Find the intervals in which the following functions are increasing or decreasing.
f(x) = 8 + 36x + 3x2 – 2x3
Given:- Function f(x) = 8 + 36x + 3x2 – 2x3
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) = 8 + 36x + 3x2 – 2x3
⇒
⇒ f’(x) = 36 + 6x – 6x2
For f(x) lets find critical point, we must have
⇒ f’(x) = 0
⇒ 36 + 6x – 6x2 = 0
⇒ 6(–x2 + x + 6) = 0
⇒ 6(–x2 + 3x – 2x + 6) = 0
⇒ –x2 + 3x – 2x + 6 = 0
⇒ x2 – 3x + 2x – 6 = 0
⇒ (x – 3)(x + 2) = 0
⇒ x = 3 , – 2
clearly, f’(x) > 0 if –2< x < 3
and f’(x) < 0 if x < –2 and x > 3
Thus, f(x) increases on x ∈ (–2,3)
and f(x) is decreasing on interval (–∞,–2) ∪ (3, ∞)