Upload
others
View
9
Download
0
Embed Size (px)
Citation preview
CHAPTER 1:
Functions
1.1: Functions
1.2: Graphs of Functions
1.3: Basic Graphs and Symmetry
1.4: Transformations
1.5: Piecewise-Defined Functions;
Limits and Continuity in Calculus
1.6: Combining Functions
1.7: Symmetry Revisited
1.8: x = f y( )
1.9: Inverses of One-to-One Functions
1.10: Difference Quotients
1.11: Limits and Derivatives in Calculus
• Functions are the building blocks of precalculus.
• In this chapter, we will investigate the general theory of functions and their graphs.
• We will study particular categories of functions in Chapters 2, 3, 4, and even 9.
(Section 1.1: Functions) 1.1.1
SECTION 1.1: FUNCTIONS
LEARNING OBJECTIVES
• Understand what relations and functions are.
• Recognize when a relation is also a function.
• Accurately use function notation and terminology.
• Know different ways to describe a function.
• Find the domains and/or ranges of some functions.
• Be able to evaluate functions.
PART A: DISCUSSION
• WARNING 1: The word “function” has different meanings in mathematics and
in common usage.
• Much of precalculus covers properties, graphs, and categorizations of functions.
• A relation relates inputs to outputs.
• A function is a relation that relates each input in its domain to exactly one output
in its range.
• We will investigate the anatomy of functions (a name such as f , a “function
rule,” a domain, and a range), look at examples of functions, find their domains
and/or ranges, and evaluate them at an input by determining the resulting output.
(Section 1.1: Functions) 1.1.2
PART B: RELATIONS
A relation is a set of ordered pairs of the form input, output( ) ,
where the input is related to (“yields”) the output.
WARNING 2: If a is related to b, then b may or may not be related to a.
Example 1 (A Relation)
Let the relation R = 1, 5( ) , , 5( ) , 5, 7( ){ } .
1, 5( ) R , so 1 is related to 5 by R.
Likewise, is related to 5, and 5 is related to 7.
R can be represented by the arrow diagram below.
§
PART C: FUNCTIONS
A relation is a function Each input is related to (“yields”) exactly one output.
A function is typically denoted by a letter, most commonly f .
Unless otherwise specified, we assume that f represents a function.
The domain of a function f is the set of all inputs.
It is the set of all first coordinates of the ordered pairs in f .
We will denote this by Dom f( ) , although this is not standard.
The range of a function f is the set of all resulting outputs.
It is the set of all second coordinates of the ordered pairs in f .
We will denote this by Range f( ) .
• We assume both sets are nonempty.
• (See Footnote 1 on terminology.)
(Section 1.1: Functions) 1.1.3
Example 2 (A Relation that is a Function; Revisiting Example 1)
Again, let the relation R = 1, 5( ) , , 5( ) , 5, 7( ){ } .
Determine whether or not the relation is also a function.
If it is a function, find its domain and its range.
§ Solution
Refer to the arrow diagram in Example 1.
Each input is related to (“yields”) exactly one output.
Therefore, this relation is a function.
• Its domain is the set of all inputs: 1, , 5{ } .
• Its range is the set of all outputs:
5, 7{ } .
•• Do not write
5, 5, 7{ } . §
WARNING 3: Although a function cannot allow one input to yield
multiple outputs, a function can allow multiple inputs (such as 1 and
in Example 2) to yield the same output (5). However, such a
function would not be one-to-one (see Section 1.9).
Example 3 (A Relation that is Not a Function)
Repeat Example 2 for the relation S, where S = 5,1( ) , 5,( ) , 7, 5( ){ } .
§ Solution
S can be represented by the arrow diagram below.
An input (5) is related to (“yields”) two different outputs (1 and ).
Therefore, this relation is not a function. §
(Section 1.1: Functions) 1.1.4
TIP 1: Think of a function button on a basic calculator such as the
x2 or button, which represent squaring and square root functions,
respectively. If a function is applied to the input 5, the calculator can never
imply, “The outputs are 1 and .”
Example 4 (Ages of People)
For a relation R,
• The set of inputs is the set of all living people.
• The outputs are ages in years.
• a, b( ) R Person a is b years old.
Is this relation a function?
§ Solution
Yes, R is a function, because a living person has only one age in
years. §
Example 5 (Colors on Paintings)
For a relation S,
• The set of inputs is the set of all existing paintings.
• The outputs are colors.
• a, b( ) S Painting a has color b.
Is this relation a function?
§ Solution
No, S is not a function, because there are paintings with more than
one color. §
(Section 1.1: Functions) 1.1.5
PART D: EVALUATING FUNCTIONS (THE BASICS)
If an input x Dom f( ) , then its output is a well-defined (i.e., single) value,
denoted by f x( ) .
• We refer to x here as the argument of f .
• We refer to f x( ) as the function value at x, or the image of x.
f x( ) is read as “ f of x” or “ f at x.”
WARNING 4: f x( ) does not mean “ f times x.”
A function can be modeled by an input-output machine such as:
x f f x( )
When we evaluate a function at an input, we determine the resulting output
and express it in simplified form.
A function is defined (or it exists) only on its domain.
If x Dom f( ) , then
f x( ) is undefined (or it does not exist).
Example 6 (Evaluating a Function; Revisiting Examples 1 and 2)
Let the function f = 1, 5( ) , , 5( ) , 5, 7( ){ } .
a) Evaluate f 5( ) . b) Evaluate
f 6( ) .
§ Solution
a) 5, 7( ) f , so
5 Dom f( ) , and
f 5( ) = 7 .
b) 6 Dom f( ) , so f 6( ) is undefined. §
(Section 1.1: Functions) 1.1.6
Example 7 (Failure to Evaluate a Non-Function; Revisiting Example 3)
Let the relation S = 5,1( ) , 5,( ) , 7, 5( ){ } . If we had erroneously
identified S as a function and renamed it f , we would see that f 5( ) is
not well-defined. It is unclear whether its value should be 1 or . §
A “function rule” describes how a function assigns an output to an input.
It is typically given by a defining formula such as f x( ) = x2 .
Example 8 (Squaring Function: Evaluation)
Let f x( ) = x2 on .
• This means that we are defining a function f by the rule
f x( ) = x2 , with
Dom f( ) = .
• The rule could have been given as, say, f u( ) = u2 .
Either way, f squares its input.
Evaluate f 3( ) .
§ Solution
We substitute
3( ) for x, and we square it.
f x( ) = x2
f 3( ) = 3( )2
= 9
3 f 9
WARNING 5: Be prepared to use grouping symbols whenever you
perform a substitution; only omit them if you are sure you do not
need them. Note that f 3( ) is not equal to 32 , which equals 9 . §
(Section 1.1: Functions) 1.1.7
WARNING 6: As a matter of convenience, some sources refer to f x( ) as a
function, but this convention is often rejected as non-rigorous.
• One advantage to the f x( ) notation is that it indicates that f is a function of one
variable. The notation f x, y( ) indicates that f is a function of two variables.
• A function can be determined by a domain and a function rule. Together, the domain
and the function rule determine the range of the function. (See Footnote 1.)
• Two functions with the same rule but different domains are considered to be different.
• Piecewise-defined functions will be discussed in Section 1.5. Such a function applies
different rules to disjoint (non-overlapping) subsets of its domain (subdomains).
For example, consider the function f , where:
f x( ) =x2 , 2 x < 1
x +1, 1 x 2
PART E: POLYNOMIAL, RATIONAL, AND ALGEBRAIC FUNCTIONS
Review Section 0.6 on polynomial, rational, and algebraic expressions.
f is a polynomial function on f can be defined by:
f x( ) = (a polynomial in x),
which implies that Dom f( ) = .
• See Footnote 2 on whether polynomial functions can be defined on another domain.
• x could be replaced by another variable.
f is a rational function f can be defined by:
f x( ) = (a rational expression in x).
f is an algebraic function f can be defined by:
f x( ) = (an algebraic expression in x).
Example 9 (Polynomial, Rational, and Algebraic Functions)
a) Let
f x( ) =x3
+ 7x5/7
x x + 53
+. Then, f is an algebraic function.
b) Let
g t( ) =5t3 1
t2+ 7t 2
. Then, g is both rational and algebraic.
c) Let h x( ) = x7+ x2 3. Then, h is polynomial, rational, and algebraic. §
(Section 1.1: Functions) 1.1.8
PART F: REPRESENTATIONS OF FUNCTIONS
Ways to Represent a Function Rule
• The domain of the function could be the implied domain (see Part G).
In Parts B and C, we determined the domain from a set of ordered pairs or an
arrow diagram.
• For our examples in 1) through 8), we will let f be our squaring function from
Example 8, with Dom f( ) = .
A function rule can be represented by …
1) a defining formula:
f x( ) = x2
2) an input-output model (machine):
x f x2
3) a verbal description:
“This function squares its input, and the result is its output.”
4) a table of input-output pairs:
Input
x
Output
f x( )
2 4
1 1
0 0
1 1
2 4
• Since Dom f( ) = , a complete table is impossible to write.
