Lesson Video: Domain of Rational Functions Mathematics

In this video, we will learn how to identify the domain of a rational function and the common domain of two or more rational functions.

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Video Transcript

In this video, we’ll learn how to identify the domain of a rational function and the common domain of two or more rational functions. We know that the domain of a function is the set of all possible inputs of that function, whilst the range is all possible outputs after we’ve substituted the domain.

Now, for polynomial functions, the domain is simply the set of real numbers. That means we can substitute any real value of π‘₯ into an equation of the form 𝑓 of π‘₯ equals π‘Ž sub zero plus π‘Ž sub one π‘₯ all the way up to π‘Ž sub 𝑛π‘₯ to the 𝑛th power and the output will be well defined. There are times though where the domain of a function would need to be restricted. And this is particularly important when working with rational functions, that is, a function of the form 𝑓 of π‘₯ equals 𝑝 of π‘₯ over π‘ž of π‘₯, where 𝑝 and π‘ž are polynomial functions and π‘ž of π‘₯ is not the zero polynomial.

Now, that restriction on π‘ž is important, and this is because dividing by zero is undefined. So we don’t want to be dividing a polynomial by zero. And that gives us a hint as to how to find the domain of a rational function. We might recall that when we find the domain of the quotient of two functions, we find the intersection of the domains of the respective functions. But we exclude any values of π‘₯ that make the function on the denominator of the expression equal to zero.

Since the domain of a polynomial is the set of real numbers and a rational function is the quotient of two polynomials then, the domain of a rational function is the set of real numbers excluding any values of π‘₯ that make the denominator equal to zero. We’ll be using this definition throughout the remainder of this video. So, with this in mind, let’s find the domain of a rational function that involves quadratics.

For which values of π‘₯ is the function 𝑛 of π‘₯ equals π‘₯ squared minus 25 over π‘₯ squared minus 12π‘₯ plus 32 not defined?

Let’s begin by inspecting the function 𝑛 of π‘₯. 𝑛 of π‘₯ is the quotient of a pair of polynomials. That is, it is a polynomial function divided by a second polynomial function. In order to identify the values of π‘₯ for which the function is not defined, we’ll begin by considering the domain of a rational function. The domain of a rational function, of course, is the set of values of π‘₯ for which the function is defined. So, if we consider the set of values of π‘₯ for which the function is defined, we’ll be able to quickly identify the values for which it is not defined.

The domain of a rational function is the set of real numbers, but we must exclude any values of π‘₯ that make the denominator of that function equal to zero. This means that our function will be defined over the set of real numbers excluding the set of numbers that make the expression on the denominator, π‘₯ squared minus 12π‘₯ plus 32, equal to zero.

To find such values of π‘₯, we will set the denominator equal to zero and solve, that is, π‘₯ squared minus 12π‘₯ plus 32 equals zero. Since we have a quadratic, we can attempt to solve by first factoring this quadratic expression. We know that we must have an π‘₯ at the beginning of each expression because π‘₯ times π‘₯ gives us the π‘₯ squared. Then, we need to find a pair of numbers whose product is 32 and whose sum is negative 12. Well, negative four times negative eight is positive 32 as required. But negative four plus negative eight is indeed negative 12.

So we rewrite our equation as shown. π‘₯ minus four times π‘₯ minus eight is equal to zero. Then, of course, for the product of these two expressions to be equal to zero, we know that either one or other of those expressions must itself be zero. So the solutions to our equation are given by the solutions to the equations π‘₯ minus four equals zero and π‘₯ minus eight equals zero.

We solve our first equation by adding four to both sides, so we get π‘₯ is equal to four. And we solve our second equation by adding eight to both sides, so π‘₯ is equal to eight. Remember, if we’re thinking about the domain of 𝑛 of π‘₯, we know it’s the set of real numbers minus the set containing the numbers that make the denominator equal to zero. So the domain of our function is the set of real numbers minus the set containing four and eight. Of course, this means our function is defined over this set. And therefore, it must be undefined when π‘₯ is equal to four or π‘₯ is equal to eight. And so we see that the function 𝑛 of π‘₯ is undefined for the set containing four and eight.

