Question

Find the interval in which the function $f\left(x\right)=(x+1{)}^3-(x-3{)}^3$ is strictly increasing or decreasing

Answer 1

We have,
$f\left(x\right)=(x+1{)}^3(x-3{)}^3$
$f^{\prime }\left(x\right)=3(x+1{)}^2(x-3{)}^3+3(x-3{)}^2(x+1{)}^3$
$=3(x+1{)}^2(x-3{)}^2[x-3+x+1]$
$=3(x+1{)}^2(x-3{)}^2(2x-2)$
$=6(x+1{)}^2(x-3{)}^2(x-1)$
Now,
$f^{\prime }\left(x\right)=0\Rightarrow x=-1,3,1$
The points $x=-1,x=1,$ and $x=3$ divide the real line into four disjoint intervals
i.e.,$(-\infty ,-1),(-1,1),(1,3)$ and $(3,\infty )$.
In intervals $(-\infty ,-1)$ and $(-1,1)$, $f^{\prime }\left(x\right)=6(x+1{)}^2(x-3{)}^2(x-1)<0$
$\therefore$ $f$ is strictly decreasing in intervals $(-\infty ,-1)$ and $(-1,1).$
In intervals $(1,3)$ and $(3,\infty )f^{\prime }\left(x\right)=6(x+1{)}^2(x-3{)}^2(x-1)>0$
$\therefore$ $f$ is strictly increasing in intervals $(1,3)$ and $(3,\infty )$.