Calculus (3rd Edition)

Minimum value is $f(6)=0$ and maximum value is $f(4)=8$
$f'(t)= 6-2t$ $f$ is differentiable everywhere. So, the only critical point we get is $c$ such that $f'(c)=0$ i.e., $6-2c=0\implies 6=2c\implies c=3$ But this point is outside the given interval. To find maximum and minimum values, we compare values at the end points only (since there are no critical points in the given interval). $f(4)= 6(4)-(4)^{2}=8$ $f(6)=6(6)-(6)^{2}=0$ Minimum value is $f(6)=0$ and maximum value is $f(4)=8$