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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
    • one year ago
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    |dw:1438564431151:dw|

  2. anonymous
    • one year ago
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    If you can help me your amaaaaaaaazing

  3. anonymous
    • one year ago
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    \[f^{-1}(x)=\frac{1}{f')f^{-1}(x)}\] is probably what you need to work with

  4. anonymous
    • one year ago
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    that makes \[f(x)=f'(f^{-1}(x))\]

  5. anonymous
    • one year ago
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    see if that is possible

  6. anonymous
    • one year ago
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    what do you mean

  7. anonymous
    • one year ago
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    i got an example f(x)=x^2 ?

  8. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    so this is proof of no example exists?

  10. anonymous
    • one year ago
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    that is both invertible and continuous

  11. anonymous
    • one year ago
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    \[y = f(x), x = g(y)\], 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
    • one year ago
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    @oldrin.bataku this is what somewrote

  13. anonymous
    • one year ago
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    the continuous argument is bs because it says "on its domain of definition", suggesting that it could be defined away from \(0\).

  14. anonymous
    • one year ago
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    @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
    • one year ago
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    That's a good point, nvm my first post then

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