Answer
Minimum value is $ f(-2)=-2$ and maximum value is $ f(2)=10$
Work Step by Step
First, we find the critical points of $ y $.
$ f'(x)=3x^{2}+2x-1$ is differentiable everywhere. However, $ f'(x)= 0$ when $ x=-1$ or $ x=\frac{1}{3}$. That is, $-1$ and $\frac{1}{3}$ are the critical points of $ y $.
Now, we compare the values of $ f(x)$ at the critical points and the end points.
$ f(-1)=(-1)^{3}+(-1)^{2}-(-1)=1$
$ f(\frac{1}{3})=(\frac{1}{3})^{3}+(\frac{1}{3})^{2}-\frac{1}{3}=-\frac{5}{27}$
$ f(-2)=(-2)^{3}+(-2)^{2}-(-2)=-2$
$ f(2)=2^{3}+2^{2}-2=10$
Minimum value is $ f(-2)=-2$ and maximum value is $ f(2)=10$
