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Math Help - Relative Max and Relative Min

  1. #1
    Junior Member arcticreaver's Avatar
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    Relative Max and Relative Min

    f(x) = (x-2)^3 / x^2

    find the relative max and min.

    so obviously i would bring the x^2 up.

    this is what i have so far,

    = x^2(x-2)^3-(x-2)^3(x^2)

    =x^2(3)(x-2)^2-(x-2)^3(2)(x)




    ​is this correct so far?
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Relative Max and Relative Min

    If you wish to use the product rule rather than the quotient rule by bringing the denominator up as a factor, then you need to negate its exponent.
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  3. #3
    Junior Member arcticreaver's Avatar
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    Re: Relative Max and Relative Min

    let me re-write this.

    x^2(x-2)^3 - (x-2)^3-(x^2) / x^4

    and the next step, i believe is the derivative right? but i am a bit confused with (x-2)^3. is the derivative of this 3(x-2)^2 or is it just 3x^2?
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Relative Max and Relative Min

    We are given the function:

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

    Using the quotient rule, we find:

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