interpretation for $f(x)=f(\frac{1}{x})$ Is there any interpretation geometric may be for a function that satisfies
$
f(x)= f(\frac{1}{x})\; \; \forall x \in \mathbb{R}^* $
Is it of kind $f(x)= f(g(x))$ ?
Thank you very much.
 A: Let us think of that property in group-theoretic terms. We have a function $f\colon \mathbb R_{>0}\to \mathbb R$ such that
$$\tag{1}
f(t)=f(t^{-1}), \qquad \forall t >0.$$
The domain $(\mathbb R_{>0}, \cdot)$ is a group, and (1) tells us that $f$ is invariant under the operation $t\mapsto t^{-1}$.
For the more familiar group $(\mathbb R, +)$, the corresponding operation would be $x\mapsto -x$. So, the property analogous to (1) for a function $g\colon \mathbb R\to \mathbb R$ would be
$$\tag{2}
g(x)=g(-x), \qquad \forall x \in \mathbb R,$$
which is a familiar one, as it expresses that $g$ is an even function.
Actually, (1) and (2) are equivalent. Indeed, the change of variable
$$t=e^x$$
is an isomorphism of the groups $(\mathbb R, +)$ and $(\mathbb R_{>0}, \cdot)$. Now, if $f$ satisfies (1), then denoting
$$
g(x):=f(e^x), $$
we see that $g$ satisfies (2). Conversely, if $g$ satisfies (2), then denoting
$$
f(t)=g(\log t) $$
we see that $f$ satisfies (1).
Conclusion. We can say that a function $f$ satisfying the property (1) is even in the group $(\mathbb R_{>0}, \cdot)$.
P.S. I hadn't read the comments before posting. I see that various users, such as Gae.S and Chrystomath, proposed this exact same idea.
