# Functional Equation $f(x)f(f(x)+\frac{1}{x})=1$

I'd like to ask how to find all solutions to the functional equation $$f(x)\cdot f(f(x)+\frac{1}{x})=1$$, where $$f: (0, +\infty)\to\mathbb{R}$$ is strictly increasing?

• – Sil Mar 31 at 18:21
• $f(x) = 1$ is a solution – George Dewhirst Mar 31 at 18:31
• Over what domains? $\mathbb{R}\to \mathbb{R}$ with $x\neq 0$? Please specify! – user574848 Mar 31 at 23:27
• "I'd like to ask how to find all solutions to the functional equation..." please include a context and your attempts/thoughts to solve this. Otherwise will not be probably considered a good question and will attract downvotes. – Sil Apr 7 at 9:19

I'll answer the problem by myself. For any $$x$$, let $$f(x)=a$$. Then, $$f(a+\frac{1}{x})=\frac{1}{a}$$. Furthermore, we have $$f(a+\frac{1}{x})f(f(a+\frac{1}{x})+\frac{1}{a+\frac{1}{x}})=1$$. Therefore, $$f(\frac{1}{a}+\frac{1}{a+\frac{1}{x}})=a$$. Since $$f$$ is strictly increasing, we have $$\frac{1}{a}+\frac{1}{a+\frac{1}{x}}=x$$. Thus, the value of $$a$$ can be obtained by solving the quadratic equation.

• Nice. Maybe the $a$ by itself obfuscates the fact that it depends on $x$, more standard way to write it would be to keep it as $f(x)$ and derive same result, but good work. – Sil Apr 7 at 9:12

Hint:

Let $$g(x)=f(x)+\dfrac{1}{x}$$ ,

Then $$f(x)=g(x)-\dfrac{1}{x}$$

$$f\left(f(x)+\dfrac{1}{x}\right)=g\left(f(x)+\dfrac{1}{x}\right)-\dfrac{1}{f(x)+\dfrac{1}{x}}=g(g(x))-\dfrac{1}{g(x)}$$

$$\therefore\left(g(x)-\dfrac{1}{x}\right)\left(g(g(x))-\dfrac{1}{g(x)}\right)=1$$

• Sorry, but I still cannot see how to derive the formula of $g(x)$. – Hang Wu Apr 2 at 15:31