MATHS 102 Course Book
Functioning in Mathematics
2021
Copyright for this work, including all diagrams and
images, is owned by the Department of Mathematics
at the University of Auckland.
2
Contents
1
Basic Algebra 1
1.1 Order of operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.1 Rules for order of operations . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Expanding brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2.1 Distributive law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2.2 Commutative law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2.3 Associative law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2.4 Examples of expanding brackets using the three laws . . . . . . . . . . . . .
1.3 Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.1 Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.2 Methods for factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4 Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4.1 Fractions by any other name . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4.2 Rule for multiplying fractions . . . . . . . . . . . . . . . . . . . . . . . . .
1.4.3 Rule for dividing with fractions . . . . . . . . . . . . . . . . . . . . . . . .
1.4.4 Cancellation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4.5 Simplifying a fraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5 More fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5.1 Rewriting fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5.2 The distributive law and fractions . . . . . . . . . . . . . . . . . . . . . . .
1.5.3 Rule for adding fractions where the denominators do not have a common factor
1.5.4 Where the denominators do have a common factor . . . . . . . . . . . . . .
1.6 Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6.1 Meaning of exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6.2 General rules of exponents . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6.3 Simplifying . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
9
10
13
13
14
14
15
17
17
18
20
20
20
21
21
22
23
23
23
24
24
27
27
28
28
2
Basic Algebra 2
2.1 Surds . . . . . . . . . . . . . . . . . . . . . . . .
2.1.1 Working with surds . . . . . . . . . . . . .
2.1.2 Simplifying surds involving only one term .
2.1.3 Surds involving more than one term . . . .
2.1.4 Adding with surds . . . . . . . . . . . . .
31
31
32
33
33
34
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4
CONTENTS
2.2
2.3
2.4
2.5
2.6
3
2.1.5 Surds in denominators . . . . . . . . . . . . . . . . . . . . . . . .
Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.1 Manipulating an equality . . . . . . . . . . . . . . . . . . . . . . .
2.2.2 What to do when . . . . . . . . . . . . . . . . . . . . . . . . . . .
Quadratic equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.1 Factorising quadratic expressions . . . . . . . . . . . . . . . . . .
2.3.2 Factoring: guessing and testing the guess . . . . . . . . . . . . . .
2.3.3 Solving quadratic equations by factoring . . . . . . . . . . . . . .
Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.1 Solving inequalities . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.2 Changing sign, and critical values . . . . . . . . . . . . . . . . . .
2.4.3 Multiplying or dividing by an unknown number . . . . . . . . . . .
2.4.4 Solving double inequalities . . . . . . . . . . . . . . . . . . . . . .
Quadratic equations 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.1 Completing the square . . . . . . . . . . . . . . . . . . . . . . . .
2.5.2 Solving a quadratic equation . . . . . . . . . . . . . . . . . . . . .
2.5.3 Quadratic formula . . . . . . . . . . . . . . . . . . . . . . . . . .
Factoring polynomials (read only) . . . . . . . . . . . . . . . . . . . . . .
2.6.1 Difference of two squares: factoring a2 − b2 . . . . . . . . . . . . .
2.6.2 Sums and differences of two cubes: factoring a3 + b3 , a3 − b3 . . .
2.6.3 Finding linear factors using the factor theorem . . . . . . . . . . .
2.6.4 Long division of polynomials . . . . . . . . . . . . . . . . . . . .
2.6.5 Simplifying polynomial fractions, and long division with remainder
2.6.6 Using the quadratic and other formulae . . . . . . . . . . . . . . .
Basic Algebra 3
3.1 Straight lines . . . . . . . . . . . . . . . . . . . . . . . .
3.1.1 Gradient of a line . . . . . . . . . . . . . . . . . .
−y1
3.1.2 Gradient of a line in algebraic form m = xy22 −x
. .
1
3.1.3 Gradient of a line in trigonometric form m = tan θ
y−y1
3.1.4 Equation of a line given a point and a gradient x−x
1
y−y1
y2 −y1
3.1.5 Equation of a line given two points: x−x
=
x2 −x1
1
3.1.6 Gradient/intercept form of line . . . . . . . . . . .
3.1.7 Parallel and perpendicular lines . . . . . . . . . .
3.1.8 Lines parallel to axes . . . . . . . . . . . . . . . .
3.2 Linear equations . . . . . . . . . . . . . . . . . . . . . . .
3.2.1 Interpretation . . . . . . . . . . . . . . . . . . . .
3.2.2 Method of solution . . . . . . . . . . . . . . . . .
3.2.3 No solution . . . . . . . . . . . . . . . . . . . . .
3.3 Relationships with graphs (read only) . . . . . . . . . . .
3.3.1 The graphs . . . . . . . . . . . . . . . . . . . . .
3.3.2 The parabola y = x2 . . . . . . . . . . . . . . . .
3.3.3 The rectangular hyperbola xy = 1 or y = x1 . . . .
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34
36
36
37
40
40
40
43
44
45
46
47
50
52
52
54
57
59
59
60
61
61
64
65
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67
67
67
69
69
70
71
72
73
75
76
76
76
78
80
80
81
81
CONTENTS
3.3.4
3.3.5
3.3.6
5
The absolute value graph y = |x| . . . . . . . . . . . . . . . . . . . . . . .
The circle x2 + y 2 = r2 . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
2
The ellipse xa2 + yb2 = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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84
85
88
88
89
Functions 1
4.1 Function notation . . . . . . . . . . . . . . . . . . . . . . . . . .
4.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . .
4.1.2 Domain and range of a function . . . . . . . . . . . . . .
4.1.3 Composite Functions . . . . . . . . . . . . . . . . . . . .
4.1.4 Inverse functions . . . . . . . . . . . . . . . . . . . . . .
4.1.5 Calculating inverse functions . . . . . . . . . . . . . . . .
4.1.6 Inverses and graphs . . . . . . . . . . . . . . . . . . . . .
4.2 Transformations (read only) . . . . . . . . . . . . . . . . . . . .
4.2.1 Shifting graphs . . . . . . . . . . . . . . . . . . . . . . .
4.2.2 Translations . . . . . . . . . . . . . . . . . . . . . . . . .
4.2.3 Applications . . . . . . . . . . . . . . . . . . . . . . . .
4.2.4 Stretches . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 Absolute value (read only) . . . . . . . . . . . . . . . . . . . . .
4.3.1 Definition of |x| . . . . . . . . . . . . . . . . . . . . . .
4.3.2 Working with the absolute value . . . . . . . . . . . . . .
4.3.3 Equalities and inequalities with the absolute value function
4.3.4 The triangle inequality . . . . . . . . . . . . . . . . . . .
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91
91
91
92
93
94
94
95
97
97
99
99
101
105
105
106
106
109
Trigonometry 1
5.1 Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.1 Angles between 0◦ and 90◦ . . . . . . . . . . . . . . . . . . . . . .
5.1.2 Angles between 90◦ and 180◦ . . . . . . . . . . . . . . . . . . . .
5.1.3 Angles between 180◦ and 270◦ . . . . . . . . . . . . . . . . . . . .
5.1.4 Angles between 270◦ and 360◦ . . . . . . . . . . . . . . . . . . . .
5.2 Radians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.1 Radians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.2 To estimate an angle given in radians in degrees . . . . . . . . . . .
5.2.3 To know the relationship between simple multiples of π and degrees
5.2.4 The formula for arc length . . . . . . . . . . . . . . . . . . . . . .
5.3 Special angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3.1 Special triangles . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.4 Graphs of trigonometric functions . . . . . . . . . . . . . . . . . . . . . .
5.4.1 Drawing trigonometric graphs . . . . . . . . . . . . . . . . . . . .
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111
111
111
112
113
114
116
117
118
118
119
120
120
122
122
3.4
4
5
2
3.3.7 The hyperbola xa2 − yb2 = 1 . . . . . . . .
3.3.8 The power (or exponential) function y = ax
Non-linear simultaneous equations (read only) . . .
3.4.1 Other possibilities for number of solutions .
3.4.2 Examples . . . . . . . . . . . . . . . . . .
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82
82
83
6
CONTENTS
5.5
5.6
6
7
5.4.2 Drawing trigonometric graphs .
Trigonometric equations . . . . . . . .
5.5.1 Inverse trigonometric functions
5.5.2 Basic sin equation (+) . . . . .
5.5.3 Basic sin equation (−) . . . . .
5.5.4 Basic cos equation (+) . . . . .
5.5.5 Basic cos equation (−) . . . . .
5.5.6 Basic tan equation (+) . . . . .
5.5.7 Basic tan equation (−) . . . . .
Harder equations . . . . . . . . . . . .
5.6.1 Domain not 0 ≤ x ≤ 2π . . . .
5.6.2 Multiple angles . . . . . . . . .
5.6.3 Composite angles . . . . . . . .
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Differentiation 1
6.1 Slopes and simple derivatives . . . . . . . . . . . . . . . . .
6.1.1 Slopes of functions . . . . . . . . . . . . . . . . . .
6.1.2 Slope using function notation . . . . . . . . . . . .
6.1.3 Easier methods of finding derivatives . . . . . . . .
6.1.4 First rules of derivatives . . . . . . . . . . . . . . .
6.1.5 Alternative notation for differentiation . . . . . . . .
6.1.6 Functions involving x in the denominator or in roots
6.2 Points and intervals of interest for a function . . . . . . . . .
6.2.1 Does the first derivative of a function always exist? .
6.2.2 Points and intervals of interest for a function . . . .
6.2.3 Critical points: derivative is zero or does not exist . .
6.3 Graphing a function . . . . . . . . . . . . . . . . . . . . . .
6.3.1 Graphing functions . . . . . . . . . . . . . . . . . .
6.3.2 Must you do all this for every graph? . . . . . . . .
6.4 Rates of change (read only) . . . . . . . . . . . . . . . . . .
6.4.1 Rate of change . . . . . . . . . . . . . . . . . . . .
6.4.2 Finding formula for rate of change . . . . . . . . . .
Trigonometry 2
7.1 Addition formulae . . . . . . . . . . . .
7.1.1 Major values . . . . . . . . . .
7.1.2 Addition formulae . . . . . . .
7.2 Multiple angles . . . . . . . . . . . . .
7.2.1 Double angle formulae . . . . .
7.2.2 Half angle formulae . . . . . .
7.3 Sums and products . . . . . . . . . . .
7.3.1 Replacing a product with a sum
7.3.2 Replacing a sum by a product .
7.4 Basic identities (read only) . . . . . . .
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124
127
127
127
128
129
130
131
131
133
133
134
134
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135
135
136
137
139
139
142
142
144
144
146
149
151
151
155
157
157
158
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161
161
161
161
164
164
165
166
166
166
168
CONTENTS
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168
168
169
169
Differentiation 2
8.1 Derivatives of trigonometric functions . . . . . . . . . . . .
8.1.1 Finding the critical points of a trigonometric function
8.2 Chain rule 1 . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2.1 Functions of functions (composite functions) . . . .
8.2.2 Chain rule . . . . . . . . . . . . . . . . . . . . . . .
8.3 Chain rule 2 . . . . . . . . . . . . . . . . . . . . . . . . . .
8.3.1 Methods in our madness . . . . . . . . . . . . . . .
8.3.2 Chain rule Formula 1 . . . . . . . . . . . . . . . . .
8.3.3 Chain rule Formula 2 . . . . . . . . . . . . . . . . .
8.3.4 Other trigonometric derivatives (read only) . . . . .
8.4 Product rule . . . . . . . . . . . . . . . . . . . . . . . . . .
8.4.1 Which rule to use when differentiating . . . . . . . .
8.4.2 Differentiating products of functions: product rule .
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171
171
175
178
178
179
182
182
182
183
185
187
187
187
Functions 2
9.1 Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.1.1 Applications to solving equations . . . . . . . . . . .
9.2 The exponential function and the natural logarithmic function
9.2.1 The exponential function . . . . . . . . . . . . . . . .
9.2.2 The natural logarithm function . . . . . . . . . . . . .
9.2.3 Converting powers of a to powers of e . . . . . . . . .
9.2.4 Converting loga to loge = ln . . . . . . . . . . . . . .
9.2.5 Working with these function . . . . . . . . . . . . . .
9.3 Differentiating exp and natural log . . . . . . . . . . . . . . .
9.3.1 Differentiating the exponential function . . . . . . . .
9.3.2 Differentiating the natural log function . . . . . . . .
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191
191
195
198
198
199
200
200
201
202
202
203
10 Differentiation 3
10.1 Quotients of functions . . . . . . . . . . . . . . . . . . . . . . .
10.2 Summary of differentiation rules . . . . . . . . . . . . . . . . .
10.2.1 Sums, products, quotients and compositions of functions
10.2.2 Summary of differentiation rules for basic functions . .
10.2.3 Shortcuts for the chain rule . . . . . . . . . . . . . . . .
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207
207
210
210
211
211
11 Summation
11.1 Sigma notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.1.1 Sequences – functions on the set of integers . . . . . . . . . . . . . . . . . .
11.1.2 Sigma notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
213
213
213
214
7.5
8
9
7.4.1 Reciprocal functions . . .
7.4.2 Pythagoras . . . . . . . .
Identities (read only) . . . . . . .
7.5.1 What to do if you are stuck
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8
CONTENTS
11.1.3 Working with sigma notation . . . . . . . . . . . . . . . . . . . . . . . . . . 215
11.1.4 Adding with sigma notation . . . . . . . . . . . . . . . . . . . . . . . . . . 217
12 Integration
12.1 Anti-differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12.1.1 What is integration and why integrate? . . . . . . . . . . . . . .
12.1.2 Integrating polynomials . . . . . . . . . . . . . . . . . . . . .
12.1.3 Integrating expressions involving fractional and negative powers
12.1.4 Turning an expression into an integrable form . . . . . . . . . .
12.2 Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12.2.1 Integration by substitution . . . . . . . . . . . . . . . . . . . .
12.3 Trigonometric functions . . . . . . . . . . . . . . . . . . . . . . . . .
12.3.1 Integrating the basic trigonometric functions . . . . . . . . . .
12.3.2 Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12.4 Exponentials and logarithms . . . . . . . . . . . . . . . . . . . . . . .
12.4.1 Integrating the basic functions . . . . . . . . . . . . . . . . . .
12.4.2 Polynomial integration and logs . . . . . . . . . . . . . . . . .
12.4.3 Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12.4.4 A short method . . . . . . . . . . . . . . . . . . . . . . . . . .
13 Answers to exercises
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221
221
221
222
223
223
225
225
228
228
228
230
230
230
230
231
233
Chapter 1
Basic Algebra 1
1.1
Order of operations
Aim. To learn how to make sense of a mathematical statement.
Vocabulary and conventions
term
Refers to a part of a mathematical statement, usually those parts
separated by +, −, or = signs. We can say that
2x + 3 = 8
has three terms: 2x, 3, and 8.
brackets ()
Used to collect several terms together. For example, the brackets in
5 − (3x − 1)
collect 3x and −1 into one term, and means 5 subtract the difference
of 3x and 1. Similarly
2(5x + 2)
means 2 times the sum of 5x and 2. When there is no sign between
a number and a bracket, the convention is that multiplication is
intended:
2(3x + 1) means 2 × (3x + 1),
(6a + 5)4 means (6a + 5) × 4.
operation
Any one of the things we do with numbers – we multiply (find a
product), divide (find a quotient), add (find a sum), subtract (find a
difference), find squares and square roots. These are all operations,
and there are others.
order
operations
of
A set of conventions (rules) about which operation gets done first,
in order to avoid confusion about meaning.
9
10
CHAPTER 1. BASIC ALGEBRA 1
1.1.1
exponent, power
Names for the operations that create squares, cubes, square roots
etc, as in
√
1
23 , 510 , 9, 25 2 .
The numbers written as superscripts are exponents (or powers).
Briefly, their meaning is:
√
1
a2 = a × a, a3 = a × a × a, a 2 = a.
√
1 √
As 9 is another way of writing 9 2 , is also an exponent.
constant
Refers to a term or part of a term which has a fixed value. Therefore
2x + 3
has constants 2 and 3. Often constants are called by a letter name,
usually a, b, or c. Their values are fixed, but unknown. For example,
2x + a
has constants 2 and a.
variable
Refers to a term or part of a term which is called by a letter name,
usually x, y, or z. Their values are not fixed, they vary. For example,
2x + 3
has a variable x. The result of adding 2x+3 will depend on the value
chosen for x, and all real values are possible. Solving the equation
2x + 3 = 8
means finding what value the variable x can take for the equation to
be true.
coefficient
Refers to the constant in a term which involves a variable. The
equation
2x + 3 = 8
has 2 as the coefficient of x and
5xy + ay 2
has 5 as the coefficient of xy and a as the coefficient of y 2 .
