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Math Help - Inverse Functions

  1. #1
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    Inverse Functions

    I'm having the hardest time finding the inverse function of



    f(x) = (x + 1) / (2x + 1)


    I know I exchange all the the x's for y's and solve for y but I'm having trouble doing so.... Help
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  2. #2
    Eater of Worlds
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    Try long dividing first.

    This gives \frac{x+1}{2x+1}=\frac{1}{2(2x+1)}+\frac{1}{2}

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

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

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

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

    y=\frac{1-x}{2x-1}
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by galactus View Post
    Try long dividing first.

    This gives \frac{x+1}{2x+1}=\frac{1}{2(2x+1)}+\frac{1}{2}

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

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

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

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

    y=\frac{1-x}{2x-1}
    i never thought of attempting a problem like this that way.

    Here's the method i usually use.

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

    For inverse, switch x and y and solve for y

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

    \Rightarrow 2xy + x = y + 1

    \Rightarrow 2xy - y = 1 - x

    \Rightarrow y(2x - 1) = 1 - x

    \Rightarrow y = \frac {1 - x}{2x - 1}
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  4. #4
    Eater of Worlds
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    Your method appears a little more efficient. Just another way to go about it.
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  5. #5
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    Thanks alot guys!!
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