## Calculus (3rd Edition)

$f(9)=\sqrt 9= 3$ $f(25)= \sqrt {25}=5$ $\frac{f(25)-f(9)}{25-9}=\frac{2}{16}=\frac{1}{8}$ 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 $f'(c)=\frac{1}{8}.$