Rules for order of operations
Operations are carried out in the following order:
1. brackets,
3. multiplication and division,
2. exponents,
4. addition and subtraction.
You can remember the order if you remember the mnemonic bedmas.
1.1. ORDER OF OPERATIONS
11
1. Brackets
Carry out any operations inside a bracket first.
5 − 3(2 + 2)2
First do any operations inside the brackets (remembering the order given above):
(2 + 2)2 = 42 .
Brackets can be removed when there is only a single term inside them, and we can begin on the
next operation:
5 − 3(2 + 2)2 = 5 − 3 × 42 .
2. Exponents
We treat a term or part of a term which has an exponent as if it has brackets around it, that is
we carry out the exponent operation before doing any other operation on the term:
5 − 3 × 42 = 5 − 3 × (4 × 4) = 5 − 3 × 16,
2 × 53 = 2 × (5 × 5 × 5) = 2 × 125,
32 − (2 + 2) = (3 × 3) − 4 = 9 − 4.
If we find an exponent in a bracket with other terms, then we apply the order of operation rules
on the terms inside the bracket, exponents now being first:
5 + (34 − 2) = 5 + (3 × 3 × 3 × 3 − 2) = 5 + (81 − 2) = 5 + 79,
√
5 − (2 + 9) = 5 − (2 + 3) = 5 − 5.
3. Multiplication and division
After dealing with brackets and exponents, we can multiply and divide:
5 − 3 × 16 = 5 − 48,
2 × 53 = 2 × 125 = 250,
(15 ÷ 5 + 4) × 2 = (3 + 4) × 2 = 7 × 2 = 14,
52 × 3 = 25 × 3 = 75.
4. Addition and subtraction
When there are no longer any brackets, any exponents, or any multiplication or division left to
do, we compute the remaining addition and subtraction.
32 − (2 + 2) = 9 − 4 = 5,
5 + (34 − 2) = 5 + (81 − 2) = 5 + 79 = 84,
√
5 − (2 + 9) = 5 − (2 + 3) = 5 − 5 = 0.
12
CHAPTER 1. BASIC ALGEBRA 1
Practice (Section 1.1). Simplify the following:
1. 2(22 + 3)2 − 4 + 2,
3. 62 ÷ 9 + 3 × 42 ,
2. 5 − (3 + 1) ÷ 2,
4. 2(18 ÷ 2 + 3) − (21 + 15) 2 .
1
1.2. EXPANDING BRACKETS
1.2
13
Expanding brackets
Aim. To learn how to eliminate brackets which contain unknown numbers.
Vocabulary and conventions
1.2.1
expand
Rewrite terms involving brackets so that the brackets are eliminated.
simplify
Rewrite a term or collection of terms in a simpler (or maybe just
shorter, or easier to read) form.
like terms
Used to describe terms which have the same unknown numbers
(with the same exponents), but may have different known
coefficients. For example, xy 2 and 2xy 2 are like terms, and so
may be added together easily. Also 3ab and 256ab are like terms.
However x2 and 3x are not like terms because the powers of x are
different. We can simplify
xy 2 + 2xy 2 = 3xy 2 and 256ab − 3ab = 253ab,
but x2 + 3x cannot be simplified in this way since its terms are not
like terms.
Distributive law
Brackets containing known numbers can easily be expanded using the order of operation rules:
(4 + 5) ÷ 3 = 9 ÷ 3 = 3.
However if the bracket contains unknown numbers or variables, that is numbers represented by letters,
then we must recall the basic laws of mathematics. The first of these is the distributive law:
a(b + c) = ab + ac,
(b + c)a = ba + ca,
Since a(b + c) means a multiplied by the contents of the bracket, every term in the bracket must be
multiplied by a when expanding.
Example.
1. x(x − 3) = x × x + x(−3),
2. −y(3 − y + z) = (−y) × (3 + (−y) + z) = (−y)3 + (−y)(−y) + (−y)z,
3. −(3 − x) = (−1) × (3 − x) = (−1) × 3 + (−1) × (−x).
14
CHAPTER 1. BASIC ALGEBRA 1
1.2.2
Commutative law
When we have expanded a bracket in this way, we often find numbers are arranged in unconventional
ways. We may arrive at
y3,
whereas the normal convention for the arrangement of terms is to have the known numbers first and
the unknown numbers (in alphabetical order) afterwards. The commutative law says:
ab = ba,
a + b = b + a,
(multiplication)
a − b = −b + a, (addition)
which enables us to write
y3 = 3y.
1.2.3
Associative law
We also find we get terms like
3(−x),
which is not in the conventional form. The associative law says that
(ab)c = a(bc),
(multiplication)
(a + b) + c = a + (b + c), (addition)
and this law and the commutative law enable us to swap things around:
3(−x) = 3(−1 × x) = (3 × −1)x = (−1 × 3)x = −1(3x) = −3x.
With known numbers you obey these laws almost instinctively:
3×5=5×3
and
3 × −5 = 5 × −3.
However when unknown numbers get involved, sometimes you need to think. Does what you have
just done obey the three laws? If you replaced known numbers in the equation and did the same
actions, would you get a correct result? If not, you have probably not used the laws correctly. We can
now finish the examples above:
1. x(x − 3) = x × x + x(−3) = x2 − 3x,
2. −y(3 − y + z) = (−y)3 + (−y)(−y) + (−y)z = −3y + y 2 − yz,
3. −(3 − x) = (−1) × (3 − x) = (−1) × 3 + (−1) × (−x) = −3 + x.
1.2. EXPANDING BRACKETS
1.2.4
15
Examples of expanding brackets using the three laws
1. −(−a + b) = (−1)(−a) + (−1)(b) = a − b,
2. x(−y + z) = x(−y) + xz = xz − xy,
3. −x(−y + z) = (−x)(−y) + (−x)z = xy − xz,
4. Expand (a + b)(c + d):
Using the distributive law, and expanding the left bracket first, we get:
(a + b)(c + d) = a(c + d) + b(c + d)
= ac + ad + bc + bd,
or expanding the right bracket first:
(a + b)(c + d) = (a + b)c + (a + b)d
= ac + bc + ad + bd
= ac + ad + bc + bd.
That last line used the commutative law for addition (bc + ad = ad + bc) to show that the two
results are the same.
5. This is a special case of the previous example:
(a + b)2 = (a + b)(a + b)
= a(a + b) + b(a + b)
= a2 + ab + ba + b2
= a2 + 2ab + b2 .
Notice that the last line resulted from using the commutative law for multiplication (ba = ab)
and then simplifying – in this case adding like terms, ab + ab = 2ab.
Do a check with known values to see if this last result is correct. We substitute each side with
a = 3 and b = 2:
left hand side: (a + b)2 = (3 + 2)2 = 52 = 25,
right hand side: a2 + 2ab + b2 = 9 + 2(6) + 4 = 25.
Since both sides give the same result using known values we can have some trust that we have
used the three laws correctly.
6. Taking the previous example a step further:
(a + b)3 = (a + b)(a + b)2
= (a + b)(a2 + 2ab + b2 )
= a × a2 + a × 2ab + a × b2 + b × a2 + b × 2ab + b × b2
= a3 + 2a2 b + ab2 + ba2 + 2ab2 + b3
= a3 + 3a2 b + 3ab2 + b3 .
16
CHAPTER 1. BASIC ALGEBRA 1
In the process of expanding the brackets, we have used all three laws again and again, and
finally simplified by collecting like terms together.
Practice (Section 1.2). Expand and simplify where possible:
1. 2(x − 3),
3. x(x − 2y),
5. −y(y + 3),
7. (x − y)2 ,
2. 5(a + 2b),
4. −(x − 1),
6. (x − y)(2x − y),
8. −(x−y)(x+2y).
1.3. FACTORS
1.3
17
Factors
Aim. To learn how to factor simple mathematical statements.
Vocabulary and conventions
product
The result of multiplying one term by another. The number 6 is a
product of 2 and 3. That is, 2 × 3 = 6.
factor
Since
2 × (some number) = 6,
we say that 2 is a factor of 6. Then 3 is also a factor of 6. Because
a × b2 = ab2 ,
2
ab is a product of a and b2 .
Equally, we can say that a and b2 are both factors of ab2 . The
number 6 has more factors than just 2 and 3, it also has 1 and 6.
Similarly, ab2 has the factors 1, b, ab and ab2 , as well as a and b2 .
1.3.1
common
factors
Two terms are said to have common factors if there is some factor
other than 1 that they both have. Since 6 and 9 both have 3 as a
factor, we say that 3 is a common factor of 6 and 9. Similarly, ab2
and abc3 both have a, b, and ab as factors, so that a, b, and ab are
common factors of ab2 and abc3 .
highest
common
factor
Of all the common factors of two numbers, the highest common
factor is the one which has all the others are factors. The numbers
8 and 12 have the common factors 2 and 4. Of these, 2 is a factor of
4, so 4 is the highest common factor of 8 and 12. ab2 and abc3 have
the common factors a, b, ab, and as a and b are both factors of ab, it
follows that ab is the highest common factor of ab2 and abc3 .
Factoring
We can easily add or subtract like terms. However if terms are unlike, we may be able to add them
by factoring. We can only do this if they have a common factor. Factoring two or more terms means
rewriting them as one term.
18
CHAPTER 1. BASIC ALGEBRA 1
1.3.2
Methods for factoring
Factoring two or more terms is a reverse action of expanding brackets. It does not change the meaning
of the expression, it merely rewrites it in perhaps a more useful form. Since by the distributive law,
a(b + c) = ab + ac and a(b − c) = ab − ac,
then
ab + ac = a(b + c)
and ab − ac = a(b − c).
The last line is just the reverse action of the first, and factors the expressions. To completely factor an
expression, say ab2 − abc3 :
1. Find the highest common factor:
the highest common factor of ab2 and abc3 is
ab.
2. Rearrange the terms (using the three laws) so that the highest common factor is at the left (or at
the right if desired) of each term:
ab2 − abc3 = ab × b − ab × c3 .
3. Then put this highest common factor outside a set of brackets:
ab2 − abc3 = ab × b − ab × c3 = ab(
).
4. Put the sum (or difference) of the remaining factors of the terms inside the brackets:
ab2 − abc3 = ab × b − ab × c3 = ab(b − c3 ).
5. Have you factored completely? Check that the terms in the bracket have no more common
factors (they will have more only if you did not find the highest common factor).
Have you factored correctly? Check by expanding your result. Is it the same as the expression
you started with? It should be.
Example.
1. x2 − 3x = x × x − x × 3 = x(x − 3),
2. rx2 + rxy − r2 xy 2 = rx × x + rx × y − rx × ry 2 = rx(x + y − ry 2 ),
3. 3x − 12 = 3 × x − 3 × 4 = 3(x − 4),
4. −2x + 2y + 2 = 2 × (−x) + 2 × y + 2 × 1 = 2(−x + y + 1),
5. 2a(b + c) + 4c(b + c) = 2(b + c) × a + 2(b + c) × 2c = 2(b + c)(a + 2c).
1.3. FACTORS
19
Practice (Section 1.3). Factor:
1. 4x − 6,
5. xy − x3 ,
2. 5ab + 20b,
6. xy 2 − 2y + y 2 ,
3. x2 + 2xy,
7. a2 − ab2 ,
4. 4ab − 6b2 ,
8. a(a + 2b)2 − (a + 2b).
20
1.4
CHAPTER 1. BASIC ALGEBRA 1
Fractions
Aim. To learn how to multiply and divide with algebraic fractions.
Vocabulary and conventions
1.4.1
algebraic
fractions
Fractions that contain unknown numbers, that is, fractions where
some numbers are represented by letters.
numerator
Top line of a fraction, that is, the number being divided.
denominator
Bottom line of a fraction, that is, the number doing the dividing.
quotient
The result of dividing one term by another.
cancellation
Elimination of common factors in numerator and denominator.
fractions and
brackets
The numerator and denominator of fractions are treated as if they
are enclosed in brackets, even though there are none showing.
Fractions by any other name
There are a number of ways of writing fractions. The following equalities illustrate how many
different expressions can have the same meaning:
1=
1
a
= ,
1
a
a
a= ,
1
1
= 1 ÷ b,
b
a
1
a 1
a÷b= =a× = × ,
b
b
1 b
a
b
b
1÷ =1× = ,
b
a
a
1.4.2
a 6= 0,
b 6= 0,
b 6= 0,
a 6= 0, b 6= 0.
Rule for multiplying fractions
Multiply the numerators to create the new numerator; multiply the denominators to create the new
denominator:
a c
ac
× = .
b d
bd
This rule comes from the three laws (see Section 1.2).
1.4. FRACTIONS
21
Example.
1.
2.
3.
1.4.3
a
a c
a×c
ac
×c= × =
= ,
b
b 1
b×1
b
3 5y
3 × 5y
15y
×
=
=
,
x
2
x×2
2x
2
1×2
2
1
×
=
=
.
z 3x
z × 3x
3xz
Rule for dividing with fractions
We can use the following rules to rewrite division of fraction as multiplication:
a c
a d
ad
÷ = × =
b d
b
c
bc
Example.
2.
3.
1.4.4
1.
or
a/b
c/d
=
a c
a d
ad
.
÷ = × =
b d
b
c
bc
a
a 1
a×1
a
÷c= × =
= ,
b
b
c
b×c
bc
3 5y
3
2
3×2
6
÷
= ×
=
=
,
x
2
x 5y
x × 5y
5xy
3
2/x
=3÷
2
3 x
3x
= × =
.
x
1 2
2
Cancellation
ac
a
The numerator and denominator of
have a common factor of a, and since = 1, we can write the
ba
a
fraction (applying the multiplication law backwards) as
ac
ac
a c
c
c
=
= × =1× = .
ba
ab
a b
b
b
See that the common factor has been eliminated from the expression. An abbreviated way of doing
the same example is
ac
= c.
b
ab
Example.
1.
a
a b
ab
a
×b= × =
= = a.
b
b 1
1
b × 1
a
ab
a
We could abbreviate this example to × b =
× b = a. But be
= a, or even further to
b
1b
b1
warned, do not abbreviate like this until you are confident and comfortable with the reasons for
your actions.
2. 3x ×
× 5
5
3x
5
3x
5
=
×
=
= .
× 2
6x
1
3x × 2 3x
2
22
CHAPTER 1. BASIC ALGEBRA 1
1.4.5
Simplifying a fraction
The examples in the above section had results in a simple form. Often it is not so easy to decide
how best to simplify a fraction, that is, in what form to leave it. In general, if the numerator and
denominator are factored, and you expect to continue working with the result, leave the factors
unexpanded. This is because it is much easier to work with an expression that is already factored.
Cancel factors that are common to both numerator and denominator.
Example. Remember that the numerators and denominators must be treated as if they were in brackets.
1. We give two ways to simplify the same expression:
(a)
2x + 3 x + 2
(2x + 3)(x + 2)
×
=
,
5+y
y
(5 + y)y
(b)
2x + 3 x + 2
(2x + 3)(x + 2)
2x2 + 7x + 6
,
×
=
=
5+y
y
(5 + y)y
5y + y 2
2.
2x + 4 x + 2
2x + 4
y
(2x + 4)y
2(x + 2)y
2y
÷
=
×
=
=
=
,
5+y
y
5+y
x+2
(5 + y)(x + 2)
(5 + y)(x + 2)
5+y
3.
x
x−y x(x − y)
x−y
×
=
=
.
x+y
x
(x + y)
x
x+y
Practice (Section 1.4). Solve and simplify where possible:
1.
x 7
× ,
3 y
4. (4x − 4y) ÷
2.
a
b
× 2 ,
1 a c
5.
3.
c
4a
×
,
2a (3c + 1)
6.
2x
,
9y
7.
(a + b) (a + b)
÷
,
3c
6b
5
y
÷
,
4x 5xz
8.
ab + abc b + bc
×
,
abc
a
2x 2x
÷
,
y
y
9.
x/4
3/2x
.
1.5. MORE FRACTIONS
1.5
23
More fractions
Aim. To learn how to add and subtract algebraic fractions.
Vocabulary and conventions
1.5.1
common
denominator
Two fractions have a common denominator when their
denominators are identical.
The fractions 21 and 32 have common denominators, as do a1 and ab .
1
However ab
and dc do not have a common denominator.
least common
multiple
A common multiple of two numbers is one of which both numbers
are factors. The numbers 5 and 3 have the common multiples 15, 30,
45, and so on. The least of these is 15, so 15 is the least common
multiple of 5 and 3.
The least common multiple of a and b is ab, that of 2x and 3y is
6xy, and that of 5a and 10b is 10ab.
