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