Answer
Minimum value is 3 at $ x=-1$ and maximum value is 21 at $ x=2$.
Work Step by Step
$ f'(x)= 4x+4$
$ f $ is differentiable everywhere. So, the only critical point we get is $ c $ such that $ f'(c)=0$
i.e., $4c+4=0\implies 4c=-4\implies c=-1$
To find maximum and minimum values, we compare values at the critical point and end points.
$ f(-1)=2(-1)^{2}+4(-1)+5=3$
$ f(-2)= 2(-2)^{2}+4(-2)+5=5$
$ f(2)=2(2)^{2}+4(2)+5=21$
Minimum value is 3 at $ x=-1$ and maximum value is 21 at $ x=2$
