Question

Find the interval in which the function $f\left(x\right)=-2x^3-9x^2-12x+1$ is strictly increasing or decreasing

Answer 1

$f\left(x\right)=-2x^3-9x^2-12x+1$
$\therefore f^{\prime }\left(x\right)=-6x^3-18x^2-12=-6(x^2+3x+2)=-6(x+1)(x+2)$
Now,
$f^{\prime }\left(x\right)=0\Rightarrow x=-1$ and $x=-2$
Points $x=-1$ and $x=-2$ divide the real line into three disjoint intervals
i.e., $(-\infty ,-2)(-2,-1)$ and $(-1,-\infty )$.
In intervals $(-\infty ,-2)$ and $(1,-\infty )$ i.e., when $x<-2$ and $x>-1,$
$f^{\prime }\left(x\right)=-6(x+1)(x+2)<0$
$\therefore$ $f$ is strictly decreasing for $x<-2$ and $x>-1$.
Now, in interval $(-2,-1)$ i.e., when $-2<x<-1,f^{\prime }\left(x\right)=-6(x+1)(x+2)>0$
$\therefore$ $f$ is strictly increasing for $-2<x<-1.$