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

f(x) = – 2x3 – 9x2 – 12x + 1

Given:- Function f(x) = –2x3 – 9x2 – 12x + 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) = –2x3 – 9x2 – 12x + 1



f’(x) = – 6x2 – 18x – 12


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


f’(x) = 0


–6x2 – 18x – 12 = 0


6x2 + 18x + 12 = 0


6(x2 + 3x + 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 x < –2 and x >–1


and f’(x) < 0 if –2 < x < –1


Thus, f(x) increases on (–, –2) (–1, ∞)


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



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