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: 19

Answer

Rolle's Theorem can be applied; $c=\frac{\pi}{6}.$

Work Step by Step

$f(x)$ is continuous for all values of $x$ and is differentiable at every value of $x.$ $f(0)=f(\frac{\pi}{3})=0.$ Since $f(x)$ is continuous over $[0 , \frac{\pi}{3}]$ and differentiable over $(0, \frac{\pi}{3})$, applying Rolle's Theorem over the interval $[0, \frac{\pi}{3}]$ guarantees the existence of at least one value $c$ such that $0\lt c\lt \frac{\pi}{3}$ and $f'(c)=0.$ $f'(x)=3\cos{3x}.$ $f'(x)=0\to3\cos{3x}=0\to3x=\frac{\pi}{2}\to c=\frac{\pi}{6}.$
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