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.

themathandlanguagetutor · Answer

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

Lubin · Answer

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.

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

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