Rational Functions are just a ratio of two polynomials (expression with constants and/or variables), and are typically thought of as having at least one variable in the denominator (which can never be 0).
Note that we talk about how to graph rationals using their asymptotes in the Graphing Rational Functions, including Asymptotes section. Also, since limits exist with Rational Functions and their asymptotes, limits are discussed here in the Limits and Continuity section. Since factoring is so important in algebra, you may want to revisit it first. Remember that we first learned factoring here in the Solving Quadratics by Factoring and Completing the Square section, and more Advanced Factoring can be found here.
Introducing Rational Expressions
Again, think of a rational expression as a ratio of two polynomials. Here are some examples of expressions that are and are not rational expressions:
Multiplying, Dividing, and Simplifying Rational Functions
Frequently, rational expressions can be simplified by factoring the numerator, denominator, or both, and crossing out factors. They can be multiplied and divided like regular fractions.
Here are some examples. Note that these look really difficult, but we’re just using a lot of steps of things we already know. That’s the fun of math! Also, note in the last example, we are dividing rationals, so we flip the second and multiply.
Remember that when you cross out factors, you can cross out from the top and bottom of the same fraction, or top and bottom from different factors that you are multiplying. You can never cross out two things on top, or two things on bottom.
Also remember that at any point in the problem, when variables are in the denominator, we’ll have domain restrictions, since denominators can’t be $ 0$.
Finding the Common Denominator
When we add or subtract two or more rationals, we need to find the least common denominator (LCD), just like when we add or subtract regular fractions. If the denominators are the same, we can just add the numerators across, leaving the denominators as they are. We then must be sure we can’t do any further factoring:
$ \require{cancel} \displaystyle \frac{2}{{3x}}+\frac{4}{{3x}}=\frac{{(2+4)}}{{3x}}=\frac{{{{{\cancel{6}}}^{2}}}}{{{{{\cancel{3}}}^{1}}x}}=\frac{2}{x};\,\,\,\,x\ne 0$.
Just like with regular fractions, we want to use the factors in the denominators in every fraction, but not repeat them across denominators. When nothing is common, just multiply the factors.
Let’s find the least common denominators for the following denominators (ignore the numerators for now).
Adding and Subtracting Rationals
Now let’s add and subtract the following rational expressions. Note that one way to look at finding the LCD is to multiply the top by what’s missing in the bottom. For example, in the first example, the LCD is $ \left( {x+3} \right)\left( {x+4} \right)$, and we need to multiply the first fraction’s numerator by $ \left( {x+4} \right)$, since that’s missing in the denominator.
Restricted Domains of Rational Functions
As we’ve noticed, since rational functions have variables in denominators, we must make sure that the denominators won’t end up as “0” at any point of solving the problem.
Thus, the domain of $ \displaystyle \frac{{x+1}}{{2x(x-2)(x+3)}}$ is $ \{x:x\ne -3,0,2\}$. This means if we ever get a solution to an equation that contains rational expressions and has variables in the denominator (which they probably will!), we must make sure that none of our answers would make any denominator in that equation “0”. These “answers” that we can’t use are called extraneous solutions. We’ll see this in the first example below.
Solving Rational Equations
When we solve rational equations, we can multiply both sides of the equations by the least common denominator, or LCD (which is $ \displaystyle \frac{{\text{least common denominator}}}{1}$ in fraction form), and not even worry about working with fractions! The denominators will cancel out and we just solve the equation using the numerators. Remember that with quadratics, we need to get everything to one side with 0 on the other side and either factor or use the Quadratic Formula.
Again, think of multiplying the top by what’s missing in the bottom from the LCD.
Notice that sometimes you’ll have to solve literal equations, which just means that you have to solve an equation for a variable, but you’ll have other variables in the answer. The last example shows this.
Rational Inequalities, including Absolute Values
Solving rational inequalities are a little more complicated since we are typically multiplying or dividing by variables, and we don’t know whether these are positive or negative. Remember that we have to change the direction of the inequality when we multiply or divide by negative numbers. When we solve these rational inequalities, our answers will typically be a range of numbers.
Rational Inequalities from a Graph
It’s not too bad to see inequalities of rational functions from a graph. Look at this graph to see where $ y<0$ and $ y\ge 0$. Notice that we have ranges of $ x$ values in the two cases:
Solving Rational Inequalities Algebraically Using a Sign Chart
The easiest way to solve rational inequalities algebraically is using the sign chart method, which we saw here in the Quadratic Inequalities section. A sign chart or sign pattern is simply a number line that is separated into partitions (intervals or regions), with boundary points (called “critical values“) that you get by setting the factors of the rational function, both in the numerator and denominator, to 0 and solving for $ x$.
With sign charts, you can pick any point in between the critical values, and see if the whole function is positive or negative. Then you just pick that interval (or intervals) by looking at the inequality. Generally, if the inequality includes the $ =$ sign, you have a closed bracket, and if it doesn’t, you have an open bracket. But any factor that’s in the denominator must have an open bracket for the values that make it 0, since you can’t have 0 in the denominator.
The first thing you have to do is get everything on the left side (if it isn’t already there) and 0 on the right side, since we can see what intervals make the inequality true. We can only have one term on the left side, so sometimes we have to find a common denominator and combine terms.
Also, it’s a good idea to put open or closed circles on the critical values to remind ourselves if we have inclusive points (inequalities with equal signs, such as $ \le $ and $ \ge $) or exclusive points (inequalities without equal signs, or factors in the denominators).
Let’s do some examples; you can always use your graphing calculator to check your answers, too. Put in both sides of the inequalities and check the zeros, and make sure your ranges are correct!