Rewriting fractions
We noted in Section 1.4 that there are many different ways of writing an expression without changing
its meaning. In that section we cancelled factors that were common to both the numerator and
denominator. In this section we will need to create common factors in order to create a common
denominator. See that
a=
1.5.2
a
a
a b
ab
= ×1= × =
1
1
1 b
b
and
a
a
a b
ab
= ×1= × = .
c
c
c b
bc
The distributive law and fractions
The distributive law applied to fractions gives
1
1
1
b
c
b+c
= (b + c) × = b × + c × = + .
a
a
a
a
a a
We need the distributive law (going from right to left) to add fractions, and the line above with two
equations shows that we need a common denominator before we can use the law. That is, we can only
add or subtract two fractions when they have a common denominator, and therefore we must find one.
The common denominator should be the least common multiple of the denominators of the original
fractions.
24
CHAPTER 1. BASIC ALGEBRA 1
1.5.3
Rule for adding fractions where the denominators do not have a common factor
a c
ad bc
ad + bc
+ =
+
=
b d
bd bd
bd
We now list the steps to add fractions:
1. Rewrite the fractions so that they have a common denominator (the least common multiple of
the denominators).
To write
3 y
+
x 4
as one fraction, note that 4 and x do not have a common factor, so 4x will be the common
denominator. We need to multiply the first fraction by 44 , which is the same as 1, and the
second by xx (again, the same as 1) to get 4x as the denominator for both:
3 y
3
y
3 4 y x
12 xy
+ = ×1+ ×1= × + × =
+
.
x 4
x
4
x 4 4 x
4x 4x
2. Add the numerators to create the new numerator and write the new denominator, the common
denominator:
12 + xy
12 xy
+
=
.
4x 4x
4x
1.5.4
Where the denominators do have a common factor
Be aware that if the denominators have factors in common, then the least common multiple will be
less than the product of the two denominators. You can use the method above, but will find that
the result is more complicated than necessary because its numerator and denominator will have a
common factor. To avoid this:
1. Rewrite the fractions so that they have the least common denominator (the least common
multiple of the denominators).
To write
5
y
+
2x 6
as a single fraction, note that 6 = 2 × 3, therefore 2x and 6 have the factor 2 in common. We
see that the least common multiple of 6 and 2x will be 3 × x × 2 = 6x (the product of the
common factor and the remaining factors):
y
5
y
5×3
y×x
5
+ =
+
=
+
.
2x 6
2×x 2×3
2×x×3 2×3×x
2. Add the numerators to create the new numerator and write the new denominator, the common
denominator:
5×3
y×x
15 xy
15 + xy
+
=
+
=
.
2×x×3 2×3×x
6x 6x
6x
1.5. MORE FRACTIONS
25
The difficulty in adding algebraic fractions is usually in creating the common denominator. If one
term is not a fraction we will have to rewrite it as one. We have to find the least common multiple of
the two (or more) denominators, and then rewrite so the fractions have a common denominator. Only
then do we actually add or subtract.
Example.
2+x
y
First we need to write x as a fraction:
1. x −
x−
2+x
x 2+x
= −
.
y
1
y
Since the least common multiple of 1 and y is y, we rewrite these fractions with the common
denominator y. The first fraction needs its denominator, (which is 1) to be multiplied by y,
so we must multiply its numerator by y as well. The second fraction already has y as the
denominator, so it needs no change:
x 2+x
x y 2+x
xy 2 + x
−
= × −
=
−
.
1
y
1 y
y
y
y
Now we add the numerators to create the new numerator, and write the common denominator
as the new denominator. Note that numerators and denominators need to be treated as if they
were in brackets:
xy 2 + x
xy − (2 + x)
xy − 2 − x
−
=
=
.
y
y
y
y
2.
x+1 x−1
−
2
6
The least common multiple of 2 and 6 is 6. Since 2×3 = 6, we need to multiply the first fraction
by 33 . That is, we multiply the first fraction top and bottom by 3 so that its new denominator
will be 6. The second fraction needs no work as its denominator is already 6:
3(x + 1) x − 1
3(x + 1) − (x − 1)
2x + 4
2(x + 2)
x+2
x+1 x−1
−
=
−
=
=
=
=
.
2
6
3×2
6
6
6
2×3
3
Remember to treat the numerators and denominators as if they were in brackets, and check if
the result can be simplified.
3.
y
1
+
x+y x−y
The denominators have no common factors, so their least common multiple must be (x + y)(x − y).
Multiply the first fraction top and bottom by (x − y) and the second fraction top and bottom by
(x + y):
y
1 × (x − y)
(x + y)y
x − y + (x + y)y
1
+
=
+
=
.
x+y x−y
(x + y)(x − y) (x + y)(x − y)
(x + y)(x − y)
26
CHAPTER 1. BASIC ALGEBRA 1
4.
2
y+1
−
xy
yz
The terms xy and yz have a factor of y in common, leaving x and z respectively that are not
common to both. A common denominator for the two fractions is x × y × z. The first fraction
needs to be multiplied top and bottom by z, the second top and bottom by x:
2
y+1
2z
x(y + 1)
2z − x(y + 1)
−
=
−
=
.
xy
yz
xyz
xyz
xyz
5.
4
3
2x
+
−
x xy
y
The common denominator of these three fractions is xy:
3
2x
4y
3
2x2
4y + 3 − 2x2
4
+
−
=
+
−
=
.
x xy
y
xy xy
xy
xy
Practice (Section 1.5). Solve, and simplify where possible:
1.
2
3
+
,
5a 10a
5.
x+1 2
+ ,
x
3
2.
b
c
− ,
3 3a
6.
5
1
−
,
3a2 b 6ab2
3.
2+x y−1
−
,
x
y
7.
4.
3+y
2z
4 + 3x
−
+
,
y
xy
z
1
1
− ,
a−b a
8. 5x +
2x
3y
+
.
y
x+y
1.6. EXPONENTS
1.6
27
Exponents
Aim. To learn how to use exponents with both known and unknown values.
Vocabulary and conventions
reciprocal
1.6.1
The multiplicative inverse, i.e. if ab = 1, then b is the reciprocal of
a, and b = a1 . Also, a is the reciprocalof b and a = 1b .
The reciprocal of 2 is 12 , since 2 × 12 = 1.
Meaning of exponents
The table below sets out the meaning of some common exponents or powers.
given any real number a,
this expression
means the same as
and in words means
a2
a×a
a multiplied by itself twice
a3
a×a×a
a multiplied by itself three times
an
a × a × ··· × a
a multiplied by itself n times
a1
a
a0
1
√
a
3
√
√ 3
a3 or ( a)
a3
1
√
3
2
3
√
3
a−1
1
a
reciprocal of a, where a 6= 0
a−2
1
a2
reciprocal of a2 , where a 6= 0
√
reciprocal of a, where a > 0
a
1
2
a2
a
1
a− 2
this holds for every a 6= 0
a
square root of a3 , where a ≥ 0, so
cube of square root of a
cube root of a
√
3
2
a2 or ( a)
√1
a
the (positive) square root of a, where
a≥0
cube root of a2 , also square of cube
root of a
28
CHAPTER 1. BASIC ALGEBRA 1
1.6.2
General rules of exponents
The forms below
mean the same as these
forms
Examples / notes
ab ac
ab+c
a2 a3 = (a × a) × (a × a × a)
= a × a × a × a × a = a5
ab
or ab ÷ ac
ac
ab−c
a2
a×a
1
= = = a−1 = a2−3
3
a
a × a × a
a
(ab)c
ac bc
(ab)2 = ab×ab = a×a×b×b = a2 b2
(ab )c
abc
(a2 )3 = a2 × a2 × a2 = a × a × a ×
a×a×a
= a6 = a2×3
a−b
ac
or ac ÷ bc
bc
√
b
b
1
a
a
a
1.6.3
1
ab
a c
(reciprocal of ab )
22
2×2
2 2
=
= × =
2
3
3×3
3 3
ab
(b-th root of a)
1
2
1
= =1
2
1
2
2
3
Simplifying
A lot of mathematics involves rewriting terms or collections of terms in simpler forms. The new form
may be easier to use, or perhaps the meaning will be clearer.
Example.
3
1. 22 = 22×3 = 26 = 64,
2. (−3)3 × 31 = (−1 × 3)3 × 31 = (−1)3 × 33 × 31 = (−1) × 33+1 = −34 = −81,
2
2x 2x
2x
2x × 2x
4x2
3.
=
×
=
= 2 ,
y
y
y
y×y
y
4.
x4
1
= x4−6 = x−2 = 2 ,
x6
x
1.6. EXPONENTS
29
5. (3x)3 = 33 x3 = 27x3 ,
p
1
1
1
1
6. 16x4 y 2 = 16x4 y 2 2 = 16 2 x4× 2 y 2× 2 = 4x2 y, where y ≥ 0,
7. We give two ways to simplify the same expression:
3a−2 b5
3
a−2
b5
b7
1−2 −2−3 5−(−2)
−1 −5 7
=
×
×
=
3
a
b
=
3
a
b
=
,
9a3 b−2
32
a3
b−2
3a5
b5−(−2)
3a−2 b5
3
a−2
b7
(b)
=
=
,
−2
3
−2
9a b
3a5
39a3−(−2)
b
2
3
8. a2 b3
ab2 = a2×2 b3×2 a3 b2×3 = a4+3 b6+6 = a7 b12 .
(a)
Practice (Section 1.6). Simplify:
√
3
1. 2x2 ,
3.
2
2. 23 y −2 ,
3
4. −3a2 b ,
36a2 b6 , where a, b ≥ 0,
5. 3x2 y × 5xy 3 ,
p
6. 3 8x3 y 6 .
30
CHAPTER 1. BASIC ALGEBRA 1
Chapter 2
Basic Algebra 2
2.1
Surds
Aim. To learn how to use surds correctly.
Vocabulary and conventions
surd
roots of negative
numbers
√
Refers to the root of a number, either a for the square root of a
√
(where a must be a positive number or 0), or n a for the n-th root
of a (where a can be negative if n is an odd number).
We are working only in real numbers. Because the square root of
a negative number is imaginary, that is, not a real number, a (real)
square root of a negative number does not exist.
The (real)√cube root of a negative number does exist and is negative.
See that 3 −8 = −2, because (−2)3 = (−1)3 23 = −8.
Similarly, the (real) cube root of a positive number is positive.
conjugate
The conjugate of the sum or difference of two square roots is created
by changing the sign between them.
√
√
√
√
The conjugate of x + y, where x, y ≥ 0 is√ x − y; √
√
√
√
√
the conjugate of x is x (see this as x + 0 and x − 0)
√
√
√
√
and the conjugate of x − y is x + y.
The conjugate is very
useful when dealing with square roots,
√
√ √
√
because
x+ y
x − y = x − y.
negative square
roots
You will often be asked to simplify an expression like
31
√
4.
32
CHAPTER 2. BASIC ALGEBRA 2
√
1
While you know that (−2)2 is 4, the expression 4, or 4 2 is defined
as the positive square root, so 2 is the only answer.
√
1
2
The negative square
root
is
written
−
√
√
√4, or −(4 ).
The expression ± 4, means 4 or − 4.
If asked to ‘solve for x when x2 = 4’, this would require you to
return all possible values of x for which the equation is true. That
is, x may be either the positive or the negative square root of 4, so
that ‘x = 2 or x = −2’ is the expected answer.
2.1.1
Working with surds
Surds are another way of writing exponents. For x ≥ 0, we can write roots in the following way:
√
√
1
1
1
1
x = x 2 = (x) 2 and n x = x n = (x) n .
All the rules of exponents apply to surds, and if an action is incorrect for exponents, it will be incorrect
for surds. Both exponents and surds are used, and you need to be able to deal with both.
Effectively the surd is a bracket, and the order of operations rules come into effect, i.e. carry out
operations inside a bracket first. We can do this for known numbers quite easily.
Consider these examples for square roots: if the number inside the surd is a square, it is easy to
remove the surd, as in
√
√
5 + 4 = 9 = 3.
When the (known) number inside the surd is not a square, as in
√
√
5 − 2 = 3,
then we have to either leave the surd, or resort to a calculator to remove it:
√
√
5 − 2 = 3 = 1.732050807569 (approximately).
When we have unknown numbers inside the surd we cannot use a calculator, and we may only be able
to add or subtract in the sense of factoring. If the term inside the square root turns out to be a square,
then the surd can be removed. As examples, for y ≥ 0 and x + 1 ≥ 0, we have
p
p
p
y 2 = y and
x2 + 2x + 1 = (x + 1)2 = x + 1.
The last example can equally be written using exponents:
1
(x2 + 2x + 1) 2 = (x + 1)2
1
1
= (x + 1)2× 2 = (x + 1)1 = x + 1.
p
If the term inside the surd is not a square, as in
x2 + 9, then you cannot remove the surd. If you
2
think x + 9 is the square of a real number, then write down that number and multiply it by itself. The
result will not be x2 + 9.
2
2.1. SURDS
33
The same ideas apply for cube roots (we need cubes inside the surd before the surd can be removed)
and so on. If we cannot remove the surd, we can often work with it and rewrite it in more useful
forms.
2.1.2
Simplifying surds involving only one term
One of the exponent rules says
(because
√
n
x×
√
n
1
√
√
xy = n x × n y
1
√
= (xy) n = n xy). Another says
√
n
1
y = xn × y n
(xa )b = xab .
We can use these rules to simplify complicated expressions.
In simplifying a surd, we need to factor the expression inside the surd. We look specifically for factors
which are squares (for square roots), or cubes (for cube roots), and so on. Some remaining factors
will not be nice squares or cubes, etc., so their roots must remain expressed as surds, or be rewritten
using exponents.
q
p
√
√
√
√
√
√
√
Example.
1. 54x6 = 9 × 6 × x6 = 9 × 6 × x6 = 32 × 6 × (x3 )2 = 3 6 × x3 ,
where x ≥ 0,
s
√
√
3 3
3
p
23 × x3
2 × x3
2x
3
3
3
−6
q
=
2.
8x y =
= 2 = 2xy −2 ,
3
y
3
(y 2 )
(y 2 )3
p
√
√
√
√
√
3. 250x3 = 25 × 10 × x × x2 = 25 × 10x × x2 = 5x 10x, where x ≥ 0.
2.1.3
Surds involving more than one term
The expression inside a surd may include a sum or a difference, that is, it contains more than one
term. To simplify, we need first to have just one term inside the surd. So we factor, if possible.
Note that in general,
√
x + y 6=
√
x+
√
y.
√
√
Why? Try putting√4 and√9 in place of x and y.
Then
the
left
hand
side
is
4
+
9
=
13, and the
√
right hand side is 4 + 9 = 2 + 3 = 5, but 13 6= 5. The two sides are not equal for some known
numbers, so that they are certainly not equal for all possible numbers x and y.
p
p
p
p
√
Example.
1. 4x2 + 8y = 4(x2 + 2y) = 4 × x2 + 2y = 2 x2 + 2y, where x2 + 2y ≥ 0,
p
p
p
√
p
√
3
3
3
3
3
27a6 + 54a3 b3 = 3 27a3 (a3 + 2b3 ) = 27 × a3 × a3 + 2b3 = 3a × a3 + 2b3 ,
2.
p
√
p
√
√
3. 4x2 + ax2 − 9bx2 = x2 (4 + a − 9b) = x2 × 4 + a − 9b = x 4 + a − 9b,
where x, 4 + a − 9b ≥ 0.
34
CHAPTER 2. BASIC ALGEBRA 2
2.1.4
Adding with surds
An expression that requires us to add surds can only be simplified if the terms have common factors,
or if the surds can be removed. If not, then we can do nothing further.
Example.
√
√
√
√
√
√
√
√
√
√
1. 4 8 − 2 18 = 4 4 × 2 − 2 9 × 2 = 4 × 4 × 2 − 2 × 9 × 2 = (8 − 6) 2 = 2 2
2. For x, y ≥ 0,
p
p
p
p
√
√
4 x5 y 4 − 2 x3 y 2 = 4 (xy)4 × x − 2 (xy)2 × x = 4(xy)2 x − 2xy x
√
√
√
= 2xy × 2xy × x + 2xy × (−1) × x = 2xy(2xy − 1) x.
√
√
3. For x, y ≥ 0, the terms in 4 x − 3 y do not have common factors.
4. For x, x + z ≥ 0 and y > 0,
r
√
√
√
√
r
x
x+z
5 x 3 x+z
5 x+3 x+z
+3
= √ + √
=
.
5
√
y
y
y
y
y
2.1.5
Surds in denominators
Fractions that have square roots in the denominators can be simplified. It is common to ‘simplify’ so
that any surds are in the numerator, not in the denominator. In order to achieve this, which is called
rationalising the denominator, we multiply both the numerator and the denominator by the conjugate
of the denominator. Multiplying the denominator by its conjugate creates a new denominator which
is not a surd.
√
√
2
2
2
2
2 2 √
Example.
