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1.2 Functions And Their Properties
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, and the set of all second components is called the
range of the relation.
Find the domain and range of the relation
{(1994,56.21), (1995,51.00), (1996,47.70), (1997,42.78), (1998,39.43)}
The domain is the set of all first components (the x-values).
domain: {1994,1995,1996,1997,1998}
The range is the set of all second components (the y-values).
range: {56.21, 51.00, 47.70, 42.78, 39.43}
Definition of a Function: A function is a relation that assigns
to each domain exactly one range.
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)}
Solution We begin by making a figure for each relation that shows the
domain and the range.
(a)
Figure (a) shows that every element in the
1
6
domain corresponds to exactly one element in
2
8
the range. No two ordered pairs in the given
3
9
relation have the same x and different y’s. For
4
every x, there is only one y. Thus, the relation is a
Domain
Range
function.
(b)
6
8
9
Domain
1
2
3
4
Range
Figure (b) shows that 6 corresponds to both 1
and 2 (6 has two y’s). This relation is not a
function; two ordered pairs have the same x and
different y’s.
Determine whether each equation is a function.
a. y = 2x + 3
b. x = y2
Consider the values of x and y. For every x is there only
one y value.
x
y
x
y
-1
1
1
-1
0
3
0
0
1
5
1
1
Function
x has two y values
Not a Function
The Vertical Line Test for Functions
•If any vertical line intersects a graph in more than one point,
the graph does not define y as a function of x.
a.
Use the vertical line test to identify graphs in which
y is a function b.of x.
y
y
x
y is not a function since
2 values of y correspond
to an x-value.
c.
y
x
y is a function of x
d.
y
x
y is a function of x
x
y is not a function since
2 values of y correspond
to an x-value.
y is a function of x for the graphs in (b) and (c).
Finding a Function’s Domain: It’s domain is the largest set of
real numbers for which the value of f(x) is a real number. 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.
x  2x
f(x)  2
x 1
3
Find the domain of the function:
x(x  2)

(x  1)(x  1)
2
domain : {x  : x  1,1}
or ,1, 1,1, 1, 
1 and -1 need to be
excluded from the
domain because they
cause division by zero.
Find the domain of the function:
g ( x)  2 x  3
Numbers that result in an even root of a negative number
must be excluded. 2x + 3 must be positive.
2x  3  0
-3 -3
2x  3
2
3 

