## Calculus (3rd Edition)

Minimum value is $f(0)=0$ and maximum value is $f(3)=9$
$f'(t)= 6-2t$ $f$ 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 $f(3)=9$