x

cot θ = = ’1

y

√

1

sec θ = =’ 2

x

1√

csc θ = = 2.

y

Similar calculations allow us to complete the following table for the values of

the trigonometric functions at the principal angles which are multiples of π/6

or π/4.

CHAPTER 1 Basics 29

Angle Sin Cos Tan Cot Sec Csc

0 0 1 0 undef 1 undef

√ √ √ √

1/2 3/2 1/ 3 3 2/ 3 2

π/6

√ √ √ √

2/2 2/2 1 1 2 2

π/4

√ √ √ √

3/2 1/2 3 1/ 3 2 2/ 3

π/3

1 0 undef 0 undef 1

π/2

√ √ √ √

’1/2 ’3 ’1/ 3 ’2

2π/3 3/2 2/ 3

√ √ √ √

’ 2/2 ’1 ’1 ’2

3π/4 2/2 2

√ √ √ √

’ 3/2 ’1/ 3 ’3 ’2/ 3

5π/6 1/2 2

’1 ’1

0 0 undef undef

π

√ √ √ √

’1/2 ’ 3/2 ’2/ 3 ’2

7π/6 1/ 3 3

√ √ √ √

’ 2/2 ’ 2/2 ’2 ’2

5π/4 1 1

√ √ √ √

’ 3/2 ’1/2 ’2 ’2/ 3

4π/3 3 1/ 3

’1 ’1

3π/2 0 undef 0 undef

√ √ √ √

’ 3/2 ’3 ’1/ 3 ’2/ 3

5π/3 1/2 2

√ √ √ √

’ 2/2 ’1 ’1 ’2

7π/4 2/2 2

√ √ √ √

’1/2 ’1/ 3 ’3 ’2

11π/6 3/2 2/ 3

Besides properties (1) and (2) above, there are certain identities which are

fundamental to our study of the trigonometric functions. Here are the principal ones:

tan2 θ + 1 = sec2 θ

(3)

cot 2 θ + 1 = csc2 θ

(4)

sin(θ + ψ) = sin θ cos ψ + cos θ sin ψ

(5)

cos(θ + ψ) = cos θ cos ψ ’ sin θ sin ψ

(6)

sin(2θ ) = 2 sin θ cos θ

(7)

cos(2θ ) = cos2 θ ’ sin2 θ

(8)

sin(’θ ) = ’ sin θ

(9)

cos(’θ ) = cos θ

(10)

1 ’ cos 2θ

(11) sin2 θ =

2

1 + cos 2θ

(12) cos2 θ =

2

EXAMPLE 1.20

Prove identity number (3).

30 CHAPTER 1 Basics

SOLUTION

We have

sin2 θ

tan θ + 1 = +1

2

cos2 θ

sin2 θ cos2 θ

= +

cos2 θ cos2 θ

sin2 θ + cos2 θ

=

cos2 θ

1

=

cos2 θ

(where we have used Property (1))

= sec2 θ.

You Try It: Use identities (11) and (12) to calculate cos(π/12) and sin(π/12).

1.8 Sets and Functions

We have seen sets and functions throughout this review chapter, but it is well to

bring out some of the ideas explicitly.

A set is a collection of objects. We denote a set with a capital roman letter, such

as S or T or U . If S is a set and s is an object in that set then we write s ∈ S and

we say that s is an element of S. If S and T are sets then the collection of elements

common to the two sets is called the intersection of S and T and is written S © T .

The set of elements that are in S or in T or in both is called the union of S and T

and is written S ∪ T .

A function from a set S to a set T is a rule that assigns to each element of S a

unique element of T . We write f : S ’ T .

EXAMPLE 1.21

Let S be the set of all people who are alive at noon on October 10, 2004

and T the set of all real numbers. Let f be the rule that assigns to each

person his or her weight in pounds at precisely noon on October 10, 2004.

Discuss whether f : S ’ T is a function.

SOLUTION

Indeed f is a function since it assigns to each element of S a unique element

of T . Notice that each person has just one weight at noon on October 10, 2004:

that is a part of the de¬nition of “function.” However two different people may

have the same weight”that is allowed.

CHAPTER 1 Basics 31

EXAMPLE 1.22

Let S be the set of all people and T be the set of all people. Let f be the

rule that assigns to each person his or her brother. Is f a function?

SOLUTION

In this case f is not a function. For many people have no brother (so the rule

makes no sense for them) and many people have several brothers (so the rule

is ambiguous for them).

EXAMPLE 1.23

Let S be the set of all people and T be the set of all strings of letters not

exceeding 1500 characters (including blank spaces). Let f be the rule that

assigns to each person his or her legal name. (Some people have rather

long names; according to the Guinness Book of World Records, the longest

has 1063 letters.) Determine whether f : S ’ T is a function.

SOLUTION

This f is a function because every person has one and only one legal name.

Notice that several people may have the same name (such as “JackArmstrong”),

but that is allowed in the de¬nition of function.

You Try It: Let f be the rule that assigns to each real number its cube root. Is this

a function?

In calculus, the set S (called the domain of the function) and the set T (called

the range of the function) will usually be sets of numbers; in fact they will often

consist of one or more intervals in R. The rule f will usually be given by one or

several formulas. Many times the domain and range will not be given explicitly.

These ideas will be illustrated in the examples below.

You Try It: Consider the rule that assigns to each real number its absolute value.

Is this a function? Why or why not? If it is a function, then what are its domain and

range?

