Function Operations - Online Tutor, Practice Problems & Exam Prep

Understanding operations with functions, such as addition, subtraction, multiplication, and division, is crucial. When adding or subtracting functions, combine like terms and determine the domain by identifying common values. For multiplication, distribute terms and find the domain based on restrictions from both functions. Division requires careful attention to avoid zero in the denominator, leading to additional domain restrictions. Mastering these concepts enhances problem-solving skills in algebra and prepares students for more complex mathematical challenges.

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 do some basic operations with functions like adding and subtracting. Thankfully, when it comes to adding and subtracting functions, the process is pretty straightforward. 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, it's 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 x² term and 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. We can also take these polynomials and represent them as functions so if we said 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 is 11, so we would end up with the same polynomial that we did before when adding these as functions. Now something that you may see though is different notation because rather than seeing f(x) and g(x), it's possible that you'll also see this written as (f+g)(x). This is another way to write this same thing. Likewise, if you saw f(x) - g(x), this could also be written as (f-g)(x). So 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 two functions together, you actually need to figure out all 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.

So 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. To start with f(x) + g(x), we need to take the f(x) function and g(x) function and add them together. I can see that f(x) is x² + 1/x, and then I can see that g(x) is x² + x + 2. To add these together what we can do is just add the like terms. Since we have 2x², we're going to have 2x². I only see that we have one, 1/x. So we're just going to have 1/x, then we have one x there, and then we have a +2. Nothing else combines, so this is what would happen when we add f and g together. Now we're also asked to find the domain of f and the domain of g, and then the domain of f+g. So we'll first find the domain of f, which is going to be this function that we have up here. To find the domain of f, we're looking for what this domain is, and I can see that we have an x _{in the denominator of a fraction. Whenever you have an 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. Now 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 over 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 over x that's causing problems. So the domain of f+g is simply gonna be that x cannot equal 0. So this is how you can find the domain and range of the functions individually and then together.}

Now let's take a look at this other example that we have where we have g(x) - h(x). To do this we're going to take our function g(x) that we see here, so we'll have x² + x + 2 and then we're going to subtract from this h(x) which is x + √(x - 8). So to subtract these two functions what I'll first 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 + x + 2, which is this entire thing, and then we have -x - √(x - 8), and then what we can do from here is combine like terms. 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 + 2 - √(x - 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 - 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 8s 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 - h, well, if I look at the original expression that we have here, it's only the square root of x - 8 that's going to cause restrictions on the domain. And if I look at our final expression, we have the same square root of x - 8. So it's this same domain that we're going to have for g - 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 - 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 fx is equal to the square root of x plus 4 plus 30, and then we have gx is equal to the square root of x plus 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 fx plus gx. So I just need to add the 2 functions above together. I can see here that fx is equal to the square root of x plus 30. And I can see here that gx is f of x, gx is equal to the square root of x plus 4 minus 2x plus 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 of x plus 4 and a square root of x plus 4, which will give us 2 square roots of x plus 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 would 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. And this right here would be the functions when we added them together. So fx plus gx 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 plugged 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 on our domain is the square root function because it is typically the 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 fx, which is the square root of x plus 4, plus 30, and this is going to be minus the gx function, which is the square root of x plus 4 minus 2x plus 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 plus 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 of x plus 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 the square root of x plus 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 going to be minus 5. So this is what the function f-gx 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 plus 4. So because of that, the inside of the square root has to stay positive, meaning that x plus 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 2 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 a 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 gonna 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 gonna 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 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, 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 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. $\frac{L}{M}\left(5\right)=\frac35$