# Composition of Functions in Precalculus

These are my notes on composition of functions in precalculus.

**Combining Functions Using Algebraic Operations**

Function composition is only one way to combine existing functions.

Another way is to carry out the usual algebraic operations on functions,

such as addition, subtraction, multiplication, and division. We do all this

by performing the operations with the function outputs, defining the

result as the output of our new function.

Suppose we need to add two columns of numbers that represent a husband and

wife's separate annual incomes over a period of years, with the result being

their total household income. We want to do this for every year, adding only

that year's incomes and then collecting all the data in a new column. If w(y)

is the wife's income and h(y) is the husband's income in year y, and we want T

to represent the total income, then we can define a new function.

\(T(y)=h(y)+w(y)\)

If this holds true for every year, then we can focus on the relation between

the functions without reference to a year and write:

\(T=h+w\)

Just as for this sum of two functions, we can define difference, product, and

ratio functions for any pair of functions that have the same kind of inputs

(not necessarily numbers) and also the same kinds of outputs (which do have to

be numbers so that the usual operations of algebra can apply to them, and which

also must have the same units or no units when we add and subtract). In this

way, we can think of adding, subtracting, multiplying, and dividing functions.

For two functions f(x) and g(x) with real number outputs, we define new

functions f+g, f-g, fg, and f/g by the relations:

\((f+g)(x)=f(x)+g(x)\)

\((f-g)(x)=f(x)-g(x)\)

\((fg)(x)=f(x)g(x)\)

\((f/g)(x)=f(x)/g(x)\)**Example 1**

Performing Algebraic Operations on Functions

Find and simplify the functions \((g-f)(x)\) and \((g/f)(x)\), given

\(f(x)=x-1\) and

\(g(x)=x^2-1\). Are they the same function?

Solution:

begin by writing the general form, and then substitute the given functions.

\((g-f)(x) = g(x)-f(x)\)

\((g-f)(x) = x^2-1-(x-1)\)

\((g-f)(x) = x^2-x\)

\((g-f)(x) = x(x-1)\)

\((g/f)(x) = \frac{g(x)}{f(x)}\)

\((g/f)(x) = \frac{x^2-1}{x-1}\)

\((g/f)(x) = \frac{(x+1)(x-1)}{x-1}\)

\((g/f)(x) = x+1\)

No, the functions are not the same.

Note, for (g/f)(x), the condition \(x \not = 1\) is necessary because when

x=1, the denominator is equal to 0, which makes the function undefined.**Create a Function by Composition of Functions**

Performing algebraic operations on functions combines them into a new function,

but we can also create functions by composing functions. When we wanted

to compute a heating cost from a day of the year, we created a new function

that takes a day as input and yields a cost as output. The process of combining

functions so that the output of one function becomes the input of another is

known as the composition of functions. The resulting function is known as a

composition function. We represent this combination by the following notation:

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

We read the left-hand side as "f composed with g at x", and the right=hand side

as "f of g of x". The two sides of the equation have the same mathematical

meaning and are equal. The open circle is called the composition operator. We

use this operator mainly when we wish to emphasize the relationship between the

functions themselves without referring to any particular input value.

Composition is a binary operation that takes two functions and forms a new

function, much as addition or multiplication takes two numbers and gives a new

number. However, it is important not to confuse function composition with

multiplication because, as we learned above, in most cases

\(f(g(x))\not=f(x)g(x)\).

It is also important to understand the order of operations in evaluating a

composite function. We follow the usual convention with parentheses by starting

with the innermost parentheses first, and then working to the outside. In the

equation above, the function g takes the input x first and yields an output

g(x). Then the function f takes g(x) as an input and yields an output f(g(x)).

g(x), the output of g is the input of f

In general, \(f \circ g\) and \(g \circ f\) are different functions. in other

words, in many cases, \(f(g(x)) \not= g(f(x))\) for all x. We will also see that

sometimes two functions can be composed only in one specific order.

For example, if \(f(x)=x^2\) and \(g(x)=x+2\), then:

\(f(g(x))=f(x+2) = (x+2)^2 = x^2+4x+4\)

but:

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

These expressions are not equal for all values of x, so the two functions are

not equal. It is irrelevant that the expressions happen to be equal for the

single input value \(x=1/2\).

Note that the range of the inside function (first function to be evaluated)

needs to be within the domain of the outside function. Less formally, the

composition has to make sense in terms of inputs and outputs.**Composition of Functions**

When the output of one function is used as the input of another function, we

call the entire operation a composition of functions. For any input x and

functions f and g, this action defines a composite function, which we write as

\(f \circ g\) such that \((f \circ g)(x)=f(g(x))\).

The domain of the composite function \(f \circ g\) is all x such that x is in

the domain of g and g(x) is in the domain of f. It is important to realize that

the product of functions fg is not the same as the function composition f(g(x)),

because , in general, \(f(x)g(x)\not=f(g(x))\).**Example 2**

Determining Whether Compositions of Functions is Commutative

Using the functions provided, find f(g(x)) and g(f(x)). Determine whether the

composition of the functions is commutative.

\(f(x)=2x+1\) and \(g(x)=3-x\)

Solution:

Let's begin by substituting g(x) into f(x).

\(f(g(x))=2(3-x)+1 = 6-2x+1 = 7-2x\)

Now, we can substitute f(x) into g(x).

\(g(f(x))=3-(2x+1) = 3-2x-1 = -2x+2\)

We find that \(g(f(x)) \not= f(g(x))\), so the operation of function composition

is not commutative.**Example 3**

Interpreting Composite Functions

The function c(s) gives the number of calories burned completing s sit-ups, and

s(t) gives the number of sit-ups a person can complete in t minutes. Interpret

c(s(3)).

Solution:

The inside expression in the composition is s(3). Because the input to the

s-function is time, t=3 represents 3 minutes, and s(3) is the number of sit-ups

completed in 3 minutes.

Using s(3) as the input to the function c(s) gives us the number of calories

burned during the number of sit-ups that can be completed in 3 minutes, or

simply the number of calories burned in 3 minutes by doing sit-ups. **Example 4**

Investigating the Order of Function Composition

Suppose f(x) gives miles that can be driven in x hours and g(y) gives the

gallons of gas used in driving y miles. Which of the expressions is meaningful,

f(g(y)) or g(f(x))?

Solution:

The function \(y=f(x)\) is a function whose output is the number of miles driven

corresponding to the number of hours driven.

number of miles=f(number of hours)

The function g(y) is a function whose output is the number of gallons used

corresponding to the number of miles driven. This means:

number of gallons = g(number of miles)

The expression g(y) takes miles as the input and a number of gallons as the

output. The function f(x) requires a number of hours as the input. Trying to

input a number of gallons does not make sense. The expression f(g(y)) is

meaningless.

The expression f(x) takes hours as the input and a number of miles driven as the

output. The function g(y) requires a number of miles as the input. Using

f(x)(miles driven) as an input value for g(y), where gallons of gas depends on

miles driven, does make sense. The expression g(f(x)) makes sense, and will

yield the number of gallons of gas used, g, driving a certain number of miles,

f(x), in x hours.**Are there any situations where f(g(y)) and g(f(x)) would both be meaningful or****useful expressions?**

Yes. For many pure mathematical functions, both compositions make sense, even

though they usually produce different new functions. In real-world problems,

functions whose inputs and outputs have the same units also may give

compositions that are meaningful in either order.**Evaluating Composite Functions**

Once we compose a new function from two existing functions, we need to be able

to evaluate it for any input in its domain. We will do this with specific

numerical inputs for functions expressed as tables, graphs, and formulas and

with variables as inputs to functions expressed as formulas. In each case, we

evaluate the inner function using the starting input and then use the inner

function's output as the input for the outer function.

When working with functions given as tables, we read input and output values

from the table entries and always work from the inside to the outside. We

evaluate the inside function first and then use the output as of the inside

function as the input to the outside function.**Example 5**

Using a Table to Evaluate a Composite Function

Evaluate f(g(3)) and g(f(3)).

x f(x) g(x)

1 6 3

2 8 5

3 3 2

4 1 7

Solution:

To evaluate f(g(3)), we start from the inside with the input value 3. We then

evaluate the inside expression g(3) using the table that defines the function

g:g(3)=2. We can then use that result as the input to the function f, sp g(3) is

replaced by 2 and we get f(2). Then using the table that defines the function f,

we find that f(2)=8.

\(g(3)=2\)

\(f(g(3))=f(2)=8\)

To evaluate g(f(3)), we first evaluate the inside expression f(3) using the

first table: f(3)=3. Then, using the table for g, we can evaluate:

\(g(f(3))=g(3)-2\)**Evaluating Composite Functions Using Graphs**

When we are given individual functions as graphs, the procedure for evaluating

composite functions is similar to the process we use for evaluating tables. We

read the input and output values, but this time, from the x and y axes of the

graphs.

Given a composite function and graphs of its individual functions, evaluate it

using the information provided by the graphs.

1. Locate the given input to the inner function on the x-axis of its graph

1. Read off the input of the inner function from the y-axis of its graph

3. Locate the inner function output on the x-axis of the graph of the outer

function.

4. Read the output of the outer function from the y-axis of its graph. This is

the output of the composite function.**Example 6**

Using a Graph to Evaluate a Composite Function

Evaluate f(g(1))

Solution:

We evaluate g(1) using the graph of g(x), finding the input of 1 on the x-axis

and finding the output value of the graph at that input. Here, g(1)=3. We

use this value as the input to the function f.

\(f(g(1))=f(3)\)

We can then evaluate the composite function by looking to the graph of f(x),

finding the input of 3 on the x-axis and reading the output value of the graph

at this input. Here, \(f(3)=6, \text{ so } f(g(1))=6\).**Evaluating Composite Functions Using Formulas**

When evaluating a composite function where we have either created or been

given formulas, the rule of working from the inside out remains the same. The

input value to the outer function will be the output of the inner function,

which may be a numerical value, a variable name, or a more complicated

expression.

