Question

In the following exercises, solve each rational inequality and write the solution in interval notation.

$\frac{1}{2}-\frac{3}{2{\mathrm{x}}^2}\ge \frac{1}{\mathrm{x}}$

Short answer

Solution for rational inequality is$\mathrm{x}\in (-\infty ,1]\cup [3,\infty ).$

Step-by-step solution

Step 1. Given Information

Given rational inequality is$\frac{1}{2}-\frac{3}{2{\mathrm{x}}^2}\ge \frac{1}{\mathrm{x}}$.

Step 2. Definition of inequality

On subtracting $\frac{1}{\mathrm{x}}$on both sides,

$$\begin{gathered} \frac{1}{2}-\frac{3}{2{\mathrm{x}}^2}-\frac{1}{\mathrm{x}}\ge \frac{1}{\mathrm{x}}-\frac{1}{\mathrm{x}} \\ \frac{1}{2}-\frac{3}{2{\mathrm{x}}^2}-\frac{1}{\mathrm{x}}\ge 0 \\ \frac{{\mathrm{x}}^2-3-2\mathrm{x}}{2{\mathrm{x}}^2}\ge 0 \end{gathered}$$

For the definition of inequality, denominator should not be zero. Thus, one of the critical points is$\mathrm{x}=0.$

Step 3. Required condition

To true the above inequality, the condition can be stated as,

$$\begin{gathered} {\mathrm{x}}^2-3-2\mathrm{x}\ge 0 \\ \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2-2\mathrm{x}-3 \end{gathered}$$

On factorizing the polynomial using AC method.

$$\begin{gathered} {\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c} \\ {\mathrm{x}}^2-3-2\mathrm{x}\ge 0 \\ {\mathrm{x}}^2+\mathrm{x}-3\mathrm{x}-3\ge 0 \\ \mathrm{x}(\mathrm{x}+1)-3(\mathrm{x}+1)\ge 0 \\ (\mathrm{x}+1)(\mathrm{x}+3)\ge 0 \end{gathered}$$

Step 4. Critical points 

To find the critical points,

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=0 \\ {\mathrm{x}}^2-2\mathrm{x}-3=0 \\ (\mathrm{x}+1)(\mathrm{x}-3)=0 \\ \mathrm{x}=-1,3 \end{gathered}$$

Thus, the critical points are$\mathrm{x}=0,-1,3.$

Step 5. Testing of critical points

To test these critical points, value of the $\frac{{\mathrm{x}}^2-3-2\mathrm{x}}{2{\mathrm{x}}^2}$at different points.

$$\begin{gathered} \mathrm{x}=-2\mathrm{is}\frac{4-3+4}{8}=\frac{5}{8} \\ \mathrm{x}=0.5\mathrm{is}\frac{0.25-3-1}{0.5}=-7.5 \\ \mathrm{x}=2\mathrm{is}\frac{4-3-4}{8}=\frac{-3}{8} \\ \mathrm{x}=4\mathrm{is}\frac{16-3-8}{32}=\frac{5}{32} \end{gathered}$$

This satisfies the condition of$\mathrm{x}\in (-\infty ,-1]\cup [3,\infty ).$