However, a partial table such as this can be useful, especially for
graphing purposes.
(Section 1.1: Functions) 1.1.9
5) a set of input x, output f x( )( ) ordered pairs:
• The table in 4) yields ordered pairs such as
2, 4( ) .
• Since Dom f( ) = , the set is an infinite set.
6) a graph consisting of points corresponding to the ordered pairs in 5);
see Section 1.2:
• The graph of f below represents the set
x, x2( ) x{ } .
• We assume that the graph extends beyond the figure “as expected.”
7) an equation:
• The equations y = x2 and y x2= 0 describe y as the same
function of x (explicitly so in the first equation; implicitly in the
second). Their common graph is in 6). See Section 1.2.
8) an arrow diagram:
• A partial arrow diagram for f is below.
9) an algorithm.
• Perhaps the output is computed by some computer code.
10) a series.
• (See Footnote 3.)
(Section 1.1: Functions) 1.1.10
PART G: IMPLIED (OR NATURAL) DOMAIN
Implied (or Natural) Domain
If f is defined by: f x( ) = (an expression in x), then
the implied (or natural) domain of f is the set of all real numbers
(x values) at which the value of the expression is a real number.
• x could be replaced by another variable.
Dom f( ) is assumed to be the implied domain of f ,
unless otherwise specified or implied by the context.
• In some applications (including geometry), we may restrict inputs to nonnegative and/or
integer values (rounding may be possible).
Implied Domain of an Algebraic Function
If a function is algebraic, then its implied domain is the set of all real
numbers except those that lead to (the equivalent of) …
1) dividing by zero, or
2) taking the even root of a negative-valued radicand.
• The list of restrictions will grow when we discuss non-algebraic functions.
Example 10 (Implied Domain of an Algebraic Function)
a) If f x( ) =1
xor x 1( ) , then the implied domain of f is
\ 0{ } , the set of
all real numbers except 0.
b) If g x( ) = x or x1/2( ) , then the implied domain of g is
0, ) , the set of
all nonnegative real numbers.
• WARNING 7: The implied domain of g includes 0. Observe that
0 = 0 , a perfectly good real number.
• WARNING 8: We will define 1 , for example, as an imaginary
number in Section 2.1. However, 1 will never be a real value. §
(Section 1.1: Functions) 1.1.11
PART H: FINDING DOMAINS AND/OR RANGES
Example 11 (Squaring Function: Finding Domain and Range)
Let f x( ) = x2 . Describe the domain and the range of f using set-builder,
graphical, and interval forms.
§ Solution
x2 is a polynomial, so we assume that Dom f( ) = .
• The symbol is used in place of set-builder form.
• The graph of is the entire real number line:
• In interval form, is
,( ) .
The resulting range of f is the set of all nonnegative real numbers, because
every such number is the square of some real number. For example, 7 is the
square of 7 : f 7( ) = 7 . Also:
WARNING 9: Squares of real numbers are never negative.
• In set-builder form, the range is: y y 0{ } , or
y : y 0{ } .
(We could have used x instead of y, but we tend to associate y with
outputs, and we should avoid confusion with the domain.)
• The graph of the range is:
• In interval form, the range is:
0, ) . §
(Section 1.1: Functions) 1.1.12
Example 12 (Finding a Domain)
Let f x( ) = x 3 . Find
Dom f( ) , the domain of f .
§ Solution
x 3 is a real output x 3 0 x 3 .
WARNING 10: We solve the weak inequality x 3 0 , not the
strict inequality x 3> 0 . Observe that 0 = 0 , a real number.
The domain of f …
… in set-builder form is: x x 3{ } , or
x : x 3{ }
… in graphical form is:
… in interval form is:
3, )
• Range f( ) = 0, ) . Ranges will be easier to determine once we learn how to graph
these functions in Section 1.4.
• If the rule for f had been given by f t( ) = t 3 , we still would have had the same
function with the same domain and range. The domain could be written as
t t 3{ } ,
x x 3{ } , etc. It’s the same set of numbers. §
Example 13 (Finding a Domain)
Let
f x( ) =1
x 3. Find Dom f( ) .
§ Solution
This is similar to Example 12, but we must avoid a zero denominator.
We solve the strict inequality x 3> 0 , which gives us x > 3 .
The domain of f …
… in set-builder form is:
x x > 3{ } , or
x : x > 3{ }
… in graphical form is:
… in interval form is: 3,( )
§
(Section 1.1: Functions) 1.1.13
Example 14 (Finding a Domain)
Let f x( ) = 3 x
4. Find
Dom f( ) .
§ Solution
Solve the weak inequality: 3 x 0 .
Method 1
3 x 0 Now subtract 3 from both sides.
x 3 Now multiply or divide both sides by 1.
WARNING 11: We must then reverse the direction of
the inequality symbol. x 3
Method 2
3 x 0 Now add x to both sides.
3 x Now switch the left side and the right side.
WARNING 12: We must then reverse the direction of
the inequality symbol. x 3
The domain of f …
… in set-builder form is: x x 3{ } , or
x : x 3{ }
… in graphical form is:
… in interval form is:
, 3(
§
Example 15 (Finding a Domain)
Let f x( ) = x 3
3. Find
Dom f( ) .
§ Solution
Dom f( ) = , because:
• The radicand, x 3, is a polynomial, and
• WARNING 13: The taking of odd roots (such as cube roots) does
not impose any new restrictions on the domain. Remember that the
cube root of a negative real number is a negative real number. §
(Section 1.1: Functions) 1.1.14
Example 16 (Finding a Domain)
Let g t( ) =
3t + 9
2t2+ 20t
. Find Dom g( ) .
WARNING 14: Don’t get too attached to f and x. Be flexible.
§ Solution
The square root operation requires:
3t + 9 0
3t 9
t 3
We forbid zero denominators, so we also require:
2t2+ 20t 0
2t t +10( ) 0
t 0 and t +10 0
t 0 and t 10
WARNING 15: We use the connective “and”
here, not “or.” (See Footnote 4.)
We already require t 3 , so we can ignore the restriction t 10 .
The domain of g …
… in set-builder form is: t t 3 and t 0{ } , or
t : t 3 and t 0{ }
… in graphical form is:
… in interval form is: 3, 0) 0,( )
§
(See Footnote 5 on our future study of domain and range.)
(Section 1.1: Functions) 1.1.15
PART I: EVALUATING FUNCTIONS (THE MECHANICS)
In practice, we often evaluate a function at an input without finding the domain.
We immediately attempt to evaluate the defining expression, such as 3t + 9
2t2+ 20t
below, at the input. As we simplify, if we obtain an expression that is clearly
undefined as a real value, we determine that the function value is undefined.
Example 17 (Evaluating a Function; Revisiting Example 16)
Let g t( ) =3t + 9
2t2+ 20t
. Evaluate g 1( ) ,
g ( ) ,
g 0( ) , and
g 4( ) .
§ Solution
We write:
g 1( ) =3 1( ) + 9
2 1( )2
+ 20 1( )
=12
22
=2 3
22
=3
11
WARNING 16: Your
answer must be in
simplified form.
g ( ) =3 + 9
2 2+ 20
, or
3 + 3( )2 +10( )
We also write
(informally, as it turns out):
g 0( ) =3 0( ) + 9
2 0( )2
+ 20 0( )
=9
0Undefined( )
g 4( ) =3 4( ) + 9
2 4( )2
+ 20 4( )
=3
48
Undefined as(a real value)
We saw in Example 16 that Dom g( ) = 3, 0) 0,( ) .
• 1 and are in Dom g( ) , so
g 1( ) and
g ( ) are defined.
• 0 and 4 are not in Dom g( ) , so g 0( ) and g 4( ) are undefined. §
(Section 1.1: Functions) 1.1.16
Example 18 (Evaluating a Function at a Non-numeric Input)
Let f x( ) = 3x2 2x + 5 . Evaluate f x + h( ) .
§ Solution
WARNING 17: f x + h( ) is often not equivalent to
f x( ) + h or
f x( ) + f h( ) . Instead, think: Substitution.
To evaluate f x + h( ) , we take 3x2 2x + 5 , and we replace all occurrences
of x with
x + h( ) . This may seem awkward, because we are replacing x with
another expression containing x.
f x( ) = 3x2 2x + 5
f x + h( ) = 3 x + h( )2
2 x + h( ) + 5
= 3 x2+ 2xh + h2( ) 2x 2h + 5
WARNING 18: Be careful when squaring
binomials and when applying the
Distributive Property when an expression
is being subtracted.
