Chapter 2.8, Problem 50E

Solution

To solve: The given inequality using sign chart.

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

Given information:

Consider the given inequality.

  $\frac{1}{x+2}-\frac{1}{x-1}>0$

Calculation:

Rewrite the given inequality:

  $\begin{array}{l}\frac{x-1-x-2}{\left(x+2\right)\left(x-1\right)}>0\\ \frac{-3}{\left(x+2\right)\left(x-1\right)}>0\end{array}$

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

  $\begin{array}{l}\left(x+2\right)\left(x-1\right)=0\\ x=-2,1\end{array}$

The boundary points are $-2,1$ .

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 three test intervals below:

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

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

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

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

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