Combining Functions - Online Tutor, Practice Problems & Exam Prep

Adding & Subtracting Functions

Welcome back, everyone. So up to this point, we've spent a lot of time talking about functions. And in this video, we're going to be looking at how we can do some basic operations with functions like adding and subtracting. Thankfully, when it comes to adding and subtracting functions, the process is pretty straightforward. But what can get confusing is the different types of notation that you're going to see, as well as finding the domain of functions once they've been added or subtracted.

Don't worry about it because in this video, we're going to be going over some different examples and scenarios that will hopefully make this crystal clear. Let's get right into it. Whenever we add and subtract functions, it is exactly how we add and subtract polynomials together. For example, if we had these 2 polynomials being added together, we know that we can just combine like terms to add them.

Here we can see that we have an x² term and then we have a 5x here, and then we have a 4 and a 7. The 4 and the 7 can combine to give us 11. This is what the fully simplified polynomial would look like. We can also take these polynomials and represent them as functions, so if f(x) was x² + 4, and g(x) was 5x + 7, we could add these together in a similar way, so we would have x² + 5x + 4 + 7 which totals 11, so we would end up with the same polynomial as before when adding these as functions. Another way to write this addition could be (f+g)(x).

Similarly, if you saw f(x) - g(x), this could also be written as (f-g)(x). It's just important to watch out for this kind of notation because you may see it show up in this course. Now, when dealing with the domain of adding 2 functions together, you actually need to figure out what all the numbers are that are common to both functions. So you need to find the individual domains of f and g, but then you also need to find them for the combined (f+g), or (f-g).

Let's see if we can solve an example. We are given 3 different functions, and we're asked to complete the operations below and find the domain and range for each new function. Starting with f(x) + g(x), we need to take the f(x) function, x² + 1/x, and the g(x) function, x² + x + 2, and add them together. By adding the like terms, we get 2x² + x + 11/x. To find the domain of f, since we have an x in the denominator, the x value cannot be equal to 0. For g, as it is a polynomial, any real number works, but for (f+g) the domain is all real numbers except x cannot equal 0.

Now let's look at g(x) - h(x). Taking our function g(x), x² + x + 2, and subtracting from it h(x), which is x + √(x-8), we distribute the negative sign and end up with x² + 2 - √(x-8). Since the inside of the square root cannot be negative, x - 8 must be greater than or equal to 0, leading to x ≥ 8 as the domain for h. The domain of g - h will also require x to be greater than or equal to 8.

That is how you can add and subtract functions and find their domain. Hopefully, you found this video helpful, and let me know if you have any questions.

Adding & Subtracting Functions Example 1

Hey, everyone, and welcome back. So let's see if we can solve this problem. In this problem, we have 2 functions. We have \( f(x) \) is equal to the square root of \( x + 4 \) plus 30, and then we have \( g(x) \) is equal to the square root of \( x+4 \) minus \( 2x \) plus 35. We're asked to complete the following operations below and determine the domain of the new functions.

So let's see if we can solve this problem. What I'm first going to do is see if I can find \( f+g \). And to do this, recall that this is the same thing as \( f(x) + g(x) \). So I just need to add the 2 functions above together. I can see here that

f ( x ) = x + 4 + 30

And I can see here that

g ( x ) = x + 4 - 2 x + 35

Now since I don't have any kind of negative signs or numbers that I need to distribute into these parentheses over here, I can kind of just drop the parentheses on all of these functions that we plugged in here. So all I need to do really is combine like terms. Now I see that we have a square root \( x + 4 \) and a square root \( x + 4 \), which will give us 2 square roots of \( x + 4 \). So we have 2 of these here, so we add them together.

And then I also see that we have a positive 30 and a positive 35. 30 plus 35 is 65. So we're going to get plus 65, and then we're going to have this minus \( 2x \). Although to do this in better order, I'm actually going to write this as minus \( 2x \) and then plus 65. Typically, we like to have the \( x \)'s come before the constants.

