Function Composition - Online Tutor, Practice Problems & Exam Prep

Function composition involves evaluating one function within another, such as finding \( f(g(x)) \). For example, if \( f(x) = x^2 + 3x - 10 \) and \( g(x) = x - 2 \), then \( f(g(x)) = (x - 2)^2 + 3(x - 2) - 10 \). Evaluating composed functions can be done by first finding the inner function's value and then substituting it into the outer function. Additionally, understanding the domain of composed functions is crucial, as it requires identifying restrictions from both functions involved. Function decomposition reverses this process, allowing for creative expressions of functions as compositions.

Function Composition

Hey, everyone, and welcome back. So we've been talking about function operations in recent videos. In this video, we're going to be looking at function compositions. When it comes to the composition of functions, this process can often be a bit difficult when you first encounter it. But in this video, we're going to take a look at some examples of things that we've already seen up to this point and see how we can build off of this to really understand this concept of composing a function. So let's get right into this.

When it comes to a function composition, it's like evaluating a function at a certain number, but instead of replacing the inside of the function with a variable, you replace the function with another function. Before getting into composition, let's remind ourselves how we can evaluate functions at certain values. For instance, we have the function fx=x2+3x-10 and we're asked to evaluate this at f7. In this case, we're going to take all of our x's and replace them with 7. So x2 will become 72 and then we'll have 3×7 minus 10. Now 72 is 49 and then this is going to be plus 3 times 7 which is 21 minus 10 and if you go ahead and combine all these numbers here you should get 60. So when you evaluate a function at a certain number, you're going to end up getting a number as your final result.

Now when it comes to composing a function, it's very similar to this process of evaluating a function except rather than replacing the x with a number we're going to replace x with another function. So in this case we have fx=x2+3x-10 and then we have gx=x-2, and if we want to find fgx, notice we have gx on the inside, so that means we take all of these x's and we replace them with gx. So in this case, we're going to have fgx=(x-2)2+3×(x-2)-10. I won't bore you with the details of trying to simplify and foil everything out, but if you were to simplify this entire expression you should end up with x2-x-12. So this is what we get when we simplify fgx, and notice that the result is a function whenever we're doing a function composition. This is the main idea when composing a function.

Now let's see at this point if we can solve an example. In this example we have the function fx=x+4 and gx=x2-3. We're asked to find the following composite functions and fully simplify our answer. In this first situation we have fgx. What this means is we need to put gx inside fx. So if we do this, x is going to be replaced with what we have here which is

Given the functions $f\left(x\right)=\sqrt{x+4}$ and $g\left(x\right)=\left(x-2\right)^2-4$ find $\left(f\circ g\right)\left(x\right)$ and $\left(g\circ f\right)\left(x\right)$

• A. $\left(f\circ g\right)\left(x\right)=\sqrt{x-2}$ ; $\left(g\circ f\right)\left(x\right)=\left(x+4\right)-4\sqrt{x+4}$
• B. $\left(f\circ g\right)\left(x\right)=x-2$ ; $\left(g\circ f\right)\left(x\right)=x\left(x+4\right)$
• C. $\left(f\circ g\right)\left(x\right)=x-2$ ; $\left(g\circ f\right)\left(x\right)=4\sqrt{x-4}$
• D. $\left(f\circ g\right)\left(x\right)=x-2$ ; $\left(g\circ f\right)\left(x\right)=\left(x+4\right)-4\sqrt{x+4}$

Given the functions $f(x)=\frac{1}{x^2-2}$ and $g(x)=\sqrt{x+2}$ find $(f∘g)(x)$ and $(g\circ f)(x)$.

• A. $(f\circ g)(x)=\frac{1}{x}$ ; $(g\circ f)(x)=\sqrt{\frac{2x^2-3}{x^2-2}}$
• B. $(f\circ g)(x)=\frac{1}{x}$ ; $(g\circ f)(x)=\sqrt{\frac{3}{x^2-2}}$
• C. $(f∘g)(x)=x$ ; $(g\circ f)(x)=\sqrt{x^2-2}$
• D. $(f∘g)(x)=x$ ; $(g\circ f)(x)=\frac{1}{\sqrt{x+2}-2}$

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. 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. 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. This is one way we can evaluate composed functions, and to understand this, let's take a look at an example.

In this example, we have the function f(x)=x2 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)=x2 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 our most simplified answers, so (x-1)2 is the same thing as (x-1)(x-1). Now I can do the FOIL method where I multiply these x's giving us x2, we'll have x−1 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 x2−2x+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 squared minus 2 times 3 plus 1, and 3 squared is the same thing as 9, and we have minus 2 times 3, and 2 times 3 is 6, 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.

