calculus

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find the absolute minimum and maximum valies of f(x)=-x^4+2x^3+5 on the interval [-2,2]

  • calculus -

    check f(-2) and f(2).

    f' = -4x^3 + 6x^2
    = -2x^2(2x-3)
    so there are local extrema at x = 3/2
    Not at x=0, since that's a double root of f'=0.

    Now pick the max/min values of f at -2,3/2,2

  • calculus -

    f ' (x) = - 4x^3 + 6x^2
    = 0 for max/min

    x^2(-4x + 6) = 0
    x = 0 or x = 3/2

    f(0) = 5
    f(3/2) = -81/64 + 54/8 + 5 = 671/64 = appr 10.48
    f(-2) = - 16 + 16 + 5 = 5
    f(2) = -16 - 16 + 5 = -27

    so the absolute max is 671/64 and the min is -27

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