Lesson Explainer: Rational Inequalities Mathematics • 10th Grade

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(π‘₯)=𝑃(π‘₯)𝑄(π‘₯), with 𝑃(π‘₯) and 𝑄(π‘₯) 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 π‘₯+3π‘₯βˆ’8π‘₯βˆ’3β‰₯2.

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 π‘₯=3 and (ii) the horizontal line 𝑦=2.

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

From the graph, it looks like the line 𝑦=2 meets the graph in points (βˆ’2,2) and (1,2), which gives the interval [βˆ’2,1]={π‘₯βˆ£βˆ’2≀π‘₯≀1} as part of the solution.

And then, to the right, we have the part of the curve for π‘₯>3, the asymptote. For all these values, it is also still true that 𝑓(π‘₯)β‰₯2. This adds the interval ]3,∞[. So, we can read off the solution to 𝑓(π‘₯)β‰₯2 as the union of intervals: [βˆ’2,1]βˆͺ]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 π‘₯+3π‘₯βˆ’8π‘₯βˆ’3β‰₯2, we subtract 2 from either side to get the equivalent inequality: π‘₯+3π‘₯βˆ’8π‘₯βˆ’3βˆ’2β‰₯0 or π‘₯+π‘₯βˆ’2π‘₯βˆ’3β‰₯0.
    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 (π‘₯βˆ’1)(π‘₯+2)π‘₯βˆ’3β‰₯0.
    So, we see that the vertical asymptote is indeed at π‘₯=3. This rational function is zero when π‘₯=βˆ’2 and π‘₯=1.
  3. List the zeros of the numerator and denominator in increasing order and consider the intervals between them, including βˆ’βˆž to the left and ∞ to the right.
    Create a table as shown below, where 𝑔(π‘₯)=(π‘₯βˆ’1)(π‘₯+2)π‘₯βˆ’3 as above, and we choose a point at random in that interval.
    The listing will be βˆ’βˆž,βˆ’2,1,3,∞. The table, which solves 𝑔(π‘₯)>0, is

    The table determines that (π‘₯βˆ’1)(π‘₯+2)π‘₯βˆ’3>0π‘₯∈]βˆ’2,1[βˆͺ]3,∞[.preciselywhen
  4. If the inequality is not strict (as is the case here), append the zeros of 𝑔(π‘₯).
    Here, these are π‘₯=βˆ’2 and π‘₯=1 so that (π‘₯βˆ’1)(π‘₯+2)π‘₯βˆ’3β‰₯0π‘₯∈[βˆ’2,1]βˆͺ]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 π‘₯+3π‘₯βˆ’8π‘₯βˆ’3β‰₯2βˆ’2≀π‘₯≀1π‘₯>3.ifandonlyifor In summary:

Solving a Rational Inequality Such As 𝑃(π‘₯)/𝑄(π‘₯) ≀ 𝐴

  1. If the inequality is 𝑃(π‘₯)𝑄(π‘₯)≀𝐴 with 𝐴≠0, replace with 𝑃(π‘₯)𝑄(π‘₯)βˆ’π΄β‰€0 instead.
  2. Rewrite the rational function in the form 𝑃(π‘₯)𝑄(π‘₯)=0 and factorize both 𝑃(π‘₯) and 𝑄(π‘₯).
  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(π‘₯) that are also the solutions to 𝑃(π‘₯)𝑄(π‘₯)=0.
  5. List the zeros of the two polynomials in increasing order.
  6. Using βˆ’βˆž 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(π‘₯)𝑄(π‘₯) 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 π‘₯ for which it is true that π‘₯+3π‘₯βˆ’1β‰₯3?


First, we need to rewrite this inequality using 0 instead of the 3 shown. We transform, simplify, and factor: π‘₯+3π‘₯βˆ’1β‰₯3π‘₯+3π‘₯βˆ’1βˆ’3β‰₯0βˆ’2π‘₯+6π‘₯βˆ’1β‰₯0βˆ’2(π‘₯βˆ’3)π‘₯βˆ’1β‰₯0.

Since the factor βˆ’2 can be divided out, we do so, being careful to switch the inequality from β‰₯ to ≀. Our inequality has the same solutions as: π‘₯βˆ’3π‘₯βˆ’1≀0.

We note that π‘₯βˆ’3π‘₯βˆ’1=0π‘₯=3.when

The zeros of the numerator and denominator, taken together, are 3 and 1. Forming intervals from these, taken with ±∞, gives us ]βˆ’βˆž,1[,]1,3[,]3,∞[ in that order.

We create the table to decide where π‘₯βˆ’3π‘₯βˆ’1<0:


We find that π‘₯βˆ’3π‘₯βˆ’1<0π‘₯∈]1,3[.when and conclude π‘₯βˆ’3π‘₯βˆ’1≀0π‘₯∈]1,3].when

In other words, the condition on π‘₯ is 1<π‘₯≀3.

The following is another example.

Example 2: Solving Inequalities of Rational Functions

Solve the inequality π‘₯βˆ’1(π‘₯+1)(π‘₯βˆ’3)≀13.


We convert to the form 𝑃(π‘₯)𝑄(π‘₯)≀0, suitably factorized, by the sequence of equivalences: π‘₯βˆ’1(π‘₯+1)(π‘₯βˆ’3)≀13π‘₯βˆ’1(π‘₯+1)(π‘₯βˆ’3)βˆ’13≀0βˆ’π‘₯+5π‘₯3(π‘₯+1)(π‘₯βˆ’3)≀0βˆ’π‘₯(π‘₯βˆ’5)3(π‘₯+1)(π‘₯βˆ’3)≀0βˆ’1π‘₯(π‘₯βˆ’5)(π‘₯+1)(π‘₯βˆ’3)β‰₯0.thendividingthroughby

This gives us the new rational function 𝑓(π‘₯)=π‘₯(π‘₯βˆ’5)(π‘₯+1)(π‘₯βˆ’3), and we note first that 𝑓(π‘₯)=π‘₯(π‘₯βˆ’5)(π‘₯+1)(π‘₯βˆ’3)=0π‘₯=0,5.when

To decide where 𝑓(π‘₯)>0, we use the list of zeros of both numerator and denominator. We need to look at the intervals: ]βˆ’βˆž,βˆ’1[,]βˆ’1,0[,]0,3[,]3,5[,]5,∞[.

Then, we look at the table of signs:


So, 𝑓(π‘₯)=π‘₯(π‘₯βˆ’5)(π‘₯+1)(π‘₯βˆ’3)>0π‘₯∈]βˆ’βˆž,βˆ’1[βˆͺ]0,3[βˆͺ]5,∞[.when

Finally, the solution to our inequality is the union with the zeros of 𝑓 above: ]βˆ’βˆž,βˆ’1[βˆͺ[0,3[βˆͺ[5,∞[.

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