2.3 Introduction to Functions

A is any set of ordered pairs.

The set of all first components of the

Definition of a Relation

relation

domainordered pairs is called the of

the relation, and the set of all second

components is called the of the

relat

range

ion.

Definition of a Function

A is a correspondence between

two sets and that assings to each

element of set exactly one ele

fu

me

ncti

nt

of .

on

X Y

x X y

Y

Domain and Range

For each element in , the

corresponding element in is called

the of the function at . The set

is called the of the function, and

the set of all function values,

value

domai

, is called

the

n

ran

x X

y Y

x X

Y

of the funcge tion.

Ex 1: Determine whether each relation is a function.

a. {(4,5), (6,7), (8,8)}

b. {(5,6), (4,7), (6,6), (6,7)}

We begin by making a figure

for each relation that shows set , the

domain, and set , t

So

he

luti

ra

o

e.

n

ng

X

Y

Solution for part (a)

468

578

X Y

Domain Range

The figure shows that every

element in the domain

corresponds to exactly one

element in the range.No two ordered pairs in the given relation have

the same first component different second

components. Thus, the relation is a function

Solution for part (b)

456

6

7

X Y

Domain Range

The figure shows that 6

corresponds to both 6 and 7.

If any element in the domain corresponds to

more than one element in the range, the

relation is not a function, Thus, the relation

is not a function.

Practice Exercises

Determine whether each relation is a

function. Give the domain and range

for each relation.

1. {( 7, 7), ( 5, 5), ( 3, 3), (0,0)}

2. {(4,1), (5,1), (4,2)}

Answers

1. Domain { 7, 5, 3,0}

Range { 7, 5, 3,0}

Given relation is a function.

2. Domain {4,5}

Range {1,2}

Given relation is not a function.

Function Notation

( )

The variable is called the

, because it can be assigned any

of the permissible numbers from the domain.

The variable is called the

independent

variable

dependent

, because

var its iable value depe

y f x

x

y

nds on .x

Function Notation

The special notation ( ), read " of "

or " at ," represents the value of the

function at the number .

The notation ( ) does not mean

" times ."

f x f x

f x

x

f x

f x

Ex 2: Determine whether each equation defines y as a function of x.

2 2

1. 25

2. 25

x y

x y

Solution

Solve each equation for in terms of

. If two or more values of can be

obtained for a given , the equation

is not a function.

y

x y

x

1. 25x y 25y x

Solution continued

From this last equation we can see that

for each value of , there is one and

only one value of . Thus, the equation

defines y as a function of .

x

y

x

Solution of 22 2 25x y

2 225y x 225y x

The in this last equation shows that for

certain values of (all values between 5 and

5), there are two values of . For this reason

the equation does not define as a function of .

x

y

y x

Practice Exercises

2

2

Determine whether each equation

difines as a function of .

1. 25

2. 4

y x

x y

x y

Answers

1. The equation defines a function.

2. The equation does not define a

function.

Ex 3: Evaluating a Function

2If ( ) 10 3, evaluate:

a. ( 1) b. ( 2) c. ( )

f x x x

f f x f x

Solution

We substitute 1, 2, and

for in the definition of .

x x

x f

Solution part a.

2

We find ( 1) by substituting 1 for

in the equation ( ) 10 3.xx x

f

f x

2( ) ( ) 10( ) 3f

1 10 3 8

Thus ( 1) 8.f

1 1 1

Solution part b.

2

We find ( 2) by substituting 2 for

in the equation ( ) 10 3.x x

x

f x

f x

x

2( ) ( ) 10( ) 3f

2( ) 4 4 10 20 3f x x x x 2x 2x 2x

2( ) 6 19f x x x

Solution part c.

2

We find ( ) by substituting for

in the equation ( ) 10 3.x

f x x

x xx f

2( ) ( ) 10( ) 3f

2( ) 10 3f x x x

x x x

Practice Exercise

3Evaluate the function ( ) 1

at the given values of the independent

variable and simplify.

a. (2) b. ( 1) c. ( ) d. (3 )

h x x x

h h h x h a

Answer

a. 7

b. 13c. 1x x 3d. 27 3 1a a

Piecewise Functions

A function defined by two (or more)

equations over a specified domain

is call piecewise functed a ion.

Ex 4: Evaluating a Piecewise Function

Evaluate the piecewise function at the

given values of the independent variable.

6 1 if 0( )

7 3 if 0

a. ( 3) b. (0) c. (4)

x xf x

x x

f f f

Solution part a

To find ( 3), we let 3.

Because 3 is less than 0, we use

the first line of the piecewise function.

f x

( ) 6 1f x x This is the function's

equation for 0.x

( 3) 6( 3) 1 19f

Solution part b

To find (0), we let 0.

Because 0 is equal to 0, we use

the second line of the piecewise function.

f x

( ) 7 3f x x This is the function's

equation for 0.x

(0) 7(0) 3 3f

Solution part c

To find (4), we let 4.

Because 4 is greater than 0, we use

the second line of the piecewise function.

f x

( ) 7 3f x x This is the function's

equation for 0.x

(4) 7(4) 3 31f

Practice Exercise

Evaluate the piecewise function at the

given values of the independent variable.

5 if 5( )

( 5) if 5

a. (0) b. ( 6) c. ( 5)

x xg x

x x

g g g

Answer

a. (0) 5g

b. ( 6) 1g

c. ( 5) 0g

Finding a Function’s Domain

If a function does not model data or

verbal conditions, its domain is the

largest set of real numbers for which the

value of ( ) is a real number.

f

f x

Finding a Function’s Domain

Exclude from a function's domain real

numbers that cause division by zero

and real numbers that result in an even

root of a negative number.

Ex 5: Find the domain of each function

2a. ( ) 8 5 2

2b. ( )

5

c. ( ) 2

f x x x

g xx

h x x

Solution part a

2The function ( ) 8 5 2 contains

neither division nor an even root.

The domain of is the set of all real numbers.

f x x x

f

Solution part b

2The function ( ) contains division.

5Because division by zero is undefined, we

must exclude from the domain values of

that cause 5 to be 0. Thus cannot equal

to 5. The domain of function is

g xx

x

x x

g

{ | 5}.x x

Solution part cThe function ( ) 2 contains an even

root. Because only non-negative numbers

have real square roots, the quantity under the

radical sign, 2 must be greater than or

equal to 0. Thus, 2 0 or 2

Th

h x x

x

x x

erefore the domain of is { | 2}

or the interval [ 2, ).

h x x

Practice Exercises

2

Find the domain of each function:

121. ( )

361

2. ( )2

xh x

x

f xx

Answers

1. { | 6, 6}x x x

2. { | 2} or ( 2, )x x