Pre-Calc Homework Solutions 136

Pre-Calc Homework Solutions 136 - 136 Section 4.2 (b) , 22....

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Unformatted text preview: 136 Section 4.2 (b) , 22. (a) Since f (x) 1, 1, 0 3 x x 1 1 f is not differentiable at x 1. (If f were differentiable at x 1, it would violate the Intermediate Value Theorem for Derivatives.) (b) [ , by [ 1, 2] (c) Note that f (x) We require f (c) For [0, 3] by [ 1, 3] sin x, cos x, f( ) ( f( 0 ) ) x x 1 0 . 1 1 ( 1) 2 . c sin 1 1 0, this occurs when 0.324 or 1 1 sin c , so (c) We require f (c) f (x) f (3) 3 f (0) 0 2 3 1 1 , but 3 c c sin 2.818. For 0 1 c , this 1.247. The 1 for all x where f (x) is defined. Therefore, occurs when cos c , so c 1 cos 1 there is no such value of c. 23. (a) Since f (x) 1, 1, 1 1 x x 0 0 , 25. f (x) 26. f (x) 27. f (x) (b) 28. f (x) 29. f (x) 30. f (x) [ 1, 1] by [ 1, 2] possible values of c are approximately 2.818, x2 2 0.324, and 1.247. C C x2 C 1) C, x C 0 x C C f is not differentiable at x 0. (If f were differentiable at x 1, it would violate the Intermediate Value Theorem for Derivatives.) 2x x 3 cos x ex ln (x 1 x 31. f (1) 1 f ( 1) ( 1) 0 2 0 f (x) f (2) 1 2 (c) We require f (c) f (x) 0, but 1 1 1 2 1 x 1 for all x where f (x) is defined. Therefore, C C f (x) there is no such value of c. 24. (a) We test for differentiability at x given in Section 3.5. Left-hand derivative: lim h0 0, using the limits 32. f (x) f (1) 11/4 C 1 C C f (x) 1 ,x 2 0 f (0 h) h h) h f (0) lim h0 cos h h (1 sin h h 1 0 Right-hand derivative: lim h0 x 1/4 2 2 2 3 1/4 x f (x) f ( 1) 2) C 0 C C f (x) x2 3 3 3 x2 x C 3 ln (x 3 3 3 3 ln (x sin x 2) C f(0 f(0) lim h0 sin h) h 1 33. ln ( 1 lim h0 1 Since the left- and right-hand derivatives are not equal, f is not differentiable at x 0. 34. f (x) f (0) 0 C C f (x) 2) C 3 x sin x 3 ...
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