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# Ch 2.6 - 92AE ISBN: 9780321570567 2

## Solution for problem 92AE Chapter 2.6

Calculus: Early Transcendentals | 1st Edition

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Problem 92AE

92AE

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STEP_BY_STEP SOLUTION Step-1 For any real nu mber x the bsolute value or modulus of x is denoted by |x|. |x| = { x , if x 0 = { -x , if x 0. The graph of |x| is ; f(x) = |x| It makes a right angle at (0,0) It is an even function. Its domain is the real numbers R, and its range is the non -negative real numbers : [ 0 , +). Step_2 Intermediate Value Theorem. Let f (x ) be a continuous function on the interval [a, b]. If d [f (a ), f (b )], then there is a ,c [a, b] such that f (c) = d. Inthe ca se where f (a) > f (b ), [f (a), f (b)] is meant to be the same as [f (b), f Step_3 Given that f(x) = |x| . x So , f(-2) = |2|= 2 = -1 , since |x| -x , if x<0. 2 2 |2| 2 f(2) = 2 = 2 = 1 , since |x| x , if x0. So, by the intermediate value theorem , then there exist a value d(-2,2) such that f(d) = 0. But , for no values of d such that f(d) = 0. Therefore , for no values of d such that f(d) = 0, because f(x) is non differentiable function . That is it cannot be differentiated at x=0. Step_4 Note: The derivative of the function f at the point...

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##### ISBN: 9780321570567

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