That's just a general preference thing. And this right here would be the functions when we've added them together. So \( f(x) + g(x) \) would look like this. Now to find the domain, what you have to do is find the combination of the domains from the functions when you initially plug them in. Now looking at these functions when I did the initial plug in, I can really see that the only thing that's going to give us restrictions for our domain is the square root function because typically it's going to be square roots or fractions that you see that will cause restrictions.

And since we have a square root, we need to take this into account. Recall that nothing underneath the square root can be negative. So what we can do is take this whole \( x + 4 \) and say this has to be greater than or equal to 0. Now what I can do from here is solve this mini equation by subtracting 4 on both sides, canceling the 4s there, giving me that \( x \) has to be greater than or equal to negative 4. So the \( x \) values can equal negative 4, but they just cannot be less than negative 4.

So that means that our domain is going to go from negative 4 to positive infinity. And since I see that it's the same quantity we have under the square root, this is the only restriction that I need to take into account, meaning this is the only domain. So this right here is going to be the domain of \( f+g \). And that's how you can solve these types of problems. But what would happen if instead we had \( f - g \) in this situation over here?

Well, to solve this situation, what I'm going to do is the same strategy I used up here, except now I'm going to subtract the 2 functions. So we're going to have

f ( x ) , which is x + 4

plus 30, and this this is going to be minus the

g ( x ) function , which is x + 4 - 2 x + 35 Now from here, what I'm going to do is since I have a negative sign out front, I need to distribute this negative sign into each of these terms. So I can drop the parenthesis on this first thing, this will give me square root of \( x + 4 \), this first quantity I should say, plus 30, and then I can take this negative sign and distribute it into each of these terms. Taking the negative sign and distributing it to here, we'll get minus square root \( x + 4 \). The two negative signs here will cancel giving me positive \( 2x \), and then the negative sign multiplied by 35 will give me minus 35. So this right here is what the function is going to look like when we distribute the negative sign. Now I can see here that because we have a positive and a negative quantity for square root of \( x + 4 \), these are going to cancel each other out. And I can also see that we have a positive 30 and a negative 35. 30 minus 35 or 30 plus negative 35 is going to give you negative 5. So, that means that what we're going to end up with is this \( 2x \) by itself and this is this is going to be minus 5. So this is what the function \( f - g(x) \) looks like. And as for the domain, you may look at this total function and think there are no restrictions on the domain because we only have an \( x \) here, and it's like we got this polynomial which is a linear equation but if you look up here notice that we have to take into account the functions before we simplify them. So when we initially plug things in we have the square root of \( x + 4 \). So because of that, the inside of the square root has to stay positive, meaning the \( x + 4 \) still has to be greater than or equal to 0 since we started with this square root. And if I go ahead and solve this mini equation, well, we already did that up here. We already know \( x \) is going to be greater than or equal to negative 4. So that means our domain for this situation is still going to be all real numbers from negative 4 to positive infinity. So this is going to be our domain when we subtract the two functions. Notice it's the same for when we added them. Even though this came out to a function that didn't have the square root in it anymore, we still needed the same domain because the square root was inside the function when we initially combined them together. So this is how you can do addition and subtraction of функций and find the domain. Hope you found this video helpful, and thanks for watching.

Multiplying & Dividing Functions

Hey, everyone. So, in the last video, we talked about adding and subtracting functions. Now, continuing on this theme of function operations, we're going to see how we can multiply and divide functions in this video. I will warn you that this process can be a bit tedious because when being asked to multiply or divide functions, you'll also likely be asked to find the domain of a resulting function that you get from this operation. This process can be a bit hard to keep track of, but don't worry about it because, in this video, we're going to be looking at some examples and scenarios that will hopefully clear up any confusion surrounding this topic.

