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Math Help - Inverse of function

  1. #1
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    Inverse of function

    y=3x-1/2x


    My attempt:

    y=(6x^2-1)/(2x)

    x= (6y^2-1)/(2y)

    2yx=6y^2-1

    2yx-6y^2=-1

    <br />
2y(x-3y)=-1

    2y=-1/(x-3y)

    ( I get stuck her what do l do with the y in the denominator ? )
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  2. #2
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    Once you have that 2yx= 6y^2- 1, rewrite it as 6y^2- 2xy- 1 and use the quadratic formula to solve for y.

    Solve that quadratic equation for x you will get a " \pm". This function does not have a true inverse.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Once you have that 2yx= 6y^2- 1, rewrite it as 6y^2- 2xy- 1 and use the quadratic formula to solve for y.

    Solve that quadratic equation for x you will get a " \pm". This function does not have a true inverse.

    I solved it thank you very much for your help
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    Once you have that 2yx= 6y^2- 1, rewrite it as 6y^2- 2xy- 1 and use the quadratic formula to solve for y.

    Solve that quadratic equation for x you will get a " \pm". This function does not have a true inverse.

    If my restricted domain is D={x|x<0} does that mean for my inverse function is going to be the one with a minus sign on the quadratic equation.
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