I don't really know where to begin – intuitively I understand that the y-axis intersection must be an extrema, so the derivative is obviously 0, but I'm having difficulty writing the proof..
We haven't learned yet that the derivative of an extrema is 0, so it's not sufficient to simply prove that $f(0)$ an extrema.
Any help would be appreciated.
I'd also appreciate a general explanation of how to approach problems like these.
Best Answer
You know that
$$f'(0)= \lim_{x \to 0} \frac{f(x)-f(0)}{x-0}$$
In particular $$f'(0)= \lim_{x \to 0^+} \frac{f(x)-f(0)}{x-0}=\lim_{x \to 0^-} \frac{f(x)-f(0)}{x-0} $$
Now we use the fact that $f$ is even
$$f'(0)= \lim_{x \to 0^-} \frac{f(x)-f(0)}{x}= \lim_{x \to 0^-} \frac{f(-x)-f(0)}{x}$$
Subbing in $y=-x$ you get
$$f'(0)=\lim_{y \to 0^+} \frac{f(y)-f(0)}{-y}=-\lim_{y \to 0^+} \frac{f(y)-f(0)}{y}=-f'(0)$$