So let's get into it. We'll start by taking a look at multiplying functions. So, if we have one function \( f(x) \), which is the square root of \( x \), and we have \( g(x) \), another function that's \( 3x - 6 \), then multiplying these two functions together would just be the square root of \( x \) multiplied by \( 3x - 6 \), and to simplify this I could take the square root of \( x \) and distribute it into these parenthesis giving us that the product of our functions is \( 3x \times \sqrt{x} - 6 \times \sqrt{x} \), and this right here would be our function when multiplying \( f \) and \( g \) together. Now, we should also take a look at how the domain is going to behave because when multiplying functions, the domain is the set of numbers common to the domains of \( f \) and \( g \). Now, we talked about in previous videos how when we have the square root of \( x \), the domain is going to be every positive number.

In interval notation, we're going to go from 0 to positive infinity. Now, as for this other function \( g(x) \), notice we have this basic polynomial of \( 3x - 6 \), and because of this, any number we plug in for \( x \) is going to be fine. There's nothing that's going to break our function. So, we have no domain restrictions, meaning that our domain is simply negative infinity to positive infinity or basically just all real numbers. Now, the domain for \( f \) multiplied by \( g \) is going to be the combination of the domains individually for \( f \) and \( g \).

Since we see that our restriction for the first domain is 0 to infinity and that we have no restrictions for our second domain, then this combination is going to be 0 to infinity because since this is the only restriction we have, the total domain is going to look like this. But now let's take a look at what happens if we divide functions. So, in this case, we have the same two functions square root of \( X \) and \( 3x - 6 \), and if we want to take \( f \) and divide it by \( g \), all we need to do is divide these two functions that we see. So, we're just going to have the square root of \( X \) divided by \( 3x - 6 \). Now we already discussed before how the domain of the square root of \( x \) is 0 to infinity and the domain of \( 3x - 6 \) is all real numbers.

But if you take a look at what happened when we divided these two functions, you may notice there's actually another restriction that we have here. We have this denominator and the denominator of a fraction can never be equal to 0. Since we said that this denominator here is \( g(x) \), that means \( g(x) \) cannot be 0. And the number that would make this denominator 0 is an \( x \) value of 2. So \( x \) cannot be equal to 2 because if you took 2 and replaced it with this \( x \), you'd have \( 3 \times 2 \) which is 6, and \( 6 - 6 \) which is 0 and you cannot divide by 0.

So, the total domain that we have is going to be a combination of the restrictions we have for our 2 functions and the restriction we get from this combined function. So our total domain is going to go from 0 to 2 because we cannot have anything below 0 underneath the square root that we have, but we also cannot have anything that is equal to 2, because that's a restriction as well. So, our domain is going to go from 0 to 2 and then we'll have another interval from 2 to positive infinity. So basically, we can have any real positive number that is not equal to 2. This is how you can find the domain after multiplying and dividing functions.

Now, before moving on to an example there is one thing I want to mention is the notation may look different depending on what problem you have. So, \( f \) multiplied by \( g \) this situation down here could also be written as \( f \times g(x) \). So both of these notations mean the exact same thing, and with division, \( f(x) \) over \( g(x) \) could also be written as \( f\) over \( g(x) \). So this is just something to keep in mind if you ever see this type of notation in problems. But now let's take a look at an example.

So in this example, we are given these two functions. We're given \( f(x) \) is equal to \( x^2 - 4 \) and \( g(x) \) is \( x + 2 \) and we're asked to complete the following operations below and find the domain of each function. So I'm first going to complete this operation which is \( f \) multiplied by \( g \). So what I'm going to do is take our function for \( f \) which is \( x^2 - 4 \) and then I'll take our function for \( g \) which is \( x + 2 \) and I'll simply multiply these together. I can use the foil method here so I'll have \( x^2 \) times \( x \) which will give us \( x^3 \), I'll have \( x^2 \) times 2 which_NC will give us 2\( x^2 \), negative 4 times \( x \) which will give us minus 4\( x \), and then I'll have negative 4 times 2 which will give us a minus 8 \).