1.8.1 EXAMPLES OF FUNCTIONS OF A REAL

VARIABLE

EXAMPLE 1.24

Let S = R, T = R, and let f (x) = x 2 . This is mathematical shorthand for

the rule “assign to each x ∈ S its square.” Determine whether f : R ’ R

is a function.

SOLUTION

We see that f is a function since it assigns to each element of S a unique

element of T ”namely its square.

32 CHAPTER 1 Basics

Math Note: Notice that, in the de¬nition of function, there is some imprecision

in the de¬nition of T . For instance, in Example 1.24, we could have let T = [0, ∞)

or T = (’6, ∞) with no signi¬cant change in the function. In the example of the

“name” function (Example 1.23), we could have let T be all strings of letters not

exceeding 5000 characters in length. Or we could have made it all strings without

regard to length. Likewise, in any of the examples we could make the set S smaller

and the function would still make sense.

It is frequently convenient not to describe S and T explicitly.

EXAMPLE 1.25

√

Let f (x) = + 1 ’ x 2 . Determine a domain and range for f which make

f a function.

SOLUTION

Notice that f makes sense for x ∈ [’1, 1] (we may not take the square root

of a negative number, so we cannot allow x > 1 or x < ’1). If we understand

f to have domain [’1, 1] and range R, then f : [’1, 1] ’ R is a function.

Math Note: When a function is given by a formula, as in Example 1.25, with no

statement about the domain, then the domain is understood to be the set of all x for

which the formula makes sense.

You Try It: Let

x

g(x) = .

x 2 + 4x + 3

What are the domain and range of this function?

EXAMPLE 1.26

Let

’3 if x ¤ 1

f (x) =

2x 2 if x > 1

Determine whether f is a function.

SOLUTION

Notice that f unambiguously assigns to each real number another real num-

ber. The rule is given in two pieces, but it is still a valid rule. Therefore it is

a function with domain equal to R and range equal to R. It is also perfectly

correct to take the range to be (’4, ∞), for example, since f only takes values

in this set.

Math Note: One point that you should learn from this example is that a function

may be speci¬ed by different formulas on different parts of the domain.

CHAPTER 1 Basics 33

You Try It: Does the expression

4 if x < 3

g(x) =

x2 ’ 7 if x ≥ 2

de¬ne a function? Why or why not?

EXAMPLE 1.27

√

Let f (x) = ± x. Discuss whether f is a function.

SOLUTION

This f can only make sense for x ≥ 0. But even then f is not a function

since it is ambiguous. For instance, it assigns to x = 1 both the numbers 1

and ’1.

1.8.2 GRAPHS OF FUNCTIONS

It is useful to be able to draw pictures which represent functions. These pictures,

or graphs, are a device for helping us to think about functions. In this book we will

only graph functions whose domains and ranges are subsets of the real numbers.

We graph functions in the x-y plane. The elements of the domain of a function

are thought of as points of the x-axis. The values of a function are measured on the

y-axis. The graph of f associates to x the unique y value that the function f assigns

to x. In other words, a point (x, y) lies on the graph of f if and only if y = f (x).

EXAMPLE 1.28

Let f (x) = (x 2 + 2)/(x ’ 1). Determine whether there are points of the

graph of f corresponding to x = 3, 4, and 1.

SOLUTION

The y value corresponding to x = 3 is y = f (3) = 11/2. Therefore the

point (3, 11/2) lies on the graph of f . Similarly, f (4) = 6 so that (4, 6) lies on

the graph. However, f is unde¬ned at x = 1, so there is no point on the graph

with x coordinate 1. The sketch in Fig. 1.38 was obtained by plotting several

points.

Math Note: Notice that for each x in the domain of the function there is one and

only one point on the graph”namely the unique point with y value equal to f (x).

If x is not in the domain of f , then there is no point on the graph that corresponds

to x.

EXAMPLE 1.29

Is the curve in Fig. 1.39 the graph of a function?

34 CHAPTER 1 Basics

Fig. 1.38

Fig. 1.39

SOLUTION

Observe that, corresponding to x = 3, for instance, there are two y values

on the curve. Therefore the curve cannot be the graph of a function.

You Try It: Graph the function y = x + |x|.

EXAMPLE 1.30

Is the curve in Fig. 1.40 the graph of a function?

CHAPTER 1 Basics 35

Fig. 1.40

SOLUTION

Notice that each x in the domain has just one y value corresponding to it.

Thus, even though we cannot give a formula for the function, the curve is the

graph of a function. The domain of this function is (’∞, 3) ∪ (5, 7).

Math Note: A nice, geometrical way to think about the condition that each x in

the domain has corresponding to it precisely one y value is this:

If every vertical line drawn through a curve intersects that curve just once,

then the curve is the graph of a function.

You Try It: Use the vertical line test to determine whether the locus x 2 + y 2 = 1

is the graph of a function.

1.8.3 PLOTTING THE GRAPH OF A FUNCTION

Until we learn some more sophisticated techniques, the basic method that we shall

use for graphing functions is to plot points and then to connect them in a plausible

manner.

EXAMPLE 1.31

Sketch the graph of f (x) = x 3 ’ x.

SOLUTION

We complete a table of values of the function f .

36 CHAPTER 1 Basics

x y = x3 ’ x

’3 ’24

’2 ’6

’1 0

0 0

1 0

2 6

3 24

We plot these points on a pair of axes and connect them in a reasonable way

(Fig. 1.41). Notice that the domain of f is all of R, so we extend the graph to

the edges of the picture.

