In this explainer, we will learn how to solve rational inequalities.

Recall the definition of a rational function.

### Definition: Rational Function

A rational function is one whose formula is a rational expression, that is, a quotient of two polynomial functions. So, with and polynomials.

Since both and are polynomials, we can use methods related to polynomials in handling rational functions.

Suppose that we are given an inequality, such as

As with polynomial inequalities, when the order is greater than one, it is useful to consider what is going on graphically. Even though this is not how we will solve the inequality. Here is the graph:

Included are (i) the vertical asymptote and (ii) the horizontal line .

The solution to is just the set of values for which the corresponding point on the graph lies, strictly, above the line . Solutions to are the -coordinates of the points on that line.

From the graph, it looks like the line meets the graph in points and , which gives the interval as part of the solution.

And then, to the right, we have the part of the curve for , the asymptote. For all these values, it is also still true that . This adds the interval . So, we can read off the solution to as the union of intervals:

How do we find the solution without recourse to the graph? We follow the steps below.

- Convert the question to a comparison against 0.

So, instead of , we subtract 2 from either side to get the equivalent inequality: or

The solutions here are the same as solutions of the original problem. - If possible, factor the rational function to identify the zeros of the
numerator and the zeros of the denominator (which give the locations of the vertical
asymptotes).

This gives us

So, we see that the vertical asymptote is indeed at . This rational function is zero when and . - List the zeros of the numerator and denominator in increasing order and
consider the intervals between them, including to the left and
to the right.

Create a table as shown below, where as above, and we choose a point at random in that interval.

The listing will be The table, which solves , isInterval Sign 0 2 4 18

The table determines that - If the inequality is not strict (as is the case here), append the
zeros of .

Here, these are and so that

We have found the same solution as suggested by the graph earlier. If we wanted to express the solution set using inequalities, this is In summary:

### Solving a Rational Inequality Such As 𝑃(𝑥)/𝑄(𝑥) ≤ 𝐴

- If the inequality is with , replace with instead.
- Rewrite the rational function in the form and factorize both and .
- We may clear constant factors from and , being careful to switch the sign of the inequality if we divide out a negative constant.
- Note the zeros of that are also the solutions to .
- List the zeros of the two polynomials in increasing order.
- Using and , place the set of open intervals created in a table as shown. Choose test values in each such interval to determine the sign of there.
- If the inequality is not strict, add the zeros as boundaries to some intervals to complete the solution.

### Example 1: Solving Inequalities of Rational Functions

What are all the values of for which it is true that ?

### Answer

First, we need to rewrite this inequality using 0 instead of the 3 shown. We transform, simplify, and factor:

Since the factor can be divided out, we do so, being careful to switch the inequality from to . Our inequality has the same solutions as:

We note that

The zeros of the numerator and denominator, taken together, are 3 and 1. Forming intervals from these, taken with , gives us in that order.

We create the table to decide where :

Interval | Sign | ||
---|---|---|---|

0 | 3 | + | |

2 | |||

4 |

We find that and conclude

In other words, the condition on is

The following is another example.

### Example 2: Solving Inequalities of Rational Functions

Solve the inequality .

### Answer

We convert to the form , suitably factorized, by the sequence of equivalences:

This gives us the new rational function , and we note first that

To decide where , we use the list of zeros of both numerator and denominator. We need to look at the intervals:

Then, we look at the table of signs:

Interval | Sign | ||
---|---|---|---|

2 | 2 | + | |

4 | |||

6 |

So,

Finally, the solution to our inequality is the union with the zeros of above: