Answer
Minimum value is $ f(0)=0$ and maximum value is $ f(3)=9$
Work Step by Step
$ f'(t)= 6-2t $
$ f $ is differentiable everywhere. So, the only critical point we get is $ c $ such that $ f'(c)=0$
i.e., $6-2c=0\implies 6=2c\implies c=3$
To find maximum and minimum values, we compare values at the critical point and end points.
$ f(3)=6(3)-(3)^{2}=9$
$ f(0)= 6(0)-(0)^{2}=0$
$ f(5)=6(5)-(5)^{2}=5$
Minimum value is $ f(0)=0$ and maximum value is $ f(3)=9$
