Question

Solve each rational function inequality and write the solution in interval notation.

Given the function$\mathrm{R}\left(\mathrm{x}\right)=\frac{(\mathrm{x}+1)}{(\mathrm{x}+3)}$, find the values of x that make the function less than or equal to 0.

Short answer

Solution for inequality is$\mathrm{x}\in (-3,-1].$

Step-by-step solution

Step 1. Given Information

Given that the function is$\mathrm{R}\left(\mathrm{x}\right)=\frac{(\mathrm{x}+1)}{(\mathrm{x}+3)}\le 0.$

Step 2. Critical point

To find the critical point,

$$\begin{gathered} (\mathrm{x}+1)=0 \\ \mathrm{x}=-1 \\ (\mathrm{x}+3)=0 \\ \mathrm{x}=-3 \\ \mathrm{At}\mathrm{x}=-1,\mathrm{R}\left(\mathrm{x}\right)=\frac{0}{2}=0 \\ \mathrm{At}\mathrm{x}=-3,\mathrm{R}\left(\mathrm{x}\right)=\frac{-2}{-0}=\infty \end{gathered}$$

The polynomial function which is undefined will not included in interval.

Step 3. Number line

It can be drawn as,

Step 4. Testing of polynomial expression

On testing the polynomial expression sign in every interval, it can be indicated.

$$\begin{gathered} \frac{(\mathrm{x}+1)}{(\mathrm{x}+3)}\le 0,(-3,-1] \\ \mathrm{Thus},\mathbf{R}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\le }\mathbf{0}\mathbf{}\mathbf{for}\mathbf{}\mathbf{x}\mathbf{\in }\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{]}\mathbf{.} \end{gathered}$$