1. √ = √ × 1 = √ × √ =
= 2,
2
2
2
2
2
√
√
√
√
√
√
x
x
x+ 2
x( x + 2)
x( x + 2)
√ =√
√ ×√
√ = √
√ √
√ =
2. √
, for x ≥ 0,
x−2
x− 2
x− 2
x+ 2
( x − 2)( x + 2)
x 6= 2,
p
√
√
√
√
√ √
y(5 x + 3 x + z)
5 xy + 3 y(x + z)
5 x+3 x+z
3.
=
=
, for x, x + z ≥ 0, y > 0.
√
√ √
y
y y
y
Practice (Section 2.1). Simplify:
√
4+ 2
√ ,
1.
3− 2
2.
√
√
√
8 − 5 20 + 3 2,
3.
p
3
−27x6 ,
4. (6 −
√ √
√
2)( 6 + 5 3),
2.1. SURDS
√
10 x
5. √
,
x+5
s
9x4
6.
, where y > 0,
4y 2
p
7. 25a3 + 100a2 b, where a, b ≥ 0,
35
p
(x − y)(x + y)
√
8.
, where x + y ≥ 0 and
x−y
x − y > 0,
p
√
9. 3 x5 + 6 x2 y 4 , where x, y ≥ 0,
10.
p
3
81b5 − 64(ab)3 .
36
CHAPTER 2. BASIC ALGEBRA 2
2.2
Equations
Aim. To learn how to solve simple equations.
Vocabulary and conventions
equation
equality
2.2.1
or
An equation is a statement declaring that one expression is equal to
another. The following are all equations:
3a = 5b + 1,
2(2 + z) = 2z − 2,
5y = 4(y − 2),
x2 + 4x + 2 = 0.
Solving an equation means finding what values the unknown
numbers must have for the equation to be true. Often more than
one value is possible, and the solution should include all possible
values. Sometimes no value at all is possible, i.e. the equation has
no solution.
Manipulating an equality
When solving an equality, we aim to have the desired unknown number on one side of the equality
sign, and all other numbers, known or unknown, on the other side. If we are solving for x, we want
to be able to write
x = something.
To achieve this we have to do some algebraic manipulation. We require that the two sides of an
equation have the same value (that is, the statement is true). So if we add (or subtract) the same
number to (or from) each side of the equality, we will have a new statement which must also be true.
Similarly if we multiply or divide each side by the same (non-zero) number, we will have a new
statement which must also be true. For example,
• If 4x − 6 = 2 is true, then 10 × (4x − 6) = 10 × (2) must also be true.
• If 4x − 6 = 2 is true, then (4x − 6) + 25 = (2) + 25 must also be true.
To preserve an equality, any operation performed on one side of the equality must be balanced by
performing the same operation on the other side.
We need to think of each side of the equality as being in brackets, in order to ensure that all operations
are performed on the entire side. The trick in finding solutions is to see which operations are needed.
2.2. EQUATIONS
2.2.2
37
What to do when
In general, we find solutions by using the order of operations rules backwards, always aiming to have
the unknown number alone on one side of the equality and all others on the other side. Often we will
have to expand brackets in order to separate terms, or factor in order to gather terms together.
Example.
1.
Solve for x
4x − 6 = 2
The unknown variable is on the left, so we aim to have
all terms not involving x on the right. Any addition or
subtraction to be done? Yes, we want to remove −6 from
the left side, and therefore we must also remove it from the
right side. We can do this by adding 6 to both sides.
(4x − 6) + 6 = (2) + 6
Simplify.
4x = 8
Any multiplication or division to be done? Yes, if we divide
the left side (4x) by 4, we will have x alone, and we must
also divide the right side by 4.
4x ÷ 4 = 8 ÷ 4
Simplify.
x=2
2.
Short version:
4x − 6 = 2 =⇒ 4x = 8 =⇒ x = 2.
Solve for y
5y = 4(y − 2)
We have the unknown variable on both sides of the equality,
and we would like to have all terms involving y on the left. If
we expand the bracket on the right hand side, we can separate
terms involving y from terms not involving y.
5y = 4y − 8
Any addition or subtraction to be done? Yes, subtracting 4y
from both sides will give us all terms involving y on the left.
5y − 4y = 4y − 8 − 4y
Simplify.
y = −8
3.
Short version:
5y = 4(y − 2) =⇒ 5y = 4y − 8 =⇒ y = −8
Solve for b
3a = 5b + 1
Let’s get all terms involving b on the right (to avoid having
negative b) and all not involving b on the left. Any addition
or subtraction?
3a − 1 = 5b + 1 − 1
Simplify.
38
CHAPTER 2. BASIC ALGEBRA 2
3a − 1 = 5b
Any multiplication or division? If we divide the right side by
5 we will be left with b alone, which is our aim. So we must
divide the left side by 5 too.
(3a − 1)
5b
=
=b
5
5
b=
4.
3a − 1
5
Solve for x
5ax + 20x + 6 = 10
Note that x is involved in two terms on the left hand side.
We can’t ‘add’ these as we did in Example 2 (5y − 4y = y)
because ax and x are not like terms. However we can factor
them to create one term involving x.
5x(a + 4) + 6 = 10
5x(a + 4) = 4
Any addition to be done? Yes, subtract 6 from each side.
5x(a + 4)
4
=
5(a + 4)
5(a + 4)
x=
5.
6.
Any multiplication or division? Dividing the left hand side
by 5(a + 4) will leave x alone at left.
4
5(a + 4)
Solve for z
2(2 + z) = 2z − 2
Expand the brackets to separate terms involving z from terms
not involving z.
4 + 2z = 2z − 2
Subtract 2z from each side.
4 = −2
no solution
We get an impossible result, because 4 6= −2. The equation
has no solution, that is, there is no number z for which the
equation is true.
Solve for x
4
16
=
3x
27
We need x to be in the numerator. If we multiply both sides
by 3x, then x will be in the numerator on the right side, and
we won’t have a fraction on the left side.
4
3x
16 3x
×
=
×
3x
1
27
1
Simplify.
4=
16x
9
Multiplying both sides by 9 will get rid of the fraction on the
right side.
2.2. EQUATIONS
4×9=
39
16x
×9
9
36 = 16x
x=
Now solve for x.
36
4×9
9
=
=
16
4×4
4
Short version:
7.
Simplify.
4
16
4
3x
=
=⇒
×
=
3x
27
3x
1
16x
=⇒ 4 × 9 =
× 9 =⇒
9
3x
16
×
27
9
1
4 × 9 16x
= x.
=
16
4 × 4
Solve for x
x−y
2
=
2x − y
3
Somehow we have to isolate x. Let’s start by trying to get all
terms involving x in the numerator. So multiply both sides
by (2x − y).
It is usually helpful to get rid of all fractions, so we also
multiply both sides by 3.
3(x − y) = 2 × (2x − y)
Separate the terms involving x by expanding the brackets.
3x − 3y = 4x − 2y
Now we are ready to collect terms involving x on one side,
and terms not involving x on the other.
−y = x
We have the solution.
Practice (Section 2.2). Solve for x:
1. 15 − 5x = 24 − 8x,
6.
9x
5
3x
=
+
,
7
21
2
7.
x+1
3x + 1
7
=
−
,
4x
2x
3x
8.
x+a−b
1
= .
3x
2
2. 3(169 − x) − (78 + x) = 29x,
3. 5ax + 10bx = 10x − 10,
4. y 2 − 3x = 3,
5.
4x
− 3 = 8,
7
40
CHAPTER 2. BASIC ALGEBRA 2
2.3
Quadratic equations
Aim. To learn how to factor and solve some quadratic equations.
Vocabulary and conventions
polynomial
linear expression or
equation
quadratic expression
or equation
2.3.1
A polynomial in x (or y or z or any other variable name) is an
expression containing the variable x whose terms include only zero
and / or positive whole number powers of x:
5x3 + 3x2 + x + 4 is a polynomial in x (since 4 = 4x0 ),
z 24 + 2z 7 − z 2 is a polynomial in z,
2y 3 + y − 2 is a polynomial in y,
ax2 + bx + c is a polynomial in x.
A polynomial whose degree (that is, the highest power of its
variable) is 1 is called a linear expression.
3x + 2 is a linear expression in x because the variable x has a
highest power of 1 (x = x1 ).
3x2 + 2x + 1 is not a linear expression because it contains a
term in which the variable x has greater power than 1.
3x = 0 is a linear equation in x.
A quadratic expression is a polynomial of degree 2, that is, the
highest power of its variable is 2:
ay 2 + by + c is a quadratic expression in y,
3x2 + 6 = 0 is a quadratic equation in x.
Factorising quadratic expressions
It is often much easier to work with quadratic expressions that are in a factored form, and it is a great
deal easier to solve quadratic equations that have been factored. The factored form of a quadratic
expression is the product of two linear terms, that is, it has a form like
(ax + b)(cx + d).
Expanding gives
acx2 + (ad + bc)x + bd.
2.3.2
Factoring: guessing and testing the guess
This method for finding the linear factors of a quadratic expression is most useful when all the
coefficients of the variable are whole numbers. It may not work, but if it does it is quite fast (with
2.3. QUADRATIC EQUATIONS
41
practice).
Notice that in the equality (ax + b)(cx + d) = acx2 + (ad + bc)x + bd:
• the coefficient of x2 at right (ac) is the product of the coefficients of x in the linear terms at left,
and
• the coefficient of x0 at right (the constant term bd) is the product of the coefficients of x0 at
left.
We will factor the expression x2 + 3x − 4. Recognise that it is the same as 1x2 + 3x − 4 and that we
are looking for some a, b, c, and d to make it fit
1x2 + 3x − 4 = acx2 + (ad + bc)x + bd.
We need the equality to be true for every x, and that will only be the case if the coefficients of like
powers of x on each side of the equality have the same value. Therefore we want
ac = 1,
ad + bc = 3,
and
bd = −4.
1. We need first two numbers a and c that are factors of 1, the coefficient of x2 . Assuming (as we
do in this method) that a, b, c, and d are whole numbers, we have only two sets of choices, 1
and 1, or −1 and −1. We try a = c = 1.
2. We need two numbers b and d that are factors of −4. In order to see what we have and what
we need, it is often convenient to use something like the boxes below. See that factors of 1,
the coefficient of x2 , are in the left hand boxes, and factors of −4, the coefficient of x0 (the
constant term), are in the right hand boxes.
a
b
1
−2
1
1
1
−1
c
d
1
2
1
−4
1
4
3. We want ad + bc = 3, that is, we want the sum of the products of each far corner to be the
coefficient of x in the quadratic 1x2 + 3x − 4.
a
b
1
−2
1
1
1
−1
c
d
1
2
1
−4
1
4
ad + bc = ?
1 × 2 + 1 × (−2) = 0,
1 × (−4) + 1 × 1 = −3,
The final box shows the correct result:
ax + b = 1x − 1
and
cx + d = 1x + 4.
1 × 4 + 1 × (−1) = 3.
42
CHAPTER 2. BASIC ALGEBRA 2
Here we needed three attempts to get the right result. The second attempt giving −3 showed
we were on the right track, and that we should succeed if we swap the negative sign. The third
one is correct. From the successful box, we read off the factors:
1x2 + 3x − 4 = (1x − 1)(1x + 4) = (x − 1)(x + 4).
It is a good idea to check your result by expanding it – does it give the original expression? If not,
there is an error somewhere. If we could not get the correct coefficient of x, we would start again
with a different set of factors of the coefficient of x2 .
Example.
1. Factorise 4x2 − 9
(= 4x2 + 0x − 9).
We try the factors of 4 on the left hand side, and the factors of −9 on the right hand side,
multiply the corner numbers and add the results. We are looking for 0, the coefficient of x in
this quadratic expression.
1
−3
2
−3
4
3
2
3
1 × 3 + 4 × (−3) = −9,
2 × 3 + 2 × (−3) = 0.
We see that the latter box is correct:
ax + b = 2x − 3
and cx + d = 2x + 3,
therefore we can read off the factors from this box:
4x2 − 9 = (2x − 3)(2x + 3).
2. Factorise 5x2 − 3x − 2.
We try the factors of 5 on the left hand side, and the factors of −2 on the right hand side,
multiply the corner numbers and add the results. We are looking for −3, the coefficient of x in
this quadratic expression.
5
−1
5
2
1
2
1
−1
5 × 2 + 1 × (−1) = 9,
5 × (−1) + 2 × 1 = −3.
We see that the latter box is correct:
ax + b = 5x − 2
and cx + d = 1x − 1,
therefore we can read off the factors from this box:
5x2 − 3x − 2 = (5x + 2)(x − 1).
2.3. QUADRATIC EQUATIONS
2.3.3
43
Solving quadratic equations by factoring
We can solve quadratic equations easily once they have been factored. The method we use to solve
them depends on the fact below:
If ab = 0, then a = 0 or b = 0,
Suppose we are asked to solve 2x2 − 3x = 2. That is, for what values of x can this equality be true?
We manipulate the equality so that 0 is on one side alone. Subtracting 2 gives 2x2 − 3x − 2 = 0, and
factoring the left side we obtain (2x + 1)(x − 2) = 0 then either (2x + 1) = 0 or (x − 2) = 0. These
are linear equations are easily solved. Note that there are two solutions: either − 21 or x = 2.
Every quadratic equation will have either two (real) solutions, one or none.
Example.
1.
x2 = 8x − 16
Manipulate to get 0 alone at one side.
x2 − 8x + 16 = 0
Factor.
(x − 4)(x − 4) = 0
One solution, x = 4.
Solve the linear equations (x−4) = 0 and (x−4) = 0. They
both have the same solution.
2. 1 + 3x2 = 4x =⇒ 3x2 − 4x + 1 = 0 =⇒ (3x − 1)(x − 1) = 0 =⇒ x =
1
3
or x = 1,
3. 3y 2 = 8y − 4 =⇒ 3y 2 − 8y + 4 = 0 =⇒ (3y − 2)(y − 2) = 0 =⇒ y =
2
3
or
3x
3
4. x +
= 0 =⇒ x x +
= 0 =⇒ x = 0
2
2
2
or x =
−3
.
2
Practice (Section 2.3). Solve:
1. x2 − 3x − 4 = 0,
5. 0 = 3x2 + 7x + 2,
2. 6x2 + 17x = −12,
6. 4y 2 + 7y = 2,
3. 3 = 2x2 − 5x,
7. 2x2 = 5 − 3x,
4. 9x2 − 25 = 0,
8. 3x2 − 5x − 8 = 0.
y = 2,
44
CHAPTER 2. BASIC ALGEBRA 2
2.4
Inequalities
Aim. To learn how to solve inequalities.
Vocabulary and conventions
inequality or
inequation
Inequalities are statements declaring that one expression is
less than (<),
less than or equal to (≤),
greater than (>), or
greater than or equal to (≥)
another. Note the pointed (smaller) end of the sign always points to
the smaller number. The following are all inequalities:
3a ≤ 5b + 1, 5y > 4(y − 2) and 3x2 + 4x + 2 < 0 ≤ x.
We can read inequalities forwards and backwards. That is x ≤ 2
means just the same as 2 ≥ x.
solving inequalities
Solving an inequality means finding what value or values the
unknown numbers must have for the statement to be true. An
inequality will sometimes have no solution, but usually it will have
a range of solutions. For example, the solution to 5y > 4(y − 2) is
y > −8, meaning that all values greater than −8 are solutions for y.
The inequality x2 < 0 has no solution because the square of any
(real) number must be either positive or zero, it cannot be negative,
so there is no real number x for which this inequality is true.
∈
R
N
“Belongs to”, or “is an element of”.
The set of all real numbers: “x ∈ R” means x is a real number.
The set of all natural numbers or positive integers: 1, 2, 3, . . . We
can say that x is a positive integer by writing x ∈ N.
The set of integers (whole numbers): . . . , −2, −1, 0, 1, 2, 3, . . . We
write x ∈ Z to mean that x is an integer. Often I is used to replace
Z.
Infinity – note that this is not a real number.
Means the same as a < x < b. This is called an open interval.
−∞ < x < ∞: Means exactly the same as x ∈ R.
a < x ≤ b A half open interval.
a ≤ x < b A half open interval.
a ≤ x ≤ b A closed interval.
“Implies”, or “therefore”.
Z
∞
x ∈ (a, b)
x ∈ (∞, ∞)
x ∈ (a, b]
x ∈ [a, b)
x ∈ [a, b]
=⇒
2.4. INEQUALITIES
2.4.1
45
Solving inequalities
Solving inequalities is very much like solving equalities, we do a lot of algebraic manipulation in
order to get the unknown number at one side of the inequality sign, and all other numbers on the other
side. There is however an important difference when multiplying or dividing:
when multiplying or dividing both sides of an inequality by a negative number, we must reverse the
direction of the inequality.
