(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.