= 3x2+ 6xh + 3h2 2x 2h + 5
• We will see much more of the notation f x + h( ) when we cover difference quotients
and derivatives in Sections 1.10, 1.11, and 5.7. §
(Section 1.1: Functions) 1.1.17
PART J: APPLICATIONS
In this chapter, we will discuss the following functions:
Function Input Output
(Function Value)
s
position or height
(in Section 1.2)
t = the time elapsed
(in seconds) after a coin
is dropped from the top
of a building
s t( ) = the height (in feet)
of the coin t seconds after it
is dropped
f
temperature conversion
(in Section 1.9)
x = the temperature
using the Celsius scale
f x( ) = the Fahrenheit
equivalent of x degrees
Celsius
P
profit
(in Sections 1.10, 1.11)
x = the number of
widgets produced and
sold by a company
P x( ) = the resulting profit
when x widgets are
produced and sold
(Section 1.1: Functions) 1.1.18
FOOTNOTES
1. Terminology.
When defining a function f , some sources require:
• a domain (a set containing all of the inputs in the ordered pairs making up the function,
but nothing else),
•• Some sources attempt to define the domain and the range of a relation, but
there is disagreement as to how to define the domain (and also the range, as
discussed below). Some sources allow the domain to include elements that are not
inputs for any of the ordered pairs in the relation.
• a codomain (a set containing all of the outputs and perhaps other elements that are not
outputs), and
• a “function rule,” perhaps obtained from the statement of f as a set of ordered pairs,
relating each (input) element of the domain to exactly one (output) element of the
codomain.
We can then write f : X Y , meaning that f maps the domain X to the codomain Y.
Let f x( ) = x2
, also denoted by f : x x2 , where the domain is and the codomain is .
We can then write f : . This is because f relates each real number input in the
domain to exactly one real number output, which is an element of the codomain.
The range, which is the set of all assigned outputs, is a subset of the codomain.
In the example above, the range is a proper subset of the codomain, because not every real
number in the codomain is assigned. Specifically, the negative reals are not assigned.
What we call the codomain some sources call the range, and what we call the range some
authors call the image of the function.
2. Definition of a polynomial function.
If f x( ) =x2
x, then
f x( ) = x x 0( ) . Is f a polynomial function? In his book Polynomials
(New York: Springer-Verlag, 1989), E.J. Barbeau implies that it is. Other sources imply
otherwise due to the fact that Dom f( ) . It depends on whether or not one views
Dom f( ) = as a defining characteristic of a polynomial function f for now. In Chapter 2,
we will see cases where Dom f( ) = .
(Section 1.1: Functions) 1.1.19
3. Series expansions of [defining expressions of] functions. Let f x( ) =
1
1 x. In Section 9.4
and in calculus, you will see that f x( ) has the infinite series expansion 1+ x + x2
+ x3+ ... ,
provided that 1< x < 1 . In calculus, you will consider series expansions for sin x , cos x ,
ex, etc.
4. The Zero Factor Property and inequalities.
• According to the Zero Factor Property, if ab = 0 for real numbers a and b, then
a = 0 or b = 0 .
• If we were solving the equation 2t t +10( ) = 0 , we could use the Zero Factor Property.
2t t +10( ) = 0 t = 0( ) or t = 10( )
• In Example 16, we essentially solved the inequality 2t t +10( ) 0 .
‘~’ denotes negation (“not”).
2t t +10( ) 0 ~ t = 0( ) or t = 10( )
~ t = 0( ) and ~ t = 10( )by DeMorgan's Laws of logic (see below)
t 0( ) and t 10( )
• By DeMorgan’s Laws of logic, ~ p or q( ) is logically equivalent to
~ p( ) and ~ q( ) .
For example: If I am an American, then (I am an Alabaman) or (I am an Alaskan) or ….
If I am not an American, then (I am not an Alabaman) and (I am not an Alaskan) and ….
• A Zero Factor Property for inequalities: If ab 0 for real numbers a and b, then
a 0 and b 0 .
5. Revisiting domain and range.
• In Section 1.2, we will relate domains and ranges to graphs.
• We will study domains and ranges of basic functions in Section 1.3; more complicated
functions in Sections 1.4, 1.5, and 1.6; inverse functions in Section 1.9 (and Section 4.10);
and various types of functions in Chapters 2, 3, and 4.
• The topic of solving nonlinear inequalities in Section 2.11 will be relevant, particularly
when finding domains of algebraic functions.
• In Section 9.2, we will study sequences, which are functions with domains consisting of
only integers.
• Ranges will be better understood as we discuss graphs in further detail.
(Section 1.2: Graphs of Functions) 1.2.1
SECTION 1.2: GRAPHS OF FUNCTIONS
LEARNING OBJECTIVES
• Know how to graph a function.
• Recognize when a curve or an equation describes y as a function of x, and apply the
Vertical Line Test (VLT) for this purpose.
• Recognize when an equation describes a function explicitly or implicitly.
• Use a graph to estimate a function’s domain, range, and specific function values.
• Find zeros of a function, and relate them to x-intercepts of its graph.
• Use a graph to determine where a function is increasing, decreasing, or constant.
PART A: DISCUSSION
• In Chapter 0, we graphed lines and circles in the Cartesian plane.
• If f is a function, then its graph in the usual Cartesian xy-plane is the graph of
the equation y = f x( ) , and it must pass the Vertical Line Test (VLT).
In Section 1.8, we will consider the graph of x = f y( ) .
• In this section, we will sketch graphs of functions. We will investigate how their
behaviors reflect the behaviors of their underlying functions, as well as the
information that they contain about those functions.
• For example, the real zeros of a function f correspond to the x-intercepts of its
graph in the xy-plane. If it exists, f 0( ) gives us the y-intercept. Also, a function
increases and decreases according to the rising and falling of its graph.
• After this section, we will specialize and focus on particular functions and
categories of functions, as well as their corresponding graphs.
(Section 1.2: Graphs of Functions) 1.2.2
PART B: THE GRAPH OF A FUNCTION
The graph of a function f in the xy-plane is the graph of the equation y = f x( ) .
It consists of all points of the form x, f x( )( ) , where x Dom f( ) .
• In Example 13, we will graph the function s in the th-plane by graphing h = s t( ) .
• Remember that, as a set of ordered pairs, f = x, f x( )( ) x Dom f( ){ } .
Here, as we typically assume, …
• x is the independent variable, because it is the input variable.
• y is the dependent variable, because it is the output variable.
Its value (the function value) typically “depends” on the value of the input x.
Then, it is customary to say that y is a function of x, even though y is a variable
here and not a function. The form y = f x( ) implies this.
• In Section 1.8, we will switch the roles of x and y.
Basic graphs, such as the ones presented in Section 1.3, and methods of
manipulating them, such as the ones presented in Section 1.4, are to be
remembered.
The Point-Plotting Method presented below will help us develop basic graphs,
and it can be used to refine our graphs by identifying particular points on them.
It is also available as a “last resort” if memory fails us.
The Point-Plotting Method for Graphing a Function f in the xy-Plane
• Choose several x values in Dom f( ) .
• For each chosen x value, find f x( ) , its corresponding function value.
• Plot the corresponding points
x, f x( )( ) in the xy-plane.
• Try to interpolate (connect the points, though often not with line
segments) and extrapolate (go beyond the scope of the points)
as necessary, ideally based on some apparent pattern.
•• Ensure that the set of x-coordinates of the points on the graph is,
in fact, Dom f( ) .
(Section 1.2: Graphs of Functions) 1.2.3
PART C: GRAPHING A SQUARE ROOT FUNCTION
Let f x( ) = x . We will sketch the graph of f in the xy-plane.
This is the graph of the equation y = f x( ) , or y = x .
TIP 1: As usual, we associate y-coordinates with function values.
When point-plotting, observe that: Dom f( ) = 0, ) .
• For instance, if we choose x = 9 , we find that f 9( ) = 9 = 3 ,
which means that the point 9, f 9( )( ) , or
9, 3( ) , lies on the graph.
• On the other hand, f 9( ) is undefined, because
9 Dom f( ) .
Therefore, there is no corresponding point on the graph with x = 9 .
A (partial) table can help:
x
f x( ) Point
0 0 0, 0( )
1 1 1,1( )
4 2 4, 2( )
9 3 9, 3( )
Below, we sketch the graph of f :
WARNING 1: Clearly indicate any endpoints on a graph, such as
the origin here.
The lack of a clearly indicated right endpoint on our sketch implies that the
graph extends beyond the edge of our figure. We want to draw graphs in
such a way that these extensions are “as one would expect.”
WARNING 2: Sketches of graphs produced by graphing utilities might not
extend as expected. The user must still understand the math involved.
Point-plotting may be insufficient.
• The x between the ‘4’ and the ‘9’ on the x-axis represents a generic x-coordinate in
Dom f( ) . We could use x0 (“x sub zero” or “x naught”) to represent a particular or fixed
x-coordinate.
(Section 1.2: Graphs of Functions) 1.2.4
PART D: THE VERTICAL LINE TEST (VLT)
The Vertical Line Test (VLT)
A curve in a coordinate plane passes the Vertical Line Test (VLT)
There is no vertical line that intersects the curve more than once.
An equation in x and y describes y as a function of x
Its graph in the xy-plane passes the VLT.
• Then, there is no input x that yields more than one output y.
• Then, we can write y = f x( ) , where f is a function.
• A “curve” could be a straight line.
Example 1 (Square Root Function and the VLT; Revisiting Part C)
The equation y = x explicitly describes y as a function of x, since it is of
the form y = f x( ) . f is the square root function from Part C.
Observe that the graph of y = x passes the VLT.
