Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 3 - Differentiation - 3.2 Exercises: 24

Answer

Rolle's Theorem can be applied $c=\dfrac{\sqrt{3}}{9}.$

Work Step by Step

$f(x)$ is continuous for all values of $x$ and is differentiable at every value of $x.$ $f(0)=f(1)=0.$ Since $f(x)$ is continuous over $[0 , 1]$ and differentiable over $(0, 1)$, applying Rolle's Theorem over the interval $[0, 1]$ guarantees the existence of at least one value $c$ such that $0\lt c\lt 1$ and $f'(c)=0.$ $f'(x)=1-\dfrac{1}{3\sqrt[3]{x^2}}=\dfrac{3\sqrt[3]{x^2}-1}{3\sqrt[3]{x^2}}$. $f'(x)=0\to \dfrac{3\sqrt[3]{x^2}-1}{3\sqrt[3]{x^2}}=0\to 3\sqrt[3]{x^2}-1=0\to c=\dfrac{\sqrt{3}}{9}.$
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