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 replace 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(3)). 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 x2 meaning that f(2) would be 22, and 2 squared 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. 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. $(f∘g)(2)=1$ ; $(g∘f)(2)=1$

Domain Restrictions of Composed Functions

Hey, everyone, and welcome back. So in the previous video, we talked about how we could evaluate composed functions. Now in this video, we're going to be taking a look at finding the domain of functions which have been composed. Now this process can often be a bit tricky because it can be difficult to recognize how the domain of the functions will behave when they're separate versus how the domain acts when the functions are composed together. But don't worry about it because in this video, we're going to be going over an example that I think will really clear this entire concept up. So without further ado, let's get right into this.

Now when finding the domain of composed functions, there are a few steps you can follow, and I think it's best to see these steps by looking at an example. So given below is an example where we have 2 functions. We have f(x)=1x-2 and g(x)=x. We're asked to determine the composition f∘g(x) and find its domain. Now keep in mind that this f∘g(x) that we see here is the same thing as f∘g(x), where g is the inside function.

Now your first step with dealing with these types of problems is to find the x values not defined for g(x). So what you first want to do is find any restrictions on the inside function. Now if I look at our g function, we can see here that g(x), the inside function, is equal to x. Now recall that we've talked about in previous videos how the square root of x does have restrictions, because we cannot have a negative number underneath the square root. You cannot take the square root of something negative. So in this case, we would say that x has to be greater than or equal to 0 because x cannot be negative, so this would be the restrictions for our function g.

Now once you've found these restrictions, your next step should be to find any x values that make g(x) not defined for f(x). And if you're not sure what this means, basically, what this is saying is if we take g and we put it inside of f, what restrictions are we going to have on that entire function when we compose the two together? So to figure this out, what I'm going to do is actually compose these functions. So we'll have f∘g(x). Now in doing this step, what I can do is take our function g and I can place it inside of the function f. So I'll just replace this x with the function we have here. So rather than having one over x minus 2, our function f∘g(x) is going to be 1x-2, because the square root of x is this function g.

Now we already determined the restrictions for the square root of x, which is that x has to be positive, but notice that when we plug g into f, we get this new fraction that we see here, this new kind of equation, and when we do this, we see that for a fraction, the denominator cannot equal 0. That's another restriction that we have. So we can say that this whole denominator, x-2, cannot be equal to 0. Now what I can do from here is solve for x in this mini equation by adding 2 on both sides; that'll get the positive and negative 2 to cancel, giving me that the square root of x cannot equal 2. From here, I can take both sides of this equation and square it. This will allow me to cancel the square and the square root, giving me that x cannot equal 22, which is 4. So this right here is the restriction for our function f∘g(x).

And notice at this point, we have found the restriction for g, which was the first step, and we found the restriction for f∘g(x), which was the second step. Since we have found these two restrictions, we can now write out what our domain is. So the domain for our composed function f∘g(x) is that we can go from 0, and we are including the 0 because x can be greater than or equal to 0, so we can go from 0 all the way up to 4, not including 4, and then we can go from 4 to infinity. So basically, we could have any number as long as it's above 0 and not equal to 4. So this right here would be the domain of our function f∘g(x). And notice how we found first the domain for g, then we found f∘g(x) domain, and then we combine these together to get the total domain. So this is how you can find the domain of composed functions. Hopefully, you found this video helpful, and thanks for watching.

Given the functions $f(x) = x^2$ and $g(x)=\sqrt{x-8}$ find $(f∘g)(x)$ and determine its domain.

• A. $(f∘g)(x)=x-8$ ; $Dom:(-\infty,\infty)$
• B. $(f\circ g)(x)=\sqrt{x^2-8}$ ; $Dom:(-∞,∞)$
• C. $(f∘g)(x)=x-8$ ; $Dom:[8,∞)$
• D. $(f\circ g)(x)=\sqrt{x^2-8}$ ; $Dom:[8,∞)$

