# Finding inflection points using the second derivative

So I have the function $y=(1+x^2) e^{-x^2}$ I find its first derivative $y'=-2x^3e^{-x^2}$ and its second derivative is $y''=e^{-x^2}(-6x^2+4x^4)$. Then I find the roots of $y''$ and they are $0, \pm \sqrt{1.5}$. Why are $\pm \sqrt{1.5}$ the only inflection points, why isn't zero an inflection point too?

• Be careful with your derivatives. Your function is a product, so you need to use the product rule. – AndrewG Jan 2 '13 at 12:33
• I have used it and they are right.. – eeweew Jan 2 '13 at 12:36
• Double check your work. You're missing a factor of x in the first derivative. – AndrewG Jan 2 '13 at 12:37
• Oh,I had done it correctly on my paper ..so the second derivative is still ok.. – eeweew Jan 2 '13 at 12:40

## 1 Answer

The inflection points occur where the second derivative changes sign. The second derivative is indeed $0$ at $x = 0$, but you need to look at neighborhoods of $x=0$ to see whether the sign changes. It doesn't: it remains negative as you pass through $x=0$. Compare $x=-1$ to $x=1$, for example; they're the same.

• Well, at the critical points of the first derivative (where the second one vanishes) and where the second derivative changes sign. – DonAntonio Jan 2 '13 at 13:18