Answer
$f$ is not continuous at $x=3$
Work Step by Step
$f(x) = (x-3)^{-2}$
Determine $f'(x)$:
$f'(x) = -2(x-3)^{-3}$
Solve the equation for $c$:
$f'(c) = \frac{f(4)-f(1)}{4-1}$
$f(4)-f(1) = f'(c)$
$\frac{1}{1^{2}}-\frac{1}{(-2)^{2}} = \frac{-2}{(c-3)^{3}}\times3$
$c-3 = -2$
$c = 1$
which is not in the open interval $(1,4)$.
The Mean Value Theorem states that if $f$ is a continuous function over the closed interval $[a,b]$ and differentiable over the open interval $(a,b)$, then there exists at least one point $c\in(a,b)$ so that
$$f′(c)=\dfrac{f(b)−f(a)}{b−a}.$$
The function $f(x)= (x-3)^{-2}$ is not continuous in $x=3$, therefore it is not continuous in $[1,4]$. That is why the Mean Value Theorem could not be applied.