Now nothing here is going to combine any farther, so this right here is the simplified expression. So, this is basically our solution for \( f \times g(x) \). Now, to find the domain of this equation, well, what I need to do is look at the domains for \( f \) and \( g \) and figure out how they're going to combine, but notice that we have 2 polynomials here, and for polynomials, all real numbers are going to be in the domain. So because of this, we could say that the total polynomial we got is going to have a domain which is simply all real numbers, so we'll just go from negative infinity to positive infinity. Nothing is going to break this function, so this would be the domain that we have.

But now let's see what happens if we divide these two functions. So I'll take \( f(x) \), which is \( x^2 - 4 \), and then I'll divide this by \( g(x) \), which is \( x + 2 \). Now I can simplify this fraction but before doing that, I'm actually going to find the domain first. We already talked about how these two polynomials that we initially plugged in are going to have a domain of all real numbers, but notice that when we divide these two together, we end up with a restriction because we have an \( X \) in the denominator. This denominator \( X + 2 \) cannot be equal to 0. If I go ahead and solve this little equation we have here, I'll get the \( X \) cannot equal negative 2. So this is going to be a restriction for our domain.

So now that I've found all the restrictions that we have here, I'm gonna go ahead and simplify this. So \( x^2 - 4 \) is the same thing as \( (x^2 - 2^2) \) because \( 2^2 \) is 4, and the reason I'm writing it like this is that this is a difference of 2 squares. We're taught that \( (x^2 - 2^2) \) can also be written as \( (x + 2) \times (x - 2) \), and then this is all divided by \( x + 2 \). Now notice how the \( x + 2 \)s are going to cancel in this equation, meaning all we're going to end up with is this right here, which is \( x - 2 \). So this right here is the solution for the division operation that we had.

Now, notice just by looking at the final expression we got, it seems like there should be no restrictions, but we do have restrictions because we initially had this \( X + 2 \) in the denominator, and this is why I went ahead and found this domain restriction first because whenever you are trying to find the domain of a function after you've done division or multiplication, you always want to determine the domain restrictions before simplifying the functions. So because we have this domain restriction at an \( x \) value of negative 2, we could say that our domain is going to go from negative infinity to negative 2 (not including this value), and then we'll have another interval from negative 2 to positive infinity; and this right here is going to be the domain of our function. So, this is how you can multiply and divide functions and find the domain of the resulting function. Hopefully, you found this video helpful, and thanks for watching.

Given the functions $h\left(x\right)=2x^3-4$ and $k\left(x\right)=x^2+2$, find and fully simplify $h\cdot k\left(x\right)$

• A. $h\cdot k\left(x\right)=2\left(x^5+2x^3-2x^2-4\right)$
• B. $h\cdot k\left(x\right)=2x^5-8$
• C. $h\cdot k\left(x\right)=2x^5+4x^3-8$
• D. $h\cdot k\left(x\right)=x^2+4x+4$

Given the functions $L\left(x\right)=x-2$ and $M\left(x\right)=x^2$, calculate $\frac{L}{M}\left(5\right)$

• A. $\frac{L}{M}\left(5\right)=\frac{25}{3}$
• B. $\frac{L}{M}\left(5\right)=\frac53$
• C. $\frac{L}{M}\left(5\right)=\frac{3}{25}$
• D. $\frac{L}{M}\left(5\right)=\frac35$

Evaluating Composed Functions

Hey, everyone, and welcome back. So in the last video, we talked about compositions of functions. And in this video, we're going to be taking a look at how we can evaluate functions which have been composed. Now, the interesting thing about evaluating composed functions is that there are actually multiple different methods you can use to do this. One of these methods is actually kind of a shortcut method, but it doesn't work every single time.

So without further ado, let's get right into this. So when dealing with composed functions, you'll often be asked to evaluate the function at a certain number. And to do this, one of the methods you can use actually builds off the last video where we composed 2 functions together. So the first method is taking 2 different functions and composing them like we saw in the last video and then taking whatever function we get from this composition and plugging in the number into the final function. So this is one way we can evaluate composed functions, and to understand this let's take a look at an example.