Each vertical line in the xy-plane either …
• … misses the graph entirely, meaning that the corresponding x value is
not in Dom f( ) , or
• … intersects the graph in exactly one point, meaning that the
corresponding x value yields exactly one y value as its output.
§
(Section 1.2: Graphs of Functions) 1.2.5
Example 2 (An Equation that Does Not Describe a Function)
Show that the equation x2+ y2
= 9 does not describe y as a function of x.
§ Solution (Method 1: VLT)
The circular graph of x2+ y2
= 9 below fails the VLT, because there exists
a vertical line that intersects the graph more than once. For example, we
can take the red line ( x = 2 ) below:
Therefore, x2+ y2
= 9 does not describe y as a function of x. §
§ Solution (Method 2: Solve for y)
This is also evident if we solve x2+ y2
= 9 for y:
x2+ y2
= 9
y2= 9 x2
y = ± 9 x2
• Any input value for x in the interval
3, 3( ) yields two different y outputs.
• For example, x = 2 yields the outputs y = 5 and y = 5 . §
(Section 1.2: Graphs of Functions) 1.2.6
PART E: IMPLICIT FUNCTIONS and CIRCLES
Example 3 (An Equation that Describes a Function Implicitly)
The equation xy = 1 implicitly describes y as a function of x.
This is because, if we solve the equation for y, we obtain: y =
1
x.
This is of the form y = f x( ) , where f is the reciprocal function. §
Example 4 (Implicit Functions and Circles; Revisiting Example 2)
As it stands, the equation x2+ y2
= 9 does not describe y as a function of x;
we saw this in Example 2. However, it does provide implicit functions if we
impose restrictions on x and/or y and consider smaller pieces of its graph.
• If we impose the restriction y 0 and solve the equation x2+ y2
= 9 for y,
we obtain y = 9 x2 . (See Example 2.) Its graph is the upper half of the
circle, and it passes the VLT, so y = 9 x2 does describe y as a function
of x.
• If we impose the restriction y 0 and solve the equation x2+ y2
= 9 for y,
we obtain y = 9 x2 . Its graph is the lower half of the circle, and it
passes the VLT, so y = 9 x2 does describe y as a function of x.
• This helps us graph entire circles on graphing utilities. §
(Section 1.2: Graphs of Functions) 1.2.7
PART F: ESTIMATING DOMAIN, RANGE, and FUNCTION VALUES
FROM A GRAPH
The domain of f is the set of all x-coordinates of points on the graph of
y = f x( ) . (Think of projecting the graph onto the x-axis.)
The range of f is the set of all y-coordinates of points on the graph of
y = f x( ) . (Think of projecting the graph onto the y-axis.)
Domain
Think: xf
Range
Think: y
Example 5 (Estimating Domain, Range, and Function Values from a Graph)
Let f x( ) = x2
+1. Estimate the domain and the range of f based on its
graph below. Also, estimate f 1( ) .
§ Solution
Apparently, Dom f( ) = , or ,( ) , and Range f( ) = 1, ) .
It also appears that the point 1, 2( ) lies on the graph and thus
f 1( ) = 2 .
• Finding the range of a function will become easier as you learn how to graph functions
in precalculus and calculus. §
WARNING 3: Graph analyses can be imprecise. The point 1, 2.001( ) ,
for example, may be hard to identify on a graph. Not all coordinates are integers.
(Section 1.2: Graphs of Functions) 1.2.8
PART G: ZEROS (OR ROOTS) and INTERCEPTS
The real zeros (or roots) of f are the real solutions of f x( ) = 0 , if any.
They correspond to the x-intercepts of the graph of y = f x( ) .
WARNING 4: The number 0 may or may not be a zero of f . In this sense, the
term “zero” may be confusing. On the other hand, the term “root” might be
confused with square roots and such.
• The graph of y = f x( ) can have any number of x-intercepts (possibly none), or
infinitely many, depending on f .
• We typically focus on real zeros, though we will discuss imaginary zeros in Chapters 2 and 6.
The y-intercept of the graph of y = f x( ) , if it exists, is given by
f 0( ) or by the
point 0, f 0( )( ) .
• The graph of y = f x( ) can have at most one y-intercept.
Example 6 (Finding Zeros and Intercepts)
Find the zeros (or roots) of f , where f x( ) = x2 9 , and
find the x-intercepts of the graph of y = f x( ) .
§ Solution
Solve f x( ) = 0 :
x2 9 = 0
x2= 9
x = ±3
The zeros of f are 3 and 3. They are both real, so they correspond to
x-intercepts of the graph of y = x2 9 . Some prefer to write the x-intercepts
as 3, 0( ) and
3, 0( ) .
WARNING 5: Do not confuse the process of finding zeros, which
involves solving the equation f x( ) = 0 , with the process of evaluating
at 0, which involves substituting 0( ) for x and finding
f 0( ) .
• Here, f 0( ) = 9 . In fact, 9 , or the point
0, 9( ) , is the y-intercept.
(Section 1.2: Graphs of Functions) 1.2.9
The graph of f is below.
§
We will informally refer to zeros of the defining expression for a function,
in particular zeros of radicals and fractions.
“Zeros of a Radical”
g x( )n = 0 g x( ) = 0 n = 2, 3, 4, ...( )
• That is, the zeros of a radical are the zeros of its radicand.
Example 7 (Finding “Zeros of a Radical”)
Find the zeros (or roots) of f , where f x( ) = x2 9 .
§ Solution
The zeros are the same as those for x2 9 , namely 3 and 3.
The graph of f is below.
Why does the graph disappear on the x-interval 3, 3( )? §
(Section 1.2: Graphs of Functions) 1.2.10
“Zeros of a Fraction”
If f x( ) is of the form numerator N x( )
denominator D x( ), then
the zeros of f are the zeros of N that are in Dom f( )
(WARNING 6).
• In particular, a zero of f cannot make any denominator undefined
or equal to 0.
Example 8 (Finding “Zeros of a Fraction”)
Find the zeros (or roots) of f , where f x( ) =x2 9
x + 7.
§ Solution
Solve f x( ) = 0 :
x2 9
x + 7= 0
x2 9 = 0 x 7( )
Again, the zeros of f are 3 and 3.
• The graph of f here has features we will discuss in Chapter 2. §
Example 9 (Finding “Zeros of a Fraction”)
If f x( ) =
x2 9
x 3, the only zero of f is 3, because 3 is not in
Dom f( ) . (3 yields a zero denominator.) §
(Section 1.2: Graphs of Functions) 1.2.11
Example 10 (Finding Zeros)
Find the zeros (or roots) of g, where g t( ) =
3t2 t 4
t 3.
§ Solution
• Observe that 3 is excluded from Dom g( ) , because it yields a zero
denominator.
• Dom g( ) also excludes values of t that yield negative values for the
radicand, 3t2 t 4 . We don’t have to worry about this, though, because we
only care about values of t that make that radicand zero in value, anyway.
Solve g t( ) = 0 :
3t2 t 4
t 3= 0
3t2 t 4 = 0 t 3( )3t2 t 4 = 0 t 3( )
Method 1: Factoring
3t 4( ) t +1( ) = 0 t 3( )
By the Zero Factor Property,
3t 4 = 0
t =4
3
or
t +1= 0
t = 1
WARNING 7: If we had obtained 3, we would have had to
eliminate it.
The zeros of g are
4
3 and 1.
(Section 1.2: Graphs of Functions) 1.2.12
Method 2: Quadratic Formula
We need to solve: 3t2 t 4 = 0 t 3( ) .
For the Quadratic Formula, a = 3, b = 1, and c = 4 .
WARNING 8: It helps to identify what a, b, and c are.
Sign mistakes are common.
Apply the Quadratic Formula.
t =b ± b2 4ac
2a
=1( ) ± 1( )
2
4 3( ) 4( )2 3( )
=1± 1+ 48
6
=1± 49
6
=1± 7
6
Using “+” : Using “ ” :
t =1+ 7
6
=8
6
=4
3
t =1 7
6
=6
6
= 1
WARNING 9: Again, if we had obtained 3, we would have
had to eliminate it.
Again, the zeros of g are
4
3 and 1. §
(Section 1.2: Graphs of Functions) 1.2.13
PART H: INTERVALS OF INCREASE, DECREASE, AND CONSTANT VALUE
We may have an intuitive sense of what it means for a function to increase
(respectively, decrease, or stay constant) on an interval. In Examples 11 and 12,
we will formalize this intuition.
Example 11 (Intervals of Increase and Intervals of Decrease from a Graph)
Let f x( ) = x
3 3x + 2. The graph of f is below.
Give the intervals of increase and the intervals of decrease for f .
• It is assumed that we give the “largest” intervals in the sense that no interval we
give is a proper subset of another appropriate interval.
§ Solution
• f increases on the interval , 1( . Why?
Graphically: If we only consider the part of the graph on the
x-interval , 1( , any point must be higher than any point to its
left. The graph rises from left to right.
