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

f(x) = 6 + 12x + 3x2 – 2x3

Given:- Function f(x) = -2x3 + 3x2 + 12x + 6


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 + 3x2 + 12x + 6


⇒


⇒ f’(x) = –6x2 + 6x + 12


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


⇒ f’(x) = 0


⇒ –6x2 + 6x + 12 = 0


⇒ 6(–x2 + x + 2) = 0


⇒ 6(–x2 + 2x – x + 2) = 0


⇒ x2 – 2x + x – 2 = 0


⇒ (x – 2)(x + 1) = 0


⇒ x = –1, 2





clearly, f’(x) > 0 if –1 < x < 2


and f’(x) < 0 if x < –1 and x > 2


Thus, f(x) increases on x ∈ (–1, 2)


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



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