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Math Help - [SOLVED] inverse of a function

  1. #1
    Junior Member
    Joined
    Sep 2009
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    74

    [SOLVED] inverse of a function

    I'm trying to take the inverse of the function \frac{2x-3}{x+4}
    I think I've got it following this procedure
    x=\frac{2y-3}{y+4}
    x(y+4) = 2y-3
    xy+4x = 2y-3
    xy-2y = -3-4x
    y(x-2) = -3-4x
    [tex]y = \frac{-3-4x}{x-2}[tex]
    y = -\frac{4x+3}{x-2}

    I worked that a few times and am pretty confident in my answer. The proof is in plugging that back into the original equation and seeing if I get x, and this is where I'm fumbling about
    \frac{2(-\frac{4x+3}{x-2})-3}{-(\frac{4x+3}{x-2})+4}
    \frac{-(\frac{6+8x-3x+6}{x-2})}{-(\frac{3+4x+4x-8}{x-2})}
    \frac{-(\frac{5x+12}{x-2})}{-(\frac{8x-5}{x-2})}
    -\frac{5x+12}{x-2} * -\frac{x-2}{8x-5}
    \frac{5x+12}{8x+5}
    which, unless I'm blind, does not equal x. I hope the above isn't too confusing.

    Assuming my initial answer is correct, what am I doing wrong when I'm trying to verify it?
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  2. #2
    Junior Member
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    Feb 2010
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    Lisbon
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    Well, your answer is correct. You've made a mistake in your proof:

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

    \not = \frac{-(\frac{6+8x-3x+6}{x-2})}{-(\frac{3+4x+4x-8}{x-2})}

    You factor by -1 but keep the -3 on the numerator (should be 3). I strongly recommedn writing the original answer as

    y = \frac{4x+3}{2-x}

    Keeping unecessary minus signs is a recipe for disaster
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  3. #3
    Junior Member
    Joined
    Sep 2009
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    Thanks a lot for the assistance. I agree, all those minus signs in there were causing me some serious difficulties. I'll try to avoid them as much as possible in the future.
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