Listen NCERT Audio Books to boost your productivity and retention power by 2X.
Find the intervals in which the following functions are increasing or decreasing.
f(x) = 5x3 – 15x2 – 120x + 3
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)