Answer
$f(x)$ satisfies the hypotheses because
$f(x)$ is continuous on $[0,3]$
$f'(x)$ is defined on $(0,3)$
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, $[a,b]=[0,3]$
Polynomials are continuous functions. The only problem might arise at $x=2.$
$\displaystyle \lim_{x\rightarrow 2^{-}}f(x)=2(2)-3=1$
$\displaystyle \lim_{x\rightarrow 2^{+}}f(x)=6(2)-(2)^{2}-7=1$
and, $f(2)=1$, so f is continuous on $[0,3]$.
$f'(x)=\left\{\begin{array}{ll}
2, & 0\leq x\leq 2\\
6-2x & 2\lt x\leq 3
\end{array}\right.$
$\displaystyle \lim_{x\rightarrow 2^{-}}f'(x)=2$
$\displaystyle \lim_{x\rightarrow 2^{+}}f'(x)=6-2(2)=2$
so $f'(x)$ is differentiable on $(0,3)$
Both conditions are satisfied.
