Does $f(x)=f(1/x) \forall x$ put any restrictions on the derivative of $f$?

Question

Consider a functions with the property that $f(x)=f(1/x) \forall x \not = 0$ and $f$ continuous.

Does this tell us anything about the derivatives of $f$?

I feel like it should, since if $f$ is increasing on $(2,3)$ then $f$ is decreasing on $(\frac{1}{3}.\frac12)$

But I feel like this (and potentially other properties) should be easy to show via application of chain rule/product rule.

• (Another property that I have a hunch may be true is that there is some relationship about concavity/convexity to the left and right of $x=1$)

I recall, however, that $f(2)=g(2)$ does not mean that $f'(2)=g'(2)$ so I'm not whether I can even take the derivative of both sides to use in proofs.

Answer 1

Let $g$ be any even function defined on $\Bbb R$.

Now let $f$ be defined as $f(x)=g(\log x)$. Then $f(1/x)=g(\log(1/x))=g(-\log x) =g(\log x)=f(x)$.

So $f$ will be differentiable, continuous, etc., as the original $g$ was.

Answer 2

If $f$ is differentiable at $x$, then by the chain rule, we get $$f'(x) = -\frac{f'(1 / x)}{x^2}.$$