2.7 - Composition of Functions

Overview

Composing functions means feeding the output of one function into another. It lets you build new functions out of simpler ones, connect quantities that aren't directly related by one formula, and break complicated functions into pieces. Composition also sets up inverse functions in Topic 2.8.

What composition means

f

and

g

are functions, the composite function

f \circ g

(read “

f

composed with

g

” or “

f

g

”) is defined by

(f \circ g)(x) = f(g(x))

Work from the inside out:

g

acts first, and its output becomes the input of

f

The domain of a composite function

For

f(g(x))

to exist, two things must be true:

$x$ is in the domain of $g$ .

$g(x)$ is in the domain of $f$ .

So the domain of

f \circ g

is restricted to the inputs of

g

whose outputs

f

can accept.

For example, let

f(x) = \sqrt{x}

and

g(x) = x - 3

. Then

f(g(x)) = \sqrt{x - 3}

. The function

g

accepts any

x

, but

f

only accepts inputs that are at least 0, so we need

x - 3 \geq 0

. The domain of

f \circ g

x \geq 3

Evaluating composite functions

You can find values of

f \circ g

from any representation of

f

and

g

: equations, tables, graphs, or words. The process is always the same: find the output of

g

, then use it as the input of

f

Example 1: From equations

Let

f(x) = x^2 - 1

and

g(x) = 2x + 3

. Find

f(g(-1))

and

g(f(-1))

First composition. Start inside, with

g

g(-1) = 2(-1) + 3 = 1

f(g(-1)) = f(1) = 1^2 - 1 = 0

Second composition. Now

f

goes first:

f(-1) = (-1)^2 - 1 = 0

g(f(-1)) = g(0) = 2(0) + 3 = 3

Answer:

f(g(-1)) = 0

and

g(f(-1)) = 3

. Same two functions, same input, different order, different answers.

Example 2: From a table

Use the table to find each value.

$x$	0	1	2	3	4
$f(x)$	2	4	1	6	0
$g(x)$	3	0	4	1	2

(a)

f(g(0))

: the table gives

g(0) = 3

, so

f(g(0)) = f(3) = 6

(b)

g(f(0))

: the table gives

f(0) = 2

, so

g(f(0)) = g(2) = 4

(c)

f(f(2))

: the table gives

f(2) = 1

, so

f(f(2)) = f(1) = 4

(d)

g(f(3))

: the table gives

f(3) = 6

, but 6 isn't an input in the table for

g

. So

g(f(3))

can't be found from this table. This is the domain rule in action: the output of the inside function has to be an input the outside function accepts.

Example 3: From graphs

The graphs of

f

and

g

are shown. Find

f(g(3))

g(f(3))

, and

f(g(6))

(a) From the graph of

g

g(3) = 2

. From the graph of

f

f(2) = 6

. So

f(g(3)) = 6

(b) From the graph of

f

f(3) = 5

. From the graph of

g

g(5) = 4

. So

g(f(3)) = 4

g

g(6) = 1

. From the graph of

f

f(1) = 4

. So

f(g(6)) = 4

Order matters

Composition is not commutative:

f \circ g

and

g \circ f

are usually different functions, so

f(g(x))

and

g(f(x))

usually give different values. Example 1 showed this (

0

versus

3

). Compare the formulas for

f(x) = x^2

and

g(x) = x + 3

f(g(x)) = (x + 3)^2 \qquad g(f(x)) = x^2 + 3

The first adds 3 and then squares. The second squares and then adds 3. At

x = 1

, they give

16

and

4

The identity function

The function

f(x) = x

is the identity function: it returns whatever you give it. Composing any function

g

with the identity changes nothing:

g(f(x)) = f(g(x)) = g(x) \quad \text{when } f(x) = x

The identity function plays the same role for composition that 0 plays for addition (

a + 0 = a

) and 1 plays for multiplication (

a \cdot 1 = a

). This idea is the heart of inverse functions in Topic 2.8: a function and its inverse compose to give the identity.

Building composite functions

Substituting one formula into another

To write

f(g(x))

as a formula, replace every

x

f

with the whole expression

g(x)

. Use parentheses so nothing gets lost.

Example 4: Writing a composite formula

Let

f(x) = 2x^2 - x + 1

and

g(x) = x - 4

. Write

f(g(x))

in standard form.

f(g(x)) = 2(x - 4)^2 - (x - 4) + 1

= 2(x^2 - 8x + 16) - x + 4 + 1

= 2x^2 - 16x + 32 - x + 5

= 2x^2 - 17x + 37

Answer:

f(g(x)) = 2x^2 - 17x + 37

. Quick check at

x = 4

g(4) = 0

and

f(0) = 1

, and the formula gives

2(16) - 68 + 37 = 1

. It matches.

