class="bi x0 y0 w0 h1" SCHAUM'S OUTLINE SERIES
Schaum's Outline of Theory and Problems of Beginning
Calculus
Second Edition
Elliott Mendelson, Ph.D.
Professor of Mathematics
Queens College
City University of New York
To the memory of my father, Joseph, and my mother, Helen
ELLIOTT MENDELSON is Professor of Mathematics at Queens College of the City University of
New York. He also has taught at the University of Chicago, Columbia University, and the University of
Pennsylvania, and was a member of the Society of Fellows of Harvard University. He is the author of
several books, including Schaum's Outline of Boolean Algebra and Switching Circuits. His principal
area of research is mathematical logic and set theory.
Schaum's Outline of Theory and Problems of

BEGINNING CALCULUS
Copyright © 1997,1985 by The McGraw-
Hill Companies, Inc. All rights reserved. Printed in the United
States of America. Except as permitted under the Copyright Act of 1976, no part of this publication
may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval
system, without the prior written permission of the publisher.
5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 PRS PRS 9 0 1 0 9
ISBN 0-07-041733-4
Sponsoring Editor: Arthur Biderman
Production Supervisor: Suzanne Rapcavage
Editing Supervisor: Maureen B. Walker
Library of Congress Cataloging-in-Publication Data

suffer seriously from that lack (except insofar as the use of a graphing calculator enhances their
understanding of the subject).
2. Treatment of several topics have been expanded:
(a) Newton's Method is now the subject of a separate section. The availability of calculators makes
it much easier to work out concrete problems by this method.
(b
) More attention and more problems are devoted to approximation techniques for integration, such
as the trapezoidal rule, Simpson's rule, and the midpoint rule.
(c) The chain rule now has a complete proof outlined in an exercise.
3. The exposition has been streamlined in many places and a substantial number of new problems have
been added.
The author wishes to thank again the editor of the First Edition, David Beckwith, as well as the editor of
the Second Edition, Arthur Biderman, and the editing supervisor, Maureen Walker.
ELLIOTT MENDELSON
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Contents
Chapter 1
Coordinate Systems on a Line
1
1.1 The Coordinates of a Point 1
1.2 Absolute Value 2
Chapter 2
Coordinate Systems in a Plane
8
2.1 The Coordinates of a Point 8
2.2 The Distance Formula 9
2.3 The Midpoint Formulas 10
Chapter 3
Graphs of Equations
14

8.2 Properties of Limits 59
8.3 Existence or Nonexistence of the Limit 61
Chapter 9
Special Limits
67
9.1 One-Sided Limits 67
9.2 Infinite Limits: Vertical Asymptotes 68
9.3 Limits at Infinity: Horizontal Asymptotes 70
Chapter 10

Continuity
78
10.1 Definition and Properties 78
10.2 One-Sided Continuity 79
10.3 Continuity over a Closed Interval 80
6.2 Symmetry about a Point 42
Chapter 11
The Slope of a Tangent Line
86
Chapter 12
The Derivative
92
Chapter 13
More on the Derivative
99
13.1 Differentiability and Continuity 99
13.2 Further Rules for Derivatives 100
Chapter 14
Maximum and Minimum Problems
104

21.1 Estimating the Value of a Function 155
21.2 The Differential 155
21.3 Newton's Method 156
Chapter 22
Higher-Order Derivatives
161
Chapter 23
Applications of the Second Derivative and Graph Sketching
167
23.1 Concavity 167
23.2 Test for Relative Extrema 169
23.3 Graph Sketching 171
Chapter 24
More Maximum and Minimum Problems
179
Chapter 25
Angle Measure
185
25.1 Arc Length and Radian Measure 185
26.2 Directed Angles 186
Chapter 26
Sine and Cosine Functions
190
26.1 General Definition 190
26.2 Properties 192
Chapter 27
Graphs and Derivatives of Sine and Cosine Functions

Applications of Integration II: Volume
257
33.1 Solids of Revolution 257

33.2 Volume Based on Cross Sections 259
Chapter 34
The Natural Logarithm
268
34.1 Definition 268
34.2 Properties 268
Chapter 35
Exponential Functions
275
35.1 Introduction 275
35.2 Properties of a
x
275
35.3 The Function e
x
275
36.1 L'Hôpital's Rule 284
36.2 Exponential Growth and Decay 285
Chapter 36
L'Hôpital's Rule; Exponential Growth and Decay
284

