Diploma Program
MA1131 Mathematics 1A
CALCULUS NOTES
CRICOS Provider No: 00098G c©2020 UNSW Sydney

iii
Preface
Please read carefully.
These Notes form the basis for the calculus strand of MA1131. However, not all of the material in
these Notes is included in the MA1131 calculus syllabus. In particular, any material marked [X]
is non-assessable. A detailed syllabus is given, commencing on page (ix) of these Notes.
In using these Notes, you should remember the following points:
1. Most courses at university present new material at a faster pace than you will have been
accustomed to in high school, so it is essential that you start working right from the beginning
of the session and continue to work steadily throughout the session. Make every effort to keep
up with the lectures and to do problems relevant to the current lectures.
2. These Notes are not intended to be a substitute for attending lectures or tutorials. The
lectures will expand on the material in the notes and help you to understand it.
3. These Notes may seem to contain a lot of material but not all of this material is equally
important. One aim of the lectures will be to give you a clearer idea of the relative importance
of the topics covered in the Notes.
4. Use the tutorials for the purpose for which they are intended, that is, to ask questions about
both the theory and the problems being covered in the current lectures.
5. Some of the material in these Notes is more difficult than the rest. This extra material is
marked with the symbol [H].
6. Some of the problems are marked [V]. These have a video solution available from Moodle.
7. It is essential for you to do problems which are given at the end of each chapter. If you
find that you do not have time to attempt all of the problems, you should at least attempt
a representative selection of them. The problems set in tests and exams will be similar to
the problems given in these notes. Further information on the problems and class tests is on
page (xiii).
8. You will be expected to use the computer algebra package Maple in tests and understand
Maple syntax and output for the end of semester examination.
Note.
We gratefully acknowledge the contributions of the School of Mathematics and Statistics towards
the creation of this resource. Copyright is vested in The University of New South Wales, c©2020.
c©2020 School of Mathematics and Statistics, UNSW Sydney
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c©2020 School of Mathematics and Statistics, UNSW Sydney
vContents
Preface iii
Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ix
Syllabus for MA1131 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ix
Homework schedule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xi
Tutorial schedule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xii
Test schedule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiii
Revision questions 1
1 Sets, inequalities and functions 11
1.1 Sets of numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.2 Solving inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.3 Absolute values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
1.4 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
1.5 Polynomials and rational functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
1.6 The trigonometric functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
1.7 The elementary functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
1.8 Implicitly defined functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
1.9 Continuous functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
1.10 Maple notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
Problems for Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
2 Limits 39
2.1 Limits of functions at infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
2.1.1 Basic rules for limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
2.1.2 The pinching theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
2.1.3 Limits of the form f(x)/g(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
2.1.4 Limits of the form
√
f(x)−
√
g(x) . . . . . . . . . . . . . . . . . . . . . . . . 42
2.1.5 Indeterminate forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
2.2 The definition of lim
x→∞ f(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
2.3 Proving that lim
x→∞ f(x) = L using the limit definition . . . . . . . . . . . . . . . . . . 47
2.4 Proofs of basic limit results [X] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
2.5 Limits of functions at a point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
2.5.1 Left-hand, right-hand and two-sided limits . . . . . . . . . . . . . . . . . . . . 51
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2.5.2 Limits and continuous functions . . . . . . . . . . . . . . . . . . . . . . . . . 53
2.5.3 Rules for limits at a point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
2.6 Maple notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
Problems for Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
3 Properties of continuous functions 63
3.1 Combining continuous functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
3.2 Continuity on intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
3.3 The intermediate value theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
3.4 The maximum-minimum theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
Problems for Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
4 Differentiable functions 75
4.1 Gradients of tangents and derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
4.2 Rules for differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
4.3 Proofs of differentiation rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
4.4 Implicit differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
4.5 Differentiation, continuity and split functions . . . . . . . . . . . . . . . . . . . . . . 86
4.6 Derivatives and function approximation . . . . . . . . . . . . . . . . . . . . . . . . . 89
4.7 Derivatives and rates of change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
4.8 Local maximum, local minimum and stationary points . . . . . . . . . . . . . . . . . 91
4.9 Maple notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
Problems for Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
5 The mean value theorem and applications 97
5.1 The mean value theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
5.2 Proof of the mean value theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
5.3 Proving inequalities using the mean value theorem . . . . . . . . . . . . . . . . . . . 100
5.4 Error bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
5.5 The sign of a derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
5.6 The second derivative and applications . . . . . . . . . . . . . . . . . . . . . . . . . . 104
5.7 Critical points, maxima and minima . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
5.8 Counting zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
5.9 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
5.10 L’Hoˆpital’s rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
Problems for Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
6 Inverse functions 123
6.1 Some preliminary examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
6.2 One-to-one functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
6.3 Inverse functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
6.4 The inverse function theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
6.5 Applications to the trigonometric functions . . . . . . . . . . . . . . . . . . . . . . . 132
6.6 Maple notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
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Problems for Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
7 Curve sketching 143
7.1 Curves defined by a Cartesian equation . . . . . . . . . . . . . . . . . . . . . . . . . 143
7.1.1 A checklist for sketching curves . . . . . . . . . . . . . . . . . . . . . . . . . . 144
7.1.2 Oblique asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
7.1.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
7.2 Parametrically defined curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
7.2.1 Parametrisation of conic sections . . . . . . . . . . . . . . . . . . . . . . . . . 151
7.2.2 Calculus and parametric curves . . . . . . . . . . . . . . . . . . . . . . . . . . 154
7.2.3 The cycloid and curve of fastest descent . . . . . . . . . . . . . . . . . . . . . 155
7.3 Curves defined by polar coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
7.3.1 Polar coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
7.3.2 Basic sketches of polar curves . . . . . . . . . . . . . . . . . . . . . . . . . . . 158
7.3.3 Sketching polar curves using calculus . . . . . . . . . . . . . . . . . . . . . . . 161
7.4 Maple notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
Problems for Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
8 Integration 171
8.1 Area and the Riemann Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
8.1.1 Area of regions with curved boundaries . . . . . . . . . . . . . . . . . . . . . 171
8.1.2 Approximations of area using Riemann sums . . . . . . . . . . . . . . . . . . 173
8.1.3 The definition of area under the graph of a function and the Riemann integral178
8.2 Integration using Riemann sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
8.3 The Riemann integral and signed area . . . . . . . . . . . . . . . . . . . . . . . . . . 183
8.4 Basic properties of the Riemann integral . . . . . . . . . . . . . . . . . . . . . . . . . 184
8.5 The first fundamental theorem of calculus . . . . . . . . . . . . . . . . . . . . . . . . 186
8.6 The second fundamental theorem of calculus . . . . . . . . . . . . . . . . . . . . . . 190
8.7 Indefinite integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
8.8 Integration by substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196
8.9 Integration by parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200
8.10 Improper integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202
8.11 Comparison tests for improper integrals . . . . . . . . . . . . . . . . . . . . . . . . . 205
8.12 Functions defined by an integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212
8.13 Maple notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
Problems for Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216
9 The logarithmic and exponential functions 223
9.1 Powers and logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224
9.2 The natural logarithm function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225
9.3 The exponential function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228
9.4 Exponentials and logarithms with other bases . . . . . . . . . . . . . . . . . . . . . . 230
9.5 Integration and the ln function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231
9.6 Logarithmic differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233
9.7 Indeterminate forms with powers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233
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Problems for Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235
10 The hyperbolic functions 239
10.1 Hyperbolic sine and cosine functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 240
10.2 Other hyperbolic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243
10.3 Hyperbolic identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245
10.4 Hyperbolic derivatives and integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 246
10.5 The inverse hyperbolic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248
10.6 Integration leading to the inverse hyperbolic functions . . . . . . . . . . . . . . . . . 251
10.7 A summary of important hyperbolic formulae . . . . . . . . . . . . . . . . . . . . . . 252
10.8 Appendix: The catenary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253
10.9 Maple notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256
Problems for Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258
Answers to selected problems 261
Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261
Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262
Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262
Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263
Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263
Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264
Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265
Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266
Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268
Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268
Past Class Tests 269
Index 283
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CALCULUS SYLLABUS FOR MA1131 MATHEMATICS 1A
The Calculus textbook is S.L. Salas & E. Hille and G.J. Etgen Calculus - One and Several Vari-
ables, any recent edition, Wiley. References to the 10th and 9th editions are shown as SH10 and
SH9. To improve your understanding of definitions, theorems and proofs, the following book is
recommended: Introduction to Proofs in Mathematics, J. Franklin & A. Daoud, Prentice-Hall.
In this syllabus the references to the textbook are not intended as a definition of what you will be
expected to know. They are just a guide to finding relevant material. Some parts of the course are
not covered in the textbook and some parts of the textbook (even in the sections mentioned in the
references below) are not included in the course. The scope of the course is defined by the content
of the lectures and problem sheets. The approximate lecture time for each section is given below.
References to the 9th and 10th editions of Salas & Hille are shown as SH9 and SH10.
SH10 SH9
1. Sets, inequalities and functions. (2 hours)
N,Z,Q,R. Open and closed intervals. Inequalities. 1.2, 1.3 1.2, 1.3
Functions: sums, products, quotients, composites.
Polynomials, rational functions, trig functions as
examples of continuous functions.
Implicitly defined functions. 1.6-1.7 1.6-1.7
2. Limits. (2 hours)
Formal definition of limit as x→ a (a finite). 2.1, 2.2 2.1, 2.2
Formal definition of limit as x→∞. pp177-178 pp222-224
pp195-198 pp243-246
Limit rules. The pinching theorem. 2.3, 2.5 2.3, 2.5
3. Properties of continuous functions. (2 hours)
Combinations of continuous functions. 2.4 2.4
Intermediate value and min-max theorems. 2.6, B1, B2 2.6, B1, B2
Relative and absolute maxima and minima. 4.3-4.5 4.3-4.5
4. Differentiable functions. (2 hours)
Definition of derivative via tangents. 3.1 3.1
Derivatives of sums, products, quotients and
composites. Rates of change. Higher derivatives. 3.2-3.5 3.2-3.5
Derivatives of polynomial, rational and trig functions. 3.5,3.6 3.5,3.6
Implicit differentiation, fractional powers. 3.7 3.7
5. The mean value theorem and applications. (2 hours)
Mean value theorem and applications. 4.1, 4.2 4.1, 4.2
L’Hoˆpital’s rule. 11.5, 11.6, 10.5, 10.6
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6. Inverse functions. (2 hours)
Domain, range, inverse functions,
the inverse function theorem. 7.1, B3 7.1, B3
Inverse trig functions, their derivatives and graphs. 7.7 7.7
7. Curve sketching. (3 hours)
Use of domain, range, intercepts, asymptotes,
even or odd, calculus. 4.7, 4.8 4.7, 4.8
Parametrically defined curves.
Relation between polar and Cartesian coordinates. 10.2 9.3
Sketching curves in polar coordinates. 10.3 9.4
8. Integration. (5 hours)
Riemann sums, the definite integral and its
algebraic properties. 5.1, B5 5.1, B5
Indefinite integrals, primitives and the
two fundamental theorems of calculus. 5.2-5.5 5.2-5.5
Integration by substitution and by parts. 5.6, 8.2 5.6, 8.2
Integrals on unbounded domains, limit form of
comparison test. 11.7 10.7
9. Logarithms and exponentials. (2 hours)
ln as primitive of 1/x, basic properties,
logarithmic differentiation. 7.2, 7.3 7.2, 7.3
Exponential function as inverse of ln, basic properties.
ax, logs to other bases. 7.4-7.6 7.4-7.6
10. Hyperbolic functions (2 hours)
Definitions, identities, derivatives, integrals
and graphs. 7.8 7.8
Inverse hyperbolic functions. 7.9 7.9
Integrals involving hyperbolic or trig substitution.
Review. (2 hour)
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CALCULUS PROBLEM SETS
The Calculus problems are located at the end of each chapter of the Calculus Notes booklet. They
are also available from the course module on the UNSW Moodle server. Some of the problems are
very easy, some are less easy but still routine and some are quite hard. To help you decide which
problems to try first, each problem is marked with an [R] or an [H]. The problems marked [R]
form a basic set of problems which you should try first. Problems marked [H] are harder and can
be left until you have done the problems marked [R]. Problems marked [V] have a video solution
available on Moodle.
You do need to make an attempt at the [H] problems because problems of this type will occur on
tests and in the exam. If you have difficulty with the [H] problems, ask for help in your tutorial.
Remember that working through a wide range of problems is the key to success in mathematics.
WEEKLY CALCULUS SCHEDULE
Solving problems and writing mathematics clearly are two separate skills that need to be devel-
oped through practice. We recommend that you keep a workbook to practice writing solutions to
mathematical problems. The following table gives the range of questions suitable for each week.
In addition it suggests specific recommended problems to do before your classroom tutorials.
The Online Tutorials will develop your problem solving skills, and give you examples of mathemat-
ical writing. Because this overlaps with the skills developed through homework, there are fewer
recommended homework in Online Tutorial weeks.
WEEKLY CALCULUS HOMEWORK SCHEDULE
Week Try to do up to Recommended Homework
Chapter Problem Problems
1 1. Sets, inequalities. . . 19 4(e), 5(d), 10(h), 12, 13(e), 15, 17
2 2. Limits 13 1(e), 2(b), 3(b), 5, 10(b)
3 3. Properties of continuous. . . 9 3, 5, 8(a), 8(c), 9(a), 9(b)
4 4. Differentiable functions 16 2(d), 6(d), 7(b), 10(a), 15
5 5. The MVT and applications 24 1(b), 4(b), 7(a), 9(b), 13, 17(c), 18(d)
6 6. Inverse functions 17 1, 5, 8(b), 8(d), 8(f), 11(b)
8 7. Curve sketching 11 2(b), 7(c), 8(b)
9 7. Curve sketching 19 14(c), 16(b)
8. Integration 2 1
10 8. Integration 15 4(b), 11(b), 14(d), 15(a)
11 8. Integration 27 16(b), 17(d), 20(b), 22(a)
12 9. Logs and exponentials 10 3(b), 4(e), 5(a), 8(c), 9(e), 9(h)
12 10. Hyperbolic functions 12 2(b), 3(a), 7(c), 8
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The main reason for having tutorials is to give you a chance to tackle and discuss problems which you
find difficult or don’t fully understand.
There are two kinds of tutorials: Online and Classroom. Calculus Online Tutorials are delivered using
MapleTA. These are optional and can be completed from home. Calculus Classroom Tutorials are
delivered in a classroom by a calculus tutor. The topics covered in a classroom tutorial are flexible,
and you can (and should) ask your tutor to cover any topics you find difficult. You may also be asked
to present solutions to homework questions to the rest of the class.
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xiii
CALCULUS CLASS TESTS
The tests will take place in class test times. The following table shows the week in which each
test will be held and the topics covered.
Topics covered
Test Week chapter sections
1 6 1, 2, 3, 4 & 5 All
2 11 6 , 7 & 8 All
It is important to note that:
• The class tests do not cover the whole syllabus.
• Questions in the exams may be very different from those in the class tests.
c©2020 School of Mathematics and Statistics, UNSW Sydney
xiv
c©2020 School of Mathematics and Statistics, UNSW Sydney
1Revision questions
Inequalities and Absolute Values
1. Sketch the set of points (x, y) which satisfy the following relations.
a) 0 ≤ y ≤ 2x and 0 ≤ x ≤ 2 b) y/2 ≤ x ≤ 2 and 0 ≤ y ≤ 4
2. Solve
a) x(x− 1) > 0 b) (x− 1)(x− 2) < 0 c) 2x2 + 3x− 2 ≥ 0
d)
1
x
> −1
2
e)
1
1− x >
1
2
f)
4
x− 1 ≤ 1
3. Solve
a)
∣∣x+ 1∣∣< 3 b) ∣∣x+ 2∣∣> 3 c) ∣∣3x+ 2∣∣≤ 1 d) ∣∣∣∣x− 1x+ 1
∣∣∣∣ < 1
Trigonometry
4. Find the exact value of each of the following:
a) cos
( π
12
)
b) sin
(
5π
12
)
c) tan
(
7π
12
)
d) sec
(
11π
12
)
5. If A and B are acute with sin(A) =
3
5
and tan(B) =
12
5
find (without the use of a calculator):
a) cos(A) b) tan(A) c) sin(B) d) cos(B)
e) sin(A+B) f) cos(A−B) g) sin(2A) h) tan(2B)
6. If A and B are acute with sin(A) =
24
25
and cos(B) =
8
17
find (without finding A and B):
a) cos(2A) b) sin(A−B) c) tan(A+B)
7. Find the period and amplitude for each of the following functions.
a) y = 3 sin
(
2x− π
4
)
b) y = −2 cos
(x
3
+
π
2
)
c©2020 School of Mathematics and Statistics, UNSW Sydney
2 REVISION QUESTIONS
8. Express each of the following in terms of a single sine function in the form R sin(x±α), where
R > 0 and α is acute.
a) sin(x) + cos(x) b) 2 sin(x) + 2
√
3 cos(x)
c)
√
3 sin(x)− cos(x) d) √8 sin(x)−√8 cos(x)
Functions
9. What is the (maximal) domain and range of the following functions?
a) f(x) =
√
5− x2 b) f(x) = √x2 − 5
c) f(x) =
√
x− 1 d) f(x) = 1√
x− 1
e) f(x) = (x− 8)−1/3 f) f(x) = √sinx
g) f(x) = 1 + tan2 x h) f(x) =


cos x if x < 0√
1− x if 0 ≤ x ≤ 1
|x| if x > 1
10. Sketch the graph of each of the functions in Problem 9.
11. Sketch each of the following functions without using calculus.
a) An odd function, f(x), defined on [−2, 2] such that
f(x) = x2(1− x) when 0 ≤ x ≤ 2.
b) An even function, f(x), defined on [−3, 3] such that
f(x) = (x− 1)2(x− 2) when 0 ≤ x ≤ 3.
12. If f(x) = x+ 5 and g(x) = x2 − 3 find
a) g(f(0)) b) g(f(x)) c) f(g(2)) d) f(g(x))
13. If f(x) = x− 1 and g(x) = 1√
x− 1, give the explicit forms of
a) f(x) + g(x) b) f(x)g(x) c)
f(x)
g(x)
d) f(g(x)) e) g(f(x))
Limits of some Rational Functions
14. Find
a) lim
x→2
x− 2
x2 − 5x+ 6 b) limx→2
x2 − 5x+ 6
2x2 − 3x− 2 c) limλ→1
λ2 − 0.8λ− 0.2
λ− 1
d) lim
x→1
1− x4
1− x e) limx→∞
2x2 − 3x+ 7
3x2 + x− 1 f) limx→∞
2x3 + 3x+ 2
−5x3 + 4x− 1
c©2020 School of Mathematics and Statistics, UNSW Sydney
REVISION QUESTIONS 3
Simple Differentiation
15. Find the derivative of each of the following functions.
a) f(x) = (2x+ 5)3 b) g(t) =
√
t2 − 4 c) h(x) = 1
(2x+ 3)3/2
d) f(x) = sin3 x e) g(x) = cos(x3) f) h(x) = sec(2x2 + 3)
g) f(x) = e−x2/2 h) g(x) = x2(2x− 1)4 i) h(θ) = θ tan θ
j) f(x) = x cos 2x k) g(x) = x3 sinx l) h(x) = x lnx
m) f(x) =
x+ e
x+ π
n) g(x) =
2x2 + 3
3x− 2 o) h(t) =
t√
t2 − 4
p) f(x) =
sinx
2x+ 5
Tangents and Normals
16. Find the equation of the tangent and the equation of the normal to each of the following
curves.
a) y = 4x+
1
x
at the point (1, 5)
b) y = x3 − 1 + 1
x2
at the point (1, 1)
c) y =
cos x
1− sinx at the point where x =
π
6
Stationary Points
17. Locate and identify the stationary points for
a) y = 2x3 − 9x2 + 12x− 3 b) y = x
1 + x2
c) y = e2x(1− x) d) y = xe−x
e) y = xne−x for n ∈ Z, n ≥ 2 f) y = lnx
x
g) y = 4x3 − x4 h) y = x+ cos x
18. The slope of the curve y = f(x) is given by
dy
dx
= x2(2x− 1)(x− 1)
Determine the nature of the stationary points.
19. The slope of the curve y = f(x) is
dy
dx
= 3(x− 1)2(x− 2)3(x− 3)4(x− 4)
c©2020 School of Mathematics and Statistics, UNSW Sydney
4 REVISION QUESTIONS
For what value or values of x does y have
a) a local maximum? b) a local minimum?
Integration
20. a) Use your answer to 15(i) to find a primitive function of θ sec2 θ.
Hint: From tables
∫
tan θdθ = ln | sec θ|+ C.
b) Use your answer to 15(j) to find a primitive function of x sin 2x.
c) Use your answer to 15(l) to find a primitive function of lnx.
21. The curve y = f(x) has
dy
dx
= 3x2 − 2x+ 1 and passes through the point (2, 3). Find f(x).
22. Find y where
a)
dy
dx
=
x2 + 1
x2
for x 6= 0 b) dy
dx
=
x+ 1√
x
for x > 0
23. Without recourse to tables find
a)
∫
ex dx b)
∫ 1
0
e3x dx
c)
∫ π
0
sin(2x) dx d)
∫
cos(3x) dx
e)
∫
(2x3 + 3x2 + 4x+ 5)dx f)
∫
1
3x+ 1
dx
g)
∫ −1
−2
1
2x− 3 dx h)
∫
(2x− 3)5 dx
For all the above indefinite integrals, check your answers by differentiating.
Integration by Substitution
24. Evaluate each of the following indefinite integrals by using the suggested substitution:
c©2020 School of Mathematics and Statistics, UNSW Sydney
REVISION QUESTIONS 5
a)
∫
x2
(
x3 + 1
)5
dx; u = x3 + 1
b)
∫
(t− 1)
√
t2 − 2t+ 4 dt; u = t2 − 2t+ 4
c)
∫
(x+ 1) ex
2+2x+3 dx; u = x2 + 2x+ 3
d)
∫
x sin
(
x2 + 1
)
dx; u = x2 + 1 e)
∫
esin 2x cos 2x dx; u = sin 2x
f)
∫
e2x cos
(
e2x
)
dx; u = e2x g)
∫
dz
z ln z
; u = ln z
h)
∫
x+ 1
x2 + 2x− 1 dx; u = x
2 + 2x− 1 i)
∫
ex
1 + ex
dx; u = 1 + ex
j)
∫
x+ 1
(x2 + 2x− 1)5 dx; u = x
2 + 2x− 1 k)
∫
sin(lnx) dx
x
; u = lnx
25. Evaluate each of the following definite integrals by using the suggested substitution:
a)
∫ 4
0
xex
2+1 dx; u = x2 + 1 b)
∫ π/4
π/6
sec2 x
tan x
dx; u = tanx
c)
∫ 1
0
3x
(3x+ 1)2
dx; u = 3x+ 1 d)
∫ 20
5
t√
t− 4 dt; u = t− 4
Area and Volume
26. For each of the following functions, find the area between the curve y = f(x) and the x-axis
over the given range of x values.
a) f(x) = 2x2 − 1 from x = 1 to 2 b) f(x) = x3 − 3x2 + 4x from x = 0 to 2
c) f(x) = 2x2 +
1
x2
from x = 1 to 2 d) f(x) = e−x/3 from x = 0 to 3
e) f(x) = 2 cos x+ 3 from x = 0 to π f) f(x) =
1
x+ 1
from x = 0 to 2
27. For each of the following functions, find the volume of the solid formed when the curve
y = f(x) over the given range of x is rotated about the x-axis.
a) f(x) = x2 + 1 from x = 0 to 1 b) f(x) = x+
2
x
from x = 1 to 2
c) f(x) = e−x/4 from x = 0 to 2 d) f(x) = sec x from x = 0 to
π
4
e) f(x) =
1
x+ 1
from x = 0 to 1
Logarithms
28. Simplify:
c©2020 School of Mathematics and Statistics, UNSW Sydney
6 REVISION QUESTIONS
a) log4 12− log4 3 b)
log2 16
log2 8
c) log1/3 729
29. Solve for x:
a) 22x+1 − (17)2x + 8 = 0 b) lnx = 3 ln 2 + 2 ln 3 c) logx 125 = −3
Remainder Theorem
30. Without division find the remainder when p(x) = x3 − 5x2 + 10x− 6 is divided by
a) x− 2 b) x− 1 c) x+ 2 d) x+ 1 which (if any) of these is a factor of
p(x)?
Binomial Theorem
31. Use Pascal’s triangle to expand the following:
a) (x+ y)5 b) (3x− 2y)4 c) (2x+ 3)6
32. Use the Binomial Theorem to find the following.
a) The coefficient of x12 in the expansion of (2x3 − 3)7.
b) The coefficient of x3 in the expansion of
(
x2 − 2
x
)3
.
c) The term independent of x in the expansion of
(
2x2 +
1
x
)9
.
c©2020 School of Mathematics and Statistics, UNSW Sydney
REVISION QUESTIONS 7
Answers for Revision Questions
1. Answer for both: the interior and boundary of the triangle with vertices at (0, 0),
(2, 0), and (2, 4).
2. a) x < 0 or x > 1 b) 1 < x < 2 c) x ≤ −2 or x ≥ 1
2
d) x < −2 or x > 0 e) −1 < x < 1 f) x < 1 or x ≥ 5
3. a) −4 < x < 2 b) x < −5 or x > 1 c) −1 ≤ x ≤ −1
3
d) x > 0
4. a) 14
√
2
(
1 +
√
3
)
b) 14
√
2
(
1 +
√
3
)
c) − (2 +√3) d) −√2 (√3− 1)
5. a)
4
5
b)
3
4
c)
12
13
d)
5
13
e)
63
65
f)
56
65
g)
6
13
h) −120
119
6. a) −527
625
b)
87
425
c) −297
304
7. a) amplitude = 3, period = π b) amplitude = 2, period = 6π
8. a)
√
2 sin
(
x+
π
4
)
b) 4 sin
(
x+
π
3
)
c) 2 sin
(
x− π
6
)
d) 4 sin
(
x− π
4
)
9. a) −√5 ≤ x ≤ √5; 0 ≤ y ≤ √5 b) x ≤ −√5 or x ≥ √5; y ≥ 0
c) x ≥ 1; y ≥ 0 d) x > 1; y > 0 e) x 6= 8; y 6= 0
f) {x : 2nπ ≤ x ≤ (2n + 1)π, n ∈ Z}; 0 ≤ y ≤ 1
g) {x : x 6= (2n + 1)π/2, n ∈ Z}; y ≥ 1 h) R; y ≥ −1
12. a) 22 b) x2 + 10x+ 22 c) 6 d) x2 + 2
13. a) x− 1 + 1√
x− 1 b)
√
x− 1 c) (x− 1)3/2 d) 1√
x− 1 − 1 e)
1√
x− 2
14. a) −1 b) −1
5
c) 1.2 d) 4 e)
2
3
f) −2
5
15. a) 6(2x+ 5)2 b)
t√
t2 − 4 c) −
3
(2x+ 3)5/2
d) 3 sin2 x cos x
e) −3x2 sin(x3) f) 4x sec(2x2 + 3) tan(2x2 + 3) g) −xe−x2/2
h) 2x(6x− 1)(2x − 1)3 i) θ sec2 θ + tan θ j) −2x sin 2x+ cos 2x
k) x2(x cos x+ 3 sin x) l) 1 + lnx m)
π − e
(x+ π)2
n)
6x2 − 8x− 9
(3x− 2)2
c©2020 School of Mathematics and Statistics, UNSW Sydney
8 REVISION QUESTIONS
o) − 4
(t2 − 4)3/2 p)
(2x+ 5) cos x− 2 sinx
(2x+ 5)2
16. a) y = 3x+ 2, x+ 3y = 16 b) y = x, x+ y = 2
c) y −√3 = 2
(
x− π
6
)
, y −√3 = −1
2
(
x− π
6
)
17. a) (1, 2) is a local maximum and (2, 1) is a local minimum
b) (1, 12 ) is a local maximum and (−1,−12 ) is a local minimum
c) (12 ,
e
2) is a local maximum
d) (1, e−1) is a local maximum
e)
(
n,
nn
en
)
is a local maximum and (0, 0) is a local minimum if n is even and a point of
inflection if n is odd
f) (e, e−1) is a local maximum
g) (3, 27) is a local maximum and (0, 0) is a point of inflection
h) (π2 + 2kπ,
π
2 + 2kπ) k ∈ Z are points of inflection
18. There is a point of inflection for x = 0, a local maximum for x =
1
2
, and a local minimum for
x = 1
19. a) x = 2 b) x = 4
20. a) θ tan θ − ln | sec θ| b) −1
2
x cos 2x+
1
4
sin 2x c) x lnx− x
21. f(x) = x3 − x2 + x− 3
22. a) y = x− 1
x
+ C b) y =
2
3
x3/2 + 2
√
x+ C
23. a) ex+C b)
1
3
(e3−1) c) 0 d) 1
3
sin(3x)+C e)
1
2
x4+x3+2x2+5x+C
f)
1
3
ln |3x+ 1|+C g) 1
2
ln
(
5
7
)
h)
1
12
(2x− 3)6 + C
24. a)
1
18
(
x3 + 1
)6
+ C b)
1
3
(
t2 − 2t+ 4) 32 + C c) 1
2
ex
2+2x+3 + C
d) −1
2
cos
(
x2 + 1
)
+ C e)
1
2
esin 2x + C f)
1
2
sin
(
e2x
)
+ C g) ln | ln z|+ C
h)
1
2
ln
∣∣x2 + 2x− 1∣∣+ C i) ln (1 + ex) + C j) − 1
8 (x2 + 2x− 1)4 + C
k) − cos(lnx) + C
25. a)
1
2
(
e17 − e) b) 1
2
ln 3 c)
2
3
ln 2− 1
4
d) 66
c©2020 School of Mathematics and Statistics, UNSW Sydney
REVISION QUESTIONS 9
26. a)
11
3
b) 4 c)
31
6
d) 3− 3
e
e) 3π f) ln 3
27. a)
28π
15
b)
25π
3
c) 2π
(
1− 1
e
)
d) π e)
π
2
28. a) 1 b)
4
3
c) −6
29. a) −1, 3 b) 72 c) 1
5
30. a) 2 b) 0 c) −54 d) −22; x− 1 is a factor.
31. a) x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
b) 81x4 − 216x3y + 216x2y2 − 96xy3 + 16y4
c) 64x6 + 576x5 + 2160x4 + 4320x3 + 4860x2 + 2916x + 729
32. a) −15120 b) −6 c) 672
c©2020 School of Mathematics and Statistics, UNSW Sydney
10 REVISION QUESTIONS
c©2020 School of Mathematics and Statistics, UNSW Sydney
11
Chapter 1
Sets, inequalities and functions
In the days of the Roman empire, the word ‘calculus’ denoted a pebble that was used for counting
and gambling. As time progressed, the word ‘calculare’ came to mean ‘to compute.’ In the second
half of the seventeenth century, two mathematicians, the Englishman Isaac Newton and the German
Gottfried Leibniz, independently invented methods for
• calculating gradients of tangents to curves,
• calculating instantaneous acceleration and velocity,
• calculating the area of regions with a curved boundary,
• calculating the volume of solids with curved boundary,
• calculating the length of a curve,
• calculating the work done by a force, and
• calculating the centre of mass of a general solid.
These methods were developed by combining algebra, geometry and trigonometry with the limiting
process and became known as the calculus.
Calculus (as it is known today) has many applications to engineering, physics, chemistry, bi-
ology, geology, surveying, sociology, economics and statistics. It includes two major branches: the
differential calculus (introduced in Chapter 4) and the integral calculus (introduced in Chapter 8).
These two branches are related by the fundamental theorem of calculus (see Theorem 8.5.1). The
underlying tool used to develop these branches is the concept of the limit (introduced in Chapter
2), which is applied to functions (introduced in Chapter 1).
The main goal of this chapter is to introduce functions. Although functions will be familiar to
students from high school, important points that deserve students’ attention are emphasised here.
Since the functions we study in this course take real numbers as inputs and give real numbers of
outputs, we begin the chapter with a review of different sets that consist of real numbers. We also
devote a little time to revising inequalities and absolute values, in preparation for our discussion
of limits in Chapter 2.
c©2020 School of Mathematics and Statistics, UNSW Sydney
12 CHAPTER 1. SETS, INEQUALITIES AND FUNCTIONS
1.1 Sets of numbers
(Ref: SH10 §1.2)
A set is a collection of distinct objects. The objects in a set are called the elements or members of
the set. Some commonly used sets of numbers are listed below.
• The set N of natural numbers is given by
N = {0, 1, 2, 3, 4, . . .}.
• The set Z of integers is given by
Z = {. . . ,−3,−2,−1, 0, 1, 2, 3, . . .}.
• The set Q of rational numbers is the collection of all numbers of the form pq , where p and q
are integers and q 6= 0.
• The set R of real numbers may be represented as the collection of points lying on the number
line.
| | | | | | |
−3 −2 −1 3210
If A is a set of numbers and the number x is a member of the set A, then we write
x ∈ A.
If x is not a member of A then we write
x /∈ A.
The next example illustrates this notation.
Example 1.1.1. The following statements are true:
2 ∈ N, −2 /∈ N, 3
4
/∈ Z, 3
4
∈ Q,
√
2 ∈ R,
√
2 /∈ Q.
Many other sets of numbers can be described using N, Z, Q, R and inequalities.
Example 1.1.2. (a) The set
{n ∈ Z : n > 0}
is the set of all numbers n belonging to the integers such that n is greater than 0. In other words,
it is the set of positive integers. (Note that the colon ‘:’ above is read as ‘such that’.)
(b) The set
{x ∈ R : −2 ≤ x < 3}
is represented by the following interval on the real line.
| | | | | | |
−3 −2 −1 3210
c©2020 School of Mathematics and Statistics, UNSW Sydney
1.1. SETS OF NUMBERS 13
(c) The set
{x ∈ R : x /∈ Q}
is the set of irrational numbers.
Sets that are represented by intervals of the real line occur so frequently that we introduce
a special notation for them. Suppose that a and b are real numbers and that a < b. Then the
intervals (a, b), [a, b], (a, b] and [a, b) are given by
• (a, b) = {x ∈ R : a < x < b}
a b
• [a, b] = {x ∈ R : a ≤ x ≤ b}
a b
• [a, b) = {x ∈ R : a ≤ x < b}
a b
• (a, b] = {x ∈ R : a < x ≤ b}.
a b
In each case, the numbers a and b are called the endpoints of the interval. An interval [a, b] that
includes its endpoints is called a closed interval, while an interval (a, b) that excludes its endpoints
is called an open interval. The intervals [a, b) and (a, b] are neither open nor closed.
Interval notation can be extended to describe rays of the real line by using the symbol ∞ for
infinity. Thus we have
• [a,∞) = {x ∈ R : a ≤ x}
a b
|
• (−∞, b) = {x ∈ R : x < b}
a b
|
• (−∞,∞) = R.
We end by introducing an important set relation.
c©2020 School of Mathematics and Statistics, UNSW Sydney
14 CHAPTER 1. SETS, INEQUALITIES AND FUNCTIONS
Definition 1.1.3. Suppose that A and B are two sets. We say that A is a subset
of B if x ∈ A implies that x ∈ B. If A is a subset of B then we also say that B
contains the set A.
The above definition is illustrated by the following examples:
• {0, 1, 2, 3} is a subset of {0, 1, 2, 3, 4},
• N is a subset of Z,
• [−1, 2] is a subset of (−3,∞),
• R is a subset of R, and
• [1, 4) is not a subset of (1, 4].
1.2 Solving inequalities
(Ref: SH10 §1.3)
When solving an inequality, one must remember that if the inequality is multiplied by a negative
number, the sign of the inequality reverses.
Two types of inequalities deserve special attention: quadratic inequalities and inequalities with
an unknown quantity in the denominator.
Example 1.2.1. Solve the inequality
x2 > 6− x.
Solution. Rearranging the inequality gives
x2 + x− 6 > 0.
The values of x for which x2 + x− 6 is positive are the same as those for which the graph of
y = x2 + x− 6
lies above the x-axis.
To graph
y = x2 + x− 6,
we factorise the right-hand side to obtain
y = (x− 2)(x+ 3).
Hence y = 0 when x = 2 and when x = −3. We can now easily sketch its graph.
c©2020 School of Mathematics and Statistics, UNSW Sydney
1.2. SOLVING INEQUALITIES 15
x
y
y = x2 + x− 6
−3 2
Since the graph lies above the x-axis when x < −3 and when x > 2, the solution to the inequality
is
x < −3 or x > 2.
Example 1.2.2. Solve the inequality
1
3
<
1
x− 1 .
Solution. We begin by noting that x 6= 1.
To solve the inequality, we cannot simply multiply the inequality through by x − 1, since we
do not know whether x− 1 is positive or negative. (Taking the reciprocal of both sides is similarly
problematic.) Instead, we multiply through by the positive number (x− 1)2 to obtain
1
3
(x− 1)2 < x− 1.
Multiplication by 3 and rearrangement gives
(x− 1)2 − 3(x− 1) < 0
and by factorisation we obtain
(x− 1)[(x− 1)− 3] = (x− 1)(x− 4) < 0.
So proceding in a manner similar to the previous example, we seek to identify the values of x for
which the graph of
y = (x− 1)(x− 4)
lies below the x-axis.
c©2020 School of Mathematics and Statistics, UNSW Sydney
16 CHAPTER 1. SETS, INEQUALITIES AND FUNCTIONS
x
y
y = (x− 1)(x − 4)
1 4
Hence the solution to the inequality is given by
1 < x < 4.
The next example is slightly more difficult, but uses techniques illustrated in the previous two
examples.
Example 1.2.3. Solve the inequality
x+ 2 ≥ 4
x− 1
Solution. First observe that x 6= 1.
Multiplying the inequality through by (x− 1)2 gives
(x− 1)2(x+ 2) ≥ 4(x− 1).
Hence
(x− 1)2(x+ 2)− 4(x− 1) ≥ 0.
Now
(x− 1)2(x+ 2)− 4(x− 1) = (x− 1)[(x − 1)(x+ 2)− 4]
= (x− 1)[x2 + x− 6]
= (x− 1)(x − 2)(x+ 3).
So the original inequality is equivalent to
(x− 1)(x− 2)(x + 3) ≥ 0, x 6= 1.
We now sketch the graph of
y = (x− 1)(x− 2)(x+ 3)
and see for which values of x the graph lies on or above the x-axis.
c©2020 School of Mathematics and Statistics, UNSW Sydney
1.3. ABSOLUTE VALUES 17
x
y
y = (x− 1)(x− 2)(x+ 3)
1 2−3
Hence the solution to the inequality is
−3 ≤ x < 1 or x ≥ 2.
Remark 1.2.4. A variety of methods (some good and others bad) exist for solving inequalities
where an unknown lies in the denominator. It is strongly recommended that students use the
method shown in this section, or that shown by their first year calculus lecturer.
1.3 Absolute values
(Ref: SH10 §§1.2, 1.3)
The magnitude, or absolute value, of a real number x is denoted by |x| and defined by
|x| =
{
x if x ≥ 0
−x if x < 0. (1.1)
Absolute values of real numbers interact in the following way.
Proposition 1.3.1. Suppose that x and y are real numbers. Then
(i) | − x| = |x|,
(ii) |xy| = |x||y|,
(iii)
∣∣∣∣xy
∣∣∣∣ = |x||y| , provided that y 6= 0, and
(iv) |x+ y| ≤ |x|+ |y|.
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18 CHAPTER 1. SETS, INEQUALITIES AND FUNCTIONS
Property (iv) is known as the triangle inequality.
Three other useful facts about absolute values are noted below.
(F1) For every real number x,
|x| =
√
x2 and |x|2 = x2. (1.2)
(F2) If a and x are real numbers then |x− a| is equal to the distance between x and a on the real
number line.
(F3) For any positive real number a,
|y| < a if and only if − a < y < a
and
|y| > a if and only if y < −a or y > a.
These facts give different methods for solving inequalities involving absolute values.
Example 1.3.2. Solve the following inequalities:
(a) |x− 3| ≤ 5
(b) |2x+ 5| > 8
(c)
∣∣∣∣x− 3x− 1
∣∣∣∣ < 1.
Solution. (a) An algebraic solution. Using (F3) we have
|x− 3| ≤ 5
−5 ≤ x− 3 ≤ 5
−2 ≤ x ≤ 8.
(a) A geometric solution. By (F2) the inequality |x− 3| ≤ 5 is equivalent to saying that
‘the distance between x and 3 is less than or equal to 5.’
The points x on the number line that have this property are shown below.
|
3−2 8
55
Hence −2 ≤ x ≤ 8.
(b) An algebraic solution. Using (F3) the inequality is equivalent to
2x+ 5 < −8 or 2x+ 5 > 8.
Hence
2x < −13 or 2x > 3,
which gives the solution
x < −13
2
or x >
3
2
.
c©2020 School of Mathematics and Statistics, UNSW Sydney
1.4. FUNCTIONS 19
(c) An algebraic solution. By Prop.1.3.1 (iii), we can write, for x 6= 1,∣∣∣∣x− 3x− 1
∣∣∣∣ = |x− 3||x− 1|
and since the denominator is positive, the inequality can be written as
|x− 3| < |x− 1|.
Squaring both (positive) sides),and using (F1), we have
(x− 3)2 < (x− 1)2.
Expanding and solving, we have x > 2.
(c) A geometric solution. As above, the inequality is equivalent to
|x− 3| < |x− 1|, x 6= 1.
By (F2) the inequality is interpreted geometrically as
‘the distance from x to 3 is less than the distance from x to 1.’
In other words, x is closer to 3 than to 1. Points that have this property are shown on the numberline
below.
| |
1 2 3
Hence x > 2.
1.4 Functions
(Ref: SH10 §1.5)
A function f : A→ B is a rule which assigns to every element x belonging to a set A exactly one
element f(x) belonging to a set B. The set A is called the domain of the function f and the set B
is called the codomain of f . In this course, A and B are always sets of real numbers.
Example 1.4.1. A function f : [0,∞)→ R is given by
f(x) =
√
x
for all x in [0,∞). Informally, this means that f takes a number x from [0,∞) as input and gives
the number
√
x as output. More formally, we say that f maps an element x of [0,∞) to √x. We
offer a few comments on terminology employed to describe f .
• The expression f : [0,∞)→ R says that [0,∞) is the domain of f and that R is the codomain
of f . We write
Dom(f) = [0,∞) and Codom(f) = R.
• The domain of f is the set of all inputs for the function.
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20 CHAPTER 1. SETS, INEQUALITIES AND FUNCTIONS
• The codomain of f is a set that contains all the output values of the function. Since the
output values are all real numbers, we say that f is a real-valued function. While all the
outputs must lie in Codom(f), not every number in Codom(f) need be an output value.
• The expression f(x) (read as ‘f of x’) is the value of f at the point x. That is, f(x) is the
unique number in R that corresponds to the input x. To emphasise the point: f(x) is a real
number, not a function.
• The statement
‘f(x) =
√
x for all x in [0,∞)’
can be abbreviated as
f(x) =
√
x ∀x ∈ [0,∞).
The symbol ∀ means ‘for all.’
In high school, functions are often described by specifying a rule without specifying a domain.
For example, the following sentence was taken from an HSC study guide:
‘Consider the function f(x) = x2.’
There are two problems with this sentence. First, neither f(x) nor f(x) = x2 is a function. As
mentioned above, f(x) is a real number, and f(x) = x2 is an equation. (This may seem picky,
but being able to distinguish between a function f and its value f(x) at the point x is crucial to
understanding key concepts in both algebra and calculus.) Second, the quoted sentence does not
specify the domain of the function. The next example gives two different functions that obey the
same squaring rule.
Example 1.4.2. Consider the functions f and g, defined by
f : R→ R
f(x) = x2 ∀x ∈ Dom(f) and
g : [0,∞)→ R
g(x) = x2 ∀x ∈ Dom(g).
Even though they are both squaring functions, f and g are not the same function. For example, f
can square negative numbers while g does not. On the other hand, g is invertible (this important
and desirable property is discussed in Chapter 6) while f is not. Their graphs, shown below, are
also noticeably different.
x
f(x)
x
g(x)
If, for whatever reason, the domain of a function is not specified, but the function rule is, then
the default domain, known as the maximal or natural domain, is the largest possible domain for
which the rule makes sense. We give two examples.
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1.4. FUNCTIONS 21
• If
f(x) =
√
x− 1
then the rule makes sense only if x− 1 ≥ 0, that is, if x ≥ 1. Therefore the maximal domain
of the function f is [1,∞).
• If
f(x) =
2
(x− 2)(x− 8)
then the rule only makes sense if x 6= 2 or x 6= 8. So the maximal domain is the set
{x ∈ R : x 6= 2, x 6= 8}.
Another set associated to a function is its range.
Definition 1.4.3. Suppose that f : A→ B is a function. The range of f , denoted
by Range(f), is defined by
Range(f) = {f(x) ∈ B : x ∈ A}.
In other words, the range of f is the set of all output values. Equivalently, if the rule of a
function is described by the equation y = f(x), then the range of f is the set of all corresponding
y-values. The range is always a subset of the codomain, but the range need not equal the codomain.
The difference between range and codomain is illustrated by the following examples.
Example 1.4.4. Suppose that f : R→ R is given by the rule
f(x) = sinx ∀x ∈ R.
Then Dom(f) = R and Codom(f) = R (this is obtained from the expression f : R → R). To
determine the range, consider the graph of f is sketched below.
x
f(x)
1
−1
The set of all output values (equivalently, the set of all corresponding y-values) is [−1, 1]. Hence
Range(f) = [−1, 1].
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22 CHAPTER 1. SETS, INEQUALITIES AND FUNCTIONS
Example 1.4.5. Suppose that f : [−1, 1]→ R is given by the rule
f(x) =
√
1− x2 ∀x ∈ [−1, 1].
Then Dom(f) = [−1, 1] and Codom(f) = R (again, we are simply reading this from f : [−1, 1] →
R). To determine the range, we observe that the graph of f is a semicircle.
x
f(x)
1
−1 1
The set of all outputs (or y-values) is [0, 1]. Hence Range(f) = [0, 1].
If f and g are two functions with the same domain, then one can combine f and g to form new
functions.
Definition 1.4.6. Suppose that f : A→ B and g : A→ B are real-valued functions.
Then the functions f + g, f − g, f.g and f/g are defined by the rules
(f + g)(x) = f(x) + g(x) ∀x ∈ A
(f − g)(x) = f(x)− g(x) ∀x ∈ A
(f.g)(x) = f(x)g(x) ∀x ∈ A
(f/g)(x) =
f(x)
g(x)
∀x ∈ A, provided that g(x) 6= 0.
Another way of constructing new functions is given below.
Definition 1.4.7. Suppose that f : C → D and g : A→ B are functions such that
Range(g) is a subset of Dom(f). Then the composition f ◦ g : A→ D is defined by
the rule
(f ◦ g)(x) = f(g(x)) ∀x ∈ A.
Example 1.4.8. Suppose that f : R→ R and g : R→ R are given by the rules
f(x) = x2 + 1 ∀x ∈ R and g(x) = x− 2 ∀x ∈ R.
Find (f + g)(5) and (f ◦ g)(x). Is 2 in the domain of the function f/g? Give reasons.
c©2020 School of Mathematics and Statistics, UNSW Sydney
1.5. POLYNOMIALS AND RATIONAL FUNCTIONS 23
Solution. First,
(f + g)(5) = f(5) + g(5)
= (52 + 1) + (5− 2)
= 29.
Second, if x in R then
(f ◦ g)(x) = f(g(x))
= (x− 2)2 + 1
= x2 − 4x+ 5.
Finally, g(2) = 0 so 2 is not in the domain of f/g.
The final example illustrates the importance of bearing in mind domain and range when com-
posing two functions.
Example 1.4.9. Suppose that f : [0,∞)→ R and g : R→ R are given by
f(x) =
√
x ∀x ∈ [0,∞) and g(x) = sinx− 2 ∀x ∈ R.
Find, if they exist, (f ◦ g)(x) and (g ◦ f)(x).
Solution. Consider first f ◦ g. For this function to exist, we require that Range(g) is a subset of
Dom(f). Now
−1 ≤ sinx ≤ 1 ∀x ∈ R
so
−3 ≤ sinx− 2 ≤ −1 ∀x ∈ R;
that is, Range(g) = [−3,−1]. Since Dom(f) = [0,∞), Range(g) is not a subset of Dom(f) and
hence f ◦ g does not exist. (The point here is that, if
(f ◦ g)(x) = f(g(x)) = √sinx− 2,
then the expression under the square root sign is always negative.)
Now consider g ◦f . Since f is a real-valued function and Dom(g) = R, it is clear that Range(f)
is a subset of Dom(g). (The fact that Range(f) = [0,∞) is not needed because Dom(g) is so large.)
Hence
(g ◦ f)(x) = g(f(x)) = sin(√x)− 2 ∀x ∈ R.
1.5 Polynomials and rational functions
(Ref: SH10 §1.6)
In the next three sections we discuss some important classes of functions. Since these were intro-
duced in high school, we give a brief survey only.
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24 CHAPTER 1. SETS, INEQUALITIES AND FUNCTIONS
Definition 1.5.1. A function f : R→ R is called a polynomial if
f(x) = anx
n + an−1xn−1 + · · · + a2x2 + a1x+ a0 ∀x ∈ R,
where n is a natural number, the constants a0, a1, . . . , an are real numbers and
an 6= 0. We call n the degree of f , and the constants a0, a1, . . . , an are called the
coefficients of f . The constant an is called the leading coefficient of f .
Polynomials of small degree have specials names and properties.
• If n = 0 then f(x) = a0. Hence f is a constant function and the graph of f is a horizontal
straight line.
• If n = 1 then f(x) = a1x+ a0. The graph of f is a straight line with gradient a1 and y-axis
intercept a0.
• If n = 2 then f is called a quadratic function and its graph is a parabola.
• If n is 3, 4 or 5 then f is called a cubic, quartic or quintic, polynomial respectively.
The general shape of the graph of a polynomial function of degree greater than 1 is indicated
by the table below.
Positive leading coefficient Negative leading coefficient
Odd degree
Even degree
The number of turning points in the graph is always strictly less than the degree of the polynomial.
Exactly how many turning points the graph has depends on the coefficients of the polynomial, and
can be determined using calculus (see, for example, Chapters 4 and 5).
By dividing two polynomials, one obtains a new type of function.
c©2020 School of Mathematics and Statistics, UNSW Sydney
1.6. THE TRIGONOMETRIC FUNCTIONS 25
Definition 1.5.2. Suppose that p and q are polynomials. A function f is called a
rational function if
Dom(f) = {x ∈ R : q(x) 6= 0}
and
f(x) =
p(x)
q(x)
, ∀x ∈ Dom(f).
Hence the expressions
f(x) =
x
x2 + 2
and f(x) =
x3 − 4x2 + 1
x− 2
both give rise to rational functions, as does
f(x) = x− 1 + 3
x2 + 3
(to see why, rewrite right-hand side with a common denominator of x2 + 3).
The graphs of two simple rational functions are shown below.
x
y
y = 1x
x
y
y = 1x2
The graphs of more sophisticated rational functions are studied in Chapter 7.
1.6 The trigonometric functions
(Ref: SH10 §1.6)
The trigonometric functions will already be familiar to you. This section gives a brief summary of
their definitions and some of their properties.
Remark 1.6.1. In this course, angles are always measured in radians rather than degrees. (Radian
measure is a more natural way of measuring angles; moreover, important formulae used in calculus,
such as
d
dθ
(sin θ) = cos θ,
only make sense geometrically when θ is interpreted as an angle measured in radians rather than
degrees.) We remind readers that
2π radians = 360 degrees.
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26 CHAPTER 1. SETS, INEQUALITIES AND FUNCTIONS
Consider an angle θ and the corresponding point P (x, y) that lies on the unit circle centred at
the origin, as shown below.
1
x
y
P (x, y)
θ
The functions cos : R→ R and sin : R→ R are defined by
cos θ = x and sin θ = y.
Other trig functions are defined in terms of the sine and cosine functions:
tan θ =
sin θ
cos θ
, provided that cos θ 6= 0
sec θ =
1
cos θ
, provided that cos θ 6= 0
cosec θ =
1
sin θ
, provided that sin θ 6= 0
cot θ =
cos θ
sin θ
, provided that sin θ 6= 0.
The graphs of sin, cos and tan are shown below.
θ
sin θ
π
2
π 3π
2
2π−2π
| | | | |
1
−1
|
|
θ
cos θ
π
2
π 3π
2
2π−2π
| | | | |
1
−1
|
|
c©2020 School of Mathematics and Statistics, UNSW Sydney
1.6. THE TRIGONOMETRIC FUNCTIONS 27
θ
tan θ
π
2
π 3π
2
2π−2π
| | | | |
1
−1
|
|
The sine and cosine functions are 2π-periodic, while the tangent function is π-periodic, which
means that
sin(θ + 2π) = sin θ, cos(θ + 2π) = cos θ and tan(θ + π) = tan θ
whenever θ is in the domain of the respective function. Sine and cosine both have an amplitude of
1.
It is not hard to obtain functions whose graphs have the same general shape as the sine (or
cosine) curve with a period other than 2π and amplitude other than 1. For example, if A and ω
are positive real numbers then the function f : R→ R, given by
f(x) = A sin(ωx) ∀x ∈ R,
has an amplitude of A and a period of 2πω . Its graph is shown below.
x
f(x)
2π
ω−2πω
| | | | |
A
−A
|
|
The six trigonometric functions are related to one another by various identities and formulae:
• complementary identities
sin
(π
2
− x
)
= cos x
cos
(π
2
− x
)
= sinx
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28 CHAPTER 1. SETS, INEQUALITIES AND FUNCTIONS
• Pythagorean identities
cos2 x+ sin2 x = 1
1 + tan2 x = sec2 x
cot2 x+ 1 = cosec2 x
• the sum and difference formulae
sin(x± y) = sinx cos y ± cos x sin y
cos(x± y) = cos x cos y ∓ sinx sin y
tan(x± y) = tanx± tan y
1∓ tanx tan y
• double-angle formulae
sin(2x) = 2 sinx cos x
cos(2x) = cos2 x− sin2 x
tan(2x) =
2 tan x
1− tan2 x.
The above formulae, especially the double-angle formulae for sin(2x) and cos(2x), should be learnt
carefully.
The inverse trigonometric functions will be discussed in Chapter 6.
1.7 The elementary functions
(Ref: SH10 §1.6)
Other important functions that have not yet been mentioned include the exponential and logarithm
functions. Their graphs are shown below.
x
ex
1
(1, e)b
x
lnx
1
(e, 1)
b
These functions will be studied in greater depth in Chapter 9.
