Problem :
Assume that two functions $f(x), g(x)$ are differentiable function.
(And, no one of them has interval which makes them constant function.)
And consider a function $f(x)$ has local maxima(or local minima) at $x=a$.
Then $f(g(x))$ also at $g(x)=a$?
I saw a function which as oscillation property works well for counterexample about increasing/decreasing and local maximum, minimum.
And, It also works if I switch the problem for 'if $g$ has local max, $f(g)$ also?'
The counterexample for switched problem is :
$$f(x)=x^2\left(1+ \sin\left(\frac{1}{x}\right)\right),\quad f(0)=0 \\ g(x)=x^2$$
Then we know $g$ has local minimum at $x = 0$ but $f(g(x))$ isn't.
So, I wonder how can I find a counterexample about the original problem?
Or, I want to know the proof about original problem if that's true.
Thanks for help.
When $f(x)$ has a local maximum at $x=a$, it means that you can choose $r>0$ so that $f(a)\geq f(x)$ whenever $x\in(a-r,a+r)$.
Suppose $g(c)=a$. Because $g$ is continuous, there is some $\delta>0$ such that $g(x)\in(a-r,a+r)$ whenever $x\in(c-\delta,c+\delta)$.
Therefore, for such $\delta$ and for $x\in(c-\delta,c+\delta)$, you have $f(a)\geq f(g(x))$.
So $f(g(x))$ has a local maximum at $x=c$.