# 1.4: Composition of Functions

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Skills to Develop

- Combine functions using algebraic operations to create a new function.
- Create a new function by composition of functions.
- Evaluate composite functions.
- Find the domain of a composite function.
- Decompose a composite function into its component functions.

Suppose we want to calculate how much it costs to heat a house on a particular day of the year. The cost to heat a house will depend on the average daily temperature, and in turn, the average daily temperature depends on the particular day of the year. Notice how we have just defined two relationships: The cost depends on the temperature, and the temperature depends on the day.

Using descriptive variables, we can notate these two functions. The function \(C(T)\) gives the cost \(C\) of heating a house for a given average daily temperature in \(T\) degrees Celsius. The function \(T(d)\) gives the average daily temperature on day d of the year. For any given day, \(Cost=C(T(d))\) means that the cost depends on the temperature, which in turns depends on the day of the year. Thus, we can evaluate the cost function at the temperature \(T(d)\). For example, we could evaluate \(T(5)\) to determine the average daily temperature on the 5th day of the year. Then, we could evaluate the **cost function** at that temperature. We would write \(C(T(5))\).

*Figure \(\PageIndex{1}\): **Explanation of \(C(T(5))\), which is the cost for the temperature and \(T(5)\) is the temperature on day 5.*

By combining these two relationships into one function, we have performed function composition, which is the focus of this section.

## 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 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.

\[T(y)=h(y)+w(y) \nonumber\]

If this holds true for every year, then we can focus on the relation between the functions without reference to a year and write

\[T=h+w \nonumber\]

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 kinds 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 \(\frac{f}{g}\) by the relations.

\[ \begin{align*} (f+g)(x) &=f(x)+g(x) \\[5pt] (f−g)(x) &=f(x)−g(x) \\[5pt] (fg)(x)&=f(x)g(x) \\[5pt] \left(\dfrac{f}{g}\right)(x) &=\dfrac{f(x)}{g(x)} \end{align*}\]

Example \(\PageIndex{1}\): Performing Algebraic Operations on Functions

Find and simplify the functions \((g−f)(x)\) and \(\left(\dfrac{g}{f}\right)(x)\), 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.

\[\begin{align*} (g−f)(x) &= g(x)−f(x) \\[5pt] &= (x^2−1)−(x−1) \\[5pt] &=x^2−x , \mbox{ or } x(x−1) \end{align*}\]

\[\begin{align*} \left(\dfrac{g}{f}\right)(x)&=\dfrac{g(x)}{f(x)} \\[5pt] &=\dfrac{x^2−1}{x−1} \\[5pt] &=\dfrac{(x+1)(x−1)}{x−1} \\[5pt]&=x-1 \end{align*}\]

No, the functions are not the same.

Note: For \(\left(\dfrac{g}{f}\right)(x)\), the condition \(x\neq1\) is necessary because when \(x=1\), the denominator is equal to 0, which makes the function undefined. Thus, the domain of \( \frac{g}{f}\) is \(D = \{x \;| \; x \neq 1 \} \).

**Note**: What do you think the graph of \( \frac{g}{f}\) would look like? Hand this graph to your instructor along with an explanation of your reasoning for *Extra Credit.*

\(\PageIndex{1}\)

Find and simplify the functions \((fg)(x)\) and \((f−g)(x)\), given \(f(x)=x−1 \nonumber\) and \(g(x)=x^2−1 \nonumber\). Are they the same function?

**Answer**-
\((fg)(x)=f(x)g(x)=(x−1)(x^2−1)=x^3−x^2−x+1 \)

\((f−g)(x)=f(x)−g(x)=(x−1)−(x^2−1)=x−x^2,\) or \(x(1-x)\)No, the functions are not the same.

## Create a Function by Composition of Functions

New functions can also be created 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 a **composition of functions**. The resulting function is known as a **composite function**. We represent this combination by the following notation:

\[f{\circ}g(x)=f(g(x))\nonumber\]

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 symbol \(\circ\) 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 will see, in most cases \(f(g(x)){\neq}f(x)g(x)\).

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 \(f(g(x))\).

