chapter 1: functions(section 1.1: functions) 1.1.4 tip 1: think of a function button on a basic...

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.


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.)


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. §


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. §


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. §


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 . §


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. §


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.


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.)


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. §


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, ) . §


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. §


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 interval form is: 3,( )

§



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 interval form is:

, 3(

§


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. §



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 interval form is: 3, 0) 0,( )

§

(See Footnote 5 on our future study of domain and range.)


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. §


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. §


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


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( ) = .


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.


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( ) .


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.


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.

§


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 . §


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. §


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.


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.


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( )? §


“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.) §


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.


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. §


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 .


• 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.


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.


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.


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( )


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. §


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,( ) .


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 .


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 .


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( )


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. §


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.


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.


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.


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.


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.


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.


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.


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.)


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.)


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.)


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.


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


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


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( ) .


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 .



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 .


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.


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.


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, )

§


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 .


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. §


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. §


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( ) .


• 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 . §


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


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


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.


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. §


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.


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

§

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