Answer:
Explanation:
f(x) = x⁴ - 8x² + 16
f'(x) = 4x³ - 16x = 4(x³ - 4x) = 4x(x² - 4) = 4x(x - 2)(x + 2)
f'(x) = 0, x = 2, -2
f"(x) = 4(3x² - 4)
f''(0) = -16 < 0, f""(2) = f''(-2) 3(12 - 4) > 0
so f(2) and f(-2) are minimum values
f(0) is maximum value.
f(0) = 16, f(2) = f(-2) = 2⁴ - 8*2² + 16 = 0
Maximum = 16, minimum value = 0
-∞ < x < -2, f'(x) < 0
-2 < x < 0, f'(x) > 0
0 < x < 2, f'(x) < 0
2 < x < ∞, f(x) > 0
increase: (-2, 0) and (2, ∞)
decrease: (-∞, -2) and (0, 2)