Question

Find the interval in which the following function is strictly incerasing or decreasing,

$(x+1{)}^3(x-3{)}^3$

Answer 1

Given, $f\left(x\right)=(x+1{)}^3(x-3{)}^3$
On differentiating, we get
$f^{\prime }\left(x\right)=(x+1{)}^3.3(x-3{)}^{2}1+(x-3{)}^3.3(x+1{)}^2.1=3(x-3{)}^2(x+1{)}^2\left\{\right(x+1)+(x-3\left)\right\}=3(x-3{)}^2(x+1{)}^2(2x-2)=6(x-3{)}^2(x+1{)}^2(x-1)$
On putting f'(x)=0, we get x=-1, 1, 3
Which divides real line into four disjoint
intervals namely $(-\infty ,-1)(-1,1)(1,3)$ and $(3,\infty )$.

$\begin{array}{ccc}Intervals& Signof{f}^{\prime }\left(x\right)& Natureoff\left(x\right)\\ -\infty <x<-1& (+)(+)(-)=-ve& Strictlydecreasing\\ -1<x<1& (+)(+)(-)=-ve& Strictlydecreasing\\ 1<x<3& (+)(+)(+)=+ve& Strictlyincreasing\\ 3<x<\infty & (+)(+)(+)=+ve& strictlyincreasing\end{array}$
Therefore, f(x) is strictly increasing in (1,3) and $(3,\infty )$ and strictly decreasing in $(-\infty ,-1)$ and $(-1,1)$