silemus.

19Absolute max on a closed interval

Find the absolute maximum value of f(x)=x3−3x2+1\displaystyle f(x) = x^3 - 3x^2 + 1 on [−1,4]\displaystyle [-1, 4].

Compare f at the critical points and endpoints. Critical points are x=0,2 (where f'=0); candidates are x=-1,0,2,4 with f-values -3, 1, -3, 17. The absolute maximum is 17, at x=4.

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