Learning Objectives
In this section, you will:
- Create a new function by composition of functions.
- Evaluate composite functions.
- Decompose a composite function into its component functions.
Optional:
- Find the domain of a composite function.
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 [latex]C\left(T\right)[/latex] gives the cost [latex]C[/latex] of heating a house for a given average daily temperature in [latex]T[/latex] degrees Celsius. The function [latex]T\left(d\right)[/latex] gives the average daily temperature on day [latex]d[/latex] of the year. For any given day, [latex]\text{Cost}=C\left(T\left(d\right)\right)[/latex] 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 [latex]T\left(d\right).[/latex] For example, we could evaluate [latex]T\left(5\right)[/latex] 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 [latex]C\left(T\left(5\right)\right).[/latex]
By combining these two relationships into one function, we have performed function composition, which is the focus of this section.
Create a Function by Composition of Functions
We can 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 a composition of functions. The resulting function is known as a composite function. We represent this combination by the following notation:
We read the left-hand side as [latex]\text{“}f[/latex] composed with [latex]g[/latex] at [latex]x,\text{”}[/latex] and the right-hand side as [latex]\text{“}f[/latex] of [latex]g[/latex] of [latex]x.\text{”}[/latex] The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol [latex]\circ [/latex] 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. It is important to realize that the product of functions [latex]fg[/latex] is not the same as the function composition [latex]f\left(g\left(x\right)\right),[/latex] because, in general, [latex]f\left(x\right)g\left(x\right)\ne f\left(g\left(x\right)\right).[/latex]
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 [latex]g[/latex] takes the input [latex]x[/latex] first and yields an output [latex]g\left(x\right).[/latex] Then the function [latex]f[/latex] takes [latex]g\left(x\right)[/latex] as an input and yields an output [latex]f\left(g\left(x\right)\right).[/latex]
In general, [latex]f\circ g[/latex] and [latex]g\circ f[/latex] are different functions. In other words, in many cases [latex]f\left(g\left(x\right)\right)\ne g\left(f\left(x\right)\right)[/latex] for all [latex]x.[/latex] We will also see that sometimes two functions can be composed only in one specific order.
For example, if [latex]f\left(x\right)={x}^{2}[/latex] and [latex]g\left(x\right)=x+2,[/latex] then
but
These expressions are not equal for all values of [latex]x,[/latex] so the two functions are not equal. It is irrelevant that the expressions happen to be equal for the single input value [latex]x=-\frac{1}{2}.[/latex]
Note that 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.
Definition
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 [latex]x[/latex] and functions [latex]f[/latex] and [latex]g,[/latex] this action defines a composite function, which we write as [latex]f\circ g[/latex] such that
The domain of the composite function [latex]f\circ g[/latex] is all [latex]x[/latex] such that [latex]x[/latex] is in the domain of [latex]g[/latex] and [latex]g\left(x\right)[/latex] is in the domain of [latex]f.[/latex]
Example 1: Determining Whether Composition of Functions is Commutative
Using the functions provided, find [latex]f\left(g\left(4\right)\right)[/latex] and [latex]g\left(f\left(4\right)\right).[/latex] Determine whether the composition of the functions is commutative.
Example 2: Interpreting Composite Functions
The function [latex]c\left(s\right)[/latex] gives the number of calories burned completing [latex]s[/latex] sit-ups, and [latex]s\left(t\right)[/latex] gives the number of sit-ups a person can complete in [latex]t[/latex] minutes. Interpret [latex]c\left(s\left(3\right)\right).[/latex]
Example 3: Investigating the Order of Function Composition
Suppose [latex]f\left(x\right)[/latex] gives the number of miles that can be driven in [latex]x[/latex] hours and [latex]g\left(y\right)[/latex] gives the number of gallons of gas used in driving [latex]y[/latex] miles. Which of these expressions is meaningful: [latex]f\left(g\left(y\right)\right)[/latex] or [latex]g\left(f\left(x\right)\right)?[/latex]
Q&A
Are there any situations where [latex]f\left(g\left(y\right)\right)[/latex] and [latex]g\left(f\left(x\right)\right)[/latex] 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.
Try it #1
The gravitational force on a planet a distance [latex]r[/latex] from the sun is given by the function [latex]G\left(r\right).[/latex] The acceleration of a planet subjected to any force [latex]F[/latex] is given by the function [latex]a\left(F\right).[/latex] Form a meaningful composition of these two functions, and explain what it means.
