NAME
1-1
DATE
PERIOD
Study Guide and Intervention
Functions
Describe Subsets of Real Numbers The set of real numbers
includes the rationals , irrationals , integers , wholes , and naturals .
Another way is to use interval notation. With interval notation, you use
brackets if an endpoint is included and parentheses if an endpoint is not
included. Use ∞ to indicate positive infinity and -∞ to indicate negative
infinity.
Example
Describe x > 18 using set-builder notation and interval notation.
The set includes all numbers that are greater than 18 but are not equal to 18.
Set-builder notation:
{x | x > 18, x ∈ }
The vertical line | means “such that.” The symbol ∈ means “is an element
of.” Read the expression as the set of all x such that x is greater than 18 and
x is an element of the set of real numbers.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Interval notation:
(18, ∞)
Use parentheses on the left because 18 is not included in the set. Use
parentheses with infinity since it never ends.
Exercises
Write each set of numbers in set-builder and interval notation, if possible.
1. {17, 18, 19, 20, …}
2. x ≤ -2
{x | x ≥ 17, x ∈ }
3. x > -8.8
{x | x ≤ -2, x ∈ }; (-∞, -2]
4. 5 < x < 15
{x | x > -8.8, x ∈ }; (-8.8, ∞)
5. x < -11 or x ≥ 1
{x | 5 < x < 15, x ∈ }; (5, 15)
6. {…, -10, -9, -8, -7}
{x | x < -11 or x ≥ 1, x ∈ };
(-∞, -11) ∪ [1, ∞)
Chapter 1
{x | x ≤ -7, x ∈ }
5
Glencoe Precalculus
Lesson 1-1
One way to describe a subset of the real numbers is to use set-builder
notation. With set-builder notation, you choose a variable, list the
properties of the variable, and tell to which set of numbers the variable
belongs.
NAME
DATE
1-1
Study Guide and Intervention
PERIOD
(continued)
Functions
Identify Functions A relation is a rule that relates, or pairs, the
elements in set A with the elements in set B. Set A contains the inputs, or
the domain, and set B contains the outputs, or the range. A function f
from set A to set B is a relation that assigns to each element x in set A
exactly one element y in set B. To evaluate a function, replace the
independent variable with the given value from the domain and simplify.
Example 1
Find each function value.
a. If f(x) = 4x3 + 6x2 + 3x, find f(-2).
f(x) = 4x3 + 6x2 + 3x
f(-2) = 4(-2)3 + 6(-2)2 + 3(-2)
= -32 + 24 - 6 or -14
Original function
Substitute -2 for x.
Simplify.
⎧ √x + 1 if x ≤ 4
⎨
b. If g(x) =
⎩
3x if 4 < x < 10, find g(6) and g(10).
2
2x - 15 if x ≥ 10
Example 2
3+x
x - 6x
.
State the domain of f(x) = −
2
3+x
x - 6x
When the denominator of −
is zero, the expression is undefined.
2
Solving x2 - 6x = 0, the excluded values in the domain are x = 0 and x = 6.
The domain is {x | x ≠ 0, 6, x ∈ }.
Exercises
Find each function value.
1. If f(x) = 5x2 - 4x - 6, find f(3).
2. If h(x) = 9x9 - 4x4 + 3x - 2, find h(t).
9t9 - 4t4 + 3t - 2
27
⎧ x + 45 if x ≤ -1
, find g(-5) and g(36). 40; 45
3. If g(x) = ⎨
⎩ 81 - x if x > -1
⎧
4. If f(x) =
⎨
⎩
Chapter 1
√2x if x < 3
2x + 10 if 3 ≤ x < 8, find f(3) and f(8.5). 16; 42
42 if x ≥ 8
6
Glencoe Precalculus
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Look at the “if ” statements to see that 6 fits into the second rule,
so g(6) = 3(6) or 18.
The value 10 fits into the third rule, so g(10) = 2(10)2 - 15 or 185.
NAME
DATE
1-2
PERIOD
Study Guide and Intervention
Analyzing Graphs of Functions and Relations
Analyzing Function Graphs
By looking at the graph of a function,
you can determine the function’s domain and range and estimate the x- and
y-intercepts. The x-intercepts of the graph of a function are also called the
zeros of the function because these input values give an output of 0.
