Question

f(x) = 2x³ – 3x² – 36x + 7

(a) For (-∞, -2) interval.

(b) For (-2, 3)

(c) For (3, ∞)

Answer 1

Given

f(x) = 2x³ – 3x² – 36x + 7

⇒ f'(x) = 6x² – 6x – 36

f'(x) = 0

⇒ 6x² – 6x – 36 = 0

⇒ 6(x² – x – 6) = 0

⇒ 6(x – 3) (x + 2) = 0

⇒ x – 3 = 0 or x + 2 = 0

⇒ x = 3 or x = -2

Hence, point x = – 2, x = 3 divides the real number line into three intervals (-∞,-2), (-2, 3) and (3,∞).

(a) For (-∞, -2) interval.

f'(x) = 6x² – 6x – 36 > 0

∵ At x = – 3

f'(x) = 6(- 3)² – 6(- 3) – 36

= 6 × 9 + 6 × 3 – 36

= 54 + 18 – 36 = 36 >0

So,f'(x) > 0 can be shown like this by taking other points. 

Hence, function is continuously increasing in interval (-∞, – 2).

(b) For (-2, 3)

f'(x) = 6x² – 6x – 36 < 0

∴ At x = 1

f'(x) = 6(1)² – 6 × 1 – 36

= 6 – 6 – 36 = -36 < 0

At x = 0, f'(x) = 6(0)² – 6 × 0 – 36

= – 36 < 0

So,f'(x) < 0 can be shown like this by taking other points.

Hence, function is continuously decreasing for x ∈ (- 2, 3).

(c) For (3, ∞)

f'(x) = 6x² – 6x – 36 > 0

∵ At x = 4,
f'(x) = 6 × (4)² – 6 × 4 – 36

= 96 – 24 – 36

= 96 – 60 = 36 > 0

At x = 5, f'(x)= 6(5)² – 6 × 5 – 36

= 6 × 25 – 30 – 36

= 150 – 30 – 36

= 84 > 0

So,f'(x) > 0 can be shown like this for other points.

Hence, function f(x) is continuous increasing in interval

(-∞, -2) ∪ (3,∞). {∵ f'(x) > 0}

Function is continuously decreasing in interval (- 2, 3).