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

f(x) = 2x^{3} – 15x^{2} + 36x + 1

Given:- Function f(x) = 2x^{3} – 15x^{2} + 36x + 1

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) = 2x^{3} – 15x^{2} + 36x + 1

⇒

⇒ f’(x) = 6x^{2} – 30x + 36

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

⇒ f’(x) = 0

⇒ 6x^{2} – 30x + 36 = 0

⇒ 6(x^{2} – 5x + 6) = 0

⇒ 3(x^{2} – 3x – 2x + 6) = 0

⇒ x^{2} – 3x – 2x + 6 = 0

⇒ (x – 3)(x – 2) = 0

⇒ x = 3, 2

clearly, f’(x) > 0 if x < 2 and x > 3

and f’(x) < 0 if 2 < x < 3

Thus, f(x) increases on (–∞, 2) ∪ (3, ∞)

and f(x) is decreasing on interval x ∈ (2,3)

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