Chapter 4: Applications of Derivatives - Section 4.4 - Concavity and Curve Sketching - Exercises 4.4 - Page 215: 110

$x=-3$ and $x=2$

Step 1. Given $y''=x^2(x-2)^3(x+3)$, we can find its zeros as $x=-3,0,2$. Step 2. Test the sign changes of $y''$ across its zeros as: $..(+)..(-3)..(-)..(0)..(-)..(2)..(+)$. Thus the inflection points are at $x=-3$ and $x=2$, but not at $x=0$. Step 3. We can not find more inflection points for this function.