Functions arising from the study of roots of polynomials are also important to mathematics,
especially those functions f of the form f(x) = n
√
x, where n is a positive integer. It is important
c©2020 School of Mathematics and Statistics, UNSW Sydney
1.7. THE ELEMENTARY FUNCTIONS 29
that students are familiar with the graphs of the square root and cube root functions, shown below.
x
√
x
(1, 1)
b
x
3
√
x
(1, 1)
(−1,−1)
b
b
The elementary functions are all those functions that can be constructed by combining a finite
number of polynomials, exponentials, logarithms, roots and trigonometric functions (including the
inverse trigonometric functions) via function composition, addition, subtraction, multiplication and
division. Hence the following expressions give rise to elementary functions:
f(x) = esinx + x2
g(x) =
lnx− tan x√
x
h(x) =
3
√
x4 − 2x2 + 5.
It also follows that every rational function is an elementary function.
The absolute values function, given by
abs : R→ R
abs(x) = |x| ∀x ∈ R,
is an elementary function by equation (1.2). It follows that if f is an elementary function then
abs ◦ f is also an elementary function. It is standard to denote function abs ◦ f by |f |. Thus
|f |(x) = (abs ◦ f)(x) = |f(x)| ∀x ∈ Dom(f).
The graph of |f | is obtained by reflecting in the x-axis any part of the graph of f that lies below
the x-axis. This is illustrated below.
x
f(x)
x
|f(x)|
Most, if not all functions studied in high school were elementary functions. In this course we
shall meet some useful functions that are not elementary functions. (These are defined by integrals;
see Section 8.12.)
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30 CHAPTER 1. SETS, INEQUALITIES AND FUNCTIONS
1.8 Implicitly defined functions
Many curves on the plane can be described as all those points (x, y) on the plane that satisfy some
equation involving x and y. For example, consider the equation
(x2 + y2 − 1)3 − x2y3 = 0. (1.3)
The set of points (x, y) satisfying this equation are shown on the graph below.
x
y
Note that the curve cannot be described by one function since some x-values have two corresponding
y-values. However, the curve can be described by combining the graphs of two functions f :
[−a, a]→ R and g : (−a, a)→ R, which are defined by the rules

y = f(x)
(x2 + y2 − 1)3 − x2y3 = 0
y ≥ b
and 

y = g(x)
(x2 + y2 − 1)3 − x2y3 = 0
y < b
(where a ≈ 1.139 and b ≈ 0.545), as illustrated below.
x
y
| b
||
a−a
y = f(x)
x
y
| b
||
a−a
y = g(x)
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1.8. IMPLICITLY DEFINED FUNCTIONS 31
We say that f and g are implicitly defined by equation (1.3).
Example 1.8.1. The set of points (x, y) that satisfy the equation
x2 + y2 = 1
describes a circle of radius 1 centred at the origin. The above equation could be used to implicitly
define many different functions. We give two examples.
• The function f : [0, 1] → R is defined by the rule

y = f(x)
x2 + y2 = 1
y ≥ 0.
Its graph is shown below.
x
y
1
1 y = f(x)
• The function g : (−1, 1)→ R is defined by the rule

y = f(x)
x2 + y2 = 1
y ≤ 0.
Its graph is shown below.
x
y
1−1
−1 y = g(x)
Many heavenly bodies (such as planets and comets) trace out paths that are roughly elliptic,
parabolic or hyperbolic. (The fact that planets trace out elliptic paths around the sun was discov-
ered by Johannes Kepler at the beginning of the sixteenth century, and was later used by Newton, in
conjunction with calculus, to establish his universal law of gravitation.) Circles, ellipses, parabolas
and hyperbolas are known as conic sections because they can be obtained by intersecting a cone
with a plane. Equations that describe conic sections, accompanied by corresponding diagrams, are
shown below.
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32 CHAPTER 1. SETS, INEQUALITIES AND FUNCTIONS
x
y
r
r
−r
−r
Circle: x2 + y2 = r2
x
y
a
b
−a
−b
Ellipse:
x2
a2
+
y2
b2
= 1
x
y y = bax
a−a
Hyperbola:
x2
a2
− y
2
b2
= 1
x
y
y = bax
b
−b
Hyperbola:
y2
b2
− x
2
a2
= 1
Students should be able to recognise such curves.
1.9 Continuous functions
The sine, cosine, exponential, logarithmic and polynomial functions are examples of continuous
functions. Intuitively, a function f is continuous on its domain if small changes in the input variable
x produce small changes in f(x) (assuming, of course, that x always remains within Dom(f)).
The intuitive notion that a function is continuous if ‘its graph can be drawn without lifting the
pencil off the page’ works well for some functions, but not for others. For example, the tangent
function is a continuous function, even though its graph (part of which is shown below) cannot be
drawn without lifting the pencil off the page.
x
tan x
π
2
π−π2−π
| | | |
The break in the graph when, say, x = −π/2 or x = π/2, is a consequence of a break in the domain
of the function rather than a discontinuity of the function. Similarly, the trigonometric functions
c©2020 School of Mathematics and Statistics, UNSW Sydney
1.10. MAPLE NOTES 33
sec, cosec and cot are continuous, as are the rational functions. The notion of continuity shall be
discussed in more precise terms in Chapters 2 and 3.
Intuitively, a function f has a discontinuity at a point a if a ∈ Dom(f) and there is a break in
the graph of f at a. It is important to realise that there are different types of discontinuities. We
compare three examples.
x
f(x)
bc
b
|
a x
g(x)
bc
b
|
a x
h(x)
b
|
a
The functions f , g and h each have a discontinuity at a.
• For f , the discontinuity can be easily removed by redefining the value of f(a) to ‘plug up the
hole’ at a. This is an example of a removable discontinuity.
• For g, the discontinuity is a result of a finite jump in the graph of g at the point a. The
discontinuity for g, which is an example of a jump discontinuity, is worse than that for f but
not nearly as bad as that for h.
• For h, the discontinuity is a result of rapid oscillation on the right-hand side of a and an
infinite jump on the left-hand side of a. This is an example of an essential discontinuity.
While students are not expected to learn or use the terms removable, jump and essential disconti-
nuity, it is expected that they will appreciate that some discontinuities are worse than others.
1.10 Maple notes
Many chapters in these notes conclude with a summary of relevant Maple commands. Other sources
of information are the First Year Computing Notes and the online Maple help command. While
Maple can relieve you of the tedium of calculations, it is not a substitute for understanding the
mathematics behind the calculations, which is what you need in order to interpret Maple output
intelligently.
The following Maple command is relevant to the material of this chapter:
plot(f(x),x=a..b); draws a (two-dimensional) plot of the graph of y = f(x) for
a ≤ x ≤ b. To find out about the many options available, use ?plot. Implicitly defined
functions can be plotted using the smartplot command. For example,
> # draw a rational function
> f:=x^5-2*x^4+3*x^3-6*x^2-4*x+8:
> g:=4*x^4-8*x^3-12*x^2+16*x+16:
> plot(f/g, x=-10..10,y=-10..10,numpoints=1000,discont=true);
c©2020 School of Mathematics and Statistics, UNSW Sydney
34 CHAPTER 1. SETS, INEQUALITIES AND FUNCTIONS
> # draw a hyperbola, via implicit definition
> smartplot(x^2/9-y^2/4=1);
> # use the Maple plot command implicitplot --- need more plotting tools
> with(plots):
> implicitplot(y^5+x^2*y^3+16,x=-10..10,y=-2..0);
c©2020 School of Mathematics and Statistics, UNSW Sydney
PROBLEMS FOR CHAPTER 1 35
Problems for Chapter 1
Questions marked with [R] are routine and with [H] harder. You should make sure that you
can do the easier questions before you tackle the more difficult questions. Questions marked with
[V] have a video solution available from Moodle.
Problems 1.1 : Sets of numbers
1. [R] Express the following sets in words. Graph the sets on the number line (if possible).
a) {x ∈ Z : −π < x < π}
b) {x ∈ R : x2 − x− 1 < 0}
c) {x ∈ Q : x2 = 2}
2. [R] Graph on the number line the following sets.
a) [3,∞), (−∞, 3), (−∞,∞), (−3, 3]
b) {x : |x− 2| < 5}
c) {x : x2 + 4x− 5 > 0}
3. [R] Sketch the set of points (x, y) which satisfy the following relations.
a) 0 ≤ y ≤ 2x and 0 ≤ x ≤ 2 b) y/2 ≤ x ≤ 2 and 0 ≤ y ≤ 4
Problems 1.2 : Solving inequalities and
1.3 : Absolute values
4. [R] Solve the following inequalities.
a) x(x− 1) > 0 b) (x− 1)(x− 2) < 0 c) 1
x
> −1
2
d)
1
1− x >
1
2
e) x ≥ 6
x− 1
5. [R] Solve the following inequalities.
a)
∣∣x+ 1∣∣< 3 b) ∣∣x+ 2∣∣> 3 c) ∣∣3x+ 2∣∣< 1 d) ∣∣∣∣x− 1x+ 1
∣∣∣∣ < 1
6. [R] [V]
a) By expanding (x− y)2, prove that x2 + y2 ≥ 2xy for all real numbers x and y.
b) Deduce that
a+ b
2
≥
√
ab for all non-negative real numbers a and b. When does equality
hold?
c) Use the result above to find the minimum value of y = x2 +
1
x2
.
c©2020 School of Mathematics and Statistics, UNSW Sydney
36 CHAPTER 1. SETS, INEQUALITIES AND FUNCTIONS
7. [R] True or false:
a) If a > b then
1
a
<
1
b
. b) If a < b then a2 < b2.
c) If 0 < a < b then a2 < b2. d) If a2 + b2 = 0 then a = b = 0.
e) If −1 < a < b then a2 < b2.
8. [H] Prove that (x+ y)2 ≥ 4xy and hence deduce that 1
x2
+
1
y2
≥ 4
x2 + y2
.
9. [H] [V]
a) Prove that f(x) = 1 + x+ x2 is positive for all real numbers x.
b) By considering cases (or otherwise) prove that 1 + x+ x2 + x3 + x4 is always positive.
c) Generalise the above results.
Problems 1.4 : Functions
10. [R] Determine the (maximal) domain and corresponding range for each function f described
below.
a) f(x) =
√
5− x2 b) f(x) = √x2 − 5
c) f(x) = (x− 8)−1/3 d) f(x) = √x− 1
e) f(x) =
1√
x− 1 f) f(x) =
√
sinx
g) f(x) =
√
1− 2 sinx h) f(x) = 1 + tan2 x
i) f(x) =


cos x if x < 0√
1− x if 0 ≤ x ≤ 1
|x| if x > 1
11. [R] If f(x) = x+ 5 and g(x) = x2 − 3 then find
a) (g ◦ f)(0) b) (g ◦ f)(x) c) (f ◦ g)(2) d) (f ◦ g)(x).
12. [R] If f(x) = x− 1 and g(x) = 1√
x− 1, then give the explicit forms of
a) (f + g)(x) b) (fg)(x) c)
(
f
g
)
(x) d) (f ◦ g)(x).
Problems 1.5 : Polynomials and rational functions,
1.6 : The trigonometric functions and
1.7 : The elementary functions
13. [R] Draw neat sketches (preferably without using calculus) of the graphs given by the fol-
lowing equations.
c©2020 School of Mathematics and Statistics, UNSW Sydney
PROBLEMS FOR CHAPTER 1 37
a) y = x2 − 5x+ 6 b) y = 2x3 − 16 c) y = 4
x− 3
d) y = 2ex−1 e) y =
1
x2 + 4
f) y = 3 sin 2x
g) y =
√
x− 1
14. [R]
a) Sketch the graph of y =
√
x+ 1 and use your graph to sketch (on the same diagram)
y =
1√
x+ 1
.
b) Repeat for y = x2 − 4x+ 3.
15. [R] Sketch the graph of y = x2 − 7x− 8 and hence sketch the graph of y = |x2 − 7x− 8|.
16. [R] What range of values will x2 + 4 take if −2 ≤ x ≤ 3?
17. [R] Use a graphical approach to solve |2x− 5| = x+ 2.
18. [R]
a) Show that if p and q are polynomials then p ◦ q is again a polynomial.
b) [H] Is the same true for rational functions?
Problems 1.8 : Implicitly defined functions
19. [R] Sketch the graphs given by the following equations.
a)
x2
9
+
y2
4
= 1 b)
x2
9
− y
2
4
= 1
c) 4x2 + 9y2 = 36 d)
y2
9
− x
2
4
= 1
c©2020 School of Mathematics and Statistics, UNSW Sydney
38 CHAPTER 1. SETS, INEQUALITIES AND FUNCTIONS
c©2020 School of Mathematics and Statistics, UNSW Sydney
39
Chapter 2
Limits
In many situations, it is desirable to know what the long term state of a system will be. For
example, suppose that an initially unpolluted lake containing 109 litres of water has a river flowing
through it at a rate of 106 litres per day. A factory is built next to the lake and discharges 104
litres of pollution per day. By making some simple assumptions, it can be shown that the amount
P (t) of pollution in the lake after t days of the factory’s operation is given by
P (t) =
109
101
(
1− e−101t/105). (2.1)
Environmental authorities want to know what the long term level of the pollution will be. To see
whether the pollution in the lake eventually stabilises, and what the level of pollution will be, one
studies the behaviour of P (t) as t→∞.
Understanding the limiting behaviour of functions at infinity is important for practical and
mathematical reasons. That is what we study in the next four sections.
2.1 Limits of functions at infinity
(Ref: SH10 §4.7)
In this section, we examine some techniques for calculating limits of the form lim
x→∞ f(x).
2.1.1 Basic rules for limits
There are a few basic things to bear in mind when working with limits.
• If f(x) gets arbitrarily close to some real number L as x tends to infinity, then we write
lim
x→∞ f(x) = L.
We can also write f(x)→ L as x→∞.
• If f(x) gets arbitrarily large (that is, approaches ∞) as x tends to ∞ then we write
f(x)→∞ as x→∞.
We do not write lim
x→∞ f(x) =∞ since∞ is not a real number. (The lim notation is only used
when the limit exists and is a real number.)
c©2020 School of Mathematics and Statistics, UNSW Sydney
40 CHAPTER 2. LIMITS
• If f(x)→∞ as x→∞ then
lim
x→∞
1
f(x)
= 0.
This is intuitively obvious and gives limits such as
lim
x→∞
1
x
= 0, lim
x→∞
5
4x6
= 0, lim
x→∞ e
−x = 0
(for the last example, recall that e−x = 1ex ).
The following proposition shows that the arithmetic of limits behaves nicely in many situations.
Proposition 2.1.1. Suppose that lim
x→∞ f(x) and limx→∞ g(x) exist and are finite real numbers. Then
(i) lim
x→∞[f(x) + g(x)] = limx→∞ f(x) + limx→∞ g(x)
(ii) lim
x→∞[f(x)− g(x)] = limx→∞ f(x)− limx→∞ g(x)
(iii) lim
x→∞[f(x)g(x)] = [ limx→∞ f(x)][ limx→∞ g(x)]
(iv) lim
x→∞
f(x)
g(x)
=
lim
x→∞ f(x)
lim
x→∞ g(x)
, provided that lim
x→∞ g(x) 6= 0.
Part of this proposition will be proved in Section 2.4. For now, we demonstrate how the
proposition is used to calculate limits.
Example 2.1.2. Calculate
lim
x→∞
3− 1/x2
2 + e−x
if the limit exists. Justify each step of your working with reference to Proposition 2.1.1.
Solution.
lim
x→∞
3− 1/x2
2 + e−x
=
lim
x→∞(3− 1/x
2)
lim
x→∞(2 + e
−x)
(rule (iv))
=
lim
x→∞ 3− limx→∞ 1/x
2
lim
x→∞ 2 + limx→∞ e
−x (rules (i) and (ii))
=
3− 0
2 + 0
=
3
2
.
In practice, you may use the rules of Proposition 2.1.1 without referring to them, and leave out
some of the steps shown above.
c©2020 School of Mathematics and Statistics, UNSW Sydney
2.1. LIMITS OF FUNCTIONS AT INFINITY 41
2.1.2 The pinching theorem
The idea behind the following theorem is intuitively obvious. Suppose, as illustrated in the diagram
below, that the graph of a function h always lies above the graph of a function f . Suppose also
that f and h have the same limit as x →∞. If the graph of a function g always lies between the
graphs of h and f (so that g is ‘pinched’ between f and h) then g has the same limit also.
x
y
|
b
L
y = f(x)
y = g(x)
y = h(x)
This idea is stated precisely below.
Theorem 2.1.3 (The pinching theorem). Suppose that f , g and h are all defined on the interval
(b,∞), where b ∈ R. If
f(x) ≤ g(x) ≤ h(x) ∀x ∈ (b,∞)
and
lim
x→∞ f(x) = limx→∞h(x) = L
then
lim
x→∞ g(x) = L.
One can modify the theorem to cover the case when x→ −∞.
Example 2.1.4. Use the pinching theorem to find lim
x→∞
sinx
x
.
Solution. We begin with the basic inequality
−1 ≤ sinx ≤ 1,
which is valid for every real number x. If we restrict x to the interval (0,∞) then we have
−1
x
≤ sinx
x
≤ 1
x
.
Now
lim
x→∞−
1
x
= lim
x→∞
1
x
= 0,
and so
lim
x→∞
sinx
x
= 0
by the pinching theorem.
c©2020 School of Mathematics and Statistics, UNSW Sydney
42 CHAPTER 2. LIMITS
2.1.3 Limits of the form f(x)/g(x)
Suppose that we want to calculate a limit of the form
lim
x→∞
f(x)
g(x)
where both f(x) and g(x) tend to infinity as x→∞. In this situation, we cannot apply the rules of
Proposition 2.1.1 directly because the numerator and denominator don’t have finite limits. Instead,
we find an equivalent form of f(x)g(x) for which Proposition 2.1.1 can be applied. The key idea is to
divide both f and g by the fastest growing term appearing in the denominator g.
Example 2.1.5. Evaluate lim
x→∞
4x2 − 5
2x2 + 3x
, if it exists.
Solution. There are two terms appearing in the denominator: 2x2 and 3x. As x → ∞, the term
which grows fastest is the one involving x2. So we divide the numerator and denominator by x2 to
evaluate the limit:
4x2 − 5
2x2 + 3x
=
4− 5/x2
2 + 3/x
→ 4− 0
2 + 0
as x→∞. Therefore
lim
x→∞
4x2 − 5
2x2 + 3x
= 2.
In general, we divide the numerator and denominator by the highest power of x in the denom-
inator.
Example 2.1.6. Find lim
x→∞
5x3 + 6x2 − 4 sinx
cos 3x+ 5x− 2x3 .
Example 2.1.7. Find lim
x→∞
x2 + 3x√
2x4 + 3− 4x .
2.1.4 Limits of the form
√
f(x)−√g(x)
Consider the limit of
√
x+ 5−√x+ 2 as x→∞. Since both √x+ 5 and √x+ 2 tend to infinity
(rather than a finite number) as x→∞, one cannot apply Proposition 2.1.1. As the next example
shows, a simple algebraic trick is used to overcome this difficulty.
Example 2.1.8. Evaluate lim
x→∞
(√
x+ 5−√x+ 2), if it exists.
c©2020 School of Mathematics and Statistics, UNSW Sydney
2.1. LIMITS OF FUNCTIONS AT INFINITY 43
Solution. The algebraic trick is to multiply both the numerator and denominator by
√
x+ 5 +√
x+ 2 and expand the numerator as a difference of squares:
√
x+ 5−√x+ 2 = (
√
x+ 5−√x+ 2)(√x+ 5 +√x+ 2)√
x+ 5 +
√
x+ 2
=
(x+ 5)− (x+ 2)√
x+ 5 +
√
x+ 2
=
3√
x+ 5 +
√
x+ 2
→ 0
as x→∞.
One should not be fooled into thinking, on the basis of the above example, that all limits of
this type are zero.
Example 2.1.9. Show that lim
x→∞
(√
x2 + x− x) = 1
2
.
2.1.5 Indeterminate forms
We have already mentioned that the rules of Proposition 2.1.1 only apply when the limits involved
are finite. In Section 2.1.3, we studied limits of the type
lim
x→∞
f(x)
g(x)
,
when f(x)→∞ and g(x)→∞ as x→∞. We say that such a limit is of the form ∞∞ . We cannot
say in advance whether or not a limit of the form ∞∞ exists, and if it does exist, what its value is.
For example, while the following limits have the form ∞∞ , each displays very different limiting
behaviour as x→∞:
• x2x →∞ as x→∞
• x
x2
→ 0 as x→∞
• 2x2
x2
→ 2 as x→∞
• x2
2x2
→ 12 as x→∞.
Because we cannot determine in advance what kind of limiting behaviour something of the form
∞
∞ has, we say that
∞
∞ is an indeterminate form.
Other types of indeterminate forms (for example, limits of the form 00 and∞−∞) arise naturally
in many applications. Techniques for evaluating such limits will be developed at various stages of
the course.
c©2020 School of Mathematics and Statistics, UNSW Sydney
44 CHAPTER 2. LIMITS
2.2 The definition of lim
x→∞
f(x)
Up to this point, we have treated limits in a rather intuitive way. It is now time to establish a
rigorous basis for our treatment of limits. That is, we seek a definition for lim
x→∞ f(x) = L that is
mathemetically precise and agrees with our intuitive notion of limits. How do we do this? A good
strategy is to think deeply about a few simple examples before considering the general case.
Consider the real-valued functions f , g and h defined by
f(x) =
1
x
, g(x) = sinx and h(x) = 0.05
whenever x > 0. Their graphs are sketched below.
x
f(x)
0
x
g(x)
0 2pi 4pi
x
h(x)
0
Intuitively, we see that f(x) tends to 0 as x tends to infinity, while neither g nor h have this
property. What is it about f that distinguishes its limiting behaviour from that of g or h?
The above graph shows that the distance between f(x) and its limit 0 is small whenever x is
large. However, to get to the core difference between f and the other two functions, we must be
more precise.
• The distance between h(x) and 0 is also small, but its limit is not 0. The reason why f(x)
tends to 0 as x → ∞ is that the distance between f(x) and 0 can be made as small as we
like whenever x is large enough.
• The distance between g(x) and 0 can also be made as small as we like, by taking x sufficiently
close to an integer multiple of π. However the limit of g(x) is not 0. The difference between
f and g is that the distance between f(x) and 0 remains small for all sufficiently large values
of x. The same is not true for g.
In summary, f(x) tends to 0 as x →∞ because the distance between f(x) and 0 can be made as
small as we like for all values of x that are sufficiently large.
The next step in crafting a good definition is to express phrases like ‘can be made as small as we
like’ and ‘sufficiently large’ in concrete mathematical language. The following list is one attempt
to do this. (As a reminder, f(x) = 1/x.)
• The distance between f(x) and 0 is less than 1 whenever x > 1.
c©2020 School of Mathematics and Statistics, UNSW Sydney
2.2. THE DEFINITION OF LIM
X→∞
F (X) 45
x
y
y = f(x)
1
0
−1
1
(1, 1)
• The distance between f(x) and 0 is less than 0.2 whenever x > 5.
x
y
y = f(x)
0.2
0
−0.2 5
(5, 0.2)
• The distance between f(x) and 0 is less than 0.1 whenever x > 10.
• The distance between f(x) and 0 is less than 0.01 whenever x > 100.
• The distance between f(x) and 0 is less than 0.0001 whenever x > 10000.
• etc.
Of course, to express the fact that ‘the distance between f(x) and 0 can be made as small as
we like’ in this way would require an infinite list. Instead, we say that, for every small positive
number ǫ, there is real number M such that
• the distance between f(x) and 0 is less than ǫ whenever x > M .
x
y
y = f(x)
ǫ
0−ǫ M
(1ǫ , ǫ)
By looking at the graph we can see that M = 1ǫ . (The symbol ǫ is the Greek letter ‘epsilon’ and is
traditionally used by mathematicians in this context.)
So far we have agreed that f(x)→ 0 as x→∞ precisely because
c©2020 School of Mathematics and Statistics, UNSW Sydney
46 CHAPTER 2. LIMITS
for every positive real number ǫ, there is a real numberM such that the distance between
f(x) and 0 is less than ǫ whenever x > M .
This statement forms a basis for defining what is meant by lim
x→∞ f(x) = 0. A general definition for
what is meant by lim
x→∞ f(x) = L is now obtained by
• replacing the limit 0 with the limit L, and then
• replacing ‘the distance between f(x) and L’ with |f(x)− L| (see Section 1.3).
We are now ready to give a general definition for the limit of a function at infinity.
Definition 2.2.1. Suppose that L is a real number and f is a real-valued function
defined on some interval (b,∞). We say that lim
x→∞ f(x) = L if
for every positive real number ǫ, there is a real number M such that if
x > M then |f(x)− L| < ǫ.
Remark 2.2.2. Suppose that lim
x→∞ f(x) = L. Definition 2.2.1 can be interpreted geometrically in
the following way. For every small number ǫ, you can find a marker M on the x-axis such that the
distance between f(x) and the limit L is less than ǫ whenever x lies to the right of M .
x
y
y = f(x)
L+ ǫ
L
L− ǫ
M
|
Remark 2.2.3. In Definition 2.2.1, the number M depends on ǫ. In general, the smaller the value
of ǫ, the larger the value of M . We note, however, that for any particular value of ǫ the choice of
M is not unique. For example, the M chosen for the above diagram is different from the M chosen
for the diagram below, but both choices guarantee that
|f(x)− L| < ǫ whenever x > M.
c©2020 School of Mathematics and Statistics, UNSW Sydney
2.3. PROVING THAT LIM
X→∞
F (X) = L USING THE LIMIT DEFINITION 47
x
y
y = f(x)
L+ ǫ
L
L− ǫ
M
|
Remark 2.2.4. Students who find Definition 2.2.1 difficult to understand can take comfort in the
fact that a rigorous formulation of limits evaded mathematicians for over 2000 years. The problems
and paradoxes involving limiting processes tabled around 450 BC by the ancient Greek philosopher
Zeno of Elea were only satisfactorily resolved in the nineteenth century, when a rigorous definition
of the limit was given by Bolzano (in 1817) and Weierstrass (in the 1850s).
2.3 Proving that lim
x→∞
f(x) = L using the limit definition
In the previous section we gave a definition for lim
x→∞ f(x) = L. In this section this definition is used
to prove limits that were previously taken for granted.
Example 2.3.1. Prove from Definition 2.2.1 that lim
x→∞
4
x2
= 0.
Proof. Suppose that f(x) = 4/x2, L = 0 and ǫ > 0. We begin by calculating the distance between
the function and its limit:
|f(x)− L| =
∣∣∣∣ 4x2 − 0
∣∣∣∣
=
4
x2
. (2.2)
We need to find a condition on x such that |f(x)− L| < ǫ. By (2.2), we require that
4
x2
< ǫ.
By rearrangement, we see that this is satisfied whenever
x >
2√
ǫ
.
So far we have shown that |f(x)− L| < ǫ whenever x > 2√
ǫ
. Hence if
M =
2√
ǫ
. (2.3)
then |f(x)− L| < ǫ whenever x > M . This completes the proof.
c©2020 School of Mathematics and Statistics, UNSW Sydney
48 CHAPTER 2. LIMITS
The above proof can be streamlined by omitting some explanatory commentary. In the next
example, explanatory comments (which are not part of the proof) are inserted inside square brack-
ets.
Example 2.3.2. Prove that
lim
x→∞
5x
x+ 3
= 5
by using Definition 2.2.1.
Proof. Suppose that ǫ > 0. For convenience, suppose also that x > 0. [This is allowed since we are
considering the behaviour of f(x) for large positive values of x.] Now
|f(x)− L| =
∣∣∣∣ 5xx+ 3 − 5
∣∣∣∣
=
∣∣∣∣5x− 5(x+ 3)x+ 3
∣∣∣∣
=
∣∣∣∣ −15x+ 3
∣∣∣∣
=
15
x+ 3
[since x > 0]
<
15
x
[to make algebra simpler later on].
In summary,
|f(x)− L| < 15
x
.
[This inequality gives an upper bound for the distance between f(x) and L.] Hence |f(x)− L| < ǫ
whenever
15
x
< ǫ.
This condition is equivalent to
x >
15
ǫ
.
Hence if M = 15ǫ then
|f(x)− L| < ǫ whenever x > M,
as required.
Before attempting one more proof, we reflect on the general strategy employed. Given ǫ, we
need to find a number M such that
|f(x)− L| < ǫ whenever x > M.
The number M can be found by following the procedure below.
1. Find a good upper bound for |f(x)− L|.
2. Find a simple condition on x such that this upper bound is less than ǫ.
c©2020 School of Mathematics and Statistics, UNSW Sydney
2.4. PROOFS OF BASIC LIMIT RESULTS [X] 49
3. Use this condition to state an appropriate value for M (in terms of ǫ).
Example 2.3.3. Prove that
lim
x→∞
sinx
9x2 + e3x
= 0
by using Definition 2.2.1.
Proof. Suppose that ǫ > 0. Now
|f(x)− L| =
∣∣∣∣ sinx9x2 + e3x − 0
∣∣∣∣
=
| sinx|
|9x2 + e3x|
≤ 1
9x2 + e3x
[using the fact that −1 ≤ sinx ≤ 1]
<
1
9x2
[decreasing the denominator increases the RHS].
In summary,
|f(x)− L| < 1
9x2
.
[This is our upper bound for |f(x)− L|.] Now
1
9x2
< ǫ
whenever
x >
1
3
√
ǫ
.
Set
M =
1
3
√
ǫ
.
Then |f(x)− L| < ǫ whenever x > M , as required.
Remark 2.3.4. As pointed out in Remark 2.2.3, the value of M is not unique. For instance, the
upper bound for |f(x) − L| in our solution to Example 2.3.3 was 1/(9x2) and the corresponding
value for M was 1/(3
√
ǫ). If instead we use the upper bound 1/e3x then the corresponding value
for M is 13 ln
1
ǫ .
2.4 Proofs of basic limit results [X]
By using Definition 2.2.1, one can now prove the limit rules stated in Section 2.1. The selection of
proofs given in this section is designed to give students a taste of how such results are proved.
Proof of Proposition 2.1.1 (i). Suppose that
lim
x→∞ f(x) = L1 and limx→∞ g(x) = L2.
c©2020 School of Mathematics and Statistics, UNSW Sydney
50 CHAPTER 2. LIMITS
We need to show, using Definition 2.2.1, that
lim
x→∞
(
f(x) + g(x)
)
= L1 + L2.
We begin by fixing a positive real number ǫ.
Since lim
x→∞ f(x) = L, there is a positive M1 such that
|f(x)− L1| < ǫ
2
whenever x > M1.
Similarly, there is a positive M2 such that
|g(x) − L2| < ǫ
2
whenever x > M2.
Hence, by the triangle inequality,∣∣(f(x) + g(x)) − (L1 + L2)∣∣ ≤ |f(x)− L1|+ |g(x) − L2|
<
ǫ
2
+
ǫ
2
= ǫ
whenever both x > M1 and x > M2. Now set M = max{M1,M2} so that∣∣(f(x) + g(x)) − (L1 + L2)∣∣ < ǫ whenever x > M.
This proves that lim
x→∞
(
f(x) + g(x)
)
= L1 + L2.
The other rules in Proposition 2.1.1 can be proved similarly, except that sometimes the desired
inequalities are harder to obtain. We move on to prove the pinching theorem.
Proof of Theorem 2.1.3. Suppose that
f(x) ≤ g(x) ≤ h(x) ∀x ∈ (b,∞)
and that
lim
x→∞ f(x) = limx→∞h(x) = L.
Fix a positive real number ǫ.
Since lim
x→∞ f(x) = limx→∞h(x) = L, there are real numbers M1 and M2, both greater than b, such
that
|f(x)− L| < ǫ whenever x > M1
and
|h(x) − L| < ǫ whenever x > M2.
In other words,
L− ǫ < f(x) < L+ ǫ whenever x > M1
and
L− ǫ < h(x) < L+ ǫ whenever x > M2.
Hence
L− ǫ < f(x) ≤ g(x) ≤ h(x) < L+ ǫ
whenever x > max{M1,M2}. This shows that
|g(x) − L| < ǫ whenever x > max{M1,M2},
completing the proof.
c©2020 School of Mathematics and Statistics, UNSW Sydney
2.5. LIMITS OF FUNCTIONS AT A POINT 51
2.5 Limits of functions at a point
(Ref: SH10 §§2.1–2.3, 2.5)
So far we have only discussed limits of the form lim
x→∞ f(x). In this section we study limits of the
form lim
x→a f(x), where a is a finite real number.
2.5.1 Left-hand, right-hand and two-sided limits
Consider the function f whose graph is shown below.
x
f(x)
| | | |
2 4 6 8
|
|
|
|
1
2
3
4
b
b
b
bc
bc
With reference to this graph, we will discuss the behaviour of f(x) when x is near the points 2, 4,
6 and 8.
• As x approaches 2 from the left-hand side, f(x) approaches 3. In this situation, we write
lim
x→2−
f(x) = 3
and say that ‘the limit of f(x) as x approaches 2 from the left is 3.’ (Note that f(2) = 1 6= 3.
One should not confuse a limit with a function value.) As x approaches 8 from the left-hand
side, f(x) approaches 1 and hence we write
lim
x→8−
f(x) = 1.
Both of these limits are called left-hand limits.
• As x approaches 2 from the right-hand side, f(x) approaches 1. In this situation, we write
lim
x→2+
f(x) = 1
and say that ‘the limit of f(x) as x approaches 2 from the right is 1.’ As x approaches 6 from
the right-hand side, f(x) approaches 2 and hence we write
lim
x→6+
f(x) = 2.
Both of these limits are called right-hand limits.
c©2020 School of Mathematics and Statistics, UNSW Sydney
52 CHAPTER 2. LIMITS
• As x approaches 4 from the left-hand side, f(x) is positive and grows large without bound. In
this case, lim
x→4−
f(x) does not exist because f(x) does not approach a real number. However,
we can write
f(x)→∞ as x→ 4−.
• As x approaches 4 from the right-hand side, f(x) is negative and grows large without bound.
In this case, lim
x→4+
f(x) does not exist but we can we write
f(x)→ −∞ as x→ 4+.
Once an understanding of left- and right-hand limits is grasped, we can talk about two-sided
limits.
Definition 2.5.1. If the left-hand limit lim
x→a−
f(x) and the right-hand limit
lim
x→a+
f(x) both exist and equal the same real number L, then we say that the limit
of f(x) as x→ a exists and is equal to L, and we write
lim
x→a f(x) = L.
If any one of these conditions fails then we say that lim
x→a f(x) does not exist.
Recall the graph of the function f introduced at the beginning of this section.
• The two-sided limit lim
x→2
f(x) does not exist because
lim
x→2−
f(x) 6= lim
x→2+
f(x).
• The two-sided limit lim
x→4
f(x) does not exist because lim
x→4−
f(x) does not exist.
• Since the left-hand and right-hand limits both exist at 6 and
lim
x→6−
f(x) = lim
x→6+
f(x) = 2,
the two-sided limit lim
x→6
f(x) exists and
lim
x→6
f(x) = 2.
Note that f(6) = 4 and so
lim
x→6
f(x) 6= f(6).
That is, the two-sided limit of a function at a point need not equal the value of the function
at that point.
c©2020 School of Mathematics and Statistics, UNSW Sydney
2.5. LIMITS OF FUNCTIONS AT A POINT 53
• Since the left-hand and right-hand limits at 8 both exist and equal 1, the two-sided limit
lim
x→8
f(x) exists and
lim
x→8
f(x) = 1.
In this case, the two-sided limit of f at 8 equals the value of the function at 8. That is,
lim
x→8
f(x) = f(8).
The last two points emphasise an important feature of limits and functions. If lim
x→a f(x) = L
then f(x) approaches L as x tends to a. The function f may not be defined at a, or if it is defined
at a, we can have the situation where f(a) = L or where f(a) 6= L. In other words, the value of
f(a) does not determine the value of lim
x→a f(x). As far as the limit is concerned, all that matters is
how f behaves when it is very close to the point a.
Remark 2.5.2. In this section we have not given a proper definition for limits of the form lim
x→a−
f(x)
or lim
x→a+
f(x). At this stage it is enough to know that such a rigorous definition exists and is similar
to the definition for limits of the form lim
x→∞ f(x) (see Definition 2.2.1). This definition can be used
to prove the basic limit results presented in the remainder of this chapter.
2.5.2 Limits and continuous functions
Using limits we can now give a definition for continuity.
Definition 2.5.3. Suppose that f is defined on some open interval containing the
point a. If lim
x→a f(x) = f(a), then we say that f is continuous at a; otherwise, we
say that f is discontinuous at a.
To illustrate, consider the function f whose graph is sketch below.
x
f(x)
| | |
2 4 6
b
b
bc
bc
The function f is continuous at 2, discontinuous at 4 and undefined at 6. (We do not say that f is
discontinuous at 6 since 6 /∈ Dom(f).)
We now extend our formal notion of continuity at a point to continuity on the real line.
c©2020 School of Mathematics and Statistics, UNSW Sydney
54 CHAPTER 2. LIMITS
Definition 2.5.4. If f : R → R is continuous at every point a in R then we say
that f is continuous everywhere.
If a function f : R→ R is continuous everywhere then it has the property that its graph can be
drawn on the Cartesian plane as an unbroken curve. This coincides with intuitive notions about
continuity gained in high school.
As mentioned in Section 1.9, polynomials, rational functions, the trigonometric functions, ex-
ponentials and logarithms are continuous at every point in the respective domains. Moreover, if
f is continuous at a point a then |f | is also continuous at a (see Section 1.7 for the definition of
|f |). The proof of these facts requires a rigorous definition of the limit at a point, and, as already
mentioned, such a definition is not presented in this course. However, we will discuss aspects of
the proof in Chapter 3.
A consequence of the above definition is that both one-sided and two-sided limits of f at a point
a are easy to evaluate provided that (a) a ∈ Dom(f) and (b) we know that f is a continuous at a.
Hence, in view the facts mentioned above,
• lim
x→2
sin(x) = sin(2)
• lim
x→3+
(x2 − x+ 1) = lim
x→3
(x2 − x+ 1) = 32 − 3 + 1 = 7
• lim
x→4
|x− 10| = |4− 10| = 6.
Remark 2.5.5. Continuity is a deep property for a function to have. Contrary to the impression
that many students form from their study of functions at school, most functions are not continuous.
Three interesting functions with discontinuities are mentioned below.
• The function f : R→ R, defined by
f(x) =
{
1 if x ∈ Q
0 if x /∈ Q,
is discontinuous at every point in its domain.
• The function f : R→ R, defined by
f(x) =
{
x if x ∈ Q
−x if x /∈ Q,
is continuous at 0 but discontinuous everywhere else.
• It is not too difficult to construct a function f : R→ R which is continuous at every irrational
point but discontinuous at every rational point.
Hence, one should not assume that the graph of a general function f : R → R is continuous at
even a majority of points in its domain. Continuous functions are quite special and deserve much
attention. We will study them in greater depth in Chapter 3.
c©2020 School of Mathematics and Statistics, UNSW Sydney
2.5. LIMITS OF FUNCTIONS AT A POINT 55
2.5.3 Rules for limits at a point
The following proposition, which is analogous to Proposition 2.1.1, shows that limits at a point
have nice arithmetic properties.
Proposition 2.5.6. Suppose that a ∈ R and that lim
x→a f(x) and limx→a g(x) exist and are finite real
numbers. Then
(i) lim
x→a(f(x) + g(x)) = limx→a f(x) + limx→a g(x)
(ii) lim
x→a(f(x)− g(x)) = limx→a f(x)− limx→a g(x)
(iii) lim
x→a f(x)g(x) = limx→a f(x) limx→a g(x)
(iv) lim
x→a
f(x)
g(x)
=
lim
x→a f(x)
lim
x→a g(x)
, provided that lim
x→a g(x) 6= 0.
A similar set of rules hold for left- and right-hand limits.
Proposition 2.5.6 says that limits interact nicely with function addition, subtraction, multiplica-
tion and division. The next proposition shows that limits interact nicely with function composition.
Proposition 2.5.7. If lim
x→a f(x) = L and g is continuous at L then
lim
x→a g(f(x)) = g
(
lim
x→a f(x)
)
.
If the functions f and g are continuous everywhere, then Proposition 2.5.7 implies that
lim
x→a g(f(x)) = g
(
lim
x→a f(x)
)
for any point a in R.
Example 2.5.8. Evaluate the limit
lim
x→
√
π/3
cos
(
x2 + π6
)
,
justifying each step of your calculation.
Solution. Since the cosine function and polynomials are continuous everywhere,
lim
x→
√
π/3
cos
(
x2 +
π
6
)
= cos
(
lim
x→
√
π/3
x2 +
π
6
)
(by Proposition 2.5.7)
= cos
(π
3
+
π
6
)
(since polynomials are continuous)
= cos
(π
2
)
= 0.
c©2020 School of Mathematics and Statistics, UNSW Sydney
56 CHAPTER 2. LIMITS
A version of the pinching theorem for two-sided limits is given below. Similar versions exist for
left- and right-hand limits.
x
y
|
a
L
y = f(x)
y = g(x)
y = h(x)
Theorem 2.5.9 (The pinching theorem). Let I be an open interval containing the point a. Suppose
that f , g and h are all defined on I except possibly at the point a. If
f(x) ≤ g(x) ≤ h(x) ∀x ∈ I
and
lim
x→a f(x) = limx→ah(x) = L
then
lim
x→a g(x) = L.
Remark 2.5.10. The pinching theorem can be used to prove the well-known limit
lim
θ→0
sin θ
θ
= 1.
See the problem set of Chapter 2 for details.
We close this chapter with four examples involving limits of the form lim
x→a f(x), where f is not
necessarily continuous at the point a. Determining whether or not the limit exists, and if so what
its value is, requires special analysis of the function f .
Example 2.5.11. Determine whether or not the limit
lim
x→0
x sin(1/x)
exists. If it exists, find its value.
Solution. We begin with the inequality
−1 ≤ sin θ ≤ 1
which is valid for all θ in R. In particular, if θ = 1/x then
−1 ≤ sin(1/x) ≤ 1 whenever x 6= 0.
c©2020 School of Mathematics and Statistics, UNSW Sydney
2.5. LIMITS OF FUNCTIONS AT A POINT 57
Case 1: Suppose that x > 0. Then we have
−x ≤ x sin(1/x) ≤ x.
Since lim
x→0+
−x = lim
x→0+
x = 0, it follows from the pinching theorem that lim
x→0+
x sin(1/x) = 0.
Case 2: Suppose that x < 0. Then
−x ≥ x sin(1/x) ≥ x.
Since lim
x→0−
−x = lim
x→0−
x = 0, it follows from the pinching theorem that lim
x→0−
x sin(1/x) = 0.
Conclusion: Since the left- and right-hand limits exist and are both equal to 0, the two-sided
limit exists and
lim
x→0
x sin(1/x) = 0.
This limit is illustrated graphically below.
x
y
y = xy = −x
y = x sin(1/x)
Example 2.5.12. Suppose that f : R→ R is defined by the formula
f(x) =


x2 − 4
x− 2 if x 6= 2
0 if x = 2.
Discuss the limiting behaviour of f(x) as x approaches 2.
Solution. Since we are discussing limiting behaviour, the value of f at 2 is irrelevant. All that
matters is how f(x) behaves near 2. So suppose that x 6= 2. Then
f(x) =
x2 − 4
x− 2
=
(x− 2)(x + 2)
x− 2
= x+ 2.
The graph of f is now easy to sketch.
c©2020 School of Mathematics and Statistics, UNSW Sydney
58 CHAPTER 2. LIMITS
x
f(x)
bc
b
2
|4
Since
lim
x→2−
f(x) = lim
x→2+
f(x) = 4,
the two-sided limit exists and we have lim
x→2
f(x) = 4. (Note that f is not continuous at 2 since
lim
x→2
f(x) 6= f(2).)
Example 2.5.13. Suppose that f : R→ R is defined by the formula
f(x) =


|x2 − 9|
x− 3 if x 6= 3
2 if x = 3.
Discuss the limiting behaviour of f(x) as x approaches 3.
Example 2.5.14. Suppose that
f(x) = sin
(π
x
)
, x 6= 0.
Discuss the limiting behaviour of f(x) as x approaches 0.
Solution. Since f is an odd function, we initially restrict our attention to positive values of x. Now
sin θ = 0 whenever θ ∈ {π, 2π, 3π, . . .}. If θ = π/x then we have sin(π/x) = 0 whenever
x =
1
1
,
1
2
,
1
3
, . . . .
Similarly, sin θ = 1 whenever θ ∈ {π/2, 5π/2, 9π/2, 13π/2, . . .} so sin(π/x) = 1 whenever
x =
2
1
,
2
5
,
2
9
,
2
13
. . . .
Consequently, no matter how close x is to 0 on the right, we can always find two closer points x1
and x2 such that f(x1) = 0 and f(x2) = 1. We conclude that f has no right-hand limit at 0. Since
f is odd, it follows that f has no left-hand limit at 0. Hence lim
x→0+
f(x) does not exist.
The graph of f near the origin is shown below.
c©2020 School of Mathematics and Statistics, UNSW Sydney
2.6. MAPLE NOTES 59
x
f(x)
−1
1
The f(x) oscillates wildly between 1 and −1 as x tends to 0, and lim
x→0+
f(x) does not exist.
2.6 Maple notes
The following Maple command is relevant to the material of this chapter:
limit(f(x),x=a); finds the (two-sided) limit of f(x) as x → a. One-sided limits can be
specified using limit(f(x),x=a,right); or limit(f(x),x=a,left);. For example,
> limit(sin(x)/x, x=0);
1
> limit(sin(x)/x, x=infinity);
0
> # the next limit needs methods from chapter 5 for hand calculations
> limit((1-cos(x))/x^2, x=0, left);
1/2
> # Maple produces results that are not in strict agreement with our definitions
as x goes to plus or minus infinity
> limit(x^2, x=-infinity);
∞
> limit(sin(x), x=-infinity);
−1 . . . 1
> limit(x^2*sin(x), x=-infinity);
undefined
c©2020 School of Mathematics and Statistics, UNSW Sydney
60 CHAPTER 2. LIMITS
Problems for Chapter 2
Problems 2.1 : Limits of functions at infinity
1. [R] [V] Evaluate the following limits if they exist.
a) lim
x→∞
x2 − 1
x2 + 1
b) lim
x→∞
2x2 + x− 1
x2 + 4x− 3
c) lim
x→∞
2x2 + 5x− 1
x3 + x
d) lim
x→∞
x5 + 5x+ 1
x4 + 3
e) lim
x→∞
5x2 − 3x+ cos 7x
4 + sin 2x+ x2
f) lim
x→∞ sinx
2. [R] Use the pinching theorem to find the following limits.
a) lim
x→∞
sinx
x
b) lim
x→∞
cosx
x2
3. [R] [V]
a) Prove that lim
x→∞(
√
x+ 1−√x) = 0.
b) Show that lim
x→∞(
√
x2 + x − x) = 1
2
.
Problems 2.2 : The definition of lim
x→∞
f(x) and
2.3 : Proving that lim
x→∞
f(x) = L using the limit definition
4. [R]
a) Write down the formal definition for the statement
lim
x→∞ f(x) = L.
b) Evaluate lim
x→∞
1
2x2
.
c) Verify from the formal definition that your answer in (b) is correct.
5. [R]
a) Evaluate lim
x→∞
x2 + 1
x2
.
b) Find a real number M such that the distance between
x2 + 1
x2
and its limit is less than
0.01 whenever x > M .
c) Suppose that ǫ > 0. Find a real number M (expressed in terms of ǫ) such that the
distance between
x2 + 1
x2
and its limit is less than ǫ whenever x > M .
c©2020 School of Mathematics and Statistics, UNSW Sydney
PROBLEMS FOR CHAPTER 2 61
6. [R] [V] For each of the following, find the limit of f(x) as x tends to infinity and prove from
the definition that your answer is correct.
a) f(x) =
4x
x+ 7
b) f(x) =
x− 3
x2 + 3
c) f(x) = e−2x
d) f(x) =
sinx
x
e) f(x) =
sin 3x
x2 + 4
7. [R] [V] A parcel is dropped from an aeroplane. A simple model, taking into account gravity
and air resistance, suggests that the parcel’s velocity v(t) (in metres per second) is given by
v(t) = 50(1 − e−t/5), where t is the number of seconds since leaving the plane.
a) Calculate the terminal velocity of the parcel (that is, find lim
t→∞ v(t)).
b) The parcel never attains its terminal velocity. How long does it take to come within
1 metre per second of its terminal velocity?
Problems 2.5 : Limits of functions at a point
8. [R] Evaluate the following limits.
a) lim
x→3
2x+ 4 b) lim
x→2
x2 − 4
x− 2 c) limx→1
x3 − 1
x− 1 d) limx→3
1
x − 13
x− 3
9. [R] [V]
a) Find the left-hand limit lim
x→2−
|x− 2|
x− 2 .
b) Find the right-hand limit lim
x→2+
|x− 2|
x− 2 .
c) Does lim
x→2
|x− 2|
x− 2 exist?
10. [R] By finding the left- and right-hand limits first, decide whether or not each of the following
limits exist and if so find their values.
a) lim
x→0
x
|x| b) limx→2
|x2 − 4|
x− 2 c) limx→4
x− 4
|x− 4| d) limx→0
4
x
11. [R] [V]
a) Use the pinching theorem to find lim
x→0
x sin 1x .
b) Repeat for lim
x→0
x2 sin
1
2x
.
c©2020 School of Mathematics and Statistics, UNSW Sydney
62 CHAPTER 2. LIMITS
12. [R] [V] Suppose that θ is a (positive) angle measured in radians and consider the diagram
below.
θ
O A B
C
D
The curve segment CB is the arc of a circle of radius 1 centre O.
a) Write down, in terms of θ, the length of arc CB and the lengths of the line segments
CA and DB.
b) By considering areas, deduce that sin θ cos θ ≤ θ ≤ tan θ whenever 0 < θ < π2 .
c) Use the pinching theorem to show that lim
θ→0+
θ
sin θ
= 1.
d) Deduce that lim
θ→0
sin θ
θ
= 1.
13. [H] Discuss the limiting behaviour of cos 1x as x→ 0.
c©2020 School of Mathematics and Statistics, UNSW Sydney
63
Chapter 3
Properties of continuous functions
In the world of functions, there is a small civilised country inhabited by the race of continuous
functions. Citizens of this country, such as the polynomials and exponentials, are studied (and
perhaps even loved) by students in high schools throughout the human world. However, outside
this small civilised country is a vast untamed universe of all sorts of discontinuous functions. From
this perspective, the continuous functions are a very rare breed.
However, continuous functions are far from endangered. They are among the most useful
functions for modelling real-life phenomena, particularly when the quantity that is being modelled
occurs in nature and changes smoothly with time. For example,
• the distance from the earth to the sun at time t,
• the height of Mount Everest at time t,
• the volume of water in a dam at time t,
• the mass of an earthworm at time t, and
• the temperature at the back of a lecture theatre at time t
can all be modelled by continuous functions of the variable t.
In addition to their usefulness for modelling naturally occurring quantities, continuous functions
have important mathematical characteristics. The combination of the usefulness of continuous
functions and their rarity among the world of functions means that continuity is a deep property.
In this chapter we begin to explore the ramifications of continuity.
3.1 Combining continuous functions
(Ref: SH10 §2.4)
If two continuous functions are combined via function addition, subtraction, multiplication, division
or composition, when is the resulting function also continuous? The two next propositions answer
this question.
Proposition 3.1.1. Suppose that the functions f and g are continuous at a point a. Then f + g,
f − g and fg are continuous at a. If g(a) 6= 0 then f/g is also continuous at a.
c©2020 School of Mathematics and Statistics, UNSW Sydney
64 CHAPTER 3. PROPERTIES OF CONTINUOUS FUNCTIONS
Proof. Suppose that f and g are continuous at a. Then, by the definition of continuity at a point
(see Definition 2.5.3), lim
x→a f(x) = f(a) and limx→a g(x) = g(a). Therefore
lim
x→a(f + g)(x) = limx→a(f(x) + g(x)) (by the definition of f + g)
= lim
x→a f(x) + limx→a g(x) (by Proposition 2.5.6)
= f(a) + g(a) (since f and g are continuous)
= (f + g)(a) (by the definition of f + g).
Hence f + g is continuous at a.
The proofs that the functions f − g, fg and f/g are continuous at a are similar.
Proposition 3.1.2. Suppose that f is continuous at a and that g is continuous at f(a). Then g ◦f
is continuous at a.
Proof. Combine Definition 2.5.3 with Proposition 2.5.7.
In Chapters 1 and 2 it was stated (without proof) that the polynomials, rational functions
and trigonometric functions are continuous at every point in their respective domains. The next
example shows that the continuity of these functions can be deduced from the continuity of
• the constant functions
• the sine function, and
• the function f : R→ R given by f(x) = x.
Example 3.1.3. Suppose that a is a point in R. By assuming that the constant functions, the sine
function and the function f : R→ R, given by f(x) = x, are continuous everywhere, show that
(a) every polynomial is continuous at a,
(b) every rational function r is continuous at a (provided that a ∈ Dom(r)), and
(c) the cosine and tangent functions are continuous at a (provided that a ∈ Dom(tan)).
Solution. (a) Every polynomial can be constructed from constant functions and the function f by
a finite number of function multiplications and additions. (For example,
x3 − 4x2 + 5 = (x× x× x) + ((−4) × x× x) + 5
for all x in R.) Since the constant functions and f are continuous at a, it follows from Proposition
3.1.1 that every polynomial is continuous at a.
(b) Every rational function r can be written in the form p/q, where p and q are polynomials.
By the result of part (a), p and q are continuous a. It follows from Proposition 3.1.1 that p/q is
also continuous at a, provided that q(a) 6= 0.
(c) Recall that
cos x = sin(π/2− x) ∀x ∈ R.
That is, cos x = h(g(x)) where h(x) = sinx and g(x) = π/2 − x. Since the polynomial g is
continuous at a and the sine function h is continuous everywhere, the cosine function must also be
continuous at a by Proposition 3.1.2.
c©2020 School of Mathematics and Statistics, UNSW Sydney
3.2. CONTINUITY ON INTERVALS 65
Finally, since
tan x =
sinx
cos x
,
the tangent function is also continuous at a (provided that a ∈ Dom(tan)) by Proposition 3.1.1.
The next example justifies an assertion made in Section 2.5.2. We continue working under
the assumption that the constant functions and the function f : R → R, given by f(x) = x, are
continuous everywhere.
Example 3.1.4. Suppose that g : R → R is continuous at the point a and let |g| denote the
function given by
|g|(x) = |g(x)| ∀x ∈ R.
Show that |g| is continuous at the point a.
Remark 3.1.5. Throughout this section we have taken for granted that the constant functions,
the sine function and the function f : R → R, given by f(x) = x, are continuous everywhere. To
prove this, one needs a rigorous definition of the limit of a function at a point. Such a definition is
similar in flavour to the definition for the limit of a function at infinity, but will not be presented
in these notes.
Example 3.1.6. Suppose that a and b are real numbers and consider the function f : R→ R given
by
f(x) =
{
eax+b if x ≥ 0
cos x if x < 0.
