#### Answer

c= 4

#### Work Step by Step

$f'(c) = -\frac{1}{c^{2}}$
MVT states that, if the interval is [a,b]
$f'(c)=\frac{f(b)-f(a)}{b-a}$.
Then in this case
$f'(c)= \frac{f(8)-f(2)}{8-2}=\frac{\frac{1}{8}-\frac{1}{2}}{6}= -\frac{1}{16}$
⇒ $-\frac{1}{c^{2}}= -\frac{1}{16}$
⇒ $c^{2}= 16$ or c= 4 $\epsilon$ (2,8) satisfying mean value theorem.