University of the Philippines Diliman
MATHEMATICS 21
Elementary Analysis I
Course Module
Institute of Mathematics
MATHEMATICS 21
Elementary Analysis I
Course Module
Institute of Mathematics
University of the Philippines Diliman
iv
c 2018 by the Institute of Mathematics, University of the Philippines Diliman.
All rights reserved.
No part of this document may be distributed in any way, shape, or form, without prior written
permission from the Institute of Mathematics, University of the Philippines Diliman.
Mathematics 21 Module Writers and Editors:
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
Carlo Francisco Adajar
Michael Baysauli
Katrina Burdeos
Lawrence Fabrero
Alip Oropeza
Contents
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
1 Limits and Continuity
1.1 Limit of a Function: An Intuitive Approach . . . . . . . . . . . . . . . . . . . . . . .
1.1.1 An Intuitive Approach to Limits . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.2 Evaluating Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.3 Other Techniques in Evaluating Limits . . . . . . . . . . . . . . . . . . . . . .
1.1.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 One-Sided Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Limits Involving Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.1 Infinite Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.2 Limits at Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4 Limit of a Function: The Formal Definition . . . . . . . . . . . . . . . . . . . . . . .
1.4.1 The Formal Definition of Limits . . . . . . . . . . . . . . . . . . . . . . . . .
1.4.2 Proving Limits using the Definition . . . . . . . . . . . . . . . . . . . . . . . .
1.4.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5 Continuity of Functions; The Intermediate Value Theorem . . . . . . . . . . . . . . .
1.5.1 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5.2 The Intermediate Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . .
1.5.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6 Trigonometric Functions: Limits and Continuity;
The Squeeze Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6.1 The Squeeze Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6.2 Continuity of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . .
1.6.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.7 New Classes of Functions: Limits and Continuity . . . . . . . . . . . . . . . . . . . .
1.7.1 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.7.2 Exponential and Logarithmic Functions . . . . . . . . . . . . . . . . . . . . .
1.7.3 Inverse Circular Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.7.4 Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.7.5 Inverse Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.7.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
v
1
1
1
4
6
8
10
16
19
19
24
29
31
31
33
37
38
38
45
48
50
50
54
57
59
59
60
65
68
72
75
vi
CONTENTS
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
2 Derivatives and Differentiation
79
2.1 Slopes, the Derivative, and Basic Differentiation Rules . . . . . . . . . . . . . . . . . 79
2.1.1 The Tangent Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
2.1.2 Definition of the Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
2.1.3 Differentiability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
2.1.4 Techniques of Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
2.1.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
2.2 The Chain Rule, and more on Differentiability . . . . . . . . . . . . . . . . . . . . . 88
2.2.1 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
2.2.2 Derivatives from the Left and from the Right . . . . . . . . . . . . . . . . . . 90
2.2.3 Differentiability and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . 91
2.2.4 Graphical Consequences of Differentiability and Non-differentiability . . . . . 93
2.2.5 Higher Order Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
2.2.6 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
2.2.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
2.3 Derivatives of Exponential and Logarithmic Functions . . . . . . . . . . . . . . . . . 100
2.3.1 Derivatives of Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . 100
2.3.2 Logarithmic Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
2.3.3 Derivatives of Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . 103
2.3.4 Derivative of f (x)g(x) , where f (x) > 0 . . . . . . . . . . . . . . . . . . . . . . 105
2.3.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
2.4 Derivatives of Other New Classes of Functions . . . . . . . . . . . . . . . . . . . . . 107
2.4.1 Derivatives of Inverse Circular Functions . . . . . . . . . . . . . . . . . . . . . 107
2.4.2 Derivatives of Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . 108
2.4.3 Derivatives of Inverse Hyperbolic Functions . . . . . . . . . . . . . . . . . . . 109
2.4.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
2.5 The Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
2.5.1 Rolle’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
2.5.2 The Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
2.5.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
2.6 Relative Extrema of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
2.6.1 Relative Extrema . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
2.6.2 Critical Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
2.6.3 Increasing/Decreasing Functions . . . . . . . . . . . . . . . . . . . . . . . . . 118
2.6.4 The First Derivative Test for Relative Extrema . . . . . . . . . . . . . . . . . 119
2.6.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
2.7 Concavity and the Second Derivative Test . . . . . . . . . . . . . . . . . . . . . . . . 124
2.7.1 Concavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
2.7.2 Point of Inflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
2.7.3 The Second Derivative Test for Relative Extrema . . . . . . . . . . . . . . . . 127
2.7.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
2.8 Graph Sketching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
2.8.1 Graphing Polynomial Funtions . . . . . . . . . . . . . . . . . . . . . . . . . . 131
CONTENTS
vii
2.8.2
2.8.3
2.8.4
2.8.5
Review of Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
Graphing Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
The Graph of f from the Graph of f 0 . . . . . . . . . . . . . . . . . . . . . . 137
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
3 Applications of Differentiation
143
3.1 Absolute Extrema of a Function on an Interval . . . . . . . . . . . . . . . . . . . . . 143
3.1.1 Absolute Extrema on Closed and Bounded Intervals . . . . . . . . . . . . . . 144
3.1.2 Absolute Extrema On Open Intervals . . . . . . . . . . . . . . . . . . . . . . 146
3.1.3 Optimization: Application of Absolute Extrema on Word Problems . . . . . . 148
3.1.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
3.2 Rates of Change, Rectilinear Motion, and Related Rates . . . . . . . . . . . . . . . . 157
3.2.1 Rates of Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157
3.2.2 Rectilinear Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
3.2.3 Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
3.2.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
3.3 Local Linear Approximation, Differentials, and Marginals . . . . . . . . . . . . . . . 169
3.3.1 Differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169
3.3.2 Local Linear Approximation and Approximating ∆y . . . . . . . . . . . . . . 170
3.3.3 Marginals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
3.3.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
3.4 Indeterminate Forms and L’Hôpital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . 179
∞
0
. . . . . . . . . . . . . . . . . . . . . 179
3.4.1 Indeterminate Forms of Type and
0
∞
3.4.2 L’Hôpital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
3.4.3 Indeterminate Forms of Type 0 · ∞ and ∞ − ∞ . . . . . . . . . . . . . . . . . 183
3.4.4 Indeterminate Forms of Type 1∞ , 00 and ∞0 . . . . . . . . . . . . . . . . . . 185
3.4.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
4 Integration and Its Applications
191
4.1 Antidifferentiation and Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . . 191
4.1.1 Antiderivatives or Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . 191
4.1.2 Particular Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
4.1.3 Integration by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
4.1.4 Rectilinear Motion Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . 197
1
4.1.5 Antiderivatives of f (x) = and of the other Circular Functions . . . . . . . 198
x
4.1.6 Antiderivatives of Exponential Functions . . . . . . . . . . . . . . . . . . . . 200
4.1.7 Antiderivatives Yielding the Inverse Circular Functions . . . . . . . . . . . . 200
4.1.8 Antiderivatives of Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . 203
4.1.9 *Antiderivatives Yielding Inverse Hyperbolic Functions . . . . . . . . . . . . 204
4.1.10 Summary of Antidifferentiation Rules . . . . . . . . . . . . . . . . . . . . . . 206
4.1.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
4.2 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
4.2.1 Area of a Plane Region: The Rectangle Method . . . . . . . . . . . . . . . . . 211
4.2.2 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
viii
CONTENTS
s
em
at
ic
at
h
of
M
e
4.7
ut
4.6
tit
4.5
In
s
4.4
UP
4.3
4.2.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219
The Fundamental Theorem of the Calculus . . . . . . . . . . . . . . . . . . . . . . . 221
4.3.1 First Fundamental Theorem of the Calculus . . . . . . . . . . . . . . . . . . . 221
4.3.2 The Second Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . 224
4.3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227
Generalization of the Area of a Plane Region . . . . . . . . . . . . . . . . . . . . . . 230
4.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245
Arc Length of Plane Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248
4.5.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252
Volumes of Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254
4.6.1 Volumes of Solids of Revolution . . . . . . . . . . . . . . . . . . . . . . . . . . 254
4.6.2 Volume of Solids by Slicing . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272
4.6.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276
Mean Value Theorem for Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280
4.7.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285
Chapter 1
Limits and Continuity
1.1
Limit of a Function: An Intuitive Approach
em
at
ic
s
We begin this course with an introduction to the core concept needed in studying calculus: the limit
of a function. We start studying the notion of limits in an informal, intuitive way. We treat limits
using a descriptive, graphical, and numerical approach. We then develop computational methods
in evaluating limits of algebraic expressions.
at
h
At the end of this section, the student will be able to:
of
M
• interpret the limit of a function through graphs and tables of values;
ut
e
• compute the limit of polynomial and rational functions using limit theorems; and
1.1.1
4
UP
In
s
tit
• evaluate limits of functions using substitution, cancellation of common factors,
and rationalization of radical expressions (for indeterminate forms 0/0).
An Intuitive Approach to Limits
f (x) = 3x − 1
4
3x2 − 4x + 1
g(x) =
x−1
(
4
3
3
3
2
2
2
1
1
1
0
−1
1
2
0
−1
1
2
3
0
−1
h(x) =
1
3x − 1,
0,
2
x 6= 1
x=1
3
Figure 1.1.1: Graphs of y = f (x), y = g(x) and y = h(x) in Illustration 1.1.1.
In this subsection, we use graphs of functions in order to develop an intuitive notion of the basic
concept of limits. We make a distinction between the value of a function at a real number a and
1
2
CHAPTER 1. LIMITS AND CONTINUITY
the function’s behavior for values very near a. A function f may be undefined at a, but it can be
described by studying the values of f when x is very close to a, but not equal to a. To illustrate
our point, let us consider the following functions:
Illustration 1.1.1.
1. Let f (x) = 3x − 1 and consider the tables below.
x
0
0.5
0.9
0.99
0.99999
f (x)
−1
0.5
1.7
1.97
1.99997
x
2
1.5
1.1
1.001
1.00001
f (x)
5
3.5
2.3
2.003
2.00003
em
at
ic
s
In the tables above, we evaluated f at values of x very close to 1. Observe that as the values
of x get closer and closer to 1, the values of f (x) get closer and closer to 2. If we continue
replacing x with values even closer to 1, the value of f (x) will get even closer to 2.
3x2 − 4x + 1
(3x − 1)(x − 1)
=
. Note that g(x) is undefined at x = 1. Observe
x−1
x−1
though that if x 6= 1, then g(x) = 3x − 1 = f (x). Thus, g is identical to f except only at
x = 1. Hence, as in the first item, if x assumes values going closer and closer to 1 but not
reaching 1, then the values of g(x) go closer and closer to 2.
(
3x − 1, x 6= 1
3. Let h(x) =
. Here, h(1) = 0. If x 6= 1, then h(x) = f (x) and as in above,
0,
x=1
h(x) goes closer and closer to 2 as x goes closer and closer to 1. (See Figure 1.1.1 for a
comparison of f , g and h.)
UP
In
s
tit
ut
e
of
M
at
h
2. Let g(x) =
In each of the above examples, we saw that as x got closer and closer to a certain number a, the
value of the function approached a particular number. This does not always happen, but in the
case that it does, the number to which the function value gets closer and closer is what we will call
the limit of the function as x approaches a.
Let f be a function defined on some open interval I containing a, except possibly at a. We say
that the limit of f (x) as x approaches a is L, where L ∈ , denoted
R
lim f (x) = L,
x→a
if we can make f (x) as close to L as we like by taking values of x sufficiently close to a (but not
necessarily equal to a).
Remark 1.1.2. Alternatively, lim f (x) = L if the values of f (x) get closer and closer to L as x
x→a
assumes values going closer and closer to a but not reaching a.
1.1. LIMIT OF A FUNCTION: AN INTUITIVE APPROACH
3
Example 1.1.3. Since the value of 3x − 1 goes closer and closer to 2 as x goes closer and closer
to 1 as shown in Illustration 1.1.1, we now write
lim (3x − 1) = 2.
x→1
Remark 1.1.4. Note that in finding the limit of f (x) as x tends to a, we only need to consider
values of x that are very close to a but not exactly a. This means that the limit may exist even if
f (a) is undefined.
3x2 − 4x + 1
is undefined at x = 1.
x−1
However, since x only approaches 1 and is not equal to 1, we conclude that x − 1 6= 0. Hence,
Example 1.1.5. In Illustration 1.1.1, we see that g(x) =
(3x − 1)(x − 1)
= lim (3x − 1) = 2.
x→1
x→1
x−1
lim g(x) = lim
x→1
s
Remark 1.1.6. If lim f (x) and f (a) both exist, their values may not be equal. In other words, it
em
at
ic
x→a
is possible that f (a) 6= lim f (x).
x→a
Example 1.1.7. Recall that
3x − 1, x 6= 1
0,
x=1
of
M
h(x) =
at
h
(
from Illustration 1.1.1. Here, h(1) = 0 but lim h(x) = 2.
e
x→1
In
s
tit
ut
Remark 1.1.8. If f (x) does not approach a real number as x tends to a, then we say that the
limit of f (x) as x approaches a does not exist (dne).
UP
Example 1.1.9. Let H(x) be defined by
H(x) =

1,
x≥0
0,
x<0
This function is called the Heaviside step function. The graph of the function is given below:
1
−3 −2 −1
−1
0
1
2
3
From the graph, we see that there is no particular value to which H(x) approaches as x approaches
0. We cannot say that the limit is 0 because if x approaches 0 through values greater than 0,
the value of H(x) approaches 1. In the same way, we cannot say that the limit is 1 because if x
approaches 0 through values less than 0, the value of H(x) approaches 0. In this case, lim H(x)
x→0
does not exist.
4
CHAPTER 1. LIMITS AND CONTINUITY
1.1.2
Evaluating Limits
In the previous subsection, we tried to compute the limit of a given function using tables of values.
However, this method only gives us an estimate of the limit, not guaranteeing that the value which
the function seems to approach is indeed the limit. In this section, we will compute limits not by
making tables of values or by graphing, but by applying the theorem below.
Theorem 1.1.10. Let f (x) and g(x) be functions defined on some open interval containing a,
except possibly at a.
1. If lim f (x) exists, then it is unique.
x→a
2. If c ∈
R, then x→a
lim c = c.
3. lim x = a
x→a
4. Suppose lim f (x) = L1 and lim g(x) = L2 where L1 , L2 ∈
R, and c ∈ R.
s
x→a
em
at
ic
x→a
(a) lim [f (x) ± g(x)] = L1 ± L2
x→a
(b) lim [cf (x)] = cL1
at
h
x→a
(c) lim [f (x)g(x)] = L1 L2
of
M
x→a
L1
, provided that g(x) 6= 0 on some open interval containing a, except
L2
possibly at a, and L2 6= 0.
=
(e) lim [f (x)]n = (L1 )n for n ∈
N.
In
s
x→a
ut
e
f (x)
x→a g(x)
tit
(d) lim
UP
p
p
(f ) lim n f (x) = n L1 for n ∈
x→a
N, n > 1 and provided that L1 > 0 when n is even.
Example 1.1.11. Determine lim (2x3 − 4x2 + 1).
x→−1
Solution.
From the theorem above,
lim (2x3 − 4x2 + 1) = lim 2x3 − lim 4x2 + lim 1
x→−1
x→−1
x→−1
x→−1
= 2 lim x3 − 4 lim x2 + 1
x→−1
x→−1
= 2(−1)3 − 4(−1)2 + 1
= −5.
(x − 3)(x2 − 2)
.
x→1
x2 + 1
Example 1.1.12. Evaluate lim
1.1. LIMIT OF A FUNCTION: AN INTUITIVE APPROACH
5
Solution.
First, note that
lim (x2 + 1) = lim x2 + lim 1 = 1 + 1 = 2 6= 0.
x→1
x→1
x→1
Using the theorem,
lim (x − 3)(x2 − 2)
(x − 3)(x2 − 2)
lim
x→1
x2 + 1
=
x→1
lim (x2 + 1)
x→1
lim (x − 3) · lim (x2 − 2)
=
x→1
x→1
=
=
x→1
lim (x2 + 1)
x→1
lim x − lim 3 lim x2 − lim 2
x→1
x→1
x→1
2
lim (x + 1)
x→1
(1 − 3)(12 − 2)
2
em
at
ic
s
= 1.
√
2x + 5
x→2 1 − 3x
at
h
Example 1.1.13. Evaluate: lim
of
M
Solution.
First, note that
lim (1 − 3x) = lim 1 − lim 3x = 1 − 6 = −5 6= 0.
x→2
x→2
ut
e
Also,
x→2
x→2
x→2
In
s
Thus, using the theorem,
tit
lim (2x + 5) = lim 2x + lim 5 = 4 + 5 = 9 > 0.
x→2
UP
q
√
√
√
lim (2x + 5)
lim
2x
+
5
3
2x + 5
9
x→2
x→2
=
=
=
=− .
lim
x→2 1 − 3x
lim 1 − 3x
lim 1 − 3x
−5
5
x→2
x→2
Theorem 1.1.14. Let f be a polynomial or rational function. If a ∈ dom f , then
lim f (x) = f (a).
x→a
Example 1.1.15. Evaluate lim (2x3 − 4x2 + 1).
x→−1
Solution.
Using Theorem 1.1.14,
lim (2x3 − 4x2 + 1) = 2(−1)3 − 4(−1)2 + 1 = −5.
x→−1
6
CHAPTER 1. LIMITS AND CONTINUITY
Example 1.1.16. Evaluate lim
x→1
1 − 5x
1 + 3x2 + 4x4
4
.
Solution. 4
1−5x
. Note that 1 ∈ dom f . By Theorem 1.1.14,
Let f (x) = 1+3x
2 +x4
lim
x→1
1.1.3
1 − 5x
1 + 3x2 + 4x4
4
=
1 − 5(1)
1 + 3(1)2 + 4(1)4
4
=
1
.
16
Other Techniques in Evaluating Limits
3x2 − 4x + 1
F (x)
= lim
x→1
x→1 G(x)
x−1
em
at
ic
lim
s
In this part, we shall see that there are functions whose limits cannot be calculated using only the
previous techniques. For instance, Theorems 1.1.10 and 1.1.14 do not apply to
(where F (x) = 3x2 − 4x + 1 and G(x) = x − 1) in Illustration 1.1.1. This is because lim G(x) = 0.
x→1
tit
ut
e
of
M
at
h
F (x)
Now, observe that lim F (x) = 0 = lim G(x). We call the limit lim
an indeterminate form.
x→1
x→1
x→1 G(x)
3x2 − 4x + 1
Such limits may or may not exist. Example 1.1.5 showed us that the limit lim
exists,
x→1
x−1
and is in fact equal to 2. We shall see that limits with indeterminate forms, if they exist, may be
determined using algebraic manipulation.
0
.
0
UP
In
s
f (x)
If lim f (x) = 0 and lim g(x) = 0, then lim
is called an indeterminate form of type
x→a
x→a
x→a g(x)
Remark 1.1.17.
f (x)
is undefined at x = a, and NOT indeterminate.
g(x)
f (x)
Remember that the term “indeterminate” only applies to the limit lim
, and not the
x→a g(x)
f (a)
function value
.
g(a)
1. If f (a) = 0 and g(a) = 0, then
2. By our intuitive notion of the limit, recall that when computing the limit, we are not concerned
with the function value when x = a.
3. A limit that is indeterminate of type 00 may exist, and to compute the limit, one may use
cancellation of common factors and rationalization of expressions (if applicable).
x2 + 2x + 1
.
x→−1
x+1
Example 1.1.18. Evaluate lim
1.1. LIMIT OF A FUNCTION: AN INTUITIVE APPROACH
7
Solution.
The limit of both the numerator and the denominator as x approaches −1 is 0. Thus, this limit in
its current form is indeterminate of type 00 . However, observe that x + 1 is a common factor of
the numerator and the denominator. Thus, we may simplify the function as
x2 + 2x + 1
(x + 1)2
=
= x + 1, provided x 6= −1.
x+1
x+1
Therefore, we obtain the limit as follows:
x2 + 2x + 1
= lim (x + 1) = 0
x→−1
x→−1
x+1
lim
x2 − 5x + 6
.
x→2
x2 − 4
Example 1.1.19. Evaluate lim
Solution.
x2 − 5x + 6
is an indeterminate form of type
x→2
x2 − 4
Note that lim
0
0
. Using the same technique,
s
x2 − 5x + 6
(x − 2)(x − 3)
x−3
1
= lim
= lim
=− .
2
x→2
x→2 (x − 2)(x + 2)
x→2 x + 2
x −4
4
of
M
Solution.
at
h
x2 − 16
√ .
x→4 2 − x
Example 1.1.20. Evaluate lim
em
at
ic
lim
x2 − 16
√ is an indeterminate form of type 00 . Observe that in its current form, the
x→4 2 − x
numerator and denominator do not have common factors. So we multiply the numerator and
√
denominator by 2 + x to get
√
√
√
x2 − 16 2 + x
(x − 4)(x + 4)(2 + x)
√ ·
√ =
= −(x + 4)(2 + x),
4−x
2− x 2+ x
In
s
tit
ut
e
Again, lim
UP
provided x 6= 4. Thus, we have
√
x2 − 16
√ = lim −(x + 4)(2 + x) = −32.
x→4 2 − x
x→4
lim
√
Example 1.1.21. Evaluate lim
x→4
x+5−3
.
x−4
Solution.
This limit is also an indeterminate form of type ( 00 ). Similar to the previous example,
√
√
x+5−3
x+5+3
(x + 5) − 9
√
lim
·√
= lim
x→4
x→4 (x − 4)( x + 5 − 3)
x−4
x+5+3
x−4
√
= lim
x→4 (x − 4)( x + 5 + 3)
1
= lim √
x→4
x+5+3
1
= .
6
8
CHAPTER 1. LIMITS AND CONTINUITY
1.1.4
Exercises
Exercises for Discussion
A. Let f be the function whose graph is shown in the figure below.
4
2
0
2
4
Evaluate f (0), f (2), and f (3). Evaluate also lim f (x), lim f (x) and lim f (x).
B. Evaluate the following limits.
1. lim x(x − 2)(x + 2)
at
h
of
M
3x2 + 2x − 1
x→−1
x3 + 1
√
t−2−4
3. lim
t→18
t − 18
ut
tit
In
s
UP
2x2 − 13x + 20
x→4
x3 − 64
5. lim
e
2. lim
2s2 − 7s + 3
s→3 s2 − 4s + 3
3
2z − z 2
6. lim
z→2
z2 − 4
√
x2 + 3 − 2
7. lim
x→−1
x2 − 1
√
√
2x − 6 − x
8. lim
x→2
4 − x2
p3 − 1
9. lim √
p→1
2p − 1 − 1
x→−1
4. lim
x→3
s
x→2
em
at
ic
x→0
6x + 2x2
√
x→−3 −1 − 3 2x + 5
10. lim
C. Do as indicated.
1. Find lim f (x) where f (x) = x2 for all x 6= 10 but f (10) = 99.
x→10
2. Determine the values of the constants a and b such that lim √
x→0
x
= 1.
ax + b − 2
1.1. LIMIT OF A FUNCTION: AN INTUITIVE APPROACH
9
Supplementary Exercises
A. Evaluate the following limits.
x3 − x2 − x + 1
x→−1
x3 − 3x − 2
t−1
10. lim √
2
t→1
6t + 3 − 3t
√
4x2 + 5x + 9 − 3
11. lim
x→0
x
√
√
8 − x − 1 − 8x
√
√
12. lim
x→−1
3 − x − 6x + 10
(x + t)3 − x3
13. lim
t→0
t
1
+1
14. lim x2 4
x→−4 x − 16
1
−1
15. lim x+t x
t→0
t
4
8
16. lim
+ 2
x→−2 x + 2
x + 2x
q3 + q2 − q − 1
q→−1
q2 − 1
2y 2 − 3y + 1
2. lim
y→1
y3 − 1
6 + x − x2
3. lim 2
x→−2 x − 4x − 12
√
2− 7−a
4. lim 2
a→3 2a − 3a − 9
x3 − x2 − x + 10
5. lim
x→−2
x2 + 3x + 2
√
√
4
x4 + 1 − x2 + 1
6. lim
x→0
x2
p
√
7+ 3x−3
7. lim
x→8
x−8
p
√
9q 2 − 4 − 17 + 12q
8. lim
q→−1
q 2 + 3q + 2
9. lim
of
M
at
h
em
at
ic
s
1. lim
bx2 + 15x + b + 15
exists. If it exists,
x→−2
x2 + x − 2
B. Find whether there exists a constant b so that lim
ut
e
determine the value of b and find the limit.
In
s
tit
C. For each of the following functions below, use a calculator to evaluate f (x) when
x = ±0.1, ±0.001, ±0.000001. Based on your results, what could the value of lim f (x) be?
UP
sin x
x
1 − cos x
2. f (x) =
x
tan x
3. f (x) =
x
1. f (x) =
x→0
10
1.2
CHAPTER 1. LIMITS AND CONTINUITY
One-Sided Limits
When we compute the limit of a function f as x approaches a, we observe the behavior of f as x
approaches a from both sides. However, there are instances when the behavior of f as x approaches
a from the right is not the same as its behavior as x approaches a from the left. This may happen
for piecewise-defined functions. Moreover, it is also possible that a function is not defined for some
open interval containing a, but defined only for values greater than a or less than a. In this case,
we can only observe the behavior of f as x approaches a from one side.
At the end of this section, the student will be able to:
• interpret the one-sided limit of a function through graphs and tables of values;
• evaluate one-sided limits of functions; and
em
at
ic
s
• determine the limit of piecewise functions using one-sided limits.
of
M
at
h
Consider the following functions.
Illustration 1.2.1. Let
x<1
4x − 3,
x≥1
.
tit
ut
e
f (x) =

3 − 5x2 ,
x→1
UP
In
s
If x is less than and very close to 1, then f (x) = 3 − 5x2 . If we let x approach 1 through these
values, then f (x) would approach −2. On the other hand, if we let x approach 1 through values
greater than 1, then we use f (x) = 4x − 3 and f (x) approaches 1. Similar to Example 1.1.9,
lim f (x) does not exist.
y = 3 − 5x2
y = 4x − 3
1
−1 0
−1
1
2
−2
Figure 1.2.1: Graph of y = f (x) in Illustration 1.2.1
√
Illustration 1.2.2. Let f (x) = x. The domain of f is [0, ∞). So we can only consider values of
x which are greater than 0 if we want to to get the “limit” of f as x goes to 0.
1.2. ONE-SIDED LIMITS
11
f (x) =
2
√
x
1
0
1
2
3
4
Figure 1.2.2: Graph of y = f (x) in Illustration 1.2.2
These cases lead us to what are called one-sided limits.
Let f be a function defined at every number in some open interval (c, a). We say that the limit of
f (x) as x approaches a from the left is L, denoted
lim f (x) = L
x→a−
em
at
ic
s
if the values of f (x) get closer and closer to L as the values of x get closer and closer to a, but are
less than a.
of
M
at
h
Similarly, let f be a function defined at every number in some open interval (a, c). We say that the
limit of f (x) as x approaches a from the right is L, denoted
lim f (x) = L
x→a+
In
s
tit
ut
e
if the values of f (x) get closer and closer to L as the values of x get closer and closer to a, but are
greater than a.
UP
Remark 1.2.3.
1. The conclusions in Theorem 1.1.10 still hold when “x → a” is replaced by “x → a− ” or
“x → a+ ”.
2. We sometimes refer to lim f (x) as the two-sided limit to distinguish it from one-sided limits.
x→a
Example 1.2.4. Let H(x) be defined by
H(x) =

1,
x≥0
0,
x<0
.
(Recall that this is the Heaviside step function.) We easily see that lim H(x) = lim 1 = 1 while
x→0+
lim H(x) = lim 0 = 0.
x→0−
x→0−
Example 1.2.5. Let
f (x) =

3 − 5x2 ,
x<1
4x − 3,
x≥1
.
x→0+
12
CHAPTER 1. LIMITS AND CONTINUITY
From Illustration 1.2.1, we have
lim f (x) = lim (3 − 5x2 ) = −2,
x→1−
x→1−
and
lim f (x) = lim (4x − 3) = 1.
x→1+
x→1+
Suppose a function f satisfies lim f (x) = 0. We shall distinguish between ways by which f
x→a
approaches 0. For example, for f (x) = 2 − x, notice that lim (2 − x) = 0. As x approaches
x→2
2 from the left, f (x) approaches 0 but passes through positive values. On the other hand, if x
approaches 2 from the right, f (x) approaches 0 but passes through negative values:
x
f (x)
1.9
0.1
1.99
0.01
1.999
0.001
2
0
2.001
−0.001
2.01
−0.01
2.1
−0.1
s
Meanwhile, consider g(x) = (2−x)2 . Similarly, lim (2−x)2 = 0, but this time, regardless of whether
em
at
ic
x→2
x approaches 2 from the left or right, g(x) passes through positive values as it approaches 0:
1.99
0.0001
1.999
0.000001
2
0
2.001
0.000001
at
h
1.9
0.01
2.01
0.0001
2.1
0.01
of
M
x
g(x)
ut
e
We now state the following definition which will be helpful in our computations later.
Suppose lim f (x) = 0. If f approaches 0 through positive values, we write
tit
x→a
In
s
f (x) → 0+ .
UP
Similarly, if f approaches 0 through negative values, we write
f (x) → 0− .
Remark 1.2.6. If f (x) → 0+ as x → a, and n is even, then
p
lim n f (x) = 0.
x→a
p
If f (x) → 0− as x → a, and n is even, then lim n f (x) does not exist.
x→a
Example 1.2.7. Let f (x) = 2 − x. Note that f (x) → 0+ as x → 2− , so lim
x→2−
√
other hand, f (x) → 0− as x → 2+ , so lim 2 − x does not exist.
√
2 − x = 0. On the
x→2+
Example 1.2.8. Consider g(x) =
•
lim
x→−1−
√
2x2 + x − 1.
p
p
2x2 + x − 1 = lim
(2x − 1)(x + 1) = 0
x→−1−
hp
√ i
(−3) · (0− ) = 0+
1.2. ONE-SIDED LIMITS
•
•
p
lim
x→−1+
13
hp
√ i
(−3) · (0+ ) = 0−
(2x − 1)(x + 1) does not exist
hq
p
(2x − 1)(x + 1) does not exist
lim
1−
x→ 2
i
(0− ) · ( 23 )
hq
i
(0+ ) · ( 32 )
p
• lim
(2x − 1)(x + 1) = 0
1+
x→ 2
Example 1.2.9. Let
 2
x −x−2


, x<0


 x+1
√
f (x) =
4 − x,
0≤x≤4




x2 − 5x + 4, x > 4
lim f (x), (b) lim f (x), (c) lim f (x), and (d) lim f (x).
x→0−
x→0+
x→4+
s
x→−1−
em
at
ic
Find (a)
lim f (x) = lim
x→−1−
(x − 2)(x + 1)
= lim (x − 2) = −3.
x+1
x→−1−
ut
e
x→−1−
of
M
at
h
Solution.
To evaluate (a), we need x to be less than −1 but very close to −1, so we use the expression
x2 − x − 2
. Thus,
x+1
In
s
tit
For (b), we take x to be less than 0 but very close to zero. By the definition of the function, we
x2 − x − 2
for such values of x. Hence, lim f (x) = −2.
should take f (x) =
x+1
x→0−
x→0+
√
UP
For (c), lim f (x) = lim
x→0+
4 − x = 2.
Finally, for (d), lim f (x) = lim (x2 − 5x + 4) = 0.
x→4+
x→4+
Some limits can be calculated by finding the one-sided limits first. In fact, we have the following
result:
Theorem 1.2.10. lim f (x) = L if and only if lim f (x) = L = lim f (x).
x→a
x→a−
Example 1.2.11. From Example 1.2.7, since lim
x→2−
√
lim 2 − x does not exist.
√
x→a+
2 − x = 0 and lim
x→2+
√
2 − x does not exist,
x→2
Example 1.2.12. If H(x) is the Heaviside step function, then lim H(x) does not exist (see Example
x→0
1.2.4).
14
CHAPTER 1. LIMITS AND CONTINUITY
Example 1.2.13. Let
f (x) =
√
 4 − x,
x≤4
x2 − 5x + 4,
x>4
Evaluate lim f (x).
x→4
Solution.
Observe that f is defined differently when x < 4 and when x > 4, so we consider one-sided limits.
Since
√
lim f (x) = lim 4 − x = 0,
x→4−
x→4−
and
lim f (x) = lim (x2 − 5x + 4) = 0,
x→4+
x→4+
we have lim f (x) = 0.
x→4
em
at
ic
at
h
|t + 4| =

t + 4,
if t ≥ −4
.
of
M
Solution.
Recall that
s
t+4
. Evaluate lim g(t).
t→−4
|t + 4|
Example 1.2.14. Let g(t) =
−(t + 4),
e
Therefore,
if t < −4
t+4
= lim −1 = −1,
−(t + 4) t→−4−
In
s
and
t→−4−
tit
t→−4−
ut
lim g(t) = lim
t+4
= lim 1 = 1.
t→−4+
t→−4+ t + 4
lim g(t) = lim
Hence, lim g(t) does not exist.
t→4
UP
t→−4+
Example 1.2.15. Evaluate: lim [[x]], where [[x]] is the greatest integer less than or equal to x.
x→−2
Solution.

..


.






−3, −3 ≤ x < −2


Note that [[x]] = −2, −2 ≤ x < −1




−1,
−1 ≤ x < 0





 ..
.
Thus,
lim [[x]] =
x→−2−
lim (−3) = −3, while
x→−2−
lim [[x]] =
x→−2+
exist.
Example 1.2.16. Evaluate: lim s + [[1 − s]]
s→2+
lim (−2) = −2, so lim [[x]] does not
x→−2+
x→−2
1.2. ONE-SIDED LIMITS
15
Solution.

..


.




−2, −2 ≤ 1 − s < −1 ⇔ 2 < s ≤ 3
Note that [[1 − s]] =

−1,
−1 ≤ 1 − s < 0 ⇔ 1 < s ≤ 2



 .

 .
.
Therefore, lim s + [[1 − s]] = lim (s − 2) = 0.
s→2+
s→2+
[[2x − 1]] − 2x
−
2x + 1
x→ 1
Example 1.2.17. Evaluate: lim
2
2
x2 − 2x + 1
x→1
x − [[x]]
e
Solution.
of
M
Example 1.2.18. Evaluate: lim
at
h
2
em
at
ic

..


.




−1, −1 ≤ 2x − 1 < 0 ⇔ 0 ≤ x < 1
2
Note that [[2x − 1]] =
1

0,
0 ≤ 2x − 1 < 1 ⇔ 2 ≤ x < 1




 .

..
[[2x − 1]] − 2x
−1 − 2x
So, lim
= lim
= −1.
1−
1 − 2x + 1
2x
+
1
x→
x→
ut
x2 − 2x + 1
(x − 1)2
= lim
=0
x − [[x]]
x→1−
x→1− x − 0
In
s
tit
• lim
x2 − 2x + 1
(x − 1)2
= lim
= lim (x − 1) = 0
x − [[x]]
x→1+
x→1+ x − 1
x→1+
UP
• lim
x2 − 2x + 1
= 0.
x→1
x − [[x]]
Therefore, lim
s
Solution.
16
CHAPTER 1. LIMITS AND CONTINUITY
1.2.1
Exercises
Exercises for Discussion
A. Evaluate the following limits.
x→3−
p
7x2 − 12x + 5
5. lim [[3x + 1]]
2
x→ 3
p
2. lim 4x2 − 12x + 9
|4x| − |x − 5|
1−x
2t + 1
7. lim
1 |2t2 − 3t − 2|
t→−
6. lim
3
x→ 2
x→1
3. lim [[3x + 1]]
2+
x→ 3
4.
2
lim [[3x + 1]]
8. lim
2−
x→ 3
x→1−
B. Let
x [[x]] − 1
[[x]] − 1
lim g(x), lim g(x)
x→−1
x→0
C. Given
h(x) =
of
M
lim g(x),
x→−2

kx − 3,
e
Evaluate:
at
h
em
at
ic
 2
x −4


,
x < −1
 2
x + 2x
g(x) =
.
x+1


√
, x > −1
2x + 3 − 1
s
1. lim
,
x > −1
tit
ut
x2 + k,
x ≤ −1
where k is a constant. Find k so that lim h(x) exists.
In
s
x→−1
UP
D. Sketch a graph of a function f (x) satisfying all of the following:
• dom f = [−4, 4]
• f (4) = 0
• f (−4) = f (−2) = 3
•
• f (0) = 1
• lim f (x) = 1
• f (2) = −1
• lim f (x) = 1
lim f (x) = 0
x→−4+
• lim f (x) = 4
x→0+
• lim f (x) = −1
x→2
x→−2
x→0−
• lim f (x) = 0
x→4−
1.2. ONE-SIDED LIMITS
17
Supplementary Exercises
A. Evaluate the following limits.
1. lim
x→3−
x2 − 9
x+2
7.
2
[[x]] − 2
x→2+ [[x]] − x
[[2 − y]]
9. lim
+
y→4 [[y + 2]] − y
2. lim [[1 − 2x]]
8. lim
x→5−
3. lim
x→2−
1
1
−
x [[x]]
x2 − [[x + 2]] x
x→2
x − [[2x]]
[[x]] + 2
11. lim
−
x→2 [[x]] − 2
4. lim [[2 + x]]
10. lim
x→2
5.
lim
x→−2+
1 − 4x2
1 − |2x2 − 3x + 1| + |2x2 − x|
x→
lim
1
1
−
[[−x]] |x|
x2 − 3x + 2
[[x]] − x
x→2−
|3x + 2| − |x − 2|
x→0
x
12. lim
em
at
ic
6. lim
s
s
B. Let
tit
lim f (x), lim f (x), lim f (x), lim f (x)
x→−1
x→0
x→1
x→2
In
s
Evaluate:
ut
e
of
M
at
h
 2
x + 4x + 3


, x ≤ −1


 x2 − x − 2
f (x) = [[x + 3]] ,
−1 < x < 1 .




√x − 1,
x≥1
UP
C. Let

|2x − x2 |



,
x<0


x


[[x + 3]] − 6
g(x) =
,
0≤x<3.

3
−
[[x]]



x2 − 6x + 5


 3
, x≥3
x − 19x − 30
Evaluate: lim g(x), lim g(x), lim g(x)
x→0
x→3
x→5
18
CHAPTER 1. LIMITS AND CONTINUITY
D. Let


a − 3x x < −2



f (x) = ax + 3b −2 ≤ x ≤ 1 ,




4x + b
x>1
where a and b are constants. Find a and b so that lim f (x) and lim f (x) both exist.
x→−2
x→1
E. Sketch a graph of a function f (x) satisfying all of the following:
• dom f = [−3, 2) ∪ (2, 5]
7
• f (−3) = f (−1) =
2
• f (4) = 3
•
lim f (x) = −2
• lim f (x) = 4
1
2
• lim f (x) = 3
• f (5) = −1
• lim f (x) = 0
x→−3+
• lim f (x) = −
x→−1
x→2+
x→4
• lim f (x) = −
x→5−
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
x→2−
5
2
1.3. LIMITS INVOLVING INFINITY
1.3
19
Limits Involving Infinity
In this section, we consider functions that may increase indefinitely either positively or negatively.
We will also be interested in the behavior of a function as x increases or decreases without bound,
or as we shall say, as x approaches positive infinity or negative infinity.
At the end of this section, the student will be able to:
• interpret infinite limits and limits at infinity of a function through graphs and
tables of values; and
1.3.1
Infinite Limits
of
M
In
s
tit
ut
f (x)
1
4
100
1000000
10000000000
UP
x
1
0.5
0.1
0.001
0.00001
1
. Note that f is undefined at x = 0.
x2
e
Illustration 1.3.1. Let f (x) =
at
h
Let us start with what are called infinite limits.
em
at
ic
s
• evaluate infinite limits and limits at infinity of functions (e.g. rational, radical).
x
−1
−0.5
−0.1
−0.001
−0.00001
f (x)
1
4
100
1000000
10000000000
Observe that as the values of x get closer and closer to 0, the values of f (x) become larger and
larger, and that there is no bound to the growth of the values of f (x).
Figure 1.3.1: Graphs of y =
1
1
and y = − 2 near x = 0
x2
x
20
CHAPTER 1. LIMITS AND CONTINUITY
Let f be a function defined on some open interval containing a, except possibly at a.
We say that the limit of f (x) as x approaches a is positive infinity, denoted
lim f (x) = +∞,
x→a
if the value of f (x) increases without bound whenever the values of x get closer and closer to a
(but does not reach a).
Also, we say that the limit of f (x) as x approaches a is negative infinity, denoted
lim f (x) = −∞,
x→a
if the value of f (x) decreases without bound whenever the values of x get closer and closer to a
(but does not reach a).
em
at
ic
s
Remark 1.3.2. The expressions above with “x → a” replaced by “x → a− ” or “x → a+ ” are
similarly defined.
lim
1
= +∞,
of
M
x→0 x2
at
h
Example 1.3.3. In Illustration 1.3.1, we have
and
ut
e
1
lim − 2 = −∞.
x→0
x
UP
In
s
tit
Remark 1.3.4. Note that ∞ is not a real number. Thus, if lim f (x) = +∞ or −∞, we do not
x→a
mean that the limit exists. Though the limit does not exist, through the symbol we are able to
describe the behavior of f near a: that it increases or decreases indefinitely as x → a.
3x
. Note that lim 3x = 3 while lim (x − 1) = 0. In this case, lim f (x) does
x→1
x→1
x→1
x−1
not exist, but observe that:
Consider f (x) =
• as x → 1− , x − 1 → 0− . Thus, the numerator approaches 3 while the numerical value of the
3x
denominator becomes smaller and smaller. As a result,
approaches negative infinity.
x−1
• as x → 1+ , x − 1 → 0+ , so this time,
3x
approaches positive infinity.
x−1
In general, we have the following theorem.
Theorem 1.3.5. Let c be a nonzero real number. Suppose lim f (x) = c and lim g(x) = 0.
x→a
1. If c > 0,
(a) and g(x) → 0+ as x → a, then lim
f (x)
x→a g(x)
= +∞.
x→a
1.3. LIMITS INVOLVING INFINITY
21
f (x)
= −∞.
x→a g(x)
(b) and g(x) → 0− as x → a, then lim
2. If c < 0,
f (x)
= −∞.
x→a g(x)
f (x)
(b) and g(x) → 0− as x → a, then lim
= +∞.
x→a g(x)
(a) and g(x) → 0+ as x → a, then lim
Example 1.3.6. Consider: g(x) =
5x
.
4 − x2
−10
(0− )(4)
5x
• lim
= +∞
x→−2− (2 + x)(2 − x)
5x
= +∞
x→2− (2 + x)(2 − x)
• lim
= −∞
of
M
at
h
5x
x→2+ (2 + x)(2 − x)
em
at
ic
−10
(0+ )(4)
10
(4)(0+ )
10
(4)(0− )
s
5x
• lim
= −∞
x→−2+ (2 + x)(2 − x)
• lim
In
s
Theorem 1.3.7.
tit
ut
e
The next theorem involves some properties about infinite limits. In particular, it involves evaluating
limits of sum and/or product of two functions where one function approaches a limit while the other
function increases/decreases without bound as x approaches a.
UP
1. If lim f (x) exists and lim g(x) = ±∞, then lim (f (x) + g(x)) = ±∞.
x→a
x→a
x→a
2. If lim f (x) exists and lim g(x) = ±∞, then lim (f (x) − g(x)) = ∓∞.
x→a
x→a
x→a
3. If lim f (x) = +∞ and lim g(x) = +∞, then lim (f (x) + g(x)) = +∞.
x→a
x→a
x→a
4. If lim f (x) = +∞ and lim g(x) = −∞, then lim (f (x) − g(x)) = +∞ and
x→a
x→a
x→a
lim (g(x) − f (x)) = −∞.
x→a
5. Let c ∈
R \ {0}. Suppose x→a
lim f (x) = c and lim g(x) = ±∞. Then
x→a
(a) lim f (x)g(x) = ±∞, if c > 0.
x→a
(b) lim f (x)g(x) = ∓∞, if c < 0.
x→a
22
CHAPTER 1. LIMITS AND CONTINUITY
Example 1.3.8. Determine lim
x→1−
x2 −
2
.
x−1
Solution.
Using the above theorem, since lim x2 = 1 and lim
2
= −∞, we have (see Figure 1.3.2)
x→1−
x→1− x − 1
2
2
= +∞.
lim x −
x−1
x→1−
x→−2+
lim 2 = 2, lim 5x = −10 and
x→−2+
x→−2+
lim
2 − 5x +
x→−2+
2
(2 + x)3
lim
x→−2+
2
= +∞. Thus,
(2 + x)3
= +∞.
s
Solution.
Using the above theorem, we get
2
.
(2 + x)3
em
at
ic
Example 1.3.9. Determine
2 − 5x +
lim
at
h
The presence of infinite limits motivates this next definition, which pertains to properties of the
graph of a function. A more detailed discussion of this definition will be given in a later chapter.
of
M
The graph of x = a is a vertical asymptote of the graph of y = f (x) if at least one of the following
is true:
• lim f (x) = −∞
ut
e
• lim f (x) = −∞
tit
x→a−
UP
x→a−
In
s
• lim f (x) = +∞
x→a+
• lim f (x) = +∞
x→a+
Example 1.3.10. From the previous example, since lim
a vertical asymptote of the graph of y = x2 −
x→1−
x2 −
2
x−1
= +∞, the line x = 1 is
2
. (See Figure 1.3.2.)
x−1
Notice that in Theorem 1.3.7, nothing is said about lim [f (x) + g(x)], where lim f (x) = +∞ while
x→a
x→a
lim g(x) = −∞. In this case, does the limit exist? If it does, is the limit simply equal to zero? The
x→a
answer is no. In fact, this is another indeterminate form: ∞ − ∞. Moreover, in Theorem 1.3.7,
limits of the form c · ∞ where c 6= 0 were dealt with. When c = 0, the limit is also indeterminate.
1. Suppose lim f (x) = +∞ and lim g(x) = +∞. Then lim [f (x) − g(x)] is called an indetermix→a
x→a
x→a
nate form of type ∞ − ∞.
2. Suppose lim f (x) = 0 and lim g(x) = ±∞. Then lim [f (x)g(x)] is called an indeterminate
x→a
x→a
x→a
form of type 0 · ∞.
1.3. LIMITS INVOLVING INFINITY
23
x=1
Figure 1.3.2: Graph of y = x2 −
lim
x→−1−
1
3
+ 2
x + 1 2x + x − 1
em
at
ic
s
Example 1.3.11. Evaluate:
2
with vertical asymptote x = 1
x−1
lim
3
1
+
x + 1 (2x − 1)(x + 1)
of
M
x→−1−
at
h
Solution.
1
3
+
−
0
(−3)(0− )
tit
(2x − 1) + 3
= lim
x→−1− (2x − 1)(x + 1)
2x + 2
= lim
x→−1− (2x − 1)(x + 1)
2
= lim
−
x→−1 2x − 1
2
=−
3
UP
x→−1−
1
3
+
x + 1 (2x − 1)(x + 1)
In
s
lim
ut
e
Thus, the limit is indeterminate of type ∞ − ∞. To compute, we combine the expressions into one:
1
Example 1.3.12. Evaluate: lim
+
t→4 4 − t
t
8
−
t−1 t+2
0
0
Solution.
1
lim
t→4+ 4 − t
t
8
−
t−1 t+2
1
0−
4 8
−
3 6
Thus, the limit is indeterminate of type 0 · ∞. To compute, we rewrite the expression as a quotient:
24
CHAPTER 1. LIMITS AND CONTINUITY
1
lim
t→4+ 4 − t
1.3.2
t
8
−
t−1 t+2
1
t(t + 2) − 8(t − 1)
= lim
(t − 1)(t + 2)
t→4+ 4 − t
2
t − 6t + 8
= lim
t→4+ (4 − t)(t − 1)(t + 2)
(t − 2)(t − 4)
= lim
t→4+ (4 − t)(t − 1)(t + 2)
−(t − 2)
= lim
t→4+ (t − 1)(t + 2)
1
=−
9
0
0
Limits at Infinity
em
at
ic
s
We now discuss limits at infinity, that is, the behavior of a function as x increases or decreases
without bound. Let us consider the following illustration.
x
−1
−100
−10000
−10000000
−1000000000
In
s
f (x)
1
0.1
0.001
0.000001
0.000000001
UP
x
1
10
1000
1000000
1000000000
tit
ut
e
of
M
at
h
1
Illustration 1.3.13. Let f (x) = . Observe from the tables below that the values of f (x) get
x
closer and closer to zero as the values of x approach positive infinity, as seen in the left table.
Similarly, the values of f (x) get closer and closer to zero as the values of x approach negative
infinity, as seen in the right table.
f (x) =
1
x
0
Figure 1.3.3: Graph of y =
1
x
f (x)
−1
−0.01
−0.0001
−0.0000001
−0.000000001
1.3. LIMITS INVOLVING INFINITY
25
Let f be a function defined at every number in some interval (a, ∞). We say that the limit of f (x)
as x approaches positive infinity is L, denoted
lim f (x) = L
x→+∞
if the values of f (x) get closer and closer to L as the values of x increase without bound.
Similarly, let f be a function defined at every number in some interval (−∞, a). We say that the
limit of f (x) as x approaches negative infinity is L, denoted
lim f (x) = L
x→−∞
if the values of f (x) get closer and closer to L as the values of x decrease without bound.
Remark 1.3.14. We have similar notions for the following symbols:
s
x→+∞
lim f (x) = +∞ or −∞
x→−∞
at
h
•
lim f (x) = +∞ or −∞
em
at
ic
•
Example 1.3.15. In the previous illustration, lim
1
= 0 and lim
of
M
x→+∞ x
1
x→−∞ x
= 0.
tit
ut
e
In general, we have the following results.
2.
3.
lim xn = +∞, if n is even.
x→±∞
UP
1.
In
s
Theorem 1.3.16. Let n be a positive integer.
lim xn = ±∞, if n is odd.
x→±∞
lim
1
x→±∞ xn
4. Let c ∈
=0
R. Suppose x→+∞
lim f (x) = c and lim g(x) = ±∞. Then
x→+∞
f (x)
= 0.
x→+∞ g(x)
lim
Remark 1.3.17. In statement 4 of the previous theorem, “x → +∞” may be replaced by
“x → −∞”, “x → a”, “x → a+ ”, and “x → a− ”. What is important is that the limit of the
numerator exists, while the denominator increases or decreases without bound.
Example 1.3.18. Evaluate:
Solution.
lim (5x4 − x3 − x + 8)
x→+∞
26
CHAPTER 1. LIMITS AND CONTINUITY
Note that since we are letting x increase without bound, we have x 6= 0. We may then write
1
1
8
5x4 − x3 − x + 8 = x4 · 5 − − 3 + 4 . By the previous theorem, lim x4 = +∞, while each
x→+∞
x x
x
1 1
8
of , 3 and 4 approach 0, as x → +∞. Statement 4 of Theorem 1.3.7 then implies that
x x
x
1
1
8
lim x4 · 5 − − 3 + 4 = +∞.
(+∞)(5 − 0 − 0 + 0)
x→+∞
x x
x
Example 1.3.19. Evaluate:
lim (3x5 − x4 + 2x − 4)
x→−∞
Solution.
1
2
4
lim (3x5 − x4 + 2x − 4) = lim x5 · 3 − + 4 − 5 = −∞
x→−∞
x→−∞
x x
x
(−∞)(3 − 0 + 0 − 0)
s
Remark 1.3.20. In general, to find lim f (x) if f is a polynomial function, it suffices to consider
em
at
ic
x→±∞
the behavior of the leading term of f (x) as x → +∞ (or as x → −∞).
1
lim
Solution.
Note that lim x3 − 4 = −∞. Thus, lim
at
h
x→−∞ x3 − 4
1
of
M
Example 1.3.21. Evaluate:
x→−∞ x3 − 4
= 0.
e
x→−∞
In
s
tit
ut
4
2
Example 1.3.22. Evaluate: lim
3x + 2x −
x→+∞
x
UP
Solution.
4
Here, since lim (3x2 + 2x) = +∞ and lim
= 0, we have
x→+∞
x→+∞ x
4
2
lim
3x + 2x −
= +∞.
x→+∞
x
Example 1.3.23. Evaluate:
3x − 1
.
x→+∞ 9x + 3
lim
Solution.
∞
Note that lim (3x − 1) = +∞ and lim (9x + 3) = +∞. Thus, the limit has the form
. Does
x→+∞
x→+∞
∞
this mean that the limit does not exist? Consider the table of values below.
x
1
10
100000
1000000000
3x−1
9x+3
0.1666667
≈ 0.311828
≈ 0.3333311
≈ 0.3333333
1.3. LIMITS INVOLVING INFINITY
27
3x − 1
approaches a particular value: 0.3333333. Let us
9x + 3
2
1
3x − 1
3
= −
. Therefore,
verify this. Using long division, we can write
9x + 3
3 3x + 1
!
2
3x − 1
1
1
1
3
lim
= −0= .
= lim
−
x→+∞ 9x + 3
x→+∞ 3
3x + 1
3
3
It seems that as x → +∞, the quotient
So the limit exists, but it is not equal to 0.3333333 as we initially guessed—it is equal to 13 .
From the above example, we see that if both the numerator and denominator have infinite limits,
then the limit of the quotient may exist. In fact, this is another indeterminate form.
Suppose
lim f (x)
x→a
∞
=
and
lim g(x)
=
x→a
∞.
Then
∞
indeterminate form of type ∞
.
f (x)
x→a g(x)
lim
is
called
an
em
at
ic
s
Remark 1.3.24. The expression “x → a” may be replaced by “x → a− ”, “x → a+ ”, “x → −∞”
and “x → +∞”.
x3 − 2x2 + 3
Example 1.3.25. Evaluate: lim
x→+∞ 4x4 − x2 + x + 1
at
h
Solution.
x3 − 2x2 + 3
∞
is an indeterminate form of type
. Since we are letting x
x→+∞ 4x4 − x2 + x + 1
∞
approach +∞, we have x 6= 0. Thus, we may divide both numerator and denominator by the
highest power of x in the denominator which is x4 :
lim
e
of
M
Note that
UP
In
s
tit
ut
1
1
2
3
x3 − 2x2 + 3
0
x − x2 + x4
x4
lim
· 1 = lim
1
1
1 = 4 =0
x→+∞ 4x4 − x2 + x + 1
x→+∞
4 − x2 + x3 + x4
x4
√
x2 − 3
Example 1.3.26. Evaluate: lim
x→+∞ x + 2
Solution.
This is an indeterminate form of type
∞
∞
. To evaluate this, we use the fact that

x,
if x ≥ 0
√
x2 = |x| =
.
−x, if x < 0
√
Since we are letting x → +∞, we have x > 0 and so, x2 = x. Hence,
√
√
1
x2 − 3
x 2 − 3 √ x2
lim
= lim
· 1
x→+∞ x + 2
x→+∞ x + 2
√
2
x
√
= lim
x→+∞
= lim
x→+∞
= 1.
1
x 2 − 3 √ x2
· 1
x+2
x
q
1 − x32
1 + x2
28
CHAPTER 1. LIMITS AND CONTINUITY
Example 1.3.27. Evaluate:
lim (
x→−∞
p
9x2 − x + 3x)
Solution.
This is an indeterminate form of type ∞ − ∞. We solve it as follows:
√
p
p
( 9x2 − x − 3x)
2
2
lim ( 9x − x + 3x) = lim ( 9x − x + 3x) · √
x→−∞
x→−∞
( 9x2 − x − 3x)
∞
−x
= lim √
x→−∞ ( 9x2 − x − 3x)
∞+∞
1
√
−x
2
= lim √
· 1x
x→−∞ ( 9x2 − x − 3x) √
x2
1
√
−x
= lim √
· −x
x→−∞ ( 9x2 − x − 3x) √1
x2
−x
−x
9x2
3x
− xx2 − −x
x2
s
x→−∞
1
= lim q
x→−∞
9 − x1 + 3
at
h
1
6
of
M
=
x2 = |x| = −x when x < 0
em
at
ic
= lim q
since
or
ut
lim f (x) = L
e
The line y = L is a horizontal asymptote of the graph of y = f (x) if
lim f (x) = L.
x→−∞
In
s
tit
x→+∞
UP
(A more detailed discussion of the previous definition will be given in a later section.)
Let us illustrate the above definition with the following examples.
Example 1.3.28. Consider lim
the graph of y =
3
x2 − 9
3
x→+∞ x2 − 9
= 0. Here, the line y = 0 is a horizontal asymptote of
.
Example 1.3.29. From a previous example,
asymptote of the graph of y =
3x − 1
.
9x + 3
3x − 1
1
= . The line y = 13 is a horizontal
x→−∞ 9x + 3
3
lim
p
1
Example 1.3.30. From a previous example, lim
9x2 − x + 3x = . The line y = 16 is a
x→−∞
6
√
horizontal asymptote of the graph of y = 9x2 − x + 3x.
y
1.3. LIMITS INVOLVING INFINITY
29
y = x23−9
y = 3x−1
9x+3
y = 13
x
y=
√
9x2 − x + 3x
y = 61
Figure 1.3.4: The graphs of some functions with their respective horizontal asymptotes
1.3.3
Exercises
em
at
ic
s
Exercises for Discussion
Evaluate the following limits.
2. lim
t→0+
3. lim
x→2−
at
h
6. lim
x→2+
1
1
√
−
t
t t+3
x−2
√
2 − 4x − x2
of
M
x→ 3
2
1 − 3x
[[x]] − 1
[[x]] − x
7.
2x3 − 6x + 5
x→+∞ 4 + 7x − 6x3
8.
4z 3 + 5
z→+∞ 1 − 2z + 3z 2
e
1−
ut
lim
tit
1.
lim
lim
lim √
UP
In
s
3 − x2
x→+∞
4x2 + 1 + x2
4
1
1
4. lim
−
−
p
(x + 2)2 x
x→−2− (x + 2)2 (2 − x)
10. lim w + w2 + 2w
w→+∞
p
x
x−2
2 + 2w
5. lim
+ 2
11.
lim
w
+
w
x +x−2
x→−2− x + 2
w→−∞
9.
Supplementary Exercises
Evaluate the following limits.
2x3 − 5x2
1. lim
x→1− x2 − 1
4. lim
4 − x2
2. lim
x→2+ 4 − 4x + x2
5. lim
3. lim
2−y
y→4 y 2 − 8y + 16
s→−1
2
1
7s
+
+
s2 − 1 s2 + 3s + 2 s3 + 8
1
t→0+ t
1
1− √
2t + 1
[[s]] − 1
s→−1 [[s]] + 1
6. lim
30
CHAPTER 1. LIMITS AND CONTINUITY
√
2x + x2 − 1
12. lim
x→+∞
x+3
2
x − [[x]]2
7. lim
x2 − 1
x→1+
8.
9.
lim
3
x→−∞ x2 − 3
13.
lim (3x4 − 3x2 + x + 4)
1+
lim
x→−∞
√
x6 + x2 + 1
2x3 + 3
x→−∞
3y 2 − 5y + 2
10. lim
y→−∞ 6y 3 − 2y 2 − 1
14.
y 2 + 3y − 8
y→∞ 2 − 5y − 3y 2
15.
p
t2 − 3 + t
lim
p
y 2 + 3y − y
t→−∞
y→+∞
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
11. lim
lim
1.4. LIMIT OF A FUNCTION: THE FORMAL DEFINITION
1.4
31
Limit of a Function: The Formal Definition
In this section, we give the formal definition of the limit of a function. This involves a more rigorous
approach on showing the limit of a function. We will then establish limits of a function using the
formal definition.
At the end of this section, the student will be able to:
• describe the limit of a function using the formal definition; and
• illustrate the formal definition of the limit of a function using graphs.
1.4.1
The Formal Definition of Limits
of
M
at
h
em
at
ic
s
We start this section with the formal definition of the limit of a function. Recall that informally,
we define limit of a function f is the real number L as x approaches a, written lim f (x) = L, if
x→a
the values of the function f get closer and closer to L as we allow the values of x to get closer and
closer to a. We will make this informal notion of ”closer and closer” mathematically precise.
Let us consider the next illustration.
Illustration 1.4.1. Consider the function f (x) = 2x − 1. Note that lim f (x) = 1.
x→1
tit
ut
e
• For which values of x do we have 0 < f (x) < 2?
Answer: For all x in 21 , 32 , or for all x such that 12 < x < 32 .
UP
In
s
• For which values of x do we have 12 < f (x) < 23 ?
Answer: For all x in 43 , 54 , or for all x such that 34 < x < 54 .
• For which values of x do we have 0.98 < f (x) < 1.02?
Answer: For all x in (0.99, 1.01), or for all x such that 0.99 < x < 1.01.
• For which values of x do we have 0.9998 < f (x) < 1.0002?
Answer: For all x in (0.9999, 1.0001), or for all x such that 0.9999 < x < 1.0001.
In general, for any positive number ε, suppose we want to find a range of values for x so that
1 − ε < f (x) < 1 + ε
or
|f (x) − 1| < ε.
Then we can choose x satisfying
1−
ε
ε
<x<1+
2
2
or
ε
0 < |x − 1| < .
2
Moreover, if ε is very, very close to zero, then the expression
|f (x) − 1| < ε
32
CHAPTER 1. LIMITS AND CONTINUITY
f (x) = 2x − 1
2
(
1+ε
ε
1
ε
)
1−ε
|{z}|{z}
δ
δ
(
)
1
1−δ
0
2
1+δ
Figure 1.4.1: Given ε > 0, what is δ?
em
at
ic
s
means that f (x) is very, very close to 1. This happens if we choose x such that
ε
0 < |x − 1| < ,
2
at
h
that is, if x is chosen to be very, very close to 1.
of
M
The illustration above summarizes to the following problem:
tit
ut
e
If ε is any positive number, what should δ be so that
|f (x) − 1| < ε whenever 0 < |x − 1| < δ?
ε
.
2
UP
In
s
The solution above implies that we may choose δ to be
Let f (x) be a function defined on some open interval containing a, except possibly at a. The limit
of f (x) as x approaches a is L, written lim f (x) = L, if and only if for every ε > 0 (no matter how
x→a
small) there exists a δ > 0 such that
|f (x) − L| < ε whenever 0 < |x − a| < δ.
This definition formalizes our intuitive notion of a limit: that lim f (x) = L if f (x) can be made as
x→a
close as possible to L by taking values of x sufficiently close to a (but not equal to a). Recall that:
|f (x) − L| < ε ⇔
−ε < f (x) − L < ε
⇔
L − ε < f (x) < L + ε
0 < |x − a| < δ ⇔ −δ < x − a < δ, x − a 6= 0 ⇔ a − δ < x < a + δ, x 6= a
That is, the absolute value inequalities represent open intervals “centered” at L and at a. The
formal definition then says this: for any open interval Iy centered at L, there is a corresponding
open interval Jx “centered” at a such that whenever x is in Jx (but x 6= a), f (x) is guaranteed to
be in Iy . Because this can be done for any open interval Iy , the interval Iy can be chosen to be as
short as we like.
1.4. LIMIT OF A FUNCTION: THE FORMAL DEFINITION
1.4.2
33
Proving Limits using the Definition
In the next examples, we illustrate how the definition is used to prove that the limit of a function
is a particular number.
Example 1.4.2. Prove that lim (3x + 2) = 5.
x→1
Solution.
The solution is composed of two parts: the first part is to systematically guess what δ would be
sufficient, and the second part is to verify whether the guessed δ works.
I. Choosing a value for δ. Let ε be a positive number. We want to find a number δ such that
0 < |x − 1| < δ,
if
then |(3x + 2) − 5| < ε.
Note that |(3x + 2) − 5| = |3x − 3| = 3|x − 1|. So, we want to satisfy the following:
if
0 < |x − 1| < δ,
then |x − 1| < ε/3.
of
M
ε
This suggests that we may take δ = .
3
3|x − 1| < ε,
at
h
or
then
s
0 < |x − 1| < δ,
em
at
ic
if
Given any ε > 0, choose δ = ε/3.
tit
ut
e
II. Proof (verification that the chosen δ works).
If 0 < |x − 1| < δ, then
That is,
UP
In
s
|(3x + 2) − 5| = |3x − 3| = 3|x − 1| < 3δ = 3
if
0 < |x − 1| < δ,
ε
3
= ε.
then |(3x + 2) − 5| < ε.
Therefore, by the definition of a limit,
lim (3x + 2) = 5.
x→1
Example 1.4.3. Prove that lim (5x + 6) = −4.
x→−2
Solution.
I. Choosing a value for δ. Let ε be a given positive number. We want to find a number δ such
that
if
0 < |x − (−2)| = |x + 2| < δ,
then |(5x + 6) − (−4)| < ε.
34
CHAPTER 1. LIMITS AND CONTINUITY
Note that |(5x + 6) − (−4)| = |5x + 10| = 5|x + 2|. So, we want to satisfy the following:
if
0 < |x + 2| < δ,
then
if
0 < |x + 2| < δ,
then |x + 2| < ε/5.
5|x + 2| < ε
or
ε
This suggests that we may take δ = .
5
II. Proof (verification that the chosen δ works).
If 0 < |x + 2| < δ, then
Given any ε > 0, choose δ = ε/5.
|(5x + 6) − (−4)| = |5x + 10| = 5|x + 2| < 5δ = 5
ε
5
= ε.
That is,
then |(5x + 6) − (−4)| < ε.
s
0 < |x − (−2)| < δ,
Therefore, by the definition of a limit,
at
h
lim (5x + 6) = −4.
of
M
x→−2
Example 1.4.4. Prove that lim (4 − 3x) = 4.
ut
e
x→0
em
at
ic
if
In
s
tit
Solution.
if
UP
I. Choosing a value for δ. Let ε be a positive number. We want to find a number δ such that
0 < |x − 0| = |x| < δ,
then |(4 − 3x) − 4| < ε.
Note that |(4 − 3x) − 4| = | − 3x| = 3|x|. So, we want to satisfy the following:
if
0 < |x| < δ,
then
if
0 < |x| < δ,
then |x| < ε/3.
3|x| < ε
or
ε
This suggests that we may take δ = .
3
II. Proof (verification that the chosen δ works).
If 0 < |x − 0| < δ, then
Given any ε > 0, choose δ = ε/3.
|(4 − 3x) − (4)| = 3|x| < 3δ = 3
That is,
ε
3
= ε.
1.4. LIMIT OF A FUNCTION: THE FORMAL DEFINITION
if
0 < |x − 0| < δ,
35
then |(4 − 3x) − 4| < ε.
Therefore, by the definition of a limit,
lim (4 − 3x) = 4.
x→0
Remark 1.4.5. The value of δ is not unique. Note that if a given δ works such that
if
0 < |x − a| < δ, then |f (x) − L| < ε
then we can take any smaller positive δ0 ≤ δ such that
0 < |x − a| < δ0 .
s
The statement then becomes
at
h
Example 1.4.6. Prove that lim (x2 − 1) = 3.
em
at
ic
0 < |x − a| < δ0 ≤ δ, then |f (x) − L| < ε.
if
of
M
x→−2
Solution.
tit
0 < |x − (−2)| = |x + 2| < δ,
then |(x2 − 1) − 3| < ε.
In
s
if
ut
e
I. Choosing a value for δ. Let ε be a positive number. We want to find a number δ such that
UP
Note that |(x2 − 1) − 3| = |x2 − 4| = |x − 2||x + 2|. So, we want to satisfy the following:
if
0 < |x − (−2)| < δ,
then |x − 2||x + 2| < ε.
Since we are just getting the limit of the function for values of x very close to −2, it is safe
to assume that
0 < |x + 2| < 1.
Thus,
−1 < x + 2 < 1 =⇒ −3 < x < −1 =⇒ −5 < x − 2 < −3 < 5 =⇒ |x − 2| < 5.
Summarizing, if |x + 2| < 1, one way to ensure that |(x2 − 1) − 3| < is to ensure that
|x + 2| < ε/5. Thus, if |x + 2| is less than both 1 and ε/5, then we arrive at the desired
conclusion. This suggests that we may take δ to be the smaller of the two numbers, written
δ = min{1, ε/5}1 .
1
By δ = min{1, ε/5}, we mean that if 1 ≤ ε/5, δ may be chosen to be 1, and when ε/5 < 1, δ may be taken to
be ε/5.
36
CHAPTER 1. LIMITS AND CONTINUITY
II. Proof (verification that the chosen δ works). Given any ε > 0, choose δ = min{1, ε/5}.
If 0 < |x − (−2)| = |x + 2| < δ, then
ε
|(x2 − 1) − 3| = |x2 − 4| = |x − 2||x + 2| < 5δ ≤ 5
= ε.
5
That is,
if
then |(x2 − 1) − 3| < ε.
0 < |x − (−2)| < δ,
Therefore, by the definition of a limit,
lim (x2 − 1) = 3.
x→−2
Remark 1.4.7. If lim f (x) = L, observe that in one way or another, the expression |x − a| can
x→a
em
at
ic
s
be factored (but it may not be easy to do so!) from the expression |f (x) − L|. This means that if
0 < |x − a| < δ, then
|f (x) − L| = |x − a||g(x)| < δ|g(x)|
UP
In
s
tit
ut
e
of
M
at
h
for some function g(x). In this case, if g(x) is a nonzero constant, say g(x) = c, then δ = ε/|c|. If
g(x) is an expression in x, we shall find an upper bound for |g(x)| by restricting values of x to a
certain interval about a.
1.4. LIMIT OF A FUNCTION: THE FORMAL DEFINITION
1.4.3
37
Exercises
Supplementary Exercises
A. Prove the following.
4. lim (x2 + 2x − 3) = 0
1. lim (5x − 3) = −3
x→0
x→1
2. lim (2x + 1) = −3
5. lim (2 − x2 ) = 1
3. lim (5 − 2x) = −1
6. lim (4 − 3x + x2 ) = 8
x→−2
x→−1
x→−1
x→3
B. Use the definition of the limit of a function to prove that lim f (x) = −6 if
x→− 34
x 6= − 34
s
x = − 34
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic

2

 16x − 9 ,
4x + 3
f (x) =

2,
.
38
CHAPTER 1. LIMITS AND CONTINUITY
1.5
Continuity of Functions; The Intermediate Value Theorem
Notice that in the previous sections, to compute the limit of polynomial or rational functions as
x approaches a, we simply calculate the value of the function at a, if possible. We will see later
on that this property is a condition for possession of other “desirable” characteristics. Functions
satisfying that property are said to be continuous at x = a.
At the end of this section, the student will be able to:
• interpret the continuity of a function at a point using graphs;
• distinguish the types of discontinuity graphically;
• find the possible points of discontinuity of a function;
• describe the continuity of a function on an interval;
em
at
ic
• redefine functions with removable discontinuities;
s
• identify the type of discontinuity as either removable, jump essential, or infinite
essential;
at
h
• evaluate limits and determine continuity of composite functions; and
Continuity
tit
1.5.1
ut
e
of
M
• interpret the Intermediate Value Theorem using graphs and real life situations
Continuity at a Point
UP
In
s
Continuity is the precise mathematical translation of an unbroken curve. A function is continuous
if its graph has no gaps, breaks, or holes.
A function f is said to be continuous at x = a if the following conditions are all satisfied:
(i) f is defined at x = a
(ii) lim f (x) exists
x→a
(iii) f (a) = lim f (x)
x→a
Otherwise, f is said to be discontinuous at x = a.
Example 1.5.1. Let f (x) = x3 + x2 − 2. Let us examine the continuity of f at x = 1. First of all,
f (1) = 0. Moreover, lim f (x) = 0. Therefore, f is continuous at x = 1.
x→1
x2 − x − 2
. Since f is not defined at x = 2, f is discontinuous at
x−2
x = 2. However, it is continuous at every other a ∈ \ {2}.
Example 1.5.2. Let f (x) =
R
1.5. CONTINUITY OF FUNCTIONS; THE INTERMEDIATE VALUE THEOREM
39
Remark 1.5.3. In general, if f is a polynomial or a rational function and a is any element in the
domain of f , then f is continuous at x = a.
Example 1.5.4. Let g be defined by
 2
 x −x−2
, x 6= 2
g(x) =
.
x−2

0,
x=2
Note that g is defined at x = 2 with g(2) = 0. However,
x2 − x − 2
= lim (x + 1) = 3.
x→2
x→2
x−2
lim g(x) = lim
x→2
Since g(2) is not equal to lim g(x), g is discontinuous at x = 2.
x→2
em
at
ic
s
Example 1.5.5. Let h be defined by
at
h
 2
 x − x − 2 , x 6= 2
x−2
h(x) =
.

3,
x=2
of
M
For this function, we have h(2) = 3 = lim h(x). Therefore, h(x) is continuous at x = 2.
f (x) = 3x − 1
4
3
3
2
In
s
4
tit
ut
e
x→2
3x2 − 4x + 1
x−1
1
2
2
1
UP
1
0
−1
g(x) =
1
0
−1
2
3
3
1
h(x) = 2
x
1
−3 −2 −1
−1
Heaviside function H
0
1
2
3
1
−2
−3 −2 −1
−1
−3
0
1
2
3
Figure 1.5.1: The figures above show graphs of various functions. The function whose graph appears
on the upper left is continuous at all a ∈ , while the other three graphs are those of functions
that are discontinuous at some a ∈ .
R
R
40
CHAPTER 1. LIMITS AND CONTINUITY
1. If lim f (x) exists but either f (a) is undefined or f (a) 6= lim f (x), then we say that f has a
x→a
x→a
removable discontinuity at x = a.
2. If lim f (x) does not exist, then we say that f has an essential discontinuity at x = a.
x→a
Moreover,
(a) if lim f (x) and lim f (x) both exist but are not equal, then f is said to have a jump
x→a−
x→a+
essential discontinuity at x = a.
(b) if lim f (x) = +∞ or −∞, or lim f (x) = +∞ or −∞, then f is said to have an infinite
x→a−
x→a+
essential discontinuity.
Example 1.5.6. Refer to Figure 1.5.1.
• For f (x) = 3x − 1, note that f (1) = 2 = lim (3x − 1), so f is continuous at x = 1.
em
at
ic
s
x→1
3x2 − 4x + 1
3x2 − 4x + 1
, g(1) is undefined but lim
= 2, so g has a removable
x→1
x−1
x−1
discontinuity at x = 1.
at
h
• For g(x) =
1
. Then h is undefined at x = 0, while lim h(x) = −∞ and lim h(x) = +∞.
x
x→0−
x→0+
Thus, h has an infinite essential discontinuity at x = 0.
of
M
• Take h(x) =
ut
e
• Finally, for the Heaviside function H, we have H(0) = 1, lim H(x) = 0, and lim H(x) = 1.
x→0−
x→0+
In
s
tit
Therefore, H has a jump essential discontinuity at x = 0.
UP
Remark 1.5.7. The discontinuity in statement 1 of the above definition is called removable,
because the discontinuity can be “removed” by redefining the value of f at a so that f (a) = lim f (x),
x→a
resulting in a function that is now continuous at x = a. On the other hand, it is not possible to do
this for essential discontinuities.
Example 1.5.8. The function g in Example 1.5.4 has a removable discontinuity at x = 2, since
g(2) = 0 while lim g(x) = 3. However, if we redefine g at x = 2 such that g(2) = 3, then the
x→2
resulting function would be continuous at x = 2.
Example 1.5.9. Determine if the function


4x − 3,
f (x) =
x−2


,
2
2x − 5x + 2
x<1
x≥1
is continuous at x = 1 and at x = 2. If discontinuous, classify the type of discontinuity as removable,
jump essential, or infinite essential.
1.5. CONTINUITY OF FUNCTIONS; THE INTERMEDIATE VALUE THEOREM
41
Solution.
• x = 1:
1−2
=1
2−5+2
(ii) lim f (x) = lim (4x − 3) = 4(1) − 3 = 1
(i) f (1) =
x→1−
x→1−
(iii) lim f (x) = lim
x−2
x→1− 2x2 − 5x + 2
x→1+
=
1−2
=1
2−5+2
Thus, lim f (x) = 1 = f (1). Hence, f is continuous at x = 1.
x→1
• x = 2:
Note that f (x) =
x−2
2x2 − 5x + 2
for values of x close enough to 2.
(i) f (2) is undefined
x−2
1
1
= lim
=
x→2 (2x − 1)(x − 2)
x→2 2x − 1
3
= lim
s
x→2
x→2 2x2 − 5x + 2
em
at
ic
x−2
(ii) lim f (x) = lim
Hence, f has a removable discontinuity at x = 2.
e
of
M
at
h
Remark 1.5.10. In general, the discontinuities of a given function f may occur at points where
f is undefined or, for the case of piecewise functions, at the endpoints of intervals. Such points are
aptly called the possible points of discontinuity of f .
3. f g
In
s
2. f − g
UP
1. f + g
tit
ut
Theorem 1.5.11. Let f and g be continuous at x = a and let c ∈
also continuous at x = a:
4.
R. Then the following are
f
provided that g(a) 6= 0
g
5. cf
Similar to limits, we define continuity from the left and continuity from the right.
1. A function f is said to be continuous from the left at x = a if
f (a) = lim f (x).
x→a−
2. A function f is said to be continuous from the right at x = a if
f (a) = lim f (x).
x→a+
√
Example 1.5.12. The function f defined by f (x) = 2 − x is continuous from the left at x = 2
but not from the right at x = 2 (since f is undefined for x > 2).
42
CHAPTER 1. LIMITS AND CONTINUITY
Example 1.5.13. Let g(x) = [[x]] be the greatest integer function. Then g is continuous from the
right at x = 1, but not continuous from the left at x = 1. In general, if n is any integer, then g is
continuous from the right at x = n, but not continuous from the left at x = n (recall the graph of
g(x) = [[x]]).
Continuity on an Interval
We can also consider continuity of functions on intervals. Being continuous on an interval means
that the graph of a function has no “gaps” or “holes” on that interval.
A function f is said to be continuous
1. everywhere if f is continuous at every real number.
2. on (a, b) if f is continuous at every point x in (a, b).
3. on [a, b) if f is continuous on (a, b) and from the right of a.
em
at
ic
5. on [a, b] if f is continuous on (a, b] and on [a, b).
at
h
6. on (a, ∞) if f is continuous at all x > a.
s
4. on (a, b] if f is continuous on (a, b) and from the left of b.
e
8. on (−∞, b) if f is continuous at all x < b.
of
M
7. on [a, ∞) if f is continuous on (a, ∞) and from the right of a.
In
s
tit
ut
9. on (−∞, b] if f is continuous on (−∞, b) and from the left of b.
UP
Remark 1.5.14.
1. Polynomial functions are continuous everywhere.
2. The absolute value function f (x) = |x| is continuous everywhere.
3. Rational functions are continuous on their respective domains.
√
4. The square root function f (x) = x is continuous on [0, ∞).
5. The greatest integer function f (x) = [[x]] is continuous on [n, n + 1), where n is any integer.
The following theorem allows us to get the limit of a composition of functions provided that certain
conditions are satisfied.
Theorem 1.5.15. If lim g(x) = b and f is continuous at b, then lim f (g(x)) = f (b). In other
x→a
x→a
1.5. CONTINUITY OF FUNCTIONS; THE INTERMEDIATE VALUE THEOREM
43
words, if f is continuous at lim g(x),
x→a
lim f (g(x)) = f lim g(x) .
x→a
x→a
x−2
x→2 x2 − 4
Example 1.5.16. Evaluate: lim
Solution.
Since the absolute value function is continuous everywhere, we can apply the previous theorem:
lim
x→2
x−2
1
x−2
1
= lim 2
= − =
x→2 x − 4
x2 − 4
4
4
Example 1.5.17. Evaluate: lim 3x2
em
at
ic
s
1
x→ 3
Solution.
1
1
Since lim 3x2 = and the greatest integer function is continuous at x = , we have
1
3
3
x→
of
M
at
h
3
lim 3x2 =
tit
ut
e
1
x→ 3
1
= 0.
3
UP
In
s
Theorem 1.5.18. If g is continuous at x = a and f is continuous at g(a), then (f ◦ g)(x) =
f (g(x)) is continuous at x = a.
Example 1.5.19. Determine the intervals at which h(x) =
√
x2 − 1 is continuous.
Solution.
√
Note that if a > 0, the square root function f (x) = x is continuous at x = a. Moreover,
√
g(x) = x2 − 1 is continuous everywhere. By the previous theorem, h(x) = x2 − 1 is continuous
at all a satisfying a2 − 1 > 0, or for all a in (−∞, −1) or (1, ∞). It is left to the reader to verify
that h is continuous from the left at x = −1 and from the right at x = 1. Therefore, the intervals
at which h is continuous are (−∞, −1] and [1, ∞).
Example 1.5.20. Determine the values of x where the function

x−2


,
x ≤ −1

2x + 6
f (x) =
2x2 + x − 6


, x > −1

|2x − 3|
is discontinuous. Classify each discontinuity as removable, jump essential, or infinite essential.
44
CHAPTER 1. LIMITS AND CONTINUITY
Solution.
The function f is undefined at x = −3 and at x = 32 , and its definition splits at x = −1. Thus, f
is continuous at each x ∈ , except possibly at x = −3, −1 and 32 . Moreover, note that
R
(
|2x − 3| =
2x − 3, 2x − 3 ≥ 0
⇔ x ≥ 23
−(2x − 3), 2x − 3 < 0
⇔ x < 32
.
• x = −3:
(i) f (−3) is undefined
x−2
(ii) lim f (x) = lim
= +∞
x→−3−
x→−3− 2(x + 3)
−5
2(0− )
x−2
(iii) lim f (x) = lim
= −∞
x→−3+
x→−3+ 2(x + 3)
−5
2(0+ )
em
at
ic
s
Thus, f has an infinite essential discontinuity at x = −3.
• x = −1:
−1 − 2
3
=−
2(−1) + 6
4
x−2
3
(ii) lim f (x) = lim
=−
4
x→−1−
x→−1− 2x + 6
of
M
2x2 + x − 6
2x2 + x − 6
2−1−6
= lim
=
= −1
+
+
|2x
−
3|
−(2x
−
3)
−(2(−1)
− 3)
x→−1
x→−1
x→−1+
e
lim f (x) = lim
ut
(iii)
at
h
(i) f (−1) =
tit
Thus, f has a jump essential discontinuity at x = −1.
(i) f 32 is undefined
(ii)
UP
In
s
• x = 23 :
2x2 + x − 6
(2x − 3)(x + 2)
7
= lim
= lim −(x + 2) = −
−
−
−
|2x
−
3|
−(2x
−
3)
2
3
3
3
x→
x→
x→
lim f (x) = lim
3−
x→ 2
2
(iii) lim f (x) = lim
3+
x→ 2
3+
x→ 2
2
2x2 + x − 6
|2x − 3|
= lim
3+
x→ 2
2
(2x − 3)(x + 2)
7
= lim (x + 2) =
−
2x − 3
2
3
x→
2
Thus, f has a jump essential discontinuity at x = 32 .
Example 1.5.21. Determine the values of x where the function

1
1

+ , x<0


2
 x −x x
g(x) =
[[x − 1]] ,
0≤x≤2



 √
x − 2,
x>2
is discontinuous. Classify each discontinuity as removable, jump essential, or infinite essential.
1.5. CONTINUITY OF FUNCTIONS; THE INTERMEDIATE VALUE THEOREM
45
Solution.
Note that between 0 and 2, x − 1 is an integer when x = 1. Thus, the definition of the function
splits at x = 0, 1 and 2. Moreover, g is defined at all x ∈ . Thus, g is continuous everywhere
except possibly at 0, 1 and 2.
R
• x = 0:
(i) g(0) = [[0 − 1]] = −1
1 + (x − 1)
1
1
1
= lim
(ii) lim g(x) = lim
+
= lim
= −1
x
x→0− x(x − 1)
x→0−
x→0− x − 1
x→0− x(x − 1)
(iii) lim g(x) = lim [[x − 1]] = lim (−1) = −1
x→0+
x→0+
x→0+
Thus, g is continuous at x = 0.
• x = 1:
x→1−
x→1−
x→1−
(iii) lim g(x) = lim [[x − 1]] = lim 0 = 0
x→1+
x→0+
at
h
x→1+
em
at
ic
(ii) lim g(x) = lim [[x − 1]] = lim (−1) = −1
s
(i) g(1) = [[1 − 1]] = 0
of
M
Thus, g has a jump essential discontinuity at x = 1.
• x = 2:
e
(i) g(2) = [[2 − 1]] = 1
x→2+
In
s
tit
ut
(ii) lim g(x) = lim [[x − 1]] = lim 0 = 0
x→2−
x→2−
x→2−
√ √
0+
(iii) lim g(x) = lim x − 2 = 0
x→2+
1.5.2
UP
Thus, f has a removable discontinuity at x = 2.
The Intermediate Value Theorem
The next theorem gives an important property of continuous functions. It states that a continuous
function f on [a, b] takes on any value between the numbers f (a) and f (b).
Theorem 1.5.22 (Intermediate Value Theorem (IVT)). Let f be continuous on a closed interval [a, b] with f (a) 6= f (b). For every k between f (a) and f (b), there exists c in (a, b) such
that f (c) = k.
Geometrically, IVT states that if f is continuous on [a, b], then the graph of y = f (x) intersects
any horizontal line y = k between y = f (a) and y = f (b).
Remark 1.5.23.
1. The continuity of the function on [a, b] in IVT is required. In general, the theorem is not true
for functions that are discontinuous on [a, b].
46
CHAPTER 1. LIMITS AND CONTINUITY
f (b)
k
(c, f (c))
a
c
b
f (a)
Figure 1.5.2: A graphical illustration for IVT
2. The number c in the conclusion of IVT may not be unique.
x−1
.
x
em
at
ic
s
Example 1.5.24. Consider the function f (x) =
1. Verify that the Intermediate Value Theorem holds for f in the interval [1, 3].
of
M
at
h
1
2. Use the Intermediate Value Theorem to justify why there exists c ∈ (1, 3) such that f (c) = ,
3
and find the value of c.
Solution.
e
R \ {0} and f is continuous at every number in its domain, f is continuous on
ut
1. Since dom f is
In
s
tit
2
[1, 3]. Moreover, f (1) = 0 and f (3) = , so f (1) 6= f (3). Thus, IVT holds for f in [1, 3].
3
UP
2
2. IVT guarantees that for every k between 0 and , there exists c ∈ (1, 3) such that f (c) = k.
3
1
2
1
Since is between 0 and , there exists c ∈ (1, 3) such that f (c) = . We solve for c:
3
3
3
1
3
c−1
1
=
c
3
3c − 3 = c
f (c) =
2c = 3
3
c =
2
3
Thus, the desired c in the interval (1, 3) is c = .
2
Remark 1.5.25. Note that in using IVT (or any theorem), one must first verify that the theorem
holds before stating the conclusion of the theorem. For instance, in IVT, we must first show that the
function f is continuous on the interval [a, b] and that k is between f (a) and f (b) where f (a) 6= f (b)
before we can conclude that there exists c ∈ (a, b) such that f (c) = k.
1.5. CONTINUITY OF FUNCTIONS; THE INTERMEDIATE VALUE THEOREM
47
The Intermediate Value Theorem can also be used to locate roots of equations, as illustrated in the
next example.
Example 1.5.26. Show that the equation
3x4 − 4x2 + 5x − 1 = 0
has a solution between 0 and 1.
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
Solution.
Let f (x) = 3x4 −4x2 +5x−1. We wish to find c ∈ (0, 1) such that c is a root of the equation f (x) = 0
(that is, f (c) = 0). We check if IVT applies for f in (0, 1) with k = 0. Note that f is a polynomial
function so f is continuous on [0, 1]. Moreover, we have f (0) = −1 and f (1) = 3 − 4 + 5 − 1 = 3.
Hence, k = 0 is between f (0) and f (1). Therefore, the assumptions of IVT are satisfied, and we can
conclude that there exists c in (0, 1) such that f (c) = 0, that is, the equation 3x4 − 4x2 + 5x − 1 = 0
has at least one root c between 0 and 1.
48
CHAPTER 1. LIMITS AND CONTINUITY
1.5.3
Exercises
Exercises for Discussion
A. Determine if f is continuous at all possible points of discontinuity. Classify each discontinuity
as removable, jump essential, or infinite essential. Redefine f so that it will be continuous at
every point of removable discontinuity.
1
,
if x < 3
x
1. f (x) =
2


, if 3 ≤ x
9−x


|x + 3|, if x ≤ 0




[[2x]] ,
if 0 < x < 1
2. f (x) =


x2 − 5x



, if x ≥ 1
x−5

3x + 3


,
if x < 1


x3√+ 1
3. f (x) =
2 + 2 − x, if 1 ≤ x ≤ 2




x2 − 6,
if x > 2
at
h
em
at
ic
s



In
s
C. Use IVT to prove the following.
tit
ut
e
of
M


 x + 2a, if x < −2
B. Find all values of a and b such that h(x) =
3ax + b, if − 2 ≤ x ≤ 1 is continuous every
 3x − 2b, if x > 1
where.
cos x
has a solution between 1 and 2.
x
2. The graph of f (x) = x3 − 4x2 + x − 3 intersects the x-axis at least once between 3 and 4.
UP
1. Show that p(x) =
Supplementary Exercises
A. Determine if f is continuous at all possible points of discontinuity. Classify each discontinuity
as removable, jump essential, or infinite essential. Redefine f so that it will be continuous at
every point of removable discontinuity.
1. f (x) =



|x + 6|,
x2 − 6x + 9







x2 − 3x
if x ≤ −1
, if x > −1
x2 + 2x
, if x < 0
x+2
2. f (x) =
|x − 1| + 1, if 0 ≤ x < 2




x2 − 2,
if x ≥ 2
1.5. CONTINUITY OF FUNCTIONS; THE INTERMEDIATE VALUE THEOREM
at
h
em
at
ic
s
 2
x −x−2


, if x < 0

√x − 1
3. f (x) =
4 − x,
if 0 ≤ x ≤ 4



2
x − 5x + 4, if x > 4
 2
x −x−2


, if x < −1

x+2
4. f (x) =
[[x]] + 1,
if − 1 ≤ x < 1



|x + 2|,
if x ≥ 1
 2
x − 3x + 2


, if x < 0


x−1
5. f (x) =
[[x]] + [[−x]] , if 0 ≤ x < 2




x2 − 5,
if x ≥ 2


x,
if x ≤ 0



3,
if 0 < x ≤ 1
B. Is the function g(x) =
continuous on [1, 4]? [0, 4)? (−∞, 0)?
2

3 − x , if 1 < x ≤ 4



x − 3, if x > 4
(
x + 2, if x ≤ m
C. Find all values of m such that g(x) =
is continuous everywhere.
x2 ,
if x > m
of
M
D. Use IVT to prove the following.
1. The equation x5 + 4x3 + 14 − 7x = 0 has at least one real solution.
UP
In
s
tit
ut
e
2. Show that f (x) = x4 − 7x2 + x + 4 has at least two real zeros.
49
50
CHAPTER 1. LIMITS AND CONTINUITY
1.6
Trigonometric Functions: Limits and Continuity;
The Squeeze Theorem
In the preceding sections, we considered the behavior of algebraic functions such as polynomial and
rational functions, and those that involve rational exponents and radicals. We now investigate the
limits and continuity of trigonometric functions.
At the end of this section, the student will be able to:
• interpret the Squeeze Theorem graphically;
• evaluate limits using Squeeze Theorem;
• apply theorems on limits and continuity to trigonometric functions; and
The Squeeze Theorem
at
h
1.6.1
em
at
ic
s
• evaluate limits of trigonometric functions using prescribed theorems
and techniques.
of
M
We begin with an important theorem that is helpful in computing limits involving trigonometric
functions. In fact, this theorem is used in the proofs of other theorems in this section.
tit
ut
e
Theorem 1.6.1 (Squeeze Theorem). Let f (x), g(x) and h(x) be defined on some open interval
I containing a except possibly at x = a such that
In
s
f (x) ≤ g(x) ≤ h(x),
UP
for all x ∈ I \ {a}. If lim f (x) and lim h(x) exist and are both equal to L ∈
x→a
x→a
L.
R, then x→a
lim g(x) =
Remark 1.6.2. With some modifications, “x → a” can be replaced by “x → a− ”, “x → a+ ”,
“x → +∞” and “x → −∞”.
The Squeeze Theorem, sometimes called the Sandwich Theorem, states that if g(x) is “squeezed”
between f (x) and h(x) near a, and if f and h have the same limit at a, then g must have the same
limit at a.
1
2
Example 1.6.3. Evaluate: lim x cos
x→0
x
Solution.
1
Note that we cannot distribute the limit over the product since lim cos
does not exist. Howx→0
x
ever, we do know that for all real numbers x 6= 0,
1
−1 ≤ cos
≤ 1.
x
1.6. TRIGONOMETRIC FUNCTIONS: LIMITS AND CONTINUITY;THE SQUEEZE THEOREM51
y = h(x)
y = g(x)
y = f (x)
Figure 1.6.1: An illustration of Squeeze Theorem
em
at
ic
s
Since x2 is nonnegative, we may multiply all sides of the inequality by x2 to obtain
1
2
2
−x ≤ x cos
≤ x2 .
x
at
h
Since
lim (−x2 ) = lim x2 = 0,
we conclude by Squeeze Theorem that
of
M
x→0
x→0
In
s
tit
ut
e
1
lim x cos
= 0.
x→0
x
2
UP
Example 1.6.4. Evaluate:
sin x
x→+∞ x
lim
Solution.
Note that lim sin x does not exist, while lim x = +∞. We will show, however, that the limit
x→+∞
x→+∞
of the quotient exists. Observe that whenever x > 0:
−1
1
−
x
≤
≤
sin x
sin x
x
≤
≤
1
1
x
1
sin x
1
Now, lim
−
= 0 = lim
. Thus, by Squeeze Theorem, lim
= 0.
x→+∞
x→+∞
x→+∞
x
x
x
52
CHAPTER 1. LIMITS AND CONTINUITY
Example 1.6.5. Evaluate:
2 [[x]] + 1
x→−∞
x
lim
Solution.
Note that lim [[x]] + 1 = −∞, while lim x = −∞. Observe that whenever x < 0:
x→−∞
x→−∞
x−1
2x − 2
2x − 1
2x − 1
x
≤
≤
≤
≤
≤
≤
x
2x
2x + 1
2x + 1
≥
≥
x
!
2 − x1
2x + 1
= 2, and lim
= lim
x→−∞
x→−∞
1
x
We now present the following special trigonometric limits.
sin x
=1
x
1 − cos x
2. lim
=0
x→0
x
!
= 2, we
at
h
Theorem 1.6.6.
2 + x1
1
s
2x − 1
Since lim
= lim
x→−∞
x→−∞
x
2 [[x]] + 1
= 2 by Squeeze Theorem.
get lim
x→+∞
x
em
at
ic
[[x]]
2 [[x]]
2 [[x]] + 1
2 [[x]] + 1
x
3. lim sin x = 0
1. lim
x→0
of
M
x→0
4. lim cos x = 1
ut
e
x→0
sin(4x)
x→0
x
In
s
tit
Example 1.6.7. Evaluate: lim
UP
Solution.
First of all, note that the argument of sine is 4x, while the denominator is simply x. To compute
the limit, we multiply both the numerator and denominator by 4. We have
sin(4x) 4
4 sin(4x)
· = lim
.
x→0
x
4 x→0
4x
lim
sin(4x)
= 1, and so,
x→0
4x
Note that x → 0 if and only if 4x → 0. Applying the theorem, we have lim
4 sin(4x)
= 4.
x→0
4x
lim
Remark 1.6.8.
1. lim
x
x→0 sin x
=1
This is because lim
x
x→0 sin x
= lim
1
x→0 sin x
x
=
1
= 1.
1
1.6. TRIGONOMETRIC FUNCTIONS: LIMITS AND CONTINUITY;THE SQUEEZE THEOREM53
2. lim
x
x→0 1 − cos x
6= 0
x
Similar to the previous item, lim
x→0 1 − cos x
= lim
1
x→0 1 − cos x
=
1
6= 0.
0
x
x2
Example 1.6.9. Evaluate: lim
x→0 sin(3x2 )
Solution.
x2
3
1
3x2
·
=
lim
·
=
x→0 sin(3x2 ) 3
x→0 3 sin(3x2 )
lim
1
1
(1) =
3
3
sin(x2 − 1)
x→−1
x+1
Example 1.6.10. Evaluate: lim
s
Solution.
em
at
ic
sin(x2 − 1) x − 1
sin(x2 − 1)
·
= lim
· (x − 1) = 1(−2) = −2
x→−1
x+1
x − 1 x→−1 x2 − 1
lim
tan x
x→0 x
of
M
at
h
Example 1.6.11. Evaluate: lim
Solution.
sin x
1
tan x
= lim
·
= 1(1) = 1
x→0 x
x→0 x
cos x
tit
ut
e
lim
In
s
sin(3x)
x→0 sin(5x)
UP
Example 1.6.12. Evaluate: lim
Solution.
sin(3x) 3x 5
3 sin(3x)
5x
·
· = lim ·
·
=
x→0 sin(5x) 3x 5
x→0 5
3x
sin(5x)
lim
3
3
(1)(1) =
5
5
tan(3x)
x→0+ 1 − cos2 (2x)
Example 1.6.13. Evaluate: lim
Solution.
lim
x→0+
tan(3x)
1 − cos2 (2x)
tan(3x)
sin2 (2x)
sin(3x)
3 2 x
= lim
· · ·
2
+
3
2 x
x→0 cos(3x) sin (2x)
3 sin(3x)
2x
1
= lim ·
·
·
3x
sin(2x)
cos(3x)
x→0+ 2
sin(2x)
3
1
= +∞
(1)(1)
2
(1)(0+ )
= lim
x→0+
54
CHAPTER 1. LIMITS AND CONTINUITY
1 − cos x
x→0
x2
Example 1.6.14. Evaluate: lim
Solution.
1 − cos x 1
1 − cos x
= lim
· , and whether one takes the limit as x → 0− or as
Note that lim
x→0
x→0
x2
x
x
x → 0+ , the form is 0 · ∞. We therefore wish to rewrite the function such that the resulting limit
is not indeterminate. The trick is to multiply the numerator and denominator by 1 + cos x:
1 − cos x 1 + cos x
1 − cos x
= lim
·
2
x→0
x→0
x
x2
1 + cos x
2
1
1 − cos x
·
= lim
2
x→0
x
1 + cos x
sin2 x
1
= lim
·
x→0 x2
1 + cos x
2
sin x
1
= lim
·
x→0
x
1 + cos x
1
= (1)2
2
1
= .
2
1.6.2
of
M
at
h
em
at
ic
s
lim
Continuity of Trigonometric Functions
tit
ut
e
The next theorem, which is a consequence of the Squeeze Theorem, deals with the continuity of
trigonometric functions.
In
s
Theorem 1.6.15.
x→a
UP
1. For all a ∈ R, lim cos x = cos a and lim sin x = sin a.
x→a
2. The trigonometric functions are continuous on their respective domains.
Remark 1.6.16. By the previous theorem, we can conclude that if f is a trigonometric function
and a ∈ dom f , then
lim f (x) = f (a).
x→a
This conclusion can also be seen by looking at the graphs of the trigonometric functions below.
Example 1.6.17. Evaluate the following limits.
1. limπ tan x
x→ 4
Solution.
limπ tan x = tan
x→ 4
π
= 1.
4
1.6. TRIGONOMETRIC FUNCTIONS: LIMITS AND CONTINUITY;THE SQUEEZE THEOREM55
f (x) = sin x
f (x) = cos x
f (x) = cot x
f (x) = sec x
f (x) = csc x
at
h
em
at
ic
s
f (x) = tan x
1 − cos x
x
e
x→π
ut
2. lim
of
M
Figure 1.6.2: Graphs of trigonometric functions
tit
Solution.
In
s
lim
1
−
lim
cos
x
1 − cos π
1 − cos x
2
=
lim
= x→π x→π
= .
x→π
x
π
π
lim x
UP
x→π
3. lim sin x2 − 1
x→1
Solution.
Since the sine function is continuous everywhere and lim (x2 − 1) = 0, we have
x→1
lim sin x2 − 1 = sin lim (x2 − 1) = sin 0 = 0.
x→1
x→1
1
4. lim cos
x→+∞
x
Solution.
Since the cosine function is continuous everywhere and lim
1
x→+∞ x
= 0, we have
1
1
lim cos
= cos
lim
= cos 0 = 1.
x→+∞
x→+∞ x
x
56
CHAPTER 1. LIMITS AND CONTINUITY
5. lim tan
x→0
1 − cos x
x
Solution.
1 − cos x
Since lim
= 0 and the tangent function is continuous at x = 0, we have
x→0
x
1 − cos x
1 − cos x
= tan lim
= tan 0 = 0.
lim tan
x→0
x→0
x
x
6. lim cot x
x→π +
Solution.
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
If f (x) = cot x, then π ∈
/ dom f , so we cannot use the preceding remark to compute this limit.
cos x
However, we may use the identity cot x =
. Since sin x → 0− as x → π from the right, we
sin x
have
cos x
−1
lim cot x = lim
= +∞.
+
+
0−
x→π
x→π sin x
1.6. TRIGONOMETRIC FUNCTIONS: LIMITS AND CONTINUITY;THE SQUEEZE THEOREM57
1.6.3
Exercises
Exercises for Discussion
A. Evaluate the following limits.
sin(x2 + x − 2)
x→1
x−1
sin 3x
2. lim
x→0 sin 4x
x
3. lim
x→0+ tan x
1 − cos 5x + tan 3x
4. lim
−
sin 2x
x→0
x tan 5x
5. lim
x→0 sin2 2x
6. lim 2x cos(2x) csc(4x)
3y − sin(4y)
y→0
y
7. lim
1. lim
x2 − 3 sin x
x→0
x
8. lim
tan3 (3x)
x→0
3x3
1
10. lim x 1 − cos
x→−∞
x
9. lim
sin(sin t)
t→0
t
11. lim
em
at
ic
s
x→0
B. Use the Squeeze Theorem to evaluate the following limits.
1. lim x sin
2.
lim
x→+∞
sin x + 3 cos x
x
of
M
x→0+
1
x2
at
h
UP
In
s
tit
ut
e
C. Test the following function for continuity at x = −3, −2, 0, and π:

x+2



 |x2 + 5x + 6| , if x < 0
x csc x
f (x) =
,
if 0 < x ≤ π


2


x + π,
if x > π
Supplementary Exercises
A. Evaluate the following limits.
1. lim x csc 3x
x→0
sin4 (2x)
x→0
4x4
sin(4x − 8)
3. lim 2
x→2 4x − 5x − 6
sin(4x)
4. lim
x→0
|x|
1
5. lim sin
x
x→0+
1
6. lim x sin
x→+∞
x2
t
7. lim
π
t→0 cos( − t)
2
2. lim
8. limπ
x→ 6
2 sin x − 1
6x − π
cos x − cos 3
x→3
x−3
sin πx
4 −1
10. lim
x→2
2x − 4
π−x
11. lim
x→π sin x
1 − cos 2x + tan2 x
12. lim
x→0
x sin 2x
(2x − π) cos x
13. limπ
1 − sin x
x→ 2
9. lim
58
CHAPTER 1. LIMITS AND CONTINUITY
B. Use the Squeeze Theorem to evaluate the following limits.
1.
2x3 + 1 + 5 cos x
x→−∞
3x3
2.
lim
C. Given:
 2
x −x−2


