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Let $f:(0,+\infty)\mapsto R$ be a strictly increasing function such that $\forall x\ge0,$ $$f(x)+\frac{1}{x}\ge0, \qquad f(x)f\left(f(x)+\frac{1}{x}\right)=1.$$ Show that $$f(1)=\frac{1-\sqrt{5}}{2}.$$

Please give an example that satisfies these conditions.

Thanks in advance.

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1 Answer 1

up vote 6 down vote accepted

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

$x=1 $ gives $ f(1)f(f(1)+1)=1$

$x=f(1)+1\ge0$ gives $ f(f(1)+1)f(f(f(1)+1)+\frac{1}{f(1)+1})=1$

By replacing $f(f(1)+1)$ by $\frac{1}{f(1)}$ you get $\frac{1}{f(1)}f(\frac{1}{f(1)}+\frac{1}{f(1)+1})=1$

Multiply both sides by $f(1)$ to get $f(\frac{1}{f(1)}+\frac{1}{f(1)+1})=f(1)$

Then you know that the function is strictly increasing and therefore injective.

So $\frac{1}{f(1)}+\frac{1}{f(1)+1}=1$

And then solve for $f(1)$.

Let $x=f(1)$

$\frac{1}{x}+\frac{1}{x+1}=1$

Multiply by $x$ and $x+1$ to get $x+1+x=x(x+1)$

$2x+1=x^2+x$

$0=x^2-x-1$

$x=\cfrac{1\pm\sqrt{5}}{2}$

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you know to take the positive solution since $$f(1)+\frac{1}{1}\ge0$$ –  user27182 Apr 2 '13 at 19:19

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