Video: Simplifying and Determining the Domain of a Product of Two Rational Functions

Given that 𝑛₁(π‘₯) = ((5π‘₯ βˆ’ 8)/(25π‘₯Β² βˆ’ 4)) Γ· ((25π‘₯Β² βˆ’ 30π‘₯ βˆ’ 16)/(125π‘₯Β³ + 8)), 𝑛₂(π‘₯) = (25π‘₯Β² βˆ’ 4)/(50π‘₯Β² βˆ’ 20π‘₯ + 8), and 𝑛(π‘₯) = 𝑛₁(π‘₯) Γ— 𝑛₂(π‘₯), simplify the function 𝑛, and determine its domain.

08:56

Video Transcript

Given that 𝑛 sub one of π‘₯ is equal to five π‘₯ minus eight over 25π‘₯ squared minus four divided by 25π‘₯ squared minus 30π‘₯ minus 16 over 125π‘₯ cubed plus eight, 𝑛 sub two of π‘₯ is equal to 25π‘₯ squared minus four over 50π‘₯ squared minus 20π‘₯ plus eight, and 𝑛 of π‘₯ is equal to 𝑛 sub one of π‘₯ times 𝑛 sub two of π‘₯, simplify the function 𝑛 and determine its domain.

Ultimately, we’re going to be looking to find the product of the two functions given to us in this question. Before we do, we’re going to need to begin by simplifying our function 𝑛 sub one of π‘₯. 𝑛 sub one of π‘₯ is the quotient of two rational functions. That’s two functions that are fractions. And so we recall that to divide by a fraction, we multiply by the reciprocal of that fraction.

The reciprocal of our second fraction is one over that fraction. But of course, to find the reciprocal, we simply reverse the numerator and the denominator. And so we can rewrite 𝑛 sub one of π‘₯ as five π‘₯ minus eight over 25π‘₯ squared minus four times 125π‘₯ cubed plus eight over 25π‘₯ squared minus 30π‘₯ minus 16. Now, whenever we’re dealing with rational functions, we should always check whether we can simplify by factoring. In fact, this function contains a few expressions that can be factored.

Let’s begin with the expression 25π‘₯ squared minus four. We should see that both 25π‘₯ squared and four are square numbers. So we’re actually dealing with the difference of two squares, meaning that 25π‘₯ squared minus four can be written as five π‘₯ minus two times five π‘₯ plus two.

And what about this expression 25π‘₯ squared minus 30π‘₯ minus 16? Well, this one could be factored as five π‘₯ minus eight times five π‘₯ plus two. In fact, whilst it might not look like it, there is one further factorable expression here. This is the sum of two cubes. And we can factor a sum of two cubes as shown. We say that π‘Ž cubed plus 𝑏 cubed is π‘Ž plus 𝑏 times π‘Ž squared minus π‘Žπ‘ plus 𝑏 squared.

We can see that if we compare our expression to the expression π‘Ž cubed plus 𝑏 cubed, we’ll need to let π‘Ž be equal to five π‘₯ and 𝑏 be equal to two. And so this factors to be five π‘₯ plus two times 25π‘₯ squared minus 10π‘₯ plus four. Now, if you’re unsure where these numbers come from, 25π‘₯ squared is the value of π‘Ž squared. So that’s five π‘₯ squared. Negative 10π‘₯ is negative π‘Žπ‘. And π‘Ž is five π‘₯, and 𝑏 is two. And five π‘₯ times two is 10π‘₯. Then 𝑏 squared is four. And now, we see we can rewrite 𝑛 sub one of π‘₯ as five π‘₯ minus eight over five π‘₯ minus two times five π‘₯ plus two times five π‘₯ plus two times 25π‘₯ squared minus 10π‘₯ plus four over five π‘₯ minus eight times five π‘₯ plus two.

Next, before we multiply, we’re going to cross cancel. We look for common factors in the numerator of the first fraction and the denominator of the second and the denominator of the first fraction and the numerator of the second. We see we have a factor of five π‘₯ plus eight. Then we could actually cancel by any five π‘₯ plus two on our denominator. Now, we multiply the numerators and separately multiply the denominators. And we have an expression for 𝑛 sub one of π‘₯.