We’ll now consider a second example which involves finding the domain of a rational function which is the quotient of a pair of quadratics.

What is the domain of the function 𝑦 equals π‘₯ squared minus one over π‘₯ squared plus one?

Remember, the domain of a function is the set of all possible inputs to that function. And if we inspect our function 𝑦 equals π‘₯ squared minus one over π‘₯ squared plus one carefully, we can see it’s a rational function. That is, it’s the quotient of a pair of polynomials. So we’ll remind ourselves what we know about the domain of a rational function. The domain of a rational function is the set of real numbers, but we exclude any values of π‘₯ that make the denominator of the function equal to zero. In this case then, the function is undefined for any values of π‘₯ which satisfy the equation π‘₯ squared plus one equals zero. To establish which values of π‘₯ this is true for, let’s solve this equation.

We can begin by subtracting one from both sides, so π‘₯ squared is equal to negative one. Then, we would look to take both the positive and negative square root of negative one. But of course the square root of a negative number is not a real number. And since we said the domain of a rational function is just the set of real numbers, there are no values of π‘₯ in this case which are going to make the denominator equal to zero. We can therefore say that the domain of our function and the set of numbers that ensure it is well defined is simply the set of real numbers.

We’ve now considered some examples where we’ve looked at rational functions and calculated the points where they’re not defined and, by extension, their domains. We might also encounter problems where we’re given the domain of a function, and we need to use this to find missing values. Let’s consider an example that uses this idea.

Given that the domain of the function 𝑛 of π‘₯ equals 36 over π‘₯ plus 20 over π‘₯ plus π‘Ž is the set of real numbers minus the set containing negative two, zero, evaluate 𝑛 of three.

𝑛 of π‘₯ is the sum of a pair of rational functions. Both 36 over π‘₯ and 20 over π‘₯ plus π‘Ž are the quotient of a pair of polynomials. We also know that we can find the domain of the sum of a pair of functions by considering the intersection of those domains. So let’s begin by looking at the domains of 36 over π‘₯ and 20 over π‘₯ plus π‘Ž to the domain that we’ve been given, the set of real numbers minus the set containing negative two, zero. This will allow us to find the value of π‘Ž, which will in turn allow us to evaluate 𝑛 of three.

Let’s begin by looking at the expression 36 over π‘₯. Remember, the domain of a rational function is the set of real numbers, but we exclude any values of π‘₯ that make the denominator equal to zero. In this case, the denominator is simply π‘₯. So we set π‘₯ equal to zero, and we see that π‘₯ equals zero is a value that we’re going to exclude from the domain of this function. The domain 𝑛 of 36 over π‘₯ is the set of real numbers minus the set containing zero. We’ll now consider the second rational function. We have 20 over π‘₯ plus π‘Ž. This time, the domain is going to be the set of real numbers excluding any values of π‘₯ that make the denominator π‘₯ plus π‘Ž equal to zero.

So let’s set π‘₯ plus π‘Ž equal to zero and solve for π‘₯. If we do, we find π‘₯ is equal to negative π‘Ž. So we can say that the domain of this second function is the set of real numbers minus the set containing negative π‘Ž. The domain of 𝑛 of π‘₯ then is the intersection of these two domains. The intersection here is the set of real numbers minus the set containing negative π‘Ž, zero. Now, if we compare this to the domain we’ve been given, we can see that negative π‘Ž must be equal to negative two. And if negative π‘Ž is equal to negative two, π‘Ž itself must be equal to two. So we can rewrite 𝑛 of π‘₯ using the value we have for π‘Ž. It’s 36 over π‘₯ plus 20 over π‘₯ plus two.

