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 3x^2, 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.
Notice that we get different answers when we compose the functions in different directions. This means that you've got to be careful to not do them in the wrong way. I limit my mistakes by always starting by writing down the outside function, and only then do I think about the inside one.
Example #2
There are a few other ways to make these problems slightly more complex. One of those is to compose a function with itself. Maybe r(r(x)): r is the outside function, so we start with -x + 1, but then r is the inside function as well, so where we saw the x, we put another -x + 1. That gives us this: -(-x + 1) + 1, which simplifies down to just plain x.
We can also evaluate a composition of functions at a specific value - maybe like s(s(3)). We start with the sfunction, 2x + 5, substitute in another s function, 2(2x+ 5) + 5, and then substitute a 3 into that (where the x used to be), giving us 2(2(3) + 5) + 5. Now, instead of just simplifying, we multiply and add it out. 2 times 3 is 6 plus 5 is 11 times 2 is 22 plus 5 is 27. So, s(s(3)) is just 27!