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Find the absolute maximum of (x^2-1)^3 in the interval [-1,2]

User Luccas Clezar
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1 Answer

22 votes
22 votes

Let f(x) = (x ² - 1)³. Find the critical points of f in the interval [-1, 2]:

f '(x) = 3 (x ² - 1)² (2x) = 6x (x ² - 1)² = 0

6x = 0 or (x ² - 1)² = 0

x = 0 or x ² = 1

x = 0 or x = 1 or x = -1

Check the value of f at each of these critical points, as well as the endpoints of the given domain:

f (-1) = 0

f (0) = -1

f (1) = 0

f (2) = 27

So maxf(x) = 27.

User Kimmy
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