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Math Help - [f(x+deltax)-f(x)]/delta x

  1. #1
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    [f(x+deltax)-f(x)]/delta x

    algebraically simplify [f(x+deltax)-f(x)]/delta x to -1/[(x+delta x-3)(x-3)] for f(x) = (x-2)/(x-3)
    Last edited by mr fantastic; September 23rd 2010 at 03:32 PM. Reason: Re-titled.
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  2. #2
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    Hello, yess!


    \text{Given: }\;f(x) \:=\:\dfrac{x-2}{x-3}

    \text{Simply }\,\dfrac{f(x+\Delta x) - f(x)}{\Delta x}\:\text{ to }\:\dfrac{-1}{(x+\Delta x - 3)(x - 3)}

    When working with a Difference Quotient, I make three steps out of it. .**

    . . [1] Find f(x+\Delta x) . . . Replace x with x + \Delta x ... and simplify.

    . . [2] Subtract f(x) . . . Subtract the original function ... and simplify.

    . . [3] Divide by \Delta x . . . Write \Delta x in the denominator ... and reduce.


    Here we go . . .


    [1]\;\;f(x+\Delta x) \;=\;\dfrac{(x+\Delta x) - 2}{(x+\Delta x) - 3}



    [2]\;\;f(x+\Delta x) - f(x) \;=\;\dfrac{x+\Delta x - 2}{x + \Delta x - 3} - \dfrac{x-2}{x -3}

    . . =\;\dfrac{(x-3)(x+\Delta x - 2) - (x-2)(x+\Delta x - 3)}{(x+\Delta x - 3)(x - 3)}

    . . =\;\dfrac{x^2 + x\,\Delta x - 2x - 3x - 3\Delta x + 6 - x^2 - x\,\Delta x + 3x + 2x + 2\Delta x - 6}{(x+\Delta x - 3)(x - 3)}

    . . =\;\dfrac{-\Delta x}{(x+\Delta x - 3)(x - 3)}



    [3]\;\;\dfrac{f(x+\Delta x) - f(x)}{\Delta x} \;=\;\dfrac{-\Delta x}{\Delta x\, (x + \Delta x - 3)(x - 3)}

    . . . . . . . . . . . . . . . . . =\;\dfrac{-1}{(x+\Delta x - 3)(x - 3)}\quad\hdots\;There!


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    **

    Am I the only teacher on the entire planet who taught it this way?
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