For what values of a and b will the function f be continuous at 0?
Solution. We require that
lim
x→0−
f(x) = f(0) = lim
x→0+
f(x).
Now
lim
x→0−
f(x) = lim
x→0−
cos x = 1, f(0) = eb and lim
x→0+
f(x) = lim
x→0+
eax+b = eb.
So we require that eb = 1, which implies that b = 0. There is no restriction on a.
3.2 Continuity on intervals
(Ref: SH10 §2.4)
In section 2.5.2, a definition for continuity at a point was given. We now extend this definition to
define what is meant by continuity on an interval.
Definition 3.2.1. Suppose that f is a real-valued function defined on an open
interval (a, b). We say that f is a continuous on (a, b) if f is continuous at every
point in the interval (a, b).
c©2020 School of Mathematics and Statistics, UNSW Sydney
66 CHAPTER 3. PROPERTIES OF CONTINUOUS FUNCTIONS
Definition 3.2.2. Suppose that f is a real-valued function defined on a closed
interval [a, b]. We say that
(a) f is continuous at the endpoint a if lim
x→a+
f(x) = f(a),
(b) f is continuous at the endpoint b if lim
x→b−
f(x) = f(b), and
(c) f is continuous on the closed interval [a, b] if f is continuous on the open
interval (a, b) and at each of the endpoints a and b.
To illustrate these definitions, consider the functions f , g and h, whose graphs are shown below.
x
f(x)
| |
a b
b
b
bc
x
g(x)
| |
a b
b
b
b
bc
x
h(x)
| |
a b
b
b
All three functions are defined on the interval [a, b]. We see that
• f is continuous on the open interval (a, b) and at the endpoint b;
• g is continuous at the endpoints a and b but not continuous on the open interval (a, b); and
• h is continuous on the closed interval [a, b] (and, by implication, on the open interval (a, b)
and at both endpoints a and b).
3.3 The intermediate value theorem
(Ref: SH10 §§2.6, B.1)
The idea of this section is very simple. Suppose that a function f is continuous on some closed
interval [a, b] and that z is any real number that lies between f(a) and f(b), as illustrated below.
x
f(x)
| | |
|
|
|
f(a)
z
f(b)
a c b
b
b
c©2020 School of Mathematics and Statistics, UNSW Sydney
3.3. THE INTERMEDIATE VALUE THEOREM 67
It is clear from the graph that one can find a real number c in [a, b] such that f(c) = z. The
intermediate value theorem, which is stated below, says that this can always be done for any
function that is continuous on [a, b].
Theorem 3.3.1 (The intermediate value theorem). Suppose that f is continuous on the closed
interval [a, b]. If z lies between f(a) and f(b) then there is at least one real number c in [a, b] such
that f(c) = z.
The intermediate value theorem is proven using the least upper bound property of the real
numbers. (It is this property that distinguishes the real numbers from the rational numbers.) We
shall not examine the proof here.
We note three important points about the theorem. First, there may be more than one number
c in [a, b] that satisfies the conclusion of the intermediate value theorem. The diagram below
illustrates this possibility.
x
f(x)
| |
|
|
|f(a)
z
f(b)
a b
b
b
c1 c2 c3
Second, the condition in Theorem 3.3.1 that f is continuous cannot be relaxed. For example,
consider the function f : [a, b]→ R whose graph is shown below.
x
f(x)
| |
|
|
|
f(a)
z
f(b)
a b
b
b
b
bc
Because of the discontinuity, z is not in the range of f , even though z lies between f(a) and f(b).
That is, there is no real number c in [a, b] such that f(c) = z.
Third, the intermediate value theorem is a theorem about continuous functions defined on the
real numbers. There is no analogous theorem for continuous functions defined on the rational
numbers. For example, consider the functions f : R→ R and g : Q→ Q, given by
f(x) = x2 − 2 and g(x) = x2 − 2.
c©2020 School of Mathematics and Statistics, UNSW Sydney
68 CHAPTER 3. PROPERTIES OF CONTINUOUS FUNCTIONS
Both functions are continuous on their domains. Also,
f(0) = g(0) = −2 and f(2) = g(2) = 2.
However, while there is a number c in [0, 2] such that f(c) = 0, there is no number c in [0, 2] such
that g(c) = 0 (since
√
2 /∈ Dom(g)).
One of the chief applications of the intermediate value theorem is determining whether an
equation has a solution. Moreover, if a solution exists, then the theorem helps us determine an
approximate location of this solution. This is particularly helpful when the equation cannot be
solved by simple algebra.
Example 3.3.2. Show that the equation
e2x = sinx+ 4 (3.1)
has at least one positive solution.
Solution. Consider the function f given by
f(x) = e2x − sinx− 4.
Now
f(0) = e0 − sin 0− 4 = −3
while
f(π/2) = eπ + 1− 4 ≈ 20.14.
Since f is continuous on the closed interval [0, π/2] and 0 lies between f(0) and f(π/2), the
intermediate value theorem implies that there is a real number c in [0, π/2] such that f(c) = 0.
That is,
e2c = sin c+ 4
for some c in [0, π/2]. Moreover, one can check that c 6= 0. Hence the equation has at least one
positive solution c.
Note that in equation (3.1), one cannot isolate x to find an explicit solution. However, now
that it is known a solution exists in the interval [0, π/2], one could use Newton’s method to find a
good approximation to this solution.
3.4 The maximum-minimum theorem
(Ref: SH10 §§2.6, B.2)
In many situations, it is desirable to know the maximum (or minimum) value of a quantity that
varies over an interval of time. Examples include the maximum temperature in a 24 hour period
or the minimum stock price over a seven day period. Depending on the quantity and the time
interval, a maximum or minimum value may not be attained. The maximum-minimum theorem
gives conditions under which a maximum and minimum is attained. To clarify discussion, we begin
with a definition.
c©2020 School of Mathematics and Statistics, UNSW Sydney
3.4. THE MAXIMUM-MINIMUM THEOREM 69
Definition 3.4.1. Suppose that f is defined on a closed interval [a, b].
(a) We say that a point c in [a, b] is an absolute minimum point for f on [a, b]
if f(c) ≤ f(x) for all x in [a, b]. The corresponding value f(c) is called the
absolute minimum value of f on [a, b]. If f has an absolute minimum point on
[a, b] then we say that f attains its minimum on [a, b].
(b) We say that a point d in [a, b] is an absolute maximum point for f on [a, b]
if f(x) ≤ f(d) for all x in [a, b]. The corresponding value f(d) is called the
absolute maximum value of f on [a, b]. If f has an absolute maximum point
on [a, b] then we say that f attains its maximum on [a, b].
An absolute maximum point and an absolute minimum point are sometimes referred to as a
global maximum point and a global minimum point.
Example 3.4.2. Consider the functions g and h, whose graphs are illustrated below.
x
g(x)
| | | | |
1 2 3 4 5
|
|
|
|
1
2
3
4
x
h(x)
| | | | |
1 2 3 4 5
|
|
|
|
1
2
3
4
b
b
The absolute minimim and maximum points of g and h on [1, 5] are recorded in the following table.
g h
Absolute minimum points none 3
Absolute minimum value n.a. 1
Absolute maximum points 2, 4 none
Absolute maximum value 4 n.a.
The above example shows that a function f : [a, b] → R need not have an absolute maximum
point (or an absolute minimum point) on a closed interval [a, b]. However, if f is continuous on
[a, b] then such points always exist.
Theorem 3.4.3 (The maximum-minimum theorem). If f is continuous on a closed interval [a, b]
then f attains its minimum and maximum on [a, b]. That is, there exist points c and d in [a, b]
such that
f(c) ≤ f(x) ≤ f(d)
for all x in [a, b].
c©2020 School of Mathematics and Statistics, UNSW Sydney
70 CHAPTER 3. PROPERTIES OF CONTINUOUS FUNCTIONS
As with the intermediate value theorem, the least upper bound property of the real numbers is
used to prove the maximum-minimum theorem. We omit the proof.
Example 3.4.4. Suppose that the function f : R→ R is given by the rule
f(x) = x2 − 10x+ 16.
Find the absolute maximum and absolute minimum values of f on the interval [1, 10].
Solution. Since f(x) = (x− 2)(x− 8), the function f has zeros at 2 and 8.
x
f(x)
| |
1 10
2 8
5
By symmetry, we know that the axis of the parabola is the line x = 5. Hence 5 is an absolute
minimum point for f on [1, 10]. Since f(5) = −9, the absolute minimum value of f on [1, 10] is −9.
It is clear from the sketch of f that at least one of the endpoints of the interval [1, 10] must be
an absolute maximum point for f on [1, 10]. Now f(1) = 7 while f(10) = 16. So 10 is an absolute
maximum point for f on [1, 10] and the corresponding maximum value of f is 16.
Remark 3.4.5. While continuity on a closed interval guarantees the existence of at least one
absolute maximum point and absolute minimum point, locating such points may not be easy. A
systematic approach to finding maximum and minimum points will be developed in Chapters 4 and
5.
Remark 3.4.6. All the hypotheses stated in the maximum-minimum theorem must be satisfied for
the conclusion of the theorem to be valid. If the interval is closed but the function is not continuous,
then an absolute maximum (or minimum) point may not exist (see Example 3.4.2). The next
example shows that if the function is continuous on the interval but the interval is not closed then
an absolute maximum (or minimum) point may not exist. It is expected that students will know
the precise hypotheses of the maximum-minimum theorem (and other important theorems, such as
the intermediate value theorem) and know how to apply the theorem correctly.
Example 3.4.7. Consider the function f : R→ R given by f(x) = x+1. Since f is a polynomial,
it is continuous on the open interval (0, 1). However, f does not have an absolute maximum point
on (0, 1). To see why, consider the graph of f on (0, 1).
x
f(x)
|
10
1
2 |
c©2020 School of Mathematics and Statistics, UNSW Sydney
3.4. THE MAXIMUM-MINIMUM THEOREM 71
As x → 1−, f(x) → 2 but there is no real number d in (0, 1) such that f(d) = 2. By a similar
observation, f has no absolute minimum point on (0, 1).
The final result of this chapter is a corollary of maximum-minimum theorem. We begin with a
definition.
Definition 3.4.8. A function f is said to be bounded on an interval I if there is
some positive number M such that |f(x)| ≤M for all x in I.
In other words, f is bounded if the y-values of its graph lie between −M and M for some
positive number M . The sine and cosine functions are obvious examples of functions that are
bounded on the whole of the real line.
Corollary 3.4.9. If f : [a, b]→ R is continuous on [a, b] then f is bounded on [a, b].
Proof. Suppose that f is continuous on [a, b]. Then the function |f | is also continuous on [a, b]
(see Example 3.1.4). By the maximum-minimum theorem, |f | attains its maximum on [a, b]. If M
denotes the absolute maximum value of |f | on [a, b] then
|f(x)| ≤M ∀x ∈ [a, b].
Hence f is bounded.
c©2020 School of Mathematics and Statistics, UNSW Sydney
72 CHAPTER 3. PROPERTIES OF CONTINUOUS FUNCTIONS
Problems for Chapter 3
Problems 3.1 : Combining continuous functions and
3.2 : Continuity on intervals
1. [R] Suppose that f : R→ R is defined by f(x) = |x|.
a) Show that f is continuous at 0.
b) Is f continuous everywhere? Give brief reasons for your answer.
2. [R] Determine at which points each function f : R→ R is continuous. Give reasons.
a) f(x) =
{
e2x if x < 0
cos x if x ≥ 0 b) f(x) =


e−2x if x < 0
sinx+ 1 if 0 ≤ x ≤ π/2
2x− π if x > π/2
3. [R] Suppose that
f(x) =
{
x2−16
x−4 if x 6= 4
k if x = 4,
where k is a real number. For which values of k (if any) will f be continuous everywhere?
4. [H] Use the pinching theorem for limits to show that if f , g and h are three functions defined
on an open interval I, such that
• f(x) ≤ g(x) ≤ h(x) for all x ∈ I,
• f(a) = g(a) = h(a) for some a ∈ I, and
• f and h are continuous at a,
then g is also continuous at a.
Problems 3.3 : The intermediate value theorem
5. [R] Show that the function f , given by f(x) = x3− 5x+3, has a zero in each of the intervals
[−3,−2], [0, 1] and [1, 2].
6. [R] [V] Use the intermediate value theorem to show that the equation ex = 2cos x has at
least one positive real solution.
7. [H] Suppose that f is continuous on [0, 1] and that Range(f) is a subset of [0, 1]. By using
g(x) = f(x)− x, prove that there is a real number c in [0, 1] such that f(c) = c.
c©2020 School of Mathematics and Statistics, UNSW Sydney
PROBLEMS FOR CHAPTER 3 73
Problems 3.4 : The maximum-minimum theorem
8. [R] In each case, determine whether or not f attains a maximum on the given interval. Give
reasons for your answer.
a) f(x) = x2 − 4 on [−3, 5] b) f(x) =
∣∣∣∣sin(ex) + lnxx2 − 1
∣∣∣∣ on [2, 4]
c) f(x) = x2 − 4 on (−3, 5) d) f(x) = −(x2 − 4) on (−3, 5)
9. [H] [V] Suppose that f is a continuous function on R and that lim
x→∞ f(x) = limx→−∞ f(x) = 0.
a) Give an example of such a function which has both a maximum value and a minimum
value.
b) Give an example of such a function which has a minimum value but no maximum value.
c©2020 School of Mathematics and Statistics, UNSW Sydney
74 CHAPTER 3. PROPERTIES OF CONTINUOUS FUNCTIONS
c©2020 School of Mathematics and Statistics, UNSW Sydney
75
Chapter 4
Differentiable functions
A cyclist accelerates from rest in such a way that her displacement s (in metres) from her starting
position after t seconds is given by
s = t2 ∀t ∈ [0, 3].
The corresponding displacement–time graph for the first three seconds is shown below.
|
|
1
4
9
| |
1 2 30
b
b
displacement (m)
time (s)
What is the cyclist’s instantaneous velocity when t = 1?
At present, we do not have a formula for calculating instantaneous velocity. However, we can
calculate the average velocity v of the cyclist over any time interval by using the formula
v =
∆s
∆t
,
where ∆s denotes the change in displacement corresponding to a change ∆t in time. For example,
the average velocity over the time interval [1, 1.5] is 2.5 metres per second, as calculated below:
v[1,1.5] =
∆s
∆t
=
(1.5)2 − (1)2
1.5− 1 = 2.5. (4.1)
Average velocities over small time intervals are recorded in the following table.
c©2020 School of Mathematics and Statistics, UNSW Sydney
76 CHAPTER 4. DIFFERENTIABLE FUNCTIONS
Time interval ∆t Average velocity
[1, 1.5] 0.5 2.5
[1, 1.4] 0.4 2.4
[1, 1.3] 0.3 2.3
[1, 1.2] 0.2 2.2
[1, 1.1] 0.1 2.1
[1, 1.01] 0.01 2.01
[1, 1.001] 0.001 2.001
The table suggests that the average velocity approaches 2 meters per second as ∆t→ 0. Hence it
appears that instantaneous velocity of the cyclist when t = 1 is 2 meters per second.
One of the goals of this chapter is put these kinds of ideas on a rigorous footing in a general
context (where they can be applied to understanding rates of change of any quantity, not just
displacement). Hence we reframe these ideas in terms of functions and analytic geometry. More
precisely, we shall now show that the problem of determining instantaneous velocity is equivalent
to the problem of finding the gradient of a tangent line to the graph of a function.
To begin, the problem of calculating average velocity is equivalent to the problem of calculating
the gradient of a secant to the graph of a function. (A secant is a straight line that intersects a
curve at least twice.) For example, compare the calculation of average velocity in (4.1) with the
calculation of the gradient of the secant shown in the following diagram.
t
s
b
b
|
|
1
2.25
| |
1 1.5
s = t2
secant
gradient of secant = ∆s∆t =
(1.5)2−(1)2
1.5−1 = 2.5
In the same way, the above table of average velocities can be reinterpreted geometrically as a table
of gradients corresponding to secants of the curve s = t2.
Intersection points ∆t Gradient of secant
t = 1, t = 1.5 0.5 2.5
t = 1, t = 1.4 0.4 2.4
t = 1, t = 1.3 0.3 2.3
t = 1, t = 1.2 0.2 2.2
t = 1, t = 1.1 0.1 2.1
c©2020 School of Mathematics and Statistics, UNSW Sydney
4.1. GRADIENTS OF TANGENTS AND DERIVATIVES 77
The secants used in the table are illustrated over the page in Figure 4.1. The figure suggests that
as ∆t→ 0, the gradient of the corresponding secant approaches the gradient of the tangent line to
the curve. Thus the problem of determining instantaneous velocity is equivalent to the problem of
determining the gradient of a tangent to the graph of a function. It is to this general problem that
we now turn.
4.1 Gradients of tangents and derivatives
(Ref: SH10 §3.1)
Suppose that a function f has a ‘smooth’ graph. In this section, we give a general approach to
calculating the gradient of tangents to f . This leads the concept of a gradient function, also known
as a derivative (since it is derived from f), whose values give the gradient of the tangent to f at
any point. We begin with an example.
Example 4.1.1. Suppose that f : R→ R is given by f(x) = x3. Find the gradient of the tangent
to the graph of f when x = 2.
Solution. The tangent to the graph of f passes through the point P (2, 23). We will approximate
the gradient of the tangent with the gradient of a secant passing through the points P (2, 23) and
Q(2 + h, (2 + h)3), where h is a very small real number. (See the illustration below for the case
when h > 0.)
x
y
| |
|23
2 2 + h
b
b
P (2, 23)
Q(2 + h, (2 + h)3)
secant
tangent
y = f(x)
Using the ‘rise over run’ formula, the gradient of the secant is
(2 + h)3 − 23
h
.
Now, as h approaches 0, the point Q moves along the curve towards P , and hence the gradient of
c©2020 School of Mathematics and Statistics, UNSW Sydney
78 CHAPTER 4. DIFFERENTIABLE FUNCTIONS
b
b
b
b
b
b
|
|
|
|
|
|
1
1.21
1.44
1.69
1.96
2.25
| | | | | |
1 1.1 1.2 1.3 1.4 1.5
t
s
tangent
s = t2
secants
Figure 4.1: Using secants to determine the gradient of a tangent.
c©2020 School of Mathematics and Statistics, UNSW Sydney
4.1. GRADIENTS OF TANGENTS AND DERIVATIVES 79
the secant approaches the gradient of the tangent. In other words,
(gradient of tangent) = lim
h→0
(2 + h)3 − 23
h
= lim
h→0
8 + 12h+ 6h2 + h3 − 8
h
= lim
h→0
12h + 6h2 + h3
h
= lim
h→0
12 + 6h+ h2
= 12.
Thus the gradient of the tangent to f at the point 2 is 12. This gives a quantitative description of
the rate of change of f at the point 2.
There is nothing special about the point 2 in the previous example; we could calculate the
gradient of the tangent to f at any point x. The calculation would go something like this:
(gradient of tangent at x) = lim
h→0
(x+ h)3 − x3
h
= lim
h→0
x3 + 3x2h+ 3xh2 + h3 − x3
h
= lim
h→0
3x2h+ 3xh2
h
= lim
h→0
3x2 + 3xh
= 3x2.
So the gradient of the tangent to f when x = 5 is 75 (since 3× 52 = 75), while the gradient of the
tangent when x = −4 is 48. The gradient function of f , denoted by f ′ and given by
f ′(x) = 3x2 ∀x ∈ R,
is called the derivative of f .
We can repeat the same procedure for functions other than f(x) = x3. The gradient function
(or derivative) may be calculated using the limit given in the next definition.
Definition 4.1.2. Suppose that f is defined on some open interval containing the
point x. We say that f is differentiable at x if
lim
h→0
f(x+ h)− f(x)
h
exists. If the limit exists, we denote it by f ′(x). We call f ′(x) the derivative of f at
x.
Other notation for f ′(x) includes dfdx(x) and
d
dxf(x). If y = f(x) then the derivative is often
denoted by y′ or dydx . The ratio
f(x+ h)− f(x)
h
c©2020 School of Mathematics and Statistics, UNSW Sydney
80 CHAPTER 4. DIFFERENTIABLE FUNCTIONS
is called the difference quotient for f at the point x.
Remark 4.1.3. The notation dydx originates from the gradient formula
∆y
∆x , where ∆y represents a
change in y corresponding to a small change ∆x in x. (The symbol ∆ represents the Greek letter
‘delta’, which is equivalent to our capital ‘D’.) Some calculus texts prefer the use of delta notation
and write the difference quotient for a function f as
f(x+∆x)− f(x)
∆x
.
Example 4.1.4. Suppose that f : [0,∞) → R is given by f(x) = √x. Find the equation of the
tangent to f when x = 16.
Solution. First we shall calculate the derivative of f (if it exists) at 16. Note that, in the working
below, the limit as h approaches 0 is difficult to evaluate immediately, so we use a standard trick
(see, for example, Subsection 2.1.4):
f(16 + h)− f(16)
h
=
√
16 + h−√16
h
=
√
16 + h−√16
h
×
√
16 + h+
√
16√
16 + h+
√
16
=
16 + h− 16
h(
√
16 + h+
√
16)
=
h
h(
√
16 + h+
√
16)
=
1√
16 + h+
√
16
→ 1
8
as h→ 0. So the gradient of the tangent is 1/8. Using the point-gradient formula, the equation of
the tangent is
y − y1 = m(x− x1)
y − 4 = 1
8
(x− 16)
8y − x− 16 = 0.
Example 4.1.5. Suppose that f : R → R is given by f(x) = |x|. Determine whether or not f is
differentiable at the point 0.
Solution. We calculate the difference quotient of f at 0:
f(0 + h)− f(0)
h
=
|h|
h
=
{
h
h if h > 0
−h
h if h < 0.
=
{
1 if h > 0
−1 if h < 0.
c©2020 School of Mathematics and Statistics, UNSW Sydney
4.2. RULES FOR DIFFERENTIATION 81
Because of the ‘split formula,’ we must consider the left- and right-hand limits as h approaches 0
separately. Now
lim
h→0−
f(0 + h)− f(0)
h
= −1
while
lim
h→0+
f(0 + h)− f(0)
h
= 1.
Since the left- and right-hand limits are different, the two-sided limit doesn’t exist and hence f is
not differentiable at 0.
Note that this conclusion makes sense geometrically, since the graph of f has a ‘vertex’ at 0
and there is no unique tangent that touches the graph at this point.
x
y
Example 4.1.6. Suppose that f : R→ R is defined by
f(x) =
{
x sin 1x if x 6= 0
0 if x = 0.
Determine whether or not f is differentiable at 0.
4.2 Rules for differentiation
(Ref: SH10 §§3.2, 3.5, 3.6)
So far we have calculated derivatives working directly from the difference quotient. In this section we
see that there are fast ways of calculating derivatives without referring to the difference quotient
and limits at all. This fast method is part of what is known as ‘the calculus.’ We begin with
the derivatives of some standard functions, and then show how derivatives of other functions are
obtained by applying some simple rules.
The following table gives the derivatives for some standard functions.
f(x) f ′(x)
C, where C is a constant 0
xn, where n is a positive integer nxn−1
sinx cos x
ex ex
c©2020 School of Mathematics and Statistics, UNSW Sydney
82 CHAPTER 4. DIFFERENTIABLE FUNCTIONS
With these basic functions, it is possible to construct many other functions via function addition,
subtraction, multiplication and division. The next theorem describes what happens to the resulting
derivatives.
Theorem 4.2.1 (Rules for differentiation). Suppose that f and g are differentiable at x. Then
f + g, f − g and fg are differentiable at x. If g(x) 6= 0 then fg is also differentiable at x. Moreover,
(i) (f + g)′(x) = f ′(x) + g′(x),
(ii) (C.f)′(x) = C.f ′(x), where C is a constant,
(iii) (fg)′(x) = f ′(x)g(x) + f(x)g′(x), and
(iv)
(
f
g
)′
(x) =
f ′(x)g(x) − f(x)g′(x)
g(x)2
, provided that g(x) 6= 0.
Rules (iii) and (iv) are usually called the product and quotient rules. A familiar way of express-
ing these rules is
d(uv)
dx
= v
du
dx
+ u
dv
dx
and
d
dx
(u
v
)
=
v dudx − u dvdx
v2
,
where u and v are both functions of x.
Given two functions f and g, a new function f ◦g may be constructed by function composition.
The next theorem describes the derivative of this new function.
Theorem 4.2.2 (The chain rule). Suppose that g is differentiable at the point x and f is differen-
tiable at the point g(x). Then f ◦ g is differentiable at x and
(f ◦ g)′(x) = f ′(g(x))g′(x). (4.2)
You are probably more familiar with the chain rule expressed in the following way: if y = f(u)
and u = g(x) then
dy
dx
=
dy
du
du
dx
. (4.3)
While formula (4.3) looks simpler than (4.2), it does not specify the relationship between the
variables or the points at which each derivative is to be evaluated. It should therefore be used with
care.
The next example illustrates how the chain rule formula (4.2) is applied.
Example 4.2.3. Suppose that y = (sinx+ 1)7. Find the derivative of y at the point x.
Solution. Suppose that f(x) = x7 and g(x) = sinx+1. Then f ′(x) = 7x6 and g′(x) = cos x. Then
y = (f ◦ g)(x). By the chain rule, the derivative of y at x is given by
(f ◦ g)′(x) = f ′(g(x))g′(x)
= 7(g(x))6 × cos x
= 7(sin x+ 1)6 cos x.
c©2020 School of Mathematics and Statistics, UNSW Sydney
4.3. PROOFS OF DIFFERENTIATION RULES 83
As mentioned at the beginning of this section, derivatives of many functions can be obtained
from those of a few basic functions by the application of simple rules. The next example illustrates
this point.
Example 4.2.4. Show that
(a)
d
dx
(
xm
)
= mxm−1, where m is an integer;
(b)
d
dx
(
cosx
)
= − sinx; and
(a)
d
dx
(
tanx
)
= sec2 x.
4.3 Proofs of differentiation rules
In Section 4.2 we stated two theorems and gave a table of derivatives. In this section we prove
some of these results. We begin with the derivative of the sine function, using the standard result
that
lim
θ→0
sin θ
θ
= 1. (4.4)
For a geometric argument for this limit, see Question 12 from the problems for Chapter 2.
Proposition 4.3.1. If f(x) = sinx then f ′(x) = cos x for all real x.
Proof. The difference quotient for the sine function and the angle sum formula for sine gives
sin(x+ h)− sinx
h
=
sinx cos h+ cos x sinh− sinx
h
= sinx
cos h− 1
h
+ cosx
sinh
h
. (4.5)
Now cos2 h+ sin2 h = 1 so
cos h− 1
h
=
cos2 h− 1
h(cos h+ 1)
=
− sin2 h
h(cos h+ 1)
= −sinh
h
sinh
cos h+ 1
→ −1× 0
= 0
as h→ 0, by (4.4). If we combine this with (4.5) then
lim
h→0
sin(x+ h)− sinx
h
= sinx lim
h→0
cos h− 1
h
+ cos x lim
h→0
sinh
h
= sinx× 0 + cos x× 1
= cos x.
Thus f ′(x) = cos x.
c©2020 School of Mathematics and Statistics, UNSW Sydney
84 CHAPTER 4. DIFFERENTIABLE FUNCTIONS
The derivative of ex will be dealt with in Chapter 9. The derivative of constant functions can
be easily computed from Definition 4.1.2.
Suppose that f is given by f(x) = xn, where n is a positive integer. The derivative of f can be
computed using the same method as Example 4.1.1, with the assistance of the binomial theorem.
Alternatively, one can use mathematical induction and the product rule for differentiation.
Proof of the product rule for differentiation. Suppose that f and g are differentiable at the point
x. The difference quotient of fg at x gives
(fg)(x+ h)− (fg)(x)
h
=
f(x+ h)g(x + h)− f(x)g(x)
h
=
f(x+ h)g(x + h)− f(x)g(x+ h) + f(x)g(x+ h)− f(x)g(x)
h
= g(x+ h)
f(x+ h)− f(x)
h
+ f(x)
g(x+ h)− g(x)
h
. (4.6)
Now g is differentiable at x, so g is also continuous at x (see Theorem 4.5.1). Hence g(x+h)→ g(x)
as h→ 0. Therefore (4.6) implies that
(fg)(x + h)− (fg)(x)
h
→ g(x)f ′(x) + f(x)g′(x)
as h→ 0.
Proofs of other differentiation rules are found in most undergraduate calculus textbooks.
4.4 Implicit differentiation
(Ref: SH10 §3.7)
In Section 1.8 we introduced functions that are defined by an equation relating the variables x and
y. Using the chain rule, one can calculate the derivative of such functions without first expressing
y explicitly as a function of x. We begin with a simple example to illustrate the principle. The
point to keep in mind is that y is a function of x.
Example 4.4.1. Suppose that y is a function of x, and that y and x are related by the formula
x2 + y2 = 1.
Calculate dydx .
Solution. We begin with the equation
x2 + y2 = 1
and differentiate both sides with respect to x to obtain
d
dx
(x2) +
d
dx
(y2) =
d
dx
(1). (4.7)
c©2020 School of Mathematics and Statistics, UNSW Sydney
4.4. IMPLICIT DIFFERENTIATION 85
Now clearly, ddx(x
2) = 2x and ddx(1) = 0. To simplify the central term of (4.7), we regard y as a
function of x and suppose that g(y) = y2. The chain rule gives
d
dx
(y2) =
dg
dx
=
dg
dy
dy
dx
= 2y
dy
dx
.
So continuing from (4.7) we have
2x+ 2y
dy
dx
= 0.
By isolating dydx we obtain
dy
dx
=
−2x
2y
= −x
y
as required.
(We note in passing that this answer coincides with the derivative obtained by first expressing
y as a function of x and then differentiating. In particular, if
y = (1− x2)1/2
then
dy
dx
=
1
2
(1− x2)−1/2(−2x) = − x
(1− x2)1/2 = −
x
y
as above.)
In general, if g is a function of y and y is a function of x then
d
dx
g(y) = g′(y)
dy
dx
(4.8)
by the chain rule. The use of this formula is illustrated below.
Example 4.4.2. Suppose that y is a function of x, implicitly related by the equation
y4 + x3 − x2e3y = 4. (4.9)
Find the equation of the tangent to the corresponding curve at the point where (x, y) = (2, 0).
In Section 4.2 we saw that ddxx
n = nxn−1 whenever n is an integer. A similar formula holds
when n is any rational number. This can be easily proved using implicit differentiation.
Proposition 4.4.3. Suppose that q is a rational number. Then
d
dx
xq = qxq−1.
c©2020 School of Mathematics and Statistics, UNSW Sydney
86 CHAPTER 4. DIFFERENTIABLE FUNCTIONS
Proof. Since q is a rational number, there are integers m and n such that q = mn and n 6= 0. Now
if y = xq then y = xm/n and taking the nth power of both sides gives
yn = xm.
Differentiating both sides with respect to x yields
nyn−1
dy
dx
= mxm−1.
Hence
dy
dx
=
mxm−1
nyn−1
=
m
n
xm−1
xq(n−1)
= qx(m−1)−qn+q
= qxq−1
as required.
4.5 Differentiation, continuity and split functions
(Ref: SH10 §3.1)
Not every function that is continuous at a point a is differentiable at a. (Consider, for example,
the function f , given by f(x) = |x|, at the point 0.) However, we do have the following result.
Theorem 4.5.1. If f is differentiable at a then f is continuous at a.
Proof. Suppose that f is differentiable at the point a. Then
lim
h→0
f(a+ h)− f(a)
h
exists. To show that f is continuous at a, we need to show that
lim
x→a f(x) = f(a),
or equivalently, that
lim
h→0
f(a+ h) = f(a).
Now
lim
h→0
f(a+ h)− f(a) = lim
h→0
(
f(a+ h)− f(a)
h
× h
)
=
(
lim
h→0
f(a+ h)− f(a)
h
)
× lim
h→0
h (4.10)
= f ′(a)× 0
= 0.
c©2020 School of Mathematics and Statistics, UNSW Sydney
4.5. DIFFERENTIATION, CONTINUITY AND SPLIT FUNCTIONS 87
(Note that we can separate the limits in (4.10) because we know that the limit of the difference
quotient exists.) Hence
lim
h→0
f(a+ h) = f(a),
proving the theorem.
The contrapositive form of the theorem is stated below.
Corollary 4.5.2. If f is not continuous at a then it is not differentiable at a.
Remark 4.5.3. Differentiability is a much stronger property than continuity. While every differen-
tiable function is continuous, there exist functions that are continuous everywhere but differentiable
nowhere. In fact, functions that are differentiable everywhere are a very rare breed, even among the
continuous functions. An example of a function that is continuous everywhere but differentiable
nowhere is the Weierstrass function whose graph is shown below.
x
y
The next function studied is an example of a ‘split function’. Whether or not the function is
differentiable at the ‘split point’ can be determined by calculating left- and right-hand limits of the
difference quotient.
Example 4.5.4. Suppose that f : R→ R is defined by
f(x) =
{
sinx if x ≥ 0
x2 + bx+ c if x < 0,
where b and c are real numbers. Find all possible values of b and c such that f is (a) continuous at
0 and (b) differentiable at 0.
Solution. (a) For f to be continuous at 0, we require that
lim
x→0−
f(x) = f(0) = lim
x→0+
f(x).
Now f(0) = sin 0 = 0 while
lim
x→0−
f(x) = c and lim
x→0+
f(x) = 0.
Hence f is continuous if and only if c = 0 and b ∈ R.
(b) If f is differentiable at 0 then it is continuous at 0 and hence c = 0 by part (a). Also, for
differentiability we require that
lim
h→0−
f(0 + h)− f(0)
h
= lim
h→0+
f(0 + h)− f(0)
h
.
c©2020 School of Mathematics and Statistics, UNSW Sydney
88 CHAPTER 4. DIFFERENTIABLE FUNCTIONS
Now,
lim
h→0+
f(0 + h)− f(0)
h
= lim
h→0+
sinh− sin 0
h
= lim
h→0+
sinh
h
= 1
while
lim
h→0−
f(0 + h)− f(0)
h
= lim
h→0−
h2 + bh− 0
h
= lim
h→0−
h+ b = b.
Hence we require that b = 1. So f is differentiable at 0 if and only if b = 1 and c = 0. The graph
of the corresponding function is sketched below.
x
f(x)
||
|
−1
1
2π
An important application of split functions is the theory of splines. A spline is a function
defined piecewise by polynomials. They are frequently used in computer graphics and elsewhere
because of their simplicity and their capacity to approximate complex curves. The following is an
example of a spline.
Example 4.5.5.
f(x) =
{
(x− 1)2 − 1 if − 2 ≤ x < 0
1− (x− 1)2 if 0 ≤ x ≤ 2.
In this example, each of the functions p(x) = (x − 1)2 − 1 and q(x) = 1 − (x − 1)2 have had
their domains restricted and then ‘glued’ together at the point 0.
If the constituent functions are continuous and differentiable in some interval containing the
point a, then the following theorem tells us when the corresponding split function is differentiable.
Theorem 4.5.6. Suppose a is a fixed real number and the function f is defined by
f(x) =
{
p(x) if x ≥ a
q(x) if x < a,
where p(x) and q(x) are continuous and differentiable in some interval containing a. Then if f is
continuous at a and p′(a) = q′(a), then f is differentiable at x = a.
This theorem in particular can be applied to splines since the constituent functions are polyno-
mials.
c©2020 School of Mathematics and Statistics, UNSW Sydney
4.6. DERIVATIVES AND FUNCTION APPROXIMATION 89
Example 4.5.7.
f(x) =
{
sinx if x ≥ 0
x2 + x if x < 0.
The functions sinx and x2 + x are continuous and differentiable everywhere. We saw in Example
4.5.4 that f is continuous at x = 0. Also, with p(x) = sinx and q(x) = x2 + x, we have p′(0) =
cos 0 = 1 and q′(0) = 1 so f is differentiable at x = 0.
4.6 Derivatives and function approximation
(Ref: SH10 §4.11)
The last three sections of this chapter are devoted to some applications of the derivative.
Suppose that a function f is differentiable at a. Consider its graph and the tangent to f at a,
shown below.
x
y
|
|f(a)
a
b
The equation of the tangent to f at a may be calculated using the point-gradient formula:
y − f(a) = f ′(a)(x − a)
y = f ′(a)(x − a) + f(a).
Observe in the above diagram that the tangent line is close to the graph of f for all points that are
close to a. In other words,
f(x) ≈ f ′(a)(x− a) + f(a) (4.11)
when x is close to a.
Formula (4.11) reflects the fact that every differentiable function can be locally approximated
by a linear function. We use this principle in the following example.
Example 4.6.1. Estimate
√
9.001 without using a calculator.
Solution. Suppose that f(x) =
√
x. Since 9.001 is close to 9 and f is a continuous function, we
could use the approximation √
9.001 = f(9.001) ≈ f(9) = 3. (4.12)
However, we can do better than this. We shall approximate f with a linear function at the
point 9 using (4.11). Since f ′(9) = 1/6 we have
f(x) ≈ (x− 9)/6 + 3
c©2020 School of Mathematics and Statistics, UNSW Sydney
90 CHAPTER 4. DIFFERENTIABLE FUNCTIONS
whenever x is close to 9. When x = 9.001 this gives
√
9.001 = f(9.001) ≈ (9.001 − 9)/6 + 3 = 3 + 1/6000. (4.13)
The table below gives the error involved in each approximation.
Approximation Error
Approximation (4.12) 1.667−4
Approximation (4.13) 4.637−9
Clearly (4.13) is the superior approximation.
4.7 Derivatives and rates of change
(Ref: SH10 §3.4)
Many physical processes involve quantities (such as temperature, volume, concentration, velocity)
that change with time. If Q is a quantity that varies with time, then the derivative dQdt gives the
rate of change of that quantity with respect to time. The chain rule is a major tool for answering
questions about rates of quantities which are related.
Example 4.7.1. A spherical balloon is being inflated and its radius is increasing at a constant
rate of 6 mm/sec. At what rate is its volume increasing when the radius of the balloon is 20 mm?
Solution. Suppose that V (t) is the volume of the balloon and r(t) is its radius at time t. We are
told that drdt = 6 and we need to find
dV
dt when r = 20. By the chain rule,
dV
dt
=
dV
dr
dr
dt
.
Now V = 43πr
3 so dVdr = 4πr
2. Hence
dV
dt
= 4πr2 × 6.
When r = 20,
dV
dt
= 24π(20)2 = 9600π.
So the volume is increasing at a rate of 9600π mm3/sec when the radius is 20 mm.
The above example illustrates an approach to solving such problems.
1. Define variables for the quantities involved.
2. Write down what is known in terms of these variables and their derivatives.
3. Write down what you need to find in terms of the these variables and their derivatives.
4. Write down anything else you know that relates the variables (for example, a volume or area
formula).
5. Use the chain rule (or implicit differentiation) to find the relevant derivative.
c©2020 School of Mathematics and Statistics, UNSW Sydney
4.8. LOCAL MAXIMUM, LOCAL MINIMUM AND STATIONARY POINTS 91
Sometimes drawing a diagram helps in the first few steps.
The next example follows this procedure.
Example 4.7.2. A point P moves to the right along the positive x-axis at a constant rate of 5
cm per second, and a point Q moves up the positive y-axis at a constant rate of 10 cm per second.
How fast is the distance between them changing when OP = 30 cm and OQ = 40 cm?
4.8 Local maximum, local minimum and stationary points
(Ref: SH10 §4.3)
In Example 3.4.4, we determined the absolute maximum and minimum points of a continuous
function over a closed interval. While the existence of these extreme points is given by the max-
min theorem, the theorem provides no systematic way of locating these points. In this section we
begin to develop a systematic approach to locating maxima and minima. A complete approach will
be presented in the next chapter.
We begin by defining what is meant by a local minimum point c. This is a point such that for
all x sufficiently close to c, f(c) ≤ f(x). A more precise statement is below.
x
cc− h c+ h
| | |
b
Local minimum point
Definition 4.8.1. Suppose that f is defined on some interval I. We say that a point
c in I is a local mininum point if there is a positive number h such that f(c) ≤ f(x)
whenever x ∈ (c − h, c + h) and x ∈ I. We say that a point d in (a, b) is a local
maximum point if there is a positive number h such that f(x) ≤ f(d) whenever
x ∈ (d− h, d + h) and x ∈ I.
The following theorem will be familiar from high school.
Theorem 4.8.2. Suppose that f is defined on (a, b) and has a local maximum or minimum point
at c for some c in (a, b). If f is differentiable at c then f ′(c) = 0.
Proof. Suppose that f is a local maximum at c. To show that f ′(c) = 0 we consider the difference
quotient
f(c+ h)− f(c)
h
.
c©2020 School of Mathematics and Statistics, UNSW Sydney
92 CHAPTER 4. DIFFERENTIABLE FUNCTIONS
Since f is a maximum at c, f(c+ h) ≤ f(c) for h sufficiently close to 0. So if h is sufficiently small
and positive,
f(c+ h)− f(c)
h
≤ 0,
while if h is sufficiently small and negative,
f(c+ h)− f(c)
h
≥ 0.
That is,
lim
h→0+
f(c+ h)− f(c)
h
≤ 0
while
lim
h→0−
f(c+ h)− f(c)
h
≥ 0.
Hence
lim
h→0
f(c+ h)− f(c)
h
= 0
which is equivalent to saying that f ′(c) = 0.
The proof when c is a local minimum is similar and will be omitted.
The set of points where the derivative of a function is zero is important. In particular, tangents
to the graph of the function at these points are horizontal lines. This motivates the following
definition.
Definition 4.8.3. If a function f is differentiable at a point c and f ′(c) = 0 then c
is called a stationary point of f .
Example 4.8.4. Suppose that f : R → R is given by f(x) = 4x5 − 5x4 − 40x3 − 2. Find all the
stationary points of f .
Solution. We begin by differentiating f and factorising the derivative:
f ′(x) = 20x4 − 20x3 − 120x2
= 20x2(x2 − x− 6)
= 20x2(x+ 2)(x− 3).
Hence the solutions to f ′(x) = 0 are x = −2, 0, 3. So the points −2, 0 and 3 are the stationary
points of f .
The graph of the function f appearing in Example 4.8.4 is shown below.
c©2020 School of Mathematics and Statistics, UNSW Sydney
4.9. MAPLE NOTES 93
x
f(x)
| |
|
|
200
−200
−2 3
It is clear that −2 is a local maximum point of f and that 3 is a local minimum point of f . However,
0 is neither a local minimum point nor local maximum point, and none of these points are global
maximum points or global minimum points. So while the local maxima and minima of a function
occur where f ′ is zero, it does not follow that any point c satisfying f ′(c) = 0 is a local maximum
point or local minimum point of f .
Tools for identifying whether a stationary point is a local maximum, local minimum or neither
are developed in the next chapter using an important result called the mean value theorem.
4.9 Maple notes
The Maple diff command is used to compute derivatives: diff(f(x),x); calculates
df
dx
and
diff(f(x),x$n); calculates
dnf
dxn
. For example,
> # Note Maple use of csc to denote function we usually call cosec
> diff(sqrt(csc(x^12-53*x^5-1)),x);
−12
√
csc (x12 − 53x5 − 1) cot (x12 − 53x5 − 1) (12x11 − 265x4)
> diff(x^3*exp(5*x),x);
3x2e5 x + 5x3e5x
> diff(%,x);
6xe5 x + 30x2e5x + 25x3e5x
> f2ndDeriv:=diff(x^3*exp(5*x),x$2);
f2ndDeriv := 6xe5 x + 30x2e5 x + 25x3e5x
c©2020 School of Mathematics and Statistics, UNSW Sydney
94 CHAPTER 4. DIFFERENTIABLE FUNCTIONS
Problems for Chapter 4
Problems 4.1 : Gradients of tangents and derivatives
1. [R] Using the definition of the derivative, show that:
a) if f(x) = x2 then f ′(x) = 2x;
b) if f(x) = x3 then f ′(x) = 3x2;
c) if f(x) = 1x then f
′(x) = −1x2 ;
d) [H] if f(x) =
√
x then f ′(x) = 1
2
√
x
.
Problems 4.2 : Rules for differentiation
2. [R] Find the derivative in each case.
a) f(x) = 5(x4 + 3x7) b) g(x) = (x4 − 2x)(4x2 + 2x+ 5)
c) h(y) =
y2
y3 + 8
d) f(x) = x(x2 − 4)1/2
e) f(t) = t/
√
t2 − 4 f) g(y) = sin 3y − 3 cos2 2y
g) g(x) = x4e−x h) f(x) = (x2 + 1) ln
√
x3 + 1
i) f(x) = ln(etan x) j) f(x) = ln(cos x)
3. [H] Suppose that f : R→ R is defined by f(x) = x|x| for all x in R.
a) If it exists, evaluate lim
h→0+
f(0 + h)− f(0)
h
.
b) If it exists, evaluate lim
h→0−
f(0 + h)− f(0)
h
.
c) State the value of f ′(0) or explain why f is not differentiable at 0.
4. [R] [V] Determine at which points each function f is (i) differentiable; (ii) continuous.
a) f(x) = |x| b) f(x) =
{
sinx if x ≤ 0
x if x > 0
c) f(x) =
x3 − 6x+ 4
x2 + 4x+ 4
5. [R] Sketch the graph of f , where f(x) = x1/3. Is f differentiable at 0? Give reasons.
6. [R] (An exercise on notation.) Suppose that f(x) = x+ cos 2x. Write down
a) f(x+ 17π) b) f ′(x+ nπ) c) f(2− x2)
d) f ′(2− x2) e) ddx f(2− x2).
c©2020 School of Mathematics and Statistics, UNSW Sydney
PROBLEMS FOR CHAPTER 4 95
Problems 4.4 : Implicit differentiation
7. [R] Find
dy
dx
in terms of x and y if
a) x3 + y3 = xy b) x2 −√xy + y2 = 6.
8. [R] Find
dy
dx
for the curve x4 + y4 = 16. Sketch the graph of the curve.
9. [R] [V] Find the equation of the line tangent to the curve x3 + y3 = 3(x + y) at the point
(1, 2).
Problems 4.5 : Differentiation, continuity and split functions
10. [R] Suppose that a and b are real numbers. Find all values of a and b (if any) such that the
functions f and g, given by
a) f(x) =
{
ax+ b if x < 0
sinx if x ≥ 0 and b) g(x) =
{
ax+ b if x < 0
e2x if x ≥ 0 ,
are (i) continuous at 0 and (ii) differentiable at 0.
11. [H] The function f : R→ R is is defined by
f(x) =
{√
x sin
√
x if x ≥ 0
ax+ b if x < 0,
where a and b are real numbers. Find all values of a and b (if any) such that f is differentiable
at 0.
Problems 4.6 : Derivatives and function approximation
12. [R] [V] Suppose that f(x) = 3
√
x.
a) Without using a calculator, give a rough estimate of f(8.01).
b) i) Find the equation of the tangent to f at the point (8, 2).
ii) Use your answer to part (i) to find a different approximation for f(8.01).
c) Using a calculator, determine the error for the approximation in (a) and in (b). Which
approximation is better?
Problems 4.7 : Derivatives and rates of change
13. [R] At a certain instant the side length of an equilateral triangle is a cm and this length is
increasing at r cm/sec. How fast is the area increasing?
c©2020 School of Mathematics and Statistics, UNSW Sydney
96 CHAPTER 4. DIFFERENTIABLE FUNCTIONS
14. [R] [V] A 5 m ladder is leaning against a vertical wall. Suppose that the bottom of the
ladder is being pulled away from the wall at a rate of 1 m/sec. How fast is the area of the
triangle underneath the ladder changing at the instant that the top of the ladder is 4 m from
the floor?
15. [R] A spherical balloon is to be filled with water so that there is a constant increase in the
rate of its surface area of 3 cm2/sec.
(The surface area A and volume V of a sphere of radius r is given by A = 4πr2 and V = 43πr
3.)
a) Find the rate of increase in the radius when the radius is 3 cm.
b) Find the volume when the volume is increasing at a rate of 10 cm3/sec.
16. [R]
a) A container in the shape of a right circular cone, of semi-vertical angle tan−1(12), is
placed vertex downwards with its axis vertical.
θ
θ = tan−1(12)
Water is poured in at the rate of 10mm3 per sec. Find the rate at which the depth,
hmm, is increasing when the depth of water in the cone is 50mm.
b) [H] The cone is filled to a depth of 100mm and pouring is then stopped. A hole is
then opened at the vertex of the cone and water flows out of the hole at the rate of
50π
√
hmm3 per second, where h is the depth at time t. Show that it takes 200 seconds
to empty the cone.
c©2020 School of Mathematics and Statistics, UNSW Sydney
97
Chapter 5
The mean value theorem and
applications
The mean value theorem is one of the most important results for establishing the theoretical
structure of calculus. Several results that you are familiar with from high school calculus are based
on the mean value theorem. Applications of the mean value theorem include
• identifying where a function is increasing or decreasing,
• identifying different types of stationary points,
• determining how many zeros a polynomial has,
• evaluating limits which are indeterminate forms of type ∞∞ and 00 ,
• proving useful inequalities and
• estimating errors in approximations.
We begin the chapter by introducing the theorem.
5.1 The mean value theorem
(Ref: SH10 §4.1)
The basic idea behind the mean value theorem is straightforward. Consider a differentiable function
f and choose an interval [a, b]. We construct a chord AB joining the points (a, f(a)) and (b, f(b))
as shown in Figure 5.1. Because the function is continuous and differentiable, there is a point c in
(a, b) such that tangent to the graph of f at c lies parallel to the chord AB.
Since the tangent and the chord are parallel, their gradients are equal. The gradient of the
tangent at c is f ′(c) while the gradient of the chord is given by
f(b)− f(a)
b− a
(using the ‘rise over run’ formula). Hence
f(b)− f(a)
b− a = f
′(c).
c©2020 School of Mathematics and Statistics, UNSW Sydney
98 CHAPTER 5. THE MEAN VALUE THEOREM AND APPLICATIONS
x
f(x)
| | |
|
|
f(a)
f(b)
a c b
A
B
C
Figure 5.1: The mean value theorem – AB is parallel to the tangent at C.
A precise statement of the theorem is given below.
Theorem 5.1.1 (The mean value theorem). Suppose that f is continuous on [a, b] and differentiable
on (a, b). Then there is at least one real number c in (a, b) such that
f(b)− f(a)
b− a = f
′(c).
The proof of the theorem will be presented in the next section. For now we illustrate the
theorem with a simple example.
Example 5.1.2. Suppose that f : [1, 8] → R is given by f(x) = 3√x. Find a number c in (1, 8)
that satisfies the conclusions of the mean value theorem for f on [1, 8].
Solution. Note that f is continuous on [1, 8] and differentiable on (1, 8). By the mean value theorem
there is a real number c in (1, 8) such that
f(8)− f(1)
8− 1 = f
′(c).
Now f ′(x) = 13x
−2/3 so the above equation becomes
2− 1
7
=
1
3
3
√
c2
.
Rearranging to find c gives
3
√
c2 =
7
3
c =
√
343
27
.
Since c ≈ 3.56 it is clear that c ∈ (1, 8). This is illustrated graphically below.
c©2020 School of Mathematics and Statistics, UNSW Sydney
5.2. PROOF OF THE MEAN VALUE THEOREM 99
x
3
√
x
| | |
|
|
1
2
1 c 8
b
b
b
5.2 Proof of the mean value theorem
(Ref: SH10 §4.1)
In this section we prove the mean value theorem. The proof is part of MATH1141 only, so
MATH1131 students may want to jump straight to Section 5.3.
To prove the mean value theorem, we begin by considering the special case when f(a) = f(b).
This case is known as Rolle’s theorem and is illustrated in Figure 5.2.
Theorem 5.2.1. Suppose that g is continuous on [a, b] and differentiable on (a, b). Suppose also
that g(a) = g(b) = 0. Then there is a real number c in (a, b) such that g′(c) = 0.
Proof. The proof considers three cases.
Case 1: Suppose that g is the constant function given by g(x) = 0. Then g′(c) = 0 for every c
in (a, b) (see Section 4.2).
Case 2: Suppose that there is a point d in (a, b) such that g(d) > 0. By the max-min theorem
(Theorem 3.4.3), g attains a maximum value at some point c in [a, b]. Moreover, c cannot be a or
b since g(d) > g(a) = g(b) = 0. Hence c lies in (a, b), and since g is differentiable on (a, b), we have
g′(c) = 0 (see Theorem 4.8.2).
Case 3: Suppose that g(x) ≤ 0 for all x in [a, b] and that g is not constant on [a, b]. Then g
attains a minimum at a point c in (a, b), and similarly to Case 2 one can show that g′(c) = 0.
c©2020 School of Mathematics and Statistics, UNSW Sydney
100 CHAPTER 5. THE MEAN VALUE THEOREM AND APPLICATIONS
x
g(x)
a c b
A B
C
Figure 5.2: Rolle’s theorem – the tangent at C is horizontal.
From here it is not hard to see how to prove the mean value theorem. Consider the graph of
the function f in Figure 5.1 and ‘subtract’ the chord AB. The result will look something like the
graph of g in Figure 5.2, to which we can apply Rolle’s theorem and obtain a horizontal tangent
at C. We then ‘add’ the chord back again to obtain the tangent to f of Figure 5.1. This rough
geometric argument is made rigorous below.
Proof of the mean value theorem. Suppose that f is continuous on [a, b] and differentiable on (a, b).
We consider the function g given by
g(x) = f(x)−
[
f(b)− f(a)
b− a (x− a) + f(a)
]
.
(The part in square brackets is taken from the equation of the chord AB in Figure 5.1.) One can
check that g is continuous on [a, b], differentiable on (a, b) and that g(a) = g(b) = 0. By Rolle’s
theorem there is a c in (a, b) such that g′(c) = 0; that is, such that
f ′(c)−
[
f(b)− f(a)
b− a
]
= 0.
If we rearrange this equation then we obtain
f ′(c) =
f(b)− f(a)
b− a
as required.
5.3 Proving inequalities using the mean value theorem
Our first application of the mean value theorem will be to prove some useful inequalities. We begin
with an example.
Example 5.3.1. By using the mean value theorem, show that lnx < x− 1 whenever x > 1.
c©2020 School of Mathematics and Statistics, UNSW Sydney
5.3. PROVING INEQUALITIES USING THE MEAN VALUE THEOREM 101
Solution. Suppose that x > 1 and consider the closed interval [1, x]. We define a function f :
[1, x] → R by f(t) = ln t. Now f is continuous on [1, x] and differentiable on (1, x). So we may
apply the mean value theorem to f on the interval [1, x]. Now f ′(t) = 1/t so
f(x)− f(1)
x− 1 =
1
c
for some c in (1, x) by the mean value theorem. That is,
lnx
x− 1 =
1
c
for some c between 1 and x. Since c > 1 we have 1/c < 1 and hence
lnx
x− 1 < 1.
Rearranging the inequality gives lnx < x− 1 as desired.
