Whats the local maximum and minimum points for x^3-2x^2-8x?

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  • DaM
    Lv 6
    1 decade ago
    Favorite Answer

    If y=x^3-2x^2-8x

    then dy/dx=3x^2-4x-8

    equate to 0 and solve for x

    3x^2-4x-8=0

    using quadratic formula, with a=3, b=-4, and c=-8

    we get x=-1.0972 and x=2.4305

    plug in those values in the origional formula to get the corresponding values of y

    first pair, x=-1.097,y=5.049

    2nd pair x=2.431, y=-16.901

    EDIT to add,

    sahsjing, you factored wrong in

    3x^2-4x-8 = (3x+4)(x-2) = 0

    (3x+4)(x-2)=3x^2-2x-8

    Source(s): some mental, confirmed by both graph and numeric solution on calculator, (and also a computer program)
  • 1 decade ago

    y = f(x) = x^3-2x^2-8x

    f'(x) = 3x^2-4x-8 = (3x+4)(x-2) = 0

    Critical points: -4/3, 2

    local maximum: f(-4/3) = 128/27

    local minimum: f(2) = -16

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