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Trying to find an inverse equation - maybe cosh/sinh

  1. Sep 17, 2007 #1
    Given [tex]f(t) =[/tex] [tex]\frac{2e^{t} + 3e^{-t}}{e^{t} + 2e^{-t}}[/tex] find [tex]f^{-1}(t)[/tex]

    My attempt:
    i worked for about half an hour but i think im not doing the right thing. i tried multiplying it out, factorizing, pretty much just playing around with it.
    OMMGG mayb got it
    Last edited: Sep 17, 2007
  2. jcsd
  3. Sep 17, 2007 #2


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    I would do this: write the equation as
    [tex]x= \frac{2e^t+ 3e^{-t}}{e^t+ 2e^{-t}}[/tex]
    Now swap x and t:
    [tex]t= \frac{2e^x+ 3e^{-x}}{e^x+ 2e^{-x}}[/tex]
    That "gives" the inverse function. Now you "only" have to solve for x!

    Multiply that denominator on both sides:
    [tex]te^x+ 2te^{-x}= 2e^x+ 3e^{-x}[/tex]
    Multiply the entire equation by ex
    [tex]te^{2x}+ 2t= 2e^{2x}+ 3[/tex]
    Let y= e^x so that you get a quadratic in y
    [tex]ty^2+ 2t= 2y^2+ 3[/tex]
    Solve that by a simple square root, then take ln of both sides to find x as a function of x.
  4. Sep 17, 2007 #3
    ooooohh thankszzz
    sooo [tex] x = ln\sqrt{\frac{3-2t}{t-2}}[/tex]
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