Answer
The minimum value is $ f(2)=-128$ and the maximum value is $ f(-2)=128$
Work Step by Step
$ f'(z)=5z^{4}-80$ is differentiable everywhere. The critical points of the function are the solutions of $5z^{4}-80=0$.
$5z^{4}-80=0$ when $ z=2$ or when $ z=-2$.
The critical points of $ y $ are 2 and -2.
We compare the values of the function at the critical points and the end points to obtain the maximum and minimum value.
$ f(2)=(2)^{5}-80(2)=-128$
$ f(-2)=(-2)^{5}-80(-2)=128$
$ f(3)=(3)^{5}-80(3)=3$
$ f(-3)=(-3)^{5}-80(-3)=-3$
The minimum value is $ f(2)=-128$ and the maximum value is $ f(-2)=128$
