Chapter 2.8, Problem 68E

Solution

To solve: The problem without using a calculator.

The solution set is equal to all real values except $x\ne -2$ .

Given information:

Consider the given inequality.

  $\frac{1}{{\left(x+2\right)}^2}\ge 0$

Calculation:

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

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

So, the boundary point is $-2$ .

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 two test intervals. The intervals are below:

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

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

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

Therefore, the intervals $\left(-\infty -2\right)$ and $\left(-2,\infty \right)$ belongs to the solution set.

Hence, the solution set is all real values except $x\ne -2$ .