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. In this video, we're going to be looking at how we can perform some basic operations with functions like adding and subtracting. Now, 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.

But 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. So let's get right into it. Whenever we add and subtract functions, this is exactly how we add and subtract polynomials together. It's the same process.

For example, if we had these two polynomials being added together, we know that we can just combine like terms to add them. We can see that we have an x squared term, then we have a 5x here, and then we have a 4 and a 7. Now the 4 and the 7 can combine to give us 11. So this is what the fully simplified polynomial would look like. Now we can also take these polynomials and represent them as functions; so if we said f(x) was x² plus 4, and g(x) was 5x plus 7, we could add these together in a similar way, so we would have x² plus 5x plus 4 plus 7 which equals 11, so we would end up with the same polynomial that we did before when adding these as functions.

This is another way to write this same thing. Now likewise, if you saw f(x) minus 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 adding two functions together, you actually need to figure out all of the numbers that are common to both functions. So, you need to find the domain that we have for the individual functions f and g, but then you also need to find them for the combined f+g, or f-g.

This can be a bit of a complicated process, but to really solidify our understanding let's see if we can solve an example. In this example, we are given three different functions up here, and we're asked to complete the operations below and find the domain and range for each new function. So we'll start with f(x) plus g(x). In this case, we need to take the f(x) function and the g(x) function and add them together. So I can see that f(x) is x² plus 1/x, and then I can see that g(x) is x² plus x plus 2.

2 x2 1 x + x + 2

To add these together, we just add the like terms. I see that we have two x² terms, so we're going to have 2x². I only see that we have one 1/x term, so we're just going to have one minus x, then we have 1x there, and then we have a plus 2. Nothing else combines, so this is what would happen when we add f and g together.

Now, for finding the domain of f, which is this function that we have up here. To find the domain of f, we're doing it because we have an x in the denominator of a fraction. Whenever you have x in the denominator, the x value cannot be equal to 0 because you can never divide by 0. So in order to make this function not break, we need to say that the x in the denominator cannot be equal to 0.

For g, we just have a polynomial, and whenever you have these kinds of polynomials, any number that you plug in should work. So for g, we can say that our numbers go from negative infinity to positive infinity because all real numbers are defined. Now as for the domain of f+g, what we need to do is first look at all the individual situations we have, and I can see this one inversed x is the only thing really causing any kind of restrictions here, and I also see that when we add these together it's the same one inversed x that's causing problems. So the domain of f+g is simply going to be that x cannot equal 0.

So this is how you can find the domain and range of the functions individually and then together. But now let's take a look at this other example that we have where we have g(x) minus h(x). Now to do this, we're going to take our function g(x) that we see here, so we'll have this x² plus x plus 2, and then we're going to subtract from this h(x), which is x plus the square root of x minus 8. So to subtract these two functions, what I'm first going to do is take this negative sign and distribute it into the function we have over here, so we're going to end up with x² plus x plus 2, which is this entire thing, and then we have minus x minus the square root of x minus 8, and then what we can do from here is combine like terms.

So I can see that we have this positive x and negative x, which are just going to cancel each other, meaning that our final polynomial is going to be x² plus 2, minus the square root of x minus 8, and since I can see that nothing else is going to combine, this is what our function is going to look like when we subtract h from g.

Now we already discussed that the domain of g is all real numbers, it goes from negative infinity to positive infinity. And as for h, notice that we have a square root right here. The inside of the square root cannot be negative, so what we need to do is say that this x minus 8 that is inside the square root has to be greater than or equal to 0, so that way it's not negative. Now if I go ahead and add this 8 to both sides of the equation, they'll get the eights to cancel there, giving us that x is greater than or equal to positive 8. Meaning that our domain for h is going to be that x has to be greater than or equal to 8.

Now as for the domain of g minus h, we'll if I look at the original expression that we have here, it's only the square root of x minus 8 that's gonna cause restrictions on the domain. And if I look at our final expression, we have the same square root of x minus 8. So it's this same domain that we're going to have for g minus h, meaning that x has to be greater than or equal to 8. So this is the domain of g, the domain of h, and the domain of g minus h.

