Find the intervals in which f (x) = (x + 1)3 (x - 3)3 is increasing or decreasing.

Asked by Topperlearning User | 6th Aug, 2014, 09:54: AM

Expert Answer:

We have,

f (x) = (x + 1)3 (x - 3)3

=> f ' (x) = 3(+ 1)2\frac{d}{dx}(+ 1)(x - 3)3 + (x + 1)33(x - 3)2.\frac{d}{dx}(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

Answered by  | 6th Aug, 2014, 11:54: AM