Find the intervals in which the function f given by f (x) = 2x3 – 3x2 – 36x + 7 is
(a) strictly increasing (b) strictly decreasing
(a) It is given that function f(x) = 2x3 – 3x2 – 36x + 7
⇒ f’(x) = 6x2 – 6x + 36
⇒ f’(x) = 6(x2 – x + 6)
⇒ f’(x) = 6(x + 2)(x – 3)
If f’(x) = 0, then we get,
⇒ x = -2, 3
So, the points x = -2 and x = 3 divides the real line into two disjoint intervals, (-∞,2), (-2,3) and (3,∞)
So, in interval 
f’(x) = 6(x + 2)(x – 3) >0
Therefore, the given function (f) is strictly increasing in interval
.
(b) It is given that function f(x) = 2x3 – 3x2 – 36x + 7
⇒ f’(x) = 6x2 – 6x + 36
⇒ f’(x) = 6(x2 – x + 6)
⇒ f’(x) = 6(x + 2)(x – 3)
If f’(x) = 0, then we get,
⇒ x = -2, 3
So, the points x = -2 and x = 3 divides the real line into two disjoint intervals, (-∞,2), (-2,3) and (3,∞)
So, in interval
f’(x) = 6(x + 2)(x – 3) < 0
Therefore, the given function (f) is strictly decreasing in interval
.