Find the intervals in which the following functions are increasing or decreasing.

f(x) = x^{3} – 6x^{2} – 36x + 2

Given:- Function f(x) = x^{3} – 6x^{2} – 36x + 2

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) = x^{3} – 6x^{2} – 36x + 2

⇒

⇒ f’(x) = 3x^{2} – 12x – 36

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

⇒ f’(x) = 0

⇒ 3x^{2} – 12x – 36 = 0

⇒ 3(x^{2} – 4x – 12) = 0

⇒ 3(x^{2} – 6x + 2x – 12) = 0

⇒ x^{2} – 6x + 2x – 12 = 0

⇒ (x – 6)(x + 2) = 0

⇒ x = 6, – 2

clearly, f’(x) > 0 if x < –2 and x > 6

and f’(x) < 0 if –2< x < 6

Thus, f(x) increases on (–∞,–2) ∪ (6, ∞)

and f(x) is decreasing on interval x ∈ (–2,6)

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