Chapter 37
Inverse Trigonometric Functions
292
37.1 One-One Functions 292

Answers to Supplementary Problem 335
Index 371
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Chapter
1
Coordinate Systems on
a
Line
1.1
THE COORDINATES
OF
A
POINT
Let
9
be a line. Choose a point
0
on the line and call this point the
origin.
Now select a direction along
9;
say, the direction from left to right on the diagram.
For every point
P
to the right
of
the origin
0,
let the
coordinate

9
to the right of the origin
0;
namely,
of
that point to the right of
0
whose distance from
0
is
r.
To every point
Q
on
9
to the left of the origin
0,
-
we assign a negative real number as its coordinate; the number
-Q0,
the negative
of
the distance
between
Q
and
0.
For example, in the diagram
the point
U

ON
A
LINE
This assignment of real numbers to the points on the line
9
is called a coordinate system on
9.
Choosing a different origin, a different direction along the line, or a different unit distance would result
in a different coordinate system.
1.2
ABSOLUTE VALUE
For any real number
b
define the absolute value
I
b
I
to be the magnitude of
b;
that is,
b ifb20
-6 ifb<O
lbl=
In other words, if b is a positive number or zero, its absolute value
I
b
I
is
6
itself. But if b is negative, its

la12
=
a‘
(1.4)
An important special case
of
(1.1)
results from choosing
r
=
U
-
U
and recalling that
-(U
-
U)
=
U
-
U,
lu
-
U1
=
lu
-
U1
If
I

Replacing
a
in
(1
4)
by ab yields
I
ab
l2
=
(ab)2
=
a2b2
=
I
U
1’
I
b
1’
=
(I
U
I I
b
I)’
whence, the absolute value being nonnegative,
lab1
=
lallbl

COORDINATE SYSTEMS ON A LINE
3
012
345
I
I.
.
,,
.

,Y
0
AI
A2
-3
-2
-1
0
1
2
3
4
I
1
U U
1
1
I
PY
A2

is the coordinate of
A,
then
I
a
I
=
distance between
A
and the origin
Notice that, for any positive number
c,
lul
I
c
is equivalent to
-c
I
u
I
c
1
1
1
*Y
1
1
1
-C
0

substitute
x
-
3 for
U
in
(1.9),
obtaining
-
5
<
x
-
3
<
5.
Adding
3,
we have
-
2
<
x
<
8.
From a geometric standpoint, note that
I
x
-
3

a
I
lal
and
-161
I
6
I
lbl
(In fact, either
a
=
I
a
I
or
a
=
-1
a
I.)
Adding the inequalities, we obtain
(-14)
+(-lbl)
I
a
+
b
2s
I4

(1.10)
The inequality
(1.10)
is known as the
triangle inequality.
In
(1.20)
the sign
c
applies if and only if
a
and
b
are of opposite signs.
EXAMPLE
13
+
(-2)(
=
11
I
=
1,
but 131
+
1-21
=
3
+
2=

I.
(d)
Why isn’t the formula
p=
x
always true?
(a)
fl=
Jd
=
3.
(d)
By part
(c),
,/?
=
I
x
I,
but
I
x
I
=
x
is false when
x
<
0.
For example,

x
I.
1.2
Solve
I
x
+
3
I
5
5;
that is, find all values of
x
for which the given relation holds.
By
(1.8),
Ix
+
31
5
5
if and only if
-5
5
x
+
3
5
5.
Subtracting

-
1
<
3x
+
2
<
1.
Subtracting
2,
we obtain the equivalent rela-
tion
-
3
<
3x
<
-
1.
This is equivalent, upon division by
3,
to
-
1
<
x
<
-
3.
I

<
-
3.
Dividing by
-
3,;
>
x
>
1.
~ ~ ~
ALGEBRA
REVIEW
Multiplying or dividing both sides of an inequality by a negative number
reuerses
the
inequality:
if
U
<
b
and
c
<
0,
then
uc
>
bc.
To see this, notice that

I
1
11
7
I
I
1.
b
0
1
23
x+4<2
x-3
We cannot simply multiply both sides by
x
-
3,
because we do not know whether
x
-
3
is positive or
negative.
Case
1:
x
-
3
>
0.

