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

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.
