## Calculus (3rd Edition)

$f'(c) = -\frac{1}{c^{2}}$ 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.