Chapter 5 Solutions.pdf

# Chapter 5 Solutions.pdf - CHAPTER 5 MORE APPLICATIONS OF...

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CHAPTER 5 MORE APPLICATIONS OF THE DERIVATIVE EXERCISE 5.1 (pp. 260–263) 1. f(x) = 1 3 x 3 – x 2 – 3x + 3 f ' (x) = x 2 – 2x – 3 = (x – 3)(x + 1) The critical numbers are x = 3 and x = –1. f ' (–2) > 0 and f ' (0) < 0. There is a local maximum at x = –1. f ' (2) < 0 and f ' (4) > 0. There is a local minimum at x = 3 –10 x 10 –10 y 10 2. f(x) = 3x 4 – 4x 3 – 6x 2 + 12x f ' (x) = 12x 3 – 12x 2 – 12x + 12 = 12 ( x 3 – x 2 – x + 1 ) = 12 [ x 2 (x – 1) – 1(x – 1) ] = 12 ( x 2 – 1 ) (x – 1) = 12(x + 1)(x – 1) 2 The critical numbers are x = ± 1. f ' (–2) < 0 and f ' (0) > 0. There is a local minimum at x = –1. f ' (0) > 0 and f ' (2) > 0. There is no local extreme value at x = 1 since the function is increasing on both sides of x = 1. –3 x 3 –12 y 8 3. f(x) = x 4 – 4x 3 f ' (x) = 4x 3 – 12x 2 = 4x 2 (x – 3) The critical numbers are x = 0 and x = 3. f ' (–1) < 0 and f ' (1) < 0. There is no local extreme value at x = 0 since f is decreasing on both sides of x = 0. f ' (2) < 0 and f ' (4) > 0. There is a local minimum at x = 3. –3 x 6 –40 y 40 4. f(x) = sin(2x) f ' (x) = 2 cos(2x) cos(2x) = 0 if 2x is an odd multiple of π 2 . That is, if 2x = ± π 2 , ± 3 π 2 , ± 5 π 2 , ± 7 π 2 , … . Then the critical numbers are the odd multiples of π 4 . That is, x = ± π 4 , ± 3 π 4 , ± 5 π 4 , ± 7 π 4 , … . We test values of f ' (x) at the even multiples of π 4 . f ' (0) > 0; f '( π 2 ) < 0; f ' ( π ) > 0; f '( 3 π 2 ) < 0; f ' (2 π ) > 0; … . There is a local maximum at x = π 4 ; a local minimum at x = 3 π 4 ; a local maximum at x = 5 π 4 ; and so on. The maximum and minimum points alternate. –2 π x 2 π –4 y 4

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168 Increasing and Decreasing Functions Section 5.1 5. f(x) = x 2/3 (x – 5) = x 5/3 – 5x 2/3 f ' (x) = 5 3 x 2/3 10 3 x –1/3 = 5 3 x –1/3 (x – 2) The critical numbers are x = 0 (where f ' (x) is undefined) and x = 2 (where f ' (x) = 0). f ' (–1) > 0 and f ' (1) < 0. There is a local maximum at x = 0. f ' (1) < 0 and f ' (3) > 0. There is a local minimum at x = 2. –10 x 10 –10 y 10 6. f(x) = 2x  3 – x f ' (x) = 2x · –1 2  3 – x +  3 – x · 2 = –x  3 – x + 2  3 – x = – x + 2(3 – x)  3 – x = 6 3x  3 – x = 3(2 – x)  3 – x The critical numbers are x = 2 and x = 3. The domain of the function is ( , 3 ] . There is a vertical tangent at x = 3, but the curve only exists at and to the left of 3. There is a horizontal tangent at x = 2. f ' (1) > 0 and f ' (2.5) < 0. Hence there is a local maximum at x = 2. We don't refer to the point at x = 3 as being a local minimum because we can't use inputs x on each side of 3. –9.4 x 9.4 –6.2 y 6.2 7. f(x) = x 4/3 + 4x 1/3 f ' (x) = 4 3 x 1/3 + 4 3 x –2/3 = 4 3 x –2/3 (x + 1) The critical numbers are x = 0 and x = –1. f ' (–2) < 0 and f ' (–.5) > 0. There is a local minimum at x = –1. f ' (–.5) > 0 and f ' (1) > 0. There is no local extreme value at x = 0, since the function is increasing on both sides of x = 0.
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