The technique used in the above example is as follows. Consider a function f that is continuous
on the interval [a, x] and differentable on the interval (a, x). The mean value theorem gives
f(x)− f(a)
x− a = f
′(c) (5.1)
for some c between a and x. If f ′(c) < M then
f(x)− f(a)
x− a < M
and hence
f(x) < M(x− a) + f(a). (5.2)
whenever x > a.
Sometimes it is desirable for the inequality to go the other way. In this case we find m such
that f ′(c) > m. Then, instead of (5.2), we obtain
f(x) > m(x− a) + f(a).
Often, the tricky part of the process is determining which function f will give the desired
inequality.
Example 5.3.2. Use the mean value theorem to show that
√
x+ 4− 2 < x
4
whenever x > 0.
Solution. Fix positive x. The first step is to identify a good choice of f . If we rewrite the inequality
as √
x+ 4 <
x
4
+ 2
c©2020 School of Mathematics and Statistics, UNSW Sydney
102 CHAPTER 5. THE MEAN VALUE THEOREM AND APPLICATIONS
then we have a similar form to inequality (5.2). This suggests that we should define f : [0, x]→ R
by the formula f(t) =
√
t+ 4. Then f is continuous on [0, x] and differentiable on (0, x). An
application of the mean value theorem gives
f(x)− f(0)
x− 0 =
1
2
√
c+ 4
for some c in (0, x). Now the right-hand side is bounded above by 1
2
√
4
, so we have
√
x+ 4−√0 + 4
x− 0 <
1
2
√
4
.
Simplifying and rearranging gives √
x+ 4− 2 < x
4
as required.
5.4 Error bounds
A second application of the mean value theorem is in calculating error bounds. Suppose that a
calculation involves
√
26. Clearly
√
26 is close to
√
25, but the latter is easier to work with because
it is equal to 5. If we use 5 instead of
√
26 then calculations are made easier with the tradeoff that
we introduce an error. How bad is this error?
Example 5.4.1. Use the mean value theorem to find an upper bound for the error involved if we
approximate
√
26 by
√
25.
Solution. The precise error in the approximation is given by
√
26−
√
25.
We apply the mean value theorem to the function f , given by f(x) =
√
x, on the interval [25, 26].
This gives √
26−√25
26− 25 =
1
2
√
c
for some c in (25, 26). Hence
error =
√
26−
√
25
=
√
26−√25
26− 25
=
1
2
√
c
<
1
2
√
25
=
1
10
,
since c > 25. Hence an upper bound for the error is 1/10.
We illustrate the process with one more example.
Example 5.4.2. Use the mean value theorem to find an upper bound for the error involved if we
approximate sin 15π21 by sin
2π
3 .
c©2020 School of Mathematics and Statistics, UNSW Sydney
5.5. THE SIGN OF A DERIVATIVE 103
5.5 The sign of a derivative
(Ref: SH10 §§4.2, 4.3)
Suppose that a function f is differentiable at 4 and that f ′(4) = −2. In high school, it was taught
that, since the derivative of f at 4 is negative, the function f is decreasing at the point 4. This
makes intuitive sense since the gradient of the tangent to f at the point 4 is −2 and so the tangent
slopes downward. In this section, these ideas are given a rigorous foundation by using the mean
value theorem.
Definition 5.5.1. Suppose that a function f is defined on an interval I. We say
that
a) f is increasing on I if for every two points x1 and x2 in I,
x1 < x2 implies that f(x1) < f(x2);
b) f is decreasing on I if for every two points x1 and x2 in I,
x1 < x2 implies that f(x1) > f(x2).
Example 5.5.2. The function f : R→ R given by f(x) = x2 is increasing on [0,∞) and decreasing
on (−∞, 0].
x
f(x)
decreasing increasing
The following theorem should be familiar to you. It is proved using the mean value theorem.
Theorem 5.5.3. Suppose that f is continuous on [a, b] and differentiable on (a, b).
(i) If f ′(x) > 0 for all x in (a, b) then f is increasing on [a, b].
(ii) If f ′(x) < 0 for all x in (a, b) then f is decreasing on [a, b].
(iii) If f ′(x) = 0 for all x in (a, b) then f is constant on [a, b].
Proof. We only prove statement (i) since the proofs of statements (ii) and (iii) are similar. Suppose
that f ′(x) > 0 for all x in (a, b) and choose two points x1 and x2 in [a, b] such that x1 < x2. By
Definition 5.5.1, it suffices to show that f(x1) < f(x2).
Since f is differentiable on I, it is continuous on [x1, x2] (see Theorem 4.5.1) and differentiable
on (x1, x2). Hence, by the mean value theorem, there is a c in (x1, x2) such that
f(x2)− f(x1)
x2 − x1 = f
′(c).
c©2020 School of Mathematics and Statistics, UNSW Sydney
104 CHAPTER 5. THE MEAN VALUE THEOREM AND APPLICATIONS
But f ′(c) > 0 and x2 − x1 > 0. This means that f(x2) − f(x1) > 0 and hence f(x1) < f(x2) as
required.
Theorem 5.5.3 has several important applications. The first of these is classifying different
types of stationary points. The basic idea is as follows. Suppose that f is differentiable on the
open interval (1, 3) and that for some small positive number h we have the following:
• f ′(x) > 0 for all x in (2− h, 2),
• f ′(2) = 0, and
• f ′(x) < 0 for all x in (2, 2 + h).
Now f is differentiable on (2 − h, 2 + h) so it is also continuous on this interval. Moreover, f
increases on the interval (2− h, 2), is stationary at 2 and then decreases on the interval (2, 2 + h).
From these facts we conclude that the stationary point 2 is a local maximum point for f .
Example 5.5.4. Find and classify all stationary points of the function f : R→ R whose derivative
is given by
f ′(x) = (x− 4)(x− 1)(x+ 5)2.
Solution. The stationary points, which are found by solving the equation f ′(x) = 0, are 4, 1 and
−5. To classify each stationary point, we examine the sign of f ′ on small intervals either side of
each stationary point. This process is concisely documented in the following table.
−5− −5 −5+ 1− 1 1+ 4− 4 4+
x− 4 − − − − − − − 0 +
x− 1 − − − − 0 + + + +
(x+ 5)2 + 0 + + + + + + +
f ′(x) + 0 + + 0 − − 0 +
Gradient upslope − upslope upslope − − upslope
For example, the notation 4− is shorthand for those points x lying in a small interval to the
immediate left of 4. For such a point x, the factor x − 4 is negative, while the factors x − 1 and
(x+ 5)2 are positive. Hence f ′(x) is negative when x is slightly to the left of 4. Other columns of
the table are filled by a similar process.
In conclusion, 4 is a local minimum point, 1 is a local maximum point and −5 is neither a
maximum nor minimum point. (In fact, −5 is called a horizontal point of inflexion since the
tangent to f at −5 is horizontal and the concavity of the function changes about this point. See
any standard undergraduate calculus text or any advanced high school calculus text for further
details.)
5.6 The second derivative and applications
(Ref: SH10 §4.3)
In this section we give another method for classifying the stationary points of a function f . To do
so, we use the second derivative of f , which is denoted by f ′′.
c©2020 School of Mathematics and Statistics, UNSW Sydney
5.6. THE SECOND DERIVATIVE AND APPLICATIONS 105
Example 5.6.1. Find the second derivative of the function f : R→ R, given by the rule
f(x) = 3x4 + 2x− sinx.
Solution. The first derivative of f is given by
f ′(x) = 12x3 + 2− cos x.
To find the second derivative f ′′ of f , we simply differentiate f ′. Hence
f ′′(x) = 36x2 + sinx.
Remark 5.6.2. If y = f(x) then the first derivative of f is often written as dydx or y
′ while the
second derivative of f is often written as d
2y
dx2
or y′′.
Remark 5.6.3. Even if a function is differentiable at a point a, the function may not have a second
derivative at a. (Consider, for example, the function f : R→ R given by
f(x) = x4/3,
at the point 0.) A function which has a second derivative at a is called twice differentiable at a.
If f is twice differentiable, the sign of f ′′ can often be used to determine whether a stationary
point is a local maximum point or a local minimum point.
Theorem 5.6.4 (The second derivative test). Suppose that a function f is twice differentiable on
(a, b) and that c ∈ (a, b).
(i) If f ′(c) = 0 and f ′′(c) > 0 then c is a local minimum point of f ;
(ii) If f ′(c) = 0 and f ′′(c) < 0 then c is a local maximum point of f .
The proof of the theorem uses the following result.
Lemma 5.6.5. Suppose that g is differentiable at a point c.
(i) If g′(c) > 0 then g(c − h) < g(c) < g(c + h) for all positive h sufficiently small.
(ii) If g′(c) < 0 then g(c + h) < g(c) < g(c − h) for all positive h sufficiently small.
Sketch proof. We give an intuitive argument for statement (i) and leave (ii) as an exercise. Suppose
that g′(c) > 0. By the definition of the derivative,
lim
h→0
g(c + h)− g(c)
h
= g′(c). (5.3)
That is,
g(c+ h)− g(c)
h
≈ g′(c)
when h is close to 0. Now, if h > 0 then
g(c + h)− g(c) ≈ hg′(c) > 0
c©2020 School of Mathematics and Statistics, UNSW Sydney
106 CHAPTER 5. THE MEAN VALUE THEOREM AND APPLICATIONS
Hence
g(c+ h)− g(c) > 0 (5.4)
provided that h > 0 and that h is sufficiently small.
On the other hand, the definition of the derivative also implies that
lim
h→0
g(c − h)− g(c)
−h = g
′(c)
(to see this, simply replace h with −h in (5.3)). Now, if h > 0 and h is close to 0 then
g(c) − g(c− h) ≈ hg′(c) > 0.
Hence
g(c)− g(c − h) > 0 (5.5)
provided that h > 0 and that h is sufficiently small.
If we combine (5.4) and (5.5) then we have
g(c − h) < g(c) < g(c + h)
for all positive h sufficiently small.
Proof of Theorem 5.6.4. We prove statement (i) and leave the proof of (ii) as an exercise. Suppose
that f ′(c) = 0, f ′′(c) > 0 and that g = f ′. Then g′(c) > 0 and
g(c − h) < g(c) < g(c + h)
for all positive h sufficiently small, by Lemma 5.6.5. Since f ′(c) = g(c) = 0 we conclude that
f ′(c− h) < 0 < f ′(c+ h)
for all positive h sufficiently small. So f ′ is positive on a small interval immediately to the right
of c, zero at x and negative on a small interval immediately to the left of c. By Theorem 5.5.3 we
now conclude that f has a local minimum at the point c.
Example 5.6.6. Find and classify the stationary points of the function f : R→ R given by
f(x) = x3 − 6x2 + 9x− 5.
Solution. First we calculate the first and second derivatives of f :
f ′(x) = 3x2 − 12x+ 9 = 3(x− 1)(x − 3)
f ′′(x) = 6x− 12.
Hence f has stationary points at 1 and 3. Now f ′′(1) = −6 < 0, so f has a local maximum point
at 1. For second stationary point, f ′′(3) = 6 > 0, so f has a local minimum stationary point at 3.
(To illustrate these conclusions, we have included the graph of f , shown below.)
c©2020 School of Mathematics and Statistics, UNSW Sydney
5.7. CRITICAL POINTS, MAXIMA AND MINIMA 107
x
f(x)
| |
1 3
−5
Remark 5.6.7. Suppose that f ′(c) = 0 and that f ′′(c) = 0. Then we cannot apply the second
derivative test. In fact, the point c may be a local maximum, a local minimum or neither of these.
The examples below illustrate various possibilities.
x
x4
x
−x4
x
x3
• If f(x) = x4 then f ′(0) = f ′′(0) = 0 and there is a local minimum at 0.
• If f(x) = −x4 then f ′(0) = f ′′(0) = 0 and there is a local maximum at 0.
• If f(x) = x3 then f ′(0) = f ′′(0) = 0 and there is neither a a local maximum and minimum
at 0 (in fact, we have a horizontal point of inflexion at 0).
Hence if f ′(c) = f ′′(c) = 0 then it is best to classify the stationary point c by examining the sign
of the derivative on either side of c, as illustrated in Section 5.5.
5.7 Critical points, maxima and minima
(Ref: SH10 §§4.3–4.5)
By combining results from Chapters 3, 4 and 5, we now give a comprehensive presentation of how
to locate global maxima and minima for any real-valued continuous function defined on a closed
interval.
Suppose that a function f : [a, b] → R is continuous on [a, b]. By the max-min theorem
(Theorem 3.4.3), f attains a global maximum and a global minimum on [a, b]. Suppose that the
global maximum occurs at the point c. Now either c is one of the endpoints a or b, or c lies in the
open interval (a, b). In the latter case, f is either differentiable at c or it is not. If it is differentiable
at c, then f ′(c) = 0 by Theorem 4.8.2.
Thus we have identified that a global maximum point c must satisfy one of the three following
properties:
c©2020 School of Mathematics and Statistics, UNSW Sydney
108 CHAPTER 5. THE MEAN VALUE THEOREM AND APPLICATIONS
• c is one of the endpoints a or b of the closed interval [a, b],
• f is differentiable at c and f ′(c) = 0, or
• f is not differentiable at c.
The set of points satisfying these properties is useful so we give it a special name.
Definition 5.7.1. Suppose that f is defined on [a, b]. We say that a point c in [a, b]
is a critical point for f on [a, b] if c satisfies one of the following properties:
(a) c is an endpoint a or b of the interval [a, b],
(b) f is not differentiable at c, or
(c) f is differentiable at c and f ′(c) = 0.
Theorem 5.7.2. Suppose that f is continuous on [a, b]. Then f has a global maximum and global
minimum on [a, b]. Moreover, the global maximum point and the global minimum point are both
critical points for f on [a, b].
The proof of the theorem follows the same reasoning given at the beginning of this section. The
theorem provides a systematic way of finding global minima and maxima of functions.
Example 5.7.3. Suppose that f : R→ R is given by
f(x) = 3(x− 1)2/3.
Find the absolute maximum and minimum values of f on [0, 2].
Solution. The derivative of f is given by
f ′(x) = 2(x− 1)−1/3 = 2
3
√
x− 1 .
Clearly f is not differentiable at 1. Morover, its derivative is never zero. So the only critical
points of f on [0, 2] are 1 and the endpoints 0 and 2. By Theorem 5.7.2, the global maximum and
minimum points must be among the points 0, 1 and 2. But
f(0) = 3, f(1) = 0 and f(2) = 3.
Hence the global minimum value of f on [0, 2] is 0 while the global maximum value of f on [0, 2] is
3. (See Figure 5.3 (a) for a sketch of f that shows its critical points.)
Example 5.7.4. Suppose that the function f : R→ R is given by the rule
f(x) = |x2 − 3x− 4|.
Find the absolute maximum and absolute minimum values of f on the interval [0, 5].
c©2020 School of Mathematics and Statistics, UNSW Sydney
5.7. CRITICAL POINTS, MAXIMA AND MINIMA 109
x
f(x)
b
b
b
|
1 2
3
(a)
x
f(x)
b
b
b
b
| |
4
3
2
4 5
(b)
Figure 5.3: Graphs of functions and their critical points.
x
y
| |
−2 3
(a)
x
g(x)
b
b
b
b
| |
−2 3
|34
(b)
Figure 5.4: Graphs for Example 5.7.5.
c©2020 School of Mathematics and Statistics, UNSW Sydney
110 CHAPTER 5. THE MEAN VALUE THEOREM AND APPLICATIONS
Example 5.7.5. Consider the parabolic arc
{(x, y) ∈ R2 : y = x2 − x− 1,−2 ≤ x ≤ 3}
(see Figure 5.4 (a)). Find where the arc is (i) closest to and (ii) furthest from the origin.
Solution. If (x, y) is a point on the arc then its distance from the origin is given by√
x2 + y2.
To find where the arc is closest to the origin, we want to minimise
√
x2 + y2 subject to the con-
straints y = x2− x− 1 and −2 ≤ x ≤ 3. In other words, we want to find a global minimum for the
function
f(x) =
√
x2 + (x2 − x− 1)2
on the interval [−2, 3]. Because calculations involving square roots can get messy, we instead
consider the function g, given by
g(x) = x2 + (x2 − x− 1)2
on [−2, 3]. (Note that g(x) may be interpreted as the square of the distance between the origin
and a point on the arc with coordinate (x, x2 − x− 1). The global minimum of both f and g will
occur at the same point x.)
We seek the critical points of g, so we compute its derivative:
g′(x) = 2x+ 2(x2 − x− 1)(2x − 1)
= 2(x+ 2x3 − 2x2 − 2x− x2 + x+ 1)
= 2(2x3 − 3x2 − x2 + 1)
= 2(x− 1)(2x2 − x− 1)
= 2(x− 1)(x − 1)(2x + 1).
Hence the critical points of g on [−2, 3] are −2, −1/2, 1 and 3. Now
g(−2) = 29, g(−1/2) = 5/16, g(1) = 2, g(3) = 34
(see Figure 5.4 (b)). Therefore the arc is closest to the origin when x = −1/2 and furthest from
the origin when x = 3.
5.8 Counting zeros
In this section we combine the intermediate value theorem with results about the sign of derivatives
to determine how many solutions there are to various equations. The main ideas are illustrated in
the next example.
Example 5.8.1. Determine how many real numbers satisfy the equation
x3 − 6x2 − 15x+ 8 = 0. (5.6)
c©2020 School of Mathematics and Statistics, UNSW Sydney
5.8. COUNTING ZEROS 111
Solution. Suppose that f(x) = x3 − 6x2 − 15x + 8 for all real x. Ideally, we would look for one
root of f , factor this out and solve a quadratic equation to determine the remaining roots (if any).
However, finding one root by inspection is not straightforward, so we need another approach.
Note that f is differentiable (and hence continuous) everywhere. We begin by identifying
intervals where f is monotonically decreasing and intervals where f is monotonically increasing.
The derivative of f is given by
f ′(x) = 3x2 − 12x− 15 = 3(x− 5)(x+ 1).
So the stationary points of f are 5 and −1. Morover, the table below shows that f is increasing on
the intervals (−∞,−1] and [5,∞), while it is decreasing on [−1, 5].
Interval (−∞,−1) (−1, 5) (5,∞)
(x− 5) − − +
(x+ 1) − + +
f ′(x) + − +
We evaluate f at the (finite) endpoints of these intervals and see that
f(−1) = 16 and f(5) = −92.
Since f is a cubic, we can now give a rough sketch of the graph of f .
x
f(x)
b
b
(−1, 16)
(5,−92)
We now argue that f has exactly one root on each of the intervals mentioned above.
• On the interval [−1, 5], the function f is decreasing. Hence f cannot have more than one
root on [−1, 5]. On the other hand, since f(−1) > 0 and f(5) < 0 there is at least one c in
[−1, 5] such that f(c) = 0 by the intermediate value theorem. Hence f has exactly one root
on [−1, 5].
• On the interval (−∞,−1], the function f is increasing. Hence f cannot have more than one
root on (−∞,−1]. On the other hand, since f(−1) > 0 and f(−5) < 0 there is at least one c
in [−5,−1] such that f(c) = 0 by the intermediate value theorem. Hence f has exactly one
root on (−∞,−1].
• On the interval [5,∞), the function f is increasing. Hence f cannot have more than one
root on [5,∞). On the other hand, since f(5) < 0 and f(10) > 0 there is at least one c in
[5, 10] such that f(c) = 0 by the intermediate value theorem. Hence f has exactly one root
on [5,∞).
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112 CHAPTER 5. THE MEAN VALUE THEOREM AND APPLICATIONS
In conclusion, equation (5.6) has exactly three real solutions, one in each of the intervals [−5,−1],
[−1, 5] and [5, 10].
Guided by our solution to Example 5.8.1, we now outline a general approach for solving problems
of this type.
Suppose that f is continuous and differentiable everywhere and that its derivative is also con-
tinuous. To determine how many real solutions the equation f(x) = 0 has, one may follow the
procedure below.
1. Calculate f ′ and solve f ′(x) = 0.
2. Determine the intervals where the derivative is positive and the intervals where it is negative.
3. By step 2 we know the intervals where the function f is monotonically increasing, and the
intervals where f is monotonically decreasing.
4. Evaluate the function f at the endpoints of each interval. If f changes sign on the interval,
there is exactly one root on that interval. If it does not change sign, there are no roots on that
interval. (This step must be modified slightly for intervals of the form (−∞, a] or [b,∞).)
Example 5.8.2. Determine how many real numbers satisfy the equation
2x3 − 9x2 + 12x− 1 = 0. (5.7)
Solution. We follow the steps outlined above. Suppose that
f(x) = 2x3 − 9x2 + 12x− 1.
By differentiating f we obtain
f ′(x) = 6x2 − 18x+ 12 = 6(x− 1)(x− 2).
The stationary points of f are 1 and 2. By studying the sign of f ′, it is clear that f is increasing
on the intervals (−∞, 1) and (2,∞) while it is decreasing on (1, 2). Now
f(1) = 4 and f(2) = 3,
as illustrated in the sketch below.
x
f(x)
b
b
(1, 4)
(2, 3)
c©2020 School of Mathematics and Statistics, UNSW Sydney
5.9. ANTIDERIVATIVES 113
So it is clear that f has no zeros on [1, 2] or on [2,∞). On (−∞, 1), the function f has no more
than one zero (since it is increasing on this interval). Moreover, f changes sign on the interval [0, 1]
(since f(0) = −1 < 0 and f(1) > 0). Hence f has exactly one root on [0, 1] by the intermediate
value theorem.
In conclusion, equation (5.7) has exactly one real solution and this solution lies somewhere in
the interval [0, 1].
5.9 Antiderivatives
We open with an example.
Example 5.9.1. While filming a movie, a stunt man jumps out of a stationary helicopter 900
meters above the ground. The scriptwriter wants the man to make a 20 second mobile phone call
before opening the parachute. Physical considerations (gravity and air resistance) suggest that the
man’s velocity f(t) (in metres per second) without an open parachute is given by
f(t) = 50(1 − e−t/5),
where t is the number of seconds after jumping from the helicopter. How far will the man fall 20
seconds?
Solution. Write F (t) for the distance fallen (in metres) after t seconds. We note that F (0) = 0.
Since velocity is the rate of change of displacement, F satisfies the equation
F ′(t) = f(t).
(Any equation like this involving differentiation is called a differential equation). The differential
equation may be rewritten as
dF
dt
= 50− 50e−t/5. (5.8)
One can check by differentiation that if
F (t) = 50t+ 250e−t/5 + C, (5.9)
where C is a real constant, then F satisfies (5.8). Let’s assume (for the moment) that all possible
solutions to the differential equation (5.8) are of the form (5.9). Imposing the condition that
F (0) = 0 allows us to evaluate the unknown constant C:
0 = F (0) = 50× 0 + 250e0 + C,
whence C = −250. Therefore F (t) = 50t+ 250e−t/5 − 250. To complete the solution,
F (20) = 50× 20 + 250e−4 − 250 ≈ 754.6.
Hence the stunt man will fall about 755 metres in the first 20 seconds.
To be certain of this solution, we need to show that every function F satisfying the differential
equation (5.8) is of the form (5.9). This can be done using the mean value theorem. We begin by
introducing the notion of an antiderivative.
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114 CHAPTER 5. THE MEAN VALUE THEOREM AND APPLICATIONS
Definition 5.9.2. Suppose that f is continuous on an open interval I. A function
F is said to be an antiderivative (or a primitive) of f on I if F ′(x) = f(x) for all x in
I. The process of finding an antiderivative of a function is called antidifferentiation.
Example 5.9.3. (a) Suppose that
f(t) = 50− 50e−t/5 ∀t ∈ R
Then the function F , given by
F (t) = 50t+ 250e−2t − 250
is an antiderivative of f on (−∞,∞).
(b) Suppose that n is a positive integer and f(x) = xn. If
F (x) =
xn+1
n+ 1
and G(x) =
xn+1
n+ 1
+ 5
then F and G are both antiderivatives of f on the interval (−∞,∞). So f has more than
one antiderivative on (−∞,∞). In fact, if C is any real number and
H(x) = xn+1/(n+ 1) + C
then H is an antiderivative of f on (−∞,∞). Thus the function f has infinitely many
antiderivatives on (−∞,∞).
Example 5.9.3 (b) illustrates the general principle that if F is an antiderivative of f on I then
F + C
is also an antiderivative of f on I, for any real constant C. The next theorem says that all
antiderivatives are of this form.
Theorem 5.9.4. Suppose that f is a continuous function on an open interval I and that F and G
are two antiderivatives of f on I. Then there is a real constant C such that
G(x) = F (x) + C
for all x in I.
Proof. Suppose that F and G are two antiderivatives of f on I and let H denote the function given
by
H(x) = G(x)− F (x)
for all x in I. Then H is differentiable on I and
H ′(x) = G′(x)− F ′(x)
= f(x)− f(x)
= 0
c©2020 School of Mathematics and Statistics, UNSW Sydney
5.10. L’HOˆPITAL’S RULE 115
for all x in I. Hence there is a constant C such that H(x) = C for all x in I (by Theorem 5.5.3
(c)). Therefore
G(x) = F (x) +H(x) = F (x) + C
for all x in I.
Theorem 5.9.4 justifies the assumption made in the solution to Example 5.9.1. It also enables
us to write down all possible antiderivatives of some well-known functions. In the table below, C
is any real constant.
Function Antiderivative
xr, where r is rational and r 6= −1 1r+1xr+1 + C
sinx − cos x+ C
cos x sinx+ C
eax 1ae
ax + C
f ′(x)
f(x)
ln |f(x)|+C
5.10 L’Hoˆpital’s rule
(Ref: SH10 §§11.5, 11.6)
Consider the limit
lim
x→0
ex − 1
5x+ x2
.
Since both the numerator and denominator approach 0 as x → 0+, the limit is an example of an
indeterminate form of type 00 . None of the limit rules of Chapter 2 can be easily applied to evaluate
the limit. However, the mean value theorem gives another rule which helps in this situation.
Suppose that f(x) = ex − 1 and g(x) = 5x+ x2. Instead of considering the quotient f(x)/g(x)
(whose limit is difficult to calculate), we consider the quotient f ′(x)/g′(x) of derivatives (whose
limit is easier to calculate):
lim
x→0
f ′(x)
g′(x)
= lim
x→0
ex
5 + 2x
=
1
5
.
Since this second limit exists, l’Hoˆpital’s rule (which is stated below) implies that
lim
x→0
f(x)
g(x)
= lim
x→0
f ′(x)
g′(x)
.
Hence
lim
x→0
ex − 1
5x+ x2
=
1
5
.
Theorem 5.10.1 (l’Hoˆpital’s rule). Suppose that f and g are both differentiable functions and a
is a real number. Suppose also that either one of the two following conditions hold:
c©2020 School of Mathematics and Statistics, UNSW Sydney
116 CHAPTER 5. THE MEAN VALUE THEOREM AND APPLICATIONS
• f(x)→ 0 and g(x)→ 0 as x→ a;
• f(x)→∞ and g(x)→∞ as x→ a.
If
lim
x→a
f ′(x)
g′(x)
exists then
lim
x→a
f(x)
g(x)
= lim
x→a
f ′(x)
g′(x)
.
Remark 5.10.2. The theorem also holds for
• limits as x→∞ or x→ −∞;
• one-sided limits (that is, as x→ a+ or as x→ a−).
L’Hoˆpital’s rule is proved using the mean value theorem (see the sketch proof at the end of this
section).
Example 5.10.3. Calculate lim
x→1
lnx
1− x .
Solution. If f(x) = lnx and g(x) = 1−x then both f(x) and g(x) approach 0 as x→ 1. Moreover,
f and g are both differentiable. So we look at the quotient of the derivatives:
f ′(x)
g′(x)
=
1/x
−1 = −
1
x
→ −1
as x→ 1. We conclude by l’Hoˆpital’s rule that
lim
x→1
f(x)
g(x)
= −1.
(Note: for short we may write
lim
x→1
lnx
1− x = limx→1
1/x
−1 (5.10)
= lim
x→1
−1
x
= −1
provided that we check that our application of l’Hoˆpital’s rule in (5.10) is valid.)
Sometimes l’Hoˆpital’s rule must be applied more than once to calculate a limit. The next
example illustrates this.
Example 5.10.4. Determine the limiting behaviour of
x2
e2x
as x→∞.
c©2020 School of Mathematics and Statistics, UNSW Sydney
5.10. L’HOˆPITAL’S RULE 117
Solution. Observe that both the numerator and denominator approach ∞ as x → ∞. Differenti-
ating the numerator and denominator gives
2x
2e2x
.
Again, as x→∞ we have an indeterminate form of type ∞∞ . If we differentiate the numerator and
denominator once more we have
2
4e2x
and this limit can be evaluated. Piecing this together, we have
lim
x→∞
x2
e2x
= lim
x→∞
2x
2e2x
= lim
x→∞
2
4e2x
= 0
by l’Hoˆpital’s rule.
As the next example shows, l’Hoˆpital’s rule can be used to solve limits that are indefinite forms
of type 0×∞.
Example 5.10.5. Determine the limiting behaviour of x lnx as x→ 0+.
The next example warns against the improper application of l’Hoˆpital’s rule.
Example 5.10.6. Determine the limiting behaviour of
2x− cos x
3x+ cos x
as x→∞.
We shall now sketch the proof of l’Hoˆpital’s rule for the case when x→ a+. Other cases are done
similarly. (A complete proof uses a generalisation of the mean value theorem known as Cauchy’s
mean value theorem.)
Sketch proof of Theorem 5.10.1. Suppose that f and g are differentiable (and hence continuous)
everywhere, that f(x)→ 0 and g(x)→ 0 as x→ a+, and that
lim
x→a+
f ′(x)
g′(x)
exists. We need to show that
lim
x→a+
f(x)
g(x)
= lim
x→a+
f ′(x)
g′(x)
. (5.11)
We begin by applying the mean value theorem to both functions f and g on the interval [a, x].
So there are real numbers c and d in (a, x) such that
f(x)− f(a)
x− a = f
′(c) and
g(x)− g(a)
x− a = g
′(d). (5.12)
Since f is continuous and f(x)→ 0 as x→ a+, we have f(a) = 0. Similarly g(a) = 0. So by (5.12)
we have
f(x) = (x− a)f ′(c) and g(x) = (x− a)g′(d)
for some c and d in (a, x). Hence
f(x)
g(x)
=
(x− a)f ′(c)
(x− a)g(d) =
f ′(c)
g′(d)
. (5.13)
Now c→ a+ and d→ a+ as x→ a+. Hence (5.13) becomes (5.11) as required.
c©2020 School of Mathematics and Statistics, UNSW Sydney
118 CHAPTER 5. THE MEAN VALUE THEOREM AND APPLICATIONS
Problems for Chapter 5
Problems 5.1 : The mean value theorem
1. [R] Find a real number c which satisfies the conclusions of the mean value theorem for each
function f on the given interval.
a) f(x) = x3 on [1, 2] b) f(x) =
√
x on [0, 2].
2. [R] Suppose that f(x) = 1/x. Show that there is no real number c in [−1, 2] such that
f ′(c) =
f(2)− f(−1)
2− (−1) .
Why does this not contradict the mean value theorem?
3. [R] Consider the function f given by f(x) = (x − 2)4 cos(x2 − 4x+ 4). Use the mean value
theorem to show that f ′ has a zero on the interval [1, 3].
Problems 5.3 : Proving inequalities using the mean value theorem
4. [R] [V] By using the mean value theorem, show that
a) ln(1 + x) < x whenever x > 0;
b) − ln(1− x) < x/(1 − x) whenever 0 < x < 1;
c) 1 + x < ex whenever x > 0.
5. [R]
a) Use the mean value theorem to show that sin t < t whenever t > 0.
b) Using the pinching theorem and part (a), evaluate the limit lim
x→∞ sin
1
x
.
6. [H] Prove that
1 +
x
2
√
1 + x
<
√
1 + x < 1 +
x
2
whenever x > 0.
Problems 5.4 : Error bounds
7. [R] [V] Use the mean value theorem to find an upper bound for the error involved if we
approximate
a)
√
17 by
√
16 = 4;
b)
(
1998
1000
)2
by 22 = 4;
c)
1
1002
by
1
1000
.
c©2020 School of Mathematics and Statistics, UNSW Sydney
PROBLEMS FOR CHAPTER 5 119
Problems 5.5 : The sign of a derivative,
5.6 : The second derivative and applications and
5.7 : Critical points, maxima and minima
8. [R] The derivative of a function f : R → R is given by f ′(x) = 3(x + 1)(x − 1)2(x − 4)3.
Locate all stationary points of f and identify any local maximum or minimum points of f .
9. [R] Find the maximum and minimum values for each function f over the given interval.
a) f(x) = 3− x3 over [−2, 4] b) f(x) = 3− x4 over [−2, 4]
c) f(x) = x3 − x4 over [−5, 5] d) f(x) = 2x(x+ 4)3 over [−2, 1]
e) f(x) = |x2 − 3x+ 2| over [0, 3]
10. [R] Find the point on the straight line 2x+ 3y = 6 which is closest to the origin.
11. a) i) [R] Show that the polynomial p3, where p3(x) = 1 + x+
x2
2!
+
x3
3!
, has at least one
real root.
ii) [H] Show that the polynomial p2, where p2(x) = 1 + x+
x2
2!
, has no real roots and
deduce that p3 has exactly one real root.
12. [R] A wire of length 100 cm is cut into two pieces of length x cm and y cm. The piece of
length x cm is bent into the shape of a square and the piece of length y cm into the shape of
a circle. Find x and y so that the sum of the areas enclosed by the shapes will be
a) a minimum b) a maximum.
Problems 5.8 : Counting zeros
13. [R] Show that x3 + x− 9 = 0 has only one real solution.
14. [R] [V] Suppose that p(x) = x3 − 12x2 + 45x − 51 whenever x ∈ R. How many real zeros
does p have?
Problems 5.9 : Antiderivatives
15. [R]
a) Find a function f that has the following properties:
f ′(t) = sin t+ t whenever t ∈ R,
f(0) = 2.
b) Are there any other functions with these properties? Explain your answer.
16. [R] [V] A particle moving along the x-axis has velocity 2t − t2 units per second after t
seconds. Find
c©2020 School of Mathematics and Statistics, UNSW Sydney
120 CHAPTER 5. THE MEAN VALUE THEOREM AND APPLICATIONS
a) the distance from the starting point after three seconds;
b) the total distance travelled after three seconds.
Problems 5.10 : L’Hoˆpital’s rule
17. [R] Calculate the following limits.
a) lim
x→0
ex − 1
x(3 + x)
b) lim
x→1
xm − 1
xn − 1 , n 6= 0 c) limx→π/2
x− π/2
cos x
d) lim
x→0
ln (1 + x)− x
x2
e) lim
x→π/2
(
1− sinx
1 + cos 2x
)
f) lim
x→0
tanx− x
x3
18. [R] Determine the limiting behaviour in the following cases.
a)
x3 + 1
x4 + 1
as x→∞ b) e
5x
x3
as x→∞
c)
e5x
x3
as x→ −∞ d) x sin(1/x) as x→∞
e)
√
x4 + 1
3
√
x6 + 1
as x→∞ f) ln(x
3 + 1)
ln(x2 + 1)
as x→∞
19. [H] Find the value of lim
t→0
(
1
ln(1 + t)
+
1
ln(1− t)
)
.
20. [H] Find (a, b) such that lim
x→0
ax− 1 + ebx
x2
= 1.
21. [R] Explain why l’Hoˆpital’s Rule cannot be used to find lim
x→∞
4x+ sinx
2x− sinx . Use another method
to find this limit.
22. [R] [V] Show that the function f , given by
f(x) =
{
e2x if x ≥ 0
2x+ 1 if x < 0,
is differentiable at 0.
23. [R]
a) Evaluate lim
h→0+
cos
√
h− 1
h
.
b) A function f is defined by
f(x) =
{
cos
√
x if x ≥ 0
ax+ b if x < 0,
where a and b are real numbers. By using the limit calculated in (a), find all possible
values of a and b such that f is differentiable at 0.
c©2020 School of Mathematics and Statistics, UNSW Sydney
PROBLEMS FOR CHAPTER 5 121
24. [H] [V]
a) Use l’Hoˆpital’s rule to show that lim
x→0+
x lnx = 0.
b) By using part (a), or otherwise, show that lim
x→0+
x2 lnx = 0.
c) A function f is defined by
f(x) =
{
x2 lnx if x > 0
ax+ b if x ≤ 0,
where a and b are real numbers. Find all possible values of a and b such that f is
differentiable at 0.
c©2020 School of Mathematics and Statistics, UNSW Sydney
122 CHAPTER 5. THE MEAN VALUE THEOREM AND APPLICATIONS
c©2020 School of Mathematics and Statistics, UNSW Sydney
123
Chapter 6
Inverse functions
We return again to the pollution example (see the introductions to Chapter 2). According to our
model, the volume of pollution P (t) in the lake after t days of the factory’s operation is given by
P (t) =
109
101
(
1− e−101t/105).
Environmental authorities, after hearing that the eventual amount of pollution in the lake will be
109/101 litres, determine that this level would be environmentally devastating. Recommendations
are made that the amount of pollution should not exceed 8 × 106 litres. If factory operations do
not change, in how many days will pollution levels in the lake exceed this amount?
Note that the function P expresses pollution as a function of time. One approach to solving
this problem is to find a function T that expresses time as a function of pollution. Such a function
is called an inverse function of P . This chapter is devoted to a study of inverse functions.
6.1 Some preliminary examples
The examples of this section give an introduction to some of the key concepts appearing in Chapter
6. (A rigorous treatment of these concepts will be given in later sections.)
Example 6.1.1. Jack and Jill are playing a number game. Jack says, ‘A positive number, when
squared, is equal to 16. What was the original number?’ Jill correctly answers that the number
was 4.
Jack goes again: ‘A positive number, when squared, is equal to 49. What was the original
number?’ This time Jill answers ‘7.’
This game may be modelled using two functions. Jack’s function f takes a positive number and
squares it. That is, f is given by
f : (0,∞)→ (0,∞), f(x) = x2.
The function g that Jill uses to answer Jack’s question is given by
g : (0,∞)→ (0,∞), g(x) = √x.
So f takes an initial number and squares it, while g takes the square root and recovers the original
number. A way of writing down the fact that g undoes (or reverses) what f does to any positive
number x is:
g(f(x)) = x ∀x ∈ Dom(f). (6.1)
c©2020 School of Mathematics and Statistics, UNSW Sydney
124 CHAPTER 6. INVERSE FUNCTIONS
x
f(x)
output
input
(a) Example 6.1.1
x
f(x)
output
input
(b) Example 6.1.2
x
f(x)
output
inputinput
(c) Example 6.1.3
Figure 6.1: Correspondence between inputs and outputs
In fact, f also undoes what g does to any positive number. This is easily verified by checking that
f(g(x)) = x ∀x ∈ Dom(g). (6.2)
Because f and g have this special relationship, they are called inverses of each other.
The relationship between any function f and its inverse function g can also be viewed in the
following way. If y = f(x) and we want to express x as a function of y, then we find that x = g(y).
For the functions of Example 6.1.1, this is easily verified by the following calculation:
y = f(x) ⇐⇒ y = x2 ⇐⇒ √y = x ⇐⇒ x = g(y).
The next example shows that changing the domain of f gives a different inverse function g.
Example 6.1.2. Jack changes the game slightly. He says, ‘A negative number, when squared, is
equal to 36. What was the original number?’ Jill correctly answers that the number was −6.
Again, we model the game with two functions. Jack’s function f is given by
f : (−∞, 0)→ (0,∞), f(x) = x2,
while Jill’s function g is given by
g : (0,∞)→ (−∞, 0), g(x) = −√x.
Note that g undoes what f does to any negative number, and vice-versa. That is, f and g satisfy
equations (6.1) and (6.2) and are therefore inverse functions of each other.
Each function f in Example 6.1.1 and Example 6.1.2 had the property that for every output
there is a unique corresponding input (See Figure 6.1 (a) and (b)). However, the function f graphed
in Figure 6.1 (c) does not have a one-to-one correspondence between its outputs and inputs. As
we see in the next example, this function f has no inverse function.
Example 6.1.3. Jack changes the game one more time. This time he says, ‘A real number, when
squared, is equal to 64. What was the original number?’ This time Jill cannot give a definite
answer to his question. The answer is either 8 or −8, but she has no way of telling which it is.
The model for this situation is the following. Jack’s function f is now given by
f : R→ [0,∞), f(x) = x2;
c©2020 School of Mathematics and Statistics, UNSW Sydney
6.2. ONE-TO-ONE FUNCTIONS 125
but there is no function g that undoes what f does to a real number x. That is, there is no function
g that satisfies the equation (6.1). Therefore f has no inverse function. The reason why this is the
case is clear: the domain of f now includes the positive and negative real numbers, and therefore
some (in fact most) outputs of f have two possible corresponding inputs. That is, there is no
one-to-one correspondence between inputs and outputs of f .
The ideas introduced through these examples are summarised below.
• If f has an inverse function g, then g undoes (or reverses) what f does to elements of Dom(f).
• Suppose f and g are inverse functions. Then y = f(x) if and only if x = g(y).
• A function f has an inverse function only if there is a one-to-one correspondence between
inputs and outputs of f .
• The domain of a function f plays an important role in determining whether f has an inverse
function g, and if so, exactly what the inverse function is.
• If a function f does not have an inverse function, then we can sometimes modify the domain
of f so that it does have an inverse function. (For example, if f(x) = x2 then f : R→ [0,∞)
does not have an inverse function while the function f : (0,∞) → (0,∞), with domain
restricted to the positive real numbers, does have an inverse function.)
In the following sections we treat these ideas more rigorously.
6.2 One-to-one functions
(Ref: SH10 §7.1)
As was demonstrated in Section 6.1, the property that a function has a one-to-one correspondence
between inputs and outputs is critical to our discussion.
Definition 6.2.1. A function f is said to be one-to-one if
f(x1) = f(x2) implies that x1 = x2
whenever x1, x2 ∈ Dom(f).
In other words, a function is one-to-one if every output has a unique input. Equivalently, f is
one-to-one if
f(x1) 6= f(x2) whenever x1 6= x2,
provided that both x1 and x2 belong to Dom(f). One-to-one functions are sometimes called
injective functions.
Example 6.2.2. Show that the function f : R→ R, given by f(x) = 2x− 3, is one-to-one.
c©2020 School of Mathematics and Statistics, UNSW Sydney
126 CHAPTER 6. INVERSE FUNCTIONS
Solution. Suppose that f(x1) = f(x2). Then
2x1 − 3 = 2x2 − 3.
If we add 3 to both sides and afterwards divide by 2 then we obtain
x1 = x2.
Hence f is one-to-one.
Example 6.2.3. Show that the function f : R→ R, given by f(x) = x2, is not one-to-one.
Proof. Clearly f(−2) = f(2). However, −2 6= 2 and so f is not one-to-one.
There are other ways of identifying one-to-one functions. The following method is a simple
geometric test.
Proposition 6.2.4 (The horizontal line test). Suppose that f is a real-valued function defined on
some subset of R. Then the following statements are equivalent:
(a) f is one-to-one;
(b) every horizontal line in the Cartesian plane intersects the graph of f at most once.
Example 6.2.5. Consider the functions graphed below.
x
f(x)
A
b
x
g(x)
x
h(x)
• f is not one-to-one because the horizontal line passing through the point A cuts the graph of
f more than once;
• g is one-to-one (in fact, since g is increasing, every horizontal line can cut the graph of g
graph no more than once);
• h is also one-to-one (even though it is not always increasing).
Although not every one-to-one function is increasing (or decreasing), it is true that every in-
creasing function is one-to-one.
Proposition 6.2.6. If a function f is either increasing or decreasing, then f is one-to-one.
Sketch proof. If f is increasing and x1, x2 ∈ Dom(f), then
x1 < x2 ⇐⇒ f(x1) < f(x2). (6.3)
Now compare (6.3) with Definition 6.2.1.
c©2020 School of Mathematics and Statistics, UNSW Sydney
6.3. INVERSE FUNCTIONS 127
Example 6.2.7. Show that the function f : R→ R, given by
f(x) = 2x5 + x3 + x− 10,
is one-to-one.
Solution. The derivative of f is given by f ′(x) = 10x4 +3x2 +1. Since f ′(x) ≥ 1 for all real x, the
function f is increasing and hence one-to-one.
Remark 6.2.8. Not every function whose derivative is only positive (or only negative) on its
domain is one-to-one. For example, on Dom(tan)
d
dx
tanx = sec2 x ≥ 1,
but tan is not one-to-one, nor is it increasing on its domain. The problem here is that the domain
of tan has gaps.
6.3 Inverse functions
(Ref: SH10 §7.1)
We have already mentioned the term inverse function in Section 6.1. In this section we define
precisely what is meant by an inverse function and explore some of the relationships between a
function and its inverse. Our starting point is the following theorem. It says that if f is one-to-one,
then there is a function g which undoes what f does to any input x (c.f. equation (6.1)).
Theorem 6.3.1. Suppose that f is a one-to-one function. Then there is a unique function g
satisfying
g(f(x)) = x ∀x ∈ Dom(f) (6.4)
and
f(g(x)) = x ∀x ∈ Range(f). (6.5)
Moreover,
Dom(g) = Range(f), Range(g) = Dom(f)
and g is one-to-one.
Proof. Suppose that f is one-to-one and let D and R denote the domain and range of f respectively.
Since f is one-to-one, for every y in R there is a unique x in D such that y = f(x).
Now define a function g : R→ D by the following rule: if y ∈ R then g(y) is the unique x in D
such that f(x) = y.
It is left as an exercise to show that g is one-to-one and that equations (6.4) and (6.5) hold.
The theorem allows us to define the term inverse function.
Definition 6.3.2. Suppose that f is a one-to-one function. Then the inverse func-
tion of f is the unique function g given by Theorem 6.3.1. The inverse function for
f is often written as f−1.
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128 CHAPTER 6. INVERSE FUNCTIONS
Remark 6.3.3. If f−1 denotes the inverse function of a one-to-one function f , then equations (6.4)
and (6.5) can be expressed as
f−1(f(x)) = x ∀x ∈ Dom(f)
and
f(f−1(x)) = x ∀x ∈ Range(f).
(Warning: This notation is potentially confusing if not understood correctly. The function f−1
is not the reciprocal function of f (that is, 1/f(x)). To denote the reciprocal of f(x) using index
notation we write [f(x)]−1 rather than f−1(x).)
Remark 6.3.4. If g is the inverse of a function f , then f is the inverse of g. This is due to the
symmetry between f and g in the statements of Theorem 6.3.1.
If f is a one-to-one function then equation (6.5) can often be used to find an explicit formula
for its inverse g.
Example 6.3.5. Suppose that f : R→ R is given by f(x) = 4− 13x3.
(a) Explain why f has an inverse function.
(b) Find the inverse function g of f .
(c) Sketch f and g on the Cartesian plane.
(d) Use the inverse function g to solve the equation f(x) = 13.
Solution. (a) Since f ′(x) = −x2 ≤ 0 for all x in R (with f ′(x) = 0 only when x = 0), the function
f is decreasing and is therefore one-to-one. Hence f has an inverse function.
(b) Let g denote the inverse function of f . Then Dom(g) = Range(f) = R and Range(g) =
Dom(f) = R. To find an explicit formula for g(x), suppose that
y = g(x).
If we apply f to both sides of the equation, and use (6.5), then
f(y) = f(g(x)) = x.
Using the definition of f , this simplifies to
4− 1
3
y3 = x,
whereupon rearrangement gives
y = 3
√
12 − 3x.
Therefore g : R→ R is given by
g(x) = 3
√
12− 3x. (6.6)
(c) The graphs of f and g are shown below. Note that each graph is the reflection the other in
the line y = x.
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6.3. INVERSE FUNCTIONS 129
x
y
y = x
y = f(x)
y = g(x)
(d) To solve the equation f(x) = 13, apply g to both sides. This gives
g(f(x)) = g(13).
By (6.4) and (6.6) this simplifies to
x = 3
√
12− 3× 13
= −3.
Remark 6.3.6. As was illustrated in Example 6.3.5 (c), the graph of a one-to-one function f and
its inverse g are reflections of each other in the line y = x. We give a proof of this fact below:
(x, y) lies on the graph of f ⇐⇒ y = f(x)
⇐⇒ g(y) = g(f(x))
⇐⇒ g(y) = x
⇐⇒ (y, x) lies on the graph of g.
If a function f is not one-to-one, then we can sometimes restrict the domain of f so that it
becomes one-to-one. The function f with restricted domain then has an inverse function.
Example 6.3.7. Consider the function f : R→ R given by
f(x) = 4− x2.
Find a restriction of f such that f becomes one-to-one. Find the inverse function g of this restriction.
Solution. A quick sketch of the graph of f (see Figure 6.2 (a)) shows that f is not one-to-one. If we
restrict the domain of f to [0,∞), then the restricted function f : [0,∞)→ R passes the horizontal
line test and is therefore one-to-one. Let g denote the inverse of f : [0,∞)→ R. Then
g : (−∞, 4]→ [0,∞)
c©2020 School of Mathematics and Statistics, UNSW Sydney
130 CHAPTER 6. INVERSE FUNCTIONS
x
y
y = f(x)
4
−2 2
(a)
x
y
y = f(x)
y = g(x) 4
4
(b)
x
y
y = f(x)
y = g(x)
4
4
(c)
Figure 6.2: Two different restrictions of the function f , and their corresponding inverses (see
Example 6.3.7).
since Dom(g) = Range(f) = (−∞, 4] and Range(g) = Dom(f) = [0,∞). We find an explicit
formula for g:
y = g(x)
f(y) = x (by equation 6.5)
4− y2 = x
y =
√
4− x (taking the positive root since Range(g) = [0,∞)).
Hence g(x) =
√
4− x. The graphs of f and g are sketched in Figure 6.2 (b).
Remark 6.3.8. In the solution to Example 6.3.7, one could have instead chosen the restriction
f : (−∞, 0]→ R. The corresponding inverse g is given by
g : (−∞, 4]→ (−∞, 0], g(x) = −
√
4− x2.
The graphs of f and g are given in Figure 6.2 (c).
Example 6.3.9. Consider the function f : R→ R, given by
f(x) = −x3 + 3x2 + 24x − 13.
Find all intervals I, each as large as possible, such that the function f : I → R (with domain
restricted to I) has an inverse function.
6.4 The inverse function theorem
(Ref: SH10 §§7.1, B.3)
If a function f is one-to-one then its inverse function g is also one-to-one (see Theorem 6.3.1).
What other properties of f does its inverse function g inherit?
First, if f is continuous, then so is g. (A rough argument for this fact would go like this: if f
is continuous then its graph can be drawn as an unbroken line. Now reflect this graph in the line
y = x. The reflected graph is also an unbroken line. Moreover it corresponds to the graph of g.
Hence g is continuous.)
c©2020 School of Mathematics and Statistics, UNSW Sydney
6.4. THE INVERSE FUNCTION THEOREM 131
What about differentiability? If a one-to-one function f is differentiable then its inverse g is
not always differentiable. For example, if f : R→ R is given by f(x) = x3, then f is differentiable
everywhere. Its inverse function g is given by
g : R→ R, g(x) = 3√x.
Since
g′(x) =
1
3
√
x2
,
g is not differentiable at 0. So while f is differentiable everywhere, its inverse g is not.
The problem with the above example occurs where the derivative of f is zero. When a horizontal
tangent to f is reflected in the line y = x, it becomes a vertical tangent to g and hence g is not
differentiable at this point.
x
y
However, if f is a differentiable function whose derivative is never zero, its inverse g will also be
differentiable.
Theorem 6.4.1 (The inverse function theorem). Suppose that I is an open interval, f : I → R is
differentiable and f ′(x) 6= 0 for all x in I. Then
(i) f is one-to-one and has an inverse function g : Range(f)→ Dom(f),
(ii) g is differentiable at all points in Range(f), and
(iii) the derivative of g is given by the formula
g′(x) =
1
f ′(g(x))
for all x in Range(f).
Some parts of the inverse function theorem are easy to prove while others are hard. Statement
(i) is true since f ′(x) 6= 0 for x in I implies that f is either increasing on I or decreasing on I (by
the mean value theorem). Hence f is one-to-one. Statement (ii) is difficult to prove and involves a
delicate limiting argument with the difference quotients of f and g.
It is important that students understand the proof of statement (iii). Suppose (by statement
(ii)) that g is differentiable. Beginning with the equation
f(g(x)) = x
(see 6.5), differentiation with respect to x gives
f ′(g(x)).g′(x) = 1
c©2020 School of Mathematics and Statistics, UNSW Sydney
132 CHAPTER 6. INVERSE FUNCTIONS
by the chain rule. Since f ′ is never zero on I, we can divide by f ′(g(x))) to obtain
g′(x) =
1
f ′(g(x))
as required.
In the next section the inverse function theorem will be applied to the trigonometric functions.
6.5 Applications to the trigonometric functions
(Ref: SH10 §7.7)
In elementary trigonometry, the sine function gives the ratio ba of two side lengths of a right-angle
triangle corresponding to the angle θ.
a
b
c
θ
However, if we know the ratio and instead want to find the corresponding angle, we use the inverse
sine function. Thus relationship between the sine and inverse sine functions is given by
sin θ =
b
c
, sin−1
b
c
= θ. (6.7)
This relationship works in elementary trigonometry where the angle θ is acute. However, if θ is
allowed to be any real number then the relationship between sin and sin−1 is not so straightforward.
This is because the function sin : R→ R is not one-to-one and therefore has no inverse (see Figure
6.3).
In the next example, we define exactly what we mean by sin−1 and compute its derivative.
Example 6.5.1 (The inverse sine function). Since the usual sine function is not a one-to-one
function, we consider instead a restricted sine function
sin : [−π2 , π2 ]→ [−1, 1]
which is one-to-one (see Figure 6.3 (a)). This restricted function has an inverse
sin−1 : [−1, 1]→ [−π2 , π2 ],
which is graphed in Figure 6.3 (b).
We note that sin−1 is an odd function, that is,
sin−1(−x) = − sin−1 x ∀x ∈ [−π2 , π2 ].
We now examine the differentiability of sin−1. If I = (−π2 , π2 ) then the sine function is differen-
tiable on I and
d
dx
(sinx) = cos x 6= 0
c©2020 School of Mathematics and Statistics, UNSW Sydney
6.5. APPLICATIONS TO THE TRIGONOMETRIC FUNCTIONS 133
x
sinx
b
b
(π2 , 1)
(−π2 ,−1)
(a)
x
sin−1 x
b
b
(1, π2 )
(−1,−π2 )
(b)
Figure 6.3: Graphs of sin and sin−1
for all x in I. So by the inverse function theorem (Theorem 6.4.1), sin−1 is differentiable on (−1, 1).
While the derivative of sin−1 could be computed using the formula of Theorem 6.4.1 (iii), we prefer
to compute it directly by implicit differentiation.
