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26 September, 01:40

F (x) = 3x^2-12x+5, find the absolute max and min value of f (x) on the interval [0,1].

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  1. 26 September, 02:01
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    F (0) = 3 (0) ^2 - 12 (0) + 5 = 5

    f (1) = 3 (1) ^2 - 12 (1) + 5 = 3 - 12 + 5 = - 4

    The absolute maximum in the interval [0, 1] is at y = 5 which occurs at x = 0.

    The absolute minimum in the interval [0, 1] is at y = - 4 which occurs at x = 1.
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