#### Answer

The minimum value is $ f(1)=5$ and the maximum value is $ f(2)=28$

#### Work Step by Step

$ 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$