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

We have,

*f* (*x*) = (*x* + 1)^{3} (*x* - 3)^{3}

*=> f ' *(*x*) = 3(*x *+ 1)^{2}(*x *+ 1)(*x* - 3)^{3} + (*x* + 1)^{3}3(*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

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

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