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anonymous
 one year ago
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:
anonymous
 one year ago
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:

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438564431151:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you can help me your amaaaaaaaazing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[f^{1}(x)=\frac{1}{f')f^{1}(x)}\] is probably what you need to work with

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that makes \[f(x)=f'(f^{1}(x))\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0see if that is possible

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i got an example f(x)=x^2 ?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1First 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/Howcouldfinverseofxbeequalto1fx

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so this is proof of no example exists?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is both invertible and continuous

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[y = f(x), x = g(y)\], 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
 one year ago
Best ResponseYou've already chosen the best response.0@oldrin.bataku this is what somewrote

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the continuous argument is bs because it says "on its domain of definition", suggesting that it could be defined away from \(0\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.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 \(\pi/2\)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1That's a good point, nvm my first post then
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