## Calculus (3rd Edition)

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