Numerically: Any x-value in the interval
, 1( yields a greater
function value f x( ) than any lesser x-value in the interval does.
f increases on an interval I
x2> x
1 implies that f x
2( ) > f x1( ) , x
1, x
2I .
(Section 1.2: Graphs of Functions) 1.2.14
• f decreases on the interval
1,1 . Why?
Graphically: If we only consider the part of the graph on the
x-interval 1,1 , any point must be lower than any point to its left.
The graph falls from left to right.
Numerically: Any x-value in the interval
1,1 yields a lesser
function value f x( ) than any lesser x-value in the interval does.
f decreases on an interval I
x
2> x
1 implies that f x
2( ) < f x1( ) , x
1, x
2I .
• f increases on the interval 1, ) . §
Example 12 (Intervals of Constant Value from a Graph)
The graph of g below implies that g is constant on the interval
1,1 ,
because the graph is flat there.
§
f is constant on an interval I
f x
2( ) = f x1( ) ,
x
1, x
2I .
In calculus, you will reverse this process. You will first determine intervals where a function is
increasing / decreasing / constant, and then you will sketch a graph.
You will locate turning points such as the ones indicated on the graph of f in Example 11.
• The point
1, 4( ) is called a local (or relative) maximum point.
• The point 1, 0( ) is called a local (or relative) minimum point.
Derivatives, which are key tools, will be previewed in Section 1.11.
(Section 1.2: Graphs of Functions) 1.2.15
PART I: USING OTHER NOTATION
WARNING 10: Don’t get too attached to y, f , and x. Be flexible.
Example 13 (Falling Coin)
You drop a coin from the top of a building.
• Let t be the time elapsed (in seconds) since you dropped the coin.
• Let h be the height (in feet) of the coin.
• Let s be a position function such that h = s t( ) .
We ignore what happens after the coin hits the ground.
Instead of graphing y = f x( ) , we graph
h = s t( ) .
• t, not x, is the independent variable.
• h, not y, is the dependent variable.
• s, not f , is the function.
• the th-plane, not the xy-plane, is the coordinate plane containing the
graph of the function s.
The graph of s, or the graph of h = s t( ) , in the th-plane is given below.
As a set of ordered pairs, s = t, s t( )( ) t Dom s( ){ } .
• WARNING 11: The horizontal and vertical axes are scaled
differently here. We typically try to avoid this unless necessary.
• The reader can analyze this graph, including the indicated points, in
the Exercises. §
(Section 1.3: Basic Graphs and Symmetry) 1.3.1
SECTION 1.3: BASIC GRAPHS and SYMMETRY
LEARNING OBJECTIVES
• Know how to graph basic functions.
• Organize categories of basic graphs and recognize common properties,
such as symmetry.
• Identify which basic functions are even / odd / neither and relate this to
symmetry in their graphs.
PART A: DISCUSSION
• We will need to know the basic functions and graphs in this section without
resorting to point-plotting.
• To help us remember them, we will organize them into categories. What are the
similarities and differences within and between categories, particularly with
respect to shape and symmetry in graphs? (We will revisit symmetry in Section 1.4
and especially in Section 1.7.)
• A power function f has a rule of the form f x( ) = xn , where the exponent or
power n is a real number.
• We will consider graphs of all power functions with integer powers, and
some power functions with non-integer powers.
• In the next few sections, we will manipulate and combine these building blocks to
form a wide variety of functions and graphs.
(Section 1.3: Basic Graphs and Symmetry) 1.3.2
PART B: CONSTANT FUNCTIONS
If f x( ) = c , where c is a real number, then f is a constant function.
• Any real input yields the same output, c.
If f x( ) = 3, for example, we have the input-output model and the flat graph of
y = 3, a horizontal line, below.
PART C: IDENTITY FUNCTIONS
If f x( ) = x , then f is an identity function.
• Its output is identical to its input.
6 f 6
10 f 10
• There are technically different identity functions on different domains.
The graph of y = x is the line below.
(Section 1.3: Basic Graphs and Symmetry) 1.3.3
PART D: LINEAR FUNCTIONS
If f x( ) = mx + b , where m and b are real numbers, and m 0 ,
then f is a linear function.
In Section 0.14, we graphed y = mx + b as a line with slope m and y-intercept b.
If f x( ) = 2x 1, for example, we graph the line with slope 2 and y-intercept 1.
PART E: SQUARING FUNCTION and EVEN FUNCTIONS
Let f x( ) = x2 . We will construct a table and graph f .
x
f x( ) Point
x
f x( ) Point
0 0 0, 0( ) 0 0
0, 0( )
1 1 1,1( )
1 1
1,1( )
2 4 2, 4( )
2 4
2, 4( )
3 9 3, 9( )
3 9
3, 9( )
(Section 1.3: Basic Graphs and Symmetry) 1.3.4
TIP 1: The graph never falls below the x-axis, because squares of real
numbers are never negative.
Look at the table. Each pair of opposite x values yields a common function
value f x( ) , or y.
• Graphically, this means that every point x, y( ) on the graph has a
“mirror image partner”
x, y( ) that is also on the graph. These
“mirror image pairs” are symmetric about the y-axis.
• We say that f is an even function. (Why?)
A function f is even f x( ) = f x( ) , x Dom f( )
The graph of y = f x( ) is
symmetric about the y -axis.
Example 1 (Even Function: Proof)
Let f x( ) = x2 . Prove that f is an even function.
§ Solution
Dom f( ) = . x ,
f x( ) = x( )2
= x2
= f x( )
Q.E.D. (Latin: Quod Erat Demonstrandum)
• This signifies the end of a proof. It means “that which was to
have been proven, shown, or demonstrated.”
TIP 2: Think: If we replace x with
x( ) as the input, we obtain
equivalent outputs. §
(Section 1.3: Basic Graphs and Symmetry) 1.3.5
PART F: POWER FUNCTIONS WITH POSITIVE, EVEN POWERS and
INTERSECTION POINTS
The term “even function” comes from the following fact:
If f x( ) = xn , where n is an even integer, then f is an even function.
• The graph of y = x2 is called a parabola (see Chapters 2 and 10).
• The graphs of y = x4 , y = x6 , etc. resemble that parabola, although
they are not called parabolas.
• We will discuss the cases with nonpositive exponents later.
How do these graphs compare?
For example, let f x( ) = x2 and
g x( ) = x4 . Compare the graphs of f and g.
Their relationship when x > 1 is unsurprising:
x
f x( )
x2
g x( )
x4
2
4 16
3
9 81
4
16 256
• As expected, x4> x2
if x > 1. As a result, the graph of y = x4 lies
above the graph of y = x2 on the x-interval 1,( ) .
(Section 1.3: Basic Graphs and Symmetry) 1.3.6
However, their relationship on the x-interval
0,1 might be surprising:
x
f x( )
x2
g x( )
x4
0
0 0
0.1
0.01 0.0001
1
3
1
9
1
81
1
2
1
4
1
16
1
1 1
• WARNING 1: As it turns out, x4< x2
on the x-interval 0,1( ) .
As a result, the graph of y = x4 lies below the graph of y = x2 on that
x-interval.
• Since the graphs have the points 0, 0( ) and 1,1( ) in common, those
points are intersection points.
Graphically, here’s what we have (so far) on the x-interval
0, ) .
Below, f x( ) = x2 and
g x( ) = x4 .
(Section 1.3: Basic Graphs and Symmetry) 1.3.7
How can we quickly get the other half of the picture? Exploit symmetry!
f and g are both even functions, so their graphs are symmetric about
the y-axis.
• Observe that
1,1( ) is our third intersection point.
• In calculus, you might find the area of one or both of those tiny regions
bounded (trapped) by the graphs.
Let h x( ) = x6 . How does the graph of h below compare?
• The graph of h rises even faster than the others as we move far away
from x = 0 , but it is even flatter than the others close to x = 0 .
(Section 1.3: Basic Graphs and Symmetry) 1.3.8
PART G: POWER FUNCTIONS WITH POSITIVE, ODD POWERS and
ODD FUNCTIONS
Let f x( ) = x3 . We will construct a table and graph f .
x
f x( ) Point
x
f x( ) Point
0 0 0, 0( ) 0 0
0, 0( )
1 1 1,1( )
1
1
1, 1( )
2 8 2, 8( )
2
8
2, 8( )
3 27 3, 27( )
3
27
3, 27( )
(Section 1.3: Basic Graphs and Symmetry) 1.3.9
Look at the table. Each pair of opposite x values yields opposite function
values. That is, f x( ) and f x( ) are always opposites.
• Graphically, this means that every point x, y( ) on the graph has a
“mirror image partner”
x, y( ) on the other side of the origin.
The two points are separated by a 180 rotation (a half revolution)
about the origin. These “mirror image pairs” are symmetric about
the origin.
• We say that f is an odd function. (Why?)
A function f is odd f x( ) = f x( ) , x Dom f( )
The graph of y = f x( ) is
symmetric about the origin.
• In other words, if the graph of f is rotated 180 about the origin,
we obtain the same graph.
Example 2 (Odd Function: Proof)
Let f x( ) = x3 . Prove that f is an odd function.