Example
Use the graph of f to find the domain
and range of the function and to approximate the
y
y-intercept and zero(s). Then confirm the estimate
(-0.75, 5.0625)
f (x) = -x 2 - 1.5x
algebraically.
The graph is not bounded on the left or right, so the
domain is the set of all real numbers.
0
x
{x | x ∈ }
The graph does not extend above 5.0625 or f(-0.75), so
the range is all real numbers less than or equal to 5.0625.
{y | y ≤ 5.0625, y ∈ }
The y-intercept is the point where the graph intersects the y-axis. It appears
to be 4.5. Likewise, the zeros are the x-coordinates of the points where the
graph crosses the x-axis. They seem to occur at -3 and 1.5.
+ 4.5
To find the y-intercept algebraically, find f(0).
f(0) = -(0)2 - 1.5(0) + 4.5 = 4.5
Exercises
Use the graph of g to find the domain and range of the function and to approximate
its y-intercept and zero(s). Then find its y-intercept and zeros algebraically.
1.
2.
y
g (x) = 8 + 2x - x 2
8
−6
0
−12
−20
4
(0, 8)
8x
−8
(2, -16)
8
−4 0
y
8x
−4
g(x) = x 2 - 4x - 12
−8
D = {x | x ∈ }, R = {y | y ≥ -16,
y ∈ }
y-intercept: -12
zeros: -2 and 6
x2 - 4x - 12 = (x + 2)(x - 6);
x = -2 or x = 6
g(0) = 02 - 4(0) - 12 = -12
Chapter 1
D = {x | x ∈ }, R = {y | y ≤ 9,
y ∈ }
y-intercept: 8
zeros: -2 and 4
8 + 2x - x2 = -(x + 2)(x - 4);
x = -2 or x = 4
g(0) = 8 + 2(0) - 02 = 8
10
Glencoe Precalculus
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
To find the zeros algebraically, let f(x) = 0 and solve for x.
-x2 - 1.5x + 4.5 = 0
-1(x + 3)(x - 1.5) = 0
x = -3 or x = 1.5
NAME
DATE
1-2
PERIOD
Study Guide and Intervention
(continued)
Analyzing Graphs of Functions and Relations
Symmetry of Graphs A graph of a relation that is symmetric to the
x-axis and/or the y-axis has line symmetry. A graph of a relation that is
symmetric to the origin has point symmetry.
Description
Algebraic Test
x-axis
For every (x, y) on the graph,
(x, -y) is also on the graph.
Replacing y with -y produces
an equivalent equation.
y-axis
For every (x, y) on the graph,
(-x, y) is also on the graph.
Replacing x with -x produces
an equivalent equation.
origin
For every (x, y) on the graph,
(-x, -y) is also on the graph.
Replacing x with -x and y with
-y produces an equivalent
equation.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Functions that are symmetric with respect to the y-axis are even functions,
and for every x in the domain, f(-x) = f(x). Functions that are symmetric
with respect to the origin are odd functions and for every x in the domain,
f(-x) = -f(x).
Example
GRAPHING CALCULATOR Graph f(x) = -x3 + 2x.
Analyze the graph to determine whether the function is even, odd, or
neither. Confirm algebraically. If odd or even, describe the
symmetry of the graph of the function.
From the graph, it appears that the function is symmetric to
the origin.
Confirm: f(-x) = -(-x)3 + 2(-x) = x3 - 2x = -f(x)
[-10, 10] scl: 1 by [-10, 10] scl: 1
The function is odd because f(-x) = -f(x).
Exercises
GRAPHING CALCULATOR Graph each function. Analyze the graph
to determine whether each function is even, odd, or neither. Confirm
algebraically. If odd or even, describe the symmetry of the graph of
the function.
1. f(x) = 4x3 + 1
2. g(x) = x4 - 10x2 + 9
neither; 4(-x)3 + 1 = -4x3 + 1
5
3. g(x) = −
4
even;
(-x)4 -10(-x)2 + 9 = x4 - 10x2 + 9
symmetric with respect to y-axis
4. g(x) = x3 - 6x
x
5
5
even; −
=−
4
4
(-x)
odd; (-x)3 - 6(-x) = -x3 + 6x
symmetric with respect to origin
x
symmetric with respect to y-axis
Chapter 1
11
Glencoe Precalculus
Lesson 1-2
Symmetric with
respect to…
NAME
DATE
1-3
PERIOD
Study Guide and Intervention
Continuity, End Behavior, and Limits
Continuity A function f(x) is continuous at x = c if it satisfies the
following conditions.
