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