Given:- Function f(x) = 2x³ – 12x² + 18x + 15

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) = 2x³ – 12x² + 18x + 15

⇒

⇒ f’(x) = 6x² – 24x + 18

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒ 6x² – 24x + 18 = 0

⇒ 6(x² – 4x + 3) = 0

⇒ 6(x² – 3x – x + 3) = 0

⇒ 6(x – 3)(x – 1) = 0

⇒ (x – 3)(x – 1) = 0

⇒ x = 3 , 1