Decomposition of Functions

Hey, everyone, and welcome back. So over the course of the past couple of videos, we've been talking about the composition of functions. Now, in this video, we're going to be taking a look at function decomposition. Decomposition is basically just the reverse of function composition, and this process can often be a bit tricky because it's hard to know when reversing this process what the functions are going to look like. We're used to starting with 2 functions and composing them into one thing, but if we start with the composed function and go backward, that step can be a bit tricky. But don't worry about it, because in this video, we're going to look at some strategies you can use to do this, and you might find that after doing this a few times, this process can be kind of fun because you can be creative with your answers here. Let's get right into this. There are many correct answers when it comes to decomposing a function. To understand this best, I think it's overall best to look at an example. In this example, we're told to express the function \( h(x) \) as:

h ( x ) = x - 2

as a composition of 2 functions \( f \) and \( g \), so that \( h \) is equal to \( f(g(x)) \). This means that the function \( h(x) \) could also be written as \( f(g(x)) \), and we are asked to find what the individual functions would look like before this composition. We need to figure out what our functions \( f(x) \) and \( g(x) \) are going to look like. One common strategy for doing this is to look at your inside function and set that equal to whatever is inside an operation. For example, if you had a square root, you could set \( g(x) \) equal to what's inside the square root, which would be \( x - 2 \). So, in this case, we could say that \( g(x) \) is equal to \( x - 2 \), and then \( f(x) \) would be the square root, which is on the outside. However, we can't just write a square root by itself because that doesn't mean anything, so we have to make this: f ( x ) = x

This is one of the ways that we could decompose this function into \( f(x) \) and \( g(x) \), and this is going to be one of the most common strategies that you see. We know that this is correct, because if you were to take the function \( g(x) \) and plug it into \( f(x) \), \( \sqrt{x - 2} \) is what we started with, so this is an accurate decomposition.

However, I mentioned that there are many correct ways that we could decompose functions. Another strategy we could use is to set \( g(x) \) equal to the entire thing, \( \sqrt{x - 2} \), and in this case, \( f(x) \) would just be equal to \( x \), because notice if we took this function and plugged it in for \( x \), we would once again get \( \sqrt{x - 2} \). This might seem too simple and a bit ridiculous, but this is technically a completely correct answer also because if we recompose these functions, we get what we started with. In fact, when you get really good at this, you can try doing some really creative things. So let's say we start again with our original function \( f(g(x)) \), which is always going to be \( \sqrt{x - 2} \) for this example, you could make it something ridiculous. You could say that \( g(x) \) is equal to \( \sqrt{x - 2} - 1000 \), and in this case, \( f(x) \), the original function, would be \( x + 1000 \). Because notice, if you were to take this entire thing and plug it in there, the \( 1000 \) would cancel, and you would just be left with \( \sqrt{x - 2} \). This is another weird but perfectly okay strategy that you could take for decomposing these functions. Once you get good at this, you can get creative with the various decompositions you do. All of these solutions that we see here are perfectly acceptable ways to decompose this function. Some of these are a little ridiculous. The most common case that you'll see is the one that we have over here, but either way, this is how you can do basic function decomposition. Hopefully, you found this video helpful. Thanks for watching.

Decomposition of Functions Example 1

So let's see how we can solve this problem. In this problem, we are given the function \( h(x) \) is equal to \( \frac{1}{x^2 + 3x - 10} \). We're told that this function is a composition of two functions \( f \) and \( g \), and to express these functions such that \( h(x) \) is equal to \( f(g(x)) \). So basically what this problem is asking us is to find individual functions \( f(x) \) and \( g(x) \) that when composed would give us this function \( h(x) \). So to figure this out, I'm actually going to write this \( h(x) \) as a composition of these two functions. So we're going to say that \( f(g(x)) \), where \( g \) would be the inside function here, is going to equal \( \frac{1}{x^2 + 3x - 10} \). Now, in order to decompose this function, we need to figure out what \( f(x) \) and \( g(x) \) are going to be. Well, to do this, I can think of some examples that I could do to split these into two functions. And one thing that I recognize is this is a fraction, and with a fraction, I can see that the denominator is the inside function and that the entire thing would be the outside function so what we would end up having is \( g(x) \) would equal everything in this denominator so we'd have \( x^2 + 3x - 10 \), this would be the \( g(x) \) function. And then for \( f(x) \), we would say that \( f(x) \) is everything that we did not set equal to \( g \). So in this case, we would have the one over this entire thing which we said was \( g \), so in this case, we just say this is \( x \). So this would be the functions for \( g \) and \( f(x) \) if we decompose this, and notice if you were to take this function and plug it back into \( f(x) \), we would get this exact same thing that we had over here. So this right here would be the solution for decomposing this function. Now there are other different things you could try with this. We discussed in the video on decompositions that you can get creative if you want to, but I think this is one of the most intuitive ways to decompose this function. So hopefully you found this helpful. Thanks for watching. Let's move on.