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Math Help - Maxima/minima/points of inflection

  1. #1
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    Maxima/minima/points of inflection

    Hi all,
    Can somebody tell me if my solution is correct?

    y= (x^3)-[(3/2)*(x^2)]-6x+9

    Find the stationary values, determine their nature, and find the point of inflection.

    Max= (-1,12.5)
    Min= (2, -1)
    Point of inflection= (1/2, 5.75)

    thanks
    John

    Solution:

    At stationary points dy/dx (slope=0)

    dy/dx=0=3x^2 -3x-6

    solve quadratic to find x values

    X=2, -1

    Substitute into expression for y gives corresponding y values of (-1,12.5)

    The second derivative=6x-3

    The second derivative for x=2 =9 which is greater than 0 so this is the minimum point
    The second derivative for x=-1 =-9 which is less than 0 so this is the maximum point

    At the point of inflection the second derivative=0

    d2y/dx2=0=6x-3 therefore x=1/2

    Substituting x=1/2 into the expression for y gives y value of 5.75.

    Therefore the point of inflection= (1/2, 5.75)
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Maxima/minima/points of inflection

    Looks fine! To determine if you're dealing with a minimum or maximum you can also make a sign table.
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  3. #3
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    Re: Maxima/minima/points of inflection

    thanks Siron
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