Given:- Function f(x) = 3x⁴ – 4x³ – 12x² + 5

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) = 3x⁴ – 4x³ – 12x² + 5

⇒

⇒ f’(x) = 12x³ – 12x² – 24x

⇒ f’(x) = 12x(x² – x – 2)

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

⇒ f’(x) > 0

⇒ 12x(x² – x – 2)> 0

⇒ x(x² – 2x + x – 2) > 0

⇒ x(x – 2)(x + 1) > 0

⇒ –1 < x < 0 and x > 2

⇒ x ∈ (–1,0)∪(2, ∞)

Thus f(x) is increasing on interval (–1,0)∪(2, ∞)

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

f’(x) < 0

⇒ 12x(x² – x – 2)< 0

⇒ x(x² – 2x + x – 2) < 0

⇒ x(x – 2)(x + 1) < 0

⇒ –∞ < x < –1 and 0 < x < 2

⇒ x ∈ (–∞, –1) ∪ (0, 2)

Thus f(x) is decreasing on interval (–∞, –1) ∪ (0, 2)