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: 17

Answer

Rolle's theorem can be applied; $c=\dfrac{\pi}{2}$ or $c=\dfrac{3\pi}{2}.$

Work Step by Step

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