Basics of Functions

Definition of a Relation

• A relation is any set of ordered pairs. • The set of all first components of the

ordered pairs is called the domain of the relation,

• The set of all second components is called the range of the relation.

Example

SolutionThe domain is the set of all first components. Thus, the domain is

{1994,1995,1996,1997,1998}.The range is the set of all second components. Thus, the range is

{56.21, 51.00, 47.70, 42.78, 39.43}.

Find the domain and range of the relation{(1994, 56.21), (1995, 51.00), (1996, 47.70),

(1997, 42.78), (1998, 39.43)}

Definition of a Function• A function is a correspondence between

two sets X and Y that assigns to each element x of set X exactly one element y of set Y.

• For each element x in X, the corresponding element y in Y is called the value of the function at x.

• The set X is called the domain of the function, and the set of all function values, Y, is called the range of the function.

Text Example

Solution We begin by making a figure for each relation that shows set X, the domain, and set Y, the range, shown below.

Determine whether each relation is a function.a. {(1, 6), (2, 6), (3, 8), (4, 9)} b. {(6,1),(6,2),(8,3),(9,4)}

1234

689

Domain Range

(a) Figure (a) 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 and different second components. Thus, the relation is a function.

689

1234

Domain Range

(b) Figure (b) shows that 6 corresponds to both 1 and 2. This relation is not a function; two ordered pairs have the same first component and different second components.

Text Example

Solution We substitute 2, x + 3, and -x for x in the definition of f. When replacing x with a variable or an algebraic expression, you might find it helpful to think of the function's equation as

f (x) = x2 + 3x + 5.

If f (x) = x2 + 3x + 5, evaluate: a. f (2) b. f (x + 3) c. f (-x)

a. We find f (2) by substituting 2 for x in the equation.

f (2) = 22 + 3 • 2 + 5 = 4 + 6 + 5 = 15

Thus, f (2) = 15.

Text Example cont.

Solutionb. We find f (x + 3) by substituting x + 3 for x in the equation.

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

If f (x) = x2 + 3x + 5, evaluate: a. f (2) b. f (x + 3) c. f (-x)

Equivalently,

f (x + 3) = (x + 3)2 + 3(x + 3) + 5= x2 + 6x + 9 + 3x + 9 + 5= x2 + 9x + 23.

Square x + 3 and distribute 3 throughout the parentheses.

Text Example cont.

Solutionc. We find f (-x) by substituting -x for x in the equation.

f (-x) = (-x)2 + 3(-x) + 5

If f (x) = x2 + 3x + 5, evaluate: a. f (2) b. f (x + 3) c. f (-x)

Equivalently,

f (-x) = (-x)2 + 3(-x) + 5 = x2 –3x + 5.

Definition of a Difference Quotient

• The expression

for h not equal to 0 is called the difference quotient.

f (x h) f (x)

h

Example

}1,1:{

)1)(1(

)2(

1

2)(

2

2

3

xx

domain

xx

xx

x

xxxf

function

Find the domain of the function.

32)( xxg

Example

• Find the domain of the function.

}2

3:{

2

3

032

xxanswer

x

so

x

Example cont.

Solution

Basics of Functions