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Math Help - Find the inverse of F(x)?

  1. #1
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    Find the inverse of F(x)?

    f(x) = (2x+3)/(4x-5)

    So, I let f(x) = y and then I get y = (2x+3) / (4x-5) and then I figure I solve for x... But I am having difficulty in this, any suggestions? Thank you
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  2. #2
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    Hi WashingtonStudent!

    Quote Originally Posted by WashingtonStudent View Post
    f(x) = (2x+3)/(4x-5)

    So, I let f(x) = y and then I get y = (2x+3) / (4x-5) and then I figure I solve for x... But I am having difficulty in this, any suggestions? Thank you
    Do you know this method: "Polynomial long division"

    You get   \frac{2x+3}{4x-5} = 0.5 + \frac{11}{2*(4x-5)}

    thus you got to solve  y = 0.5 + \frac{11}{2*(4x-5)}

    I
    Yours
    Rapha
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  3. #3
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    Could you go over the steps in a bit more detail? I am familiar with polynomial long divison.. but when I make 2x + 3 the divided and 4x - 5 the divisor then I end up with..

    0.5*(4x-5) + 5.5
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  4. #4
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    I found

     \frac{2x+3}{4x-5}= \frac{1}{2(4x-5)}+\frac{1}{2}

    Now y = \frac{1}{2(4x-5)}+\frac{1}{2}

    To find the inverse, swap y and x and solve for y.

    x = \frac{1}{2(4y-5)}+\frac{1}{2}

    x-\frac{1}{2} = \frac{1}{2(4y-5)}

    4y-5 = \frac{1}{2(x-\frac{1}{2})}

    4y = \frac{1}{2(x-\frac{1}{2})}+5

    y = \frac{1}{8(x-\frac{1}{2})}+\frac{5}{4}

    y = \frac{1}{8x-4}+\frac{5}{4}
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  5. #5
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    to find inverse, i'm pretty sure you have to exchange the Y with the X, and then solve for Y.

    original equation:
    y=\frac{2x+3}{4x-5}

    exchange variables:
    x=\frac{2y+3}{4y-5}

    solve for y:
    x(4y-5)=2y+3 <----move the denominator on the right to the left
    4xy-5x=2y+3 <----factor
    4xy-2y=5x+3 <----move all the y's to one side and the single x to the other
    2y(2x-1)=5x+3 <----factor out the 2y from the left side
    2y=\frac{5x+3}{2x-1} <----get the 2y by itself
    y=\frac{5x+3}{4x-2} <----get the y by itself

    EDIT: I think my way is easier, pickslides...
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  6. #6
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    yes, i like it.

    I was only making note of the previous poster's reference to long division.
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