Lesson Video: Combining Functions Mathematics

In this video, we will learn how to add, subtract, multiply, or divide two given functions to create a new function and how to identify the domain of the new function.

15:46

Video Transcript

In this video, we’ll learn how to add, subtract, multiply, or divide two given functions to create a new function and how to identify the domain of the new function. We are not going to be looking at a function composition, which is a slightly different concept.

Now, combining functions in this manner is actually incredibly intuitive. Let’s say we have two functions 𝑓 of π‘₯ and 𝑔 of π‘₯. And we have the following operations and associated notation. The function 𝑓 plus 𝑔 of π‘₯ is equal to the sum of the functions, 𝑓 of π‘₯ plus 𝑔 of π‘₯. We can say the same for subtraction: 𝑓 minus 𝑔 of π‘₯ is equal to the difference of the functions 𝑓 of π‘₯ minus 𝑔 of π‘₯. Then the function 𝑓𝑔 of π‘₯ is 𝑓 of π‘₯ times 𝑔 of π‘₯. And then 𝑓 over 𝑔 of π‘₯ is 𝑓 of π‘₯ over 𝑔 of π‘₯. And so these are all fairly straightforward.

But we do need to be really careful when combining functions and thinking about the domain of the new function. The domain of the new function will be the intersection or the overlap of the domain of 𝑓 and the domain of 𝑔. In other words, both functions must be defined at a given point for the combination of the functions to also be defined. And, of course, if we’re working with the division of functions, an additional requirement is that the denominator, here that’s 𝑔 of π‘₯, cannot be equal to zero. So, now that we have some of these definitions, let’s see how we might answer a question.

If 𝑓 and 𝑔 are two real functions where 𝑓 of π‘₯ is equal to π‘₯ plus nine over π‘₯ squared plus 15π‘₯ plus 54 and 𝑔 of π‘₯ is equal to π‘₯ plus eight, determine the value of 𝑓 minus 𝑔 of negative six if possible.

Let’s begin by considering what we mean by the combination function 𝑓 minus 𝑔. 𝑓 minus 𝑔 of π‘₯ is equal to the difference between the functions 𝑓 of π‘₯ and 𝑔 of π‘₯; it’s 𝑓 of π‘₯ minus 𝑔 of π‘₯. Given that our functions are defined as 𝑓 of π‘₯ equals π‘₯ plus nine over π‘₯ squared plus 15π‘₯ plus 54 and 𝑔 of π‘₯ is π‘₯ plus eight, we see that 𝑓 minus 𝑔 of π‘₯ here is π‘₯ plus nine over π‘₯ squared plus 15π‘₯ plus 54 minus π‘₯ plus eight. And we could go ahead and simplify this, but actually it’s not entirely necessary since we’re looking to work out whether we can evaluate the function at π‘₯ equals negative six.

And so we need to consider what the domain of our function 𝑓 minus 𝑔 of π‘₯ is. Well, the domain of a combination of functions is equal to the intersection of the domains of the respective functions. So here, we need to find the domain of 𝑓 of π‘₯ and 𝑔 of π‘₯. Firstly, 𝑔 of π‘₯ is the function π‘₯ plus eight, and that’s a polynomial. Now we know the domain of a polynomial is simply the set of real numbers. And so we can say that the domain of 𝑔 of π‘₯ is the set of real numbers; π‘₯ can be any of these. But what about the function 𝑓 of π‘₯? Well, 𝑓 of π‘₯ is the quotient of two polynomials. And when we think about these quotients, we must make sure that the denominator is not equal to zero.

And so what we’re going to do is set the denominator of 𝑓 of π‘₯ equal to zero and solve for π‘₯. And these will tell us the values of π‘₯ that we need to exclude from our function since they will give us a result of zero. So π‘₯ squared plus 15π‘₯ plus 54 equals zero. We’ll factor the expression on the left-hand side. It’s a quadratic, so we’re going to have a pair of parentheses with π‘₯ at the front of each binomial. Then, we want two numbers that multiply to make 54, they have a product of 54, and they add to make 15. Those numbers are six and nine, so our expression factors to π‘₯ plus six times π‘₯ plus nine.

And now we know that since the product of these binomials is equal to zero, either one or other of them must itself be equal to zero. That is, π‘₯ plus six is equal to zero or π‘₯ plus nine is equal to zero. If we solve this first equation by subtracting six from both sides, we get π‘₯ equals negative six. And when we solve the second equation, we get π‘₯ equals negative nine. And so the values of π‘₯ that make the denominator of our function 𝑓 of π‘₯ equal to zero are negative six and negative nine. Now, since 𝑓 of π‘₯ is itself the quotient of two polynomials, we know its domain is the set of real numbers not including the set containing the elements that make the denominator equal to zero.

