Assume that $f(x)$ is a function which is infinite many times differentiable and also continuous for every derivative. The function graph is in general not known.
How is it possible to detect extremas?
Till now i thought $f'(x)=0$ and $f''(x)\neq 0$ just works, but then i saw the following function:
\begin{cases} 0 & x= 0 \\ \exp\left(-\frac{1}{x^2}\right)&x\neq 0 \end{cases}
For this function it seems to be more tricky to calculate the extrema at $x=0$. Probably there are more functions which are even more "complex".
This brings me to my question: What is a general way calculate the extrema of a given function $f(x)$ for which it is only known that it is infinite many times differentiable and that it is continuous?
Thank you
Fermat's test still works for extrema of your function, you just have to calculate the derivative properly. You have probably noticed that for $x\ne 0$ the derivative is always nonzero. However, computing the derivative at 0 using the difference quotient we find $$ f'(0)=\lim_{x\rightarrow 0}\frac{e^{-1/x^2}}{x}=0 $$ Which tells you that this is the only extremum candidate for the smooth function. The first derivative test is what I would then use to conclude that it is a minimum.
Edit: To avoid any potential ambiguity, for a critical point $x_0$, the first derivative test requires examining the sign $f'(x_0-\epsilon)$ and $f'(x_0+\epsilon)$. $f'(x_0-\epsilon)< 0$ and $f'(x_0+\epsilon)> 0$ yields that $x_0$ is a minimum. Vice versa for a local maximum.
In this case, $$ f'(x)=\frac{2}{x^3}e^{-1/x^2} $$ Which is less than 0 for $x<0$ and greater than 0 for $x>0$, allowing our conclusion.