Given:- Function

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

⇒

⇒ f’(x) = 6x³ – 12x² – 90x

⇒ f’(x) = 6x(x² – 2x – 15)

⇒ f’(x) = 6x(x² – 5x + 3x – 15)

⇒ f’(x) = 6x(x – 5)(x + 3)

For f(x) to be increasing, we must have

⇒ f’(x) > 0

⇒ 6x(x – 5)(x + 3)> 0

⇒ x(x – 5)(x + 3) > 0

⇒ –3 < x < 0 or 5 < x < ∞

⇒ x ∈ (–3,0)∪(5, ∞)

Thus f(x) is increasing on interval (–3,0)∪(5, ∞)

Again, For f(x) to be decreasing, we must have

f’(x) < 0

⇒ 6x(x – 5)(x + 3)> 0

⇒ x(x – 5)(x + 3) > 0

⇒ –∞ < x < –3 or 0 < x < 5

⇒ x ∈ (–∞, –3)∪(0, 5)

Thus f(x) is decreasing on interval (–∞, –3)∪(0, 5)