**Figure \(\PageIndex{2}\): **Explanation of the composite function.

In general, \(f{\circ}g\) and \(g{\circ}f\) are different functions. In other words, in many cases \(f(g(x)){\neq}g(f(x))\) for all \(x\).

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

\[\begin{align*} f(g(x))&= f(x+2) \\[5pt]&=(x+2)^2 \\[5pt] &=x^2+4x+4 \end{align*}\]

but

\[\begin{align*} g(f(x))&= g(x^2) \\[5pt]&=x^2+2 \end{align*}\]

These expressions are not equal for all values of \(x\), so the two functions are not equal. Although the expressions happen to be equal for the single input value \(x=−\frac{1}{2}\), this is irrelevant.

** Extra Credit**: Given the formulas for each composite function, show how to find all \(x\)-values that make the functions equal, and hand your work in to your instructor.

Composition of Functions

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

\[(f{\circ}g)(x)=f(g(x))\nonumber\]

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 \(f\circ g\).

Example \(\PageIndex{2}\): Determining whether Composition 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**.

\[f(x)=x^2 \;\;\;\; g(x)=\sqrt{x} \nonumber\]

**Solution**

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

\[\begin{align*} f(g(x))&= (\sqrt{x})^2 \\[5pt] &= x ,\; D=[0,\infty) \end{align*}\]

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

\[\begin{align*} g(f(x))&= \sqrt{(x^2)} \\[5pt]&=\vert x \vert ,\; D=(-\infty,\infty) \end{align*}\]

We find that \(g(f(x)){\neq}f(g(x))\), so the operation of function composition is not commutative. Note that not only are the formulas different, but the domains are also different.

Example \(\PageIndex{3}\): Comparing Multiplication of Functions with Composition of Functions

Let \(f(x)=2x-3\), and let \(g(x)=x^2+1\). Find two new functions by combining \(f(x)\) and \(g(x)\) in the following ways: \(f\circ g\) and \(fg\). Are they the same?

**Solution**

\[f\circ g = f(g(x)) = 2(x^2+1) - 3 = 2x^2-1 \nonumber\]

\[fg = (2x-3)(x^2+1) = 2x^3-3x^2+2x-3 \nonumber\]

The two new functions are not the same.

Example \(\PageIndex{4}\): 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).

Sometimes two functions can be composed only in one specific order. The range of the inside function (the 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. In a mathematical model, this can be very important. See Example \(\PageIndex{5}\).

Example \(\PageIndex{5}\): 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 these 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.

\[\text{number of miles } =f (\text{number of hours}) \nonumber\]

The function \(g(y)\) is a function whose output is the number of gallons used corresponding to the number of miles driven. This means:

\[\text{number of gallons } =g(\text{number of miles}) \nonumber\]

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 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.*

\(\PageIndex{2}\)

The gravitational force on a planet a distance \(r\) from the sun is given by the function \(G(r)\). The acceleration of a planet subjected to any force \(F\) is given by the function \(a(F)\). Form a meaningful composition of these two functions, and explain what it means.

**Answer**-
A gravitational force is still a force, so \(a(G(r))\) makes sense as the acceleration of a planet at a distance \(r\) from the Sun (due to gravity), but \(G(a(F))\) does not make sense.

## 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.

### Evaluating Composite Functions Using Tables

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 of the inside function as the input to the outside function.

Example \(\PageIndex{6}\): Using a Table to Evaluate a Composite Function

Using Table \(\PageIndex{1}\), 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 \(f\), so \(g(3)\) is replaced by 2 and we get \(f(2)\). Then, using the table that defines the function \(f\), we find that \(f(2)=8\).

\[g(3)=2 \nonumber\]

\[f(g(3))=f(2)=8 \nonumber\]

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

\[g(f(3))=g(3)=2 \nonumber\]

Table \(\PageIndex{2}\) shows the composite functions \(f{\circ}g\) and \(g{\circ}f\) as tables.

\(x\) | \(g(x)\) | \(f(g(x))\) | \(f(x)\) | \(g(f(x))\) |
---|---|---|---|---|

3 | 2 | 8 | 3 | 2 |

\(\PageIndex{3}\)

Using Table \(\PageIndex{1}\), evaluate \(f(g(1))\) and \(g(f(4))\).

