Q. 1 G5.0( 2 Votes )

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

f(x) = 5x3 – 15x2 – 120x + 3

Answer :

Given:– Function f(x) = 5x3 – 15x2 – 120x + 3


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) = 5x3 – 15x2 – 120x + 3



f’(x) = 15x2 – 30x – 120


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


f’(x) = 0


15x2 – 30x – 120 = 0


15(x2 – 2x – 8) = 0


15(x2 – 4x + 2x – 8) = 0


x2 – 4x + 2x – 8 = 0


(x – 4)(x + 2) = 0


x = 4 , – 2





clearly, f’(x) > 0 if x < –2 and x > 4


and f’(x) < 0 if –2< x < 4


Thus, f(x) increases on (–,–2) (4, ∞)


and f(x) is decreasing on interval x (–2,4)



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