,


x2 − 1
sin 4x
f (x) =
,



 2 2x
x + 3x − 1,
[[2x − 1]]
x→+∞
x
lim
if x ≤ 0
if 0 < x ≤ 1
if x > 1
Discuss the continuity of f at x = −1, 0, and 1. Classify each discontinuity as jump essential,
infinite essential, or removable.
D. Miscellaneous Problems.
1. If 2x ≤ f (x) ≤ 3x2 − 16 for all x ≤ 0, find lim f (x).
s
x→−2
em
at
ic
2. Use the Squeeze Theorem to find lim g(x) given that
x→5
|g(x) − 8| < 3(x − 5)2 ,
R \ {5}.
at
h
for all x ∈
of
M
3. Prove that lim f (x) = 0 if and only if lim |f (x)| = 0.
x→0
x→0
4. Let f and g be functions such that |f (x)| ≤ 2 and lim |g(x)| = 0. Use the Squeeze
ut
e
Theorem to verify that lim f (x)g(x) = 0.
UP
In
s
tit
x→a
x→a
1.7. NEW CLASSES OF FUNCTIONS: LIMITS AND CONTINUITY
1.7
59
New Classes of Functions: Limits and Continuity
In this section, we study new classes of transcendental functions. We recall inverse functions and in
particular, exponential and logarithmic functions. We also revisit the inverse trigonometric functions. We review the properties and graphs of these functions.
After, we introduce new types of functions, the hyperbolic functions and inverse hyperbolic functions, and examine their properties and graphs. Our goal is to investigate the limits and continuity
of all aforementioned functions.
At the end of this section, the student will be able to:
• recall the basic concepts of inverse functions;
• recall the properties and graphs of exponential and logarithmic functions;
em
at
ic
s
• define and identify the properties of the inverse circular functions, the hyperbolic
functions, and the inverse hyperbolic functions;
at
h
• familiarize themselves with the graphs of the inverse circular functions, the hyperbolic functions, and the inverse hyperbolic functions;
Inverse Functions
In
s
1.7.1
tit
ut
e
of
M
• apply theorems on limits and continuity to exponential, logarithmic, inverse
circular, hyperbolic, and inverse hyperbolic function.
UP
Let us revisit the definition of inverse functions and their properties. This will prepare us in our
study of the exponential and logarithmic functions in the next subsection,
If the functions f and g satisfy
(f ◦ g)(x) = x for all x ∈ dom g, and
(g ◦ f )(x) = x for all x ∈ dom f ,
then we say f and g are inverse functions of each other. We write,
g = f −1 or g = f −1 .
Example 1.7.1. The functions f (x) = −2x + 4 and g(x) = − 12 x + 2 are inverse functions of each
other. Indeed,
(f ◦ g)(x) = f (− 12 x + 2) = −2(− 21 x + 2) + 4 = x, and
(g ◦ f )(x) = g(−2x + 4) = − 12 (−2x + 4) + 2 = x.
Hence, we can write f −1 (x) = 21 x + 2.
60
CHAPTER 1. LIMITS AND CONTINUITY
We note that not every function has an inverse. We define a class of functions that always possesses
an inverse.
A function f is a one-to-one function if for all x1 , x2 ∈ dom f with x1 6= x2 , f (x1 ) 6= f (x2 ).
Remark 1.7.2. (Horizontal Line Test) A function is one-to-one if and only if no horizontal line
intersects its graph more than once.
We present the following important theorem that will guarantee the existence of an inverse.
Theorem 1.7.3. A function f has an inverse if and and only if it is one-to-one.
Example 1.7.4. The function g(x) = x2 is not one-to-one. Indeed, g(1) = g(−1) = 1, so it violates
the definition of one-to-one function. Further, the horizontal line y = 1 intersects the graph of g in
two points. Based on the theorem above, g has no inverse function.
1. We have y = f (x) if and only if x = f −1 (y).
dom f −1 = ran f
ran f −1 = dom f .
of
M
and
at
h
2. The domains and range of f and f −1 are related. Indeed,
em
at
ic
s
Remark 1.7.5. Let f be a one-to-one function.
3. The graph of f −1 is obtained by reflecting the graph of f about the line y = x.
tit
ut
e
4. We have the following cancellation equations:
In
s
f −1 (f (x)) = x for all x ∈ dom f , f (f −1 (x)) = x for all x ∈ dom f −1 .
UP
5. To find the inverse function f −1 , we write y = f (x) and then solve for x in terms of y to obtain
x = f −1 (y). Finally, interchange x and y to get y = f −1 (x).
Example 1.7.6. Find the inverse of f (x) =
4x − 1
.
2x + 3
Solution.
4x − 1
. We then solve for x in terms of y. We get
Following the remark above, we write y =
2x + 3
3y + 1
3x + 1
f −1 (y) = x =
. Hence, f −1 (x) =
.
4 − 2y
4 − 2x
1.7.2
Exponential and Logarithmic Functions
In this section, we recall the exponential and logarithmic functions and review their properties.
These functions are important in studying different phenomena in the natural sciences, such as
population growth and decay, radioactive decay, acidity of chemicals and earthquakes.
Let a > 0 and a 6= 1. The exponential function with base a is given by f (x) = ax , where
x∈ .
R
1.7. NEW CLASSES OF FUNCTIONS: LIMITS AND CONTINUITY
61
Exponential functions can be classified into two types: those with base greater than 1, and those
with base between 0 and 1. The fundamental distinction between the two classes is that if a > 1,
then p < q implies ap < aq ; whereas if 0 < a < 1, then p < q implies ap > aq . The graphs of these
functions are found in Figure 1.7.1. Note that exponential functions are continuous everywhere.
It is good to remember the graphs of these functions as they serve as visual aids in remembering
their properties, such as their behavior as x → +∞ or as x → −∞, and where the functions are
increasing or decreasing.
f (x) = ax
0<a<1
f (x) = ax
(0, 1)
a>1
(0, 1)
em
at
ic
s
Figure 1.7.1: Graphs of Exponential Functions
1. dom f =
R and ran f = (0, +∞).
R.
of
M
2. f is one-to-one and continuous on
at
h
Remark 1.7.7. Let f (x) = ax with a > 0 and a 6= 1. Then we have the following:
R, x→+∞
lim ax = +∞ and lim ax = 0.
x→−∞
4. If 0 < a < 1, then f is decreasing on R, lim ax = 0 and lim ax = +∞.
x→+∞
x→−∞
tit
ut
e
3. If a > 1, then f is increasing on
UP
In
s
Because the exponential function with base a is one-to-one, it has an inverse function, called the
logarithmic function with base a.
Let a > 0 and a 6= 1. The logarithmic function with base a, denoted by loga , is the inverse
function of the exponential function with base a; that is,
y = loga x if and only if x = ay .
Hence, the logarithm of x to the base a is the exponent to which a must be raised to obtain x.
Following the inverse relationship of exponentials and logarithms, we have loga (ax ) = x for all
x ∈ and aloga x = x for all x > 0.
R
As with exponential functions, logarithmic functions can be classified into those with base greater
than 1 and those with base between 0 and 1. The graphs of these functions are given in Figure 1.7.2.
Again, note the continuity of these functions on their domains.
Remark 1.7.8. Let f (x) = loga x with a > 0 and a 6= 1. Then we have the following:
1. dom f = (0, +∞) and ran f =
R.
62
CHAPTER 1. LIMITS AND CONTINUITY
f (x) = loga x
a>1
f (x) = loga x
0<a<1
(1, 0)
(1, 0)
Figure 1.7.2: Graphs of Logarithmic Functions
2. f is one-to-one and continuous on (0, +∞).
3. If a > 1, then f is increasing on (0, +∞), lim loga x = +∞ and lim loga x = −∞.
x→+∞
x→0+
4. If 0 < a < 1, then f is decreasing on (0, +∞), lim loga x = −∞ and lim loga x = +∞.
x→+∞
em
at
ic
s
x→0+
Table 1.7.1 gives additional properties of exponentials and logarithms.
R a, b, m, n > 0, a, b 6= 1, c ∈ R
of
M
loga 1 = 0
loga a = 1
UP
ut
e
loga mn = loga m + loga n
m
loga
= loga m − loga n
n
c
log
a m = c loga m
1
loga
= − loga m
m
logb m
1
loga m =
=
logb a
logm a
tit
In
s
(ab)x = ax bx
a x ax
= x
b
b
1
−r
a = r
a
at
h
a, b > 0, r, s ∈
a0 = 1, a1 = a
ar as = ar+s
ar
= ar−s
as
(ar )s = ars
Table 1.7.1: Laws of Exponents and Logarithms
Example 1.7.9. Rewrite 22x · 5x as an exponential expression with a single base.
Solution.
By the laws of exponents, we have
22x · 5x = (22 )x · 5x = 4x · 5x = (4 · 5)x = 20x .
Example 1.7.10. Rewrite
272x
as an exponential expression in base 3.
(9x+2 )2−x
Solution.
Using laws of exponents, we get
272x
(9x+2 )2−x
=
(33 )2x
(3)6x
2
2
=
= 36x−(8−2x ) = 32x +6x−8 .
(32 )(2+x)(2−x)
(3)2(4−x2 )
1.7. NEW CLASSES OF FUNCTIONS: LIMITS AND CONTINUITY
63
Example 1.7.11. Find the exact value of log2 6 − log2 15 + log2 20.
Solution.
Because of the laws of logarithms, we have
log2 6 − log2 15 + log2 20 = log2
6
(6)(20)
+ log2 20 = log2
= log2 8 = 3.
15
15
Example 1.7.12. Solve for x: 32 log3 2 − 3x+2 = 2.
Solution.
Because of the properties of logarithms, inverse functions and laws of exponents, we have
2
32 log3 2 − 3x+2 = 3log3 2 − 32 · 3x = 4 − 9 · 3x .
Therefore, the equation becomes
9 · 3x = 2
2x + 3x
.
x→+∞
6x
⇐⇒
3x =
2
9
s
⇐⇒
em
at
ic
4 − 9 · 3x = 2
x = log3
2
.
9
at
h
Example 1.7.13. Evaluate lim
⇐⇒
of
M
Solution.
We recall that lim ax = 0, whenever 0 < a < 1. Thus, we get
x→+∞
x
x x 2
3x
1
1
+ x = lim
+
= 0.
x
x→+∞
6
6
3
2
In
s
tit
ut
e
2x + 3x
lim
= lim
x→+∞
x→+∞
6x
h→0
UP
In the next chapter, we will compute the derivative of logarithmic functions. In doing so, the
1
existence of lim (1 + h) h is crucial. This limit exists, but advanced concepts beyond the scope
of this course are needed to establish this fact rigorously. Instead, we estimate its value to a few
1
decimal places by computing (1 + h) h for values of h very near 0. The limit shall be called Euler’s
number and denoted by e.
h
0.5
0.1
0.01
0.0001
0.00000001
↓
0+
1
(1 + h) h
2.25
2.593742460
2.704813829
2.718145927
2.718281815
↓
e−
h
-0.5
-0.1
-0.01
-0.0001
-0.00000001
↓
0−
1
Euler’s number e is defined as e = lim (1 + h) h .
h→0
1
(1 + h) h
4
2.867971991
2.731999026
2.718417728
2.718281842
↓
e+
64
CHAPTER 1. LIMITS AND CONTINUITY
Remark 1.7.14. The number e is irrational, and to the first 15 decimal places,
e = 2.718281828459045 . . . .
1. The natural exponential function is given by f (x) = ex .
2. The natural logarithmic function is given by f (x) = ln x = loge x.
Remark 1.7.15. Let a, b > 0 and a, b 6= 1. Then we have the following:
1. ln ex = x for all x ∈
R.
x
4. ax = eln a = ex ln a = (ex )ln a for all x ∈
R.
2. eln x = x for all x > 0.
5. loga x =
3. ln e = 1.
ln x
logb x
=
for all x > 0.
logb a
ln a
at
h
1
.
1 − ex
em
at
ic
s
Items 4 and 5 in the previous remark show that any exponential or logarithmic function can
be expressed in terms of the natural exponential function or the natural logarithmic function,
respectively.
of
M
Example 1.7.16. Determine the domain of f (x) =
ut
e
Solution.
For the function to be defined, the denominator should not be zero. Now, the denominator is zero
exactly when ex = 1, that is, x = ln 1 = 0. Hence, dom f = \ {0}.
R
In
s
tit
Example 1.7.17. Express 3x+2 in terms of the natural exponential function.
UP
Solution.
Using the remark above, we have
3x+2 = eln (3
x+2 )
= (ex+2 )ln 3 .
Example 1.7.18. Express as a single logarithm: ln 5 + 2 ln 3.
Solution.
Applying the laws of logarithms, we have
ln 5 + 2 ln 3 = ln 5 + ln 32 = ln(5)(32 ) = ln 45.
Example 1.7.19. Solve for x: ln(x + 1) − ln(x − 2) = 2.
Solution.
We have ln(x + 1) − ln(x − 2) = ln
ln
x+1
x−2
=2
⇐⇒
x+1
. Thus,
x−2
x+1
= e2
x−2
⇐⇒
x + 1 = e2 x − 2e2
⇐⇒
x=
2e2 + 1
.
e2 − 1
1.7. NEW CLASSES OF FUNCTIONS: LIMITS AND CONTINUITY
65
e−x − 2
.
x→−∞ e−x + 2
Example 1.7.20. Evaluate lim
Solution.
Recall that since e > 1 we have lim ex = 0. Thus, we have
x→−∞
e−x − 2
e−x − 2 ex
1 − 2ex
1−0
=
lim
·
=
lim
=
= 1.
x→−∞ e−x + 2
x→−∞ e−x + 2 ex
x→−∞ 1 + 2ex
1+0
lim
1.7.3
Inverse Circular Functions
We now apply techniques of basic calculus to another familiar class of functions: the inverses of the
circular functions. Recall that the circular functions are periodic; a fortiori they are not one-to-one.
In order to define the inverse of each such function, we restrict the domain of each to a set over
which the corresponding circular function is one-to-one.
em
at
ic
s
1. The inverse sine function, denoted sin−1 , is defined as follows:
y = sin−1 x if and only if x = sin y, y ∈ − π2 , π2 .
at
h
2. The inverse cosine function, denoted cos−1 , is defined as follows:
of
M
y = cos−1 x if and only if x = cos y, y ∈ [0, π] .
In
s
tit
ut
e
3. The inverse tangent function, denoted tan−1 , is defined as follows:
π π
y = tan−1 x if and only if x = tan y, y ∈ − ,
.
2 2
UP
4. The inverse cotangent function, denoted cot−1 , is defined as follows:
y = cot−1 x if and only if x = cot y, y ∈ (0, π) .
5. The inverse secant function, denoted sec−1 , is defined as follows:
h π 3π −1
y = sec x if and only if x = sec y, y ∈ 0,
∪ π,
.
2
2
6. The inverse cosecant function, denoted csc−1 , is defined as follows:
πi πi
y = csc−1 x if and only if x = csc y, y ∈ −π, −
∪ 0,
.
2
2
66
CHAPTER 1. LIMITS AND CONTINUITY
Example 1.7.21. Evaluate the following.
√ !
3
2
1. sin−1
3. tan−1
√ 2. sec−1 − 2
√ 4. csc−1 − 2
Solution.
1. sin−1
√ !
3
−
3
3. tan−1
√ !
π
3
=
2
3
√ 5π
2. sec−1 − 2 =
4
√ !
3
π
−
=−
3
6
√ 3π
4. csc−1 − 2 = −
4
Example 1.7.22.
s
1. Evaluate: sin
9π
.
sin
7
em
at
ic
−1
e
of
M
at
h
Solution.
9π
2π
2π
−1
−1
sin
sin
= sin
sin −
=− .
7
7
7
1
.
2. Evaluate: sin cos−1
4
ut
Solution.
tit
UP
given as
1
1
. Then, cos k = where k ∈ ran cos−1 = [0, π] . We may now rewrite the
4
4
−1 1
sin cos
= sin k.
4
In
s
Let k = cos−1
Solving for sin k, we have that
sin2 k = 1 − cos2 k,
and so
sin k =
p
r
1 − cos2 k =
1
1−
=
16
since sin k ≥ 0 for k ∈ [0, π]. Thus,
√
15
−1 1
sin cos
=
.
4
4
"
−1
3. Evaluate: sec sin
Solution.
√ !#
3
.
5
√
15
4
1.7. NEW CLASSES OF FUNCTIONS: LIMITS AND CONTINUITY
67
√
√ !
h π πi
3
3
. Then, sin k =
Let k = sin
where k ∈ ran sin−1 = − , . We may now
5
5
2 2
rewrite the given as
"
√ !#
3
−1
sec sin
= sec k.
5
√
3
Before solving for sec k, observe that sin k =
≥ 0 indicates that k ∈ [0, π]. Thus,
5
h π πi
h πi
k∈ − ,
∩ [0, π] = 0, .
2 2
2
Now, solving for sec k, we have that
−1
sec2 k = 1 + tan2 k,
so that
em
at
ic
s
r
p
sin2 k
sec k = 1 + tan2 k = 1 +
2
v cos k
s
2
u
√
u
3/5
sin2 k
= t1 +
= 1+
2
√
1 − sin2 k
1 − 3/5
of
M
at
h
5
=√
22
h πi
since sec k > 0 for k ∈ 0, . Thus,
2
e
"
In
s
tit
ut
sec sin−1
√ !#
3
5
=√
5
22
UP
The graphs of the inverse circular functions defined above are shown in Figure 1.7.3. From these
graphs, one can predict the behavior of each inverse circular function and deduce some limit statements involving them.
Example 1.7.23.
1. From the graph of tan−1 x, we can infer the following statements involving limits at infinity:
lim tan−1 x =
x→ +∞
π
2
and
π
lim tan−1 x = − .
x→ −∞
2
Similarly, we infer the following from the graphs of sec−1 x and csc−1 x:
lim sec−1 x =
x→ +∞
lim csc−1 x = 0
x→ +∞
π
,
2
lim sec−1 x =
x→ −∞
and,
3π
,
2
lim csc−1 x = −π.
x→ −∞
68
CHAPTER 1. LIMITS AND CONTINUITY
y = 3π
2
(1, π2 )
y = π2
(−1, π)
f (x) = sec−1 x
f (x) = tan−1 x
y = π2
f (x) = sin−1 x
y = − π2
(−1, − π2 )
(1, 0)
(1, π2 )
(−1, π)
f (x) = cos−1 x
y=π
(−1, − π2 )
f (x) = cot−1 x
f (x) = csc−1 x
y = −π
s
(1, 0)
x→ +∞
1
π.
2
tan−1 x −
at
h
lim
of
M
2. Evaluate:
em
at
ic
Figure 1.7.3: Graphs of inverse circular functions
Solution.
π
π
from values less than . Thus,
2
2
1
1
lim
= −∞
π
−1
x→ +∞ tan
0−
x− 2
lim tan−1 (ex ).
x→ −∞
UP
3. Evaluate:
In
s
tit
ut
e
Note that as x → +∞, tan−1 x approaches
Solution.
Observe from the graph of ex that
lim ex = 0. Now,
x→ −∞
−1
lim tan
x→ −∞
x
(e ) = tan
−1
lim e
x
x→ −∞
= tan−1 (0) = 0.
Passing the limit to the inner function is permissible in this situation due to the continuity of
tan−1 x at x = 0.
1.7.4
Hyperbolic Functions
We now introduce another new class of functions, arising from the natural exponential function.
They are called hyperbolic functions and have properties very similar to the circular functions.
Each circular function has a hyperbolic analogue, and we will see in Theorems 1.7.25 that identities
1.7. NEW CLASSES OF FUNCTIONS: LIMITS AND CONTINUITY
69
involving these new functions will look very much like those involving their corresponding circular
analogues.
1. The hyperbolic sine function, denoted sinh, is defined by
sinh x =
for any x ∈
R.
ex − e−x
,
2
2. The hyperbolic cosine function, denoted cosh, is defined by
cosh x =
for any x ∈
R.
ex + e−x
,
2
3. The hyperbolic tangent function, denoted tanh, is defined by
R.
s
at
h
for any x ∈
ex − e−x
,
ex + e−x
em
at
ic
tanh x =
of
M
4. The hyperbolic cotangent function, denoted coth, is defined by
coth x =
tit
ut
e
for any x 6= 0.
ex + e−x
,
ex − e−x
In
s
5. The hyperbolic secant function, denoted sech, is defined by
UP
for any x ∈
sech x =
2
ex + e−x
,
R.
6. The hyperbolic cosecant function, denoted csch, is defined by
csch x =
2
ex − e−x
for any x 6= 0.
Example 1.7.24.
1. sinh 0 =
e0 − e−0
1−1
=
=0
2
2
1
2. cosh(ln 2) =
eln 2 + eln 2
2 + 12
eln 2 + e− ln 2
5
=
=
=
2
2
2
4
,
70
CHAPTER 1. LIMITS AND CONTINUITY
y=1
f (x) = sinh x
f (x) = tanh x
f (x) = sech x
(0, 1)
y = −1
f (x) = coth x
f (x) = csch x
y=1
y = −1
em
at
ic
s
(0, 1) f (x) = cosh x
Figure 1.7.4: Graphs of hyperbolic functions
tit
ut
Identities Involving Hyperbolic Functions
of
M
at
h
1
25 − 25
eln 25 − e− ln 25
624
eln 25 − eln(1/25)
312
=
=
=
=
.
1/25)
1
ln
25
−
ln
25
ln(
ln
25
626
313
e
+e
e
+e
25 + 25
e
3. tanh(2 ln 5) = tanh(ln 25) =
1
1
1
, csch x =
, coth x =
cosh x
sinh x
tanh x
UP
1. sech x =
In
s
Theorem 1.7.25.
5. cosh2 x − sinh2 x = 1
sinh x
cosh x
, coth x =
cosh x
sinh x
x
3. cosh x + sinh x = e
6. 1 − tanh2 x = sech2 x
4. cosh x − sinh x = e−x
7. 1 − coth2 x = − csch2 x
2. tanh x =
8. sinh(x ± y) = sinh x cosh y ± cosh x sinh y
9. cosh(x ± y) = cosh x cosh y ± sinh x sinh y
10. sinh 2x = 2 sinh x cosh x
11. cosh 2x = cosh2 x + sinh2 x = 1 + 2 sinh2 x = 2 cosh2 x − 1
Proof. We shall prove statements 3, 5 and 8 only.
1.7. NEW CLASSES OF FUNCTIONS: LIMITS AND CONTINUITY
cosh x + sinh x =
71
ex + e−x ex − e−x
ex + e−x + ex − e−x
+
=
= ex
2
2
2
cosh2 x − sinh2 x = (cosh x + sinh x)(cosh x − sinh x) = ex · e−x = 1
x
y
x
y
e − e−x
e + e−y
e + e−x
e − e−y
sinh x cosh y + cosh x sinh y =
+
2
2
2
2
x+y
x−y
−x+y
−x−y
x+y
x−y
e
+e
−e
−e
+e
−e
+ e−x+y − e−x−y
=
4
x+y
−x−y
x+y
−(x+y)
2e
− 2e
e
−e
=
=
= sinh(x + y)
4
2
Example 1.7.26. If tanh x =
12
, find the values of the other hyperbolic functions of x.
13
at
h
em
at
ic
s
Solution.
13
We readily have coth x = . Now, since sech x > 0 for all x, using the identities above we obtain
12
s
2
p
5
12
2
= .
sech x = 1 − tanh x = 1 −
13
13
of
M
13
. Using another identity, we get
5
e
This implies that cosh x =
In
s
5
.
12
UP
Finally, we get csch x =
12
12 13
·
= .
13 5
5
tit
ut
sinh x = tanh x cosh x =
Why they are called Hyperbolic Functions
In the same way that points with coordinates (cos t, sin t) are on the unit circle, the points with
coordinates (cosh t, sinh t) are on the unit hyperbola, which has equation x2 − y 2 = 1. In particular,
they are on the right “branch” of the hyperbola. Those on the other branch have coordinates
(− cosh t, sinh t). See Figure 1.7.5.
(cos t, sin t)
(cosh t, sinh t)
x2 − y 2 = 1
x2 + y 2 = 1
Figure 1.7.5: Circular functions versus Hyperbolic functions
72
CHAPTER 1. LIMITS AND CONTINUITY
Applications of Hyperbolic Functions
As mentioned earlier, hyperbolic functions have several applications, such as understanding the
behavior of hanging cables, electric current, and waves. A telephone or electrical wire suspended
between fixed ends at the same height forms a curve described by a function involving the hyperbolic
cosine function. We clarify that the graph of y = a cosh[b(x − c)] + d is not a parabola, but is
a catenary, from the Latin word catena for chain. In fact, it can be shown that a catenary
always outgrows a parabola having the same vertex and opening in the same direction. Hyperbolic
functions are also used in describing current flow in electrical wires. Similarly, the hyperbolic
tangent function is used in models describing the velocity of (idealized) ocean waves.
1.7.5
Inverse Hyperbolic Functions
em
at
ic
s
We recall that the inverse of a function can be defined only when the function is one-to-one. Among
the hyperbolic functions, only the hyperbolic cosine and secant functions are not one-to-one. We
restrict the domain of these functions to [0, +∞), on which the two functions become one-to-one.
f (x) = tanh−1 x
x = −1
x=1
f (x) = sech−1 x
In
s
UP
f (x) = sinh−1 x
tit
ut
e
of
M
at
h
Now, we may define an inverse function for each hyperbolic function. The graphs of the inverse
hyperbolic functions, obtained by reflecting about the line y = x the graphs of the corresponding
hyperbolic functions, are given in Figure 1.7.6.
(1, 0)
f (x) = csch−1 x
f (x) = coth−1 x
f (x) = cosh−1 x
(1, 0)
x = −1
x=1
Figure 1.7.6: Graphs of inverse hyperbolic functions
Since the hyperbolic functions are constructed using exponential functions, we expect that their
inverses can be written in terms of logarithms. In fact, the following hold.
1.7. NEW CLASSES OF FUNCTIONS: LIMITS AND CONTINUITY
Theorem 1.7.27.
√
1. sinh−1 x = ln x + x2 + 1 for all x ∈
R
√
2. cosh−1 x = ln x + x2 − 1 for all x ∈ [1, +∞)
1+x
3. tanh
for all x ∈ (−1, 1)
1−x
1
x+1
−1
4. coth x = ln
for all x ∈ (−∞, −1) ∪ (1, +∞)
2
x−1
!
√
1 + 1 − x2
−1
for all x ∈ (0, 1)
5. sech x = ln
x
x = ln
1
+
x
√
1 + x2
|x|
!
for all x ∈
R\{0}
√
Example 1.7.28. Prove that sinh−1 x = ln(x +
x2 + 1).
at
h
Proof. Let y = sinh−1 x. Then sinh y = x, and
s
6. csch
−1
1
x = ln
2
em
at
ic
−1
of
M
ey − e−y
2
2xey = e2y − 1
e
x=
UP
In
s
tit
ut
0 = (ey )2 − 2x (ey ) − 1
p
2x ± (−2x)2 − 4(1)(−1)
y
e =
2
p
y
2
e =x± x +1
p
p
ey = x + x2 + 1,
since ey > 0, whereas x − x2 + 1 < 0 .
√
Finally, taking the natural logarithm of both sides yields y = ln x + x2 + 1 .
Example 1.7.29. Find the numerical value of the following.
1. cosh−1 1
−1 5
2. coth
4
3. sinh−1 1
−1 3
4. sech
5
Solution.
√
1. cosh−1 (1) = ln 1 + 1 − 12 = ln 1 = 0
2. coth
−1
5
1
= ln
4
2
5
4 +1
5
4 −1
!
1
= ln
2
9
4
1
4
!
=
1
ln 9 = ln 3
2
73
74
CHAPTER 1. LIMITS AND CONTINUITY
√
√ 3. sinh−1 (1) = ln 1 + 1 + 12 = ln 1 + 2
q

1
+
1−
3
4. sech−1
= ln 
3
5
5

3 2
5

1 + 45
= ln
!
3
5
= ln
9
5
3
5
!
= ln 3
Example 1.7.30.
1. Observe from the graph of tanh−1 that the following hold:
lim tanh−1 x = +∞,
and
x→ 1−
2. Evaluate:
lim ecoth
x→ −∞
−1
x
lim tanh−1 x = −∞.
x→ 1+
.
Solution.
lim ex = 0. Furthermore, ex is continuous on R. Thus,
s
x→ −∞
we have:
lim e
x→ −∞
lim coth−1 x
= ex→ −∞
= e0
UP
In
s
tit
ut
e
of
M
= 1.
at
h
coth−1 x
em
at
ic
Observe from the graph of ex that
1.7. NEW CLASSES OF FUNCTIONS: LIMITS AND CONTINUITY
1.7.6
75
Exercises
Exercises for Discussion
Evaluate the following limits.
A.
1. lim
x→0+
2.
3.
4.
3e1/x + 2sin x
4
10.
2
11. lim cot−1 (ln x)
lim ex −x +1
x→−∞
lim
x→+∞
lim
x→−∞
e
4x
−e
x→0+
2x
+ 2e
−2x
e6x + 2e−2x − 8e
12.
−3x
13.
2ex − 1
x→+∞ ex + 2
5x
6. lim x
x→+∞ 3 + 4x
4x + 2 x
7. lim x
x→−∞ 8 − 2x
4e4x − e−2x
8. lim
x→+∞ 6e4x − e2x + 3e−x
4e4x − e−2x
9. lim
x→−∞ 6e4x − e2x + 3e−x
cosh x
x→+∞
ex
x
2. lim e sinh x
lim
5. lim sec−1 (csch x)
x→0+
6.
−1
lim tan
x→−∞
(cosh x)
lim sin
x→+∞
sec−1
π
lim csc−1 ex + x2
x→−∞
s
em
at
ic
at
h
16.
17.
e
3 csch(3x)
x→−∞ 2 sech(2x)
−1
x→0−
tit
x→+∞
In
s
4.
lim sinh(2x) sech(4x)
lim cos−1 (x − 1)
x→+∞
15. lim sec−1 (1/x)
lim
x→−∞
3.
14.
ut
1.
UP
B.
lim
of
M
5.
lim ln(7x3 − x2 )
x→+∞
7.
lim sec
x→+∞
2 tan−1 2x
3
1
x→−∞ π + 2 tan−1 x
lim
lim coth(x3 − x + 2)
x→−∞
8. lim tanh csch−1 (sinh−1 x)
x→0−
9. lim 2csch
−1
x
x→0−
10. lim ln(sech−1 x)
x→0+
csch−1 x
1
11. lim
+
2
x→0
76
CHAPTER 1. LIMITS AND CONTINUITY
Supplementary Exercises
A. Find the exact value of the following.
1. log3 100 − log3 18 − log3 50
5. sech(ln 2)
2. 22 log2 3 + ln(− ln ee ) − e−2 ln 2
6. sinh(3 ln 2 − ln 4)
csc−1 (−4)
3. tan
2
1
3
−1
−1
4. sin tan
−
− sec
4
2
7. cosh(ln 5 + ln 6)
8. tanh−1 0.5
B. Given below is a value of the hyperbolic function of a positive number x. Find the exact
value of the remaining five hyperbolic functions of x.
7
25
4
4. sech x =
5
1. sinh x = 2
tanh x + tanh y
1 + tanh x tanh y
1
+
tanh
x
2. e2x =
1 − tanh x
of
M
at
h
C. Establish the following identities.
3. cosh 3x = 4 cosh3 x − 3 cosh x
r
x
1 + cosh x
4. cosh =
2
2
tit
ut
e
1. tanh(x + y) =
UP
In
s
D. Do as indicated.
1. Find the domain of
s
15
8
em
at
ic
2. cosh x =
3. coth x =
a. f (x) = sin(e−x )
b. g(x) =
√
2 − 2x
h
f (x + h) − f (x)
5 −1
x
=5
.
h
h
3. Find the exponential function f (x) = Cax whose graphs passes (1, 6) and (3, 24).
2. If f (x) = 5x , show that
4. Show that the horizontal lines y = 1 and y = −1 are asymptotes of y = tanh x and
y = coth x.
5. Show that the graph of y = sech x is asymptotic to the x-axis.
6. If x = ln(sec s + tan s), show that sec s = cosh x.
1.7. NEW CLASSES OF FUNCTIONS: LIMITS AND CONTINUITY
77
Reviewer
I. Evaluate the following limits.
√
5 − x2 − 1
1. lim 2
x→2 x − 3x + 2
√
2x2 − 7 − 1
2. lim
x→2 x2 − x − 2
8x + 3
√
3. lim
x→−∞ 2x − 4x2 + 1
1 − 3x
4. lim √
x→−∞
4x2 − 3 − 3x
5.
6.
x−9
x
+
2
2x + 3x − 2 x + 2
1
3
− 2
2
x − 9 3x + 8x − 9
lim
x→−2−
lim
x→−3+
tan(x2 − x − 2)
x→2 (x − 2) cosh(x − 2)
7. lim
sin(y 2 − y)
x→0+
y(21/y )
8. lim
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
II. Determine if the following functions are continuous or discontinuous at the given values of
x. If discontinuous, classify the type of discontinuity as removable, jump essential, or infinite
essential.
 2
x −1


if x ≤ 0

 |x + 1| ,
1. f (x) = [[2x − 1]], if 0 < x ≤ 1 at x = −1, 0, and 1.




sin(ln x), if x > 1
√ 2

 x + 5,
if x < 0



x
if 0 ≤ x ≤ 1 at x = 0, 12 , and 1.
2. g(x) = [[2x − 2]],


 2x2 − 5x + 3


, if x > 1
|1 − x|
 2
x + 2x


if x ≤ −1

 |x + 2| ,
3. h(x) = [[2x + 1]],
if − 1 < x ≤ 0 at x = −2, −1, and 0.




