Find the interval

It is given that function f(x) = –(x + 1)³ (x – 3)³

⇒ f’(x) = 3(x + 1)² (x – 3)³ +3(x + 1)³ (x – 3)²

⇒ f’(x) = 3(x + 1)² (x – 3)²[x – 3 + x + 1]

⇒ f’(x) = 6(x + 1)² (x – 3)²(x-1)

If f’(x) = 0, then we get,

⇒ x = -1,3 and 1

So, the points x = -1, x =1 and x = 3 divides the real line into four disjoint intervals,

So, in interval

f’(x) = 6(x + 1)² (x – 3)²(x-1) < 0

Therefore, the given function (f) is strictly decreasing in intervals .

So, in interval

f’(x) = 6(x + 1)² (x – 3)²(x-1) > 0

Therefore, the given function (f) is strictly increasing in intervals.