Chapter 4: Applications of Derivatives - Section 4.4 - Concavity and Curve Sketching - Exercises 4.4 - Page 215: 109

$x=-1$ and $x=2$

Step 1. Given $y''=(x+1)(x-2)$, we can find the zeros for $y''$ as $x=-1$ and $x=2$. Step 2. Test the sign changes of $y''$ across these zeros as: $..(+)..(-1)..(-)..(2)..(+)..$; thus $x=-1$ and $x=2$ are inflection points of the function. Step 3. We can not find any other inflection points for the function.