So this is how you could 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 \( f(x) = \sqrt{x+4} + 30 \), and then we have \( g(x) = \sqrt{x+4} - 2x + 35 \). 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 \( f(x) = \sqrt{x+4} + 30 \).

And I can see here that \( g(x) \), this is \( f(x) \), the \( g(x) \) is equal to \( \sqrt{x+4} - 2x + 35 \). 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 kinda 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 \(\sqrt{x+4}\) and a \(\sqrt{x+4}\), which will give us \(2\sqrt{x+4}\). 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. \(30+35 = 65\). So we're gonna get \(+ 65\), and then we're gonna have this \(-2x\). Although to do this in better order, I'm actually going to write this as \(-2x + 65\). 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, cancelling the 4s there, giving me that \(x\) has to be greater than or equal to \(-4\). So the \(x\) values can equal \(-4\), but they just cannot be less than \(-4\).

So that means that our domain is going to go from \(-4\) 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 the square root of \(x + 4\), plus 30, and this this is going to be minus the \(g(x)\) function, which is the square root of \(x + 4\) minus \(2x + 35\). 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 \(\sqrt{x + 4}\), 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 \(-\sqrt{x + 4}\).

The two negative signs here will cancel giving me \(+2x\), and then the negative sign multiplied by 35 will give me \(-35\). 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. \(30 - 35\) or \(30 + (-35)\) is going to give you \(-5\).

So that means that what we're going to end up with is this \(2x\) by itself, and this is going to be \(-5\). 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 \(-4\). So that means our domain for this situation is still going to be all real numbers from \(-4\) 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 functions 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. Now, 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 the resulting function that you get from this operation. And this process can be a little 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 of x, which is the square root of x, and we have g of x, another function that's 3x minus 6, then multiplying these two functions together would just be the square root of x multiplied by 3x minus 6, and to simplify this I could take the square root of x and distribute it into these parentheses giving us that the product of our functions is 3x times the square root of x minus 6 times the square root of 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.

So in interval notation, we're going to go from 0 to positive infinity. Now as for this other function g of x, notice we have this basic polynomial of 3x minus 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 minus 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 minus 6. Now we already discussed before how the domain of the square root of x is 0 to infinity and the domain of 3x minus 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 of x, that means the g of 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 minus 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. So 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 in this situation down here could also be written as f times g of x. So both of these notations mean the exact same thing, and with division, f of x over g of x could also be written as f over g of 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 of x is equal to x squared minus 4 and g of x is x plus 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 gonna do is take our function for f which is x squared minus 4 and then I'll take our function for g which is x plus 2 and I'll simply multiply these together. I can use the FOIL method here so I'll have x squared times x which will give us x cubed, I'll have x squared times 2 which will give us 2x squared, negative 4 times x which will give us minus 4x, 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 of 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 of x, which is x squared minus 4, and then I'll divide this by g of x which is x plus 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 2 polynomials_that we initially plugged_in are going to have a domain of all real numbers, but notice that when we divide these 2 together we end up_with a restriction because we have an X in the denominator. This denominator X plus 2 cannot be equal to 0, and 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 squared minus 4 is the same thing as x squared minus 2 squared because 2 squared is 4, and the reason I'm writing it like this is because this is a_difference of 2 squares. We're taught that x squared minus 2 squared can also be written as x plus 2 times x minus 2, and_then this is all divided by x plus 2. Now notice how the x plus twos are going to cancel in this equation, meaning all we're going to_end up with is this right here which is x minus 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 plus 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 that 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. 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 two functions together. So the first method is taking two 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 \( f(g(3)) \). 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 \( (x - 1)^2 \) 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 \( x \times (-1) \), which is \(-x\), we'll have \(-1 \times x\), which is also \(-x\) and the \(-1 \times (-1)\) which is \(+1\). 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 \(-2 \times 3 \) and \( 2 \times 3 \) is 6 plus 1. \( 9 - 6 \) is 3, so we end up with 3 + 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 \( 3 - 1 \) because we just replaced this \( x \) with a 3 and \( 3 - 1 \) 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 \( f(x) = x^2 \), 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. $(f∘g)(2)=1$ ; $(g∘f)(2)=1$