While we can compose the functions for each individual input value, it is

sometimes helpful to find a single formula that will calculate the result of a

composition \(f9g(x))\). To do this, we will extend our idea of the function

evaluation. Recall that, when we evaluate a function like \(f(t)=t^2-t\), we

substitute the value inside the parentheses into the formula wherever we see

the input variable.

Given a formula for a composite function, evaluate the function.

1. Evaluate the inside out function using the input value or variable

provided.

2. use the resulting output as the input to the outside function.**Example 7**

Evaluating a Composition of Functions Expressed as Formulas with a numerical

Input

Given \(f(t)=t^2-t\) and \(h(x)=3x+2\), evaluate f(h(1)).

Solution:

Because the inside expression is h(1), we start by evaluating h(x) at 1.

\(h(1)=3(1)+2 = 5\)

Then \(f(h(1))=f(5)\), so we evaluate f9t) at an input of 5.

\(f(h(1))= f(5) = 5^2-5 = 20\)

It makes no difference what the input variables t and x were called because we

evaluated for specific numerical values.**Finding the Domain of a Composite Function**

As we discussed previously, the domain of a composite function such as \(f

\circ g\) is dependent on the domain of g and the domain of f. It is

important to know when we can apply a composite function and when we

cannot, that is, to know the domain of a function such as \(f \circ g\). Let us

assume we know the domains of the functions f and g separately. If we write

the composite function for an input x as f(g(x)), we can see right away that x

must be a member of the domain of g in order for the expression to be

meaningful, because otherwise we cannot complete the inner function

evaluation. However, we can see right away that x must be a member of the domain

of f, otherwise the second function evaluation in f(g(x)) cannot be

completed, and the expression is still undefined. Thus the domain of \(f \circ

g\) consists of only those inputs in the domain of g that produce outputs from

g belonging to the domain of f. Note that the domain of f composed with g is

the set of all x such that x is in the domain of f composed with g is the

set of all x such that x is in the domain of g and g9x) is in the domain of

f.**Domain of a Composite Function**

The domain of a composite function f(g(x)) is the set of those inputs x in the

domain of g for which g(x) is in the domain of f.

Given a function composition f(g(x)), determine its domain

1. Find the domain of g

2. Find the domain of f

3. Find those inputs x in the domain of g for which g(x) is in the domain of

f. That is, exclude those inputs x from the domain of g for which g(x) is not

in the domain of f. The resulting set is the domain of \(f \circ g\).**Example 8**

Finding the Domain of a Composite Function

Find the domain of \((f \circ g)(x)\) where \(f(x)=5/(x-1) \text{ and } g(x) =

4/(3x-2)\).

Solution:

The domain of g(x) consists of all real numbers except \(x=2/3\), since that

input value would cause us to divide by 0. Likewise, the domain of f consists

of all real numbers except 1. So we need to exclude from the domain of g(x)

that value of x for which g9x)=1.

\(\frac{4}{3x-2} = 1\)

\(4=3x-2\)

\(6=3x = 2\)

So the domain of \(f \circ g\) is the set of all real numbers except 2/3 and 2.

This means that :

\(x \not= \frac{2}{3} \text{ or } x\not=2\)

We can write this in interval notation as:

\((-\infty,\frac{2}{3}) \cup (\frac{2}{3},2) \cup (2,\infty)\)**Example 9**

Finding the Domain of a Composite Function Involving Radicals

Find the domain of :

\((f \circ g)(x) \text{ where } f(x) = \sqrt{x+2} \text{ and } g(x) =

\sqrt{3-x}\)

Solution:

Because we cannot take the square root of a negative number, the domain of g

is \(-\infty,3]. Now we can check the domain of the composite function:

\((f \circ g)(x) = \sqrt{\sqrt{3-x}+2}\)

This expression must \(\geq 0\), since the radicand of a square root must be

positive. Since square roots are positive, \(\sqrt{3-x} \geq 0\) which gives

the domain of \((-\infty,3].

This example shows that knowledge of the range of functions (specifically

the inner function) can also be helpful in finding the domain of a composite

function. It also shows that the domain of \(f \circ g\) can contain values

that are not in the domain of f, though they must be in the domain of g.

Decomposing a Composite Function into its Component Functions

In some cases, it is necessary to decompose a complicated function. In

other words, we can write it as a composition of two simpler functions. There

may be more than one way to decompose a composite function, so we may

choose the decomposition that appears to be most expedient.**Example 10**

Decomposing a Function

Write \(f(x)=\sqrt{5-x^2}\) as the composition of two functions.

We are looking for two functions, g and h, so f(x)=g(h(x)). To do this,

we look for a function inside a function in the formula for f(x). As one

possibility, we might notice that the expression \(5-x^2\) is the inside of

the square root. We could then decompose the function as \(h(x)=5-x^2 \text{

and } g(x) = \sqrt{x}\)

We can check out answer by recomposing the functions.

\(g(h(x))=g(5-x^2)=\sqrt{5-x^2}\)