composition of functions assignment

1.4 Composition of Functions

Learning objectives.

In this section, you will:

• Combine functions using algebraic operations.
• 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 ) C ( T ) gives the cost C C of heating a house for a given average daily temperature in T T degrees Celsius. The function T ( d ) T ( d ) gives the average daily temperature on day d d of the year. For any given day, Cost = C ( T ( d ) ) 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 ) . T ( d ) . For example, we could evaluate T ( 5 ) 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 ) ) . C ( T ( 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 ) w ( y ) is the wife’s income and h ( y ) h ( y ) is the husband’s income in year y , y , and we want T T to represent the total income, then we can define a new function.

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

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 ) f ( x ) and g ( x ) g ( x ) with real number outputs, we define new functions f + g , f − g , f g , f + g , f − g , f g , and f g f g by the relations

Performing Algebraic Operations on Functions

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

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

No, the functions are not the same.

Note: For ( g f ) ( x ) , ( g f ) ( x ) , the condition x ≠ 1 x ≠ 1 is necessary because when x = 1 , x = 1 , the denominator is equal to 0, which makes the function undefined.

Find and simplify the functions ( f g ) ( x ) ( f g ) ( x ) and ( f − g ) ( x ) . ( f − g ) ( x ) .

Are they the same function?

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 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 “ f “ f composed with g g at x ,” x ,” and the right-hand side as “ f “ f of g g of x . ” x . ” The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol ∘ ∘ 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 ) ) ≠ f ( x ) g ( x ) . f ( g ( x ) ) ≠ 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 g takes the input x x first and yields an output g ( x ) . g ( x ) . Then the function f f takes g ( x ) g ( x ) as an input and yields an output f ( g ( x ) ) . f ( g ( x ) ) .

In general, f ∘ g f ∘ g and g ∘ f g ∘ f are different functions. In other words, in many cases f ( g ( x ) ) ≠ g ( f ( x ) ) f ( g ( x ) ) ≠ g ( f ( x ) ) for all x . x . We will also see that sometimes two functions can be composed only in one specific order.

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

These expressions are not equal for all values of x , 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 . x = − 1 2 .

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.

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 x and functions f f and g , g , this action defines a composite function , which we write as f ∘ g f ∘ g such that

The domain of the composite function f ∘ g f ∘ g is all x x such that x x is in the domain of g g and g ( x ) g ( x ) is in the domain of f . f .

It is important to realize that the product of functions f g f g is not the same as the function composition f ( g ( x ) ) , f ( g ( x ) ) , because, in general, f ( x ) g ( x ) ≠ f ( g ( x ) ) . f ( x ) g ( x ) ≠ f ( g ( x ) ) .

Determining whether Composition of Functions is Commutative

Using the functions provided, find f ( g ( x ) ) f ( g ( x ) ) and g ( f ( x ) ) . g ( f ( x ) ) . Determine whether the composition of the functions is commutative .

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

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

We find that g ( f ( x ) ) ≠ f ( g ( x ) ) , g ( f ( x ) ) ≠ f ( g ( x ) ) , so the operation of function composition is not commutative.

Interpreting Composite Functions

The function c ( s ) c ( s ) gives the number of calories burned completing s s sit-ups, and s ( t ) s ( t ) gives the number of sit-ups a person can complete in t t minutes. Interpret c ( s ( 3 ) ) . c ( s ( 3 ) ) .

The inside expression in the composition is s ( 3 ) . s ( 3 ) . Because the input to the s -function is time, t = 3 t = 3 represents 3 minutes, and s ( 3 ) s ( 3 ) is the number of sit-ups completed in 3 minutes.

Using s ( 3 ) s ( 3 ) as the input to the function c ( s ) 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).

Investigating the Order of Function Composition

Suppose f ( x ) f ( x ) gives miles that can be driven in x x hours and g ( y ) g ( y ) gives the gallons of gas used in driving y y miles. Which of these expressions is meaningful: f ( g ( y ) ) f ( g ( y ) ) or g ( f ( x ) ) ? g ( f ( x ) ) ?

The function y = f ( x ) y = f ( x ) is a function whose output is the number of miles driven corresponding to the number of hours driven.

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

The expression g ( y ) g ( y ) takes miles as the input and a number of gallons as the output. The function f ( x ) 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 ) ) f ( g ( y ) ) is meaningless.

The expression f ( x ) f ( x ) takes hours as input and a number of miles driven as the output. The function g ( y ) g ( y ) requires a number of miles as the input. Using f ( x ) f ( x ) (miles driven) as an input value for g ( y ) , g ( y ) , where gallons of gas depends on miles driven, does make sense. The expression g ( f ( x ) ) g ( f ( x ) ) makes sense, and will yield the number of gallons of gas used, g , g , driving a certain number of miles, f ( x ) , f ( x ) , in x x hours.

Are there any situations where f ( g ( y ) ) f ( g ( y ) ) and g ( f ( x ) ) 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.

The gravitational force on a planet a distance r from the sun is given by the function G ( r ) . G ( r ) . The acceleration of a planet subjected to any force F F is given by the function a ( F ) . a ( F ) . 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.

Using a Table to Evaluate a Composite Function

Using Table 1 , evaluate f ( g ( 3 ) ) f ( g ( 3 ) ) and g ( f ( 3 ) ) . g ( f ( 3 ) ) .

To evaluate f ( g ( 3 ) ), f ( g ( 3 ) ), we start from the inside with the input value 3. We then evaluate the inside expression g ( 3 ) g ( 3 ) using the table that defines the function g : g : g ( 3 ) = 2. g ( 3 ) = 2. We can then use that result as the input to the function f , f , so g ( 3 ) g ( 3 ) is replaced by 2 and we get f ( 2 ) . f ( 2 ) . Then, using the table that defines the function f , f , we find that f ( 2 ) = 8. f ( 2 ) = 8.

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

Table 2 shows the composite functions f ∘ g f ∘ g and g ∘ f g ∘ f as tables.

Using Table 1 , evaluate f ( g ( 1 ) ) f ( g ( 1 ) ) and g ( f ( 4 ) ) . g ( f ( 4 ) ) .

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 - x - and y - 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 - x - axis of its graph.
• Read off the output of the inner function from the y - y - axis of its graph.
• Locate the inner function output on the x - x - axis of the graph of the outer function.
• Read the output of the outer function from the y - y - axis of its graph. This is the output of the composite function.

Using a Graph to Evaluate a Composite Function

Using Figure 1 , evaluate f ( g ( 1 ) ) . f ( g ( 1 ) ) .

To evaluate f ( g ( 1 ) ) , f ( g ( 1 ) ) , we start with the inside evaluation. See Figure 2 .

We evaluate g ( 1 ) g ( 1 ) using the graph of g ( x ) , g ( x ) , finding the input of 1 on the x - x - axis and finding the output value of the graph at that input. Here, g ( 1 ) = 3. g ( 1 ) = 3. We use this value as the input to the function f . f .