10.
Case
2:
x
-
3
<
0.
Multiplying
(1)
by this negative quantity reverses the inequality:
x
+
4
>
2~
-
6
4
>
x
-
6
10
>
x
[add
61
[subtract
x]

1
and
2,
(1)
holds for
x
>
10
and for
x
<
3.
0
3
10
1.6
Solve
(x
-
2Xx
+
3)
>
0.
A
product is positive if and only if both factors are of like sign.
Case
1:
x
-

Then
x<2
and
x<-3,
x
<
-3
implies
x
<
2.
Thus,
(x
-
2)(x
+
3)
>
0
holds when either
x
>
2
or
x
<
-3.
these are equivalent to
which are equivalent
x

1.
By
(2.9),
-1<3x-2<1
1<3x<3
[add
21
+<X<l
[divide by
31
Therefore, the solution of
I3x
-
2
I
2
1
is
x
5
4
or
x
2
1.
I
3
1
1.8
Solve

(x
-
3)
changes from positive to negative, and
so
the product will be negative between
1
and
3.
As we pass
from right to left through
x
=
1,
the factor
(x
-
1)
changes from positive to negative, and
so
the product
changes back from negative to positive throughout the interval between
x
=
-2
and
x
=
1.
Finally, as we

x
>
3
or
-2
<
x
<
1.
Supplementary Problems
1.9
(a)
For what kind of number
u
is
I
u
I
=
-U?
(b)
For what values of
x
does
I3
-
x
I
equal
x

COORDINATE SYSTEMS
ON
A LINE
(a)
Solve I2x
+
3
I
=
4.
(b)
Solve I5x
-
7
I
=
1.
(c)
m
Solve part
(a)
by graphing
y,
=
(2x
+
3
I
and
y,

(a)-(f)
by graphing.
(4
11+:1>2
(e)
1<3-2x<5
(f)
312x+1<4
(9)
IE3
Check your answers to parts
(a)-(f)
by graphing.
Solve:
(a)
x(x
+
2)
>
0
(b)
(X
-
1)(x
+
4)
<
0
(c)
X’

(h)
Check your answers to parts
(a)-@)
by graphing.
[Hints:
In part
(c),
factor; in part
(f),
use the method of Problem 1.8.1
Show that if
b
#
0,
then
-
=
-
1;l
It;
Prove:
(a)
1a21
=
lalZ
(b)
Solve:
(a)
12x
-

12x
-
31
<
Ix
+
21
(b)
I
3x
-
2
I
5
I
x
-
1
I
[Hint:
Consider the threecasesx
2
3,
-2
I
x
<
3,x
<
-2.1

61
I
lal
+
161.
[Hint:
Use the triangle inequality to prove that
lal
5
la
-
bl
+
lbl
Determine whether
fl=
a’
holds for all real numbers
a.
Does
fl
<
always imply that
a
<
b?
Let
0,
I,
A,

and
b.
Find
b
if:
(a)
a
=
7,
B
is to the right
of
A,
and
Ib
-
a1
=
3;
(b)
a
=
-1,
Bis to the left of
A,
and
Ib
-
a/
=

b
+
c.
7
ALGEBRA
a
<
b
means that
b
-
a
is positive. The sum and the product
of
two positive numbers are posi-
tive, the product
of
two negative numbers is positive, and the product
of
a positive and a negative number
is negative.
ab
(b)
If
0
<
c,
then
a
<

A,
on opposite sides
of
the origin.]
Chapter
2
Coordinate Systems in a Plane
2.1
THE
COORDINATES
OF
A
POINT
We shall establish a correspondence between the points of a plane and pairs of real numbers.
Choose two perpendicular lines in the plane of Fig.
2-1.
Let us assume for the sake of simplicity that
one of the lines is horizontal and the other vertical. The horizontal line will
be
called the x-axis and the
vertical line will be called the y-axis.
c
:L
‘I
1
-lt
Fig.
2-1
Next choose a coordinate system on the x-axis and one on the y-axis. The origin for both coordi-
nate systems is taken to be the point

In Fig. 2-2, the coordinates
of
several points have been indicated. We have limited ourselves
to
integer coordinates only for simplicity.
Conversely, every pair (a,
6)
of real numbers is associated with a unique point in the plane.
EXAMPLES
In the coordinate system
of
Fig. 2-3,
to
find the point having coordinates
(3,
2), start at the origin
0,
move three units to the
right
and then two units
upward.
To
find the point with coordinates (-2,
4),
start at the
origin
0,
move two units to the
left
and then four units

