What functions satisfies : $f^{-1}(x)+f(\frac{1}{x})=x+\frac{1}{x}$?

Question

let $f$ be a real function defined as :$f:\mathbb{R^*}\to \mathbb{R^*}$, My question here is:

Question: What functions satisfies this property: $f^{-1}(x)+f(\frac{1}{x})=x+\frac{1}{x}$ with $f$ is injective ?

Note: $f^{-1}$ is the inverse compositional of $f$ .

Edit: I have edited the question according to the definition of $f$ to make sense

Answer 1

View $f$ as $f = \{(x,f(x))|x \in \mathbb R^*\} \subset \mathbb {R^*}^2$

Let $w, u=f^{-1}(w) \in \mathbb R^*$ then

$f^{-1}(w) + f(\frac 1w) = u + f(\frac 1w) = w + 1/w$

So $f(\frac 1w) = w + 1/w - f^{-1}(w) $

And if $w = 1$ we have $f^{-1}(1) + f(1) = 2$

So simply let $g:(1,\infty)\rightarrow R^*$ be any injective function so that $g^{-1}(1) < 2$ Define $f(x) = g(x)$ if $x > 1$ and $f(x) = x + 1/x + g^{-1} (1/x)$ if $x < 1$ and let $g(1) = 2 -g^{-1} (1)$.