Question

let $f\left(x\right)=\frac{{(x-1)}^3{(x+2)}^4{(x-3)}^5(x+6)}{x^2{(x-7)}^3}$ solve the following inequilaties :
(1) $f\left(x\right)>0$
(2) $f\left(x\right)⩾0$
(3) $f\left(x\right)<0$
(4) $f\left(x\right)⩽0$

Answer 1

$f\left(x\right)=\frac{{(x-1)}^3{(x+2)}^4{(x-3)}^5(x+6)}{x^2{(x-7)}^3}$

The critical points are $-6,-2,0,1,3,7$

Plotting these points on a number line in ascending order.

Changing the sign from right as shown in the number line.

The sign of the positive powers must be repeated.

$\left(1\right)f\left(x\right)>0$

Considering the positive part on the number line,

$\Rightarrow x\in (-\infty ,-6)\cup (1,3)\cup (7,\infty )$

$\left(2\right)f\left(x\right)\ge 0$

Considering the positive part on the number line,

$\Rightarrow x\in (-\infty ,-6]\cup [1,3]\cup [7,\infty )$

$\left(3\right)f\left(x\right)<0$

Considering the negative part on the number line,

$\Rightarrow x\in (-6,-2)\cup (-2,0)\cup (0,1)\cup (3,7)$

$\therefore x\in (-6,1)\cup (3,7)$

$\left(4\right)f\left(x\right)\le 0$

Considering the negative part on the number line,

$\Rightarrow x\in [-6,-2]\cup (-2,0]\cup (0,1]\cup [3,7]$

$\therefore x\in (-6,1]\cup [3,7]$