So in this example, we have the function \( f(x) = x^2 \) and \( g(x) = x - 1 \). We're asked to find \( f(g(x)) \) and then evaluate \( f(g(3)) \). So to solve this what I'm first going to do is find \( f(g(x)) \), and this would require me to take our inside function \( g(x) \) and plug it into \( f(x) \). So in this case, since we see that \( g(x) = x - 1 \) and \( f(x) = x^2 \) we're going to end up with \( (x - 1)^2 \) as our composed function. Now I do need to simplify this a bit because we typically want to get our most simplified answer so \( x - 1 \) squared is the same thing as \( x - 1 \) times \( x - 1 \).

Now I can do the foil method where I multiply these x's giving us \( x^2 \), we'll have \( x \) times negative one, which is negative \( x \), we'll have negative one times \( x \), which is also negative \( x \) and the negative one times negative one which is positive one. Now from here I can combine the two negative \( x \)'s giving us \( x^2 - 2x + 1 \). So this right here is what we get when we do the composition \( f(g(x)) \). Now we're asked to take this function here and evaluate it at 3 because we're trying to figure out what \( f(g(3)) \) is. So to find \( f(g(3)) \) notice how I can just take this x that we have here and replace it with 3, meaning all these x's can be replaced with 3.

So we're going to end up having \( 3^2 - 2 \times 3 + 1 \), and \( 3^2 \) is the same thing as 9, and we have minus 2 times 3 and 2 times 3 is 6 plus 1. 9 minus 6 is 3, so we end up with 3 plus 1 which is 4. So this right here is the solution for \( f(g(3)) \), and this is one of the methods we can use for evaluating composed functions. Now you may recall at the start of this video, I mentioned that there's a potential shortcut method we could use, so let's take a look at what this method would look like if we were solving the same problem. So for this method, what you want to do is first figure out what your inside function is and then you want to evaluate that inside function at the number you're looking at.

Once you've found what number this is equal to, you take that number and plug it into your final function to evaluate the composed function. Now that might sound a bit confusing, but let's take a look at this example here and see if we can do this step by step. So we have the same two functions that we had before where we're trying to figure out what \( f(g(3)) \) is. Now rather than finding \( f(g(x)) \) first, instead, I'm going to figure out what \( g(3) \) is first. So \( g(3) \) is going to give us 3 minus 1 because we just replaced this x with a 3 and 3 minus 1 is equal to 2.

So now that we found \( g(3) \), we can evaluate \( f(g(3)) \) because notice that I did this first step where we found \( g \) of our number, so now I just need to evaluate \( f(g) \) of that number. And \( f(g(3)) \), well, since \( g(3) \) is 2, this is the same thing as \( f(2) \). And for \( f(2) \), all I need to do is take 2 and plug it into our function \( f(x) \). So \( f(x) \) is \( x^2 \), meaning that \( f(2) \) would be \( 2^2 \), and \( 2^2 \) is simply equal to 4. So notice when using the second method, we got the same answer that we did for the first method, except we did this much faster.

So the question becomes at this point, why don't we just use this method every single time? Well, the reason for this is because oftentimes when solving these problems, you'll be asked to find \( f(g(x)) \) first before evaluating, and whenever this happens, you have to find the original composition before you can plug in your number. So this is a shortcut method that only works for some problems, but be very wary of whether or not you're asked to find \( f(g(x)) \) first. So this is how you can evaluate composed functions. Hopefully, you found this video helpful.

Let me know if you have any questions.

Given the functions $f(x)=x+3$ and $g(x)= x^2$ find $(f∘g)(2)$ and $(g∘f)(2)$.

• A. $(f∘g)(2)=5$ ; $(g∘f)(2)=25$
• B. $(f\circ g)(2)=7;(g\circ f)(2)=4$
• C. $(f∘g)(2)=7$ ; $(g∘f)(2)=25$
• D. $(f∘g)(2)=1$ ; $(g∘f)(2)=1$