copyright © 2010 pearson education, inc. all rights reserved sec 4.5 - 1 3.5/3.6 introduction to...
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Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 4.5 - 1
3.5/3.6
Introduction to Functions
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 4.5 - 2
Introduction to Functions
Define and identify relations and functions.
Relation
A relation is any set of ordered pairs.
Function
A function is a relation in which, for each value of the first component
of the ordered pairs, there is exactly one value of the second component.
A special kind of relation, called a function, is very important in mathemat-
ics and its applications.
Introduction to Functions
EXAMPLE 1 Determining Whether Relations Are Functions
Tell whether each relation defines a function.
L = { (2, 3), (–5, 8), (4, 10) }
M = { (–3, 0), (–1, 4), (1, 7), (3, 7) }
N = { (6, 2), (–4, 4), (6, 5) }
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Introduction to Functions
Mapping Relations
1
F is a function.
–3
4
2
5
3
F
–1
G is not a function.
–2
6
0
G
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Introduction to Functions
Tables and Graphs
Graph of the function, F
x
y
Table of the
function, F
x y
–2 6
0 0
2 –6
O
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4.5 Introduction to Functions
Domain and Range
In a relation, the set of all values of the independent variable (x) is the
domain. The set of all values of the dependent variable (y) is the range.
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4.5 Introduction to Functions
EXAMPLE 2 Finding Domains and Ranges of Relations
Give the domain and range of each relation. Tell whether the relation defines
a function.
(a) { (3, –8), (5, 9), (5, 11), (8, 15) }
The domain, the set of x-values, is {3, 5, 8}; the range, the set of y-values,
is {–8, 9, 11, 15}. This relation is not a function because the same x-value 5 is
paired with two different y-values, 9 and 11.
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4.5 Introduction to Functions
EXAMPLE 2 Finding Domains and Ranges of Relations
Give the domain and range of each relation. Tell whether the relation defines
a function.
(b)
The domain of this relation is {6, 1, –9}. The range is {M, N}.
This mapping defines a function – each x-value corresponds to exactly one
y-value.
6
1
–9
M
N
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Introduction to Functions
EXAMPLE 2 Finding Domains and Ranges of Relations
Give the domain and range of each relation. Tell whether the relation defines
a function.
(c)
This is a table of ordered pairs, so the domain is the set of x-values,
{–2, 1, 2}, and the range is the set of y-values, {3}.
x y
–2 3
1 3
2 3
The table defines a
function because each different x-value corresponds to exactly one y-value
(even though it is the same y-value).
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 4.5 - 10
Introduction to Functions
EXAMPLE 3 Finding Domains and Ranges from Graphs
Give the domain and range of each relation.
(a)
The domain is the set of x-values,
{–3, 0, 2 , 4}. The range, the set of
y-values, is {–3, –1, 1, 2}.
x
y
(–3, 2)
(0, –3)
(2, 1)
(4, –1)O
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 4.5 - 11
Introduction to Functions
EXAMPLE 3 Finding Domains and Ranges from Graphs
Give the domain and range of each relation.
(b) The x-values of the points on the
graph include all numbers between
–7 and 2, inclusive.
x
y
Domain
Range
The y-valuesinclude all numbers between –2 and
2, inclusive. Using interval notation,
the domain is [–7, 2];
the range is [–2, 2].
O
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Introduction to Functions
EXAMPLE 3 Finding Domains and Ranges from Graphs
Give the domain and range of each relation.
(c) The arrowheads indicate that the
line extends indefinitely left and right,
as well as up and down. Therefore,
both the domain and range include
all real numbers, written (-∞, ∞).
x
y
O
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Introduction to Functions
EXAMPLE 3 Finding Domains and Ranges from Graphs
Give the domain and range of each relation.
(d) The arrowheads indicate that the
graph extends indefinitely left and
right, as well as upward. The domain
is (-∞, ∞).
x
y
Because there is a least y-
value, –1, the range includes all
numbers greater than or equal to –1,
written [–1, ∞).O
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4.5 Introduction to Functions
Agreement on Domain
The domain of a relation is assumed to be all real numbers that produce
real numbers when substituted for the independent variable.
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4.5 Introduction to Functions
Vertical Line Test
If every vertical line intersects the graph of a relation in no more than
one point, then the relation represents a function.
y
x
y
x
Not a function – the same
x-value corresponds to two
different y-values.
Function – each x-value
corresponds to only one
y-value.
(a) (b)
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4.5 Introduction to Functions
EXAMPLE 4 Using the Vertical Line Test
Use the vertical line test to determine whether each relation is a function.
(a)
This relation is a function.
x
y
(–3, 2)
(0, –3)
(2, 1)
(4, –1)O
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 4.5 - 17
Introduction to Functions
EXAMPLE 4 Using the Vertical Line Test
Use the vertical line test to determine whether each relation is a function.
(b)This graph fails the vertical line test
since the same x-value corresponds
to two different y-values; therefore,
it is not the graph of a function.
x
y
O
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 4.5 - 18
Introduction to Functions
EXAMPLE 4 Using the Vertical Line Test
Use the vertical line test to determine whether each relation is a function.