Reversing the direction of the inequality means changing less than (<) to greater than (>) or changing
greater than or equal to (≥) to less than or equal to (≤) and visa versa. Why do we do this?
Let’s do a simple example. Take the true statement
−3 < 1.
Multiplying both sides of this inequality by −1 and not
reversing the direction of the inequality, we get
(−1)(−3) < (−1)1
and simplifying gives
3 < −1
But that is nonsense, an untrue statement, since 3 is not less
than −1.
Let’s do it again, this time reversing the inequality.
(−1)(−3) > (−1)1
and simplifying, we get
3 > −1,
a true statement.
Look at this situation on a number line:
−4
−3
−2
−1
0
−4
−3
−2
−1
0
1
1
2
2
3
3
4
4
Now solve −x < 1 for x using addition and subtraction only:
−x < 1
0<1+x
−1 < x.
Add x to both sides.
Subtract 1 from both sides.
Solve −x < 1 for x using multiplication:
−3 is to the left of −1 on the number
line, i.e. −3 < 1
multiplying −3 by −1 gives a shift right
(with zero as the centre) to 3; multiplying 1 by −1 gives a shift left to −1. The
swap about zero means that the resulting
numbers, 3 and −1, have a different order
from the originals, −3 and 1.
46
CHAPTER 2. BASIC ALGEBRA 2
Multiply both sides by −1 and reverse the direction of the
inequality.
−x < 1
x > −1.
Example. Solve for x:
1.
2x − 2 ≥ 3
2x ≥ 5
x≥
2.
Add 2 to each side.
Divide by 2 (which is positive so there is no need to reverse
the inequality).
5
2
3(1 − 2x) < 4
1 − 2x <
Divide by 3.
4
3
Subtract 1.
1
−2x <
3
−1
x>
6
2.4.2
Divide by −2 and reverse inequality direction.
Changing sign, and critical values
If we solve 1 − 2x > 0 for x, we get x < 21 . That is, the term 1 − 2x is positive as long as x < 12 .
Also, 1 − 2x is negative as long as x > 12 . We say that 1 − 2x changes sign as it passes through 0,
and this happens at x = 21 .
We can picture this on a number line:
If x > 12 , then 1 − 2x < 0
−4
−3
−2
−1
If x < 12 , then 1 − 2x > 0
0
1
2
3
4
If x = 12 , then 1 − 2x = 0. It is at
this point that the sign of 1 − 2x
changes.
A point at which a term changes sign is a critical value. For 1 − 2x, the only critical value is x = 12 .
1
changes sign when its denominator is
A critical value can also occur where a term is undefined. x+1
1
zero, so that x = −1 is a critical value for x+1 .
Example. Find the critical points for the following expressions:
1.
(1 − 2x)(1 + x)
Find where this term is 0 or undefined.
Set (1 − 2x)(1 + x) = 0 and solve. (1 − 2x)(1 + x) is always
defined.
2.4. INEQUALITIES
47
1
1 − 2x = 0 when x = .
2
1 + x = 0 when x = −1.
Solve for x, 1 − 2x = 0 and 1 + x = 0
1
The critical values are x =
2
and −1.
1+x
.
2x + 6
2.
Find where this term is zero or undefined.
1+x
Set
= 0 and solve. This happens where the
2x + 6
numerator is zero. So solve for x, 1 + x = 0.
1+x
is undefined when the denominator is zero.
2x + 6
So solve for x, 2x + 6 = 0.
1 + x = 0 when x = −1.
2x + 6 = 0 when x = −3.
1+x
The critical values of
2x + 6
are x = −1 and −3.
2.4.3
Multiplying or dividing by an unknown number
Sometimes it seems we must multiply by an unknown number in order to solve an inequality. Because
we do not know whether it is positive or negative, we do not know what to do with the inequality sign.
The most straightforward method of solving such problems is to rearrange the inequality by addition
and subtraction only in order to have zero on one side of the inequality. Then we find what values the
unknown number must have for this new inequality to be true.
Example.
1.
1
≤2
x
Suppose x = 0. Then the expression on the left does not
exist (division by zero). So x 6= 0.
Now subtract 2 from both sides: we aim to have zero on one
side of the inequality.
1
−2≤0
x
We want just one term on the other side from zero, so we
must add these fractions together.
1 − 2x
≤0
x
Now we have one term with two factors, 1 − 2x and x1 .
We find the critical values of 1−2x
x , that is, the values of x at which it can change sign, which is where
it is zero or undefined. The two critical values are 12 and 0. Then we draw a number line and do some
thinking. We need to find if 1−2x
x is positive or negative on the intervals between the critical values
and if it is zero at the critical values.
It cannot change sign or be zero at any other point.
48
CHAPTER 2. BASIC ALGEBRA 2
−2
1
2
0<x<
x<0
−1
1
2
0
x>
1
1
2
2
It suffices to check a point from each of the three intervals and the two endpoints separating the
intervals. We see that the endpoints are x = 0, 12 and the intervals are x < 0, 0 < x < 21 and x > 12 .
• For the interval x < 0, we test with x = −1. Substituting this into the inequality gives us
1 − (−2)
=
−1
+ ve
− ve
=− ve < 0,
so every x < 0 is a solution.
• Substituting x = 0 makes the left hand side of the inequality undefined, so x = 0 is not a
solution.
• For the interval 0 < x < 12 , we test with x = 41 . Substituting this into the inequality gives us
1 − (1/2)
=
1/4
+ ve
+ ve
=+ ve > 0,
i.e. no values in this interval are solutions.
• Substituting x =
1
2
gives
1−1
1/2
≤ 0, which is true. So x =
1
2
is a solution.
• For the interval x > 1, we test with x = 1:
1 − (2)
=
1
so every x >
1
2
− ve
+ ve
=− ve < 0,
is a solution.
We have tested each of the intervals and endpoints and see that the full solution is ‘x < 0 or x ≥ 12 ’.
Actually, most of the working can be done on the number line itself. We show how this is done below.
1 − 2x
≤0
x
Critical values are
Therefore
1
and 0.
2
1
Test between the points x = −1, , 1, and at the points, as
4
shown below.
Show that 0 is not a solution by putting a hollow circle there.
1
Show that is a solution by putting a filled circle there.
2
Then check where the results are negative or zero (shown by
the heavy line below).
1
1 − 2x
1
≤ 2 when
≤ 0, which is when x < 0 or x ≥ .
x
x
2
2.4. INEQUALITIES
49
1−1/2
1/4
1−(−2)
−1
=
+
ve
5
1
≥
3x
3
5
1
− ≥0
3x 3
1
4
−
=
ve
= − ve < 0
+ ve
1
2
1
1−1
1/2
=0
Suppose x = 0. Then the expression on the left does not
exist (division by 0).
1
Subtract to get 0 at right.
3
Combine fractions to get one term.
1
We have factors, 5 − x and
. For what values of x are
3x
these zero or undefined?
5 and 0 are the critical values.
5−1
3
5−(−1)
−1
=
5−0
0
+
ve
− ve
=
+
ve
+ ve
= + ve > 0
= − ve < 0
−1
0
5−6
18
1
=
−
ve
+ ve
5
= − ve < 0
6
5−5
15
is not defined
=0
5
1
5−x
≥ when
≥ 0, which is when 0 < x ≤ 5.
3x
3
3x
2
>1
x−3
Suppose x = 3. Then the expression on the left does not
exist (division by zero). So x 6= 3. Subtract 1 from both
sides.
2
1
5−x
− =
>0
x−3 1
x−3
We have two factors, 5 − x and
are 5 and 3.
5−4
4−3
5−2
2−3
=
+
ve
− ve
5−3
3−3
=
+
ve
+ ve
is not defined
3
1
so the critical values
x−3
= + ve > 0
5−6
6−3
= − ve < 0
2
Therefore
= + ve > 0
1−2
1
is not defined
5−x
≥0
3x
3.
ve
+ ve
0
1−0
0
Therefore
+
= − ve < 0
− ve
−1
2.
=
4
=
−
ve
+ ve
5
= − ve < 0
6
5−5
5−3
2
5−x
> 1 when
> 0, which is when 3 < x < 5.
x−3
x−3
=0
50
2.4.4
CHAPTER 2. BASIC ALGEBRA 2
Solving double inequalities
Often a statement will include two inequalities. An example is
3 < 2 + x ≤ 4.
We solve these in the same manner as before, but we may need to solve the parts of the statement
separately: that is solve 3 < 2 + x and 2 + x ≤ 4 before obtaining the solution. The solution will be
only those values of x for which both parts of the statement are true.
Examples
1.
2.
3<2+x≤4
Subtract 2 from each part of the inequality.
1<x≤2
This is the solution. There was no need to solve separately.
3x < 2 + x ≤ 4
Solve the right hand part first.
2+x≤4
Subtract 2 from each side.
x≤2
If the right hand part is true, then x must be less than or equal
to 2.
Now solve the left hand part.
3.
3x < 2 + x
Subtract x, then divide by 2.
x<1
If the left hand side is true, x must be less than 1.
Solution: x < 1
We must have both parts true, that is, if x is a solution of one
part, but not the other, then it is not a solution of the whole
inequality. Therefore the solution must be x < 1, which is
the interval for which both solutions are true.
3x < 2 ≤ 4 + x
Solve the right hand part.
2≤4+x
Subtract 4.
−2 ≤ x
Solve the left hand part.
3x < 2
Divide by 3.
x<
2
3
Solution: −2 ≤ x <
We need both parts to be true.
2
3
2.4. INEQUALITIES
51
Practice (Section 2.4). Solve for x:
7.
3x 4
− > 0,
5
9
3.
8.
3x + 2
≤ 2,
x−5
5x
5.
< x,
4
1+x
2x + 3
≥
,
6.
3
3
9.
x+1
3x + 1
7
<
−
,
4x
2x
3x
1. 2(9 − 2x) − (8 + x) < −3x,
2. 4x − 6b2 + 2(x − b) > 0,
5
3x
9x
>
+
,
7
21
2
4. 15 − 5x ≥ 24 − 8x,
10.
x
4
<4≤
.
3
x+1
52
CHAPTER 2. BASIC ALGEBRA 2
2.5
Quadratic equations 2
Aim. To learn how to complete the square and so solve quadratic equations.
Vocabulary and conventions
completing
square
the
irreducible quadratic
equation
2.5.1
Rewriting a quadratic expression as the sum or difference of two
squares. One term should be the square of a linear expression, the
other should be a constant.
A quadratic equation that has no (real) solutions.
Completing the square
Consider the quadratic expression x2 + 4x + 3. Notice that
(x + 2)2 − 1 = x2 + 4x + 4 − 1 = x2 + 4x + 3.
that is we can rewrite the original expression as the difference of the two squares (x + 2)2 and 12 . We
have two different ways of writing the same expression,
x2 + 4x + 3 = (x + 2)2 − 1
Now suppose we have any quadratic expression x2 + bx + c. We can always rewrite it as the sum or
difference of two squares because
b 2 b2
x + bx + c = x +
−
+c
2
4
2
Writing a quadratic expression in this form is known as completing the square. You should remember
this formula.
Check for yourself that the formula in the box is true by expanding the brackets on the right hand
side.
If the result is the difference of two squares, that is, if
b2
− c ≥ 0,
4
then we can use it to solve the equation x2 + bx + c = 0. If however the result gives us the sum of
two squares, that is, if
b2
− c < 0,
4
then the quadratic does not have real factors, and it is called irreducible.
2.5. QUADRATIC EQUATIONS 2
53
Example. Complete the square:
1.
x2 + 3x − 4
Use the formula with
b = 3, c = −4: b 2 b2
2
x + bx + c = x +
−
+c
2
4
2
x + 3x − 4 =
3
x+
2
2
−
9
−4
4
Simplify.
3 2 25
= x+
−
2
4
Alternative way to remember
3 2 9
2
x + 3x − 4 = x +
− −4
2
4
2
25
3
−
= x+
2
4
2.
x2 − 8x + 16
x2 − 8x + 16 = (x − 4)2 − 16 + 16
= (x − 4)2
Alternatively
x2 − 8x + 16 = (x − 4)2 − 16 + 16
= (x − 4)2
3.
4.
x2 + 2x + 2
x2 + 2x + 2 = (x + 1)2 − 1 + 2
= (x + 1)2 + 1
5x2 − 3x − 2
The coefficient of x is 3.
We halve
this,
3
write the square of the linear term x +
,
2
subtract the square of the constant in this
term, and add the constant in the original
term.
Use the formula with b = −8, c = 16:
b 2 b2
2
x + bx + c = x +
−
+ c.
2
4
The coefficient of x is −8. We halve this to
get −4, write the square of the linear term
(x − 4), subtract the square of the constant in
this term, and add the constant in the original
term.
Notice that this is the sum of two squares.
This is slightly different from the examples
we have discussed before. The problem here
is that the coefficient of x2 is not 1. A
method for dealing with this is to rewrite the
expression by factoring out the coefficient of
x2 .
.
54
CHAPTER 2. BASIC ALGEBRA 2
2
3
2
5 x − x−
5
5
Then complete the square of the expression in
brackets.
3 2
9
2
x−
−
−
10
100 5
2
3
49
= x−
−
10
100
To complete the square for the original
expression, we multiply this by the factor 5
5x2 − 3x − 2
!
3 2
49
=5
x−
−
10
100
2
3
245
=5 x−
−
10
100
2.5.2
Solving a quadratic equation
Now that we can write a quadratic expression as the sum or difference of two squares, we can solve a
quadratic equation from this form without factoring. Suppose that we wanted to solve x2 +3x−4 = 0.
We complete the square, finding that
3 2 25
x + 3x − 4 = x +
− ,
2
4
2
the difference of two squares. It follows that
3 2 25
x+
= .
2
4
Since one side equals the other, then the square root of one side must equal the square root of the
5
5
other. Because 25
4 has both a positive and negative square root ( 2 and − 2 ), the unknown square root
3 2
3
of (x + 2 ) , which is x + 2 could be either positive or negative. That is, x + 32 is the positive or
negative square root of 25
4 . This is written
r
3
25
5
x+ =±
=± .
2
4
2
We now have two linear equations to solve:
x+
3
5
=
2
2
or x +
3
−5
=
.
2
2
From these equations, we find that either x = 1 or x = −4.
2.5. QUADRATIC EQUATIONS 2
55
Example.
1.
4x2 − 9 = 0
4x2 = 9
9
x2 =
4
r
x=±
so x =
2.
The expression at left is already the difference
of two squares, that is, there is no term
involving x1 . We can solve for x2 .
Now take square roots both sides, which
solves for x.
9
4
3
−3
or x =
.
2
2
1 + 3x2 = 4x
Rearrange to get 0 alone at the right.
3x2 − 4x + 1 = 0
Take out factor of 3.
4x 1
2
+
= 0.
3 x −
3
3
4x 1
+ .
We have two factors, 3 and x2 −
3
3
Since 3
6=
0, it must be that
4x 1
x2 −
+ = 0. Complete the square.
3
3
2 2
Solve for x −
.
3
4x 1
x2 −
+
3 32
2
4 1
= x−
− + =0
3
9 3
2
x−
3
2
=
1
9
r
1
2
1
x− =±
=±
3
9
3
x=
2 1
±
3 3
1
x = 1 or x = .
3
2
Solve for x − .
3
Solve for x.
56
3.
CHAPTER 2. BASIC ALGEBRA 2
3y 2 = 8y + 6
Rearrange.
3y 2 − 8y − 6 = 0
Factor out 3.
8y
2
3 y −
−2 =0
3
8y
3 6= 0, so y 2 −
− 2 = 0. Complete the
3
−8 −4
square: half of
is
.
3
3
4
y−
3
2
4
3
2
y−
−
16
−2=0
9
=
34
9
√
34
4
y− =±
3
3
√
4
34
y= ±
3
3
√
√
4 + 34
4 − 34
or y =
.
y=
3
3
4.
4
Solve for y − .
3
Solve for y.
3x
=0
2
3 2
9
x+
−
=0
4
16
x2 +
3
x+
4
x=
2
=
9
16
−3 3
±
4
4
x = 0 or x =
−3
.
2
Notice that it would have been a lot quicker
to factor this example (x is common to both
terms) than to solve by this method.
2.5. QUADRATIC EQUATIONS 2
5z 2 + 9 = 0
5.
Since the left hand side is the sum of
two squares, this equation does not have a
solution. To see this, solve for z 2 .
5z 2 = −9z
z2 =
−9
5
z 2 has to be either positive or zero. It cannot
be negative. Our final expression cannot be
true for any (real) value of z, so the original
expression cannot be true for any (real) value
of z either.
No solution.
