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Math Help - Inverse of equation involving e^(-x)

  1. #1
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    Inverse of equation involving e^(-x)

    Hi,

    My problem is to find the inverse of y = f(x) = \frac{1}{2}(e^x - \frac{1}{e^x}) .

    I can get up to using v for e^x, and that v > 0

    2yv = v^2 -1

    0 = v^2 - 2yv - 1

    The using the quadratic formula I get v = \frac{2y \pm \sqrt{(-2y)^2 + 4}}{2}.

    Here I am stuck


    Sorry for bad format.
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  2. #2
    Super Member Deadstar's Avatar
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    Following on from where you left off...

    v = \frac{2y}{2} \pm \frac{\sqrt{4(y^2 + 1)}}{2} (since (-2y)^2 + 4 = 4y^2 + 4 = 4(y^2 + 1))

    Note that 2 = \sqrt{4} so we get

    v = y \pm \sqrt{(y^2 + 1)}

    Now since v = e^x, take natural logs of both sides to get...

    x = \ln(v) = ..? Can you finish?
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  3. #3
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    Yes.

    I kept going around in circles about what to do with the square root in the quadratic formula.

    All is good now. Thanks
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  4. #4
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    Quote Originally Posted by Deadstar View Post
    Following on from where you left off...

    v = \frac{2y}{2} \pm \frac{\sqrt{4(y^2 + 1)}}{2} (since (-2y)^2 + 4 = 4y^2 + 4 = 4(y^2 + 1))

    Note that 2 = \sqrt{4} so we get

    v = y \pm \sqrt{(y^2 + 1)}

    Now since v = e^x, take natural logs of both sides to get...

    x = \ln(v) = ..? Can you finish?
    I'll add a small note: One of the solutions (either the positive or the negative root) must be discarded ....
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