And so, the domain of 𝑓 of π‘₯ is the set of real numbers minus the set including the elements negative six and negative nine. The domain of our combination function 𝑓 minus 𝑔 of π‘₯ is the intersection of our two domains, the domain of 𝑓 of π‘₯ and 𝑔 of π‘₯. The overlap of the domain of 𝑓 of π‘₯ and 𝑔 of π‘₯ is the set of real numbers minus the set including negative six and negative nine. And so, if we go back to what our question is asking us, to evaluate 𝑓 minus 𝑔 of negative six, we can see that negative six is not in the domain of this function. And so, we can’t actually evaluate it. The function 𝑓 minus 𝑔 of π‘₯ is undefined at that point. And so we say that the value of 𝑓 minus 𝑔 of negative six is undefined.

In our next example, we’ll look at how the inclusion of a root function affects the domain of the sum of two functions.

If 𝑓 and 𝑔 are two real functions where 𝑓 of π‘₯ is π‘₯ squared minus five π‘₯ and 𝑔 of π‘₯ is equal to the square root of π‘₯ plus one, find the domain of the function 𝑓 plus 𝑔.

Firstly, we recall that the combination function 𝑓 plus 𝑔 is simply the sum of the functions 𝑓 and 𝑔. Now, we’re looking to find the domain of this combination function. And so, we recall that the domain of 𝑓 plus 𝑔, that’s the set of inputs that will yield real outputs, is the intersections of the domains of 𝑓 and 𝑔. So, let’s find the domains of 𝑓 and 𝑔. We begin with the function 𝑓 of π‘₯. It’s π‘₯ squared minus five π‘₯. It’s simply a polynomial, and we know that the domain of a polynomial function is the set of real numbers. So, the domain of 𝑓 of π‘₯ is the set of real numbers.

And what about the function 𝑔 of π‘₯? Well, with a root function, we know that to yield a real output, the number inside the square root must be bigger than or equal to zero. In 𝑔 of π‘₯, we have a function inside the square root, so π‘₯ plus one must be bigger than or equal to zero. And this means to find the domain of 𝑔 of π‘₯, we need to solve the inequality π‘₯ plus one is greater than or equal to zero. We’ll do this by subtracting one from both sides. And that tells us that π‘₯ must be greater than or equal to negative one. We can use interval notation to represent the domain of 𝑔 of π‘₯. π‘₯ must be greater than or equal to negative one, so we say that the domain of 𝑔 of π‘₯ is the left-closed, right-open interval from negative one to ∞.

Notice that we can’t really define ∞, hence why we can’t have a square bracket on the right-hand side of this interval. And so we know that the domain of our function 𝑓 plus 𝑔 is the intersection, the overlap, between these two domains. If we consider the domain of 𝑔 of π‘₯ to be a subset of the set of real numbers, then we see that the domain of 𝑓 plus 𝑔, the overlap, is in fact the left-closed, right-open interval from negative one to ∞. And so, that interval, that set of values for π‘₯, is the domain of the function 𝑓 plus 𝑔.

Let’s now consider the product of two polynomials.

Given that 𝑓 sub one maps numbers from the set of positive real numbers onto the set of real numbers such that 𝑓 sub one of π‘₯ is equal to π‘₯ minus four and 𝑓 sub two maps numbers from the left-open, right-closed interval from negative nine to one onto the set of real numbers such that 𝑓 sub two of π‘₯ is equal to five π‘₯ minus two, find 𝑓 one times 𝑓 two of π‘₯ and state its domain.

Firstly, we recall that the product of 𝑓 one and 𝑓 two of π‘₯ is actually equal to the product of the two functions. It’s 𝑓 one of π‘₯ times 𝑓 two of π‘₯. And we know that when we combine functions, the domain of the resulting function will be the intersection of the domains of the respective functions. The only time we need to take into account additional information is if we’re finding the quotient and we need to make sure the denominator is not equal to zero. But of course, that’s not applicable here, so let’s begin by finding 𝑓 one times 𝑓 two of π‘₯. 𝑓 one is π‘₯ minus four and 𝑓 sub two is five π‘₯ minus two, so the product is π‘₯ minus four times five π‘₯ minus two.

Let’s distribute these parentheses by using the FOIL method. We’re going to multiply the first term in each binomial. π‘₯ times five π‘₯ is five π‘₯ squared. Next, we multiply the outer terms, which gives us negative two π‘₯. And we multiply the inner terms, giving us negative 20π‘₯. Finally, we multiply the last term in each binomial. Negative four times negative two is eight. And so we see 𝑓 one times 𝑓 two of π‘₯ is five π‘₯ squared minus 22π‘₯ plus eight. But what is the domain of this function?

Well, both 𝑓 one of π‘₯ and 𝑓 sub two of π‘₯ are, in fact, polynomials, and usually the domain of a polynomial is the set of real numbers. But we’re actually told that 𝑓 sub one maps numbers from the set of positive real numbers. And so that’s the domain of 𝑓 sub one. It’s the positive real numbers. Then we’re told that 𝑓 sub two maps numbers from the left-open, right-closed interval from negative nine to one. In other words, π‘₯ can be greater than negative nine and less than or equal to one. And so, that’s the domain of 𝑓 sub two. The domain of the product of our functions will be the intersection, the overlap, of these two domains. And so, let’s use a number line to establish where this overlap lies.