We can then evaluate the composite function by looking to the graph of f ( x ) , f ( x ) , finding the input of 3 on the x - x - axis and reading the output value of the graph at this input. Here, f ( 3 ) = 6 , f ( 3 ) = 6 , so f ( g ( 1 ) ) = 6. f ( g ( 1 ) ) = 6.

Figure 3 shows how we can mark the graphs with arrows to trace the path from the input value to the output value.

Using Figure 1 , evaluate g ( f ( 2 ) ) . g ( f ( 2 ) ) .

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 ) ) . 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 , 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.

Evaluating a Composition of Functions Expressed as Formulas with a Numerical Input

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

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

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

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

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

• ⓐ h ( f ( 2 ) ) h ( f ( 2 ) )
• ⓑ h ( f ( − 2 ) ) h ( f ( − 2 ) )

Finding the Domain of a Composite Function

As we discussed previously, the domain of a composite function such as f ∘ g f ∘ g is dependent on the domain of g g and the domain of f . 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 ∘ g . f ∘ g . Let us assume we know the domains of the functions f f and g g separately. If we write the composite function for an input x x as f ( g ( x ) ) , f ( g ( x ) ) , we can see right away that x x must be a member of the domain of g 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 ) g ( x ) must be a member of the domain of f , f , otherwise the second function evaluation in f ( g ( x ) ) f ( g ( x ) ) cannot be completed, and the expression is still undefined. Thus the domain of f ∘ g f ∘ g consists of only those inputs in the domain of g g that produce outputs from g g belonging to the domain of f . f . Note that the domain of f f composed with g g is the set of all x x such that x x is in the domain of g g and g ( x ) g ( x ) is in the domain of f . f .

Domain of a Composite Function

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

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

• Find the domain of g . g .
• Find the domain of f . f .
• Find those inputs x x in the domain of g g for which g ( x ) g ( x ) is in the domain of f . f . That is, exclude those inputs x x from the domain of g g for which g ( x ) g ( x ) is not in the domain of f . f . The resulting set is the domain of f ∘ g . f ∘ g .

Find the domain of

The domain of g ( x ) g ( x ) consists of all real numbers except x = 2 3 , x = 2 3 , since that input value would cause us to divide by 0. Likewise, the domain of f f consists of all real numbers except 1. So we need to exclude from the domain of g ( x ) g ( x ) that value of x x for which g ( x ) = 1. g ( x ) = 1.

So the domain of f ∘ g f ∘ g is the set of all real numbers except 2 3 2 3 and 2. 2. This means that

We can write this in interval notation as

Finding the Domain of a Composite Function Involving Radicals

Because we cannot take the square root of a negative number, the domain of g g is ( − ∞ , 3 ] . ( − ∞ , 3 ] . Now we check the domain of the composite function

For ( f ∘ g ) ( x ) = 3 − x + 2 , 3 − x + 2 ≥ 0 , ( f ∘ g ) ( x ) = 3 − x + 2 , 3 − x + 2 ≥ 0 , since the radicand of a square root must be positive. Since square roots are positive, 3 − x ≥ 0 , 3 − x ≥ 0 , or, 3 − x ≥ 0 , 3 − x ≥ 0 , which gives a domain of ( -∞ , 3 ] ( -∞ , 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 ∘ g f ∘ g can contain values that are not in the domain of f , f , though they must be in the domain of g . 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.

Decomposing a Function

Write f ( x ) = 5 − x 2 f ( x ) = 5 − x 2 as the composition of two functions.

We are looking for two functions, g g and h , h , so f ( x ) = g ( h ( x ) ) . f ( x ) = g ( h ( x ) ) . To do this, we look for a function inside a function in the formula for f ( x ) . f ( x ) . As one possibility, we might notice that the expression 5 − x 2 5 − x 2 is the inside of the square root. We could then decompose the function as

We can check our answer by recomposing the functions.

Write f ( x ) = 4 3 − 4 + x 2 f ( x ) = 4 3 − 4 + x 2 as the composition of two functions.

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

• Composite Functions
• Composite Function Notation Application
• Composite Functions Using Graphs
• Decompose Functions
• Composite Function Values

1.4 Section Exercises

How does one find the domain of the quotient of two functions, f g ? f g ?

What is the composition of two functions, f ∘ g ? f ∘ g ?

If the order is reversed when composing two functions, can the result ever be the same as the answer in the original order of the composition? If yes, give an example. If no, explain why not.

How do you find the domain for the composition of two functions, f ∘ g ? f ∘ g ?

Given f ( x ) = x 2 + 2 x f ( x ) = x 2 + 2 x and g ( x ) = 6 − x 2 , g ( x ) = 6 − x 2 , find f + g , f − g , f g , f + g , f − g , f g , and f g . f g . Determine the domain for each function in interval notation.

Given f ( x ) = − 3 x 2 + x f ( x ) = − 3 x 2 + x and g ( x ) = 5 , g ( x ) = 5 , find f + g , f − g , f g , f + g , f − g , f g , and f g . f g . Determine the domain for each function in interval notation.

Given f ( x ) = 2 x 2 + 4 x f ( x ) = 2 x 2 + 4 x and g ( x ) = 1 2 x , g ( x ) = 1 2 x , find f + g , f − g , f g , f + g , f − g , f g , and f g . f g . Determine the domain for each function in interval notation.

Given f ( x ) = 1 x − 4 f ( x ) = 1 x − 4 and g ( x ) = 1 6 − x , g ( x ) = 1 6 − x , find f + g , f − g , f g , f + g , f − g , f g , and f g . f g . Determine the domain for each function in interval notation.

Given f ( x ) = 3 x 2 f ( x ) = 3 x 2 and g ( x ) = x − 5 , g ( x ) = x − 5 , find f + g , f − g , f g , f + g , f − g , f g , and f g . f g . Determine the domain for each function in interval notation.

Given f ( x ) = x f ( x ) = x and g ( x ) = | x − 3 | , g ( x ) = | x − 3 | , find g f . g f . Determine the domain of the function in interval notation.