Tables and graphs of a composite function

Without formulas, you can still build a table or graph of

f \circ g

: for each input

x

, find

f(g(x))

and record the pair

(x, f(g(x)))

. Using the graphs from Example 3:

$x$	0	1	2	3	4	5	6
$g(x)$	5	4	3	2	3	4	1
$f(g(x))$	2	4	5	6	5	4	4

Plotting the points

(x, f(g(x)))

gives a graph of

f \circ g

Connecting quantities through a middle quantity

Composition is useful when two quantities aren't linked by one formula but are both linked to a third quantity. If

A

depends on

B

and

B

depends on

C

, composing gives

A

in terms of

C

Example 5: Factory costs over time (AP-style)

A factory produces

n(t) = 40t

items in

t

hours. The cost of producing

n

items is

C(n) = 500 + 12n

dollars.

(a) Write the cost as a function of time. (b) Find and interpret

C(n(5))

. (c) Does

n(C(t))

make sense?

Part (a). Substitute

n(t)

into

C

C(n(t)) = 500 + 12(40t) = 500 + 480t

Part (b).

C(n(5)) = 500 + 480(5) = 2900

In 5 hours the factory makes

n(5) = 200

items, and producing them costs $2,900.

Part (c). No. The output of

C

is a cost in dollars, but

n

expects an input in hours. The output of the inside function has to be the kind of input the outside function uses. This is another reason order matters.

Decomposing functions

Going the other way, a complicated function can often be written as a composition of simpler ones. Find the “inside” expression that is being plugged into something and call it

g(x)

. The operation done to it becomes

f

. When a function is properly decomposed, the variable of

f

replaces every copy of

g(x)

$h(x)$	Inside: $g(x)$	Outside: $f(x)$
$\sqrt{5x - 2}$	$5x - 2$	$\sqrt{x}$
$3^{x^2}$	$x^2$	$3^x$
$\frac{1}{2^x + 1}$	$2^x + 1$	$\frac{1}{x}$
$(x^2 + 1)^3 - 4(x^2 + 1)$	$x^2 + 1$	$x^3 - 4x$

The last row shows the rule: both copies of

x^2 + 1

become

x

f

. Decompositions aren't unique. For example,

\sqrt{5x - 2}

is also

f(g(x))

with

f(x) = \sqrt{x - 2}

and

g(x) = 5x

. Any pair that works is correct unless a question asks for a specific form.

Transformations as compositions

The transformations from Unit 1 are compositions with simple linear functions:

An additive transformation (a translation) is a composition with $g(x) = x + k$ .

A multiplicative transformation (a dilation) is a composition with $g(x) = kx$ .

Whether

g

goes inside or outside decides the direction:

Composition	Result	Effect on the graph of $f$
$f(g(x))$ with $g(x) = x + k$	$f(x + k)$	Horizontal translation by $-k$
$g(f(x))$ with $g(x) = x + k$	$f(x) + k$	Vertical translation by $k$
$f(g(x))$ with $g(x) = kx$	$f(kx)$	Horizontal dilation by a factor of $\frac{1}{\|k\|}$
$g(f(x))$ with $g(x) = kx$	$kf(x)$	Vertical dilation by a factor of $\|k\|$

k < 0

, the dilations also reflect the graph: over the

y

-axis for

f(kx)

and over the

x

-axis for

kf(x)

. The pattern to remember: inside changes are horizontal, outside changes are vertical.

Example 6: A transformation as a chain of compositions

Write

h(x) = 5 \cdot 2^{x - 1} + 4

as a composition of

f(x) = 2^x

with simple linear functions.

Follow what happens to

x

, in order:

Subtract 1: $p(x) = x - 1$ . This shifts the graph right 1.

Apply $f$ : $f(p(x)) = 2^{x - 1}$ .

Multiply by 5: $q(x) = 5x$ . This stretches the graph vertically by a factor of 5.

Add 4: $r(x) = x + 4$ . This shifts the graph up 4.

h(x) = r(q(f(p(x)))) = 5 \cdot 2^{x - 1} + 4

Common mistakes

Working outside in. In $f(g(x))$ , $g$ goes first.

Treating composition as multiplication. $f(g(x))$ is not $f(x) \cdot g(x)$ .

Dropping parentheses. With $f(x) = x^2$ and $g(x) = x - 4$ , $f(g(x)) = (x - 4)^2$ , not $x^2 - 4$ .

Ignoring the domain. The output of the inside function must be in the domain of the outside function.

Assuming order doesn't matter. $f(g(x))$ and $g(f(x))$ are usually different.

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