Chapter 2.8, Problem 33E

Solution

To solve: The given inequality using sign chart.

The solution set are equal to $\left(-\infty ,-2\right)\cup \left(1,2\right)$ .

Given information:

Consider the given inequality.

  $\frac{x-1}{x^2-4}<0$

Calculation:

Rewrite the given equality below:

  $\begin{array}{c}\frac{x-1}{x^2-2^2}<0\\ \frac{x-1}{\left(x+2\right)\left(x-2\right)}<0\end{array}$

To set numerator and the denominator equal to zero. These are the values that make the quotient zero or undefined and solve for $x$ :

  $\begin{array}{c}x-1=0\\ x=1\end{array}$

And

  $\begin{array}{c}\left(x+2\right)\left(x-2\right)=0\\ x=-2,2\end{array}$

The boundary points are $-2,1,2$ .

Now, locate these boundary points on the number line (or sign chart) and dividing the number line into intervals and check the function $f\left(x\right)$ is positive or negative at each interval below:

  

The boundary points divide the number line into four test intervals below:

  $\left(-\infty ,-2\right),\left(-2,1\right),\left(1,2\right),\left(2,\infty \right)$

Now, take one representative number within each test interval and substitute that number into the inequality below:

  $\begin{array}{l}\left(-\infty ,-2\right),\left(-2,1\right),\left(1,2\right),\left(2,\infty \right)\\ \frac{-3-1}{{\left(-3\right)}^2-4}<0\frac{0-1}{{\left(0\right)}^2-4}<0\frac{\frac{3}{2}-1}{{\left(\frac{3}{2}\right)}^2-4}<0\frac{3-1}{{\left(3\right)}^2-4}<0\\ -\frac{4}{5}<0\text{True }\frac{1}{4}<0\text{False}-\frac{2}{7}<0\text{True}\frac{2}{5}<0\text{False}\end{array}$

Therefore, the intervals $\left(-\infty ,-2\right)$ and $\left(1,2\right)$ belongs to the solution set and the intervals $\left(-2,1\right)$ and $\left(2,\infty \right)$ does not belong to the solution set.

Hence, the solution set are $\left(-\infty ,-2\right)\cup \left(1,2\right)$ .