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Math Help - Inverse help!

  1. #1
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    Inverse help!

    If f= x+1/x+3 , find g, it's inverse, if it exists.

    Thanks in advance!
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  2. #2
    Senior Member topher0805's Avatar
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    Finding the Inverse of a Function

    This site gives a good explanation.

    y=\frac {x+1}{x+3}

    Solve for x:

    y(x+3)=x+1

    <br />
yx+3y=x+1

    3y-1=x-yx

    3y-1=x(1-y)

    x=\frac {3y-1}{1-y}

    Now switch the x and y:

    y = \frac {3x-1}{1-x}

    And y is your inverse.
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  3. #3
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    Quote Originally Posted by antz215 View Post
    If f= x+1/x+3 , find g, it's inverse, if it exists.

    Thanks in advance!
    y = f(x) = \frac{x+1}{x+3}~,~x\neq -3

    Now change the variables:

    g: x = \frac{y+1}{y+3}~\iff~x(y+3)=y+1

    g: xy+3x = y+1~\iff~ 3x-1=y-xy~\iff~3x-1=y(1-x)

    g(x)=y=\frac{3x-1}{1-x}~,~x\neq 1
    Attached Thumbnails Attached Thumbnails Inverse help!-graph_inverse.jpg  
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  4. #4
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    Thank you both very much!!
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  5. #5
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    for these inverse problems, simply switch all x's with y's and all y's with x's. Note that for f(x), just consider it to mean y. Ok, so you switched the x's and y's. Now solve for y. That's it.
    -Andy

    Edit: And I almost forgot, check your restriction(s). Only God can divide by zero.
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