2.5.3
57
Quadratic formula
The third method of solving quadratic equations is a generalisation of the method of completing the
square. We suppose our quadratic equation is
ax2 + bx + c = 0,
where a 6= 0.
Completing the square, we get
"
#
b
c
b2
c
b 2
2
ax + bx + c = a x + x +
=a x+
− 2+
= 0.
a
a
2a
4a
a
2
Since a 6= 0, we have
b2
c
b 2
− 2 + = 0.
x+
2a
4a
a
Solving for x, we find
2
b 2
b
c
x+
=
−
,
2a
4a2 a
then
s
r
√
b
b2
c
b2 − 4ac
b2 − 4ac
=±
−
=±
=±
.
x+
2
2
2a
4a
a
4a
2a
So finally,
if
ax2
+ bx + c = 0 for some a 6= 0, then x =
−b ±
√
b2 − 4ac
.
2a
This is called the quadratic formula and you should memorise it.
Notice that if b2 − 4ac < 0, then x is not a real number, it is complex because the square root of a
negative number is an imaginary (complex) number. The expression b2 − 4ac is frequently called the
discriminant of the quadratic expression ax2 + bx + c.
58
CHAPTER 2. BASIC ALGEBRA 2
Example.
1.
x2 + 4x − 3 = 0
We have a = 1, b = 4, and c = −3. Replace
these numbers in the quadratic formula.
p
42 − 4 × 1 × (−3)
x=
2×1
√
√
−4 ± 2 7
−4 ± 28
=
=
2
2
√
√
x = −2 + 7 or x = −2 − 7.
−4 ±
2.
2x2 + 2x +√3 = 0
−2 ± 22 − 4 × 2 × 3
x=
2×2
√
−2 ± −20
=
4
No (real) solution.
Simplify.
√
−20 is an imaginary number. See that we
could have noticed this early by choosing to
inspect this discriminant, b2 − 4ac which is
4 − 4 × 2 × 3 = −20, a negative number, so
that we must expect no solution.
Practice (Section 2.5). Solve the following by completing the square – do not use the quadratic
formula. While it is usually an easier method for solving quadratic equations, being able to complete
the square is an essential skill for problems you will meet later.
1. x2 − 3x = 4,
4. 9x2 − 25 = 0,
7. x2 = 2x + 24,
2. x2 = 6x − 7,
5. −x2 + 6x = 9,
8. x2 − 5x = 14,
3. 2x2 − 5x = 3,
6. 2x2 = 3x − 5,
9. 3x2 − 5x = 8.
Then solve the same problems using the quadratic formula.
2.6. FACTORING POLYNOMIALS (READ ONLY)
2.6
59
Factoring polynomials (read only)
Aim. To learn how to factor and solve polynomial equations.
Vocabulary and conventions
cubic expression
2.6.1
A polynomial of degree 3, that is, the highest power of its variable
is 3. An example is 2y 3 − 5y.
Difference of two squares: factoring a2 − b2 .
In Section 2.5, we wrote quadratic expressions as the sum or difference of two squares and so solved
quadratic equations without factoring. However we can easily factor the difference of two squares.
√
√ √
√
Remember the conjugate used in Section 2.1: for x, y ≥ 0, x − y = ( x − y)( x + y). This
idea gives us
a2 − b2 = (a − b)(a + b),
and you should memorise this result.
Example.
Factor and hence solve:
5x2 − 3x − 2 = 0.
5(x2 −
3x 2
− )=0
5
5
49
3 2
5 (x − ) −
=0
10
100
First factor for 5.
Complete the square of the term inside the
brackets.
Factor, using a2 −rb2 = (a − b)(a + b), with
49
7
3
a = x − ,b =
= .
10
100
10
3
7
3
7
5 x−
−
x−
+
= 0 Simplify.
10 10
10 10
2
5 (x − 1) x +
=0
5
2
x = 1 or x = − .
5
Solve.
60
2.6.2
CHAPTER 2. BASIC ALGEBRA 2
Sums and differences of two cubes: factoring a3 + b3 , a3 − b3
a3 + b3 = (a + b)(a2 − ab + b2 )
a3 − b3 = (a − b)(a2 + ab + b2 )
It is useful to memorise these results. Satisfy yourself that they are true by expanding the right hand
sides.
Example.
1.
Factor 8x3 − 27.
See that this is the difference of two cubes.
= (2x)3 − (3)3
Factor, using
a3 −b3 = (a−b)(a2 +ab+b2 ), where a = 2x,
b = 3.
= (2x − 3) (2x)2 + 2x × 3 + 32
Simplify.
= (2x − 3)(4x2 + 6x + 9)
Now we try to factor the quadratic part. But notice the discriminant:
62 − 4 × 4 × 9 = −108 < 0
so that the quadratic is irreducible and we have already factored as much as possible.
2.
Factor x3 + y 6 .
= (x)3 + y 2
Rewrite as the sum of two cubes.
3
= x + y2
(x)2 − x × y 2 + (y 2 )2
= x + y2
x2 − xy 2 + y 4 .
Factor, using
a3 −b3 = (a−b)(a2 +ab+b2 ), where a = x,
b = y2.
Simplify.
The quadratic part is irreducible - in fact this is always the case with sums or differences of two cubes.
You can see this by using the quadratic formula to try to solve for a in either a2 − ab + b2 = 0, or
a2 + ab + b2 = 0.
2.6. FACTORING POLYNOMIALS (READ ONLY)
2.6.3
61
Finding linear factors using the factor theorem
Suppose we have a polynomial already factored, say
(x − a)(x − b)(x − c).
If we put x = a (or b or c), the polynomial has the value zero. This is because
(a − a)(a − b)(a − c) = 0 × (a − b)(a − c) = 0.
Now consider any polynomial, for example
x3 − x2 + x − 1.
We find (by trying out some values) that a certain value for x (in this case x = 1) makes the
polynomial have the value zero (13 − 12 + 1 − 1 = 0). Then we can be certain that (x − 1) is
a factor of the polynomial.
If a polynomial in x has the value zero
when x is replaced by a, then
(x − a) is a factor of the polynomial.
This result is known as the factor theorem.
Example.
Find a factor of
3x3 + x2 − 13x − 2.
Test some values: try x = 1.
3 × 13 + 12 − 13 × 1 − 2 = −11
(x − 1) is not a factor. Try x = 2.
3 × 23 + 22 − 13 × 2 − 2 = 0
(x − 2) is a factor.
Then 3x3 + x2 − 13x − 2 = (x − 2) × (something).
The next difficulty is to find what the something is. We can test other numbers to try and find more
linear factors, but you will see that these aren’t immediately obvious.
2.6.4
Long division of polynomials
This method comes directly from long division with known numbers. It is very useful when one factor
is known, but others are not.
Example.
1. Given that (x − 1) is a factor of x3 − 3x2 + x + 1, find the remaining factors of this cubic.
Start by dividing x3 − 3x2 + x + 1 by (x − 1).
62
CHAPTER 2. BASIC ALGEBRA 2
(x − 1) x3 − 3x2 + x + 1
x2
(x − 1) x3 − 3x2 + x + 1
x2
(x − 1) x3 − 3x2 + x + 1
−(x3 − x2 )
2x2 + x + 1
We ask, how many times will x (the term with
the highest power of the variable in x − 1),
“go into” x3 (the term with the highest power
of the variable in x3 − 3x2 + x + 1. Since
x3 ÷ x = x2 , we can say it “goes” x2 times
and write x2 on the line above.
Now multiply: x2 × (x − 1) = x3 − x2 , write
this below, and subtract it from x3 − 3x2 +
x + 1.
x2 − 2x − 1
(x − 1) x3 − 3x2 + x + 1
−(x3 − x2 )
2x2 + x + 1
−(−2x2 + 2x)
−x + 1
−(−x + 1)
0
Look next at the term with the highest power
of x? in the remainder, −2x2 + x + 1, and
ask, how many times will x “go into” −2x2 ?
Since −2x2 ÷ x = −2x, we can write −2x
on the line above.
Since
Multiply: −2x × (x − 1) = −2x2 + 2x, write
this below, and subtract it from −2x2 + x + 1.
The process goes on until we have zero
remainder, which must happen since (x − 1)
is a factor of x3 − 3x2 + x + 1.
x3 − 3x2 + x + 1
= x2 − 2x − 1, we have
x−1
x3 − 3x2 + x + 1 = (x − 1)(x2 − 2x − 1).
Now we can factor the quadratic expression remaining: since by the quadratic
formula, if x2 − 2x√
− 1 = 0, then (with a = 1, b = −2, and c = −1)
√
2± 4+4
x=
= 1 ± 2.
2
√
√
It follows that when x = 1 + 2, or x = 1√− 2, the quadratic
√ has the value
zero. Then by the factor theorem, (x − (1 + 2)) and (x − √
(1 − 2)) are √
factors
of the quadratic. So x3 − 3x2 + x + 1 = (x − 1)(x − 1 − 2)(x − 1 + 2).
2.6. FACTORING POLYNOMIALS (READ ONLY)
63
2. Given that (x − 2) is a factor of 3x3 + x2 − 13x − 1, find the remaining factors of this cubic.
3x2 + 7x + 1
(x − 2) 3x3 + x2 − 13x − 2
−(3x3 − 6x2 )
7x2 − 13x − 2
−(7x2 − 14x)
x−2
−(x − 2)
0
We ask how many times x (the term with the
highest power of the variable in (x − 2)) will
“go into” 3x3 . Since 3x3 ÷ x = 3x2 , we can
say it “goes” 3x2 times and write 3x2 on the
line above.
Since 3x2 × (x − 2) = 3x3 − 6x2 , we need
to subtract this last from 3x3 + x2 − 13x − 2,
getting the remainder 7x2 − 13x − 2.
Carry on the process until the remainder
0 is achieved.
Therefore 3x3 + x2 − 13x − 2 = (x − 2)(3x2 + 7x + 1). Now we can factor (if
possible) the remaining quadratic factor:
√ !
√ !
7
−
7
+
37
37
3x3 + x2 − 13x − 2 = 3(x − 2) x +
x+
.
6
6
3. Given that (x + 1) is a factor of x3 − 2x − 1, find the remaining factors of this cubic.
x2 − x − 1
(x + 1) x3 + 0x2 − 2x − 1
−(x3 + x2 )
−x2 − 2x − 1
−(−x2 − x)
−x − 1
−(−x − 1)
0
The major problem here is that one of the
coefficients of the cubic is zero. In this
case, the coefficient for x2 is zero. It is
easy to forget that there will some non-zero
coefficient of x2 in the quadratic we are
looking for. To avoid mistakes created in this
way, we put the zero in as at left.
Therefore x3 − 2x − 1 = (x + 1)(x2 − x − 1). Now factor (if possible) the
remaining quadratic factor:
√ !
√ !
1− 5
1+ 5
3
x − 2x − 1 = (x + 1) x −
x−
.
2
2
64
CHAPTER 2. BASIC ALGEBRA 2
2.6.5
Simplifying polynomial fractions, and long division with remainder
It often happens that we have a fraction in which both numerator and denominator are polynomial
expressions. If the numerator and denominator have common factors we may be able to simplify the
expression by cancelling the common factors.
Example.
1.
x2 − 3x − 4
(x − 4)(x + 1)
=
= x + 1 whenever x 6= 4.
x−4
x−4
(The original expression is not defined when x = 4.)
More often such expressions do not have common factors. If the numerator
has a power of x that is higher than or equal to the highest power of x in the
denominator, we can rewrite the expression using long division.
2.
x2 − 3x − 2
x−4
x+1
(x − 4)
− 3x − 2
−(x2 − 4x)
x−2
−(x − 4)
2
Therefore
3.
x2
Since x − 4 is not a factor of the numerator,
there is a remainder. But a remainder is just a
number which has not yet been divided by the
divisor. Therefore we can rewrite the original
expression as a sum of terms, one of which is
a fraction.
x2 − 3x − 2
2
=x+1+
, if x 6= 4.
x−4
x−4
3x2 − 3x + 4
x2 − 2x
(x2
Therefore
Numerator and denominator do not have a
common factor. Simplify by long division.
3
− 2x) 3x2 − 3x + 4
−(3x2 − 6x)
3x + 4
Simplify by long division.
The remainder is 3x + 4: we cannot use long
division to divide this by x2 − 2x because the
degree of the denominator is higher than the
degree of the numerator.
3x + 4
3x2 − 3x + 4
=3+ 2
, if x2 − 2x 6= 0.
2
x − 2x
x − 2x
2.6. FACTORING POLYNOMIALS (READ ONLY)
2.6.6
65
Using the quadratic and other formulae
1. Factor y 4 + y 2 − 6.
Since y 4 = (y 2 )2 , we can see this as a quadratic expression in the variable y 2 ,√and use the
−1 ± 1 + 24
, so
quadratic formula on (y 2 )2 + (y 2 ) − 6 = y 4 + y 2 − 6 = 0 to get y 2 =
2
2 + 3) and (y 2 − 2) are factors so that y 4 + y 2 − 6 = (y 2 + 3)(y 2 − 2) =
y 2 = −3 or 2.√
Hence (y√
2
(y + 3)(y − 2)(y + 2).
2.
Factor and solve:
(x + 1)3 + 18 = 10
Rearrange.
(x + 1)3 + 8 = 0
This is the sum of two cubes.
(x + 1)3 + 23 = 0
Use the formula a3 + b3 = (a + b)(a2 − ab +
b2 ).
((x + 1) + 2)((x + 1)2 − (x + 1) × 2 + 22 ) = 0
Simplify.
(x + 3)(x2 + 3) = 0
x = −3 is the only (real) solution.
Practice (Section 2.6).
1. Factor and solve:
(a) x3 − 3x2 − x + 3 = 0,
(d) 27y 3 + 64 = 0,
(c) z 3 − z 2 − z − 2 = 0,
(f) x4 + 10x3 + 35x2 + 50x + 24 = 0.
(b) x4 − 8x2 + 16 = 0,
(e) (x2 − 3)2 = 25,
2. Simplify:
(a)
2x2 − 5x + 2
,
x+1
(b)
x4 − 3x2 + 2x − 1
.
x2 − 2x − 2
66
CHAPTER 2. BASIC ALGEBRA 2
Chapter 3
Basic Algebra 3
3.1
Straight lines
Aim. To know the graphs of straight lines.
Vocabulary and conventions
3.1.1
linear equation
An equation with a line as its graph. It has the form
ax + by + c = 0.
δ
Delta: a (small) increase in something. For example δx means a
small increase in x.
x-intercept
y-intercept
The x-value (y-value) at the point where the graph cuts the x-axis
(y-axis).
gradient
A measure of the slope of a straight line (see below).
Gradient of a line
We want a measure of the steepness of a line.
Construct a right angled triangle starting from this point by moving to the right in the positive xdirection, then moving up or down to meet the curve.
67
68
CHAPTER 3. BASIC ALGEBRA 3
the gradient of this
line is positive
the y-increase
δy
the rise
the x-increase
δx
the run
The gradient of the line is defined as
gradient =
y-increase
δy
rise
=
=
.
x-increase
δx
run
Since all the right angle triangles on this line are similar (the ratios
of the sides are all the same), the gradient is the same at any point
on the line and for any sized triangle.
Remark. If the line slopes down, then the x-increase remains positive but the y-increase is negative.
the x-increase
δx
the run
the y-increase
δy
the rise
the gradient of this
line is positive
3.1. STRAIGHT LINES
3.1.2
69
Gradient of a line in algebraic form m =
y2 −y1
x2 −x1
We often use the name m for the gradient of a straight line.
(x2 , y2 )
m=
δy
= difference in y-values
= y2 − y1
y2 − y1
x2 − x1
(x1 , y1 )
δx
= difference in x-values
= x2 − x1
Remark. It doesn’t matter what order we take the points in, as long as we keep the same order in the
x and y values. That is,
y2 − y1
y1 − y2
=
.
m=
x2 − x1
x1 − x2
This is because
y1 − y2
−(y2 − y1 )
.
=
x1 − x2
−(x2 − x1 )
3.1.3
Gradient of a line in trigonometric form m = tan θ
If θ is the angle the straight line makes with the positive x-axis (measured in an anticlockwise
direction from the positive x-axis) then the gradient is m = tan θ. (We will learn more about this
when we cover trigonometry).
δy
θ
δx
Simple trigonometry tells us that
opposite
δy
=
.
adjacent
δx
Indeed, this is just the definition of the gradient.
tan θ =
70
CHAPTER 3. BASIC ALGEBRA 3
Example.
1. Sketch lines with gradients (a) 2 and (b) −1.
If we let the run be 1, i.e. δx = 1, then δy = m. Now draw the right angled triangles, then we
can extend this to the desired line as below.
