Answer:
Increasing on (-∞, -5) and (2.5, ∞)
Explanation:
ƒ(x) = 4x³ + 15x² - 150x + 9
Step 1.Find the zeros of the first derivative of the function
ƒ'(x) = 12x² + 30x -150 = 0
2x² + 5x - 25 = 0
(x + 5)(2x - 5) = 0
x = -5 or x = 2.5
These are your critical points.
Step 2. Apply the first derivative test.
Test all intervals to the left and to the right of these values to determine if the derivative is positive or negative.
(a) x = -10
ƒ'(-10) = 12(-10)² + 30(-10) - 150 = 1200 - 300 - 150 = 750
ƒ'(x) > 0 so the function is increasing on (-∞, -5).
(b) x = 0
ƒ'(0) = 12(0)² + 30(0) -150 = -150
ƒ'(x) < 0 so the function is decreasing.
(c) x = 10
ƒ'(0) = 12(10)² + 30(10) -150 = 1200 + 300 - 150 = 1350
ƒ'(x) > 0 so the function is increasing on (2.5, ∞).
The function is increasing on (-∞, -5) and (2.5, ∞).