Answer
0 and $\frac{1}{12}$
Work Step by Step
$ f'(t)$ exists for all t. Therefore, the only critical points are the solutions of $ f'(t)=0$.
$ f'(t)=24t^{2}-2t=0$
$\implies t=0$ or
$24t^{2}=2t\implies 12t=1\implies t=\frac{1}{12}$
Thus, the critical points are the roots $ c= 0$ and $ c=\frac{1}{12}$
