Find the intervals in which the functions are increasing or decreasin

1.	Find the intervals in which the functions are increasing or decreasing. f(x) = 5x3 – 15x2 – 120x + 3
Answer» Given as f(x) = 5x³ – 15x² – 120x + 3 Differentiate the above equation with respect x, we get f'(x) = (d/dx)(5x³ - 15x² - 120x + 3) ⇒ f’(x) = 15x² – 30x – 120 For the function f(x) we have to find critical point, we must have ⇒ f’(x) = 0 ⇒ 15x² – 30x – 120 = 0 ⇒ 15(x² – 2x – 8) = 0 ⇒ 15(x² – 4x + 2x – 8) = 0 ⇒ x² – 4x + 2x – 8 = 0 ⇒ (x – 4) (x + 2) = 0 ⇒ x = 4, – 2 It is clear, f’(x) > 0 if x < –2 and x > 4 and f’(x) < 0 if –2 < x < 4 Hence, f(x) increases on (–∞,–2) ∪ (4, ∞) and f(x) is decreasing on interval x ∈ (–2, 4)

Find the intervals in which the functions are increasing or decreasing. f(x) = 5x3 – 15x2 – 120x + 3

Answer»

Given as f(x) = 5x³ – 15x² – 120x + 3

Differentiate the above equation with respect x, we get

f'(x) = (d/dx)(5x³ - 15x² - 120x + 3)

⇒ f’(x) = 15x² – 30x – 120

For the function f(x) we have to find critical point, we must have

⇒ f’(x) = 0

⇒ 15x² – 30x – 120 = 0

⇒ 15(x² – 2x – 8) = 0

⇒ 15(x² – 4x + 2x – 8) = 0

⇒ x² – 4x + 2x – 8 = 0

⇒ (x – 4) (x + 2) = 0

⇒ x = 4, – 2

It is clear, f’(x) > 0 if x < –2 and x > 4 and f’(x) < 0 if –2 < x < 4

Hence, f(x) increases on (–∞,–2) ∪ (4, ∞) and f(x) is decreasing on interval x ∈ (–2, 4)