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 4: Using a Table to Evaluate a Composite Function
Using Table 1, evaluate [latex]f\left(g\left(3\right)\right)[/latex] and [latex]g\left(f\left(3\right)\right).[/latex]
| [latex]x[/latex] | [latex]f\left(x\right)[/latex] | [latex]g\left(x\right)[/latex] |
|---|---|---|
| 1 | 6 | 3 |
| 2 | 8 | 5 |
| 3 | 3 | 2 |
| 4 | 1 | 7 |
Try it #2
Using Table 1, evaluate [latex]f\left(g\left(1\right)\right)[/latex] and [latex]g\left(f\left(4\right)\right).[/latex]
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 [latex]x\text{-}[/latex] and [latex]y\text{-}[/latex]axes of the graphs.
How To
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 [latex]x\text{-}[/latex]axis of its graph.
- Read off the output of the inner function from the [latex]y\text{-}[/latex]axis of its graph.
- Locate the inner function output on the [latex]x\text{-}[/latex]axis of the graph of the outer function.
- Read the output of the outer function from the [latex]y\text{-}[/latex]axis of its graph. This is the output of the composite function.
Example 5: Using a Graph to Evaluate a Composite Function
Using Figure 1, evaluate [latex]f\left(g\left(1\right)\right).[/latex]
Figure 1
Try it #3
Using Figure 1, evaluate [latex]g\left(f\left(2\right)\right).[/latex]
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 [latex]f\left(g\left(x\right)\right).[/latex] To do this, we will extend our idea of function evaluation. Recall that, when we evaluate a function like [latex]f\left(t\right)={t}^{2}-t,[/latex] we substitute the value inside the parentheses into the formula wherever we see the input variable.
How To
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 6: Evaluating a Composition of Functions Expressed as Formulas
Given [latex]f\left(t\right)={t}^{2}-t[/latex] and [latex]h\left(x\right)=3x+2,[/latex] evaluate
- [latex]f\left(h\left(1\right)\right).[/latex]
- [latex]f\left(h\left(x\right)\right).[/latex]
- [latex]h\left(f\left(t\right)\right).[/latex]
Example 7: Determining Whether Composition of Functions is Commutative
Using the functions provided, find [latex]f\left(g\left(x\right)\right)[/latex] and [latex]g\left(f\left(x\right)\right).[/latex] Determine whether the composition of the functions is commutative.
Try it #4
Given [latex]f\left(t\right)={t}^{2}-t[/latex] and [latex]h\left(x\right)=3x+2,[/latex] evaluate
- [latex]h\left(f\left(2\right)\right)[/latex]
- [latex]h\left(f\left(-2\right)\right)[/latex]
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. However, in calculus, you will be studying the chain rule in order to find the derivative of a composite function. While we can’t begin to try to define the concept of a derivative at this point, we can help you begin to think about an “inner” and “outer” function in a composition. When you study the chain rule, you will think of the outer function as [latex]f\left(x\right)[/latex] and the inner function as [latex]g\left(x\right).[/latex]
Let’s consider the function [latex]h\left(x\right)={\left(x+10\right)}^{5}.[/latex] This reminds us of the function [latex]f\left(x\right)={x}^{5}[/latex] where the input [latex]x[/latex] has been replaced by [latex]x+10.[/latex] We can see this as though we have a place holder where we would normally see the [latex]x.[/latex] This would look like [latex]{\left(\text{__}\right)}^{5},[/latex] where we are thinking about the “outer” function as being the function that raises the “inside” to the 5th power. We would let the outside function be [latex]f\left(x\right)={x}^{5}[/latex] and the “inside” function be [latex]g\left(x\right)=x+10.[/latex] Can you see that [latex]f\left(g\left(x\right)\right)={\left(x+10\right)}^{5}?[/latex]
Now let’s try a function that isn’t as obvious with outer and inner functions. Consider [latex]k\left(x\right)={2}^{3x+4}.[/latex] This reminds us of the function [latex]f\left(x\right)={2}^{x}[/latex] where the [latex]x[/latex] has been replaced by [latex]3x+4.[/latex] Let’s let our outer function be [latex]f\left(x\right)={2}^{x},[/latex] and our inner function be [latex]g\left(x\right)=3x+4.[/latex] Now form the composite function [latex]f\left(g\left(x\right)\right).[/latex] You get [latex]k\left(x\right)={2}^{3x+4}.[/latex]
Sometimes the easiest way to see the “inside” and “outside” is to consider what a simpler function would look like if the input were simply x and not some more complex expression.
