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) = 2x3 – 24x + 7
Given:- Function f(x) = 2x3 – 24x + 7
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 – 24x + 7
⇒
⇒ f’(x) = 6x2 – 24
For f(x) to be increasing, we must have
⇒ f’(x) > 0
⇒ 6x2 – 24 > 0
⇒
⇒ x2 < 4
⇒ x < –2, +2
⇒ x ∈ (–∞,–2) and x ∈ (2,∞)
Thus f(x) is increasing on interval (–∞, –2) ∪ (2, ∞)
Again, For f(x) to be increasing, we must have
f’(x) < 0
⇒ 6x2 – 24< 0
⇒
⇒ x2 < 4
⇒ x> –1
⇒ x ∈ (–1,∞)
Thus f(x) is decreasing on interval x ∈ (–1, ∞)