## Calculus (3rd Edition)

The minimum value is $f(1)=5$ and the maximum value is $f(2)=28$
$f'(t)=6t^{2}+6t$ is differentiable everywhere. The critical points of the function are the solutions of $6t^{2}+6t=0$ $6t^{2}+6t=0\implies t=0,-1$ Both critical points are outside the given interval. Therefore, we compare the values of $f(t)$ at the end points only to find the maximum and minimum. $f(1)=2(1)^{3}+3(1)^{2}=5$ $f(2)=2(2)^{3}+3(2)^{2}=28$ The minimum value is $f(1)=5$ and the maximum value is $f(2)=28$