The domain of 𝑓 sub one is the set of positive real numbers. So that’s any number greater than zero as shown. The domain of 𝑓 sub two is values of π‘₯ greater than negative nine and less than or equal to one. The overlap lies here. It’s values of π‘₯ greater than zero and less than or equal to one. We can represent this domain using interval notation as shown. And so we see that 𝑓 one times 𝑓 two of π‘₯ is five π‘₯ squared minus 22π‘₯ plus eight, and the domain is π‘₯ is an element of the left-open, right-closed interval from zero to one.

We’ll now look at how we need to amend our process somewhat when finding the quotient of two functions.

Given that 𝑓 and 𝑔 are two real functions where 𝑓 of π‘₯ is equal to π‘₯ squared minus one and 𝑔 of π‘₯ is equal to the square root of π‘₯ plus five, find the value of 𝑔 over 𝑓 of negative two if possible.

Firstly, we recall that 𝑔 over 𝑓 of π‘₯ is simply the quotient of the functions. It’s 𝑔 of π‘₯ over 𝑓 of π‘₯. And the reason this question says to find the value of 𝑔 over 𝑓 of negative two β€œif possible” is because when we think about the domain of our combined function, that’s 𝑔 over 𝑓 of π‘₯, we think about the intersection of the domains of 𝑓 and 𝑔. But we want to make sure the denominator, here that’s 𝑓 of π‘₯, is not equal to zero. So, let’s begin by looking for the domains of our respective functions. 𝑓 of π‘₯ is π‘₯ squared minus one. Now that’s just a polynomial, and the domain of a polynomial is simply the set of all real numbers. So, the domain of 𝑓 of π‘₯ is indeed the set of all real numbers.

𝑔 of π‘₯ is a little more complicated though. We have the square root of π‘₯ plus five. And we know that for the square root of a number to be real, that number must be greater than or equal to zero. So here, we know that π‘₯ plus five must be greater than or equal to zero. Let’s solve this inequality by subtracting five from both sides. And when we do, we find that π‘₯ must be greater than or equal to negative five. And therefore, the domain of our function 𝑔 of π‘₯ is π‘₯ is an element of the left-closed, right-open interval from negative five to ∞. And so we move on to finding the domain of 𝑔 over 𝑓 of π‘₯. It’s the intersection of the domains of 𝑓 of π‘₯ and 𝑔 of π‘₯, which is the left-closed, right-open interval from negative five to ∞.

But we need to exclude the values of π‘₯ that make the denominator equal to zero. In other words, we need to exclude the values of π‘₯ that make the expression π‘₯ squared minus one, that’s the function 𝑓 of π‘₯, equal to zero. This time we’ll solve this equation by adding one to both sides, which gives us π‘₯ squared is equal to one. Next, we take the square root of both sides, remembering, of course, to take both the positive and negative square root of one. And when we do, we find π‘₯ is equal to positive or negative one. And so, this means that the domain of 𝑔 over 𝑓 of π‘₯ is the left-closed, right-open interval from negative five to ∞ minus the set including the elements negative one and one.

So, what does this mean for the value of 𝑔 over 𝑓 of negative two? π‘₯ is equal to negative two is within the domain of the function, and therefore it can be evaluated. And so, we’re now going to move on to finding the function 𝑔 over 𝑓 of π‘₯. We take the function 𝑔 of π‘₯, and we divide it by the function 𝑓 of π‘₯, so that’s the square root of π‘₯ plus five over π‘₯ squared minus one. Then, 𝑔 over 𝑓 of negative two is found by replacing π‘₯ with negative two, so we get the square root of negative two plus five over negative two squared minus one, which simplifies to the square root of three over three. And so, given the functions 𝑓 and 𝑔, we’ve shown that the value of 𝑔 over 𝑓 of negative two is the square root of three over three.

We’ll now move on and establish the key points from this lesson. In this video, we saw that given two functions 𝑓 of π‘₯ and 𝑔 of π‘₯, the combined function 𝑓 plus 𝑔 of π‘₯ is the sum of the functions. It’s 𝑓 of π‘₯ plus 𝑔 of π‘₯. We saw that 𝑓 minus 𝑔 of π‘₯ is 𝑓 of π‘₯ minus 𝑔 of π‘₯. And we also saw that 𝑓 times 𝑔 of π‘₯ is simply 𝑓 of π‘₯ times 𝑔 of π‘₯. We learned that 𝑓 over 𝑔 of π‘₯ is 𝑓 of π‘₯ over 𝑔 of π‘₯. Finally, we learned that the domain of each of our combinations is the intersection of the domains of 𝑓 and 𝑔 of π‘₯. In other words, both functions must be defined at a point for the combination to be defined.

We also saw that an additional requirement that applies only to the division of functions is that the denominator cannot be equal to zero. So, the domain of 𝑓 over 𝑔 of π‘₯ is the intersection of the domains of 𝑓 and 𝑔 minus the values of π‘₯ that make the denominator equal to zero.

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