domain :  x  : x  
2

 3 
or  ,  

2
2
3
x
For these values an even root will
2 be positive.
Confirm this graphically.
Find the range graphically.
range : y  : y  0
or 0,  
2
Find the domain and range of the function: h(x) 
x
Find the domain and range graphically.
domain : x  : x  0 range : y  : y  0
or , 0 , 0,  
or , 0 , 0,  
Confirm the domain and range algebraically.
2
x = 0 is not in the domain because
h(x) 
a denominator cannot be zero.
0
y = 0 is not in the range by trying to solve when y = 0
2
0
x
20
y = 0 is not in the range
Graph f (x) = x2 + 1. Use integer values of x from the set {-3, -2, -1, 0, 1, 2, 3}.
Use the graph to specify the function's domain and range.
Solution We begin by setting up a partial table of coordinates.
f (x) = x2 + 1
(x, y) or (x, f (x))
-3
f (-3) = (-3)2 + 1 = 10
(-3, 10)
-2
f (-2) =
(-2)2
+1=5
(-2, 5)
-1
f (-1) = (-1)2 + 1 = 2
(-1, 2)
0
f (0) = 02 + 1 = 1
(0, 1)
1
f (1) = 12 + 1 = 2
(1, 2)
2
f (2) = (2)2 + 1 = 5
(2, 5)
3
(3)2
Plot the points.
Range: [1, oo)
x
10
9
8
7
6
5
4
3
2
1
-4 -3 -2 -1
1
2
3
4
Domain: all Reals
f (3) =
+ 1 = 10
(3, 10)
The points on the graph of f have x-coordinates
that extend indefinitely far to the left and to the
right. Thus, the domain consists of all real
numbers, represented by (-∞, ∞).
The points on the graph have y-coordinates that
start at 1 and extend indefinitely upward. Thus,
the range consists of all real numbers greater
than or equal to 1, represented by [1, ∞).
Complete Student Checkpoint
Determine whether each relation is a function:
a. {(1,2),(3,4),(5,6),(5,8)}
Not a function
b. {(1,2),(3,4),(6,5),(8,5)}
Function
Solve each equation for y and then determine whether the
equation defines y as a function of x.
a. 2x  y  6
2x
 2x
y  2x  6
x y
-1 8
0 6
1 4
Function
b. x 2  y 2  1
x 2
 x2
y 2  x 2  1
x y y   x 2  1
-1 0
0 1
0 -1
Not a function
Smooth, Continuous Graphs
Two important features of the graphs of polynomial functions are that
they are smooth and continuous. By smooth, we mean that the graph
contains only rounded
y
y
Smooth
curves with no sharp
rounded
Smooth
corner
rounded
corners. Bycontinuous,
corner
we mean that the graph
has no breaks and can
be drawn without
x
lifting your pencil from
the rectangular
Smooth
coordinate system.
Smooth
rounded
rounded
corner
corner
These ideas are
illustrated in the figure.
x
Discontinuities
These graphs have discontinuity. They are not continuous everywhere.
a.
b.
5
4
3
5
4
3
2
1
-5 -4 -3 -2 -1
-1
-2
-3
-4
-5
1
1 2 3 4 5
This graph has removable
discontinuity at x = 1.
It is removable because we could
define f(1) in such a way as to
plug the hole and make f(x)
continuous as 1.
-5 -4 -3 -2 -1
-1
-2
-3
-4
-5
1 2 3 4 5
This graph has jump
discontinuity.
This discontinuity is not
removable because there is
more than just a hole at x = 0;
the gap makes it impossible to
plug.
Discontinuities
This graph has discontinuity. It is not continuous everywhere.
This graph has infinite
discontinuity.
This discontinuity is definitely
not removable at x = -2.
5
4
3
2
1
-5 -4 -3 -2 -1
-1
-2
-3
-4
-5
1 2 3 4 5
Definitions of Local Maximum
and Local Minimum
A function value f(a) is a
-5 -4 -3 -2
local maximum of f if there
exists an open interval about a
such that f(a) > f(x) for all x in
the open interval. It occurs at (1, 4),
5
4
3
2
1
-1
-1
-2
-3
-4
-5
1 2 3 4 5
the maximum value is 4
A function value f(b) is a local minimum of f if there
exists an open interval about b such that f(b) < f(x) for
all x in the open interval. It occurs at (-3, -4),
the minimum value is -4
Local is also referred to as Relative or Absolute .
Increasing, Decreasing, and Constant Functions
A function is increasing on an interval if for any x1, and x2
in the interval, where x1 < x2, then f (x1) < f (x2).
A function is decreasing on an interval if for any x1, and x2
in the interval, where x1 < x2, then f (x1) > f (x2).
A function is constant on an interval if for any x1, and x2 in
the interval, where x1 < x2, then f (x1) = f (x2).
(x2, f (x2))
(x1, f (x1))
(x1, f (x1))
(x2, f (x2))
(x1, f (x1))
(x2, f (x2))
Increasing
f (x1) < f (x2)
Decreasing
f (x1) > f (x2)
Constant
f (x1) = f (x2)
Example
Describe the increasing, decreasing, or constant behavior of each
function whose graph is shown.
a.
b.
5
4
3
5
4
3
2
1
-5 -4 -3 -2 -1
-1
-2
-3
-4
-5
1
1 2 3 4 5
decreasing on the interval (-∞, 0)
increasing on the interval (0, 2)
decreasing on the interval (2, ∞)
-5 -4 -3 -2 -1
-1
-2
-3
-4
-5
1 2 3 4 5
The function is defined in two
pieces.
constant on the interval (-∞, 0)
increasing on the interval [0, ∞)
Lower Bound, Upper Bound, and Bounded
The function is bounded below if
there is some number b that is
less than or equal to every
number in the range of f. Any
such number b is called a
lower bound of f.
y
The function is bounded above if
there is some number b that is
less than or equal to every
number in the range of f. Any
such number b is called a
upper bound of f.
y
x
not bounded above
bounded below
x
bounded above
not bounded below
Lower Bound, Upper Bound, and Bounded
The function is bounded if it is
bounded both above and below.
y
y
x
bounded
The function is:
x
not bounded above
not bounded below
Definition of Even and Odd
Functions
The function f is an even function if f (-x) = f (x)
for all x in the domain of f.
The right side of the equation of an even function
does not change if x is replaced with -x.
The function f is an odd function if f (-x) = -f (x)
for all x in the domain of f.
Every term in the right side of the equation of an
odd function changes sign if x is replaced by -x.
Identify the following function as even, odd,
or neither: f(x) = 3x2 - 2.
Use the given function equation to find f(-x).
f(-x) = 3(-x) 2-2
= 3x2 - 2
The right side of the equation of the given
function did not change when we replaced x
with -x. Algebraically:
Because f(-x) = f(x)
f is an even function.
Even Functions and
y-Axis Symmetry
y
x
The graph of an even function in which
f (-x) = f (x)
is symmetric with respect to the y-axis.
Odd Functions and
Origin Symmetry
y
The graph of an odd function in which
f (-x) = - f (x)
is symmetric with respect to the origin.
x
Identify the following function as even, odd,
or neither: f(x) = 3x2 - 2.
Identify graphically.
Because f(x) looks the same on
either side of the y-axis
f is an even function.
Determine algebraically whether each of the following functions
is even, odd, or neither.
a. f (x)  x 2  6
f (x)  (x)2  6
 x2  6
does
not
change
b. g(x)  7x 3  x
all terms
g(x)  7(x)3  (x) change
signs
 7x 3  x
ODD
EVEN
c. h(x)  x 5  1
h(x)  (x)5  1
 x 5  1
NEITHER
all terms
do not
change
signs
Arrow Notation
Symbol
Meaning
xa
xa
x
x approaches a from the right.
x approaches a from the left.
x approaches infinity; that is, x
increases without bound.
x
x approaches negative infinity;
that is, x decreases without bound.
Definition of a Vertical Asymptote
The line x  a is a vertical asymptote of the graph of a
function f if f (x) increases or decreases without bound
as x approaches a.
f (x)   as x  a 
f (x)   as x  a 
y
y
f
f
x
a
a
x=a
x=a
lim f (x)   or lim f (x)  
xa
x
xa
Definition of a Vertical Asymptote
The line x  a is a vertical asymptote of the graph of a
function f if f (x) increases or decreases without bound
as x approaches a.
f (x)  as x  a 
f (x)    as x  a 
y
y
x=a
x=a
x
a
f
a
f
lim f (x)   or lim f (x)  
xa
xa
x
Locating Vertical Asymptotes
If f(x) = p(x) / q(x) is a rational function in
which p(x) and q(x) have no common factors
and a is a zero of q(x) , the denominator,
then x = a is a vertical asymptote of the graph
of f.
Find the vertical asymptote
of this function:
x7
f (x) 
x5
5 is a zero of x - 5, therefore
x = 5 is a vertical asymptote of f
Find the vertical asymptotes, if any, of the graph of each rational
function.
x 1
a.
b.
x 1
h(x)  2
g(x)  2
x 1
x 1
x 1
g(x) 
no vertical asymptotes
x 1 x 1
  
