Answer
At $x=-2$, $y=17$ local maximum,
At $x=\frac{4}{3}$, $y=-\frac{41}{27}$ local minimum.
Work Step by Step
Step 1. Given the function $y=x^3+x^2−8x+5$, take its derivative to get $y′=3x^2+2x−8=(3x-4)(x+2)$
Step 2. The critical points can be found when $y'=0$ or undefined; we have $(3x-4)(x+2)=0$ which gives $x=-2, \frac{4}{3}$
Step 3. At $x=-2$, $y=(-2)^3+(-2)^2−8(-2)+5=17$ and we can identify this as a local maximum (use local test points) as shown in the figure.
Step 4. At $x=\frac{4}{3}$, $y=(\frac{4}{3})^3+(\frac{4}{3})^2−8(\frac{4}{3})+5=\frac{64}{27}+\frac{16}{9}-\frac{32}{3}+5=\frac{64+48-288+135}{27}=-\frac{41}{27}$ and we can identify this as a local minimum (use local test points) as shown in the figure.