Remember, we were told that 𝑛 of π‘₯ is equal to 𝑛 sub one of π‘₯ times 𝑛 sub two of π‘₯. So let’s find the product of the two functions in our question. 𝑛 of π‘₯ is the product of 25π‘₯ squared minus 10π‘₯ plus four over five π‘₯ minus two times five π‘₯ plus two and 25π‘₯ squared minus four over 50π‘₯ squared minus 20π‘₯ plus eight. Once again, we see if we can cross cancel. Well, we know that 25π‘₯ squared minus four is the same as five π‘₯ minus two over five π‘₯ plus two. And so we divide through by this common factor.

Similarly, the denominator of our second fraction can be written as shown. We take out a constant factor of two. And this means we can then divide through by 25π‘₯ squared minus 10π‘₯ plus four. This leaves us with one times one on our numerator and one times two on our denominator. So 𝑛 of π‘₯ is simply one-half.

But we’re not quite finished. We were told to determine the domain of this function. The domain of a function is the set of possible inputs that will yield a real output. When we deal with the product or quotient of two functions, the domain is the intersection of the domains of the two functions. If our resultant function is also a rational function, then we need to ensure that we exclude any values of π‘₯ that make the function in the denominator equal to zero. In this case, though, our function is simply one-half, so there’s no way of making the denominator equal to zero. So we’ll consider the domains of our two original functions, 𝑛 sub one of π‘₯ and 𝑛 sub two of π‘₯.

Let’s begin by looking at the domain of 𝑛 sub one of π‘₯. There are a couple of things we need to consider when looking at the domain. Firstly, if we have any square roots, then we know the number inside that square root must be positive. We don’t actually have any square roots. But what we do have is fractional functions, where the denominator itself could be equal to zero. We don’t want to be dividing by zero, so we’re going to set the denominators equal to zero to see which values of π‘₯ we need to exclude in our domain.

We’ll begin by looking at the denominator of our simplified expression. That’s five π‘₯ minus two times five π‘₯ plus two. We set this equal to zero and find the values of π‘₯ that we want to exclude. Now, for the expression five π‘₯ minus two times five π‘₯ plus two to be equal to zero, either one or other of these expressions must itself be equal to zero. So five π‘₯ minus two is equal to zero or five π‘₯ plus two is equal to zero.

By adding two to both sides of this first equation and then dividing by five, we find π‘₯ is equal to two-fifths. Similarly, we also find π‘₯ is equal to negative two-fifths to be a solution to this equation. And so these are two of the values we’re going to exclude from our domain of 𝑛 sub one of π‘₯. And similarly, since we’re excluding these from the domain of 𝑛 sub one, we exclude them for the domain of 𝑛.

But let’s go back to our unsimplified version of 𝑛 sub one of π‘₯. If we look at that second fraction, we remember that we multiplied by the reciprocal of that fraction. When multiplying by its reciprocal, we had a denominator of 25π‘₯ squared minus 30π‘₯ minus 16. So we’re going to find the values of π‘₯ such that this expression is equal to zero.

When we factored, we got five π‘₯ minus eight times five π‘₯ plus two. And so the two equations that we need to solve will be five π‘₯ minus eight equals zero and five π‘₯ plus two is equal to zero. But of course, this second equation is a duplicate. And so we solve the first equation by adding eight to both sides and dividing by five. And we find π‘₯ is equal to eight-fifths. So currently we have three values of π‘₯ that we cannot include in the domain of our function 𝑛 sub one of π‘₯ and, therefore, in the domain of the function 𝑛.

Of course, we said that the domain of 𝑛 will be the intersection of our two domains. So we need to consider the domain of 𝑛 sub two of π‘₯. In this case, we know it will be all real numbers except those values of π‘₯ such that the denominator is equal to zero. So we want to try and solve this equation. We could begin by dividing through by two to get 25π‘₯ squared minus 10π‘₯ plus four equals zero. The problem is this expression here, 25π‘₯ squared minus 10π‘₯ plus four, is not factorable. One way we can check for this is to look for the discriminant.

The discriminant of an equation in the form π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 equals zero is 𝑏 squared minus four π‘Žπ‘. So here that’s negative 10 squared minus four times 25 times four. That gives us a value of negative 300. Now, negative 300 is less than zero. And this tells us there are no real solutions to our equation 25π‘₯ squared minus 10π‘₯ plus four equals zero. In other words, there are no values of π‘₯ that will make this expression equal to zero.

And so the domain of our second function, 𝑛 sub two of π‘₯, must be the set of real numbers. The intersection of our two domains and, therefore, the domain of the function 𝑛 can be represented using set notation as shown. It’s the set of all real numbers minus the set including the numbers negative two-fifths, two-fifths, and eight-fifths.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.