EXAMPLE 1.32

Sketch the graph of

’1 if x ¤ 2

f (x) =

if x > 2

x

SOLUTION

We again start with a table of values.

x y = f (x)

’3 ’1

’2 ’1

’1 ’1

’1

0

’1

1

’1

2

3 3

4 4

5 5

We plot these on a pair of axes (Fig. 1.42).

Since the de¬nition of the function changes at x = 2, we would be mistaken

to connect these dots blindly. First notice that, for x ¤ 2, the function is

identically constant. Its graph is a horizontal line. For x > 2, the function is a

line of slope 1. Now we can sketch the graph accurately (Fig. 1.43).

√

You Try It: Sketch the graph of h(x) = |x| · 3 x.

CHAPTER 1 Basics 37

Fig. 1.41

EXAMPLE 1.33

√

Sketch the graph of f (x) = x +1.

SOLUTION

We begin by noticing that the domain of f , that is the values of x for which

the function makes sense, is {x: x ≥ ’1}. The square root is understood to

be the positive square root. Now we compute a table of values and plot some

points.

38 CHAPTER 1 Basics

y

x

Fig. 1.42

y

Y

FL

x

AM

Fig. 1.43

TE

√

x y= x+1

’1 0

0 1

√

1 √2

2 3

3 2

√

4 √5

5 √6

6 7

Connecting the points in a plausible way gives a sketch for the graph of f

(Fig. 1.44).

EXAMPLE 1.34

Sketch the graph of x = y 2 .

CHAPTER 1 Basics 39

Fig. 1.44

SOLUTION

The sketch in Fig. 1.45 is obtained by plotting points. This curve is not the

graph of a function.

Fig. 1.45

A curve that is the plot of an equation but which is not necessarily the graph

of a function is sometimes called the locus of the equation. When the curve is

the graph of a function we usually emphasize this fact by writing the equation

in the form y = f (x).

You Try It: Sketch the locus x = y 2 + y.

40 CHAPTER 1 Basics

1.8.4 COMPOSITION OF FUNCTIONS

Suppose that f and g are functions and that the domain of g contains the range of f .

This means that if x is in the domain of f then f (x) makes sense but also g may

be applied to f (x) (Fig. 1.46). The result of these two operations, one following

the other, is called g composed with f or the composition of g with f . We write

(g —¦ f )(x) = g(f (x)).

x f (x) g ( f (x))

Fig. 1.46

EXAMPLE 1.35

Let f (x) = x 2 ’ 1 and g(x) = 3x + 4. Calculate g —¦ f .

SOLUTION

We have

(g —¦ f )(x) = g(f (x)) = g(x 2 ’ 1). (—)

Notice that we have started to work inside the parentheses: the ¬rst step was

to substitute the de¬nition of f , namely x 2 ’ 1, into our equation.

Now the de¬nition of g says that we take g of any argument by multiplying

that argument by 3 and then adding 4. In the present case we are applying g to

x 2 ’ 1. Therefore the right side of equation (—) equals

3 · (x 2 ’ 1) + 4.

This easily simpli¬es to 3x 2 + 1. In conclusion,

g —¦ f (x) = 3x 2 + 1.

EXAMPLE 1.36

Let f (t) = (t 2 ’ 2 )/(t + 1 ) and g(t) = 2t + 1. Calculate g —¦ f and f —¦ g.

SOLUTION

We calculate that

t2 ’ 2

(g —¦ f )(t) = g(f (t)) = g (——)

.

t +1

CHAPTER 1 Basics 41

We compute g of any argument by doubling it and adding 1. Thus equation

(——) equals

t2 ’ 2

+1

2

t +1

2t 2 ’ 4

= +1

t +1

2t 2 + t ’ 3

= .

t +1

One of the main points of this example is to see that f —¦ g is different from

g —¦ f . We compute f —¦ g:

(f —¦ g)(t) = f (g(t))

= f (2t + 1)

(2t + 1)2 ’ 2

=

(2t + 1) + 1

4t 2 + 4t ’ 1

= .

2t + 2

So f —¦ g and g —¦ f are different functions.

√

You Try It: Let f (x) = |x| and g(x) = x/x. Calculate f —¦ g(x) and g —¦ f (x).

We say a few words about recognizing compositions of functions.

EXAMPLE 1.37

How can we write the function k(x) = (2x + 3)2 as the composition of two

functions g and f ?

SOLUTION

Notice that the function k can be thought of as two operations applied in

sequence. First we double and add three, then we square. Thus de¬ne f (x) =

2x + 3 and g(x) = x 2 . Then k(x) = (g —¦ f )(x).

We can also compose three (or more) functions: De¬ne

(h —¦ g —¦ f )(x) = h(g(f (x))).

EXAMPLE 1.38

Write the function k from the last example as the composition of three

functions (instead of just two).

42 CHAPTER 1 Basics

SOLUTION

First we double, then we add 3, then we square. So let f (x) = 2x, g(x) =

x + 3, h(x) = x 2 . Then k(x) = (h —¦ g —¦ f )(x).

EXAMPLE 1.39

Write the function

2

r(t) =

t2 + 3

as the composition of two functions.

SOLUTION

First we square t and add 3, then we divide 2 by the quantity just obtained.As a

result, we de¬ne f (t) = t 2 +3 and g(t) = 2/t. It follows that r(t) = (g —¦f )(t).

You Try It: Express the function g(x) = 3/(x 2 + 5) as the composition of two

functions. Can you express it as the composition of three functions?

