Pre-Calc Homework Solutions 138

Pre-Calc Homework Solutions 138 - 138 Section 4.3 52(a...

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45. Because the function is not continuous on [0, 1]. The function does not satisfy the hypotheses of the Mean Value Theorem, and so it need not satisfy the conclusion of the Mean Value Theorem. 46. Because the Mean Value Theorem applies to the function y 5 sin x on any interval, and y 5 cos x is the derivative of sin x . So, between any two zeros of sin x , its derivative, cos x , must be zero at least once. 47. f ( x ) must be zero at least once between a and b by the Intermediate Value Theorem. Now suppose that f ( x ) is zero twice between a and b . Then by the Mean Value Theorem, f 9 ( x ) would have to be zero at least once between the two zeros of f ( x ), but this can’t be true since we are given that f 9 ( x ) ± 0 on this interval. Therefore, f ( x ) is zero once and only once between a and b . 48. Let f ( x ) 5 x 4 1 3 x 1 1. Then f ( x ) is continuous and differentiable everywhere. f 9 ( x ) 5 4 x 3 1 3, which is never zero between x 52 2 and x 1. Since f ( 2 2) 5 11 and f ( 2 1) 1, exercise 47 applies, and f ( x ) has exactly one zero between x 2 and x 1.
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Fall '08 term at University of Florida.

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