(c)This relation is a function.
x
y
O
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 4.5 - 19
Introduction to Functions
EXAMPLE 4 Using the Vertical Line Test
Use the vertical line test to determine whether each relation is a function.
(d)This relation is a function.
x
y
O
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 4.5 - 20
Introduction to Functions
EXAMPLE 5 Identifying Functions from Their Equations
Decide whether each relation defines a function and give the domain.
(a) y = x – 5
Introduction to Functions
EXAMPLE 5 Identifying Functions from Their Equations
Decide whether each relation defines a function and give the domain.
(b) y = 3x – 1
Introduction to Functions
EXAMPLE 5 Identifying Functions from Their Equations
Decide whether each relation defines a function and give the domain.
(e) y = 3x + 4
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 4.5 - 23
Introduction to Functions
Variations of the Definition of Function
1. A function is a relation in which, for each value of the first component
of the ordered pairs, there is exactly one value of the second
component.
2. A function is a set of ordered pairs in which no first component is
repeated.
3. A function is a rule or correspondence that assigns exactly one range
value to each domain value.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 4.5 - 24
Introduction to Functions
Function Notation
When a function f is defined with a rule or an equation using x and y for the
independent and dependent variables, we say “y is a function of x” to
emphasize that y depends on x. We use the notation
y = f (x),
called function notation, to express this and read f (x), as “f of x”.
The letter f stands for function. For example, if y = 5x – 2, we can name
this function f and write
f (x) = 5x – 2.
Note that f (x) is just another name for the dependent variable y.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 4.5 - 25
CAUTION
The symbol f (x) does not indicate “f times x,” but represents the y-valuefor the indicated x-value. As shown below, f (3) is the y-value that corresponds to the x-value 3.
Introduction to Functions
Function Notation
y = f (x) = 5x – 2
y = f (3) = 5(3) – 2 = 13
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Introduction to Functions
EXAMPLE 6 Using Function Notation
(a) f (4)
Let f (x) = x + 2x – 1. Find the following.2
f (x) = x + 2x – 12
f (4) = 4 + 2 • 4 – 12
f (4) = 16 + 8 – 1
f (4) = 23
Since f (4) = 23, the ordered pair (4, 23) belongs to f.
Replace x with 4.
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Introduction to Functions
EXAMPLE 6 Using Function Notation
(b) f (w)
f (x) = x + 2x – 12
f (w) = w + 2w – 12
The replacement of one variable with another is important in later courses.
Replace x with w.
Let f (x) = x + 2x – 1. Find the following.2
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Introduction to Functions
EXAMPLE 7 Using Function Notation
Let g(x) = 5x + 6. Find and simplify g(n + 2).
g(x) = 5x + 6
g(n + 2) = 5(n + 2) + 6
= 5n + 10 + 6
= 5n + 16
Replace x with n + 2.
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Introduction to Functions
EXAMPLE 8 Using Function Notation
For each function, find f (7).
f (x) = –x + 2
f (7) = –7 + 2
= –5
Replace x with 7.
(a) f (x) = –x + 2
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 4.5 - 30
4.5 Introduction to Functions
EXAMPLE 8 Using Function Notation
For each function, find f (7).
(b) f = {(–5, –9), (–1, –1), (3, 7), (7, 15), (11, 23)}
We want f (7), the y-value of the ordered pair
where x = 7. As indicated by the ordered pair
(7, 15), when x = 7, y = 15, so f (7) = 15.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 4.5 - 31
4.5 Introduction to Functions
EXAMPLE 8 Using Function Notation
For each function, find f (7).
(c)
The domain element 7 is paired with 17
in the range, so f (7) = 17.
4
7
10
11
17
23
f
Domain Range
7 17
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 4.5 - 32
x
y
O
4.5 Introduction to Functions
EXAMPLE 8 Using Function Notation
For each function, find f (7).
(d)To evaluate f (7), find 7 on
the x-axis.
1 3 5
Then move up
until the graph of f is
reached.
tally to the y-axis gives 3
for the corresponding
y-value. Thus, f (7) = 3.
Moving horizon-
7
1
3
5
7
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Introduction to Functions
Finding an Expression for f (x)
Step 1 Solve the equation for y.
Step 2 Replace y with f (x).
Introduction to Functions
EXAMPLE 9 Writing Equations Using Function Notation
Rewrite each equation using function notation. Then find f (–3) and f (n).
This equation is already solved for y. Since y = f (x),
(a) y = x – 12
Introduction to Functions
EXAMPLE 9 Writing Equations Using Function Notation
Rewrite each equation using function notation. Then find f (–3) and f (n).
First solve x – 5y = 3 for y. Then replace y with f (x).
(b) x – 5y = 3
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 4.5 - 36
Introduction to Functions
Linear Function
A function that can be defined by
f (x) = ax + b,
for real numbers a and b is a linear function. The value of a is the slope
of m of the graph of the function.