Answer
$x=\pm 2,$
relative maximum at $(-2,16), $ relative minimum at $(2,-16).$
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)
$g(x)=x^{3}-12x$
$g'(x)=3x^{2}-12\qquad $defined for all x.
$g'(x)=0$
$3x-12=0$
$3x^{2}=12$
$x^{2}=4$
$ x=\pm 2\qquad$ ... critical points: $(2,-16),(-2,16)$
Second-derivative test:
$g''(x)=6x$
$ g''(-2)=-12 \lt 0$, so f has a relative maximum at $(-2,16)$
$ g''(2)=12 \gt 0$, so f has a relative minimum at $(2,-16)$
