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Math Help - Compute f'(x) and finding critical numbers

  1. #1
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    Compute f'(x) and finding critical numbers

    Hey guys

    Here's the solution to a question i had on a test



    ok so i don't understand how they got from:
    -x^3/(3x-2)(2+3x)
    to
    {3x^2(4-3x^2)}/(2-3x)^2(2+3x)^2

    and i don't get how they find the "set numerator to 0" and "set denominator to 0" answers.

    any help REALLY appreciated, thanks!
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  2. #2
    MHF Contributor harish21's Avatar
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    This looks like a part of a test or assignment, and it is against MHF Policy to post these..


    Read Rule No 6.
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  3. #3
    MHF Contributor Unknown008's Avatar
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    When you differentiate this:

    f(x) = \dfrac{-x^3}{(3x-2)(2+3x)}

    you get:

    f'(x) = \dfrac{3x^2(4-3x^2)}{(2-3x)^2(2+3x)^2}

    using the quotient rule.

    The denominator can be written as:

    (3x-2)(2+3x) = (3x-2)(3x+2) = 9x^2 -4

    and the quotient rule applied:
    If
    y = \dfrac{u}{v}

    then;

    y' = \dfrac{v u' - uv'}{v^2}

    When you set the numerator to zero, you are looking for the maxima and minima (and/or inflexion)
    When you set the denominator to zero, you are looking for the vertical asymptotes.
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