Chapter 1.6, Problem 25PPS

To determine

To express: the given graph as absolute value inequality.

Answer to Problem 25PPS

In terms of absolute value inequality the graph can be expressed as $\left|x+9\right|\le 3$ .

Explanation of Solution

Given information:

A graph is given as

Concept used:

Absolute value of a function a is given by

  $\left|a\right|=\{\begin{array}{l}aa>0\\ -aa<0\\ 0a=0\end{array}$

And

  $\begin{array}{l}\left|-a\right|=a\\ \left|a\right|=a\end{array}$

More over

  $\begin{array}{l}\left|x+a\right|\le b\\ \Rightarrow -b\le x+a\le b\end{array}$

If the terminal number or point is included in the interval it can be represented by closed ball on the number line otherwise it will be represented by an open ball.

Calculation:

Consider the given graph.

Here, for both $-12$ and $-6$ closed balls are placed on the number line, so both will be included to the interval.

So, the interval will be

  $-12\le x\le -6$

Now, express the interval as an absolute value inequality as shown:

  $\begin{array}{c}-12\le x\le -6\\ -12+9\le x+9\le -6+9\\ -3\le x+9\le 3\end{array}$

So, the absolute value inequality the graph can be expressed as $\left|x+9\right|\le 3$ .