Given f ( x ) = 2 x 2 + 1 f ( x ) = 2 x 2 + 1 and g ( x ) = 3 x − 5 , g ( x ) = 3 x − 5 , find the following:

• ⓐ f ( g ( 2 ) ) f ( g ( 2 ) )
• ⓑ f ( g ( x ) ) f ( g ( x ) )
• ⓒ g ( f ( x ) ) g ( f ( x ) )
• ⓓ ( g ∘ g ) ( x ) ( g ∘ g ) ( x )
• ⓔ ( f ∘ f ) ( − 2 ) ( f ∘ f ) ( − 2 )

For the following exercises, use each pair of functions to find f ( g ( x ) ) f ( g ( x ) ) and g ( f ( x ) ) . g ( f ( x ) ) . Simplify your answers.

f ( x ) = x 2 + 1 , g ( x ) = x + 2 f ( x ) = x 2 + 1 , g ( x ) = x + 2

f ( x ) = x + 2 , g ( x ) = x 2 + 3 f ( x ) = x + 2 , g ( x ) = x 2 + 3

f ( x ) = | x | , g ( x ) = 5 x + 1 f ( x ) = | x | , g ( x ) = 5 x + 1

f ( x ) = x 3 , g ( x ) = x + 1 x 3 f ( x ) = x 3 , g ( x ) = x + 1 x 3

f ( x ) = 1 x − 6 , g ( x ) = 7 x + 6 f ( x ) = 1 x − 6 , g ( x ) = 7 x + 6

f ( x ) = 1 x − 4 , g ( x ) = 2 x + 4 f ( x ) = 1 x − 4 , g ( x ) = 2 x + 4

For the following exercises, use each set of functions to find f ( g ( h ( x ) ) ) . f ( g ( h ( x ) ) ) . Simplify your answers.

f ( x ) = x 4 + 6 , f ( x ) = x 4 + 6 , g ( x ) = x − 6 , g ( x ) = x − 6 , and h ( x ) = x h ( x ) = x

f ( x ) = x 2 + 1 , f ( x ) = x 2 + 1 , g ( x ) = 1 x , g ( x ) = 1 x , and h ( x ) = x + 3 h ( x ) = x + 3

Given f ( x ) = 1 x f ( x ) = 1 x and g ( x ) = x − 3 , g ( x ) = x − 3 , find the following:

• ⓐ ( f ∘ g ) ( x ) ( f ∘ g ) ( x )
• ⓑ the domain of ( f ∘ g ) ( x ) ( f ∘ g ) ( x ) in interval notation
• ⓒ ( g ∘ f ) ( x ) ( g ∘ f ) ( x )
• ⓓ the domain of ( g ∘ f ) ( x ) ( g ∘ f ) ( x )
• ⓔ ( f g ) x ( f g ) x

Given f ( x ) = 2 − 4 x f ( x ) = 2 − 4 x and g ( x ) = − 3 x , g ( x ) = − 3 x , find the following:

• ⓐ ( g ∘ f ) ( x ) ( g ∘ f ) ( x )
• ⓑ the domain of ( g ∘ f ) ( x ) ( g ∘ f ) ( x ) in interval notation

Given the functions f ( x ) = 1 − x x and g ( x ) = 1 1 + x 2 , f ( x ) = 1 − x x and g ( x ) = 1 1 + x 2 , find the following:

• ⓑ ( g ∘ f ) ( 2 ) ( g ∘ f ) ( 2 )

Given functions p ( x ) = 1 x p ( x ) = 1 x and m ( x ) = x 2 − 4 , m ( x ) = x 2 − 4 , state the domain of each of the following functions using interval notation:

• ⓐ p ( x ) m ( x ) p ( x ) m ( x )
• ⓑ p ( m ( x ) ) p ( m ( x ) )
• ⓒ m ( p ( x ) ) m ( p ( x ) )

Given functions q ( x ) = 1 x q ( x ) = 1 x and h ( x ) = x 2 − 9 , h ( x ) = x 2 − 9 , state the domain of each of the following functions using interval notation.

• ⓐ q ( x ) h ( x ) q ( x ) h ( x )
• ⓑ q ( h ( x ) ) q ( h ( x ) )
• ⓒ h ( q ( x ) ) h ( q ( x ) )

For f ( x ) = 1 x f ( x ) = 1 x and g ( x ) = x − 1 , g ( x ) = x − 1 , write the domain of ( f ∘ g ) ( x ) ( f ∘ g ) ( x ) in interval notation.

For the following exercises, find functions f ( x ) f ( x ) and g ( x ) g ( x ) so the given function can be expressed as h ( x ) = f ( g ( x ) ) . h ( x ) = f ( g ( x ) ) .

h ( x ) = ( x + 2 ) 2 h ( x ) = ( x + 2 ) 2

h ( x ) = ( x − 5 ) 3 h ( x ) = ( x − 5 ) 3

h ( x ) = 3 x − 5 h ( x ) = 3 x − 5

h ( x ) = 4 ( x + 2 ) 2 h ( x ) = 4 ( x + 2 ) 2

h ( x ) = 4 + x 3 h ( x ) = 4 + x 3

h ( x ) = 1 2 x − 3 3 h ( x ) = 1 2 x − 3 3

h ( x ) = 1 ( 3 x 2 − 4 ) − 3 h ( x ) = 1 ( 3 x 2 − 4 ) − 3

h ( x ) = 3 x − 2 x + 5 4 h ( x ) = 3 x − 2 x + 5 4

h ( x ) = ( 8 + x 3 8 − x 3 ) 4 h ( x ) = ( 8 + x 3 8 − x 3 ) 4

h ( x ) = 2 x + 6 h ( x ) = 2 x + 6

h ( x ) = ( 5 x − 1 ) 3 h ( x ) = ( 5 x − 1 ) 3

h ( x ) = x − 1 3 h ( x ) = x − 1 3

h ( x ) = | x 2 + 7 | h ( x ) = | x 2 + 7 |

h ( x ) = 1 ( x − 2 ) 3 h ( x ) = 1 ( x − 2 ) 3

h ( x ) = ( 1 2 x − 3 ) 2 h ( x ) = ( 1 2 x − 3 ) 2

h ( x ) = 2 x − 1 3 x + 4 h ( x ) = 2 x − 1 3 x + 4

For the following exercises, use the graphs of f , f , shown in Figure 4 , and g , g , shown in Figure 5 , to evaluate the expressions.

f ( g ( 3 ) ) f ( g ( 3 ) )

f ( g ( 1 ) ) f ( g ( 1 ) )

g ( f ( 1 ) ) g ( f ( 1 ) )

g ( f ( 0 ) ) g ( f ( 0 ) )

f ( f ( 5 ) ) f ( f ( 5 ) )

f ( f ( 4 ) ) f ( f ( 4 ) )

g ( g ( 2 ) ) g ( g ( 2 ) )

g ( g ( 0 ) ) g ( g ( 0 ) )

For the following exercises, use graphs of f ( x ) , f ( x ) , shown in Figure 6 , g ( x ) , g ( x ) , shown in Figure 7 , and h ( x ) , h ( x ) , shown in Figure 8 , to evaluate the expressions.

g ( f ( 2 ) ) g ( f ( 2 ) )

f ( g ( 4 ) ) f ( g ( 4 ) )

f ( h ( 2 ) ) f ( h ( 2 ) )

h ( f ( 2 ) ) h ( f ( 2 ) )

f ( g ( h ( 4 ) ) ) f ( g ( h ( 4 ) ) )

f ( g ( f ( − 2 ) ) ) f ( g ( f ( − 2 ) ) )

For the following exercises, use the function values for f  and  g f  and  g shown in Table 3 to evaluate each expression.

f ( g ( 8 ) ) f ( g ( 8 ) )

f ( g ( 5 ) ) f ( g ( 5 ) )

g ( f ( 5 ) ) g ( f ( 5 ) )

g ( f ( 3 ) ) g ( f ( 3 ) )

f ( f ( 1 ) ) f ( f ( 1 ) )

g ( g ( 6 ) ) g ( g ( 6 ) )

For the following exercises, use the function values for f  and  g f  and  g shown in Table 4 to evaluate the expressions.

( f ∘ g ) ( 1 ) ( f ∘ g ) ( 1 )

( f ∘ g ) ( 2 ) ( f ∘ g ) ( 2 )

( g ∘ f ) ( 2 ) ( g ∘ f ) ( 2 )

( g ∘ f ) ( 3 ) ( g ∘ f ) ( 3 )

( g ∘ g ) ( 1 ) ( g ∘ g ) ( 1 )

( f ∘ f ) ( 3 ) ( f ∘ f ) ( 3 )

For the following exercises, use each pair of functions to find f ( g ( 0 ) ) f ( g ( 0 ) ) and g ( f ( 0 ) ) . g ( f ( 0 ) ) .

f ( x ) = 4 x + 8 , g ( x ) = 7 − x 2 f ( x ) = 4 x + 8 , g ( x ) = 7 − x 2

f ( x ) = 5 x + 7 , g ( x ) = 4 − 2 x 2 f ( x ) = 5 x + 7 , g ( x ) = 4 − 2 x 2

f ( x ) = x + 4 , g ( x ) = 12 − x 3 f ( x ) = x + 4 , g ( x ) = 12 − x 3

f ( x ) = 1 x + 2 , g ( x ) = 4 x + 3 f ( x ) = 1 x + 2 , g ( x ) = 4 x + 3

For the following exercises, use the functions f ( x ) = 2 x 2 + 1 f ( x ) = 2 x 2 + 1 and g ( x ) = 3 x + 5 g ( x ) = 3 x + 5 to evaluate or find the composite function as indicated.

f ( g ( 2 ) ) f ( g ( 2 ) )

f ( g ( x ) ) f ( g ( x ) )

g ( f ( − 3 ) ) g ( f ( − 3 ) )

( g ∘ g ) ( x ) ( g ∘ g ) ( x )

For the following exercises, use f ( x ) = x 3 + 1 f ( x ) = x 3 + 1 and g ( x ) = x − 1 3 . g ( x ) = x − 1 3 .

Find ( f ∘ g ) ( x ) ( f ∘ g ) ( x ) and ( g ∘ f ) ( x ) . ( g ∘ f ) ( x ) . Compare the two answers.

Find ( f ∘ g ) ( 2 ) ( f ∘ g ) ( 2 ) and ( g ∘ f ) ( 2 ) . ( g ∘ f ) ( 2 ) .

What is the domain of ( g ∘ f ) ( x ) ? ( g ∘ f ) ( x ) ?

What is the domain of ( f ∘ g ) ( x ) ? ( f ∘ g ) ( x ) ?

Let f ( x ) = 1 x . f ( x ) = 1 x .

• ⓐ Find ( f ∘ f ) ( x ) . ( f ∘ f ) ( x ) .
• ⓑ Is ( f ∘ f ) ( x ) ( f ∘ f ) ( x ) for any function f f the same result as the answer to part (a) for any function? Explain.

For the following exercises, let F ( x ) = ( x + 1 ) 5 , F ( x ) = ( x + 1 ) 5 , f ( x ) = x 5 , f ( x ) = x 5 , and g ( x ) = x + 1. g ( x ) = x + 1.

True or False: ( g ∘ f ) ( x ) = F ( x ) . ( g ∘ f ) ( x ) = F ( x ) .

True or False: ( f ∘ g ) ( x ) = F ( x ) . ( f ∘ g ) ( x ) = F ( x ) .

For the following exercises, find the composition when f ( x ) = x 2 + 2 f ( x ) = x 2 + 2 for all x ≥ 0 x ≥ 0 and g ( x ) = x − 2 . g ( x ) = x − 2 .

( f ∘ g ) ( 6 ) ; ( g ∘ f ) ( 6 ) ( f ∘ g ) ( 6 ) ; ( g ∘ f ) ( 6 )

( g ∘ f ) ( a ) ; ( f ∘ g ) ( a ) ( g ∘ f ) ( a ) ; ( f ∘ g ) ( a )

( f ∘ g ) ( 11 ) ; ( g ∘ f ) ( 11 ) ( f ∘ g ) ( 11 ) ; ( g ∘ f ) ( 11 )

Real-World Applications

The function D ( p ) D ( p ) gives the number of items that will be demanded when the price is p . p . The production cost C ( x ) C ( x ) is the cost of producing x x items. To determine the cost of production when the price is $6, you would do which of the following?

• ⓐ Evaluate D ( C ( 6 ) ) . D ( C ( 6 ) ) .
• ⓑ Evaluate C ( D ( 6 ) ) . C ( D ( 6 ) ) .
• ⓒ Solve D ( C ( x ) ) = 6. D ( C ( x ) ) = 6.
• ⓓ Solve C ( D ( p ) ) = 6. C ( D ( p ) ) = 6.

The function A ( d ) A ( d ) gives the pain level on a scale of 0 to 10 experienced by a patient with d d milligrams of a pain-reducing drug in her system. The milligrams of the drug in the patient’s system after t t minutes is modeled by m ( t ) . m ( t ) . Which of the following would you do in order to determine when the patient will be at a pain level of 4?

• ⓐ Evaluate A ( m ( 4 ) ) . A ( m ( 4 ) ) .
• ⓑ Evaluate m ( A ( 4 ) ) . m ( A ( 4 ) ) .
• ⓒ Solve A ( m ( t ) ) = 4. A ( m ( t ) ) = 4.
• ⓓ Solve m ( A ( d ) ) = 4. m ( A ( d ) ) = 4.

A store offers customers a 30% discount on the price x x of selected items. Then, the store takes off an additional 15% at the cash register. Write a price function P ( x ) P ( x ) that computes the final price of the item in terms of the original price x . x . (Hint: Use function composition to find your answer.)

A rain drop hitting a lake makes a circular ripple. If the radius, in inches, grows as a function of time in minutes according to r ( t ) = 25 t + 2 , r ( t ) = 25 t + 2 , find the area of the ripple as a function of time. Find the area of the ripple at t = 2. t = 2.

A forest fire leaves behind an area of grass burned in an expanding circular pattern. If the radius of the circle of burning grass is increasing with time according to the formula r ( t ) = 2 t + 1 , r ( t ) = 2 t + 1 , express the area burned as a function of time, t t (minutes).

Use the function you found in the previous exercise to find the total area burned after 5 minutes.

The radius r , r , in inches, of a spherical balloon is related to the volume, V , V , by r ( V ) = 3 V 4 π 3 . r ( V ) = 3 V 4 π 3 . Air is pumped into the balloon, so the volume after t t seconds is given by V ( t ) = 10 + 20 t . V ( t ) = 10 + 20 t .

• ⓐ Find the composite function r ( V ( t ) ) . r ( V ( t ) ) .
• ⓑ Find the exact time when the radius reaches 10 inches.

The number of bacteria in a refrigerated food product is given by N ( T ) = 23 T 2 − 56 T + 1 , N ( T ) = 23 T 2 − 56 T + 1 , 3 < T < 33 , 3 < T < 33 , where T T is the temperature of the food. When the food is removed from the refrigerator, the temperature is given by T ( t ) = 5 t + 1.5 , T ( t ) = 5 t + 1.5 , where t t is the time in hours.

• ⓐ Find the composite function N ( T ( t ) ) . N ( T ( t ) ) .
• ⓑ Find the time (round to two decimal places) when the bacteria count reaches 6752.

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• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: Precalculus
• Publication date: Oct 23, 2014
• Location: Houston, Texas
• Book URL: https://openstax.org/books/precalculus/pages/1-introduction-to-functions
• Section URL: https://openstax.org/books/precalculus/pages/1-4-composition-of-functions

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1.6.1: Composition of Functions

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Composition of Functions

If f(x) = x + 2, and g(x) = 2x + 4, what is f(g(x))?

A function can be conceptualized as a 'black box'. The input , or x value is placed into the box, and the box performs a specific set of operations on it. Once the operations are complete, the output (the " f(x) " or " y " value) is retrieved. Once the output is retrieved, the box is ready to work on the next input.

Using this idea, function composition can be seen as a box inside of a box. The input x value goes into the inner box, and then the output of the inner box is used as the input of the outer box.

Functions are often described in terms of “input” and “output.” For example, consider the function f(x) = 2x + 3. When we input an x value, we output a y value, or a function value. We find the output by taking the input x, multiplying by 2 , and adding 3. We can do this for any value of x. Now consider a second function g(x) = 5x. For this function too, we can take an x value, input the x into g(x), and obtain an output. What happens if we take the output of g and use it as the input of f?

Earlier, you were given a problem about finding a composite function .

f(g(x)) = f(2x + 4) = (2x + 4) + 2 = 2x + 6

Given the function definition above, g(x) = 5x. Therefore if x = 4, then we have g(4) = 5(4) = 20. What happens if we then take the output of 20 and use it as the input of f?

Substituting 20 in for x in f(x) = 2x + 3 gives: f(20) = 2(20) + 3 = 43.

The table below shows several examples of this same process:

Examining the values in the table, we can see a pattern: all of the final output values from f are 3 more than 10 times the initial input. We have created a new function called h(x) out of f(x) = 2x + 3 in which g(x) = 5x is the input:

h(x) = f(5x) = 2(5x) + 3 = 10x + 3

When we input one function into another, we call this the composition of the two functions. Formally, we write the composed function as f(g(x)) = 10x + 3 or write it as (f o g)x = 10x + 3

Find f(g(x)) and g(f(x)):

• f(x) = 3x + 1 and g(x) = x 2
• f(x) = 2x + 4 and g(x) = (1/2) x - 2

g(f(x)) = g(3x + 1) = (3x + 1) 2 = 9x 2 + 6x + 1

In both cases, the resulting function is quadratic.

g(f(x)) = g(2x + 4) = (1/2)(2x + 4) - 2 = x+ 2 - 2 = x.

In this case, the composites were equal to each other, and they both equal x, the original input into the function. This means that there is a special relationship between these two functions. We will examine this relationship in Chapter 3. It is important to note, however, that f(g(x) is not necessarily equal to g(f(x)).

Decompose the function f(x) = (3x - 1) 2 - 5 into a quadratic function g(x) and a linear function h(x).

When we compose functions, we are combining two (or more) functions by inputting the output of one function into another. We can also decompose a function. Consider the function f(x) = (2x + 1) 2 . We can decompose this function into an “inside” and an “outside” function. For example, we can construct f(x) = (2x+ 1) 2 with a linear function and a quadratic function. If g(x) = x 2 and h(x) = (2x + 1), then f(x) = g(h(x)). The linear function h(x) = (2x + 1) is the inside function, and the quadratic function g(x) = x 2 is the outside function.

Let h(x) = 3x - 1 and g(x) = x 2 - 5. Then f(x) = g(h(x)) because g(h(x)) = g(3x - 1) = (3x - 1) 2 - 5.

The decomposition of a function is not necessarily unique. For example, there are many ways that we could express a linear function as the composition of other linear functions.

Find: f(g(4))

To find f(g(4)), we need to know what g(4) is, so we know what to substitute into f(x):

Substitute 4 for x for the function g(x), giving: 3⋅4 2

Simplify: 3⋅16=48

∴ g(4)=48

Substitute 48 for the x in the function f(x) giving: 5(48)+3

Simplify: 240+3=243

∴ f(g(4))=243

h(n)=7n+1+4(g(n))

g(t)=−t

f(x)=−2x+g(x)

Find: f(h(−5))

First, let's solve for the value of the inner function, h(−5). Then we'll know what to plug into the outer function.

h(−5)=(7)(−5)+1+4(g(−5))

To solve for the value of h , we need to solve g(−5)

g(−5)=−(−5)

∴ g(−5)=5

Now we have: h(−5)=(7)(−5)+1+(4)(5)

Simplify to get: h(−5)=−14

Now we know that h(−5)=−14. That tells us that f(h(−5)) is f(−14)

Find f(−14)=(−2)(−14)+g(−14)

So to solve for the value of f(−14), we need to solve for the value of g(−14)

g(−14)=−(−14)

∴ g(−14)=14

Now we can finish up!

f(−14)=(−2)(−14)+14

∴ f(−14)=42

For problems 1-4:

f(x)=2x−1 g(x)=3x h(x)=x 2 +1

• Find: f(g(−3))
• Find: f(h(7))
• Find: h(g(−4))
• Find: f(g(h(2)))

Evaluate each composition below:

• Given: \(\ f(x)=-5 x+2 \text { and } g(x)=\frac{1}{2} x+4 . \text { Find } f(g(12))\).
• Given: \(\ g(x)=-3 x+6 \text { and } h(x)=9 x+3 . \text { Find } g\left(h\left(\frac{1}{3}\right)\right)\).
• Given: \(\ f(x)=-\frac{1}{5} x+4 \text { and } g(x)=4 x^{2} . \text { Find } f(g(10))\).
• Given: \(\ g(x)=3|x-4|+6 \text { and } h(x)=-x^{3} . \text { Find } h(g(4))\).
• Given: \(\ f(x)=\sqrt{x+2} \text { and } g(x)=|2 x| \text { . Find } g(f(-7))\).
• Given: \(\ f(x)=-3 x+2 \text { and given } g(x)=2 x^{2} \text { and given } h(x)=4|7-x|+6 \text { . Find }f(g(h(1)))\).
• Given: \(\ f(x)=(-3) \text { and given } g(x)=\sqrt{2 x} \text { and given } h(x)=|4 x|-12 . \text { Find } f(h(g(18)))\).
• Are compositions commutative? In other words, does \(\ f(g(x))=g(f(x))\)?
• Given: \(\ f(x)=-2^{2}-5 x \text { and } h(x)=3 x+2 . \text { Find } f(h(x))\).
• Two functions are inverses of each other if \(\ f(g(x))=x \text { and } g(f(x))=x \text { If } f(x)=x+3\), find its inverse: \(\ g(x)\)
• A toy manufacturer has a new product to sell. The number of units to be sold, \(\ n\), is a function of the price p such that: \(\ n(p)=30-25 p\). The revenue r earned from the sales is a function of the number of units sold n such that: \(\ r(n)=1000-\frac{1}{4} x^{2}\). Find the function for revenue in terms of price, \(\ p\).

Review (Answers)

To see the Review answers, open this PDF file and look for section 1.16.

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Chapter 3 – Functions

Topic 3.7 – The Composition of Functions

The Composition of Functions establishes the technique for combining two functions, using one as the input to the other. Several examples are performed.

Slideshow Full – 4 per page – 9 per page

Algebra Copyright © 2022 by Mike Weimerskirch and the University of Minnesota Board of Regents is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License , except where otherwise noted.

• Math Article
• Composition Of Functions

Composition of Functions

In Maths, the composition of a function is an operation where two functions say f and g generate a new function say h in such a way that h(x) = g(f(x)). It means here function g is applied to the function of x. So, basically, a function is applied to the result of another function.

Let’s have a look at the definition of a composite function .

Composite Functions Definition

Let f : A → B and g : B → C be two functions. Then the composition of f and g, denoted by g ∘ f, is defined as the function g ∘ f : A → C given by g ∘ f (x) = g(f (x)), ∀ x ∈ A.

The below figure shows the representation of composite functions .

The order of function is an important thing while dealing with the composition of functions since (f ∘ g) (x) is not equal to (g ∘ f) (x).

The relation and function is an important concept of Class 11 and 12. See below the function composition symbol and domain with example.

Symbol: It is also denoted as (g∘f)(x), where ∘ is a small circle symbol. We cannot replace ∘ with a dot (.), because it will show as the product of two functions, such as (g.f)(x).

Domain: f(g(x)) is read as f of g of x. In the composition of (f o g) (x) the domain of function f becomes g(x). The domain is a set of all values which go into the function.

Example: If f(x) = 3x+1 and g(x) = x 2 , then f of g of x, f(g(x)) = f(x 2 ) = 3x 2 +1.

If we reverse the function operation, such as f of f of x, g(f(x)) = g(3x+1) = (3x+1) 2

Also, read:

Properties of Function Compositions

Associative Property: As per the associative property of function composition, if there are three functions f, g and h, then they are said to be associative if and only if;

f ∘ (g ∘ h) = (f ∘ g) ∘ h

Commutative Property: Two functions f and g are said to be commute with each other, if and only if;

g ∘ f = f ∘ g

Few more properties are:

• The function composition of one-to-one function is always one to one.
• The function composition of two onto function is always onto
• The inverse of the composition of two functions f and g is equal to the composition of the inverse of both the functions, such as (f ∘ g) -1 = ( g -1 ∘ f -1 ).

How to Solve Composite Functions

In maths, solving a composite function signifies getting the composition of two functions. A small circle (∘) is used to denote the composition of a function.  

Go through the below-given steps to understand how to solve the given composite function.

Step 1: First write the given composition in a different way.

Consider f(x) = x 2 and g(x) = 3x

(f ∘ g) (x) can be written as f[g(x)].

Step 2: Substitute the variable x that is there in the outside function with the inside function by taking the individual functions as a reference.

That means,

(f ∘ g)(x) = f(3x) {since g(x) = 3x}

Step 3: Finally, simplify the obtained function.

(f ∘ g)(x) = f(3x) = (3x) 2  {since f(x) = x 2 }

Function Composition With Itself

It is possible to compose a function with itself. Suppose f is a function, then the composition of function f with itself will be

(f∘f)(x) = f(f(x))

Let us understand this with an example:

Example: If f(x) = 3x 2 , then find (f∘f)(x).

Solution: Given: f(x) = 3x 2

= f (3x 2 )

Example of Composition of Functions

Q.1: If f (x) = 2x and g(x) = x+1, then find (f∘g)(x) if x = 1.

Solution: Given, f(x) = 2x

g(x) = x+ 1

Therefore, the composition of f from g will be;

(f∘g)(x) = f(g(x)) = f(x+1) = 2(x+1)

Now putting the value of x = 1

f(g(1)) = 2(1+1) = 2 (2) = 4

Q.2: If f(x) = 2x +1 and g(x) = -x 2 , then find (g∘f)(x) for x = 2.

Solution: Given,

f(x) = 2x+1

g(x) = -x 2

To find: g(f(x))

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

Now put x =2 to get;

g(f(2)) = -(2.2+1) 2

Q.3: If there are three functions, such as f(x) = x, g(x) = 2x and h(x) = 3x. Then find the composition of these functions such as [f ∘ (g ∘ h)] (x) for x = -1.

To find: [f ∘ (g ∘ h)] (x)

= f ∘ g(3x)

If x = -1, then;

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Composition of Functions

Introduction to composition of functions, learning objectives.

By the end of this lesson, you will be able to:

• Combine functions using algebraic operations.
• 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 [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.

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Suppose we wanted 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 the average daily temperature depends on the particular day of the year. Notice how we have just defined two relationships: The temperature depends on the day, and the cost depends on the temperature. Using descriptive variables, we can notate these two functions.

The first function, \(C(T)\), gives the cost \(C\) of heating a house when the average daily temperature is \(T\) degrees Celsius, and the second, \(T(d)\), gives the average daily temperature on day \(d\) of the year in some city. If we wanted to determine the cost of heating the house on the 5 \({}^{th}\) day of the year, we could do this by linking our two functions together, an idea called composition of functions. Using the function \(T(d)\), we could evaluate \(T(5)\) to determine the average daily temperature on the 5 \({}^{th}\) day of the year. We could then use that temperature as the input to the \(C(T)\) function to find the cost to heat the house on the 5 \({}^{th}\) day of the year: \(C(T(5))\).

Definition: 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 . We write \(f(g(x))\), and read this as “\(f\) of \(g\) of \(x\)” or “\(f\) composed with \(g\) at \(x\)”.

An alternate notation for composition uses the composition operator: \(\circ\)

\((f \circ g)(x)\) is read “\(f\) of \(g\) of \(x\)” or “\(f\) composed with \(g\) at \(x\)”, just like \(f(g(x))\).

Example \(\PageIndex{1}\)

Suppose \(c(s)\) gives the number of calories burned doing s sit-ups, and \(s(t)\) gives the number of sit-ups a person can do in t minutes. Interpret \(c(s(3))\).

When we are asked to interpret, we are being asked to explain the meaning of the expression in words. The inside expression in the composition is \(s(3)\). Since the input to the s function is time, the 3 is representing 3 minutes, and \(s(3)\) is the number of sit-ups that can be done in 3 minutes. Taking this output and using it as the input to the \(c(s)\) function will gives us the calories that can be burned by the number of sit-ups that can be done in 3 minutes.

Note that it is not important that the same variable be used for the output of the inside function and the input to the outside function. However, it is essential that the units on the output of the inside function match the units on the input to the outside function, if the units are specified.

Example \(\PageIndex{2}\)

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 \(g(f(x))\)?

The expression \(g(y)\) takes miles as the input and outputs a number of gallons. The function \(f(x)\) is expecting a number of hours as the input; trying to give it a number of gallons as input does not make sense. Remember the units must match, and number of gallons does not match number of hours, so the expression \(f(g(y))\) is meaningless.

The expression \(f(x)\) takes hours as input and outputs a number of miles driven. The function \(g(y)\) is expecting a number of miles as the input, so giving the output of the \(f(x)\) function (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 give the number of gallons of gas used, \(g\), driving a certain number of miles, \(f(x)\), in \(x\) hours.

Exercise \(\PageIndex{1}\)

In a department store you see a sign that says 50% off clearance merchandise, so final cost \(C\) depends on the clearance price, \(p\), according to the function \(C(p)\). Clearance price, \(p\), depends on the original discount, \(d\), given to the clearance item, \(p(d)\). Interpret \(C(p(d))\).

The final cost, \(C\), depends on the clearance price, \(p\), which is based on the original discount, \(d\). (Or the original discount \(d\), determines the clearance price and the final cost is half of the clearance price.)

Composition of Functions using Tables and Graphs

When working with functions given as tables and graphs, we can look up values for the functions using a provided table or graph, as discussed in section 1.1. We start evaluation from the provided input, and first evaluate the inside function. We can then use the output of the inside function as the input to the outside function. To remember this, always work from the inside out.

Example \(\PageIndex{3}\)

Using the tables below, evaluate \(f(g(3))\)and \(g(f(4))\)

To evaluate \(f(g(3))\), we start from the inside with the 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 \(f\) function, so \(g(3)\) is replaced by the equivalent value 2 and we can evaluate \(f(2)\). Then using the table that defines the function \(f\), we find that \(f(2) = 8\).

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

To evaluate \(g(f(4))\), we first evaluate the inside expression \(f(4)\)using the first table: \(f(4) = 1\). Then using the table for \(g\) we can evaluate:

\[g(f(4))=g(1)=3. \nonumber\]

Exercise \(\PageIndex{2}\)

Using the tables from the example above, evaluate \(f(g(1))\) and \(g(f(3))\).

\(f(g(1)) = f(3) = 3\) and \(g(f(3)) = g(3) = 2\)

Example \(\PageIndex{4}\)

Using the graphs below, evaluate \(f(g(1))\).

To evaluate \(f(g(1))\), we again start with the inside evaluation. We evaluate \(g(1)\) using the graph of the g(x) function, finding the input of 1 on the horizontal axis and finding the output value of the graph at that input. Here, \(g(1) = 3\).

Using this value as the input to the f function, \(f(g(1))=f(3)\). We can then evaluate this by looking to the graph of the \(f(x)\) function, finding the input of 3 on the horizontal axis, and reading the output value of the graph at this input.

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

Exercise \(\PageIndex{3}\)

Using the graphs from the previous example, evaluate \(g(f(2))\).

\(g(f(2)) = g(5) = 3\)

Composition using Formulas

When evaluating a composition of functions where we have either created or been given formulas, the concept of working from the inside out remains the same. First, we evaluate the inside function using the input value provided, then use the resulting output as the input to the outside function.

Example \(\PageIndex{5}\)

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

Since the inside evaluation is \(h(1)\) we start by evaluating the \(h(x)\) function at 1:

\[h(1) = 3(1) + 2 = 5\nonumber \]

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

\[f(h(1)) = f(5) = 5^{2} - 5 = 20\nonumber \]

Exercise \(\PageIndex{4}\)

Using the functions from the example above, evaluate \(h(f(-2))\).

\(h(f(-2)) = h(6) = 20\) ( did you remember to insert your input values using parentheses? )

While we can compose the functions as above for each individual input value, sometimes it would be really helpful to find a single formula which 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 put whatever value is inside the parentheses after the function name into the formula wherever we see the input variable.

Example \(\PageIndex{6}\)

Given \(f(t) = t^{2} - t\), evaluate \(f(3)\) and \(f(-2)\).

\[ \begin{align*} f(3) &= 3^2 - 3 \\[4pt] f(-2) &= (-2)^2 - (-2)\end{align*} \]

We could simplify the results above if we wanted to

\[ \begin{align*} f(3) &= 3^2 - 3 = 9 - 3 = 6 \\[4pt] f(-2) &= (-2)^2 - (-2) = 4 + 2 = 6 \end{align*}\]

We are not limited, however, to using a numerical value as the input to the function. We can put anything into the function: a value, a different variable, or even an algebraic expression, provided we use the input expression everywhere we see the input variable.

Example \(\PageIndex{7}\)

Using the function from the previous example, evaluate f(a).

This means that the input value for \(t\) is some unknown quantity \(a\). As before, we evaluate by replacing the input variable \(t\) with the input quantity, in this case \(a\).

\[f(a)=a^{2} - a\nonumber \]

The same idea can then be applied to expressions more complicated than a single letter.

Example \(\PageIndex{8}\)

Using the same \(f(t)\) function from above, evaluate \(f(x + 2)\).

Everywhere in the formula for f where there was a t , we would replace it with the input \((x+2)\). Since in the original formula the input t was squared in the first term, the entire input \(x+2\) needs to be squared when we substitute, so we need to use grouping parentheses . To avoid problems, it is advisable to always use parentheses around inputs.

\[f(x+2)=(x+2)^{2} -(x+2)\nonumber \]

We could simplify this expression further to \(f(x+2)=x^{2} +3x+2\) if we wanted to:

\[f(x+2)=(x+2)(x+2)-(x+2)\nonumber \]

Use the “FOIL” technique (first, outside, inside, last)

\[f(x+2)=x^{2} +2x+2x+4-(x+2)\nonumber \]

distribute the negative sign

\[f(x+2)=x^{2} +2x+2x+4-x-2\nonumber \]

combine like terms

\[f(x+2)=x^{2} +3x+2\nonumber \]

Example \(\PageIndex{9}\)

Using the same function, evaluate \(f(t^{3} )\).

Note that in this example, the same variable is used in the input expression and as the input variable of the function. This doesn’t matter – we still replace the original input t in the formula with the new input expression, \(t^{3}\).

\[f(t^{3} )=(t^{3} )^{2} -(t^{3} )=t^{6} - t^{3}\nonumber \]

Exercise \(\PageIndex{5}\)

Given \(g(x) = 3x - \sqrt{x}\), evaluate \(g(t - 2)\).

\[g(t - 2) = 3(t - 2) - \sqrt{(t - 2)}\nonumber \]

This now allows us to find an expression for a composition of functions. If we want to find a formula for \(f(g(x))\), we can start by writing out the formula for \(g(x)\). We can then evaluate the function \(f(x)\) at that expression, as in the examples above.

Example \(\PageIndex{10}\)

Let \(f(x)=x^{2}\) and \(g(x)=\dfrac{1}{x} -2x\), find \(f(g(x))\) and \(g(f(x))\).

To find \(f(g(x))\), we start by evaluating the inside, writing out the formula for \(g(x)\).

\[g(x)=\dfrac{1}{x} -2x\nonumber \]

We then use the expression \((\dfrac{1}{x} -2x)\) as input for the function \(f\).

\[f(g(x)) = f(\dfrac{1}{x} - 2x)\nonumber \]

We then evaluate the function \(f(x)\) using the formula for \(g(x)\) as the input.

Since \(f(x)=x^{2}\),

\[f(\dfrac{1}{x} -2x) = (\dfrac{1}{x} - 2x)^{2}\nonumber \]

This gives us the formula for the composition:

\[f(g(x)) = (\dfrac{1}{x} - 2x)^{2}\nonumber \]

Likewise, to find g(f(x)), we evaluate the inside, writing out the formula for f(x)

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

Now we evaluate the function g(x) using \(x^{2}\) as the input.

\[g(f(x))=\dfrac{1}{x^{2} } -2x^{2}\nonumber \]

Exercise \(\PageIndex{6}\)

Let \(f(x) = x^{3} +3x\) and \(g(x)=\sqrt{x}\), find \(f(g(x))\) and \(g(f(x))\).

\[f(g(x) = f(\sqrt{x}) = (\sqrt{x})^3 + 3(\sqrt{x})\nonumber \]

\[g(f(x) = g(x^3 + 3x) = \sqrt{(x^3 + 3x)}\nonumber \]

Example \(\PageIndex{11}\)

A city manager determines that the tax revenue, \(R\), in millions of dollars collected on a population of \(p\) thousand people is given by the formula \(R(p) = 0.03p + \sqrt{p}\), and that the city’s population, in thousands, is predicted to follow the formula \(p(t) = 60 + 2t + 0.3t^{2}\), where t is measured in years after 2010. Find a formula for the tax revenue as a function of the year.

Since we want tax revenue as a function of the year, we want year to be our initial input, and revenue to be our final output. To find revenue, we will first have to predict the city population, and then use that result as the input to the tax function. So we need to find \(R(p(t))\). Evaluating this,

\[R(p(t)) = R(60 + 2t + 0.3t^{2} ) = 0.03(60+2t+0.3t^{2}) + \sqrt{60 + 2t + 0.3t^{2} }\nonumber \]

This composition gives us a single formula which can be used to predict the tax revenue during a given year, without needing to find the intermediary population value.

For example, to predict the tax revenue in 2017, when t = 7 (because t is measured in years after 2010),

\[R(p(7))=0.03(60 + 2(7) + 0.3(7)^{2})+ \sqrt{60 + 2(7) + 0.3(7)^{2} } \approx 12.079\text{ million dollars}\nonumber \]

• Domain of Compositions

When we think about the domain of a composition \(h(x)=f(g(x))\), we must consider both the domain of the inner function and the domain of the composition itself. While it is tempting to only look at the resulting composite function, if the inner function were undefined at a value of \(x\), the composition would not be possible.

Example \(\PageIndex{12}\)

Let \(f(x)=\dfrac{1}{x^{2} -1}\) and \(g(x)=\sqrt{x-2}\). Find the domain of \(f(g(x))\).

Since we want to avoid the square root of negative numbers, the domain of \(g(x)\) is the set of values where \(x - 2 \ge 0\). The domain is \(x \ge 2\).

The composition is \[f(g(x))=\dfrac{1}{(\sqrt{x-2})^{2} - 1} = \dfrac{1}{(x - 2)-1} = \dfrac{1}{x - 3}\nonumber \].

The composition is undefined when \(x = 3\), so that value must also be excluded from the domain. Notice that the composition doesn’t involve a square root, but we still have to consider the domain limitation from the inside function.

Combining the two restrictions, the domain is all values of x greater than or equal to 2, except \(x = 3\).

In inequalities, the domain is: \(2 \le x<3\) or \(x > 3\)

In interval notation, the domain is: \([2, 3) \cup (3, \infty)\).

Exercise \(\PageIndex{7}\)

Let \(f(x) = \dfrac{1}{x-2}\) and \(g(x)=\dfrac{1}{x}\). Find the domain of \(f(g(x))\).

\(g(x) = \dfrac{1}{x}\) is undefined at \(x = 0\).

The composition, \[f(g(x)) = f(\dfrac{1}{x}) = \dfrac{1}{\dfrac{1}{x} - 2} = \dfrac{1}{\dfrac{1}{x} - \dfrac{2x}{x}} = \dfrac{1}{\dfrac{1 - 2x}{x}} = \dfrac{x}{1 - 2x}\nonumber \] is undefined when \(1 - 2x = 0\), when \(x = \dfrac{1}{2}\).

Restricting these two values, the domain is \((-\infty, 0) \cup (0, \dfrac{1}{2}) \cup (\dfrac{1}{2}, \infty)\)

Decomposing Functions

In some cases, it is desirable to decompose a function – to write it as a composition of two simpler functions.

Example \(\PageIndex{13}\)

Write \(f(x)=3+\sqrt{5 - x^{2} }\) 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 \(5 - x^{2}\) is the inside of the square root. We could then decompose the function as:

\[h(x)=5-x^{2}\nonumber \]

\[g(x)=3+\sqrt{x}\nonumber \]

We can check our answer by recomposing the functions:

\[g(h(x)) = g(5 - x^{2})=3 + \sqrt{5-x^2 }\nonumber \]

Note that this is not the only solution to the problem. Another non-trivial decomposition would be \(h(x) = x^{2}\) and \(g(x) = 3 + \sqrt{5-x}\)

Important Topics of this Section

• Definition of Composition of Functions
• Compositions using:
• Decomposition of Functions