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October 31, 2014

October 31, 2014

Posted by **Math** on Thursday, November 22, 2012 at 8:08pm.

Thank you so much!!

- Jessica -
**Steve**, Thursday, November 22, 2012 at 11:45pmf = 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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