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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?

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1 Answer

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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.

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