If $f$ is even and $f′(0)$ exists, what is $f′(0)$?

Question

This is a homework question, and I was wondering if anyone could critique my answer?

We know that the derivative of every even function is an odd function.

We also know that every odd function is rotationally symmetric about the point $(0,0)$. So the function $f'(x)$ must contain the point $(0,0)$, therefore $f'(0)=0$.

Answer 1

I think your solution works.

If you want to use the hint then if you add the two expressions for the derivative then you get $$2f'(x)=\lim \limits_{h \to o}\frac{f(x+h)-f(x-h)}{h} $$ then let $x=0$ then we get. $$2f'(0)=\lim \limits_{h \to o}\frac{f(h)-f(-h)}{h} $$ and as $f$ is even then $f(-x)=f(x)$, which impliees $f(-h)-f(h)=0$ therefore $$2f'(0)=\lim \limits_{h \to o}\frac{f(h)-f(-h)}{h} =\lim \limits_{h \to o}\frac{0}{h}=0$$ therefore $f'(0)=0$

Answer 2

We have

$(1) \quad f'(0)= \lim_{h \to 0}\frac{f(h)-f(0)}{h}$

and

$(2) \quad f'(0)= \lim_{h \to 0}\frac{f(-h)-f(0)}{h}$.

Now add $(1)$ and $(2)$ and use that $f(-h)=f(h).$

Can you take it from here ?