If y = sin−1 x then
sin y = x (by applying sin to both sides)
cos y
dy
dx
= 1 (by implicit differentiation)
dy
dx
=
1
cos y
(since cos y > 0 when −π2 < y < π2 ). (6.8)
We seek a simple expression for cos y in terms of x. By the trigonometric identity sin2 y+cos2 y = 1,
and the fact that y = sin−1 x, we have
cos y =
√
1− sin2 y =
√
1− (sin(sin−1 x))2 =
√
1− x2, (6.9)
where the positive square root was taken since cos y > 0. Substituting (6.9) into (6.8) gives
dy
dx
=
1√
1− x2
and hence
d
dx
(sin−1) =
1√
1− x2
whenever −1 < x < 1.
As a corollary, note that the derivative of sin−1 is positive, so sin−1 is an increasing function.
The following table summarises the domain, range and derivatives of the restricted (and hence
one-to-one) trigonometric functions and their inverses. The derivatives exist on the largest open
intervals contained in the domain of each function. Proofs of the derivatives for the inverse trigono-
metric functions are done in much the same way as illustrated in Example 6.5.1. It is expected that
students will be able to derive these derivatives, using the method of Example 6.5.1, if required.
The graphs of cos−1 and tan−1 are shown in Figure 6.4.
c©2020 School of Mathematics and Statistics, UNSW Sydney
134 CHAPTER 6. INVERSE FUNCTIONS
x
cos−1 x
b
b
b
(1, 0)
(−1, π)
(0, π2 )
x
tan−1 x
π
2
−π2
b
b
Figure 6.4: Graphs of cos−1 and tan−1
Function Domain Range Derivative
sin [−π2 , π2 ] [−1, 1]
d
dx
(sinx) = cos x
sin−1 [−1, 1] [−π2 , π2 ]
d
dx
(sin−1 x) =
1√
1− x2
cos [0, π] [−1, 1] d
dx
(cos x) = − sinx
cos−1 [−1, 1] [0, π] d
dx
(cos−1 x) = − 1√
1− x2
tan (−π2 , π2 ) (−∞,∞)
d
dx
(tanx) = sec2 x
tan−1 (−∞,∞) (−π2 , π2 )
d
dx
(tan−1 x) =
1
1 + x2
We note also that sin−1 and tan−1 are both odd functions, but that
cos−1(−x) = π − cos−1 x ∀x ∈ [−1, 1].
The next example illustrates that one must be careful with the notation sin and sin−1. It is
not always true that sin−1(sinx) = x. This is because sin−1 is not the inverse of sin : R→ R. It is
only the inverse of sin : [−π2 , π2 ]→ R. In particular,
sin−1(sinx) = x
only when x ∈ [−π2 , π2 ]. Similar comments apply to the other trigonometric functions.
Example 6.5.2. Evaluate the following numbers without using a calculator:
(a) sin−1(sin 5π4 ),
(b) sin(cos−1 35 ).
Solution. (a) We cannot write sin−1(sin 5π4 ) =
5π
4 , since
5π
4 is not in the interval [−π2 , π2 ]. However,
since sin−1 is an odd function,
sin−1
(
sin
5π
4
)
= sin−1
(
− 1√
2
)
= − sin−1
(
1√
2
)
= −π
4
.
c©2020 School of Mathematics and Statistics, UNSW Sydney
6.5. APPLICATIONS TO THE TRIGONOMETRIC FUNCTIONS 135
(b) Let θ denote cos−1 35 . Then we seek sin θ. By the trigonometric identity
sin2 θ + cos2 θ = 1,
we have
sin θ = ±
√
1− cos2 θ,
which simplifies to
sin θ = ±
√
1−
(
3
5
)2
= ±4
5
.
Since θ = cos−1 35 > 0 we have sin θ > 0 and hence
sin θ = +
4
5
,
completing the question.
An alternate solution to (b) is the following. Once again, let θ denote the angle cos−1 35 . This
means that cos θ = 35 , whereupon we construct the right-angled triangle shown below.
4
5
θ
A C
B
3
By Pythagoras’ theorem, the length of side AB is 4. Hence sin θ = 45 .
The final example uses theory from all chapters covered in the course so far.
Example 6.5.3. A man and a woman sit on opposite edges of a canyon which is 50 metres wide
and 100 metres deep. The man drops a small stone from rest into the canyon. The woman watches
the stone fall to the bottom of the canyon through a telescope. At what time during the stone’s
fall will the angle that the telescope makes with the horizontal change the fastest? (That is, when
will the angular velocity of the telescope be greatest?) Assume that the air resistance experienced
by the stone is negligible.
Solution. Let x(t) denote the distance (in metres) that the stone has travelled exactly t seconds
after the stone was dropped. Let θ(t) denote the angle that the telescope makes with the horizontal
at t seconds. Let g denote acceleration due to gravity (g ≈ 9.8 ms−2).
c©2020 School of Mathematics and Statistics, UNSW Sydney
136 CHAPTER 6. INVERSE FUNCTIONS
r
man woman
stone
50 m
100 m
x(t)
θ(t)
If v(t) denotes the velocity of the stone at time t, then we have
dx
dt
= v, x(0) = 0, (6.10)
dv
dt
= g, v(0) = 0. (6.11)
From (6.11) we obtain v(t) = gt+C1. By imposing the initial condition that v(0) = 0 we see that
C1 = 0. Hence v(t) = gt and substituting this into (6.10) gives
dx
dt
= gt, x(0) = 0.
It follows that x(t) = 12gt
2 + C2 and one easily shows that C2 = 0.
To find out when the stone hits the canyon floor, we solve the equation
100 =
1
2
gt2
for t. Hence t =
√
200/g ≈ 4.5. So it takes about 4.5 seconds for the stone to hit the canyon floor.
By examining the triangle in the above diagram, we see that
tan θ =
x(t)
50
=
gt2
100
.
To solve the problem, we need to determine the value of t when dθdt is a maximum. Now
θ = tan−1
(
gt2
100
)
and so
dθ
dt
=
1
1 +
(
gt2
100
)2 × gt50
by the derivative of tan−1 and the chain rule. This simplifies to
dθ
dt
=
200gt
104 + g2t4
.
c©2020 School of Mathematics and Statistics, UNSW Sydney
6.6. MAPLE NOTES 137
For simplicity, denote dθdt by ω. We need to find when ω attains its maximum on the interval
[0,
√
200/g] (that is, over the time interval when the stone is falling). To locate the critical points
of ω, we calculate dωdt :
dω
dt
=
d2θ
dt2
=
(104 + g2t4)200g − 200gt(4g2t3)
(104 + g2t4)2
=
(104)200g − 600g3t4
(104 + g2t4)2
.
The stationary points of ω occur when
(104)200g − 600g3t4 = 0,
that is, when
t4 =
104
3g2
.
Since t ≥ 0, we obtain t = 10/ 4
√
3g2 ≈ 2.42. So the critical points of ω on [0,
√
200/g] are
0, 10/ 4
√
3g2 and
√
200/g.
The value of ω at each of these points is calculated below:
ω(0) = 0, ω(10/ 4
√
3g2) ≈ 1.10, ω(
√
200/g) ≈ 0.26.
Clearly ω(t) attains its maximum when t = 10/ 4
√
3g2.
In conclusion, the angular velocity of the telescope is greatest 10/ 4
√
3g2 (or about 2.26) seconds
after the stone is dropped. At this time, the angular velocity is about 1.10 radians per second (that
is, about 62.9 degrees per second).
6.6 Maple notes
Maple knows about the inverse trigonometric functions. The functions sin−1, cos−1, tan−1, cosec−1, sec−1, cot−1
are written respectively as: arcsin, arccos, arctan, arccsc, arcsec and arccot. For example,
> plot([arctan(x), arccsc(x)], x=-10..10, y=-2..2, linestyle=[dash,solid]);
c©2020 School of Mathematics and Statistics, UNSW Sydney
138 CHAPTER 6. INVERSE FUNCTIONS
> # Note the gap around zero in the graph of arccsc(x) --- please explain!
c©2020 School of Mathematics and Statistics, UNSW Sydney
PROBLEMS FOR CHAPTER 6 139
Problems for Chapter 6
Problems 6.1 : Some preliminary examples
1. [R] [V] Suppose that the functions f : [0,∞)→ [1,∞) and g : [1,∞) → [0,∞) are given by
f(x) =
√
1 + x2 and g(x) =
√
x2 − 1.
a) By calculating (f ◦ g)(x) and (g ◦ f)(x), verify that g is the inverse function to f .
b) What are the domains of f ◦ g and g ◦ f?
2. [R]
a) Suppose that f : R→ R is given by f(x) = 3x+ 1. Find f−1(x). Sketch the graph of f
and the graph of its inverse function, f−1, on the same diagram.
b) The function g : (−∞, 0]→ R is defined by g(x) = x2 + 1. Write down the domain and
range of the inverse function g−1 and find a formula for g−1(x). Find the derivative of
g−1.
Problems 6.2 : One-to-one functions and
6.3 : Inverse functions
3. [R] Show that the function f : R→ R, given by f(x) = x3 + 3x+ 1, has an inverse function
whose domain is R.
4. [R] [V] Suppose that f : R→ R is given by f(x) = 4x+ cos x.
a) Show that f has an inverse function g.
b) By using the inverse function theorem, find g′(2π).
5. [R] Suppose that f : R→ R is defined by f(x) = x3 − 3x+ 1.
a) Show that f : R→ R is not a one-to-one function.
b) Find all possible intervals I of R, each as large as possible, such that the restricted
function f : I → R has an inverse. What is the domain of each of corresponding inverse
function?
6. [H]
a) Can you find a quadratic function from R to R which is one-to-one?
b) Can you find a cubic function from R to R which is not one-to-one?
Problems 6.4 : The inverse function theorem
7. [H] For each function f : R→ R given below, find all possible intervals I of R, each as large
as possible, such that the restricted function f : I → R is one-to-one. State the range of
each restricted function f : I → R. What can you say about existence, domain of definition,
continuity and differentiability of the corresponding inverse functions?
c©2020 School of Mathematics and Statistics, UNSW Sydney
140 CHAPTER 6. INVERSE FUNCTIONS
a) f(x) = x(x2 − 1)(x + 2)
b) f(x) = (x+ 1)17
c) f(x) = |x| |x+ 1|
Problems 6.5 : Applications to the trigonometric functions
8. [R] Simplify each expression without using a calculator.
a) sin−1(
√
3/2) b) cos(cos−1(2/5)) c) sin−1(sin(5π/3))
d) cos−1(cos(−π/3)) e) cos(sin−1(3/5)) f) sin(tan−1(3/5))
g) sec−1(2) h) sin−1(sinx) when π2 ≤ x ≤ 3π2
9. [R] Sketch the graph of f : [1, 3]→ R, where f(x) = cos−1(x− 2).
10. [R] Show that
a)
d
dx
(
cos−1 x
)
=
−1√
1− x2 b)
d
dx
(
tan−1 x
)
=
1
1 + x2
.
11. [R] Differentiate
a) cos−1(2x) b) sin−1
√
x c) tan−1(2x− 3).
12. [R] Prove that sin−1 x+ cos−1 x is constant. For what values of x is this valid and what is
the constant?
13. [H] [V] Suppose that f(x) = tan−1 x+ tan−1(1/x) whenever x 6= 0.
a) Show that f ′(x) = 0 whenever x 6= 0.
b) Hence evaluate f on the intervals (0,∞) and (−∞, 0).
c) How do you account for this result geometrically?
14. [H]
a) Draw the graph of cosec x.
b) Show that cosec restricted to the interval (0, π2 ] has an inverse function. Sketch the graph
of the inverse and calculate its derivative.
15. [H] A function f : R→ R is defined by
f(x) =
{
x tan−1
(
1√
x
)
if x > 0
ax+ b if x ≤ 0,
where a and b are real numbers. Find all values of a and b such that f is differentiable at 0.
c©2020 School of Mathematics and Statistics, UNSW Sydney
PROBLEMS FOR CHAPTER 6 141
16. [H] A lighthouse containing a revolving beacon is located 3 km from P , the nearest point
on a straight shoreline. The beacon revolves with a constant rotation rate of 4 revolutions
per minute and throws a spot of light onto the shoreline. How fast is the spot of light moving
when it is (a) at P and (b) at a point on the shoreline 2 km from P?
17. [H] [V] A picture 2 metres high is hung on a wall with its bottom edge 6 metres above the
eye of the viewer. How far from the wall should the viewer stand for the picture to subtend
the largest possible vertical angle with her eye?
c©2020 School of Mathematics and Statistics, UNSW Sydney
142 CHAPTER 6. INVERSE FUNCTIONS
c©2020 School of Mathematics and Statistics, UNSW Sydney
143
Chapter 7
Curve sketching
There are a variety of ways to describe curves that lie in the plane. In this chapter we study curves
which are described by using
• a Cartesian equation (for example, y = x2 or y = √1− x2),
• a parameter, and
• polar coordinates.
Each section of this chapter is devoted to one of these methods. The use of Cartesian equations
will already be familiar. However, many curves cannot be easily described by Cartesian equations.
The following diagrams illustrate two curves that are better described by a parameter or with polar
coordinates.
x
y
An example of a parametric curve:

x(t) = sin t cos t ln |t|
y(t) =
√
|t| cos t
t ∈ [−1, 1], t 6= 0
x
y
An example of a polar curve:
r = | cos(4θ)|
7.1 Curves defined by a Cartesian equation
(Ref: SH10 §§4.7, 4.8)
c©2020 School of Mathematics and Statistics, UNSW Sydney
144 CHAPTER 7. CURVE SKETCHING
In this section we survey techniques for sketching curves that are described by a Cartesian equation
of the form y = f(x).
7.1.1 A checklist for sketching curves
Students will already be familiar with techniques for sketching the graph of a function f . For
convenience we summarise some of the most useful here.
• Identify the domain of f . (In particular, Dom(f) should not include any points that lead to
division by 0 or the square root of a negative number.)
• Identify any symmetries:
f is odd if f(−x)= −f(x) ∀x ∈ Dom(f)
f is even if f(−x) = f(x) ∀x ∈ Dom(f).
• Find x- and y-axis intercepts.
• Identify vertical asymptotes.
• Examine the behaviour of f(x) as x → ±∞. (This will identify any other asymptotes that
exist).
• Use calculus to identify stationary points and other features, if necessary.
7.1.2 Oblique asymptotes
Consider the graph of the function f that is sketched below.
x
y
y = f(x)
y = ax+ b
It appears that the graph of f approaches the line y = ax+ b as x→∞. If this is indeed the case,
we say that f is asymptotic to the line. Since this line is neither vertical, nor horizontal, we call it
an oblique asymptote to the function f .
c©2020 School of Mathematics and Statistics, UNSW Sydney
7.1. CURVES DEFINED BY A CARTESIAN EQUATION 145
Definition 7.1.1. Suppose that a and b are real numbers and that a 6= 0. We say
that a straight line, given by the equation
y = ax+ b,
is an oblique asymptote for a function f if
lim
x→∞
(
f(x)− (ax+ b)) = 0.
Example 7.1.2. Suppose that f is defined by the equation
f(x) = x+ 1 +
2
x− 2
whenever x 6= 2. Identify all asymptotes to f and hence sketch its graph.
Solution. Clearly there is a vertical asymptote when x = 2. To identify the other asymptotes (if
any), we examine the behaviour of f(x) as x→∞ and as x→ −∞.
Now as x → ∞, 2x−2 → 0 from above. Hence we conclude that f(x) approaches x + 1 from
above. Therefore the line y = x+ 1 is an oblique asymptote to f . (The calculation
lim
x→∞
(
f(x)− (x+ 1)) = lim
x→∞
2
x− 2 = 0
demonstrates from Definition 7.1.1 that this is the case.)
On the other hand, as x → −∞, 2x−2 → 0 from below. Therefore f(x) approaches x + 1 from
below.
The asymptotes for f help us construct the sketch (which is pictured below). To complete the
sketch, we identify the axes intercepts. For the y-axis, we have
f(0) = 0.
For the x-axis,
f(x) = 0 ⇒ (x+ 1)(x− 2) + 2 = 0 ⇒ x2 − x = 0 ⇒ x = 0, 1.
Thus we have the following sketch:
x
y
1
x = 2
y = f(x)
y = x+ 1
c©2020 School of Mathematics and Statistics, UNSW Sydney
146 CHAPTER 7. CURVE SKETCHING
(Note that we have obtained sufficient information for sketching the graph without resorting to
calculus.)
If f is a rational function of the form p/q, where p and q are polynomials and deg(p) > deg(q),
then the asymptotic behaviour of f may be determined by polynomial division.
Example 7.1.3. Suppose that f is defined by the equation
f(x) =
x2 − 4
x+ 1
whenever x 6= −1. Identify all asymptotes to f and hence sketch its graph.
Solution. Clearly there is a vertical asymptote when x = −1. It is also clear that f(x) → ∞
as x → ∞, but the precise asymptotic behaviour of f(x) as x → ∞ is still not clear. However,
polynomial division gives
x− 1
x+ 1 )x2 − 4
x2 + x
−x− 4
−x− 1
−3.
Hence
f(x) = x− 1− 3
x+ 1
. (7.1)
(See Remark 7.1.4 for an alternative method of establishing (7.1).)
Since 3x+1 → 0 from above as x → ∞, we conclude that f(x) approaches x − 1 from below as
x →∞. A similar observation shows that f(x) approaches x− 1 from above as x→ −∞. So the
line y = x− 1 is an oblique asymptote to f at both ∞ and −∞.
Finally, we calculate the axes intercepts:
f(0) = −4
and
f(x) = 0 ⇒ x2 − 4 = 0 ⇒ x = ±2.
Hence we obtain the following sketch.
c©2020 School of Mathematics and Statistics, UNSW Sydney
7.1. CURVES DEFINED BY A CARTESIAN EQUATION 147
x
y
x = −1
2−2
−4y = f(x)
y = x− 1
(Again, we have obtained sufficient information to sketch the graph without the use of calculus.)
Remark 7.1.4. We give a method for establishing equation (7.1) that uses an algebraic ‘trick’
rather than polynomial ‘long division’:
f(x) =
x2 − 4
x+ 1
=
x2 − 1− 3
x+ 1
=
(x− 1)(x+ 1)− 3
x+ 1
= (x− 1)− 3
x+ 1
.
7.1.3 Examples
We complete this section with three examples.
Example 7.1.5. Sketch the graph of f , where f is defined by
f(x) = xe−x
2
.
Solution. It is clear that Dom(f) = R. Moreover,
f(−x) = −xe−x2 = −f(x) ∀x ∈ R,
so f is an odd function. Therefore, it suffices to consider x ≥ 0; the rest of the graph can be
obtained by symmetry.
c©2020 School of Mathematics and Statistics, UNSW Sydney
148 CHAPTER 7. CURVE SKETCHING
Since e−x
2
> 0, it is easy to see that the only axis intercept occurs at (0, 0). We look now at
the behaviour of f(x) as x→∞. By l’Hoˆpital’s rule,
lim
x→∞ f(x) = limx→∞
x
ex2
= lim
x→∞
1
2xex2
= 0.
So y = 0 is an horizontal asymptote.
Finally, we use calculus to determine the shape of the curve. By the product rule,
f ′(x) = x
(− 2xe−x2)+ e−x2(1)
= (1− 2x2)e−x2 .
Since e−x
2
> 0 for all x in R, we conclude that
f ′(x) = 0 when x =
1√
2
,
while
f ′(x) > 0 when 0 ≤ x < 1√
2
and f ′(x) < 0 when x >
1√
2
.
So f is increasing on the interval (0, 1√
2
), stationary at 1√
2
and decreasing on ( 1√
2
,∞). Now
f(1/
√
2) =
1√
2
e−1/2 =
1√
2e
≈ 0.429.
Moreover, f ′(0) = 1, implying that the tangent line to the graph at the origin has a gradient of 1.
This information can be used to sketch the graph when x ≥ 0, and the fact that f is odd allows
us to deduce the shape of the graph when x < 0.
x
f(x)
−2
21√
2
− 1√
2
− 1√
2e
1√
2e
| | | |
|
|
Example 7.1.6. Sketch the graph of f , where f is defined by
f(x) = x2 − 4 + 3
x2
.
The next function is important since it is later used to define the Si function (see Example
8.12.1). The Si function is used by electrical engineers for digital signal processing and by surveyors
using GPS (Global Positioning System).
Example 7.1.7. Sketch the graph of f , where f is defined by
f(x) =


sinx
x
if x 6= 0
1 if x = 0.
c©2020 School of Mathematics and Statistics, UNSW Sydney
7.1. CURVES DEFINED BY A CARTESIAN EQUATION 149
x
y
1
√
3− 4
√
3
−4
2
√
3− 4
y = f(x)
y = x2 − 4
|
Figure 7.1: Sketch for Example 7.1.6.
c©2020 School of Mathematics and Statistics, UNSW Sydney
150 CHAPTER 7. CURVE SKETCHING
7.2 Parametrically defined curves
(Ref: SH10 §§10.5, 10.6)
Many moving bodies trace out a path that lies in a plane. An important problem is to describe
the curve of the trajectory. We give an example.
Example 7.2.1. Suppose that a stone is thrown horizontally from the deck of Sydney Harbour
Bridge at 20 metres per second, and that the air resistance experienced by the stone is negligible.
Let x(t) denote the horizontal distance (in metres) travelled by the stone t seconds after being
thrown and let y(t) denote its height (in metres) above the water. The deck is 50 metres above sea
level.
water
deck
b
stone
50m
x(t)
y(t)
If we take the acceleration due to gravity to be 10 m/s2, then Newton’s laws of motion show that
x(t) = 20t and y(t) = 50− 5t2 (7.2)
when 0 < t <
√
10 (that is, after the stone is thrown and just before it hits the water). Sketch the
trajectory of the stone when 0 < t <
√
10.
Solution. We want to express y as a function of x. By (7.2) we have
t =
x(t)
20
and y(t) = 50 − 5t2.
If we substitute the equation on the left into the equation on the right then
y(t) = 50 − 5
(
x(t)
20
)2
which becomes
y = 50− x
2
80
.
So the trajectory is a parabola. It is easy to see that y = 50 when x = 0 and that x = 20
√
10 when
y = 0. We therefore obtain the following sketch.
c©2020 School of Mathematics and Statistics, UNSW Sydney
7.2. PARAMETRICALLY DEFINED CURVES 151
x
y
50
20
√
10
Remark 7.2.2. The equation(
x(t), y(t)
)
=
(
20t, 50 − 5t2), t ∈ R
is called a parametrisation of the parabola
y = 50− x
2
80
.
The variable t is called a parameter.
Parametrisations of curves are important to study for several reasons.
• They arise naturally with moving bodies, as illustrated in Example 7.2.1. In this context they
also contain more information than a corresponding Cartesian equation (since a Cartesian
equation only records the trajectory of the body, not the position of the body at any given
time).
• Many curves that cannot be described by an equation of the form y = f(x) can be described
parametrically.
• Even if a curve is the graph of a function, it may be much easier to describe the curve
parametrically than to write down an equation describing the function. (This is the case with
the cycloid, which will be studied in Example 7.2.7.)
• Curves in three (or higher) dimensional space are most easily described using parameters.
7.2.1 Parametrisation of conic sections
The trajectories of many heavenly bodies (such as planets and comets) can be modelled as conic
sections. In this section we will give some standard parametrisations of conic sections.
Example 7.2.3. Show that the curve given by the functions
x(t) = a cos t, y(t) = b sin t, 0 ≤ t ≤ 2π (7.3)
is an ellipse. Sketch the curve, showing how the point (x(t), y(t)) moves as t varies from 0 to 2π.
Solution. Suppose that t ∈ [0, 2π]. From (7.3) we see that
x(t)2
a2
= cos2 t,
y(t)2
b2
= sin2 t.
c©2020 School of Mathematics and Statistics, UNSW Sydney
152 CHAPTER 7. CURVE SKETCHING
Since cos2 t+ sin2 t = 1,
x(t)2
a2
+
y(t)2
b2
= cos2 t+ sin2 t = 1.
Hence the point (x(t), y(t)) lies on the ellipse
x2
a2
+
y2
b2
= 1.
(Conversely, any point (x, y) that satisfies this equation can be expressed as x = a cos t, y = b sin t
for some t in R.)
In fact, it is easy to see that, as t moves from 0 to 2π, (x(t), y(t)) traces out the entire ellipse.
This is illustrated in the diagram below.
x(t)
y(t)
b
−b
a−a bb
b
b
t = 0, 2π
t = π2
t = π
t = 3π2
0 < t < π2
π
2 < t < π
π < t < 3π2
3π
2 < t < 2π
The table below lists some commonly used parametrisations of conic sections.
Conic section Cartesian equation Parametric equation
Parabola 4ay = x2 x(t) = 2at
y(t) = at2
Circle x2 + y2 = a2 x(t) = a cos t
y(t) = a sin t
Ellipse
x2
a2
+
y2
b2
= 1 x(t) = a cos t
y(t) = b sin t
Hyperbola
x2
a2
− y
2
b2
= 1 x(t) = a sec t
y(t) = b tan t
c©2020 School of Mathematics and Statistics, UNSW Sydney
7.2. PARAMETRICALLY DEFINED CURVES 153
There are also other parametrisations for each of these curves. (We give a second parametrisation
of the hyperbola in Chapter 10.)
In Examples 7.2.1 and 7.2.3, we sketched a parametrically defined curve by first eliminating the
parameter t. It is not always possible to do this, as the next example illustrates. Such curves can
be sketched with the aid of a table of values.
Example 7.2.4. Sketch the curve defined parametrically by(
x(t), y(t)
)
= (t2 cos t, t2 sin t), 0 ≤ t ≤ 3π. (7.4)
Solution. We begin by noting that the above parametric equation is similar to the parametrisation(
x(t), y(t)
)
= (a cos t, a sin t) (7.5)
of the circle. The only difference is that the points in (7.5) have a fixed distance a from the origin,
whereas the points in (7.4) have a variable distance t2 from the origin. So intuitively, as t increases,
the point (x(t), y(t)) in (7.4) rotates about origin, while at the same time getting further away from
the origin. Therefore it is likely that the curve is a spiral.
The following table of values helps verify this reasoning.
t 0 π2 π
3π
2 2π
5π
2 3π
x 0 0 −π2 0 4π2 0 −9π2
y 0 π
2
4 0
9π2
4 0
25π2
4 0
We plot these points on the Cartesian plane and interpolate between them to produce the following
sketch.
x(t)
y(t)
−π2
−9π24
4π2
25π2
4
−9π2
π2
4
b
b
b
b
b
b
b
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154 CHAPTER 7. CURVE SKETCHING
7.2.2 Calculus and parametric curves
Given a smooth curve defined using a parameter t, one would like to calculate its gradient at any
point along the curve. The following proposition enables us to do this without having to eliminate
t.
Proposition 7.2.5. Suppose that x and y are both differentiable functions of t and that y is a
function of x. If x′(t) 6= 0 then
dy
dx
=
y′(t)
x′(t)
. (7.6)
Proof. Since y is a function of x, we write y = f(x). Hence
y(t) = f(x(t)).
Differentiating both sides with respect to t gives
y′(t) =
df
dx
(x(t))x′(t)
by the chain rule. Since x′(t) 6= 0,
df
dx
(x(t)) =
y′(t)
x′(t)
which is equivalent to (7.6).
Example 7.2.6. A curve is described parametrically by(
x(t), y(t)
)
= (t2 − 1, t3 + 2t), t ∈ R.
Find the equation of the tangent to the curve at the point (0, 3).
Solution. By Proposition 7.2.5,
dy
dx
=
y′(t)
x′(t)
=
3t2 + 2
2t
(7.7)
provided that t 6= 0.
It is easy to see that
t2 − 1 = 0 and t3 + 2t = 3
only when t = 1. By substituting this into (7.7), we find that the gradient of the curve at (0, 3) is
5/2. Hence the equation of the tangent at (0, 3) is given by
y =
5x
2
+ 3.
c©2020 School of Mathematics and Statistics, UNSW Sydney
7.2. PARAMETRICALLY DEFINED CURVES 155
7.2.3 The cycloid and curve of fastest descent
Suppose that a point A is higher than a point B. What curve, starting at A and ending at B,
would enable a particle moving under the influence of gravity (and ignoring friction) to move from
A to B in the shortest time?
b
b
A
B
Curve of fastest descent.
Such a curve is known as a curve of fastest descent or a brachistochrone (which, in Greek, means
‘shortest time’).
In 1638 Galileo stated that the curve of fastest descent is the arc of a circle. This was shown to
be false by Johann Bernoulli who found the correct description of the curve. In 1696 he issued a
public challenge for others to find this description. Four mathematicians responded with a correct
answer: Jakob Bernoulli (Johann’s brother), Gottfried Leibniz, Isaac Newton and Guillaume de
l’Hoˆpital. The curve is the arc of a cycloid.
Example 7.2.7 (The cycloid). A circle of radius r rolls along the x-axis, starting from the origin
as shown below. Describe the locus (x(t), y(t)) of the point P on the edge of the circle that satisfies
(x(0), y(0)) = (0, 0)
x
y
P
P
P
P
The cycloid.
Solution. The solution will be built up over three steps, of which the first two correspond to the
diagrams below.
x
y
θ
P (x, y)
t
rt
Step 1.
x
y
t
rt
rt
y = −r
Step 2.
P (x, y)
c©2020 School of Mathematics and Statistics, UNSW Sydney
156 CHAPTER 7. CURVE SKETCHING
Step 1. First we consider a circle of radius r centred at the origin. If the circle rotates about
its centre by t radians in a clockwise direction (as shown) then
x(t) = r cos θ = r cos(3π2 − t) = r
(
cos 3π2 cos t+ sin
3π
2 sin t
)
= −r sin t,
y(t) = r sin θ = r sin(3π2 − t) = r
(
sin 3π2 cos t− cos 3π2 sin t
)
= −r cos t
and the point P sweeps out a distance of rt.
Step 2. We now consider what happens when the circle rolls along the line y = −r. If the circle
rotates by t radians, then the centre of the circle moves to the right by rt (corresponding to the
distance that P sweeps out – see Step 1). Hence
x(t) = −r sin t+ rt.
By Step 1 we still have
y(t) = −r cos t.
Step 3. We retain the same situation as in Step 2, except that we shift the circle up by r (so
that the circle rolls along the line y = 0 rather than along the line y = −r. Thus
y(t) = −r cos t+ r.
By Step 2 we still have
x(t) = −r sin t+ rt.
Final solution. The locus of the curve is given by
x(t) = r(t− sin t),
y(t) = r(1− cos t)
where t ≥ 0.
Remark 7.2.8. The curve of fastest descent from A to B is the unique arc of an (inverted) cycloid
whose tangent at A is vertical.
7.3 Curves defined by polar coordinates
(Ref: SH10 §§10.2, 10.3)
In this final section, we introduce a different coordinate system for the plane. Many equations in
this system correspond to beautiful curves.
7.3.1 Polar coordinates
Every point P in plane can be specified by
• (x, y), where x is the horizontal distance of P from the origin, and y is the vertical distance,
or
• (r, θ), where r is the distance of P from the origin and θ is the angle (taken in the anticlockwise
direction) between OP and the positive horizontal axis.
c©2020 School of Mathematics and Statistics, UNSW Sydney
7.3. CURVES DEFINED BY POLAR COORDINATES 157
b
P
O
x
y
b
P
O
θ
r
The pair (x, y) is called the Cartesian coordinate of P and the pair (r, θ) are called a polar coordinate
of P . (If P is the origin then r = 0 and θ is not defined.)
A polar coordinate (r, θ) of P is related to the Cartesian coordinate (x, y) of P by the formulae
x = r cos θ
y = r sin θ
r =
√
x2 + y2
tan θ =
y
x
, provided x 6= 0. (7.8)
This is easily seen using trigonometry, Pythagoras’ theorem and the diagram below.
b
P (x, y)
θ
r
r cos θ
r sin θ
When converting between the polar coordinates and Cartesian coordinates, it is best to draw a
diagram.
Example 7.3.1. Suppose that the polar coordinates of P and Q are given by (2, π4 ) and (3,
7π
6 )
respectively. Find the Cartesian coordinates for P and Q.
Solution. First we plot these points on the Cartesian plane.
x
y
P (2, π4 )
π
4
2
Q(3, 7π6 )
13π
6
3
π
6
From the diagram, it is clear that the Cartesian coordinate of P is given by(
2 cos
π
4
, 2 sin
π
4
)
=
(
2√
2
,
2√
2
)
= (
√
2,
√
2),
while the Cartesian coordinate of Q is given by(
−3 cos π
6
,−3 sin π
6
)
=
(
−
√
3,−3
2
)
.
c©2020 School of Mathematics and Statistics, UNSW Sydney
158 CHAPTER 7. CURVE SKETCHING
Example 7.3.2. Suppose that the Cartesian coordinate of R is given by (−2, 2√3). Find a polar
coordinate for R.
Solution. We begin with the following diagram.
x
yR(−2, 2√3)
2
2
√
3
r
θ
α
In the diagram, tanα = 2
√
3
2 =
√
3, so α = π3 . Hence
θ = π − α = 2π
3
.
By Pythagoras’ theorem,
r2 = 22 + (2
√
3)2 = 16.
Hence in polar coordinates, R can be written as (4, 2π3 ).
Remark 7.3.3. Note that the point R has no unique polar coordinate. For example
(4, 2π3 + 2π) and (4,
2π
3 − 2π)
are also polar coordinates for R. In practice, we often choose the value for θ such that −π < θ ≤ π.
7.3.2 Basic sketches of polar curves
Many curves can be described by an equation relating the polar variables r and θ. We begin with
the simplest case: when either r or θ are constant.
Example 7.3.4. Sketch the curves corresponding to the the equations
(a) r = 3
(b) θ = 3π4 .
Solution. (a) Since r = 3, we seek all points in the plane whose distance from the origin is 3. This
is a circle of radius three centred at (0, 0).
(b) This is the collection of all points P such that the angle between OP and the positive x-axis
is 3π4 . Hence the curve is the ray sketched below with the origin excluded.
c©2020 School of Mathematics and Statistics, UNSW Sydney
7.3. CURVES DEFINED BY POLAR COORDINATES 159
x
y
r = 3
3
3 x
y
θ = 3π4
3π
4
To sketch a curve, described by the polar equation r = f(θ), in the xy-plane, it may be helpful
to first sketch r = f(θ) in the rθ-plane.
Example 7.3.5. Sketch the curve whose polar equation is
r = 2− cos θ.
Solution. First we graph r against θ.
θ
r
r = 2− cos θ
|
|
|
|
| | | | | | | |
3
2 + 1√
2
2
2− 1√
2
1
π
4
π
2
3π
4
π 5π
4
3π
2
7π
4
2π
b
b
b
b
b
b
b b
b
a
b
c
d
e
f
g
h
i
The r-value of each point a, b, c, ..., i in the rθ-plane corresponds to the distance from the origin to
each point a′, b′, ..., i′ in the xy-plane.
x
y
θ = 0
θ = π4
θ = π2
θ = 3π4
θ = π
θ = 5π4
θ = 3π2
θ = 7π4
b
b
b
b
b
b
b
b
r = 1
r = 2
r = 3
r = 2
a′
b′
c′
d′
e′
f ′ g′
h′
i′
c©2020 School of Mathematics and Statistics, UNSW Sydney
160 CHAPTER 7. CURVE SKETCHING
Our task now is to interpolate between the points a′, b′, ..., i′ to obtain a sketch of the curve.
Consider what happens to r as θ moves from 0 to 2π.
• On the rθ-plane, as θ moves from 0 to π, r increases from 1 to 3. So on the xy-plane, if we
travel along the curve in an anticlockwise direction from the positive to the negative x-axis,
then the distance r from the origin to points on the curve increases from 1 to 3.
• On the rθ-plane, as θ moves from π to 2π, r decreases from 3 to 1. So on the xy-plane, if we
travel along the curve in an anticlockwise direction from the negative to the positive x-axis,
then the distance r from the origin to points on the curve decreases from 3 to 1.
These considerations lead to the final sketch.
x
y
θ = π4θ =
3π
4
θ = 5π4 θ =
7π
4
r = 2− cos θ
b
b
b
b
b
b
b
b
1
2
−3
−2
We list a few more helpful tips for sketching a curve described by the polar equation r = f(θ).
• If f is an even function (that is, f(−θ) = f(θ)) then the polar curve is symmetric about the
x-axis.
• If f(π − θ) = f(θ) then the polar curve is symmetric about the y-axis.
• If f is 2π–periodic then it suffices to consider θ in the range 0 ≤ θ < 2π.
• Sometimes it is possible to rewrite the equation r = f(θ) in terms of x and y only.
The use of calculus will be discussed in the next subsection.
Example 7.3.6. Sketch the curve that is described by the polar equation
r = 2 sin θ, 0 ≤ θ ≤ π.
Solution. Graphing r against θ, followed by y against x leads to the following two diagrams.
c©2020 School of Mathematics and Statistics, UNSW Sydney
7.3. CURVES DEFINED BY POLAR COORDINATES 161
θ
r
r = 2 sin θ
|
|
| | | |
2
2√
2
π
4
π
2
3π
4
π
b
b
b
b
b
x
y
θ = pi
4
θ = 3pi
4
r = 2 sin θ
b
b
b b
r = 0
r = 2
r =
√
2r =
√
2
The curve looks as though it is a circle, an observation that can be confirmed by rewriting r = 2 sin θ
in terms of x and y:
r = 2 sin θ
r2 = 2r sin θ (multiply by r)
x2 + y2 = 2y (by (7.8))
x2 + y2 − 2y + 1 = 1 (completing the square)
x2 + (y − 1)2 = 1.
Therefore the polar curve is a circle with Cartesian centre (0, 1) and radius 1.
Example 7.3.7. Sketch the curve that is described by the polar equation
r = 4| sin(3θ)|.
7.3.3 Sketching polar curves using calculus
Suppose that a curve can be expressed in polar form as r = f(θ). Since x = r cos θ and y = r sin θ,
the curve can also be written in parametric form (with parameter θ) as
x(θ) = f(θ) cos θ and y(θ) = f(θ) sin θ.
Hence,
dy
dx
=
y′(θ)
x′(θ)
=
f(θ) cos θ + f ′(θ) sin θ
−f(θ) sin θ + f ′(θ) cos θ
by the product rule. Since r = f(θ), this is usually written as
dy
dx
=
r cos θ + drdθ sin θ
−r sin θ + drdθ cos θ
.
(Rather than memorising this formula, we recommend that students understand how it is obtained.)
Knowing the value of derivative of the curve at various points allows for greater accuracy in
sketching. We illustrate with one example.
Example 7.3.8. Sketch the curve described by the polar equation r = 1− sin θ.
c©2020 School of Mathematics and Statistics, UNSW Sydney
162 CHAPTER 7. CURVE SKETCHING
Solution. First we graph r against θ and then give a preliminary sketch of y against x.
θ
r
r = 1− sin θ
|
|
|
|
2
1 + 1√
2
1
1− 1√
2
| | | | | | | |
π
4
π
2
3π
4
π 5π
4
3π
2
7π
4
2π
b
b
b
b
b
b
b
b
b
x
y θ = π4
θ = 5π4
θ = 3π4
θ = 7π4
r = 1− sin θ
1−1
−2
b
b
b
b
b
b
b
b
To obtain a clearer idea of the behaviour of the curve, we calculate dydx . We have
x = r cos θ = (1− sin θ) cos θ
and
y = r sin θ = (1 − sin θ) sin θ.
Hence
x′(θ) = cos θ(− cos θ) + (1− sin θ)(− sin θ)
= − cos2 θ − sin θ + sin2 θ
= 2 sin2 θ − sin θ − 1
= (2 sin θ + 1)(sin θ − 1)
and
y′(θ) = cos θ − 2 sin θ cos θ
= cos θ(1− 2 sin θ).
Thus
dy
dx
=
y′(θ)
x′(θ)
=
cos θ(1− 2 sin θ)
2 sin2 θ − sin θ − 1 .
To locate horizontal and vertical tangents, we examine where
x′(θ) = 0 and y′(θ) = 0.
Solving y′(θ) = 0 gives
cos θ = 0 or sin θ =
1
2
,
that is,
θ =
π
2
,
3π
2
,
π
6
,
5π
6
.
Solving x′(θ) = 0 gives
sin θ = 1 or sin θ = −1
2
,
c©2020 School of Mathematics and Statistics, UNSW Sydney
7.3. CURVES DEFINED BY POLAR COORDINATES 163
that is
θ =
π
2
,
7π
6
,
11π
6
.
So horizontal tangents occur when
θ =
3π
2
,
π
6
,
5π
6
and vertical tangents occur when
θ =
7π
6
,
11π
6
.
(Understanding how the derivative behaves when θ is close to π2 is harder since
x′(π2 ) = y
′(π2 ) = 0.
A full analysis is given in Remark 7.3.9.)
Finally, we observe that dydx = −1 when θ = 0 and that dydx = 1 when θ = π.
Hence we obtain the following sketch. (The abbreviations HT and VT, which appear in the
diagram, stand for ‘horizontal tangent’ and ‘vertical tangent.’)
x
y
θ = π6
θ = 7π6
θ = 5π6
θ = 11π6
HTHT
HT
VT VT
gradient of tangent is −1gradient of tangent is 1
r = 1− sin θ
1−1
−2
Remark 7.3.9. [X] In the above example, the curve has a ‘vertical cusp’ at the origin. This is
seen by analysing dydx when θ is close to
π
2 . By l’Hoˆpital’s rule we see that
lim
θ→pi
2
−
x′(θ)
y′(θ)
= lim
θ→pi
2
−
2 sin2 θ − sin θ − 1
cos θ − 2 sin θ cos θ
= lim
θ→pi
2
−
4 sin θ cos θ − cos θ
− sin θ − 2(cos θ cos θ − sin θ sin θ)
= lim
θ→pi
2
−
0
1
= 0.
c©2020 School of Mathematics and Statistics, UNSW Sydney
164 CHAPTER 7. CURVE SKETCHING
Hence
dy
dx
=
y′(θ)
x′(θ)
→∞ as θ → π
2
−
.
By symmetry,
dy
dx
=
y′(θ)
x′(θ)
→ −∞ as θ → π
2
+
.
Therefore the derivative has an infinite discontinuity when θ = π2 . Such a point is called a ‘cusp
point.’
Remark 7.3.10. In most cases, a reasonable sketch of a polar curve can be made without appealing
to calculus. As seen in the example above, while employing calculus gives a more accurate sketch,
it also dramatically increases the amount of work and time required.
7.4 Maple notes
It is easy to draw parametric curves using Maple. Maple also knows about polar coordinates, but
you should be aware that Maple allows r < 0. For example,
> # Draw a hyperbola via parametric specification
> plot([3*sec(t), 2*tan(t), t=-Pi/2..3*Pi/2], x=-10..10, y=-8..8, numpoints=100);
> # What is the reason for the straight lines through the origin in graph above?
> plot([sin(5*theta), theta, theta=-Pi..Pi], coords=polar);
c©2020 School of Mathematics and Statistics, UNSW Sydney
7.4. MAPLE NOTES 165
> plot([2+cos(5*t), t, t=-Pi..Pi], coords=polar);
c©2020 School of Mathematics and Statistics, UNSW Sydney
166 CHAPTER 7. CURVE SKETCHING
Problems for Chapter 7
Problems 7.1 : Curves defined by a Cartesian equation
1. [R] Find the maximal domain and range of the function f , given by f(x) =
√
5 + 4x− x2,
and sketch its graph.
2. [R] Write down the period of each of the following functions f (where possible). Determine
which are odd or even. Sketch the graph of each function.
a) f(x) = sin 3x b) f(x) = 1 + sin(2x/3) c) f(x) = x sinx
d) f(x) = tan 3x e) f(x) = cos2 x f) f(x) = sinx+ cosx
3. [R] Suppose that f is an odd function (not everywhere zero). Determine whether each
function g below is odd, even or neither.
a) g(x) = x2f(x) b) g(x) = x3f(x) c) g(x) = x2 + f(x)
d) g(x) = x3 + f(x) e) g(x) = sin(f(x)) f) g(x) = f(cos x)
4. [R] For each function f , identify any vertical and oblique asymptotes and hence sketch the
graph. (Do not use calculus.)
a) f(x) = x+ 2 +
1
x− 3 b) f(x) =
x2 − 2
x+ 1
c) [H] f(x) =
x3 − 7x+ 8
x2 + x− 6
5. [R] Sketch the following curves, showing their main features.
a) y = x2 +
1
x2
b) [V] y =
x− 1
x− 2 c) y = e
−x2/2
d) y = xe−x e) y2 = x(x− 4)2 f) y = x
2
x− 2
g) y =
x2 − 1
x2 − 2x h) y = x cos
−1 x
6. [H] (Longer rather than difficult) Suppose that y =
3x2 − 10x+ 3
3x2 + 10x+ 3
.
a) Find the values of x for which y ≥ 0. b) Find the asymptotes.
c) Find the turning points. d) Find the domain and range.
e) Sketch the graph.
Problems 7.2 : Parametrically defined curves
7. [R] Sketch the curves given by the following parametric equations. Also find, where possible,
a Cartesian equation for the curve.
c©2020 School of Mathematics and Statistics, UNSW Sydney
PROBLEMS FOR CHAPTER 7 167
a) x = 4cos t, y = 5 sin t
b) x = 3 sec t, y = 2 tan t
c) x = t3, y = t2
d) x = et cos t, y = et sin t
.
8. [R] For each of the curves given in parametric form by
a)
{
x = 1− t
y = 1 + t
b)
{
x = 3t+ 2
y = t4 − 1 c)
{
x = cos t
y = sin t,
i) find the points on the curve corresponding to t = −1, 0, 1, and 2;
ii) find any point on the curve where y = 0;
iii) find
dy
dx
as a function of t.
9. [R] [V]
a) Find the equation of the normal to the curve x =
t
t+ 1
, y =
t
t− 1 at the point P when
t = 2.
b) Eliminate t from the above equations and find the gradient of the normal at P using the
Cartesian form.
10. [H] Consider a fixed circle of radius 1 centred at the origin and a smaller circle of radius 14
initially centred at (34 , 0). The smaller circle rolls (without slipping) around the inside rim
of the larger circle such that the centre Q of the smaller circle moves in an anticlockwise
direction. A point P , fixed on the rim of the smaller circle and initially with coordinates
(1, 0), traces out a curve as the smaller circle moves inside the larger circle.
x
y
O Q P
Initial configuration
x
y
O
θ
Q
P
Configuration after motion has begun
The goal of this question is to find the Cartesian form of the trajectory of P . Let θ denote
the angle (in radians) between OQ and the positive x-axis, as shown in the above diagram.
a) Explain why
⇀
OQ = 34 (cos θ, sin θ).
c©2020 School of Mathematics and Statistics, UNSW Sydney
168 CHAPTER 7. CURVE SKETCHING
b) [X] Explain why
⇀
QP = 14 (cos(−3θ), sin(−3θ)).
c) Show that
⇀
OP = (cos3 θ, sin3 θ).
(You may find techniques from MA1131 Algebra useful here.)
d) Hence the trajectory of P is given by
x = cos3 θ, y = sin3 θ, 0 ≤ θ ≤ 2π.
By using an appropriate trigonometric identity, eliminate θ to find the cartesian equation
of the trajectory of P .
e) Sketch the curve corresponding to this equation. (This curve is called an astroid after
the Greek word for ‘star’.)
11. [R] [V] In the 1960s two French car engineers, Paul de Casteljau and Pierre Bezier, inde-
pendently discovered a remarkable new approach to parameterising curves. Let’s introduce
their approach by finding a quadratic curve determined by the control points A(1, 1), B(2, 2)
and C(3, 1) with respecitve coordinate vectors a,b and c.
a) The linear de Casteljau Bezier curve determined by A and B is given parametrically as
p(t) = (1− t)a+ tb.
This is called the linear interpolation of the vectors a and b. Find a parametric vector
equation for p(t) and evaluate p(0) and p(1).
b) Similarly, the linear de Casteljau Bezier determined by B and C is
q(t) = (1− t)b+ tc.
Find a Cartesian equation for this line, and evaluate q(1/2).
c) The quadratic de Casteljau Bezier curve r(t) determined A,B and C, is the linear inter-
polation of the vectors p(t) and q(t):
r(t) = (1− t)p(t) + tq(t).
Find an explicit expression for r(t) and find the polynomials p0, p1 and p2 such that
r(t) = p0(t)a+ p1(t)b+ p2(t)c.
Problems 7.3 : Curves defined by polar coordinates
12. [R] The following points are given in polar coordinate form. Plot them on a diagram and
find their Cartesian coordinates.
a) (3, 0) b) (6, 7π/6) c) (2, 7π/4)
13. [R] Convert these Cartesian coordinates into polar forms with r ≥ 0 and −π < θ ≤ π.
a) (−3, 0) b) (−1,−1) c) (−2, 2√3)
d) (0, 1) e) (−2√3, 2) f) (−2√3,−2)
c©2020 School of Mathematics and Statistics, UNSW Sydney
PROBLEMS FOR CHAPTER 7 169
14. [R] Sketch the graph corresponding to each polar equation.
a) r = 4 b) θ = 2 c) r = 3θ, for θ ≥ 0.
15. [R]
a) Express r = 6 sin θ, where 0 ≤ θ ≤ π, in Cartesian form and hence draw its graph.
b) Repeat this for r = 2cos θ, where −π/2 ≤ θ ≤ π/2.
16. [R] Sketch the graph corresponding to each polar equation.
a) r = 2 + sin θ b) r = 3 + cos θ c) [V] r = 2− 2 cos θ
d) r = 2| cos θ| e) r = 3| sin 6θ| f) r = | tan θ2 | (−π < θ < π)
17. [H] [V] The hyperbolic spiral is described by the equation rθ = a whenever θ > 0, where
a is a positive constant. Using the fact that lim
θ→0
sin θ
θ
= 1, show that the line y = a is a
horizontal asymptote to the spiral. Sketch the spiral.
18. [H] Show that r =
5
3− 2 cos θ is the polar equation of an ellipse by finding the Cartesian
equation of the curve (and completing the square).
19. [H] Find dydx for the curves in Q16a,b,c.
c©2020 School of Mathematics and Statistics, UNSW Sydney
170 CHAPTER 7. CURVE SKETCHING
c©2020 School of Mathematics and Statistics, UNSW Sydney
171
Chapter 8
Integration
The problem of understanding and calculating the area of regions in the plane has a long history.
Some of the main contributors to this field are Archimedes (the great Greek mathematician of
antiquity), Isaac Barrow (Isaac Newton’s mentor), Isaac Newton (one of the greatest mathemati-
cians and physicists of all time), Gottfried Leibniz (Newton’s contemporary), Bernhard Riemann
(a German mathematician of the nineteenth century) and Henri Lebesgue (a French mathemati-
cian of the early twentieth century). In this chapter we see a method for calculating the area of
regions with curved boundaries known as Riemann integration. We then give a startling result,
known as the fundamental theorem of calculus, which connects the problem of calculating areas
with antidifferentiation. This theorem was known to Barrow and its implications were developed
by Newton, Leibniz and their disciples. A consequence of this theorem and other related results,
is that the problem of calculating area is much the same as that of calculating mass, volume, work
and probability. The unifying feature is a body of theory known as the integral calculus.
8.1 Area and the Riemann Integral
(Ref: SH10 §§5.1, 5.2, B.5)
In this section we examine the problem of defining the area of regions with curved boundaries. The
main technique introduced is called integration, which leads to defining ‘the area under the graph
of a function’ via the Riemann integral.
8.1.1 Area of regions with curved boundaries
We all know how to calculate the area of a rectangle or a triangle. The area of more exotic polygons
can be computed by partitioning the polygon into triangles and then summing the areas of these
triangles (see Figure 8.1 (a)). Such calculations are based on an intuitive understanding of area
that takes for granted that
• the area of a rectangle is the product of its length and height,
• areas of congruent regions are equal, and
• the area of a whole region is the sum of the areas of its parts.
c©2020 School of Mathematics and Statistics, UNSW Sydney
172 CHAPTER 8. INTEGRATION
(a) Partition of a polygon
b
h
(b) A = 1
2
bh
b
b1 b2
h
(c) A = 1
2
bh
Figure 8.1: Calculating the area of polygons
When precisely formulated, these points could be taken together as a definition of area. With this
definition, one can derive a formula for the area of a right-angled triangle (see Figure 8.1 (b)), then
for an arbitrary triangle (see Figure 8.1 (c)) and then calculate the area of an arbitrary polygon
(as in Figure 8.1 (a)).
When it comes to calculating the area of a region with curved boundaries (see, for example,
Figures 8.2 and 8.3), a more sophisticated definition of area is needed. Before asking what shape
this definition might take, it is worthwhile asking the question, What is area?
First, we recognise that the area of a region is a non-negative real number. Also, our intuition
of area informs us that it should satisfy the following properties (see Figure 8.2 for a diagram
illustrating each property). We assume throughout that the regions considered are bounded.
(A1) If Ω is a region of the plane then
area(Ω) ≥ 0.
(A2) If one region Ω1 is contained in another region Ω2, then
area(Ω1) ≤ area(Ω2).
(A3) If the area of a region Ω is partitioned into two smaller disjoint regions Ω1 and Ω2, then
area(Ω) = area(Ω1) + area(Ω2).
(A4) If Ω1 and Ω2 are congruent regions then
area(Ω1) = area(Ω2).
(A5) If Ω is a rectangle of length a and height b then
area(Ω) = ab.
These properties are axioms that any definition of area should satisfy.
Rather than give a definition for the area of any bounded region of the plane, we shall only
give a definition for ‘the area under the graph of a function.’ The following example gives some
justification for this decision.
c©2020 School of Mathematics and Statistics, UNSW Sydney
8.1. AREA AND THE RIEMANN INTEGRAL 173
Ω
(a) (A1)
Ω2
Ω1
(b) (A2)
Ω
Ω2Ω1
(c) (A3)
Ω1
Ω2
(d) (A4)
Ω
a
b
(e) (A5)
Figure 8.2: Properties of area
Example 8.1.1. Consider the region Ω of Figure 8.3. By partitioning Ω with straight lines as
shown, one can see by axiom (A3) that
area(Ω) = area(Ω1) + area(Ω2) + area(Ω3).
By rotating and translating each subregion, we see by axiom (A4) that each of area(Ω1), area(Ω2)
and area(Ω3) is equal to the area under the graph of a function.
This procedure can be done for any region in the plane with a ‘reasonable’ boundary. Hence
the problem of defining the area of a region with a curved boundary can be reduced to defining
what is meant by ‘the area under the graph of a function.’
8.1.2 Approximations of area using Riemann sums
To guide us towards a definition for ‘the area under the graph of a function,’ we study the following
example.
Consider the function f : R→ R, given by the rule f(x) = x2. Let Ω denote the region bounded
by the graph of f , the x-axis and the lines x = 0 and x = 1. The region Ω is shaded in Figure
8.4 (a). The region Ω may be ‘approximated’ with rectangles as in Figure 8.4 (b) and (c). Since
the regions Ω1 and Ω2 are composed of disjoint rectangles, their areas are easily calculated using
axioms (A3) and (A5) (see Subsection 8.1.1). Moreover, however we eventually choose to define
area(Ω), it is clear by axiom (A2) that we want
area(Ω1) ≤ area(Ω) ≤ area(Ω2).
