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Prove that the real roots of equation $$f(x)=f^{-1}(x);x\in R$$ always lie on $y=x$.

I know that $y=f(x)$ and $y=f^{-1}(x)$ are symmetric about $y=x$ so, I do have some intuitions on this but I can't seem to be able to prove it rigorously.

  • This doesn't quite work for $f(x) = -x$. – Arthur Jan 25 '14 at 13:20
  • I think the only thing you can get from that equation is $f(f(x)) = x$. – ploosu2 Jan 25 '14 at 13:44
  • My earlier comment about $-cos^{-1}(x)$ was wrong. It doesn't intersect with it's inverse $cos(x)$, $x\in [-\pi, 0]$. – ploosu2 Jan 25 '14 at 13:52
  • I graphed the functions about 5 mins. ago only to see what you said. And you were also write about the first thing, no matter what I try to do I eventually end up with $f(f(x))=x$ and $f(f(y))=y$ and similarly $f^{-1}(f^{-1}(x))=x$ and $f^{-1}(f^{-1}(y))=y$. –  Jan 25 '14 at 16:48

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I have at least three hints for that problem:

  1. Well, if $f(x)$ and $f^{-1}(x)$ are symmetric about $y=x$, then when exactly is $f(x)=f^{-1}(x)$?
  2. Probably it's helpful to look at this the other way round: They are symmetric, so they differ whenever the points are not on the symmetry axis.
  3. $x=f^{-1}(y)$ for some $y$ by definition of the inverse. Apply this to the left side. Then go on...

The last one is probably the most rigorous one but the first two match your intuition maybe a little bit more.

Edit: After looking at the comments, I have to say that I understood the problem to be "at least one solution of the equation [...] lies on [...]". Probably my hints are not very helpful then, I have to check that...

Piwi
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