Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 4 - Applications of the Derivative - 4.3 The Mean Value Theorem and Monotonicity - Exercises - Page 189: 8

Answer

$$c=\frac{1}{2},\ \ \frac{-1}{2} $$

Work Step by Step

Given $$y=x- \sin(\pi x), \quad\left[-1,1\right]$$ Since $$f'(x)=1-\pi \cos (\pi x) $$ Then by MVT, there exists a constant $c\in (-1,1)$ such that \begin{align*} f'(c)&= \frac{f(b)-f(a) }{b-a}\\ 1-\pi \cos (\pi c)&=\frac{ 1+1 }{2}\\ -\pi \cos (\pi c) &=0 \end{align*} Then $$\pi c= \pi/2, \ -\pi/2$$ Hence $$c=\frac{1}{2},\ \ \frac{-1}{2} $$
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