`f(x) = 2x^3 - 3x^2 - 12x + 1, [-2, 3]` Find the absolute maximum and minimum values of f on the given interval

Textbook Question

Chapter 4, 4.1 - Problem 49 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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This function is differentiable everywhere. Therefore it reaches its absolute minimum and maximum either at the endpoints or where its derivative is zero.

f(-2) = -16 - 12 + 24 + 1 = -3, f(3) = 54 - 27 - 36 + 1 = -8.

`f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2),` f'(x) = 0 at `x_1=-1` and `x_2=2` (both in [-2, 3]).

`f(x_1) = -2 - 3 +12 + 1 = 8,` `f(x_2) = 16 - 12 - 24 + 1 = -19.`

The answer: the absolute minimum is -19, the absolute maximum is 8.

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