**Answer**-
\(f(g(1))=f(3)=3\) and \(g(f(4))=g(1)=3\)

### 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

- Locate the given input to the inner function on the \(x\)-axis of its graph.
- Read off the output of the inner function from the \(y\)-axis of its graph.
- Locate the inner function output on the \(x\)-axis of the graph of the outer function.
- Read the output of the outer function from the \(y\)-axis of its graph. This is the output of the composite function.

Example \(\PageIndex{7}\): Using a Graph to Evaluate a Composite Function

Using Figure \(\PageIndex{3}\), evaluate \(f(g(1))\).

**Figure \(\PageIndex{3}\): **Two graphs of a positive parabola \(y=g(x)\) and negative parabola \(y=f(x)\).

**Solution**

To evaluate \(f(g(1))\), we start with the inside evaluation. See Figure \(\PageIndex{4}\).

**Figure \(\PageIndex{4}\): **Two graphs of a positive parabola \(g(x)\) and a negative parabola \(f(x)\). The following points are plotted: \(g(1)=3\) and \(f(3)=6\).

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 point on the graph at that input. Here, \(g(1)=3\). We use this value as the input to the function \(f\).

\[f(g(1))=f(3) \nonumber\]

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, \(f(3)=6\), so \(f(g(1))=6\).

**Analysis**

Figure \(\PageIndex{5}\) shows how we can mark the graphs with arrows to trace the path from the input value to the output value.

**Figure \(\PageIndex{5}\)**: *Two graphs of a positive parabola \(y=g(x)\) and negative parabola \(y=f(x)\).*

**Solution**

\(\PageIndex{4}\)

Using Figure \(\PageIndex{3}\), evaluate \(g(f(2))\).

**Answer**-
\(g(f(2))=g(5)=3\)

### 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 \(f(g(x))\). To do this, we will extend our idea of 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

- Evaluate the inside function using the input value or variable provided.
- Use the resulting output as the input to the outside function.

Example \(\PageIndex{8}\): 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.

\[ \begin{align*} h(1)=3(1)+2 \\[5pt] h(1)=5 \end{align*} \]

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

\[ \begin{align*} f(h(1)) =f(5) \\[5pt] f(h(1))=5^2−5 \\[5pt] f(h(1))=20 \end{align*} \]

**Analysis**

It makes no difference what the input variables \(t\) and \(x\) were called in this problem because we evaluated for specific numerical values.

\(\PageIndex{5}\)

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

a. \(h(f(2))\)

b. \(h(f(−2))\)

**Answer**-
a. \(h(f(2))=h(2)=8\)

b. \(h(f(-2))=h(6)=20\)

## 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 also see that \(g(x)\) must be a member of the domain of \(f\), otherwise the outer 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 \(g\) and \(g(x)\) is in the domain of \(f\).

Definition: 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

- Find the domain of \(g\).
- Find the domain of \(f\).
- Find those inputs \(x\) in the domain of \(g\) for which \(g(x)\) is in the domain of \(f\). That is, exclude input values \(x\) from the domain of \(g\) if \(g(x)\) is not in the domain of \(f\). The resulting set is the domain of \(f \circ g\).

Example \(\PageIndex{9A}\): Finding the Domain of a Composite Function Involving Quotients

Find the domain of

\[(f∘g)(x) \text{ where } f(x)=\dfrac{5}{x−1} \text{ and } g(x)=\dfrac{4}{3x−2} \nonumber\]

**Solution**

The domain of \(g(x)\) consists of all real numbers except \(x=\frac{2}{3}\), 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 \(g(x)\) any value of \(x\) for which \(g(x)=1\).

\[\begin{align*} \dfrac{4}{3x-2}&= 1 \\[5pt] 4 &=3x-2 \\[5pt] 6&=3x \\[5pt] x&= 2 \end{align*}\]

The domain of \(f{\circ}g\) is the set of all real numbers except \(\frac{2}{3}\) and \(2\). This means that

\[x{\neq} \dfrac{2}{3} \text{ and } x\neq2 \nonumber\]

