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

Textbook Question

Chapter 4, 4.1 - Problem 52 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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f(x) is continuous and differentiable everywhere, so it reaches its maximum and minimum either at the endpoint or where f'(x)=0.

f(-1) = 0, f(2) = 27.

`f'(x) = 3(x^2-1)^2*2x` . This is zero at x=0, x=-1 and x=1.

f(1) = 0, f(0) = -1, f(-1) = 0.

The answer: the absolute minimum is -1, the absolute maximum is 27.

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