Hence, a list of local extrema will automatically include absolute extrema if there are any. Finding Extreme Values The interior domain points where the function in Figure 5.4 has local extreme values are points where either f  is zero or f does not exist. This is generally the case, as we see from the following theorem. THEOREM 2 Local Extreme Values If a function f has a local maximum value or a local minimum value at an interior point c of its domain, and if f  exists at c, then f¿(c) = 0. 7535_Ch05_pp190-229.qxd 1/10/11 10:08 AM Page 194 194 Chapter 5 Applications of Derivatives Because of Theorem 2, we usually need to look at only a few points to find a function’s extrema. These consist of the interior domain points where f¿ = 0 or f¿ does not exist (the domain points covered by the theorem) and the domain endpoints (the domain points not covered by the theorem). At all other domain points, f¿ 7 0 or f¿ 6 0. The following definition helps us summarize these findings. DEFINITION Critical Point A point in the interior of the domain of a function f at which f¿ = 0 or f¿ does not exist is a critical point of f. Thus, in summary, extreme values occur only at critical points and endpoints. DEFINITION Stationary Point A point in the interior of the domain of a function f at which f¿ = 0 is called a stationary point of f. A stationary point can be a minimum, a maximum, or an inflection point. Note that crit- ical points and stationary points of a function f are not necessarily the same. See Examples 3 and 5. EXAMPLE 3 Finding Absolute Extrema Find the absolute maximum and minimum values of f(x) = x2/3 on the interval [-2, 3]. SOLUTION Solve Analytically We evaluate the function at the critical points and endpoints and take the largest and smallest of the resulting values. The first derivative 2 - 1/3 2 f¿(x) = x = 3 3 x 32 has no zeros but is undefined at x = 0. The values of f at this one critical point and at the endpoints are Critical point value: f(0) = 0; Endpoint values: f( -2) = (-2)2>3 = 2 3 4; f(3) = (3) = 2 2>3 3 9. 3 9 « 2.08, and 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 occurs at the interior point x = 0. y  x 2/3 Support Graphically The graph in Figure 5.5 suggests that f has an absolute max- imum value of about 2 at x = 3 and an absolute minimum value of 0 at x = 0. The critical point (0, 0) is not a stationary point. Now Try Exercise 11. In Example 4, we investigate the reciprocal of the function whose graph was drawn in Example 3 of Section 1.2 to illustrate “grapher failure.” EXAMPLE 4 Finding Extreme Values 1 [–2, 3] by [–1, 2.5] Find the extreme values of f(x) =. 24 - x2 Figure 5.5 (Example 3) continued 7535_Ch05_pp190-229.qxd 1/10/11 10:08 AM Page 195 Section 5.1 Extreme Values of Functions 195 SOLUTION Solve Graphically Figure 5.6 suggests that f has an absolute minimum of about 0.5 at x = 0. There also appear to be local maxima at x = - 2 and x = 2. However, f is not defined at these points and there do not appear to be maxima anywhere else. Confirm Analytically The function f is defined only for 4 - x2 7 0, so its domain is the open interval (- 2, 2). The domain has no endpoints, so all the extreme values must occur at critical points. We rewrite the formula for f to find f¿ : 1 [–4, 4] by [–2, 4] f(x) = = (4 - x2) - 1>2 Figure 5.6 The graph of 24 - x 2 1 Thus, f(x) =. 24 - x2 1 x (Example 4) f¿(x) = - (4 - x2) - 3>2 (-2x) =. 4 (4 - x2)3>2 The only critical point in the domain (- 2, 2) is x = 0. The value 1 1 f(0) = = 24 - 0 2 2 is therefore the sole candidate for an extreme value. To determine whether 1> 2 is an extreme value of f, we examine the formula 1 f(x) =. 24 - x2 As x moves away from 0 on either side, the denominator gets smaller, the values of f increase, and the graph rises. We have a minimum value at x = 0, and the minimum is absolute. The function has no maxima, either local or absolute. This does not violate Theorem 1 (The Extreme Value Theorem) because here f is defined on an open interval. To invoke Theorem 1’s guarantee of extreme points, the interval must be closed. Now Try Exercise 25. 