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Solved: Explain why or why not Determine whether the

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

Solution for problem 23E 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 23E

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The continuous? fu ? nction ?? ) = 1 ? |? | satisfies the conditions of the Mean Value Theorem on the interval [? 1, 1]. b. Two differentiable functions, that differ by a constant always have the same derivative. ? ? c. If? ?? ) = 0, then ?f(?x) = 10.

Step-by-Step Solution:

Solution 23E Step 1 In this problem we have to check whether the given statements are true and we have to give a brief explanation about our conclusion. a. The continuous fu nction ) = 1 | | satisfies the conditions of the Mean Value Theorem on the interval [ 1, 1]. Here we have to check whether f (x) = 1 | satisfies the conditions of Mean value theorem on [-1,1] The given statement is false. Reas on: f ) = 1 | | is not differentiable on [-1,1] Consider lim f(x)f(0= lim 1|x|(1|0|) x0 x0 x0 x0 1|x|1 = lx0 x |x| = lx0 x Since we have the limit as x 0 , the value of xvaries from left of 0. That is x < 0. Therefore |x| = x. Hence we get, = lim (x) x0 x = lim x x0 x = lim 1 x0 = 1 f(x)f(0) Hence lim x0 = 1 …. (1) x0 f(x)f(0) 1|x|(1|0|) Now consider lim + x0 = lim + x0 x0 x0 1|x|1 = lim + x x0 |x| = lim + x x0 + Since we have the limit as x 0 , the value of xvaries from right of 0. That is x > 0. Therefore |x| = x. Hence we get, = lim (x) x0+ x = lim + xx x0 = lim(+1) x0 = 1 f(x)f(0) Hence lix0 + x0 = 1…. (2) From (1) and (2) we get, lim f(x)f(0)=/ lim f(x)f(0) x0 + x0 x0 x0 f is not differentiable at x = 0. Hence f is not differentiable in (-1,1). Ther efore Mean Value theorem is not applicable for f ) = 1 | on [-1,1] Step 2 b. Two differentiable functions, that differ by a constant always have the same derivative. The given statement is t rue. Let f(x)and g(x)be two functions such that f (x) = g (x)for all xin an interval I. To prove: f(x)g(x) = C on I where C is any constant. Given f (x) = g (x)for all x I f (x) g (x) = 0for all x I By using (ab) = a b we get, (f g) (x) = 0for all x I Integrating both sides we get, (f g)(x) = C for all x I f(x) g(x) = C for all x I Thus f and g differ by a constant. Hence two differentiable functions, that differ by a constant always have the same derivative.

Step 3 of 3

Chapter 4.6, Problem 23E is Solved
Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

This full solution covers the following key subjects: always, conditions, constant, Continuous, Counterexample. This expansive textbook survival guide covers 85 chapters, and 5218 solutions. Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567. The full step-by-step solution to problem: 23E from chapter: 4.6 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM. The answer to “Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The continuous? fu ? nction ?? ) = 1 ? |? | satisfies the conditions of the Mean Value Theorem on the interval [? 1, 1]. b. Two differentiable functions, that differ by a constant always have the same derivative. ? ? c. If? ?? ) = 0, then ?f(?x) = 10.” is broken down into a number of easy to follow steps, and 71 words. Since the solution to 23E from 4.6 chapter was answered, more than 298 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1.

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