#### Answer

c= 16.

#### Work Step by Step

$f(9)=\sqrt 9= 3$
$f(25)= \sqrt {25}=5$
$\frac{f(25)-f(9)}{25-9}=\frac{2}{16}=\frac{1}{8}$
MVT states that there is a point c$\in$(9,25) such that $f'(c)=\frac{1}{8}$. But $f'(x)= \frac{1}{2\sqrt x}$ which implies c=16. Thus at c=16$\,\in$(9,25), we have $f'(c)=\frac{1}{8}.$