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Let $f(x):\mathbb{R}\to\mathbb{R}$ be a continuously differentiable function infinite times $(f\in C^\infty)$ such that for every $n\in\mathbb{N^+}$:

$$f^{(n)}(0)=0$$

Prove or disprove that if $f(x)$ is not constant at a neighborhood of $x=0$, then $x=0$ is an extremum of $f(x)$.


Because I don't know how to prove this, I tried to find a counterexample. However, the only function I could think of, which was not constant and could be manipulated to follow the requirements, is $f(x)=e^{-\frac{1}{x^2}}$. Problem is, of course, that $x=0$ is indeed an extremum (minimum) of the function.

I would glad to hear your opinions. Thank you!

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1 Answer 1

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You are quite close: instead of $e^{-\frac{1}{x^2}}$, consider $xe^{-\frac{1}{x^2}}$. This function satisfies your hypothesis, is odd and non-constant, and so $0$ is neither a minimum nor a maximum.

Another way of obtaining the same result, is to note that if $f(x)$ satisfies your hypotesis and it's even, $f'$ satisfies the hypotesis and it's odd, and thus cannot have an extremum at $0$

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  • $\begingroup$ Oh that's right. Thank you very much! $\endgroup$
    – Amit Zach
    Nov 28, 2019 at 13:27
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    $\begingroup$ @AmitZach note that you could have reached this result considering the derivative of your function. Since $f(x)$ is even, $\forall_n f^{(n)}(0)=0$ , $f'$ satisfies your hypothesis and it is odd (since the derivative of an even function it's odd), so the origin cannot be an extremum $\endgroup$
    – Caffeine
    Nov 28, 2019 at 13:29

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