1.8.5 THE INVERSE OF A FUNCTION

Let f be the function which assigns to each working adult American his or her

Social Security Number (a 9-digit string of integers). Let g be the function which

assigns to each working adult American his or her age in years (an integer between

0 and 150). Both functions have the same domain, and both take values in the non-

negative integers. But there is a fundamental difference between f and g. If you

are given a Social Security number, then you can determine the person to whom it

belongs. There will be one and only one person with that number. But if you are

given a number between 0 and 150, then there will probably be millions of people

with that age. You cannot identify a person by his/her age. In summary, if you know

g(x) then you generally cannot determine what x is. But if you know f (x) then you

can determine what (or who) x is. This leads to the main idea of this subsection.

Let f : S ’ T be a function. We say that f has an inverse (is invertible) if

there is a function f ’1 : T ’ S such that (f —¦ f ’1 )(t) = t for all t ∈ T and

(f ’1 —¦ f )(s) = s for all s ∈ S. Notice that the symbol f ’1 denotes a new function

which we call the inverse of f .

Basic Rule for Finding Inverses To ¬nd the inverse of a function f , we

solve the equation

(f —¦ f ’1 )(t) = t

for the function f ’1 (t).

EXAMPLE 1.40

Find the inverse of the function f (s) = 3s.

CHAPTER 1 Basics 43

SOLUTION

We solve the equation

(f —¦ f ’1 )(t) = t.

This is the same as

f (f ’1 (t)) = t.

We can rewrite the last line as

3 · f ’1 (t) = t

or

t

f ’1 (t) = .

3

Thus f ’1 (t) = t/3.

EXAMPLE 1.41

Let f : R ’ R be de¬ned by f (s) = 3s 5 . Find f ’1 .

SOLUTION

We solve

(f —¦ f ’1 )(t) = t

or

f (f ’1 (t)) = t

or

3[f ’1 (t)]5 = t

or

t

[f ’1 (t)]5 =

3

or

1/5

t

’1

(t) =

f .

3

√

You Try It: Find the inverse of the function g(x) = x ’ 5.

3

It is important to understand that some functions do not have inverses.

EXAMPLE 1.42

Let f : R ’ {t : t ≥ 0} be de¬ned by f (s) = s 2 . If possible, ¬nd f ’1 .

44 CHAPTER 1 Basics

SOLUTION

Using the Basic Rule, we attempt to solve

(f —¦ f ’1 )(t) = t.

Writing this out, we have

[f ’1 (t)]2 = t.

uniquely for f ’1 (t).

But now there is a problem: we cannot solve this equation √

√

We do not know whether f ’1 (t) = + t or f ’1 (t) = ’ t. Thus f ’1 is not

a well de¬ned function. Therefore f is not invertible and f ’1 does not exist.

Math Note: There is a simple device which often enables us to obtain an inverse”

even in situations like Example 1.42. We change the domain of the function. This

idea is illustrated in the next example.

EXAMPLE 1.43

De¬ne f : {s : s ≥ 0} ’ {t : t ≥ 0} by the formula f (s) = s 2 . Find f ’1 .

SOLUTION

We attempt to solve

(f —¦ f ’1 )(t) = t.

Writing this out, we have

f (f ’1 (t)) = t

or

[f ’1 (t)]2 = t.

This looks like the same situation we had in Example 1.42. But in fact things

√

have improved. Now we know that f ’1 (t) must be + t, because f ’1 must

have range S = {s : s ≥ 0}. Thus f ’1 : {t : t ≥ 0} ’ {s : s ≥ 0} is given by

√

f ’1 (t) = + t.

You Try It: The equation y = x 2 +3x does not describe the graph of an invertible

function. Find a way to restrict the domain so that it is invertible.

Now we consider the graph of the inverse function. Suppose that f : S ’ T

is invertible and that (s, t) is a point on the graph of f . Then t = f (s) hence

s = f ’1 (t) so that (t, s) is on the graph of f ’1 . The geometrical connection

between the points (s, t) and (t, s) is exhibited in Fig. 1.47: they are re¬‚ections of

each other in the line y = x. We have discovered the following important principle:

The graph of f ’1 is the re¬‚ection in the line y = x of the graph of f .

Refer to Fig. 1.48.

CHAPTER 1 Basics 45

y

x

Fig. 1.47

y

x

Fig. 1.48

EXAMPLE 1.44

Sketch the graph of the inverse of the function f whose graph is shown in

Fig. 1.49.

SOLUTION

By inspection of the graph we see that f is one-to-one (i.e., takes different

domain values to different range values) and onto (i.e., takes on all values in

the range) from S = [’2, 3] to T = [1, 5]. Therefore f has an inverse. The

graph of f ’1 is exhibited in Fig. 1.50.

46 CHAPTER 1 Basics

Fig. 1.49

Fig. 1.50

You Try It: Sketch f (x) = x 3 + x and its inverse.

Another useful fact is this: Since an invertible function must be one-to-one, two

different x values cannot correspond to (that is, be “sent by the function to”) the

same y value. Looking at Figs. 1.51 and 1.52, we see that this means

In order for f to be invertible, no horizontal line can

intersect the graph of f more than once.

Fig. 1.51

CHAPTER 1 Basics 47

In Fig. 1.51, the fact that the line y = 2 intersects the graph twice means that

the function f takes the value 2 at two different points of its domain (namely at

x = ’2 and x = 6). Thus f is not one-to-one so it cannot be invertible. Figure 1.52

shows what happens if we try to invert f : the resulting curve is not the graph of a

function.

Fig. 1.52

EXAMPLE 1.45

Look at Figs. 1.53 and 1.55. Are the functions whose graphs are shown in

parts (a) and (b) of each ¬gure invertible?

