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Composite Functions

Module by: Kenny M. Felder. E-mail the author

Summary: This module describes composite functions in Algebra.

You are working in the school cafeteria, making peanut butter sandwiches for today’s lunch.

  • The more classes the school has, the more children there are.
  • The more children there are, the more sandwiches you have to make.
  • The more sandwiches you have to make, the more pounds (lbs) of peanut butter you will use.
  • The more peanut butter you use, the more money you need to budget for peanut butter.

...and so on. Each sentence in this little story is a function. Mathematically, if cc size 12{c} {} is the number of classes and hh size 12{h} {} is the number of children, then the first sentence asserts the existence of a function h(c)h(c) size 12{h \( c \) } {}.

The principal walks up to you at the beginning of the year and says “We’re considering expanding the school. If we expand to 70 classes, how much money do we need to budget? What if we expand to 75? How about 80?” For each of these numbers, you have to calculate each number from the previous one, until you find the final budget number.

Table 1
# Classes # Classes Gearbox # Children # Children Gearbox # Sandwiches # Sandwiches Gearbox lb. lb. Gearbox $$ $$

But going through this process each time is tedious. What you want is one function that puts the entire chain together: “You tell me the number of classes, and I will tell you the budget.”

Table 2
# Classes # Classes Gearbox $$ $$

This is a composite function—a function that represents in one function, the results of an entire chain of dependent functions. Since such chains are very common in real life, finding composite functions is a very important skill.

How do you make a composite Function?

We can consider how to build composite functions into the function game that we played on the first day. Suppose Susan takes any number you give her, quadruples it, and adds 6. Al takes any number you give him and divides it by 2. Mathematically, we can represent the two functions like this:

S ( x ) = 4x + 6 S ( x ) = 4x + 6 size 12{S \( x \) =4x+6} {}
(1)
A ( x ) = x 2 A ( x ) = x 2
(2)

To create a chain like the one above, we give a number to Susan; she acts on it, and gives the resulting number to Al; and he then acts on it and hands back a third number.

3 Susan S ( 3 ) = 18 Al A ( 18 ) = 9 3 Susan S ( 3 ) = 18 Al A ( 18 ) = 9 size 12{3 rightarrow ital "Susan" rightarrow S \( 3 \) ="18" rightarrow ital "Al" rightarrow A \( "18" \) =9} {}

In this example, we are plugging S(3)S(3) size 12{S \( 3 \) } {}—in other words, 18— into Al’s function. In general, for any xx size 12{x} {} that comes in, we are plugging S(x)S(x) size 12{S \( x \) } {} into A(x)A(x) size 12{A \( x \) } {}. So we could represent the entire process as A(S(x))A(S(x)) size 12{A \( S \( x \) \) } {}. This notation for composite functions is really nothing new: it means that you are plugging S(x)S(x) size 12{S \( x \) } {} into the AA function.

But in this case, recall that S ( x ) = 4x + 6 S ( x ) = 4x + 6 size 12{S \( x \) =4x+6} {} . So we can write:

A ( S ( x ) ) = S ( x ) 2 = 4x + 6 2 = 2x + 3 A ( S ( x ) ) = S ( x ) 2 = 4x + 6 2 = 2x + 3
(3)

What happened? We’ve just discovered a shortcut for the entire process. When you perform the operation A(S(x))A(S(x)) size 12{A \( S \( x \) \) } {}—that is, when you perform the Al function on the result of the Susan function—you are, in effect, doubling and adding 3. For instance, we saw earlier that when we started with a 3, we ended with a 9. Our composite function does this in one step:

3 2x + 3 9 3 2x + 3 9 size 12{3 rightarrow 2x+3 rightarrow 9} {}

Understanding the meaning of composite functions requires real thought. It requires understanding the idea that this variable depends on that variable, which in turn depends on the other variable; and how that idea is translated into mathematics. Finding composite functions, on the other hand, is a purely mechanical process—it requires practice, but no creativity. Whenever you are asked for f(g(x))f(g(x)) size 12{f \( g \( x \) \) } {}, just plug the g(x)g(x) size 12{g \( x \) } {} function into the f(x)f(x) size 12{f \( x \) } {} function and then simplify.

Example 1: Building and Testing a Composite Function

f(x)=x24xf(x)=x24x size 12{f \( x \) =x rSup { size 8{2} } - 4x} {}

g(x)=x+2g(x)=x+2 size 12{g \( x \) =x+2} {}

What is f(g(x))f(g(x)) size 12{f \( g \( x \) \) } {}?

  • To find the composite, plug g(x)g(x) size 12{g \( x \) } {} into f(x)f(x) size 12{f \( x \) } {}, just as you would with any number.

f(g(x))=(x+2)24(x+2)f(g(x))=(x+2)24(x+2) size 12{f \( g \( x \) \) = \( x+2 \) rSup { size 8{2} } - 4 \( x+2 \) } {}

  • Then simplify.

f(g(x))=(x2+4x+4)(4x+8)f(g(x))=(x2+4x+4)(4x+8) size 12{f \( g \( x \) \) = \( x"" lSup { size 8{2} } +4x+4 \) - \( 4x+8 \) } {}

f(g(x))=x24f(g(x))=x24 size 12{f \( g \( x \) \) =x rSup { size 8{2} } - 4} {}

  • Let’s test it. f(g(x))f(g(x)) size 12{f \( g \( x \) \) } {} means do gg size 12{g} {}, then ff size 12{f} {}. What happens if we start with x=9x=9 size 12{x=9} {}?

7 g ( x ) 7 + 2 = 9 f ( x ) ( 9 ) 2 4 ( 9 ) = 45 7 g ( x ) 7 + 2 = 9 f ( x ) ( 9 ) 2 4 ( 9 ) = 45 size 12{7 rightarrow g \( x \) rightarrow 7+2=9 rightarrow f \( x \) rightarrow \( 9 \) rSup { size 8{2} } - 4 \( 9 \) ="45"} {}

  • So, if it worked, our composite function should do all of that in one step.

7x24=(7)24=457x24=(7)24=45 size 12{7 rightarrow x rSup { size 8{2} } - 4= \( 7 \) rSup { size 8{2} } - 4="45"} {} It worked!

There is a different notation that is sometimes used for composite functions. This book will consistently use f(g(x))f(g(x)) size 12{f \( g \( x \) \) } {} which very naturally conveys the idea of “plugging g(x)g(x) size 12{g \( x \) } {} into f(x)f(x) size 12{f \( x \) } {}.” However, you will sometimes see the same thing written as f°g(x)f°g(x) size 12{f circ g \( x \) } {}, which more naturally conveys the idea of “doing one function, and then the other, in sequence.” The two notations mean the same thing.

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