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Can an even function have an inflection point? If yes, then give an example while if not then give a proof. If needed you may assume that $f$ is two times differentiable .

Intution says that even functions don't have inflection points, but I cannot settle down an appropriate proof.

On the contrary for an odd function this is possible. For instance $f(x)=x^3$ has an inflection point at $x=0$.

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    $\begingroup$ Does it have to be at $x=0$? $\endgroup$
    – mickep
    Commented Sep 15, 2015 at 18:22
  • $\begingroup$ Thank you all for your answers. $\endgroup$
    – Tolaso
    Commented Sep 15, 2015 at 18:58

3 Answers 3

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$\cos\colon\mathbb{R}\to\mathbb{R}$ is even and it has infinitely many inflection points (at its zeros, which are found at $\{n\pi\}_{n\in\mathbb{Z}}$).

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Consider $$f(x)=\max\bigl((x-1)^3,-(x+1)^3\bigr).$$ Or, for a nicer example, consider $$g(x)=(x-1)^3(x+1)^3.$$

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the function $y=x^4-4x^2$ has at least one inflection point WA plot

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