Fig. 1.53

SOLUTION

Graphs (a) and (b) in Fig. 1.53 are the graphs of invertible functions since

no horizontal line intersects each graph more than once. Of course we must

choose the domain and range appropriately. For (a) we take S = [’4, 4] and

T = [’2, 3]; for (b) we take S = (’3, 4) and T = (0, 5). Graphs (a) and (b)

48 CHAPTER 1 Basics

in Fig. 1.54 are the graphs of the inverse functions corresponding to (a) and (b)

of Fig. 1.53 respectively. They are obtained by re¬‚ection in the line y = x.

Fig. 1.54

Y

FL

AM

Fig. 1.55

TE

In Fig. 1.55, graphs (a) and (b) are not the graphs of invertible functions.

For each there is exhibited a horizontal line which intersects the graph twice.

However graphs (a) and (b) in Fig. 1.56 exhibit a way to restrict the domains

of the functions in (a) and (b) of Fig. 1.55 to make them invertible. Graphs (a)

and (b) in Fig. 1.57 show their respective inverses.

Fig. 1.56

You Try It: Give an example of a function from R to R that is not invertible, even

when it is restricted to any interval of length 2.

CHAPTER 1 Basics 49

Fig. 1.57

1.9 A Few Words About Logarithms

and Exponentials

We will give a more thorough treatment of the logarithm and exponential functions

in Chapter 6. For the moment we record a few simple facts so that we may use

these functions in the sections that immediately follow.

The logarithm is a function that is characterized by the property that

log(x · y) = log x + log y.

It follows from this property that

log(x/y) = log x ’ log y

and

log(x n ) = n · log x.

It is useful to think of loga b as the power to which we raise a to get b, for any

a, b > 0. For example, log2 8 = 3 and log3 (1/27) = ’3. This introduces the idea

of the logarithm to a base.

You Try It: Calculate log5 125, log3 (1/81), log2 16.

The most important base for the logarithm is Euler™s number e ≈ 2.71828 . . . .

Then we write ln x = loge x. For the moment we take the logarithm to the base e, or

the natural logarithm, to be given. It is characterized among all logarithm functions

by the fact that its graph has tangent line with slope 1 at x = 1. See Fig. 1.58. Then

we set

ln x

loga x = .

ln a

Note that this formula gives immediately that loge x = ln x, once we accept that

loge e = 1.

50 CHAPTER 1 Basics

Fig. 1.58

Fig. 1.59

Math Note: In mathematics, we commonly write log x to mean the natural log-

arithm. Thus you will sometimes encounter ln x and sometimes encounter log x

(without any subscript); they are both understood to mean loge x, the natural

logarithm.

The exponential function exp x is de¬ned to be the inverse function to ln x.

Figure 1.59 shows the graph of y = exp x. In fact we will see later that exp x = ex .

More generally, the function a x is the inverse function to loga x. The exponential

has these properties:

a b+c = a b · a c ;

(a)

CHAPTER 1 Basics 51

(a b )c = a b·c ;

(b)

ab

= c.

b’c

(c) a

a

These are really just restatements of properties of the logarithm function that we

have already considered.

You Try It: Simplify the expressions 32 · 54 /(15)3 and 24 · 63 · 12’4 .

Exercises

1. Each of the following is a rational number. Write it as the quotient of two

integers.

2/3 ’ 7/8

(a)

(b) 43.219445

’37 ’4

·

(c)

533 ’6

2

(d)

3.45969696 . . .

’73.235677677677 . . .

(e)

3

5

(f) ’17

4 +9

3

’4

9 +5

2

(g) ’11

3 +7

6

(h) 3.2147569569569 . . .

√ √

2. Plot the numbers 3.4, ’π/2, 2π , ’ 2 + 1, 3 · 4, 9/2, ’29/10 on a real

number line. Label each plotted point.

3. Sketch each of the following sets on a separate real number line.

S = {x ∈ R: |x ’ 2| < 4}

(a)

T = {t ∈ R: t 2 + 1 = 5}

(b)

U = {s ∈ R: 2s ’ 5 ¤ 3}

(c)

V = {y ∈ R: |6y + 1| > 2}

(d)

S = {x ∈ R: x 2 + 3 < 6}

(e)

T = {s ∈ R: |s| = |s + 1|}

(f )

52 CHAPTER 1 Basics

√√ √

4. Plot each of the points (2, ’4), (’6, 3), (π, π 2 ), (’ 5, 8), ( 2π, ’3),

(1/3, ’19/4) on a pair of cartesian coordinate axes. Label each point.

5. Plot each of these planar loci on a separate set of axes.

{(x, y): y = 2x 2 ’ 3}

(a)

{(x, y): x 2 + y 2 = 9}

(b)

y = x3 + x

(c)

x = y3 + y

(d)

x = y2 ’ y3

(e)

x2 + y4 = 3

(f)

6. Plot each of these regions in the plane.

{(x, y): x 2 + y 2 < 4}

(a)

{(x, y): y > x 2 }

(b)

{(x, y): y < x 3 }

(c)

{(x, y): x ≥ 2y + 3}

(d)

{(x, y): y ¤ x + 1}

(e)

{(x, y): 2x + y ≥ 1}

(f)

7. Calculate the slope of each of the following lines:

(a) The line through the points (’5, 6) and (2, 4)

(b) The line perpendicular to the line through (1, 2) and (3, 4)

The line 2y + 3x = 6

(c)

x ’ 4y

=6

(d) The line

x+y

(e) The line through the points (1, 1) and (’8, 9)

The line x ’ y = 4

(f)

8. Write the equation of each of the following lines.

The line parallel to 3x + 8y = ’9 and passing through the point

(a)

(4, ’9).

