## Precalculus (6th Edition) Blitzer

The solution of rational inequality $\frac{-x-2}{x-4}\ge 0$ is $\left\{ \left[ -2,4 \right) \right\}$ or $\left\{ x|-2\le x<4 \right\}$.
Firstly, write the inequality in the form $f\left( x \right)\ge 0$ The provided expression is already in the form $f\left( x \right)\ge 0$ Here, $f\left( x \right)=\frac{-x-2}{x-4}$ Now, put the numerator and denominator of the rational inequality $f\left( x \right)=\frac{-x-2}{x-4}$ equal to zero. That is, \begin{align} & -x-2=0 \\ & x=-2 \\ \end{align} and \begin{align} & x-4=0 \\ & x=4 \\ \end{align} Now, solve the above expressions for x. Thus, $x=-2,4$ Now, locate the point $x=-2,4$ on the number line. There are three intervals $\left( -\infty ,-2 \right],\left[ -2,4 \right]\text{ and }\left[ 4,\infty \right)$. Now, test any value in the interval and evaluate $f$ at that point. Now, take value $-3$ from the interval $\left( -\infty ,-2 \right]$ Put $-3$ in place of x in the equation of $f\left( x \right)=\frac{-x-2}{x-4}$ , \begin{align} & f\left( -3 \right)=\frac{-\left( -3 \right)-2}{-3-4} \\ & =\frac{3-2}{-3-4} \\ & =\frac{1}{-7} \\ & =-\frac{1}{7} \end{align} Thus, $f\left( x \right)<0$ Now, take value $1$ from the interval $\left[ -2,4 \right]$ Put $1$ in place of x in the equation of $f\left( x \right)=\frac{-x-2}{x-4}$ , \begin{align} & f\left( -1 \right)=\frac{-\left( -1 \right)-2}{-1-4} \\ & =\frac{1-2}{-1-4} \\ & =\frac{-1}{-5} \\ & =\frac{1}{5} \end{align} Thus, $f\left( x \right)>0$ Now, take value $5$ from the interval $\left[ 4,\infty \right)$ Put $5$ in place of x in the equation of $f\left( x \right)=\frac{-x-2}{x-4}$ , \begin{align} & f\left( 5 \right)=\frac{-\left( 5 \right)-2}{5-4} \\ & =\frac{-7}{1} \\ & =-7 \end{align} Thus, $f\left( x \right)<0$ Since it is provided that the function is greater than or equal to zero, include the solution of $f\left( x \right)=0$ , which is obtained on simplifying the numerator. So, the only interval is: $\left[ -2,4 \right)$ or $\left\{ x|-2\le x<4 \right\}$