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Thread: Help analyzing inverse of an abs function

  1. #1
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    Help analyzing inverse of an abs function

    I am trying to find an expression for $\displaystyle f^{-1}$

    $\displaystyle f(x)=ln\lvert\frac{x}{x-1}\rvert$


    And then the algebra

    $\displaystyle x=ln\lvert\frac{y}{y-1}\rvert$

    $\displaystyle e^{x}=\lvert\frac{y}{y-1}\rvert$

    First solution:
    $\displaystyle e^{x}=\frac{y}{y-1}$
    $\displaystyle ye^{x}-e^{x}=y$
    $\displaystyle ye^{x}-y=e^{x}$
    $\displaystyle y(e^{x}-1)=e^{x}$
    $\displaystyle y=\frac{e^{x}}{e^{x}-1}$

    Second Solution:
    $\displaystyle -e^{x}=\frac{y}{y-1}$
    $\displaystyle -ye^{x}+e^{x}=y$
    $\displaystyle e^{x}=y+e^{x}y$
    $\displaystyle e^{x}=(1+e^{x})y$
    $\displaystyle y=\frac{e^x}{e^{x}+1}$

    I think my calculator is telling me that the first solution I did is true for $\displaystyle y<0$ and $\displaystyle y\ge1$ and the second solution is true for all reals.

    If I am understanding my calculator correctly, how do I analytically determine that the second solution is the inverse of f for all real values of x? And if this is incorrect, how do I analyze the functions I found for $\displaystyle f^{-1}$?

    Thanks a lot if you can help!
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by mikedwd View Post
    I am trying to find an expression for $\displaystyle f^{-1}$

    $\displaystyle f(x)=ln\lvert\frac{x}{x-1}\rvert$


    And then the algebra

    $\displaystyle x=ln\lvert\frac{y}{y-1}\rvert$

    $\displaystyle e^{x}=\lvert\frac{y}{y-1}\rvert$

    First solution:
    $\displaystyle e^{x}=\frac{y}{y-1}$
    $\displaystyle ye^{x}-e^{x}=y$
    $\displaystyle ye^{x}-y=e^{x}$
    $\displaystyle y(e^{x}-1)=e^{x}$
    $\displaystyle y=\frac{e^{x}}{e^{x}-1}$

    Second Solution:
    $\displaystyle -e^{x}=\frac{y}{y-1}$
    $\displaystyle -ye^{x}+e^{x}=y$
    $\displaystyle e^{x}=y+e^{x}y$
    $\displaystyle e^{x}=(1+e^{x})y$
    $\displaystyle y=\frac{e^x}{e^{x}+1}$

    I think my calculator is telling me that the first solution I did is true for $\displaystyle y<0$ and $\displaystyle y\ge1$ and the second solution is true for all reals.

    If I am understanding my calculator correctly, how do I analytically determine that the second solution is the inverse of f for all real values of x? And if this is incorrect, how do I analyze the functions I found for $\displaystyle f^{-1}$?

    Thanks a lot if you can help!
    we can divide the function into two functions if x<0 and x>1

    $\displaystyle f(x) = ln \frac{x}{x-1} $

    $\displaystyle e^x = \frac{y}{y-1} \Rightarrow y = \frac{e^x}{e^x-1} $ this for y<0 and y>1 this is the range, the domain here is all real numbers

    if 0<x<1

    $\displaystyle f(x) = ln \frac{-x}{x-1} $

    $\displaystyle e^x = \frac{-y}{y-1} \Rightarrow y= \frac{-e^x}{e^x-1} $

    this when 0<y<1 this is the range, that means x will take the values when 1-e^x is larger than e^x

    $\displaystyle \frac{e^x}{1-e^x} $

    $\displaystyle 0<2e^x <1 \Rightarrow 0<e^x < \frac{1}{2} $

    $\displaystyle \ln 0 < x < \ln \frac{1}{2} \Rightarrow -\infty < x< \ln 1/2 $

    0,1 the function is not defined so these points are not included in the domain of f(x)( the range of $\displaystyle f^{-1}$ )
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  3. #3
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    Quote Originally Posted by Amer View Post
    we can divide the function into two functions if x<0 and x>1

    $\displaystyle f(x) = ln \frac{x}{x-1} $

    $\displaystyle e^x = \frac{y}{y-1} \Rightarrow y = \frac{e^x}{e^x-1} $ this for y<0 and y>1 this is the range, the domain here is all real numbers

    if 0<x<1

    $\displaystyle f(x) = ln \frac{-x}{x-1} $

    $\displaystyle e^x = \frac{-y}{y-1} \Rightarrow y= \frac{-e^x}{e^x-1} $

    this when 0<y<1 this is the range, that means x will take the values when 1-e^x is larger than e^x

    $\displaystyle \frac{e^x}{1-e^x} $

    $\displaystyle 0<2e^x <1 \Rightarrow 0<e^x < \frac{1}{2} $

    $\displaystyle \ln 0 < x < \ln \frac{1}{2} \Rightarrow -\infty < x< \ln 1/2 $

    0,1 the function is not defined so these points are not included in the domain of f(x)( the range of $\displaystyle f^{-1}$ )
    Yes, except that $\displaystyle e^{x}=\frac{-y}{y-1} \Rightarrow y=\frac{e^{x}}{e^{x}+1}$

    So then $\displaystyle e^{x}+1>e^{x} \Rightarrow 1>0$ ?? So does this then mean that the inverse is $\displaystyle y=\frac{e^{x}}{e^{x}+1}$ for all values (because this cannot equal 0 or 1)?

    But the output of that is never negative, while the original function can certainly have negative input...I'm still working on this...all help is appreciated!

    When I tried that inverse out, indeed it only worked for y values between 0 and 1 (I just tested to see if it would undo the original function).

    Hmm...

    I wrote a piecewise:

    denominator e^x-1 for x<0 and x>1
    denominator e^x+1 for 0<x<1

    Is that correct?
    Last edited by mikedwd; Dec 6th 2009 at 01:28 PM.
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  4. #4
    MHF Contributor Amer's Avatar
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    the problem is

    we can write the function f(x) like this

    $\displaystyle ln \frac{x}{x-1} \;\;\; \; 1<x , x<0 $ the range of this peace is all real number

    $\displaystyle ln \frac{-x}{x-1} \;\;\;\; 0<x<1 $ the range is all real numbers

    note that:-f(x) is not one-one function

    and the inverses

    $\displaystyle y =\frac{e^x}{e^x -1 } $ the inverse of the first function the domain is all real numbers the range is 1<y, y<0

    $\displaystyle y= \frac{e^x }{1+e^x} $ the inverse of the second one the domain is all real numbers the range 0<y<1

    there exist a real number have a two different images and this contradict with function conditions

    as you know the inverse function take the domain of the original function as a range and take the range as a domain if the function we have is one-one

    see this is the image of f(x)

    Help analyzing inverse of an abs function-12.jpg
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  5. #5
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    Thanks a lot for your help, Amer.

    I understand what you're saying and I really appreciate it!

    -Mike
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