ln(2 − cos x), if x > 0
III. Suppose that the graph of y = f (x) is given by the figure below.
1. Evaluate lim f (x).
x→2−
4
2. Identify the type of discontinuity
(removable, jump essential or infinite
essential) at x = 2.
3
2
3. Evaluate lim f (x).
x→+∞
1
−1
1
−1
2
3
4
4. TRUE or FALSE: The line y = 3 is
a horizontal asymptote of the graph of
y = f (x).
78
CHAPTER 1. LIMITS AND CONTINUITY
IV. Use Intermediate Value Theorem to show that each of the following functions has a root on
the given interval.
1. f (x) = x3 − 3x − 1 on [0, 2]
1
on (−∞, 0)
2. g(x) = x6 − 6x +
x−6
x √
3. h(x) = 2 log3
+ 3x on (0, 3)
3
V. Use Squeeze Theorem to evaluate the following limits.
1.
lim sin x csch x
x→+∞
ex (sin x + cos x)
x→−∞
x
sin x cos x
3. lim
x→−∞ x(4/5)x
2.
lim
x→0
x
e
2
lim 3tanh x
6. lim cot π 1/x
x→0−
1
7. lim cosh
ln x
x→0+
5.
2. lim tan−1 (log5 x)
of
M
x→1
3.
lim log 1 (x2 − 9)
x→−3−
x→+∞
at
h
1. lim cos
−1
em
at
ic
s
VI. Evaluate the following limits.
2
e
4. lim ln(sinh x)
UP
In
s
tit
ut
x→0+
8.
lim tan−1 cosh−1 x
x→+∞
Chapter 2
Derivatives and Differentiation
2.1
Slopes, the Derivative, and Basic Differentiation Rules
at
h
em
at
ic
s
In this section, we extend the notion of a tangent line to a circle to other curves at a point. Recall
from Euclidean Geometry that a line tangent to a circle is a line that intersects the circle at exactly
one point. However, in a more general sense, a line tangent to a curve may intersect the curve at
points other than the point of tangency.
of
M
At the end of this section, the student will be able to:
• find the equation of the tangent line to the graph of a function at a given point;
ut
e
• define and interpret the derivative of a function at a given point;
In
s
tit
• compute the derivatives of algebraic functions using power rule, sum/difference
rule, product rule, quotient rule; and
UP
• differentiate functions involving trigonometric functions;
2.1.1
The Tangent Line
`
y = f (x)
Given a function f (x), we want to define the equation of
the tangent line ` at a point, say P (x0 , f (x0 )) on the graph
of y = f (x).
P
Figure 2.1.1
To do this, we need to find the slope of this tangent line. Consider another point Q(x1 , f (x1 )) on
←→
the graph of y = f (x). Form the secant line P Q.
79
80
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
←→
Then P Q has slope
Q
f (x1 )
y = f (x)
)
∆y
f (x0 )
→ =
m←
PQ
P
|
{z
∆x
f (x1 ) − f (x0 )
.
x1 − x0
}
Letting ∆x = x1 − x0 , the above expression is
equivalent to
x0
x1
Figure 2.1.2
→ =
m←
PQ
f (x0 + ∆x) − f (x0 )
.
∆x
`
y = f (x)
Now, imagine the point Q moving along the curve y = f (x)
toward P .
s
em
at
ic
As Q approaches P , the secant line approaches an object in
the plane. This object coincides with the line `.
P
at
h
Figure 2.1.3
of
M
→ of the secant line as Q gets
We define the slope m of the line ` to be the limit of the slope m←
PQ
closer and closer to P and call ` the tangent line to the graph of f at P . We have
→.
m = lim m←
PQ
ut
e
Q→P
tit
Observe that as Q → P , x1 → x0 and so ∆x = x1 − x0 → 0. Hence,
In
s
f (x0 + ∆x) − f (x0 )
.
∆x→0
∆x
UP
m = lim
If the function f is defined on an open interval containing x = x0 , then the tangent line to the
graph of f at the point P (x0 , f (x0 )) is the line
1. passing through P whose slope is given by
m = lim
∆x→0
f (x0 + ∆x) − f (x0 )
,
∆x
provided that this limit exists.
2. with equation x = x0 if
f (x0 + ∆x) − f (x0 )
∆x
∆x→0−
f (x0 + ∆x) − f (x0 )
lim
∆x
∆x→0+
lim
= +∞ or − ∞ and
= +∞ or − ∞.
Otherwise, there is no tangent line to the graph of f at P .
2.1. SLOPES, THE DERIVATIVE, AND BASIC DIFFERENTIATION RULES
81
Remark 2.1.1.
1. The slope of the tangent line to the graph of f at P gives us an idea of the “flatness” or
“steepness” of the graph of f at P and whether the graph of f rises or falls at P . See Figure
2.1.4.
2. The tangent line to the graph of a function may intersect the graph at points other than the
point of tangency as shown in Figure 2.1.5.
P
`
TL
NL
P
0
Figure 2.1.5
0
Figure 2.1.6
Figure 2.1.4
em
at
ic
s
The normal line to the graph of f at the point P is the line perpendicular to the tangent line at
P . (See Figure 2.1.6.)
of
M
at
h
1
Example 2.1.2. Give equations of the tangent line and the normal line to the graph of f (x) =
x
at x = 1.
ut
e
Solution.
Let mT L denote the slope of the tangent line to the graph of f (x) at x = 1. We have
UP
In
s
tit
mT L
=
=
=
=
=
=
f (1 + ∆x) − f (1)
∆x
1
−
1
lim 1+∆x
∆x→0
∆x
1 − (1 + ∆x)
lim
∆x→0 ∆x(1 + ∆x) −∆x
0
lim
∆x→0 ∆x(1 + ∆x)
0
−1
lim
∆x→0 1 + ∆x
−1
lim
∆x→0
Thus, the equation of the tangent line to the graph of f (x) = x1 at (1, 1) is y − 1 = −(x − 1). Now,
since mT L = −1, the slope of the normal line is 1 and hence, the equation of the normal line to the
graph of f (x) = x1 at x = 1 is y − 1 = (x − 1), or simply y = x.
Example 2.1.3. Find the slope of the tangent line to the graph of g(x) = x2 + 2 at x = 1 and at
x = 2.
Solution.
At x = 1 the slope of the tangent line to the graph of g(x) is
g(1 + ∆x) − g(1)
mT L = lim
∆x→0
∆x
82
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
[(1 + ∆x)2 + 2] − (12 + 2)
∆x→0
∆x
[1 + 2∆x + (∆x)2 + 2] − 3
= lim
∆x→0
∆x
2∆x + (∆x)2
= lim
∆x→0
∆x
∆x(2 + ∆x)
= lim
∆x→0
∆x
= lim (2 + ∆x)
= lim
∆x→0
= 2.
of
M
at
h
em
at
ic
s
On the other hand, at x = 2 the slope of the tangent line to the graph of g(x) is
g(2 + ∆x) − g(2)
mT L = lim
∆x→0
∆x
[(2 + ∆x)2 + 2] − (22 + 2)
= lim
∆x→0
∆x
[4 + 4∆x + (∆x)2 + 2] − 6
= lim
∆x→0
∆x
4∆x + (∆x)2
= lim
∆x→0
∆x
= lim (4 + ∆x)
∆x→0
ut
e
= 4.
UP
In
s
tit
In fact, we can compute for the slope of the tangent line to the graph of g(x) for any value of x.
We have
g(x + ∆x) − g(x)
mT L = lim
∆x→0
∆x
[(x + ∆x)2 + 2] − (x2 + 2)
= lim
∆x→0
∆x
[x2 + 2x∆x + (∆x)2 + 2] − (x2 + 2)
= lim
∆x→0
∆x
2
2x∆x + (∆x)
= lim
∆x→0
∆x
= lim (2x + ∆x)
∆x→0
= 2x.
Using this formula, we see that the slope at x = 1 is 2 · 1 = 2 while the slope at x = 2 is 2 · 2 = 4,
which agree with the preceding calculations.
2.1.2
Definition of the Derivative
In the previous example, we obtained a function that gives the slope of the tangent line to the
graph of a function g(x) at any value of x. This function shall be called the derivative of g(x).
2.1. SLOPES, THE DERIVATIVE, AND BASIC DIFFERENTIATION RULES
83
The derivative of a function f (x), denoted f 0 (x), is the function
f (x + ∆x) − f (x)
.
∆x→0
∆x
f 0 (x) = lim
It is defined at all points x in the domain of f where the limit exists.
Remark 2.1.4.
1. Hence, from the definition, dom f 0 ⊆ dom f since there may be points x0 ∈ dom f at which
f 0 (x0 ) does not exist.
2. The definition also tells us that f 0 (x0 ) is the slope of the tangent line to the graph of the function
at the point P (x0 , f (x0 )).
3. To get the derivative of f at x = x0 , we use
f (x0 + ∆x) − f (x0 )
.
∆x→0
∆x
em
at
ic
s
f 0 (x0 ) = lim
Alternatively, by setting ∆x = x − x0 , we have
f (x) − f (x0 )
.
x − x0
at
h
f 0 (x0 ) = lim
dy
d
,
[f (x)] , Dx [f (x)].
dx dx
ut
e
4. Other notations: y 0 if y = f (x),
of
M
x→x0
tit
5. The process of computing the derivative is called differentiation.
UP
In
s
Example 2.1.5. If g(x) = x2 + 2, then, following the computations in the previous example, we
have g 0 (x) = 2x.
Example 2.1.6. Find the derivative of f (x) =
√
x.
Solution.
Using the definition of the derivative, we have
√
√
√
√
√
√
x + ∆x − x
x + ∆x − x
x + ∆x + x
0
f (x) = lim
= lim
·√
√
∆x→0
∆x→0
∆x
∆x
x + ∆x + x
(x + ∆x) − x
1
1
√
= lim
√ = lim √
√ = √
∆x→0 ∆x( x + ∆x + x)
∆x→0
2
x
x + ∆x + x
1
Thus, f 0 (x) = √ . Note that dom f = [0, +∞) while dom f 0 = (0, +∞).
2 x
2.1.3
Differentiability
This part discusses what it means for a function to be differentiable. We start with the following
definition.
84
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
1. A function f is said to be differentiable at x = x0 if the derivative f 0 (x0 ) exists.
2. A function f is differentiable on (a, b) if f is differentiable at every real number in (a, b).
3. A function f is differentiable everywhere if it is differentiable at every real number.
√
√
1
Example 2.1.7. From the preceding example, Dx [ x ] = √ and so f (x) = x is differentiable
2 x
at any positive real number x. Also, f is not differentiable at x = 0.
Example 2.1.8. It will be shown later that if f (x) is a polynomial, a rational, or a trigonometric
function, then f (x) is differentiable on its domain.
2.1.4
Techniques of Differentiation
of
M
at
h
em
at
ic
s
The previous section introduced a method of differentiation using Definition 2.1.2. The drawback
of such a method is that the process involved is tedious, even for the differentiation of relatively
simple functions such as in Example 2.1.2. The theorems introduced in this chapter will greatly
simplify the process of finding the derivative of a function, by providing formulas for the derivatives
of algebraic and some transcendental functions.
e
Differentiation Rules
tit
ut
The following theorem introduces basic rules in finding the derivative of a function.
R, then f 0(x) = 0.
UP
1. If f (x) = c ∈
In
s
Theorem 2.1.9. Let f and g be functions and c ∈
2. (Power Rule) If f (x) = xn , where n ∈
R.
Q, then f 0(x) = nxn−1.
3. If f (x) = c · g(x), then f 0 (x) = c · g 0 (x) if g 0 (x) exists.
4. (Sum Rule) If h(x) = f (x) ± g(x), then h0 (x) = f 0 (x) ± g 0 (x), provided both f 0 (x) and
g 0 (x) exist.
5. (Product Rule) If h(x) = f (x)g(x), then h0 (x) = f 0 (x)g(x)+f (x)g 0 (x), provided f 0 (x)
and g 0 (x) both exist.
f (x)
g(x)f 0 (x) − f (x)g 0 (x)
, where g(x) 6= 0, then h0 (x) =
,
g(x)
[g(x)]2
provided f 0 (x) and g 0 (x) both exist.
6. (Quotient Rule) If h(x) =
Proof. We show the proof of the first two statements only.
c−c
0
= lim
= lim 0 = 0.
∆x→0 ∆x
∆x→0 ∆x
∆x→0
1. Dx [c] = lim
2.1. SLOPES, THE DERIVATIVE, AND BASIC DIFFERENTIATION RULES
85
2. We make use of the Binomial Theorem1 in the proof of statement 2.
Dx [xn ] =
=
(x + ∆x)n − xn
∆x→0
∆x
lim
n−2 ∆x2 + . . . + nx∆xn−1 + ∆xn − xn
xn + nxn−1 ∆x + n(n−1)
x
2
lim
∆x→0
=
lim
∆x→0
∆x
n(n
−
1)
n−1
n−2
n−2
n−1
nx
+
x
∆x + . . . + nx∆x
+ ∆x
2
= nxn−1
Example 2.1.10.
em
at
ic
s
1. Dx [x5 ] = 5x5−1 = 5x4
2. Dx [(2x)5 ] = Dx [32x5 ] = 32 · 5x4 = 160x4 (Note that the derivative is not 5(2x)4 which is 80x4 .)
√
d
d 4√ d
d
[2x4 − 45 x + 7] =
[2x4 ] −
[7] = 8x3 − 54 · 2√1 x + 0 = 8x3 − 5√2 x .
x +
dx
dx
dx 5
dx
at
h
3.
of
M
4. If y = (x − 3)(2x2 − 3), then y = 2x3 − 6x2 − 3x + 9 and so y 0 = 6x2 − 12x − 3.
ut
e
5. Alternatively, we may also use the Product Rule. Thus,
Dx [(x − 3)(2x2 − 3)] = (1)(2x2 − 3) + (x − 3)(4x) = 2x2 − 3 + 4x2 − 12x = 6x2 − 12x − 3.
tit
√
4
d
[(3x5 + 6x 3 − 2)(4x3 − 5 3 x)]
dx
√
1
4
2
= (15x4 + 8x 3 )(4x3 − 5 3 x) + (3x5 + 6x 3 − 2)(12x2 − 5 · 31 x− 3 ).
7. If f (x) =
UP
In
s
6.
2x3 − x−3 + 4
(3x − 5)(6x2 + 3x−4 ) − (2x3 − x−3 + 4)(3)
, then f 0 (x) =
.
3x − 5
(3x − 5)2
(2x3 − 4√x − 3x)(8x3 − √1 ) − (2x4 − √x + 2)(6x2 − √2 − 3)
√
d
2x4 − x + 2
2 x
x
√
√
8.
=
.
dx 2x3 − 4 x − 3x
(2x3 − 4 x − 3x)2
Remark 2.1.11.
1. Using items 1 to 4 of Theorem 2.1.9, one can show that the derivative of a polynomial function is
also a polynomial function. This means that a polynomial function is differentiable everywhere.
2. From the first item and the Quotient Rule, one can deduce that a rational function is differentiable on its domain.
1
n
Binomial Theorem: If n is a positive integer, then (x + y) =
n
P
k=0
n
k
!
xn−k y k
86
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
Derivatives of Trigonometric Functions
At this point, we know how to differentiate polynomial functions, rational functions, and functions
involving radicals. We now present formulas for the derivatives of trigonometric functions.
Theorem 2.1.12.
1. Dx [sin x] = cos x
4. Dx [cot x] = − csc2 x
2. Dx [cos x] = − sin x
5. Dx [sec x] = sec x tan x
3. Dx [tan x] = sec2 x
6. Dx [csc x] = − csc x cot x
Proof. We will prove statement 1 only. Statement 2 can be proved similarly, while items 3–6 can
be proved using items 1 and 2, and the product and quotient rules.
sin(x + ∆x) − sin x
∆x→0
∆x
sin x cos(∆x) + cos x sin(∆x) − sin x
= lim
∆x→0
∆x
1 − cos(∆x)
sin(∆x)
− sin x ·
= lim cos x ·
∆x→0
∆x
∆x
of
M
= (cos x)(1) − (sin x)(0)
at
h
em
at
ic
s
Dx [sin x] = lim
ut
e
= cos x
In
s
tit
Remark 2.1.13.
UP
1. The formulas in the previous theorems consider trigonometric functions as real–valued functions.
Thus, whenever these formulas are applied to problems where trigonometric functions are viewed
as functions on angles, the measure of an angle must be in radians.
2. Observe that the derivative of a trigonometric function is either another trigonometric function
or a product of trigonometric functions. That means that a trigonometric function is differentiable where its derivative is defined. Moreover, observe that the domains of a trigonometric
function and its derivative are the same. Hence, a trigonometric function is differentiable on its
domain.
Example 2.1.14.
1.
d
[3 sin x − 7 cos x] = 3 cos x + 7 sin x
dx
2. Dx [sec x csc x] = (sec x tan x)(csc x) + (sec x)(− csc x cot x)
3. If f (x) =
cot x − x
(1 + tan x)(− csc2 x − 1) − (cot x − x)(sec2 x)
, then f 0 (x) =
1 + tan x
(1 + tan x)2
4. Dx [sin(2x)] = Dx [2 sin x cos x] = 2·(cos x·cos x+sin x·(− sin x)) = 2(cos2 x−sin2 x) = 2 cos(2x)
2.1. SLOPES, THE DERIVATIVE, AND BASIC DIFFERENTIATION RULES
2.1.5
87
Exercises
Exercises for Discussion
dy
A. Find
. There is no need to simplify.
dx
1. y = 2x4 − 3x2 + x − 1
√
1
8
2. y = 6 3 x − 2 + √
x
x
3. y = x3 cos x
5
4. y = − sec x csc x
x
√
2
2
5. y = x − 2 ( 3 x − cos x)
x
4 − x2
6. y = √
3 x + tan x
√
cos x + π
7. y =
2x − x33
√
x x
8. y = 5
x + sin x
B. Miscellaneous Exercises.
em
at
ic
s
1. Determine the equation of the normal line to the graph of g(x) = 2 sin x + tan x at the
point where x = π3 .
of
M
at
h
2. Find all the points on the graph of y = (x − 2)2 at which the tangent line is perpendicular
to the line with equation 2x − y + 2 = 0.
Supplementary Exercises
dy
. There is no need to simplify.
dx
tit
1. y = 5 sin x − 2 cot x
ut
e
A. Find
UP
In
s
2. y = (2x2 + 5x − 2)(3x − 7)
2x + 1
3. y =
csc x
x sin x − x
4. y = 5
x cot x − 5
B. Given f (5) = 5 , f 0 (5) = −
1. (f − g)0 (5)
2. (f · g)0 (5)
√
2x3 csc x + x − 2
5. y =
(sin x + 1)(x cos x − 1)
√
6. y = (x2 − 2x + 2)(sin x − x)(2 + cos x)
√
1
5
2
7. y =
x3 − x4 + 5x5 (sin x + csc x)
x − 3
x
7
3
, g(5) = , and g 0 (5) = −8 , determine
2
2
3.
f 0
(5)
g
g 0
(5)
4.
f
C. Miscellaneous Exercises.
1. Find the equation of the tangent line to the graph of f (x) = 2x3 + 1 at x = −1.
2. Determine the values of a and b so that the line with equation 2x + y = b is tangent to the
graph of y = ax2 when x = 2.
88
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
2.2
The Chain Rule, and more on Differentiability
The previous section introduced formulas for differentiating a large scope of algebraic functions.
But there are many types of functions which do not neatly fall into one category. For instance, how
√
can the function f (x) = x2 − 1 be differentiated? The formulas provided in the previous section
do not cover function compositions.
This section provides techniques for differentiation of compositions of functions. We also tackle
the geometric implications of differentiability and non-differentiability, and discuss the notion of a
higher-order derivative.
By the end of this section, the student will be able to
• differentiate using the chain rule;
em
at
ic
• discuss the differentiability and continuity of a function; and
at
h
• find higher order derivatives of a function.
The Chain Rule
of
M
2.2.1
s
• differentiate implicitly defined functions of one variable;
In
s
tit
ut
e
Recall that in the previous section, to differentiate sin 2x, an identity was used to write it in terms
of sin x and cos x. The following theorem allows us to differentiate a given function, a composite
function in particular, without having to write it as a product or quotient of basic functions.
UP
Theorem 2.2.1 (Chain Rule). If the function g is differentiable at x = x0 and the function f
is differentiable at g(x0 ), then (f ◦ g)(x) is differentiable at x = x0 and
(f ◦ g)0 (x0 ) = f 0 (g(x0 )) · g 0 (x0 ).
Remark 2.2.2. The chain rule can also be stated in the following manner:
dy
dy du
=
·
or Dx [f (u)] = f 0 (u)Dx [u] .
If y = f (u) and u = g(x) , then
dx
du dx
Example 2.2.3.
1. Find Dx [(2x)5 ].
Solution.
Note that (2x)5 = (f ◦ g)(x) where f (x) = x5 and g(x) = 2x. We have f 0 (x) = 5x4 and
g 0 (x) = 2. Using chain the rule,
Dx [(2x)5 ] = f 0 (g(x)) · g 0 (x) = 5[g(x)]4 · 2 = 10(2x)4 .
2. Find h0 (x) if h(x) =
√
x2 − 1.
2.2. THE CHAIN RULE, AND MORE ON DIFFERENTIABILITY
89
Solution.
Note that
√
x2 − 1 = (f ◦ g)(x) where f (x) =
√
and g 0 (x) = 2x. Using the chain rule,
1
x and g(x) = x2 − 1. We have that f 0 (x) = √
2 x
1
2x
h0 (x) = f 0 (g(x)) · g 0 (x) = √
· 2x = √
.
2 x2 − 1
2 x2 − 1
3. Find Dx [sin(2x)].
Solution.
We have sin(2x) = (f ◦ g)(x) where f (x) = sin x and g(x) = 2x. So, f 0 (x) = cos x and g 0 (x) = 2
and by the chain rule,
Dx [sin(2x)] = f 0 (g(x)) · g 0 (x) = (cos[g(x)]) · (2) = 2 cos(2x).
em
at
ic
s
4. Find h0 (x) if h(x) = 2 cos(x2 ).
Solution.
of
M
at
h
Note that h(x) = (f ◦ g)(x) where f (x) = 2 cos x and g(x) = x2 . Now, f 0 (x) = 2(− sin x) =
−2 sin x and g 0 (x) = 2x. Using chain the rule,
u and u = x3 − tan x. Thus,
UP
√
In
s
Solution.
We let y =
ut
√
dy
if y = x3 − tan x.
dx
tit
5. Find
e
h0 (x) = f 0 (g(x)) · g 0 (x) = (−2 sin g(x)) · (2x) = −4x sin(x2 ).
dy du
1
dy
3x2 − sec2 x
=
·
= √ · (3x2 − sec2 x) = √
.
dx
du dx
2 u
2 x3 − tan x
Remark 2.2.4.
1. The chain rule can also be extended to a finite composition of functions. To illustrate,
(f1 ◦ f2 ◦ f3 ◦ f4 )0 (x) = f10 ((f2 ◦ f3 ◦ f4 )(x)) · f20 ((f3 ◦ f4 )(x)) · f30 (f4 (x)) · f40 (x)
2. When computing derivatives using chain rule, we don’t actually write out the functions f and
g as in the previous examples, but we bear them in mind. It may help to keep in mind:
“The derivative of f (g(x)) is the derivative of the outside function evaluated at the inside function times the derivative of the inside function.“
Example 2.2.5.
1. Dx [(3x2 − x − 6)27 ] = 27(3x2 − x − 6)26 · Dx [3x2 − x − 6] = 27(3x2 − x − 6)26 · (6x − 1)
90
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
2. If y = (2x − 1)100 (2 − x)200 then
dy
dx
= 100(2x − 1)99 · (2) · (2 − x)200 + (2x − 1)100 · 200(2 − x)199 · (−1)
= (2x − 1)99 (2 − x)199 [200(2 − x) − 200(2x − 1)]
= (2x − 1)99 (2 − x)199 [400 − 200x − 400x + 200)]
= (2x − 1)99 (2 − x)199 (600 − 600x)
= 600(1 − x)(2x − 1)99 (2 − x)199
"
3. Dx
4.
√
3
5
5
√1 · 2 + csc x cot x
2 + 2)3 · (5x 2 ) − (2x 2 + 2)4 ·
(
2x
−
csc
x)
·
4(2x
(2x
2 2x
√
√
=
2x − csc x
( 2x − csc x)2
5
2
+ 2)4
#
d
d
[sec(cos x)] = sec(cos x) tan(cos x) ·
[cos x] = sec(cos x) tan(cos x) · (− sin x)
dx
dx
3
3
3
3
1
Derivatives from the Left and from the Right
em
at
ic
2.2.2
s
5. Dx [cos2 (cot 4x 2 )] = 2 cos(cot 4x 2 ) · (− sin(cot 4x 2 )) · (− csc2 4x 2 ) · (6x 2 )
at
h
In testing for continuity of piecewise-defined functions at the endpoints, it is necessary to check
one-sided limits. Similarly, testing differentiability may involve one-sided derivatives defined below.
Let the function f (x) be defined at x = x0 .
of
M
1. The derivative from the left of f (x) at x = x0 , denoted by f−0 (x0 ), is given by
f (x) − f (x0 )
.
x − x0
x→x−
0
tit
ut
e
f−0 (x0 ) = lim
In
s
2. The derivative from the right of f (x) at x = x0 , denoted by f+0 (x0 ), is given by
UP
f+0 (x0 ) = lim
x→x+
0
f (x) − f (x0 )
.
x − x0
Remark 2.2.6.
1. In Definition 2.2.2, it is necessary that the function f is defined at x0 . Otherwise, the limit
expressions do not make sense.
2. The derivative from the left [right] is also referred to as the left-hand derivative [right-hand
derivative], or simply left derivative [right derivative].
3. The function f is differentiable at x = x0 if and only if f−0 (x0 ) and f+0 (x0 ) exist and
f−0 (x0 ) = f+0 (x0 ) = f 0 (x0 ).
Example 2.2.7. Determine if f (x) = |x| is differentiable at x = 0.
Solution.
We compute the derivatives from the left and from the right at x = 0.
2.2. THE CHAIN RULE, AND MORE ON DIFFERENTIABILITY
f (x) − f (0)
x
x→0−
(−x) − 0
= lim
x
x→0−
= −1
f−0 (0) =
91
f (x) − f (0)
x
x−0
= lim
x
x→0+
= 1
f+0 (0) =
lim
lim
x→0+
Since f−0 (0) 6= f+0 (0), the function f (x) is not differentiable at x = 0.
(
1 2
24 , x < 4
2 x +√
Example 2.2.8. Determine if f (x) =
is differentiable at x = 4.
16 x , x ≥ 4
Solution.
We compute the derivatives from the left and from the right at x = 4.
=
=
s
x→4+
=
=
=
=
e
=
=
f (x) − f (4)
√x − 4
16 x − 32
lim
x√− 4
x→4+
16( x − 2)
lim
x−4
x→4+
x−4
√
lim 16
(x − 4) x + 2
x→4+
16
lim √
x+2
x→4+
4
lim
em
at
ic
=
f+0 (4) =
x→4−
at
h
=
f (x) − f (4)
x−4
( 12 x2 + 24) − 32
lim
x−4
x→4−
1 2
x
−
8
lim 2
x→4− x − 4
1 (x + 4)(x − 4)
lim
x−4
x→4− 2
1
lim (x + 4)
x→4− 2
4
lim
of
M
f−0 (4) =
Differentiability and Continuity
In
s
2.2.3
tit
ut
Since f−0 (4) = f+0 (4) = 4, the function f (x) is differentiable at x = 4. Moreover, f 0 (4) = 4.
UP
In applications that we will tackle later, functions that are used to model quantities and relationships
are assumed to be continuous and/or differentiable on certain intervals. The following presents a
link between continuity and differentiability.
Theorem 2.2.9. If f is differentiable at x = x0 , then f is continuous at x = x0 .
Remark 2.2.10.
1. If f is discontinuous at x = x0 , then f is not differentiable at x = x0 .
2. If f is continuous at x = x0 , it does not mean that f is differentiable at x = x0 .
3. If f is not differentiable at x = x0 , it does not mean that f is not continuous at x = x0 .
Example 2.2.11. The function f (x) = |x| is continuous at x = 0 but is not differentiable at x = 0.
We present a theorem that is especially useful in determining whether a piecewise-defined function
f is differentiable at x = x0 .
92
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
Theorem 2.2.12.
1. If f is continuous at x = x0 from the left and lim f 0 (x) exists, then
x→x−
0
f−0 (x0 ) = lim f 0 (x).
x→x−
0
2. If f is continuous at x = x0 from the right and lim f 0 (x) exists, then
x→x+
0
f+0 (x0 ) = lim f 0 (x).
x→x+
0
Example 2.2.13. Determine if the following function is differentiable at x = −1:
x2 , x < −1
−1 − 2x , x ≥ −1
em
at
ic
f (x) =
s
(
at
h
Solution.
First, we check the continuity of f (x) at x = −1. Note that f (−1) = 1 and
lim f (x) =
x→−1−
of
M
lim f (x) = 1, and thus, f (x) is continuous at x = −1. Next, we determine the derivative from
x→−1+
2x , x < −1
−2 , x > −1
ut
(
e
the left and derivative from the right at x = −1. We obtain
In
s
tit
f 0 (x) =
and thus, f−0 (−1) = lim f 0 (x) = 2(−1) = −2, while f+0 (−1) = lim f 0 (x) = −2. Finally, since
x→−1−
x→−1+
UP
f−0 (−1) = f+0 (−1), f (x) is differentiable at x = −1.
Example 2.2.14. Determine if the following function is differentiable at x = 1:
(
g(x) =
x2 + x + 2 , x ≤ 1
3x , x > 1
Solution.
It can be shown that g is discontinuous at x = 1. However, note that
(
g 0 (x) =
2x + 1 , x < 1
,
3 , x>1
and thus, lim g 0 (x) = 3 and lim g 0 (x) = 3. Although lim g 0 (x) = lim g 0 (x), we cannot
x→1−
x→1+
x→1−
x→1+
conclude that the function is differentiable at x = 1. (Why?) In fact, g is not differentiable at
x = 1. (Why?)
2.2. THE CHAIN RULE, AND MORE ON DIFFERENTIABILITY
2.2.4
93
Graphical Consequences of Differentiability and Non-differentiability
Geometrically, if f is differentiable at x = x0 , then the graph of f has a non-vertical tangent line
at x = x0 . The following remark gives the most commonly encountered circumstances for which f
fails to be differentiable at a value x = x0 .
Remark 2.2.15. A function f is not differentiable at x = x0 if one of the following is true:
1. f is discontinuous at x = x0 (see Figure 2.2.1)
2. the graph of f has a vertical tangent line at x = x0 (see Figure 2.2.2)
Higher Order Derivatives
x0
Figure 2.2.3
of
M
2.2.5
x0
Figure 2.2.2
at
h
x0
Figure 2.2.1
em
at
ic
s
3. the graph of f has no well–defined tangent line at x = x0 , i.e., the graph of f has a corner, edge
or cusp at x = x0 (see Figure 2.2.3).
tit
ut
e
If the derivative f 0 of a function f is itself differentiable, then the derivative of f 0 is called the
second derivative of f and is denoted f 00 . We can continue to obtain the third derivative, f 000 ,
the fourth derivative, f (4) , and even higher derivatives of f as long as we have differentiability.
UP
In
s
The n-th derivative of the function f , denoted by f (n) , is the derivative of the (n − 1)-th
derivative of f , that is,
f (n−1) (x + ∆x) − f (n−1) (x)
∆x→0
∆x
f (n) (x) = lim
Remarks.
1. The n in f (n) is called the order of the derivative.
2. The derivative of a function f is sometimes called the first derivative of f .
3. The function f is sometimes written as f (0) (x).
4. Other notations:
Dx n [f (x)] ,
dn y
dn
,
[f (x)] , y (n)
dxn
dxn
Example 2.2.16.
1. Find f (n) (x) for all n ∈
Solution.
N where f (x) = x6 − x4 − 3x3 + 2x2 − 4.
94
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
We differentiate repeatedly and obtain
f 0 (x) = 6x5 − 4x3 − 9x2 + 4x,
f 00 (x) = 30x4 − 12x2 − 18x + 4,
f 000 (x) = 120x3 − 24x − 18,
f (4) (x) = 360x2 − 24,
f (5) (x) = 720x,
f (6) (x) = 720,
f (n) (x) = 0
2. Find f (4) (x) if f (x) =
√
for all n ≥ 7.
2x − 3.
Implicit Differentiation
em
at
ic
e
2.2.6
of
M
at
h
We differentiate repeatedly and obtain
1
1
1
f 0 (x) =
(2x − 3)− 2 · (2)
= (2x − 3)− 2
2
3
3
1
f 00 (x) = − (2x − 3)− 2 · (2) = − (2x − 2)− 2
2
5
5
3
000
(2x − 3)− 2 · (2)
f (x) =
= 3 (2x − 2)− 2
2
7
7
15
f (4) (x) = − (2x − 3)− 2 · (2) = −15 (2x − 2)− 2
2
s
Solution.
UP
In
s
tit
ut
Equations in two variables are used to define a function explicitly or implicitly. The equation
y = x2 − 1 defines the function f (x) = x2 − 1 explicitly. On the other hand, the equation y 2 = x + 1
defines two functions of x implicitly, namely:
√
√
f1 (x) = x + 1 and f2 (x) = − x + 1.
Suppose y = f (x) is a function defined implicitly by the equation x3 − 2xy − 3y − 6 = 0, how do
dy
we find its derivative
?
dx
Solution.
x3 − 6
From x3 − 2xy − 3y − 6 = 0, we can define y explicitly in terms of x as y =
. Thus,
2x + 3
dy
(2x + 3)(3x2 ) − (x3 − 6)(2)
4x3 + 9x2 + 12
=
=
.
dx
(2x + 3)2
(2x + 3)2
However, suppose a function is defined implictly by a function for which y cannot be isolated easily,
dy
such as x4 y 3 − 7xy = 7 or tan(x2 − 2xy) = y. How do we find
?
dx
dy
To obtain
without solving for y explicitly in terms of x, we use the method called implicit
dx
dy
differentiation. To find
using implicit differentiation, we
dx
1. think of the variable y as a differentiable function of the variable x,
2.2. THE CHAIN RULE, AND MORE ON DIFFERENTIABILITY
2. differentiate both sides of the equation, using the chain rule where necessary, and
3. solve for
dy
.
dx
Example 2.2.17. Find
dy
using implicit differentiation.
dx
1. x3 − 2xy − 3y − 6 = 0
Solution.
tit
ut
e
of
M
at
h
em
at
ic
s
Dx [x3 − 2xy − 3y − 6] = Dx [0]
dy dy
3x2 − 2 1 · y + x ·
=0
−3·
dx
dx
dy
dy
−3·
=0
3x2 − 2y − 2x ·
dx
dx
dy
dy
3x2 − 2y = 2x ·
+3·
dx
dx
dy
(2x + 3) = 3x2 − 2y
dx
3x2 − 2y
dy
=
dx
2x + 3
x3 −6
3x2 − 2 · 2x+3
2x + 3
=
·
2x + 3
2x + 3
2
3x · (2x + 3) − 2(x3 − 6)
=
(2x + 3)2
4x3 + 9x2 + 12
=
(2x + 3)2
UP
Solution.
In
s
2. x4 y 3 − 7xy = 7
Dx [x4 y 3 − 7xy] = Dx [7]
dy
dy −7 y+x·
=0
4x3 y 3 + 3x4 y 2 ·
dx
dx
dy
−4x3 y 3 + 7y
=
dx
3x4 y 2 − 7x
3. tan(x2 − 2xy) = y
Solution.
Dx [tan(x2 − 2xy)] = Dx [y]
h
dy i dy
sec2 (x2 − 2xy) · 2x − 2 y + x ·
=
dx
dx
dy
2x sec2 (x2 − 2xy) − 2y sec2 (x2 − 2xy)
=
dx
1 + 2x sec2 (x2 − 2xy)
95
96
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
Example 2.2.18. Determine
d2 y
if xy 2 = y − 2.
dx2
Solution.
dy
We use implicit differentiation to get
.
dx
Dx [xy 2 ] = Dx [y − 2]
dy
dy
y 2 + x · 2y
=
dx
dx
y2
dy
=
dx
1 − 2xy
Thus,
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
dy
dy
(1 − 2xy)(2y dx
) − (y 2 )[−2(y + x dx
)]
d2 y
=
2
2
dx
(1 − 2xy)
h
i
h i
y2
y2
− (y 2 ) −2 y + x · 1−2xy
(1 − 2xy) 2y · 1−2xy
=
.
(1 − 2xy)2
2.2. THE CHAIN RULE, AND MORE ON DIFFERENTIABILITY
2.2.7
97
Exercises
Exercises for Discussion
A. The Chain Rule.
dy
1. Find
. There is no need to simplify.
dx
a. y = cos 3x
√
2
b. y = 6x 3h(x2 − x)5 i
4
c. y = csc 2x3 + 1
d. y = cot4 (5x)
5 cos x
4
e. y = sec
tan 2x
sin x −
f. y =
1
x2 + 1
cot 3x −
√
2π
B. Differentiability.
1. Determine if
(
f (x) =
2
3x 3 − 1 , x ≤ 1
3
x −x+2 , x>1
em
at
ic
s
is differentiable at x = 1.
2. Determine if
(
x2 − 4 , x < 3
√
x−2 , x≥3
at
h
g(x) =
of
M
is differentiable at x = 3.
3. Determine if
(
(x − 1)2 , x ≤ 1
√
x−1 , x>1
In
s
tit
is differentiable at x = 1.
ut
e
h(x) =
UP
C. Implicit Differentiation.
1. Use implicit differentiation to find
a. cos(x2 + y 2 ) = x −
√
y
dy
. There is need to simplify.
dx
2
2
b. x3 y 2 = x 3 + y 3 + 1
√
ds
if (3s + t)4 = 3 s − cos t.
dt
D. Higher Order Derivatives.
2. Find
N
1. Find f (n) (x) for all n ∈ if f (x) = 6x5 − 5x4 − 4x3 + 3x2 − 2x + 1.
√
2. Evaluate Dx2 x 4 − x2
E. Do as indicated.
1. If g(5) = 3, g 0 (5) = 4, f (3) = 5 and f 0 (3) = 6, find (f ◦ g)0 (5) and (g ◦ f )0 (3).
2. Determine the point/s on the graph of xy = (1 − x − y)2 where the tangent line/s is/are
parallel to the x−axis.
3. Determine Dx103 (cos 2x).
4. Find y 00 at the point with coordinate (2, 1) if 2x2 y − 4y 3 = 4.
98
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
Supplementary Exercises
A. The Chain Rule.
1. Find
dy
. There is no need to simplify.
dx
√
√
e. y = cos3 [cot (csc x)]
r
9 − x2
1
b. y = tan
x
a. y =
1 + cot x
c. y =
x cos x
3
d. y = sin 2x cos2 3x
3
f. y =
(1 − x2 )5 cos x
√
3x3 − tan x
2. Given the following table of values:
x
2
3
4
f 0 (x)
1
-3
-4
f (x)
4
-1
5
g(x)
3
4
4
g 0 (x)
-2
-5
1
a. (f ◦ g)0 (2)
em
at
ic
s
Evaluate:
b. (g ◦ f )0 (2)
c. (g ◦ g)0 (2)
at
h
B. Differentiability.
d. (f ◦ g ◦ f )0 (2)
of
M
1. Find the values of m and n so that
(
x2 , x < 1
mx + n , x ≥ 1
ut
e
f (x) =
tit
is differentiable at x = 1.
UP
In
s
2. Find all the values of a and b such that
(
g(x) =
ax + b , x < 2
2x2 − 1 , x ≥ 2
is differentiable at x = 2.
C. Implicit Differentiation.
1. Use implicit differentiation to find
dy
. There is need to simplify.
dx
a. 2x sin y = (x + 2y)6
√
x4
b. sec(2x − y) − 3 = sin2 y −
4
3
3
5
c. cot4 (xy) = 4x 2 − sin(x 5 + y 3 )
2. Give the equation of the normal line to the graph of y 3 − xy 2 + cos(xy) = 2 at (0, 1).
D. Higher Order Derivatives.
1. Evaluate Dx3 sin 2x − x3 + cos x2
2. Determine
d2 y
if (x + y)3 = xy 2 − 2y.
dx2
2.2. THE CHAIN RULE, AND MORE ON DIFFERENTIABILITY
99
E. Do as indicated.
1. If f (t) = at2 + bt + c , f (1) = 5, f 0 (1) = 3 and f 00 (1) = −4 , find f (3).
2. Determine the values of m and n for the curve x2 y + my 2 = n if the point (1, 1) is on its
graph and the tangent line at this point has equation 4x + 3y − 7 = 0.
d2 y
4
3. If x2 + 9y 2 = 36, show that
= − 3.
2
dx
9y
4. Find equations for the two tangent lines through the origin that are tangent to the curve
x2 − 4x + y 2 + 3 = 0.
5. Let
(
f (x) =
x2 , x ≤ 0
.
x3 , x > 0
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
Show that f 0 (0) exists but f 00 (0) does not.
100
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
2.3
Derivatives of Exponential and Logarithmic Functions
In this section, we compute for the derivatives of functions that involve logarithmic and exponential
functions.
At the end of this section, the student will be able to:
• perform differentiation involving logarithmic functions;
• find the derivative using logarithmic differentiation;
• differentiate functions of the form f (x)g(x) ; and
Derivatives of Logarithmic Functions
em
at
ic
2.3.1
s
• perform differentiation involving exponential functions.
at
h
Our goal is to find the derivatives of logarithmic functions. We begin with the natural logarithmic
function and will find later that its derivative appears simplest among all logarithmic functions.
Let x > 0. Then
of
M
ln(x + ∆x) − ln x
∆x
1
x + ∆x
= lim
ln
∆x→0 ∆x
x
1 x
∆x
= lim
ln 1 +
∆x→0 x ∆x
x
"
x #
1
∆x ∆x
=
lim ln 1 +
x ∆x→0
x
"
x #
1
∆x ∆x
= ln lim 1 +
, by continuity of ln .
∆x→0
x
x
Dx (ln x) = lim
UP
In
s
tit
ut
e
∆x→0
Let h =
∆x
. Then h → 0 if and only if ∆x → 0. Thus,
x
1
1
1
1
h
Dx (ln x) = ln lim (1 + h) = ln e = .
h→0
x
x
x
Interestingly, the derivative of the ln function, one that cannot be expressed using a finite number
of operations (including extraction of nth roots) on polynomials, is a simple algebraic expression.
In contrast, the derivatives of circular functions are products of their fellow circular functions.
Now, since any logarithmic function can be expressed as the ln function times a constant, one easily
obtains its derivative. Indeed,
ln x
1
1 1
1
Dx (loga x) = Dx
=
Dx (ln x) =
· =
.
ln a
ln a
ln a x
x ln a
2.3. DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Now, let f (x) = ln |x| on
chain rule,
R \ {0}. Since x 6= 0, |x| > 0, so ln |x| is defined. Moreover, using the
√
1
1
2x
1
Dx (ln |x|) = Dx (ln x2 ) = √ · √ · 2x = 2 = .
2
2
2x
x
x 2 x
We summarize the preceding computations into the following theorem.
Theorem 2.3.1.
1. Dx (ln x) =
1
for all x > 0.
x
1
for all x > 0.
x ln a
3. Dx (ln |x|) =
1
for all x 6= 0.
x
s
2. Dx (loga x) =
dy
.
dx
em
at
ic
Example 2.3.2. Find
101
at
h
1. y = log2 x
e
of
M
Solution.
1
dy
=
dx
x ln 2
tit
ut
2. y = ln(3x2 + 2)
UP
In
s
Solution.
dy
1
6x
= 2
· 6x = 2
dx
3x + 2
3x + 2
3. y = ln |7 − cos(2x)|
Solution.
dy
1
2 sin(2x)
=
· sin(2x) · 2 =
dx
7 − cos(2x)
7 − cos(2x)
4. y = ln [sin (log5 x)]
Solution.
1
dy
1
=
· cos (log5 x) ·
dx
sin (log5 x)
x ln 5
5. y = x2 log x
Solution.
dy
1
x
= 2x · log x + x2 ·
= 2x log x +
dx
x ln 10
ln 10
102
2.3.2
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
Logarithmic Differentiation
Consider the function given by the equation
√
(x2 − 1)3 ( 3 1 + x + x3 )(sec4 x)
y=
.
√
( csc x)( x45 )
dy
. One could imagine the excessive computations that would
Suppose you are asked to find
dx
be brought about by repeated application of the product, quotient, power and chain rules. We
shall therefore introduce a process called logarithmic differentiation, a technique helpful in
differentiating an expression involving many products and quotients. In using this technique given
an equation in x and y:
1. Take the absolute value of both sides of the equation and apply properties of the absolute
value.
em
at
ic
s
2. Take the natural logarithm of both sides and apply properties of logarithms to obtain a sum.
dy
.
dx
at
h
3. Take the derivative of both sides implicitly with respect to x and solve for
Example 2.3.3.
ut
e
of
M
√
3
dy
x+1
√
1. Find
if y =
.
5
dx
csc (x) 1 − x2
tit
Solution.
UP
In
s
We begin by taking the absolute value of both sides of the equation. Simplifying the result, we
get
√
3
|y| =
x+1
√
1 − x2
csc5 (x)
1
=
|x + 1| 3
1
| csc x|5 |1 − x2 | 2
Next, we take the natural logarithm of both sides of the function. Using the properties of
logarithms, we rewrite the equation as a sum.
!
1
|x + 1| 3
ln |y| = ln
1
| csc x|5 |1 − x2 | 2
1
1
= ln |x + 1| 3 − ln | csc x|5 − ln |1 − x2 | 2
1
1
= ln |x + 1| − 5 ln | csc x| − ln |1 − x2 |
3
2
2.3. DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS
103
Finally, we take the derivatives of both sides of the equation implicitly with respect to x and
dy
solve for
,
dx
1
1
2
Dx (ln |y|) = Dx
ln |x + 1| − 5 ln | csc x| − ln |1 − x |
3
2
1 dy
1
1
5
1
1
·
= ·
·1−
· (− csc x cot x) − ·
· (−2x)
y dx
3 x+1
csc x
2 1 − x2
dy
1
x
=y
+ 5 cot x +
dx
3(x + 1)
1 − x2
√
3
x+1
x
1
√
=
+ 5 cot x +
3(x + 1)
1 − x2
csc5 (x) 1 − x2
√
3
dy
x2 tan4 x
2. Find
if y =
.
dx
(2x2 + 1)3 log x
em
at
ic
s
Solution.
Once again, we begin by taking the absolute value of both sides of the equation. Thus,
2
of
M
at
h
|x| 3 | tan x|4
|y| =
.
|2x2 + 1|3 | log x|
tit
ut
e
Then, we take the natural logarithm of both sides and rewrite the equation as a sum using the
properties of logarithms to obtain
In
s
2
ln |x| + 4 ln | tan x| − 3 ln |2x2 + 1| − ln | log x|.
3
UP
ln |y| =
Finally, we take the derivative of both sides implicitly with respect to x and solve for
dy
,
dx
1 dy
2 1
1
1
1
1
= · +4·
· sec2 x − 3 · 2
· (4x) −
·
y dx
3 x
tan x
2x + 1
log x x ln 10
√
3
4
2
2
dy
x tan x
2
4 sec x
12x
1
=
+
− 2
−
.
dx
(2x2 + 1)3 log x 3x
tan x
2x + 1 (log x)(x ln 10)
2.3.3
Derivatives of Exponential Functions
Next, we determine the derivatives of exponential functions, which we give in the following theorem.
Theorem 2.3.4.
1. Dx (ax ) = ax ln a (a > 0 and a 6= 1)
2. Dx (ex ) = ex
104
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
Proof. Consider the exponential function of base a,
y = ax .
This is equivalent to the logarithmic equation
x = loga y,
and taking the derivative of both sides of the equation implicitly with respect to x, we obtain
Dx (x) = Dx (loga y)
1
dy
1=
·
.
y ln a dx
Solving for
dy
, we obtain
dx
em
at
ic
s
dy
= y ln a = ax ln a.
dx
at
h
dy
= ex ln e = ex ,
This proves the first statement of Theorem 2.3.4. Observe that if a = e, then
dx
which proves the second statement of Theorem 2.3.4.
dy
.
dx
e
Example 2.3.5. Find
of
M
Thus, we obtain a function whose derivative is itself. (Is this the only function with this property?)
tit
ut
1. y = 4x
UP
2. y = ex
In
s
Solution.
dy
= 4x ln 4
dx
3
Solution.
dy
3
= ex · 3x2
dx
3. y = 24x csc(ex )
Solution.
dy
= 24x ln 2 · 4 csc(ex ) + 24x [− csc (ex ) cot (ex ) · ex ]
dx
As a consequence also, if r ∈
R and x > 0, then
1
Dx (xr ) = Dx er ln x = er ln x · r ·
= xr · rx−1 = rxr−1 .
x
Hence, the power rule holds even for irrational exponents. We state this as a theorem:
2.3. DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Theorem 2.3.6. (Power Rule) If f (x) = xr where r ∈
√
For example, Dx (xπ ) = πxπ−1 and Dx ((cos x) 2 ) =
2.3.4
√
105
R, then f 0(x) = rxr−1.
√
2(cos x) 2−1 (− sin x).
Derivative of f (x)g(x) , where f (x) > 0
To differentiate expressions of the form f (x)g(x) , we either use logarithmic differentiation or rewrite
f (x)g(x) as eg(x) ln f (x) .
Example 2.3.7. Find
dy
.
dx
1. y = xx , x > 0
Solution.
Using logarithmic differentiation, we have
em
at
ic
s
ln y = ln (xx ) (no need to take absolute values since x > 0)
of
M
e
ut
tit
y = ex ln x
at
h
ln y = x ln x
1 dy
1
= 1 · ln x + x ·
y dx
x
dy
= y(ln x + 1)
dx
dy
= xx (ln x + 1)
dx
Alternatively, we have
UP
In
s
dy
1
x ln x
=e
1 · ln x + x ·
dx
x
dy
= xx (ln x + 1)
dx
2. y = (sin x)cos x , sin x > 0
Solution.
ln y = cos x ln(sin x)
1 dy
1
= (− sin x) · ln(sin x) + cos x ·
· cos x
y dx
sin x
dy
cos2 x
cos x
= (sin x)
− sin x ln(sin x) +
dx
sin x
106
2.3.5
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
Exercises
Exercises for Discussion
dy
A. Find
. Use logarithmic differentiation whenever appropriate.
dx
6. y = x2 · 35x−3
1. y = log5 (2x5 − 1) sec x
2. y = cos(ln |4 − x|)
√
3. y = x5 (x2 − 1)4 sin x
(1 + x − x2 )3
4. y = √
sec 4x(x − log x)6
x2 cot4 x
5. y = √
3
5x − 2(log4 x)10
7. ln(x2 + y 2 ) = ex+y − ex−y
8. ln 2 = (x2 − 1)π + 5(e
√
9. y = (log4 x)e
xy )
x
2
10. y = (tan 3x)1−x
em
at
ic
s
B. Find the equation of the normal line to the graph of f (x) = 2x log2 x at the point where
x = 1.
Supplementary Exercises
dy
. Use logarithmic differentiation whenever appropriate.
dx
at
h
A. Find
8. y = 2x + 3x − 5x
4 − 14x
9. y = log8
1 − x2
e
of
M
csc(log3 x)
1. y = √
x − ln(5x)
2. sin x2 − y 2 = tan log 1 x
7. ln(x − y) = (tan x)y
tit
In
s
UP
6. y x = xy
ut
4
3. y = logx2 +4 3
√
esin 2x + ln(1 + x)
4. y =
2log5 x
2
5. y = (x + 2x )sin 5x
10. y = e4x−1 + ln x2 + e
11. y =
1 + x − log12 x
log2 2 − x + log5 (2x + 1)
12. ln |x − cos y| = y 2 + x − 2
13. x tan y − sin [(x − 1) ln y] = x
2.4. DERIVATIVES OF OTHER NEW CLASSES OF FUNCTIONS
2.4
107
Derivatives of Other New Classes of Functions
We revisit the new classes of transcendental functions we defined in the last chapter. Our aim is
to differentiate expressions involving these functions.
At the end of this section, the student will be able to:
• perform differentiation involving inverse trigonometric functions;
• differentiate expressions involving hyperbolic functions; and
• find derivatives of functions involving inverse hyperbolic functions.
2.4.1
Derivatives of Inverse Circular Functions
em
at
ic
s
We now do calculus on these functions. We begin with their derivatives, which we enumerate in
the following theorem.
at
h
Theorem 2.4.1.
e
ut
1
1 + x2
In
s
3. Dx tan−1 x =
tit
1
2. Dx cos−1 x = − √
1 − x2
of
M
1
1. Dx sin−1 x = √
1 − x2
4. Dx cot−1 x = −
1
1 + x2
1
5. Dx sec−1 x = √
x x2 − 1
1
6. Dx csc−1 x = − √
x x2 − 1
x = sin y
UP
Proof. We show the proof of statement 1 only. The rest can be proved similarly.
y = sin−1 x
Dx (x) = Dx (sin y)
dy
1 = cos y ·
dx
dy
1
=
dx
cos y
p
√
dy
Note that cos y = ± 1 − sin2 y = ± 1 − x2 . Since y ∈ − π2 , π2 , cos y ≥ 0. Therefore,
=
dx
1
√
.
1 − x2
Example 2.4.2. Find
1. y = sin−1 (3x)
Solution.
dy
.
dx
108
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
dy
3
1
· (3) = √
=p
dx
1 − 9x2
1 − (3x)2
2. y = cot−1 (ln x)
Solution.
1
dy
1
·
= −
2
dx
1 + (ln x)
x
y
2
1
3.
= log2 (x − y) − csc−1 (3x2 )
5
e
Derivatives of Hyperbolic Functions
tit
ut
2.4.2
of
M
at
h
em
at
ic
We perform implicit differentiation.
y 1
dy
1
1 dy
=
· 1−
ln
5
5 dx
(x − y) ln 2
dx
1
−1
2
√
− 2 csc
3x · −
· 6x
3x2 9x4 − 1
1
6x
−1
2
√
− 2 csc
3x · −
dy
(x − y) ln 2
3x2 9x4 − 1
y =
dx
1
1
1
ln
+
5
5
(x − y) ln 2
s
Solution.
2. Dx (cosh x) = sinh x
UP
1. Dx (sinh x) = cosh x
In
s
Theorem 2.4.3.
3. Dx (tanh x) = sech2 x
4. Dx (coth x) = − csch2 x
5. Dx (sech x) = − sech x tanh x
6. Dx (csch x) = − csch x coth x
Proof. Again, we shall prove the first statement only.
If y = sinh x =
ex − e−x
dy
ex − (e−x ) · (−1)
ex + e−x
, then
=
=
= cosh x.
2
dx
2
2
Example 2.4.4. Find
1. y = csch x5 − 2
dy
.
dx
Solution.
dy
= − csch x5 − 2 coth x5 − 2 · (5x4 )
dx
2.4. DERIVATIVES OF OTHER NEW CLASSES OF FUNCTIONS
2. y = tanh3
√
x+3
109
Solution.
√
√
dy
1
x + 3 · sech2
x+3 · √
= 3 tanh2
dx
2 x+3
3. y = (cosh x)sinh x , cosh x > 0
Solution.
y = esinh x ln(cosh x)
1
dy
sinh x ln(cosh x)
cosh x · ln(cosh x) + sinh x ·
=e
· sinh x
dx
cosh x
2.4.3
Derivatives of Inverse Hyperbolic Functions
em
at
ic
s
We will find that the derivative of each inverse hyperbolic function resembles that of the corresponding inverse circular function. The proofs of the statements in the next theorem are left as
exercises.
Theorem 2.4.5.
at
h
1
4. Dx (coth−1 x) =
x2 + 1
1
x2 − 1
1
, |x| > 1
1 − x2
1
5. Dx (sech−1 x) = − √
x 1 − x2
e
2. Dx (cosh−1 x) = √
of
M
1. Dx (sinh−1 x) = √
ut
1
, |x| < 1
1 − x2
dy
Example 2.4.6. Find
.
dx
6. Dx (csch−1 x) = −
UP
In
s
tit
3. Dx (tanh−1 x) =
1. y = sinh−1 (1 − x)
Solution.
dy
1
−1
=p
· (−1) = p
dx
(1 − x)2 + 1
(1 − x)2 + 1
x 1
−1 4
−1
2. y = coth (x ) − sech
4
Solution.
dy
1
=
· 4x3 +
8
dx
1−x
1
1 x
4
q
1−
x 1
1
·
ln
4
4
1 2x
4
3. y = log10 cosh−1 (2x)
Solution.
dy
1
1
=
·√
·2
−1
dx
cosh (2x) ln 10
4x2 − 1
1
√
|x| x2 + 1
110
2.4.4
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
Exercises
Exercises for Discussion
dy
A. Find
.
dx
√
sec−1 (2x ) − e
4. y =
[log (3x)] tan−1 x
−1 x
1. y = 5cos
√
2. y = cos−1 6 cos x4 + ecot x − 4π 2
5. sin−1 (xy) = cos−1 (x − y)
tan−1 √x
6. y = sin−1 x2
3. y = sec ln csc−1 x
B. Find
dy
.
dx
3
1. y = cosh(ln x)
√
2. y = tanh3 3 2x
log5 (sech(ex ))
3. y =
sinh (x − tan−1 x)
em
at
ic
s
5. cosh(x − y) = log3 x − tanh(x2 + y 2 )
√
6. sec−1 ( x) = tanh(x2 y) + y sinh 2
at
h
dy
.
dx
of
M
C. Find
4. y = 10csch(x ) sec−1 (cosh 4x)
1. y = cosh−1 x7
e
ut
tit
UP
A. Find
In
s
2. y = tanh−1 (5x2 ) csch−1 (ln x)
√
tanh−1 x
3. y =
1 + sec x2
Supplementary Exercises
sin−1 (ex sinh 3x) + coth−1 (x2 )
√
sech−1 x
√
5. sinh−1 (x + y) = cosh−1 (x − y) + x
3x2 +1
6. cosh x = tanh−1 y
4. y =
dy
. There is no need to simplify.
dx
x+1
√
1. y = sin
2
−1
2. y = tan (sech x) + sech tan−1 x
−1
3. y = csch(cos−1 (x2 ))
2
x −1
2
4. y = tanh log7
x+1
tanh
x
e
5. y = 2
x − sech x
6. y = sinh−1 (ln x) + sin−1 (ln x)
7. y =
√
log5 x
+sech(4x−2 −sec−1 ( x))
−1
coth (2x)
8. sec−1 (2y cos x) = xy
9. 3y = y tanh(x) + 2
10. 4y = y 2 − cos−1 (x) + 1
2.5. THE MEAN VALUE THEOREM
2.5
111
The Mean Value Theorem
In this section, we discuss one of the most important theorems in calculus, namely, the Mean
Value Theorem. It is used to prove several other theorems including many of the ones that will be
discussed in this course.
At the end of this section, the student will be able to:
• interpret Rolle’s Theorem and Mean Value Theorem using graphs
• solve problems by applying Rolle’s Theorem and Mean Value Theorem
2.5.1
Rolle’s Theorem
em
at
ic
s
The following theorem is a specific case of the Mean Value Theorem and it was first proved by the
French mathematician Michel Rolle in 1691.
Theorem 2.5.1 (Rolle’s Theorem). Let f be a function such that
at
h
(i) f is continuous on the closed interval [a, b]
of
M
(ii) f is differentiable on the open interval (a, b) and
(iii) f (a) = 0 = f (b).
In
s
tit
ut
e
Then there exists a number c in the open interval (a, b) such that f 0 (c) = 0.
y
y
y
UP
c
a
a
a
bx
Figure 2.5.1
b1
Figure 2.5.2
b2
b
x
x
Figure 2.5.3
Remarks.
1. Note that condition (iii) of Rolle’s Theorem implies that the line passing through the points
(a, f (a)) and (b, f (b)) is horizontal. On the other hand, the conclusion implies that there is
a horizontal tangent line to the graph of f and the point of tangency has x-coordinate lying
between a and b. Refer to Figure 2.5.1.
2. Continuity on [a, b] is important because there are functions that satisfy only conditions (ii) and
(iii) but do not satisfy the conclusion. Refer to Figure 2.5.2 and consider the function on the
interval [a, b1 ]. It may also be the case that the conclusion is satisfied even if one of premises is
not satisfied. Refer to Figure 2.5.2 and consider the function on the interval [a, b2 ].
112
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
3. Notice that f need not be differentiable at the endpoints a and b. Refer to Figure 2.5.1.
4. The number c ∈ (a, b) in the conclusion need not be unique. Refer to Figure 2.5.3.
2.5.2
The Mean Value Theorem
Theorem 2.5.2 (The Mean Value Theorem). Let f be a function such that
(i) f is continuous on the closed interval [a, b]
(ii) f is differentiable on the open interval (a, b).
Then there is a number c on the open interval (a, b) such that f 0 (c) =
f (b) − f (a)
.
b−a
em
at
ic
s
Remark 2.5.3. The Mean Value Theorem is the generalization of Rolle’s Theorem where the line
`, passing through (a, f (a)) and (b, f (b)) is not necessarily horizontal. The conclusion says that
there is a tangent line to the graph of f that is parallel to ` and whose point of tangency has
x-coordinate between a and b. Refer to Figure 2.5.4 and 2.5.5.
y
at
h
y
c
x
b
tit
ut
e
a
of
M
`
Figure 2.5.5
UP
Example 2.5.4.
x
In
s
Figure 2.5.4
c2 b
a c1
1. Determine if Rolle’s Theorem is applicable to the given functions on the indicated intervals:
a. f (x) = x3 − 4x2 + 5x − 2 on [1, 2]
Solution.
Note that f (x) is continuous on [1, 2] and differentiable on (1, 2) because it is a polynomial.
Moreover, f (1) = 0 = f (2). Hence Rolle’s Theorem can be applied. We therefore conclude
that there exists a c such that 1 < c < 2 and f 0 (c) = 0. In fact, we can solve for c. Since
f 0 (x) = 3x2 − 8x + 5 = (3x − 5)(x − 1),
then the number c = 53 is in the interval (1, 2) and satisfies f 0 (c) = 0.
(
x2
, x ≤ 21
b. f (x) =
on [0, 1]
x − 1 , x > 12
Solution.
2.5. THE MEAN VALUE THEOREM
113
Notice that the only possible point of discontinuity of f (x) is at x = 21 . Checking conditions
for continuity at x = 21 , we have
1
1
1
1
(i) f
=
(ii) lim f (x) = lim (x − 1) = −
(iii) lim f (x) = lim x2 =
1+
1+
1−
1−
2
4
2
4
x→
x→
x→
x→
2
2
2
2
Hence, f is not continuous at x = 21 and Rolle’s Theorem cannot be applied.
2. Apply Rolle’s Theorem to f (x) = 2x3 − 3x2 + x to show that 6x2 + 1 = 6x has at least one real
root between 0 and 1.
Solution.
Note that f 0 (x) = 6x2 − 6x + 1. Since f is a polynomial and f (0) = 0 = f (1), then f satisfies
the assumptions of Rolle’s Theorem on [0, 1]. Hence there is a c ∈ (0, 1) such that f 0 (c) = 0.
This c is a root of the equation 6x2 + 1 = 6x.
s
1
x+2
. Show that there is a c ∈ (1, 2) such that if f 0 (c) = − .
x+1
6
em
at
ic
3. Let f (x) =
Solution.
of
M
at
h
Since −1 ∈
/ [1, 2] and f is a rational function, f is continuous on [1, 2] and differentiable on (1, 2).
4 3
1
f (2) − f (1)
= − =− .
By the Mean Value Theorem, there is a c ∈ (1, 2) such that f 0 (c) =
2−1
3 2
6
ut
e
4. Suppose that f (x) is continuous on [6, 15] and differentiable on the interval (6, 15) and f 0 (x) ≤ 10
for all x. If f (6) = −2, what is the largest possible value for f (15)?
tit
Solution.
UP
In
s
Note that f satisfies the hypotheses of the Mean Value Theorem. Hence there is a c ∈ (6, 15)
such that
f (15) − f (6)
f (15) + 2
f 0 (c) =
=
≤ 10.
15 − 6
9
Therefore, f (15) ≤ 88.
114
2.5.3
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
Exercises
Exercises for Discussion
A. Determine if Rolle’s Theorem applies to the following functions on the given intervals, and if
so, find all values of c satisfying the conclusion of the theorem.
1. f (x) = x2 − x − 2 on [−1, 2]
4. f (x) =
2. f (x) = x3/4 − 2x1/4 on [0, 4]
3. f (x) = 3 cos2 x on [ π2 , 3π
2 ]
x2 − x − 12
on [−3, 4]
x−3
5. f (x) = 1 − |x| on [−1, 1]
B. Determine if the Mean Value Theorem applies to the following functions on the given intervals,
and if so, find all values of c satisfying the conclusion of the theorem.
1. f (x) = x2 + 2x − 1 on [0, 1]
2. f (x) =
em
at
ic
s
3. f (x) = 3(x − 4)2/3 on [−4, 5]
√
4. f (x) = 1 + cos x on [− π2 , π2 ]
4
on [ 25 , 4]
9 − 2x
5. f (x) = (1 + sin x)2 on [−π, 0]
at
h
C. Do as indicated.
of
M
1. Using the function f (x) = x4 − 2x2 + 4x, show that the Rolle’s Theorem confirms there
exist x ∈ [−2, 0] such that x is a root of the equation x3 − x + 1 = 0.
ut
e
2. Prove that the equation x3 + 2x + 5 = 0 cannot have more than one real root.
In
s
tit
3. Let g be a function such that g 0 (x) ≥ −6 for all x ∈ [0, 2]. If g(0) = 0, determine the least
possible value of g(2).
UP
4. Show that sin b − sin a ≤ b − a whenever b ≥ a. In particular, show that sin x < x for all
x > 0.
5. Let f (x) be differentiable at everywhere and suppose that f (1) = 1, f 0 (x) < 0 on (−∞, 1),
and f 0 (x) = 1 on (1, ∞).
a. Show that f (x) ≥ 1 for all x.
b. Must f 0 (1) = 0? Explain your answer.
6. Suppose two runners in a 100m dash finish a tie. Show that there is at least one instant
during the race where they had the same velocity.
Supplementary Exercises
A. Determine if Rolle’s Theorem applies to the following functions on the given intervals, and if
so, find all values of c satisfying the conclusion of the theorem.
1. f (x) = x2 − 4x + 3 on [1, 3]
2. f (x) = sin πx − x2 + 2x on [0, 2]
3. f (x) = sin3 x cos x on [0, 2π]
4. f (x) = |x − 2| − 1 on [1, 3]
(
x2 − 4 if x < 1
5. f (x) =
on [−2, 58 ]
5x − 8 if 1 ≤ x
2.5. THE MEAN VALUE THEOREM
115
B. Determine if the Mean Value Theorem applies to the following functions on the given intervals,
and if so, find all values of c satisfying the conclusion of the theorem.
1. f (x) =
−2
on [−2, −1]
3x + 7
3. f (x) = ex + e4−x on [0, 4]
4. f (x) = ln 3x on [1, e]
5. f (x) = csc−1 (x − 1) on [1, 3]
2. f (x) = (2x − 11)−2 on [4, 6]
C. Do as indicated.
1. Suppose f (x) is continuous and differentiable everywhere. Suppose also that f (x) has at
least two distinct zeros. Show that f 0 (x) has at least one zero.
at
h
em
at
ic
s
2. Use the Mean Value Theorem to show that the graph of f (x) = x5 − 3x3 − x + 5 has a
tangent line on (0, 2) which is parallel to 3x − y = 2.
18
3. Let f (x) = cos(2x) − 5 sin(5x). Show that there exists c ∈ (0, π6 ) such that f 0 (c) = − .
π
√
y−x
√
4. Use the Mean Value Theorem to prove that if 0 < x < y, then y − x < √ . In
2 x
particular, show that the geometric mean of x and y is less than their arithmetic mean,
√
i.e. xy < 12 (x + y).
UP
In
s
tit
ut
e
of
M
5. Suppose f is continuous on an interval I. Use the Mean Value Theorem to show that if
f 0 (x) = 0 for all x ∈ I, then f is constant in I.
π
6. Use the previous item to prove the identity sin−1 x + cos−1 x = on [−1, 1].
2
116
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
2.6
Relative Extrema of a Function
In this chapter, we wish to analyze a function’s behavior, which is best described using its graph.
We aim to sketch the graph of a function without the help of graphing software. To this end, we
will first develop tools to determine the turning points of the graph of a function.
At the end of this section, the student will be able to:
• interpret the notion of increasing functions, decreasing functions, and relative extrema of a function graphically
• determine if a function is increasing or decreasing on an interval using the first
derivative
2.6.1
em
at
ic
s
• find the relative extrema using the First Derivative Test
Relative Extrema
at
h
We define below the instances when a function attains a relative extremum at a given point.
of
M
1. A function f is said to have a relative maximum at x = c if there is an open interval I,
containing c, such that f (x) is defined for all x ∈ I and f (x) ≤ f (c) for all x ∈ I.
tit
ut
e
2. A function f is said to have a relative minimum at x = c if there is an open interval I,
containing c, such that f (x) is defined for all x ∈ I and f (x) ≥ f (c) for all x ∈ I.
In
s
3. We say f has a relative extremum at x = c if f has either a relative maximum at x = c or a
relative minimum at x = c.
UP
4. If f has a relative extremum at x = c, then it is equivalent to saying that (c, f (c)) is a relative
extremum point of f , or f (c) is a relative extremum value of f .
Example 2.6.1. Let f be a function whose graph is illustrated in Figure 2.6.1.
c1
c2
c3
c4
Figure 2.6.1
• f has a relative maximum at x = c3 and at x = c5 .
c5
c6
2.6. RELATIVE EXTREMA OF A FUNCTION
117
• f has a relative minimum at x = c4 .
• f has neither a relative maximum at x = c1 nor a relative minimum at x = c6 because there is
no open interval for which the point (c1 , f (c1 )) is the highest, or (c6 , f (c6 )) is the lowest.
• f has no relative minimum at x = c2 because f (c2 ) is not defined.
2.6.2
Critical Numbers
To make it easier to locate the relative extrema of a function, we narrow down our search to a few
candidates.
A number c ∈ domf is said to be a critical number of f if either f 0 (c) = 0 or f 0 (x) is undefined at
x = c.
Example 2.6.2. Find all the critical numbers of the following functions.
s
x3
− 3x
9
em
at
ic
1. f (x) =
Solution.
R. Differentiating f , we get f 0(x) = x3 − 3 = 13 (x2 − 9) = 13 (x − 3)(x + 3).
2
Note that dom f =
of
M
at
h
The derivative is always defined and f 0 (x) = 0 when x = 3 or x = −3. Therefore the critical
numbers are 3 and −3.
The domain of f is
R and its derivative is f 0(x) = 4x3 + 6x2 = 2x2(2x + 3). Hence, the critical
tit
Solution.
ut
e
2. f (x) = x4 + 2x3
3. f (x) =
x2
9 − x2
UP
In
s
3
numbers of f are 0 and − .
2
Solution.
Note that dom f =
R \ {±3}. The derivative of f is f 0(x) = (9 −18xx2)2 . Note that f 0(x) is
not defined at x = ±3 but ±3 are not critical numbers of f because ±3 ∈
/ domf . Meanwhile,
0
f (x) = 0 when x = 0. Thus, the only critical number of f is 0.
4. f (x) = −x4/3 + 4x1/3
Solution.
First, dom f =
− 1)
R. Differentiating f , we get f 0(x) = − 34 x1/3 + 43 x−2/3 = −4(x
. Note that
3x2/3
f 0 (x) = 0 if and only if x = 1 and that f 0 (x) is undefined at x = 0. Moreover, 0 ∈ domf . Thus,
the critical numbers of f are 0 and 1.
The following theorem tells us that finding the critical numbers of a function takes us one step
closer to finding its relative extremum points.
118
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
Theorem 2.6.3. If f has a relative extremum at x = c, then c is a critical number of f .
Illustration 2.6.4.
1. The graph of f (x) = x2 given in Figure 2.6.2 has a relative maximum point at its vertex
(0, 0). Note that f 0 (x) = 2x. Indeed f 0 (0) = 0.
2. Take f (x) = |x − 1|. From its graph shown in Figure 2.6.3, it is clear that f has a relative
minimum at x = 1. Since the graph has a corner at x = 1, we know that f 0 (1) is undefined.
y = −x2
s
y = |x − 1|
Figure 2.6.3
em
at
ic
Figure 2.6.2
Increasing/Decreasing Functions
tit
2.6.3
ut
e
of
M
at
h
Remark 2.6.5. The preceding theorem says that relative extrema can only be attained at critical
numbers of f , that is, if a is not a critical number, then the point (a, f (a)) is not a relative extremum
point of f . However, these critical numbers are only candidates for relative extrema. If c is a critical
number, f may or may not have a relative extremum at x = c. This will be illustrated in the next
sections.
UP
In
s
Our next goal is to determine when the graph of a function rises or falls. We will see then that this
is closely related to finding the relative extremum points of a function.
Let f be a function defined on an interval I.
1. f is said to be (strictly) increasing on I if f (a) < f (b) for all a, b ∈ I such that a < b.
2. f is said to be (strictly) decreasing on I if f (a) > f (b) for all a, b ∈ I such that a < b.
Example 2.6.6. Let f be a function whose graph is illustrated in Figure 2.6.1. Then
• f is decreasing on [c1 , c2 ), [c3 , c4 ] and [c5 , c6 ]. But note that it is incorrect to say f is decreasing
on [c1 , c2 ) ∪ [c3 , c4 ] ∪ [c5 , c6 ].
• f is increasing on (c2 , c3 ] and [c4 , c5 ].
The following theorem gives us an analytical (as opposed to graphical) method of showing that the
function is increasing or decreasing. It should be noted that the Mean Value Theorem is a key in
proving this theorem.
2.6. RELATIVE EXTREMA OF A FUNCTION
119
Theorem 2.6.7. Let f be a function that is continuous on the closed interval [a, b] and differentiable on the open interval (a, b).
1. If f 0 (x) > 0 for all x ∈ (a, b), then f is increasing on [a, b].
2. If f 0 (x) < 0 for all x ∈ (a, b), then f is decreasing on [a, b].
3. If f 0 (x) = 0 for all x ∈ (a, b), then f is constant on [a, b].
Example 2.6.8.
A. The derivative of f (x) = x2 is f 0 (x) = 2x. Note that f 0 (x) < 0 for all x ∈ (−∞, 0) and f 0 (x) > 0
for all x ∈ (0, ∞). Indeed, as seen in Figure 2.6.4, the graph of f is decreasing on the interval
(−∞, 0] and is increasing on the interval [0, ∞).
em
at
ic
s
B. If f (x) = x3 , then f 0 (x) = 3x2 which is never negative. Observe the graph of f (x) = x3 in
Figure 2.6.5. Indeed it is always increasing.
y = x2
ut
The First Derivative Test for Relative Extrema
In
s
2.6.4
Figure 2.6.5
tit
Figure 2.6.4
e
of
M
at
h
y = x3
UP
Note that if we want to reach the peak of a hill, we first have to climb up a slope and when we
finally reach the top, we would go downhill. Analogously, when a continuous function y = f (x)
attains a relative extremum, say a relative maximum, the behavior of the graph of f transitions
from increasing to decreasing as it proceeds from the left of the point to its right.
Theorem 2.6.9 (First Derivative Test for Relative Extrema). Let f be a function continuous
on the open interval (a, b) which contains the number c. Suppose that f is also differentiable
on the interval (a, b), except possibly at c.
1. If f 0 (x) > 0 for all x ∈ (a, c) and f 0 (x) < 0 for all x ∈ (c, b), then f has a relative maximum
at x = c.
2. If f 0 (x) < 0 for all x ∈ (a, c) and f 0 (x) > 0 for all x ∈ (c, b), then f has a relative minimum
at x = c.
3. If f 0 (x) does not change signs from (a, c) to (c, b), then there is no relative extremum at
x = c.
120
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
We will use Theorem 2.6.9 to verify whether a function has relative extrema at its critical numbers.
We enumerate below some suggested steps in finding the relative extrema of a given function.
Locating relative extremum values of a function
• Determine the critical numbers of f .
• Determine the sign of f 0 on the left and right of each critical number. (Note that in creating
the table of signs of f 0 , one should consider all values of x that make f 0 either zero or undefined,
regardless of whether they are critical numbers or not.)
• Conclude accordingly using the First Derivative Test.
Example 2.6.10. Determine the intervals where the given function is increasing or decreasing and
find all of its relative extrema.
s
x3
− 3x
9
em
at
ic
1. f (x) =
Solution.
x2 − 9
and the critical numbers of f are ±3.
3
−3
3
−
+
of
M
f 0 (x)
at
h
From Example 2.6.2.1, we know f 0 (x) =
+
tit
ut
e
Then f is increasing on (−∞, −3] and on [3, ∞), while it is decreasing on [−3, 3]. By the first
derivative test, (−3, 6) is a relative maximum point of the graph of f and (3, −6) is a relative
minimum point of the graph of f .
UP
Solution.
In
s
2. f (x) = x4 + 2x3
From Example 2.6.2.2, f 0 (x) = 4x3 + 6x2 = 2x2 (2x + 3) and the critical numbers of f are 0 and
− 23 .
− 32
0
f 0 (x) −
+
+
Hence f is decreasing on (−∞, − 23 ] and increasing on [− 32 , ∞). Therefore, f has a relative
minimum at x = − 23 . It is good to observe that 0 is a critical number of f yet f does not have
a relative extremum at 0.
3. f (x) =
x2
9 − x2
Solution.
18x
and the only critical number of f is 0.
(9 − x2 )2
0
−
3
+
999
−3
f 0 (x) −
999
From Example 2.6.2.3, f 0 (x) =
+
2.6. RELATIVE EXTREMA OF A FUNCTION
121
The graph of f is decreasing on (−∞, −3) and on (−3, 0]. It is increasing on [0, 3) and on (3, ∞).
Therefore, f has a relative minimum at x = 0.
4. f (x) = −x4/3 + 4x1/3
Solution.
From Example 2.6.2.4, f 0 (x) =
−4(x − 1)
and the critical numbers of f are 0 and 1.
3x2/3
0
f 0 (x)
1
+
−
+
Then the graph of f is increasing on (−∞, 1] and decreasing on [1, ∞). We conclude that f has
a relative maximum at x = 1.
5. f (x) =
x2 + 4x + 3
2(x − 1)2
e
f 0 (x) −
− 53
+
1
999
of
M
5
of f is − . We form the following table:
3
−(3x + 5)
. Note that the only critical number
(x − 1)3
at
h
Differentiating and simplifying, we get f 0 (x) =
em
at
ic
s
Solution.
−
UP
In
s
tit
ut
We conclude that f is increasing on [− 35 , 1); decreasing on the intervals (−∞, − 53 ] and (1, ∞);
and has a relative minimum at x = − 35 . Notice that even if there was a change of sign at x = 1,
we don’t say that f has a relative maximum at x = 1 since f is undefined when x = 1. In the
first place, 1 is not a critical number of f .
122
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
2.6.5
Exercises
Exercises for Discussion
A. Determine the intervals on which the graph of the following functions is increasing or decreasing. Use the first derivative test to find its relative extremum points.
√
6. f (x) = 3 x + 2
√
7. f (x) = 2x 3 − x
1. f (x) = 1 − 4x − x2
2. f (x) = x2 (x − 1)3
3. f (x) = x2/3 (x − 1)2
8. f (x) = x2 − sin x
x2
4. f (x) = 2
x +2
√
1+ x
√
5. f (x) =
1− x
9. f (x) =
e2x + 16e−x
2
s
10. f (x) = ln(x2 + 1)
2.
of
M
1.
3
1
ut
tit
(−2, 0)
−2
UP
In
s
1
−1(−1, 0)
2
e
(0, 2)
2
−2
at
h
em
at
ic
B. Given the graph of the derivative of f and assuming f is continuous everywhere, find the
intervals where f is increasing, decreasing and the x-coordinates of the relative extrema of f .
1
(2, 0)
−1
1
2
3
−1
−2
2
−1
−3
C. Do as indicated.
1. Suppose f (x) in increasing on (0, 1]. Show that f
1
2
x +1
is decreasing on
R.
2. The population P of a new species of frogs in a conservation facility is modelled by
P (t) = 50 +
2500t2
,
25 + t2
where t is the number of years after the introduction of the new species.
a. How many frogs will populate the area in the long run?
b. Describe the behavior of the population. Is this a good model? Why or why not?
2.6. RELATIVE EXTREMA OF A FUNCTION
123
Supplementary Exercises
A. Determine the intervals on which the graph of the following functions is increasing or decreasing. Use the first derivative test to find its relative extremum points.
1. f (x) = 4x3 + 3x2 − 18x
(
6. f (x) =
2. f (x) = x3 − 9x2 + 24x − 16
3. f (x) = x3 − 12x2 + 21x
x2 − x − 2, if x < 2
(x − 2)3 ,
if x ≥ 2
7. f (x) = sin2 x − 2 cos x
√
8. f (x) = 2x − 2 cos x
x2 + 4
4. f (x) =
x
x
5. f (x) =
(3x + 1)2
9. f (x) = ln cosh x
em
at
ic
s
B. Given the graph of the derivative of f and assuming f is continuous everywhere, find the
intervals where f is increasing, decreasiing and the x-coordinates of the relative extrema of
f.
1
at
h
2
2
of
M
(0, 1)
−1
(0, −1)
2
3
(−3, 0)
−4
(−1, 0)
−3
−2
(2, 0)
−1
(0, 0)1
2
−1
(−2, −1)
(2, −2)
−2
UP
−2
1
ut
−1
(1, 32 )
(− 21 , 34 ) 1
tit
−2
In
s
−3
e
1
2
C. Do as indicated.
1
x3
2. Use a first derivative to show that sin x < x for all x > 0.
1. Show that if f is decreasing on (0, +∞), the g(x) = f
is increasing on (0, +∞).
3. Find a polynomial with critical numbers 0 and 3 with relative maximum at x = 0 and no
relative extremum at x = 3.
4. Find the value of c such that f (x) = x + ln(x2 + 1) + c tan−1 x has a relative maximum at
x = −5 and a relative minimum at x = 3.
124
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
2.7
Concavity and the Second Derivative Test
The shape of the graph of a function provides many useful insights on the physical quantity that
it represents. Thus, we study the concavity of a graph.
At the end of this section, the student will be able to:
• interpret the notion of concavity, and points of inflection of a function graphically
• determine if a function is concave up or concave down on an interval using the
second derivative
• find the relative extrema using the Second Derivative Test
Concavity
em
at
ic
s
2.7.1
We define below the concept of concavity at a point, and on an interval.
of
M
at
h
1. The graph of a function f is said to be concave up at the point P (c, f (c)) if f 0 (c) exists and
there is an open interval I containing c such that for all x ∈ I \ {c}, the point (x, f (x)) is
above the tangent line to the graph of f at P . We say that the graph of f is concave up on
an interval I if it is concave up at (c, f (c)) for all c ∈ I.
UP
In
s
tit
ut
e
2. The graph of a function f is said to be concave down at the point P (c, f (c)) if f 0 (c) exists
and there is an open interval I containing c such that for all x ∈ I \ {c}, the point (x, f (x))
is below the tangent line to the graph of f at P . We say that the graph of f is concave down
on an interval I if it is concave down at (c, f (c)) for all c ∈ I.
Example 2.7.1. Let f be a function whose graph is given below.
P
c1
c2
c3
c4
c5
c6
c7
Figure 2.7.1
• The graph of f is concave up at the point P .
• The graph of f is concave up on (c1 , c2 ) ∪ (c2 , c3 ) ∪ (c3 , c5 ) and concave down on (c5 , c7 ).
2.7. CONCAVITY AND THE SECOND DERIVATIVE TEST
125
• The graph of f is neither concave up nor concave down at the point (c5 , f (c5 )).
• Concavity of the graph of f at x = c1 and x = c3 cannot be determined because f 0 (c1 ) and
f 0 (c3 ) are not defined.
• Concavity of the graph of f at x = c2 and x = c7 cannot be determined because f is not defined
at these values of x.
em
at
ic
s
We wish to find another characterization of concavity aside from its graphical definition. Observe
the behavior of the tangent lines to the concave up curve in Figure 2.7.2a. As a point goes from
left to right, the slope increases. Meanwhile, the slope of the tangent line decreases as a point goes
from left to right along the concave down curve in Figure 2.7.2b. We shall state this as a theorem.
Figure 2.7.2b
of
M
at
h
Figure 2.7.2a
tit
ut
e
Recall that f 00 (x) = Dx (f 0 (x)). Hence if f 00 (x) > 0 for all x ∈ (a, b), then f 0 (x) is increasing on
(a, b) and the graph of f is concave up on (a, b). If f 00 (x) < 0, then f 0 (x) is decreasing on (a, b) and
the graph of f is concave down on (a, b).
In
s
Theorem 2.7.2 (Test for Concavity). Let f be a function such that f 00 exists on (a, b).
UP
1. If f 00 (x) > 0 for all x ∈ (a, b), then the graph of f is concave up on (a, b).
2. If f 00 (x) < 0 for all x ∈ (a, b), then the graph of f is concave down on (a, b).
2.7.2
Point of Inflection
Example 2.7.3. If f (x) = x3 , then f 00 (x) = 6x. Note that f 00 (x) > 0 when x > 0 and f 00 (x) < 0
when x < 0. Hence, the graph of f is concave up on (0, ∞) and concave down on (−∞, 0) (refer to
Figure 2.7.3a). Observe that that the graph of f changed concavity at x = 0. In this case, we call
the point (0, f (0)) = (0, 0) a point of inflection.
g(x) = x2/3
f (x) = x3
√
f (x) = 3 x
f (x) = x4
Figure 2.7.3a
Figure 2.7.3b
Figure 2.7.3c
Figure 2.7.3d
126
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
The graph of f (x) has a point of inflection at P (c, f (c)) if f is continuous at x = c and the graph
of f changes concavity at P , i.e. there is an open interval (a, b) containing c such that
1. f 00 (x) > 0 for all x ∈ (a, c) and f 00 (x) < 0 for all x ∈ (c, b) or;
2. f 00 (x) < 0 for all x ∈ (a, c) and f 00 (x) > 0 for all x ∈ (c, b).
Theorem 2.7.4. If P (c, f (c)) is a point of inflection of the graph of the function f , then
f 00 (c) = 0 or f 00 (c) does not exist.
Example 2.7.5.
of
M
at
h
em
at
ic
s
1. The graph of f (x) = x3 has a point of inflection when x = 0. Indeed, f 00 (0) = 0.
√
2. The graph of f (x) = 3 x (refer to Figure 2.7.3b) has a point of inflection at x = 0. Note that
2
f 00 (x) = − 5/3 . Indeed, f 00 (0) is undefined.
9x
Remark 2.7.6. The converse of the preceding theorem is not true. That is, if f 00 (c) = 0 or f 00 (c)
dne, the graph of f may or may not have a point of inflection at x = c. Take for example the
graphs of f (x) = x4 and g(x) = x2/3 illustrated in Figures 2.7.3c and 2.7.3d. The graph of both
functions have no points of inflection but f 00 (x) = 0 and g 00 (x) does not exist when x = 0. From
now on, we will refer to the values of x in the domain of f that satisfy either f 00 (x) = 0 or f 00 (x)
does not exist, as possible points of inflection.
Solution.
Note that f 00 (x) =
In
s
x3
− 3x
9
UP
1. f (x) =
tit
ut
e
Example 2.7.7. Determine all points of inflection of the graph of the following functions and
discuss their concavity.
2x
= 0 when x = 0.
3
0
f 00 (x) −
+
Then the graph of f has a point of inflection at (0, 0) and the graph of f is concave down on
(−∞, 0) and concave up on (0, ∞).
2. f (x) = x4 + 2x3
Solution.
Observe that f 00 (x) = 12x2 + 12x = 12x(x + 1) = 0 when x = 0 or x = −1.
−1
f 00 (x)
+
0
−
+
Then (−1, −1) and (0, 0) are points of inflection of the graph f . The graph of f is concave up
on (−∞, −1) and (0, ∞). The graph of f is concave down on (−1, 0).
2.7. CONCAVITY AND THE SECOND DERIVATIVE TEST
3. f (x) =
127
x2
9 − x2
Solution.
(9 − x2 )2 18 − 18x(2)(9 − x2 )(−2x)
54(3 + x2 )
=
which is never zero and is
(9 − x2 )4
(9 − x2 )3
undefined when x = ±3.
−3
3
00
+
−
f (x)
−
999
999
Note that f 00 (x) =
Note that the graph of f cannot have points of inflection at x = ±3 because ±3 ∈
/ dom f . Also,
the graph of f is concave up on (−∞, −3) and (3, ∞), while f is concave down on (−3, 3).
4. f (x) = −x4/3 + 4x1/3
Solution.
−4(x + 2)
= 0 when x = −2 and f 00 (x) is undefined when x = 0.
9x5/3
−2
0
em
at
ic
f 00 (x) −
s
Verify that f 00 (x) =
+
−
2.7.3
of
M
at
h
√
Then (−2, 2 3 2) and (0, 0) are points of inflection of the graph of f . The graph of f is concave
down on (−∞, −2) and (0, ∞) and concave up on (−2, 0).
The Second Derivative Test for Relative Extrema
tit
ut
e
The following is another test for relative extrema that uses the second derivative of a function.
Compare this test with the first derivative test for relative extrema.
UP
In
s
Theorem 2.7.8 (Second Derivative Test for Relative Extrema). Let f be a function such that
f 0 and f 00 exist for all values of x on some open interval containing x = c and f 0 (c) = 0.
1. If f 00 (c) < 0, then f has a relative maximum value at x = c.
2. If f 00 (c) > 0, then f has a relative minimum value at x = c.
3. If f 00 (x) = 0, we have no conclusion (f may or may not have a relative extrema at x = c).
Remark 2.7.9. Note that the second derivative test is only applicable to the first type of critical
numbers: those that make the first derivative zero (x-coordinates of so-called stationary points).
Example 2.7.10. Given the following functions, determine if the second derivative test is applicable to their critical numbers. If yes, what can you conclude using the second derivative test?
1. f (x) =
x3
− 3x with C.N. : −3, 3
9
Solution.
x2 − 9
2x
and f 00 (x) =
. The second derivative test is applicable to both
3 00
3
−3 and 3. Observe that f (−3) = −2 < 0 and thus, f has a relative maximum at x = −3.
Similarly, f 00 (3) = 2 > 0 and thus, f has a relative minimum at x = 3.
Recall that f 0 (x) =
128
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
2. f (x) = x4 + 2x3 with C.N. : 0, − 32
Solution.
Recall that f 0 (x) = 4x3 + 6x2 and f 00 (x) = 12x2 + 12x. The second derivative test is applicable
to both 0 and − 23 . Note that f 00 (0) = 0 and this gives us an inconclusive case. On the other
hand, f 00 − 32 = 9 > 0 and so f has a relative minimum at x = − 32 .
3. f (x) =
x2
with C.N. : 0
9 − x2
Solution.
2
18x
00 (x) = 54(3 + x ) . The second derivative test is
and
f
(9 − x2 )2
(9 − x2 )3
54(3)
applicable to 0. Since f 00 (0) = 93 > 0, f has a relative minimum at x = 0.
We have seen that f 0 (x) =
4. f (x) = −x4/3 + 4x1/3 with C.N. : 0, 1
at
h
em
at
ic
s
Solution.
−4(x − 1)
00 (x) = −4(x + 2) . The second derivative test is applicable to 1 but
and
f
f 0 (x) =
3x2/3
9x5/3
not to 0. Because f 00 (1) = − 43 , then f has a relative maximum at x = 1.
of
M
b
5. f (x) = ax2 + bx + c, where a 6= 0. C.N. : − 2a
.
Solution.
UP
In
s
tit
ut
e
b
Note that f 0 (x) = 2ax+b and f 00 (x) = 2a. The second derivative test is applicable to − 2a
. From
b
b
00
f (− 2a ) = 2a, we conclude that if a > 0, f has a relative minimum at x = − 2a . Meanwhile,
b
a < 0, then f has a relative maximum at x = − 2a
.
2.7. CONCAVITY AND THE SECOND DERIVATIVE TEST
2.7.4
129
Exercises
Exercises for Discussion
A. Identify the possible points of inflection of the graph of each function. Determine which of
these are indeed points of inflection using a table of signs. Determine the intervals on which
the graph of the function is concave up or concave down.
x
x2 + 1
√
1
6. f (x) = x + √
x
7. f (x) = sin x cos x
sin x
8. f (x) =
2 − cos x
1. f (x) = (x + 2)3
5. f (x) = √
2. f (x) = x2/3
x
3. f (x) = 2
x +2
4. f (x) =
x2 − 3
x2 + 1
em
at
ic
s
B. Determine which critical numbers satisfy the hypotheses of the second derivative test and
apply the test to these numbers.
1. f (x) = (x2 − 1)4
4. f (x) = 2x + cot x
5. f (x) = cos2 x − 2 sin x
√
6. f (x) = 3x + 2 sin x
at
h
2. f (x) = x1/3 (x + 4)
of
M
3. f (x) = x4/3 − 6x1/3
In
s
tit
ut
e
C. Given the graph of the derivative of f and assuming f is continuous everywhere, find the
intervals where f is concave up, concave down and the x-coordinates of the points of inflection
of f .
2
UP
1
(0, 0)
2
−2
(0, 1)
1
−2
−1
1
2
3
−2
(−1, 0)
−3
−1
(3, 0)
−1
−1
1
2
−3
3
(0, −1)
(−2, −4)
−4
−5
−2
(2, −2)
−6
Supplementary Exercises
A. Identify the possible points of inflection of the graph of each function. Determine which of
these are indeed points of inflection using a table of signs. Determine the intervals on which
the graph of the function is concave up or concave down.
130
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
1. f (x) = −x4 + 10x3 − 36x2 − 90x + 200
5. f (x) =
2. f (x) = x2 (2x2 − 4x + 3)
4. f (x) =
x2 + 2x + x
9−x
, if x ≤ 3
1 − 2x2 e2x
e2x
7. f (x) = ln(x3 − 1)
x2 − 3
, if x > 3
8. f (x) = 2x2 + ln x
6. f (x) =
3. f (x) = (x + 2)5 + x4 + 8x3 + 24x2
(
√
B. Determine which critical numbers satisfy the hypotheses of the second derivative test and
apply the test to these numbers.
√
√
3
4. f (x) = 2 3 x − x2
16000
x
200
2. f (x) = (x − 4)
−2
x
√
3. f (x) = x − 2 x + 1 + 1
1. f (x) = x2 +
5. f (x) = cos(π sin x)
6. f (x) = e−x
2
em
at
ic
s
7. f (x) = x − ln sech x
2.
D. Do as indicated.
UP
In
s
tit
ut
e
of
M
1.
at
h
C. Suppose that water is flowing at a constant rate into the container below. If f (t) is the water
level (height) in the container after time t, how would you describe the graph of y = f (t)?
1. If f is a function concave down everywhere, show that e−f (x) is concave up everywhere.
2. Suppose n is a positive integer such that n ≥ 2. Let g(x) = xn+1 − (n + 1)xn .
a. Prove that g has a relative minimum at x = n.
b. Explain why the Second Derivative Test cannot be used to show that g has a relative
maximum at x = 0 when n is even.
c. Show that g has a point of inflection at x = n − 1 and at x = 0 when n is odd.
2.8. GRAPH SKETCHING
2.8
131
Graph Sketching
We have laid out the tools we need to analyze the behavior of a function. We are now ready to
sketch the graph of functions using its derivatives.
At the end of this section, the student will be able to:
• find the vertical, horizontal, and oblique asymptotes of a function
• using derivatives, graph polynomial functions (of degree at most 5) and rational
functions
Here are some guidelines we can follow.
s
Guidelines for Graphing a Function f Analytically:
em
at
ic
1. Find the domain and intercepts of f .
2. Determine vertical, horizontal and oblique asymptotes of f .
at
h
3. Compute f 0 (x) and f 00 (x), and determine the critical numbers.
ut
e
of
M
4. Divide the real number line using the numbers that make f 0 zero or undefined and the
numbers that make f 00 zero or undefined. Determine the sign of f 0 (x) and f 00 (x) on each
interval. It would be easier if you make a table(s) of signs.
In
s
tit
5. Determine intervals on which the function is increasing,decreasing, concave up, concave down.
Determine the coordinates of the relative extremum points and points of inflection.
UP
6. Plot important points (intercepts, holes, extrema, points of inflection) and asymptotes first.
7. Use the table of signs of f 0 and f 00 to graph the rest of the function.
2.8.1
Graphing Polynomial Funtions
Example 2.8.1. Sketch the graph of f (x) =
x3
− 3x.
9
Solution.
R
domf :
√
x-intercept: x = 0, ±3 3
y-intercept: y = 0
x2 − 9
f 0 (x) =
CN: −3, 3
3
2x
PPOI : 0
f 00 (x) =
3
interval
(−∞, −3)
−3
(−3, 0)
0
(0, 3)
3
(3, ∞)
f
6
0
−6
f0
+
0
0
+
f 00
0
+
+
+
conclusion
inc., cd
rel. max.
dec, cd
POI
dec, cu
rel. min.
inc, cu
132
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
(−3, 6)
√
(−3 3, 0)
√
(3 3, 0)
(0, 0)
(3, −6)
Example 2.8.2. Sketch the graph of f (x) = x4 + 2x3 using derivatives.
domf :
em
at
ic
s
Solution.
R
f 0 (x) = 4x3 + 6x2
x-intercept: x = 0, −2
PPOI: 0, −1
at
h
f 00 (x) = 12x2 + 12x
CN: 0, − 23
ut
tit
−27
16
In
s
−1
0
f0
0
+
+
+
0
+
e
f
UP
interval
(−∞, − 32 )
− 32
(− 32 , −1)
−1
(−1, 0)
0
(0, ∞)
of
M
y-intercept: y = 0
(0, 0)
(−1, −1)
( −3
, −27
)
2
16
f 00
+
+
+
0
0
+
conclusion
dec., cu
rel. min.
inc, cu
POI
inc, cd
POI
inc, cu
2.8. GRAPH SKETCHING
2.8.2
133
Review of Asymptotes
In graphing functions accurately, it is also important to take note of its asymptotes. In this course,
we will discuss linear asymptotes of a function. The formal definition of an asymptote involves
limits.
The line x = a is a vertical asymptote of the graph of f (x) if at least one of the following is true:
(a) lim f (x) = ∞
x→a−
(b) lim f (x) = −∞
x→a−
(c) lim f (x) = ∞
(d) lim f (x) = −∞
x=a
x=a
x→a+
x=a
em
at
ic
s
x=a
x→a+
of
M
at
h
Remark 2.8.3. To look for the vertical asymptotes of the graph of a function, we have to think
of a real number a such that the limit of f as x approaches a is infinite. For a rational polynomial
P (x)
, x = a is a vertical asymptote if (x − a) is a factor of Q but not of P (or the multiplicity of
Q(x)
x − a as a factor of Q is greater than the multiplicity of x − a as a factor of P ).
ut
e
Example 2.8.4.
tit
x2 − 4
(x − 2)(x + 2)
=
has a vertical asymptote at x = 3 but has none at x = 2.
2
x − 5x + 6
(x − 3)(x − 2)
In
s
1. f (x) =
UP
To verify this using limits,
lim
(x − 2)(x + 2)
x→2 (x − 3)(x − 2)
2. f (x) =
= lim
(x + 2)
x→2 (x − 3)
= −4
(x − 2)(x + 2)
lim
= −∞
−
x→3 (x − 3)(x − 2)
1·5
(0− ) · 1
(x − 2)(x + 2)
lim
=∞
x→3+ (x − 3)(x − 2)
1·5
(0+ ) · 1
4 − x2
(2 − x)(2 + x)
=
has a vertical asymptote at x = 2.
2
x − 4x + 4
(x − 2)2
(2 − x)(2 + x)
−(x + 2)
lim
= lim
= ±∞
2
(x − 2)
x→2∓
x→2∓ (x − 2)
−4
0∓
134
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
The line y = b is a horizontal asymptote of the graph of f if at least one of the following is true:
y=b
(a) lim f (x) = b
x→∞
y=b
y=b
(b)
lim f (x) = b
x→−∞
y=b
at
h
em
at
ic
s
Remark 2.8.5. To find all horizontal asymptotes of a polynomial or rational function, we only
need to evaluate the limit of the function as x → ∞ and as x → −∞. If the limit is finite, then there
is a horizontal asymptote. Notice that in a rational function, whenever the degree of the numerator
is less than or equal to the degree of the denominator, then there is a horizontal asymptote.
e
x+3
has a horizontal asymptote at y = 0.
x2 − 1
tit
ut
1. f (x) =
of
M
Example 2.8.6.
UP
In
s
1
3
x + 3 x12
x + x2
· 1 = lim
=0
lim
x→∞ 1 − 12
x→∞ x2 − 1
x2
x
1
3
x + 3 x12
x + x2
·
=
lim
=0
1
x→∞ 1 − 12
x→−∞ x2 − 1
x2
x
lim
2. f (x) =
2x + 1
2
2
has a horizontal asymptote at y = and y = − .
|5x − 2|
5
5
2x + 1
2x + 1
2
= lim
=
x→∞ |5x − 2|
x→∞ 5x − 2
5
lim
2x + 1
2x + 1
2
= lim
=−
x→−∞ −5x + 2
x→−∞ |5x − 2|
5
lim
2.8. GRAPH SKETCHING
135
The graph of f has the line y = mx + b, m 6= 0, as an oblique asymptote if at least one of the
following is true:
(a) lim f (x) − (mx + b) = 0
x→∞
y = mx + b
lim f (x) − (mx + b) = 0.
x→−∞
em
at
ic
s
(b)
y = mx + b
y = mx + b
y = mx + b
y = mx + b
y = mx + b
y = mx + b
of
M
at
h
y = mx + b
Remark 2.8.7. If P (x) and Q(x) are polynomial functions with deg(P ) = deg(Q) + 1, then
e
has an oblique asymptote.
P (x)
Q(x)
ut
x2 + 3
has an oblique asymptote because the dex−1
gree of the numerator is 2, which is one degree greater than that of the denominator. To find the
equation of the oblique asymptote, we use long division.
UP
In
s
tit
Example 2.8.8. The rational function f (x) =
x+1
x−1
x2
+3
− x2 + x
x+3
−x+1
4
Thus
4
.
x−1
From here it can be shown that y = x + 1 is an oblique asymptote since
f (x) = (x + 1) +
4
= 0.
x→±∞ x + 1
lim f (x) − (x + 1) = lim
x→±∞
136
Graphing Rational Functions
Example 2.8.9. Sketch the graph of f (x) =
x2
using derivatives.
9 − x2
R
Solution.
domf : \ {±3}
x-int: x = 0 y-int: y = 0
Vertical Asymptotes: x =
−3and x = 3
9
lim f (x) = −∞
−
−
0
x→−3
9
lim f (x) = ∞
+
0
x→−3+
9
lim f (x) = ∞
0+ x→3−
9
lim f (x) = −∞
+
0−
x→3
Horizontal Asymptotes: y = −1
x2
1
lim
= lim 9
= −1
x→±∞ 9 − x2
x→±∞ 2 − 1
x
Oblique Asymptotes: none
18x
CN: 0
f 0 (x) =
(9 − x2 )2
54(3 + x2 )
f 00 (x) =
PPOI: none
(9 − x2 )3
interval
(−∞, −3)
−3
(−3, 0)
0
(0, 3)
3
(3, ∞)
f0
und
0
+
und
+
f
und
0
und
f 00
und
+
+
+
und
-
f (x) =
x = −3
conclusion
dec., cd
V.A.
dec, cu
rel. min.
inc, cu
V.A.
inc, cd
x2
9 − x2
x=3
em
at
ic
s
(0, 0)
at
h
y = −1
4x
x2 + 1
using derivatives.
Solution.
domf :
x-int: x = 0
y-int: y = 0
Vertical Asymptote.: none
Horizontal Asymptote: y = 0
ut
e
Example 2.8.10. Sketch the graph of f (x) =
of
M
2.8.3
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
UP
In
s
tit
R
4
x
lim
= 01 = 0
x→±∞ 1 + 12
x
Oblique Asymptote: none
4 − 4x2
f 0 (x) = 2
CN: −1, 1
(x + 1)2
√
8x(x2 − 3)
f 00 (x) =
PPOI: 0, ± 3
2
3
(x + 1)
interval
√
(−∞, − 3)
√
− 3
√
(− 3, −1)
−1
(−1, 0)
0
(0, 1)
1
√
(1, 3)
√
3
√
( 3, ∞)
(1, 2)
f
√
− 3
−2
0
2
√
3
√ √
( 3, 3)
(0, 0)
y=0
f (x) =
√
√
(− 3, − 3)
(−1, −2)
4x
x2 + 1
f0
0
+
+
+
0
-
f 00
0
+
+
+
0
0
+
conclusion
dec., cd
POI
dec, cu
rel. min.
inc, cu
POI
inc, cd
rel. max.
dec, cd
POI
dec, cu
2.8. GRAPH SKETCHING
137
Example 2.8.11. Sketch the graph of f (x) = 4 + x +
x−1
using derivatives.
x2 − 1
Solution.
R
x = −1
domf : \ {−1, 1} √
x-intercept: x = −5±2 5
1
x2 + 5x + 5
f (x) = 4 + x +
=
, x 6= 1
x+1
x+1
y-intercept: y = 5
11
V.A.: x = −1
hole: 1,
2
H.A.: none
lim f (x) = ∞
lim f (x) = −∞
x→∞
x−1
x2 − 1
(1, 11
)
2
y =x+4
(0, 5)
x→−∞
O.A.: y = x + 4
x2 + 2x
f 0 (x) =
(x + 1)2
2
f 00 (x) =
(x + 1)3
CN: 0, −2
(−2,1)
PPOI: none
5
5
, 0)
of
M
at
h
conclusion
inc, cd
rel. max.
dec, cd
V.A.
dec, cu
rel. min.
inc, cu
√
( −5+
2
e
und
f
und
+
+
+
ut
1
f
+
0
und
0
+
00
tit
f
0
em
at
ic
s
√
5
( −5−
, 0)
2
In
s
interval
(−∞, −2)
−2
(−2, −1)
−1
(−1, 0)
0
(0, ∞)
2.8.4
f (x) = 4 + x +
The Graph of f from the Graph of f 0
UP
We know that we can gain insight on the shape of a function f if we know the behavior of its
derivative. From the graph of the derivative of f , what can we deduce about the graph of f ?
Example 2.8.12. Given the graph of f 0 below and assuming that f is continuous everywhere,
sketch a possible graph of f .
−1
1
−1
Solution.
2
138
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
(a) First, we need to determine the critical numbers of f . Remember that these are the numbers
that make f 0 zero or undefined. From the graph of f 0 , we see that the critical numbers are −1,
0 and 2.
(b) Next, we need to determine where the graph of f has points of inflection. These are the points
where the graph of f 0 changes from increasing to decreasing, or vice versa. The possible points
of inflection of the graph of f are attained at values of x where the graph of f 0 has a horizontal
tangent line or a vertical tangent line or has no tangent line. Thus, the x-coordinates of the
possible points of inflection of f are 0 and 1.
(c) Recall that the graph of f is increasing on [a, b] if f 0 (x) > 0 for all x ∈ (a, b). This means that
the graph of f 0 is above the x-axis on the interval (a, b). Similarly, the graph of f is decreasing
on [a, b] if the graph of f 0 is below the x-axis on the interval (a, b).
e
of
M
at
h
em
at
ic
conclusions on the graph of f
increasing, concave down
relative maximum
decreasing, concave down
relative minimum, POI
increasing, concave up
POI
increasing, concave down
relative maximum
decreasing, concave down
ut
f 00
−
−
−
und
+
0
−
−
−
tit
f0
+
0
−
und
+
+
+
0
−
In
s
interval
(−∞, −1)
−1
(−1, 0)
0
(0, 1)
1
(1, 2)
2
(2, ∞)
s
(d) Meanwhile, the graph of f is concave up on an interval (a, b) if the graph of f 0 is increasing on
(a, b). It is concave down on (a, b) if the graph of f 0 is decreasing on (a, b).
UP
From this table, we can construct a possible graph of f .
−1
0
1
2
2.8. GRAPH SKETCHING
2.8.5
139
Exercises
Exercises for Discussion
A. Explain why the following functions do not have a linear asymptote and sketch their graphs.
1. y = x3 − 3x2 + 1
3. y = x1/3 (x + 4)
2. y = −x4 + 8x3 − 18x2 + 10
4. y = sin x cos x
B. Find all vertical, horizontal and oblique asymptotes of the following functions using limits.
√
x2 + 1
1. f (x) =
x+1
sin x
2. f (x) =
x
x2 − 1
3. f (x) = 2
5x + 1
x+1−1
x
2
x − 4x + 4
5. f (x) = 2
x +x−6
4
6. f (x) = x + 1 +
x−4
em
at
ic
s
4. f (x) =
x
(2x + 1)2
2. f (x) =
12x2
(x − 2)2
3. f (x) =
2x3
x2 − 4
x2 + 1
x+1
x3 − 1
5. f (x) = 2
x −1
(x + 1)3
6. f (x) =
x2
4. f (x) =
UP
In
s
tit
ut
e
1. f (x) =
of
M
at
h
C. Sketch the graph of the following functions analytically.
D. Sketch a possible graph of the function f given the graph of f 0 . Assume that f is continuous
everywhere.
1.
2.
(−1, 2)
3
2
(0, 2.6)
2
1
(3 −
(−2, 0)
−3
−2
−1(0, 0)
√
(−2, 1)
1
−4
−3
−2
−1
−1
(1, −2)
(3, 0)
(−1, −0.6)
2
−1
−2
(0, 1)
1
2, 0)
−2
1
2
140
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
Supplementary Exercises
A. Explain why the following functions do not have a linear asymptote and sketch their graphs.
1. y = 3x4 − 4x3
3. y = x4/3 − x1/3
2. y = x4 − 8x2 + 16
4. y = cosh x
B. Find all vertical, horizontal and oblique asymptotes of the following functions using limits.
2x + 3
x+4
3. f (x) =
1
|x − 3|
4. f (x) = ln(1 +
x−9
2. f (x) = √
x−3
1
)
x2
C. Sketch the graph of the following functions analytically.
1. f (x) = 4x3 + 12x2
s
5. f (x) = coth x
em
at
ic
1. f (x) =
(x + 1)3
x2
(x − 2)(3x + 2)
6. f (x) =
x+1
(2x − 1)2
7. f (x) =
(x − 2)2
3. f (x) =
4(2x + 3)(x + 3)
(x + 2)2
4. f (x) =
x2 + 3x + 2
3x + 2
e
(x − 1)2
x2
8. f (x) =
x3
x2 − 4
In
s
tit
ut
2. f (x) =
of
M
at
h
5. f (x) =
UP
D. Sketch a possible graph of the function f given the table of signs of f 0 and f 00 . Assume that
f is continuous everywhere
1.
2.
x
x = −1
x=0
x=1
f 0 (x)
−
und
−
0
+
+
+
f 00 (x)
−
und
+
+
+
0
−
x
x=0
x=1
x=4
f 0 (x)
+
0
+
und
−
0
+
f 00 (x)
−
0
+
und
+
+
+
2.8. GRAPH SKETCHING
141
Reviewer
I. Find
dy
. No need to simplify your final answers.
dx
x
21
2
1. y = (2 −1)s sec (3x+4) (use logarithmic differentiation)
1
3
5−
x2
cosh(πx)
2. y = cot−1 (7x )
3. y = f 0 (x), where f (x) = tan−1 (5x2 + 3)
√
4. log3 (x5 + y 3 ) = e + sin(x2 − y)
5. 12 y 6 − π 3 = cosh(xy 2 ) + log2 (x3 + 7)
II. Do as indicated.
s
1. Find the point-slope form of the equation of the normal line to the graph of y = cot(πx)
at x = 21 .
em
at
ic
2. Find the equation of the tangent line to the graph of
1
y
at
h
π + x3 sinh(y 4 ) = 3x +
1
at the point 0,
.
π
3. Determine if the function
of
M
x≤3
x>3
In
s
tit
ut
e

