$f(f(\sqrt{2}))=\sqrt{2}$ then f has a fixed point

$f(x)$ is continuous $f:\mathbb{R}\rightarrow\mathbb{R}$

$f(f(\sqrt{2}))=\sqrt{2}$

Prove that $f$ has a fixed point

in other words prove the there is $x_1$ such that $f(x_1)=x_1$

I tried using $g(x)=f(x)-x$ and tried to use the Intermediate value theorem but did not succeed. and it's obvious that $x=\sqrt{2}$ is the answer

• I tried using g(x)=f(x)-x and tried to use the mean value theorem but did not succeed. – Was Fr Feb 7 '14 at 21:47
• Is this function continuous on a closed interval and differentiable on a open interval? Otherwise you can't use the mean value theorem. – Felipe Jacob Feb 7 '14 at 21:50
• sorry I meant Intermediate value theorem – Was Fr Feb 7 '14 at 21:51
• You'd still have to make sure the function is defined at a closed bounded interval to use any theorem that deals with compactness. Were you given any information on the domain of your function, or wether its invertible (it seems to be...)? – Felipe Jacob Feb 7 '14 at 21:54
• "it's obvious that $x=\sqrt2$ is the answer" Not really. Suppose $f(x)=10-x$. – Rahul Feb 7 '14 at 21:57

Hint: Consider $\alpha=f(\sqrt2)$. Show that $f(\alpha)=\sqrt2$. Inspect $f(x)-x$ at $\sqrt2$ and $\alpha$.
For further enjoyment: When you grok this question, try showing that if $$\overbrace{f\circ f\circ\cdots\circ f}^{n\text{ compositions}}(x_0)=x_0$$ then $f$ has a fixed point.
• @Sawarnik: we are given that $f(\alpha)=f(f(\sqrt2))=\sqrt2$. – robjohn Feb 12 '14 at 18:53