23
4Y
(-2.4)
T
4
I
4
4-
(-2.3)
3-
0
(3.3)
2-
0
(2.2)
(5.2)
+
I
(3.0)
1
-I
+
1-1-
I-
(-4,I)
1
111
1
IAI
1

4Y
4-
3-
2-
9
(3*2)
1-
3
4
5x
-
9
11
3-
(-1.2)
0
2
(
+)
I
1
I
-3
-2
-1
0
0
(-3,-1)
-1
4Y

Thus, one might say: “The point
(1,O)
lies on the x-axis.”
2.2
THE
DISTANCE
FORMULA
Let
P1
and
P,
be points with coordinates
(x,,
y,)
and
(x,,
y,)
in
a given coordinate system (Fig.
2-5).
We wish to find a formula for the distance
PIP2.
Let
R
be the point where the vertical line through
P,
intersects the horizontal line through
P,.
Clearly, the x-coordinate of
R

P,R
=
A,&
But
A,A2
=
Ix,
-
x,
I
by
(1.6).
Thus,

=
I
x,
-
x2
I.
Similarly,
P,
R
=
I
y1
-
y,
I.
Consequently,

whence
[Equation
(2.1)
is called the
distance
formula.]
The reader should check that this formula
also
holds
when
P,
and
P,
lie on the same horizontal line or on the same vertical line.
(2.1)
10
COORDINATE
SYSTEMS
IN
A
PLANE [CHAP.
2
1’
I
I
I
I
I
I
X

-
11)’
=
,/(-4)’
+
(-3)’
=
JW
=
JZ
=
5
(b)
The distance between
(4,
-
3)
and
(2,7)
is
,/(4
-
2)’
+
(-3
-
7)’
=
Jm
=

b)
and the origin
(0,O)
is
Jm.
2.3
THE
MIDPOINT
FORMULAS
Again considering two arbitrary points
Pl(xl,
y,)
and
P2(x2,
y2),
we shall find the coordinates
(x,
y)
of
the midpoint
M
of
the segment
P,P2
(Fig.
2-6).
Let
A,
B,
C

hence
AB
=
E.
Since
AB
=
x
-
x1
and

=
x2
-
x,
x
-
x1=
x2
-
x
2x
=
XI
+
x2
x1
+x2
2

and
y=-
x1+
x2
x=-
2
2
In words, the coordinates
of
the midpoint are the averages of the coordinates
of
the endpoints.
CHAP.
21
COORDINATE SYSTEMS IN
A
PLANE
11
EXAMPLES
(a)
The midpoint of the segment connecting
(1, 7)
and
(3,
5)
is
(F,
y)
=
(2,6).

-
4)2
+(2
-
7)2
=
J(-5)2 +(-5)2
=
,/ET%
=
Jso
AC
=
J[
-
1
-
(
-3)12
+
(2
-
6)2
=
Jw'
=
JGZ
=
JZ
BC

7,
3), C(2,6) is a right triangle.
Use
(2.2)
to find the squares
of
the sides,
-
AB2
=
(-5
+
7)2
+
(-3
-
3)2
=
22
+
(-6)2
=
4
+
36
=
40
BC2
=
(-7

AB2
+
m2
=
E2,
AABC
is a right triangle with right angle at
B.
GEOMETRY
The converse of the Pythagorean theorem is also true: If
x2
=
AB2
+
m2
in
AABC,
then
<ABC
is a right angle.
2.3
Prove by use
of
coordinates that the midpoint of the hypotenuse of a right triangle is equidistant
from the three vertices.
Let the origin of a coordinate system be located at the right angle
C;
let the positive x-axis contain leg
CA
and the positive y-axis leg

-
(a
)
Fig.
2-7
12
COORDINATE SYSTEMS IN
A
PLANE
Now by the Pythagorean theorem,
and by the distance formula
(24,
[CHAP.
2
ALGEBRA
For any positive numbers
U,
U,

Hence,
MA
=
MC.
[For a simpler, geometrical proof, see Fig.
2-7(b);
MD
and
BC
are parallel.]
Supplementary Problems

3)
and
(2,
8);
(b)
(3,
1)
and
(3,
-4);
(c)
(4,
1)
and
(2,
1);
(4
(
-
3,4)
and
(54).
2.7
Draw the triangle with vertices
A(4, 7), B(4,
-
3),
and C(
-
1,