3 + x,
x
f (x) = 4 − √
x + 2x + 3,
UP
is differentiable at x = 3.
4. Find all possible values for a and b such that the function
(
tan(πx), x < 3
p(x) =
ax2 − b, x ≥ 3
is differentiable at x = 3.
5. Verify that the Mean Value Theorem is applicable to the following functions on the given
interval. Find all values of c satisfying the conclusion of the Mean Value Theorem on
the given interval.
2x
1. f (x) =
on [0, 1]
x+1
2. g(x) = 3sin x on [0, π]
3. h(x) = x − ex on [0, 2]
142
CHAPTER 2. DERIVATIVES AND DIFFERENTIATION
III. Let f (x) =
x2 − 5x + 4
18(4 − x)
(x − 1)(x − 4)
9(x − 2)
=
and f 00 (x) =
.
, with f 0 (x) =
(x + 2)2
x2 − 4x + 4
(x + 2)3
(x + 2)4
1. Determine the horizontal and the vertical asymptotes to the graph of f by computing
all necessary limits.
2. Determine the x-intercepts, critical numbers and the x-coordinates of the possible points
of inflection of the graph of f .
3. Construct a table of signs for f 0 , and f 00 . Using this table, identify all intervals in which
f is increasing, decreasing, concave upward, or concave downward. Identify the relative
extremum point/s and point/s of inflection.
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
4. Sketch the graph of f . Label all intercepts, relative extremum points, points of inflection,
and linear asymptotes of f .
Chapter 3
Applications of Differentiation
3.1
Absolute Extrema of a Function on an Interval
em
at
ic
s
At the end of this section, the student will be able to:
• determine the absolute extrema of a function on an interval
of
M
at
h
• solve optimization problems using techniques for finding absolute extrema
UP
In
s
tit
ut
e
Suppose a businessman wishes to know how many toy robots he needs to produce to maximize his
profit. If he produces too many toys, some of these may not be bought because of limited demand.
Using market research, he finds out that if he produces x toy robots, his profit will be given by the
function f (x). In deciding how many he will produce, he also has to consider some constraints, like
his budget or his factory’s production speed. Hence, he only considers values of x on a given set,
say, an interval I. How will the businessman determine what value of x on the interval I will give
the highest value of f (x)?
A function f is said to have an absolute maximum value on an interval I at x0 if f (x0 ) ≥ f (x)
for any x ∈ I. Similarly, f is said to have an absolute minimum value on an interval I at x0 if
f (x0 ) ≤ f (x) for any x ∈ I. If f has either an absolute maximum or absolute minimum value on I
at x0 , then we say f has an absolute extremum on I at x0 .
Remarks.
1. Unlike with relative extrema, the interval I need not be an open interval.
2. Note that the absolute maximum (or minimum) value of a function on an interval is unique, if
it exists. However, the graph of f may have more than one absolute maximum (or minimum)
point (same y-values but different x-values).
143
144
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
(−2, 4)
f (x) = cot x
f (x) = x2
π
2
− π2
(0, 3)
(1, 1)
f (x) = 2x + 3
(−1, 1)
−2
0
1
−1
0
Example 3.1.1.
• f (x) = cot x has no absolute extrema on the interval [− π2 , π2 ].
em
at
ic
s
• f (x) = x2 has an absolute minimum on the interval (−2, 1) at x = 0 but has no absolute
maximum on (−2, 1).
of
M
at
h
• f (x) = 2x+3 has an absolute minimum on the interval [−1, ∞) at x = −1 but has no absolute
maximum on [−1, ∞).
tit
Absolute Extrema on Closed and Bounded Intervals
In
s
3.1.1
ut
e
• f (x) = 2x + 3 has an absolute minimum on the interval [−1, 0] at x = −1 and an absolute
maximum on [−1, 0] at x = 0.
UP
Ultimately, we want to find the absolute extrema of a function on a given interval. It would also be
helpful if we can easily determine the existence of an absolute minimum or an absolute maximum
of a function, so that we are at least guaranteed that our efforts in finding them are not wasted.
Theorem 3.1.2 (Extreme Value Theorem). If f is continuous on [a, b], then f has both an
absolute maximum and an absolute minimum on [a, b].
Remark 3.1.3. If you replace [a, b] by one of the following intervals (a, b), [a, b), (a, b], (a, ∞),
(−∞, b), [a, ∞), (−∞, b], then the conclusion of the Extreme Value Theorem does not follow.
The Extreme Value Theorem assures us that a continuous function on [a, b] will attain both an
absolute maximum and minimum on [a, b]. Hence, an absolute extremum may be attained at either
an interior point, that is x ∈ (a, b), or at an endpoint, that is x = a or x = b.
Theorem 3.1.4. If f has an absolute extremum on the open interval (a, b) at x = c, then c is
a critical number of f on (a, b), that is, either f 0 (c) = 0 or f 0 (c) is undefined.
3.1. ABSOLUTE EXTREMA OF A FUNCTION ON AN INTERVAL
145
Finding absolute extrema of f on [a, b]:
• find critical numbers of f between a and b, say c1 , . . . , cm
• evaluate f at each critical number and at the endpoints a and b
• compare f (c1 ), . . . , f (cm ), f (a) and f (b)
• the number that gives the highest (lowest) value of f gives the absolute maximum
(minimum).
Note: Always check if the function is continuous on I. One way to do this is to check if the interval
in consideration is a subset of dom f .
Example 3.1.5. Find the absolute maximum and the absolute minimum values of f (x) on the
given closed interval. Specify the values of x at which these absolute extrema are attained.
em
at
ic
s
1. f (x) = x3 − 3x2 + 1 on I = [− 21 , 4]
Solution.
R. Thus, f is continuous on any interval;
of
M
at
h
We note that f is a polynomial, which is continuous on
in particular, it is continuous on I.
Now, we solve for the critical numbers of f :
f (2) = −3
f (4) = 17
In
s
tit
ut
e
f 0 (x) = 3x2 − 6x = 3x(x − 2)
C.N.s on I: 0, 2
1
1
f −
=
f (0) = 1
2
8
2. f (x) =
UP
Therefore f has an absolute minimum at x = 2 and an absolute maximum at x = 4.
1
on I = [−2, 1]
1 + x2
Solution.
Since the denominator of f is nonzero, then dom f =
[−2, 1]. Solving for the CNs of f , we get:
f 0 (x) =
−2x
(x2 + 1)2
R. Thus, f is defined and continuous on
C.N.s on I: 0
f (−2) =
1
5
f (0) = 1
f (1) =
1
2
Therefore f has an absolute minimum at x = −2 and an absolute maximum at x = 0.
3. f (x) = x2/3 on I = [−8, 1]
Solution.
146
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
Since taking cube roots is possible for any real number, f is defined, and thus continuous, on
I. We solve for the CNs of f :
f 0 (x) =
2
3x1/3
C.N.s on I: 0
f (−8) = 4
f (0) = 0
f (1) = 1
Therefore f has an absolute minimum at x = 0 and an absolute maximum at x = −8.
4. f (x) = x − tan x on I = [− π4 , π4 ]
Solution.
The function f1 (x) = x is continuous everywhere. From the graph of f2 (x) = tan x, it should
make sense to say that f2 is continuous on I. Thus, f (x) = f1 (x) − f2 (x) is continuous on I.
We solve for the CNs of f :
s
C.N.s on I: 0
π
π
=− +1
f −
4
4
f (0) = 0
f
em
at
ic
f 0 (x) = 1 − sec2 x
π 4
=
π
−1
4
5. f (x) =
x2 − 4 , x < 3
on I = [0, 5]
8−x , x≥3
of
M
(
at
h
Therefore f has an absolute minimum at x = π4 and an absolute maximum at x = − π4 .
ut
e
Solution.
In
s
tit
Each component function is defined and continuous on
Since
lim f (x) =
UP
x→3−
lim f (x) =
x→3+
R. We now check for continuity at x = 3.
lim (x2 − 4) = 5
x→3−
lim (8 − x) = 5
x→3+
we have lim f (x) = 5 = f (3). Hence f is continuous at x = 3 and so on I. Solving for the CNs
x→3
of f , we get:
(
2x , x < 3
f 0 (x) =
−1 , x > 3
C.N.s on I: 0, 3
f (0) = −4
f (3) = 5
f (5) = 3
Therefore f has an absolute minimum at x = 0 and an absolute maximum at x = 3.
3.1.2
Absolute Extrema On Open Intervals
Even if a function is continuous, we are not always guaranteed that it attains an absolute extremum
on an interval like (a, b), (a, b], (a, b] or infinite intervals. In those cases, the following theorem will
be useful.
3.1. ABSOLUTE EXTREMA OF A FUNCTION ON AN INTERVAL
147
Theorem 3.1.6. Suppose the function f is continuous on an interval I containing x0 and x0
is the only number in I for which f has a relative extremum.
(a) If f has a relative maximum at x0 , then f has an absolute maximum on I at x0 .
(b) If f has a relative minimum at x0 , then f has an absolute minimum on I at x0 .
Example 3.1.7. Find the value(s) of x where the given function attains its absolute extrema, if
any, on the given interval.
1. f (x) =
1
on I = (−2, 1)
1 + x2
Solution.
From item 2 of Example 3.1.5, f is continuous on (−2, 1).
−2x
s
C.N.: 0
(x2 + 1)2
−4x2 + 2
and f 00 (0) = 2 > 0
(x2 + 1)3
at
h
f 00 (x) =
em
at
ic
f 0 (x) =
e
R
ut
2. f (x) = x4 − 4x on
of
M
By the second derivative test, f has a relative minimum at x = 0. Since the point (0, 1) is the
only relative extremum of f on I, then f has an absolute minimum at x = 0.
tit
Solution.
In
s
Since f is a polynomial, then f is continuous on
R.
UP
f 0 (x) = 4x3 − 4 C.N.: 1
f 00 (x) = 12x2 and f 00 (1) = 12 > 0
By the second derivative test, f has a relative minimum at x = 1. It is the only relative
extremum of f in . Hence, f has an absolute minimum at x = 1.
R
3. f (x) = x +
16
on (1, ∞)
x
Solution.
Since dom f =
f 0 (x) = 1 −
R \ {0}, then f is continuous on (1, +∞).
16
x2 − 16
=
x2
x2
C.N. on (1, ∞): 4
f 00 (x) =
1
32
and f 00 (4) = > 0
3
x
2
By the second derivative test, f has a relative minimum at x = 4. Since it is the only relative
extremum of f on (1, ∞), then f has an absolute minimum at x = 4.
148
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
Consider a function f continuous on an open interval I with more than one relative extremum on
I. We can also determine the absolute extrema of f by observing the limit of f as x approaches
the endpoints/tails of the interval.
Example 3.1.8. Find the value(s) of x where f (x) = x4 − 5x2 + 4 attains its absolute extrema on
I = (−3, ∞).
Solution.
Note that f is continuous on the open interval (−3, ∞).
√
√
10
2
s
f 0 (x) = 4x3 − 10x = 2x(2x2 − 5) and C.N. : 0, − 210 ,
em
at
ic
f 00 (x) = 12x2 − 10
√ √ and f 00 − 210 = 20 = f 00 210
at
h
f 00 (0) = −10
of
M
By the√ second derivative
test, f has a relative maximum at x = 0 and relative minimum at
√
10
10
x = − 2 and x = 2 . Now,
ut
e
√ √ f − 210 = − 94 = f 210
tit
f (0) = 4,
and
lim f (x) = +∞.
x→+∞
UP
In
s
lim f (x) = 40
x→−3+
Make a rough sketch of the graph of f on (−3, ∞).
Clearly, f has no absolute maximum. Note that the interval (−3, ∞) is open. Thus, if f has
an absolute minimum on (−3, ∞), then it must be a relative minimum. From the behavior of
the function towards the left√ endpoint and towards ∞, we can conclude that f has an absolute
minimum attained at x = ± 210 .
3.1.3
Optimization: Application of Absolute Extrema on Word Problems
Many real-life situations require us to find a value that best suits our needs. If we are given
several options for the value of a variable x, how do we choose the “best value”? Such a problem is
classified as an optimization problem. We wish to apply our previous discussion on finding absolute
extremum values of a function to solve some optimization problems.
3.1. ABSOLUTE EXTREMA OF A FUNCTION ON AN INTERVAL
149
Suggestions on Solving Optimization Problems
1. If possible, draw a diagram of the problem corresponding to a general situation.
2. Assign variables to all quantities involved.
3. Identify the objective function.
(a) Identify the quantity, say q, to be maximized or minimized.
(b) Formulate an equation involving q and other quantities. Express q in terms of a single
variable, say x. If necessary, use the information given and relationships between
quantities to eliminate some variables.
(c) The objective function to be maximized/minimized: q = f (x)
4. Determine the domain or constraint of q from the physical restrictions of the problem.
em
at
ic
s
5. Use appropriate theorems involving absolute extrema to solve the problem. Make sure to
give the exact answer to the question (indicate units).
of
M
at
h
Example 3.1.9.
ut
e
1. Find the number in the interval [−2, 2] so that the difference of the number from its square is
maximized.
In
s
tit
Solution.
UP
Let x be the desired number. We want to maximize
f (x) = x2 − x
where x ∈ [−2, 2]. Note that f is continuous on [−2, 2] and thus, we can apply the Extreme
Value Theorem. We first find the critical numbers of f in the interval (−2, 2).
f 0 (x) = 2x − 1
Hence we have one critical number in (−2, 2), that is x = 21 . Then we compare the function
value at the critical number and the endpoints.
f (−2) = 6
f (2) = 2
1
1
f
=−
2
4
From this, we conclude that f attains an absolute maximum on I at x = −2. Hence, the number
we are looking for is −2.
150
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
s
s
s
s
9 in
2. A rectangular box is to be made from a piece of
cardboard 24 inches long and 9 inches wide by
cutting out identical squares from the four corners
and turning up the sides. Find the volume of the
largest rectangular box that can be formed.
s
s
s
s
24 in
Solution.
If s is the length of the side of the squares to be cut out, then the volume of the open-topped
box we can construct is
s
V (s) = (24 − 2s)(9 − 2s)s = 2(108s − 33s2 + 2s3 ).
em
at
ic
We wish to maximize V (s) but note that s is a positive real number which has to be less than
half the width of the cardboard. That is, s ∈ (0, 4.5). Now
at
h
V 0 (s) = 216 − 132s + 12s2 = 12(18 − 11s + s2 ) = 12(s − 2)(s − 9)
tit
ut
e
of
M
and hence the only critical number of V in (0, 4.5) is 2. Using the second derivative test, we
see that V 00 (s) = −132 + 24s and V 00 (2) < 0. Therefore V has a relative maximum at s = 2.
Since it is the only relative extremum of V on the open interval (0, 4.5), then V also attains
its absolute maximum at s = 2. We conclude that the volume of the largest box that can be
formed is V (2) = 200 in3 .
UP
In
s
3. A 1000 square yard rectangular lot along a highway is to be fenced off such that one side of the
fence is on the highway. The fencing on the highway costs Php 80 per yard and the fencing on
the other sides costs Php 20 per yard. Determine the dimensions of the lot that will minimize
the total cost of the fencing. (Assume that materials can be bought in any fractional parts.)
Solution.
Let s1 be the length of the side of the lot facing the highway and s2 be the length of the other
side. Since the area of the lot is 1000 yd2 , then we can write s2 = 1000
s1 . If C is the cost of
fencing the lot, then
40, 000
C(s1 ) = 100s1 + 40s2 = 100s1 +
.
s1
We want to minimize this function. But first, we need to ask: what are the feasible values for
s1 ? Clearly, s1 is a positive real number. Aside from this, we do not have any other constraint
for s1 . Thus, the domain is (0, ∞). Note that
C 0 (s1 ) = 100 −
40, 000
s21
and so C has a critical number at s1 = 20. Since C 00 (20) = 2(40,000)
> 0, then C has a relative
203
minimum at s1 = 20. Using similar arguments as the previous examples, we conclude that the
3.1. ABSOLUTE EXTREMA OF A FUNCTION ON AN INTERVAL
151
side facing the highway must be 20 yards and the other side must be 50 yards to minimize the
cost of fencing the lot.
4. Determine the dimensions of the right circular cylinder of greatest volume that can be inscribed
in a right circular cone of radius 6 inches and height 9 inches.
r
6 in
h
em
at
ic
s
9 in
Solution.
e
of
M
at
h
Let h and r respectively denote the height and radius of the cylinder. The volume of the cylinder
is πr2 h. Looking at the middle cross-section of the cylinder and cone, we can see similar triangles
and so
6
9
=
6−r
h
In
s
tit
ut
We can now write our objective function
3
V (r) = 9πr2 − πr3
2
UP
to be maximized. Clearly, r ∈ (0, 6). Now, V 0 (r) = 18πr − 29 πr2 = 9πr(2 − 2r ) and hence, the
only critical number is 4. Applying the second derivative test, we get V 00 (r) = 18π − 9πr and
V 00 (4) = −18π < 0. Since V has only one relative extremum on the interval (0, 6) and it is a
relative maximum, then it is also an absolute maximum. Therefore, the inscribed right circular
cylinder will have the greatest volume if its dimensions are r = 4 inches and h = 3 inches.
5. Angelo, who is in a rowboat 2 miles from a straight shoreline, notices smoke billowing from his
house, which is on the shoreline and 6 miles from the point on the shoreline nearest Angelo. If
he can row at 6 miles per hour and run at 10 miles per hour, how should he proceed in order to
get to his house in the least amount of time?
Starting Point
2 mi
Angelo’s house
P
c
6 mi
shoreline
152
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
Solution.
Let c be the distance between the house and the point P on the shore from which Angelo will
start to run. Using the Pythagorean theorem, we see that the distance
he will travel by boat is
p
p
4
+
(6
− c)2
4 + (6 − c)2 . Note that speed= distance
.
Thus
he
will
sail
for
hours and walk
time
6
c
for 10 hours. We wish to minimize
p
4 + (6 − c)2
c
+
T (c) =
6
10
We can assume that c ∈ [0, 6]. Solving for the critical numbers of f on (0, 6),
p
5c
−
30
+
3
4 + (6 − c)2
p
T 0 (c) =
,
30 4 + (6 − c)2
we get c = 92 . Comparing function values at the endpoints and the critical number,
9
13
=
2
15
T (6) =
14
,
15
s
T
em
at
ic
√
2 10
T (0) =
6
ut
e
of
M
at
h
we see that the absolute minimum of T is attained at c = 92 . Angelo must row up to the point
P on the shore 92 miles from his house and 32 miles from the point on the shore nearest him.
Then he must run straight to his house.
r
tit
6. A Norman window consists of a rectangle surmounted by
a semicircle. If the perimeter of a Norman window is to
be 20ft., what should the radius of the semicircle and the
height of the rectangle be so that the window will admit the
most light?
UP
In
s
h
Solution.
Let r denote the radius of the semicircle and h denote the height of the rectangular as seen in
the illustration. Since the perimeter is 20 ft., we get the equation 20 = 2h + 2r + rπ. From this,
we can solve for h in terms of r and vice versa. We have
1
h = 10 − (2 + π)r
2
and
r=
20 − 2h
.
2+π
From these two equations, notice that since h and r are positive real numbers, then h cannot
exceed 10 and
20
r<
.
2+π
Thus, we want to maximize the surface area
1
πr2
A(r) = 2rh + πr2 = −
− 2r2 + 20r,
2
2
3.1. ABSOLUTE EXTREMA OF A FUNCTION ON AN INTERVAL
153
20
20
where r ∈ (0, 2+π
). Since A0 (r) = −πr − 4r + 20, then the only critical number is r = π+4
. If
we can show that A attains a relative maximum at this critical number, then we can conclude
that A also attains its absolute maximum at this critical number. Indeed,
20
00
00
<0
A (r) = −π − 4 = A
π+4
and so by the second derivative test we obtain the desired conclusion. Therefore, the window
20
20
admits most light when r = π+4
feet and h = π+4
feet.
7. Find all points on the parabola with equation x = 2y 2 that are closest to the point (10, 0).
x = 2y 2
(x, y)
em
at
ic
s
(10, 0)
Solution.
of
M
at
h
Since the point/s we are looking for is/are on the graph of x = 2y 2 , then we let (2y 2 , y) be the
coordinates of the desired point/s. We wish to minimize
p
p
p
d := (x − 10)2 + y 2 = (2y 2 − 10)2 + y 2 = 4y 4 − 39y 2 + 100
R
D(y) := 4y 4 − 39y 2 + 100
In
s
tit
ut
e
where y ∈ . Note however that a value of y minimizing d also minimizes d2 because d is a
positive value. For simplicity, we minimize
UP
q
instead. Note that D0 (y) = 16y 3 − 78y and so the critical numbers of D are 0, ± 39
8 . From the
following observations,
r !
r !
39
39
D00 (y) = 48y 2 − 78 and D00 (0) = −78 < 0 and D00
= 156 = D00 −
8
8
q
we see that D has a relative maximum at y = 0 and relative minimum at y = ± 39
8 . Moreover,
lim D(y) = +∞
y→−∞
and
lim D(y) = +∞.
y→+∞
Thus D has no absolute
q maximum value but has an absolute minimum attained at two
q values
39
39
of y, namely, y = ± 39
.
Therefore,
the
points
in
the
graph
nearest
(10,
0)
are
,
and
8
4
8
q 39
39
4 ,−
8 .
154
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
3.1.4
Exercises
Exercises for Discussion
A. Find the absolute extrema of the function on the given interval, if there are any, and determine
the values of x at which the absolute extrema occur.
√
1. f (x) = x 4 − x2 on [−1, 2]
(
4x − 2,
x<1
2. f (x) =
(x − 2)(x − 3), x ≥ 1
3. f (x) = 2 sec x − tan x on 0, π4
on [ 12 , 72 ]
em
at
ic
s
4. f (x) = 4x3 − 3x2 − 6x + 3 on (0, ∞)
1
5. f (x) = x + on (−∞, 0)
x
√
6. f (x) = 4 − x2 on (−2, 2)
4
on [2, 5)
7. f (x) =
(x − 3)2
at
h
8. f (x) = [[x]] − x2 on (−1, 1)
√ √
9. f (x) = x2 + cos(x2 ) on (− π, π)
of
M
B. Answer the following optimization problems systematically.
1. Find two numbers such that their difference is 50 and their product is as small as possible.
In
s
tit
ut
e
2. A closed box with a square base is to have a volume of 2000 in3 . The material for the top
and bottom of the box cost 30 pesos per square inch amd the material for the sides cost
15 pesos per square inch. Find the dimensions of the box so that the total cost of material
is least.
UP
3. Find the area of the largest rectangle having two vertices on the x-axis and two vertices
above the x-axis and on the parabola with equation y = 9 − x2 .
4. Find an equation of the tangent line to the curve y = x3 − 3x2 + 5x that has the least slope.
5. Find the radius and height of the right circular cylinder of largest volume that can be
inscribed in a right circular cone with radius 6 inches and height 10 inches.
6. A closed cylindrical can is to hold 1 liter of liquid. Assuming there is no waste or overlap,
how should we choose the height and radius to minimize the amount of material needed to
manufacture the can?
7. Find a point on the curve y = x2 that is closest to the point (18, 0).
8. The product of two positive numbers x and y is equal to 9. Find the largest possible value
of xy .
3.1. ABSOLUTE EXTREMA OF A FUNCTION ON AN INTERVAL
155
Supplementary Exercises
A. Find the absolute extrema of the function on the given interval, if there are any, and determine
the values of x at which the absolute extrema occur.
1. f (x) =
x
x2 + 2
sin x
√ on [0, 2π]
cos x + 2
7. f (x) = sin(cos x) on [0, 2π]
on [−1, 4]
6. f (x) =
2. f (x) = 4x3 − 3x2 − 6x + 3 on (−1, ∞)
1
3. f (x) = on [−2, 3]
x
4. f (x) = |x − 4| + 1 on (0, 6)
8. f (x) = x ln x on (0, +∞)
√
9. f (x) = ln(1 + 3x2 ) + 2 tan−1 ( 3x) on
(−∞, +∞)
π
5. f (x) = 2 cos x on [ −2π
3 , 3)
B. Answer the following optimization problems systematically.
em
at
ic
s
1. Express the number 11 as a sum of two nonnegative numbers whose product is as large as
possible.
at
h
2. A garden is to be laid out in a rectangular area and protected by a chicken wire fence.
What is the largest possible area of the garden if only 100 running feet of chicken wire is
available for the fence?
of
M
3. An open box is to be made from a 16-inch by 30-inch piece of cardboard by cutting out
squares of equal size from the four corners and bending up the sides. What size should the
squares be to obtain a box with the largest volume?
UP
In
s
tit
ut
e
4. An offshore oil well located at a point W that is 5 km from the closest point A on a straight
shoreline. Oil is to be piped from W to a shore point B that is 8 km from A by piping it
on a straight line underwater from W to some shore point P between A and B and then
on to B via pipe along the shoreline. If the cost of laying pipe is around P10,000,000/km
underwater and P5,000,000/km over land, where should the point P be located to minimize
the cost of laying the pipe?
5. Find the area of the largest rectangle that can be inscribed in a semicircle of radius 5.
6. Paper Packaging and Artwork Philippines, a paper manufacturing company, makes opentopped rectangular boxes as follows. First, they cut out a large rectangular sheet of cardboard. Then, from the corners of that rectangular sheet, they cut out small squares, and
fold up the sides of the resulting shape, as shown below. The company wishes to make a
rectangular box of volume 2000 cm3 with a square base. What dimensions will minimize
the amount of cardboard used in the production of the box, i.e., the area of the sheet from
which the box is cut?
7. If P (a, a2 ) is a point on the parabola y = x2 except for the origin, let Q be the point where
the normal line intersects the parabola again. Find the value of a for which P Q is shortest.
8. If a projectile is fired with an initial velocity v at an angle of elevation θ above the horizontal,
then its trajectory, neglecting air resistance, is the parabola
y = x tan θ −
5
v 2 cos2 θ
x2 ,
0≤θ≤
π
.
2
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
Suppose the projectile is fired from the base of a plane that is inclined at an angle of π4 from
the horizontal. Determine θ so that the range of the projectile (i.e., its net displacement
from its point of origin before it lands on the plane) is maximized.
9. The bottom of a projector screen 1 m high is 0.8 m above eye level. How far should one
stand from the screen in order to get the best view? (A viewer gets the best view when
the angle subtended by the screen whose vertex is at the viewer’s eyes is maximized.)
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
10. A rectangular sheet of paper 12 cm × 6 cm is folded so that one corner meets the longer
of the opposite edges. Find the shortest possible length of the crease.
UP
156
3.2. RATES OF CHANGE, RECTILINEAR MOTION, AND RELATED RATES
3.2
157
Rates of Change, Rectilinear Motion, and Related Rates
At the end of this section, the student will be able to:
• use derivatives to compute the instantaneous rates of change of certain quantities
with respect to one or more other quantities
• use the techniques of differentiations in problems of rectilinear motion
3.2.1
Rates of Change
Rates of change occur in many real-world applications and phenomena.
em
at
ic
s
• A microbiologist might be interested in the rate at which the number of bacteria in a culture
changes over time.
of
M
at
h
• An engineer might be interested in the rate at which the length of a metal rod changes with
temperature.
tit
ut
e
• An economist might be interested in the rate at which production cost changes with the quantity
of a product that is manufactured.
UP
In
s
• A medical researcher might be interested in the rate at which the radius of an artery changes
with the concentration of alcohol in the bloodstream.
Suppose f is a function and y = f (x).
1. The average rate of change of y with respect to x on [x0 , x] is
f (x) − f (x0 )
∆f
=
.
∆x
x − x0
2. The instantaneous rate of change of y with respect to x at x = x0 is
∆f
f (x) − f (x0 )
= lim
= f 0 (x0 ).
x→x
∆x→0 ∆x
x − x0
0
lim
Remark 3.2.1.
158
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
Q
1. Graphically, the average rate of change of y with
respect to x on [x0 , x] is the slope of the secant
line passing through P (x0 , f (x0 )) and Q(x, f (x)).
f (x)
)
y = f (x)
∆f
P
f (x0 )
|
{z
∆x
} R
x = x0 + dx
x0
Figure 3.3.1
2. The derivative of f at x = x0 , f 0 (x0 ) can be interpreted as the instantaneous rate of change of
y with respect at x at x = x0 . That is, f 0 (x0 ) is the rate of change of y per unit change in x at
the instant when x = x0 .
em
at
ic
s
3. Let y be a function of x.
dy
> 0 on an interval I, then y increases as x increases, and y decreases as x decreases.
dx
dy
< 0 on an interval I, then y decreases as x increases, and y increases as x decreases.
(b) If
dx
dy
(c) If
= 0 on an interval I, then y does not change with respect to x.
dx
of
M
at
h
(a) If
tit
ut
e
Example 3.2.2. A right circular cylinder has a fixed height of 6 units. Find the instantaneous
rate of change of its volume with respect to the radius of its base.
UP
In
s
Solution.
Let r and h be the radius and height, respectively, of the cylinder. Then the volume of the cylinder
is given by V = πr2 h = 6πr2 . Thus, the rate of change of the volume with respect to the radius is
dV
= 12πr. This means that a unit change in the radius of the sphere produces a change of 12πr
dr
cubic units in the volume.
Example 3.2.3. A bactericide was introduced to a nutrient broth in which bacteria were growing.
The bacterium population continued to grow for some time but then stopped growing and began
to decline. The size of the population at time t (hours) was
P = 106 + 104 t − 103 t2 .
Determine the growth rates at t = 0, t = 5 and t = 10 hours.
Solution.
The growth rate or the rate of change of the bacterium population P with respect to time t is given
by P 0 (t). We have
P 0 (t) = 104 − 2(103 )t = 2(103 )(5 − t).
3.2. RATES OF CHANGE, RECTILINEAR MOTION, AND RELATED RATES
159
The growth rates are P 0 (0) = 104 , P 0 (5) = 104 − 2(103 )(5) = 0 and P 0 (10) = 104 − 2(103 )(10) =
−104 .
Also, notice that P 0 (t) > 0 on [0, 5), P 0 (5) = 0, and P 0 (t) < 0 on (5, +∞). This shows that
the bacterium population was increasing until t = 5 hours, then it stopped growing and began to
decline.
Example 3.2.4. A ladder 24 ft. long rests against a vertical wall. Let θ be the angle between the
top of the ladder and the wall and let x be the distance from the bottom of the ladder to the wall.
If the bottom of the ladder slides away from the wall, how fast does x change with respect to θ
π
when θ = ?
3
Solution.
From the figure, the equation that relates x and
θ is
em
at
ic
s
θ
x
24
x = 24 sin θ.
x
at
h
sin θ =
24 ft.
of
M
Thus, the rate of change of x with respect to θ is
dx
π
= 24 cos θ. At θ = , we have
dθ
3
Rectilinear Motion
UP
3.2.2
In
s
tit
ut
e
π dx
=
24
cos
= 12 ft/radian.
dθ θ= π3
3
Suppose a particle is moving along a horizontal line, which we shall refer to as the s-axis. Suppose
the position of the particle at time t is given by the function s(t), called the position function of
s(t) − s(t0 )
∆s
=
.
the particle. The average velocity of the particle on [t0 , t] is vave =
t − t0
∆t
Let s(t) be the position function of a particle moving along s-axis.
1. The instantaneous velocity of the particle at time t is
v(t) = lim
∆s
∆t→0 ∆t
=
ds
= s0 (t).
dt
2. The instantaneous speed of the particle at time t is |v(t)|.
3. The instantaneous acceleration of the particle at time t is
a(t) =
dv
= v 0 (t)
dt
or
a(t) =
d2 s
= s00 (t).
dt2
160
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
Remark 3.2.5. Let s(t) be the position function of a particle moving along the s-axis. Note that
∆t is always positive. The signs of v(t) and a(t) give us information about the motion of the
particle.
1.
a. If v(t) > 0, then the particle is moving in the positive direction of s (usually to the right
or upward) at time t.
b. If v(t) < 0, then the particle is moving in the negative direction of s (usually to the left or
downward) at time t.
c. If v(t) = 0, either the particle is not moving or is changing direction at time t.
2.
a. If a(t) > 0, then the velocity of the particle is increasing at time t. In addition,
i. if v(t) > 0, then the speed of the particle is increasing at time t (speeding up).
ii. if v(t) < 0, then the speed of the particle is decreasing at time t (slowing down).
b. If a(t) < 0, then the velocity of the particle is decreasing at time t. In addition,
em
at
ic
s
i. if v(t) > 0, then the speed of the particle is decreasing at time t.
ii. if v(t) < 0, then the speed of the particle is increasing at time t.
at
h
c. If a(t) = 0, then the velocity of the particle is constant.
(This does not mean that the particle is NOT moving!)
of
M
Example 3.2.6. A particle moves along a horizontal coordinate line in such a way that its position
at time t is specified by s(t) = t3 − 12t2 + 36t − 30 where s is measured in feet and t in seconds.
ut
e
1. Find the instantaneous velocity and the instantaneous acceleration in terms of t.
UP
In
s
tit
2. Describe the position and motion of the particle in a table that includes the intervals of time
when the particle is moving to the left or to the right, when the velocity is increasing or
decreasing, when the speed is increasing or decreasing, and the particle’s position with respect
to the origin during these intervals of time.
3. Show the motion of the particle schematically.
4. Determine the total distance traveled by the particle during the first 7 seconds.
Solution.
1. We have s(t) = t3 − 12t2 + 36t − 30 so v(t) =
ds
dv
= 3t2 − 24t + 36 and a(t) =
= 6t − 24.
dt
dt
2. First, we find t ≥ 0 such that v(t) = 0 and a(t) = 0.
v(t) = 0
3t2 − 24t + 36 = 0
2
a(t) = 0
t − 8t + 12 = 0
6t − 24 = 0
(t − 6)(t − 2) = 0
t = 4
t = 6, t = 2
3.2. RATES OF CHANGE, RECTILINEAR MOTION, AND RELATED RATES
161
Thus we have the following table.
t=0
0<t<2
t=2
2<t<4
t=4
4<t<6
t=6
t>6
s(t)
−30
2
−14
−30
v(t)
36
+
0
−
−12
−
0
+
a(t)
−24
−
−12
−
0
+
12
+
Conclusions
left of origin, towards right, decreasing velocity, slowing down
towards right, decreasing velocity, slowing down
right of origin, changing direction, decreasing velocity
towards left, decreasing velocity, speeding up
left of origin, towards left, constant velocity
towards left, increasing velocity, slowing down
left of origin, changing direction, increasing velocity
towards right, increasing velocity, speeding up
3. Schematically, the particle moved in the following way:
t=6
t=2
t=0
−30
em
at
ic
s
2
at
h
4. Note that the particle changed directions at t = 2 and t = 6. Thus the total distance traveled
is |s(2) − s(0)| + |s(6) − s(2)| + |s(7) − s(6)| = 32 + 32 + 7 = 71 feet.
of
M
Example 3.2.7. A ball is thrown upwards from the top of a high building. The height of the ball
from the ground at any time t (in seconds) is given by the equation s(t) = −16t2 + 128t + 320,
where distance is measured in feet.
ut
e
1. At what height will the ball start its motion?
In
s
tit
2. Find the velocity and acceleration of the ball at any time t.
UP
3. Find the maximum height reached by the ball.
4. Find the time when the ball crosses the top of the building again.
5. Find the velocity of the ball upon impact.
Solution.
1. The ball starts its motion at t = 0, giving us the initial height
h = s(0) = −16(0)2 + 128(0) = 320 = 320 feet
2. We have s(t) = −16t2 + 128t + 320, so that v(t) = −32t + 128 and a(t) = −32.
3. At the instant when the ball reaches its maximum height, the ball is actually at rest; that is,
its velocity equals zero. So we equate v(t) with 0 and solve for t:
v(t) = −32t + 128 = 0
32t = 128
t=4
162
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
Thus, the ball reaches its maximum height at t = 4 seconds. This gives a maximum height
of s(4) = −16(42 ) + 128(4) + 320 = 576 feet.
4. The ball will cross the top of the building again when s(t) = 320:
s(t) = −16t2 + 128t + 320 = 320
16t2 − 128t = 0
t = 0, 8
But the time t = 0 corresponds to the time when the ball is thrown upward. Hence, it crosses
the top of the building again after 8 seconds.
5. First, we get the value of t when s(t) = 0:
s(t) = −16t2 + 128t + 320 = 0
em
at
ic
t = −2, 10
s
−16(t − 10)(t + 2) = 0
at
h
But since we only consider positive values of time, we conclude that the ball hits the ground
at t = 10 seconds. Hence, its velocity upon impact is
Related Rates
e
3.2.3
of
M
v(10) = −32(10) + 128 = −192 ft/s
UP
In
s
tit
ut
Consider a still pond. If you drop a stone on the pond, the water is disturbed, thus creating ripples.
The ripples are composed of circles increasing in size as time passes. The areas and the radii of the
circles both increase and are related to each other. Thus, if we know how fast one increases, say,
how fast the radius increases, we can determine how fast the areas of the circles are changing.
dx
Let x be a quantity that is a function of time t, meaning the quantity changes over time. Then
dt
is the rate of change of x with respect to t.
A problem on related rates is a problem involving rates of change of several variables where one
variable is dependent on another. In particular, if y is dependent on x, then the rate of change of y
dy
is dependent
with respect to t is dependent on the rate of change of x with respect to t, that is,
dt
dx
on
.
dt
Suggestions in solving problems involving related rates:
1. Let t denote the elapsed time. If possible, draw a diagram of the problem that is valid for any
time t > 0.
2. Select variables to represent the quantities that change with respect to time. Label those
quantities whose values do not depend on t with their given constant values.
3.2. RATES OF CHANGE, RECTILINEAR MOTION, AND RELATED RATES
163
3. Write down any numerical facts known about the variables. Interpret each rate of change as
the derivative of a variable with respect to time. Remember that if a quantity decreases over
time, then its rate of change is negative.
4. Identify what is being asked.
5. Write an equation relating the variables that is valid at all time t > 0.
6. Differentiate the equation in (5) implicitly with respect to t.
7. Substitute in the equation obtained in (6) all values that are valid at the particular time of
interest. Solve for what is being asked.
8. Write a conclusion that answers the question of the problem. Do not forget to include the
correct units of measurement.
em
at
ic
s
Example 3.2.8. A ladder 10 meters long is leaning against a wall. If the bottom of the ladder is
being pulled horizontally towards the wall at 2 m/s, how fast is the top of the ladder moving when
the bottom is 6 meters from the wall?
of
M
at
h
Solution.
Let x be the distance of the bottom of the ladder from the wall, and y
be the distance of the top of the ladder from the ground. Then we have
dx
= −2 m/s, negative since x decreases as time goes on. We want to find
dt
dy
when x = 6 m. Using the Pythagorean Theorem, we obtain
dt
y
ladder
x
UP
In
s
tit
ut
e
x2 + y 2 = 100.
Differentiating both sides with respect to time t,
wall
Dt [x2 + y 2 ] = Dt [100]
dy
dx
+ 2y
= 0
2x
dt
dt
dy
x dx
= −
.
dt
y dt
Now, if x = 6 then y 2 = 100 − 62 or y = 8. Hence,
dy
=
dt x=6,y=8
−6
3
· (−2) = m/s.
8
2
Example 3.2.9. Water is pouring into an inverted cone at the rate of 8 cubic feet per minute. If
the height of the cone is 12 feet and the radius of its base is 6 feet, how fast is the water level rising
when the water is 4 feet deep?
Solution.
Let V be the volume of the water inside the cone at any time t. Let h, r be the height and radius,
dV
respectively, of the cone formed by the volume of water at any time t. We have
= 8 ft3 /min
dt
dh
and we wish to find
when h = 4 ft.
dt
164
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
Now, using similar triangles, we obtain
6
r
6
=
h
12
or r =
r
h
at any time t. Thus,
2
12
h
2
1
1
h
1
V = πr2 h = π
h = πh3 .
3
3
2
12
Differentiating both sides with respect to t,
at
h
Thus,
s
1
Dt [V ] = Dt
πh3
12
dV
1 2 dh
=
πh
dt
4
dt
4 dV
dh
=
.
dt
πh2 dt
em
at
ic
of
M
dh
4
2
=
· 8 = ft/min.
2
dt h=4 π(4)
π
Solution.
UP
In
s
tit
ut
e
Example 3.2.10. An automobile traveling at a rate of 20 ft/s is approaching an intersection.
When the automobile is 100 ft. from the intersection, a truck traveling at the rate of 40 ft/s
crosses the intersection. The automobile and the truck are on roads that are at right angles to each
other. How fast are the truck and the automobile separating 2 seconds after the truck leaves the
intersection?
Let x denote the distance of the automobile from
the intersection, y denote the distance of the
truck from the intersection and z denote the
distance between the truck and the automobile.
dx
dy
Then we have
= −20 ft/s,
= 40 ft/s.
dt
dt
After 2 seconds, note that x = 60 ft and y = 80
dz
ft. So we want to find
when x = 60 ft and
dt
y = 80 ft.
Observe that at any time t, we have
x2 + y 2 = z 2 .
x
y
z
truck
automobile
3.2. RATES OF CHANGE, RECTILINEAR MOTION, AND RELATED RATES
Differentiating both sides with respect to t,
Dt x2 + y 2 = Dt z 2
dx
dy
dz
2x
+ 2y
= 2z
dt
dt
dt
dz
1 dx
dy =
x
+y
.
dt
z
dt
dt
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
Now, if x = 60, y = 80, then 602 + 802 = z 2 or z = 100. Hence,
dz
1 =
60(−20) + 80 · 40 = 20 ft/s.
dt x=60,y=80,z=100 100
165
166
3.2.4
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
Exercises
Exercises for Discussion
A. 1. An object moves on a horizontal coordinate line. Its directed distance s from the origin
at the end of t seconds is s(t) = (t3 − 6t2 + 9t) feet.
a. when is the object moving to the left?
b. what is its acceleration when its velocity is equal to zero?
c. when is the acceleration positive?
d. when is its speed increasing?
2. Suppose that a ball was thrown upward from the top of a building. If the equation of the
ball is s = −16t2 + 64t + 160 where s is the directed distance from the ground in feet and
t is in seconds,
a. when did the ball reach its maximum height?
s
b. what was the ball’s maximum height?
em
at
ic
c. when did the ball hit the ground?
d. with what speed did the ball hit the ground?
at
h
e. what was the acceleration of the ball after 2 seconds?
of
M
f. what is the height of the building?
ut
e
3. A stone is thrown vertically upward from the top of a building. If the equation of the
motion of the stone is s = −5t2 + 30t + 200, where s is the directed distance from the
ground in meters and t is in seconds,
tit
a. find the acceleration of the stone when the velocity is 10 meters per second.
In
s
b. after how many seconds will the stone reach its maximum height?
c. what is the height of the building?
UP
d. what is the maximum height the stone will reach?
e. what is the velocity of the stone upon impact?
B. 1. Find how fast the volume of a sphere increases as the radius increases.
2. If water is being drained from a swimming pool and V liters is the volume of the water
at the time t minutes after the draining starts, where V = 250(1600 − 80t + t2 ), how fast
is the water flowing out of the water pool 5 minutes after the draining starts?
3. Find the slope of the tangent line at each point of the graph of y = x4 + x3 − 3x2 where
the rate of change of the slope is equal to zero.
4. An airplane is flying on a horizontal path at a height of 3800 ft. At what rate is the angle
of elevation θ of the airplane from P changing with respect to the distance between the
airplane and a fixed point P on the ground if θ = 30◦ ?
5. If two resistors with resistance R1 and R2 are connected in parallel, the total resistance
1
1
1
R in ohms is given by
=
+
. If R1 and R2 are increasing at 0.4 ohms/s and 0.25
R
R1 R2
ohms/s, respectively, how fast is R changing when R1 = 600 ohms and R2 = 400 ohms?
3.2. RATES OF CHANGE, RECTILINEAR MOTION, AND RELATED RATES
167
Supplementary Exercises
A. 1. A particle is moving along a horizontal line. Its directed distance s from the origin at the
end of t seconds is s = t3 − 9t2 + 24t − 16 feet.
1. when is the particle moving to the left?
2. what is the acceleration when its velocity is equal to zero?
3. when is the acceleration positive?
4. when is the particle slowing down?
5. what is the total distance travelled by the particle after 4 seconds?
2. An apple is thrown vertically upward from the top of a tree. If the directed distance of
the apple from the ground after t seconds is s = −5t2 + 6t + 8 meters,
a. when did the apple hit the ground?
b. with what speed did the apple hit the ground?
c. what was the apple’s maximum height?
em
at
ic
s
d. if a person stands directly below the path of the apple to catch it and he can extend
his arm to get a reach of 2 meters from the ground, at what time will he be able to
catch the apple?
of
M
at
h
3. An object’s position along a horizontal axis after t seconds is given by s = −t3 +4t2 −4t+5
feet.
a. When will the object change direction?
b. When is the object speeding up?
ut
e
c. What is the total distance travelled by the object afer 3 seconds?
In
s
tit
d. Suppose another object is on the same horizontal axis with position functon r (t) =
−t3 + 5t2 − 3t + 6. Will the two objects collide?
5
ft/s while Neil starts
2
running towards the south at 6 ft/s. How fast is the distance between Reden and Neil
increasing after 2 s?
UP
B. 1. Starting from the same point, Reden starts walking eastward at
2. A woman standing on a cliff is watching a motor boat through a telescope as the boat
approaches the shoreline directly below her. If the telescope is 250 feet above the water
level and if the boat is approaching the cliff at 20 ft/s, at what rate is the acute angle made
by the telescope with the vertical changing when the boat is 250 feet from the shore?
3. A right circular cylindrical balloon is being inflated in such a way that the radius and
height are both increasing at the rate of 1 in/s and 3 in/s, respectively. What is the rate of
change of its total surface area when its radius and height are 20 in and 50 in, respectively?
4. A right circular cone with radius 3 cm and height 6 cm is submerged point first into a tall
cylinder of radius 5 cm that is partially filled with water. How fast is the water level rising
when the cylinder is completely submerged?
5. A baseball diamond has the shape of a square with sides 90 ft long. A player 60 ft from
the second base is running towards the third plate at a speed of 28 ft/min. At what rate
is the player’s distance from the home plate changing?
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
6. At 12:10:00 p.m., a bug starts crawling away from the center of a clock, which is 2 meters
above the ground. It moves along the minute hand at a constant rate of 0.1 cm/s. Find
the initial rate at which its height above the ground changes. (For this problem, assume
that all hands of the clock have negligible thickness and move at a constant rate.)
7. Shan, who is 5 feet tall, is approaching a post that holds a lamp 15 feet above the ground.
If he is walking at a speed of 4 ft/s, how fast is the end of his shadow moving when he is
17 feet away from the base of the lamp post?
8. Water is being poured at the rate of 2π ft3 /min. into an inverted conical tank that is 12
ft deep and having radius of 6 ft at the top. If the water level is rising at the rate of 61
ft/min and there is a leak at the bottom of the tank, how fast is the water leaking when
the water is 6 ft deep?
9. Newton is traveling eastward at 20 km/hr while Leibniz is traveling northward at 10 km/hr
At 1:00 PM, Leibniz is located 50 km south of Newton. At what rate is the distance between
Newton and Leibniz changing at 3:00 PM?
at
h
em
at
ic
s
10. The apparent brightness of a star is equal to the rate at which it emits energy (in watts)
divided by the square of its distance from Earth (in meters). Currently, the sun is 1.5 × 108
−16
m away, while it emits energy at a rate of 3.15 × 1019−10 t W t seconds from now. If
the Earth is moving away from the sun at a constant rate of 0.05 m/s, at what rate is the
apparent brightness of the sun changing?
of
M
11. A particle moves along the curve y = 2x−2 , with its x-coordinate increasing at a constant
rate of 1 unit/s. At what rate is its distance from the origin changing when it is at (3, 4)?
In
s
tit
ut
e
12. A point moves along the upper half of the curve y 2 − y = x3 − x, with its x-coordinate
increasing at a constant rate of 1 unit/s. At what rate is the slope of the tangent line to
the point changing when it is at (2, 3)?
UP
168
3.3. LOCAL LINEAR APPROXIMATION, DIFFERENTIALS, AND MARGINALS
3.3
169
Local Linear Approximation, Differentials, and Marginals
At the end of this section, the student will be able to:
• use differentials to approximate the net change of quantities
• approximate quantities using local linear differentiation more other quantities
• estimate the change in cost, profit, revenue for an additional unit produced using
marginals
3.3.1
Differentials
Recall:
Q
f (x) − f (x0 )
x→x0
x − x0
∆y
.
= lim
∆x→0 ∆x
R
lim
f (x0 )
of
M
at
h
∆y
If ∆x is small enough, then
is approximately
∆x
0
equal to f (x0 ). Thus, ∆y may be approximated
by f 0 (x0 )∆x.
s
f (x)
em
at
ic
f 0 (x0 ) =
∆y
P
|
{z
dx=∆x
dy
}S
x0
y = f (x)
x = x0 + dx
Figure 2.3.1
e
Let the function y = f (x) be differentiable at x.
tit
ut
1. The differential dx of the independent variable x denotes an arbitrary increment of x.
UP
In
s
2. The differential dy of the dependent variable y associated with x is given by dy =
f 0 (x)dx.
dy
Observe that if dx 6= 0, then dy = f 0 (x)dx implies that
= f 0 (x). This serves as motivation for
dx
the following remark.
dy
Remark 3.3.1. Symbolically, we may interpret
as either the derivative of y = f (x) with respect
dx
to x or the quotient of the differential of y by the differential of x. Another interpretation of this
quantity may be derived from the figure above:
• The value of dy closely resembles that of the actual change in function value, which we called
∆y.
• The value of dx corresponds with the increment of x, ∆x.
dy
as an approximation of the slope of the secant line through the
dx
points P (x0 , f (x0 )) and Q(x, f (x)).
• Thus, we can think of
170
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
The following theorems follow immediately from the differentiation rules discussed previously.
Theorem 3.3.2. Let u and v be differentiable functions of x and c a constant.
1. d(c) = 0
4. d(uv) = udv + vdu
2. d(xn ) = nxn−1 dx
5. d( uv ) =
3. d(cu) = cdu
6. d(un ) = nun−1 du
vdu − udv
v2
Example 3.3.3.
1. Let y = x5 − x3 + 2x. Then dy = (5x4 − 3x2 + 2) dx .
1
· (3x2 + 6x) dx .
x3 + 3x2 , then dy = √
3
2 x + 3x2
dy
if xy 2 = y + x. Using implicit differentiation,
dx
s
3. Find
√
em
at
ic
2. If y =
x(2ydy) + y 2 dx = dy + dx
Local Linear Approximation and Approximating ∆y
In
s
tit
ut
e
3.3.2
1 − y2
dx.
2xy − 1
of
M
dy =
at
h
and therefore
∆y ≈ f 0 (x0 )∆x.
UP
In the previous discussion, we have shown
that if ∆x is very small, we have
Equivalently, we have
y = f (x)
`
f (x)
f (x0 ) + f 0 (x0 )(x − x0 )
f (x) ≈ f (x0 ) + f 0 (x0 )(x − x0 ).
Now, the equation of the tangent line to the
graph of y = f (x) at the point (x0 , f (x0 ))
is given by
y − f (x0 ) = f 0 (x0 )(x − x0 )
⇐⇒
y = f (x0 ) + f 0 (x0 )(x − x0 ).
x0
Figure 2.3.2
x
3.3. LOCAL LINEAR APPROXIMATION, DIFFERENTIALS, AND MARGINALS
171
Remark 3.3.4.
1. The function L(x) = f (x0 ) + f 0 (x0 )(x − x0 ) is called the local linear approximation of f (x)
at x0 . It follows that f (x) ≈ L(x) at points close to (x0 , f (x0 )). Moreover, from the discussion
above, the tangent line to the graph of f at x0 approximates the graph of f when x is near x0 .
2. Note that there are many ways to approximate any function using a linear function. However,
the local linear approximation turns out to be the ”best” linear approximation of f near x0 .
3. If dx = ∆x = x − x0 , then x = x0 + dx. Since f (x) ≈ f (x0 ) + f 0 (x0 )(x − x0 ), we have
f (x0 + dx) ≈ f (x0 ) + f 0 (x0 )dx
4. If dx ≈ 0 , then dy = f 0 (x)dx = f 0 (x)∆x ≈ ∆y. Hence, for sufficiently small values of dx,
dy ≈ ∆y.
em
at
ic
s
5. Because dy is easier to compute than ∆y , dy is used to approximate ∆y when dx is
sufficiently small (depending on the required accuracy of the increment).
at
h
Example 3.3.5.
of
M
√
√
1. Find the local linear approximation of f (x) = 3 x at x0 = 8 and use this to approximate 3 8.03.
e
Solution.
UP
In
s
tit
ut
1
We have f 0 (x) = √
, thus at x0 = 8,
3
3 x2
Therefore,
L(x) = f (8) + f 0 (8)(x − 8)
1
L(x) = 2 + (x − 8).
12
√
1
8.03 = f (8.03) ≈ L(8.03) = 2 + (8.03 − 8) = 2.0025.
12
2. Approximate the following:
(a)
√
3
27.027
Solution.
√
1
Let f (x) = 3 x. Then f 0 (x) = √
. Thus,
3
3 x2
√
3
27.027 = f (27 + 0.027)
≈ f (27) + f 0 (27) · (0.027)
1
= 3+
· (0.027)
3·9
= 3.01.
172
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
(b)
√
15.96
Solution.
Let f (x) =
√
1
x. Then f 0 (x) = √ . Thus,
2 x
√
15.96 = f (16 − 0.04)
≈ f (16) + f 0 (16) · (−0.04)
1
= 4+
· (−0.04)
2·4
= 4 − 0.005
= 3.995.
1
3. A ball 10 inches in diameter is to be covered by a rubber material which is 16
inch thick. Use
differentials to estimate the volume of the rubber material that will be used.
Solution.
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
4
The volume of the ball with radius r is given by V (r) = πr3 . So dV = 4πr2 dr . The volume
3
of the rubber material is
1
V 5+
− V (5) = ∆V
16
≈ dV
1
= 4πr2 dr, where r = 5, dr =
16
1
2
= 4π (5) ·
16
25π 3
in .
=
4
UP
4. A metal rod 15 cm long and 8 cm in diameter is to be insulated, except for the ends, with a
material 0.001 cm thick. Use differentials to estimate the volume of the insulation.
Solution.
Note that since the ends are not to be insulated, the height h of the rod remains the same.
Thus, the volume of the rod with radius r at the ends is given by V = πr2 h = 15πr2 . We have
dV = 30πrdr. The volume of the insulation is
V (4 + 0.001) − V (4) = ∆V
≈ dV
= 30πrdr, where r = 4, dr = 0.001
= 30π (4) · (0.001)
= 0.12π cm3 .
5. Suppose that the side of a square is measured with a ruler to be 8 inches with a measurement
1
error of at most ± 64
of an inch. Estimate the error in the computed area of the square.
3.3. LOCAL LINEAR APPROXIMATION, DIFFERENTIALS, AND MARGINALS
173
Solution.
The area of a square with side x is A(x) = x2 . Thus, dA = 2xdx. The measurement error of
1
1
at most ± 64
implies that |dx| = 64
. We can approximate the error in the computed area ∆A
using dA. We have
1
1
=4
|∆A| ≈ |dA| = 2x|dx| = 2(8) 64
Therefore, the propagated error in the computed area is at most ± 14 of a square inch.
3.3.3
Marginals
Rates of change are also used in the field of economics. Economists refer to marginal profit, marginal
revenue, and marginal cost as the rates of change of the profit, revenue and cost, respectively, with
respect to the number of units produced or sold.
s
1. Cost function C(x) - cost of producing x units of a product
em
at
ic
2. Revenue function R(x) - revenue in the sale of x units of a product
3. Profit function P (x) - profit earned in the sale of x units of a product
of
M
at
h
4. Price demand funciton p(x) - price of product x if there are x demands
e
Remark 3.3.6.
tit
ut
1. The revenue and price demand functions are related by the equation
In
s
R(x) = x · p(x)
UP
2. The profit, revenue and cost functions are related by the equation
P (x) = R(x) − C(x)
= x · p(x) − C(x)
3. C(x) ≥ 0 and C(0) is the overhead or fixed cost
4. R(x) ≥ 0 and R(0) = 0
5. P (x) may be positive, negative or zero
Differentiating each of these quantities with respect to the number of units x gives a term in
economics called Marginal. That is
dP
= marginal profit,
dx
dR
= marginal revenue,
dx
dC
= marginal cost.
dx
Note that the marginal function value at a number n is the rate of change of the function when
x = n. This value is an approximation of F (n + 1) − F (n), where F = C, R, or P .
174
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
Example 3.3.7. A manufacturer determines that the profit derived from selling x units of a certain
item is given by
P (x) = 0.0002x3 + 10x.
1. Find the marginal profit for a production level of 50 units.
Solution.
The marginal profit is
dP
= 0.0006x2 + 10. When x = 50, we have
dx
dP
= 0.0006(50)2 + 10 = 11.50 per unit.
dx
2. Compare this to the actual profit obtained by increasing the production level from 50 units to
51 units.
em
at
ic
P (51) = 0.0002(51)3 + 10(51) = 536.53
s
Solution.
P (50) = 0.0002(50)3 + 10(50) = 525
at
h
The actual gain in profit is given by
of
M
P (51) − P (50) = 536.53 − 525 = 11.53.
e
Notice that the actual gain in profit can be approximated by the marginal profit.
In
s
tit
ut
Example 3.3.8. A company manufactures fuel tanks for automobiles. The total weekly cost (in
dollars) of producing x tanks is given by
UP
C(x) = 10, 000 + 90x − 0.05x2 .
1. Find the marginal cost function.
Solution.
dC
= 90 − 0.1x
dx
2. Find C 0 (500) and the actual additional cost of producing the 501st tank.
Solution.
C 0 (500) = 90 − 0.1(500) = 40 dollars.
To compute the actual additional cost of producing the 501st tank, we need
C(501) = 10, 000 + 90(501) − 0.05(501)2 = 42, 539.95 dollars
C(500) = 10, 000 + 90(500) − 0.05(500)2 = 42, 500 dollars
Hence, the actual additional cost is given by
C(501) − C(500) = 42, 539.95 − 42.500 = 39.95 dollars
3.3. LOCAL LINEAR APPROXIMATION, DIFFERENTIALS, AND MARGINALS
175
Example 3.3.9. A fastfood restaurant has determined that the monthly price-demand function
for their hamburgers is given by
60, 000 − x
p(x) =
,
20, 000
where x is the number of hamburgers sold. Find the increase in revenue per hamburger for monthly
sales of 20,000 hamburgers.
Solution.
The revenue function R(x) is equal to the product x · p(x) where p(x) is the price-demand function.
x2
60, 000 − x
= 3x −
R(x) = x ·
20, 000
20, 000
So
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
dR
= 1. Hence the marginal revenue for the sale of 20,000 hamburgers is 1.
dx
UP
If x = 20, 000,
dR
x
=3−
dx
10, 000
176
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
3.3.4
Exercises
Exercises for Discussion
A. Find the local linear approximation of f at x0 .
√
1. f (x) = x2 2x + 3; x0 = −1
2. f (x) = ln(2 − sec x); x0 = 0
B. Approximate the following using local linear approximation.
1. (2.001)5
√
2. 2.99 3.99
3. e0.01 + 0.01 ln 1.02
s
C. Solve the following completely.
em
at
ic
1. Use differentials to approximate the increase in the surface area of a soap bubble when its
radius increases from 3 inches to 3.025 inches.
of
M
at
h
2. A metal box in the form of a cube is to have an interior volume of 1000 cm3 . The six sides
are to be made of metal 12 cm thick. If the cost of the metal to be used is 0.20 pesos per
cubic centimeter, use differentials to find the approximate cost of the metal to be used in the
manufacture of the box.
ut
e
3. A cylindrical roller is exactly 12 inches long and its diameter is measured as 6 ± 0.05 inches.
Approximate its volume with an estimate for the error.
UP
In
s
tit
1
D. 1. Suppose that R(x) = 300x − x2 dollars is the total revenue from the sales of x tables. Find:
2
a. the marginal revenue function
b. the marginal revenue when x = 40
c. the actual additional revenue from the sale of the 41st table
2. The research department of a certain company found out that the price-deman equation and
cost functions of a particular product are given by
p(x) = 10 − 0.001x and C(x) = 7, 000 + 2x,
respectively, where x is the number of units produced and sold.
a. Determine the marginal profit
b. Use the marginal profit to approximate the profit gained in producing and selling 1001
units.
3. The profit and revenue functions, measured in PhP, for selling x items are given by P (x) =
2x2 + 46x − 200 and R(x) = x3 + x2 + 6x, respectively. Find the minimum cost.
4. The cost of producing x necklaces is given by C(x) = x2 − 20x + 100. If each necklace is
sold for P 80, how many necklaces should you sell to get maximum profit?
3.3. LOCAL LINEAR APPROXIMATION, DIFFERENTIALS, AND MARGINALS
177
Supplementary Exercises
A. Find the local linear approximation of f at x0 .
1. f (x) = sech(π − x) + e1+cos x ; x0 = π
√
2. f (x) = tan−1 ( x); x0 = 1
B. Approximate the following using local linear approximation.
1
64.12
2. sec 0.2 tan 0.2
1. √
6
3. sin(tan−1 0.04)
C. Solve the following completely.
em
at
ic
s
1. A ball of malt, with diameter 1.5 cm, is coated by a layer of chocolate 0.02 cm thick to
form a candy. Use differentials to approximate the amount of chocolate (in mL) each candy
contains.
2. A burn on a person’s skin is in the shape of a circle. Estimate the decrease in the area of
the burn when the radius decreases from 1 cm to 0.8 cm.
of
M
at
h
3. The pH of an aqueous solution is given by
pH = − log10 (aH+ ),
tit
ut
e
where aH+ is the concentration of hydrogen ions in the solution. A solution with initial
pH = 3 decreases in volume from 1 L to 0.99 L, but the amount of hydrogen ions in the
solution remains unchanged. Estimate the net change in its pH.
UP
In
s
4. The population of a bacterium in a nutrient broth is given by P = 10000e0.001t t hours after
it was introduced. Estimate the net change in the population within the first six minutes.
D. 1. The total cost of producing x food processors is
C(x) = 2, 000 + 50x − 0.5x2
a Find the actual additional cost of producing the 21st food processor.
b Use the marginal cost to approximate the cost of producing the 21st food processor.
2. The price-demand equation and the cost function for the production of a certain product
are given by
x = 6, 000 − 30p and C(x) = 72, 000 + 60x,
respectively, where x is the number of units that can be sold monthly at a price p (Pesos)
per piece.
a. Determine the marginal cost.
b. Determine the revenue function and the break-even point(s), i.e. the production level
when the revenue is equal to the cost.
c. Determine R0 (1500).
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
d. Find the marginal profit at x = 1500.
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
3. A company is known for producing high quality calculators. The price-demand function for
selling x calculators is given by p(x) = 4x2 − 60x, while the profit in selling x is determined
by P (x) = 3x3 − 54x2 + 15x − 150. Determine the number of calculators that must be sold
in order to minimize cost.
UP
178
3.4. INDETERMINATE FORMS AND L’HÔPITAL’S RULE
3.4
179
Indeterminate Forms and L’Hôpital’s Rule
We began this course with the concept of the limit: the behavior of a function as the independent variable approaches a certain value, or as it increases or decreases without bound. The new
techniques of differentiation we have learned so far will aid us in evaluating a wider range of limits, including new types of indeterminate forms which we could not previously evaluate with the
techniques learned in Chapter 1.
By the end of this section, the student will be able to:
• evaluate limits with indeterminate form 0/0, ∞/∞ using L’Hôpital’s Rule; and
Indeterminate Forms of Type
∞
0
and
0
∞
em
at
ic
3.4.1
s
• evaluate limits with indeterminate forms 0 · ∞, ∞ − ∞, 00 , 1∞ , ∞0 .
of
M
at
h
With the aid of derivatives, certain limits can be evaluated more conveniently. We first recall
some terminology defined in the early part of this course. We also recall here some techniques in
evaluating limits we have previously encountered.
f (x)
The lim
is an indeterminate form of type
x→a g(x)
0
if lim f (x) = lim g(x) = 0.
x→a
0 x→a
∞
2.
if lim f (x) and lim g(x) are both +∞ or −∞.
x→a
∞ x→a
In
s
tit
ut
e
1.
UP
Of course, ”x → a” may be replaced by ”x → a+ ”, ”x → a− ”, ”x → +∞” or ”x → −∞”.
Example 3.4.1. Evaluate the following limits.
x2 − 3x
0
1. lim 2
x→0 2x + x
0
Solution.
x2 − 3x
x(x − 3)
x−3
= lim
= lim
= −3
x→0 2x2 + x
x→0 x(2x + 1)
x→0 2x + 1
lim
sin 5x
x→0 sin 3x
2. lim
0
0
Solution.
sin 5x
lim
= lim
x→0 sin 3x
x→0
sin 5x
5x
3x
sin 3x
5
5
5
=1·1· =
3
3
3
180
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
x2 + 3x − 10
3. lim 2
x→2− x − 4x + 4
0
0
Solution.
x2 + 3x − 10
x→2− x2 − 4x + 4
lim
3x − 1
4. lim
x→+∞ 7 − 6x
+∞
−∞
(x + 5)(x − 2)
(x − 2)2
7
x+5
= lim
0−
x→2− x − 2
= −∞
=
lim
x→2−
Solution.
em
at
ic
s
3− 1
1
3x − 1 x1
3x − 1
= lim
· 1 = lim 7 x = −
x→+∞ 7 − 6x
x→+∞
x→+∞ 7 − 6x
2
x
x −6
lim
L’Hôpital’s Rule
at
h
3.4.2
ut
e
of
M
The following theorem tells us how derivatives can be used to evaluate limits that are indeterminate
∞
0
or
. It is usually referred to as L’Hôpital’s Rule, after the French mathematician
of type
0
∞
Guillaume François Marquis de L’Hôpital.
UP
In
s
tit
Theorem 3.4.2. Let f and g be functions differentiable on an open interval I containing a
f (x)
0
except possibly at a and g 0 (x) 6= 0 for all x ∈ I \ {a}. If lim
is indeterminate of type
x→a g(x)
0
∞
, then
or
∞
f (x)
f 0 (x)
lim
= lim 0
,
x→a g(x)
x→a g (x)
f 0 (x)
f 0 (x)
exists or lim 0
= ±∞.
0
x→a g (x)
x→a g (x)
provided lim
Remark 3.4.3. L’Hôpital’s Rule, with suitable modifications, is valid if “x → a” is replaced by
“x → a+ ”, “x → a− ”, “x → +∞” or “x → −∞”.
Example 3.4.4. Evaluate the following limits.
x2 − 3x
0
1. lim 2
x→0 2x + x
0
Solution.
Dx x2 − 3x
x2 − 3x
2x − 3
= lim
lim 2
= lim
= −3
2
x→0 2x + x
x→0 Dx 2x + x
x→0 4x + 1
3.4. INDETERMINATE FORMS AND L’HÔPITAL’S RULE
sin 5x
x→0 sin 3x
2. lim
181
0
0
Solution.
5 cos 5x
5
sin 5x
= lim
=
x→0 3 cos 3x
x→0 sin 3x
3
lim
x2 + 3x − 10
3. lim 2
x→2− x − 4x + 4
0
0
+∞
−∞
s
em
at
ic
2x + 3
= lim
−
x→2 2x − 4
= −∞
7
0−
of
M
3x − 1
4. lim
x→+∞ 7 − 6x
x2 + 3x − 10
lim 2
x→2− x − 4x + 4
at
h
Solution.
tit
ut
e
Solution.
In
s
lim
= lim
3
x→+∞ −6
=−
1
2
0
0
UP
x3 − 3x + 2
5. lim
x→1 1 − x + ln x
3x − 1
x→+∞ 7 − 6x
Solution.
x3 − 3x + 2
lim
x→1 1 − x + ln x
csc x
6. lim
x→0− 1 − cot x
Solution.
−∞
+∞
3x2 − 3
= lim
x→1 −1 + 1
x
6x
= lim 1
x→1 − 2
x
= −6
0
0
182
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
csc x
x→0− 1 − cot x
lim
− csc x cot x
csc2 x
x→0−
− cot x
lim
x→0− csc x
csc2 x
lim
x→0− − csc x cot x
csc x
lim
x→0− − cot x
− cot x
lim
x→0− csc x
=
lim
=
=
=
=
+∞
−∞
−∞
+∞
em
at
ic
s
Observe that the expression in the last line above is exactly the same as the expression in the
second line. Hence, continued application of L’Hôpital’s Rule here will just lead us to an infinite
string of equations and will not help us evaluate the limit. This example should make you realize
that L’Hôpital’s Rule is not always helpful. Sometimes, we must fall back to old-fashioned tricks.
of
M
at
h
For instance, we evaluate this limit by simply manipulating the given expression to obtain
a simpler expression:
e
1
sin
x
= lim
cos x
x→ 0− 1 −
sin x
1
= lim
−
sin
x
−
cos x
x→ 0
= −1
UP
In
s
tit
ut
csc x
lim
x→ 0− 1 − cot x
It is imperative to remember the behavior of each function introduced in this course; doing so
will help us in computing new limits. Recalling the graphs of our new functions will be helpful in
remembering their behavior. For instance, using the graph of f (x) = loga x, where 0 < a < 1, one
sees that lim loga x = +∞. Thus, if f (x) approaches 0 through positive values as x approaches
x→0+
k, then
lim loga [f (x)] = lim loga y = +∞.
y→0+
x→k
Example 3.4.5. Evaluate:
lim
x→+∞
x + sin x
x
Solution.
Note that since x + sin x ≥ x − 1 for any x ∈
R and x→+∞
lim x − 1 = +∞, then lim x + sin x = +∞
x→+∞
as well. Thus, the limit is indeterminate of type
∞
∞
.
3.4. INDETERMINATE FORMS AND L’HÔPITAL’S RULE
183
Dx (x + sin x)
1 + cos x
1 + cos x
=
and lim
does not exist since cos x does not apx→+∞
Dx (x)
1
1
proach any particular value as x → +∞. Neither does 1 + cos x grow without bound. Thus,
L’Hôpital’s Rule does not apply.
However,
We therefore need to employ other techniques. In particular, notice that for any x > 0, the following
hold:
x − 1 ≤ x + sin x ≤ x + 1
x−1
x + sin x
x+1
⇔
≤
≤
x
x
x
x+1
x + sin x
x−1
= 1 = lim
, so by the Squeeze Theorem, lim
= 1.
Also, lim
x→+∞
x→+∞
x→+∞
x
x
x
3.4.3
Indeterminate Forms of Type 0 · ∞ and ∞ − ∞
We revisit the indeterminate forms of type 0 · ∞ and ∞ − ∞. We recall their definitions below.
x→a
em
at
ic
s
1. The limit lim f (x)g(x) is an indeterminate form of type 0 · ∞ if either
lim f (x) = 0 and lim g(x) = +∞ or − ∞, or
x→a
x→a
at
h
lim f (x) = +∞ or − ∞ and lim g(x) = 0.
x→a
of
M
x→a
2. The limit lim (f (x) + g(x)) is an indeterminate form of type ∞ − ∞ if either
x→a
e
lim f (x) = +∞ and lim g(x) = −∞, or
x→a
ut
x→a
tit
lim f (x) = −∞ and lim g(x) = +∞.
x→a
UP
In
s
x→a
0
∞
Remark 3.4.6. L’Hôpital’s Rule works only for indeterminate forms of type
and
. Any
0
∞
other indeterminate form must be expressed equivalently in one of these two forms if we wish to
apply L’Hôpital’s Rule. For the new indeterminate forms described above, these conversions can
be performed as described below.
1. If lim f (x) = 0 and lim g(x) = +∞ or −∞, write lim f (x)g(x) as:
x→a
(a) lim
x→a
x→a
f (x)
0
g(x)
∞
, which is indeterminate of type , or
x→a 1
0
g(x)
(b) lim
, which is indeterminate of type
x→a 1
∞
f (x)
and apply L’Hôpital’s Rule.
2. If lim f (x) + g(x) is indeterminate of type ∞ − ∞, rewrite f (x) + g(x) as a single expression to
x→a
0
∞
obtain an indeterminate form of type or
and apply L’Hôpital’s Rule.
0
∞
184
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
Example 3.4.7. Evaluate the following limits.
1. lim sin−1 (2x) csc x
(0 · +∞)
x→0+
Solution.
sin−1 (2x)
sin x
x→0+
1
√
· (2)
1 − 4x2
= lim
cos x
x→0+
2
=
1
= 2
lim sin−1 (2x) csc x =
x→0+
lim
0
0
(+∞ · 0)
2. lim tan θ ln (sin θ)
em
at
ic
s
θ→ π2 −
at
h
Solution.
0
0
ut
e
of
M
ln(sin θ)
lim tan θ ln (sin θ) = lim
π−
π−
cot θ
θ→ 2
θ→ 2
1
cos θ
= lim sin θ 2
π−
− csc θ
θ→ 2
lim − sin θ cos θ
3. lim
x→1+
x
1
−
x − 1 ln x
UP
In
s
tit
=
1
1
− +
+
0
0
θ→ π2 −
= −1(0)
= 0
Solution.
lim
x→1+
x
1
−
x − 1 ln x
x ln x − (x − 1)
= lim
+
(x − 1) ln x
x→1
x · x1 + ln x − 1
= lim
x→1+ (x − 1) · 1 + ln x
x
ln x
= lim
x→1+ 1 − 1 + ln x
x
=
=
lim
1
x
x→1+ 12 + 1
x
x
1
2
0
0
0
0
3.4. INDETERMINATE FORMS AND L’HÔPITAL’S RULE
3.4.4
185
Indeterminate Forms of Type 1∞ , 00 and ∞0
We now define new indeterminate forms of exponential type.
Let f be a nonconstant function. The lim f (x)g(x) is an indeterminate form of type
x→a
1. 1∞ if lim f (x) = 1 and lim g(x) = +∞ or −∞.
x→a
x→a
2. 00 if lim f (x) = 0, through positive values, and lim g(x) = 0.
x→a
x→a
3. ∞0 if lim f (x) = +∞ and lim g(x) = 0.
x→a
x→a
Remark 3.4.8. If lim f (x)g(x) is indeterminate of type 1∞ , 00 or ∞0 , we write
x→a
lim f (x)g(x) = lim eg(x) ln[f (x)]
x→a
x→a
s
and evaluate lim g(x) ln[f (x)] first. Then, if
1. lim g(x) ln[f (x)] = L ∈
x→a
em
at
ic
x→a
R, then x→a
lim f (x)g(x) = eL .
x→a
of
M
x→a
at
h
2. lim g(x) ln[f (x)] = +∞, then lim f (x)g(x) = +∞.
3. lim g(x) ln[f (x)] = −∞, then lim f (x)g(x) = 0.
x→a
x→a
tit
ut
e
Example 3.4.9. Evaluate the following limits.
1. lim xsin x
00
In
s
x→0+
Solution.
UP
First, write xsin x = esin x ln x . Evaluate first lim sin x ln x.
x→0+
lim sin x ln x
x→0+
(0 · (−∞)) =
=
ln x
x→0+ csc x
lim
lim
−∞
+∞
1
x
x→0+ − csc x cot x
2
− sin x
0
= lim
+
x cos x
0
x→0
−2 sin x cos x
= lim
x→0+ x(− sin x) + cos x
= 0
Hence, lim xsin x = e0 = 1.
x→0+
3 2x
2. lim
1−
x→+∞
x
(1∞ )
186
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
Solution.
2x
3
= e2x ln(1− x )
1 − x3
3
lim 2x ln 1 −
x→+∞
x
(+∞ · 0) =
=
=
em
at
ic
s
=
3
ln 1 −
x
lim
1
x→+∞
2x 1
3
·
3
x2
1−
x
lim
1
x→+∞
− 2
2x
−6
lim
3
x→+∞
1−
x
−6
3 2x
= e−6 .
1−
x→+∞
x
UP
In
s
tit
ut
e
of
M
at
h
Hence, lim
0
0
3.4. INDETERMINATE FORMS AND L’HÔPITAL’S RULE
3.4.5
187
Exercises
Exercises for Discussion
A. Evaluate the following limits.
3−x
1. lim x
x→3 2 − 8
ln x
2. lim 1
x→0+ e x
sec x
3. lim
π − 1 + tan x
x→ 2
x
4. lim √
x→−∞
x2 + 1
ln (3 + ex )
5. lim
x→+∞
7x
sin x − x
6. lim
x→0
x3
1 − x + ln x
7. lim
x→1 1 + cos πx
1
8. lim x2 e x
12.
13.
x→+∞
x−
p
x2 − x + 2
lim (tan x)cos x
x→ π2 −
1
14. lim x + e2x x
x→0
16.
x→+∞
1
x + e2x x
√
x
lim
x→+∞
cosh 3x
lim x ln
em
at
ic
17.
lim
s
15.
x→+∞
x−1
x+1
sin x + tan x
ex + e−x − 2
cos x
lim x − π2
18. lim
x→0−
at
h
x→0+
19.
9. lim tan( x) ln x
of
M
x→0+
20.
21.
x→ π2 +
lim logx (x + 10)
x→+∞
lim ln(e3x + 1) − 3x
x→+∞
In
s
tit
ut
e
10. lim tan−1 (x) ln x
x→0+
1
11. lim csc x −
x
x→0−
B. Do as indicated.
lim
UP
1. If an electrostatic field E acts on a gaseous polar dielectric, the net dipole moment P per
1
unit volume is P (E) = coth E − . Evaluate lim P (E).
E
E→0+
x
x+m
2. For what values of m is lim
= e?
x→+∞ x − m
188
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
Supplementary Exercises
A. Evaluate the following limits.
2x3 + x2 + 1
x→−∞
x2 − x
2x
e −4
2. lim
x→ln 2 x − ln 2
cos x
3. limπ
x→ 2 x − π 3
2
tanh x
4. lim
x→0+ sech x − 1
11.
lim
lim (ln(x + 1) − ln(x − 1))
x→+∞
12. lim cot(3x) ln [1 + tan(5x)]
x→0+
13. lim ex csch x
x→0
14. lim x(ln 2)/(1+ln x)
x→∞
1 3x
15. lim
1+
x→+∞
x
2
1 − ex
x→0 log3 (1 − x)
2x3 + ln 5x
6. lim
x→+∞
7 + ex
sinh−1 x4
7. lim
1
x→+∞
3x − 1
8. lim x (ln(−x))
x→0−
2
8 x
9. lim
1−
−
x→−∞
5x 5x2
1
4
10. lim 2
−
ln (2x − 5)
x→3+ x − 9
5. lim
16. lim [1 + tan(11x)]cot(2x)
x→0+
lim
x→+∞
18.
x3
x3 + 4
x3
lim (2x + 5)csch(2x+4)
x→−2+
of
M
19.
lim
x→+∞
2x+1
em
at
ic
17.
2x − 3
2x + 5
20. lim tanh
In
s
tit
ut
e
x→2−
UP
s
at
h
1.
√
9
16 − 8x
x−2
3.4. INDETERMINATE FORMS AND L’HÔPITAL’S RULE
189
Reviewer
I. A particle is moving along a horizontal line with its position function given by
s(t) = t3 − 6t2 + 9t − 4 = (t − 1)2 (t − 4),
where s is in meters and t ≥ 0 is in seconds. Assume that the positive direction is to the
right of the origin.
1. Determine the interval(s) when the particle is to the right of the origin.
2. Find the functions v(t) and a(t) of the particle at any time t ≥ 0.
3. Determine the time interval(s) when the particle is moving to the left.
4. Find all time interval(s) when the particle is speeding up.
II. Evaluate the following limits.
4 cot−1 (x + 3) − π
x→−2
x2 + 3x + 2
4
2
2. lim
−
cos2 x
x→ π2 + 2x − π
csc x
cos x 3. lim
−
x
x sin x
x→0+
1−x
4. lim (x − 1)
x→1+
x
x
5. lim
x→+∞ x + 2
ut
e
of
M
at
h
em
at
ic
s
1. lim
In
s
tit
III. Find the local linear approximation of f (x) = e−x cosh(2x) at x = 0 and use it to estimate
the value of e−0.01 cosh(0.02).
3x − 5
. Determine the x-values at which the absolute extrema of f occur on
e3x
[0, 3]. (Note e ≈ 2.718)
UP
IV. Given g(x) =
V. A new product, Happy Shalala, is now available. Preliminary market studies show that its
x
x2
ideal price-demand and cost functions are p(x) = 100 +
and C(x) = 3, 000 + 50x +
100
50
respectively, where x is the number of Happy Shalala sold and p is the price in pesos.
1. Find the marginal revenue for a sale of 150 Happy Shalala.
2. Find the profit function for the sale of x Happy Shalala.
3. When the profit is maximum, how many Happy Shalala are sold?
VI. Solve the following problems completely.
1. A hemispherical bowl with a radius of 10 cm is to be coated with red paint 0.05 cm
thick. Use differentials to estimate the amount of paint needed for the bowl.
2. The surface area of a closed cylinder is 54π square meters. Determine the base radius of
this cylinder if it is to have the largest volume. (A right circular cylinder with height h
and base radius r has volume πr2 h, top area πr2 , bottom area πr2 , and side area 2πrh.)
CHAPTER 3. APPLICATIONS OF DIFFERENTIATION
3. Find the volume of the largest box with a square base that can be constructed from 24
square inches of a flat, re-shapable sheet of metal.
4. Find the largest possible volume of a cylindrical can if it is to be made such that the
sum of its radius and height is 21 cm.
5. Vin’s car is traveling westward at a rate of 60 mi/hr while Al’s car is traveling due north
at a rate of 50 mi/hr. Both are headed for the intersection of the two roads. Determine
the rate at which the two cars are approaching each other when Vin is 0.4 mi and Al is
0.3 mi away from the intersection.
6. Vin is climbing up a 20-ft ladder leaning on a vertical wall. If the bottom of the ladder
is moving away from the wall at a rate of 3 ft/s, determine how fast the top of the ladder
is falling at the instant when the foot of the ladder is 3 ft away from the wall.
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
7. Having their friendship ended, Vin left Al by running southwards at a rate of 2 ft/s and
at the same time, Al decided to walk eastwards at a rate of 32 ft/s. Determine how fast
the distance between Vin and Al changing after four seconds.
UP
190
Chapter 4
Integration and Its Applications
4.1
Antidifferentiation and Indefinite Integrals
em
at
ic
s
In Chapters 1 and 2, we developed tools for differentiation, and saw various applications of the
derivative as a rate of change. Now we introduce antidifferentiation, the process which reverses
differentiation. Given a function f , can we find a function F whose derivative is f ? We will see
that this process is related to the computation of the area of a region in the plane.
at
h
At the end of this section, the student will be able to:
of
M
• recognize antidifferentiation as the inverse process of differentiation
ut
e
• integrate algebraic functions and functions involving the six circular functions
tit
• integrate functions using substitution rule
In
s
• find a particular antiderivative given an initial condition
UP
• solve problems involving rates of change given initial conditions (e.g. rectilinear
motion)
4.1.1
Antiderivatives or Indefinite Integrals
A function F is an antiderivative of the function f on an interval I if F 0 (x) = f (x) for every value
of x in I.
Example 4.1.1.
1. An antiderivative of f (x) = 12x2 + 2x is F (x) = 4x3 + x2 .
2. An antiderivative of g(x) = cos x is G(x) = sin x.
3. Another antiderivative of f (x) = 12x2 + 2x is F1 (x) = 4x3 + x2 − 1.
4. Another antiderivative of g(x) = cos x is G1 (x) = sin x + π.
191
192
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
Remark 4.1.2. If an antiderivative of f exists, then it is not unique.
Theorem 4.1.3. If F is an antiderivative of f on an interval I, then every antiderivative of
f on I is given by F (x) + C, where C is an arbitrary constant.
Remark 4.1.4. By the theorem above, we can conclude that if F1 and F2 are antiderivatives of f ,
then F2 (x) = F1 (x) + C. That is, F2 and F1 differ only by a constant.
Terms and Notations:
• Antidifferentiation is the process of finding antiderivatives.
Z
• If F is an antiderivative of f , we write f (x) dx = F (x) + C.
Z
s
, called the integral sign, denotes the operation of antidifferentiation.
em
at
ic
• The symbol
• The function f is called the integrand.
Z
cos x dx = sin x + C
e
ut
tit
UP
2.
In
s
Example 4.1.5.
Z
1.
12x2 + 2x dx = 4x3 + x2 + C
of
M
at
h
• The expression F (x)+C is called the general antiderivative of f . Meanwhile, each antiderivative of f is called a particular antiderivative of f .
Theorem 4.1.6 (Theorems on Antidifferentiation).
Z
1.
dx = x + C
2. If a is a constant, then
Z
Z
af (x) dx = a f (x) dx.
3. If f and g are defined on the same interval, then
Z
Z
Z
f (x) ± g(x) dx = f (x) dx ± g(x) dx.
4.1. ANTIDIFFERENTIATION AND INDEFINITE INTEGRALS
4. If n is any real number and n 6= −1, then
Z
xn+1
xn dx =
+ C.
n+1
Example 4.1.7. Find the following antiderivatives.
Z
x6
1.
x5 dx =
+C
6
Z
Z
2.
2 dx = 2 dx = 2(x + C1 ) = 2x + 2C1 = 2x + C
Z
3.
3
dx =
x5
Z
3x
−5
Z
dx = 3
x−5 dx =
3x−4
3
+C =− 4 +C
−4
4x
16u7/4
4u7/4
+C =
+C
7/4
7
Z
Z
Z
12x3
2x2
2
2
5.
12x + 2x dx = 12x dx + 2x dx =
+ C1 +
+ C2 = 4x3 + x2 + C
3
2
Z
Z
Z Z √
2t3 6t5/2
3
3
2t2 − 3t /2 dt = 2t2 dt − 3t /2 dt =
−
+C
6.
t 2t − 3 t dt =
3
5
Z 2
Z
x +1
x−1
1
−2
7.
dx
=
1
+
x
dx
=
x
+
+C =x− +C
2
x
−1
x
Z
4u /4 du =
3
tit
ut
e
of
M
at
h
em
at
ic
s
4.
Z
UP
In
s
Theorem 4.1.8 (Antiderivatives of Trigonometric Functions).
sin x dx = − cos x + C
1.
Z
4.
Z
2.
Z
3.
csc2 x dx = − cot x + C
Z
cos x dx = sin x + C
5.
sec2 x dx = tan x + C
6.
sec x tan x dx = sec x + C
Z
csc x cot x dx = − csc x + C
Example 4.1.9. Find the following antiderivatives.
Z
1.
(sin x + cos x) dx
= − cos x + sin x + C
Z
2.
cos x
dx
sin2 x
Z
Z
1
cos x
=
·
dx = csc x cot x dx = − csc x + C
sin x sin x
193
194
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
Z
tan2 x dx
Z
sec2 x − 1 dx = tan x − x + C
=
Z
dx
1 + sin x
Z
Z
1
1 − sin x
1 − sin x
=
·
dx =
dx
1 + sin x 1 − sin x
cos2 x
Z
=
sec2 x − sec x tan x dx = tan x − sec x + C
3.
4.
4.1.2
Particular Antiderivatives
em
at
ic
s
Now suppose that given a function f (x), we wish to find a particular antiderivative F (x) of f (x)
that satisfies a given condition. Such a condition is called an initial or boundary condition.
Example 4.1.10.
at
h
1. Given that F 0 (x) = 2x and F (2) = 6, find F (x).
Solution.
of
M
Since F 0 (x) = 2x, we have
Z
2x dx = x2 + C.
e
F (x) =
In
s
tit
ut
The initial condition F (2) = 6 implies that F (2) = 22 + C = 6. we get C = 2. Therefore,
the particular antiderivative that we wish to find is
UP
F (x) = x2 + 2.
√
2. The slope of the tangent line at any point (x, y) on a curve is given by 3 x. Find an equation
of the curve if the point (9, 4) is on the curve.
Solution.
Let y = F (x) be an equation of the curve. The slope of the tangent line mT L at a point
√
(x, y) on the graph of the curve is given by F 0 (x) = 3 x. We have
Z
1
3
F (x) = 3x /2 dx = 2x /2 + C.
The initial condition that (9, 4) is on the curve implies that F (9) = 2 · 93/2 + C = 4. We obtain
C = −50. Thus, an equation of the curve is
y = 2x /2 − 50.
3
4.1. ANTIDIFFERENTIATION AND INDEFINITE INTEGRALS
4.1.3
195
Integration by Substitution
Suppose F (x) is an antiderivative of f (x), and g is a differentiable function of x whose range is
contained in the domain of F . Recall the chain rule:
Dx [F (g(x))] = F 0 (g(x)) · g 0 (x).
Hence, F 0 (g(x))g 0 (x) is an antiderivative of g 0 (x). If we take u = g(x), then du = g 0 (x) dx.
The following theorem thus allows us to take the antiderivatives of more complex functions:
Theorem 4.1.11 (Substitution Rule). If u = g(x) is a differentiable function whose range
is an interval I and f is continuous on I, then
Z
Z
0
f (g(x)) · g (x) dx = f (u) du.
em
at
ic
s
Example 4.1.12.
Z
1
1.
(1 − 4x) /2 dx
1/2
(1 − 4x)
Z
dx =
1/2
(1 − 4x)
−4
dx =
·
−4
Z
of
M
Z
at
h
If we let u = 1 − 4x, then du = −4 dx. We multiply the integrand by
1/2
u
−4
. Thus,
−4
Z
du
1
1 2u3/2
1
−
=−
u /2 du = − ·
+ C.
4
4
4
3
UP
In
s
tit
ut
e
We put the final answer in terms of x by substituting u = 1 − 4x. Therefore,
Z
(1 − 4x)3/2
1/2
(1 − 4x) dx = −
+ C.
6
Z
2.
x2 (x3 − 1)10 dx
du
= x2 dx. By substitution,
3
11
Z
Z
Z
x3 − 1
1
u11
10
2 3
10
10 du
x (x − 1) dx = u ·
=
u du =
+C =
+ C.
3
3
33
33
Let u = x3 − 1. Then du = 3x2 dx, or
Z
3.
x
(x2 + 1)3
dx
du
Let u = x2 + 1. Then du = 2x dx, or
= x dx. By substitution,
2
Z
Z
x
1
1 u−2
1
−3
dx
=
u
du
=
·
+C =−
+ C.
3
2
2
2 −2
4(x + 1)2
(x2 + 1)
Z
4.
cos4 x sin x dx
Let u = cos x. Then du = − sin x dx, or −du = sin x dx. By substitution,
Z
Z
u5
cos5 x
cos4 x sin x dx = − u4 du = − + C = −
+ C.
5
5
196
Z
5.
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
x sec3 x2 tan x2 dx
du
Let u = sec x2 . Then du = sec x2 tan x2 · 2x dx, or
= sec x2 tan x2 · x dx. By
2
substitution,
Z
Z
3
2
2
x sec x tan x dx = sec2 x2 sec x2 tan x2 · x dx
Z
1 u3
1
u2 du = ·
=
+C
2
2 3
3
2
sec x
+ C.
=
6
6.
tan 1s + tan 1s sin 1s
ds
s2 cos 1s
of
M
at
h
em
at
ic
1
1
ds
Let u = . Then du = − 2 ds or −du = 2 . By substitution,
s
s
s
Z
Z
1
1
1
tan s + tan s sin s
tan u + tan u sin u
du
ds = −
1
2
cos u
s cos s
Z
=−
sec u tan u + tan2 u du
Z
=−
sec u tan u + sec2 u − 1 du
s
Z
ut
√
t t − 1 dt
tit
Z
7.
e
= − (sec u + tan u − u) + C
= − sec 1s − tan 1s + 1s + C.
UP
In
s
Let u = t − 1. Then u = dt. Also, t = u + 1. By substitution,
Z Z
Z
√
2u5/2 2u3/2
1
3
1
t t − 1 dt = (u + 1) u /2 du =
+
+C
u /2 + u /2 du =
5
3
5
3
2 (t − 1) /2 2 (t − 1) /2
=
+
+ C.
5
3
Z
t3
√
8.
dt
t2 + 3
du
Let u = t2 + 3. Then du = 2t dt, or
= t dt. Also, t2 = u − 3. By substitution,
2
Z
Z
Z
t3
t2 · t
du
−1
√
√
dt =
dt = u /2 (u − 3)
2
2
2
t +3
t +3
!
Z
1 1/2
1 2u3/2
−1/2
1/2
+C
=
u − 3u
du =
− 6u
2
2
3
3/2
1/2
t2 + 3
=
− 3 t2 + 3
+ C.
3
Z q
√
9.
4 + x dx
4.1. ANTIDIFFERENTIATION AND INDEFINITE INTEGRALS
Let u = 4 +
√
197
1
dx
x. Then du = √ dx or 2 du = √ . By substitution,
2 x
x
√
Z q
Z q
√
√
x
4 + x dx =
4 + x · √ dx
x
Z q
√ √
√
dx
=
4+ x· x· √
( x = u − 4)
x
Z
1
= u /2 · (u − 4) · 2 du
Z 3
1
=
2u /2 − 8u /2 du
2 · 2u5/2 2 · 8u3/2
−
+C
5
3
√ 5/2
√ 3/2
4 (4 + x)
16 (4 + x)
=
−
+ C.
5
3
4.1.4
em
at
ic
s
=
Rectilinear Motion Revisited
at
h
Suppose that a particle is traveling along a straight line and s (t), v (t) and a (t) are equations of
motion, velocity and acceleration, respectively, of the particle. Recall that
of
M
v(t) = s0 (t) and a(t) = v 0 (t).
ut
e
Therefore, s(t) is a particular antiderivative of v(t) while v(t) is a particular antiderivative of a(t).
In
s
tit
Example 4.1.13.
UP
1. A heavy projectile is fired straight up from a platform 3 meters above the ground with an
initial velocity of 160 m/s. Find an equation of motion of the particle. (Use −10 m/s2 for
acceleration due to gravity.)
Solution.
Let the acceleration and velocity of the particle at time t be given by a (t) and v (t).
Since a(t) = −10 at any time t and v (t) is a particular antiderivative of a (t), we have
Z
v(t) = −10 dt = −10t + C.
The initial velocity is 160 m/s, so v(0) = 160. Thus, 160 = −10 · 0 + C, or C = 160, and we
obtain
v(t) = −10t + 160.
Let s (t) be the position of the particle from the ground at time t. Now s (t) is a particular
antiderivative of v (t), so
Z
s (t) = (−10t + 160) dt = −5t2 + 160t + C.
198
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
The initial condition s(0) = 3 yields C = 3. Finally,
s(t) = −5t2 + 160t + 3.
2. The acceleration of a particle moving along a line at t seconds is given by a (t) = t2 + 2t.
Find an equation of motion of the particle if the particle is one unit to the right of the origin
when t = 0, and 3 units to the left of the origin when t = 2.
Solution.
Since v(t) is an antiderivative of a(t), we have
Z
t3
v(t) =
t2 + 2t dt =
+ t2 + C1 .
3
at
h
em
at
ic
s
Because there is no initial condition involving velocity, we antidifferentiate the previous expression to obtain a general expression for s(t) first:
Z 3
t
t4
t3
s(t) =
+ t2 + C1 dt =
+ + C1 t + C2 .
3
12
3
Antiderivatives of f (x) =
Recall: Du (ln |u|) =
UP
4.1.5
In
s
tit
ut
e
of
M
The initial condition s(0) = 1 yields C2 = 1. Meanwhile, the other initial condition s(2) = −3
4
8
implies that + + 2C1 + 1 = −3, or C1 = −4. Therefore, an equation of motion of the
3
3
particle is
t3
t4
+ − 4t + 1.
s(t) =
12
3
1
for all u 6= 0. As an immediate consequence, we have the following theorem:
u
Z
Theorem 4.1.14.
1
and of the other Circular Functions
x
1
du = ln |u| + C
u
3x2 + 5
Example 4.1.15.
1.
dx
4x
Z Z
Z
3x
5
3
5 1
3
5
=
+
dx =
x dx +
· dx = x2 + ln |x| + C
4
4x
4
4 x
8
4
Z
At this point, among the six circular functions, we know the antiderivatives of only the sine and
the cosine functions. We will see that the antiderivatives of the other circular functions involve the
natural logarithmic function.
4.1. ANTIDIFFERENTIATION AND INDEFINITE INTEGRALS
199
Theorem 4.1.16.
Z
1.
tan x dx = ln | sec x| + C
Z
cot x dx = ln | sin x| + C
2.
Z
sec x dx = ln | sec x + tan x| + C
3.
Z
4.
csc x dx = ln | csc x − cot x| + C
em
at
ic
s
Proof. We shall prove items 1 and 3 only. Statements 2 and 4 can be proved similarly.
Z
Z
sin x
dx. Let u = cos x. Then du = (− sin x) dx. Hence,
First, note that tan x dx =
cos x
Z
Z
1
tan x dx = − du = − ln |u| + C = − ln | cos x| + C = ln | sec x| + C.
u
Z
Z
sec x(sec x + tan x)
dx =
sec x + tan x
Z
sec2 x + sec x tan x
dx.
sec x + tan x
ut
e
sec x dx =
of
M
Now, for the secant function, note that
at
h
That proves the statement for the tangent function.
UP
In
s
tit
Let u = sec x + tan x. Then du = (sec x tan x + sec2 x) dx. Therefore,
Z
Z
du
= ln |u| + C = ln | sec x + tan x| + C.
sec x dx =
u
Example 4.1.17. Find the following antiderivatives.
Z
1.
x3 csc x4 dx
Solution.
Let u = x4 . Then du = 4x3 dx. Therefore,
Z
Z
1
1
1
3
4
csc u du = ln | csc u − cot u| + C = ln csc x3 − cot x3 + C.
x csc x dx =
4
4
4
Z
sin( x2 ) + 4
dx
2.
cos( x2 )
Solution.
Let u = x2 . Then du = 21 dx. Hence,
200
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
Z
sin( x2 ) + 4
dx = 2
cos( x2 )
Z
sin u + 4
du
cos u
Z
=2
(tan u + 4 sec u) du
= 2 (ln | sec u| + 4 ln |sec u + tan u|) + C
= 2 ln | sec( x2 )| + 4 ln sec( x2 ) + tan( x2 ) + C.
4.1.6
Antiderivatives of Exponential Functions
at
h
of
M
Example 4.1.19. Find the following antiderivatives.
Z
1.
7x dx
em
at
ic
Theorem 4.1.18 (Antiderivatives of Exponential Functions).
Z
ax
+ C (a > 0 and a 6= 1)
1.
ax dx =
ln a
Z
2.
ex dx = ex + C
s
The following theorem is a direct consequence of the derivatives of exponential functions.
Solution.
Solution.
UP
In
s
tit
ut
e
Direct application of the theorem above gives us,
Z
7x
7x dx =
+ C.
ln 7
Z
2.
22x · 5x dx
Applying laws of exponents to express the integrand into an equivalent exponential function in
a single base, we have
Z
Z
Z
20x
2x
x
x
x
+ C.
2 · 5 dx = 4 · 5 dx = 20x dx =
ln 20
4.1.7
Antiderivatives Yielding the Inverse Circular Functions
The next theorem presents the antiderivatives yielding the inverse circular functions.
Theorem 4.1.20 (Antiderivatives Yielding the Inverse Circular Functions).
Z
dx
√
1.
= sin−1 x + C
2
1−x
4.1. ANTIDIFFERENTIATION AND INDEFINITE INTEGRALS
Z
dx
= tan−1 x + C
1 + x2
Z
dx
= sec−1 x + C
x x2 − 1
2.
3.
201
√
One can generalize Theorem 4.1.20 to the case when the constant 1 in the integrand is replaced by
some other positive number, as shown in the next theorem.
em
at
ic
s
Theorem 4.1.21. Let a > 0.
Z
u
du
√
= sin−1
1.
+C
a
a2 − u2
Z
du
1
−1 u
2.
=
tan
+C
a2 + u2
a
a
Z
u
du
1
√
+C
3.
= sec−1
a
a
u u2 − a2
Proof. We will prove the first statement only and leave the proofs of the rest as exercises.
Z
du
=
√ q
2
a2 1 − ua2
of
M
du
√
=
2
a − u2
Z
du
q
2 , since a > 0.
a 1 − ua
e
Z
at
h
Let a > 0. Then,
ut
u
1
. Then dv = du and hence
a
a
Z
Z
u
du
dv
√
√
=
= sin−1 v + C = sin−1
+ C.
a
a2 − u2
1 − v2
UP
In
s
tit
Let v =
Example 4.1.22. Find the following antiderivatives.
Z
x4
1.
dx
x2 + 1
Solution.
Z Z
x4
x3
1
2
dx
=
x
−
1
+
dx
=
− x + tan−1 x + C.
x2 + 1
x2 + 1
3
Z
1
√
2.
dx
4 − x2
Solution.
Z
x
1
√
dx = sin−1
+C
2
4 − x2
202
Z
3.
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
dx
9x2 + 25
Solution.
Let u = 3x. Then du = 3dx and hence
Z
Z
1
du
1 1
1
dx
−1 u
−1 3x
+ C.
=
= · tan
+C =
tan
9x2 + 25
3
u2 + 25
3 5
5
15
5
Z
dx
p
4.
x ln x (ln x)2 − 8
Solution.
1
dx and hence
x
Z
Z
du
1
u
1
ln x
dx
−1
−1
p
√
√
√
=
= √ sec
+ C = √ sec
+ C.
2 2
2 2
2 2
2 2
u u2 − 8
x ln x (ln x)2 − 8
Z
5.
em
at
ic
s
Let u = ln x. Then du =
1
√ x
dx
4 −4
at
h
Solution.
R
UP
In
s
tit
ut
e
of
M
Note that 4x = (2x )2 for any x ∈ . Let u = 2x . Then du = 2x ln 2 dx and hence
Z
Z
Z
u
2x ln 2
1
1
1
1
√
√ x
q
dx =
du =
sec−1
+C
dx =
2−4
ln 2
2 ln 2
2
2
4 −4
u
u
x
x
2 ln 2 (2 ) − 4
x
1
1
−1 2
=
sec
+C =
sec−1 2x−1 + C.
2 ln 2
2
2 ln 2
Z
dx
√
6.
7 + 6x − x2
Solution.
First, note that
Z
Z
Z
Z
dx
dx
dx
dx
√
p
p
p
=
=
=
.
7 + 6x − x2
7 − (x2 − 6x)
7 − (x2 − 6x + 9 − 9)
16 − (x − 3)2
Let u = x − 3. Then du = dx and hence
Z
Z
dx
du
u
x−3
√
√
=
= sin−1 + C = sin−1
+ C.
4
4
7 + 6x − x2
16 − u2
Z
8x + 1
7.
dx
2
4x + 12x + 10
Solution.
Let u = 4x2 + 12x + 10. Then du = (8x + 12) dx and hence
4.1. ANTIDIFFERENTIATION AND INDEFINITE INTEGRALS
Z
203
Z
8x + 12
11
dx −
dx
2
2
4x + 12x + 10
4x + 12x + 10
Z
Z
dx
du
− 11
=
2
u
4x + 12x + 9 − 9 + 10
Z
dx
= ln |u| − 11
.
(2x + 3)2 + 1
8x + 1
dx =
2
4x + 12x + 10
Z
Antiderivatives of Hyperbolic Functions
em
at
ic
4.1.8
s
Let v = 2x + 3. Then dv = 2dx and hence
Z
Z
1
8x + 1
11
dx
=
ln
|u|
−
dv
2
2
4x + 12x + 10
2
v +1
11
= ln |u| −
tan−1 v + C
2
11
= ln |4x2 + 12x + 10| −
tan−1 (2x + 3) + C.
2
at
h
The succeeding theorem gives the antiderivatives of hyperbolic functions. Some of these antiderivative formulas are direct consequence of the derivatives of hyperbolic functions.
of
M
Theorem 4.1.23 (Antiderivatives of Hyperbolic Functions).
In
s
Z
sinh x dx = cosh x + C
Z
3.
UP
2.
ut
cosh x dx = sinh x + C
tit
1.
e
Z
sech2 x dx = tanh x + C
Z
csch x coth x du = − csch x + C
6.
Z
7.
tanh x dx = ln(cosh x) + C
Z
coth x dx = ln | sinh x| + C
8.
Z
Z
4.
9.
csch2 x dx = − coth x + C
= tan−1 (sinh x) + C
Z
5.
sech x dx = 2 tan−1 (ex ) + C
Z
sech x tanh x dx = − sech x + C
10.
csch x dx = ln |csch x − coth x| + C
Example 4.1.24.
Z
1.
2x cosh (2x ) dx
Solution.
Let u = 2x . Then du = 2x ln 2 dx and
Z
Z
1
1
1
cosh u du =
sinh (u) + C =
sinh (2x ) + C.
2x cosh (2x ) dx =
ln 2
ln 2
ln 2
204
Z
2.
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
sech(ln x) tanh(ln x)
dx
x
Solution.
1
Let u = ln x. Then du = dx and
x
Z
Z
sech(ln x) tanh(ln x)
dx = sech u tanh u du = − sech (u) + C = − sech (ln x) + C.
x
Z
csch x − tanh3 x
3.
dx
tanh x
Solution.
Z
Z
csch x − tanh3 x
dx = (csch x coth x − tanh2 x) dx
tanh x
Z
= (csch x coth x + sech2 x − 1) dx
4.1.9
em
at
ic
s
= − csch x + tanh x − x + C.
*Antiderivatives Yielding Inverse Hyperbolic Functions
at
h
We now give without proof the antiderivatives yielding the inverse hyperbolic functions.
UP
In
s
tit
ut
e
of
M
Theorem 4.1.25 (Antiderivatives Yielding the Inverse Hyperbolic Functions). Let
a > 0.
Z
u
p
1
√
du = sinh−1
1.
+ C = ln(u + u2 + a2 ) + C
a
u2 + a2
Z
u
p
1
√
2.
du = cosh−1
+ C = ln(u + u2 − a2 ) + C, u > a
a
u2 − a2

