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Math Help - Two Equations (solving for inverses)

  1. #1
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    Two Equations (solving for inverses)

    Find the inverse equations of the following:

    1. f(x) = 3x + ln(x)

    2. g(x) = \frac{x+1}{2x+1}

    So the equations become...

    1. x = 3y + ln(y)

    2. x = \frac{y + 1}{2y + 1}

    Sovle for y.

    I've played with the algebra a few different ways, especially on the rational function, but I still can't solve for 'y' in either equation! I've also graphed both functions, and they both appear to have an inverse (geometrically anyway).

    Step-by-step algebra would be helpful - thanks!
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  2. #2
    ADY
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    [quote=TaylorM0192;446936]Find the inverse equations of the following:

    1. f(x) = 3x + ln(x)

    2. g(x) = \frac{x+1}{2x+1}

    ___________________________________________


    2.

    g^{-1}(x) = \frac{1-x}{2 x-1}
    Last edited by ADY; January 28th 2010 at 01:53 PM.
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  3. #3
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    [quote=ADY;446956]
    Quote Originally Posted by TaylorM0192 View Post
    Find the inverse equations of the following:

    1. f(x) = 3x + ln(x)

    2. g(x) = \frac{x+1}{2x+1}

    ___________________________________________

    1.

    f^{-1}(x) = 1/3 W(3 e^x)


    2.

    g^{-1}(x) = \frac{1-x}{2 x-1}
    Your g^{-1}(x)=\frac{1-x}{2x-1} cannot be the inverse of my g(x)=\frac{x+1}{2x+1} (confirmed by the graphs of each & algebraically by g(g^-1(x)) not being equal to 'x').

    I don't know what f^{-1}(x) = 1/3 W(3 e^x) is...
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  4. #4
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    Figured out g^-1(x) = \frac{x-1}{1-2x}.

    Still don't know f^-1(x) though.
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  5. #5
    ADY
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    Quote Originally Posted by TaylorM0192 View Post
    Figured out g^-1(x) = \frac{x-1}{1-2x}.

    Still don't know f^-1(x) though.
    f^{-1}(x) = \frac {1}{3} W(3 e^x)

    W is the product log function
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