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Find the intervals in which the following functions are increasing or decreasing.
f(x) = 3x4 – 4x3 – 12x2 + 5
Given:- Function f(x) = 3x4 – 4x3 – 12x2 + 5
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) = 3x4 – 4x3 – 12x2 + 5
⇒
⇒ f’(x) = 12x3 – 12x2 – 24x
⇒ f’(x) = 12x(x2 – x – 2)
For f(x) to be increasing, we must have
⇒ f’(x) > 0
⇒ 12x(x2 – x – 2)> 0
⇒ x(x2 – 2x + x – 2) > 0
⇒ x(x – 2)(x + 1) > 0
⇒ –1 < x < 0 and x > 2
⇒ x ∈ (–1,0)∪(2, ∞)
Thus f(x) is increasing on interval (–1,0)∪(2, ∞)
Again, For f(x) to be decreasing, we must have
f’(x) < 0
⇒ 12x(x2 – x – 2)< 0
⇒ x(x2 – 2x + x – 2) < 0
⇒ x(x – 2)(x + 1) < 0
⇒ –∞ < x < –1 and 0 < x < 2
⇒ x ∈ (–∞, –1) ∪ (0, 2)
Thus f(x) is decreasing on interval (–∞, –1) ∪ (0, 2)