Answer
Use the Mean Value Theorem
Work Step by Step
Since $f$ is continuous on $[32,33]$ and differentiable on $(32,33)$, then by the Mean Value Theorem there exists a number $c$ in $(32,33)$ such that
$f'(c)$ = $\frac{1}{5}c^{-\frac{4}{5}}$ = $\frac{\sqrt[5] {33}-\sqrt[5] {32}}{33-32}$ = $\sqrt[5] {33}-{2}>0$
The function $f'(x)=\frac{1}{5}x^{-4/5}$ is decreasing, so it follows that $f'(c)$ $\lt$ $f'(32)$ = $\frac{1}{5}(32)^{-\frac{4}{5}}$ = $0.0125$, so
$0.0125$ $\gt$ $f'(c)$
$0.0125$ $\gt$ $\sqrt[5] {33}-{2}$
Therefore
$2$ $\lt$ $\sqrt[5] {33}$ $\lt$ $2.0125$
