2
$\begingroup$

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

  • 1
    Should be assumed that $f$ is injective and that $f(\Bbb R^*)=\Bbb R^*$? If not, for what values of $x$ does the equality hold? – 2017-02-10
  • 1
    Apart of $f(x)=x$? – 2017-02-10
  • 0
    @N74 and $f(x) = 1/x$? – 2017-02-10
  • 0
    How about using $f(x) = \frac{1}{x}$? It works simply. – 2017-02-10
  • 1
    If there is no assumption about continuity, there are infinitely many such functions. For each pair $\left\{x,\frac{1}{x}\right\}$ you can choose $f(x)=x$ and $f\left(\frac{1}{x}\right)=\frac{1}{x}$, or $f(x)=\frac{1}{x}$ and $f\left(\frac{1}{x}\right)=x$. For instance, we can make $$f(x)=\left\{\begin{array}{ll} x & \text{ if }x \neq a, \frac{1}{a} \\ \frac{1}{x} & \text{ otherwise}\end{array} \right.$$ for some $a \in \mathbb{R}^*$, and there are infinitely many other variations. – 2017-02-10
  • 0
    @O.VonSeckendorff *it gives $2x=x+1/x$.* Nope, composition... then $f(x)=x$ also works. – 2017-02-10
  • 1
    @Wolfram $f^{-1}(x) = \frac{1}{x}$, so it should work. – 2017-02-10

1 Answers 1

0

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 $g(1) = 2 -g^{-1} (1)$.