# If $f$ is an even function, and $f$ is differentiable at $x=0$, prove that $f'(0)=0$?

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.

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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

$$f'(0)=\lim_{y \to 0^+} \frac{f(y)-f(0)}{-y}=-\lim_{y \to 0^+} \frac{f(y)-f(0)}{y}=-f'(0)$$

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Thank you!! This might be sort of a silly question but can you explain what led you to separate the limit into one-sided limits and what 'inspired' you to substitute $y=-x$? I think that even if I had gotten to the third line of your proof I still wouldn't have thought of that. – Danny Dec 11 '12 at 15:29
The substitution $y=-x$ should be pretty natural because you need to use $f(-x)=f(x)$. Splitting it into one-sided limits is not needed, I just think it makes the proof a little easier to follow. – N. S. Dec 11 '12 at 15:51

We have $f(-x)=f(x)$ and so differentiating, $$f^{\prime}(-x)(-x)^{\prime}=f^{\prime}(x)\Rightarrow -f^{\prime}(-x)=f^{\prime}(x)$$ for the $x$ that $f$ is differentiable on $x$ and $-x$ (one such $x$ is $0$). Plugging $x=0$ yields the result.

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Suppose that $f\,'(0)=\alpha>0$. Then

$$\lim_{h\to 0}\frac{f(h)-f(0)}h=\alpha\;,$$

so there is a $\delta>0$ such that $$\frac{f(h)-f(0)}h>\frac{\alpha}2\quad\text{whenever}\quad |h|<\delta\;.$$

Let $x=\dfrac{\delta}2$. Then $$\frac{f(x)-f(0)}x>\frac{\alpha}2\quad\text{and}\quad\frac{f(-x)-f(0)}{-x}>\frac{\alpha}2\;,$$ so

$$f(x)>\frac{\alpha}2x+f(0)\quad\text{and}\quad f(-x)<-\frac{\alpha}2+f(0)\;.$$

But $f$ is even, so $f(-x)=f(x)$, and we have

$$f(x)<-\frac{\alpha}2+f(0)<\frac{\alpha}2+f(0)<f(x)\;,$$

which is absurd. A similar argument shows that $f\,'(0)$ cannot be negative, and we conclude that it must be $0$.

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