Question

Find the complete solution set of the inequality $x(2^x-1)(x+2)(x-3{)}^2\le 0$.

Answer 1

We are given,

$x(2^x-1)(x+2)(x-3{)}^2\le 0$

let all quantity tends to zero

so $x=0$

$2^x-1=0\Rightarrow x=0$

$x+2=0\Rightarrow x=-2$

$(x-3{)}^2=0\Rightarrow x=3,3$

at $x=0,-2$ & $3$ we get zero

Let quantity $=x(2^x-1)(x+2)(x-3{)}^2$

$\left(i\right)$ Let $x>3$

we get, total quantity $>0$

$\left(ii\right)$ Let $x\in (0,3)$

as $x>0,2^x-1>0$ & $(x-3{)}^2>0$

we get, total quantity $>0$

$\left(iii\right)$ Let $x\in (-2,0)$

as $x<0,2^x-1<0,x+2>0$ & $(x-3{)}^2>0$

we get, total quantity $>0$

$\left(iv\right)$ Let $x<-2$

as $x<0,2^x-1<0,x+2<0,(x-3{)}^2>0$

we get, total quantity $<0$

so, $x\in (-\infty ,-2)\cup \{0,3\}$