Solving $f(x)=f(\frac{1}{x}),\,\,\,x>0$

Question

What are the non-trivial solutions of the following functional equations: $$f(x)=f(\frac{1}{x}),\,\,\,x>0$$ given $f$ if differentiable on $(0,\infty)$.

$\textbf{Edit:}$ Do all the solutions attain their absolute max/min at $x=1$?

Answer 1

Take any even function $g$ with whatever conditions you need (continuous, differentiable, etc.). Then let $f(x)=g\mathopen{}\left(\ln(x)\right)\mathclose{}$.

Answer 2

Differentiating the functional equation gives

$$ f'(x)=-\frac{1}{x^2} f'(1/x) $$

and so, in particular, $f'(1)=0$. So any such $f$ must be of the form $$ \begin{cases}g(x), & x \in (0, 1] \\ g(1/x), & x \in (1, \infty)\end{cases} $$ for some differentiable $g$ defined on $(0,1]$ whose left derivative at $1$ vanishes.

Conversely, if $g$ is any such function, then we can define $f$ by the above formula and it will satisfy the functional equation; it will be differentiable at $1$ because the value and derivative of $g(x)$ match up with the value and derivative of $g(1/x)$ there, and it will be differentiable everywhere else by the chain rule.