## Calculus (3rd Edition)

Minimum value is $f(-2)=-2$ and maximum value is $f(2)=10$
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 $f(2)=10$