And here’s another one where we have to do a lot of “organizing” first, including moving terms to one side, and then factoring:
Sometimes we get a funny interval with a single $ x$ value as part of the interval:
And here’s one where we have a removable discontinuity in the rational inequality, so we have to make sure we skip over that point:
Here are a couple that involves solving radical inequalities with absolute values. (You might want to review Solving Absolute Value Equations and Inequalities before continuing on to this topic.)
Here are more complicated ones, where the absolute value may need to be multiplied by other variables (think of if you had to cross multiply). Notice how it’s best to separate the inequality into two separate inequalities: one case when $ x$ is positive, and the other when $ x$ is negative. Notice also how we had to use the Quadratic Formula to get the critical points when $ x$ is negative:
Here’s one more that’s a bit tricky, since we have two expressions with absolute value in it. In this case, we have to separate in four cases, just to be sure we cover all the possibilities.
There’s one more that we did here in the Compositions of Functions, Even and Odd, and Increasing and Decreasing section, when we worked on domains of composites.
Applications of Rationals
There are certain types of word problems that typically use rational expressions. These tend to deal with rates, since rates are typically fractions (such as distance over time). We also see problems dealing with plain fractions or percentages in fraction form.
Here is a rational “fraction” problem:
Here is a rational “percentage” problem:
Distance/Rate/Time Problems
With rational rate problems, we must always remember: $ \text{Distance}=\text{Rate }\times \,\,\text{Time}$. It seems most of the problems deal with comparing times or adding times.
Problem:
Shalini can run 3 miles per hour faster than her sister Meena can walk. If Shalini ran 12 miles in the same time it took Meena to walk 8 miles, what is the speed of each sister in this case?
Solution:
This is a “$ \text{Distance}=\text{Rate }\times \,\,\text{Time}$” problem, and let’s use a table to organize this information like we did in the Algebra Word Problems and Systems of Linear Equations and Word Problems sections. Let $ x=$ Meena’s speed, since Shalini runs faster and it’s easier to add than subtract:
An Upstream/Downstream Rational Problem:
Work Problems:
Work problems typically have to do with different people or things working together and alone, at different rates. Instead of distance, we work with jobs (typically, 1 complete job).
I find that usually the easiest way to work these problems is to remember:
$ \displaystyle \frac{{\operatorname{time} \,\text{together}}}{{\text{time alone}}}\,\,+\,…\,\,+\,\,\frac{{\text{time together}}}{{\text{time alone}}}\,\,=\,\,1$ (or whatever part of the job or jobs is done; if they do half the job, this equals $ \displaystyle \frac{1}{2}$; if they do a job twice, this equals 2).
(Use a “$ +$” between the terms if working towards the same goal, such as painting a room, and “$ -$” if working towards opposite goals, such as filling and emptying a pool.)
“Proof”: For work problems, $ \text{Rate }\times \,\text{Time = }1\text{ Job}$. Add up individuals’ portions of a job with this formula, using the time working with others (time together):
$ \displaystyle \begin{array}{c}\left( {\text{individual rate }\times \text{ time }} \right)\text{+}…\text{+}\left( {\text{individual rate }\times \text{ time }} \right)=1\\\left( {\displaystyle \frac{1}{{\text{individual time – 1 job}}}\,\,\times \,\,\text{time}} \right)\text{+}…\text{+}\left( {\displaystyle \frac{1}{{\text{individual time – 1 job}}}\,\,\times \,\,\text{time}} \right)=1\\\left( {\displaystyle \frac{{\text{time working with others}}}{{\text{individual time – 1 job}}}} \right)\text{+}…\text{+}\left( {\displaystyle \frac{{\text{time working with others}}}{{\text{individual time – 1 job}}}} \right)=1\end{array}$
Also, as explained after the first example below, often you see this formula as $ \displaystyle \frac{\text{1}}{{\text{time alone}}}\,+…+\,\frac{\text{1}}{{\text{time alone}}}\,=\,\frac{\text{1}}{{\text{time together}}}$.
Let’s do some problems:
Note: The formula above can also be derived by using the concept that you can figure out how much of the job the girls do per hour (or whatever the time unit is), both together and alone. Then you can add the individual “rates” to get the “rate” of their painting together.
We are actually adding the Work they complete (together and alone) using formula $ \text{Rate }\times \text{ }\text{Time }=\text{ Work}$, where the Time is 1 hour (or whatever the unit is).
In this example, Erica’s rate per hour is $ \displaystyle \frac{1}{5}$ (she can do $ \displaystyle \frac{1}{5}$ of the job in 1 hour); Rachel’s rate per hour is $ \displaystyle \frac{1}{R}$; we can add their rates to get the rate of their painting together: $ \displaystyle \frac{1}{5}+\frac{1}{R}=\frac{1}{3}$. If we multiply all the terms by 3, we come up with the equation above! (Sometimes you see this as $ \displaystyle \frac{\text{1}}{{\text{time alone}}}\,\,+…+\,\frac{\text{1}}{{\text{time alone}}}\,=\,\frac{\text{1}}{{\text{time together}}}$.)
Here is a systems problem and also a work problem at the same time. We did this problem without using rationals here in the Systems of Linear Equations and Word Problems section (and be careful, since the variables we assigned were different).
Cost Problems
A third type of problem you might get while studying rationals has to do with average cost, or possibly costs per person (or unit cost) problems. Both of these types of costs can still be thought as rates, as they are an amount over time. Here are some examples:
Rational Inequality Word Problem
Here is a rational inequality word problem that we saw in the Algebra Word Problems section.
Note that there is a Rational Asymptote Application Problem here in the Graphing Rational Functions, including Asymptotes section.
Understand these problems, and practice, practice, practice!
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On to Graphing Rational Functions, including Asymptotes – you’re ready!