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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:

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1. anonymous

|dw:1438564431151:dw|

2. anonymous

If you can help me your amaaaaaaaazing

3. anonymous

$f^{-1}(x)=\frac{1}{f')f^{-1}(x)}$ is probably what you need to work with

4. anonymous

that makes $f(x)=f'(f^{-1}(x))$

5. anonymous

see if that is possible

6. anonymous

what do you mean

7. anonymous

i got an example f(x)=x^2 ?

8. 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

9. anonymous

so this is proof of no example exists?

10. anonymous

that is both invertible and continuous

11. 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

12. anonymous

@oldrin.bataku this is what somewrote

13. anonymous

the continuous argument is bs because it says "on its domain of definition", suggesting that it could be defined away from $$0$$.

14. 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 $$\pi/2$$

15. jim_thompson5910

That's a good point, nvm my first post then

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