#### Answer

The minimum value is $ f(-2)=-44$ and the maximum value is $ f(2)=84$

#### Work Step by Step

$ f'(x)=10x^{4}+10x $ is defined everywhere. The critical points are the solutions of $ f'(x)=0$
$10x^{4}+10x= 10x(x^{3}+1)=0$
$\implies x=0$ or $ x=-1$
The critical points are $0$ and $-1$. In order to find the maximum and minimum values, we compare the values of the function at the critical points and at the end points.
$ f(0)=0$
$ f(-1)= 2(-1)^{5}+5(-1)^{2}=3$
$ f(-2)= 2(-2)^{5}+5(-2)^{2}=-44$
$ f(2)= 2(2)^{5}+5(2)^{2}=84$
The minimum value is $ f(-2)=-44$ and the maximum value is $ f(2)=84$