# How to Compose Functions

An error occurred trying to load this video.

Try refreshing the page, or contact customer support.

Coming up next: Inverse Functions

### You're on a roll. Keep up the good work!

Replay
Your next lesson will play in 10 seconds
• 0:06 Function Composition
• 0:34 Function Notation
• 2:04 Composition of Functions
• 3:34 Example #1
• 4:50 Example #2
• 6:06 Lesson Summary

Want to watch this again later?

Timeline
Autoplay
Autoplay
Create an account to start this course today
Try it free for 5 days!

#### Recommended Lessons and Courses for You

Lesson Transcript
Instructor: Luke Winspur

Luke has taught high school algebra and geometry, college calculus, and has a master's degree in education.

Function composition is the process of putting two or more functions together. This video lesson will explain how this process works and also show you how to evaluate functions that have been composed.

## Function Composition

It's really easy for mathematicians to make things seem much harder than they actually are. This often comes down to either confusing vocabulary or confusing notation. While these words or symbols will always have a purpose and will end up making life easier, when you're first learning them it can be hard to keep it all straight.

The topic that this lesson is on, function composition, is one of those topics. It can seem complicated at first, so let's start small and ease you into it.

## Function Notation

We'll begin by reviewing what function notation is.

Basically, it's just another way of writing an equation. Instead of saying y = 4x - 1, we can say f(x) = 4x - 1. This notation now gives this function a name, f, and allows us to substitute anything we want into it.

Instead of f(x), what if it was f(w)? That means f(w) is just 4w - 1.

We don't just have to use symbols, either. How about f(6)? Now we just put a 6 in that spot: 4(6) - 1 = 23.

We could even use random shapes if we want! How about f(:))? I just plug that smiley face right in, which means f(:)) = 4:) - 1.

Let's up the difficulty a little bit. Instead of substituting in a single term, what if we tried an expression with multiple terms? Maybe f(-2m+3)? Just because it's a bigger expression doesn't mean we do anything different. Where there used to be an x (or a smiley, or a 6, or a w), now I put -2m + 3. That gives us this: 4(-2m + 3) - 1, which we can then simplify with the distributive property and combining like terms to end up with our answer: -8m + 11.

## Composition of Functions

So, as you can see, we can substitute any old thing into a function. So, why not another function? That's exactly what a composition of functions is - we take one function and plug it into another one. If we defined another function, let's say g(x) to be 3x2, we can then evaluate f(g(x)) by doing exactly what we have been doing for the last few minutes and just plug one function into another!

We start with the outside function, f: 4 times something - 1, but everywhere that we would normally have put an x, we now substitute in the function g(x). So instead of 4x - 1, or 4w- 1, or 4 :) - 1, we have 4(g(x)) - 1. But since we know that g(x) is just 3x2, we can substitute that in as well, which makes f(g(x)) equal to 4(3x2) - 1. Simplifying again gives us our final answer as 12x2 - 1.

And that's it! But composing functions can be difficult because seeing all those letters - f and g and x - can be daunting. Even when you get that part, it can be easy to do the problem backwards and substitute the functions into each other the wrong way. So, let's look at an example or two, and see if we can address those two common mistakes and prevent them from happening to you.

## Example #1

Let's set up some new functions - maybe r(x) = -x + 1 and s(x) = 2x + 5 - and run through the different ways we could compose them.

How about r(s(x))? Well, r is the outside function, so we start with that: negative something plus 1. But instead of an x, we're substituting in s(x). That turns what we have, -x + 1, into -(2x + 5) + 1. Again, distributing and simplifying gives us r(s(x)) = -2x- 4.

How about the other way: s(r(x))? This time the outside function is s, which means we'll start with 2x + 5, but then substitute the r function where the x used to be. That gives us 2(-x + 1) + 5, and our simplified answer is -2x + 7.

To unlock this lesson you must be a Study.com Member.

### Register for a free trial

Are you a student or a teacher?
Back

Back

### Earning College Credit

Did you know… We have over 95 college courses that prepare you to earn credit by exam that is accepted by over 2,000 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.