(1) f(x) is defined at c; in other words, f(c) exists.
(2) f(x) approaches the same function value to the left and right of c; in other
words, lim f(x) exists.
x→c
(3) The function value that f(x) approaches from each side of c is f(c); in
other words, lim f(x) = f(c).
x→c
Functions that are not continuous are discontinuous. Graphs that are
discontinuous can exhibit infinite discontinuity, jump discontinuity,
or removable discontinuity (also called point discontinuity).
Example
Determine whether each function is continuous at the given
x-value. Justify using the continuity test. If discontinuous, identify the type of
discontinuity as infinite, jump, or removable.
2x
;x=1
b. f(x) = −
2
a. f(x) = 2|x| + 3; x = 2
x -1
(1) f(2) = 7, so f(2) exists.
(2) Construct a table that shows values for
f(x) for x-values approaching 2 from the
left and from the right.
y = f(x)
x
y = f(x)
1.9
6.8
2.1
7.2
1.99
6.98
2.01
7.02
1.999
6.998
2.001
7.002
The tables show that y approaches 7
as x approaches 2 from both sides.
It appears that lim f(x) = 7.
x
y = f(x)
x
y = f(x)
0.9
-9.5
1.1
10.5
0.99
-99.5
1.01
100.5
0.999
-999.5
1.001
1000.5
The function has infinite discontinuity
at x = 1.
x→2
(3) lim f(x) = 7 and f(2) = 7.
x→2
The function is continuous at x = 2.
Exercises
Determine whether each function is continuous at the given x-value.
Justify your answer using the continuity test. If discontinuous,
identify the type of discontinuity as infinite, jump, or removable.
⎧ 2x + 1 if x > 2
;x=2
2. f(x) = x2 + 5x + 3; x = 4 f(4) = 39
1. f(x) = ⎨
⎩ x - 1 if x ≤ 2
lim f(x) = 1 and lim f(x) = 5 ,
x → 2–
lim f(x) = 39 and lim f(x) = 39,
x → 4-
x → 2+
so the function is not continuous;
it has jump discontinuity.
Chapter 1
x → 4+
so the function is continuous.
16
Glencoe Precalculus
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
x
The function is not defined at x = 1
because it results in a denominator of 0.
The tables show that for values of x
approaching 1 from the left, f(x)
becomes increasingly more negative. For
values approaching 1 from the right,
f(x) becomes increasingly more positive.
NAME
DATE
1-3
PERIOD
Study Guide and Intervention
(continued)
Continuity, End Behavior, and Limits
End Behavior The end behavior of a function describes how the function behaves at
either end of the graph, or what happens to the value of f(x) as x increases or decreases
without bound. You can use the concept of a limit to describe end behavior.
Left-End Behavior (as x becomes more and more negative):
lim
x → -∞
f(x)
Right-End Behavior (as x becomes more and more positive): lim f(x)
x→∞
The f(x) values may approach negative infinity, positive infinity, or a specific value.
Example
Use the graph of f(x) = x3 + 2 to describe
its end behavior. Support the conjecture numerically.
As x decreases without bound, the y-values also
decrease without bound. It appears the limit is negative
infinity: lim f(x) = -∞.
8
y
4
−4
−2 0
f(x) = x 3 + 2
2
4x
−4
x → -∞
As x increases without bound, the y-values increase
without bound. It appears the limit is positive infinity:
lim f(x) = ∞.
−8
x→∞
x
-1000
-100
-10
0
10
100
1000
f(x)
-999,999,998
-999,998
-998
2
1002
1,000,002
1,000,000,002
As x −∞, f(x) -∞. As x ∞, f(x) ∞. This supports the conjecture.
Exercises
Use the graph of each function to describe its end behavior. Support
the conjecture numerically.
1.
8
4
−4
0
−2
y
2.
8
4
f (x ) = -x 4 - 2x
2
4x
−16 −8 0
−4
−4
−8
−8
y
f (x ) =
8
5x
x -2
16x
lim
f(x) = -∞; lim f(x) = -∞
lim
f(x) = 5; lim f(x) = 5
x → -∞
x→∞
x → -∞
x→∞
See students’ work.
Chapter 1
See students’ work.
17
Glencoe Precalculus
Lesson 1-3
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Construct a table of values to investigate function values as |x| increases.