# Prove or disprove: $\forall n\in\mathbb{N^{+}}: f^{(n)}(0)=0$, $f$ is not constant $\implies x=0$ is an extremum of $f$

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.

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 $$0$$
• @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 Nov 28, 2019 at 13:29