Answer
$f(x)$ satisfies the hypotheses
(see reasons below)
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 $x=2.$
$\displaystyle \lim_{x\rightarrow 2^{-}}f(x)=\lim_{x\rightarrow 2^{-}}(2x-3)=1$
$\displaystyle \lim_{x\rightarrow 2^{+}}f(x)=\lim_{x\rightarrow 2^{+}}(6x-x^{2}-7)=12-4-7=1$
So $\displaystyle \lim_{x\rightarrow 2}f(x)$ exists and equals $f(2)=1$.
f is continuous at $x=2 \Rightarrow$ hypothesis (1) is satisfied.
$(2) $ Observe what happens with the derivative at $x=2.$
$\displaystyle \left.\frac{d}{dx}(2x-3)\right|_{x=2}=\left.(2)\right|_{x=2}=2$
$\displaystyle \left.\frac{d}{dx}(6x-x^{2}-7)\right|_{x=2}=\left.(6-2x)\right|_{x=2}=2$
This means that $\displaystyle \lim_{h\rightarrow 0}\frac{f(x+h)-f(-1)}{h}$ exists (one sided limits exist and are equal)
$\Rightarrow f$ is differentiable at $x=2.$
$f(x)$ satisfies ($2$).
