# $f(x)$ such that $f(x)*f(f(x)+\frac{1}{x})=1$ and $f(x) > - \frac{1}{x}$ $\forall$ $x > 0$ and $f(x)$ is an injective function

I don't even know where to begin in solving this functional equation. I got this in a multiple choice question exam and was able to solve it by substituting all of the given options into the equation. The one that worked is $$f(x) = \frac{1 -\sqrt 5}{2x}$$

I can't even prove that this is a unique solution. However, I was able to show that $f(1)=\frac{1-\sqrt 5}{2}$

• Why can't we just take $f(x)=1\;\;\forall x>0$?
– lulu
May 5 '18 at 13:58
• That is a solution as well I guess. Still doesn't really solve the question? I assumed the function is one-one mapped which is why I got that particular value of f(1). I have the edited the question accordingly. May 5 '18 at 15:11

Let's for a moment assume that $f(x)$ is invertible with $g(x)$ being its inverse. We can always discard our assumption later on if things don't turn out the way we want them to.

So, $f(x)f\biggl(f(x) + \frac{1}{x}\biggr) = 1$ can be modified as $x.f\biggl(x + \frac{1}{g(x)}\biggr) = 1$

$$\Rightarrow x + \frac{1}{g(x)} = g(\frac{1}{x})$$

$$\Rightarrow g(\frac{1}{x}) - \frac{1}{g(x)} = x$$

Substituting $x$ for $\frac{1}{x}$, we get $g({x}) - \frac{1}{g(\frac{1}{x})} = \frac{1}{x}$

Eliminating $g(\frac{1}{x})$ from the above equations, we get

$$g(x)\bigl(g(x)-\frac{1}{x}\bigr) = \frac{1}{x^2}$$

$$\Rightarrow x^2g^2(x) - xg(x) -1 =0$$ This can be modified as $$f^2(x)x^2 - f(x)x - 1=0$$

which is simply a quadratic equation leading to the solution: $$f(x) = \frac{1 \pm \sqrt{5}}{2x}$$

Now, to decide between the two possible solutions, one could substitute and check for $f(x)$ in the question. Both $f(x) = \frac{1 \pm \sqrt{5}}{2x}$ satisfy the conditions of the question. Also, our original assumption remains true that $f(x)$ is an invertible function, hence no contradictions here.