Question

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

$\frac{1}{2}-\frac{4}{{\mathrm{x}}^2}\le \frac{1}{\mathrm{x}}$

Short answer

Solution in interval notation is$\mathrm{x}\in [-2,0)\cup (0,4].$

Step-by-step solution

Step 1. Given Information

The rational inequality is given as$\frac{1}{2}-\frac{4}{{\mathrm{x}}^2}\le \frac{1}{\mathrm{x}}$.

Step 2. Inequality definition

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

$$\begin{gathered} \frac{1}{2}-\frac{4}{{\mathrm{x}}^2}-\frac{1}{\mathrm{x}}\le \frac{1}{\mathrm{x}}-\frac{1}{\mathrm{x}} \\ \frac{1}{2}-\frac{4}{{\mathrm{x}}^2}-\frac{1}{\mathrm{x}}\le 0 \\ \frac{{\mathrm{x}}^2-8-2\mathrm{x}}{2{\mathrm{x}}^2}\le 0 \end{gathered}$$

To define the inequality, the denominator should not be zero. One of the critical points are $\mathrm{x}=0.$

Step 3. Polynomial function

The required condition for true inequality can be,

$$\begin{gathered} {\mathrm{x}}^2-8-2\mathrm{x}\le 0 \\ {\mathrm{x}}^2-2\mathrm{x}-8\le 0 \\ \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2-2\mathrm{x}-8 \end{gathered}$$

On factorizing the polynomial function using AC method.

$$\begin{gathered} {\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c} \\ {\mathrm{x}}^2-2\mathrm{x}-8\le 0 \\ {\mathrm{x}}^2-4\mathrm{x}+2\mathrm{x}-8\le 0 \\ \mathrm{x}(\mathrm{x}-4)+2(\mathrm{x}-4)\le 0 \\ (\mathrm{x}-4)(\mathrm{x}+2)\le 0 \end{gathered}$$

Step 4. Critical points

The critical point for given polynomial function,

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=0 \\ (\mathrm{x}-4)(\mathrm{x}+2)=0 \\ \mathrm{x}=4,-2 \end{gathered}$$

Critical points are$\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{4}\mathbf{.}$

Step 5. Testing of critical points

The value of $\frac{{\mathrm{x}}^2-8-2\mathrm{x}}{2{\mathrm{x}}^2}$can be tested,

$$\begin{gathered} \mathrm{x}=-3\mathrm{is}\frac{\left(9\right)-8+6}{18}=\frac{7}{18} \\ \mathrm{x}=-1\mathrm{is}\frac{1-8+2}{2}=\frac{-5}{2} \\ \mathrm{x}=1\mathrm{is}\frac{1-8-2}{2}=\frac{-9}{2} \\ \mathrm{x}=5\mathrm{is}\frac{25-8-10}{50}=\frac{6}{50} \end{gathered}$$

Hence,$\mathbf{x}\mathbf{\in }\mathbf{[}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{\cup }\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{4}\mathbf{]}\mathbf{.}$