The line perpendicular to x + y = 2 and passing through the point

(b)

(’4, ’8).

The line passing through the point (4, 6) and having slope ’8.

(c)

(d) The line passing through (’6, 4) and (2, 3).

(e) The line passing through the origin and having slope 6.

The line perpendicular to x = 3y ’ 7 and passing through (’4, 7).

(f)

9. Graph each of the lines in Exercise 8 on its own set of axes. Label your

graphs.

10. Which of the following is a function and which is not? Give a reason in

each case.

CHAPTER 1 Basics 53

(a) f assigns to each person his biological father

(b) g assigns to each man his dog

(c) h assigns to each real number its square root

(d) f assigns to each positive integer its cube

(e) g assigns to each car its driver

(f) h assigns to each toe its foot

(g) f assigns to each rational number the greatest integer that does not

exceed it

(h) g assigns to each integer the next integer

(i) h assigns to each real number its square plus six

11. Graph each of these functions on a separate set of axes. Label your graph.

f (x) = 3x 2 ’ x

(a)

x+2

g(x) =

(b)

x

h(x) = x 3 ’ x2

(c)

f (x) = 3x + 2

(d)

g(x) = √ ’ 2x

x2

(e)

h(x) = x + 3

(f)

12. Calculate each of the following trigonometric quantities.

(a) sin(8π/3)

(b) tan(’5π/6)

(c) sec(7π/4)

(d) csc(13π/4)

(e) cot(’15π/4)

(f) cos(’3π/4)

13. Calculate the left and right sides of the twelve fundamental trigonometric

identities for the values θ = π/3 and ψ = ’π/6, thus con¬rming the

identities for these particular values.

14. Sketch the graphs of each of the following trigonometric functions.

f (x) = sin 2x

(a)

g(x) = cos(x + π/2)

(b)

h(x) = tan(’x + π )

(c)

f (x) = cot(3x + π )

(d)

g(x) = sin(x/3)

(e)

h(x) = cos(’π + [x/2])

(f)

15. Convert each of the following angles from radian measure to degree

measure.

54 CHAPTER 1 Basics

= π/24

(a) θ

= ’π/3

(b) θ

= 27π/12

(c) θ

= 9π/16

(d) θ

=3

(e) θ

= ’5

(f) θ

16. Convert each of the following angles from degree measure to radian

measure.

= 65—¦

(a) θ

= 10—¦

(b) θ

= ’75—¦

(c) θ

= ’120—¦

(d) θ

= π—¦

(e) θ

= 3.14—¦

(f) θ

17. For each of the following pairs of functions, calculate f —¦ g and g —¦ f .

f (x) = x 2 + 2x + 3 g(x) = (x ’ 1)2

(a) √

√

f (x) = x + 1 g(x) = x 2 ’ 2

3

(b)

f (x) = sin(x + 3x 2 ) g(x) = cos(x 2 ’ x)

(c)

f (x) = ex+2 g(x) = ln(x ’ 5)

(d)

f (x) = sin(x 2 + x) g(x) = ln(x 2 ’ x)

(e)

g(x) = e’x

2 2

f (x) = ex

(f)

f (x) = x(x + 1)(x + 2) g(x) = (2x ’ 3)(x + 4)

(g)

18. Consider each of the following as functions from R to R and say whether

the function is invertible. If it is, ¬nd the inverse with an explicit formula.

f (x) = x 3 + 5

(a)

g(x) = x 2 ’ x √

(b)

h(x) = (sgn x) · |x|, where sgn x is +1 if x is positive, ’1 if x is

(c)

negative, 0 if x is 0.

f (x) = x 5 + 8

(d)

g(x) = e’3x

(e)

h(x) = sin x

(f)

f (x) = tan x

(g)

g(x) = (sgn x) · x 2 , where sgn x is +1 if x is positive, ’1 if x is

(h)

negative, 0 if x is 0.

19. For each of the functions in Exercise 18, graph both the function and its

inverse in the same set of axes.

CHAPTER 1 Basics 55

20. Determine whether each of the following functions, on the given domain

S, is invertible. If it is, then ¬nd the inverse explicitly.

f (x) = x 2 , S = [2, 7]

(a)

g(x) = ln x, S = [1, ∞)

(b)

h(x) = sin x, S = [0, π/2]

(c)

f (x) = cos x, S = [0, π]

(d)

g(x) = tan x, S = (’π/2, π/2)

(e)

h(x) = x 2 , S = [’2, 5]

(f)

f (x) = x 2 ’ 3x, S = [4, 7]

(g)

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CHAPTER 2

Foundations of

Calculus

2.1 Limits

The single most important idea in calculus is the idea of limit. More than 2000 years

ago, the ancient Greeks wrestled with the limit concept, and they did not succeed. It

is only in the past 200 years that we have ¬nally come up with a ¬rm understanding

of limits. Here we give a brief sketch of the essential parts of the limit notion.

Suppose that f is a function whose domain contains two neighboring intervals:

f : (a, c) ∪ (c, b) ’ R. We wish to consider the behavior of f as the variable x

approaches c. If f (x) approaches a particular ¬nite value as x approaches c, then

we say that the function f has the limit as x approaches c. We write

lim f (x) = .

x’c

The rigorous mathematical de¬nition of limit is this:

De¬nition 2.1 Let a < c < b and let f be a function whose domain contains

(a, c) ∪ (c, b). We say that f has limit at c, and we write limx’c f (x) = when

this condition holds: For each > 0 there is a δ > 0 such that

|f (x) ’ | <

whenever 0 < |x ’ c| < δ.

