## Calculus (3rd Edition)

0 and $\frac{1}{12}$
$f'(t)$ 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 $c=\frac{1}{12}$