Let’s do one more. Consider [latex]r\left(x\right)=\frac{3}{\sqrt[\leftroot{1}\uproot{2} ]{2x+5}}.[/latex] We can see that this fits the form of a simpler function [latex]f\left(x\right)=\frac{3}{\sqrt[\leftroot{1}\uproot{2} ]{x}}[/latex] where the [latex]x[/latex] has been replaced by [latex]2x+5.[/latex] Our outer function would be [latex]f\left(x\right)=\frac{3}{\sqrt[\leftroot{1}\uproot{2} ]{x}}[/latex] and our inner function would be [latex]g\left(x\right)=2x+5.[/latex] If you form the composite function [latex]f\left(g\left(x\right)\right),[/latex] you will get [latex]r\left(x\right)=\frac{3}{\sqrt[\leftroot{1}\uproot{2} ]{2x+5}}.[/latex]
Example 8: Decomposing a Function
Write [latex]f\left(x\right)=\sqrt{5-{x}^{2}}[/latex] as the composition of two functions.
Try it #5
Write [latex]f\left(x\right)=\frac{4}{3-\sqrt{4+{x}^{2}}}[/latex] as the composition of two functions.
Finding the Domain of a Composite Function (Optional)
As we discussed previously, the domain of a composite function such as [latex]f\circ g[/latex] is dependent on the domain of [latex]g[/latex] and the domain of [latex]f.[/latex] 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 [latex]f\circ g.[/latex] Let us assume we know the domains of the functions [latex]f[/latex] and [latex]g[/latex] separately. If we write the composite function for an input [latex]x[/latex] as [latex]f\left(g\left(x\right)\right),[/latex] we can see right away that [latex]x[/latex] must be a member of the domain of [latex]g[/latex] in order for the expression to be meaningful, because otherwise we cannot complete the inner function evaluation. However, we also see that [latex]g\left(x\right)[/latex] must be a member of the domain of [latex]f,[/latex] otherwise the second function evaluation in [latex]f\left(g\left(x\right)\right)[/latex] cannot be completed, and the expression is still undefined. Thus the domain of [latex]f\circ g[/latex] consists of only those inputs in the domain of [latex]g[/latex] that produce outputs from [latex]g[/latex] belonging to the domain of [latex]f.[/latex] Note that the domain of [latex]f[/latex] composed with [latex]g[/latex] is the set of all [latex]x[/latex] such that [latex]x[/latex] is in the domain of [latex]g[/latex] and [latex]g\left(x\right)[/latex] is in the domain of [latex]f.[/latex]
Definition
The domain of a composite function [latex]f\left(g\left(x\right)\right)[/latex] is the set of those inputs [latex]x[/latex] in the domain of [latex]g[/latex] for which [latex]g\left(x\right)[/latex] is in the domain of [latex]f.[/latex]
How To
Given a function composition [latex]f\left(g\left(x\right)\right),[/latex] determine its domain.
- Find the domain of [latex]g.[/latex]
- Find the domain of [latex]f.[/latex]
- Find those inputs [latex]x[/latex] in the domain of [latex]g[/latex] for which [latex]g\left(x\right)[/latex] is in the domain of [latex]f.[/latex] That is, exclude those inputs [latex]x[/latex] from the domain of [latex]g[/latex] for which [latex]g\left(x\right)[/latex] is not in the domain of [latex]f.[/latex] The resulting set is the domain of [latex]f\circ g.[/latex]
Example 9: Finding the Domain of a Composite Function
Find the domain of
Example 10: Finding the Domain of a Composite Function Involving Radicals
Find the domain of [latex]\left(f\circ g\right)\left(x\right)[/latex] where [latex]f\left(x\right)=\sqrt[\leftroot{1}\uproot{2} ]{x+2}[/latex] and [latex]g\left(x\right)=\sqrt[\leftroot{1}\uproot{2} ]{3-x}.[/latex]
Try it #6
Find the domain of [latex]\left(f\circ g\right)\left(x\right)[/latex] where [latex]f\left(x\right)=\frac{1}{x-2}[/latex] and [latex]g\left(x\right)=\sqrt[\leftroot{1}\uproot{2} ]{x+4}.[/latex]
Access these online resources for additional instruction and practice with composite functions.
Key Equation
| Composite function | [latex]\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)[/latex] |
Key Concepts
- 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.
- 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.
- A composite function can be evaluated from a graph.
- A composite function can be evaluated from a formula.
- Just as functions can be combined to form a composite function, composite functions can be decomposed into simpler functions.
- Functions can often be decomposed in more than one way.
- (Optional)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.
Glossary
- commutative property
- the order of the operations being preformed does not matter if the commutative property holds
- composite function
- the new function formed by function composition, when the output of one function is used as the input of another
