Find the intervals in which the following functions are increasing or decreasing.
f(x) = 2x3 – 24x + 107
Given:- Function f(x) = 2x3 – 24x + 107
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 + 107
⇒
⇒ f’(x) = 6x2 – 24
For f(x) lets find critical point, we must have
⇒ f’(x) = 0
⇒ 6x2 – 24 = 0
⇒ 6(x2 – 4) = 0
⇒ (x – 2)(x + 2) = 0
⇒ x = –2, 2
clearly, f’(x) > 0 if x < –2 and x > 2
and f’(x) < 0 if –2 < x < 2
Thus, f(x) increases on (–∞, –2) ∪ (2, ∞)
and f(x) is decreasing on interval x ∈ (–2,2)