Introduction

Dabin and Melina are having a walking race. Dabin can walk 1 meter per second, but Melina can walk 2 meters per second. Since Melina is the faster walker, she gives Dabin a head start of 5 meters. At this point, we can ask a few questions about the race. Two questions we’ll focus on are “When is Dabin in the lead?” and “When is Melina in the lead?” both of which can be answered by considering inequalities.

To start, let’s define some relevant functions. The function defined by represents how far (in meters) Dabin has walked seconds after the start of the race. Similarly, the function defined by represents how far Melina has walked seconds after the start of the race. Here, asking when Dabin is in the lead is the same as asking for all such that . In the vocabulary that we’ll use, we want to solve the inequality .

Note that we define a solution to an inequality as a set. We will often write the sets in interval notation.

Solving inequalities graphically

Find a solution to the inequality .

Now let’s turn our attention to inequalities involving absolute values, which are often a source of confusion. The following theorem provides the complete story. As you read through the theorem, use the interactive figure below to help you interpret and make sense of each bullet point statement.

In light of what we have developed in this section, we can understand these statements graphically. For instance, if , the graph of is a horizontal line which lies above the -axis through . To solve , we are looking for the values where the graph of is below the graph of . We know that the graphs intersect when , which we know happens when or . Graphing, we get

We see that the graph of is below for between and , and hence we get is equivalent to . The other properties in the theorem can be shown similarly. You can try changing the value of using Desmos.

The above example illustrates a common technique. Rather than considering two functions and and asking when one is greater than, less than, or equal to the other, we can move one function to the other side, and consider the function . Now, the problem becomes one of finding when the function is positive, negative, or zero.