Answer
$f(x)$ does not satisfy the hypotheses because
$f(x)$ is not continuous on $[-\pi,0]$
Work Step by Step
The Mean Value Theorem requires
- that $f(x)$ is continuous on $[a,b]$
- that $f'(x)$ is defined on $(a,b)$
Here, $[a,b]=[-\pi,0]$
Observe the left-sided limit at x=0:
By Th.7 in section 2.4
$\displaystyle \quad \lim_{t\rightarrow 0}\frac{\sin t}{t}=1,\quad$
so
$\displaystyle \lim_{x\rightarrow 0^{-}}f(x)=1\qquad $,
But, at the right endpoint, $f(0)=0$,
so f is not left-continuous at x=0, meaning that
f is not continuous over the closed interval $[-\pi,0]$
