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Math Help - inverse function question..

  1. #1
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    inverse function question..

    i need to find the inverse function of these:

    ImageShack - Image Hosting :: 63348338nf8.gif

    i know that if y=sinx
    then its inverse whould be x=arcsin y

    but the first case has two = signs
    and the second one is a split function

    what to do in these cases?
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    i need to find the inverse function of these:

    ImageShack - Image Hosting :: 63348338nf8.gif

    i know that if y=sinx
    then its inverse whould be x=arcsin y

    but the first case has two = signs
    and the second one is a split function

    what to do in these cases?
    In the first one you just have to find the inverse of tanh(x), given that tanh(x) = all those e things. Given that natural logarithm (ln(x)) is the inverse of (e^x), I suspect that the inverse of tanh will be a function of ln(x).

    In the 2nd piecewise function you just have to find the inverse of each piece as if it were a separate function.

    For example f(x) = x^2 1 < x < 4 would lead to the inverse, g(x) being:

    g(x) = sqrt(x) 1<x<4
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  3. #3
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    regarding the first "hiperbolic" question

    there are two "=" signs
    thats not a equation

    i dont know how to get one equation with one "=" sign out of it?
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  4. #4
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    Quote Originally Posted by transgalactic View Post
    regarding the first "hiperbolic" question

    there are two "=" signs
    thats not a equation

    i dont know how to get one equation with one "=" sign out of it?
    Lol there's nothing funny going on, I promise you!

    All it's saying it this:

     f(x) = tanh(x)

    AND

     tanh(x) = \frac{e^x - e^{-x}}{e^x+e^{-x}}

    THEREFORE:

     f(x) = \frac{e^x - e^{-x}}{e^x+e^{-x}}

    What is f^{-1}(x) ?

    An "equation" means, two eqate to things. Or to state them to be equal. Therefore, an equation can have as many equal signs as it wants. As long as what it's saying is TRUE, then it's still an equation. This is an equation:

     5 = 4+1=3+2=2+2+1=1+1+1+2=3+1+1
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mush View Post
     f(x) = \frac{e^x - e^{-x}}{e^x+e^{-x}}

    What is f^{-1}(x) ?
    \tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^x-\frac{1}{e^x}}{e^x+\frac{1}{e^x}}. Let z=e^x and this becomes y=\frac{z-\frac{1}{z}}{z+\frac{1}{z}} multiplying both sides by z+\frac{1}{z} gives y\left(z+\tfrac{1}{z}\right)=z-\tfrac{1}{z}. Multiplying both sides by z gives y(z^2+1)=(z^2-1) or equivalently yz^2+y-z^2+1=0. Factoring gives (y-1)z^2+(y+1)=0. Solving gives z=\pm\sqrt{\frac{y+1}{y-1}}. Which implies that e^x=\pm\sqrt{\frac{y+1}{y-1}} or x=\ln\left(\pm\sqrt{\frac{y+1}{y-1}}\right)=\text{arctanh}(y)
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  6. #6
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    Quote Originally Posted by Mathstud28 View Post
    \tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^x-\frac{1}{e^x}}{e^x+\frac{1}{e^x}}. Let z=e^x and this becomes y=\frac{z-\frac{1}{z}}{z+\frac{1}{z}} multiplying both sides by z+\frac{1}{z} gives y\left(z+\tfrac{1}{z}\right)=z-\tfrac{1}{z}. Multiplying both sides by z gives y(z^2+1)=(z^2-1) or equivalently yz^2+y-z^2+1=0. Factoring gives (y-1)z^2+(y+1)=0.

    Solving gives z=\pm\sqrt{\frac{y+1}{y-1}}. Mr F says: No. {\color{red}z=\pm\sqrt{\frac{y+1}{{\color{blue}1-y}}}}. But z > 0 since {\color{red}z = e^x} and x is real. Therefore {\color{red}z = \sqrt{\frac{y+1}{1-y}}}.

    [snip]
    There some errors (see red). Once these corrections are made, the solution will continue in a similar way to that already shown.
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