This gives a lower and upper bound for area(Ω).
The regions Ω1 and Ω2 each consist of ten rectangles. By changing the number of rectangles
used, we obtain different upper and lower bounds for area(Ω).
Example 8.1.2. Find different upper and lower bounds for the area of the shaded region Ω of
Figure 8.4 (a).
c©2020 School of Mathematics and Statistics, UNSW Sydney
174 CHAPTER 8. INTEGRATION
Ω
(a)
Ω1
Ω2
Ω3
(b)
x
y
f1
x
y
f2
x
y
f3
(c)
Figure 8.3: Partitioning of a region with curved boundary
x
f(x)
1
Ω
(a)
x
f(x)
1
Ω1
(b)
x
f(x)
1
Ω2
(c)
Figure 8.4: Area under the graph of f(x) = x2.
c©2020 School of Mathematics and Statistics, UNSW Sydney
8.1. AREA AND THE RIEMANN INTEGRAL 175
x
y
1
n
2
n
3
n
4
n
· · · k−1
n
k
n
· · · n−1
n
n
n
y = f(x)
Rk
· · · · · ·
Figure 8.5: An upper Riemann sum for Example 8.1.2.
The following formula will help our solution:
n∑
k=1
k2 = 12 + 22 + 32 + · · ·+ n2 = 1
6
n(n+ 1)(2n + 1). (8.1)
(This may be proved using mathematical induction. If students need to use such a formula in an
exam, they will be given it).
Solution to Example 8.1.2. Let A denote the (as yet undefined) area of the region Ω. Rather than
calculating an upper bound for A by using 10 rectangles (as in Figure 8.4 (c)), we use n rectangles,
where n is a positive integer (see Figure 8.5). The bases of the rectangles are constructed by
subdividing the interval [0, 1] into n subintervals
[0, 1n ], [
1
n ,
2
n ], [
2
n ,
3
n ], . . . , [
n−1
n , 1].
The set Pn, given by
Pn = {0, 1n , 2n , 3n , . . . , n−1n , 1},
that divides the interval [0, 1] into these subintervals is called a partition of [0, 1].
Let Rk denote the area of the kth rectangle in Figure 8.5. Then
Rk = width× height
=
1
n
× f( kn)
=
1
n
×
(
k
n
)2
=
1
n3
× k2.
c©2020 School of Mathematics and Statistics, UNSW Sydney
176 CHAPTER 8. INTEGRATION
If SPn(f) denotes the total area of the shaded region in Figure 8.5, then
SPn(f) =
n∑
k=1
Rk
=
n∑
k=1
1
n3
× k2
=
1
n3
n∑
k=1
k2
=
1
n3
(
12 + 22 + 32 + · · ·+ (n− 1)2 + n2)
=
1
n3
× 1
6
n(n+ 1)(2n + 1) (by formula (8.1))
=
1
6
(
1 +
1
n
)(
2 +
1
n
)
=
1
3
+
1
2n
+
1
6n2
.
The quantity SPn(f) is called the upper Riemann sum of f with respect to the partition Pn. Axiom
(A2) imples that
A ≤ SPn(f) =
1
3
+
1
2n
+
1
6n2
,
and by taking different values of n we obtain different upper bounds for A.
To compute a lower bound corresponding to n rectangles, consider Figure 8.6. The area Rk of
the kth rectangle is given by
Rk =
1
n
× f(k−1n ) =
1
n3
(k − 1)2.
The sum of all the areas of the rectangles is called the lower Riemann sum for the function f over
the partition Pn and is denoted by SPn(f). We see that
SPn(f) =
n∑
k=1
Rk
=
n∑
k=1
1
n3
× (k − 1)2
=
1
n3
n∑
k=1
(k − 1)2
=
1
n3
(
12 + 22 + 32 + · · · + (n− 1)2)
=
1
n3
× 1
6
(n− 1)n(2n − 1) (by formula (8.1))
=
1
6
(
1− 1
n
)(
2− 1
n
)
=
1
3
− 1
2n
+
1
6n2
.
c©2020 School of Mathematics and Statistics, UNSW Sydney
8.1. AREA AND THE RIEMANN INTEGRAL 177
x
y
1
n
2
n
3
n
4
n
· · · k−1
n
k
n
· · · n−1
n
n
n
y = f(x)
Rk
· · · · · ·
Figure 8.6: A lower Riemann sum for Example 8.1.2.
By axiom (A2),
A ≥ SPn(f) =
1
3
− 1
2n
+
1
6n2
.
Hence for each positive integer n, the inequality
1
3
− 1
2n
+
1
6n2
≤ A ≤ 1
3
+
1
2n
+
1
6n2
(8.2)
gives an upper and lower bound for A.
From inequality (8.2) we see that, however the area A of Ω is eventually defined, it must be
squeezed between
1
3
− 1
2n
+
1
6n2
and
1
3
+
1
2n
+
1
6n2
for every value of n. But as n→∞,
1
3
− 1
2n
+
1
6n2
→ 1
3
and
1
3
+
1
2n
+
1
6n2
→ 1
3
. (8.3)
So there is only one number A that satisfies (8.2) for every positive integer n, namely A = 13 . Our
definition of the area under the graph of a function must agree with this calculation.
Remark 8.1.3. The process of calculating upper and lower Riemann sums and taking a limit as
in (8.3) is called integration.
Guided by the ideas introduced in this subsection, we are now ready to give a definition for the
area under the graph of a function.
c©2020 School of Mathematics and Statistics, UNSW Sydney
178 CHAPTER 8. INTEGRATION
x
f(x)
a0 a1 a2 a3 ak−1 ak an−2 an−1 an· · ·
· · ·
· · ·
· · ·
Figure 8.7: An example of an upper Riemann sum
8.1.3 The definition of area under the graph of a function and the Riemann
integral
Suppose that f is a bounded function on [a, b] and that f(x) ≥ 0 for all x in [a, b]. In this subsection
we define what is meant by ‘the area under the graph of f from a to b’. This is done by constructing
upper and lower Riemann sums with respect to partitions of [a, b].
Definition 8.1.4. A finite set P of points in R is said to be a partition of [a, b] if
P = {a0, a1, a2, . . . , an}
and
a = a0 < a1 < a2 < . . . < an = b.
Suppose that P is a partition of [a, b]. In a manner similar to Example 8.1.2, P is used
to construct rectangles that approximate the region under the graph of f (see Figure 8.7). As
illustrated in the diagram, the points of P need not be evenly spaced.
The area of the kth rectangle in Figure 8.7 is
width× height = (ak − ak−1)× fk,
where fk is the maximum value of the function f on the subinterval [ak, ak−1]. The upper Riemann
sum SP(f) for f with respect to the partition P is defined by the formula
SP(f) =
n∑
k=1
(ak − ak−1)fk. (8.4)
c©2020 School of Mathematics and Statistics, UNSW Sydney
8.1. AREA AND THE RIEMANN INTEGRAL 179
Hence the upper Riemann sum corresponds to the total area of the rectangles in Figure 8.4. Like-
wise, the lower Riemann sum SP(f) for f with respect to the partition P is defined by the formula
SP(f) =
n∑
k=1
(ak − ak−1)fk, (8.5)
where f
k
is the minimum value of f on the subinterval [ak, ak−1].
We are now ready to give a definition of the area under the graph of a function f .
Definition 8.1.5 (Definition of the area under the graph of a function). Suppose
that a function f is bounded on [a, b] and that f(x) ≥ 0 for all x in [a, b]. If there
exists a unique real number A such that
SP(f) ≤ A ≤ SP(f) for every partition P of [a, b], (8.6)
then we say that A is the area under the graph of f from a to b.
It may not seem obvious now but, as we shall see in Section 8.2, not every bounded function f
has an area under its graph in the sense of Definition 8.1.5. The following definition gives a special
name to those functions that do have a well-defined area under their graph. At this point, we lift
the restriction that f(x) ≥ 0 for all x in [a, b].
Definition 8.1.6 (Definition of the Riemann integral). Suppose that a function f
is bounded on [a, b]. If there exists a unique real number I such that
SP(f) ≤ I ≤ SP(f) for every partition P of [a, b], (8.7)
then we say that f is Riemann integrable on the interval [a, b]. If f is Riemann
integrable, then the unique real number I satisfying (8.7) is called the definite integral
of f from a to b and we write
I =
∫ b
a
f(x) dx.
The function f is called the integrand of the definite integral, while the points a and
b are called the limits of the definite integral.
Remark 8.1.7. The notation
∫ b
a f(x) dx is due to Leibniz. It evolved from a slightly different way
of writing down lower and upper Riemann sums. For example, SP(f) may be written as
SP(f) =
n∑
k=1
f(xk)∆xk,
where ∆xk = ak − ak−1 and f attains its maximum value on [ak−1, ak] at the point xk. When
taking a limit, as in (8.3), ∆xk was replaced with dx and the symbol
∑
was replaced with an
elongated ‘S’ (‘S’ stands for ‘sum’).
c©2020 School of Mathematics and Statistics, UNSW Sydney
180 CHAPTER 8. INTEGRATION
f(x)
x
C
a b
(a)
x
f(x)
1
(b)
f(x)
x1 2
(c)
f(x)
x1
1
2
|
(d)
Figure 8.8: Diagrams for examples in Section 8.2.
Remark 8.1.8. If f is Riemann integrable on [a, b], then the real number
∫ b
a f(x) dx is the area
under the graph of f from a to b, provided that f(x) ≥ 0 for all x in [a, b]. A geometric interpretation
of the real number
∫ b
a f(x) dx when f takes negative values will be discussed in Section 8.3.
Remark 8.1.9. [X] If f is bounded but not continuous, then f may not attain a maximum or
minimum value on the closed interval [ak, ak−1]. This technical difficulty can be overcome by
defining the upper and lower Riemann sums in terms of suprema and infima rather than maxima
and minima. The concept of a supremum and infimum is introduced only in higher Mathematics
courses.
Remark 8.1.10. One should think about why Definition 8.1.5 is a good definition for the area
under the graph of a function. The following questions need to be answered:
• If the area under the graph of f is well-defined by Definition 8.1.5, would it agree with any
other reasonable definition for the area under the graph of f?
• Is Definition 8.1.5 consistent with the axioms of area listed in Subsection 8.1.1?
The answer to each of these questions is ‘yes,’ but we leave it to the reader to ponder why this
might be so. Some of the results given over the next few sections do confirm that our definition of
area under a graph is consistent with the axioms of area.
8.2 Integration using Riemann sums
(Ref: SH10 §5.2)
In this section we calculate the area under the graph of functions by computing Riemann sums.
Our notation for the upper and lower Riemann sums follows that of equations (8.4) and (8.5).
Example 8.2.1. Suppose that f(x) = C where C is a positive constant. Show that f is Riemann
integrable on [a, b] and calculate the area under the graph of f from a to b (see Figure 8.8 (a)).
Solution. Suppose that P is a partition of [a, b] and that P = {a0, a1, . . . , an}. Since f is constant,
c©2020 School of Mathematics and Statistics, UNSW Sydney
8.2. INTEGRATION USING RIEMANN SUMS 181
x
f(x)
a b
| |
b
x
g(x)
a b
| |
b
b b
bc
Figure 8.9: Piecewise continuous functions
fk = fk = C for every k between 1 and n. Therefore
SP (f) =
n∑
k=1
fk(ak − ak−1)
=
n∑
k=1
C(ak − ak−1)
= C(a1 − a0 + a2 − a1 + a3 − a2 + · · ·+ an−1 − an−2 + an − an−1)
= C(ak − a0)
= C(b− a).
The same calculation gives SP(f) = C(b− a). Since P is an arbitrary partition of [a, b], we have
SP(f) = C(b− a) = SP for every partition P of [a, b].
Hence there is a unique number A satisfying (8.7), namely A = C(b− a). Therefore f is Riemann
integrable, ∫ b
a
f(x) dx = C(b− a)
and the area under the graph of f is C(b− a).
Remark 8.2.2. The area under the graph of f is consistent with axiom (A5).
There are many other kinds of functions which are Riemann integrable. One of the largest
classes consists of those functions that are bounded and piecewise continuous.
Definition 8.2.3. A function f : [a, b] → R is said to be piecewise continuous if it
is continuous on [a, b] at all except perhaps a finite number of points.
Figure 8.9 illustrates the graphs of two functions f and g that are piecewise continuous on [a, b].
The function f is bounded on [a, b] while the function g is not. According to the next theorem, the
area of the shaded region under the graph of f is well-defined.
Theorem 8.2.4. If f is bounded and piecewise continuous on [a, b] then f is integrable on [a, b].
c©2020 School of Mathematics and Statistics, UNSW Sydney
182 CHAPTER 8. INTEGRATION
Since the proof of this theorem is difficult, we omit it.
In practice, it often hard to prove from first principles that a function f is Riemann integrable.
However, if we know in advance that f is integrable on [a, b] then it is much easier to calculate its
definite integral from a to b. All one needs to do is show that
lim
n→∞SPn(f) = limn→∞SPn(f)
for some sequence of partitions Pn of [a, b]. If I denotes this common limit, then∫ b
a
f(x) dx = I.
Example 8.2.5. If f(x) = x2, find the area under the graph of f from 0 to 1 (see Figure 8.8 (b)).
Solution. We first note that f is integrable on [0, 1] since it is bounded and continuous on [0, 1].
Suppose that
Pn = {0, 1n , 2n , 3n , . . . , n−1n , nn .}
Then
SPn(f) =
1
3
+
1
2n
+
1
6n2
and SPn(f) =
1
3
− 1
2n
+
1
6n2
(see Example 8.1.2 for these calculations). Since
lim
n→∞SPn(f) =
1
3
= lim
n→∞SPn(f),
we conclude that ∫ 1
0
f(x) dx =
1
3
.
So the area underneath the graph of f from 0 to 1 is 13 .
Example 8.2.6. Suppose that f(x) = 3− x. Calculate ∫ 21 f(x) dx (see Figure 8.8 (c)).
Our final example shows that some functions are not Riemann integrable. For such a function
f , the area under the graph of f is not defined according to Riemann integration.
Example 8.2.7. Suppose that f : [0, 1]→ R is defined by
f(x) =
{
2 if x ∈ Q
1 if x /∈ Q.
(see Figure 8.8 (d)). Show that f is not Riemann integrable.
Solution. Suppose that Pn is an arbitrary partition of [0, 1]. Then fk = 2 while fk = 1 for every k
between 1 and n. Hence
SPn(f) = 1 and SPn(f) = 2.
(Exercise: show that this is the case by following a method similar to Example 8.2.1). Therefore
there is no unique real number I satisfying (8.7). We conclude that f is not Riemann integrable.
c©2020 School of Mathematics and Statistics, UNSW Sydney
8.3. THE RIEMANN INTEGRAL AND SIGNED AREA 183
x
f(x)
a b
c
Ω1
Ω2
(a)
x
f(x)
a b
c
(b)
Figure 8.10: Signed area and the definite integral
Remark 8.2.8. A more sophisticated definition of the integral, known as the Lebesgue integral, was
developed at the beginning of the twentieth century by the French mathematician Henri Lebesgue.
Lebesgue’s definition of the integral allows us to integrate some functions which are not Riemann
integrable. For example, if f is the function of Example 8.2.7 then
• f is not Riemann integrable and the Riemann integral
∫ 1
0
f(x) dx is not defined;
• f is Lebesgue integrable and the Lebesgue integral
∫ 1
0
f(x) dx is equal to 1.
The Lebesgue integral is studied in some third year mathematics courses.
8.3 The Riemann integral and signed area
(Ref: SH10 §5.5)
The definite integral ∫ b
a
f(x) dx
can only be interpreted as the area under the graph of f from a to b if f(x) ≥ 0 for all x in [a, b].
How should we interpret the definite integral if the integrand f has negative values on [a, b]?
Consider the function f graphed in Figure 8.10 (a). The upper sum from a to b, illusrated in
Figure 8.10 (b), approximates area(Ω1). However, since f(x) < 0 when b < x < c, each term in the
upper sum from b to c will be negative. Hence the upper sum from b to c approximates −area(Ω2).
By increasing the number of subdivisions of [a, c] and decreasing the size of each subdivison, these
approximations get better and better. It follows that∫ c
a
f(x) dx = area(Ω1)− area(Ω2).
We call this quantity the signed area under the graph of f from a to c.
Thus there are two ways to calculate the (unsigned or absolute) area of the entire shaded region
Ω of Figure 8.10 (a):
c©2020 School of Mathematics and Statistics, UNSW Sydney
184 CHAPTER 8. INTEGRATION
• area(Ω) =
∫ c
a
|f(x)| dx; or
• area(Ω) =
∫ b
a
f(x) dx−
∫ c
b
f(x) dx.
This kind of calculation will be illustrated in Section 8.6.
8.4 Basic properties of the Riemann integral
(Ref: SH10 §5.8)
The following proposition gives a list of some of the basic properties of the Riemann integral.
Proposition 8.4.1 (Basic properties of the Riemann integral). Suppose that f and g are integrable
functions over [a, b].
(i) If α and β are real numbers then αf + βg is integrable and
∫ b
a
(αf + βg)(x) dx = α
∫ b
a
f(x) dx+ β
∫ b
a
g(x) dx.
(ii) If a < c < b then ∫ b
a
f(x) dx =
∫ c
a
f(x) dx+
∫ b
c
f(x) dx.
(iii) If f(x) ≥ 0 for all x in [a, b] then
∫ b
a
f(x) dx ≥ 0.
(iv) If f(x) ≤ g(x) for all x in [a, b] then
∫ b
a
f(x) dx ≤
∫ b
a
g(x) dx.
(v) If m ≤ f(x) ≤M for all x in [a, b] then
m(b− a) ≤
∫ b
a
f(x) dx ≤M(b− a).
(vi) If |f | is integrable on [a, b] then
∣∣∣∣
∫ b
a
f(x) dx
∣∣∣∣ ≤
∫ b
a
|f(x)| dx.
Each of these properties can be interpreted geometrically (see Figure 8.11). For example, under
the assumption that f(x) ≥ 0, property (iv) says that if the graph of f lies beneath the graph of g
then the area under the graph of f will be less than the area under the graph of g. Except for (i),
each of these properties is easy to prove.
c©2020 School of Mathematics and Statistics, UNSW Sydney
8.4. BASIC PROPERTIES OF THE RIEMANN INTEGRAL 185
x
f(x)
a c b
(a) Prop 8.4.1(ii)
x
y
a b
g
f
(b) Prop 8.4.1 (iv)
x
y
a b
M
m
f
(c) Prop 8.4.1 (v)
Figure 8.11: Geometric interpretation of Proposition 8.4.1.
Proof. We omit the proof of (i) and (ii).
(iii) If f(x) ≥ 0 for all x in [a, b], then each term in lower Riemann sum (8.5) is non-negative
for any partition P of [a, b]. Hence
0 ≤ SP(f) ≤
∫ b
a
f(x) dx
by Definition 8.1.6.
(iv) Suppose that h = g − f . Then h(x) ≥ 0 for all x in [a, b] and by applying (iii) and (i) we
have
0 ≤
∫ b
a
h(x) dx =
∫ b
a
(
g(x)− f(x)) dx = ∫ b
a
g(x) dx−
∫ b
a
f(x) dx.
Rearranging the inequality gives the result.
(v) By (iv) we have ∫ b
a
mdx ≤
∫ b
a
f(x) dx ≤
∫ b
a
M dx.
By Example 8.2.1, the integral on the left is m(b − a) and the integral on the right is M(b − a),
completing the proof.
(vi) Now
−|f(x)| ≤ f(x) ≤ |f(x)| ∀x ∈ [a, b],
so ∫ b
a
−|f(x)| dx ≤
∫ b
a
f(x) dx ≤
∫ b
a
|f(x)| dx
by (iv). By (i),
∫ b
a −|f(x)| dx = −
∫ b
a |f(x)| dx and so
−
∫ b
a
|f(x)| dx ≤
∫ b
a
f(x) dx ≤
∫ b
a
|f(x)| dx
which is equivalent to ∣∣∣∣
∫ b
a
f(x) dx
∣∣∣∣ ≤
∫ b
a
|f(x)| dx
by a property of the absolute value function.
c©2020 School of Mathematics and Statistics, UNSW Sydney
186 CHAPTER 8. INTEGRATION
So far the definite integral ∫ b
a
f(x) dx
has only been defined when a < b. We now extend this definition.
Definition 8.4.2. Suppose that b < a and that f in integrable on [b, a]. Then we
define ∫ b
a
f(x) dx
to be the real number
−
∫ a
b
f dx
and we define ∫ a
a
f(x) dx
to be 0.
This definition gives a more general version of Proposition 8.4.1 (ii). We omit the proof.
Proposition 8.4.3. Suppose that a, b and c are real numbers and that f is integrable over some
interval containing a, b and c. Then∫ b
a
f(x) dx =
∫ c
a
f(x) dx+
∫ b
c
f(x) dx.
8.5 The first fundamental theorem of calculus
In this section we present a surprising theoretical result that indicates a strong connection between
integration (calculating areas) and differentiation (calculating gradients of tangents). This result
has enormous ramifications for the problem of calculating area.
We begin by studying the area under the graph of a function. Suppose that f is a continuous
function on the interval [a, b]. Since f is integrable on [a, b] (see Theorem 8.2.4) we can define the
function F : [a, b]→ R by the formula
F (x) =
∫ x
a
f(t) dt ∀x ∈ [a, b]. (8.8)
For x in [a, b], the number F (x) represents the (signed) area captured underneath the graph of x
on the interval [a, x] (see Figure 8.12 (a)). Thus F is an ‘area function’.
(Note that the limit x in (8.8) is a different symbol from the ‘dummy variable’ t. It would be
incorrect and confusing to write
F (x) =
∫ x
a
f(x) dx
for formula (8.8).)
c©2020 School of Mathematics and Statistics, UNSW Sydney
8.5. THE FIRST FUNDAMENTAL THEOREM OF CALCULUS 187
y = f(t)
F (x)
t
y
| | |
a x b
(a)
y = f(t)
t
y
| | | |
a x x+ h b
(b)
Figure 8.12: The area function F
We investigate some of the properties of F . First, it is easy to see that if f(t) ≥ 0 for all t in
[a, b] then the shaded area under f increases as x increases. That is, if f is a non-negative function
then F is an increasing function.
Second, we ask whether F continuous on [a, b] and differentiable on (a, b). Since differentiability
on (a, b) implies continuity on (a, b), we address the question of differentiability first. Consider the
difference quotient
F (x+ h)− F (x)
h
when x ∈ (a, b) and h is small. Now, if h is a small positive number then Figure 8.12 (b) shows
that
F (x+ h)− F (x) = (area under f on the interval [x, x+ h])
≈ f(x)× h,
since f(t) ≈ f(x) for all t in [x, x+ h] by the continuity of f . Dividing both sides by h gives
F (x+ h)− F (x)
h
≈ f(x).
As h gets smaller, this approximation gets better and so we expect that
lim
h→0+
F (x+ h)− F (x)
h
= f(x).
A similar argument may be repeated when h is a small negative number to obtain the corresponding
left-hand limit. Therefore it seems true that
F ′(x) = lim
h→0
F (x+ h)− F (x)
h
= f(x).
This intuitive argument can be made rigorous (see the proof at the end of this section). Hence we
conclude that F is continuous and differentable on (a, b) and that F ′(x) = f(x) for all x in (a, b).
The fact that a continuous function f and its ‘area function’ F are related by the differential
equation
F ′(x) = f(x)
has far-reaching consequences. This result is so important that it has come to be known as the
fundamental theorem of calculus.
c©2020 School of Mathematics and Statistics, UNSW Sydney
188 CHAPTER 8. INTEGRATION
Theorem 8.5.1 (The first fundamental theorem of calculus). If f is continuous function defined
on [a, b] then the function F : [a, b]→ R, defined by
F (x) =
∫ x
a
f(t) dt, (8.9)
is continuous on [a, b], differentiable on (a, b) and has derivative F ′ given by
F ′(x) = f(x)
for all x in (a, b).
The first fundamental theorem of calculus is a deep mathematical result which relates differen-
tiation to integration. From the statement of the theorem we note the following points.
• The fact that F satisfies the equation F ′ = f on (a, b) says that F is an antiderivative of f
on (a, b). Hence every continuous function f has an antiderivative F , given by formula (8.9).
• Since any two antiderivatives of f differ by a constant (see Theorem 5.9.4), every antiderivative
of f is of the form
F + constant,
where F is given by the integral formula (8.9). Hence the process of integration and of
antidifferentiation are essentially the same. (This is surprising from a geometric point of
view, where integration is used to calculate areas while differentiation is used to calculate
gradients of tangents.)
• The above point suggests that integration and differentiation are inverse processes. If one
takes a function f , integrates it and differentiates the result, then one obtains f again. That is,
differentiation undoes what integration does to f . This is precisely expressed by the formula
f(x) =
d
dx
(∫ x
a
f(t) dt
)
.
Whether or not the converse statement is true (that is, that integration undoes what differ-
entiation does to a function f) will be discussed in the next section.
Thus the first fundamental theorem of calculus suggests that there may be a different method for
calculating the area under the graph of f ; rather than integrating via limits of Riemann sums, try
integrating via antidifferentiation. This approach is investigated in the next section.
We end this section by presenting a proof of the first fundamental theorem of calculus. The
proof of the differentiability of F is based on the intuitive argument given immediately prior to the
statement of Theorem 8.5.1.
Proof of Theorem 8.5.1. We begin by showing that F is differentiable on (a, b). Suppose that
x ∈ (a, b) and consider a small positive number h. By Proposition 8.4.1 (ii) one can show that
F (x+ h)− F (x) =
∫ x+h
x
f(t) dt. (8.10)
c©2020 School of Mathematics and Statistics, UNSW Sydney
8.5. THE FIRST FUNDAMENTAL THEOREM OF CALCULUS 189
x
y
x x+ h
Mh
mh
y = f(x)
Figure 8.13: Diagram for the proof of Theorem 8.5.1.
Since f is continuous on [a, b], it attains a minimum valuemh and a maximum valueMh on [x, x+h];
that is,
mh ≤ f(t) ≤Mh ∀t ∈ [x, x+ h].
By Proposition 8.4.1 (v) (see also Figure 8.13),
mhh ≤
∫ x+h
x
f(t) dt ≤Mhh
and hence
mhh ≤ F (x+ h)− F (x) ≤Mhh.
by (8.10). Now h > 0 so
mh ≤ F (x+ h)− F (x)
h
≤Mh.
Since f is continuous on [x, x+ h],
lim
h→0+
mh = f(x) = lim
h→0+
Mh
and hence
lim
h→0+
F (x+ h)− F (x)
h
= f(x)
by the pinching theorem (Theorem 2.5.9).
In a similar manner, one can show that
lim
h→0−
F (x+ h)− F (x)
h
= f(x).
Hence F is differentiable on (a, b) and F ′(x) = f(x) for all x in (a, b).
To complete the proof, we need to show that F is continuous on [a, b]. Since F is differentiable
on (a, b), it is also continuous on (a, b) by Theorem 4.5.1. So it suffices to show that F is continuous
at the endpoints of [a, b].
c©2020 School of Mathematics and Statistics, UNSW Sydney
190 CHAPTER 8. INTEGRATION
We will show that F is continuous at a. Since f is continuous on [a, b] there is a positive
constant M such that |f(x)| ≤ M for all x in [a, b] (see Corollary 3.4.9). Suppose that x ∈ (a, b).
By using the properties of Proposition 8.4.1,
|F (x)− F (a)| =
∣∣∣∣
∫ x
a
f(t) dt−
∫ a
a
f(t) dt
∣∣∣∣
=
∣∣∣∣
∫ x
a
f(t) dt
∣∣∣∣ (Definition 8.4.2)
≤
∫ x
a
|f(t)| dt (Proposition 8.4.1 (vi))
≤
∫ x
a
M dt (f is bounded)
=M |x− a| (Example 8.2.1).
Hence
|F (x) − F (a)| ≤M |x− a| → 0
as x→ a+, or in other words, lim
x→a+
F (x) = F (a). So F is continuous at a.
The proof of continuity at the endpoint b is similar.
8.6 The second fundamental theorem of calculus
(Ref: SH10 §5.4)
The first fundamental theorem of calculus says that the processes of integration and antidifferen-
tiation are essentially the same. The second fundamental theorem of calculus gives a fast way of
calculating integrals by exploiting this similarity.
Theorem 8.6.1 (The second fundamental theorem of calculus). Suppose that f is a continuous
function on [a, b]. If F is an antiderivative of f on [a, b] then
∫ b
a
f(t) dt = F (b)− F (a). (8.11)
The second fundamental theorem of calculus is an astonishing result. To calculate the signed
area under the graph of a continuous function f , all one needs to know is the value of any an-
tiderivative of f at the endpoints a and b. To illustrate the power of this result, compare the
lengthy calculation of Riemann sums in Example 8.1.2 with the example below.
Example 8.6.2. Suppose that f(x) = x2 for all real numbers x. Calculate the area of the region
bounded by the x-axis, the graph of f and lines x = 0 and x = 1.
Solution. Essentially we are being asked to evaluate
∫ 1
0
f(t) dt
c©2020 School of Mathematics and Statistics, UNSW Sydney
8.6. THE SECOND FUNDAMENTAL THEOREM OF CALCULUS 191
where f(t) = t2 (see Figure 8.4 (a)). An antiderivative F of f is given by F (t) = t
3
3 . Hence, by the
second fundamental theorem of calculus,
∫ 1
0
f(t) dt = F (1) − F (0) = 1
3
3
− 0
3
3
=
1
3
.
So the area of the region is 1/3.
Remark 8.6.3. The expression F (b) − F (a) appearing in (8.11) is used so frequently that it is
often abbreviated to
F (x)
∣∣∣b
a
or
[
F (x)
]b
a
.
Thus the calculation of Example 8.6.2 may be written as
∫ 1
0
t2 dt =
[
t3
3
]1
0
=
1
3
− 0 = 1
3
.
When calculating areas, it is important to bear in mind that the integral in (8.11) represents
the signed area.
Example 8.6.4. Calculate the area of the region bounded by the x-axis, the lines x = 0 and x = 4
and the function f given by
f(x) = (x− 2)(x− 4).
y = f(x)
x
y
2 4
8
Solution. The above sketch shows that part of the region lies beneath the x-axis. So the area we
c©2020 School of Mathematics and Statistics, UNSW Sydney
192 CHAPTER 8. INTEGRATION
seek corresponds to the integral∫ 4
0
|f(x)| dx =
∫ 2
0
f(x) dx+
∫ 4
2
(−f(x)) dx
=
∫ 2
0
x2 − 6x+ 8 dx−
∫ 4
2
x2 − 6x+ 8 dx
=
[
x3
3
− 3x2 + 8x
]2
0
−
[
x3
3
− 3x2 + 8x
]4
2
=
[(
8
3
− 12 + 16
)
− 0
]
−
[(
64
3
− 48 + 32
)
−
(
8
3
− 12 + 16
)]
= 8.
So the area is 8 square units.
Note that this is a different problem to: Evaluate
∫ 4
0
f(x) dx, whose solution is:
∫ 4
0
f(x) dx =
∫ 4
0
x2 − 6x+ 8 dx
=
[
x3
3
− 3x2 + 8x
]4
0
=
16
3
.
We now present a proof of the second fundamental theorem of calculus. The tools used in the
proof are the first fundamental theorem of calculus and a corollary of the mean value theorem.
Proof of Theorem 8.6.1. Suppose that f is continuous on [a, b] and that F is an antiderivative of
f on [a, b]. Define a function G : [a, b]→ R by the formula
G(x) =
∫ x
a
f(t) dt ∀x ∈ [a, b].
By the first fundamental theorem of calculus (see Theorem 8.5.1), G is continuous on [a, b] and
G′(x) = f(x) for all x in (a, b). Hence G is also an antiderivative of f on [a, b]. By Theorem 5.9.4,
there is a constant C such that
G(x) = F (x) + C (8.12)
for all x in [a, b]. Note that
G(a) =
∫ a
a
f(t) dt = 0.
So if x = a then equation (8.12) becomes
0 = F (a) + C.
c©2020 School of Mathematics and Statistics, UNSW Sydney
8.6. THE SECOND FUNDAMENTAL THEOREM OF CALCULUS 193
Hence C = −F (a). Therefore
∫ b
a
f(t) dt = G(b)
= F (b) + C (by (8.12))
= F (b)− F (a)
as required.
We return to the question of whether integration and differentiation are inverse processes.
Suppose that f is continuous on [a, b]. The first fundamental theorem of calculus states that if one
integrates a function f and differentiates the result, then one obtains f again. This is expressed by
the formula
f(x) =
d
dx
(∫ x
a
f(t) dt
)
∀x ∈ (a, b).
To be able to say that integration and differentiation are inverses operations of each other, we must
ask whether the converse is true. That is, if one differentiates a function f and integrates the result,
then does one obtain f again? In precise terms, we would like the formula
f(x) =
∫ x
a
f ′(t) dt ∀x ∈ (a, b) (8.13)
to hold. Unfortunately, this converse is false. The following corollary to the second fundamental
theorem of calculus says that one obtains f again only when f(a) = 0. Otherwise formula (8.13) is
‘out by a constant.’
Corollary 8.6.5. Suppose that f is continuous on [a, b] and has a continuous derivative on (a, b).
Then ∫ x
a
f ′(t) dt = f(x)− f(a)
for all x in [a, b]
Proof. Apply the second fundamental theorem of calculus to the function f ′ on the interval [a, x].
The second fundamental theorem of calculus allows a fruitful interaction between Riemann
sums and antidifferentiation.
Example 8.6.6. Suppose that f(x) =
1
1 + x2
. By considering the lower Riemann sum of f with
respect to the partition
{0, 1n , 2n , . . . , n−1n , nn}
of [0, 1], show that
π
4
= lim
n→∞
n∑
k=1
n
n2 + k2
= lim
n→∞
(
n
n2 + 12
+
n
n2 + 22
+
n
n2 + 32
+ · · ·+ n
n2 + n2
)
.
c©2020 School of Mathematics and Statistics, UNSW Sydney
194 CHAPTER 8. INTEGRATION
Solution. Let Pn denote the above partition of [0, 1]. Since f is a decreasing function on [0, 1], the
lower Riemann sum is given by
SPn(f) =
n∑
k=1
1
n
1
1 + (k/n)2
=
n∑
k=1
n
n2 + k2
.
Since f is bounded and continuous, it is integrable on [0, 1] and hence
lim
n→∞SPn(f) =
∫ 1
0
1
1 + x2
dx
(see Remark 8.6.7). But by the second fundamental theorem of calculus,∫ 1
0
1
1 + x2
dx =
[
tan−1(x)
]1
0
=
π
4
.
Hence
π
4
= lim
n→∞
n∑
k=1
n
n2 + k2
as desired.
Remark 8.6.7. The solution to Example 8.6.6 uses the following result. Suppose that f is Riemann
integrable on [a, b] and that {Pn}∞n=1 is a sequence of partitions of [a, b]. Let sn denote the maximum
size of the subintervals generated by Pn. If lim
n→0
sn = 0 then
lim
n→∞SPn(f) =
∫ b
a
f(x) dx = lim
n→∞SPn(f).
8.7 Indefinite integrals
(Ref: SH10 §5.6)
The fundamental theorems of calculus say that integration and antidifferentiation are, in a certain
sense, the same process. In particular, if a F is an antiderivative of f then∫ x
a
f(t) dt = F (x) + C
for some suitable constant C. If we have no particular interest in the interval [a, x] and merely
want to indicate that F is an antiderivative of f , then we write∫
f(t) dt = F (t) + C. (8.14)
An integral expressed in this way, without limits, is called an indefinite integral. The constant C
in (8.14) is called the constant of integration.
The table of antiderivatives given in Section 5.9 is equivalent to the following table of indefinite
integrals.
c©2020 School of Mathematics and Statistics, UNSW Sydney
8.7. INDEFINITE INTEGRALS 195
Indefinite integrals∫
xr dx =
1
r + 1
xr+1 + C, where r is a rational number and r 6= −1∫
sinx dx = − cos x+ C∫
cos x dx = sinx+ C∫
eax dx =
1
a
eax + C∫
f ′(x)
f(x)
dx = ln |f(x)|+ C
Since differentiation and indefinite integration (or antidifferentiation) are inverse processes, they
have many analogous properties. For example, differentiation behaves linearly, which means that
(f + g)′(x) = f ′(x) + g′(x)
and
(αf)′(x) = α× f ′(x)
for every real number α and every differentiable function f and g. The next proposition says that
indefinite integration is also behaves linearly. (We have already shown in Proposition 8.4.1 that
definite integration is linear.)
Proposition 8.7.1. Suppose that f and g are integrable functions and that α is a real constant.
Then ∫
(f + g)(x) dx =
∫
f(x) dx+
∫
g(x) dx (8.15)
and ∫
(αf)(x) dx = α
∫
f(x) dx. (8.16)
Proof. We prove (8.15) only and leave the proof of 8.16 as an exercise. Suppose that F , G and H
are antiderivatives of f , g and f + g respectively. Then
H ′(x) = (f + g)(x) = f(x) + g(x) = F ′(x) +G′(x)
by the linearity of differentiation. By expressing the equation H ′ = F ′ +G′ in terms of indefinite
integrals we obtain (8.15) as desired.
Example 8.7.2. Find ∫
(10x4 + 8 sin x− 9ex) dx.
Proof. By Proposition 8.7.1,∫
(10x4 + 8 sinx− 9ex) dx = 10
∫
x4 dx+ 8
∫
sinx dx− 9
∫
ex dx
= 10
(
1
5
x5 + C1
)
+ 8 (− cos x+ C2)− 9 (ex + C3)
= 2x5 − 8 cos x− 9ex + C
c©2020 School of Mathematics and Statistics, UNSW Sydney
196 CHAPTER 8. INTEGRATION
where C1, C2 and C3 are constants of integration and C = 10C1 + 8C2 − 9C3.
Remark 8.7.3. It is standard practice to combine of all constants of integration and simply write
‘+C’. Thus the calculation in Example 8.7.2 may be written as∫
(10x4 + 8 sin x− 9ex) dx = 10
∫
x4 dx+ 8
∫
sinx dx− 9
∫
ex dx
= 10× 1
5
x5 + 8(− cos x)− 9ex + C
= 2x5 − 8 cos x− 9ex + C.
8.8 Integration by substitution
(Ref: SH10 §5.7)
When differentiating the composition of two functions, we apply the chain rule. By reversing this
process with integration, we obtain a technique known as integration by substitution.
The basic idea is the following. Suppose that F is an antiderivative for f . Then∫
f(g(x)).g′(x) dx = F (g(x)) + C, (8.17)
as can be easily verified by differentiating the right-hand side via the chain rule. On the other
hand, ∫
f(u) du = F (u) +C. (8.18)
If we make the substitution u = g(x) then the right-hand sides of (8.17) and (8.18) are the same.
This suggests the substitution du = g′(x) dx, so that the left-hand side of (8.17) is transformed
into the left-hand side of (8.18). These mechanical substitutions can be used to transform the
complicated integral into a simpler one.
Example 8.8.1. Calculate the indefinite integral∫
(2x− 3) cos(x2 − 3x+ 4) dx
by the method of substitution.
Solution. We begin with the substitution
u = x2 − 3x+ 4.
Since
du
dx
= 2x− 3,
we also make the substitution
du = (2x− 3) dx.
c©2020 School of Mathematics and Statistics, UNSW Sydney
8.8. INTEGRATION BY SUBSTITUTION 197
Then ∫
(2x− 3) cos(x2 − 3x+ 4) dx =
∫
cos(u) du
= sinu+ C
= sin(x2 − 3x+ 4) + C.
We can verify our answer by differentiating sin(x2 − 3x+ 4) via the chain rule.
A similar technique can be used to evaluate definite integrals.
Example 8.8.2. Evaluate the definite integral∫ 2
0
3x2
√
1 + x3 dx.
Solution. If u = 1 + x3 then dudx = 3x
2. Hence we make the following substitutions:
u = 1 + x3
du = 3x2 dx
u = 1 (when x = 0)
u = 9 (when x = 2).
Thus ∫ 2
0
3x2
√
1 + x3 dx =
∫ 9
1
√
u du
=
[
2
3
u3/2
]9
1
=
2
3
(93/2 − 13/2)
=
52
3
.
In general, the reversal of the chain rule via integration gives rise to the change of variables
formula ∫ b
a
f(g(x))g′(x) dx =
∫ g(b)
g(a)
f(u) du, (8.19)
The substitutions
u = g(x), du = g′(x) dx,
are only to be understood within this context. Precise conditions under which formula (8.19) holds
is given by the following theorem.
Theorem 8.8.3 (Change of variables formula). Suppose that g is a differentiable function such
that g′ is continuous on [a, b]. If f is continuous on any interval I containing g(a) and g(b) then
the change of variables formula (8.19) holds.
c©2020 School of Mathematics and Statistics, UNSW Sydney
198 CHAPTER 8. INTEGRATION
Proof. Since f is continuous on I it has an antiderivative F : I → R by the first fundamental
theorem of calculus. By the chain rule (Theorem 4.2.2),
(F ◦ g)′(x) = F ′(g(x))g′(x)
= f(g(x))g′(x).
So by two applications of the second fundamental theorem of calculus,∫ b
a
f(g(x))g′(x) dx = (F ◦ g)(b)− (F ◦ g)(a)
= F (g(b)) − F (g(a))
=
∫ g(b)
g(a)
f(u) du,
completing the proof.
Finding a fruitful substitution is not always easy (and sometimes not even possible). In principle,
one looks for a function g in the integrand whose derivative g′ is also in the integrand. Once such
a function g is identified, try the substitution u = g(x). As illustrated below, it may be necessary
to manipulate the integrand to implement this strategy.
Example 8.8.4. Evaluate ∫
e
√
x
√
x
dx.
Solution. Note that
d
dx
(√
x
)
=
1
2
√
x
and so if g(x) =
√
x then g′(x) appears in the integrand (up to the constant factor 1/2). So the
substitution we use is
u =
√
x, du =
1
2
√
x
dx.
Hence ∫
e
√
x
√
x
dx = 2
∫
e
√
x
2
√
x
dx
= 2
∫
eu du
= 2eu + C
= 2e
√
x + C.
There are occasions when we make a substitution of the form x = g(u) rather than u = g(x).
Example 8.8.5. Evaluate the integral I, where
I =
∫ 9
4
dx
2 +
√
x
.
c©2020 School of Mathematics and Statistics, UNSW Sydney
8.8. INTEGRATION BY SUBSTITUTION 199
x
f(x)
−a
a
| |
Figure 8.14: Integration with an odd function f over [−a, a]
Solution. If the integral were ∫ 9
4
dx
2 + x
,
then it would be easy to evaluate. The difficulty with I clearly lies with the square root appearing
in the integrand. To remove the square root, we use the substitution x = u2. This gives
x = u2
dx = 2u du
u = 2 when x = 4
u = 3 when x = 9.
(Note that we are implicitly assuming that u > 0. It would be incorrect to take the limits in the
variable u to be −2 and 3, for example.) Hence
I =
∫ 3
2
2u du
2 + u
= 2
∫ 3
2
u
2 + u
du
= 2
∫ 3
2
2 + u− 2
2 + u
du
= 2
∫ 3
2
1− 2
2 + u
du
= 2
[
u− 2 ln |2 + u|
]3
2
= 2 + 4 ln
(
4
5
)
.
If an integrand possesses symmetry, then the corresponding definite integral may be easy to
evaluate. For example, if f is an odd function and we integrate f over a balanced interval [−a, a],
then the ‘negative area’ cancels out the ‘positive area’ and consequently the integral is zero (see
Figure 8.14). This result is proved using integration by substitution.
Proposition 8.8.6. Suppose that f is a continuous function and a is a real number.
c©2020 School of Mathematics and Statistics, UNSW Sydney
200 CHAPTER 8. INTEGRATION
(i) If f is even then ∫ a
−a
f(x) dx = 2
∫ a
0
f(x) dx.
(ii) If f is odd then ∫ a
−a
f(x) dx = 0.
(iii) If f is periodic with period T then∫ a+T
a
f(x) dx =
∫ T
0
f(x) dx.
Proof. We prove (ii) only and leave the proofs of (i) and (iii) as an exercise. Suppose that f is an
odd function. By breaking up the integral and using a substitution we have∫ a
−a
f(x) dx =
∫ 0
−a
f(x) dx+
∫ a
0
f(x) dx
= −
∫ 0
a
f(−u) du+
∫ a
0
f(x) dx (substituting u = −x)
=
∫ a
0
f(−u) du+
∫ a
0
f(x) dx
= −
∫ a
0
f(u) du+
∫ a
0
f(x) dx (since f(−u) = f(u))
= 0,
completing the proof of (ii).
Example 8.8.7. Find ∫ pi
17
+2π
pi
17
sin5 x dx.
Solution. If f(x) = sin5 x then f is both odd and periodic with period 2π. Hence
∫ pi
17
+2π
pi
17
sin5 x dx =
∫ 2π
0
sin5 x dx (by Proposition 8.8.6 (iii) when a = π17 )
=
∫ π
−π
sin5 x dx (by Proposition 8.8.6 (iii) when a = −π)
= 0 (by Proposition 8.8.6 (ii)).
8.9 Integration by parts
(Ref: SH10 §8.2)
c©2020 School of Mathematics and Statistics, UNSW Sydney
8.9. INTEGRATION BY PARTS 201
Suppose that u and v are two functions of a variable x. The product rule for differentiation gives
d
dx
(uv) = v
du
dx
+ u
dv
dx
.
By integrating both sides with respect to x we obtain
uv =
∫
v
du
dx
dx+
∫
u
dv
dx
dx.
Rearranging this equation gives the integration by parts formula∫
u
dv
dx
dx = uv −
∫
v
du
dx
dx. (8.20)
For definite integrals the formula becomes
∫ b
a
u
dv
dx
dx =
[
uv
]b
a
−
∫ b
a
v
du
dx
dx. (8.21)
by the second fundamental theorem of calculus.
Integration by parts allows one to evaluate the integral of a product of two functions. Its success
depends on choosing u and v such that∫ b
a
v
du
dx
dx is easier to calculate than
∫ b
a
u
dv
dx
dx.
Once u and v are chosen, simply apply formula (8.20) or (8.21).
Remark 8.9.1. The shorthand version∫
uv′ = uv −
∫
vu′
of formulae (8.20) and (8.21) is easier to remember.
Example 8.9.2. Evaluate the integral ∫
xe2x dx.
Solution. We set
u = x v = 12e
2x
u′ = 1 v′ = e2x.
Then integration by parts (see formula (8.20)) gives∫
xe2x dx = 12xe
2x −
∫
1
2e
2x × 1 dx
= 12xe
2x − 14e2x +C,
completing the solution.
c©2020 School of Mathematics and Statistics, UNSW Sydney
202 CHAPTER 8. INTEGRATION
(Note that if we had instead set
u = e2x v = 12x
2
u′ = 2e2x v′ = x.
then integration by parts gives ∫
xe2x dx = 12x
2e2x −
∫
x2e2x dx
and the integral on the right-hand side is harder to evaluate than the one we started with.)
Sometimes integration by parts must be applied several times to evaluate the integral.
Example 8.9.3. Find the integral I, where
I =
∫ π
0
x2 sinx dx.
Example 8.9.4. Evaluate the indefinite integral I, where
I =
∫
ex sinx dx
Integration by parts can also be used to evaluate some integrals of the form
∫
f(x) dx. The
trick is to rewrite the integral as ∫
f(x)× 1 dx.
Example 8.9.5. Evaluate the integral ∫
cos−1 x dx.
8.10 Improper integrals
(Ref: SH10 §11.7)
So far we have defined definite integrals for bounded functions over finite intervals. In this section
we look at integrals over infinite intervals. Such integrals have many applications in probability,
statistics, physics and engineering.
To give meaning to the improper integral ,∫ ∞
a
f(x) dx,
we examine the behaviour of
∫ R
a f(x) dx as R→∞, as illustrated below.
x
f(x)
a R
| |
c©2020 School of Mathematics and Statistics, UNSW Sydney
8.10. IMPROPER INTEGRALS 203
Definition 8.10.1. (a) Suppose that there is a real number L such that∫ R
a
f(x) dx→ L
as R → ∞. Then f is said to be integrable over [a,∞) and the integral∫∞
a f(x) dx is said to be convergent. Morever, we write∫ ∞
a
f(x) dx = L.
(b) Suppose that ∫ R
a
f(x) dx
does not have a limit as R → ∞. Then f we say that f not integrable over
(a,∞) and we say that the integral ∫∞a f(x) dx is divergent.
One can write down a similar definition for improper integrals of the form
∫ b
−∞ f(x) dx.
Example 8.10.2. Evaluate the following improper integrals or show that they diverge:
(a)
∫ ∞
0
1
x2 + 1
dx;
(b)
∫ ∞
1
1√
x
dx;
(c)
∫ 0
−∞
e2x dx.
Solution. (a) If R is a real number then∫ R
0
1
x2 + 1
dx =
[
tan−1
]R
0
= tan−1R− 0
→ π
2
as R→∞. Hence the improper integral converges and∫ ∞
0
1
x2 + 1
dx =
π
2
.
(b) If R is a real number then ∫ R
1
x−1/2 dx =
[
2x1/2
]R
1
= 2
√
R− 2
→∞
as R→∞. Hence the improper integral diverges.
c©2020 School of Mathematics and Statistics, UNSW Sydney
204 CHAPTER 8. INTEGRATION
We now consider improper integrals whose interval of integration is the entire real line.
Definition 8.10.3. We say f is integrable over (−∞,∞) if f is integrable over both
(−∞, 0) and (0,∞). In this case we write∫ ∞
−∞
f(x) dx =
∫ 0
−∞
f(x) dx+
∫ ∞
0
f(x) dx.
If f is not integrable on either of the intervals (−∞, 0) or (0,∞), then we say that
the improper integral ∫ ∞
−∞
f(x) dx
diverges.
The following example shows that the improper integral
∫ ∞
−∞
f(x) dx may diverge even if
lim
R→∞
∫ R
−R
f(x) dx exists.
Example 8.10.4. Consider the improper integral∫ ∞
−∞
x dx.
If R is a positive number then ∫ R
−R
x dx = 0
since the function f , given by f(x) = x, is odd (see Proposition 8.8.6). Hence
lim
R→∞
∫ R
−R
x dx = 0.
However ∫ R
0
x dx =
[
1
2x
2
]R
0
=
R2
2
→∞
as R→∞. Hence ∫ ∞
0
x dx
diverges, and by Definition 8.10.3 ∫ ∞
−∞
x dx
diverges also.
The following proposition determines the convergence or divergence of improper integrals whose
integrand is a power of x. Such integrals will be used frequently in Section 8.11
c©2020 School of Mathematics and Statistics, UNSW Sydney
8.11. COMPARISON TESTS FOR IMPROPER INTEGRALS 205
Proposition 8.10.5 (Convergence and divergence of p-integrals). The improper integral∫ ∞
1
1
xp
dx
is convergent if p > 1 and divergent if p ≤ 1.
Proof. If p 6= 1 then
∫ R
1
x−p dx =
[
x1−p
1− p
]R
1
=
R1−p − 1
1− p
→
{
1
p−1 when 1− p < 0
∞ when 1− p > 0
as R→∞. Hence the integral converges when 1− p < 0 (that is, when p > 1) and diverges when
1− p > 0 (that is, when p < 1).
In the case when p = 1, we have
∫ R
1
1
x
dx =
[
lnx
]R
1
= lnR− ln 1→∞ as R→∞.
Hence the integral diverges when p = 1.
Remark 8.10.6. While you may be familiar with the fact that the ln function is unbounded, the
proof that lnR→∞ as R→∞ requires the use of Riemann sums and is given later in §9.2.
8.11 Comparison tests for improper integrals
(Ref: SH10 §11.7)
Some of the most important improper integrals, such as∫ ∞
0
e−x
2
dx
have integrands without any antiderivative in the elementary functions. In these cases, determining
convergence via a computation like
∫ ∞
a
f(x) dx = lim
R→∞
∫ R
a
f(x) dx = lim
R→∞
F (R)− F (a),
(where F is an antiderivative of f) is not a viable strategy. Instead, we compare the integral of f
with another integral
∫∞
a g(x) dx whose behaviour is known. The basic idea is as follows.
Suppose that 0 ≤ f(x) ≤ g(x) whenever x > a, as illustrated below.
c©2020 School of Mathematics and Statistics, UNSW Sydney
206 CHAPTER 8. INTEGRATION
g
f
x
y
a
Obviously the area under the graph of g is greater than that under the graph of f . Hence if the
area under g is finite then the area under f is also finite. If the area under f is infinite then so too
is the area under g. The next theorem expresses this conclusion in terms of improper integrals.
Theorem 8.11.1 (The comparison test). Suppose that f and g are integrable functions and that
0 ≤ f(x) ≤ g(x) whenever x > a.
(i) If
∫ ∞
a
g(x) dx converges then
∫ ∞
a
f(x) dx converges.
(ii) If
∫ ∞
a
f(x) dx diverges then
∫ ∞
a
g(x) dx diverges.
Proof. We begin by proving (i). Suppose that f and g are integrable functions, that 0 ≤ f(x) ≤ g(x)
for all x in [a,∞) and that ∫∞0 g(x) dx converges. Then
0 ≤
∫ R
a
f(x) dx ≤
∫ R
a
g(x) dx
whenever R > 0 by Proposition 8.4.1. Since the integral∫ R
a
f(x) dx
increases as R increases and is bounded above by
∫∞
0 g(x) dx, it has a limit as R → ∞. Hence∫∞
a f(x) dx is also convergent.
The proof of (ii) is left as an exercise.
When applying the comparison test, we often compare an improper integral I with∫ ∞
1
1
xp
dx,
since the behaviour of this second integral is known (see Proposition 8.10.5). The value of p is chosen
by analysing the dominant terms appearing in the integrand of I. The next example illustrates
this procedure.
Example 8.11.2. Determine whether or not the following improper integrals converge.
c©2020 School of Mathematics and Statistics, UNSW Sydney
8.11. COMPARISON TESTS FOR IMPROPER INTEGRALS 207
(a)
∫ ∞
1
1
x3 + 1
dx
(b)
∫ ∞
2
x+ 2
x3/2 − 1 dx.
Solution. (a) When x is large, the dominant term in the denominator of
1
x3 + 1
is x3. So intuitively,
1
x3 + 1
≈ 1
x3
for large x. Consequently we aim to compare∫ ∞
1
1
x3 + 1
dx with
∫ ∞
1
1
x3
dx.
To make this comparison rigorous, we seek an appropriate inequality. Note that x3 + 1 > x3
for all x > 1 and so
1
x3 + 1
<
1
x3
for all x > 1. Since ∫ ∞
1
1
x3
dx
converges (see Proposition 8.10.5), ∫ ∞
1
1
x3 + 1
dx
also converges by the comparison test.
There are many examples where the ‘dominant term analysis’ is straightforward but it is difficult
to obtain an appropriate inequality for successful use of the comparison test. This difficulty can
sometimes be overcome by multiplying one of the integrands by a ‘fudge factor.’
Example 8.11.3. Determine whether or not the the improper integral∫ ∞
2
1
x3/2 − 1 dx.
converges.
Solution. Dominant term analysis suggests that
1
x3/2 − 1 ≈
1
x3/2
when x is large. So we compare∫ ∞
2
1
x3/2 − 1 dx with
∫ ∞
2
1
x3/2
dx.
c©2020 School of Mathematics and Statistics, UNSW Sydney
208 CHAPTER 8. INTEGRATION
Now the integral to the right converges (by Proposition 8.10.5) but it is clear that
1
x3/2 − 1 >
1
x3/2
when x > 2, so we cannot immediately apply the comparison test.
However, by multiplying the integrand on the right with the ‘fudge factor’ 2, we see that
1
x3/2 − 1 <
2
x3/2
(8.22)
when x > 2. (To see why, observe that when x > 2,
x3/2 − 1 > x3/2 − 12x3/2 = 12x3/2
whence (8.22) follows.) Since ∫ ∞
2
2
x3/2
dx
converges, ∫ ∞
2
1
x3/2 − 1 dx.
must also.
Later we give a method for determining the convergence of this integral that does not rely on
the use of fudge factors.
Dominant term analysis is not always straightforward, and at times may not even be useful.
One must develop an intuition of how functions decay at infinity and of what comparisons to use.
Example 8.11.4. Determine whether or not the following improper integrals converge.
(a)
∫ ∞
1
e−
√
x dx
(b)
∫ ∞
1
sinx+ 2√
x+ 1
dx.
Solution. (a) Dominant term analysis does not apply since there is only one term appearing in the
integrand. Instead, we recognise that the decay of the exponential e−
√
x is very rapid; it is certainly
c©2020 School of Mathematics and Statistics, UNSW Sydney
8.11. COMPARISON TESTS FOR IMPROPER INTEGRALS 209
much faster than the decay of 1x2 . This can be verified by the following limit calculation:
lim
x→∞
e−
√
x
1
x2
= lim
x→∞
x2
ex
1/2
= lim
x→∞
2x
1
2x
−1/2ex1/2
(by l’Hoˆpital’s rule)
= lim
x→∞
4x3/2
ex1/2
= lim
x→∞
4.32x
1/2
1
2x
−1/2ex1/2
(by l’Hoˆpital’s rule)
= lim
x→∞
12x
ex
1/2
... (two more applications of l’Hoˆpital’s rule)
= lim
x→∞
24
ex
1/2
= 0.
Hence we conclude that
e−
√
x <
1
x2
whenever x is sufficiently large. Since ∫ ∞
1
1
x2
dx
converges, ∫ ∞
1
e−
√
x dx
also converges by the comparison test.
(b) While the numerator of
sinx+ 2√
x+ 1
has no dominant term, intuitively we see that, since the numerator is bounded and the dominant
term of the denominator is
√
x, we should compare the integrand with
1√
x
.
Since ∫ ∞
1
1√
x
dx
diverges, we seek an inequality of the form
f(x) ≤ sinx+ 2√
x+ 1
where f(x) is 1/
√
x with an appropriate fudge factor.
c©2020 School of Mathematics and Statistics, UNSW Sydney
210 CHAPTER 8. INTEGRATION
Now, −1 ≤ sinx, and so
sinx+ 2√
x+ 1
≥ 1√
x+ 1
≥ 1
2
√
x
for all x ≥ 1. Since ∫ ∞
1
1
2
√
x
dx
diverges (see Proposition 8.10.5), ∫ ∞
1
sinx+ 2√
x+ 1
dx
also diverges by the comparison test.
The next theorem provides an alternative approach for constructing a comparison of integrals.
It is particularly useful if dominant term analysis is straightforward but one wants to avoid the use
of fudge factors.
Theorem 8.11.5 (The limit form of the comparison test). Suppose that f and g are non-negative
and bounded on [a,∞). If
lim
x→∞
f(x)
g(x)
= L
and 0 < L <∞ then either∫ ∞
a
f(x) dx and
∫ ∞
a
g(x) dx both converge
or ∫ ∞
a
f(x) dx and
∫ ∞
a
g(x) dx both diverge.
The hypothesis
lim
x→∞
f(x)
g(x)
= L
indicates that f and g exhibit similar behaviour at infinity. The proof of the theorem will be given
at the end of this section.
Example 8.11.6. Determine whether or not the following improper integrals converge.
(a)
∫ ∞
1
1√
1 + x+ x2
dx
(b)
∫ ∞
2
x2 + 3x
5x4 − 2 dx.
Solution. (a) Suppose that f(x) =
1√
1 + x+ x2
. We need to find a suitable function g to compare
f with. As x gets large, the dominant term under the square root is x2. Therefore it seems that
1√
1 + x+ x2
≈ 1√
x2
=
1
x
c©2020 School of Mathematics and Statistics, UNSW Sydney
8.11. COMPARISON TESTS FOR IMPROPER INTEGRALS 211
when x is large. So define g by g(x) = 1x . The following calculation verifies that our choice of g is
suitable:
lim
x→∞
f(x)
g(x)
= lim
x→∞
x√
1 + x+ x2
= lim
x→∞
√
x2
1 + x+ x2
= lim
x→∞
√
1
1
x2
+ 1x + 1
= 1.
Since this limit is a positive real number, we can apply the limit form of the comparison test. By
Proposition 8.10.5,
∫∞
1 g(x) dx diverges. Hence
∫∞
1 f(x) dx also diverges.
Remark 8.11.7. As a general rule, students will find the limit form of the comparison test easier
to use than the inequality form of the comparison test when the integrand is a rational function,
or is a ratio of roots of polynomials.
Proof of Theorem 8.11.1 [X] . Suppose that f and g are non-negative and bounded on [a,∞). (It
follows, for later use, that if a < M < R and R is increasing then the integrals∫ R
M
f(x) dx and
∫ R
M
g(x) dx,
are also increasing.) Suppose also that
lim
x→∞
f(x)
g(x)
= L
where 0 < L < ∞. Choose a positive number ǫ such that 0 < ǫ < L. By Definition 2.2.1, there is
a positive real number M such that ∣∣∣∣f(x)g(x) − L
∣∣∣∣ < ǫ
whenever x > M . Hence
−ǫ < f(x)
g(x)
− L < ǫ ∀x > M
and rearranging gives
0 < (L− ǫ)g(x) < f(x) < (L+ ǫ)g(x) ∀x > M.
By integrating with respect to x we obtain
0 < (L− ǫ)
∫ R
M
g(x) dx <
∫ R
M
f(x) dx < (L+ ǫ)
∫ R
M
g(x) dx
whenever R > M . By letting R approach infinity it is seen that
• if ∫∞M g(x) dx is convergent then ∫ RM f(x) is also convergent (since ∫ RM f(x) increases as R
increases and it is bounded above by (L+ ǫ)
∫∞
M g(x) dx);
• if ∫∞M f(x) dx is convergent then ∫ RM g(x) is also convergent (since ∫ RM g(x) is increases as R
increases and is bounded above by 1L−ǫ
∫∞
M f(x) dx);
• if ∫∞M f(x) dx is divergent then ∫ RM g(x) dx→∞ as R→∞; and
c©2020 School of Mathematics and Statistics, UNSW Sydney
212 CHAPTER 8. INTEGRATION
• if
∫ ∞
M
g(x) dx is divergent then
∫ R
M f(x) dx→∞ as R→∞.
Hence either ∫ ∞
M
f(x) dx and
∫ ∞
M
g(x) dx both converge
or ∫ ∞
M
f(x) dx and
∫ ∞
M
g(x) dx both diverge.
The proof of the theorem now easily follows by first observing that∫ R
a
f(x) dx =
∫ M
a
f(x) dx+
∫ R
M
f(x) dx
and ∫ R
a
g(x) dx =
∫ M
a
g(x) dx+
∫ R
M
g(x) dx
whenever a < M < R, and then letting R approach infinity.
8.12 Functions defined by an integral
Many important functions are defined by an integral. If f is a continuous function then we may
define a function F by
F (x) =
∫ x
a
f(t) dt.
We note the following points.
• The derivative of F may be calculated by the first fundamental theorem of calculus.
• The value of F (x) may be approximated by computing Riemann sums.
• Symmetries of f can help in understanding the behaviour of F (see, for example, Proposition
8.8.6).
• The behaviour of F (x) as x→∞ or as x→ −∞ may be ascertained by studying an improper
integral of f .
We give one example. Others may be found in the problem set for Chapter 8 and at the beginning
of Chapter 9.
Consider the function f : R→ R given by
f(t) =
{
sin t
t if t 6= 0
1 if t = 0.
Since lim
t→0
sin t
t
= 1, the function f is continuous everywhere. We can now define the function
Si : R→ R by
Si(x) =
∫ x
0
f(t) dt ∀x ∈ R.
The name Si comes from ‘sine integral.’ This function is used in electrical engineering for signal
processing and in surveying for the Global Positioning System (GPS). Its graph is plotted in Figure
8.15.
c©2020 School of Mathematics and Statistics, UNSW Sydney
8.12. FUNCTIONS DEFINED BY AN INTEGRAL 213
x
Si(x)
π
2
−π2
10π−10π
||
|
|
Figure 8.15: The Si function.
Example 8.12.1. Consider the functions f and Si defined above.
(a) Find the value of Si(0).
(b) Show that Si is an odd function.
(c) Find and classify the critical points of Si on (0, 3π).
(d) A function G : R→ R is defined by the formula
G(x) =
∫ x2
0
f(t) dt.
Calculate G′(x).
Solution. (a) Si(0) =
∫ 0
0
f(t) dt = 0.
(b) We need to show that Si(−x) = − Si(x). Now
Si(−x) =
∫ −x
0
sin t
t
dt
=
∫ x
0
sin(−u)
−u (−du) (using the substitution t = −u)
= −
∫ x
0
− sinu
−u du
= −
∫ x
0
sinu
u
du
= − Si(x).
Hence Si is an odd function.
(c) By the first fundamental theorem of calculus,
Si′(x) =
sinx
x
c©2020 School of Mathematics and Statistics, UNSW Sydney
214 CHAPTER 8. INTEGRATION
for all x in (0, 3π). Critical points occur when Si′(x) = 0, that is, when x = π or when x = 2π. To
classify the critical points, we calculate the second derivative:
Si′′(x) =
x cos x− sinx
x2
.
Now Si′′(π) = − 1π and Si′′(2π) = 12π . Therefore Si has a local maximum point when x = π and a
local minimum point when x = 2π.
(d) Note that G(x) = Si(x2), so to calculate G′(x) we use the chain rule:
G′(x) =
d
dx
(
Si(x2)
)
= 2x Si′(x2) = 2x× f(x2) = 2x sin(x
2)
x2
=
2 sin(x2)
x
.
(If this calculation is too dense, then consider
y = G(x), u = x2
so that
y =
∫ u
0
f(t) dt and
dy
du
= f(u)
by the first fundamental theorem of calculus. Then by the chain rule,
dy
dx
=
dy
du
du
dx
= f(u)× 2x = 2xf(x2) = 2x sin(x
2)
x2
=
2 sin(x2)
x
.
Hence G′(x) = 2 sin(x
2)
x .)
c©2020 School of Mathematics and Statistics, UNSW Sydney
8.13. MAPLE NOTES 215
8.13 Maple notes
The command int(f(x),x); will cause Maple to attempt to find the indefinite integral
∫
f(x) dx,
and the command int(f(x),x=a..b); is used to find definite integrals
∫ b
a
f(x) dx. If Maple is
unable to find a primitive, then numerical integration can be carried out by combining the int and
evalf commands.
For example,
> int(1/(x*log(x)^2),x);
− (ln (x))−1
> int(1/(x*log(x)^2),x=2..infinity);
(ln (2))−1
> int( exp(-x^2)*ln(x), x=0..1 );∫ 1
0 e
−x2 ln (x) dx
> evalf(%);
−0.9059404763
c©2020 School of Mathematics and Statistics, UNSW Sydney
216 CHAPTER 8. INTEGRATION
Problems for Chapter 8
Problems 8.1 : Area and the Riemann Integral and
8.2 : Integration using Riemann sums
1. [R] [V]
a) By taking the partition Pn =
{
0,
1
n
,
2
n
, . . . , 1
}
of the interval [0,1], calculate the lower
sum SPn(f) and the upper sum SPn(f) for each function f .
i) f(x) = 1
ii) f(x) = x
iii) f(x) = x2
[You may need
n∑
k=1
k2 = 16n(n+ 1)(2n + 1).]
iv) f(x) = x3
[You may need
n∑
k=1
k3 = 14n
2(n+ 1)2.]
v) f(x) =
{
1 if x ∈ Q
0 if x /∈ Q
b) By taking the limit as n→∞ for each sum SPn(f) and SPn(f) calculated in (a), either
calculate
∫ 1
0
f(x) dx, or show that f is not Riemann integrable.
2. [R] An electrical signal S(t) has its amplitude |S(t)| tested (sampled) every 110 of a second.
It is desired to estimate the energy over a period of half a second, given exactly by
(∫ 1
2
0
|S(t)|2 dt
) 1
2
.
The results of the measurement are shown in the following table:
t .1 .2 .3 .4 .5
|S(t)| 60 50 50 45 55
e(t) 5 3 7 4 10
a) Using the above data for S(t), set up an appropriate Riemann sum and compute an
approximate value for the energy.
b) It is known that the signal varies by an amount of at most ±e(t), as shown above, in
each 110 second period. Calculate upper and lower bounds for the energy.
c©2020 School of Mathematics and Statistics, UNSW Sydney
PROBLEMS FOR CHAPTER 8 217
Problems 8.3 : The Riemann integral and signed area and
8.4 : Basic properties of the Riemann integral
3. [R] Find the area of the region bounded by the line y = x and the parabola y = x2 − 2.
4. [R] Find
a)
∫ 9
4
x3 − x
x3/2
dx b)
∫ 2
−4
|x| dx.
5. [H] Find a function f which satisfies the integral equation∫ x
0
tf(t) dt =
∫ 0
x
(t2 + 1)f(t) dt+ x.
6. [R] Explain why
∫ 1
−1
1
x2
dx =
[
−1
x
]1
−1
= −1− 1 = −2 is not valid.
7. [H]
a) Suppose that f is a continuous increasing (and hence invertible) function on [a, b]. If
c = f(a), d = f(b) and a, b, c, d ≥ 0, then explain why∫ d
c
f−1(t) dt = bd− ac−
∫ b
a
f(x) dx.
b) Use this to find
∫ 1
1/2
sin−1 x dx.
8. [H] Suppose that U ′(x) = u(x).
a) Find V ′(x) if V (x) = (a− x)U(x) +
∫ x
0
U(t) dt where a is a constant.
b) Hence show that
∫ a
0
U(x) dx = aU(0) +
∫ a
0
(a− x)u(x) dx.
Problems 8.5 : The first fundamental theorem of calculus
9. [H] Suppose that f(t) = ⌊t⌋ and F (x) =
∫ x
0
f(t) dt, where ⌊t⌋ is the greatest integer less
than or equal to t. Use a graph of f to sketch F on the interval [−1, 3]. Is F continuous?
Where is F differentiable?
10. [H] Suppose that f(t) = sin(t2). Sketch the graph of f on the interval [0, 3]. Use this to
sketch the graph of F on the interval [0, 3], where F (x) =
∫ x
0
f(t) dt. Indicate where F has
local maxima and minima.
c©2020 School of Mathematics and Statistics, UNSW Sydney
218 CHAPTER 8. INTEGRATION
11. [R] [V] Find F ′(x) for each function F : R→ R given below.
a) F (x) =
∫ x
0
sin(t2) dt b) F (x) =
∫ x3
0
sin(t2) dt
c) F (x) =
∫ 1
x3
sin(t2) dt d) F (x) =
∫ x3
x
sin(t2) dt
12. [R] Find
d
dx
∫ 4
x
(5− 4t)5 dt.
Problems 8.6 : The second fundamental theorem of calculus
13. [R]
a) Suppose that f(x) =
1
x
. By considering the lower Riemann sum for f with respect to
the partition {
n
n
,
n+ 1
n
,
n+ 2
n
, . . . ,
2n
n
}
of [1, 2], show that
ln 2 = lim
n→∞
(
1
n+ 1
+
1
n+ 2
+ · · ·+ 1
2n
)
.
b) Suppose that f(x) =
1√
1− x2 .
i) Show that f is increasing on the interval [0, 12 ].
ii) Find the upper Riemann sum for f with respect to the partition{
0
2n
,
1
2n
,
2
2n
,
3
2n
, . . . ,
n
2n
}
of [0, 12 ].
iii) Hence evaluate
lim
n→∞
(
1√
4n2 − 12 +
1√
4n2 − 22 +
1√
4n2 − 32 + · · · +
1√
4n2 − n2
)
.
Problems 8.7 : Indefinite integrals
14. [R] Evaluate the following integrals by inspection.
a)
∫
x ex
2
dx b)
∫
sin
√
x√
x
dx
c)
∫ 1
0
2x(1 + x2)3 dx d)
∫ a
−a
x2
√
a3 − x3 dx (a > 0)
e)
∫ π/2
0
cos3 x sin x dx f) [H]
∫ 0
−1
√
t2 + t4 dt
c©2020 School of Mathematics and Statistics, UNSW Sydney
PROBLEMS FOR CHAPTER 8 219
Problems 8.8 : Integration by substitution
15. [R] Use a substitution to evaluate the following integrals.
a)
∫
dx
1 +
√
x
b)
∫
x(5x− 1)19 dx
c) [V]
∫
1− x
(1 + x)3
dx d) [V]
∫ 4
0
dx
5 +
√
x
Problems 8.9 : Integration by parts
16. [R] Use integration by parts to evaluate the following integrals.
a)
∫ 1
0
x e5x dx b)
∫
x2 cos x dx c)
∫
ln x dx
d)
∫ 0.5
0
sin−1 x dx e)
∫ e
1
x7 ln x dx f)
∫ π
0
x2 cos 2x dx
g) [V]
∫
ex cos x dx h)
∫
tan−1 x dx i) [H]
∫ π/4
0
sec3 θ dθ
Problems 8.10 : Improper integrals
17. [R] [V] Evaluate the following improper integrals or show that they diverge.
a)
∫ ∞
0
e−5xdx b)
∫ 1
−∞
e−0.01xdx c)
∫ ∞
0
dx
4 + x2
d)
∫ ∞
−∞
x3 e−x
4
dx e)
∫ ∞
2
dx
(x− 1)3/2 f)
∫ ∞
e
dx
x ln x
18. [H] Prove that
∫ ∞
0
xne−xdx = n! whenever n = 0, 1, 2, . . . .
19. [H]
a) Find lim
R→∞
∫ R
−R
x
1 + x2
dx.
b) Find lim
R→∞
∫ 2R
−R
x
1 + x2
dx.
c) Does
∫ ∞
−∞
x
1 + x2
dx converge? Explain.
Problems 8.11 : Comparison tests for improper integrals
20. [R] Use the inequality form of the comparison test to determine whether or not the following
improper integrals converge.
c©2020 School of Mathematics and Statistics, UNSW Sydney
220 CHAPTER 8. INTEGRATION
a) [V]
∫ ∞
1
1√
1 + x4
dx b)
∫ ∞
2
1
3
√
x2 − x dx c)
∫ ∞
2
1
lnx
dx
21. [R] Use the limit form of the comparison test to determine whether or not the following
improper integrals converge.
a)
∫ ∞
2
x
2x3 − 1 dx b)
∫ ∞
1
2x− 1
x2 + 2
dx c)
∫ ∞
2
1√
x6 − 1 dx
22. [R] Use a comparison test to determine whether or not the following improper integrals
converge.
a)
∫ ∞
1
3x+ sinx+ 2
2x3 − x+ 8 dx b)
∫ ∞
4
4x3 − x+ 5
x4 − x2 + 1 dx c) [H]
∫ ∞
2
ln t
t3/2
dt
23. [H] Find all real numbers s such that the improper integral∫ ∞
1
xs
1 + x
dx
is convergent.
24. [H] Find all real numbers p such that
∫ ∞
2
1
x(lnx)p
dx converges.
25. [H] For which pairs of numbers (a, b) does the improper integral
∫ ∞
1
xb
(1 + x2)a
dx converge?
Problems 8.12 : Functions defined by an integral
26. [R] [V] Given a positive real number x, let π(x) denote the number of primes less than or
equal to x. The function Li with domain (1,∞) is given by
Li(x) =
∫ x
2
1
ln t
dt
and is known as the ‘logarithmic integral function’. It has the property that
Li(x)
π(x)
≈ 1
when x is sufficiently large.
a) Evaluate π(10), π(20) and π(3.14159).
b) Suppose that x > 0. What does
π(x)
x
represent?
c) Find ddx Li(x) and Li(2).
d) By applying the mean value theorem to Li on the interval [2, 106], find a lower bound
for Li(106).
c©2020 School of Mathematics and Statistics, UNSW Sydney
PROBLEMS FOR CHAPTER 8 221
e) If x is large then
π(x)
x
≈ π(x)
x
Li(x)
π(x)
=
Li(x)
x
.
Using this approximation and your answer to part (d), find an approximate lower bound
for
π(106)
106
.
Note: There are 78, 498 primes less than one million so the actual value of
π(106)
106
is
0.078498.
27. [R] The function erf : R→ R is defined by the formula
erf(x) =
2√
π
∫ x
0
e−t
2
dt.
The function erf is an error function and can be used to calculate the probability that a
measurement has an error in a given range of values.
a) Calculate erf ′(x).
b) Explain why erf is an increasing function on R.
c) [H] Show that erf is an odd function.
d) i) By calculating Riemann sums with respect to the partition {0, 14 , 12 , 34 , 1}, find upper
and lower bounds for erf(1).
ii) Explain why e−t2 < e−t whenever t > 1.
iii) Hence show that
∫ ∞
1
e−t
2
dt converges and find an upper bound for this improper
integral.
iv) Using your answers to (i) and (iii), find an upper bound for lim
x→∞ erf(x). (In fact,
lim
x→∞ erf(x) = 1 but this is not so easy to prove.)
e) Sketch the graph of erf.
f) Explain why erf has an inverse function erf−1 and sketch its graph.
c©2020 School of Mathematics and Statistics, UNSW Sydney
222 CHAPTER 8. INTEGRATION
c©2020 School of Mathematics and Statistics, UNSW Sydney
223
Chapter 9
The logarithmic and exponential
functions
The volume of a colony of bacteria on an agar plate is 1 cubic millimetre and doubles every day.
Let V (t) denote the volume of bacteria after t days. It is clear that
V (0) = 1, V (1) = 2, V (2) = 4, V (3) = 8, V (4) = 16,
and so on. In general
V (t) = 2t
if t is a nonnegative integer. The corresponding points are plotted in Figure 9.1 (a).
If one wanted to determine the volume of the colony part-way through a day, it would be natural
to use the formula
V (t) = 2t,
where t is a nonnegative real number. This formula makes sense when t is a rational number. If
t = p/q where p and q are integers and q > 0, then 2t is defined to be the unique positive qth root
of 2p. However, if t is an irrational number, this definition does not apply. So what do we mean
by 2
√
3 or 2π? We have not yet seen a definition for such numbers.
That this shortcoming in our current definition of 2t is a serious problem can be illustrated
graphically. The graph corresponding to V (t) = 2t, where t is confined to the nonnegative rational
numbers, has an infinite number of ‘gaps’ (see Figure 9.1 (b)). Clearly this is not a satisfactory
state of affairs. One of the aims of this chapter is to rectify this deficiency and show that, when
t is an irrational number, 2t can be defined in such a way that the graph of V : [0,∞) → R is
continuous (see in Figure 9.1 (c)).
Our plan of attack is the following.
1. In Section 9.2, we define the function ln as an integral (by the fundamental theorem of
calculus).
2. In Section 9.3, we define the function exp as the inverse of ln (by the inverse function theorem).
3. In Section 9.4, we define the number bt, where b > 0 and t is an irrational number, by a
formula involving both ln and exp.
c©2020 School of Mathematics and Statistics, UNSW Sydney
224 CHAPTER 9. THE LOGARITHMIC AND EXPONENTIAL FUNCTIONS
t
V (t)
b
b
b
b
b
|
|
4
16
(a)
t
V (t)
|
|
4
16
(b)
t
V (t)
|
|
4
16
(c)
Figure 9.1: Growth in volume of bacteria on an agar plate.
We shall see that when the definition of bt (for rational t) is combined with the definition of bt (for
irrational t), the resulting function f : R→ R, given by
f(t) = bt ∀t ∈ R,
is continuous.
9.1 Powers and logarithms
The content of this section is revision, and is placed here primarily as reference. It summarises a
few known definitions and facts about powers and logarithms.
Definition 9.1.1. Suppose that b is a positive real number. If p and q are integers
and q > 0 then bp/q is defined to be the unique positive qth root of bp.
The following is an elementary definition of the logarithm to the base b.
Definition 9.1.2. Suppose that b is a positive real number not equal to one, c is a
rational number and a = bc. Then logb a is defined by the formula
c = logb a.
Note that, according to Definition 9.1.2, one cannot write down logb a for any positive number
a (even if a is rational); logb a is only defined if a = b
c for some rational number c. (This short-
coming in the domain of the function logb will be rectified later in the chapter when Definition
9.1.2 is superseded by a superior definition.)
c©2020 School of Mathematics and Statistics, UNSW Sydney
9.2. THE NATURAL LOGARITHM FUNCTION 225
The elementary properties of logarithms are summarised below: if x > 0, y > 0 and b > 0 with
b 6= 1 then
logb 1 = 0, logb b = 1,
logb(xy) = logb x+ logb y, logb
(
x
y
)
= logb x− logb y,
logb(b
r) = r, logb(x
r) = r logb x. (9.1)
Remark 9.1.3. In the properties listed above, it is implicitly assumed that r is a rational number,
and that x and y are of the form bc for some rational number c. One of the aims of this chapter is
to remove these awkward assumptions.
9.2 The natural logarithm function
(Ref: SH10 §7.2)
The ln function will already be familiar to you from high school. However, it was probably never
defined rigorously. In order to address the problems identified in the introduction to Chapter 9, it
will be necessary to start from scratch. Fortunately, by the fundamental theorem of calculus, this
is not difficult.
Definition 9.2.1. The function ln : (0,∞)→ R is defined by the formula
ln(x) =
∫ x
1
1
t
dt.
We read ln as either ‘ell en’ or as ‘log’.
Since the function ln is defined by an integral, lnx is simply the area of the shaded region shown
below in the case when x > 1.
t
1/t
1 x
One can also immediately see, by the first fundamental theorem of calculus, that
d
dx
(
lnx
)
=
1
x
.
This and other properties of ln are stated in the proposition below.
c©2020 School of Mathematics and Statistics, UNSW Sydney
226 CHAPTER 9. THE LOGARITHMIC AND EXPONENTIAL FUNCTIONS
Proposition 9.2.2. The function ln has the following properties:
(i) ln is differentiable on (0,∞) and d
dx
(
lnx
)
=
1
x
;
(ii) lnx > 0 if x > 1, ln 1 = 0 and lnx < 0 if 0 < x < 1;
(iii) lnx→ −∞ as x→ 0+ and lnx→∞ as x→∞;
(iv) ln(xy) = ln(x) + ln(y) for all positive real numbers x and y;
(v) ln
(
x
y
)
= ln(x)− ln(y) for all positive real numbers x and y; and
(vi) ln(xr) = r ln(x) whenever r is a rational number and x is a positive real number.
Proof. (i) Apply the first fundamental theorem of calculus (Theorem 8.5.1) to the definition of ln.
(ii) This follows from Definition 8.4.2 and the fact that 1t > 0 when t > 0.
(iii) We need to show that the improper integral
∫∞
1
1
t dt diverges to infinity. The diagram
below shows that ∫ 2
1
dt
t
≥ 1× 1
2
,
∫ 4
2
dt
t
≥ 2× 1
4
,
∫ 8
4
dt
t
≥ 4× 1
8
and so on.
t
y
1 2 4 8
y =
1
t
Hence ∫ 2n
1
dt
t
≥ 1
2
+
1
2
+ · · ·+ 1
2︸ ︷︷ ︸
n terms
=
n
2
→∞
as n→∞. Hence the integral diverges and therefore lnx→∞ as x→∞.
This argument can be adapted to show that∫ 1
2−n
dt
t
→ −∞
as n→∞. Hence lnx→ −∞ as x→ 0+.
(iv) Suppose that y is some fixed positive real number and that x ∈ (0,∞). Then, by the chain
rule and part (i),
d
dx
(
ln(xy)
)
= y ln′(xy) = y × 1
xy
=
1
x
=
d
dx
(
ln(x)
)
.
c©2020 School of Mathematics and Statistics, UNSW Sydney
9.2. THE NATURAL LOGARITHM FUNCTION 227
Hence
ln(xy) = ln(x) + C
for some constant C. When x = 1 we obtain
ln(y) = ln(1) + C = 0 + C = C
by part (ii). Hence
ln(xy) = ln(x) + C = ln(x) + ln(y)
as required.
(v) The proof uses the same technique as the proof of (vi) and is left to the reader.
(vi) This proof also uses the same technique as the proof of (iv).
The fact that ln(1) = 0 and that ln satisfies properties (iv), (v) and (vi) suggests that ln coincides
with one of the logarithm functions logb for some base b (with the advantage that Dom(ln) =
(0,∞)). We now aim to identify the base.
The function ln is increasing and continuous with Range(ln) = R (see Proposition 9.2.2 (i) and
(ii)). Hence, by the intermediate value theorem, there is exactly one real number x that satisfies
the equation
ln(x) = 1.
We call this solution e (after Leonard Euler, who was probably the greatest mathematician of the
eighteenth century).
Definition 9.2.3. The real number e is defined to be the unique number x that
satisfies the equation ∫ x
1
1
t
dt = 1.
Thus ln(e) = 1. One can show that e is irrational and that e ≈ 2.71828. From Proposition
9.2.2, we see now that
ln(er) = r
for every rational number r. This equation shows that ln coincides with the logarithm to the base
e. Since the number e arises naturally in many contexts, ln is usually called the natural logarithm.
By using the properties of ln given by Propoistion 9.2.2, one can draw the graph of ln.
c©2020 School of Mathematics and Statistics, UNSW Sydney
228 CHAPTER 9. THE LOGARITHMIC AND EXPONENTIAL FUNCTIONS
x
lnx
1 e
1
We note that, although ln(x) → ∞ as x → ∞, we have ddx ln(x) = 1x → 0 as x → ∞. This means
the graph of ln becomes flatter towards infinity and hence approaches infinity very slowly.
9.3 The exponential function
(Ref: SH10 §7.4)
In Section 9.2 it was shown that ddx(lnx) > 0 for all x > 0. Therefore, by the inverse function
theorem (Theorem 6.4.1), ln has an inverse function.
Definition 9.3.1. The function exp : R → (0,∞) is defined to be the inverse
function of ln : (0,∞)→ R.
By reflecting the graph of ln in the line y = x, one obtains the graph of exp.
x
y
1
(e, 1)
b
(1, e)
b
1
y = lnx
y = expx
Proposition 9.3.2. The function exp : R→ (0,∞) has the following properties:
c©2020 School of Mathematics and Statistics, UNSW Sydney
9.3. THE EXPONENTIAL FUNCTION 229
(i) exp(ln x) = x for all x in (0,∞) and ln(exp x) = x for all x in R;
(ii) exp(1) = e and exp(0) = 1;
(iii) expx→∞ as x→∞ and expx→ 0 as x→ −∞.
(iv) exp is differentiable on R and ddx(expx) = expx for all x in R;
(v) exp(x+ y) = exp(x) exp(y) for all x and y in R; and
(vi) exp(rx) = (expx)r for every real number x and every rational number r.
Proof. The proof of (i) follows immediately from Definition 9.3.1 and the proof of (ii) and (iii) then
follow from (i) and the continuity of ln and exp.
That exp is differentiable on R follows from the inverse function theorem. To complete the
proof of (iii), we differentiate both sides of the equation
ln(expx) = x
with respect to x to obtain
1
expx
× d
dx
(exp x) = 1
by the chain rule. Rearranging gives ddx(expx) = expx as required.
If x and y are real numbers then
exp(x+ y) = exp
(
ln(exp x) + ln(exp y)
)
(by part (i))
= exp
(
ln(exp x× exp y)
)
(by Proposition 9.2.2 (iv))
= expx× exp y (by part (i)).
This completes the proof of (iv)
To prove (v), suppose that r is a rational number and x is a real number. Then
exp(rx) = exp
(
r ln(exp x)
)
(by part (i))
= exp
(
ln
(
(exp x)r
))
(by Proposition 9.2.2 (v))
(exp x)r (by part (i)).
This completes the proof of the proposition.
Note that by Proposition 9.3.2 (ii) and (v) we have
exp(r) = (exp 1)r = er (9.2)
for every rational number r. This inspires the following definition.
Definition 9.3.3. Suppose that x is an irrational number. Then we define the real
number ex by the formula
ex = expx.
c©2020 School of Mathematics and Statistics, UNSW Sydney
230 CHAPTER 9. THE LOGARITHMIC AND EXPONENTIAL FUNCTIONS
Thus Definition 9.3.3 extends Definition 9.1.1, in the case when b = e, to include irrational
powers. Moreover, since expx = ex and exp is a continuous function, the function f : R→ R, given
by f(x) = ex, is also continuous. We have thus made significant progress towards the problem
outlined at the beginning of the chapter. Finally, since expx = ex, we call exp the exponential
function.
9.4 Exponentials and logarithms with other bases
(Ref: SH10 §7.5)
At the end of the last section, we gave a definition of ex, where x is an irrational number. Moreover,
if f : R → R is given by f(x) = ex, then f is a continuous and differentiable function. In this
section, we do the same for bx, where b is any positive real number.
To begin, consider a rational number r and any positive real number b. Then
br = exp
(
ln(br)
)
= exp
(
r ln b
)
= er ln b (9.3)
by the definitions and properties established in the last two sections. This inspires the following
definition.
Definition 9.4.1. Suppose that b is a positive real number and x is an irrational
number. We define the number bx by the formula
bx = ex ln b.
By combining equation (9.3) and Definition 9.4.1 we see that
bx = ex ln b ∀x ∈ R. (9.4)
Since exp and ln are continuous functions, the function f : R→ R, given by
f(x) = bx ∀x ∈ R,
is also continuous (see Proposition 3.1.2). This solves the problem articulated at the beginning of
the chapter. Thus
2
√
3 = e
√
3 ln 2 ≃ 1.492106 . . . .
Moreover, we can discard our old definition of logb (see Definition 9.1.2), together with its
technical difficulties (see Remark 9.1.3), and replace it with a simpler and more powerful definition.
Definition 9.4.2. Suppose that b is a positive real number, b 6= 1. Then the function
logb : (0,∞)→ R is defined to be the inverse of the function f : R→ (0,∞), where
f(x) = bx
for all x in R.
c©2020 School of Mathematics and Statistics, UNSW Sydney
9.5. INTEGRATION AND THE LN FUNCTION 231
We leave it as an exercise to show that the function f of Definition 9.4.2 is one-to-one and
hence invertible. The graphs of various exponential functions and their corresponding logarithmic
functions are shown below.
x
y
1
1
y = log8 x
y = log3 x
y = log2 x
y = 2x
y = 3xy = 8x
The following proposition gives a formula for logb in terms of ln. Its proof is left as an exercise.
Proposition 9.4.3. Suppose that b is a positive real number, b 6= 1. Then
logb x =
lnx
ln b
∀x ∈ (0,∞). (9.5)
Using equations (9.4) and (9.5), one can easily establish familiar properties for the exponentials
and logarithms to the base b, with the advantage that we no longer need the technical restrictions
of Remark 9.1.3.
9.5 Integration and the ln function
We have already seen that
d
dx
(ln x) =
1
x
provided that x > 0. It follows from the chain rule that
d
dx
(
ln(f(x))
)
=
1
f(x)
× f ′(x) = f
′(x)
f(x)
,
provided that f is differentiable and f(x) > 0. Hence
∫
f ′(x)
f(x)
dx = ln(f(x)) +C.
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232 CHAPTER 9. THE LOGARITHMIC AND EXPONENTIAL FUNCTIONS
It is not difficult to remove the restriction that f(x) > 0. To take a simple example, suppose
that x 6= 0. Then
d
dx
(
ln |x|) =
{
d
dx
(
lnx
)
if x > 0
d
dx
(
ln(−x)) if x < 0
=
{
1
x if x > 0
1
−x × (−1) if x < 0
=
1
x
.
Thus ∫
1
x
dx = ln |x|+ C
provided x is restricted to an interval not containing zero. By generalising this calculation we
obtain the formula ∫
f ′(x)
f(x)
dx = ln |f(x)|+ C,
provided that f is never zero over the interval of integration.
Example 9.5.1. Find
(a)
∫
tan x dx
(b)
∫
sec x dx.
Solution. (a) By using the fact that tan x = sinxcos x we have∫
tanx dx = −
∫ − sinx
cos x
dx
= − ln | cos x|+ C.
(The indefinite integral of cot can be found similarly.)
(b) This example involves a special trick:
∫
sec x dx =
∫
secx
tan x+ sec x
tan x+ sec x
dx
=
∫
tanx sec x+ sec2 x
sec x+ tanx
dx
= ln | sec x+ tanx|+ C.
(A similar trick can be used to find the indefinite integral of cosec.)
c©2020 School of Mathematics and Statistics, UNSW Sydney
9.6. LOGARITHMIC DIFFERENTIATION 233
9.6 Logarithmic differentiation
(Ref: SH10 §7.3)
Logarithms are powerful because they transform powers into products, products into sums and
quotients into differences. In this section, we illustrate the use of the logarithm for differentiating
functions that are defined by (combinations of) powers, products or quotients.
Example 9.6.1. Find dydx in each case:
(a) y = 10x
(b) y =
(
(3x2 + 4)(x+ 2)
x3 + 5x
)3/5
(c) y = (sinx)cos x.
Solution. (a) If we take ln of both sides of the equation y = 10x then
ln y = ln 10x
= x ln 10.
By differentiating with respect to x, we obtain
1
y
dy
dx
= ln 10
and hence
dy
dx
= (ln 10) × y
= (ln 10)10x.
Alternatively: ddx(10
x) = ddx(e
x ln 10) = ln 10ex ln 10 = (ln 10)10x.
Remark 9.6.2. Recall that one can only take the logarithm of a positive number. Hence our
solution to Example 9.6.1 (b) and (c) is only valid when y > 0.
9.7 Indeterminate forms with powers
(Ref: SH10 §11.6)
Consider the limits
lim
x→0+
xx and lim
x→∞x
1/x. (9.6)
The first limit is of the form 00 while the second is of the form ∞0. Both of these forms are
examples of indeterminate forms and hence we cannot say what the corresponding limits are in
each case without further calculation. Since each limit involves a power, it is natural to first take
the logarithm of the limit and then bring l’Hoˆpital’s rule into play.
Example 9.7.1. Evaluate the limit
lim
x→0+
x2x.
c©2020 School of Mathematics and Statistics, UNSW Sydney
234 CHAPTER 9. THE LOGARITHMIC AND EXPONENTIAL FUNCTIONS
Solution. The limit is an indeterminate form of the type 00. By taking the natural logarithm, we
can transform the limit into an indeterminate form of the type ∞∞ :
lim
x→0+
x2x = lim
x→0+
exp
(
lnx2x
)
(since ln and exp are inverses)
= lim
x→0+
exp
(
2x ln x
)
= exp
(
lim
x→0+
2x lnx
)
(since exp is continuous)
= exp
(
lim
x→0+
lnx
1/(2x)
)
.
We can now apply l’Hoˆpital’s rule to the problem. By differentiating the numerator and denomi-
nator and then simplifying we obtain
lim
x→0+
x2x = exp
(
lim
x→0+
1/x
−1/(2x2)
)
= exp
(
lim
x→0+
−2x
)
= exp(0)
= 1.
Each of the limits in (9.6) can be evaluated by using the ideas in this example.
Remark 9.7.2. The limit of Example 9.7.1 is of the form 00 and has the value 1. The limit
lim
x→0+
(
e−
1
x
)x
is also of the form 00. However, its value is not 1:
lim
x→0+
(
e−
1
x
)x
= lim
x→0+
e−1 =
1
e
.
The fact that limits of the type 00 can have different values explains why 00 is an indeterminate
form.
c©2020 School of Mathematics and Statistics, UNSW Sydney
PROBLEMS FOR CHAPTER 9 235
Problems for Chapter 9
Problems 9.1 : Powers and logarithms and
9.2 : The natural logarithm function
1. [R]
a) Write down the definition of lnx, where x > 0.
b) Explain why
d
dx
lnx =
1
x
whenever x > 0.
c) Suppose that r is a rational number and that x and y are positive real numbers.
i) By first differentiating ln(xy) with respect to x, show that ln(xy) = lnx+ ln y.
ii) Use the same technique to show that
ln
(
x
y
)
= lnx− ln y and ln(xr) = r lnx.
2. [R]
a) Prove, using upper and lower Riemann sums and the definition of lnx, that ln 2 < 1 <
ln 4, and hence that 2 < e < 4.
b) [H] Use Maple and the method of part (a) to prove that 52 < e < 3. How many partition
points do you need?
3. [R] Find the derivatives of
a) f(x) = ln
√
x3 + 1 b) g(x) = e|x|
c) h(x) = ln(ln(ln x)) d) q(x) = eln(x
5+6)
Problems 9.3 : The exponential function and
9.5 : Integration and the ln function
4. [R] Find
a)
∫
e2x
1 + e2x
dx b)
∫
e1/x
x2
dx c)
∫
3xdx
d)
∫
e
√
x
8
√
x
dx e)
∫
ln x
x
dx f)
∫
cot x dx.
(Hint for part (f): express cot in terms of sin and cos.)
5. [R] Sketch the curves
a) y = ln(1 + ex) b) y =
(ex + x)
(ex − x) .
c©2020 School of Mathematics and Statistics, UNSW Sydney
236 CHAPTER 9. THE LOGARITHMIC AND EXPONENTIAL FUNCTIONS
6. [R]
a) Sketch the curve y =
lnx
x
, noting any turning points and asymptotes.
b) By using (a) or otherwise, prove that πe < eπ.
7. [R] [V]
x
y
y = 1x
1
t
1+t
1 1 + 1t
a) From the graph, explain why
1
1 + t
≤ ln
(
1 +
1
t
)
≤ 1
t
whenever t ≥ 0.
b) Deduce that lim
t→∞ ln
(
1 +
1
t
)t
= 1 and hence find the value of lim
t→∞
(
1 +
1
t
)t
.
Problems 9.6 : Logarithmic differentiation
8. [R] Use logarithmic differentiation to find
dy
dx
if
a) y = 3x b) y =
(
x3 − 3
1 + x2
)1/5
c) y = (sinx)sinx d) y = sin(xsinx).
Problems 9.7 : Indeterminate forms with powers
9. [R] Calculate the following limits:
a) [V] lim
x→∞
lnx
xa
, a > 0 b) lim
x→0+
xa lnx, a > 0
c) lim
x→0+
xx d) lim
x→0+
x2/ ln x
e) lim
x→∞ x
1/x f) lim
x→∞ a
1/x, a > 0
g) lim
x→∞
(
1 +
a
x
)x
h) lim
x→∞ x
100 e−x
i) lim
x→∞ p(x) e
−x ,where p is any polynomial.
c©2020 School of Mathematics and Statistics, UNSW Sydney
PROBLEMS FOR CHAPTER 9 237
10. [H] Prove that the functions f : (−1,∞)→ R and g : (−1,∞)→ R, given by
f(x) = ln(1 + x)−
(
x− x
2
2
)
and g(x) =
(
x− x
2
2
+
x3
3
)
− ln(1 + x),
are increasing on (0,∞). Deduce that
x− x
2
2
< ln(1 + x) < x− x
2
2
+
x3
3
whenever x > 0.
c©2020 School of Mathematics and Statistics, UNSW Sydney
238 CHAPTER 9. THE LOGARITHMIC AND EXPONENTIAL FUNCTIONS
c©2020 School of Mathematics and Statistics, UNSW Sydney
239
Chapter 10
The hyperbolic functions
A telecommunications company wishes to suspend a cable from a sequence of telegraph poles in
a cost effective manner. There are at least two competing considerations. First, it is desirable to
minimise the length of cable used. Second, it is desirable to minimise the amount of tension in
the cable (since an increase in tension requires an increase in cable strength). The configuration
in Figure 10.1 (a) requires a short length of strong cable while that in Figure 10.1 (b) reduces the
tension in the cable at the cost of increasing its length. Is there an optimal configuration for this
problem?
In order to solve this problem, one needs an equation which describes the curve of a suspended
cable. Such a curve is called a catenary, after the Latin word for ‘chain.’ Galileo claimed that
every catenary was a parabola, but this was later disproved in 1669. Two decades later, the Swiss
mathematician Jakob Bernoulli issued a public challenge to find the general equation that describes
a catenary. Three mathematicians Gottfried Leibniz, Christiaan Huygens and Johann Bernoulli
(Jakob’s brother) independently derived the correct equation:
y =
a
2
(
ex/a + e−x/a
)
(10.1)
(see Figure 10.1 (c)). In this chapter we study the function described by equation (10.1) and
other functions related to it. Such functions are important to mathematics, engineering, physics
(especially the theory of relativity) and architecture.
We note from the outset that a large number of new formulae will be introduced in this chapter.
However, students do not need to memorise all of them. Those which should be memorised are
listed in Section 10.7.
(a) (b)
x
y
a
(c)
Figure 10.1: Catenaries.
c©2020 School of Mathematics and Statistics, UNSW Sydney
240 CHAPTER 10. THE HYPERBOLIC FUNCTIONS
10.1 Hyperbolic sine and cosine functions
(Ref: SH10 §7.8)
We mentioned in the introduction that the equation describing a freely suspended chain is given
by y = f(x), where
f(x) =
a
2
(
ex/a + e−x/a
)
.
To begin we study the simplest case: when a = 1.
Definition 10.1.1. The hyperbolic cosine function cosh : R → R is defined by the
formula
coshx = 12(e
x + e−x) ∀x ∈ R.
The reasoning behind the name ‘hyperbolic cosine’ will gradually become apparent. Since cosh
is a linear combination of exponential functions, it is differentiable and
d
dx
(
coshx
)
=
1
2
d
dx
(
ex + e−x
)
=
1
2
(ex − e−x).
The derivative of cosh is important in its own right and is given a special name.
Definition 10.1.2. The hyperbolic sine function sinh : R → R is defined by the
formula
sinhx = 12(e
x − e−x) ∀x ∈ R.
Some people pronounce sinh as ‘shine.’ One can easily show that the derivative of sinh is cosh.
Thus we have
d
dx
(sinhx) = coshx and
d
dx
(cosh x) = sinhx. (10.2)
We now consider the graphs of these functions.
Example 10.1.3. Sketch the graph of sinh.
Solution. We begin by identifying a few key features of the sinh function.
• Since
sinh(−x) = 12(e−x − ex) = −12(ex − e−x) = − sinhx,
we see that sinh is an odd function. As a corollary, sinh(0) = 0.
• Since
d
dx
(sinhx) = cosh x = 12(e
x + e−x) > 0,
sinh is an increasing function.
c©2020 School of Mathematics and Statistics, UNSW Sydney
10.1. HYPERBOLIC SINE AND COSINE FUNCTIONS 241
x
y
y = sinhx
y = 12e
x
(a)
x
y y = cosh x
y = 12e
x
1
(b)
Figure 10.2: Graphs of sinh and cosh.
• Since lim
x→∞ e
−x = 0, sinhx gets arbitrarily close to 12e
x as x→∞.
From these observations we obtain a sketch of the graph of sinh (see Figure 10.2).
Example 10.1.4. Sketch the graph of cosh.
Solution. Some properties of the cosh function are listed below:
• cosh is an even function;
• cosh(0) = 1;
• cosh is decreasing on (−∞, 0), stationary at 0 and increasing on (0,∞);
• coshx ≥ 1 for all x in R;
• coshx gets arbitrarily close to 12ex as x→∞.
The reader should verify each of these properties by appealing to Definition 10.1.1. From these
observations, we obtain a sketch of the graph of cosh (see Figure 10.2).
The functions cosh and sinh are related by the following important identity.
Proposition 10.1.5. If x is a real number then
cosh2 x− sinh2 x = 1.
Proof. If x ∈ R then
cosh2 x− sinh2 x =
(
1
2(e
x + e−x)
)2
−
(
1
2 (e
x − e−x)
)2
=
1
4
× 2ex × 2e−x difference of two squares
= 1,
completing the proof.
c©2020 School of Mathematics and Statistics, UNSW Sydney
242 CHAPTER 10. THE HYPERBOLIC FUNCTIONS
We are now in a position to appreciate why these functions are called ‘hyperbolic cosine’ and
‘hyperbolic sine.’ First, we examine their similarity to the trigonometric cosine and sine functions.
The sine and cosine functions have the following properties:
d
dx
(cos x) = − sinx d
dx
(sinx) = cos x
cos is an even function sin is an odd function
cos2 x+sin2 x = 1.
The functions sinh and cosh have analogous properties (with the occasional adjustment of a negative
sign):
d
dx
(cosh x) = sinhx
d
dx
(sinhx) = coshx
cosh is an even function sinh is an odd function
cosh2 x− sinh2 x = 1.
The next example explains the origin of the name hyperbolic in ‘hyperbolic cosine.’
Example 10.1.6. Sketch the curve defined by the parametric equation{
x(t) = cosh t
y(t) = sinh t,
(10.3)
where t ∈ R.
Solution. The parameter t in equation (10.3) can be eliminated by Proposition 10.1.5 to obtain
1 = cosh2 t− sinh2 t
=
(
x(t)
)2 − (y(t))2,
or simply
x2 − y2 = 1.
This equation describes an hyperbola. Since cosh t > 0 for every real number t, it follows that
x > 0 and hence the curve is the branch of the hyperbola that lies in the right-half plane (see the
black curve in Figure 10.3 (b)). Its asymptotes are the lines y = x and y = −x.
(The other branch of the hyperbola is parameterised by{
x(t) = − cosh t
y(t) = sinh t
where t ∈ R, and is shown in Figure 10.3 (b) in gray.)
Hence we use trigonometric sine and cosine to paramaterise a circle or ellipse, and hyperbolic
sine and cosine to parameterise an hyperbola.
c©2020 School of Mathematics and Statistics, UNSW Sydney
10.2. OTHER HYPERBOLIC FUNCTIONS 243
x
y
1
Figure 10.3: Sketch for Example 10.1.6.
10.2 Other hyperbolic functions
(Ref: SH10 §7.9)
Other hyperbolic functions are defined in analogy to the trigonometric functions. Thus we have
tanhx =
sinhx
cosh x
, coth x =
cosh x
sinhx
,
sech x =
1
cosh x
, cosech x =
1
sinhx
.
Some people pronounce tanh as ‘than.’
The derivative of each of these functions is easy to compute.
Example 10.2.1. Compute the derivative of tanh.
Solution. By the definition of tanh,
d
dx
(tanh x) =
d
dx
(
sinhx
coshx
)
=
cosh x ddx(sinhx)− sinhx ddx(cosh x)
cosh2 x
(by the quotient rule)
=
cosh2 x− sinh2 x
cosh2 x
(by (10.2))
=
1
cosh2 x
(by Proposition 10.1.5)
= sech2 x.
A complete list of derivatives will be given in Section 10.4.
c©2020 School of Mathematics and Statistics, UNSW Sydney
244 CHAPTER 10. THE HYPERBOLIC FUNCTIONS
x
tanhx
1
−1
Figure 10.4: The graph of tanh.
Example 10.2.2. Sketch the graph of the function tanh.
Solution. Since tanhx = sinhxcosh x and coshx > 0 for all x in R, we see that Dom(tanh) = R. Some
properties of tanh are listed below.
• By using the fact that cosh is even and sinh is odd, we see that
tanh(−x) = sinh(−x)
cosh(−x) =
− sinhx
coshx
= − tanh(x).
Hence tanh is an odd function. It follows that tanh(0) = 0.
• Since ddx(tanhx) = sech2 x > 0 for all x in R, the function tanh is increasing everywhere.
• By writting tanhx in terms of exponentials we find that
tanhx =
sinhx
cosh x
=
ex − e−x
ex + e−x
=
1− e−2x
1 + e−x
→ 1
as x→∞.
• Finally, the slope of the graph of tanh at the origin is
d
dx
(tanhx)
∣∣∣
x=0
= sech2 0 = 1.
These features are shown in the following sketch in Figure 10.4.
c©2020 School of Mathematics and Statistics, UNSW Sydney
10.3. HYPERBOLIC IDENTITIES 245
10.3 Hyperbolic identities
(Ref: SH10 §7.8)
It has already been observed that some identities involving cosh and sinh are almost analogous to
those for cos and sin. This analogy can also be seen in the following larger list of identities. There
are the ‘difference of squares’ identities
cosh2 x− sinh2 x = 1,
1− tanh2 x = sech2 x,
coth2 x− 1 = cosech2 x,
the ‘sum and difference’ formulae
sinh(x± y) = sinhx cosh y ± cosh x sinh y,
cosh(x± y) = coshx cosh y ± sinhx sinh y,
tanh(x± y) = tanhx± tanh y
1± tanhx tanh y ,
and the ‘double-angle’ formulae
sinh(2x) = 2 sinhx coshx,
cosh(2x) = cosh2 x+ sinh2 x,
tanh(2x) =
2 tanh x
1 + tanh2 x
.
The following remark gives an easy method for remembering these identities.
Remark 10.3.1. There is a convenient mnemonic for these hyperbolic identities, provided that
you know the corresponding trigonometric identities. Simply replace cos with cosh and sin with
i sinh. For example,
cos2 x+ sin2 x = 1 −→ cosh2 x+ i2 sinh2 x = 1
−→ cosh2 x− sinh2 x = 1,
and
sin(x+ y) = sinx cos y + cos x sin y
−→ i sinh(x+ y) = i sinhx cosh y + i cosh x sinh y
−→ sinh(x+ y) = sinhx cosh y + cosh x sinh y.
Since tan = sincos we replace tan with
i sinh
cosh = i tanh. Similarly sec is replaced with sech. Thus
1 + tan2 x = sec2 x −→ 1 + i2 tanh2 x = sech2 x
−→ 1− tanh2 x = sech2 x.
We emphasise that Remark 10.3.1 provides a way to remember each formula, not to prove it.
Proofs are constructed using techniques that are illustrated in the following example.
c©2020 School of Mathematics and Statistics, UNSW Sydney
246 CHAPTER 10. THE HYPERBOLIC FUNCTIONS
Example 10.3.2. (a) Prove the identity
cosh(x+ y) = cosh x cosh y + sinhx sinh y.
(b) Hence show that
cosh(x− y) = cosh x cosh y − sinhx sinh y.
(c) By using the result of (a), prove the double-angle formula for cosh.
(d) By assuming that
sinh(x+ y) = sinhx cosh y + cosh x sinh y, (10.4)
use the result of (a) to show that
tanh(x+ y) =
tanhx+ tanh y
1 + tanhx tanh y
.
Solution. (a) By the definition of cosh,
LHS = 12
(
ex+y + e−(x+y)
)
while
RHS = 12
(
ex + e−x
)
1
2
(
ey + e−y
)
+ 12
(
ex − e−x)12(ey − e−y)
= 14
(
ex+y + ex−y + e−x+y + e−x−y
)
+ 14
(
ex+y − ex−y − e−x+y + e−x−y)
= 12e
x+y + 12e
−x−y
= LHS.
This proves the identity.
Remark 10.3.3. Note that in Example 10.3.2, each proof started by simplifying either the left-
hand side of the equation, or the right-hand side of the equation. It would be wrong to begin the
proof of (a) by
cosh(x+ y) = coshx cosh y + sinhx sinh y
1
2
(
ex+y + e−(x+y)
)
= 12
(
ex + e−x
)
1
2
(
ey + e−y
)
+ 12
(
ex − e−x)12(ey − e−y)
· · · = · · ·
Such a method is flawed and can be used to ‘prove’ erroneous statements such as ‘1 = 0’.
10.4 Hyperbolic derivatives and integrals
(Ref: SH10 §§7.8, 7.9)
Below is a list of derivatives for the hyperbolic functions:
d
dx
(
sinhx
)
= cosh x
d
dx
(
cosh x
)
= sinhx
d
dx
(
tanhx
)
= sech2 x
d
dx
(
coth x
)
= − cosech2 x
d
dx
(
sech x
)
= − sech x tanhx d
dx
(
cosech x
)
= − cosech x coth x.
c©2020 School of Mathematics and Statistics, UNSW Sydney
10.4. HYPERBOLIC DERIVATIVES AND INTEGRALS 247
These derivatives can be established by using the definition of each function, as in Example 10.2.1.
The derivatives of sinh, cosh and tanh are the most important to remember.
Corresponding to these derivatives are the indefinite integrals∫
sinhx dx = coshx+ C,
∫
sech2 x dx = tanhx+ C
and so on.
Example 10.4.1. Calculate the derivative of the function f given by
f(x) = ln(sech(2x)).
Solution. First we have
d
dx
(
sech 2x
)
= −2 sech(2x) tanh(2x).
Hence the chain rules gives
f ′(x) =
−2 sech(2x) tanh(2x)
sech(2x)
= −2 tanh(2x).
Example 10.4.2. Find the following integrals:
(a)
∫ 1
5
ln 3
0
sinh 5x dx
(b)
∫
x coshx dx
(c)
∫
sech2(
√
x)√
x
dx
(d)
∫
ex sinhx dx.
Solution. (a) We have
∫ 1
5
ln 3
0
sinh 5x dx =
[
1
5 cosh 5x
] 1
5
ln 3
0
= 110
[
e5x + e−5x
] 1
5
ln 3
0
= 110
(
eln 3 + e− ln 3 − e0 − e0
)
= 110
(
3 + 13 − 1− 1
)
= 215 .
c©2020 School of Mathematics and Statistics, UNSW Sydney
248 CHAPTER 10. THE HYPERBOLIC FUNCTIONS
x
y
(a) sinh and sinh−1
x
y
1
1
(b) cosh and cosh−1
x
y
−1
1
(c) tanh and tanh−1
Figure 10.5: Graphs of the hyperbolic functions (in gray) and their inverses (in black).
10.5 The inverse hyperbolic functions
(Ref: SH10 §7.9)
The functions sinh : R → R and tanh : R → (−1, 1) are both increasing on R and hence have
well-defined inverses. However, cosh : R → [1,∞) is not one-to-one on R. To obtain an inverse,
one must restrict the domain of cosh. The following definition gives the conventional restriction.
Definition 10.5.1. The function cosh−1 : [1,∞) → [0,∞) is defined to be the
inverse function of the restricted hyperbolic cosine function
cosh : [0,∞)→ [1,∞).
Graphs of the functions sinh−1, cosh−1 and tanh−1 are illustrated in Figure 10.5.
Expressions involving the hyperbolic functions and their inverses can be simplified using the
identities of Section 10.3. It is important to note that
cosh−1(coshx) = x
only when x ≥ 0.
Example 10.5.2. Simplify
(a) sinh
(
cosh−1 43
)
,
(b) cosh−1(cosh(−7)), and
(c) cosh(2 sinh−1 3).
Solution. (a) The identity
cosh2 t− sinh2 t = 1
can be rearranged to give
sinh2 t = cosh2 t− 1.
By substituting cosh−1 43 for t, we obtain
sinh2
(
cosh−1 43
)
=
(
4
3
)2 − 1 = 79 .
c©2020 School of Mathematics and Statistics, UNSW Sydney
10.5. THE INVERSE HYPERBOLIC FUNCTIONS 249
We need to decide whether to take the positive or negative square root. Since cosh−1 is a nonneg-
ative function, we have cosh−1 43 > 0 and hence
sinh(cosh−1 43 ) > 0.
Therefore
sinh
(
cosh−1 43
)
= +
√
7
3 .
(b) We cannot write cosh−1(cosh(−7)) = −7 because cosh−1 is only the inverse function of cosh
with domain restricted to [0,∞). However, since cosh is an even function,
cosh−1(cosh(−7)) = cosh−1(cosh(7)) = 7.
(c) We use the double-angle formula
cosh(2t) = 1 + 2 sinh2 t
for cosh:
cosh(2 sinh−1 3) = 1 + 2
(
sinh(sinh−1 3)
)2
= 1 + 2× 32
= 19.
Each of the functions sinh, cosh and tanh can be expressed in terms of the exponential function.
It is not surprising, then, that their inverses can be expressed in terms of the natural logarithm.
Proposition 10.5.3. The following identities hold:
sinh−1 x = ln(x+
√
x2 + 1) ∀x ∈ R,
cosh−1 x = ln(x+
√
x2 − 1) ∀x ∈ [1,∞),
tanh−1 x =
1
2
ln
(
1 + x
1− x
)
∀x ∈ (−1, 1).
Proof. We prove the formula for sinh−1 only; the others can be proved similarly.
Suppose that y = sinh−1 x. Then
sinh y = x,
which means that
1
2(e
y − e−y) = x.
If we multiply this equation through by 2ey then we obtain
e2y − 1 = 2xey,
which is a quadratic equation in ey. Further rearrangement gives
(ey)2 − 2x(ey) = 1,
c©2020 School of Mathematics and Statistics, UNSW Sydney
250 CHAPTER 10. THE HYPERBOLIC FUNCTIONS
to which we complete the square to obtain
(ey − x)2 = 1 + x2.
Since ey > 0, we take the positive square root. Thus
ey = x+
√
x2 + 1.
(Note that this may also be obtained by using the quadratic formula rather than completing the
square). Taking the natural logarithm of both sides of this equation gives
y = ln(x+
√
x2 + 1)
as required.
By the inverse function theorem, the functions
sinh−1 : R→ R, cosh−1 : (1,∞)→ (0,∞) and tanh−1 : (−1, 1)→ R
are differentiable on their respective domains. A formula for each derivative is given in the next
proposition.
Proposition 10.5.4. The derivatives of sinh−1, cosh−1 and tanh−1 are given by
d
dx
(
sinh−1 x
)
=
1√
x2 + 1
,
d
dx
(
cosh−1 x
)
=
1√
x2 − 1 ,
d
dx
(
tanh−1 x
)
=
1
1− x2 .
Proof. We prove the formula for ddx(sinh
−1 x) only; the others can be proved similarly.
Suppose that y = sinh−1 x, Then
sinh y = x
and differentiating with respect to x gives
cosh y
dy
dx
= 1.
Hence
dy
dx
=
1
cosh y
=
1
cosh(sinh−1 x)
. (10.5)
The identity cosh2 t− sinh2 t = 1 gives
cosh t =
√
1 + sinh2 t
(where we have taken the positive square root since cosh t > 0 for all t). Hence
cosh(sinh−1 x) =
√
1 + x2.
If we substitute this into (10.5) then we obtain
d
dx
(
sinh−1 x
)
=
1√
1 + x2
as required.
Remark 10.5.5. Students should be familiar with the proofs of Propositions 10.5.3 and 10.5.4.
c©2020 School of Mathematics and Statistics, UNSW Sydney
10.6. INTEGRATION LEADING TO THE INVERSE HYPERBOLIC FUNCTIONS 251
10.6 Integration leading to the inverse hyperbolic functions
When propositions 10.5.3 and 10.5.4 for sinh−1 are restated in terms of indefinite integration we
obtain ∫
dx√
x2 + 1
= sinh−1 x+ C = ln(x+
√
x2 + 1) + C.
A more general version may be obtained by observing that, when a > 0,
d
dx
(
sinh−1
x
a
)
=
1
a
1√(
x
a
)2
+ 1
=
1√
x2 + a2
.
In this manner, one is able to prove the formulae∫
dx√
x2 + a2
= sinh−1
x
a
+ C
= ln(x+
√
x2 + a2) + (C − ln a), a > 0,
∫
dx√
x2 − a2 = cosh
−1 x
a
+ C
= ln(x+
√
x2 − a2) + (C − ln a), x ≥ a > 0,
∫
dx
a2 − x2 =
{
1
a tanh
−1 x
a + C, |x| < a
1
a coth
−1 x
a + C, |x| > a > 0
=
1
2a
ln
∣∣∣∣a+ xa− x
∣∣∣∣+ C, x2 6= a2.
These formulae are included in the table of standard integrals that is issued at the final examination.
Example 10.6.1. Find
(a)
∫
dx√
1 + 9x2
(b)
∫
dx√
x2 − 2x+ 10.
Solution. (a) A little algebraic manipulation shows that
1√
1 + 9x2
=
1
3
√
(1/3)2 + x2
.
Hence ∫
dx√
1 + 9x2
=
1
3
∫
dx√
(1/3)2 + x2
=
1
3
sinh−1
x
1/3
+ C
=
1
3
sinh−1(3x) + C.
c©2020 School of Mathematics and Statistics, UNSW Sydney
252 CHAPTER 10. THE HYPERBOLIC FUNCTIONS
Alternatively, one could approach this problem by using the substitution u = 3x.
(b) The technique we apply to this example is known as ‘completing the square’:
x2 − 2x+ 10 = x2 − 2x+ 1 + 9 = (x− 1)2 + 32 = u2 + 32,
where u = x− 1. Hence ∫
dx√
x2 − 2x+ 10 =
∫
du√
u2 + 32
= sinh−1
u
3
+ C
= sinh−1
x− 1
3
+ C.
10.7 A summary of important hyperbolic formulae
A large number of formulae was introduced in this chapter. Below is a list of formulae, with the
more important appearing earlier in the list, that students are expected to know.
• The definitions of sinh and cosh:
sinhx = 12(e
x − e−x) coshx = 12 (ex + e−x)
• The derivatives of sin and cosh:
d
dx
(sinhx) = cosh x
d
dx
(cosh x) = sinhx
• The definition of tanh:
tanhx =
sinhx
coshx
• The hyperbolic identity
cosh2 x− sinh2 x = 1
• The derivative of sinh−1:
d
dx
(sinh−1 x) =
1√
1 + x2
• The derivative of tanh:
d
dx
(tanh x) = sech2 x.
Other formulae need not be memorised, though students will do well to note the following
points.
• Derivatives for the inverse hyperbolic functions can be read from the table of standard inte-
grals.
• Identities which express the inverse hyperbolic functions in terms of the natural logarithm
(see Proposition 10.5.3) can be read from the table of standard integrals.
c©2020 School of Mathematics and Statistics, UNSW Sydney
10.8. APPENDIX: THE CATENARY 253
• The definition of coth, sech and cosech are in exact analogy with the definition of cot, sec
and cosec.
• The hyperbolic identities listed in Section 10.3 are easily remembered from their trigonometric
counterparts (see Remark 10.3.1).
• It is not expected that students will memorise the derivatives of coth, sech and cosech. How-
ever, they can be easily derived from first principles provided that the derivatives of sinh and
cosh are known.
A table of standard integrals will be issued at the final examination.
10.8 Appendix: The catenary
We end this chapter showing that the cosh function describes the curve of a suspended cable.
Suppose that a cable of uniform mass is suspended from two fixed points. To describe the
equation of the resulting curve in terms of x and y, we fix a coordinate system such that vertex
of the curve passes through the point A(0, y0) as in Figure 10.6. (The ordinate y0 can be any real
number; its choice only fixes the coordinate system and does not alter the shape of the catenary.)
Consider a point P (x, y) that lies on the curve and suppose that
• T is the tension in the cable at P (x, y),
• H is the tension in the cable at A(0, y0),
• ℓ(x) is the length of the cable from A(0, y0) to P (x, y), and
• w is the weight per unit length of the cable.
(see Figure 10.6). Thus the gravitational force acting on the cable along the segment AP is given
by w ℓ(x). We note that the quantities T , θ and ℓ depend on x but that w and H do not. We also
make the natural assumption that the catenary is described by a differentiable function.
Theorem 10.8.1. There is a choice of coordinate system such that the catenary in Figure 10.6 is
described by the equation
y =
H
w
cosh
(wx
H
)
.
To prove the theorem, it suffices to show that if the catenary is described by y = f(x), where
f is a differentiable function, then
f(x) =
H
w
cosh
(wx
H
)
for a particular choice of y0.
We begin with a general formula that relates a function f to its arc length function ℓ by a
differential equation.
Lemma 10.8.2. Suppose that f is a differentiable function. Then the arc length function ℓ is also
differentiable and
ℓ′(x) =
√
1 +
(
f ′(x)
)2
(10.6)
c©2020 School of Mathematics and Statistics, UNSW Sydney
254 CHAPTER 10. THE HYPERBOLIC FUNCTIONS
x
y
b
b
H
w ℓ(x)
T
T cos θ
T sin θ
θP
A
suspended cable
Figure 10.6: Forces acting on a suspended cable.
Sketch proof. Suppose that h is a small nonzero real number and consider the diagram below.
t
f(t)
b
b
(
x, f(x)
)
(
x+ h, f(x+ h)
)
h
f(x+ h)− f(x)ℓ(x+ h)− ℓ(x)
By Pythagoras’ theorem,
ℓ(x+ h)− ℓ(x) ≈
√
h2 +
(
f(x+ h)− f(x))2.
If we divide both sides by h then
ℓ(x+ h)− ℓ(x)
h
≈
√
1 +
(
f(x+ h)− f(x)
h
)2
.
As h→ 0 the approximation gets better and we obtain
ℓ′(x) =
√
1 +
(
f ′(x)
)2
as required.
c©2020 School of Mathematics and Statistics, UNSW Sydney
10.8. APPENDIX: THE CATENARY 255
The next lemma uses facts specific to the catenary.
Lemma 10.8.3. If the equation of the catenary in Figure 10.6 is given by y = f(x), then f satisfies
the following conditions:
f(0) = y0, (10.7)
f ′(0) = 0, (10.8)
f ′′(x) =
w
H
√
1 +
(
f ′(x)
)2
. (10.9)
Proof. The first two conditions are a result of our choice of coordinate system and the assumption
that f is differentiable. It remains to establish the third condition.
Since the cable is stationary, the forces w ℓ(x) and T sin θ that act vertically cancel each other
out. That is,
T sin θ = w ℓ(x).
Similarly, the horizontal forces H and T cos θ also cancel each other out and so
T cos θ = H.
Using these equations, and the fact that f ′(x) = tan θ at the point P (x, y), gives
f ′(x) = tan θ =
T sin θ
T cos θ
=
w ℓ(x)
H
.
So far we have
f ′(x) =
w ℓ(x)
H
.
Since w and H are constants, differentiating this equation with respect to x gives
f ′′(x) =
w
H
ℓ′(x).
Therefore
f ′′(x) =
w
H
√
1 +
(
f ′(x)
)2
.
by equation (10.6).
Proof of Theorem 10.8.1. It remains to find all functions f that satisfy the conditions of Lemma
10.8.3. If we let u denote f ′(x) and b denote w/H, then equation (10.9) becomes
du
dx
= b
√
1 + u2.
Hence
dx
du
=
1
b
√
1 + u2
.
Every solution to this differential equation is of the form
x =
1
b
sinh−1 u+C1.
c©2020 School of Mathematics and Statistics, UNSW Sydney
256 CHAPTER 10. THE HYPERBOLIC FUNCTIONS
Hence we have
f ′(x) = u = sinh(bx− bC1).
Enforcing condition (10.8) shows that either b = 0 or C1 = 0. Now if b = 0 then f
′(x) = 0 and f
is simply a constant function (that is, the cable is pulled tight and exhibits no sag). On the other
hand, if b 6= 0 then C1 = 0 and hence
f ′(x) = sinh(bx).
Therefore
f(x) =
cosh(bx)
b
+ C2
and enforcing the condition (10.7) yields C2 =
1
b − y0. If we choose y0 to be 1/b then C2 = 0 and
the function f that describes the catenary is given by
f(x) =
cosh(bx)
b
=
H
w
cosh
(wx
H
)
,
proving the theorem.
10.9 Maple notes
Maple knows about the hyperbolic functions sinh, cosh, tanh, csch, sech, and coth, and inverse
hyperbolic functions arcsinh, arccosh, arctanh, arccsch, arcsech, and arccoth. For example,
> int(arctanh(x),x);
x arctanh (x) +
1
2
ln
(
1− x2)
> # which is closely analogous to
> int(arctan(x),x);
x arctan (x)− 1
2
ln
(
1 + x2
)
> # BUT the following result does not follow the analogy between trig and hyperbolic
functions
> int(sech(x),x);
arctan (sinh (x))
> plot(arccsch(x),x=-6..6, y=-6..6, discont=true, numpoints=100);
c©2020 School of Mathematics and Statistics, UNSW Sydney
10.9. MAPLE NOTES 257
c©2020 School of Mathematics and Statistics, UNSW Sydney
258 CHAPTER 10. THE HYPERBOLIC FUNCTIONS
Problems for Chapter 10
Problems 10.1 : Hyperbolic sine and cosine functions and
10.2 : Other hyperbolic functions
1. [R] [V] Define sinhx and coshx. Hence show that
a)
d
dx
(cosh 6x) = 6 sinh 6x;
b) ln(sinh x) < x− ln 2 whenever x > 0.
2. [R] By expressing the following hyperbolic functions in terms of sinhx and coshx, find the
derivative of each function f given below.
a) f(x) = tanhx b) f(x) = sech x c) f(x) = coth x
3. [R] In each case, find f ′(x).
a) f(x) = sinh(3x2) b) f(x) = cosh( 1x) c) f(x) = sinh(ln x)
Problems 10.3 : Hyperbolic identities
4. [R]
a) Given the formula sinh(A+B) = sinhA coshB+coshA sinhB, find a formula for sinh 2x.
By differentiation or otherwise, find a formula for cosh 2x.
b) [H] Using the results of part (a), express sinh 3x as a cubic polynomial in sinhx. Hence,
or otherwise, find
∫
sinh3 x dx.
5. [R] Show that coshx+ sinhx = ex. Deduce that (cosh x+ sinhx)n = coshnx+ sinhnx.
6. [H] [V] Consider the hyperbola x2 − y2 = 1, where x ≥ 1.
A(t)
x
y
(cosh t, sinh t)
0 1
c©2020 School of Mathematics and Statistics, UNSW Sydney
PROBLEMS FOR CHAPTER 10 259
a) Using the definitions of cosh and sinh, prove that, for every real number t, the point
(cosh t, sinh t) lies on the hyperbola.
b) When t > 0, let A(t) denote the shaded region in the diagram. Explain why
A(t) =
1
2
cosh t sinh t−
∫ cosh t
1
√
x2 − 1 dx.
c) By first calculating A′(t), prove that A(t) =
t
2
.
Problems 10.4 : Hyperbolic derivatives and integrals
7. [R] Evaluate the following integrals.
a)
∫
cosh(4x) dx b)
∫ 1
3
ln 2
0
sinh 3x dx
c)
∫
cosh2 x dx d)
∫
sinh(
√
x)√
x
dx
Problems 10.5 : The inverse hyperbolic functions
8. [R] Simplify cosh(sinh−1(3/4)), cosh−1(cosh(−3)) and sinh(tanh−1(5/13)).
9. [R] Show that
a)
d
dx
(
cosh−1 x
)
=
1√
x2 − 1 , for x > 1 b)
d
dx
(
tanh−1 x
)
=
1
1− x2 .
10. [R] Show that
a) cosh−1 x = ln(x+
√
x2 − 1) ∀x ∈ [1,∞)
b) [V] tanh−1 x =
1
2
ln
(
1 + x
1− x
)
∀x ∈ (−1, 1).
11. [R] Find
dy
dx
if
a) y = sinh−1(2x)
b) y = tanh−1(1/x)
c) y = cosh−1(sec x) whenever 0 < x < π/2.
Problems 10.6 : Integration leading to the inverse hyperbolic functions
12. [R] Find
a) [V]
∫
dx√
1 + 4x2
b)
∫ 1/2
0
dx
1− x2 c)
∫
dx√
x2 + 4x+ 13
.
c©2020 School of Mathematics and Statistics, UNSW Sydney
260 CHAPTER 10. THE HYPERBOLIC FUNCTIONS
c©2020 School of Mathematics and Statistics, UNSW Sydney
261
Answers to selected problems
Chapter 1
1. a) The set of integers between −π and π.
c) The empty set.
3. Answer for both: the interior and boundary of the triangle with vertices at (0, 0), (2, 0) and (2, 4).
4. a) x < 0 or x > 1 b) 1 < x < 2 c) x < −2 or x > 0
d) −1 < x < 1 e) −2 ≤ x < 1 or x ≥ 3
5. a) −4 < x < 2 b) x < −5 or x > 1
c) −1 < x < −1/3 d) x > 0
6. c) From (a) we have x2 +
1
x2
≥ 2 with equality if and only if x = ±1.
7. a) F b) F c) T d) T e) F
8. Hint: (x2 + y2)2 ≥ 4x2y2.
10. a) −√5 ≤ x ≤ √5; 0 ≤ y ≤ √5
b) x ≤ −√5 or x ≥ √5; y ≥ 0
c) x 6= 8; y 6= 0
d) [1,∞); [0,∞)
e) (1, ∞); (0, ∞)
f) {x ∈ R : 2nπ ≤ x ≤ (2n+ 1)π; n ∈ Z}; [0, 1]
g) The union of the intervals [− 7pi6 + 2kπ, pi6 + 2kπ] where k ∈ Z; 0 ≤ y ≤
√
3
h) {x ∈ R : x 6= (2n+ 1)π/2, n an integer}; [1, ∞)
i) R; [−1, ∞)
11. a) 22 b) x2 + 10x+ 22 c) 6 d) x2 + 2
12. a) x− 1 + 1/√x− 1 b) √x− 1 c) (x − 1)3/2 d) (1/√x− 1)− 1
16. [4, 13]
17. x = 1, 7
c©2020 School of Mathematics and Statistics, UNSW Sydney
262 CHAPTER 3
18. a) If p(x) = a0+a1x+ · · ·+anxn then p(q(x)) = a0+a1q(x)+a2(q(x))2+ · · ·+an(q(x))n. Products
and sums of polynomials are again polynomials.
b) Yes.
Chapter 2
1. a) 1 b) 2 c) 0
d) Doesn’t exist (→∞). e) 5 f) Doesn’t exist.
2. a) 0 b) 0
4. b) 0
5. a) 1 b) M = 10 (best possible) c) M = 1/
√
ǫ will do.
6. a) 4 b) 0 c) 0 d) 0 e) 0
7. a) 50 metres per second b) 5 ln 50 ≈ 19.56 seconds after leaving the plane.
8. a) 10 b) 4 c) 3 d) −1/9
9. a) −1 b) 1 c) No
10. a) Doesn’t exist. b) Doesn’t exist. c) Doesn’t exist. d) Doesn’t exist.
11. a) 0 b) 0
12. a) |CB| = θ, |CA| = sin θ, |DB| = tan θ.
13. Neither the left-hand nor right-hand limits exist due to wild oscillatory behaviour.
Chapter 3
1. b) Yes
2. a) Continuous everywhere. b) Continuous everywhere except at π/2.
3. k = 8
5. Use the intermediate value theorem.
8. a) Yes b) Yes c) No d) Yes
c©2020 School of Mathematics and Statistics, UNSW Sydney
ANSWERS 263
Chapter 4
2. a) 5(4x3 + 21x6) b) (4x3−2)(4x2+2x+4)+(x4−2x)(8x+2)
c) (16y − y4)/(y3 + 8)2 d) (2x2 − 4)/(x2 − 4)1/2
e) −4/(t2 − 4)3/2 f) 3 cos 3y + 12 cos 2y sin 2y
g) (4x3 − x4)e−x h) x ln(x3 + 1) + 3x2(x2 + 1)/2(x3 + 1)
i) sec2 x j) − tanx
3. a) 0 b) 0 c) f ′(0) = 0
4. a) i) x 6= 0 ii) all x b) i) all x ii) all x
c) i) x 6= −2 ii) x 6= −2
6. a) x+ 17π + cos 2x b) 1− 2 sin 2x c) 2−x2+cos 2(2−x2)
d) 1− 2 sin 2(2− x2) e) −2x(1− 2 sin 2(2− x2))
7. a)
dy
dx
=
3x2 − y
x− 3y2 b)
dy
dx
= (y − 4x√xy)/(4y√xy − x)
9. y = 2
10. a) (i) b = 0 (ii) a = 1, b = 0
b) (i) b = 1 (ii) a = 2, b = 1.
11. a = 1, b = 0
12. a) f(8.01) ≈ f(8) = 2
b) i) y = (x− 8)/12 + 2
ii) f(8.01) ≈ (8.01− 8)/12 + 2 = 2 + 11200
c) The approximation in (b) is much better.
13.
√
3 ar/2
14. 7/8
15. a) 18pi
b) 32000pi81 cm
3
16. a) dhdt =
2
125pi when h = 50.
Chapter 5
1.
√
7
3 b)
1
2
5. b) 0
7. a) By the Mean Value Theorem, for some c with 16 < c < 17,
√
17−√16 = 1
2
√
c
< 1
2
√
16
= 0.125.
b) 0.008
c©2020 School of Mathematics and Statistics, UNSW Sydney
264 CHAPTER 6
c) 2× 10−6.
8. −1, 1 and 4 are stationary points; 4 is a local minimum point;−1 is a local maximum point.
9. a) 11,−61 b) 3,−253 c) 27/256,−750
d) 250,−54 e) 2, 0
10. (12/13, 18/13)
11.
12. a) (400)/(4 + π), 100π/(4 + π) b) 0, 100
14. Three real zeros
15. a) f(t) = − cos t+ t2/2 + 3 b) No
16. a) 0 b) 8/3
17. a) 13 b)
m
n c) −1 d) − 12 e) 14 f) 13
18. a) → 0 b) →∞ c) → 0
d) → 1 e) → 1 f) → 32
19. Combine the two fractions and apply l’Hoˆpital twice only. You will need to simplify the quotient
obtained after the first application of l’Hoˆpital. Maple can confirm your answer.
20. (a, b) = (−√2,√2) or (√2,−√2)
23. a) −1/2
b) a = −1/2, b = 1
24. c) a = b = 0
Chapter 6
2. a) f−1(x) =
1
3
(x− 1)
b) g−1(x) = −√x− 1, Dom(g−1) = [1,∞),
Range(g−1) = (−∞, 0], (g−1)′(x) = −1
2
√
x−1
4. b) 1/3
5. b) The restriction of f to (−∞,−1] has an inverse with domain (−∞, 3],
the restriction of f to [−1, 1] has an inverse with domain [−1, 3], and
the restriction of f to [1,∞) has an inverse with domain [−1,∞).
6. a) No b) Yes
c©2020 School of Mathematics and Statistics, UNSW Sydney
ANSWERS 265
7. a) The graph is symmetric about x = − 12 , which surely gives a local maximum of f(x). There
will be four (maximal) intervals where f will have an inverse. Try this exercise on Maple. The
commands plot, diff and solve should suffice.
b) f is one-to-one; f−1(x) = x1/17 − 1 is not differentiable when x = 0.
c) I can be one of four intervals.
8. a) π/3 b) 2/5 c) −π/3 d) π/3
e) 4/5 f) 3/
√
34 g) π/3 h) π − x
11. a) −2/√1− 4x2 b) 1/(2√x− x2 ) c) 2/(4x2 − 12x+ 10)
12. Differentiate; −1 ≤ x ≤ 1; π/2.
13. b) f(x) = π/2 when x > 0 and f(x) = −π/2 when x < 0.
14. b) The derivative of the inverse is −1/x√x2 − 1 when x > 1.
15. a = π/2, b = 0
16. a) 24π km/min b) 104π/3 km/min
17.
√
48 metres
Chapter 7
1. [−1, 5], [0, 3], upper half of circle.
2. a) period 2π/3, odd b) period 3π, neither
c) not periodic, even d) period π/3, odd
e) period π, even f) 2π
3. odd, even, neither, odd, odd, even.
4. The asymptotes are
a) x = 3, y = x+ 2 b) x = −1, y = x− 1 c) x = −3, x = 2, y = x− 1.
6. a) x ≥ 3, − 13 < x ≤ 13 b) x = − 13 , x = −3, y = 1 c) (1,− 14 ), (−1,−4) d) Domain:
x 6= 3,− 13 , Range: (−∞,−4], [− 14 ,∞).
7. a)
x2
16
+
y2
25
= 1, ellipse b)
x2
9
− y
2
4
= 1, hyperbola
c) y = x2/3 d) spiral
8. a) ii) (2, 0) iii) −1
b) ii) (5, 0), (−1, 0) iii) 4t3/3
c) ii) (1, 0), (−1, 0) iii) − cot t
9. a) 3x− 27y + 52 = 0 b) 19
c©2020 School of Mathematics and Statistics, UNSW Sydney
266 CHAPTER 8
10. b) Hint: the length of one particular arc of the larger circle equals the length of one arc on the
smaller circle.
d) x2/3 + y2/3 = 1
11. a) p(t) = a + t(b − a),p(0) = a,p(1) = b b) y = 4 − x,q(1/2) is the coodinate vector for the
midpoint of B and C c) p0(t) = (1 − t)2, p1(t) = 2t(1− t), p2(t) = t2
12. a) (3, 0) b) (−3√3, −3) c) (√2,−√2)
13. a) (3, π) b) (
√
2,−3π/4) c) (4, 2π/3)
d) (1, π/2) e) (4, 5π/6) f) (4,−5π/6)
14. a) Circle, centre (0,0), radius 4
b) A ray in the second quadrant
c) A spiral of Archimedes
15. a) Circle, centre (0,3), radius 3
b) Circle, centre (1,0), radius 1
16. The following sketches are a guide to shape only.
a)
x
y
b)
x
y
c)
x
y
d)
x
y
e)
x
y
f)
x
y
18.
(x− 2)2
9
+
y2
5
= 1
Chapter 8
1. a) i) SPn(f) = SPn(f) = 1
ii) SPn(f) =
1
2
(
1− 1n
)
, SPn(f) =
1
2
(
1 + 1n
)
iii) SPn(f) =
1
6
(
1− 1n
) (
2− 1n
)
, SPn(f) =
1
6
(
1 + 1n
) (
2 + 1n
)
v) SPn(f) = 1, SPn(f) = 0
b) i) 1 ii) 12 iii)
1
3 iv)
1
4 (v) Not Riemann integrable
2. a)
√
1365 = 36.95
b)
√
1690.9 = 41.12 and the lower bound is
√
1078.9 = 32.85
3. 4.5
4. a) 82.4 b) 10
c©2020 School of Mathematics and Statistics, UNSW Sydney
ANSWERS 267
5. f(x) =
1
x2 + x+ 1
6. 1x is not differentiable on all of [−1, 1] so the FTC doesn’t apply.
7. a) Draw a picture! b) 5π/12−√3/2
9. F is continuous everywhere, but not differentiable at the integers.
11. a) sinx2 b) 3x2 sinx6 c) −3x2 sinx6 d) 3x2 sinx6 − sinx2
12. −(5− 4x)5
13. biii) pi6 .
14. a) 12 e
x2 + C b) −2 cos√x+ C c) 15/4
d) 4
√
2 a9/2
/
9 e) 1/4 f) (2
√
2− 1)/3
15. a) 2
√
x− 2 ln(1 +√x) + C b) 125
(
1
21 (5x− 1)21 + 120 (5x− 1)20
)
+ C
c) x/(x + 1)2 + C d) 4− 10 ln(7/5)
16. a) 4e
5+1
25
b) x2 sinx+ 2x cosx− 2 sinx+ C
c) x(ln(x) − 1) + C d) pi12 +
√
3
2 − 1
e) 7e
8+1
64
f) pi2
g) e
x
2 (cos x+ sinx) h) x tan
−1 x− ln√1 + x2 + C
i)
√
2
2
+ 1
2
ln(1 +
√
2)
17. a) 1/5 b) diverges c) π/4
d) 0 e) 2 f) diverges
19. a) 0 b) ln 2 c) No
20. a) convergent b) divergent c) divergent
21. a) convergent b) divergent c) convergent
22. a) convergent b) divergent c) convergent
23. s < 0
24. p > 1
25. The integral converges whenever 2a− b > 1.
26. a) 4, 8, 2
c) Li′(x) = 1ln x > 0 so Li is an increasing function; Li(2) = 0.
d) Li(106) ≥ 106−26 ln 10 .
e) pi(10
6)
x ' 0.07238.
27. a) 2√
pi
e−x
2
c©2020 School of Mathematics and Statistics, UNSW Sydney
268 CHAPTER 10
d) (i) 0.749 < erf(1) < 0.928 (iii) 1/e (iv) 1.344
Chapter 9
2. a) A partition into 7 equal parts will suffice
3. a) 3x2/2(x3 + 1) b) ex for x > 0, −e−x for x < 0
c) 1(ln(ln x))(ln x)x d) 5x
4 (where x > −61/5)
4. a) 12 ln(1 + e
2x) b) −e1/x
c) 3x/ ln 3 d) e
√
x
4
e) (ln x)
2
2
f) ln | sinx|
7. b) e
8. a) 3
x ln 3 b)
(
x3 − 3
x2 + 1
)1/5(
3x2
5(x3 − 3) −
2x
5(1 + x2)
)
c) (sinx)sin x cosx (1+ln(sinx)) d) cos(xsin x)xsin x
(
cosx lnx+
sinx
x
)
9. a) 0 b) 0 c) 1 d) e2 e) 1
f) 1 g) ea h) 0 i) 0
Chapter 10
2. a) sech2 x b) −sechx tanhx c) −cosech2x
3. a) 6x cosh(3x2) b) − sinh(1/x)x2 c)
1
2 +
1
2x2
4. a) sinh 2x = 2 coshx sinhx ; cosh 2x = cosh2 x+ sinh2 x
b) 14 (
1
3 cosh 3x− 3 coshx) or 13 cosh3 x− coshx
7. a) sinh 4x4 b)
1
12 c) (2x+ sinh 2x)/4 d) 2 cosh
√
x
8. 5/4, 3, 5/12
11. a) 2/
√
1 + 4x2 b) 11−x2 for |x| > 1 c) secx
12. a) 12 sinh
−1 2x b) tanh−1 12 =
1
2 ln 3 c) sinh
−1 (x+2
3
)
c©2020 School of Mathematics and Statistics, UNSW Sydney
269
Past class tests
The following selection of past MATH1131 class tests can be used as a guide to the degree of
difficulty of calculus class tests in this course. However, due to variations in timing and frequency,
the material examined in each class test differs from the tests given here. Thus students must
consult the Information booklet for MA1131, or page xiii of these notes, to ascertain the precise
topics that may be examined in each calculus class test.
c©2020 School of Mathematics and Statistics, UNSW Sydney
UNIVERSITY OF NEW SOUTH WALES
SCHOOL OF MATHEMATICS AND STATISTICS
MATH1131 Mathematics 1A Calculus S1 2008
TEST 1 VERSION 5a
This sheet must be filled in and stapled to the front of your answers
Student’s Family Name Initials Student Number
Tutorial Code Tutor’s Name Mark
Note: The use of a calculator is NOT permitted in this test
QUESTIONS (Time allowed: 20 minutes)
1. (2 marks)
Solve |2− 3x| ≤ 1.
2. (2 marks)
Find the (maximal) domain and the range of the function f(x) =
1√
3− x.
3. (2 marks)
Sketch the graph y = x2 − 3x− 10, and hence sketch the graph y = 1
x2 − 3x− 10.
4. (2 marks)
For f(x) =
|x2 − 9|
x− 3 and a = 3, discuss the limiting behaviour of f(x) as x→ a
+, as x→ a−
and as x→ a.
5. (2 marks)
Determine the limiting behaviour of f(x) =
2x+ 3x2 + e−x
2x2 + cos x
as x→∞.
UNIVERSITY OF NEW SOUTH WALES
SCHOOL OF MATHEMATICS AND STATISTICS
MATH1131 Mathematics 1A Calculus S2 2008
TEST 1 VERSION 2b
This sheet must be filled in and stapled to the front of your answers
Student’s Family Name Initials Student Number
Tutorial Code Tutor’s Name Mark
Note: The use of a calculator is NOT permitted in this test
QUESTIONS (Time allowed: 20 minutes)
1. (2 marks)
Let f(x) = x2 + 4, and g(x) =
1√
x+ 1
. Give the explicit forms of (f ◦ g)(x) and (g ◦ f)(x).
2. (2 marks)
Find the limiting value of f(x) =
x2 − 5x+ 6
2x2 − 5x+ 2 as x tends to 2.
3. (2 marks)
For f(x) =
|x2 − 4x+ 3|
x− 1 and a = 1, discuss the limiting behaviour of f(x) as x → a
+, as
x→ a− and as x→ a.
4. (2 marks)
Solve |2− 3x| ≤ 1.
5. (2 marks)
Find the (maximal) domain and the range of the function f(x) = ln(x2 − 5).
UNIVERSITY OF NEW SOUTH WALES
SCHOOL OF MATHEMATICS AND STATISTICS
MATH1131/1141 Calculus S1 2009
TEST 1 VERSION 8a
This sheet must be filled in and stapled to the front of your answers
Student’s Family Name Initials Student Number
Tutorial Code Tutor’s Name Mark
Note: The use of a calculator is NOT permitted in this test
QUESTIONS (Time allowed: 20 minutes)
1. (2 marks)
Sketch the graph y =
√
x+ 2, and hence sketch the graph y =
1√
x+ 2
.
2. (2 marks)
Solve |3x+ 2| ≥ 1.
3. (2 marks)
Find the (maximal) domain and range of the function f(x) =
1√
9− x2 .
4. (2 marks)
Determine the limiting behaviour of f(x) =
e−x + 3x2 − 2
4x2 + 3x+ sinx
as x→∞.
5. (2 marks)
For f(x) =
x− 2
|x2 − 4| and a = 2, discuss the limiting behaviour of f(x) as x→ a
+, as x→ a−
and as x→ a.
UNIVERSITY OF NEW SOUTH WALES
SCHOOL OF MATHEMATICS AND STATISTICS
MATH1131 Mathematics 1A Calculus S1 2009
TEST 1 VERSION 6a
This sheet must be filled in and stapled to the front of your answers
Student’s Family Name Initials Student Number
Tutorial Code Tutor’s Name Mark
Note: The use of a calculator is NOT permitted in this test
QUESTIONS (Time allowed: 20 minutes)
1. (2 marks)
Solve
1
x+ 1
≤ −1
2
.
2. (2 marks)
Find the (maximal) domain and the range of the function f(x) =
√
2− e−x.
3. (2 marks)
Let f(x) = 3x+ 4, and g(x) =
1√
x− 2. Give the explicit forms of (f ◦ g)(x) and (g ◦ f)(x).
4. (2 marks)
Find the limiting value of f(x) =
2x2 − x− 6
3x2 − 2x− 8 as x tends to 2.
5. (2 marks)
For f(x) =
|x2 + 3x− 18|
x− 3 and a = 3, discuss the limiting behaviour of f(x) as x→ a
+, as
x→ a− and as x→ a.
UNIVERSITY OF NEW SOUTH WALES
SCHOOL OF MATHEMATICS AND STATISTICS
MATH1131 Mathematics 1A Calculus S2 2009
TEST 1 VERSION 1a
This sheet must be filled in and stapled to the front of your answers
Student’s Family Name Initials Student Number
Tutorial Code Tutor’s Name Mark
Note: The use of a calculator is NOT permitted in this test
QUESTIONS (Time allowed: 20 minutes)
1. (2 marks)
Sketch the set of points in the (x, y) plane satisfying 0 < x < 3y and 0 < y < 2.
2. (2 marks)
Solve
∣∣∣∣3x+ 12
∣∣∣∣ ≤ 2.
3. (2 marks)
Find the (maximal) domain and the range of the function f(x) =
1√
x− 1 .
4. (2 marks)
For f(x) =
|x2 − 4x+ 3|
3− x and a = 3, discuss the limiting behaviour of f(x) as x → a
+, as
x→ a− and as x→ a.
5. (2 marks)
Consider the function
f(x) = − cos x
on the interval (−π/2, π/2). Determine whether f attains a maximum value on the interval.
Give reasons for your answer.
UNIVERSITY OF NEW SOUTH WALES
SCHOOL OF MATHEMATICS AND STATISTICS
MATH1131/MATH1141 Calculus S1 2010
TEST 1 VERSION 7b
This sheet must be filled in and stapled to the front of your answers
Student’s Family Name Initials Student Number
Tutorial Code Tutor’s Name Mark
Note: The use of a calculator is NOT permitted in this test
QUESTIONS (Time allowed: 20 minutes)
1. (2 marks)
Sketch the graph y =
√
x− 1, and hence sketch the graph y = 1√
x− 1.
2. (2 marks)
For f(x) =
|x2 + x− 2|
x− 1 and a = 1, discuss the limiting behaviour of f(x) as x → a
+, as
x→ a− and as x→ a.
3. (2 marks)
Let p(x) = x3− 3x2− 4x+2. Use the Intermediate Value Theorem to show that p has a root
between −2 and 0.
4. (2 marks)
Solve
1
x+ 1
> −1
2
.
5. (2 marks)
Find the (maximal) domain and the range of the function f(x) =
√
3 + x.
UNIVERSITY OF NEW SOUTH WALES
SCHOOL OF MATHEMATICS AND STATISTICS
MATH1131 Mathematics 1A Calculus S1 2008
TEST 2 VERSION 8a
This sheet must be filled in and stapled to the front of your answers
Student’s Family Name Initials Student Number
Tutorial Code Tutor’s Name Mark
Note: The use of a calculator is NOT permitted in this test
QUESTIONS (Time allowed: 20 minutes)
1. (2 marks)
Show that the function f given by f(x) = x3−2x2−3x+3 has a zero in each of the intervals
[0, 1] and [2, 3].
2. (2 marks)
Using the definition of the derivative, show that if f(x) = −2x2 + x then f ′(x) = −4x+ 1.
3. (2 marks)
The length L of a rectangle is decreasing at the rate of 2 cm per second, and the width W
is increasing at the rate of 4 cm per second. Find the rate of change of the area when L=13
cm and W=10 cm.
4. (2 marks)
State the Mean Value Theorem. Find a point which satisfies the conclusions of the Mean
Value Theorem for f(x) = x3 − 2x2 + 5 on the interval [0, 2].
5. (2 marks)
Find lim
x→0
x sinx
1− cos x.
UNIVERSITY OF NEW SOUTH WALES
SCHOOL OF MATHEMATICS AND STATISTICS
MATH1131 Mathematics 1A Calculus S2 2008
TEST 2 VERSION 4b
This sheet must be filled in and stapled to the front of your answers
Student’s Family Name Initials Student Number
Tutorial Code Tutor’s Name Mark
Note: The use of a calculator is NOT permitted in this test
QUESTIONS (Time allowed: 20 minutes)
1. (2 marks)
State the Mean Value Theorem and find a point which satisfies the conclusions of the Mean
Value Theorem for f(x) =
√
x− 1 on the interval [1, 3].
2. (2 marks)
Find lim
x→0
1− cos 3x
x2
.
3. (2 marks)
Show that the function f given by f(x) = x3−3x2−2x+5 has a zero in each of the intervals
[−2,−1] and [1, 2].
4. (2 marks)
Using the definition of the derivative, show that if f(x) = −x3 then f ′(x) = −3x2.
5. (2 marks)
Find the equation of the line tangent to x2 + y3 − x2y = 1 at (1, 1).
UNIVERSITY OF NEW SOUTH WALES
SCHOOL OF MATHEMATICS AND STATISTICS
MATH1131 Mathematics 1A Calculus S1 2009
TEST 2 VERSION 8b
This sheet must be filled in and stapled to the front of your answers
Student’s Family Name Initials Student Number
Tutorial Code Tutor’s Name Mark
Note: The use of a calculator is NOT permitted in this test
QUESTIONS (Time allowed: 20 minutes)
1. (2 marks)
Carefully state the Mean Value Theorem. Find a point which satisfies the conclusions of the
Mean Value Theorem for f(x) = x3 − x2 + 3 on the interval [0, 1].
2. (2 marks)
Find lim
x→1
3x3 − 5x2 + x+ 1
x2 − 2x+ 1 .
3. (2 marks)
Determine the values of x at which the function
f(x) =
{
x3 for x < 1
(x− 1)3 + 2 for x ≥ 1
is continuous. Give reasons for your answer.
4. (2 marks)
Using the definition of the derivative, show that if f(x) = −x3 then f ′(x) = −3x2.
UNIVERSITY OF NEW SOUTH WALES
SCHOOL OF MATHEMATICS AND STATISTICS
MATH1131/1141 Calculus S1 2009
TEST 2 VERSION 1a
This sheet must be filled in and stapled to the front of your answers
Student’s Family Name Initials Student Number
Tutorial Code Tutor’s Name Mark
Note: The use of a calculator is NOT permitted in this test
QUESTIONS (Time allowed: 20 minutes)
1. (2 marks)
The function
f(x) =
x− 2
x2 − 3x+ 2
is not defined for x = 2. Find a value to be given to f(2) that will make f continuous at 2.
2. (3 marks)
Determine all real values of a and b such that the function
f(x) =

ax+ b for x ≤ 1tan πx
4
for 1 < x < 2
is differentiable at x = 1.
3. (3 marks)
Let f(x) = |x2 − 2x|.
(i) Giving reasons for your answer, find all critical points of f on the interval [0, 5].
(ii) Find the absolute maximum and absolute minimum values of f(x) on the given interval.
4. (2 marks)
Find lim
x→1
2x4 − 3x3 + x
(x− 1)2 .
UNIVERSITY OF NEW SOUTH WALES
SCHOOL OF MATHEMATICS AND STATISTICS
MATH1131 Mathematics 1A Calculus S2 2009
TEST 2 VERSION 3a
This sheet must be filled in and stapled to the front of your answers
Student’s Family Name Initials Student Number
Tutorial Code Tutor’s Name Mark
Note: The use of a calculator is NOT permitted in this test
QUESTIONS (Time allowed: 20 minutes)
1. (2 marks)
Find the equation of the line tangent to x+ lnx = y + 2 ln y at (1, 1).
2. (2 marks)
A ladder of length 3 metres is leaning against a vertical wall. The foot of the ladder is pulled
away from the wall at the rate of 0.5 metres per second. How fast is the top of the ladder
moving down the wall when the foot is 1 metre away from the wall? (Leave your answer in
surds.)
3. (2 marks)
Carefully state the Mean Value Theorem and find a point which satisfies the conclusions of
the Mean Value Theorem for f(x) =
√
x− 1 on the interval [1, 3].
4. (3 marks)
Let f(x) = (x− 1)2/3.
(i) Giving reasons for your answer, find all critical points of f on the interval [0, 2].
(ii) Find the absolute maximum and absolute minimum values of f(x) on tthe given interval.
5. (1 mark)
Find lim
x→0
tanx
x
.
UNIVERSITY OF NEW SOUTH WALES
SCHOOL OF MATHEMATICS AND STATISTICS
MATH1131/MATH1141 Calculus S1 2010
TEST 1 VERSION 2a
This sheet must be filled in and stapled to the front of your answers
Student’s Family Name Initials Student Number
Tutorial Code Tutor’s Name Mark
Note: The use of a calculator is NOT permitted in this test
QUESTIONS (Time allowed: 20 minutes)
1. (3 marks)
Determine all real values of a and b such that the function
f(x) =
{
aex + b for x < 0,
sinx for x ≥ 0
is differentiable at x = 0.
2. (1 mark)
Find lim
x→0
tanx
e3x − 1 .
3. (2 marks)
Find the equation of the line tangent to x3 + y3 − x− y2 = 0 at (1, 1).
4. (3 marks)
Determine how many real numbers satisfy the equation x3−6x2+1 = 0. Give reasons for your
answer, naming any theorems you use. (Hint: find the stationary points of the polynomial
function.)
5. (1 mark)
Differentiate tan−1(4x+ 1).
282
c©2020 School of Mathematics and Statistics, UNSW Sydney
283
Index
absolute maximum point, 69
absolute maximum value, 69
absolute minimum point, 69
absolute minimum value, 69
absolute value, 17
antiderivative, 114
antidifferentiation, 114, 193
area, 172
absolute, 183
signed, 183
under the graph of f , 179
unsigned, 183
astroid, 168
asymptote, 144
oblique, 144
vertical, 144
Barrow, Isaac, 171
Bernoulli, Jakob, 155, 239
Bernoulli, Johann, 155, 239
Bolzano, 47
bounded function, 71
brachistochrone, 155
Cartesian coordinate, 157
catenary, 239, 253
chain rule, 82, 90
change of variables formula, 197
circle, 31, 152
closed interval, 13
codomain, 19
composition (of functions), 22
conic sections, 31
contains, 14
continuity, 53, 87
and differentiability, 87
at an end point, 66
everywhere, 54
on an interval, 65, 66
continuous, 86
continuous functions, 32
and limits, 53
combination of, 63
piecewise, 181
critical point, 108
cubic, 24
curve of fastest descent, 155
cusp point, 164
cycloid, 155
decreasing function, 103
definite integral, see integral(s), definite
derivative(s), 77, 79
definition of, 79
difference of squares identities, 245
difference quotient, 80
differentiability, 87
and continuity, 87
local extrema, 91
differentiable, 86
at a point, 79
local approximation, 89
maximum point, 91
mean value theorem, 97
minimum point, 91
twice, 105
differential calculus, 11
differential equation, 113
differentiation, 81
chain rule, 82
product rule, 84
quotient rule, 82
discontinuity, 33
distance, 18
domain, 19
maximal, 20
c©2020 School of Mathematics and Statistics, UNSW Sydney
284
natural, 20
dominant term analysis, 206, 210
double-angle formulae, 245
dummy variable, 186
e, 227
elementary functions, 29
elements (of set), 12
ellipse, 31, 152
endpoints
critical points, 108
endpoints of intervals, 13
equations
solutions, 110
erf, 221
error bounds, 102
error function, 221
even function, 144
exp, 228
exponential function, 231
fastest growing term, 42
function(s), 11, 19
composition of, 22, 82
decreasing, 103
error function, 221
even, 144
exponential, 230
implicitly defined, 31
increasing, 103
inverse function, 127
limit at infinity, 46
local approximation, 89
natural logarithm, 225
odd, 144
quadratic, 24
range of, 21
rational, 25
real-valued, 20
sine integral, 212
value of, 20
fundamental theorem of calculus, 11, 212
first fundamental theorem, 188
second fundamental theorem, 190
Galileo, 155, 239
horizontal line test, 126
horizontal point of inflexion, 104
Huygens, Christiaan, 239
hyperbola, 31, 152, 242
hyperbolic functions, 243
double-angle formulae, 245
hyperbolic cosine, 240
hyperbolic sine, 240
sum and difference formulae, 245
improper integral
p-integrals, 205
convergence of, 203
indeterminate form(s), 43, 115, 233, 234
inequalities, 14
and absolute value, 18
triangle inequality, 18
inequality, 14
infinity, symbol for, 13
injective, 125
integers, 12
integral calculus, 11, 171
integral(s), 179
definite
limits of, 179
improper
p-integrals, 202
comparison test, 206
convergence of, 202
limit comparison test, 210
indefinite, 194
over (a,∞), 203
Riemann sum, 176, 179
table of, 253
integrand, 179
integration, 177
and differentiation, 193
by parts, 200
by substitution, 196
constant of, 194
dummy variable, 186
intermediate value theorem, 67
number of solutions, 112
intervals, 13
closed, 13
open, 13
inverse function(s), 127
hyperbolic functions, 248
inverses of, 248
restriction, 130
trigonometric functions, 132, 133
irrational numbers, 13
c©2020 School of Mathematics and Statistics, UNSW Sydney
285
l’Hoˆpital’s rule, 115, 233
l’Hoˆpital, Guillaume, 155
least upper bound property, 67, 70
Lebesgue, Henri, 171, 183
left-hand limits, 51
Leibniz, Gottfried, 11, 155, 171, 239
Leonard Euler, 227
limit(s), 11, 39
at infinity, 46
composition of function, 55
left-hand, 51
of continuous functions, 53
of the form lim
x→∞ f(x), 39
pinching theorem, 50, 56
right-hand, 51
two-sided, 52
ln, 225
logarithmic functions, 231
logarithms, 225
maximum
global, 107
local, 91, 105
second derivative test, 105
maximum-minimum theorem, 68, 69
mean value theorem, 97, 98
members (of set), 12
minimum
global, 107
local, 105
second derivative test, 105
natural domain, 20
natural logarithm, 227, 249
natural numbers, 12
Newton, Isaac, 11, 155, 171
oblique asymptote, 144, 145
odd function, 144
one-to-one correspondence, 124, 125
one-to-one function, 125
open interval, 13
parabola(s), 31, 152
parametric curves, 150
conic sections, 151
parameter, 151
parametrisation, 151
polar curves, 158
partition, 175, 178
piecewise continuous, 181
pinching theorem, 41, 50, 56
polar coordinate, 157
polar curves, 158
polynomial, 24
coefficients of, 24
degree of, 24
leading coefficient, 24
polynomial division, 146
primitive, 114
product rule, 84
quadratic function, 24
quartic, 24
quintic, 24
radians, 25
range of a function, 21
rates of change, 90
rational function, 25
rational numbers, 12, 67
real numbers, 12, 67
Riemann integrable, 179
Riemann integral, 177
area under the graph, 177
basic properties of, 184
Riemann sum
upper, 176, 178
Riemann sums, 193, 212
lower, 176, 179
Riemann, Bernhard, 171
right-hand limits, 51
Rolle’s theorem, 99
secant, 76
second derivative, 104, 107
set(s), 12
sine integral, 212
stationary point, 92
classification of, 104
subset, 14
sum and difference formulae, 245
tangent to a curve, 77
triangle inequality, 18
trigonometric functions, 25
inverse functions, 133
vertical asymptotes, 144
Weierstrass, 47
Zeno of Elea, 47
c©2020 School of Mathematics and Statistics, UNSW Sydney

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