NOTES 4.6– FORMALIZING RELATIONS
AND FUNCTIONS
A set of ordered pairs is called a RELATION.
The set of values that can be used for the first number in the ordered
pair is called the DOMAIN.
The set of values for the second number in an ordered pair is called
the RANGE.
Given an ordered pair (x, y)
The set of x-VALUES is the domain.
The set of y-VALUES is the range
A function is a RELATION where each member of the domain has 1
and only 1 corresponding member of the range. In other words,
“A function has NO repeats in the domain!”
Every function is a RELATION but not every relation is a function.
Example: Given the relation { ( 1, 2), ( 5, 6), ( 12, -2), ( -2, 8) ( 0, 6) }
find the domain and range. Is this a function?
DOMAIN:
1, 5, 12, - 2, 0
Or in order:
-2, 0, 1, 5, 12
RANGE:
2, 6, - 2, 8, 6
Or in order:
-2, 2, 6, 8
Any repeats in the domain???
NO! This IS a FUNCTION.
Mapping Notation:
Given the set of points: {(2, 4), (3, 2), (8, 5), (3, -1)}
What is the domain?
What is the range?
x
2, 3, 8
-1, 2, 4, 5
y
2
3
8
-1
2
4
5
Is y a function of x? Why or why not?
-1 and 2 in the range.
NO. 3 in the domain maps to
The Vertical Line Test:
Another way to determine if a relation is a function is to graph the
relation and apply the VERTICAL LINE Test to the graph.
If a vertical line passes through more than 1 point at any time on a
graph, the relation is NOT a function.
Example: Use the Vertical line test to determine if the relation is a function.
1.
NO! At x = 3, y has 2 values
2.
YES! It never crosses the
graph more than once
Using f (x) is a different way of writing the y-VARIABLE in an equation.
f indicates the NAME of the function. x indicates the INDEPENDENT
variable.
The domain is the set of INPUT values for the function.
The domain values are associated with the INDEPENDENT variable.
The range is the set of OUTPUT values for the function.
The range values are associated with the DEPENDENT variable.
Example: Find the range of each function for the given domain.
1) f (x) = –3x + 2; {–2, –1, 0, 1, 2}
x
-2
-1
0
1
2
f  x
f  -2  = -3  -2   2 = 8
f  -1  = -3  -1   2 = 5
f  0  = -3  0   2 = 2
f  1  = -3  1   2 = -1
f  2  = -3  2   2 = -4
RANGE:-4, - 1, 2, 5 , 8
2  2; {0, 1 , 1 , 3 ,1}
g
(
x
)

x
2)
4 2 4
x
0
14
12
34
1
g  x
g 0  = 0   2 = 2
2
g  1 4  =  1 4   2 = 2 1 16
2
g 1 2 = 1 2   2 = 2 1 4
2
g  3 4  =  3 4   2 = 2 9 16
2
g  1 =  1  2 = 3
2
RANGE:2, 2 1 16 , 2 1 4 , 2 9 16 , 3
Identifying a Reasonable Domain and Range:
In real world situations the domain and range may be LIMITED.
Example:
1. A tenth grade class is selling granola bars for a fundraiser.
They earn $0.75 for every granola bar that they sell. They have
ordered 300 granola bars for the sale. The function P(b) = 0.75b
represents the profit P the class earns for each bar b they sell. Find
a reasonable domain and range for the function.
How many granola bars CAN they sell?
DOMAIN:
0  b  300
Where b is an integer!
How much MONEY CAN they make?
RANGE:
0  P  b   225
Evaluating Functions:
2. The function t(x) = 150x represents the number of words t(x)
you can speak in x minutes. How many words can you speak in 20
minutes? How many minutes would it take to speak 10,000 words?
How many WORDS can you speak in 20 MINUTES?
t  20   150  20   3000 words
How many MINUTES will it take to speak in 10,000 WORDS?
t  x   10,000  150x
150 150
x  66 2 3 minutes
HOMEWORK: 4.6
pages 290 #’s 8-16 and
pages 293-294 #’s 3-19