Answer
$${\text{No relative extrema}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^3} + 5x - 2 \cr
& {\text{Calculate the first derivative }} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^3} + 5x - 2} \right] \cr
& f'\left( x \right) = 3{x^2} + 5 \cr
& {\text{Find the critical points}}{\text{, set }}g'\left( x \right) = 0 \cr
& 3{x^2} + 5 = 0 \cr
& {\text{There are no real values for which }}f'\left( x \right) = 0. \cr
& {\text{Then the function does not have relative extrema}}{\text{.}} \cr} $$
