Answer
At $x=0, $f has neither a minimum nor maximum
At $x=3$, 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: $f(x)=x^{4}-4x^{3}$
$f'(x)=4x^{3}-12x^{2}, \qquad f''(x)=12x^{2}-24x$
Stationary points:
$f'(x)=0$
$4x^{3}-12x^{2}=0$
$4x^{2}(x-3)=0$
$x=0,\quad x=3$
Second-derivative test:
$f''(0)=12(0)^{2}-24(0)=0$, inconclusive, so we inspect the signs of $f' $around x=0.
Left of x=0, $f'(x)=4x^{2}(x-3)$ is negative
Between 0 and 3, $f'(x)$ is also negative, so f has neither a minimum nor maximum at $x=0$
$f''(3)=12(3)^{2}-24(3)$ is positive, so
at $x=3$, f has a relative minimum.
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