This section discusses how to graph linear and non-linear inequalities

By their nature, inequalities almost never have a single solution, but rather a set of solutions. As a result, it can sometimes be difficult to determine or display all the solutions - at least in an intuitive way. This is where graphing the solution set can be helpful and that is what we cover here.

Video lecture

_

Text and Supplemental Content

One Dimensional Inequalities

Let’s consider a relatively basic inequality by way of motivation - In particular the inequality .

This inequality is expressing that can be any value at least as big as negative three. This means that is a valid answer, as is , , , and . This is often called a “lower bound”, because it is giving the lowest (or most negative) acceptable value. We can record this solution set in a number of ways, other than the inequality given. For example, interval notation works here, which would be . But our goal here is graphing.

Since there is only one variable, we are going to display this using a one-dimensional graph, i.e. a a number line.

Since is included as a solution (due to the square bracket, the line under the inequality in , and the “at least” part in the phrase “at least as big as”), we start with a filled circle at .

Now, since the values on the positive side (large values) are also solutions (due to the fact that our inequality says values greater than are also solutions by the part of the , as well as the being on the left of the interval notation , and/or the “as big as” part of “at least as big as”), we want to include them in our graph, which we represent by extending a line from the filled circle to the right.

Finally, since the acceptable answers don’t have any upper bound (since there is no value given as a maximum acceptable value) we include an arrow on the right part of the line to indicate that the answer set continues on infinitely far.

Let’s see a few more single variable graphs by way of example, each of which follows the same general process as above:

(a)
, “ is at most ”.

(b)
, “ is less than ”.

(c)
, “ is greater than and at most ”.

Two Dimensional Inequalities

In the one dimensional case, it may seem a bit silly to try and use graphing, since the inequality itself is pretty easy to understand, as is the vocabulary used to explain it. But inequalities with more than one variable - indeed even inequalities with two variables, get markedly more complicated to explain with visuals.

Let’s start with one of the most basic two variable example to demonstrate the process, .

We start by graphing the function as if it were an equality - in this case we graph the function .

Now that we have the function graphed (something that may be a bit of a process all on its own - depending on how complicated the function is), we have effectively cut the - plane into two parts, the part above the function, and the part below the function. One of these parts is the solution set, and one of them is not - the key is to indicate which one is which. To do this, we shade in the part that represents the solution set. To determine which side works we want to return to the original inequality, . This tells us that the values must be less than the values - in other words if we look at a given value, we want to shade in the values that are smaller than that value. A good way to visualize this is to think of a vertical line (remember how the vertical line test represents all the values corresponding to a specific value? We’re doing the same thing!) somewhere along the curve, like this:

Notice that the vertical line for intersects our function at the point , so if we pull out this “segment” out and rotate it ninety degrees to the right to lay it sideways we get:

Where the blue dot represents where the function crossed the vertical line. This numberline now represents all the values for when . But, since we want and we can now shade the correct side of this line to get:

Rotating and putting this vertical line back shows us which side of the original function is being colored:

So, we can see, that we want the values below the function shaded, which we can shade to get the final graph:

The above process is the theoretical way to go about this, but there is an easier way when we are looking at only two (or some finite number) possible regions for our potential solution set. Let’s return to our original graph:

Now, we know either above the line or below the line is our solution set - but importantly either the entire region above or the entire region below satisfies out inequality . Since the entire region will work, we can just test a point of our choice on either side of the function’s graph to see which one satisfies our inequality. We can choose any two points, so let’s choose these:

For the region above the line, we have the point , which corresponds to and . Plugging those values into the inequality gives which is obviously not true. This means that this isn’t the solution region. Since there are only two options, we know that the other one must be the solution region - but it is always a good idea to test the other point to make sure no errors were made.

For the other region we have the point , which corresponds to and . Plugging this into the inequality gives , which is true! This tells us that the region that has this point is our solution region, so we want to shade the side of the line that has the point that worked - in this case the area below the line. This is typically a method that is lot easier than the theoretical method we initially used.

Again, our motivating example doesn’t seem compelling as to why we would go through this much work - after all, the inequality is quite straight forward. The power of this methodology is a bit more impressive when we have a more complicated problem.

To this end, let’s consider the following problem: Graph the solution set to the inequalities and .

As before, we start by graphing the two equalities. This gives us the following:

Now, there are three different approaches we could take:

(a)
We could try the theoretical approach
(b)
We could try testing each region that our functions created on the graph
(c)
We could look at each function individually and figure out what solution set works for each individual function, then see where those solution sets overlap

The first option is a bit messy, especially with multiple inequalities. The second one is a bit messy too, since our region is cut up into several regions, which means we’d need something like the following:

That’s quite a few points to check! Moreover, we no longer know that there is exactly one region that works because we have multiple inequalities graphed at once - which means we must check every point, regardless of when we find a valid region.

So let’s try option 3!

Let’s consider the graph of first.

As before, this function is splitting our region into two pieces, above and below. So let’s consider a point above and below and check! Since we can pick any two points, let’s pick nice points, like these:

The point is equivalent to the claim that and (i.e. ), So plugging these values into the inequality gives: . Since is true, we know this region is indeed part of our solution set. To be safe, we can check the other point, which corresponds to and . Plugging this in gets us and since is not true, this region is not in the solution set. So we have the following solution set for :

We do the same for the inequality, graphing to get:

Then we find two points to test, again picking nice points like these:

Checking these inequalities gives for the point (which works) and for the point (which doesn’t work) so we can graph our solution set as the side that has the point in it:

Now comes the tricky part - we want to graph both of these regions on the same - axis, preferably with different colors so we can see where they overlap. Where they overlap will be our final solution region:

So, now we can restrict to the regions where these two shaded areas overlap to get our final solution set:

We’ve seen that inequalities, especially those involving two or more variables, can generate very complicated solution sets - complicated enough that trying to express them analytically can be quite difficult. Instead, displaying them geometrically by determining the solution region for each inequality, and finding anywhere that all of the individual solution sets overlap. Graphing these regions of complete overlap gives us our solutioon set, as well as a nice graphical representation which is easy to visualize and understand.