We’re now ready to evaluate 𝑛 of three, and we can do so by substituting three into this equation. When we do, we get 36 over three plus 20 over three plus two. And that becomes 12 plus four, which is of course equal to 16. So, given information about the domain of our function, 𝑛 of three must be equal to 16.

In this example, we saw that when we added functions, we had to take into account the domain of both functions. Now, a similar process holds if we want to find the common domain of a pair or more of functions. In particular, we can take any number of functions, and the common domain is simply the intersections of the domains of the respective functions. So all we have to do is work out the domain of each function individually and identify the regions where they overlap. And then we can give that in set notation. Let’s look at an example that covers this concept.

Find the common domain between the functions 𝑓 sub one of π‘₯ equals negative nine over π‘₯ plus nine, 𝑓 sub two of π‘₯ equals eight over π‘₯ plus three, and 𝑓 sub three of π‘₯ equals seven π‘₯ over π‘₯ cubed minus four π‘₯.

Remember, given any number of functions, the common domain is the intersection of the domains of the respective functions. In this case then, we need to find the domain of 𝑓 sub one of π‘₯, 𝑓 sub two of π‘₯, and 𝑓 sub three of π‘₯. Then, we can find their intersection. So, next, we remind ourselves how we find the domain of a rational function. The domain of a rational function is the set of real numbers, but we exclude any values of π‘₯ that make the denominator equal to zero. So let’s take function 𝑓 sub one of π‘₯. Its domain is going to be the set of real numbers, but we need to exclude values of π‘₯ that make π‘₯ plus nine equal to zero. To solve for π‘₯, we’ll subtract nine from both sides, and we find the value of π‘₯ that satisfies this equation is negative nine. So the domain of 𝑓 sub one of π‘₯ is the set of real numbers minus the set containing negative nine.

Let’s now consider 𝑓 sub two of π‘₯. This time, we need to exclude values of π‘₯ that make π‘₯ plus three, the denominator of 𝑓 sub two of π‘₯, equal to zero. The value of π‘₯ that satisfies this equation is π‘₯ equals negative three. And so the domain of this function is the set of real numbers minus the set containing negative three. Finally, we’ll move on to 𝑓 sub three of π‘₯. The denominator here is π‘₯ cubed minus four π‘₯. So we know we need to exclude any values of π‘₯ that make this equal to zero. So we have the equation π‘₯ cubed minus four π‘₯ equals zero. And how do we solve it?

Well, we might first look to factor the expression on the left-hand side. The first step to this is to take out a common factor of π‘₯. Then, we can further factor the expression π‘₯ squared minus four using the difference of two squares. So π‘₯ cubed minus four π‘₯ can be written as π‘₯ times π‘₯ plus two times π‘₯ minus two.

The first solution to this equation is when π‘₯ is equal to zero, so it’s π‘₯ equals zero. The second solution is found by setting π‘₯ plus two equal to zero. And when we solve that equation, we get π‘₯ equals negative two. Finally, we solve the equation π‘₯ minus two equals zero, and we get π‘₯ equals two. So, finally, we have found that the domain of 𝑓 sub three of π‘₯ is the set of real numbers minus the set containing these values of π‘₯. The common domain then is the intersection of these three domains. So we’re going to have to take the set of real numbers and exclude the following values of π‘₯: negative nine, negative three, negative two, zero, and two. Hence, the common domain between the three functions we’re given is the set of real numbers minus the set containing negative nine, negative three, negative two, zero, and two.

We’re now going to recap the key points from this lesson. In this lesson, we learnt that a rational function is of the form 𝑝 of π‘₯ over π‘ž of π‘₯. These two functions, 𝑝 of π‘₯ and π‘ž of π‘₯, are themselves polynomials and π‘ž of π‘₯ is not the zero polynomial. With this definition, we can then identify the domain of a rational function. It’s the set of real numbers but we exclude any values of π‘₯ that make the denominator, π‘ž of π‘₯, equal to zero. Finally, we learnt that we can take any number of functions, and then their common domain is simply the intersection of their respective domains.

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