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Math Help - Finding inverse using hyperbolic function

  1. #1
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    Finding inverse using hyperbolic function

    I'm getting spoonfed instructions and I still don't get it. >_>

    I am to find the inverse of f(x) = x^3 + x, by using properties of sinh(x).
    The question tells me to let y = x^3 + x, let z = \frac{\sqrt{3}}{2}x and then prove that \frac{3\sqrt{3}}{2}y = 4z^3 + 3z.

    After I do that, it tells me to set z = sinh(θ), solve for θ and then reverse the changes to the variables to find x as a function of y.

    I managed to do most of it, up until the 'reverse the changes' part. My θ equals something ridiculously long and confusing, and I have no idea how to use it to my benefit.
    \theta = \frac{sinh^{-1}(\frac{3\sqrt{3}}{2}y)}{3}

    Just need a bit of guiding light, it feels like I overlooked something...

    EDIT: All right, well, I solved for θ in terms of x now too.
    \theta = sinh^{-1}(\frac{\sqrt{3}}{2}x)

    And then I equate the two.
    sinh^{-1}(\frac{\sqrt{3}}{2}x) = \frac{sinh^{-1}(\frac{3\sqrt{3}}{2}y)}{3}

    And I'm not exactly sure how to solve...
    Last edited by BlackBlaze; October 8th 2009 at 04:16 PM.
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  2. #2
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    Quote Originally Posted by BlackBlaze View Post
    I'm getting spoonfed instructions and I still don't get it. >_>

    I am to find the inverse of f(x) = x^3 + x, by using properties of sinh(x).
    The question tells me to let y = x^3 + x, let z = \frac{3\sqrt{3}}{2}x and then prove that \frac{3\sqrt{3}}{2}y = 4z^3 + 3z.

    After I do that, it tells me to set z = sinh(θ), solve for θ and then reverse the changes to the variables to find x as a function of y.

    I managed to do most of it, up until the 'reverse the changes' part. My θ equals something ridiculously long and confusing, and I have no idea how to use it to my benefit.
    \theta = \frac{sinh^{-1}(\frac{3\sqrt{3}}{2}y)}{3}

    Just need a bit of guiding light, it feels like I overlooked something...

    EDIT: All right, well, I solved for θ in terms of x now too.
    \theta = sinh^{-1}(\frac{\sqrt{3}}{2}x)

    And then I equate the two.
    sinh^{-1}(\frac{\sqrt{3}}{2}x) = \frac{sinh^{-1}(\frac{3\sqrt{3}}{2}y)}{3}

    And I'm not exactly sure how to solve...
    If y = x^3 + x and z = \frac{3\sqrt{3}}{2}x then \frac{3\sqrt{3}}{2}y = 4z^3 + 3z does not seem correct to me.
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  3. #3
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    Right you are, it has been acknowledged.

    However, I managed to complete the question. I first isolated for y in the 'θ in terms of x = θ in terms of y' equation. Then I swapped around the x and y and isolated for y a second time.

    Rather crude, but it got the job done.
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