Please help! Calculus problem!!!?

Question

6. Find the absolute extrema of g(x) = x^(4) − 32x^(2) − 7 on [−5, 6] (if it exists).

Answer 1

The function is g(x) = x⁴ - 32x² - 7 and interval [-5, 6].

Differentiate with respect to x.

g'(x) = 4x³ - 64x.

To find the critical numbers g'(x) = 0 or g'(x) does not exist.

4x³ - 64x = 0

4x(x² - 16) = 0

4x = 0 and x² - 16 = 0

The critical numbers are x = 0 and x = -4 and x = 4.

Left Endpoint (x = - 5): g(-5) = (-5)⁴ - 32(-5)² - 7 = -182.

Critical Number (x = -4): g(-4) = (-4)⁴ - 32(-4)² - 7 = -263 (Minimum).

Critical Number (x = 0): g(0) = (0)⁴ - 32(0)² - 7 = -7.

Critical Number (x = 4): g(4) = (4)⁴ - 32(4)² - 7 = -263 (Minimum).

Right Endpoint (x = 6): g(6) = (6)⁴ - 32(6)² - 7 = 137 (Maximum).

The minimum points are (-4, -263) and (4, -263) and the maximum point is (6, 137).

Graph: