problem solving relations and functions

3.5 Relations and Functions

Learning objectives.

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

• Find the domain and range of a relation
• Determine if a relation is a function
• Find the value of a function

Be Prepared 3.13

Before you get started, take this readiness quiz.

Evaluate 3 x − 5 3 x − 5 when x = −2 x = −2 . If you missed this problem, review Example 1.6 .

Be Prepared 3.14

Evaluate 2 x 2 − x − 3 2 x 2 − x − 3 when x = a . x = a . If you missed this problem, review Example 1.6 .

Be Prepared 3.15

Simplify: 7 x − 1 − 4 x + 5. 7 x − 1 − 4 x + 5. If you missed this problem, review Example 1.7 .

Find the Domain and Range of a Relation

As we go about our daily lives, we have many data items or quantities that are paired to our names. Our social security number, student ID number, email address, phone number and our birthday are matched to our name. There is a relationship between our name and each of those items.

When your professor gets her class roster, the names of all the students in the class are listed in one column and then the student ID number is likely to be in the next column. If we think of the correspondence as a set of ordered pairs, where the first element is a student name and the second element is that student’s ID number, we call this a relation .

The set of all the names of the students in the class is called the domain of the relation and the set of all student ID numbers paired with these students is the range of the relation.

There are many similar situations where one variable is paired or matched with another. The set of ordered pairs that records this matching is a relation.

A relation is any set of ordered pairs, ( x , y ) . ( x , y ) . All the x -values in the ordered pairs together make up the domain . All the y -values in the ordered pairs together make up the range .

Example 3.42

For the relation { ( 1 , 1 ) , ( 2 , 4 ) , ( 3 , 9 ) , ( 4 , 16 ) , ( 5 , 25 ) } : { ( 1 , 1 ) , ( 2 , 4 ) , ( 3 , 9 ) , ( 4 , 16 ) , ( 5 , 25 ) } :

ⓐ Find the domain of the relation.

ⓑ Find the range of the relation.

{ ( 1 , 1 ) , ( 2 , 4 ) , ( 3 , 9 ) , ( 4 , 16 ) , ( 5 , 25 ) } { ( 1 , 1 ) , ( 2 , 4 ) , ( 3 , 9 ) , ( 4 , 16 ) , ( 5 , 25 ) }

ⓐ The domain is the set of all x -values of the relation. { 1 , 2 , 3 , 4 , 5 } { 1 , 2 , 3 , 4 , 5 }

ⓑ The range is the set of all y -values of the relation. { 1 , 4 , 9 , 16 , 25 } { 1 , 4 , 9 , 16 , 25 }

Try It 3.83

For the relation { ( 1 , 1 ) , ( 2 , 8 ) , ( 3 , 27 ) , ( 4 , 64 ) , ( 5 , 125 ) } : { ( 1 , 1 ) , ( 2 , 8 ) , ( 3 , 27 ) , ( 4 , 64 ) , ( 5 , 125 ) } :

Try It 3.84

For the relation { ( 1 , 3 ) , ( 2 , 6 ) , ( 3 , 9 ) , ( 4 , 12 ) , ( 5 , 15 ) } : { ( 1 , 3 ) , ( 2 , 6 ) , ( 3 , 9 ) , ( 4 , 12 ) , ( 5 , 15 ) } :

A mapping is sometimes used to show a relation. The arrows show the pairing of the elements of the domain with the elements of the range.

Example 3.43

Use the mapping of the relation shown to ⓐ list the ordered pairs of the relation, ⓑ find the domain of the relation, and ⓒ find the range of the relation.

ⓐ The arrow shows the matching of the person to their birthday. We create ordered pairs with the person’s name as the x -value and their birthday as the y -value.

{(Alison, April 25), (Penelope, May 23), (June, August 2), (Gregory, September 15), (Geoffrey, January 12), (Lauren, May 10), (Stephen, July 24), (Alice, February 3), (Liz, August 2), (Danny, July 24)}

ⓑ The domain is the set of all x -values of the relation.

{Alison, Penelope, June, Gregory, Geoffrey, Lauren, Stephen, Alice, Liz, Danny}

ⓒ The range is the set of all y -values of the relation.

{January 12, February 3, April 25, May 10, May 23, July 24, August 2, September 15}

Try It 3.85

Use the mapping of the relation shown to ⓐ list the ordered pairs of the relation ⓑ find the domain of the relation ⓒ find the range of the relation.

Try It 3.86

A graph is yet another way that a relation can be represented. The set of ordered pairs of all the points plotted is the relation. The set of all x -coordinates is the domain of the relation and the set of all y -coordinates is the range. Generally we write the numbers in ascending order for both the domain and range.

Example 3.44

Use the graph of the relation to ⓐ list the ordered pairs of the relation ⓑ find the domain of the relation ⓒ find the range of the relation.

ⓐ The ordered pairs of the relation are: { ( 1 , 5 ) , ( −3 , −1 ) , ( 4 , −2 ) , ( 0 , 3 ) , ( 2 , −2 ) , ( −3 , 4 ) } . { ( 1 , 5 ) , ( −3 , −1 ) , ( 4 , −2 ) , ( 0 , 3 ) , ( 2 , −2 ) , ( −3 , 4 ) } .

ⓑ The domain is the set of all x -values of the relation: { −3 , 0 , 1 , 2 , 4 } . { −3 , 0 , 1 , 2 , 4 } .

Notice that while −3 −3 repeats, it is only listed once.

ⓒ The range is the set of all y -values of the relation: { −2 , −1 , 3 , 4 , 5 } . { −2 , −1 , 3 , 4 , 5 } .

Notice that while −2 −2 repeats, it is only listed once.

Try It 3.87

Try it 3.88.

Determine if a Relation is a Function

A special type of relation, called a function , occurs extensively in mathematics. A function is a relation that assigns to each element in its domain exactly one element in the range. For each ordered pair in the relation, each x -value is matched with only one y -value.

A function is a relation that assigns to each element in its domain exactly one element in the range.

The birthday example from Example 3.43 helps us understand this definition. Every person has a birthday but no one has two birthdays. It is okay for two people to share a birthday. It is okay that Danny and Stephen share July 24 th as their birthday and that June and Liz share August 2 nd . Since each person has exactly one birthday, the relation in Example 3.43 is a function.

The relation shown by the graph in Example 3.44 includes the ordered pairs ( −3 , −1 ) ( −3 , −1 ) and ( −3 , 4 ) . ( −3 , 4 ) . Is that okay in a function? No, as this is like one person having two different birthdays.

Example 3.45

Use the set of ordered pairs to (i) determine whether the relation is a function (ii) find the domain of the relation (iii) find the range of the relation.

ⓐ { ( −3 , 27 ) , ( −2 , 8 ) , ( −1 , 1 ) , ( 0 , 0 ) , ( 1 , 1 ) , ( 2 , 8 ) , ( 3 , 27 ) } { ( −3 , 27 ) , ( −2 , 8 ) , ( −1 , 1 ) , ( 0 , 0 ) , ( 1 , 1 ) , ( 2 , 8 ) , ( 3 , 27 ) }

ⓑ { ( 9 , −3 ) , ( 4 , −2 ) , ( 1 , −1 ) , ( 0 , 0 ) , ( 1 , 1 ) , ( 4 , 2 ) , ( 9 , 3 ) } { ( 9 , −3 ) , ( 4 , −2 ) , ( 1 , −1 ) , ( 0 , 0 ) , ( 1 , 1 ) , ( 4 , 2 ) , ( 9 , 3 ) }

(i) Each x -value is matched with only one y -value. So this relation is a function.

(ii) The domain is the set of all x -values in the relation. The domain is: { −3 , −2 , −1 , 0 , 1 , 2 , 3 } . { −3 , −2 , −1 , 0 , 1 , 2 , 3 } .

(iii) The range is the set of all y -values in the relation. Notice we do not list range values twice. The range is: { 27 , 8 , 1 , 0 } . { 27 , 8 , 1 , 0 } .

(i) The x -value 9 is matched with two y -values, both 3 and −3 . −3 . So this relation is not a function.

(ii) The domain is the set of all x -values in the relation. Notice we do not list domain values twice. The domain is: { 0 , 1 , 2 , 4 , 9 } . { 0 , 1 , 2 , 4 , 9 } .

(iii) The range is the set of all y -values in the relation. The range is: { −3 , −2 , −1 , 0 , 1 , 2 , 3 } . { −3 , −2 , −1 , 0 , 1 , 2 , 3 } .

Try It 3.89

Use the set of ordered pairs to (i) determine whether the relation is a function (ii) find the domain of the relation (iii) find the range of the function.

ⓐ { ( −3 , −6 ) , ( −2 , −4 ) , ( −1 , −2 ) , ( 0 , 0 ) , ( 1 , 2 ) , ( 2 , 4 ) , ( 3 , 6 ) } { ( −3 , −6 ) , ( −2 , −4 ) , ( −1 , −2 ) , ( 0 , 0 ) , ( 1 , 2 ) , ( 2 , 4 ) , ( 3 , 6 ) }

ⓑ { ( 8 , −4 ) , ( 4 , −2 ) , ( 2 , −1 ) , ( 0 , 0 ) , ( 2 , 1 ) , ( 4 , 2 ) , ( 8 , 4 ) } { ( 8 , −4 ) , ( 4 , −2 ) , ( 2 , −1 ) , ( 0 , 0 ) , ( 2 , 1 ) , ( 4 , 2 ) , ( 8 , 4 ) }

Try It 3.90

ⓐ { ( 27 , −3 ) , ( 8 , −2 ) , ( 1 , −1 ) , ( 0 , 0 ) , ( 1 , 1 ) , ( 8 , 2 ) , ( 27 , 3 ) } { ( 27 , −3 ) , ( 8 , −2 ) , ( 1 , −1 ) , ( 0 , 0 ) , ( 1 , 1 ) , ( 8 , 2 ) , ( 27 , 3 ) }

ⓑ { ( 7 , −3 ) , ( −5 , −4 ) , ( 8 , 0 ) , ( 0 , 0 ) , ( −6 , 4 ) , ( −2 , 2 ) , ( −1 , 3 ) } { ( 7 , −3 ) , ( −5 , −4 ) , ( 8 , 0 ) , ( 0 , 0 ) , ( −6 , 4 ) , ( −2 , 2 ) , ( −1 , 3 ) }

Example 3.46

Use the mapping to ⓐ determine whether the relation is a function ⓑ find the domain of the relation ⓒ find the range of the relation.

ⓐ Both Lydia and Marty have two phone numbers. So each x -value is not matched with only one y -value. So this relation is not a function.

ⓑ The domain is the set of all x -values in the relation. The domain is: {Lydia, Eugene, Janet, Rick, Marty}

ⓒ The range is the set of all y -values in the relation. The range is:

{ 321-549-3327 , { 321-549-3327 , 427-658-2314 , 427-658-2314 , 321-964-7324 , 321-964-7324 , 684-358-7961 , 684-358-7961 , 684-369-7231 , 684-369-7231 , 798-367-8541 } 798-367-8541 }

Try It 3.91

Try it 3.92.

In algebra, more often than not, functions will be represented by an equation. It is easiest to see if the equation is a function when it is solved for y . If each value of x results in only one value of y , then the equation defines a function.

Example 3.47

Determine whether each equation is a function. Assume x x is the independent variable.

ⓐ 2 x + y = 7 2 x + y = 7 ⓑ y = x 2 + 1 y = x 2 + 1 ⓒ x + y 2 = 3 x + y 2 = 3

ⓐ 2 x + y = 7 2 x + y = 7

For each value of x , we multiply it by −2 −2 and then add 7 to get the y -value

We have that when x = 3 , x = 3 , then y = 1 . y = 1 . It would work similarly for any value of x . Since each value of x , corresponds to only one value of y the equation defines a function.

ⓑ y = x 2 + 1 y = x 2 + 1

For each value of x , we square it and then add 1 to get the y -value.

We have that when x = 2 , x = 2 , then y = 5 . y = 5 . It would work similarly for any value of x . Since each value of x , corresponds to only one value of y the equation defines a function.

We have shown that when x = 2 , x = 2 , then y = 1 y = 1 and y = −1 . y = −1 . It would work similarly for any value of x . Since each value of x does not corresponds to only one value of y the equation does not define a function.

Try It 3.93

Determine whether each equation is a function.

ⓐ 4 x + y = −3 4 x + y = −3 ⓑ x + y 2 = 1 x + y 2 = 1 ⓒ y − x 2 = 2 y − x 2 = 2

Try It 3.94

ⓐ x + y 2 = 4 x + y 2 = 4 ⓑ y = x 2 − 7 y = x 2 − 7 ⓒ y = 5 x − 4 y = 5 x − 4

Find the Value of a Function

It is very convenient to name a function and most often we name it f , g , h , F , G , or H . In any function, for each x -value from the domain we get a corresponding y -value in the range. For the function f , we write this range value y as f ( x ) . f ( x ) . This is called function notation and is read f of x or the value of f at x . In this case the parentheses does not indicate multiplication.

Function Notation

For the function y = f ( x ) y = f ( x )

We read f ( x ) f ( x ) as f of x or the value of f at x .

We call x the independent variable as it can be any value in the domain. We call y the dependent variable as its value depends on x .

Independent and Dependent Variables

For the function y = f ( x ) , y = f ( x ) ,

Much as when you first encountered the variable x , function notation may be rather unsettling. It seems strange because it is new. You will feel more comfortable with the notation as you use it.

Let’s look at the equation y = 4 x − 5 . y = 4 x − 5 . To find the value of y when x = 2 , x = 2 , we know to substitute x = 2 x = 2 into the equation and then simplify.

The value of the function at x = 2 x = 2 is 3.

We do the same thing using function notation, the equation y = 4 x − 5 y = 4 x − 5 can be written as f ( x ) = 4 x − 5 . f ( x ) = 4 x − 5 . To find the value when x = 2 , x = 2 , we write:

This process of finding the value of f ( x ) f ( x ) for a given value of x is called evaluating the function.

Example 3.48

For the function f ( x ) = 2 x 2 + 3 x − 1 , f ( x ) = 2 x 2 + 3 x − 1 , evaluate the function.

ⓐ f ( 3 ) f ( 3 ) ⓑ f ( −2 ) f ( −2 ) ⓒ f ( a ) f ( a )

Try It 3.95

For the function f ( x ) = 3 x 2 − 2 x + 1 , f ( x ) = 3 x 2 − 2 x + 1 , evaluate the function.

ⓐ f ( 3 ) f ( 3 ) ⓑ f ( −1 ) f ( −1 ) ⓒ f ( t ) f ( t )

Try It 3.96

For the function f ( x ) = 2 x 2 + 4 x − 3 , f ( x ) = 2 x 2 + 4 x − 3 , evaluate the function.

ⓐ f ( 2 ) f ( 2 ) ⓑ f ( −3 ) f ( −3 ) ⓒ f ( h ) f ( h )

In the last example, we found f ( x ) f ( x ) for a constant value of x . In the next example, we are asked to find g ( x ) g ( x ) with values of x that are variables. We still follow the same procedure and substitute the variables in for the x .

Example 3.49

For the function g ( x ) = 3 x − 5 , g ( x ) = 3 x − 5 , evaluate the function.

ⓐ g ( h 2 ) g ( h 2 ) ⓑ g ( x + 2 ) g ( x + 2 ) ⓒ g ( x ) + g ( 2 ) g ( x ) + g ( 2 )

Notice the difference between part ⓑ and ⓒ . We get g ( x + 2 ) = 3 x + 1 g ( x + 2 ) = 3 x + 1 and g ( x ) + g ( 2 ) = 3 x − 4 . g ( x ) + g ( 2 ) = 3 x − 4 . So we see that g ( x + 2 ) ≠ g ( x ) + g ( 2 ) . g ( x + 2 ) ≠ g ( x ) + g ( 2 ) .

Try It 3.97

For the function g ( x ) = 4 x − 7 , g ( x ) = 4 x − 7 , evaluate the function.

ⓐ g ( m 2 ) g ( m 2 ) ⓑ g ( x − 3 ) g ( x − 3 ) ⓒ g ( x ) − g ( 3 ) g ( x ) − g ( 3 )

Try It 3.98

For the function h ( x ) = 2 x + 1 , h ( x ) = 2 x + 1 , evaluate the function.

ⓐ h ( k 2 ) h ( k 2 ) ⓑ h ( x + 1 ) h ( x + 1 ) ⓒ h ( x ) + h ( 1 ) h ( x ) + h ( 1 )

Many everyday situations can be modeled using functions.

Example 3.50

The number of unread emails in Sylvia’s account is 75. This number grows by 10 unread emails a day. The function N ( t ) = 75 + 10 t N ( t ) = 75 + 10 t represents the relation between the number of emails, N , and the time, t , measured in days.

ⓐ Determine the independent and dependent variable.

ⓑ Find N ( 5 ) . N ( 5 ) . Explain what this result means.

ⓐ The number of unread emails is a function of the number of days. The number of unread emails, N , depends on the number of days, t . Therefore, the variable N , is the dependent variable and the variable t t is the independent variable.

Since 5 is the number of days, N ( 5 ) , N ( 5 ) , is the number of unread emails after 5 days. After 5 days, there are 125 unread emails in the account.

Try It 3.99

The number of unread emails in Bryan’s account is 100. This number grows by 15 unread emails a day. The function N ( t ) = 100 + 15 t N ( t ) = 100 + 15 t represents the relation between the number of emails, N , and the time, t , measured in days.

ⓑ Find N ( 7 ) . N ( 7 ) . Explain what this result means.

Try It 3.100

The number of unread emails in Anthony’s account is 110. This number grows by 25 unread emails a day. The function N ( t ) = 110 + 25 t N ( t ) = 110 + 25 t represents the relation between the number of emails, N , and the time, t , measured in days.

ⓑ Find N ( 14 ) . N ( 14 ) . Explain what this result means.

Access this online resource for additional instruction and practice with relations and functions.

• Introduction to Functions

Section 3.5 Exercises

Practice makes perfect.

In the following exercises, for each relation ⓐ find the domain of the relation ⓑ find the range of the relation.

{ ( 1 , 4 ) , ( 2 , 8 ) , ( 3 , 12 ) , ( 4 , 16 ) , ( 5 , 20 ) } { ( 1 , 4 ) , ( 2 , 8 ) , ( 3 , 12 ) , ( 4 , 16 ) , ( 5 , 20 ) }

{ ( 1 , −2 ) , ( 2 , −4 ) , ( 3 , −6 ) , ( 4 , −8 ) , ( 5 , −10 ) } { ( 1 , −2 ) , ( 2 , −4 ) , ( 3 , −6 ) , ( 4 , −8 ) , ( 5 , −10 ) }

{ ( 1 , 7 ) , ( 5 , 3 ) , ( 7 , 9 ) , ( −2 , −3 ) , ( −2 , 8 ) } { ( 1 , 7 ) , ( 5 , 3 ) , ( 7 , 9 ) , ( −2 , −3 ) , ( −2 , 8 ) }

{ ( 11 , 3 ) , ( −2 , −7 ) , ( 4 , −8 ) , ( 4 , 17 ) , ( −6 , 9 ) } { ( 11 , 3 ) , ( −2 , −7 ) , ( 4 , −8 ) , ( 4 , 17 ) , ( −6 , 9 ) }

In the following exercises, use the mapping of the relation to ⓐ list the ordered pairs of the relation, ⓑ find the domain of the relation, and ⓒ find the range of the relation.

For a woman of height 5 ′ 4 ″ 5 ′ 4 ″ the mapping below shows the corresponding Body Mass Index (BMI). The body mass index is a measurement of body fat based on height and weight. A BMI of 18.5 – 24.9 18.5 – 24.9 is considered healthy.

For a man of height 5 ′ 11 ′ ′ 5 ′ 11 ′ ′ the mapping below shows the corresponding Body Mass Index (BMI). The body mass index is a measurement of body fat based on height and weight. A BMI of 18.5 – 24.9 18.5 – 24.9 is considered healthy.

In the following exercises, use the graph of the relation to ⓐ list the ordered pairs of the relation ⓑ find the domain of the relation ⓒ find the range of the relation.

In the following exercises, use the set of ordered pairs to ⓐ determine whether the relation is a function, ⓑ find the domain of the relation, and ⓒ find the range of the relation.

{ ( −3 , 9 ) , ( −2 , 4 ) , ( −1 , 1 ) , { ( −3 , 9 ) , ( −2 , 4 ) , ( −1 , 1 ) , ( 0 , 0 ) , ( 1 , 1 ) , ( 2 , 4 ) , ( 3 , 9 ) } ( 0 , 0 ) , ( 1 , 1 ) , ( 2 , 4 ) , ( 3 , 9 ) }

{ ( 9 , −3 ) , ( 4 , −2 ) , ( 1 , −1 ) , { ( 9 , −3 ) , ( 4 , −2 ) , ( 1 , −1 ) , ( 0 , 0 ) , ( 1 , 1 ) , ( 4 , 2 ) , ( 9 , 3 ) } ( 0 , 0 ) , ( 1 , 1 ) , ( 4 , 2 ) , ( 9 , 3 ) }

{ ( −3 , 27 ) , ( −2 , 8 ) , ( −1 , 1 ) , { ( −3 , 27 ) , ( −2 , 8 ) , ( −1 , 1 ) , ( 0 , 0 ) , ( 1 , 1 ) , ( 2 , 8 ) , ( 3 , 27 ) } ( 0 , 0 ) , ( 1 , 1 ) , ( 2 , 8 ) , ( 3 , 27 ) }

{ ( −3 , −27 ) , ( −2 , −8 ) , ( −1 , −1 ) , { ( −3 , −27 ) , ( −2 , −8 ) , ( −1 , −1 ) , ( 0 , 0 ) , ( 1 , 1 ) , ( 2 , 8 ) , ( 3 , 27 ) } ( 0 , 0 ) , ( 1 , 1 ) , ( 2 , 8 ) , ( 3 , 27 ) }

In the following exercises, use the mapping to ⓐ determine whether the relation is a function, ⓑ find the domain of the function, and ⓒ find the range of the function.

In the following exercises, determine whether each equation is a function.

ⓐ 2 x + y = −3 2 x + y = −3 ⓑ y = x 2 y = x 2 ⓒ x + y 2 = −5 x + y 2 = −5

ⓐ y = 3 x − 5 y = 3 x − 5 ⓑ y = x 3 y = x 3 ⓒ 2 x + y 2 = 4 2 x + y 2 = 4

ⓐ y − 3 x 3 = 2 y − 3 x 3 = 2 ⓑ x + y 2 = 3 x + y 2 = 3 ⓒ 3 x − 2 y = 6 3 x − 2 y = 6

ⓐ 2 x − 4 y = 8 2 x − 4 y = 8 ⓑ −4 = x 2 − y −4 = x 2 − y ⓒ y 2 = − x + 5 y 2 = − x + 5

In the following exercises, evaluate the function: ⓐ f ( 2 ) f ( 2 ) ⓑ f ( −1 ) f ( −1 ) ⓒ f ( a ) . f ( a ) .

f ( x ) = 5 x − 3 f ( x ) = 5 x − 3

f ( x ) = 3 x + 4 f ( x ) = 3 x + 4

f ( x ) = −4 x + 2 f ( x ) = −4 x + 2

f ( x ) = −6 x − 3 f ( x ) = −6 x − 3

f ( x ) = x 2 − x + 3 f ( x ) = x 2 − x + 3

f ( x ) = x 2 + x − 2 f ( x ) = x 2 + x − 2

f ( x ) = 2 x 2 − x + 3 f ( x ) = 2 x 2 − x + 3

f ( x ) = 3 x 2 + x − 2 f ( x ) = 3 x 2 + x − 2

In the following exercises, evaluate the function: ⓐ g ( h 2 ) g ( h 2 ) ⓑ g ( x + 2 ) g ( x + 2 ) ⓒ g ( x ) + g ( 2 ) . g ( x ) + g ( 2 ) .

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

g ( x ) = 5 x − 8 g ( x ) = 5 x − 8

g ( x ) = −3 x − 2 g ( x ) = −3 x − 2

g ( x ) = −8 x + 2 g ( x ) = −8 x + 2

g ( x ) = 3 − x g ( x ) = 3 − x

g ( x ) = 7 − 5 x g ( x ) = 7 − 5 x

In the following exercises, evaluate the function.

f ( x ) = 3 x 2 − 5 x ; f ( x ) = 3 x 2 − 5 x ; f ( 2 ) f ( 2 )

g ( x ) = 4 x 2 − 3 x ; g ( x ) = 4 x 2 − 3 x ; g ( 3 ) g ( 3 )

F ( x ) = 2 x 2 − 3 x + 1 ; F ( x ) = 2 x 2 − 3 x + 1 ; F ( −1 ) F ( −1 )

G ( x ) = 3 x 2 − 5 x + 2 ; G ( x ) = 3 x 2 − 5 x + 2 ; G ( −2 ) G ( −2 )

h ( t ) = 2 | t − 5 | + 4 ; h ( t ) = 2 | t − 5 | + 4 ; h ( −4 ) h ( −4 )

h ( y ) = 3 | y − 1 | − 3 ; h ( y ) = 3 | y − 1 | − 3 ; h ( −4 ) h ( −4 )

f ( x ) = x + 2 x − 1 ; f ( x ) = x + 2 x − 1 ; f ( 2 ) f ( 2 )

g ( x ) = x − 2 x + 2 ; g ( x ) = x − 2 x + 2 ; g ( 4 ) g ( 4 )

In the following exercises, solve.

The number of unwatched shows in Sylvia’s DVR is 85. This number grows by 20 unwatched shows per week. The function N ( t ) = 85 + 20 t N ( t ) = 85 + 20 t represents the relation between the number of unwatched shows, N , and the time, t , measured in weeks.

ⓑ Find N ( 4 ) . N ( 4 ) . Explain what this result means

Every day a new puzzle is downloaded into Ken’s account. Right now he has 43 puzzles in his account. The function N ( t ) = 43 + t N ( t ) = 43 + t represents the relation between the number of puzzles, N , and the time, t , measured in days.

ⓑ Find N ( 30 ) . N ( 30 ) . Explain what this result means.

The daily cost to the printing company to print a book is modeled by the function C ( x ) = 3.25 x + 1500 C ( x ) = 3.25 x + 1500 where C is the total daily cost in dollars and x is the number of books printed.

ⓑ Find C ( 0 ) . C ( 0 ) . Explain what this result means.

ⓒ Find C ( 1000 ) . C ( 1000 ) . Explain what this result means.

The daily cost to the manufacturing company is modeled by the function C ( x ) = 7.25 x + 2500 C ( x ) = 7.25 x + 2500 where C ( x ) C ( x ) is the total daily cost and x is the number of items manufactured.

Writing Exercises

In your own words, explain the difference between a relation and a function.

In your own words, explain what is meant by domain and range.

Is every relation a function? Is every function a relation?

How do you find the value of a function?

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

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• Authors: Lynn Marecek, Andrea Honeycutt Mathis
• Publisher/website: OpenStax
• Book title: Intermediate Algebra 2e
• Publication date: May 6, 2020
• Location: Houston, Texas
• Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
• Section URL: https://openstax.org/books/intermediate-algebra-2e/pages/3-5-relations-and-functions

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• Relations and Functions – Explanation & Examples

Relations and Functions – Explanation & Examples

In this article, we will define and elaborate on how you can identify if a relation is a function . Before we go deeper, let’s look at a brief history of functions.

The concept of function was brought to light by mathematicians in the 17 th century. In 1637, a mathematician and the first modern philosopher, Rene Descartes, talked about many mathematical relationships in his book Geometry. Still, the term “function” was officially first used by German mathematician Gottfried Wilhelm Leibniz after about fifty years. He invented a notation y = x to denote a function, dy/dx, to denote a function’s derivative. The notation y = f (x) was introduced by a Swiss mathematician Leonhard Euler in 1734.

Let’s now review some key concepts as used in functions and relations.

• What is a set?

A set is a collection of distinct or well-defined members or elements . In mathematics, members of a set are written within curly braces or brackets {}. Members of assets can be anything such as; numbers, people, or alphabetical letters, etc.

For example,

{a, b, c, …, x, y, z} is a set of alphabet letters.

{…, −4, −2, 0, 2, 4, …} is a set of even numbers.

{2, 3, 5, 7, 11, 13, 17, …} is a set of prime numbers

Two sets are said to be equal; they contain the same members. Consider two sets, A = {1, 2, 3} and B = {3, 1, 2}. Regardless of the members’ position in sets A and B, the two sets are equal because they contain similar members.

• What are ordered-pair numbers?

• What is a domain?

A domain is a set of all input or first values of a function . Input values are generally ‘x’ values of a function.

• What is a range?

• What is a function?

In mathematics, a function can be defined as a rule that relates every element in one set , called the domain, to exactly one element in another set, called the range. For example, y = x + 3 and y = x 2  – 1 are functions because every x-value produces a different y-value.

Types of Functions

Functions can be classified in terms of relations as follows:

• Injective or one-to-one function: The injective function f: P → Q implies that there is a distinct element of Q for each element of P.
• Many to one : The many to one function maps two or more P’s elements to the same element of set Q.
• The Surjective or onto function: This is a function for which every element of set Q there is a pre-image in set P
• Bijective function.

The common functions in algebra include:

• Linear Function
• Inverse Functions
• Constant Function
• Identity Function
• Absolute Value Function

How to Determine if a Relation is a Function?

We can check if a relation is a function either graphically or by following the steps below.

• Examine the x or input values.
• Examine also the y or output values.
• If all the input values are different, then the relation becomes a function, and if the values are repeated, the relation is not a function.

Note: if there is a repetition of the first members with an associated repetition of the second members, the relation becomes a function.

Identify the range and domain the relation below:

{(-2, 3), {4, 5), (6, -5), (-2, 3)}

Since the x values are the domain, the answer is, therefore,

⟹ {-2, 4, 6}

The range is {-5, 3, 5}.

Check whether the following relation is a function:

B = {(1, 5), (1, 5), (3, -8), (3, -8), (3, -8)}

Though a relation is not classified as a function if there is a repetition of x – values, this problem is a bit tricky because x values are repeated with their corresponding y-values.

Determine the domain and range of the following function: Z = {(1, 120), (2, 100), (3, 150), (4, 130)}.

Domain of z = {1, 2, 3, 4 and the range is {120, 100, 150, 130}

Check if the following ordered pairs are functions:

• W= {(1, 2), (2, 3), (3, 4), (4, 5)
• Y = {(1, 6), (2, 5), (1, 9), (4, 3)}
• All the first values in W = {(1, 2), (2, 3), (3, 4), (4, 5)} are not repeated, therefore, this is a function.
• Y = {(1, 6), (2, 5), (1, 9), (4, 3)} is not a function because, the first value 1 has been repeated twice.

Determine whether the following ordered pairs of numbers are a function.

R = (1,1); (2,2); (3,1); (4,2); (5,1); (6,7)

There is no repetition of x values in the given set of ordered pairs of numbers.

Therefore, R = (1,1); (2,2); (3,1); (4,2); (5,1); (6,7) is a function.

Practice Questions

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Relations And Functions

Related Pages More On Relations And Functions Graphs Of Functions Algebra Lessons

In these lessons, we will look at ordered-pair numbers, relations, and functions. We will also discuss the difference between a relation and a function, and how to use the vertical line test.

Ordered-Pair Numbers

An ordered-pair number is a pair of numbers that go together. The numbers are written within a set of parentheses and separated by a comma.

For example, (4, 7) is an ordered-pair number; the order is designated by the first element 4 and the second element 7. The pair (7, 4) is not the same as (4, 7) because of the different ordering. Sets of ordered-pair numbers can represent relations or functions.

A relation is any set of ordered-pair numbers.

The following diagram shows some examples of relations and functions. Scroll down the page for more examples and solutions on how to determine if a relation is a function.

Suppose the weights of four students are shown in the following table.

The pairing of the student number and his corresponding weight is a relation and can be written as a set of ordered-pair numbers. W = {(1, 120), (2, 100), (3, 150), (4, 130)}

The set of all first elements is called the domain of the relation. The domain of W = {1, 2, 3, 4}

The set of second elements is called the range of the relation. The range of W = {120, 100, 150, 130}

All functions are relations. A function is a relation in which no two ordered pairs have the same first element.

A function associates each element in its domain with one and only one element in its range.

Example: Determine whether the following are functions a) A = {(1, 2), (2, 3), (3, 4), (4, 5)} b) B = {(1, 3), (0, 3), (2, 1), (4, 2)} c) C = {(1, 6), (2, 5), (1, 9), (4, 3)}

Solution: a) A = {(1, 2), (2, 3), (3, 4), (4, 5)} is a function because all the first elements are different.

b) B = {(1, 3), (0, 3), (2, 1), (4, 2)} is a function because all the first elements are different. (The second element does not need to be unique)

c) C = {(1, 6), (2, 5), (1, 9), (4, 3)} is not a function because the first element, 1, is repeated.

Vertical Line Test

A function can be identified from a graph. If any vertical line drawn through the graph cuts the graph at more than one point, then the relation is not a function. This is called the vertical line test .

Determining Whether A Relation Is A Function

Understanding relations (defined as a set of inputs and corresponding outputs) is an important step to learning what makes a function. A function is a specific relation, and determining whether a relation is a function is a skill necessary for knowing what we can graph. Determining whether a relation is a function involves making sure that for every input there is only one output.

How To Determine If A Relation Is A Function?

A function is a correspondence between a first set, called the domain, and a second set, called the range, such that each member of the domain corresponds to exactly one member of the range.

The graph of a function f is a drawing hat represents all the input-output pairs, (x, f(x)). In cases where the function is given by an equation, the graph of a function is the graph of the equation y = f(x).

The vertical line test - a graph represents a function if it is impossible to draw a vertical line that intersects the graph more than once.

This Algebra 1 level math video tutorial

• defines a relation as a set of ordered pairs and a function as a relation with one to one correspondence.
• models how to determine if a relation is a function with two different methods.
• shows how to use a mapping and the vertical line test.
• discusses how to work with function notation. It is defined as replacing y in an equation that is a function.
• Using a mapping diagram, determine whether each relation is a function.
• Using a vertical line test, determine whether the relation is a function.
• Make a table for f(t) = 0.5x + 1. Use 1, 2, 3, and 4 as domain values.
• Evaluate the function rule f(g) = -2g + 4 to find the range for the domain (-1, 3, 5).

Determine If A Relation Is A Function

This video explains the concepts behind mapping a relation and the vertical line test.

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Relations and Functions

Relations and Functions in real life give us the link between any two entities. In our daily life, we come across many patterns and links that characterize relations such as a relation between a father and a son, brother and sister, etc. In mathematics also, we come across many relations between numbers such as a number x is less than y, line l is parallel to line m, etc. Relation and function map elements of one set (domain) to the elements of another set (codomain).

Functions are nothing but special types of relations that define the precise correspondence between one quantity with the other. In this article, we will study how to link pairs of elements from two sets and then define a relation between them, different types of relations and functions, and the difference between relation and function.

What are Relations and Functions?

Relations and functions define a mapping between two sets (Inputs and Outputs) such that they have ordered pairs of the form (Input, Output). Relation and function are very important concepts in algebra and calculus . They are used widely in mathematics as well as in real life. Let us define each of these terms of relation and function to understand their meaning.

Relation and Function Definition

Relation and function individually are defined as:

• Relations - A relation R from a non-empty A to a non-empty set B is a subset of the cartesian product A × B. The subset is derived by describing a relationship between the first element and the second element of the ordered pairs in A × B.
• Functions - A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B. In other words, no two distinct elements of B have the same pre-image.

Note: Please note that all functions are relations but all relations are not functions.

Representation of Relation and Function

Relations and functions can be represented in different forms such as arrow representation, algebraic form, set-builder form , graphical form, roster form , and tabular form. Define a function f: A = {1, 2, 3} → B = {1, 4, 9} such that f(1) = 1, f(2) = 4, f(3) = 9. Now, let us represent this function in different forms.

• Set-builder form - {(x, y): y = x 2 , x ∈ A, y ∈ B}
• Roster form - {(1, 1), (2, 4), (3, 9)}

• Table Representation -

Difference Between Relation and Function

The basic difference between a relation and a function is that in a relation, a single input may have multiple outputs. Whereas in a function, each input has a single output. The table given below highlights the differences between relations and functions .

Note: Look at the example of the relation above: {(1, x), (1, y), (4, z)}. This is NOT a function as a single element (1) is related to multiple elements (x and y). Hence the statement "every relation is a function" is incorrect.

Terms Related to Relations and Functions

Now that we have understood the meaning of relation and function, let us understand the meanings of a few terms related to relations and functions that will help to understand the concept in a better way:

• Cartesian Product - Given two non-empty sets P and Q, the cartesian product P × Q is the set of all ordered pairs of elements from P and Q, that is, P × Q = {(p, q) : p ∈ P, q ∈ Q}
• Domain - The set of all first elements of the ordered pairs in a relation R from a set A to a set B is called the domain of the relation R. It is called the set of inputs or pre-images.
• Range - The set of all second elements of the ordered pairs in a relation R from a set A to a set B is called the range of the relation R. It is called the set of outputs or images.
• Codomain - The whole set B in a relation R from a set A to a set B is called the codomain of the relation R. Note that range is a subset of codomain. i.e., Range ⊆ Codomain

Types of Relations and Functions

Different types of relations and functions have specific properties which make them different and unique. Let us go through the list of types of relations and functions given below:

Types of Relations

Given below is a list of different types of relations:

• Empty Relation - A relation is an empty relation if it has no elements, that is, no element of set A is mapped or linked to any element of A. It is denoted by R = ∅.
• Universal Relation - A relation R in a set A is a universal relation if each element of A is related to every element of A, i.e., R = A × A. It is called the full relation.
• Identity Relation - A relation R on A is said to be an identity relation if each element of A is related to itself, that is, R = {(a, a) : for all a ∈ A}
• Inverse Relation - Define R to be a relation from set P to set Q i.e., R ∈ P × Q. The relation R -1 is said to be an Inverse relation if R -1 from set Q to P is denoted by R -1 = {(q, p): (p, q) ∈ R}.
• Reflexive Relation - A binary relation R defined on a set A is said to be reflexive if, for every element a ∈ A, we have aRa, that is, (a, a) ∈ R.
• Symmetric Relation - A binary relation R defined on a set A is said to be symmetric if and only if, for elements a, b ∈ A, we have aRb, that is, (a, b) ∈ R, then we must have bRa, that is, (b, a) ∈ R.
• Transitive Relation - A relation R is transitive if and only if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for a, b, c ∈ A
• Equivalence Relation - A relation R defined on a set A is said to be an equivalence relation if and only if it is reflexive, symmetric and transitive.
• Antisymmetric Relation - A relation R on a set A is said to be antisymmetric if (a, b) ∈ R and (b , a) ∈ R ⇒ a = b.

Types of Functions

Given below is a list of different types of functions:

• One-to-One Function - A function f: A → B is said to be one-to-one if each element of A is mapped to a distinct element of B. It is also known as Injective Function .
• Onto Function - A function f: A → B is said to be onto, if every element of B is the image of some element of A under f, i.e, for every b ∈ B, there exists an element a in A such that f(a) = b. A function is onto if and only if the range of the function = B.
• Many to One Function - A many to one function is defined by the function f: A → B, such that more than one element of the set A are connected to the same element in the set B.
• Bijective Function - A function that is both one-to-one and onto function is called a bijective function.
• Constant Function - The constant function is of the form f(x) = K, where K is a real number. For the different values of the domain(x value), the same range value of K is obtained for a constant function.
• Identity Function - An identity function is a function where each element in a set B gives the image of itself as the same element i.e., g (b) = b ∀ b ∈ B. Thus, it is of the form g(x) = x.
• linear function
• quadratic function
• cubic function
• polynomial function
• objective functions

Important Notes on Relation and Function:

• Relations and functions define the relation between the elements of two sets and are represented as a set of ordered pairs.
• Relations and functions can be represented in different forms such as arrow representation, algebraic form, set-builder form, graph form, roster form, and tabular form.
• A relation may not be a function but every function is a relation.

☛ Related Topics:

• NCERT Solutions Class 12 Maths Chapter 1 Relations and Functions
• NCERT Solutions Class 11 Maths Chapter 2 Relations and Functions
• Relations and Functions Worksheets
• Domain and Range of a Function

Relations and Functions Examples

Example 1: Given three relations R, S, T from A = {x, y, z} to B = {u, v, w} defined as: 1) R = {(x, u), (z, v)}, 2) S = {(x, u), (y, v), (z, w)}, 3) T = {(x, u), (x, v), (z, w)}. Identify which of the given relations is/are function(s) using relations and functions definition.

Solution: Let us check each part one by one.

1) For R = {(x, u), (z, v)}, each element of A is not mapped to an element of B which violates the definition of a function. Hence, R is not a function.

2) For S = {(x, u), (y, v), (z, w)}, each element of A is mapped to a unique element of B which satisfies the definition of a function. Hence, S is a function.

3) For T = {(x, u), (x, v), (z, w)}, element x of A is mapped to two different elements of B which violates the definition of a function. Hence, T is not a function.

Answer: S = {(x, u), (y, v), (z, w)} is a function.

Example 2: Define a relation R from A to A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y = x + 1}. Determine the domain, codomain and range of R.

Solution: We can see that A = {1, 2, 3, 4, 5, 6} is the domain and codomain of R.

To determine the range, we determine the values of y for each value of x, that is, when x = 1, 2, 3, 4, 5, 6

• x = 1, y = 1 + 1 = 2;
• x = 2, y = 2 + 1 = 3;
• x = 3, y = 3 + 1 = 4;
• x = 4, y = 4 + 1 = 5;
• x = 5, y = 5 + 1 = 6;
• x = 6, y = 6 + 1 = 7.

Since 7 does not belong to A and the relation R is defined on A, hence, x = 6 has no image in A.

Therefore range of R = {2, 3, 4, 5, 6}

Answer: Domain = Codomain = {1, 2, 3, 4, 5, 6}, Range = {2, 3, 4, 5, 6}

Example 3: Is the function f(x) = x 2 one-one from ℝ to ℝ? Justify your answer.

The given function is not one-one. To justify this, note the following examples:

• f(-1) = (-1) 2 = 1
• f(1) = (1) 2 = 1

Here, each of -1 and 1 are mapped to the same element 1 and hence f(x) is NOT one-one.

Answer: The given function is not one to one.

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Practice Questions on Relation and Function

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FAQs on Relations and Functions

What are relation and function in math.

In Math, Relations and functions are defined as follows:

• Relation: A relation from set A to set B is the set of ordered pairs from A to B.
• Function: A function from set A to set B is a relation such that every element of A is mapped to exactly one element of B.

What is the Difference between Relation and Function?

The difference between relations and functions is that relations define any relationship between inputs and outputs whereas a function defines a relation such that each input has only one output. Every function is a relation but a relation doesn't need to be a function.

What is the Range of Relation and Function?

The set of all outputs obtained from a relation or a function is called its range.

What is an Inverse Relation?

An inverse relation of a relation R is denoted by R -1 and is obtained by interchanging x and y coordinates of every ordered pair of R. For example, if R = {(1, 2), (2, 4), (3, 6)} then R -1 = {(2, 1), (4, 2), (6, 3)}.

How Do you Represent Relation and Function?

Relations and functions can be represented in different forms such as

• arrow diagram
• algebraic form
• set-builder form
• graphical form
• roster form
• tabular form.

What are the Types of Relations and Functions?

There are different types of relations and functions.

• Types of Relations: Empty relation, universal relation, reflexive relation, symmetric relation, transitive relation, equivalence relation, etc
• Types of Functions: Constant function, polynomial function, identity function, on-to-one function, onto function, bijective function, etc.

What are Ordered Pairs in Relations and Functions?

Relations and functions have ordered pairs of the form (Input, Output). The input belongs to the domain and the output belongs to the range of the relation/function.

How to Determine if a Relation is a Function?

If in a relation, each element in the domain is mapped to a unique element in the codomain, then it is said to be a function.

What is the Relation Between Sets Functions and Relations?

Functions and relations are subsets of the cartesian product of two sets. For example, if A = {1, 2, 3} and B = {a, b, c} are two sets then:

• R = {(1, a), (1, b), (2, c)} is a relation.
• F = {(1, b), (2, a), (3, c)} is a function.

How to Identify Relations and Functions?

A binary operation defining a link between a set of elements with another set of elements is a relation. But if each element of the first set is mapped to one and only one element of the second set, then it is a function.

Where can I Learn about Relations and Function of Class 11 and Class 12?

Relations and functions is introduced in class 11 and is taken into depth in class 12. The following links are very helpful in learning about relation and function in each of these classes.

• Relations and functions class 11
• Relation and function class 12

Math Functions, Relations, Domain & Range

So, what is a 'relation'.

In math, a relation is just a set of ordered pairs .

Note: {} are the symbol for "set"

• {(0, 1) , (55, 22), (3, -50)}
• {(0, 1) , (5, 2), (-3, 9)}
• {(-1, 7) , (1, 7), (33, 7), (32, 7)}
• { 3, 1, 2 }
• {(0, 1, 2 ) , (3,4,5)} ( these numbers are grouped as 3's so not ordered and therefore not a relation )
• {-1, 7, 3,4,5,5}

One more time: A relation is just a set of ordered pairs. There is absolutely nothing special at all about the numbers that are in a relation. In other words, any bunch of numbers is a relation so long as these numbers come in pairs .

Video Lesson

What is the domain and range of a 'relation'?

Is the set of all the first numbers of the ordered pairs. In other words, the domain is all of the x-values.

Is the set of the second numbers in each pair, or the y-values.

In the relation above the domain is { 5, 1 , 3 } .( highlight ) And the range is {10, 20, 22} ( highlight ).

Domain and range of a relation

In the relation above, the domain is {2, 4, 11, -21} the range is is {-5, 31, -11, 3} .

Arrow Chart

Relations are often represented using arrow charts connecting the domain and range elements.

I. Practice Identifying Domain and Range

What is the domain and range of the following relation ?

{(-1, 2), (2, 51), (1, 3), (8, 22), (9, 51)}

Domain : -1, 2, 1, 8, 9

Range : 2, 51, 3, 22, 51

What is the domain and range of the following relation?

{(-5, 6), (21, -51), (11, 93), (81, 202), (19, 51)}

Domain : -5, 21, 11, 81, 19

Range : 6, -51, 93, 202, 51

What makes a relation a function ?

Functions are a special kind of relation .

• A function is a set of ordered pairs such as {(0, 1) , (5, 22), (11, 9)}.
• Like a relation , a function has a domain and range made up of the x and y values of ordered pairs .

In mathematics, what distinguishes a function from a relation is that each x value in a function has one and only ONE y-value .

Since relation #1 has ONLY ONE y value for each x value, this relation is a function .

On the other hand, relation #2 has TWO distinct y values 'a' and 'c' for the same x value of '5' . Therefore, relation #2 does not satisfy the definition of a mathematical function .

Teachers has multiple students

If we put teachers into the domain and students into the range, we do not have a function because the same teacher, like Mr. Gino below, has more than 1 student in a classroom.

Mothers and Daughters Analogy

A way to try to understand this concept is to think of how mothers and their daughters could be represented as a function. Each element in the domain, each daughter , can only have 1 mother (element in the range).

Some people find it helpful to think of the domain and range as people in romantic relationships. If each number in the domain is a person and each number in the range is a different person, then a function is when all of the people in the domain have 1 and only 1 boyfriend/girlfriend in the range.

Compare the two relations on the below. They differ by just one number, but only one is a function.

What's an easy way to do this?

Look for repeated elements in the domain . As soon as an element in the domain repeats, watch out!

II. Practice Identifying Functions

Which relations below are functions ?

• Relation #1 {(-1, 2), (-4, 51), (1, 2), (8, -51)}
• Relation #2 { (13, 14) , (13, 5) , (16, 7), (18, 13)}
• Relation #3 {(3, 90), (4, 54), (6, 71), (8, 90)}

Relation #1 and Relation #3 are both functions. Remember if domain element repeats then it's not a function.

• Relation #1 {(3, 4), (4, 5), (6, 7), (8, 9)}
• Relation #2 { (3, 4) , (4, 5), (6, 7), (3, 9) }
• Relation #3 {(-3, 4), (4, -5), (0, 0), (8, 9)}
• Relation #4 { (8, 11) , (34, 5), (6, 17), (8, 19) }

Relation #1 and Relation #3 are functions because each x value, each element in the domain, has one and only only one y value, or one and only number in the range.

Remember if a domain element repeats then it's not a function.

For the following relation to be a function, X can not be what values?

• {( 8 , 11), ( 34 ,5), ( 6 ,17), (X ,22)}

X cannot be 8, 34, or 6.

If x were 8 for instance, the relation would be:

{( 8, 11 ), (34, 5), (6, 17), ( 8 ,22 )}

In this relation, the x-value of 8 has two distinct y values. Therefore this relation would NOT be a function since each element in the domain must have 1 and only value in the range.

For the relation below to be a function, X cannot be what values?

• {( 12 , 13), ( -11 , 22), ( 33 , 101), (X, 22)}

X cannot be 12 or 33.

If x were 12 for instance, the relation would be: {( 12 , 13), (-11, 22), ( 33 , 101), ( 12 ,22}

In this problem, x could be -11 . Since (-11, 22) is already a pair in our relation, -11 can again go with a range element of 22 without creating a problem (We would just have two copies of 1 ordered pair).

If x were -11 , the relation would still be a function: {(12, 13), (-11, 22) , (33, 101), (-11, 22) }

The all important rule for a function in math -- that each value in the domain has only 1 value in the range -- would still be true if we had a second copy of 1 ordered pair.

• {( 12 ,14), ( 13 ,5) , ( -2 ,7), (X,13)}

X cannot be 12, 13, or -2.

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Relations and Functions Worksheet

What is the Relation?

Relations can be described as a set of ordered pairs. Let’s look at some examples of relation, such as:

• {(1,0) , (25,50)}
• {(Mon, Sun), (Tue, Sat)}

Where { } denotes the set symbol.

A relation is a correspondence among two or sets (known as the domain and range) such that there are one or more elements assigned to every element or member of the domain.

(2, 4), (2, 3), (3, 7), (5, 2) is a relation of

Domain {2, 3, 5}

Range {2, 3, 4, 7}

What is Function?

A relation “f” from set “X” to set “y” is said to be a function, if every element of set X has only one image in set Y. The function is symbolically represented as  f : X →Y. It means that the f is a function from set X to set Y , X is called the domain of function “f’ and Y is called the codoamin of the function “f”.

Worksheet on Relations and Functions

Solve the problems on relations and functions given below:

• If  the function is f(x)= 6x-27, and the domain is {-5, 3, 15, 17}, then find the range.
• Calculate the range for the function f(x)= (3x-2)/5, if its domain is {-6, -1, 4, 9, 19}
• f(x)= -|3x-7|
• f(x) = 4/(x+1)
• f(x)= 4x 2 – 2
• {(12, -18), (15, 1), (12, 5), (0, 9), (-5, -17)} – _____
• {(15, -3), (-6, 9), (-3, 0), (-1, 16)} – _____
• 9y 9 = 11+8x

(a) 4+3x = y 8   (b) y 5 = -12-x   (c) -2y 6 =-5+9x  (d)(7x 2 +15)/4 = y 2 Find the domain and range for the given relation:

     Domain = _____

     Range = _____ Which of the following statement is true:

• All the relations are functions.
• In every relation, each input value has exactly one output value.
• A relation is defined as a set of input and output values which are related in some way.
• All the above-given statements are true.

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WORD PROBLEMS ON RELATIONS AND FUNCTIONS

Problem 1 :

The total cost of airfare on a given route is comprised of the base cost C and the fuel surcharge S in rupee. Both C and S are functions of the mileage m; C(m) = 0.4m + 50 and S(m) = 0.03m. Determine a function for the total cost of a ticket in terms of the mileage and find the airfare for flying 1600 miles.

Total cost   =  C(m) + S(m)

  =  0.4m + 50 +  0.03m

T(m)  =  0.43 m + 50

To  find the airfare for flying 1600 miles, we have to apply 1600 instead of m.

T(m)  =  0.43 (1600) + 50

  =  688 + 50

T(m)  =  738

So, the  total cost of airfare for flying 1600 miles is 738.

Problem 2 :

A salesperson whose annual earnings can be represented by the function A(x) = 30, 000 + 0.04x, where x is the rupee value of the merchandise he sells. His son is also in sales and his earnings are represented by the function S(x) = 25, 000 + 0.05x. Find (A + S)(x) and determine the total family income if they each sell Rupees 1, 50, 00, 000 worth of merchandise.

(A + S)(x)   =  A(x) + S(x)

A(x) = 30, 000 + 0.04x    -------(1)

S(x) = 25, 000 + 0.05x     -------(2)

   =   30, 000 + 0.04x +  25, 000 + 0.05x  

    A(x) + S(x)   =  55000 + 0.09x

Here x = 15000000

    A(x) + S(x)   =  55000 + 0.09(15000000)

  =  55000 +  1350000

Total income  =  1405000

So, the required income is  1405000.

Problem 3 :

The function for exchanging American dollars for Singapore Dollar on a given day is f(x) = 1.23x, where x represents the number of American dollars. On the same day the function for exchanging Singapore Dollar to Indian Rupee is g(y) = 50.50y, where y represents the number of Singapore dollars. Write a function which will give the exchange rate of American dollars in terms of Indian rupee.

The function for exchanging American dollars for Singapore Dollar :

f(x) = 1.23x

S.D  =   1.23 (A.D)

Here "x" stands for American dollar and f(x) stands for Singapore dollar.

A.D  = S.D/1.23  ----(1)

Exchanging Singapore Dollar to Indian Rupee is

g(y) = 50.50y

I.R  =  50.50 (S.D)

S.D  =  I.R/50.50

Applying  S.D  =  I.R/50.50 in(1), we get

A.D  = ( I.R/50.50) /1.23

A.D =  I.R/62.115

I.R  =  62.115 AD

So, the  exchange rate of American dollars in terms of Indian rupee is  I.R  =  62.115 AD.

Problem 4 :

The owner of a small restaurant can prepare a particular meal at a cost of Rupees 100. He estimates that if the menu price of the meal is x rupees, then the number of customers who will order that meal at that price in an evening is given by the function D(x) = 200−x. Express his day revenue, total cost and profit on this meal as functions of x.

Cost price of the meal  =  100

Selling price  =  x 

Number of customers  =  200 - x

1 day revenue  =   No of customers  ⋅ x

  =  (200 - x)  ⋅ x

  1 day revenue   =  200 x - x 2

Total cost  =  Cost of meal  ⋅ No of customers

  =  100  ⋅ ( 200 - x)

  = 20000 - 100x 

Profit  =  Total cost - 1 day revenue

(200 - x)  ⋅ x -  100  ⋅ ( 200 - x) 

Problem 5 :

The formula for converting from Fahrenheit to Celsius temperatures is y = (5x/9) − (160/9). Find the inverse of this function and determine whether the inverse is also a function

 y = (5x - 160)/9

9y  = 5x - 160

5x  =  9y + 160

x  =  (1/5) (9y + 160)

f -1 (x)  =  (1/5) (9x + 160)

f -1  (x)  =  (9x/5) + (160/5)

f -1  (x)  =  (9x/5) + 32

Inverse is also a function.

Problem 6 :

A simple cipher takes a number and codes it, using the function f(x) = 3x−4. Find the inverse of this function, determine whether the inverse is also a function and verify the symmetrical property about the line y = x (by drawing the lines).

f(x) = 3x−4

Let y = 3x - 4

3x  =  y + 4

x  =   (y + 4)/3

f -1 (x)  =  (x + 4)/3

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2.1: Relations and Functions

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Learning Objectives

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

• Find the domain and range of a relation
• Determine if a relation is a function
• Find the value of a function

Prerequisite Skills

Before you get started, take this prerequisite quiz.

1. Evaluate \(2x^2-5x+3\) when \(x=4\).

If you missed this problem, review here . (Note that this will open a different textbook in a new window.)

2. Evaluate \(2x^2-5x+3\) when \(x=-4\).

3. Solve \(y^2=9\).

\(y=3\) and \(y=-3\)

4. Express these inequalities in interval notation:

a. \(x>3\)

b. \(x\geq3\)

c. \(x\leq3\)

d. \(-2<x\leq5\)

a. \((3, \infty)\)

b. \([3, \infty)\)

c. \((-\infty, 3]\)

d. \((-2, 5]\)

If you missed any part of this problem, review Section 1.1 . (Note that this will open in a new window.)

Find the Domain and Range of a Relation

As we go about our daily lives, we have many data items or quantities that are paired to our names. Our social security number, student ID number, email address, phone number and our birthday are matched to our name. There is a relationship between our name and each of those items.

When your professor gets her class roster, the names of all the students in the class are listed in one column and then the student ID number is likely to be in the next column. If we think of the correspondence as a set of ordered pairs, where the first element is a student name and the second element is that student’s ID number, we call this a relation .

\[(\text{Student name}, \text{ Student ID #})\nonumber \]

The set of all the names of the students in the class is called the domain of the relation and the set of all student ID numbers paired with these students is the range of the relation.

There are many similar situations where one variable is paired or matched with another. The set of ordered pairs that records this matching is a relation.

Definition: Relation

A relation is any set of ordered pairs, \((x,y)\). All the x -values in the ordered pairs together make up the domain . All the y -values in the ordered pairs together make up the range .

Note that if the relation consists of a finite number of ordered pairs, then the domain and range will be listed in set notation, with {braces} around the list. If the relation consists of an infinite number of ordered pairs, then the domain and range will be expressed in interval notation, described in Section 1.1 .

Example \(\PageIndex{1}\)

For the relation \(\{(1,1),(2,4),(3,9),(4,16),(5,25)\}\):

• Find the domain of the relation.
• Find the range of the relation.

\[\begin{array} {ll} {} &{ {\{(1,1), (2,4), (3,9), (4,16), (5,25) }\} } \\ {ⓐ\text{ The domain is the set of all x-values of the relation.}} &{ {\{1,2,3,4,5}\} } \\ {ⓑ\text{ The range is the set of all y-values of the relation.}} &{ {\{1,4,9,16,25}\} } \\ \nonumber \end{array}\]

Example \(\PageIndex{2}\)

For the relation \({\{(1,1),(2,8),(3,27),(4,64),(5,125)}\}\):

\({\{1,2,3,4,5}\}\)

\({\{1,8,27,64,125}\}\)

Example \(\PageIndex{3}\)

For the relation \({\{(1,3),(2,6),(3,9),(4,12),(5,15)}\}\):

\({\{3,6,9,12,15}\}\)

A mapping is sometimes used to show a relation. The arrows show the pairing of the elements of the domain with the elements of the range.

Example \(\PageIndex{4}\)

Use the mapping of the relation shown to

• list the ordered pairs of the relation,
• find the domain of the relation, and
• find the range of the relation.

ⓐ The arrow shows the matching of the person to their birthday. We create ordered pairs with the person’s name as the x -value and their birthday as the y -value.

{(Alison, April 25), (Penelope, May 23), (June, August 2), (Gregory, September 15), (Geoffrey, January 12), (Lauren, May 10), (Stephen, July 24), (Alice, February 3), (Liz, August 2), (Danny, July 24)}

ⓑ The domain is the set of all x -values of the relation.

{Alison, Penelope, June, Gregory, Geoffrey, Lauren, Stephen, Alice, Liz, Danny}

ⓒ The range is the set of all y -values of the relation.

{January 12, February 3, April 25, May 10, May 23, July 24, August 2, September 15}

Example \(\PageIndex{5}\)

• list the ordered pairs of the relation
• find the domain of the relation

ⓐ (Khanh Nguyen, kn68413), (Abigail Brown, ab56781), (Sumantha Mishal, sm32479), (Jose Hernandez, jh47983)

ⓑ {Khanh Nguyen, Abigail Brown, Sumantha Mishal, Jose Hernandez}

ⓒ {kn68413, ab56781, sm32479, jh47983}

Example \(\PageIndex{6}\)

ⓐ (Maria, November 6), (Armando, January 18), (Cynthia, December 8), (Kelly, March 15), (Rachel, November 6)

ⓑ {Maria, Armando, Cynthia, Kelly, Rachel}

ⓒ{November 6, January 18, December 8, March 15}

A graph is yet another way that a relation can be represented. The set of ordered pairs of all the points plotted is the relation. The set of all x -coordinates is the domain of the relation and the set of all y -coordinates is the range. Generally we write the numbers in ascending order for both the domain and range.

Example \(\PageIndex{7}\)

Use the graph of the relation to

ⓐ The ordered pairs of the relation are: \[{\{(1,5),(−3,−1),(4,−2),(0,3),(2,−2),(−3,4)}\}.\nonumber\]

ⓑ The domain is the set of all x -values of the relation: \(\quad {\{−3,0,1,2,4}\}\).

Notice that while \(−3\) repeats, it is only listed once.

ⓒ The range is the set of all y -values of the relation: \(\quad {\{−2,−1,3,4,5}\}\).

Notice that while \(−2\) repeats, it is only listed once.

Example \(\PageIndex{8}\)

ⓐ The ordered pairs of the relation are: \((−4,4),(−3,3),(−2,2),(−1,1),(0,0),(1,1),(2,2),(3,3),(4,4)\) ⓑ The domain is the set of all x -values of the relation: \(\quad {\{−4,−3,−2,−1,0,1,2,3,4}\}\). ⓒ The range is the set of all y -values of the relation: \(\quad {\{0,1,2,3,4}\}\)

Example \(\PageIndex{9}\)

ⓐ The domain is the set of all x -values of the relation.  Since this includes all values from the previous graph and also all the infinite values between them, as well as those farther left and farther right, the domain is all real numbers, or \((−\infty,\infty)\). ⓑ The range is the set of all y -values of the relation.  Since this includes all values from the previous graph and also all the infinite values between them, as well as those above them, the range is all real numbers where \(y \geq 0\), or \([0,\infty)\).

Determine if a Relation is a Function

A special type of relation, called a function , occurs extensively in mathematics. A function is a relation that assigns to each element in its domain exactly one element in the range. For each ordered pair in the relation, each x -value is matched with only one y -value.

Definition: Function

A function is a relation that assigns to each element in its domain exactly one element in the range.

The birthday example from Example 2.1.6 helps us understand this definition. Every person has a birthday but no one has two birthdays. It is okay for two people to share a birthday. It is okay that Danny and Stephen share July 24 th as their birthday and that June and Liz share August 2 nd . Since each person has exactly one birthday, the relation in Example 2.1.6 is a function.

The relation shown by the graph in Example 2.1.7 includes the ordered pairs \((−3,−1)\) and \((−3,4)\). Is that okay in a function? No, as this is like one person having two different birthdays.

Example \(\PageIndex{10}\)

Use the set of ordered pairs to (i) determine whether the relation is a function (ii) find the domain of the relation (iii) find the range of the relation.

• \({\{(−3,27),(−2,8),(−1,1),(0,0),(1,1),(2,8),(3,27)}\}\)
• \({\{(9,−3),(4,−2),(1,−1),(0,0),(1,1),(4,2),(9,3)}\}\)

ⓐ \({\{(−3,27),(−2,8),(−1,1),(0,0),(1,1),(2,8),(3,27)}\}\)

(i) Each x -value is matched with only one y -value. So this relation is a function.

(ii) The domain is the set of all x -values in the relation. The domain is: \({\{−3,−2,−1,0,1,2,3}\}\).

(iii) The range is the set of all y -values in the relation. Notice we do not list range values twice. The range is: \({\{27,8,1,0}\}\).

ⓑ \({\{(9,−3),(4,−2),(1,−1),(0,0),(1,1),(4,2),(9,3)}\}\)

(i) The x -value 9 is matched with two y -values, both 3 and \(−3\). So this relation is not a function.

(ii) The domain is the set of all x -values in the relation. Notice we do not list domain values twice. The domain is: \({\{0,1,2,4,9}\}\).

(iii) The range is the set of all y -values in the relation. The range is: \({\{−3,−2,−1,0,1,2,3}\}\).

Example \(\PageIndex{11}\)

Use the set of ordered pairs to (i) determine whether the relation is a function (ii) find the domain of the relation (iii) find the range of the function.

• \({\{(−3,−6),(−2,−4),(−1,−2),(0,0),(1,2),(2,4),(3,6)}\}\)
• \({\{(8,−4),(4,−2),(2,−1),(0,0),(2,1),(4,2),(8,4)}\}\)

ⓐ Yes; \({\{−3,−2,−1,0,1,2,3}\}\); \({\{−6,−4,−2,0,2,4,6}\}\) ⓑ No; \({\{0,2,4,8}\}\); \({\{−4,−2,−1,0,1,2,4}\}\)

Example \(\PageIndex{12}\)

• \({\{(27,−3),(8,−2),(1,−1),(0,0),(1,1),(8,2),(27,3)}\}\)
• \({\{(7,−3),(−5,−4),(8,0),(0,0),(−6,4),(−2,2),(−1,3)}\}\)

ⓐ No; \({\{0,1,8,27}\}\); \({\{−3,−2,−1,0,2,2,3}\}\) ⓑ Yes; \({\{7,−5,8,0,−6,−2,−1}\}\); \({\{−3,−4,0,4,2,3}\}\)

Example \(\PageIndex{13}\)

Use the mapping to

• determine whether the relation is a function

ⓐ Both Lydia and Marty have two phone numbers. So each x -value is not matched with only one y -value. So this relation is not a function.

ⓑ The domain is the set of all x -values in the relation. The domain is: {Lydia, Eugene, Janet, Rick, Marty}

ⓒ The range is the set of all y -values in the relation. The range is:

\({\{321-549-3327, 427-658-2314, 321-964-7324, 684-358-7961, 684-369-7231, 798-367-8541}\}\)

Example \(\PageIndex{14}\)

Use the mapping to ⓐ determine whether the relation is a function ⓑ find the domain of the relation ⓒ find the range of the relation.

ⓐ no ⓑ {NBC, HGTV, HBO} ⓒ {Ellen Degeneres Show, Law and Order, Tonight Show, Property Brothers, House Hunters, Love it or List it, Game of Thrones, True Detective, Sesame Street}

Example \(\PageIndex{15}\)

ⓐ No ⓑ {Neal, Krystal, Kelvin, George, Christa, Mike} ⓒ {123-567-4839 work, 231-378-5941 cell, 743-469-9731 cell, 567-534-2970 work, 684-369-7231 cell, 798-367-8541 cell, 639-847-6971 cell}

In algebra, more often than not, functions will be represented by an equation. It is easiest to see if the equation is a function when it is solved for y . If each value of x results in only one value of y , then the equation defines a function.

Example \(\PageIndex{16}\)

Determine whether each equation is a function.  We will explore domain and range of these relations in the upcoming sections.

• \(y=x^2+1\)
• \(x+y^2=3\)

ⓐ \(2x+y=7\)

For each value of x , we multiply it by \(−2\) and then add 7 to get the y -value

We have that when \(x=3\), then \(y=1\). It would work similarly for any value of x . Since each value of x , corresponds to only one value of y the equation defines a function.

ⓑ \(y=x^2+1\)

For each value of x , we square it and then add 1 to get the y -value.

We have that when \(x=2\), then \(y=5\). It would work similarly for any value of x . Since each value of x , corresponds to only one value of y the equation defines a function.

We have shown that when \(x=2\), then \(y=1\) and \(y=−1\). It would work similarly for any value of x . Since each value of x does not corresponds to only one value of y the equation does not define a function.

Example \(\PageIndex{17}\)

Determine whether each equation is a function.

• \(4x+y=−3\)
• \(x+y^2=1\)
• \(y−x^2=2\)

ⓐ yes ⓑ no ⓒ yes

Example \(\PageIndex{18}\)

• \(x+y^2=4\)
• \(y=x^2−7\)
• \(y=5x−4\)

ⓐ no ⓑ yes ⓒ yes

Find the Value of a Function

It is very convenient to name a function and most often we name it f , g , h , F , G , or H . In any function, for each x -value from the domain we get a corresponding y -value in the range. For the function \(f\), we write this range value \(y\) as \(f(x)\). This is called function notation and is read \(f\) of \(x\) or the value of \(f\) at \(x\). In this case the parentheses does not indicate multiplication.

Definition: Function Notation

For the function \(y=f(x)\)

\[\begin{array} {l} {f\text{ is the name of the function}} \\{x \text{ is the domain value}} \\ {f(x) \text{ is the range value } y \text{ corresponding to the value } x} \\ \nonumber \end{array}\]

We read \(f(x)\) as \(f\) of \(x\) or the value of \(f\) at \(x\).

We call x the independent variable as it can be any value in the domain. We call y the dependent variable as its value depends on x .

INDEPENDENT AND DEPENDENT VARIABLES

For the function \(y=f(x)\),

\[\begin{array} {l} {x \text{ is the independent variable as it can be any value in the domain}} \\ {y \text{ the dependent variable as its value depends on } x} \\ \nonumber \end{array}\]

Much as when you first encountered the variable x , function notation may be rather unsettling. It seems strange because it is new. You will feel more comfortable with the notation as you use it.

Let’s look at the equation \(y=4x−5\). To find the value of y when \(x=2\), we know to substitute \(x=2\) into the equation and then simplify.

The value of the function at \(x=2\) is 3.

We do the same thing using function notation, the equation \(y=4x−5\) can be written as \(f(x)=4x−5\). To find the value when \(x=2\), we write:

This process of finding the value of \(f(x)\) for a given value of x is called evaluating the function.

Example \(\PageIndex{19}\)

For the function \(f(x)=2x^2+3x−1\), evaluate the function.

• \(f(−2)\)

Example \(\PageIndex{20}\)

For the function \(f(x)=3x^2−2x+1\), evaluate the function.

• \(f(−1)\)

ⓐ \(f(3)=22\) ⓑ \(f(−1)=6\) ⓒ \(f(t)=3t^2−2t+1\)

Example \(\PageIndex{21}\)

For the function \(f(x)=2x^2+4x−3\), evaluate the function.

• \(f(−3)\)

ⓐ \((2)=13\) ⓑ \(f(−3)=3\) ⓒ \(f(h)=2h2+4h−3\)

In the last example, we found \(f(x)\) for a constant value of x . In the next example, we are asked to find \(g(x)\) with values of x that are variables. We still follow the same procedure and substitute the variables in for the x .

Example \(\PageIndex{22}\)

For the function \(g(x)=3x−5\), evaluate the function.

• \(g(x)+g(2)\)

Notice the difference between part ⓑ and ⓒ. We get \(g(x+2)=3x+1\) and \(g(x)+g(2)=3x−4\). So we see that \(g(x+2)\neq g(x)+g(2)\).

Example \(\PageIndex{23}\)

For the function \(g(x)=4x−7\), evaluate the function.

• \(g(x−3)\)
• \(g(x)−g(3)\)

ⓐ \(4m^2−7\) ⓑ \(4x−19\) ⓒ \(x−12\)

Example \(\PageIndex{24}\)

For the function \(h(x)=2x+1\), evaluate the function.

• \(h(x)+h(1)\)

ⓐ \(2k^2+1\) ⓑ \(2x+3\) ⓒ \(2x+4\)

Many everyday situations can be modeled using functions.

Example \(\PageIndex{25}\)

The number of unread emails in Sylvia’s account is 75. This number grows by 10 unread emails a day. The function \(N(t)=75+10t\) represents the relation between the number of emails, N , and the time, t , measured in days.

• Determine the independent and dependent variable.
• Find \(N(5)\). Explain what this result means.

ⓐ The number of unread emails is a function of the number of days. The number of unread emails, N , depends on the number of days, t . Therefore, the variable N , is the dependent variable and the variable tt is the independent variable.

ⓑ Find \(N(5)\). Explain what this result means.

Since 5 is the number of days, \(N(5)\), is the number of unread emails after 5 days. After 5 days, there are 125 unread emails in the account.

Example \(\PageIndex{26}\)

The number of unread emails in Bryan’s account is 100. This number grows by 15 unread emails a day. The function \(N(t)=100+15t\) represents the relation between the number of emails, N , and the time, t , measured in days.

• Find \(N(7)\). Explain what this result means.

ⓐ t IND; N DEP ⓑ 205; the number of unread emails in Bryan’s account on the seventh day.

Access this online resource for additional instruction and practice with relations and functions.

• Introduction to Functions

Key Concepts

• f is the name of the function
• x is the domain value
• \(f(x)\) is the range value y corresponding to the value x We read \(f(x)\) as f of x or the value of f at x .
• x is the independent variable as it can be any value in the domain
• y is the dependent variable as its value depends on x

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HANNAH BATES: Welcome to HBR on Leadership , case studies and conversations with the world’s top business and management experts, hand-selected to help you unlock the best in those around you.

Problem solving skills are invaluable in any job. But even the most experienced among us can fall into the trap of solving the wrong problem.

Thomas Wedell-Wedellsborg says that all too often, we jump to find solutions to a problem – without taking time to really understand what we’re facing.

He’s an expert in innovation, and he’s the author of the book, What’s Your Problem?: To Solve Your Toughest Problems, Change the Problems You Solve .

  In this episode, you’ll learn how to reframe tough problems, by asking questions that reveal all the factors and assumptions that contribute to the situation. You’ll also learn why searching for one root cause can be misleading. And you’ll learn how to use experimentation and rapid prototyping as problem-solving tools.

This episode originally aired on HBR IdeaCast in December 2016. Here it is.

SARAH GREEN CARMICHAEL: Welcome to the HBR IdeaCast from Harvard Business Review. I’m Sarah Green Carmichael.

Problem solving is popular. People put it on their resumes. Managers believe they excel at it. Companies count it as a key proficiency. We solve customers’ problems.

The problem is we often solve the wrong problems. Albert Einstein and Peter Drucker alike have discussed the difficulty of effective diagnosis. There are great frameworks for getting teams to attack true problems, but they’re often hard to do daily and on the fly. That’s where our guest comes in.

Thomas Wedell-Wedellsborg is a consultant who helps companies and managers reframe their problems so they can come up with an effective solution faster. He asks the question “Are You Solving The Right Problems?” in the January-February 2017 issue of Harvard Business Review. Thomas, thank you so much for coming on the HBR IdeaCast .

THOMAS WEDELL-WEDELLSBORG: Thanks for inviting me.

SARAH GREEN CARMICHAEL: So, I thought maybe we could start by talking about the problem of talking about problem reframing. What is that exactly?

THOMAS WEDELL-WEDELLSBORG: Basically, when people face a problem, they tend to jump into solution mode to rapidly, and very often that means that they don’t really understand, necessarily, the problem they’re trying to solve. And so, reframing is really a– at heart, it’s a method that helps you avoid that by taking a second to go in and ask two questions, basically saying, first of all, wait. What is the problem we’re trying to solve? And then crucially asking, is there a different way to think about what the problem actually is?

SARAH GREEN CARMICHAEL: So, I feel like so often when this comes up in meetings, you know, someone says that, and maybe they throw out the Einstein quote about you spend an hour of problem solving, you spend 55 minutes to find the problem. And then everyone else in the room kind of gets irritated. So, maybe just give us an example of maybe how this would work in practice in a way that would not, sort of, set people’s teeth on edge, like oh, here Sarah goes again, reframing the whole problem instead of just solving it.

THOMAS WEDELL-WEDELLSBORG: I mean, you’re bringing up something that’s, I think is crucial, which is to create legitimacy for the method. So, one of the reasons why I put out the article is to give people a tool to say actually, this thing is still important, and we need to do it. But I think the really critical thing in order to make this work in a meeting is actually to learn how to do it fast, because if you have the idea that you need to spend 30 minutes in a meeting delving deeply into the problem, I mean, that’s going to be uphill for most problems. So, the critical thing here is really to try to make it a practice you can implement very, very rapidly.

There’s an example that I would suggest memorizing. This is the example that I use to explain very rapidly what it is. And it’s basically, I call it the slow elevator problem. You imagine that you are the owner of an office building, and that your tenants are complaining that the elevator’s slow.

Now, if you take that problem framing for granted, you’re going to start thinking creatively around how do we make the elevator faster. Do we install a new motor? Do we have to buy a new lift somewhere?

The thing is, though, if you ask people who actually work with facilities management, well, they’re going to have a different solution for you, which is put up a mirror next to the elevator. That’s what happens is, of course, that people go oh, I’m busy. I’m busy. I’m– oh, a mirror. Oh, that’s beautiful.

And then they forget time. What’s interesting about that example is that the idea with a mirror is actually a solution to a different problem than the one you first proposed. And so, the whole idea here is once you get good at using reframing, you can quickly identify other aspects of the problem that might be much better to try to solve than the original one you found. It’s not necessarily that the first one is wrong. It’s just that there might be better problems out there to attack that we can, means we can do things much faster, cheaper, or better.

SARAH GREEN CARMICHAEL: So, in that example, I can understand how A, it’s probably expensive to make the elevator faster, so it’s much cheaper just to put up a mirror. And B, maybe the real problem people are actually feeling, even though they’re not articulating it right, is like, I hate waiting for the elevator. But if you let them sort of fix their hair or check their teeth, they’re suddenly distracted and don’t notice.

But if you have, this is sort of a pedestrian example, but say you have a roommate or a spouse who doesn’t clean up the kitchen. Facing that problem and not having your elegant solution already there to highlight the contrast between the perceived problem and the real problem, how would you take a problem like that and attack it using this method so that you can see what some of the other options might be?

THOMAS WEDELL-WEDELLSBORG: Right. So, I mean, let’s say it’s you who have that problem. I would go in and say, first of all, what would you say the problem is? Like, if you were to describe your view of the problem, what would that be?

SARAH GREEN CARMICHAEL: I hate cleaning the kitchen, and I want someone else to clean it up.

THOMAS WEDELL-WEDELLSBORG: OK. So, my first observation, you know, that somebody else might not necessarily be your spouse. So, already there, there’s an inbuilt assumption in your question around oh, it has to be my husband who does the cleaning. So, it might actually be worth, already there to say, is that really the only problem you have? That you hate cleaning the kitchen, and you want to avoid it? Or might there be something around, as well, getting a better relationship in terms of how you solve problems in general or establishing a better way to handle small problems when dealing with your spouse?

SARAH GREEN CARMICHAEL: Or maybe, now that I’m thinking that, maybe the problem is that you just can’t find the stuff in the kitchen when you need to find it.

THOMAS WEDELL-WEDELLSBORG: Right, and so that’s an example of a reframing, that actually why is it a problem that the kitchen is not clean? Is it only because you hate the act of cleaning, or does it actually mean that it just takes you a lot longer and gets a lot messier to actually use the kitchen, which is a different problem. The way you describe this problem now, is there anything that’s missing from that description?

SARAH GREEN CARMICHAEL: That is a really good question.

THOMAS WEDELL-WEDELLSBORG: Other, basically asking other factors that we are not talking about right now, and I say those because people tend to, when given a problem, they tend to delve deeper into the detail. What often is missing is actually an element outside of the initial description of the problem that might be really relevant to what’s going on. Like, why does the kitchen get messy in the first place? Is it something about the way you use it or your cooking habits? Is it because the neighbor’s kids, kind of, use it all the time?

There might, very often, there might be issues that you’re not really thinking about when you first describe the problem that actually has a big effect on it.

SARAH GREEN CARMICHAEL: I think at this point it would be helpful to maybe get another business example, and I’m wondering if you could tell us the story of the dog adoption problem.

THOMAS WEDELL-WEDELLSBORG: Yeah. This is a big problem in the US. If you work in the shelter industry, basically because dogs are so popular, more than 3 million dogs every year enter a shelter, and currently only about half of those actually find a new home and get adopted. And so, this is a problem that has persisted. It’s been, like, a structural problem for decades in this space. In the last three years, where people found new ways to address it.

So a woman called Lori Weise who runs a rescue organization in South LA, and she actually went in and challenged the very idea of what we were trying to do. She said, no, no. The problem we’re trying to solve is not about how to get more people to adopt dogs. It is about keeping the dogs with their first family so they never enter the shelter system in the first place.

In 2013, she started what’s called a Shelter Intervention Program that basically works like this. If a family comes and wants to hand over their dog, these are called owner surrenders. It’s about 30% of all dogs that come into a shelter. All they would do is go up and ask, if you could, would you like to keep your animal? And if they said yes, they would try to fix whatever helped them fix the problem, but that made them turn over this.

And sometimes that might be that they moved into a new building. The landlord required a deposit, and they simply didn’t have the money to put down a deposit. Or the dog might need a $10 rabies shot, but they didn’t know how to get access to a vet.

And so, by instigating that program, just in the first year, she took her, basically the amount of dollars they spent per animal they helped went from something like $85 down to around $60. Just an immediate impact, and her program now is being rolled out, is being supported by the ASPCA, which is one of the big animal welfare stations, and it’s being rolled out to various other places.

And I think what really struck me with that example was this was not dependent on having the internet. This was not, oh, we needed to have everybody mobile before we could come up with this. This, conceivably, we could have done 20 years ago. Only, it only happened when somebody, like in this case Lori, went in and actually rethought what the problem they were trying to solve was in the first place.

SARAH GREEN CARMICHAEL: So, what I also think is so interesting about that example is that when you talk about it, it doesn’t sound like the kind of thing that would have been thought of through other kinds of problem solving methods. There wasn’t necessarily an After Action Review or a 5 Whys exercise or a Six Sigma type intervention. I don’t want to throw those other methods under the bus, but how can you get such powerful results with such a very simple way of thinking about something?

THOMAS WEDELL-WEDELLSBORG: That was something that struck me as well. This, in a way, reframing and the idea of the problem diagnosis is important is something we’ve known for a long, long time. And we’ve actually have built some tools to help out. If you worked with us professionally, you are familiar with, like, Six Sigma, TRIZ, and so on. You mentioned 5 Whys. A root cause analysis is another one that a lot of people are familiar with.

Those are our good tools, and they’re definitely better than nothing. But what I notice when I work with the companies applying those was those tools tend to make you dig deeper into the first understanding of the problem we have. If it’s the elevator example, people start asking, well, is that the cable strength, or is the capacity of the elevator? That they kind of get caught by the details.

That, in a way, is a bad way to work on problems because it really assumes that there’s like a, you can almost hear it, a root cause. That you have to dig down and find the one true problem, and everything else was just symptoms. That’s a bad way to think about problems because problems tend to be multicausal.

There tend to be lots of causes or levers you can potentially press to address a problem. And if you think there’s only one, if that’s the right problem, that’s actually a dangerous way. And so I think that’s why, that this is a method I’ve worked with over the last five years, trying to basically refine how to make people better at this, and the key tends to be this thing about shifting out and saying, is there a totally different way of thinking about the problem versus getting too caught up in the mechanistic details of what happens.

SARAH GREEN CARMICHAEL: What about experimentation? Because that’s another method that’s become really popular with the rise of Lean Startup and lots of other innovation methodologies. Why wouldn’t it have worked to, say, experiment with many different types of fixing the dog adoption problem, and then just pick the one that works the best?

THOMAS WEDELL-WEDELLSBORG: You could say in the dog space, that’s what’s been going on. I mean, there is, in this industry and a lot of, it’s largely volunteer driven. People have experimented, and they found different ways of trying to cope. And that has definitely made the problem better. So, I wouldn’t say that experimentation is bad, quite the contrary. Rapid prototyping, quickly putting something out into the world and learning from it, that’s a fantastic way to learn more and to move forward.

My point is, though, that I feel we’ve come to rely too much on that. There’s like, if you look at the start up space, the wisdom is now just to put something quickly into the market, and then if it doesn’t work, pivot and just do more stuff. What reframing really is, I think of it as the cognitive counterpoint to prototyping. So, this is really a way of seeing very quickly, like not just working on the solution, but also working on our understanding of the problem and trying to see is there a different way to think about that.

If you only stick with experimentation, again, you tend to sometimes stay too much in the same space trying minute variations of something instead of taking a step back and saying, wait a minute. What is this telling us about what the real issue is?

SARAH GREEN CARMICHAEL: So, to go back to something that we touched on earlier, when we were talking about the completely hypothetical example of a spouse who does not clean the kitchen–

THOMAS WEDELL-WEDELLSBORG: Completely, completely hypothetical.

SARAH GREEN CARMICHAEL: Yes. For the record, my husband is a great kitchen cleaner.

You started asking me some questions that I could see immediately were helping me rethink that problem. Is that kind of the key, just having a checklist of questions to ask yourself? How do you really start to put this into practice?

THOMAS WEDELL-WEDELLSBORG: I think there are two steps in that. The first one is just to make yourself better at the method. Yes, you should kind of work with a checklist. In the article, I kind of outlined seven practices that you can use to do this.

But importantly, I would say you have to consider that as, basically, a set of training wheels. I think there’s a big, big danger in getting caught in a checklist. This is something I work with.

My co-author Paddy Miller, it’s one of his insights. That if you start giving people a checklist for things like this, they start following it. And that’s actually a problem, because what you really want them to do is start challenging their thinking.

So the way to handle this is to get some practice using it. Do use the checklist initially, but then try to step away from it and try to see if you can organically make– it’s almost a habit of mind. When you run into a colleague in the hallway and she has a problem and you have five minutes, like, delving in and just starting asking some of those questions and using your intuition to say, wait, how is she talking about this problem? And is there a question or two I can ask her about the problem that can help her rethink it?

SARAH GREEN CARMICHAEL: Well, that is also just a very different approach, because I think in that situation, most of us can’t go 30 seconds without jumping in and offering solutions.

THOMAS WEDELL-WEDELLSBORG: Very true. The drive toward solutions is very strong. And to be clear, I mean, there’s nothing wrong with that if the solutions work. So, many problems are just solved by oh, you know, oh, here’s the way to do that. Great.

But this is really a powerful method for those problems where either it’s something we’ve been banging our heads against tons of times without making progress, or when you need to come up with a really creative solution. When you’re facing a competitor with a much bigger budget, and you know, if you solve the same problem later, you’re not going to win. So, that basic idea of taking that approach to problems can often help you move forward in a different way than just like, oh, I have a solution.

I would say there’s also, there’s some interesting psychological stuff going on, right? Where you may have tried this, but if somebody tries to serve up a solution to a problem I have, I’m often resistant towards them. Kind if like, no, no, no, no, no, no. That solution is not going to work in my world. Whereas if you get them to discuss and analyze what the problem really is, you might actually dig something up.

Let’s go back to the kitchen example. One powerful question is just to say, what’s your own part in creating this problem? It’s very often, like, people, they describe problems as if it’s something that’s inflicted upon them from the external world, and they are innocent bystanders in that.

SARAH GREEN CARMICHAEL: Right, or crazy customers with unreasonable demands.

THOMAS WEDELL-WEDELLSBORG: Exactly, right. I don’t think I’ve ever met an agency or consultancy that didn’t, like, gossip about their customers. Oh, my god, they’re horrible. That, you know, classic thing, why don’t they want to take more risk? Well, risk is bad.

It’s their business that’s on the line, not the consultancy’s, right? So, absolutely, that’s one of the things when you step into a different mindset and kind of, wait. Oh yeah, maybe I actually am part of creating this problem in a sense, as well. That tends to open some new doors for you to move forward, in a way, with stuff that you may have been struggling with for years.

SARAH GREEN CARMICHAEL: So, we’ve surfaced a couple of questions that are useful. I’m curious to know, what are some of the other questions that you find yourself asking in these situations, given that you have made this sort of mental habit that you do? What are the questions that people seem to find really useful?

THOMAS WEDELL-WEDELLSBORG: One easy one is just to ask if there are any positive exceptions to the problem. So, was there day where your kitchen was actually spotlessly clean? And then asking, what was different about that day? Like, what happened there that didn’t happen the other days? That can very often point people towards a factor that they hadn’t considered previously.

SARAH GREEN CARMICHAEL: We got take-out.

THOMAS WEDELL-WEDELLSBORG: S,o that is your solution. Take-out from [INAUDIBLE]. That might have other problems.

Another good question, and this is a little bit more high level. It’s actually more making an observation about labeling how that person thinks about the problem. And what I mean with that is, we have problem categories in our head. So, if I say, let’s say that you describe a problem to me and say, well, we have a really great product and are, it’s much better than our previous product, but people aren’t buying it. I think we need to put more marketing dollars into this.

Now you can go in and say, that’s interesting. This sounds like you’re thinking of this as a communications problem. Is there a different way of thinking about that? Because you can almost tell how, when the second you say communications, there are some ideas about how do you solve a communications problem. Typically with more communication.

And what you might do is go in and suggest, well, have you considered that it might be, say, an incentive problem? Are there incentives on behalf of the purchasing manager at your clients that are obstructing you? Might there be incentive issues with your own sales force that makes them want to sell the old product instead of the new one?

So literally, just identifying what type of problem does this person think about, and is there different potential way of thinking about it? Might it be an emotional problem, a timing problem, an expectations management problem? Thinking about what label of what type of problem that person is kind of thinking as it of.

SARAH GREEN CARMICHAEL: That’s really interesting, too, because I think so many of us get requests for advice that we’re really not qualified to give. So, maybe the next time that happens, instead of muddying my way through, I will just ask some of those questions that we talked about instead.

THOMAS WEDELL-WEDELLSBORG: That sounds like a good idea.

SARAH GREEN CARMICHAEL: So, Thomas, this has really helped me reframe the way I think about a couple of problems in my own life, and I’m just wondering. I know you do this professionally, but is there a problem in your life that thinking this way has helped you solve?

THOMAS WEDELL-WEDELLSBORG: I’ve, of course, I’ve been swallowing my own medicine on this, too, and I think I have, well, maybe two different examples, and in one case somebody else did the reframing for me. But in one case, when I was younger, I often kind of struggled a little bit. I mean, this is my teenage years, kind of hanging out with my parents. I thought they were pretty annoying people. That’s not really fair, because they’re quite wonderful, but that’s what life is when you’re a teenager.

And one of the things that struck me, suddenly, and this was kind of the positive exception was, there was actually an evening where we really had a good time, and there wasn’t a conflict. And the core thing was, I wasn’t just seeing them in their old house where I grew up. It was, actually, we were at a restaurant. And it suddenly struck me that so much of the sometimes, kind of, a little bit, you love them but they’re annoying kind of dynamic, is tied to the place, is tied to the setting you are in.

And of course, if– you know, I live abroad now, if I visit my parents and I stay in my old bedroom, you know, my mother comes in and wants to wake me up in the morning. Stuff like that, right? And it just struck me so, so clearly that it’s– when I change this setting, if I go out and have dinner with them at a different place, that the dynamic, just that dynamic disappears.

SARAH GREEN CARMICHAEL: Well, Thomas, this has been really, really helpful. Thank you for talking with me today.

THOMAS WEDELL-WEDELLSBORG: Thank you, Sarah.  

HANNAH BATES: That was Thomas Wedell-Wedellsborg in conversation with Sarah Green Carmichael on the HBR IdeaCast. He’s an expert in problem solving and innovation, and he’s the author of the book, What’s Your Problem?: To Solve Your Toughest Problems, Change the Problems You Solve .

We’ll be back next Wednesday with another hand-picked conversation about leadership from the Harvard Business Review. If you found this episode helpful, share it with your friends and colleagues, and follow our show on Apple Podcasts, Spotify, or wherever you get your podcasts. While you’re there, be sure to leave us a review.

We’re a production of Harvard Business Review. If you want more podcasts, articles, case studies, books, and videos like this, find it all at HBR dot org.

This episode was produced by Anne Saini, and me, Hannah Bates. Ian Fox is our editor. Music by Coma Media. Special thanks to Maureen Hoch, Adi Ignatius, Karen Player, Ramsey Khabbaz, Nicole Smith, Anne Bartholomew, and you – our listener.

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Title: gradient descent for unbounded convex functions on hadamard manifolds and its applications to scaling problems.

Abstract: In this paper, we study asymptotic behaviors of continuous-time and discrete-time gradient flows of a ``lower-unbounded" convex function $f$ on a Hadamard manifold $M$, particularly, their convergence properties to the boundary $M^{\infty}$ at infinity of $M$. We establish a duality theorem that the infimum of the gradient-norm $\|\nabla f(x)\|$ of $f$ over $M$ is equal to the supremum of the negative of the recession function $f^{\infty}$ of $f$ over the boundary $M^{\infty}$, provided the infimum is positive. Further, the infimum and the supremum are obtained by the limits of the gradient flows of $f$, Our results feature convex-optimization ingredients of the moment-weight inequality for reductive group actions by Georgoulas, Robbin, and Salamon,and are applied to noncommutative optimization by Bürgisser et al. FOCS 2019. We show that the gradient descent of the Kempf-Ness function for an unstable orbit converges to a 1-parameter subgroup in the Hilbert-Mumford criterion, and the associated moment-map sequence converges to the mimimum-norm point of the moment polytope. We show further refinements for operator scaling -- the left-right action on a matrix tuple $A= (A_1,A_2,\ldots,A_N)$. We characterize the gradient-flow limit of operator scaling via a vector-space generalization of the classical Dulmage-Mendelsohn decomposition of a bipartite graph. Also, for a special case of $N = 2$, we reveal that this limit determines the Kronecker canonical form of matrix pencils $s A_1+A_2$.

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