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) = 2x3 – 15x2 + 36x + 1
Answer» Given as f(x) = 2x³ – 15x² + 36x + 1 Differentiate the above equation with respect x, we get f'(x) = (d/dx)(2x³ – 15x² + 36x + 1) ⇒ f’(x) = 6x² – 30x + 36 For the function f(x) we have to find critical point, we must have ⇒ f’(x) = 0 ⇒ 6x² – 30x + 36 = 0 ⇒ 6 (x² – 5x + 6) = 0 ⇒ 3(x² – 3x – 2x + 6) = 0 ⇒ x² – 3x – 2x + 6 = 0 ⇒ (x – 3) (x – 2) = 0 ⇒ x = 3, 2 It is clear, f’(x) > 0 if x < 2 and x > 3 and f’(x) < 0 if 2 < x < 3 Hence, f(x) increases on (–∞, 2) ∪ (3, ∞) and f(x) is decreasing on interval x ∈ (2, 3)

Find the intervals in which the functions are increasing or decreasing. f(x) = 2x3 – 15x2 + 36x + 1

Answer»

Given as f(x) = 2x³ – 15x² + 36x + 1

Differentiate the above equation with respect x, we get

f'(x) = (d/dx)(2x³ – 15x² + 36x + 1)

⇒ f’(x) = 6x² – 30x + 36

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

⇒ f’(x) = 0

⇒ 6x² – 30x + 36 = 0

⇒ 6 (x² – 5x + 6) = 0

⇒ 3(x² – 3x – 2x + 6) = 0

⇒ x² – 3x – 2x + 6 = 0

⇒ (x – 3) (x – 2) = 0

⇒ x = 3, 2

It is clear, f’(x) > 0 if x < 2 and x > 3 and f’(x) < 0 if 2 < x < 3

Hence, f(x) increases on (–∞, 2) ∪ (3, ∞) and f(x) is decreasing on interval x ∈ (2, 3)