(a)
(b)
1
−1
2
1
2. Find the gradient of the line through the points (3, −2) and (1, 4).
We find this using the formula for the gradient given above:
m=
4 − (−2)
6
y2 − y1
=
=
= −3.
x2 − x1
1−3
−2
3. Find the acute angle that the line of gradient
Using the
tan−1
makes with the positive x-axis.
key on a calculator, we find
2
tan θ = m = ,
3
3.1.4
2
3
therefore θ ≈ 33.7◦ .
Equation of a line given a point and a gradient
gradient m
y−y1
x−x1
=m
(x, y)
(x1 , y1 )
Let the known point be (x1 , y1 ) and the gradient be m as shown by the short line segment. Also let
(x, y) be a general point on the line, that is, it can be any point on the line.
3.1. STRAIGHT LINES
71
If (x, y) is on the line, the gradient of the segment connecting (x1 , y1 ) to (x, y) must be m.
If the gradient of the segment connecting (x1 , y1 ) to (x, y) was not m, then (x, y) could not be on the
line: the gradient would be wrong. It might lie on one of the dotted lines, or somewhere else.
Using the algebraic form of the gradient gives
y − y1
= m.
x − x1
Example. Find the equation of the line passing through (1, 4) with gradient 32 .
3
y−4
=
x−1
2
2(y − 4) = 3(x − 1)
2y − 8 = 3x − 3
−3x + 2y − 5 = 0.
3.1.5
Remove the denominators by multiplying through by 2(x − 1) or
by cross multiplying.
Be careful to use brackets in this step.
Equation of a line given two points:
y−y1
x−x1
=
y2 −y1
x2 −x1
Let the known points be (x1 , y1 ) and (x2 , y2 ). Let (x, y) be a general point on the line.
(x2 , y2 )
(x, y)
(x, y)
(x1 , y1 )
If (x, y) lies on the line, the gradient of the line segment connecting (x1 , y1 ) to (x, y) must be the
same as the gradient of the line segment connecting (x1 , y1 ) to (x2 , y2 ). The (x, y) on the dotted line
shows that the two gradients differ if (x, y) lies on any other line than the one joining the two points.
Using the algebraic form of the gradient gives
gradient of segment joining
(x1 , y1 ) to (x, y)
y − y1
y2 − y1
=
x − x1
x2 − x1
gradient of segment joining
(x1 , y1 ) to (x2 , y2 )
Example. Find the equation of the line passing through (1, −4) and (−3, 2).
y − (−4)
2 − (−4)
y+4
6
3
=
=⇒
=
= − =⇒ 2(y + 4) = −3(x − 1)
x−1
−3 − 1
x−1
−4
2
=⇒ 2y + 8 = −3x + 3 =⇒ 3x + 2y + 5 = 0.
72
3.1.6
CHAPTER 3. BASIC ALGEBRA 3
Gradient/intercept form of line
Up to now we have given the equation of a line in general form. A very useful form is
y = mx + c,
where m is the gradient of the line (as x increases by 1, y increases by m) and
c is the y-intercept, i.e. the y-value at which the line cuts the y-axis (when x = 0, y = c).
This form is useful for sketching a line and for reading off a gradient.
Example. Sketch the line 2x − 3y = 9.
First we write the line in gradient/intercept form:
−3y = −2x + 9
3y = 2x − 9
y = 23 x − 3.
As shown here, a handy trick is to change all signs to their
opposites. This is fine as it is just the same as multiplying both
sides by −1.
2
3
Because c = −3, this line cuts the y-axis at (x, y) = (0, −3). Starting from this point, you can count
out the gradient triangle and sketch the line.
c
c is the value of y
when x = 0
The x-intercept is also easy to find from this form: it is the value of x at which the line crosses the
x-axis. The x-axis is defined by y = 0, so we set y = 0 in the equation given and find out what value
x will have in that case:
2
9
2
0 = x − 3 =⇒ 3 = x =⇒
= x.
3
3
2
3.1. STRAIGHT LINES
73
9
2
This tells us the line cuts the x-axis at the point ( 29 , 0) and gives you an alternative method of sketching
the line: find the y-intercept and the x-intercept, mark them on your grid, and then join them by a
straight line.
3.1.7
Parallel and perpendicular lines
If two lines are parallel, they have the same gradient, i.e. m1 = m2 .
If they are perpendicular, the product of the gradients = −1, i.e. m1 m2 = −1. The proof of this
result requires trigonometry and will not be given here. This means that m1 = − m12 : so that a useful
way to find a perpendicular gradient is to invert (i.e. find the reciprocal of) the original and change
the sign.
Example.
1. Show that the line y = 2x + 3 is perpendicular to y = − 12 x + 2.
This is because the lines have gradients 2 and − 21 , whose product is −1.
74
CHAPTER 3. BASIC ALGEBRA 3
y = 2x + 3
5
4
3
2
1
−5
−4
−3
−2
−1
1
2
3
4
5
−1
y = − 12 x + 2
−2
−3
2. Find the equation of the lines passing through (1, −4) which are (a) parallel to and (b) perpendicular
to the line 2x − 3y + 4 = 0, giving the answers in general form.
Write the equation of the line in gradient/intercept form:
2
4
−3y = −2x − 4 =⇒ 3y = 2x + 4 =⇒ y = x + ,
3
3
hence the gradient is 32 .
(a) Parallel lines have identical gradients, so
y+4
2
=
x−1
3
3(y + 4) = 2(x − 1)
3y + 12 = 2x − 2
−2x + 3y + 14 = 0.
(b) For the perpendicular line, we first find the new gradient:
2
3
3
−1
= −1 ÷ = −1 × = − .
2/3
3
2
2
3.1. STRAIGHT LINES
75
Therefore
y+4
3
=−
x−1
2
2(y + 4) = −3(x − 1)
2y + 8 = −3x + 3
3x + 2y + 5 = 0.
3.1.8
Lines parallel to axes
These are best illustrated by example.
Example.
1. Sketch the line with equation x = 3.
The value of x is restricted to 3 while the y-values are unrestricted. Hence we have a vertical
line through x = 3 (on the x-axis). Vertical lines have an infinite gradient because the run (i.e.
x-increase) is always zero. Note that this means that we cannot write their equations in the
form y = mx + c.
2. Sketch the line with equation y = −1.
The x-value is unrestricted while y is restricted to −1. Hence we have a horizontal line through
y = −1. A horizontal line has slope zero because the rise (i.e. y-increase) is always zero.
x
x=3
y
y = −1
Practice (Section 3.1).
1. Find the gradient of the line joining:
(a) (1, 3) and (−1, −4),
(b) (1, 3) and (1, 4).
2. Find the equation of the line:
3
(a) Through (1, −1) with gradient ,
2
(b) Joining (1, 3) and (4, −4).
3. Find the equation of the line through (1, 3)
which is:
(a) Parallel to the x-axis,
(b) Parallel to the y-axis.
4. Find the equation of the line through (4, 1)
which is:
(a) Parallel to x + 3y − 1 = 0,
(b) Perpendicular to x + 3y − 1 = 0.
76
3.2
CHAPTER 3. BASIC ALGEBRA 3
Linear equations
Aim. To be able to solve linear simultaneous equations in two variables.
3.2.1
Interpretation
A pair of equations such as
2x + 3y − 3 = 0,
3x − y + 4 = 0
represents a pair of lines. Writing their equations in gradient/intercept form gives y = − 23 x + 1 and
y = 3x + 4. We plot their graphs below.
y = 3x + 4
4
− 34
3
2
2x + 3y − 3 = 0
The algebraic solution of the pair of equations is the pair (x, y) of values for which both the equations
9
true. That is, it is the point where the two lines meet. In fact, solving these equations gives x = − 11
17
and y = 11 as the only values that are true for both of these equations.
3.2.2
Method of solution
You need a fast and error free method of solving a pair of linear equations. We begin by supposing
the two lines meet at some point. We then need to find the x- and y- values at that point. The method
of elimination is best.
Example.
1. Find the solution to the system
2x − y − 9 = 0,
2x − 4y + 12 = 0.
3.2. LINEAR EQUATIONS
77
2x − y − 9 = 0
2x − 4y + 12 = 0
The terms involving x both have 2 as a
coefficient.
2x − y − 9 = 0
(−)2x − 4y + 12 = 0
3y − 21 = 0
This means we can eliminate the terms
involving x, by subtracting one equation from
the other. Then solve for y using the result.
y=7
We have found the y-value of the point where
they meet.
2x − y − 9 = 0
y=7
Write down one of the original equations
again. Use this and your y-value to find x.
If y = 7 and 2x − y − 9 = 0 are both true,
then 2x − 7 − 9 = 0 is also true.
Now solve for x.
2x − 7 − 9 = 0
2x = 16
x=8
The two lines meet at the point (8, 7). The solution of this pair of equations is x = 8, y = 7.
Often this method will give you a lot of fractions, so that you are more likely to make errors. A
way to avoid problems with fractions is to eliminate both variables. The step by step explanation
for the following problem is given after it.
2. Find the solution to the system
2x + 3y − 3 = 0,
3x − y + 4 = 0.
6x − 9y − 9 = 0
(−)(6x − 2y + 8 = 0)
11y − 17 = 0
17
y=
11
2x − +3y − 3 = 0
(+)(9x − 3y + 12 = 0)
11x + 9 = 0
−9
x=
11
We eliminate x.
Note that the
coefficients of x are 2 and 3.
We eliminate y. The coefficients of y are 3
and −1 (ignoring signs).
The least common multiple of 2 and 3
is 6, so multiply the rows by 3 and 2
respectively.
The least common multiple of 3 and 1 is 3.
So multiply the rows by 1 and 3.
78
CHAPTER 3. BASIC ALGEBRA 3
The coefficients of x are now both 6.
Subtract one equation from the other to
eliminate x.
The coefficients of y are 3 and −3 Add the
equations to eliminate y.
Check
It is a good idea to check the answer by substituting back the values for x and y into the original
equations to see if they are true with these values.
2x + 3y − 3 = 0:
2×
−9
17
−18 + 51
+3×
−3=
− 3 = 3 − 3 = 0,
11
11
11
3x − y + 4 = 0:
3×
−9 17
−27 − 17
−
+4=
+ 4 = −4 + 4 = 0.
11
11
11
Remark. This method requires that the terms in the two equations line up. For example if the
equations had been written in the form
2x + 3y = 3,
y = 3x + 4,
we would first need to rearrange them so the terms lined up:
2x + 3y − 3 = 0,
3x − y + 4 = 0.
3.2.3
No solution
Consider the pair of equations
y = 3x − 4
6x − 2y = 11
We assume that they have a solution.
Rearrange to get the same terms lined up.
3x − y − 4 = 0
6x − 2y − 11 = 0
Eliminate x (the least common multiple of 3
and 6 is 6).
6x − 2y − 8 = 0
−(6x − 2y − 11 = 0)
−8 + 11 = 0
3=0
Notice that y gets eliminated too.
But 3 is not equal to 0! Our assumption that a solution exists must be wrong, since it leads to an
3.2. LINEAR EQUATIONS
79
untrue statement. Hence there is no solution.
We can see why by writing both equations in gradient/intercept form:
y = 3x − 4,
11
y = 3x − .
2
The two lines have the same gradient and hence are parallel. However they have different y-intercepts,
so they never meet.
If the lines had the same gradient and the same y-intercepts, then they would be the same line. That
is, they meet at every point.
Practice (Section 3.2). Find the point where the following pairs of lines meet:
1. 3x + 5y = 7 and
2x − y = 3,
2. 2x + y = 4 and 4x − 3y = 2,
3. 2y = 3x − 7 and
4. y = 2x − 1
9x + 4y = 8,
and 6x − 3y = 1.
80
CHAPTER 3. BASIC ALGEBRA 3
3.3
Relationships with graphs (read only)
Aim. To know the basic graphs.
Vocabulary and conventions
absolute value
asymptote
3.3.1
The absolute value |x| is just the positive value of x, e.g. |2| = 2
and | − 2| = 2.
A line which a curve approaches as x or y approaches infinity.
Note that a curve may cut a (non-vertical) asymptote many times
before approaching it.
For example the x-axis is an asymptote in the graph below.
The graphs
The following are the basic graphs which we will be considering:
y = x2
xy = 1
y = |x|
x2 + y 2 = r 2
x2 y 2
+ 2 =1
a22
b
x
y2
− 2 =1
a2
b
y = ax
parabola,
rectangular hyperbola,
absolute value,
circle,
ellipse,
hyperbola,
power function.
The primary way we will obtain these graphs is by plotting them.
Remark. In order to see what graphs look like we often plot points. This means we choose a set of
x-values and calculate the corresponding y-values, using enough values to give the main details. If
you still can’t see the pattern, choose some more x-values. The first three examples following show
this method, then we move on to more general investigations.
3.3. RELATIONSHIPS WITH GRAPHS (READ ONLY)
3.3.2
81
The parabola y = x2
Choose x-values, calculate y-values and plot.
x
y = x2
−3 −2 −1 0 1 2 3
9
4
1 0 1 4 9
The divisions on the graph are 1 unit each, and the points plotted are shown as dots.
3.3.3
The rectangular hyperbola xy = 1 or y =
1
x
Choose x-values, calculate y-values and plot. The divisions on the graph are 1 unit each.
x
−5
−4
−3
−2
−1
−0.5
−0.25
−0.1
0.1
0.25
0.5
1
2
3
4
5
1
x=
x
−0.2
−0.25
−0.33
−0.5
−1
−2
−4
−10
10
4
2
1
0.5
0.33
0.25
0.2
82
CHAPTER 3. BASIC ALGEBRA 3
Note that as x approaches positive or negative infinity, written x → ±∞, the graph approaches y = 0.
As x approaches 0, written x → 0, the graph approaches y = ±∞. In other words the axes are the
asymptotes of y = x1 .
3.3.4
The absolute value graph y = |x|
Choose x-values, calculate y-values and plot. The divisions on the graph are 1 unit each.
x
y = |x|
3.3.5
−3 −2 −1 0 1 2 3
3
2
1 0 1 2 3
The circle x2 + y 2 = r2
Note that r is the radius of the circle.
This expression comes from Pythagoras’ Theorem applied to the right angled triangle shown.
r
y
x
3.3. RELATIONSHIPS WITH GRAPHS (READ ONLY)
3.3.6
The ellipse
x2
a2
+
y2
b2
83
=1
Note that a and b are positive numbers. In order to make y the subject, we first rearrange the equation
to
x2
a2 − x2
y2
=
1
−
=
.
b2
a2
a2
Therefore
2
a − x2
b2
bp 2
y 2 = b2
= 2 (a2 − x2 ) or y = ±
a − x2 .
2
a
a
a
Remark.
1. This formula is only defined for −a ≤ x ≤ a (otherwise the contents of the square
root are negative, so that when x > a or x < −a, y is undefined).
2. If y = 0, then x = ±a.
√
3. If x = 0, then y = ± ab a2 = ±b.
Hence x lies between −a and a; recall that we can also write −a ≤ x ≤ a. Similarly, −b ≤ y ≤ b.
b
a
−a
−b
2
2
Example. Plot x9 + y4 = 1.
For this graph a = 3 and b = 2, so −3 ≤ x ≤ 3 and −2 ≤ y ≤ 2. First we set up a table of values
2p
for y =
9 − x2 and choose x-values between -3 and 3. Now calculate the corresponding y-values
3
and plot both ±y. The divisions on the graph below are 1 unit each.
y=
2
3
x
√
9 − x2
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3
0.0 1.1 1.5 1.7 1.9 2.0 2.0 2.0 1.9 1.7 1.5 1.1 0.0
2
−3
3
−2
84
CHAPTER 3. BASIC ALGEBRA 3
3.3.7
x2
a2
The hyperbola
−
y2
b2
=1
Note that a, b are positive numbers. Making y the subject:
x2
x2 − a2
y2
=
−
1
=
.
b2
a2
a2
So
2
2
y =b
x2 − a2
a2
=
b2 2
(x − a2 ) or
a2
y=±
bp 2
x − a2 .
a
Remark.
1. This formula is only defined for x ≥ a or x ≤ −a (otherwise the contents of the
square root are negative, so that when x lies in the open interval (−a, a), y is undefined).
2. When y = 0, we find that x = ±a.
3. Hyperbolas have asymptotes at y = ± bx
a . First note that
bp 2
b
y=±
x − a2 = ±
a
a
s
x2
r
bx
a2
a2
1− 2 =±
1 − 2.
x
a
x
The reason for writing the expression in this way is that, as x → ±∞,
a2
a2
→
0
=⇒
1
−
→ 1 =⇒
x2
x2
r
1−
a2
bx
→ 1 =⇒ y → ± .
2
x
a
We will illustrate all this in the example following.
Example. Plot
−
y2
4
y=
2
3
x2
9
= 1.
