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