Answer
$f(x)$ does not satisfy the hypotheses
($f$ is not differentiable at $x=-1$)
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) - the only issue with continuity could arise at $x=-1.$
$\displaystyle \lim_{x\rightarrow-1^{-}}f(x)=\lim_{x\rightarrow-1^{-}}(x^{2}-x)=1+1=2$
$\displaystyle \lim_{x\rightarrow-1^{+}}f(x)=\lim_{x\rightarrow-1^{+}}(2x^{2}-3x-3)=2+3-3=2$
So $\displaystyle \lim_{x\rightarrow-1}f(x)$ exists and equals $f(-1)=2$.
f is continuous at $x=-1 \Rightarrow$ hypothesis (1) is satisfied.
$(2) $ Observe what happens with the derivative at $x=-1.$
$\displaystyle \left.\frac{d}{dx}(x^{2}-x)\right|_{x=-1}=(\left.2x-1)\right|_{x=-1}=-3$
$\displaystyle \left.\frac{d}{dx}(2x^{2}-3x-3)\right|_{x=-1}=\left. (4x-3)\right|_{x=-1}=-7$
This means that $\displaystyle \lim_{h\rightarrow 0}\frac{f(x+h)-f(-1)}{h}$ does not exist (the one-sided limits are different)
$\Rightarrow f$ is not differentiable at $x=-1.$
$f(x)$ does not satisfy (2).
