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Condition for non differentiability Suppose f (x) < 0 < f

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett ISBN: 9780321570567 2

Solution for problem 38AE Chapter 4.6

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Calculus: Early Transcendentals | 1st Edition

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Problem 38AE

Condition for non differentiability Suppose f (x) < 0 < f (x)for x < a and f ?(x) > 0 > f ?(x) for x > a. Prove that f is not differentiable at x = a. (Hint: Assume f is differentiable at a, and apply the Mean Value Theorem to f .) More generally, show that if f ?and f ?change sign at the same point, then f is not differentiable at that point.

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Solution 38AE In this problem we have to discuss about the condition of non-differentiability. Given: 1. I f x < a then f (x) < 0 < f (x) 2. If x > a then f (x) > 0 > f (x) To prove: f is not differentiable at x = a Consider a very small interval near a , say [a ,a ] + Consider the first condition. “If x < a then f (x) < 0 < f (x)” Now take a very small deviation from a, say x = a Finding the derivative of a , by using mean value theorem. The statement of mean value theorem is given by Mean Value theorem: If f is defined and continuous on the closed interval [a,b]and differentiable on the open interval (a,b)then there is at least one point cin (a,b)that is f(b) f(a) a < c < bsuch that f (c) = ba . Thus f (a )is given by f(a ) = f(aa a f(a) a refers to the value of a that are less than a.So a a < 0. That is a < a Thus using 1. We get, f(a ) < 0 < f (a ) That is f (a ) < 0 and f (a ) > 0 … A Now consider x > a Now take a very small deviation from a, say x = a + Finding...

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Chapter 4.6, Problem 38AE is Solved
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Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

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Condition for non differentiability Suppose f (x) < 0 < f

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