§ Solution
Dom f( ) = . x ,
f x( ) = x( )3
= x3
= x3( )= f x( )
Q.E.D.
TIP 3: Think: If we replace x with
x( ) as the input, we obtain
opposite outputs. §
(Section 1.3: Basic Graphs and Symmetry) 1.3.10
The term “odd function” comes from the following fact:
If f x( ) = xn , where n is an odd integer, then f is an odd function.
• The graphs of y = x5 , y = x7 , etc. resemble the graph of y = x3 .
• In Part C, we saw that the graph of y = x is a line.
• We will discuss the cases with negative exponents later.
How do these graphs compare?
For example, let f x( ) = x3 and
g x( ) = x5 . Compare the graphs of f and g.
Based on our experience from Part F, we expect that the graph of g
rises or falls even faster than the graph of f as we move far away
from x = 0 , but it is even flatter than the graph of f close to x = 0 .
WARNING 2: Zero functions are functions that only output 0 (Think: f x( ) = 0 ).
Zero functions on domains that are symmetric about 0 on the real number line are
the only functions that are both even and odd. (Can you show this?)
WARNING 3: Many functions are neither even nor odd.
(Section 1.3: Basic Graphs and Symmetry) 1.3.11
PART H : f x( ) = x
0
Let f x( ) = x0 . What is f 0( ) ? It is agreed that 02= 0 and 2
0= 1 , but what is 0
0?
Different sources handle the expression 00 differently.
• If 00 is undefined, then f x( ) = 1 x 0( ) , and f has the graph below.
•• There is a hole at the point 0,1( ) .
• There are many reasons to define 00 to be 1. For example, when analyzing
polynomials, it is convenient to have x0= 1 for all real x without having to
consider x = 0 as a special case. Also, this will be assumed when we discuss
the Binomial Theorem in Section 9.6.
•• Then, f x( ) = 1 on , and f has the graph below.
• In calculus, 00 is an indeterminate limit form. An expression consisting of a base
approaching 0 raised to an exponent approaching 0 may, itself, approach a real number
(not necessarily 0 or 1) or not. The expression 00 is called indeterminate by some
sources.
In any case, f is an even function.
(Section 1.3: Basic Graphs and Symmetry) 1.3.12
PART I: RECIPROCAL FUNCTION and
POWER FUNCTIONS WITH NEGATIVE, ODD POWERS
Let f x( ) =
1
xor x 1( ) .
We call f a reciprocal function, because its output is the reciprocal
(or multiplicative inverse) of the input.
We will carefully construct the graph of f .
Let’s construct a table for x 1.
x 1 10 100
f x( ) , or 1
x 1
1
10
1
100 0+
• The 0+
notation indicates an approach to 0 from greater numbers,
without reaching 0.
The table suggests the following graph for x 1:
The x-axis is a horizontal asymptote (“HA”) of the graph.
An asymptote is a line that a curve approaches in a “long-run” or
“explosive” sense. The distance between them approaches 0.
• Asymptotes are often graphed as dashed lines, although some
sources avoid dashing the x- and y-axes.
• Horizontal and vertical asymptotes will be formally defined in Section 2.9.
(Section 1.3: Basic Graphs and Symmetry) 1.3.13
Let’s now construct a table for 0 < x 1 .
x 0
+
1
100
1
10 1
f x( ) , or 1
x 100 10 1
• We write: 1
x as x 0+
(“1
x approaches infinity as
x approaches 0 from the right, or from greater numbers”).
•• In the previous table,
1
x0+ as x . Graphically,
1
x
approaches 0 “from above,” though we say “from the right.”
• We will revisit this notation and terminology when we discuss limits
in calculus in Section 1.5.
We now have the following graph for x > 0 :
The y-axis is a vertical asymptote (“VA”) of the graph.
How can we quickly get the other half of the picture? Exploit symmetry!
f is an odd function, so its graph is symmetric about the origin.
TIP 4: Reciprocals of negative real numbers are negative real
numbers. 0 has no real reciprocal.
(Section 1.3: Basic Graphs and Symmetry) 1.3.14
The graph exhibits opposing behaviors about the vertical asymptote (“VA”).
• The function values increase without bound from the right of the VA,
and they decrease without bound from the left of the VA.
The graph of y =1
xor x 1( ) , or xy = 1, is called a hyperbola (see Chapter 10).
The graphs of y =1
x3or x 3( ) , y =
1
x5or x 5( ) , etc. resemble that hyperbola, but
they are not called hyperbolas.
Below, f x( ) =
1
x yields the blue graph;
g x( ) =
1
x3 yields the red graph.
• The graph of g approaches the x-axis more rapidly as x and as x .
• The graph of g approaches the y-axis more slowly as x 0+
and as x 0 (“as x approaches 0 from the left, or from lesser numbers”).
This is actually because the values of g “explode” more rapidly.
• When we investigate the graph of y =
1
x2 in Part J, we will understand these
behaviors better.
(Section 1.3: Basic Graphs and Symmetry) 1.3.15
PART J: POWER FUNCTIONS WITH NEGATIVE, EVEN POWERS
Let h x( ) =
1
x2or x 2( ) . We will compare the graph of h to the graph of y =
1
x.
Let’s construct a table for x 1.
x 1 10 100
f x( ) , or 1
x 1
1
10
1
100 0
h x( ) , or 1
x2 1
1
100
1
10,000 0
• This suggests that the graph of h approaches the x-axis more rapidly as x .
The table suggests the following graphs for x 1:
The x-axis is a horizontal asymptote (“HA”) of the graph of h.
Let’s now construct a table for 0 < x 1 .
x 0
+
1
100
1
10 1
f x( ) , or
1
x 100 10 1
h x( ) , or 1
x2 10,000 100 1
• This suggests that the graph of h approaches the y-axis more slowly as x 0+ .
This is actually because the values of h “explode” more rapidly.
(Section 1.3: Basic Graphs and Symmetry) 1.3.16
We now have the following graphs for x > 0 :
The y-axis is a vertical asymptote (“VA”) of the graph of h.
How can we quickly get the other half of the graph of h? Exploit symmetry!
h is an even function, so its graph is symmetric about the y-axis.
TIP 5: This graph lies entirely above the x-axis, because
1
x2 is always
positive in value for nonzero values of x.
The graph exhibits symmetric behaviors about the vertical asymptote (“VA”).
• The function values increase without bound from the left and from the
right of the VA.
The graphs of y = x 4 or 1
x4, y = x 6 or
1
x6, etc. resemble the graph above.
(Section 1.3: Basic Graphs and Symmetry) 1.3.17
PART K: SQUARE ROOT FUNCTION
Let f x( ) = x or x1/2( ) . We discussed the graph of f in Section 1.2.
WARNING 4: f is not an even function, because it is undefined for x < 0 .
The graphs of y = x
4or x1/4( ) ,
y = x
6or x1/6( ) , etc. resemble this graph, as do
the graphs of y = x34or x3/4( ) , y = x58
or x5/8( ) , etc. (See Footnote 1.)
PART L: CUBE ROOT FUNCTION
Let f x( ) = x
3or x1/3( ) . The graph of f resembles the graph in Part K for x 0 .
WARNING 5: The cube root of a negative real number is a negative real number.
Dom f( ) = .
f is an odd function; its graph is symmetric about the origin.
The graphs of y = x
5or x1/5( ) ,
y = x
7or x1/7( ) , etc. resemble this graph, as do
the graphs of y = x35
or x3/5( ) , y = x59
or x5/9( ) , etc. (See Footnote 2.)
(Section 1.3: Basic Graphs and Symmetry) 1.3.18
PART M : f x( ) = x
2/3
Let f x( ) = x23or x2/3( ) . The graph of f resembles the graphs in Parts K and L
for x 0 .
f is an even function; its graph below is symmetric about the y-axis.
• WARNING 6: Some graphing utilities omit the part of the graph to the left
of the y-axis.
• In calculus, we will call the point at 0, 0( ) a cusp, because:
•• it is a sharp turning point for the graph, and
•• as we approach the point from either side, we approach
±( ) “infinite steepness.”
The graphs of y = x25
or x2/5( ) , y = x47
or x4/7( ) , etc. resemble the graph above.
(See Footnote 3.)
(Section 1.3: Basic Graphs and Symmetry) 1.3.19
PART N: ABSOLUTE VALUE FUNCTION
We discussed the absolute value operation in Section 0.4.
The piecewise definition of the absolute value function (on ) is given by:
f x( ) = x =x, if x 0
x, if x < 0
• We will discuss more piecewise-defined functions in Section 1.5.
f is an algebraic function, because we can write: f x( ) = x = x2 .
WARNING 7: Writing x2 as x2/2
would be inappropriate if it is
construed as x, which would not be equivalent for x < 0 , or as
x( )2
,
which has domain
0, ) . (See Footnote 4.)
f is an even function, so its graph will be symmetric about the y-axis.
The graph of y = x for x 0 has a mirror image in the graph of y = x for x 0 .
• In calculus, we will call the point at 0, 0( ) a corner, because:
•• the graph makes a sharp turn there, and
•• the point is not a cusp.
