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Homework Help: Inverse Relationship

  1. Feb 23, 2006 #1
    Ok, rookie question.. but I have no one to verify it other than you folks.

    Please help me out if you can.

    Problem: Find the inverse of [tex]y = \frac{1-e^-x}{e^x+1}[/tex]

    The question is... can you do this:

    [tex]y = \frac{1-e^-x}{e^x+1} = y = \frac{1}{e^x + 1 - e^x} = \frac{1}{1} = 1 [/tex]

    I thought this was illegal because the [tex]e^-x[/tex] was tied to the 1 through the minus sign. Can the [tex]e^-x[/tex] be moved to the bottom of the denominator?


    ****EDIT****
    I cant get tex to work right. it should be e^-x "e to the power of negative x" in the numerator.

    Thanks
     
    Last edited: Feb 23, 2006
  2. jcsd
  3. Feb 23, 2006 #2
    You can't do that.
     
  4. Feb 23, 2006 #3

    AKG

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    No, you can't do that. Besides, you'd end up getting y = 1, which is clearly not equal to the original y, and which also has no inverse.
     
  5. Feb 23, 2006 #4

    AKG

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    Actually, this given y doesn't have an inverse either. It does on [itex](-\infty,\,\ln(2)][/itex] and on [itex][\ln(2),\,\infty)[/itex], but not on both simultaneously.
     
  6. Feb 23, 2006 #5
    Well ya see, this was a test question tonight on a calculus test I had and I came home racking my brain trying to figure out where I went wrong.

    AKG, are you saying that this does not have an inverse unless its in the domain you specified above?
     
  7. Feb 23, 2006 #6
    What was the test question, because I highly doubt that it was to find the inverse of that function.
     
  8. Feb 23, 2006 #7
    No, this was a calc 1 exam that covered some review of old stuff as well as lmits and beginning derivatives. This was the problem.

    If this was a positve x in the exponent it would be easy to solve.
     
  9. Feb 23, 2006 #8
    The test question was the original question.

    Find the inverse of: [tex]y = \frac{1-e^-x}{e^x+1}[/tex]

    (again, that should be negative x exponent in the numerator)
     
  10. Feb 23, 2006 #9

    AKG

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    Yeah, it has an inverse only if the domain is some subset of [itex](-\infty,\,\ln(2)][/itex] or a subset of [itex][\ln(2),\,\infty)[/itex].

    You want to find the inverse of the function g = R o exp, where exp is the exponential function, and R is the rational function defined by:

    R(x) = (x-1)/[x(x+1)]

    so you know that g-1 = exp-1 o R-1 = ln o R-1. You can find the inverse of R by rearranging the equation above so that it takes the form:

    ax² + bx + c = 0

    where the coefficients a, b, and c might contain some occurence of R(x), then solve for x using the quadratic formula. This gives you an expression for x in terms of R(x), from which you can figure out the inverse of R.
     
    Last edited: Feb 23, 2006
  11. Feb 23, 2006 #10

    AKG

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    To get the LaTeX to work right, click on the image below to see what to do:

    [itex]e^{-x}[/itex], not [itex]e^-x[/tex]
     
  12. Feb 23, 2006 #11
    Thank you for the LaTex help.

    When you say "o R.." what does the 'o' stand for? Are you saying 'of' ? R of exp? I just want to make sure I'm on the same page as you.

    Thanks again.
     
  13. Feb 23, 2006 #12
    Testing: [tex]y = \frac{1-e^{-x}}{e^x+1}[/tex]
     
  14. Feb 23, 2006 #13

    arildno

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    You should follow AKG's advice:
    Rewrite your expression as:
    [tex](e^{x}+1)y=1-e^{-x}\to{e}^{x}(e^{x}+1})y=e^{x}-1[/tex]
    Introduce [itex]t=e^{x}[/itex], and rewrite the quadratic equation in t to standard form and solve it, remembering that t>0
     
  15. Feb 23, 2006 #14

    AKG

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    The 'o' denotes function composition. So f o g is the function defined by:

    (f o g)(x) = f(g(x))
     
  16. Feb 23, 2006 #15
    Cool Thanks guys. I'll check this out later when I get home from work. I appreciate it.
     
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