Answer
At $t=-\displaystyle \frac{\sqrt{2}}{2}$, f has a relative minimum.
At $t=\displaystyle \frac{\sqrt{2}}{2}$, f has a relative maximum
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: $f(t)=-2t^{3}+3t$
$f'(t)=-6t^{2}+3, \qquad f''(t)=-12t$
Stationary points:
$f'(t)=0$
$-6t^{2}+3=0$
$-6t^{2}=-3$
$t^{2}=\displaystyle \frac{1}{2}$
$t=\displaystyle \pm\frac{\sqrt{2}}{2}$
Second-derivative test:
$f''(-\displaystyle \frac{\sqrt{2}}{2})=-12\cdot(-\frac{\sqrt{2}}{2})$ is positive, so
at $t=-\displaystyle \frac{\sqrt{2}}{2}$, f has a relative minimum.
$f''(\displaystyle \frac{\sqrt{2}}{2})=-12\cdot(\frac{\sqrt{2}}{2})$ is negative, so
at $t=\displaystyle \frac{\sqrt{2}}{2}$, f has a relative maximum.
