Be The Professor (Composite Functions)

Hello my name is Sara Sindi and I will be explaining the concepts and some examples of composite functions.


Until now, given a function  f(x), you would plug a number or another variable in for x. You could even get fancy and plug in an entire expression for x. For example, given  f(x) = 2x + 3, you could find f(y² – 1) by plugging y² – 1 in for x to get f(y² – 1) = 2(y² – 1) + 3 = 2y² – 2 + 3 = 2y² + 1.
In function composition, you're plugging entire functions in for the x. In other words, you're always getting "fancy". But let's start simple. Instead of dealing with functions as formulas, let's deal with functions as sets of (x, y) points:

        Let f = {(–2, 3), (–1, 1), (0, 0), (1, –1), (2, –3)} and

• let g = {(–3, 1), (–1, –2), (0, 2), (2, 2), (3, 1)}.
  
  Find (i) f (1), (ii) g(–1), and (iii) (g o f )(1).

(i) This type of  exercise is meant to emphasize that the (x, y) points are really (x, f (x)) points. To find  f (1), I need to find the (x, y) point in the set of (x, f (x)) points that has a first coordinate of x = 1. Then f (1) is the y-value of that point. In this case, the point with x = 1 is(1, –1), so:

f (1) = –1

(ii) The point in the g(x) set of point withx = –1 is the point (–1, –2), so:

g(–1) = –2

(iii) What is "(g o f )(1)"? This is read as "g-compose-f of 1", and means "plug 1into f, evaluate, and then plug the result into g". The computation can feel a lot easier if I use the following, more intuitive, formatting:


             (g o f )(1) = g( f(1))


Now I'll work in steps, keeping in mind that, while I may be used to doing things from the left to the right (because that's how we read), composition works from the right to the left (or, if you prefer, from the inside out). So I'll start with the x = 1. I am plugging this into f(x), so I look in the set of f(x) points for a point with x = 1. The point is (1, –1). This tells me that f(1) = –1, so now I have:  

(g o f )(1) = g( f(1)) = g(–1)


Working from the right back toward the left, I am now plugging x = –1 (from "f(1) = –1") intog(x), so I look in the set of g(x) points for a point with x = –1. That point is (–1, –2). This tells me that g(–1) = –2, so now I have my answer:

(g o f )(1) = g( f(1)) = g(–1) = –2


Note that they never told us what were the formulas, if any, for f(x) or g(x); we were only given a list of points. But this list was sufficient for answering the question, as long as we keep track of our x- and y-values.


• Let f = {(–2, 3), (–1, 1), (0, 0), (1, –1), (2, –3)} and
  let g = {(–3, 1), (–1, –2), (0, 2), (2, 2), (3, 1)}.  
• Find (i) ( f o g)(0), (ii) ( f o g)(–1), and (iii) (g o f )(–1).
(i) To find ( f o g)(0), ("f-compose- g of zero"), I'll rewrite the expression as:

( f o g)(0) = f(g(0))


This tells me that I'm going to plug zero into g(x), simplify, and then plug the result into f(x). Looking at the list of g(x) points, I find (0, 2), so g(0) = 2, and I need now to find f(2). Looking at the list of f(x) points, I find (2, –3), so f(2) = –3. Then:

( f o g)(0) = f(g(0)) = f(2) = –3


(ii) The second part works the same way:

( f o g)(–1) = f(g(–1)) = f(–2) = 3


(iii) I can rewrite the composition as (g o f )(–1) = g( f(–1)) = g(1).
Uh-oh; there is no g(x) point with x = 1, so it is nonsense to try to find the value of g(1). In math-speak, g(1) is "not defined"; that is, it is nonsense.Then (g o f )(–1) is also nonsense, so the answer is:

(g o f )(–1) is undefined.



Part (iii) of the above example points out an important consideration regarding domains and ranges. It may be that your composed function (the result you get after composing two other functions) will have a restricted domain, or at least a domain that is more restricted than you might otherwise have expected. This will be more important when we deal with composing functions symbolically later.
Another exercise of this type gives you two graphs, rather than two sets of points, and has you read the points (the function values) from these graphs.


• Given f(x) and g(x) as shown below, find ( f o g)(–1).
In this case, I will read the points from the graph. I've been asked to find ( f o g)(–1) = f(g(–1)). This means that I first need to find g(–1). So I look on the graph of g(x), and find x = –1. Tracing up from x = –1 to the graph of g(x), I arrive at y = 3. Then the point (–1, 3) is on the graph of g(x), and g(–1) = 3.
Now I plug this value, x = 3, into f(x). To do this, I look at the graph of f(x) and find x = 3. Tracing up from x = 3 to the graph of  f(x), I arrive at y = 3. Then the point (3, 3) is on the graph of  f(x), and f(3) = 3.

Then ( f o g)(–1) = f(g(–1)) = f(3)=3

•   
• 
  

  • f(x):	g(x):

• • You can also evaluate compositions symbolically. It is simpler to evaluate a composition at a point because you can simplify as you go, since you'll always just be plugging in numbers and simplifying. Evaluating a symbolic compositon, where you're first plugging x into some function and then plugging that function into some other function, can be much messier. But the process works just as the at-a-number composition does, and using parentheses to be carefully explicit at each step will be even more helpful.
    
    
    • Given f(x) = 2x + 3 and g(x) = –x² + 5, find ( f o g)(x).
    In this case, I am not trying to find a certain numerical value. Instead, I am trying to find the formula that results from plugging the formula for g(x) into the formula for f(x). I will write the formulas at each step, using parentheses to indicate where the inputs should go:
    
    ( f o g)(x) = f (g(x))     = f (–x² + 5)     = 2(             ) + 3     
        = 2(–x² + 5) + 3     = –2x² + 10 + 3     = –2x² + 13
    
    If you plug in "1" for the x in the above, you will get ( f o g)(1) = –2(1)² + 13 = –2 + 13 = 11, which is the same answer we got before. Previously, we'd plugged a number into g(x), found a new value, plugged that value into f(x), and simplified the result. This time, we plugged a formula into f(x), simplified the formula, plugged the same number in as before, and simplified the result. The final numerical answers were the same. If you've done the symbolic composition (the composition with the formulas) correctly, you'll get the same values either way, regardless of the value you pick for x. This can be a handy way of checking your work.