We can write this in interval notation as

\[D=(−\infty,\dfrac{2}{3})\cup(\dfrac{2}{3},2)\cup(2,\infty) \nonumber\]

Example \(\PageIndex{9B}\): Finding the Domain of a Composite Function Involving Radicals

Find the domain of

\[(f{\circ}g)(x) \text{ where } f(x)=\sqrt{x+2} \text{ and } g(x)=\sqrt{3−x}\nonumber\]

Because we cannot take the square root of a negative number, the domain of \(g\) is \(\left(−\infty,3\right]\). The composite function is

\[(f{\circ}g)(x)=\sqrt{\sqrt{3−x}+2}. \nonumber\]

Again, because we cannot take the square root of a negative number, we exclude any negative numbers in the radicand; thus, we want only \(x\)-values such that \(\sqrt{3−x}+2≥0,\) or \(\sqrt{3-x}\geq -2.\) Since the square root symbol \(\sqrt{}\) is defined to be the non-negative square root of a number, \(\sqrt{3−x}\) is always greater than -2; hence we are left with the restriction on the domain of \(g\), which gives \(f \circ g\) a domain of \((-∞,3]\).

**Analysis**

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 \(g\).

\(\PageIndex{6}\)

Find the domain of

\[(f{\circ}g)(x), \text{ where } f(x)=\dfrac{1}{x−2} \text{ and } g(x)=\sqrt{x+4} \nonumber\]

**Answer**-
\([−4,2)∪(2,∞)\)

## Decomposing a Composite Function into its Component Functions

In some cases, it is necessary to decompose a complicated function. In other words, we want to write a complicated function as a composition of two simpler functions. There may be more than one way to **decompose a composite function**, so we choose the decomposition that seems best.

Example \(\PageIndex{10}\): Decomposing a Function

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

**Solution**

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

\[h(x)=5−x^2 \text{ and } g(x)=\sqrt{x} \nonumber\]

We can check our answer by recomposing the functions.

\[g(h(x))=g(5−x^2)=\sqrt{5−x^2} \nonumber\]

\(\PageIndex{7}\)

Write \(f(x)=\dfrac{4}{3−\sqrt{4+x^2}}\) as the composition of two functions.

**Answer**-
Possible answer:

\(g(x)=\sqrt{4+x^2}, \; h(x)=\dfrac{4}{3−x}\)

\(f=h{\circ}g=h(g(x))\)

Access these online resources for additional instruction and practice with composite functions.

- Composite Functions (http://openstaxcollege.org/l/compfunction)
- Composite Function Notation Application (http://openstaxcollege.org/l/compfuncnot)
- Composite Functions Using Graphs (http://openstaxcollege.org/l/compfuncgraph)
- Decompose Functions (http://openstaxcollege.org/l/decompfunction)
- Composite Function Values (http://openstaxcollege.org/l/compfuncvalue)

## Key Equation

- Composite function \((f{\circ}g)(x)=f(g(x))\)

## Key Concepts

- We can perform algebraic operations on functions. See Example \(\PageIndex{1}\).
- When functions are combined, the output of the first (inner) function becomes the input of the second (outer) function.
- The function produced by combining two functions is a composite function.
- The order of function composition must be considered when interpreting the meaning of composite functions. See Examples \(\PageIndex{4}\) and \(\PageIndex{5}\).
- A composite function can be evaluated by evaluating the inner function using the given input value and then evaluating the outer function taking as its input the output of the inner function.
- A composite function can be evaluated from a table. See Example \(\PageIndex{6}\).
- A composite function can be evaluated from a graph. See Example \(\PageIndex{7}\).
- A composite function can be evaluated from a formula. See Example \(\PageIndex{8}\).
- The domain of a composite function consists of those inputs in the domain of the inner function that correspond to outputs of the inner function that are in the domain of the outer function. See Examples \(\PageIndex{9A}\) and \(\PageIndex{9B}\).
- Just as functions can be combined to form a composite function, composite functions can be decomposed into simpler functions. See Example \(\PageIndex{10}\).
- Functions can often be decomposed in more than one way.

## Glossary

**composite function**-
the new function formed by function composition, when the output of one function is used as the input of another

## Contributors

Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus.