Answer
$f(x)$ does not satisfy the hypotheses
(f is not continuous on $[-\pi,0]$)
Work Step by Step
The two hypotheses of the Mean Value Theorem are:
(1) $y=f(x)$ is continuous over a closed interval $[a, b]$ and
(2) $f(x)$ is differentiable on the interval's interior $(a, b)$ .
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In this problem,
(1)
$\displaystyle \lim_{x\rightarrow 0^{-}}f(x)=\lim_{x\rightarrow 0^{-}}\frac{\sin x}{x}=1$
Since the left limit at $x=0$ does not equal $f(0)=0,$
$f(x)$ is not left-continuous at the right border of $[-\pi,0]$. That is, f is not continuous on $[-\pi,0]$.
Thus, hypothesis (1) is not satisfied.