1

−1 u

tanh
+ C , if |u| < a

Z
 a
a
1
1
a+u
3.
du =
=
ln
+C
2
2
u

a −u
2a
a−u

1

−1
 coth
+ C , if |u| > a
a
a
It is recommended that the concise formula for the integral of
complicated piecewise function.
Z
2t4 + 3t2 + 4
Example 4.1.26. 1.
dt
t2 − 1
Solution.
Z
Z 2t4 + 3t2 + 4
9
2
dt =
2t + 5 + 2
dt .
t2 − 1
t −1
Z 9
2
=
2t + 5 −
dt
1 − t2
2t3
9
1+t
=
+ 5t − ln
+C
3
2
1−t
1
a2 − u2
be used rather than the
4.1. ANTIDIFFERENTIATION AND INDEFINITE INTEGRALS
Z
2.
√
1
dx
9x2 + 1
Solution.
Let u = 3x. Then du = 3dx and
Z
Z
1
1
1
1
1
√
√
=
du = sinh−1 u + C = sinh−1 (3x) + C
2
2
3
3
3
9x + 1
u +1
Z
x2
3.
dx
x6 − 25
Solution.
em
at
ic
s
Let u = x3 . Then du = 3x2 dx and
Z
Z
Z
1
1
1
1
x2
=
du
=
−
du
6
2
x − 25
3
u − 25
3
25 − u2
5+u
1
5 + x3
1 1
+ C = − ln
+C
ln
=− ·
3 10
5−u
30
5 − x3
Z
2x − 1
√
4.
dx
2
x + 4x − 5
at
h
Solution.
UP
In
s
tit
ut
e
of
M
Let u = x2 + 4x − 5. Then du = (2x + 4)dx and
Z
Z
2x − 1
2x + 4 − 4 − 1
√
√
dx =
dx
2 + 4x − 5
x2 + 4x − 5
x
Z
Z
5
2x + 4
√
dx − √
dx
=
2
2
x
+
4x
−
5
x
+
4x
−
5
Z
Z
1
5
√ du − √
=
dx
2
u
x + 4x + 4 − 4 − 5
Z
√
5 dx
.
=2 u− p
(x + 2)2 − 9
Let v = x + 2. Then dv = dx and
Z
Z
p
2x − 1
5
√
dx = 2 x2 + 4x − 5 − √
dv
x2 + 4x − 5
v2 − 9
v p
= 2 x2 + 4x − 5 − 5 cosh−1
+C
3
p
−1 x + 2
2
= 2 x + 4x − 5 − 5 cosh
+ C,
3
where v > 3, i.e. x > 1.
205
206
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
Summary of Antidifferentiation Rules
 n+1
u


+ C, n 6= −1

R n
n+1
1. u du =



ln |u| + C, n = −1
8.
R
sec2 u du = tan u + C
csc2 u du = − cot u + C
R
tan u du = ln | sec u| + C
11.
R
cot u du = ln | sin u| + C
Z
15.
20.
R
csch2 u du = − coth u + C
21.
R
sech u tanh u du = − sech u + C
22.
R
csch u coth u du = − csch u + C
23.
R
tanh u du = ln cosh u + C
24.
R
coth u du = ln | sinh u| + C
Z
sech u du = 2 tan−1 eu + C
26.
R
Z
sec u du = ln | sec u + tan u| + C
csc u du = ln | csc u − cot u| + C
√
1
a2 − u2
du = sin−1
27.
In
s
14.
sech2 u du = tanh u + C
25.
u
+ C, a > 0
a
UP
Z
R
sec u tan u du = sec u + C
10.
R
19.
cos u du = sin u + C
csc u cot u du = − csc u + C
13.
cosh u du = sinh u + C
sin u du = − cos u + C
R
R
R
eu du = eu + C
9.
12.
18.
s
7.
R
au
au du =
+ C, a > 0, a 6= 1
ln a
at
h
6.
R
sinh u du = cosh u + C
of
M
5.
R
R
e
4.
R
17.
1
1
u
du = tan−1 + C, a 6= 0
a2 + u2
a
a
Z
28.
Z
29.
= tan−1 (sinh u) + C
csch u du = ln | csch u − coth u| + C
√
ut
3.
R
1
1
u
√
du = sec−1 + C, a > 0
2
2
a
a
u u −a
16.
tit
2.
R
Z
em
at
ic
4.1.10
√
1
u2 + a2
1
u2 − a2
du = sinh−1
u
+ C, a > 0
a
du = cosh−1
u
+ C, a ∈ (0, u)
a
1
1
u+a
du =
ln
+ C,
a2 − u2
2a
u−a
u 6= a, a 6= 0
4.1. ANTIDIFFERENTIATION AND INDEFINITE INTEGRALS
4.1.11
207
Exercises
Exercises for Discussion
A. Find the following antiderivatives.
Z
Z 6
3/4
4
x − 5x − 4 dx
1.
x
Z
2. (4s + 3)(s − 5) ds
Z
4z 2 − 6z + 3
√
3.
dz
z
Z
4.
(5 cos x − 4 sin x) dx
Z
4 csc x cot x + 2 sec2 x dx
5.
2 cot2 θ − 3 tan2 θ dθ
6.
Z
1
dx
1 − cos x
Z
106x + 1
8.
dx
103x
Z
3 − 2x
dx
9.
3x
Z x
3 −7
10.
dx
e2x
em
at
ic
s
7.
B. Find the following antiderivatives using substitution.
Z
coth2 x csch2 x dx
Z
sec2 x
dx
1 + 16 tan2 x
Z
1
dt
(t + 3) t2 + 6t − 7
tit
ut
e
of
M
at
h
7.
UP
In
s
Z p
3
1.
16n2 − 8n + 1 dn
Z
p
4
2.
z 8 z 3 + 3 dz
√
Z
sec2 (3 t)
√
3.
dt
t 1
Z 4 cos
x2
dx
4.
1
3
3
x sin
x2
Z
2 ln x + 1
dx
5.
x(ln2 x + ln x)
Z
2 + ln2 x
6.
dx
x − x ln x
8.
9.
Z
10.
Z
11.
Z
12.
√
√
√
√
x
dx
8x − x2
ex
e2x − 4
dx
x
x4 + 4x2 + 5
dx
C. Solve the following problems completely.
1. The point (3, 2) is on a curve, and at any point (x, y) on the curve, the tangent line has
slope equal to 2x − 3. Find an equation of the curve.
d2 y
2. The points (1, 3) and (0, 2) are on a curve, and at any point (x, y) on the curve, 2 = 2−4x.
dx
Find an equation of the curve.
D. Solve the following problems completely. (When appropriate, assume acceleration due to
gravity is −10 m/s2 or −32 ft/s2 .)
1. A ball is thrown vertically upwards from a platform 8 feet above the ground. After 2
seconds, the ball is 2 feet below the platform. Find the initial velocity of the ball.
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
2. If the brakes on a car can give the car a constant negative acceleration of 8 m/s2 , what
is the maximum speed it can go to be able to stop within 25 meters after the brake is
applied?
UP
208
4.1. ANTIDIFFERENTIATION AND INDEFINITE INTEGRALS
Supplementary Exercises
A. Verify that F is an antiderivative for f .
x2 + 1
x
;
f
(x)
=
4(4x2 + 5x − 4)
(4x2 + 5x − 4)2
2. F (x) = x ln x − x; f (x) = ln x
1. F (x) = −
3. F (x) = x sin x + cos x; f (x) = x cos x
2
2x + 1
3x4 + 2x3 − 2x + 1
3
2
−1
√
√
4. F (x) = ln |x − x + 1| +
tan
; f (x) =
x5 + x + 1
3
3
B. Find the following antiderivatives.
2
2
x (x − 1) dx
Z
1
√ √
3
2.
dy
y
y+ √
4 y
Z √
1 2
3
3.
2 t+ √
dt
3
t
Z
4.
3 csc2 t − 5 sec t tan t dt
Z
5. (tan x + cot x)2 dx
2 cosh t sech t + csch2 t dt
Z
cosh x
p
dx
8 − cosh2 x
6.
7.
em
at
ic
Z
sec θ
dθ
sec θ − tan θ
Z
1 − t2
dt
1 + t2
at
h
8.
of
M
1.
Z
s
Z
9.
tit
ut
e
C. Find the following antiderivatives using substitution.
(2y − 1)6 dy
Z
2x2 (4 + 3x3 )4 dx
UP
2.
In
s
Z
1.
s3
ds
3s2 + 1
Z
t
4.
dt
4
t + 2t2 + 1
Z
5.
4x[3x + cos(2x2 + 1)] dx
Z
cos x
6.
dx
(1 + sin x)5
Z
7.
csc3 (2x + 1) cot (2x + 1) dx
Z
8.
sin θ cos θ cos (cos 2θ) dθ
Z
9.
sec2 θ − cos2 θ tan θ dθ
Z
10.
sin y − tan2 y sec2 y dy
Z
3.
√
tan2 x 1 + sin2 x
dx
11.
cos2 x
√
√
Z
cot 3 x csc2 3 x
√
12.
√ 2 dx
3
x2 (4 + cot2 3 x)
Z q
√
4 − 4 − x dx
13.
Z
csch cot−1 x
14.
dx
1 + x2
Z
15.
sech x 2 + coth2 x dx
Z
x
16.
5x ecsch(5 ) csch(5x ) coth(5x ) dx
Z
17.
sinh x tanh x dx
Z
Z
tanh2 (ey )
dy
cosh y − sinh y
Z
tan(ln x) sec(ln x)
dx
πx − x sec(ln x)
18.
19.
209
210
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
Z
Z
dx
√
20.
4x2 + 8x + 13
Z
dx
√
21.
16 − e−2x
Z
2x + 10
22.
dx
x2 + 2x − 3
Z
1
√
23.
dx
e2x − 4
Z
1
√
dx
24.
25 + 16x2
Z
x
√
25.
dx
4
x −1
26.
dx
52−x − 5x
Z
27.
cosh−1 (log5 x)
q
dx
x (log5 x)2 − 1
Z
sin 2x + 2 sin x
dx
sin2 x + 1
Z x
3 +1
29.
dx
9x + 1
Z
1
√
30.
dx
2x
e +4
Z
sech r
p
31.
dr
15 coth2 r + 1
28.
D. Solve the following problems completely.
at
h
em
at
ic
s
1. The point (3, 2) is on a curve, and at any point (x, y) on the curve, the tangent line has
slope equal to 2x − 3. Find an equation of the curve.
d2 y
2. The points (1, 3) and (0, 2) are on a curve, and at any point (x, y) on the curve, 2 = 2−4x.
dx
Find an equation of the curve.
e
of
M
3. An equation of the tangent to a curve at the point (1, 3) is y = x + 2. If at any point (x, y)
d2 y
= 6x, find an equation of the curve.
on the curve,
dx2
In
s
tit
ut
E. Solve the following problems completely. (When appropriate, assume acceleration due to
gravity is −10 m/s2 or −32 ft/s2 .)
UP
1. A sandbag is released from a balloon rising vertically with a velocity of 16 ft/s at the
instant when the balloon is 64 feet from the ground. How many seconds after its release
will the bag hit the ground?
2. If a particle starts from rest, what constant acceleration is required to move the particle
50 meters in 5 seconds along a straight line?
3. Two stones are thrown vertically upward at the same time, one with an initial velocity of
32 ft/s from a height of 128 feet, the other with an initial velocity of 64 ft/s from a height
of 80 ft. If the stones travel along parallel paths, when will the second stone overtake the
first?
4.2. THE DEFINITE INTEGRAL
4.2
211
The Definite Integral
We have probably known of some formulas regarding the area of specific shapes, or polygons for a
better term.
√ For instance, the area of an equilateral triangle of side length s is give by the function
2
s 3
A(s) =
. A square of side length s has area A(s) = s2 , and so on. In general, the area of any
4
polygon is the sum of the areas of the triangles into which it can be decomposed. It can be seen
that this computation is independent of the way in whiich we divide the polygon into the needed
triangles.
Now, we deconstruct this notion of area for us to be able to find the area of a region bounded by a
curve and some vertical lines.
At the end of this section, the student will be able to:
• generalize the notion of area of a polygonal region to the area of any plane region;
em
at
ic
s
• derive an expression that gives such a formula, given any function f (x);
UP
In
s
tit
ut
e
of
M
at
h
• use the properties of summation to obtain an exact number that gives the desired area.
4.2.1
Area of a Plane Region: The Rectangle Method
We first introduce the sigma notation which will be often used when approximating areas of a plane
region by a sum of areas of rectangles.
Definition 4.2.1. (Sigma Notation) Let n be a positive integer, and F be a function such that
{1, 2, . . . , n} is in the domain of F . We define:
n
X
F (i) := F (1) + F (2) + . . . + F (n)
i=1
We read the left hand side as “the summation of F (i), with i from 1 to n.
212
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
Remark 4.2.2.
1. The variable i is called a dummy variable. It can be changed arbitrarily, especially when two
or more summations are involved in a single calculation.
Z
2. In general, the summation need not start from 1. As long as we have k ∈ , , where k < n, we
can define the summation of F (i), with i from k to n:
n
X
F (i) := F (k) + F (k + 1) + . . . + F (n)
i=k
However, for convenience of calculations, we start from i = 1.
n
X
c = cn
em
at
ic
1.
s
Theorem 4.2.3. (Properties of Summation) Let n be a positive integer, c be a real number,
and F and G be functions defined on the set {1, 2, . . . , n}.
i=1
i=1
n
X
n
n
X
X
F (i) ± G(i) =
F (i) ±
G(i)
i=1
i=1
i=1
5.
n
X
e
i=
ut
n
X
i=1
n(n + 1)
2
i2 =
i=1
tit
4.
F (i)
i=1
n(n + 1)(2n + 1)
6
In
s
3.
n
X
at
h
cF (i) = c
of
M
n
X
UP
2.
Example 4.2.4.
1.
30
X
(6i − 1) =
i=1
2.
10
X
i=1
3.
n
X
i=1
30
X
i=1
6i +
30
X
1=6
i=1
30
X
i+
i=1
30
X
1=6·
i=1
30(31)
+ 1 · 20 = 2750
2
10
10
10
10
X
X
X
X
10(11)(21)
10(11)
2
2
(i − 1) =
(i − 2i + 1) =
i −2
i+
1=
−2·
+ 10 = 285
6
2
2
i=1
(i − 1)2 =
n
X
i=1
i=1
i2 − 2
n
X
i=1
i+
n
X
i=1
1=
i=1
i=1
n(n + 1)(2n + 1) n(n + 1)
n(2n2 − 3n + 1)
+
+n=
6
2
6
The Area of a Plane Region
Suppose we wish to find the area of the region R in the first quadrant bounded by the curve
f (x) = x2 , the x-axis and the vertical line x = 2.
4.2. THE DEFINITE INTEGRAL
213
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
We calculate the area AR of the region R in the following manner:
UP
1. First, we divide the interval [0, 2] into n subintervals of equal length ∆x. Note that
∆x =
2
2−0
= .
n
n
2. Let ci be the right endpoint of the ith subinterval, i = 1, 2, .., n.
3. Next, we cover the region with n circumscribed rectangles of width ∆x and height f (ci ).
4. Let Ai be the area of the ith rectangle. We get
Ai = f (ci ) · ∆x.
5. The area of the region can be approximated by taking the sum of the areas of the n circumscribed rectangles. Thus,
n
X
AR ≈
Ai
i=1
or
AR ≈
n
X
i=1
f (ci ) · ∆x
214
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
But ∆x =
2
and f (ci ) = c2i . So,
n
AR ≈
n
X
2
n
c2i ·
i=1
Moreover ci = i · ∆x =
2i
. So,
n
AR ≈
n 2
X
2
2i
·
n
n
i=1
6. We simplify the right hand side using summation formulas and obtain
AR ≈
n
X
n
X
8i2
Ai =
i=1
n3
i=1
i=1
em
at
ic
=
s
n
8X2
i
n3
8 n(n + 1)(2n + 1)
·
n3
6
4(n + 1)(2n + 1)
3n2
at
h
=
of
M
=
tit
ut
e
7. Note that if we increase the number of rectangles, we get a better approximation of the area.
In fact, we define the area of the region to be
We obtain
UP
In
s
AR := lim
AR := lim
n→∞
n
X
n
X
n→∞
Ai = lim
n→∞
i=1
Ai .
i=1
8
4(n + 1)(2n + 1)
= .
3n2
3
In the previous illustration, we calculated the area AR of a plane region R bounded above by a
continuous function y = f (x) and below by the interval [a, b] by covering R with n circumscribed
rectangles of equal width ∆x. We then let n → ∞ to calculate the area
AR := lim
n
X
n→∞
f (ci ) · ∆x
i=1
where ci is the right endpoint of ith subinterval and ∆x is the width of the ith rectangle, i =
1, 2, ..., n.
The following alternative method can be used to calculate AR .
4.2. THE DEFINITE INTEGRAL
215
The Definite Integral
of
M
4.2.2
i=1
at
h
n→∞
em
at
ic
s
Instead of taking the right endpoint ci of the ith subinterval, we can take any number x∗i in the
ith subinterval and let the height of the ith rectangle be h = f (x∗i ). The area of the region is now
given by
n
X
AR = lim
f (x∗i ) · ∆x.
UP
In
s
tit
ut
e
To find the area of some plane regions R bounded above by y = f (x) and below by the interval
[a, b], it is sometimes more convenient to use rectangles of different widths.
Given an interval [a, b], we define a partition of the interval to be a sequence of numbers
a = x0 < x1 < x2 < ... < xn−1 < xn = b
the divides interval into n subintervals where the length of the ith interval is given by ∆xi =
xi − xi−1 , where i = 1, ..., n. When the subintervals are of equal length we say that the partition is
a regular partition.
216
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
For a regular partition, the length of each subinterval tends to 0 as n → ∞. However, if the
partition is not regular, this is not always true. Instead, we define max∆xi to be the length of the
longest subinterval in the partition. This length is called the mesh size of the partition. The area
of the region is now given by
n
X
AR =
lim
f (x∗i ) · ∆xi .
max∆xi →0
i=1
Definition 4.2.5. Let f be defined on [a, b]. The definite integral of f from a to b is
Z b
n
X
f (x) dx =
lim
f (x∗i ) · ∆xi
max∆xi →0
a
i=1
if the limit exists and does not depend on the choice of partitions or on the choice of numbers x∗i
in the subintervals. If the limit exists, the function is said to be integrable on [a, b].
em
at
ic
• The process of calculating the integral is called integration.
Z b
• In the notation for the definite integral f (x) dx:
s
Terminologies and Notations:
at
h
a
of
M
Z
– The integral sign resembles the letter S because an integral is the limit of a sum.
– The function f (x) is called the integrand.
tit
ut
e
– The numbers a and b are called the limits of integration: a is the lower limit of integration, while b is the upper limit of integration.
• The sum
n
P
UP
In
s
– We use the same symbol as the antiderivative because the definite integral is closely
linked to the antiderivative, as shall be seen in the following section.
f (x∗i )·∆xk is called a Riemann Sum, after the mathematician Bernhard Riemann
i=1
who formulated much of the concepts in integral calculus.
Remark 4.2.6. We note the following:
1. The definite integral is a number which does not depend on the variable used. The value of the
definite integral does not change if x is replaced by any other variable. For example,
Z b
Z b
f (x) dx =
f (t) dt.
a
a
Z b
2. Geometrically, the definite integral
f (x) dx gives the net-signed area between the graph of
a
the curve y = f (x) and the interval [a, b]. In particular, if the graph of y = f (x) lies above
Z b
the x-axis in the interval [a, b],
f (x) dx gives the area of the region bounded by the curve
a
y = f (x), the x-axis and the lines x = a and x = b.
4.2. THE DEFINITE INTEGRAL
217
3. If the function f is continuous, a regular partition can be used to compute the definite integral:
Z b
f (x) dx = lim
n→∞
a
n
X
f (x∗i ) · ∆x
i=1
4. The function f must be defined on the interval [a, b] to define the definite integral of f from [a, b].
Z 1
1
1
For instance
dx is not a definite integral, since the 0 is not in the domain of f (x) = 2 .
2
x
0 x
This type of integral is called an improper integral, which is not in the scope of this course.
Example 4.2.7. Evaluate the following definite integrals:
Z 2
x2 dx
1.
0
s
Solution.
at
h
em
at
ic
We have seen that the area of the region bounded by the curve f (x) = x2 , the x-axis and the
Z 2
8
8
x2 dx = .
vertical line x = 0 and x = 2 is square units. Therefore,
3
3
0
Z 6
2.
2x dx
of
M
−2
e
Solution.
UP
In
s
tit
ut
The value of the given definite integral is the net-signed area between the curve y = 2x and the
x-axis over the interval [−2, 6].
Z 6
We have that
2x dx =
−2
1
1
· 6 · 12 − · 2 · 4 = 32.
2
2
Theorem 4.2.8. If a function is continuous on [a, b], then it is integrable on [a, b].
Example 4.2.9.
218
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
1. Polynomial functions are integrable on any closed interval.
2. The function f (x) = |x| is integrable on the interval [−1, 2], although it is not differentiable on
(−1, 2).
3. The sine and cosine functions are integrable on any closed interval.
Theorem 4.2.10 (Properties of the Definite Integral). Let f and g be integrable on [a, b].
Z a
Z b
1.
f (x) dx = − f (x) dx
b
a
Z a
f (x) dx = 0
2.
a
Z b
c dx = c(b − a)
3.
a
Z b
[f (x) ± g(x)] dx =
5.
a
em
at
ic
a
Z b
Z b
f (x) dx ±
a
g(x) dx
a
at
h
4.
Z b
c f (x)dx = c f (x) dx
of
M
Z b
s
a
6. If f is integrable on a closed interval I containing the three numbers a, b and c,
Z c
f (x) dx =
a
tit
ut
a
UP
In
s
regardless of the order of a, b and c.
Z b
f (x) dx +
e
Z b
f (x) dx
c
4.2. THE DEFINITE INTEGRAL
4.2.3
219
Exercises
Exercises for Discussion
A. Let R be the region bounded by y = 4 − x2 and the x-axis in the interval [0, 2].
1. Use circumscribed rectangles of equal width to find the area of the region.
2. Use inscribed rectangles of equal width to find the area of the region.
B. Find the value of the following definite integrals.
Z 8
1.
10 − 3x dx
−2
Z 1
2.
x2 cos x dx
1
−4
Z 2
16 − x2 dx
3−
√
s
4.
√
4 − t2 dt
em
at
ic
Z 4
3.
0
−1
−1
of
M
−1
at
h
C. Do as indicated.
Z 6
Z 6
Z 6
1. If
f (x) dx = 3 and
g(x) dx = 8, find
[3f (x) − 2g(x)] dx.
Z ln 2
15
+ ln 2, find
16 sinh2 (x) + 16 dx.
16
ln 0.5
ln 0.5
Z 5
Z 5
Z 5
3. Find the value of
[f (x) − 5]2 dx, given
f (x) dx = −17 and
[f (x)]2 dx = 32.
Z ln 2
cosh2 (x) dx =
ut
e
2. Given that
tit
−1
Z 5
In
s
Z 2
−1
−1
Z 5
f (x) dx = 21, and
−5
−5
2
Z 2
5. If
Z 1
g(x) dx = 4 and
−1
Z 1
3g(x) dx = 6, evaluate
−1
f (x) dx = 7, find
−2
UP
4. If
f (x) dx =
f (x) dx and
−5
−2
Z −2
Z 5
f (x) dx +
f (x) dx.
Z 2
g(x) dx.
2
220
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
Supplementary Exercises
A. Using the theorems on summation, prove the following:
n
X
i3 =
i=1
n2 (n + 1)2
4
B. Let R be the region bounded by y = x3 and the x-axis in the interval [0, 2].
1. Use circumscribed rectangles of equal width to find the area of the region.
2. Use inscribed rectangles of equal width to find the area of the region.
C. Find the value of the following definite integrals.
Z 7
Z 12
4 dx
−3
Z 6
2
Z 3
5.
15 + 4x dx
−4
3
cosh−1 x dx
Z 6 Z −10
Z 5
6.
|2x − 5| dx
3.
s
2.
−|12 − 3x| dx
4.
em
at
ic
1.
4
−10
at
h
0
of
M
D. Do as indicated.
csc−1 w
dw
csc w
√
dx
16 − x2 , on the following
ut
e
1. Find the area enclosed by the x-axis and the semicircle y =
intervals: (a) [0, 4], (b) [2, 4], (c) [0, 2], (d) [−2, 4].
−3
UP
In
s
tit
2. Find the area of the region enclosed by the graph of f (x) = 2 − |x − 5| , the two
coordinate axes, and the line x = 10.
Z 6
Z 6
Z 6
3. If
f (x) dx = 12,
g(x) dx = −2 and
h(x) dx = −7, find the value of
−3
−3
Z −3 6
f (x)
3h(x)
+ 3g(x) −
dx
4
7
4. Let k be any real number, and a, b be real numbers such that a 6= b. Given that
Z b
Z b
Z b
2
2
g(x) dx = k and
[g(x)] dx = k + 2k − 1, find the value of
[3g(x) − 2]2 dx.
a
a
a
4.3. THE FUNDAMENTAL THEOREM OF THE CALCULUS
4.3
221
The Fundamental Theorem of the Calculus
Now that we have the necessary background about antiderivatives and definite integrals, we are
ready to state the two statements that link differential and integral calculus.
At the end of this section, the student will be able to do the following:
• state the First and Second Fundamental Theorems of the Calculus,
• provide a proof using the concepts we had discussed so far, and
• apply these theorems to solve complicated integrals more conveniently.
s
First Fundamental Theorem of the Calculus
em
at
ic
4.3.1
a
Z 2
8
x dx =
3
0
Z 2
8
y dy =
3
0
2
Z 2
8
r dr =
3
0
2
Z 2
0
u2 du =
8
3
tit
ut
e
2
of
M
choice of the variable in the function. That is,
at
h
Let f be a continuous function on the closed interval [a, b]. Then the value of the definite integral
Z b
f (x) dx depends only on the function f and the values of a and b, and is independent of our
Z x
In
s
We consider a function defined by F (x) =
UP
x is any number in [a, b].
f (t) dt, where f is a continuous function on [a, b] and
a
Example 4.3.1.
Z x
1. Let F (x) =
t2 dt. Evaluate F at x = 0 and x = 2.
0
Z 0
a. F (0) =
t2 dt = 0
0
Z 2
b. F (2) =
t2 dt =
0
8
3
Z x
2. Let F (x) =
2t dt.
0
Notice that F (x) gives the areaZ of the triangle under the curve y = 2t in the interval [0, x].
x
Therefore, we get that F (x) =
0
F 0 (x) = f (x). Refer to the figure.
2t dt = 12 · x · 2x = x2 . In this case, F 0 (x) = 2x. Note that
222
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
As was illustrated inZthe previous example, the next theorem states that in general, for a function
x
of the form F (x) =
s
em
at
ic
[a, b], F 0 (x) = f (x).
f (t) dt, where f is a continuous function on [a, b] and x is any number in
a
at
h
Theorem 4.3.2 (The First Fundamental Theorem of Calculus). Let f be a function
continuous
on [a, b] and let x be any number in [a, b]. If F is the function defined by F (x) =
Z
of
M
x
f (t) dt, then
a
tit
ut
e
F 0 (x) = f (x)
0
Solution.
UP
In
s
Example 4.3.3. Find the derivative of the following functions.
Z x
1. F (x) =
t2 dt
Let f (t) = t2 . Applying the First Fundamental Theorem of Calculus, we obtain F 0 (x) = x2 .
Z x
2. F (x) =
cos t2 + 1 dt
−π
Solution.
In this case, let f (t) = cos t2 + 1 . Thus, F 0 (x) = cos(x2 + 1).
Z 3
3. F (x) =
sin 2t dt
x
Solution.
To apply the First Fundamental Theorem of Calculus, we first need to switch the upper and
lower
Z limits of integration. Using the first property of the definite integral, we have F (x) =
x
sin 2t dt. Now let f (t) = sin 2t. Finally, we get that F 0 (x) = − sin 2x.
−
3
4.3. THE FUNDAMENTAL THEOREM OF THE CALCULUS
223
Z g(x)
f (t) dt, where f is a function continuous on [a, b] and let
Remark 4.3.4. Suppose F (x) =
Z xa
f (t) dt, then F (x) = H(g(x)). Using the chain rule, we get
g (x) ∈ [a, b]. If we let H(x) =
a
F 0 (x) = H 0 (g(x)) · g 0 (x). By the First Fundamental Theorem of Calculus, H 0 (x) = f (x). So we
have
F 0 (x) = f (g(x)) · g 0 (x)
Example 4.3.5. Find the derivative of the following functions.
Z x2
1. F (x) =
t2 dt
−1
2 2
−x−1
Z −x−1
=−
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
F 0 (x) = (x ) · 2x = 2x5
Z cos x
dt
2. F (x) =
t
1
1
· (− sin x) = − tan x
F 0 (x) =
cos x
Z 0
t
3. F (x) =
dt
2+1
t
2x+1
Z 2x+1
t
F (x) = −
dt
2+1
t
0
2x + 1
2x + 1
2x + 1
F 0 (x) = −
·2=− 2
·2=− 2
(2x + 1)2 + 1
4x + 4x + 2
2x + 2x + 1
Z x
4. F (x) =
sin2 t dt
−x−1
Z 0
Z x
2
F (x) =
sin t dt +
sin2 t dt
2
0
Z x
sin t dt +
0
sin2 t dt
0
F 0 (x) = − sin2 (−x − 1) (−1) + sin2 x = − sin2 (x + 1) + sin2 x
Z tanh(x)
2 t−1
5. F (x) =
sinh
dt
3
x3
Z 1
Z tanh(x)
2 t−1
2 t−1
=
sinh
dt +
sinh
dt
3
3
x3
1
Z x3
Z tanh(x)
2 t−1
2 t−1
=−
sinh
dt +
sinh
dt
3
3
1
1
3
0
2 x −1
2
2 tanh(x) − 1
F (x) = sinh
· 3x + sinh
· sech2 (x)
3
3
We reserve the proof of the First Fundamental Theorem of the Calculus in Section 4.5, since it
involves the Mean Value Theorem for Integrals.
224
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
4.3.2
The Second Fundamental Theorem of Calculus
The next result provides a link between computing antiderivatives of a function f and finding the
definite integral of f on the interval [a, b].
Theorem 4.3.6 (The Second Fundamental Theorem of Calculus). Let f be a
function continuous on [a, b]. If F is any antiderivative of f on [a, b], then
Z b
f (x) dx = F (x)
ix=b
a
x=a
= F (b) − F (a)
em
at
ic
−2
s
Example 4.3.7. Evaluate the following definite integrals.
3
x=2 Z 2
x
8
8
2
1.
x dx =
+C
+ C − (0 + C) =
=
3
3
3
0
x=0
Z 6
x=6
= 36 − 4 = 32
2x dx = x2
2.
x=−2
Z 3
at
h
dx
3
−1 (x + 2)
of
M
3.
Solution.
UP
In
s
tit
ut
e
We use integration by substitution to get the desired antiderivative. Let u = x + 2. Then
du = dx. We get that
Z
Z
dx
1
1
u−2
−3
+C =− 2 +C =−
=
u
du
=
+C
3
−2
2u
(x + 2)
2 (x + 2)2
We now have
Z 3
dx
3 =
−1 (x + 2)
1
−
2(x + 2)2
x=3
=−
x=−1
1
1
24
12
+ =
=
50 2
50
25
Remark 4.3.8. By the Second Fundamental Theorem of Calculus and the Substitution Rule,
Z b
x=b
f (g(x))g 0 (x) dx = F (g(x))
= F (g(b)) − F (g(a))
a
x=a
If we let u = g(x), we have that
u=g(b)
Z b
0
f (g(x))g (x) dx = F (u)
a
u=g(a)
Therefore,
Z b
0
Z g(b)
f (g(x))g (x) dx =
a
f (u) du
g(a)
4.3. THE FUNDAMENTAL THEOREM OF THE CALCULUS
225
Example 4.3.9. Evaluate the following definite integrals using the previous remark.
Z 3
dx
1.
3
−1 (x + 2)
Solution.
Again, we let u = x + 2. Then du = dx. We use the previous remark and put the limits of
integration in terms of the new variable u. If x = −1, then u = 1. If x = 3, then u = 5. We
now have
Z 3
Z 5
dx
u−2 u=5
1
12
1
−3
=
=− + =
u
du
=
3
−2 u=1
50 2
25
−1 (x + 2)
1
Z √11
2.
1
2x(x2 − 3) 3 dx
2
s
Solution.
2x(x
2
1
+ −3) 3 dx =
Z 8
1
4
√
sin t
√ dt
t
e
π2
4
4
u=1
45
4
ut
3.
4
=
√
11, then
of
M
2
Z π2
4
4
4
1
3u 3 u=8 3(8) 3
3(1) 3
u 3 du =
=
−
at
h
Z √11
em
at
ic
Let u = x2 − 3. Then du = 2x dx. Moreover, if x = 2, then u = 1. If x =
√ 2
u=
11 − 3 = 8. Thus,
UP
In
s
tit
Solution.
√
dt
dt
π
2
Let u = t. Then du = √ , or 2du = √ . In addition, if x = π4 , then u = . Meanwhile, if
2
2 t
t
2
x = π , then u = π. Thus,
√
Z π2
Z π
u=π
sin t
√
sin
u
du
=
2(−
cos
u)
=2−0=2
dt
=
2
π
π
π2
t
u=
4
2
2
Z 2
|x| dx
4.
−1
Solution.
(
Recall that |x| =
x,
−x,
if x ≥ 0
. We use the 6th property of the definite integral and obtain
if x < 0
Z 2
Z 0
|x| dx =
−1
Z 2
(−x) dx +
−1
x2 x=0
x dx
0
x2 x=2
=−
+
2 x=−1
2
x=0
1
5
= 0− −
+ (2 − 0) =
2
2
226
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
Remark 4.3.10.
1. The Fundamental Theorems of CalculusZ establish a close connection between antiderivatives
and definite integrals. For this reason,
f (x) dx is also referred to as an indefinite integral,
and the process of antidifferentiation as integration. However, note that in Advanced Calculus,
the indefinite integral is defined independently of the antiderivative and does not always
coincide with the antiderivative of a function.
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
2. To use the Second Fundamental
Theorem of Calculus, the function f must be continuous on
Z 2
1
1 ix=2
1
1
[a, b]. For instance,
dx
=
6
−
= . In fact, the definite integral of f (x) = 2 on
2
x x=−1 2
x
−1 x
[−1, 2] cannot even be defined since the interval is not contained in the domain of f .
4.3. THE FUNDAMENTAL THEOREM OF THE CALCULUS
4.3.3
227
Exercises
Exercises for Discussion
A. Find the derivative of the following functions.
Z 2
Z 7
t3
1. F (x) =
dt
3
1−3x 1 + t
Z 1
x√
2. F (x) =
1 + m3 dm
3. F (x) =
t2 + 1 dt
csc x
Z cos x
4. F (x) =
y 2 dy
sin x
1
B. Find an equation of the tangent line to the graph of the following functions at the given value
of x:
√
3
Z x
Z x2 −5x+6
t
1. h(x) =
dt ; x = −8
2
−8 t + 8t + 1
Z 3x2 −10
z sinh(z)
dz ; x = −2
2. g(x) =
cos2 z − 4
2
Z π
sin x dx
e
1.
0
Z 64
3.
1
Z π
√
1+ 3y
dy
√
3 y
In
s
(1 + 3t − t2 ) dt
UP
2.
tit
0
Z 4
1 + cos2 θ
dθ
cos2 θ
0
Z π
x
x
5.
sin2
cos
dx
π
2
2
3
Z π
tan u
6.
du
2π sec u − tan u
3
Z π
2
7.
cos x cos(π sin x) dx
4
4.
s
em
at
ic
4. g(x) =
Z π
1
2
9.
sin 2y
2we−w
dw ; x = 0
tan−1 w
p
1 − sin ydy
0
Z ln 3
10.
tanh x sech3 x dx
ln 2
Z 4
1
√
3 dw
w ( w + 2)
1
Z 2
1 2
1
12.
1+
dm
2
m
1 m
Z √3−1
4
√
dr
13.
3 − 2r − r2
−1
Z 4
1
√
14.
dt
2
1 (t + 2) t + 4t − 5
Z 2
15.
|x + 1| dx
11.
√
−2
0
Z 2 p
8.
y y − 1 dy
3r3 + sech r
dr ; x = 3
3r3 − sech r
at
h
2x+1
ut
2
9−x2
Z cos(x2 )
of
M
C. Evaluate the following definite integrals.
3. f (x) =
Z 4
16.
0
|s2 − 5s + 6| ds
228
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
Supplementary Exercises
A. Find the derivative of the following functions.
Z x
1. F (x) =
Z √
3x
t2 − 1
dt
√
3
sec x 1 + t
!#
Z 3s2
Z 2x "
√
d
4. F (x) =
t3 + 1dt
ds
ds
1
5
t2 sin t dt
3. F (x) =
0
Z 2x3
2. F (x) =
x
1
ds
1 + s2
B. Find an equation of the tangent line and normal line to the graph of the following functions at
the given value of x:
Z 1−4x √ 2
5 s +1
4. h(x) =
ds ; x = 1
tan−1 (s)
−3
Z 2x
π
sinh−1 v
π
5. g(x) =
dv ; x =
2
csc(x) sinh v
Z cos(x2 )
2we−w
6. g(x) =
dw ; x = 0
tan−1 w
2x+1
Z x2
sec(t − 1) dt ; x = −2
1. h(x) =
s
2−x
em
at
ic
Z 0
1+u
du ; x = 1
2
x5 −1 1 + u + u
Z 0
4 ln(cosh u) + 1
π
3. f (x) =
du ; x =
2
u +1
2
cos x
of
M
at
h
2. f (x) =
Z 3
3
ut
e
C. Evaluate the following definite integrals.
2
tit
x − 4x + 5x − 8 dx
1.
4
tan3 (y)(1 + tan2 y) dy
π
6
Z 6
UP
2.
2 5x
x sin
dx
3.
6
1
Z π
6
3x
2 3x
4.
sec
tan
dx
π
4
4
−
3
√
Z
5.
3
3 tan−1 x
dx
1 + x2
0
Z 1
csch cot−1 x
6.
dx
1 + x2
0
Z e
coth2 (ln x + x)(x + 1)
7.
dx
x
1
Z 0
8.
|3x + 1 |dx
−1
Z 3
(|3x − 6| − 1) dx
9.
0
In
s
−2
Z π
Z 4
10.
|s2 − 4s + 3| ds
0
Z π
1
dx
2
0
Z ln 4
√
12.
sinh x cosh x dx
sin x −
11.
ln 0.5
Z ln 3
13.
tanh x sech3 x dx
ln 2
(
Z π
14.
f (x) dx if f (x) =
−π
Z ln 5
15.
e−x tanh(e−x ) dx
ln 0.2
Z e4
1
p
dx
x ln2 x − 1
2
Z 1 −1
2 sech (x)
√
17.
dx
x 1 − x2
0
Z 1
1
√
18.
dx
1 + e4x
0
16.
e2
x
sin x
;
;
x≤0
x>0
4.3. THE FUNDAMENTAL THEOREM OF THE CALCULUS
20.
3
em
at
ic
at
h
of
M
e
ut
2
2log3 (x +2x )
√
dx
x · 4log3 x+2
s
1
tit
1
In
s
19.
Z 3
sinh x
dx
x
e − e−x
UP
Z ln 5
229
230
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
4.4
Generalization of the Area of a Plane Region
Now that we know the most important properties of the definite integral, we are able to perform
some geometric calculations. One such computation generalizes the concept of a plane region
we discussed earlier.
At the end of the lesson, the student will be able to do the following:
• express the area bounded by arbitrary curves as a definite integral;
• use the Second Fundamental Theorem of the Calculus to compute such areas; and
• use the concept of vertical and horizontal strips to perform convenient calculations for
area.
1.
f (x)dx = lim
f (x∗i ) ∆x.
n→∞i=1
at
h
a
n
P
em
at
ic
Z b
s
Let f be a continuous function. Recall the following:
Z b
of
M
2. If the graph of y = f (x) lies entirely above the x-axis in the interval [a, b],
f (x) dx gives
a
UP
In
s
tit
ut
e
the area of the region bounded by the curves y = f (x), the x-axis and the vertical lines
x = a and x = b. This is illustrated in the figure below.
We generalize the problem to finding the area of a plane region bounded by several curves such
as the one shown below.
231
em
at
ic
s
4.4. GENERALIZATION OF THE AREA OF A PLANE REGION
of
M
at
h
Some Preliminaries:
tit
ut
e
1. The coordinates of an arbitrary point on a curve y = f (x) in terms of x are given by the
ordered pair (x, f (x)).
b. y = x3
UP
a. y = x2
In
s
Example 4.4.1.
232
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
The height h of a curve at x with reference to the x-axis is given by:
a. h = f (x), if the curve is above the x-axis.
b. h = −f (x), if the curve is below the x-axis.
em
at
ic
s
Example 4.4.2.
b. y = x3
UP
In
s
tit
ut
e
of
M
at
h
a. y = x2
The height (or distance) between two curves at x is given by
h = (y-coordinate of the upper curve) − (y-coordinateof the lower curve)
Example 4.4.3.
a. y = x2 and y = x + 1
4.4. GENERALIZATION OF THE AREA OF A PLANE REGION
233
h = (x + 1) − x2 = −x2 + x + 1
h = x2 − (−1) = x2 + 1
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
b. y = x2 and y = −1
Area between Two Curves: Vertical Approach
Let y = f (x) and y = g (x) be the equations of two curves that are continuous on [a, b] such
that g (x) ≤ f (x) for all x ∈ [a, b]. To find the area between the curves y = f (x) and y = g (x)
in the interval [a, b], we modify the method used for finding the area of a region bounded above
by a curve and below by the x-axis in a given closed interval:
em
at
ic
s
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
1. We divide the interval [a, b] into n subintervals of equal length ∆x. We assign the width
of the ith rectangle to be ∆x. Note that
at
h
b−a
n
of
M
∆x =
ut
e
2. Let i = 1, 2, ..., n. Let x∗i be an arbitrary point in the ith subinterval. We assign the height
hi of the ith rectangle to be the distance between the two curves at x∗i :
In
s
tit
hi = f (x∗i ) − g(x∗i )
3. The area of the ith rectangle is now given by
UP
234
Ai = hi ∆x = f (x∗i ) − g(x∗i ) ∆x
4. We approximate the area AR of the region R by getting the sum of the areas of the n
rectangle:
n
X
AR ≈
n
X
Ai =
f (x∗i ) − g(x∗i ) ∆x
i=1
i=1
5. We let n → ∞ and define this to be the area of the region:
AR = lim
n
X
n→∞
Ai = lim
n→∞
i=1
n
X
f (x∗i ) − g(x∗i ) ∆x
i=1
6. By the definition of the definite integral, we get
Z b
(f (x) − g(x)) dx
AR =
a
4.4. GENERALIZATION OF THE AREA OF A PLANE REGION
235
Formula for the Area of a Plane Region
If f and g are continuous functions on the interval [a, b] and f (x) ≥ g (x) for all x ∈ [a, b],
then the area of the region R bounded above by y = f (x) , below by y = g (x) and the
vertical lines x = a and x = b is
Z b
(f (x) − g(x)) dx
AR =
a
Suggestion: We can think of the formula for the area of a plane region in this way:
Z b
AR =
h dx
a
em
at
ic
s
where:
• [a, b] is the interval I covered by the region along the x-axis
at
h
• h is the height of the rectangle at an arbitrary point x in the interval I
of
M
• dx is the width of the rectangle at an arbitrary point x in the interval I
e
Example 4.4.4.
UP
In
s
tit
ut
1. Find the area of the region bounded by y = x3 , the x-axis and x = −2.
The interval covered by the region along the x-axis is
I = [−2, 0]
Meanwhile, the height of the rectangle at an arbitrary point x in the interval I is given by
236
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
h = 0 − x3 = −x3
Therefore, the area of given plane region is given by
Z 0
AR =
−x3 dx
−2
We evaluate the definite integral and obtain
Z 0
AR =
−x3 dx = −
−2
x4 x=0
= 0 − (−4) = 4 square units
4 x=−2
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
2. Find the area of the region bounded by y = x2 + 2, y = −1, x = −1 and x = 2.
We have the following:
I = [−1, 2] ; h = (x2 + 2) − (−1) = x2 + 3
Therefore, the area of the desired region is given by
x=2
x3
(x2 + 3) dx =
+ 3x
3
−1
x=−1
8
1
= AR =
+ 6 − − − 3 = 12 square units
3
3
Z 2
AR =
3. Find the area of the region bounded by y = 2 − x2 and y = −x.
4.4. GENERALIZATION OF THE AREA OF A PLANE REGION
237
em
at
ic
s
We solve for the x-coordinates of the points of intersection to get I.
2 − x2
= −x
x2 − x − 2 = 0
at
h
(x + 1) (x − 2) = 0
of
M
x = −1 or x = 2
e
Thus, I = [−1, 2]. Meanwhile,
tit
ut
h = 2 − x2 − (−x) = −x2 + x + 2
UP
In
s
The area of the plane region is
x=2
x3 x2
+ 2x
(−x2 + x + 2) dx = − +
3
2
−1
x=−1
8
1 1
9
= − +2+4 −
+ −2
square units
3
3 2
2
Z 2
AR =
Notice that the formula for the area of a plane region assumes that the upper curve y = f (x)
is always above the lower curve y = g (x) in [a, b]. That is, the formula for the height h of a
rectangle at an arbitrary x is the same throughout the region. In the case that this is not so,
we have to divide the region into an appropriate number of subregions to apply the formula.
Example 4.4.5.
1. Find the area of the region bounded by y = cos x, y = sin x and the lines x = 0 and x = 2π.
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
em
at
ic
s
238
of
M
= cos x
1
=
· cos x
cos x
= 1
5π
or x =
4
UP
We now have the following:
In
s
tit
ut
e
sin x
1
· sin x
cos x
tan x
π
x=
4
at
h
We divide the interval into three subregions R1 , R2 and R3 . To determine the appropriate
x-interval covered by each of these three subregions, we solve for the x-coordinates of the
points of intersection of y = cos x and y = sin x in the interval [0, 2π]:
For R1 :
I = [0, π4 ]
h = cos x − sin x
For R2 :
I = [ π4 , 5π
4 ]
h = sin x − cos x
For R3 :
I = [ 5π
4 , 2π]
h = cos x − sin x
The required area is given by:
AR = AR1 + AR2 + AR3
Z π
Z π
Z 2π
4
4
=
cos x − sin x dx +
cos x − sin x dx +
cos x − sin x dx
0
√
= 4 2 square units
0
5π
4
2. Find the area of the region shown below. The boundaries are y = x2 , y = 2x and 7x+4y = 15.
4.4. GENERALIZATION OF THE AREA OF A PLANE REGION
239
em
at
ic
s
We divide the interval into two subregions R1 and R2 and solve for the points of intersection:
x2
=
4x2 + 7x − 15
at
h
=
15 − 7x
4
0
of
M
Intersection of y = x2 and 7x + 4y = 15:
=
0
or
x = −3
ut
e
(4x − 5)(x + 3)
5
x=
4
7x + 4 (2x) = 15
x = 1
UP
In
s
tit
Intersection of y = 2x and 7x + 4y = 15:
Intersection of y = x2 and y = 2x:
x2 − 2x
=
0
x (x − 2)
=
0
x=0
or
x=2
We now have
For R1 :
I = [−3, 1]
15 − 7x
− x2
h=
4
For R2 :
I = [1, 2]
h = 2x − x2
Finally,
AR = AR 1 + AR 2
Z 1
Z 2
15 − 7x
40
2
=
−x
dx +
2x − x2 dx =
square units
4
3
−3
1
240
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
An Alternative Method Using Horizontal Rectangles
ut
e
of
M
at
h
em
at
ic
s
In the method we employed to find the area of a plane region, the plane region is covered using
rectangles oriented vertically. Now consider the case of the plane region below. If rectangles
oriented vertically are used to find the area of the plane region, the region would have to be
divided into two subregions because the formula for the height of the rectangle is not consistent
throughout the region. In one subregion, the upper curve is the line, while the lower curve is
the lower branch of the parabola. In the other subregion, the upper curve is the upper branch
of the parabola, while the lower curve is the lower branch of the parabola.
UP
In
s
tit
Notice, however, that if we use rectangles oriented horizontally, the expression for the length of
the rectangle is the same throughout the given region. This is in contrast to the approach using
vertical rectangles, where the expression for the height of the rectangle is the same throughout
the region.
The length of a rectangle oriented horizontally is given by
4.4. GENERALIZATION OF THE AREA OF A PLANE REGION
241
l =(x-coordinate of the right curve) − (x-coordinate of the left curve)
Throughout the region, the curve on the right is always the parabola, while the curve of the left
is always the line.
The following gives an alternative method for finding the area of a plane region using horizontal
rectangles, which would be more convenient to use in finding the area of some regions such as
the one shown above.
Formula for the Area of a Plane Region
If u and v are continuous functions in y on the interval [c, d] and v (y) ≥ u (y) for all
y ∈ [c, d], then the area of the region R bounded on the left by x = u (y) , on the right
by x = v (y) and the horizontal lines y = c and y = d is
Z d
(v(y) − u(y)) dy
AR =
em
at
ic
s
c
of
M
at
h
Suggestion: Again, we can think of the formula for the area of a plane region in this way:
Z d
l dy
AR =
ut
e
where:
c
tit
• [c, d] is the interval I covered by the region along the y-axis;
In
s
• l is the length of the rectangle at an arbitrary point y in the interval I;
UP
• dy is the width of the rectangle at an arbitrary point y in the interval I.
Some Preliminaries.
1. The coordinates of a point on the curve x = u (y) in terms of y are given by (u(y), y)
Example.
242
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
2. The length of the rectangle at an arbitrary y is given by
l =(x-coordinate of the right curve) − (x-coordinate of the left curve)
Example. In the previous example,
h = (3 − y 2 ) − (y + 1) = −y 2 − y + 2
Example 4.4.6.
1. Find the area of the region bounded by x = 3 − y 2 and the line y = x − 1. (Refer to the
figure in the previous example.)
3 − y2
em
at
ic
s
To find the y-interval covered by the region above, we solve for the y-coordinates of the
points of intersection:
= y+1
at
h
y2 + y − 2 = 0
of
M
(y + 2)(y − 1) = 0
e
y = −2 or y = 1
In
s
tit
ut
Therefore, I = [−2, 1]. We have already seen that the length of a horizontal rectangle at an
arbitrary y is given by
UP
l = −y 2 − y + 2
We get that the area of the region is given by the definite integral
Z 1
AR =
− y 2 − y + 2 dy
−2
Evaluating this integral, we obtain
AR = −
y=1
9
y3 y2
−
+ 2y
= square units
3
2
2
y=−2
2. Find the area of the region bounded by the parabolas x = 4 − y 2 and x = 2 −
y2
.
2
4.4. GENERALIZATION OF THE AREA OF A PLANE REGION
243
em
at
ic
s
We solve for the y-coordinates of the points of intersection:
4 − y2
2
y2 − 4 = 0
=
at
h
4 − y2
of
M
y = −2 or y = 2
−
4 − y2
2
=
4 − y2
. The required area is
2
ut
e
Therefore, we have I = [−2, 2] and l =
4 − y2
In
s
tit
Z 2
y2
y 3 y=2
16
AR =
2−
dy = 2y −
=
square units
2
6 y=−2
3
−2
UP
3. Find the area of the region bounded by the parabolas x2 = 4y and the parabola y 2 = −4x.
244
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
The curve on the right is the upper branch of the parabola y 2 = −4x. We isolate x and
y2
get x = − . On the other hand, the curve on the left is the left branch of the parabola
4
√
x2 = 4y. We solve for x and obtain x = ±2 y. Since the curve of interest is the left branch
√
of the parabola, the equation of the curve is x = −2 y.
We solve for the y-coordinates of the points of intersection and obtain
√
= −4 (−2 y)
y2
y 4 − 64y
= 0
y = 0 or y = 4
Thus, we get I = [0, 4] and h = −
s
em
at
ic
at
h
of
M
e
ut
tit
0
In
s
AR =
3
y2
y3
4y 2 y=4 16
√
−
+ 2 y dy = − +
=
square units
4
12
3 y=0
3
UP
Z 4
y2
√
+ 2 y. Finally,
4
4.4. GENERALIZATION OF THE AREA OF A PLANE REGION
4.4.1
245
Exercises
Exercises for Discussion
A. Find the area of the following regions.
1. Triangle whose vertices are the points (−1, 4), (2, −2), and (5, 1)
2. Trapezoid whose vertices are the points A(−1, −1), B(2, 2), C(6, 2) and D(7, −1)
3. Region bounded by the upper branch of the parabola x = y 2 , the tangent line to the parabola
at (1, 1) and the x-axis
4. Region bounded by y = x2 − 2x + 3 and y = x + 7.
5. Region bounded by y = 6x − x2 and 2y = 5x + 6.
6. R is bounded above by the graph of y = |2x|, and below by the parabola y = x2 − 8.
tit
ut
e
of
M
at
h
em
at
ic
s
7. Region shown below. Boundaries are y = 5 − x2 , y = 2x + 2, y = −x − 1
In
s
8. Region bounded below by the curves y = 3−x and y = 3x , and the line y = 9
UP
9. Region bounded by the curve y = tanh2 (x), the x-axis, and the line x = ln 14
10. Region bounded by the curve y = 2x , the coordinate axes, and the lines x = 4 and y = 8
246
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
Supplementary Exercises
A. Find the area of the following regions.
1. Region bounded by y = x2 − 2x + 1 and y = 7 − x
2. Triangle whose vertices are the points (−1, 4), (2, −2), and (5, 1)
3. Triangle with vertices at A(−2, 2), B(5, 8), and C(1, −4).
4. Triangle with vertices at A(0, 0), B(4, −8), and C(−2, −4).
5. Quadrilateral with vertices A(−5, −3), B(1, 6), C(6, 6), and D(12, −3).
6. Region bounded by the line y = −9 and the curve y = 16 − x2 .
7. Region bounded below by the x-axis, and above by the parabola x = y 2 and the tangent
line to the parabola at (2, 4).
8. R is bounded below by the graph of y = 3 sin x, above by the graph of y = −2 cos2 x,
and between x = 0 and x = 2π
ut
e
of
M
at
h
em
at
ic
s
9. Region shown below. Boundaries are x = y 2 − 2y, x = 4 − y 2 and x + 2y = 4
UP
In
s
tit
10. Set up the integral that gives the area of the region shown below. The parabola and the
circle intersect at the point (2, 0).
11. Set up the definite integral(s) equal to the area of the region with the given boundaries.
a. y = tan x, y = cot x, y = x2 − x, x = 1
b. y = ln x, y = −x2 − x + 2, 20x − 4y = 5, 4x = 1
√
c. y = x, y = 0, y = 6 − x
√
d. y = 2x2 , y = − x, 4x + 3y = 10
√
e. y = x, y = −3 x + 1, y= −x + 1, y = 0
x
x
f. y = 2 cos
, y = ln
, 4x = π
2
π
4.4. GENERALIZATION OF THE AREA OF A PLANE REGION
247
√
g. xy = 2, y = 2 x, x − 3y = 1, y = 0
B. Do as indicated.
em
at
ic
s
1. The parabola y = 10 − x2 , and the lines y = −2x + 2 and y = 3x + 6, serve as boundaries
of three finite regions in the Cartesian plane. Find the area of each region.
1
2. Find the area of the region bounded by the graph of y = √
, the x-axis, and the
x 4x2 − 1
√
3
lines x =
and x = 1.
3
x2 y 2
3. Derive the formula for the area of an ellipse with equation 2 + 2 = 1.
a
b
x
4. Derive the formula for the area of the region R bounded by the catenary y = a cosh
,
a
the coordinate axes, and the line x = a, where a > 0.
1
5. Let R be the region under the graph of y = 2 over [1, 4].
x
A. Find a such that the line with equation x = a divides R into two regions having equal
area.
B. Find b such that the line with equation y = b divides R into two regions having equal
area.
at
h
C. Do as indicated.
of
M
1. TRUE OR FALSE: Any region, whose boundaries can be expressed in terms of functions
and vertical lines, can be divided into two regions of equal area.
2. For the statement above to be always true, what condition(s) must be satisfied?
In
s
tit
ut
e
3. Prove the following statement: If a region is bounded above by y = f (x), below by y =
g(x), and the lines x = a and x = b, then there is a unique line dividing the region into
two regions of equal area.
UP
4. Construct a similar statement for regions bounded on the left by x = u(y), on the right
by x = v(y), and the lines y = c and y = d.
5. Consider the region R bounded above by the graph of f (x) = 4x − 4x3 and below by the
x-axis. What value of m will make the graph of g(x) = m divide R into two regions of
equal area?
248
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
4.5
Arc Length of Plane Curves
We proceed with another geometric computation which turns out to make use of derivatives and
integrals. We only need a fact from Euclidean geometry, which states that the distance between
p
two points on the Cartesian plane, P (x1 , y1 ) and Q(x2 , y2 ), is d(P Q) = (x1 − x2 )2 + (y1 − y2 )2 .
At the end of the lesson, the student will be able to do the following:
• characterize a smooth function;
• derive an expression for the length of any smooth function; and
em
at
ic
s
• use a particular orientation of the curve to be able to construct a convenient
expression for its length.
UP
In
s
tit
ut
e
of
M
at
h
First, we define a smooth curve. A curve with equation y = f (x) is said to be smooth on [a, b] if
f 0 is continuous on [a, b]. Suppose we want to find the length of the arc of a smooth curve over a
the interval [a, b].
We use the previous method of dividing the interval [a, b] into n subintervals of equal length. Let
i = 1, 2, ..., n.
1. Let (xi , yi ) and (xi+1 , yi+1 ) be the endpoints of the arc over the ith subinterval.
2. The distance between the endpoints of the ith subarc is given by
di =
p
(xi+1 − xi )2 + (yi+1 − yi )2 , for i = 1, 2, . . . , n
4.5. ARC LENGTH OF PLANE CURVES
249
3. Let ∆xi = xi+1 − xi and ∆yi = yi+1 − yi . So we have
p
(∆xi )2 + (∆yi )2
s
(∆xi )2 + (∆yi )2
=
· (∆xi )2
(∆xi )2
s
∆yi 2
· ∆xi
1+
=
∆xi
di =
4. We approximate the length L of the arc by getting the sum of the distances between the
endpoints of all subarcs:
s
n
n
X
X
∆yi 2
L≈
di =
1+
· ∆xi
∆xi
i=1
i=1
i=1
at
h
i=1
em
at
ic
s
5. We define the length of the arc to be the limit of the sum as n → ∞:
s
n
n
X
X
∆yi 2
L = lim
di =
1+
· ∆xi
n→∞
∆xi
In
s
tit
ut
e
of
M
6. Therefore, length of the arc is given by the definite integral
s
2
Z b
dy
L=
1+
dx
dx
a
Formula for the Length of an Arc
UP
If y = f (x) is a smooth curve on the interval [a, b], then the arc length L of this curve
from x = a to x = b is
s
2
Z b
Z bq
2
dy
1+
1 + f 0 (x) dx
L=
dx =
dx
a
a
If x = u(y) is a smooth curve on the interval [c, d], then the arc length L of this curve
from y = c to y = d is
s
2
Z d
Z dq
2
dx
1+
dy =
1 + u0 (y) dy
L=
dy
c
c
Example 4.5.1.
1. Find the length of arc of the curve y =
3/2
1 2
x +2
from x = 1 to x = 2.
3
250
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
First, we compute
p
dy
1p 2
x + 2 · 2x = x x2 + 2
=
dx
2
Z 2
L=
s
p
2
Z 2p
2
1+ x x +2
dx =
x4 + 2x2 + 1 dx
1
(x2 + 1) dx =
1
3
x=2
+x
=
x=1
10
units
3
of
M
=
x3
at
h
1
Z 2
em
at
ic
s
Therefore,
UP
In
s
tit
ut
e
2. Set up the definite integral required to find the arc length of the curve x = (y − 1)2 − 4 from
the point (−3, 2) to (0, −1).
Here, it is more convenient to use the horizontal orientation of the curve. We have that
dx
= 2(y − 1)
dy
The definite integral that gives the length of arc is
Z 2p
L=
1 + 4(y − 1)2 dy
−1
4.5. ARC LENGTH OF PLANE CURVES
251
3. Set up the definite integral required to find the perimeter of the region enclosed by the curves
3
1
1 2
3
and y = .
y=x + ,x=− y−
2
2
2
at
h
em
at
ic
s
1
If L1 is the length of the segment from −1, 32 and 1, 32 , L2 is the length of the arc of y = x3 +
2
2
1
from −1, 23 to 0, 12 ,
from 0, 12 to 1, 32 , and L3 is the length of the arc of x = − y −
2
then the perimeter P of the enclosed region is given by
of
M
P = L1 + L2 + L3
L1 = 1 − (−1) = 2
tit
ut
e
Obviously,
UP
In
s
We set up the integral to find L2 in terms of x and get that
Z 1p
L2 =
1 + 9x4 dx
0
We set up the integral to find L3 in terms of y and get that
s
Z 3
2
1 2
L3 =
1+4 y−
dy
1
2
2
Therefore,
Z 1p
Z 3
2
4
P =2+
1 + 9x dx +
0
1
2
s
1 2
1+4 y−
dy
2
Remark 4.5.2. Note in the previous example that our expression for the perimeter involves x and
y. This is sometimes the case, especially for the perimeter of regions with varying boundaries. The
thing to take away here is that each integral must contain only one variable of integration; that is,
x and y cannot appear in a single integrand.
252
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
4.5.1
Exercises
Exercises for Discussion
A. Find the arc length of the following curves in the specified interval:
2
3
(x − 5) /2 from the point where x = 6 to the point where x = 8.
3
2. 2x − 4y + 6 = 0 between y = 0 and y = 2
π
3. y = ln(cos x) from x = 0 to x =
4
√
4. 9y 2 = 4x3 from the origin to the point (3, 2 3).
1. y =
5. 8y = x4 + 2x−2 from the point where x = 1 and x = 2.
x
6. y = a cosh
, from the point where x = 0 to the point where x = a
a
7. y = sinh−1 x from the point where x = 0 to the point where x = 4
em
at
ic
s
B. Do as indicated.
UP
In
s
tit
ut
e
of
M
at
h
1. Set up a definite integral that gives the length of the arcsine function throughout its domain.
√
2. Set up a definite integral that gives the arc length of y = 3 x − 1 from x = 0 to x = 1. (Note
that this integral has to be set up in terms of y. Why not x?)
4.5. ARC LENGTH OF PLANE CURVES
253
Supplementary Exercises
A. Find the perimeter of each region in Exercises 4.4.1, Part A, Items 1-7, and Exercises 4.4.2,
Part A, 1-5.
B. Set up the (sum of) definite integral(s) equal to the perimeter of each region in Exercises 4.4.1,
Part A, Items 8-10, and Exercises 4.4.2, Part A, 6-11.
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
C. Set up the (sum of) definite integral(s) equal to the perimeter of each of the regions described
in Exercises 4.4.2, Part B, Items 1-4.
254
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
4.6
Volumes of Solids
Another application of the definite integral comes in when we consider the volume of solids formed
using special procedures. We consider first right cylinders, the normal type of cylinders we see in
everyday life. Then we consider solids formed by rotating a plane region about a particular line.
At the end of the lesson, the student will be able to do the following:
• understand the different procedures used in computing the volume of a given solid;
• generate expressions for the volume of any solid using cross-sectional area; and
4.6.1
em
at
ic
s
• solve for the volume of solids generated by revolving a plane region about an indicated
axis, using the methods of disks, washers, and cylindrical shells.
Volumes of Solids of Revolution1
at
h
Solids of Revolution
ut
e
of
M
A solid of revolution is a solid obtained when a plane region is revolved about a line called the
axis of revolution. For simplicity, we will only consider axes of revolution which are vertical
(parallel to the y-axis) or horizontal (parallel to the x-axis).
In
s
tit
Example 4.6.1.
UP
1. The solid of revolution generated when the region bounded by x = y 2 and y = x2 is revolved
about the x-axis:
2. The solid of revolution generated when the region bounded by y = sin x and the x-axis in the
interval [0, π] is revolved about the y-axis:
1
Some of figures in this section are taken from [5].
4.6. VOLUMES OF SOLIDS
255
Remark 4.6.2. Two methods allow us to find the volume of a solid of revolution:
em
at
ic
s
1. Disk or Washer Method. Use rectangles that are perpendicular to the axis of revolution.
of
M
at
h
2. Cylindrical Shell Method. Use rectangles that are parallel to the axis of revolution.
ut
e
In both methods, we will need to determine the distance d from an arbitrary point x on a curve
to the axis of revolution. To this end, we need the following observations:
In
s
tit
1. If the axis of revolution is a vertical line, the horizontal distance from the axis of revolution to
an arbitrary point on the curve is given by
UP
d = (x-coordinate of the right curve) − (x-coordinate of the left curve)
2. If the axis of revolution is a horizontal line, the vertical distance from the axis of revolution to
an arbitrary point on the curve is given by
d = (y-coordinate of the upper curve) − (y-coordinate of the lower curve)
Example. Find the distance from an arbitrary point x on the given curve to the given axis of
revolution.
1. y = x2 , x ≥ 0
a. Axis of revolution: x = 5
256
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
In terms of x: d = 5 − x
In terms of y: d = 5 −
√
y
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
b. Axis of revolution: x = −2
2. x = 4 − y 2
UP
In terms of x: d = x + 2
In terms of y: d =
√
y+2
a. Axis of revolution: y = 3
In terms of y: d = 3 − y
In terms of x: d = 3 −
√
4−x
4.6. VOLUMES OF SOLIDS
257
b. Axis of revolution: y = −1
In terms of y: d = y − (−1) = y + 1
In terms of x: d =
√
√
4 − x − (−1) = 4 − x + 1
em
at
ic
s
The Disk or Washer Method
ut
e
of
M
at
h
A disk and a washer with the formulas for their respective volumes are shown below.
Washer: V = πh(r22 − r12 )
r = radius; h = height
r2 = outer radius; r1 = inner radius
In
s
tit
Disk: V = πr2 h
UP
In the disk or washer method, the rectangles are oriented so that they are perpendicular to the axis
of revolution.
When a segment of the axis of revolution is a boundary of the region, as is the case when the
region below is revolved about the y-axis, disks are obtained.
258
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
If the axis of revolution does not intersect the region, or intersects the region at only one point, we
get washers. This is illustrated in the following example.
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
Let us consider the simple case when we only obtain disks. Refer to the region below, bounded
above by the curve y = f (x), in the interval [a, b]. To get the volume of the solid generated when
the region is revolved about the x-axis, we do the following:
1. We divide the interval [a, b] into n subintervals of equal length ∆x.
2. Let x∗i be an arbitrary point in the ith subinterval. Let ri = f (x∗i ) be the height of the ith
rectangle.
3. When we revolve the ith rectangle about the axis of revolution, we get the ith disk whose radius
is given by ri and whose height is given by ∆x.
4. The volume of the ith disk is then given by
Vi = πri2 ∆x = π [f (x∗i )]2 ∆x
4.6. VOLUMES OF SOLIDS
259
5. The volume of the solid of revolution can be approximated by adding the volumes of the disks
obtained when the rectangles are revolved about the axis of revolution. The expression becomes
V ≈
n
X
Vi =
i=1
n
X
π [f (x∗i )]2 ∆x
i=1
6. We define the volume of the solid to be the limit of the sum above as n → ∞:
n→∞
π [f (x∗i )]2 ∆x
i=1
Z b
V =
π [f (x)]2 dx
of
M
at
h
a
em
at
ic
7. By the definition of the definite integral, we have
s
V = lim
n
X
ut
e
The formula below generalizes what is obtained above.
In
s
tit
Formula for the Volume of a Solid of Revolution using Disks or Washers
Vertical Rectangles: Suppose R is the region bounded above by y = f (x), below by
UP
y = g (x), and the vertical lines x = a and x = b such that f and g are continuous
functions on [a, b]. If the line y = y0 does not intersect the interior of R, then the volume
of the solid of revolution obtained when R is revolved about the line y = y0 is given by:
1. If only disks are obtained (that is, a boundary of R lies on the axis of revolution),
then
Z b
V =
π [r(x)]2 dx
a
where r (x) is the radius of a disk at an arbitrary x in [a, b].
2. If washers are obtained (that is, a boundary of R does not lie on the axis of revolution),
then
Z b V =
π [r2 (x)]2 − [r1 (x)]2 dx
a
where r2 (x) and r1 (x) are the outer radius and inner radius, respectively, of a washer
at an arbitrary x in [a, b].
260
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
Formula for the Volume of a Solid of Revolution using Disks or Washers
Horizontal Rectangles: Suppose R is the region bounded on the left by x = u (y) , on
the right by x = v (y) and the horizontal lines y = c and y = d such that u and v are
continuous functions on [c, d]. If the line x = x0 does not intersect the interior of R, then
the volume of the solid of revolution obtained when R is revolved about the line x = x0
is given by:
1. If only disks are obtained, then
Z d
V =
π [r(y)]2 dy
c
where r (y) is the radius of a disk at an arbitrary y in [c, d].
em
at
ic
Z d V =
π [r2 (y)]2 − [r1 (y)]2 dy
s
2. If washers are obtained, then
c
of
M
at
h
where r2 (y) and r1 (y) are the outer radius and inner radius, respectively, of a washer
at an arbitrary y in [c, d].
Suggestion:
ut
e
A. We can think of the volume of a solid of revolution in this way:
where:
Z b
V =
πr2 dw
a
UP
In
s
tit
1. If only disks are obtained:
• [a, b] is the interval I covered by the region
• r is given by the distance of the farther tip of an arbitrary rectangle to the axis of
revolution
• dw is the width (shorter side) of an arbitrary rectangle
2. If washers are obtained:
Z d
V =
π r22 − r12 dw
c
where:
• [a, b] is the interval I covered by the region
• r1 is given by the distance of the nearer tip of an arbitrary rectangle to the axis of
revolution
• r2 is given by the distance of the farther tip of an arbitrary rectangle to the axis of
revolution
4.6. VOLUMES OF SOLIDS
261
• dw is the width (shorter side) of an arbitrary rectangle
B. If vertical rectangles are used, dw = dx, so set up the integral in terms of x. If horizontal
rectangles are used, dw = dy, so set up the integral in terms of y.
Example 4.6.3. Let R be the region bounded by y = x2 , x = 0 and x = 2.
at
h
em
at
ic
s
1. Find the volume of the solid generated when R is revolved about the x-axis.
ut
e
of
M
Since the rectangles must be perpendicular to the axis of revolution, we use vertical rectangles
and set up the integral in terms of x. In this case, a boundary of the region lies on the axis of
revolution so only disks are obtained.
UP
In
s
tit
The x-interval covered by the region is [0, 2] and the distance from the farther tip of an arbitrary rectangle to the axis of revolution is r = x2 − 0 = x2 . Therefore, the volume is given
by
Z 2
πx5 2 32π
4
V =
πx dx =
=
cubic units
5 0
5
0
2. Find the volume of the solid generated when R is revolved about the line x = 2.
262
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
We use horizontal rectangles and set up the integral in terms of y. Again only washers are
obtained, so we have:
I = [0, 4]
r =2−
;
√
y
The volume is given by


3
2
2
8y
y
8π
√
π(4 − 4 y + y) dy = π 4y −
V =
+ =
cubic units
3
2
3
0
Z 4
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
3. Find the volume of the solid generated when R is revolved about the line y = 4.
We use vertical rectangles and set up the integral in terms of x. This time we obtain washers.
We get the following:
I = [0, 2]
;
r1 = 4 − x2
;
r2 = 4 − 0 = 4
The volume is given by
Z 2
2
π 8x − x
V =
0
4
dx = π
8x3 x5
−
3
5
2
=
0
224π
cubic units
15
cubic units
4. Find the volume of the solid generated when R is revolved about the y-axis.
4.6. VOLUMES OF SOLIDS
263
I = [0, 4]
;
r1 =
√
y
;
r2 = 2 − 0 = 2
of
M
y2 4
π (4 − y) dy = π 4y −
= 8π cubic units
2
0
0
Z 4
tit
ut
e
V =
y−0=
at
h
Therefore, the volume required is
√
em
at
ic
s
We use horizontal rectangles and set up the integral in terms of y. We use the formula for
washers and get
In
s
The Cylindrical Shell Method
UP
A cylindrical shell and the formula for its volume are shown below.
Cylindrical Shell: V = 2πrh ∆r
2
r = r1 +r
2 ; r2 = outer radius; r1 = inner radius; ∆r = r2 − r1
Suppose we want to get the volume of the solid generated when the region below is revolved about
the y-axis. In the cylindrical shell method, we orient the rectangles so that they are parallel to the
axis of revolution.
264
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
of
M
at
h
em
at
ic
s
When we revolve the rectangles about the axis of revolution, we obtain cylindrical shells.
UP
In
s
tit
ut
e
We obtain the formula for the volume of a solid of revolution using cylindrical shells in the following
manner:
1. We divide the interval [a, b] into n subintervals of equal length ∆x.
2. Let x∗i be the midpoint of the ith subinterval.
3. The height of the ith rectangle is given by hi = f (x∗i ).
4. For the ith cylindrical shell, r = x∗i , h = f (x∗i ), ∆r = ∆x.
5. Thus, the volume of the ith cylindrical shell is
Vi = 2π(x∗i )f (x∗i )∆x
4.6. VOLUMES OF SOLIDS
265
6. The volume of the solid is approximated by adding the volumes of the n cylindrical shells:
V ≈
n
X
Vi =
i=1
n
X
2π(x∗i )f (x∗i )∆x
i=1
7. The volume of the solid is defined to be the limit of the above as n → ∞:
V = lim
n→∞
n
X
2π(x∗i )f (x∗i )∆x
i=1
8. By the definition of the definite integral,
Z b
2πxf (x) dx
V =
a
s
The formulas below generalize that which was derived above.
em
at
ic
Formula for the Volume of a Solid of Revolution using Cylindrical Shells
of
M
at
h
Vertical Rectangles: Suppose R is the region bounded above by y = f (x) , below by
y = g (x), and the vertical lines x = a and x = b such that f and g are continuous
functions on [a, b]. If the line x = x0 does not intersect the interior of R, then the volume
of the solid of revolution obtained when R is revolved about the line x = x0 is given by
Z b
2πr(x)h(x) dx
a
ut
e
V =
In
s
tit
where r (x) and h (x) are the radius and height, respectively, of a cylindrical shell at an
arbitrary x in [a, b].
UP
Horizontal Rectangles: Suppose R is the region bounded on the left by x = u (y) , on
the right by x = v (y) and the horizontal lines y = c and y = d such that u and v are
continuous functions on [c, d]. If the line y = y0 does not intersect the interior of R, then
the volume of the solid of revolution obtained when R is revolved about the line y = y0
is given by
Z d
V =
2πr(y)h(y) dy
c
where r (y) and h (y) are the radius and height, respectively, of a cylindrical shell at an
arbitrary y in [c, d].
Suggestion:
A. We can think of the formula for the volume using cylindrical shells in this way:
Z b
V =
2πrh dw
a
where:
266
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
• [a, b] is the interval I covered by the region
• r is the distance of an arbitrary rectangle to the axis of revolution
• h is the height of an arbitrary rectangle
• dw is the width of an arbitrary rectangle
B. If vertical rectangles are used, dw = dx, so set up the integral in terms of x. If horizontal
rectangles are used, dw = dy, so set up the integral in terms of y.
s
Example 4.6.4. Let R be the region bounded by y = x2 , x = 0 and x = 2.
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
1. Find the volume of the solid generated when R is revolved about the y-axis.
Since the rectangles must be parallel to the axis of revolution, we use vertical rectangles and
set up the integral in terms of x. The x-interval covered by the region is I = [0, 2]. Meanwhile
the distance of an arbitrary rectangle to the axis of revolution is r = x − 0 = x and the height
of an arbitrary rectangle is h = x2 . Therefore, we get that
Z 2
V =
0
2πx3 dx =
πx4 2
= 8π cubic units
2 0
2. Find the volume of the solid generated when R is revolved about the line x = 2.
4.6. VOLUMES OF SOLIDS
267
We use vertical rectangles and set up the integral in terms of x. We get the following:
r =2−x
;
h = x2
s
;
em
at
ic
I = [0, 2]
at
h
The volume is therefore
3
2
Z 2
2x
x4
8π
2
3
V =
2π(2x − x ) dx = 2π
−
cubic units
=
3
4
3
0
of
M
0
UP
In
s
tit
ut
e
3. Find the volume of the solid generated when R is revolved about the x-axis.
In this case, the axis of revolution is horizontal, so use horizontal rectangles and set up the
integral in terms of y. We have the following:
√
I = [0, 4] ; r = y − 0 = y ; h = 2 − y
The required volume is given by
Z 4
V =
2π
0
3
2y − y 2

dy = 2π y 2 −
5
2y 2

5

4
=
0
32π
cubic units
5
268
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
4. Find the volume of the solid generated when R is revolved about the line y = 4.
em
at
ic
s
We use horizontal rectangles and set-up the integral in terms of y We get the following
√
I = [0, 4] ; r = 4 − y ; h = 2 − y
224π
cubic units
15
0
ut
e
=
of
M
at
h
The volume of the resulting solid of revolution is


3
5
4
Z 4 1
3
8y 2
2y 2 
2

V =
2π 8 − 2y − 4y 2 + y 2 dx = 2π 8y − y −
+
3
3
0
tit
Remarks.
UP
In
s
1. If the formula for the height of the rectangle is not the same throughout the region, divide the
region into the appropriate number of subregions.
2. If you are free to choose the method, the best method is that for which the formula for the
height of the rectangle changes least throughout the region.
Example 4.6.5. Let R be the region bounded by x = 3y 2 and x = 4 − y 2 .
4.6. VOLUMES OF SOLIDS
269
For this particular region, it is better to use horizontal rectangles, so we set up all required integrals
in terms of y.
1. Set up the integral that gives the volume of the solid generated when R is revolved about the
line y = 1.
Since we are using horizontal rectangles and the axis of revolution is also horizontal, we use
the Cylindrical Shell Method.
We find the y-coordinates of the points of intersection to determine the y-interval covered by
the region.
3y 2 = 4 − y 2
4y 2 = 4
em
at
ic
s
y = ±1
r =1−y
;
;
h = (4 − y 2 ) − (3y 2 ) = 4 − 4y 2
of
M
I = [−1, 1]
at
h
We have the following:
Z 1
2π(1 − y)(4 − 4y 2 ) dy
−1
UP
In
s
tit
ut
e
The required volume is given by the integral
2. Set up the integral that gives the volume of the solid generated when R is revolved about the
line x = −1.
This time the axis of revolution is vertical so we use the Washer Method. We obtain the
following:
I = [−1, 1]
;
r = (4 − y 2 ) − (−1) = 5 − y 2
;
h = (3y 2 ) − (−1) = 3y 2 + 1
The required volume is given by the integral
Z 1
V =
π
h
−y 2 + 5
2
− 3y 2 + 1
2 i
dy
−1
Example 4.6.6. Let R be the region bounded by y = −3x + 8, y = 3x + 2 and y = x2 − 2.
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
em
at
ic
s
270
at
h
In this case, it is better to use vertical rectangles so we set-up the integral in terms of x. We divide
the region R into two subregions R1 and R2 .
of
M
1. Set up the integral that gives the volume of the solid generated when R is revolved about the
line x = 2.
In
s
tit
ut
e
Vertical rectangles are parallel to the axis of revolution so we use Cylindrical Shell Method.
First we find the x-coordinates of the points of intersection to determine the x-interval covered
by the region.
UP
Points of intersection of y = 3x + 2 and y = x2 − 2:
3x + 2 = x2 − 2
x2 − 3x − 4 = 0
x = 4 or x = −1
Points of intersection of y = 3x + 2 and y = −3x + 8:
3x + 2 = −3x + 8
x = 1
Points of intersection y = −3x + 8 and y = x2 − 2:
x2 − 2 = −3x + 8
x2 + 3x − 10 = 0
(x + 5)(x − 2) = 0
x = −5 or x = 2
4.6. VOLUMES OF SOLIDS
271
For R1 :
I = [−1, 1]
r =2−x
h = (3x + 2) − (x2 − 2)
= −x2 + 3x + 4
For R2 :
I = [1, 2]
r =2−x
h = (−3x + 8) − (x2 − 2)
= −x2 − 3x + 10
The volume of the resulting solid when R1 is revolved about the axis of revolution is the following:
Z 1
2π(2 − x)(−x2 + 3x + 4) dx
V1 =
−1
The volume of the resulting solid when R2 is revolved about the axis of revolution is:
Z 2
V2 =
2π(2 − x)(−x2 − 3x + 10) dx
1
em
at
ic
s
The required volume is given by:
V = V1 + V2
Z 1
Z 2
=
2π(2 − x)(−x2 + 3x + 4) dx +
2π(2 − x)(−x2 − 3x + 10) dx
1
of
M
at
h
−1
ut
e
2. Set up the integral that gives the volume of the solid generated when R is revolved about the
line y = −3.
tit
This time we use the Washer Method.
UP
In
s
For R1 :
I = [−1, 1]
r2 = (3x + 2) − (−3) = 3x + 5
r1 = (x2 − 2) − (−3) = x2 + 1
For R2 :
I = [1, 2]
r2 = (−3x + 8) − (−3) = −3x + 11
r1 = (x2 − 2) − (−3) = x2 + 1
The volume of the solid obtained when R1 is revolved about the axis of revolution is:
Z 1
V1 =
h
i 2
π (3x + 5)2 − x2 + 1
dx
−1
The volume of the solid obtained when R2 is revolved about the axis of revolution is
Z 2 h
i2
V2 =
π (−3x + 11)2 − x2 + 1
dx
1
The required volume is given by:
V = V1 + V2
Z 1 h
Z 2 h
i 2
i2
2
2
=
π (3x + 5) − x + 1
dx +
π (−3x + 11)2 − x2 + 1
dx
−1
1
272
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
4.6.2
Volume of Solids by Slicing2
We recall that the volume of a cylinder is given by
em
at
ic
s
Volume of a cylinder = Area of a cross-section × Height
UP
In
s
tit
ut
e
of
M
at
h
Let S be a solid bounded by two parallel planes perpendicular to the x-axis at x = a and x = b
such that the cross-sectional area of S in the plane perpendicular the x-axis at an arbitrary x in
[a, b] is given by a continuous function A (x).
We obtain the volume of the solid by ”slicing” the solid into vertical cylinders:
2
Some figures in this section are taken from [5].
4.6. VOLUMES OF SOLIDS
273
1. Divide the interval [a, b] into n subintervals of equal length ∆x.
2. Let x∗i be an arbitrary point in the ith subinterval.
3. The volume of the ith cylinder is given by
Vi = A(x∗i )∆x
4. The volume of the solid is approximated by
V ≈
n
X
Vi =
i=1
n
X
A(x∗i )∆x
i=1
5. The volume of the solid is defined to be
i=1
6. The volume of the solid is given by
Z b
A(x) dx
at
h
V =
A(x∗i )∆x
em
at
ic
n→∞
s
n
X
V = lim
of
M
a
Similarly, we can also find the volume of a solid by slicing the solid into horizontal cylinders.
ut
e
Formula for the Volume of a Solid by Slicing
UP
In
s
tit
Vertical Cylinders: Let S be a solid bounded by two parallel planes perpendicular to
the x-axis at x = a and x = b. If the cross-sectional area of S in the plane perpendicular
the x-axis at an arbitrary x in [a, b] is given by a continuous function A (x), then the
volume of the solid is
Z b
V =
A(x) dx
a
Horizontal Cylinders: Let S be a solid bounded by two parallel planes perpendicular to
the y-axis at y = c and y = d. If the cross-sectional area of S in the plane perpendicular
the y-axis at an arbitrary y in [c, d] is given by a continuous function A (y), then the
volume of the solid is
Z d
V =
A(y) dy
c
Example 4.6.7.
1. Find the volume of the solid shown below.
274
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
s
Note: Cross-sections are circles.
of
M
at
h
em
at
ic
The given solid is bounded on the left by the plane x = 0 and on the right by the plane x = 1.
Note that the cross-sectional area of S in a plane perpendicular to the x-axis at any x ∈ [0, 1]
is a circle. If r(x) is the radius of the cross-section at x, then
√
√
√
x
2 x− x
r(x) =
=
2
2
Thus,
x=1
πx
πx2
π
dx =
= cubic units.
4
8 x=0
8
UP
V =
In
s
The volume is therefore given by
Z 1
πx
4
tit
ut
e
A(x) = π [r(x)]2 =
0
2. Derive the formula for the volume of a right circular cone of radius r and height h.
4.6. VOLUMES OF SOLIDS
275
The solid is bounded on the left by the plane x = 0 and on the right by the plane x = h. Again,
an arbitrary cross-sectional area in a plane perpendicular to the x-axis is a circle. Let r(x) be
the radius of the cross-section at x. Using similar triangles, we get
r (x)
=
x
r (x) =
r
h
rx
h
The area of a cross-section is given by
A(x) = π [r(x)]2 =
πr2 x2
h2
We get the desired formula for the volume of a cone
Z h
V =
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
0
πr2 x2
πr2 x3 x=h πr2 h
dx
=
=
h2
3h2 x=0
3
276
4.6.3
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
Exercises
Exercises for Discussion
A. Find the volume of the following solids by slicing:
em
at
ic
s
1. The figure shown below by slicing. Cross-sections are circles.
at
h
2. The solid whose base is the triangular region with vertices (0, 0), (1, 0) and (0, 1). Crosssections perpendicular to the y-axis are squares.
In
s
b. about x = −2
UP
a. about y = −1
tit
ut
e
of
M
B. Find the volume of the solid generated when the indicated plane region is revolved about the
given axis of revolution.
π
π
1. Region bounded by y = cos x, x = , x = and y = 0; about the x-axis.
4
2
2. Region bounded by y = x2 − 2x + 1, y = 7 − x
3. Region bounded by x = y 2 , x = 0, y = 2;
a. about the x-axis
b. about x = −1
4. Region in Exercises 4.4.1, Part A, Item 15,
a. about the x-axis
b. about x = 1
5. Region in Exercises 4.4.1, Part A, Item 16,
a. about y = −1
b. about x = 4
C. Let R be the region bounded by y = (x − 2)2 , y = x3 and x = −1.
1. Find the volume of the solid generated when R is revolved about x = −2 using the Washer
Method.
4.6. VOLUMES OF SOLIDS
277
2. Find the volume of the solid generated when R is revolved about y = 10 using the Cylindrical Shell Method.
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
3. Find the volume of the solid generated when R is revolved about x = 1.
278
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
Supplementary Exercises
A. Find the volume of the following solids by slicing:
1. The volume of a sphere of radius 3 using vertical cylinders.
2. The volume of a pyramid with a square base using horizontal cylinders.
3. The solid whose base of is an elliptical region with boundary curve 9x2 + 4y 2 = 36. Crosssections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base.
B. Find the volume of the solid generated by revolving the given plane region about the indicated
axis:
1. Region bounded by the curve y = 9 − x2 and the two coordinate axes
a. x = 0
c. y = −4
b. x = −4
c. x = −4
b. y = −1
at
h
a. x = 3
em
at
ic
s
2. Region bounded by the curve y = 4x − x2 , the y-axis, and the lines y = 6 and x = 2,
b. x = π
c. x = −π
e
a. y = 1
of
M
3. Region bounded by the curve y = cos x, and the two coordinate axes
b. x = −3
In
s
a. y = 0
tit
ut
4. Region bounded by the curve x = y 3 , and the lines x = 8 and y = 1
c. x = −1
UP
5. Region bounded by the curve 1 − x = (y − 2)2 , and the lines x = 3, y = 0 and y = 2
a. y = 3
b. x = 5
c. x = −5
6. Region bounded by the curve y = 4 − |x| and the x-axis
a. y = 0
b. y = 5
c. x = 10
7. Region bounded by the curve x − 1 = (y − 2)2 , y-axis, and y = 10
a. x = 0
b. x = −5
C. Let R be the region bounded by y =
c. y = 0
1
, x = 4 and y = 1.
x
1. Find the volume of the solid generated when R is revolved about x = −1 using the Washer
Method.
4.6. VOLUMES OF SOLIDS
279
2. Find the volume of the solid generated when R is revolved about y = −3 using the Cylindrical
Shell Method.
3. Find the volume of the solid generated when R is revolved about x = 6.
D. Let R be the region in the first quadrant bounded by x2 + y 2 = 3, y 2 = −x + 3 and y = 0.
1. Set up the integral that gives the volume of the solid generated by revolving R about x = 4:
a. using the method of washers
b. using the method of cylindrical shells
2. Set up the integral that gives the volume of the solid generated by revolving R about y = −5:
a. using the method of washers
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
b. using the method of cylindrical shells
280
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
4.7
Mean Value Theorem for Integrals
Our main goal for this section is to develop more useful properties of the definite integral. In
particular, we want to direct our discussion in such a way that our complicated computations from
the previous section can be simplified.
At the end of this section, the student will be able to do the following:
• compare integrals of distinct functions,
• put upper and lower estimates on the definite integral of a function,
• provide a statement parallel to the Mean Value Theorem for Derivatives, and
• define a concept similar to that of the average of a collection of numbers.
em
at
ic
s
In particular, we need the third goal - that is, to state the Mean Value Theorem for Integrals
- to be able to provide a proof of the First Fundamental Theorem of the Calculus. The
proof will be given in this section.
of
M
at
h
We begin with the following result, which compares the definite integrals of two regions in the same
closed interval [a, b]:
tit
In
s
a
ut
e
Theorem 4.7.1. If the functions f and g are integrable on [a, b], and if f (x) ≥ g(x) for all x
in [a, b], then
Z b
Z b
f (x) dx ≥
g(x) dx.
a
UP
The geometric interpretation of the previous theorem is illustrated in the following figure. If the
graph of y = f (x) lies above the graph of y = g (x) in the interval [a, b], then the net-signed area
between the the curve y = f (x) and [a, b] must be greater than the net-signed area between the
the curve y = g (x) and [a, b].
4.7. MEAN VALUE THEOREM FOR INTEGRALS
281
Theorem 4.7.2. Suppose f is continuous on the closed interval [a, b]. If m and M are the
absolute minimum function value and absolute maximum function value, respectively, of f in
[a, b], then
Z b
m(b − a) ≤
f (x) dx ≤ M (b − a).
a
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
The previous theorem is illustrated in the following figure. By the Extreme Value Theorem, we
know that if f is continuous on [a, b], then f achieves an absolute maximum value M and absolute
minimum value m on [a, b]. Therefore, the area between y = f (x) and the interval [a, b] is greater
than the area of the rectangle with width (b − a) and height m, but less than the area of the
rectangle with the same width, but with height M .
Theorem 4.7.3 (Mean Value Theorem for Integrals). If the function f is continuous
on the closed interval [a, b], then there exists a number c in [a, b] such that
Z b
f (x) dx = f (c) (b − a).
a
Geometrically, this means that if a function f is nonnegative and continuous on a closed interval
[a, b], we can find a number c in [a, b] such that area of a rectangle with base equal to the length
of the interval [a, b] and height equal to the height of the curve at c is the same the the area under
the curve y = f (x) from a to b.
282
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
em
at
ic
s
Example 4.7.4. Since f (x) = x2 is continuous on [0, 2], there exists c ∈ [−2, 0] satisfying the
conclusion of the Mean Value Theorem. That is,
Z 2
x2 dx = c2 (2 − 0)
0
at
h
√
2 3
. But c ∈ [0, 2], so c =
.
3
3
3
In the previous example, there is only one value of c in the interval satisfying the conclusion of the
Mean Value Theorem. In general, however, the value of c is not unique.
3
e
of
M
2
8
We get that 2c2 = so c = ±
√
In
s
tit
ut
The value of f (c) in the Mean Value Theorem is called the average value of f in the interval [a, b].
It is a generalization of the arithmetic mean of a discrete and finite set of numbers. We thus have
the following definition.
UP
Definition 4.7.5. If the function f is integrable on [a, b], the average value of f on [a, b] is
Rb
f (x) dx
fave = a
.
b−a
Example 4.7.6. The average value of f (x) = x on [a, b], where a < b, is
Z b
b2 a2
x dx
−
a
2 = b+a
= 2
b−a
b−a
2
We recover the usual average of a and b. This should motivate the following statement.
Remark 4.7.7. The average value of a function f on [a, b] works in a similar way with taking the
usual average of a (finite) set of numbers.
Example 4.7.8. The average value of f (x) = x2 on [0, 2] is
Z 2
x2 dx
8
4
0
= 3 = .
2−0
2
3
4.7. MEAN VALUE THEOREM FOR INTEGRALS
283
Example 4.7.9. A can of beans is being sold at P20 per piece. The cost of producing each can
follows the function C(x) = −x2 + 10x + 375. Find the average profit from selling 30 cans of beans.
Solution.
From the given information, we have p(x) = 20; furthermore, from the relation R(x) = xp(x), we
get R(x) = 20x. Thus,
P (x) = R(x) − C(x)
= 20x − (−x2 + 10x + 375)
= x2 + 10x − 375
em
at
ic
s
Thus, the average profit from selling 30 cans of beans, which is the average value of P (x) on [0, 30],
is
R 30
P (x) dx
Pave = 0
30 − 0
30
3
x
+ 5x2 − 375x
3
0
= 75
=
30
e
of
M
at
h
Thus, an average profit of P75 is gained from selling 30 cans of beans.
We end this part with a proof of the First Fundamental Theorem of the Calculus.
Z x
d
Proof. Let f be continuous on [a, b]. We wish to show that
f (t) dt = f (x).
dx a
Z x
Let F (x) =
f (t) dt. We have
tit
ut
a
F (x + ∆x) − F (x)
∆x→0
∆x
F (x + ∆x) − F (x)
. We have
∆x
Z x+∆x
UP
Consider the expression
In
s
F 0 (x) = lim
F (x + ∆x) − F (x)
=
∆x
Z x
f (t) dt −
a
f (t) dt
a
∆x
Z x+∆x
f (t) dx
=
x
∆x
Now, consider the function f (t) on [x, x + ∆x]. Since f is continuous, we use the Mean Value
Theorem for Integrals to conclude that there exists c ∈ [x, x + ∆x] such that
Z x+∆x
f (t) dt = f (c) [(x + ∆x) − x] = f (c)∆x
x
Thus,
Z x+∆x
F (x + ∆x) − F (x)
= lim
∆x→0
∆x→0
∆x
F 0 (x) = lim
f (t) dx
x
∆x
= lim
∆x→0
f (c)∆x
= f (c)
∆x
284
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
Note that c depends on x and ∆x. Finally, because x ≤ c ≤ x + ∆x and lim x = lim (x + ∆x) =
∆x→0
∆x→0
x, we conclude by the Squeeze Theorem that lim c = x. Hence, by continuity of f ,
∆x→0
0
F (x) = lim c = f
∆x→0
UP
In
s
tit
ut
e
of
M
at
h
em
at
ic
s
∆x→0
lim c = f (x)
4.7. MEAN VALUE THEOREM FOR INTEGRALS
4.7.1
285
Exercises
Exercises for Discussion
A. Do as indicated.
Z 2
3
x dx = , find the average value of f (x) = x in the interval [−1, 2]. Find
1. Given that
2
−1
the value of x ∈ [−1, 2] at which the average value occurs.
Z 3
f (x) dx = 4, show that f takes on the value of 2 at least once
2. If f is continuous and
1
on the interval.
3. Use Theorem 4.3.2 to prove the following inequalities:
Z 2
1
1
a.
dx ≤
2+4
x
2
0
Z 2
1
b. 0 ≤
sin πx dx ≤ 2
2
0
1
c. ≤
2
Z 3
4
dx ≤ 4
3+5
x
−1
em
at
ic
s
Z ln 9+1
d. 0 ≤
sinh3
1
x2 − 1
64
dx ≤
2
27 ln 9
at
h
4. Find the value of b such that the average value of g(x) = 3x2 − 4x − 1 on [0, b] is equal to
2.
e
of
M
B. Find a closed interval containing the definite integral of the following functions on the corresponding closed interval:
UP
In
s
tit
ut
1. f (x) = 4x3 + 3x2 − 2x − 1 on [−1, 2]
6
2. g(x) =
on [−4, −2]
(x + 1)2
3. f (x) = 3x2 + sin(x) on [0, π]
1
4. g(x) = 2
on [0, 1]
x +1
1
on [3, 9]
x−2
√
6. h(x) = 16 − x2 on [−2, 4]
x2 + 5
1 7
7. h(x) =
on
,
x−4
2 2
5. g(x) =
C. Find the average value of the following functions on the corresponding interval, and find
a value of c in the interval which satisfies the conclusion of the Mean Value Theorem for
Integrals:
1. f (x) = 6x2 − 4x + 1 on [−2, 1]
2. f (x) = x3 − 4x − 6 on [1, 5]
x+2
3. f (x) =
on [4, 6]
x−3
1
4. f (x) = 2
on [1, 3]
x + 2x + 1 1+π
1
5. f (x) = sin(3x − 1) on
,π +
3
3
6. f (x) = sec2 3x on
π
0,
18
√
√
csc(π x) cot(π x)
1 1
√
7. f (x) =
on
,
16 9
x
1
8. f (x) = p
√ on [16, 64]
9− x
286
CHAPTER 4. INTEGRATION AND ITS APPLICATIONS
Supplementary Exercises
A. Find a closed interval containing the definite integral of the following functions on the corresponding closed interval:
1. f (x) = 12 + 3x2 + x3 on [−1, 5]
5. g(x) =
2. g(x) = ln(x + 1) on [0, e3 − 1]
1
x2 − 3
on [3, 9]
√
16 − x2
on [2, 4]
x
2 − 3x
7. h(x) =
on [−3, 3]
x+4
π
3. f (x) = sin−1 (x) + on [−1, 1]
2
6. h(x) =
4. g(x) = cosh−1 (x) on [1, 3]
B. Find the average value of the following functions on the corresponding interval, and find
a value of c in the interval which satisfies the conclusion of the Mean Value Theorem for
Integrals:
9. f (x) = ex on [−3, 1]
2. f (x) = x2 − 3x − 9 on [−3, 3]
10. f (x) = 4x − 2x on [−3, log2 5]
3. f (x) = 2x2 − 4x + 3 on [−2, 4]
11. f (x) = 2x + 2−x on [−4, −1]
4. f (x) = 2x3 + 3x2 − 3 on [−2, 0]
12. f (x) = 4x − 3 on [−2, 2]
5. f (x) = 4x3 + 6x2 + 4x − 1 on [−3, 2]
x
on [1, 5]
6. f (x) = √
24 + x2
1
on [0, 2]
7. f (x) = √
16 − x2
1
8. f (x) = √
on [−4, 4]
9 + x2
13. f (x) = sinh(x) on [0, 1]
of
M
at
h
em
at
ic
s
1. f (x) = 3x − 12 on [−3, 7]
14. f (x) = x cosh(x2 ) on [0, 1]
16. f (x) = sech(2x) on [− ln 3, ln 5]
sinh−1 (x)
17. f (x) = √
on [0, 1]
x2 + 1
UP
In
s
tit
ut
e
15. f (x) = π tanh2 x on [0, ln 3]
C. Solve the following problems completely.
1. A container containing 1,000L of some liquid loses half its contents every 8 hours. Find
the average volume of the liquid during a 24-hr period.
2. On a given day on Planet E, which has a 24-hr day, the heat index (in ◦ C) was determined
to be modeled by the equation
T (x) = −0.15x2 + 0.4x + 37
where x is the number of hours from 12nn. Calculate the average heat index on Planet E
from 2 am to 10pm.
(4 pts)
4.7. MEAN VALUE THEOREM FOR INTEGRALS
287
Reviewer
I. Evaluate the following integrals.
Z 2
Z
|4x − 1| dx
1.
4.
0
√
Z
2.
√
sech x tanh x + 6 sinh x
dx
cosh x
Z e 23
2sec x
√
√
√ dx
x cot( x) cos( x)
5.
1
e2
Z 3π
(x + 1)2
√
dx
1−x
Z ln(x)+1 p
II. Let F (x) =
1 + t4 dt.
Z
3.
2
6.
π
1
p
dx
x 2 ln x − (ln x)2
sin θ
dθ
cos2 θ + 2 cos θ + 2
1
1. Use the First Fundamental Theorem of Calculus to determine F 0 (x).
em
at
ic
s
2. Find an equation of the tangent line to the graph of F at x = 1.
of
M
at
h
III. After ending their friendship, Al threw Vin’s gift vertically from an elevation of 27 meters
atop a building. Unbeknownst to Al, Vin is at the foot of the building and reaches out to
catch the gift so he could return it ot Al. If Vin’s outsretched arms are 2 meters high and
the gift reached its maximum height after 2 seconds, when will Vin catch the gift? (Use -10
m/s2 for the acceleration due to gravity.)
y
(2, 4)
In
s
tit
ut
e
IV. Let R be the region bounded by x = log2 y, y = 4, and y = (x − 1)2 as shown below.
UP
(3, 4)
R
(0, 1)
x
SET-UP a definite integral or a sum of definite integrals equal to:
1. the area of R using vertical rectangles
2. the perimeter of R
3. the volume of the solid generated when R is revolved abot the line x = 3 using the
method of washers
Bibliography
[1] H. Anton, I. Bivens, S. Davis, Calculus: Early Transcendentals, John Wiley and
Sons, 7th Edition, 2002.
[2] M. Castillo, F. Reyes, F. Cejalvo, J. Tangco, College Algebra and Trigonometry, National Book Store, 2008.
em
at
ic
s
[3] L. Leithold, College Algebra and Trigonometry, Addison Wesley Longman Inc.,
1989, reprinted-Pearson Education Asia Pte. Ltd, 2002.
[4] L. Leithold, The Calculus 7, Harpercollins College Div., 7th edition, December 1995.
UP
In
s
tit
ut
e
of
M
at
h
[5] J. Stewart, The Calculus: Early Transcendentals, Brooks/Cole, 6th Edition, 2008.
288
Institute of Mathematics, College of Science
University of the Philippines Diliman