Give an example (or prove that none exists) of a real function f(x) which is continuous, invertible, and satisfies the following identity everywhere on its domain of definition:

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Give an example (or prove that none exists) of a real function f(x) which is continuous, invertible, and satisfies the following identity everywhere on its domain of definition:

Mathematics
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If you can help me your amaaaaaaaazing
\[f^{-1}(x)=\frac{1}{f')f^{-1}(x)}\] is probably what you need to work with

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that makes \[f(x)=f'(f^{-1}(x))\]
see if that is possible
what do you mean
i got an example f(x)=x^2 ?
First replace every x with f(x) as shown in red. Then isolate f(x) \[\Large f^{-1}(x) = \frac{1}{f(x)}\] \[\Large f^{-1}({\color{red}{f(x)}}) = \frac{1}{f({\color{red}{f(x)}})}\] \[\Large f^{-1}(f(x)) = \frac{1}{f(f(x))}\] \[\Large x = \frac{1}{f(f(x))}\] \[\Large f(f(x)) = \frac{1}{x}\] \[\Large f(x) = f^{-1}\left(\frac{1}{x}\right)\] We don't need to worry about what f(x) is. When x = 0, the right hand side will be undefined. So this function isn't continuous everywhere. This proves there is no such function that meets every condition stated above. If you drop the continuity requirement, then you can set up strange functions as described in the link below http://www.quora.com/How-could-f-inverse-of-x-be-equal-to-1-f-x
so this is proof of no example exists?
that is both invertible and continuous
\[y = f(x), x = g(y)\], g is f inverse, so x.y = 1, but, (x, y) and (y, x) are reflecting the graph across the line y = x. so the scalar product (x, y) and (y, x) \[\frac{x.y + y.x }{ x ^{2}+y ^{2} } = \cos \frac{ 3\pi }{ 4 } \] \[\frac{ 2 }{ x^2 + y^2 } = -\frac{ \sqrt{2} }{ 2 }\] is impossíble
@oldrin.bataku this is what somewrote
the continuous argument is bs because it says "on its domain of definition", suggesting that it could be defined away from \(0\).
@ginochen's answer is interesting but I don't think it's valid that the angle between \((x,y),(y,x)\) always be \(3\pi/4\) -- consider \((0,1),(1,0)\) for which it is \(\pi/2\)
That's a good point, nvm my first post then

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