## Calculus (3rd Edition)

Minimum value is 5 at $x=0$ and maximum value is 21 at $x=2$
$f'(x)= 4x+4$ $f$ is differentiable everywhere. So, the only critical point we get is $c$ such that $f'(c)=0$ i.e., $4c+4=0\implies 4c=-4\implies c=-1$ But, this point is outside the given interval. To find the maximum and minimum value, we compare values at the end points only (since there are no critical points inside the given interval). $f(0)= 2(0)^{2}+4(0)+5=5$ $f(2)=2(2)^{2}+4(2)+5=21$ The minimum value is 5 at $x=0$ and maximum value is 21 at $x=2$