(A corner may or may not be a turning point where the graph changes from rising
to falling, or vice-versa.)
(Section 1.3: Basic Graphs and Symmetry) 1.3.20
PART O: UPPER SEMICIRCLES
In Section 1.2, we saw that the graph of x2+ y2
= 9 y 0( ) is the upper half of the
circle of radius 3 centered at 0, 0( ) .
• Solving for y, we obtain: y = 9 x2 .
More generally, the graph of x2+ y2
= a2 y 0( ) , where a > 0 , is an upper
semicircle of radius a.
• Solving for y, we obtain: y = a2 x2 .
Let f x( ) = a2 x2 . f is an even function, so its upper semicircular graph
below is symmetric about the y-axis.
(Section 1.3: Basic Graphs and Symmetry) 1.3.21
PART P: A GALLERY OF GRAPHS
TIP 6: If you know the graphs well, you don’t have to memorize the domains,
ranges, and symmetries. They can be inferred from the graphs.
• In the Domain and Range column,
\ 0{ } denotes the set of nonzero real
numbers. In interval form,
\ 0{ } is
, 0( ) 0,( ) .
Function
Rule
Type of
Function
(Sample)
Graph
Domain;
Range
Even/Odd;
Symmetry
f x( ) = c Constant
;
c{ } Even;
y-axis
f x( ) = x
Identity
(Type of
Linear)
;
Odd;
origin
f x( ) = mx + b
m 0( )
Linear
;
Odd
b = 0 ;
then, origin
f x( ) = x2
xn : n 2, even( )
Power
;
0, )
Even;
y-axis
f x( ) = x3
xn : n 3, odd( )
Power
;
Odd;
origin
f x( ) = x0 Power
See
Part H
See
Part H
Even;
y-axis
f x( ) = x 1 or
1
x
xn : n < 0, odd( )
Power
\ 0{ } ;
\ 0{ }
Odd;
origin
f x( ) = x 2 or1
x2
xn : n < 0, even( )
Power
\ 0{ } ;
0,( )
Even;
y-axis
(Section 1.3: Basic Graphs and Symmetry) 1.3.22
Function
Rule
Type of
Function
(Sample)
Graph
Domain;
Range
Even/Odd;
Symmetry
f x( ) = x1/2 or x
x
n: n 2, even( )
Power
0, ) ;
0, ) Neither
f x( ) = x1/3 or x
3
x
n: n 3, odd( )
Power
;
Odd;
origin
f x( ) = x2/3 Power
;
0, ) Even;
y-axis
f x( ) = x Absolute
Value
(Algebraic)
;
0, ) Even;
y-axis
f x( ) = a2 x2
a > 0( )
(Type of
Algebraic)
a, a ;
0, a
Even;
y-axis
(Section 1.3: Basic Graphs and Symmetry) 1.3.23
FOOTNOTES
1. Power functions with rational powers of the form
odd
even.
Let f x( ) = xN / D
, where N is an odd and positive integer, and D is an even and positive integer.
f x( ) = x1/ 2 f x( ) = x3/ 2
• If
N
D is a proper fraction (where N < D ), then the graph of f is concave down and
resembles the graph on the left. Examples:
f x( ) = x or x1/ 2( ) ,
f x( ) = x34
or x3/ 4( ) .
• If
N
D is an improper fraction (where N > D ), then the graph of f is concave up and
resembles the graph on the right. Examples:
f x( ) = x3 or x3/ 2( ) ,
f x( ) = x74
or x7 / 4( ) .
2. Power functions with rational powers of the form
odd
odd.
Let f x( ) = xN / D, where N and D are both odd and positive integers.
f x( ) = x1/3
f x( ) = x3/3
= x f x( ) = x9/3
= x3
• If
N
D is a proper fraction, then the graph of f resembles the leftmost graph.
Examples: f x( ) = x
3or x1/3( ) , f x( ) = x35
or x3/5( ) .
• If
N
D is an improper fraction where N > D , then the graph of f resembles the
rightmost graph. For example, f x( ) = x9/3
= x3.
• If N = D (
N
D is still improper), then we obtain the line y = x (see the middle graph)
as a “borderline” case. For example, f x( ) = x3/3
= x .
(Section 1.3: Basic Graphs and Symmetry) 1.3.24
3. Power functions with rational powers of the form
even
odd.
Let f x( ) = xN / D
, where N is an even and positive integer, and D is an odd and positive integer.
f x( ) = x2/3
f x( ) = x6/3
= x2
• If
N
D is a proper fraction, then the graph of f resembles the graph on the left.
Examples: f x( ) = x23
or x2/3( ) , f x( ) = x47
or x4/7( ) .
• If
N
D is an improper fraction, then the graph of f resembles the graph on the right,
where f x( ) = x63
= x6/3= x2 .
4. Power functions with rational powers of the form
even
even.
Let f x( ) = xN / D
, where N and D are both even and positive integers.
Different interpretations of xN / D lead to different approaches to Dom f( ) .
• For example, let f x( ) = x2/6
.
•• If x 0 , then f x( ) = x2/6
= x1/3, or x3
.
•• If x2/6 is interpreted as x26
, then x2/6 is real-valued, even if x < 0 .
Under this interpretation, Dom f( ) = .
•• If x2/6
is interpreted as
x6( )
2
, then x2/6
is not real-valued when x < 0 .
Under this interpretation, Dom f( ) = 0, ) .
(Section 1.4: Transformations) 1.4.1
SECTION 1.4: TRANSFORMATIONS
LEARNING OBJECTIVES
• Know how to graph transformations of functions.
• Know how to find an equation for a transformed basic graph.
• Use graphs to determine domains and ranges of transformed functions.
PART A: DISCUSSION
• Variations of the basic functions from Section 1.3 correspond to variations of the
basic graphs. These variations are called transformations.
• Graphical transformations include rigid transformations such as translations
(“shifts”), reflections, and rotations, and nonrigid transformations such as
vertical and horizontal “stretching and squeezing.”
• Sequences of transformations correspond to compositions of functions, which we
will discuss in Section 1.6.
• After this section, we will be able to graph a vast repertoire of functions, and we
will be able to find equations for many transformations of basic graphs.
• We will relate these ideas to the standard form of the equation of a circle with
center h, k( ) , which we saw in Section 0.13. In the Exercises, the reader can revisit
the Slope-Intercept Form of the equation of a line, which we saw in Section 0.14.
• We will use these ideas to graph parabolas in Section 2.2 and conic sections in
general in Chapter 10, as well as trigonometric graphs in Chapter 4.
• Thus far, y and f x( ) have typically been interchangeable. This will no longer be
the case in many of our examples.
(Section 1.4: Transformations) 1.4.2
PART B: TRANSLATIONS (“SHIFTS”)
Translations (“shifts”) are transformations that move a graph without changing its
shape or orientation.
Let G be the graph of y = f x( ) .
Let c be a positive real number.
Vertical Translations (“Shifts”)
The graph of y = f x( )+ c is G shifted up by c units.
• We are increasing the y-coordinates.
The graph of y = f x( ) c is G shifted down by c units.
Horizontal Translations (“Shifts”)
The graph of y = f x c( ) is G shifted right by c units.
The graph of y = f x + c( ) is G shifted left by c units.
Example 1 (Translations)
Let f x( ) = x . Its graph, G, is the center graph in purple below.
(Section 1.4: Transformations) 1.4.3
A table can help explain how these translations work.
In the table, “und.” means “undefined.”
x
f x( )
x
f x( ) + 2
x + 2
f x( ) 2
x 2
f x 2( )
x 2
f x + 2( )
x + 2
3 und. und. und. und. und.
2 und. und. und. und. 0
1 und. und. und. und. 1
0 0 2 2 und. 2
1 1 3 1 und. 3
2 2 2 + 2 2 2 0 2
3 3 3 + 2 3 2 1 5
How points
change
y-coords.
increase
2 units
y-coords.
decrease
2 units
x-coords.
increase
2 units
x-coords.
decrease
2 units
G moves …
UP DOWN RIGHT LEFT
§
Example 2 (Finding Domain and Range; Revisiting Example 1)
We can infer domains and ranges of the transformed functions in Example 1
from the graphs and the table in Example 1.
Let f x( ) = x . Then, Dom f( ) = Range f( ) = 0, ) .
Let g x( ) = x + 2 x 2 x 2 x + 2
Dom g( )
Think: x 0, ) 0, ) 2, ) 2, )
Range g( )
Think: y
2, )
2, )
0, )
0, )
§
(Section 1.4: Transformations) 1.4.4
WARNING 1: Many people confuse the horizontal shifts.
• Compare the x-intercepts of the graphs of y = x and y = x 2 .
The x-intercept is at x = 0 for the first graph, while it is at x = 2 for the
second graph. The fact that the point 0, 0( ) lies on the first graph implies
that the point 2, 0( ) lies on the second graph.
• More generally: The point a, b( ) lies on the first graph the point
a + 2, b( ) lies on the second graph. Therefore, the second graph is obtained
by shifting the first graph to the right by 2 units.