It is important to know that there is a rigorous de¬nition of the limit concept, and

any development of mathematical theory relies in an essential way on this rigorous

de¬nition. However, in the present book we may make good use of an intuitive

57

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58 CHAPTER 2 Foundations of Calculus

understanding of limit. We now develop that understanding with some carefully

chosen examples.

EXAMPLE 2.1

De¬ne

3’x if x < 1

f (x) =

x2 + 1 if x > 1

See Fig. 2.1. Calculate limx’1 f (x).

Y

FL

Fig. 2.1

AM

SOLUTION

Observe that, when x is to the left of 1 and very near to 1 then f (x) = 3 ’ x

is very near to 2. Likewise, when x is to the right of 1 and very near to 1 then

TE

f (x) = x 2 + 1 is very near to 2. We conclude that

lim f (x) = 2.

x’1

We have successfully calculated our ¬rst limit. Figure 2.1 con¬rms the conclusion

that our calculations derived.

EXAMPLE 2.2

De¬ne

x2 ’ 4

g(x) = .

x’2

Calculate limx’2 g(x).

SOLUTION

We observe that both the numerator and the denominator of the fraction

de¬ning g tend to 0 as x ’ 2 (i.e., as x tends to 2). Thus the question seems

to be indeterminate.

However, we may factor the numerator as x 2 ’ 4 = (x ’ 2)(x + 2).

As long as x = 2 (and these are the only x that we examine when we

59

CHAPTER 2 Foundations of Calculus

calculate limx’2 ), we can then divide the denominator of the expression

de¬ning g into the numerator. Thus

g(x) = x + 2 for x = 2.

Now

lim g(x) = lim x + 2 = 4.

x’2 x’2

Fig. 2.2

The graph of the function g is shown in Fig. 2.2. We encourage the reader to

use a pocket calculator to calculate values of g for x near 2 but unequal to 2 to

check the validity of our answer. For example,

x g(x) = [x 2 ’ 4]/[x ’ 2]

1.8 3.8

1.9 3.9

1.99 3.99

1.999 3.999

2.001 4.001

2.01 4.01

2.1 4.1

2.2 4.2

We see that, when x is close to 2 (but unequal to 2), then g(x) is close (indeed,

as close as we please) to 4.

x 3 ’ 3x 2 + x ’ 3

You Try It: Calculate the limit limx’3 .

x’3

Math Note: It must be stressed that, when we calculate limx’c f (x), we do not

evaluate f at c. In the last example it would have been impossible to do so. We want

to determine what we anticipate f will do as x approaches c, not what value (if any)

f actually takes at c. The next example illustrates this point rather dramatically.

60 CHAPTER 2 Foundations of Calculus

Fig. 2.3

EXAMPLE 2.3

De¬ne

if x = 7

3

h(x) =

if x = 7

1

Calculate limx’7 h(x).

SOLUTION

It would be incorrect to simply plug the value 7 into the function h and

thereby to conclude that the limit is 1. In fact when x is near to 7 but unequal

to 7, we see that h takes the value 3. This statement is true no matter how close

x is to 7. We conclude that limx’7 h(x) = 3.

You Try It: Calculate limx’4 [x 2 ’ x ’ 12]/[x ’ 4].

2.1.1 ONE-SIDED LIMITS

There is also a concept of one-sided limit. We say that

lim f (x) =

x’c’

if the values of f become closer and closer to when x is near to c but on the left.

In other words, in studying limx’c’ f (x), we only consider values of x that are

less than c.

Likewise, we say that

lim f (x) =

x’c+

if the values of f become closer and closer to when x is near to c but on the right.

In other words, in studying limx’c+ f (x), we only consider values of x that are

greater than c.

61

CHAPTER 2 Foundations of Calculus

EXAMPLE 2.4

Discuss the limits of the function

2x ’ 4 if x < 2

f (x) =

if x ≥ 2

x2

at c = 2.

SOLUTION

As x approaches 2 from the left, f (x) = 2x ’ 4 approaches 0. As x

approaches 2 from the right, f (x) = x 2 approaches 4. Thus we see that f

has left limit 0 at c = 2, written

lim f (x) = 0,

x’2’

and f has right limit 4 at c = 2, written

lim f (x) = 4.

x’2+

Note that the full limit limx’2 f (x) does not exist (because the left and right

limits are unequal).

You Try It: Discuss one-sided limits at c = 3 for the function

±

x 3 ’ x if x < 3

f (x) = 24 if x = 3

4x + 1 if x > 3

All the properties of limits that will be developed in this chapter, as well as the

rest of the book, apply equally well to one-sided limits as to two-sided (or standard)

limits.

2.2 Properties of Limits

To increase our facility in manipulating limits, we have certain arithmetical and

functional rules about limits. Any of these may be veri¬ed using the rigorous de¬-

nition of limit that was provided at the beginning of the last section. We shall state

the rules and get right to the examples.

If f and g are two functions, c is a real number, and limx’c f (x) and

limx’c g(x) exist, then

Theorem 2.1

(a) limx’c (f ± g)(x) = limx’c f (x) ± limx’c g(x);

62 CHAPTER 2 Foundations of Calculus

(b) limx’c (f · g) (x) = (limx’c f (x)) · (limx’c g(x)) ;

limx’c f (x)

f

(x) = provided that limx’c g(x) = 0;

(c) lim

limx’c g(x)

g

x’c

(d) limx’c (± · f (x)) = ± · (limx’c f (x)) for any constant ±.

