Given:– Function f(x) = 5x³ – 15x² – 120x + 3

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) = 5x³ – 15x² – 120x + 3

⇒

⇒ f’(x) = 15x² – 30x – 120

For f(x) lets 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