x  1 and x  1
Definition of a Horizontal Asymptote
The line y = b is a horizontal asymptote of the graph of a
function f if f (x) approaches b as x increases or
decreases without bound.
y
y
y=b
y
f
f
f
y=b
x
y=b
x
x
f(x)b as x
f(x)b as x
lim f (x)  b
x
f(x)b as x
Locating Horizontal Asymptotes
Let f be the rational function given by
an x n  an 1 x n1  ...  a1 x  a0
f (x) 
, an  0,bm  0
m
m 1
bm x  bm1 x  ...  b1 x  b0
The degree of the numerator is n. The degree of
the denominator is m.
1. If n<m, the x-axis, or y=0, is the horizontal
asymptote of the graph of f.
2. If n=m, the line y = an/bm is the horizontal
asymptote of the graph of f.
3. If n>m, the graph of f has no horizontal
asymptote.
Find the horizontal asymptotes, if any, of the graph of each rational
function.
a.
b.
9x
9x 3
g(x)  2
h(x)  2
3x  1
3x  1
If n<m,
the x-axis, or y = 0
c. g(x) 
9x
3x  1
If n=m,
then y = a/b, or y = 3
If n>m,
no horizontal asymptote
Strategy for Graphing a Rational Function
Suppose that f(x) = p(x) / q(x) , where p(x) and q(x) are polynomial functions
with no common factors.
1.
Determine whether the graph of f has symmetry.
f (x)  f (x): y-axis symmetry
f (x)  f (x): origin symmetry
2.
Find the y-intercept (if there is one) by evaluating f (0).
3.
Find the x-intercepts (if there are any) by solving the equation p(x)  0.
4.
Find any vertical asymptote(s) by solving the equation q (x)  0.
5.
Find the horizontal asymptote (if there is one) using the rule for determining
the horizontal asymptote of a rational function.
6.
Plot at least one point between and beyond each x-intercept and vertical
asymptote.
Use the information obtained previously to graph the function between
and beyond the vertical asymptotes.
7.
Sketch the graph of
2x  3
f ( x) 
5 x  10
•
•
•
•
•
•
5
4
3
2
1
-5 -4 -3 -2 -1
-1
-2
-3
-4
-5
1 2 3 4 5
the graph has no symmetry
The y-intercept is (0,-3/10)
The x-intercept is (3/2, 0)
The vertical asymptote is x = -2
The horizontal asymptote is y = 2/5
Test points include (-5, 13/15), (-4, 11/10),
(-3, 9/5), (-7/2, 16/5), (-1, -1), (2, 1/20)
Identify the intervals on which the function is increasing,
decreasing, and constant.
5 if x  3