Some theoretical results, which will prove useful throughout our study of

calculus, are these:

Theorem 2.2

Let a < c < b. A function f on the interval {x : a < x < b} cannot have two

distinct limits at c.

Theorem 2.3

If

lim g(x) = 0

x’c

and

lim f (x) either does not exist or exists and is not zero

x’c

then

f (x)

lim

x’c g(x)

does not exist.

Theorem 2.4 (The Pinching Theorem)

Suppose that f, g, and h are functions whose domains each contain S = (a, c) ∪

(c, b). Assume further that

g(x) ¤ f (x) ¤ h(x)

for all x ∈ S. Refer to Fig. 2.4.

y = h(x)

y = f (x)

y = g (x)

a c b

Fig. 2.4

If

lim g(x) =

x’c

63

CHAPTER 2 Foundations of Calculus

and

lim h(x) =

x’c

then

lim f (x) = .

x’c

EXAMPLE 2.5

Calculate limx’3 4x 3 ’ 7x 2 + 5x ’ 9.

SOLUTION

We may apply Theorem 2.1(a) repeatedly to see that

lim 4x 3 ’ 7x 2 + 5x ’ 9 = lim 4x 3 ’ lim 7x 2 + lim 5x ’ lim 9. (—)

x’3 x’3 x’3 x’3 x’3

We next observe that limx’3 x = 3. This assertion is self-evident, for when x

is near to 3 then x is near to 3. Applying Theorem 2.1(d) and Theorem 2.1(b)

repeatedly, we now see that

lim 4x 3 = 4 · lim x · lim x · lim x = 4 · 3 · 3 · 3 = 108.

x’3 x’3 x’3 x’3

Also

lim 7x 2 = 7 · lim x · lim x = 7 · 3 · 3 = 63,

x’3 x’3 x’3

lim 5x = 5 · lim x = 5 · 3 = 15.

x’3 x’3

Of course limx’3 9 = 9.

Putting all this information into equation (—) gives

lim 4x 3 ’ 7x 2 + 5x ’ 9 = 108 ’ 63 + 15 ’ 9 = 51.

x’3

EXAMPLE 2.6

Use the Pinching Theorem to analyze the limit

lim x sin x.

x’0

SOLUTION

We observe that

’|x| ≡ g(x) ¤ f (x) = x sin x ¤ h(x) ≡ |x|.

Thus we may apply the Pinching Theorem. Obviously

lim g(x) = lim h(x) = 0.

x’0 x’0

We conclude that limx’0 f (x) = 0.

64 CHAPTER 2 Foundations of Calculus

EXAMPLE 2.7

Analyze the limit

x2 + 4

lim .

x+2

x’’2

SOLUTION

The denominator tends to 0 while the numerator does not. According to

Theorem 2.3, the limit cannot exist.

You Try It: Use the Pinching Theorem to calculate limx’0 x 2 sin x.

x2

You Try It: What can you say about limx’’1 ?

x2 ’ 1

2.3 Continuity

Let f be a function whose domain contains the interval (a, b). Assume that c is a

point of (a, b). We say that the function f is continuous at c if

lim f (x) = f (c).

x’c

Conceptually, f is continuous at c if the expected value of f at c equals the actual

value of f at c.

EXAMPLE 2.8

Is the function

2x 2 ’ x if x < 2

f (x) =

if x ≥ 2

3x

continuous at x = 2?

SOLUTION

We easily check that limx’2 f (x) = 6. Also the actual value of f at 2, given

by the second part of the formula, is equal to 6. By the de¬nition of continuity,

we may conclude that f is continuous at x = 2. See Fig. 2.5.

EXAMPLE 2.9

Where is the function

±

1 if x < 4

g(x) = x ’ 3

2x + 3 if x ≥ 4

continuous?

65

CHAPTER 2 Foundations of Calculus

Fig. 2.5

SOLUTION

If x < 3 then the function is plainly continuous. The function is unde¬ned

at x = 3 so we may not even speak of continuity at x = 3. The function is also

obviously continuous for 3 < x < 4. At x = 4 the limit of g does not exist”it

is 1 from the left and 11 from the right. So the function is not continuous (we

sometimes say that it is discontinuous) at x = 4. By inspection, the function is

continuous for x > 4.

You Try It: Discuss continuity of the function

±

x ’ x 2 if x < ’2

g(x) = 10 if x = ’2

’5x if x > ’2

We note that Theorem 2.1 guarantees that the collection of continuous functions

is closed under addition, subtraction, multiplication, division (as long as we do not

divide by 0), and scalar multiplication.

Math Note: If f —¦ g makes sense, if limx’c g(x) = , and if lims’ f (s) = m,

then it does not necessarily follow that limx’c f —¦ g(x) = m. [We invite the reader

to ¬nd an example.] One must assume, in addition, that f is continuous at . This

point will come up from time to time in our later studies.

In the next section we will learn the concept of the derivative. It will turn out

that a function that possesses the derivative is also continuous.

66 CHAPTER 2 Foundations of Calculus

2.4 The Derivative

Suppose that f is a function whose domain contains the interval (a, b). Let c be a

point of (a, b). If the limit

f (c + h) ’ f (c)

lim (—)

h

h’0

exists then we say that f is differentiable at c and we call the limit the derivative

of f at c.

EXAMPLE 2.10

Is the function f (x) = x 2 + x differentiable at x = 2? If it is, calculate the