CBSE Class 12-science Answered

We have,

f (x) = (x + 1)³ (x - 3)³

=> f ' (x) = 3(x + 1)²(x + 1)(x - 3)³ + (x + 1)³3(x - 3)^2.(x - 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 + 1 + x - 3)

=> f ' (x) = 3(x + 1)² (x - 3)² (x + 1 + x - 3)

=> f ' (x) = 6(x + 1)² (x - 3)² (x - 1)

For f (x) to be increasing, we must have

f ' (x) > 0

=> 6(x + 1)² (x - 3)² (x - 1) > 0  x - 1 > 0          [ 6(x + 1)² (x - 3)² > 0]

=> x > 1 & x belongs to (1, ¥)

So, f (x) is increasing on (1, ¥)

For f (x) to  be decreasing, we must have

f ' (x) < 0

=> 6(x + 1)² (x - 3)² (x - 1) < 0

=> x - 1 < 0       [ 6(x + 1)² (x - 3)² > 0]

=> x < 1 & x (-¥, 1)

So, f (x) is decreasing on