Answer
$${\text{relative minimum of 0 at }}x = 0.$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^{4/5}} \cr
& {\text{Calculate the first derivative }} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^{4/5}}} \right] \cr
& f'\left( x \right) = \frac{4}{5}{x^{ - 1/5}} \cr
& f'\left( x \right) = \frac{4}{{5\root 5 \of x }} \cr
& {\text{Find the critical points}}{\text{, set }}f'\left( x \right) = 0 \cr
& \frac{4}{{5\root 5 \of x }} = 0 \cr
& {\text{There are no real values for which }}f'\left( x \right) = 0. \cr
& {\text{The function is not defined at }}x = 0. \cr
& {\text{Then the critical point is }}x = 0 \cr
& {\text{Calculate }}f'\left( { - 1} \right){\text{ and }}f'\left( 1 \right) \cr
& f'\left( { - 1} \right) = \frac{4}{{5\root 5 \of { - 1} }} = - \frac{4}{5} < 0 \cr
& f'\left( 1 \right) = \frac{4}{{5\root 5 \of 1 }} = \frac{4}{5} > 0 \cr
& {\text{The first derivative changes negative to positive}}{\text{, then }}f\left( x \right){\text{ has a}} \cr
& {\text{relative minimum of 0 at }}x = 0. \cr} $$
