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March 30, 2017

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Find absolute maximum and minimum for f(x)=(3x^3)−(18x^2)+27x−10 over the closed interval [2,4].


Thank you so much!!

  • Jessica - ,

    f = 3x^3−18x^2+27x−10
    f' = 9x^2 - 36x + 27 = 9(x^2-4x+3) = 9(x-1)(x-3)

    so, f has a relative min/max at x=3
    From what you know about the general shape of cubics, it should be clear that (3,-10) is a relative min.

    since
    f(2) = -4
    f(4) = 2,

    f(3) = -10 is the absolute min within the interval
    f(4) = 2 is the absolute max in the interval

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