anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1438564431151:dw|
anonymous
  • anonymous
If you can help me your amaaaaaaaazing
anonymous
  • anonymous
\[f^{-1}(x)=\frac{1}{f')f^{-1}(x)}\] is probably what you need to work with

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anonymous
  • anonymous
that makes \[f(x)=f'(f^{-1}(x))\]
anonymous
  • anonymous
see if that is possible
anonymous
  • anonymous
what do you mean
anonymous
  • anonymous
i got an example f(x)=x^2 ?
jim_thompson5910
  • jim_thompson5910
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
anonymous
  • anonymous
so this is proof of no example exists?
anonymous
  • anonymous
that is both invertible and continuous
anonymous
  • anonymous
\[y = f(x), x = g(y)\], 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
anonymous
  • anonymous
@oldrin.bataku this is what somewrote
anonymous
  • anonymous
the continuous argument is bs because it says "on its domain of definition", suggesting that it could be defined away from \(0\).
anonymous
  • anonymous
@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 \(\pi/2\)
jim_thompson5910
  • jim_thompson5910
That's a good point, nvm my first post then

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