PART C: REFLECTIONS
Reflections
Let G be the graph of y = f x( ) .
The graph of y = f x( ) is G reflected about the x-axis.
The graph of y = f x( ) is G reflected about the y-axis.
The graph of y = f x( ) is G reflected about the origin.
• This corresponds to a 180 rotation (half revolution) about the
origin. It combines both transformations above, in either order.
Example 3 (Reflections)
Again, let f x( ) = x .
(Section 1.4: Transformations) 1.4.5
A table can help explain how these reflections work.
In the table, “und.” means “undefined.”
x
f x( )
x
f x( )
x
f x( )
x
f x( )
x
3 und. und. 3 3
2 und. und. 2 2
1 und. und. 1
1
0
0
0 0 0
1 1 1 und. und.
2 2 2 und. und.
3 3 3 und. und.
Points are
reflected about x-axis y-axis
Both, or
origin
§
Example 4 (Finding Domain and Range; Revisiting Example 3)
We can infer domains and ranges of the transformed functions in Example 3
from the graphs and the table in Example 3.
Let f x( ) = x . Then,
Dom f( ) = Range f( ) = 0, ) .
Let g x( ) = x x x
Dom g( )
Think: x
0, )
, 0(
, 0(
Range g( )
Think: y
, 0(
0, )
, 0(
WARNING 2: x is defined as a real value for nonpositive real values
of x, because the opposite of a nonpositive real value is a nonnegative real
value. §
(Section 1.4: Transformations) 1.4.6
Example 5 (Reflections and Symmetry)
Let f x( ) = x2 . The graph of f is below.
The graph is its own reflection about the y-axis, because f is an even
function. The graphs of y = f x( ) and
y = f x( ) are the same:
f x( ) = x( )
2
= x2 . Thus, the graph is symmetric about the y-axis. §
Example 6 (Reflections and Symmetry)
Let f x( ) = x3 . The graph of f is below.
• The graph is its own reflection about the origin, because f is an
odd function. The graphs of y = f x( ) and
y = f x( ) are the same:
f x( ) = x( )3
= x3 . The graph is symmetric about the origin. §
(Section 1.4: Transformations) 1.4.7
PART D: NONRIGID TRANSFORMATIONS; STRETCHING AND SQUEEZING
Nonrigid transformations can change the shape of a graph beyond a mere
reorientation, perhaps by stretching or squeezing, unlike rigid transformations
such as translations, reflections, and rotations.
If f is a function, and c is a real number, then cf is called a constant multiple of f .
The graph of y = cf x( ) is:
a vertically stretched version of G if c > 1
a vertically squeezed version of G if 0 < c < 1
The graph of y = f cx( ) is:
a horizontally squeezed version of G if c > 1
a horizontally stretched version of G if 0 < c < 1
If c < 0 , then perform the corresponding reflection either before or after the
vertical or horizontal stretching or squeezing.
WARNING 3: Just as for horizontal translations (“shifts”), the cases involving
horizontal stretching and squeezing may be confusing. Think of c as an
“aging factor.”
Example 7 (Vertical Stretching and Squeezing)
Let f x( ) = x . First consider the form
y = cf x( ) .
(Section 1.4: Transformations) 1.4.8
• For any x-value in
0, ) , such as 1, the corresponding y-coordinate for
the y = x graph is doubled to obtain the y-coordinate for the y = 2 x
graph. This is why there is vertical stretching.
• Similarly, the graph of y =
1
2x exhibits vertical squeezing , because the
y-coordinates have been halved. §
Example 8 (Horizontal Stretching and Squeezing; Revisiting Example 7)
Again, let f x( ) = x . Now consider the form
y = f cx( ) .
The graph of y = f 4x( ) is the graph of y = 2 x in blue, because:
f 4x( ) = 4x = 2 x . The vertical stretching we described in Example 7
may now be interpreted as a horizontal squeezing.
(This is not true of all functions.)
• The function value we got at x = 1 we now get at x =
1
4.
The graph of y = f1
4x is the graph of
y =
1
2x in red, because:
f1
4x =
1
4x =
1
2x . The vertical squeezing we described in Example 7
may now be interpreted as a horizontal stretching.
• The function value we got at x = 1 we now get at x = 4 . §
(Section 1.4: Transformations) 1.4.9
PART E: SEQUENCES OF TRANSFORMATIONS
Example 9 (Graphing a Transformed Function)
Graph y = 2 x + 3 .
§ Solution
• We may want to rewrite the equation as y = x + 3 + 2 to more clearly indicate the vertical shift.
• We will “build up” the right-hand side step-by-step. Along the way, we
transform the corresponding function and its graph.
• We begin with a basic function with a known graph. (Point-plotting should
be a last resort.) Here, it is a square root function. Let f1
x( ) = x .
Basic Graph: y = x Graph: y = x + 3
Begin with: f1
x( ) = x Transformation: f
2x( ) = f
1x + 3( )
Effect: Shifts graph left by 3 units
Graph: y = x + 3 Graph: y = x + 3 + 2
Transformation: f
3x( ) = f
2x( ) Transformation:
f
4x( ) = f
3x( ) + 2
Effect: Reflects graph about x-axis Effect: Shifts graph up by 2 units
(Section 1.4: Transformations) 1.4.10
WARNING 4: We are expected to carefully trace the movements of any
“key points” on the developing graphs. Here, we want to at least trace the
movements of the endpoint. We may want to identify intercepts, as well.
Why is the y-intercept of our final graph at 2 3 , or at 0, 2 3( )?
Why is the x-intercept at 1, or at 1, 0( )? (Left as exercises for the reader.) §
Example 10 (Finding an Equation for a Transformed Graph)
Find an equation for the transformed basic graph below.
§ Solution
The graph appears to be a transformation of the graph of the absolute value
function from Section 1.3, Part N.
Basic graph: y = x
Begin with: f1
x( ) = x
(Section 1.4: Transformations) 1.4.11
There are different strategies that can lead to a correct equation.
Strategy 1 (Raise, then reflect)
Effect: Shifts graph up by 1 unit Effect: Reflects graph about x-axis
Transformation: f
2x( ) = f
1x( ) +1 Transformation:
f x( ) = f
2x( )
Graph: y = x +1 Graph:
y = x +1( )
• WARNING 5: It may help to write f x( ) = f
2x( ) , since it
reminds us to insert grouping symbols.
Possible answers: f x( ) = x +1( ) , or f x( ) = x 1.
Strategy 2 (Reflect, then drop)
Effect: Reflects graph about x-axis Effect: Shifts graph down by 1 unit
Transformation: f
2x( ) = f
1x( ) Transformation:
f x( ) = f
2x( ) 1
Graph: y = x Graph:
y = x 1
Possible answer: f x( ) = x 1, which we saw in Strategy 1.
(Section 1.4: Transformations) 1.4.12
Strategy 3 (Switches the order in Strategy 2, but this fails!)
Basic graph: y = x
Begin with: f1
x( ) = x
Effect: Shifts graph down by 1 unit Effect: Reflects graph about x-axis
Transformation: f
2x( ) = f
1x( ) 1 Transformation:
f
3x( ) = f
2x( )
Graph: y = x 1 Graph: y = x 1( )
Observe that y = x 1( ) is not equivalent to our previous
answers.
WARNING 6: The order in which transformations are applied can
matter, particularly when we mix different types of transformations. §
(Section 1.4: Transformations) 1.4.13
PART F: TRANSLATIONS THROUGH COORDINATE SHIFTS
Translations through Coordinate Shifts
A graph G in the xy-plane is shifted h units horizontally
and k units vertically.
• If h < 0 , then G is shifted left by h units.
• If k < 0 , then G is shifted down by k units.
To obtain an equation for the new graph, take an equation for G and:
• Replace all occurrences of x with x h( ) , and
• Replace all occurrences of y with
y k( ) .
Example 11 (Translating a Circle through Coordinate Shifts;
Revisiting Section 0.13)
We want to translate the circle in the xy-plane with radius 3 and center 0, 0( )
so that its new center is at
2,1( ) . Find the standard form of the equation of
the new circle.
§ Solution
We take the equation x2+ y2
= 9 for the old black circle and:
• Replace x with
x 2( )( ) , or x + 2( ) , and
• Replace y with y 1( ) .
This is because we need to shift the black circle left 2 units and up 1 unit to
obtain the new red circle.
(Section 1.4: Transformations) 1.4.14
Answer: x + 2( )2
+ y 1( )2
= 9 . §
• We will use this technique in Section 2.2 and Chapter 10 on conic sections.
Equivalence of Translation Methods for Functions
Consider the graph of y = f x( ) . A coordinate shift of h units horizontally
and k units vertically yields an equation that is equivalent to one we
would have obtained from our previous approach:
y k = f x h( )y = f x h( ) + k
• Think: h , k , if h and k are positive numbers.
• We will revisit this form when we study parabolas in Section 2.2.
Example 12 (Equivalence of Translation Methods for Functions)
We will shift the first graph to the right by 2 units and up 1 unit.
Graph of y = x2 Graph of y 1= x 2( )
2
, or
y = x 2( )
2
+1
§