Answer
There is no number $c$ in the interval $(1,4)$ such that $f(4)- f(1) = f'(c)(4-1)$
The derivative does not exist when $x = 3$. Thus, the function is not differentiable on the interval $(1,4)$.
Work Step by Step
$f(x) = (x-3)^{-2}$
$f'(x) = -2(x-3)^{-3} = -\frac{2}{(x-3)^3}$
We can find $c$ such that $f(4)- f(1) = f'(c)(4-1)$:
$f(4)- f(1) = f'(c)(4-1)$
$\frac{1}{(4-3)^2}- \frac{1}{(1-3)^2} = -2(c-3)^{-3}(3)$
$1- \frac{1}{4} = -\frac{6}{(c-3)^3}$
$\frac{3}{4} = -\frac{6}{(c-3)^3}$
$(c-3)^3 = -8$
$c-3 = -2$
$c = 1$
Thus there is no number $c$ in the interval $(1,4)$ such that $f(4)- f(1) = f'(c)(4-1)$
The derivative does not exist when $x = 3$. Thus, the function is not differentiable on the interval $(1,4)$. The hypotheses of the Mean Value Theorem are not satisfied.
Therefore, the fact that there is no number $c$ in the interval $(1,4)$ such that $f(4)- f(1) = f'(c)(4-1)$ does not contradict the Mean Value Theorem.