Answer
$f(x)$ does not satisfy the hypotheses because
$f'(x)$ is not defined on the whole interval $(-2,0)$
(f is not differentiable at $x=-1$).
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, $[a,b]=[-2,0]$
Polynomials are continuous functions. The only problem might arise at $x=-1.$
$\displaystyle \lim_{x\rightarrow 0^{-}}f(x)=(-1)^{2}-(-1)=2$
$\displaystyle \lim_{x\rightarrow 0^{+}}f(x)=2(-1)^{2}-3(-1)-3=2$
and, $f(-1)=2$, so f is continuous on $[-2,0]$.
$f'(x)=\left\{\begin{array}{ll}
2x-1, & -2\leq x\leq-1\\
4x-3 & -1\lt x\leq 0
\end{array}\right.$
$\displaystyle \lim_{x\rightarrow 0^{-}}f'(x)=2(-1)-1=-3$
$\displaystyle \lim_{x\rightarrow 0^{+}}f'(x)=4(-1)-3=-7$
So, the derivative is not defined at $x=-1$.
