Answer
The theorem applies.
$c=2$
Work Step by Step
The Mean Value Theorem
If $f$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ ,
then there exists a number $c$ in $(a, b)$ such that
$f^{\prime}(c)=\displaystyle \frac{f(b)-f(a)}{b-a}.$
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f is not defined for x=0, but 0 is not in the interval $[1,4]$
... so, f is continuous on the closed interval $[1,4]$
$f^{\prime}(x)=-\displaystyle \frac{1}{x^{2}}$
...and f is differentiable on the open interval $(1,4)$
(not defined for x=0, but 0 is not in the interval)
The theorem applies.
There is a $ c\in (1,4)$ such that
$f^{\prime}(c)=\displaystyle \frac{f(b)-f(a)}{b-a}=\frac{(1/4)-1}{4-1}=\frac{-3/4}{3}=-\frac{1}{4}$
$\displaystyle \frac{-1}{c^{2}}=-\frac{1}{4}$
$c^{2}=4$
$c=\pm 2$
we discard $-2 $ as it is not in the interval $(1,4)$
$c=2$
