Find Absolute maximum and minimum values

Question

Find Absolute maximum and minimum values of each function over the indicated interval, and indicate the x-values at which they occur.

f(x)=x^3-x^2-8x+8 ;[-2,0]

Answer 1

The absolute max or min for a continuous function on closed interval occurs at one of the end points of the closed interval or where a critical number exists (first derivative equals zero).

f(-2) = (-2)³ - (-2)² -8(-2) + 8 = -8 - 4 +16 + 8 = 12

f(0) = 8

f'(x) = 3x² - 2x - 8, which equals zero:

3x² - 2x - 8 = 0

3x² - 6x + 4x - 8 = 0

3x(x - 2) + 4(x - 2) = 0

(3x + 4) (x - 2) = 0

x = -4/3 or x = 2 (ignore x = 2 because it is not in [-2, 0].

f(-4/3) = (-4/3)³ - (-4/3)² - 8(-4/3) + 8

= -64/27 - 16/9 + 32/3 + 8

= -64/27 - 48/27 + 288/27 + 216/27

= 392/27 ≈ 14.5

So the absolute max occurs at x = -4/3 with a max value of 392/27.

The absolute min occurs at x = 0 with a min value of 8.

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