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$f(x)$ 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

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  • $\begingroup$ I tried using g(x)=f(x)-x and tried to use the mean value theorem but did not succeed. $\endgroup$ – Was Fr Feb 7 '14 at 21:47
  • $\begingroup$ Is this function continuous on a closed interval and differentiable on a open interval? Otherwise you can't use the mean value theorem. $\endgroup$ – Felipe Jacob Feb 7 '14 at 21:50
  • $\begingroup$ sorry I meant Intermediate value theorem $\endgroup$ – Was Fr Feb 7 '14 at 21:51
  • $\begingroup$ 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...)? $\endgroup$ – Felipe Jacob Feb 7 '14 at 21:54
  • $\begingroup$ "it's obvious that $x=\sqrt2$ is the answer" Not really. Suppose $f(x)=10-x$. $\endgroup$ – Rahul Feb 7 '14 at 21:57
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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.

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  • $\begingroup$ Thank you for your help. and for solving the enjoyment do I need to write an algorithm to show that ? $\endgroup$ – Was Fr Feb 7 '14 at 22:34
  • $\begingroup$ @WasFr: Not really. It should only require the methods we have used for this question. $\endgroup$ – robjohn Feb 7 '14 at 22:37
  • $\begingroup$ okay :D got the idea $\endgroup$ – Was Fr Feb 7 '14 at 23:16
  • $\begingroup$ @Sawarnik: we are given that $f(\alpha)=f(f(\sqrt2))=\sqrt2$. $\endgroup$ – robjohn Feb 12 '14 at 18:53

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