Explainer: Rational Inequalities

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

Recall the definition of a rational function.

Definition: Rational Function

A rational function 𝑓 is one whose formula is a rational expression, that is, a quotient of two polynomial functions. So, 𝑓(𝑥)=𝑃(𝑥)𝑄(𝑥), with 𝑃(𝑥) and 𝑄(𝑥) polynomials.

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

Suppose that we are given an inequality, such as 𝑥+3𝑥8𝑥32.

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 𝑥=3 and (ii) the horizontal line 𝑦=2.

The solution to 𝑓(𝑥)>2 is just the set of 𝑥 values for which the corresponding point (𝑥,𝑓(𝑥)) on the graph lies, strictly, above the line 𝑦=2. Solutions to 𝑓(𝑥)=2 are the 𝑥-coordinates of the points on that line.

From the graph, it looks like the line 𝑦=2 meets the graph in points (2,2) and (1,2), which gives the interval [2,1]={𝑥2𝑥1} as part of the solution.

And then, to the right, we have the part of the curve for 𝑥>3, the asymptote. For all these values, it is also still true that 𝑓(𝑥)2. This adds the interval (3,). So, we can read off the solution to 𝑓(𝑥)2 as the union of intervals: [2,1](3,).

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

  1. Convert the question to a comparison against 0.
    So, instead of 𝑥+3𝑥8𝑥32, we subtract 2 from either side to get the equivalent inequality: 𝑥+3𝑥8𝑥320 or 𝑥+𝑥2𝑥30.
    The solutions here are the same as solutions of the original problem.
  2. 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 (𝑥1)(𝑥+2)𝑥30.
    So, we see that the vertical asymptote is indeed at 𝑥=3. This rational function is zero when 𝑥=2 and 𝑥=1.
  3. 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 𝑔(𝑥)=(𝑥1)(𝑥+2)𝑥3 as above, and we choose a point at random in that interval.
    The listing will be ,2,1,3,. The table, which solves 𝑔(𝑥)>0, is
    Interval𝑥𝑔(𝑥)Sign
    (,2)323
    (2,1)023+
    (1,3)24
    (3,)418+
    The table determines that (𝑥1)(𝑥+2)𝑥3>0𝑥(2,1)(3,).preciselywhen
  4. If the inequality is not strict (as is the case here), append the zeros of 𝑔(𝑥).
    Here, these are 𝑥=2 and 𝑥=1 so that (𝑥1)(𝑥+2)𝑥30𝑥[2,1](3,).preciselywhen

We have found the same solution as suggested by the graph earlier. If we wanted to express the solution set using inequalities, this is 𝑥+3𝑥8𝑥322𝑥1𝑥>3.ifandonlyifor In summary:

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

  1. If the inequality is 𝑃(𝑥)𝑄(𝑥)𝐴 with 𝐴0, replace with 𝑃(𝑥)𝑄(𝑥)𝐴0 instead.
  2. Rewrite the rational function in the form 𝑃(𝑥)𝑄(𝑥)=0 and factorize both 𝑃(𝑥) and 𝑄(𝑥).
  3. We may clear constant factors from 𝑃 and 𝑄, being careful to switch the sign of the inequality if we divide out a negative constant.
  4. Note the zeros of 𝑃(𝑥) that are also the solutions to 𝑃(𝑥)𝑄(𝑥)=0.
  5. List the zeros of the two polynomials in increasing order.
  6. 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 𝑃(𝑥)𝑄(𝑥) there.
  7. 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 𝑥 for which it is true that 𝑥+3𝑥13?

Answer

First, we need to rewrite this inequality using 0 instead of the 3 shown. We transform, simplify, and factor: 𝑥+3𝑥13𝑥+3𝑥1302𝑥+6𝑥102(𝑥3)𝑥10.

Since the factor 2 can be divided out, we do so, being careful to switch the inequality from to . Our inequality has the same solutions as: 𝑥3𝑥10.

We note that 𝑥3𝑥1=0𝑥=3.when

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

We create the table to decide where 𝑥3𝑥1<0:

Interval𝑥𝑥3𝑥1Sign
(,1)03+
(1,3)21
(3,)413+

We find that 𝑥3𝑥1<0𝑥(1,3).when and conclude 𝑥3𝑥10𝑥(1,3].when

In other words, the condition on 𝑥 is 1<𝑥3.

The following is another example.

Example 2: Solving Inequalities of Rational Functions

Solve the inequality 𝑥1(𝑥+1)(𝑥3)13.

Answer

We convert to the form 𝑃(𝑥)𝑄(𝑥)0, suitably factorized, by the sequence of equivalences: 𝑥1(𝑥+1)(𝑥3)13𝑥1(𝑥+1)(𝑥3)130𝑥+5𝑥3(𝑥+1)(𝑥3)0𝑥(𝑥5)3(𝑥+1)(𝑥3)01𝑥(𝑥5)(𝑥+1)(𝑥3)0.thendividingthroughby

This gives us the new rational function 𝑓(𝑥)=𝑥(𝑥5)(𝑥+1)(𝑥3), and we note first that 𝑓(𝑥)=𝑥(𝑥5)(𝑥+1)(𝑥3)=0𝑥=0,5.when

To decide where 𝑓(𝑥)>0, we use the list of zeros of both numerator and denominator. We need to look at the intervals: (,1),(1,0),(0,3),(3,5),(5,).

Then, we look at the table of signs:

Interval𝑥𝑓(𝑥)Sign
(,1)2145+
(1,0)0.51.571
(0,3)22+
(3,5)445
(5,)627+

So, 𝑓(𝑥)=𝑥(𝑥5)(𝑥+1)(𝑥3)>0𝑥(,1)(0,3)(5,).when

Finally, the solution to our inequality is the union with the zeros of 𝑓 above: (,1)[0,3)[5,).

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