I am trying to solve the functional equation
$$f(x)=f\left(\frac{1}{x}\right),\quad x>0$$
for a real-valued function $f$, with $f(1)=2$. We may assume $f$ is continuous or infinitely differentiable. There is, obviously, the solution $f(x)=2$ for all $x>0$, but I have been unable to show that this is the only solution. By differentiating the original equation, one easily that $f^{(2n+1)}(1)=0$, where $f^{(k)}$ denotes the $k$th derivative of $f$, and $n$ is a non-negative integer. I cannot see this helping, though. By evaluating $f$ on $f(x)$ and applying the equation which we want it to satisfy, one gets that $f$ must satisfy
$$f\left(\frac{1}{f(x)}\right)=f\left(f\left(\frac{1}{x}\right)\right).$$
This doesn't seem to help either. Substituting values on the original equation yields nothing. Another observation is that
$$\lim_{x\to0+}f(x)=\lim_{x\to0+}f\left(\frac{1}{x}\right)=\lim_{x\to+\infty}f(x),$$
if the limit even exists. I know these are very superficial observations, but I am inexperienced with functional equations. Any hints?
Solving the functional equation $f(x)=f(1/x)$ for $x>0$
functional-equations
Related Solutions
For every $x$,
$$a[f(x)]^2+bf(x)+c-g(x)=0$$
is an ordinary quadratic equation and
$$f(x)=\frac{-b\pm\sqrt{b^2-4ac+4ag(x)}}{2a}$$ indeed holds.
There are real roots when
$$b^2-4ac+4ag(x)\ge0,$$ a quartic inequation.
If there are no constraints on $f$, except that it must be a function, then for any $x$ such that there are distinct real roots, you can choose either of them. So there are infinitely, non-countably many solutions.
If $f$ is restricted to be continuous, then the choice of the sign must be consistent across the intervals of the domain.
For the case $(1,1,0)$, the roots are
$$\frac{-1\pm\sqrt{1+4(x^4+2x^3+2x^2+x)}}{2}=\frac{-1\pm(2x^2+2x+1)}2=x^2+x,-x^2-x-1.$$
A solution, among many many many others:
Best Answer
Consider any function $g : [1,+\infty) \rightarrow \mathbb{R}$ such that $g(1)=2$.
Then the function $f : \mathbb{R}_+^* \rightarrow \mathbb{R}$ defined by $$f(x)=g(1/x) \quad \text{if } 0<x<1 \quad \quad \quad \text{and} \quad \quad f(x)=g(x) \quad\text{if } x\geq 1 $$
is a solution to your problem. (So there are many solutions)
(If you want continuous solutions, just ask that $g$ is continuous, it will give a continuous $f$. If you want differentiable solutions, then try to see at which condition on $g$ the constructed function will be differentiable - hint : you will have a condition about the derivative of $g$ at $1$)