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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. §

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