Answer:
(-∞, -2) and (2, ∞)
Step-by-step explanation:
ƒ(x) = x³ -12x + 10
f'(x) = 3x² - 12
Set 3x² - 12 = 0
x² - 4 = 0
x² = 4
x = ±2
The points x = -2 and x = + 2 divide the number line into three intervals:
(-∞, -2), (-2, +2), and (+2, ∞).
a. Interval (-∞, -2)
x < -2, so f'(x) > 0 when -∞ < x < -2
The function is increasing in (-∞, -2).
b. Interval (-2, 2)
|x| < 2, so f'(x) <0
The function is decreasing in (-2, 2).
c. Interval (2, ∞)
x > 2, so f'(x) > 0 when 2 < x < ∞
The function is increasing in (2, ∞).
Thus, the function is increasing in the intervals (-∞, -2) and (2, ∞).
