<<

. 2
( 11)



>>

x
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

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