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22. (a) Since f 9 ( x ) 5 h , f is not differentiable at x 5 1. (If f were differentiable at x 5 1, it would violate the Intermediate Value Theorem for Derivatives.) (b) [0, 3] by [ 2 1, 3] (c) We require f 9 ( c ) 5 } f (3 3 ) 2 2 f 0 (0) }5} 2 2 3 1 } 5 } 1 3 } , but f 9 ( x ) 56 1 for all x where f 9 ( x ) is defined. Therefore, there is no such value of c . 23. (a) Since f 9 ( x ) 5 h , f is not differentiable at x 5 0. (If f were differentiable at x 5 1, it would violate the Intermediate Value Theorem for Derivatives.) (b) [ 2 1, 1] by [ 2 1, 2] (c) We require f 9 ( c ) 5 } f (1 1 ) 2 2 ( f 2 ( 2 1) 1) 0 2 2 0 } 5 0, but f 9 ( x ) 1 for all x where f 9 ( x ) is defined. Therefore, there is no such value of c . 24. (a) We test for differentiability at x 5 0, using the limits given in Section 3.5. Left-hand derivative: lim h 0 2 } f (0 1 h h ) 2 f (0) }5 lim h 0 2 } cos h h 2 1 0 Right-hand derivative: lim h 0 1 } f (0 1 h h ) 2 f (0) lim h 0 1 } (1 1 sin h h ) 2 1 } 5 lim h 0 1 } sin h h } 5 1 Since the left- and right-hand derivatives are not equal, f is not differentiable at x 5 0. (b) [ 2 p , p ] by [ 2 1, 2] (c) Note that f 9 ( x ) 5 h . We require f 9 ( c ) 5 } f ( p p ) 2 2 ( f 2 ( 2 p ) p ) } 1 2 2 p ( 2 1) p 1 } . For 2 p , c , 0, this occurs when 2 sin c 5 } p 1 } , so c 52 sin 2 1 1 } p 1 } 2 < 2 0.324 or c p 1 sin 2 1 1 } p 1 } 2 < 2 2.818. For 0 , c , p , this occurs when cos c 5 } p 1 } , so c 5 cos 2 1 1 } p 1 } 2 < 1.247. The possible values of c are approximately 2 2.818, 2 0.324, and 1.247. 25. f ( x ) 5 } x 2 2 } 1 C 26. f ( x ) 5 2 x 1 C 27. f ( x ) 5 x 3 2 x 2 1 x 1 C 28. f ( x ) cos x 1 C 29. f ( x ) 5 e x 1 C 30. f ( x ) 5 ln ( x 2 1) 1 C 31. f ( x ) 5 } 1 x } 1 C , x . 0 f (2) 5 1 } 1 2 } 1 C 5 1 C 5 } 1 2 } f ( x ) 5 } 1 x } 1 } 1 2 } , x . 0 32. f ( x ) 5 x 1/4 1 C f (1) 2 1 1/4 1 C 2 1 1 C 2 C 3 f ( x ) 5 x 1/4 2 3 33. f ( x ) 5 ln ( x 1 2) 1 C f ( 2 1) 5 3 ln ( 2 1 1 2) 1 C 5 3 0 1 C 5 3 C 5 3 f ( x ) 5 ln ( x 1 2) 1 3 34. f ( x ) 5 x 2 1 x 2 sin x 1 C f (0) 5 3 0 1 C 5 3 C 5 3 f ( x ) 5 x 2 1 x 2 sin x 1 3 2 p # x , 0 0 , x # p 2 sin x , cos x , 2 1 # x , 0 1 \$ x . 0 1, 2 1, 0 # x , 1 3 \$ x . 1 2 1, 1, 136 Section 4.2

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35. Possible answers: (a) (b) [ 2 2, 4] by [ 2 2, 4] [ 2 1, 4] by [0, 3.5] (c) [ 2 1, 4] by [0, 3.5] 36. Possible answers: (a) (b) [ 2 1, 5] by [ 2 2, 4] [ 2 1, 5] by [ 2 1, 8] (c) (d) [ 2 1, 5] by [ 2 1, 8] [ 2 1, 5] by [ 2 1, 8] 37. One possible answer: [ 2 3, 3] by [ 2 15, 15] 38. One possible answer: [ 2 3, 3] by [ 2 70, 70] 39. Because the trucker’s average speed was 79.5 mph, and by the Mean Value Theorem, the trucker must have been going that speed at least once during the trip. 40. Let f ( t ) denote the temperature indicated after t seconds. We assume that f 9 ( t ) is defined and continuous for 0 # t # 20. The average rate of change is 10.6 8 F/sec. Therefore, by the Mean Value Theorem, f 9 ( c ) 5 10.6 8 F/sec for some value of c in [0, 20]. Since the temperature was constant before t 5 0, we also know that f 9 (0) 5 0 8 F/min. But f 9 is continuous, so by the Intermediate Value Theorem, the rate of change f 9 (t) must have been 10.1 8 F/sec at some moment during the interval. 41. Because its average speed was approximately 7.667 knots, and by the Mean Value Theorem, it must have been going that speed at least once during the trip.
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