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Challenge question to the community - - Determine the inverse function of

  1. Jan 30, 2012 #1
    [itex] f(x) \ = \ \dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}[/itex]


    [itex]Edit: \ \ I \ sent \ a \ PM \ to \ a \ mentor.[/itex]
     
    Last edited: Jan 30, 2012
  2. jcsd
  3. Jan 30, 2012 #2

    micromass

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    [STRIKE]Can you PM the solution to one of the mentors please?? This way we can verify that it's a challenge question and not a disguised homework problem.[/STRIKE]

    OK, this is verified.
     
    Last edited: Jan 30, 2012
  4. Jan 30, 2012 #3

    mathman

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    Finding √x is trivial. Just square it afterward.
     
  5. Jan 30, 2012 #4
    Thanks for posting this. The inverse of f(x) is:

    [tex]

    f^{-1}(x) = \frac{(1-x)^{2}}{(1+x)^{2}}

    [/tex]

    and x>=0 to prevent singularities

    If you'd like to see my work, then just ask. It's easy to show that f(f^1(x)) = x.

    Edit:

    But I forgot to check whether f^-1 satisfied the requirements of inverse, I'll repost later, as I made a mistake :(
     
    Last edited: Jan 30, 2012
  6. Jan 30, 2012 #5
    [tex] \text{And what about any possible restrictions on a domain?}[/tex]
     
  7. Jan 30, 2012 #6
    Hint: What you have typed is not a one-to-one function.
     
  8. Jan 30, 2012 #7
    If I restrict x>=0, then my inverse is correct (I think...for now..). So I would say for x>=0, f^-1 is

    [tex]
    f^{-1}(x) = \frac{(1-x)^{2}}{(1+x)^{2}}

    [/tex]
     
    Last edited: Jan 30, 2012
  9. Jan 30, 2012 #8
    Mine (meaning the original function) is one-to-one.

    Recommendation:

    Graph/sketch my function and see.
     
  10. Jan 30, 2012 #9
    Yeah I had made a slight typo, I looked at the graph of a completely different function. My answer stands,

    [tex]

    f^{-1}(x) = \frac{(1-x)^{2}}{(1+x)^{2}}

    [/tex]

    on x>=0 only.
     
  11. Jan 30, 2012 #10

    No to your restriction.


    Hint: The range of f is the domain of f inverse, numerically speaking
    (as when given in interval notation).


    I must go away from the computer for a couple of hours starting now,
    but I will check back for any updates.


    --------------------------------------------------------


    [itex]Edit: \ \ What \ is \ the \ range \ of \ f(x) \ ?[/itex]


    [itex]2nd \ edit: \ \ \ I \ \ am \ \ agreeing \ \ with \ \ Curious3141, [/itex]

    [itex]but \ \ I \ \ am \ \ stating \ \ it \ \ in \ \ this \ \ edit \ \ so \ \ [/itex]

    [itex]that \ \ Curious3141's \ \ post \ \ will \ \ show \ \ the [/itex]

    [itex] answer \ \ more \ \ recently \ \ than \ \ my \ \ own \ \ recent \ \ post.[/itex]
     
    Last edited: Jan 30, 2012
  12. Jan 30, 2012 #11

    Curious3141

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    The inverse is:


    [tex]

    f^{-1}(x) = \frac{(1-x)^{2}}{(1+x)^{2}}

    [/tex]

    with a domain of (-1,1] and a range of [0,∞).
     
  13. Jan 30, 2012 #12

    chiro

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    Just for future reference for checkitagain and other interested readers, you can use identities for derivatives of inverse functions and generate a taylor series that corresponds to the inverse function.

    The key thing to be aware of is if the derivative (of the original function) is zero which indicates a branch cut in the function which means that you have to treat each branch as an individual function in its own right (if it is to have an inverse)
     
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