Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.2 - The Mean Value Theorem - 4.2 Exercises: 7

Answer

$c = \pi$

Work Step by Step

$f(x) = sin(\frac{x}{2}),[\frac{\pi}{2},\frac{3\pi}{2}]$ $f(\frac{\pi}{2}) = sin(\frac{\pi/2}{2})=\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$ $f(\frac{3\pi}{2}) = sin(\frac{3\pi/2}{2})=\sin(\frac{3\pi}{4})=\frac{\sqrt{2}}{2}$ Differentiate using chain rule: $f'(x)=\frac{1}{2}cos(\frac{x}{2})$ Since $f'(x)$ exists, $f(x)$ is differentiable which implies continuity, and because $f(\frac{\pi}{2})=f(\frac{3\pi}{2})$, Rolle's theorem's prerequisites are met and is applicable. Find zeroes of $f'(x)$: $cos(\frac{x}{2})$ has to equal 0, since $\frac{1}{2}\times0=0$ Using the unit circle, we know that $cos(\frac{\pi}{2})$ and $cos(\frac{3\pi}{2})$ both equal 0. Therefore $x = \pi,3\pi$ The function is continuous on $[\frac{\pi}{2},\frac{3\pi}{2}]$ and differentiable on $(\frac{\pi}{2},\frac{3\pi}{2})$. Only $\pi$ is on the interval $(\frac{\pi}{2},\frac{3\pi}{2})$ Therefore $c=\pi$
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