Answer
At $x=0, $f has neither a minimum nor maximum
At $x=2$, f has a relative minimum.
Work Step by Step
Critical points of $f$ are points that are either stationary or singular
(for which either $f'(c)=0$ or $f'(c)$ is undefined, but $f(c)$ is)
Given: $f(x)=3x^{4}-2x^{3}$
$f'(x)=12x^{3}-6x^{2}, \qquad f''(x)=36x^{2}-12x$
Stationary points:
$f'(x)=0$
$12x^{3}-6x^{2}=0$
$6x^{2}(x-2)=0$
$x=0,\quad x=2$
Second-derivative test:
$f''(0)=36(0)^{2}-12(0)=0$, inconclusive, so we inspect the signs of $f' $around x=0.
Left of x=0, $f'(x)=6x^{2}(x-2)$ is negative
Between $0$ and $2$,
$f'(x)$ is also negative, so f has neither a minimum nor maximum at $x=0$
$f''(2)=36(2)^{2}-12(2)$ is positive, so
at $x=2$, f has a relative minimum.
