Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 3 - Applications of Differentiation - 3.2 The Mean Value Theorem - 3.2 Exercises - Page 220: 13

Answer

$c$ = $\frac{\sqrt 3}{9}$

Work Step by Step

The Mean Value Theorem states that if $f$ is a continuous function over the closed interval $[a,b]$ and differentiable over the open interval $(a,b)$, then there exists at least one point $c\in(a,b)$ so that $$f′(c)=\dfrac{f(b)−f(a)}{b−a}.$$ The function $f(x)$ = $\sqrt[3]x$ is continuous on $[0,1]$ and differentiable on $(0,1)$. According the the Mean Value Theorem, there is at least one point $c\in(0,1)$ so that we have: $f'(c)=\dfrac{f(1)-f(0)}{1-0}$ $f'(c)=\dfrac{\sqrt[3]1-\sqrt[3]0}{1}$ $f'(c)=1$ But $f'(x)=\frac{1}{3}x^{-2/3}$, so we have: $\dfrac{1}{3}c^{-2/3}=1$ $c^2=\frac{1}{27}$ $c=\pm\dfrac{1}{3\sqrt 3}=\pm\dfrac{\sqrt 3}{9}$
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