Is it a necessary condition for an even function to have a local extremum (for $f(x)=k,$ derivative${}=0$) at $x=0$ Let $f(x)$ be an even function ($f(-x)=f(x)$) if $f(x)$ is continuous and differentiable at $x = 0$ will it be necessary for it to have a local extremum? Or more generally, have it's derivative $=0$ at $x=0$?
I thought this as:$$f(x+h)-f(x)=f(x-h)-f(x) \\ \text{(for $x=0 , h>0$)}$$ so the derivative should also be zero. Am I correct or is there a counter example?
 A: The absolute value function is an even function about which it is not true that its derivative at $0$ is $0,$ since its deriative is not defined at $0$ at all. However, all even functions $f$ that are differentiable at $0$ satisfy $f'(0)=0,$ thus:
$$
f'(0) = \lim_{h\to0} \frac{f(h)} h = \lim_{h\to0} \frac{f(-h)} h = \lim_{h\to0} \frac{-f(-h)} {-h} = -f'(0). 
$$
Since $f'(0) = -f'(0),$ we have $f'(0)=0.$
Now suppose
$$
f(x) = \begin{cases} x^2\cos(1/x) & \text{if } x\ne0, \\ 0 & \text{if } x=0. \end{cases}
$$
Then $f$ is an even function that is differentiable at $0,$ but it does not have a local extremum at $0.$
A: Derivative changes the parity (eveness/oddness) of a function. An odd function  if it is continuous vanishes at $x=0$. So  $$f(-x)=f(x) \implies  -f'(-x) =f'(x) \implies f'(0)=0.$$
A: Let $f(x)$ be a function differentiable at $x=0$.
Case 1:
$f$ is an even function.
$$ f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{f(x)-f(x-h)}{h}.$$
Now find $f'(-x)$.
$$f'(-x)= \lim_{h \to 0} \frac{f(-x+h)-f(-x)}{h} $$
Since $f(-x)=f(x)$,
We conclude that $f'(-x)=-f'(x)$
$\implies$ $f'(x)$ is an odd function, hence $f'(0)=0$.
Case 2: (Not really required for your question)
$f$ is an odd function.
$$ f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{f(x)-f(x-h)}{h}. $$
Now find $f'(-x)$.
$$f'(-x)= \lim_{h \to 0} \frac{f(-x+h)-f(-x)}{h} $$
Since $f(-x)=-f(x)$,
We conclude that $f'(-x)=f'(x)$
$\implies$ $f'(x)$ is an even function.
A: We have $f(x)$ as an even function ⇒
$f(-x)=f(x)$
Taking derivative with respect to x.
$-f'(-x)=f'(x)$ ⇒ $f'(x)$ is odd.
Now $f(x)$ is differentiable at $x=0$
$-f'(0)=f'(0)$ ⇒ $f'(0)=0.$