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 5.7 illustrates this for interior points. Exercise 55 describes a function that fails to assume an extreme value at an endpoint of its domain. y y  x3 y 1 1 y  x1/3 x –1 0 1 x –1 1 –1 –1 (a) (b) Figure 5.7 Critical points without extreme values. (a) y¿ = 3x2 is 0 at x = 0, but y = x3 has no extremum there. (b) y¿ = (1>3)x - 2>3 is undefined at x = 0, but y = x1>3 has no extremum there. 7535_Ch05_pp190-229.qxd 1/10/11 10:08 AM Page 196 196 Chapter 5 Applications of Derivatives EXAMPLE 5 Finding Extreme Values Find the extreme values of 5 - 2x2, f(x) = e x … 1 x + 2, x 7 1. SOLUTION Solve Analytically For x Z 1, the derivative is d (5 - 2x2) = - 4x, x 6 1 dx f¿(x) = d d (x + 2) = 1, x 7 1. dx The only point where f¿ = 0 is x = 0. What happens at x = 1? At x = 1, the right- and left-hand derivatives are, respectively, f(1 + h) - f(1) (1 + h) + 2 - 3 h lim+ = lim+ = lim+ = 1. h:0 h h:0 h h:0 h 2 f(1 + h) - f(1) 5 - 2(1 + h) - 3 lim- = lim- h:0 h h:0 h -2h(2 + h) = lim- = - 4. h:0 h Since these one-sided derivatives differ, f has no derivative at x = 1, and 1 is a second critical point of f. The domain (- q , q ) has no endpoints, so the only values of f that might be local extrema are those at the critical points: f(0) = 5 and f(1) = 3 From the formula for f, we see that the values of f immediately to either side of x = 0 are less than 5, so 5 is a local maximum. Similarly, the values of f immediately to either side of x = 1 are greater than 3, so 3 is a local minimum. Support Graphically The graph in Figure 5.8 suggests that f¿(0) = 0 and that f¿(1) does not exist. There appears to be a local maximum value of 5 at x = 0 [–5, 5] by [–5, 10] and a local minimum value of 3 at x = 1. The point (0, 50) is the only stationary Figure 5.8 The function in Example 5. point. Now Try Exercise 41. Most graphing calculators have built-in methods to find the coordinates of points where extreme values occur. We must, of course, be sure that we use correct graphs to find these values. The calculus that you learn in this chapter should make you feel more confident about working with graphs. EXAMPLE 6 Using Graphical Methods Find the extreme values of f(x) = In ` `. x 1 + x2 SOLUTION Solve Graphically The domain of f is the set of all nonzero real numbers. Figure 5.9 suggests that f is an even function with a maximum value at two points. The coordi- Maximum X =.9999988 Y = –.6931472 nates found in this window suggest an extreme value of about -0.69 at approximately x = 1. Because f is even, there is another extreme of the same value at approximately [–4.5, 4.5] by [–4, 2] x = - 1. The figure also suggests a minimum value at x = 0, but f is not defined there. Figure 5.9 The function in Example 6. continued 7535_Ch05_pp190-229.qxd 1/10/11 10:08 AM Page 197 Section 5.1 Extreme Values of Functions 197 Confirm Analytically The derivative 1 - x2 f¿(x) = x(1 + x2) is defined at every point of the function’s domain. The critical points where f¿(x) = 0 are x = 1 and x = - 1. The corresponding values of f are both In (1>2) = -In 2 « - 0.69. Now Try Exercise 37. EXPLORATION 1 Finding Extreme Values Let f(x) = ` ` , -2 … x … 2. x x2 + 1 1. Determine graphically the extreme values of f and where they occur. Find f  at these values of x. 2. Graph f and f¿ (or NDER ( f(x), x, x) ) in the same viewing window. Comment on the relationship between the graphs. 3. Find a formula for f¿(x). Quick Review 5.1 (For help, go to Sections 1.2, 2.1, 3.5, and 4.1.) Exercise numbers with a gray background indicate problems that the authors have designed to be solved without a calculator. In Exercises 1–4, find the first derivative of the function. 1. f(x) = 14 - x a b c a b c 2 2. f(x) = (c) (d) 19 - x2 3. g(x) = cos (In x) In Exercises 9 and 10, find the limit for 4. h(x) = e2x 2 f(x) =. 29 - x2 In Exercises 5–8, match the table with a graph of f(x). 5. x f¿(x) 6. x f¿(x) 9. lim- f(x) 10. lim f(x) x:3 x: -3 + a 0 a 0 b 0 b 0 In Exercises 11 and 12, let c 5 c f(x) = u -5 x 3 - 2x, x … 2 x + 2, x 7 2. 7. x f¿(x) 8. x f¿(x) a does not exist a does not exist 11. Find (a) f¿(1), (b) f¿(3), (c) f¿(2). b 0 b does not exist 12. (a) Find the domain of f¿. c -2 c -1.7 (b) Write a formula for f¿(x). a b c a b c (a) (b) 7535_Ch05_pp190-229.qxd 1/10/11 10:08 AM Page 198 198 Chapter 5 Applications of Derivatives Section 5.1 Exercises In Exercises 1–4, find the extreme values and where they occur. 13. h(x) = In (x + 1), 0 … x … 3 1. y 2. y - x2 14. k(x) = e , - q 5, - 3 … x5, - 2