x y
f (x)  x  3  x  2  2x  1 if -3  x  2
5 if x  2
-4 -5

need to find out
-3 -5
value it begins or
-3 -5
ends but is not
-2 -3
included.
-1 -1
0 1
1 3
Constant:
2 5
, 3 and 2,  
2 5
Increasing: 3, 2 
3 5
graphing calculator
interpretation
Suppose f is an odd function. For each statement below,
explain why the statement is possible, necessarily true, or
impossible.
a. f(3) = 4 and f(-4) = -3
Possible
For odd functions f (-x) = - f (x), so if (-3,-4) and
(4,3) are in the function this statement is possible.
b. f(3) = 4 and f(-3) = -4
Necessarily true
For odd functions f (-x) = - f (x), so f is an odd
function in this statement.
c. f(0) = -2 and f(-0) = 2
Impossible
The graph of an odd function in which f (-x) = - f (x) is
symmetric with respect to the origin. f(0) = 0 for this
to be an odd function
Matching functions using end behavior
Match the functions to their graphs on page 97.
4
2
3
3x
3x
3x
3x
d. y  2
a. y  2
b. y  2
c. y  2
x 1
x 1
x 1
x 1
When x is very large, the denominator x2+1 is almost the
same as x2, replace this to make simpler functions and
compare end behaviors to2 match with graphs.
3x
3x
3x 4
3x 3
b. y  2
a. y  2
d. y  2
c. y  2
x
x
x
x
2
3
b.
y

3
c.
y

3x
d.
y

3x
a. y 
x
c. matches
d. matches
b. matches
a. matches
graph (ii)
graph (i)
graph (iii)
graph (iv)
end behavior,
end behavior, end behavior,
ending behavior,
it approaches
y=0
it approaches
y=3
it approaches
y = 3x
it approaches
y = 3x2
Functions and Their Properties