For this graph a = 3 and b = 2 so x ≤ −3 or x ≥ 3.
x
√
x2 − 9
−7 −6 −5 −4 ±3 4
5
6
7
4.2 3.5 2.7 1.8 0.0 1.8 2.7 3.5 4.2
3.3. RELATIONSHIPS WITH GRAPHS (READ ONLY)
85
As x → ∞, the curve approaches the asymptotes y = ± 23 x.
3.3.8
The power (or exponential) function y = ax
We will plot y = 2x .
Choose x-values, calculate y-values and plot. The divisions on the graph are 1 unit each.
x
y = 2x
2
2
−3 −2 −1 0 1 2 3
0.1 0.3 0.5 1 2 4 8
Example. Sketch x25 + y36 = 1.
Mark the intercepts: a = 5 so the x intercepts are (±5, 0), and b = 6 so the y intercepts are (0, ±6).
Then sketch the ellipse.
86
Example. Sketch
CHAPTER 3. BASIC ALGEBRA 3
x2
25
−
y2
36
= 1.
Draw intercepts and asymptotes: a = 5 so the x-intercepts are (±5, 0). The asymptotes are y = ± 65 x.
Then sketch the hyperbola.
3.3. RELATIONSHIPS WITH GRAPHS (READ ONLY)
Practice (Section 3.3). Sketch:
1. x2 + y 2 = 25,
2.
x2 y 2
+
= 1,
16
9
4.
x2
y2
+
= 1,
9
16
5.
x2
y2
−
= 1,
9
16
6. y = 3(2)x ,
3.
x2 y 2
−
= 1,
16
9
7. y = (0.5)x .
87
88
3.4
CHAPTER 3. BASIC ALGEBRA 3
Non-linear simultaneous equations (read only)
Aim. To be able to solve simultaneous non linear equations in two variables.
3.4.1
Other possibilities for number of solutions
A non linear equation has a graph which is not a straight line.
Consider the following three graphs each showing the intersection of two named curves.
y = x2 − 1
y = −x2 + 1
y = x2
y = −x2
y = x2 + 1
y = −x2 − 1
This illustrates the fact that two curves can intersect, be tangent to each other (touch) or fail to intersect
(miss). We find out whether or not they intersect by substituting a result from one equation into the
other equation.
We now find solutions for each of these three pairs of equations.
(a)
y = x2 − 1
y = −x2 + 1
(b)
y = x2
y = −x2
(c)
y = x2 + 1
y = −x2 − 1
Substitute the value for y from the first equation into the second equation, and solve for x.
x2 − 1 = −x2 + 1
2x2 − 2 = 0
2(x2 − 1) = 0
2(x − 1)(x + 1) = 0
x = −1, 1
x2 = −x2
2x2 = 0
x=0
x2 +1 = −x2 −1
2x2 = −2
x2 = −1
no solution
We substitute the results back into the first equation in order to find y:
y = 12 − 1 = 0
y = (−1)2 − 1 = 0
y = 02 = 0
no solution
The curves:
intersect at (1, 0) and
(−1, 0).
are tangent at (0, 0)
but don’t cross, ie they
touch at (0, 0).
don’t intersect.
3.4. NON-LINEAR SIMULTANEOUS EQUATIONS (READ ONLY)
3.4.2
89
Examples
In the first example, we intersect a straight line and a curve.
Example. Find the intersection of y = x2 − 2x − 12 and x + y + 6 = 0.
Note that the first curve is a parabola and the second a straight line with negative gradient so the
curves will look something like this:
Since y = −x − 6 and y = x2 − 2x − 12 are both true at any point (x, y) where the lines meet, then
at these points:
−x − 6 = x2 − 2x − 12
0 = x2 − x − 6
0 = (x − 3)(x + 2)
x = −2, 3.
To find y it is usually easier substitute these x values into the line equation y = −x − 6:
y = −(−2) − 6 = −4,
y = −3 − 6 = −9,
so the intersections are at (−2, −4) and (3, −9).
Now we have a pair of equations with no common points.
Example. Find the intersection of y = x2 − 2x − 12 and x + y + 18 = 0.
Rewrite the line equation gives y = −x − 18. So
−x − 18 = x2 − 2x − 12
0 = −x2 + x − 6
0 = x2 − x + 6.
The quadratic formula then gives
√
1 − 24
,
2
which is not a real number. So there is no intersection of the line and the curve, i.e. they miss each
other entirely as in the picture below.
x=
1±
90
CHAPTER 3. BASIC ALGEBRA 3
Practice (Section 3.4).
1. Find the intersection of y = x2 − 5x − 9 and y = −x2 + 3x + 1.
2. Find the intersection of y = x2 + 4x − 1 and y = x2 + x + 2.
3. Show y = 6x − 4 is tangent to y = x2 + 4x − 3 and find the point of tangency.
Hint: solving as usual gives a single value of x, which means that y = 6x − 4 must be a tangent
line to the parabola.
4. Show that y = x2 − 3x + 2 and y = x − 5 do not meet.
Chapter 4
Functions 1
4.1
Function notation
Aim. To understand functions.
Vocabulary and conventions
function
argument
4.1.1
A rule that associates pairs of numbers. The statement f (x) = x2
tells us that f is a function which operates on any number x and
√
outputs the square of x. g(x) = x tells us that g operates on any
number x ≥ 0 and outputs its square root.
The number that we put into a function. For example in f (x) = x2 ,
the argument is x (the value in the brackets of the function). If
we want to operate this function on the number 3, we write f (3),
that is, the argument is 3. Replacing every x with 3 in the general
expression f (x) = x2 gives f (3) = 32 .
Introduction
A function is a rule that associates pairs of values in a particular way. Consider, for example, the
function f (x) = x2 . You can think of a function as being like a button on a calculator. In this case f
is like the x2 key. So f (2) = 22 = 4. f associates the number 2 with 4, that is if the input is 2, the
output is 4. If y = f (x), then for x = 2, y = f (2) = 4, so the point (2, 4) is on the graph of f .
A function has another property. If you get one answer on one occasion by putting a number into a
calculator and pressing a function key, and another answer another time by entering the same number
and pressing the same function key, then you know the calculator is not operating properly. In the
same way, a function must only give exactly one output for a given input. If there is more than one
91
92
CHAPTER 4. FUNCTIONS 1
output for a single input, then the graph will have one point directly above another point, that is, it
has two or more points on the same vertical line. It cannot be called a function if it does this.
Notice however that different inputs can give the same output. For example f (x) = x2 is a function
even though f (1) = f (−1) = 1. The graph has two points along the same horizontal line, and f
is still a function. Think of a function as being a general rule. The f (x) = x2 function will square
anything that it operates on. This means a function can take numbers, algebraic expressions or other
functions as arguments.
For example, if f (x) = x2 , then f (−3) = (−3)2 = 9, i.e. we input −3 and the function outputs
(−3)2 . Similarly f (2x + 1) = (2x + 1)2 = 4x2 + 4x + 1, i.e. we input 2x + 1 and the function
outputs (2x + 1)2 .
Example. Let g(x) = 3x − 1. Evaluate:
1. g(2),
2. g(2x − 5),
3. g(x + h) − g(x).
1. g(2) = 3 × 2 − 1 = 5,
2. g(2x − 5) = 3 × (2x − 5) − 1 = 6x − 16,
3. g(x + h) − g(x) = (3 × (x + h) − 1) − (3x − 1) = 3x + 3h − 1 − 3x + 1 = 3h.
When we are referring to graphs of functions, we tend to use y and x as in y = x2 − 1. But we can
also write y = f (x), where f (x) = x2 − 1. If y = f (x), then (x, y) = (x, f (x)) is a point on the
graph of f . Usually we swap back and forth from y to f (x) as needed.
Remark. We say f (x) when we mean one of the outputs of a function, namely the output corresponding
to x. Therefore f (x) is some number. But when we are talking about the function itself, most often
we say f , once we have given its rule.
4.1.2
Domain and range of a function
Consider a function y = f (x). The domain of that function is a set of x-values for which the function
is defined. Functions require a domain to be properly defined. If no domain is given, then it is
assumed to be the set of all x-values for which the function is defined (this is often called the natural
domain).
√
For example in the function g(x) = x we cannot have an argument that is a negative number. This
is because the square root of a negative number is not defined. However we can have 0 and all positive
numbers as arguments, so we say the (natural) domain of the function g is [0, ∞).
1
The domain of the function h(x) = x−2
is every real number except 2. This is because h(x) is not
defined for x = 2, while it is defined for every other real number.
The range of a function is the set of all y-values the function takes on, i.e. the output from the function.
The range of a function depends upon its domain. In the function h above we cannot get zero as an
output. This is because we cannot use infinity as an input (it is not a real number). Therefore the
range of h is every real number except zero. The range of g is [0, ∞).
4.1. FUNCTION NOTATION
93
√
Example. The function f (x) = x − 1 has domain x − 1 ≥ 0, i.e. x ≥ 1 (otherwise we
√ are taking
the square root of a negative quantity). Also its range is y ≥ 0 (for x ≥ 1, y = f (x) = x − 1 must
be zero or greater).
4.1.3
Composite Functions
We have already looked at f (2x + 1) = (2x + 1)2 = 4x2 + 4x + 1, where f (x) = x2 . If we define
g(x) = 2x + 1, we could instead write this as
f (g(x)) = (g(x))2 = (2x + 1)2 = 4x2 + 4x + 1.
In other words we can use another function as√an input (argument).
For example,
if f (x) = 2x2 − 1 and g(x) = x + 1 (for x ≥ −1), then f (g(x)) = 2(g(x))2 − 1 =
√
2 × ( x + 1)2 − 1 = 2x + 1 (for x ≥ −1).
Notice the order in this last example of f (g(x)):
1. Start with x,
2. apply g to get g(x) =
√
x + 1,
3. then apply f to g(x) to get f (g(x)) = 2(g(x))2 − 1 = 2x + 1.
We can also illustrate this diagrammatically as below.
x
g
f ◦g
√
x+1
f
2x + 1
We can get from x to 2x + 1 in one go by defining the single function (f ◦ g)(x) = 2x + 1, as long
as x ≥ −1. We call f ◦ g a composite function and it is defined as (f ◦ g)(x) = f (g(x)).
What about g ◦ f ? We now compute this function:
p
p
√
(g ◦ f )(x) = g(f (x)) = f (x) + 1 = (2x2 − 1) + 1 = 2x, for x ≥ 0,
√
so (g ◦ f )(x) = 2x, for x ≥ 0.
Remark.
1. The order of the functions within the composite function represents the way in which
the functions operate.
2. We can define h(x) = 2x + 1 as a function in its own right. To show its relationship to the
functions f and g above, we could write h(x) = (f ◦ g)(x), for x ≥ −1.
94
CHAPTER 4. FUNCTIONS 1
Example. Let f (x) = 3x − 1 and g(x) =
1. g ◦ f ,
1. (g ◦ f )(x) = g(f (x)) =
1
when x 6= −1. Find:
x+1
2. f ◦ g,
1
1
1
=
=
, where f (x) 6= −1,
f (x) + 1
3x − 1 + 1
3x
2. (f ◦ g)(x) = f (g(x)) = 3g(x) − 1 =
3. (g ◦ g)(x) = g(g(x)) =
but
4.1.4
3. g ◦ g.
1
=
g(x) + 1
3
3 − (x + 1)
−x + 2
−1=
=
,
x+1
x+1
x+1
1
, x 6= −1, g(x) 6= −1,
+1
1
x+1
1+x+1
x+2
x+1
1
+1=
=
, so (g ◦ g)(x) =
when x 6= −1 and g(x) 6= −1.
x+1
x+1
x+1
x+2
Inverse functions
Consider the function f , where f (x) = x + 1.
The inverse function f −1 is the function that “undoes” the work of f ; likewise f “undoes” the work
of f −1 . For this to happen, we need f −1 (x) = x − 1.
f
x
f −1
x+1
Notice that (f −1 ◦ f )(x) = x and (f ◦ f −1 )(x) = x (check this).
That is, the functions f −1 ◦ f and f ◦ f −1 do not change anything, the output is the same as the input.
4.1.5
Calculating inverse functions
Consider a function y = f (x).
x
f
f −1
y
If we write the original function in terms of x and y we can find the inverse function by swapping x
and y over and solving for the new y to get f −1 (x) = y.
Example. Find the inverse of f (x) = 2x − 1 and show it is the inverse.
First write y = 2x − 1, then swap x and y to get x = 2y − 1. Now make y the subject:
2y = x + 1
1
y = (x + 1).
2
4.1. FUNCTION NOTATION
95
We claim that f −1 (x) = y = 21 (x + 1). As a check,
1
(f ◦ f −1 )(x) = f (f −1 (x)) = 2f −1 (x) − 1 = 2 × (x + 1) − 1 = x,
2
so we really have found the correct inverse function.
Remark. If y = f (x) then x = f −1 (y). This is because
y = f (x) =⇒ f −1 (y) = f −1 (f (x)) by applying the function f −1 to each side
=⇒ f −1 (y) = x.
4.1.6
Inverses and graphs
If y = f (x), the inverse of f is found by swapping x and y and solving for the new y to get y =
f −1 (x). This means the graph of f −1 is the reflection of the graph of f in the line y = x. Thus the
graph of the inverse function f −1 can be obtained by reflection.
Example. If f (x) = 2x then f −1 (x) = x2 .
f (x) = 2x
y=x
f (x) =
x
2
The following example shows that, although we have spoken of inverse functions up to now, the
inverse of a function may need some restrictions in order to be a function itself.
Example. Find the inverse of y = x2 .
√
To solve this algebraically, we first swap x and y to get x = y 2 . Now make y the subject: y = ± x.
Otherwise we may do this graphically (below) by reflecting in the line y = x.
y = x2
y=x
x = y2
96
CHAPTER 4. FUNCTIONS 1
√
Either from the equation or the graph it can be seen that the result y = ± x has two y-values for each
√
possible x-value. This means that y = ± x is not a function. However it can be made a function by
restricting the allowable y-values (the range) of the inverse to (say) positive values and zero only.
√
Thus, the inverse defined by y = x is a function, and it is the inverse of y = x2 for x ≥ 0.
Practice (Section 4.1). Given f (x) = 4x − 1, g(x) =
1
x
and h(x) = x2 − 4, find:
1. f (2),
3. h(1),
5. (g ◦ g)(x),
7. f −1 (x),
2. g(−1),
4. (f ◦ g)(x),
6. (h ◦ f )(x),
8. g −1 (x).
Find the domain and range of:
9. f ,
10. g,
11. h.
12. Sketch h and h−1 on the same set of axes (where h−1 is restricted to make it a function).
4.2. TRANSFORMATIONS (READ ONLY)
4.2
97
Transformations (read only)
Aim. To learn how to draw the graphs of transformed functions.
4.2.1
Shifting graphs
Consider the graph of the function f (x) = x2 . It passes through the points (0, 0), (−1, 1), (1, 1), as
well as many others. Generally we can say it passes through the point (x0 , y0 ), where y0 = x20 .
(−2, 4)
(1, 1)
(0, 0)
Suppose we shifted the graph: say 1 unit to the left, and 2 units up.
(−3, 6)
(x, y) = (x0 − 1, y0 + 2)
(0, 3)
(−1, 2)
It now passes through the points (0, 3), (−1, 2), (−3, 6) and others. Shifting the point (x0 , y0 ) on
the first graph 1 unit to the left and 2 units up tells us that the second curve passes through the point
(x, y), where (x, y) = (x0 −1, y0 +2), that is, 1 unit to the left and 2 above (x0 , y0 ). Then x = x0 −1
and y = y0 + 2, so that x0 = x + 1 and y0 = y − 2. Also y0 = x20 . This means that
y0 = (y − 2) = x20 = (x + 1)2 , where (x, y) is a point on the second graph.
98
CHAPTER 4. FUNCTIONS 1
Expanding gives
y − 2 = (x + 1)2 = x2 + 2x + 1,
and so
y = x2 + 2x + 3.
Then the second graph is that of the function y = g(x) = x2 + 2x + 3.
If a function can be rewritten as
y − b = (x − a)2
for some constants a and b, then it is the graph of y = x2 shifted a units to the right and b units up.
We can see this as:
• x → x − a shifts the graph to the right by a, and
• y → y − b shifts the graph up by b.
y − b = (x − a)2
y = x2
(a, b)
(0, 0)
Example. Quickly sketch the graph of y = (x − 2)2 − 1.
We can rewrite this equation as y + 1 = (x − 2)2 . Therefore the graph is y = x2 , with x → x − 2
and y → y + 1 = y − (−1). That is, the graph is the parabola y = x2 shifted to the right by 2 and up
by −1, i.e. down by 1.
y = (x − 2)2 − 1
4.2. TRANSFORMATIONS (READ ONLY)
4.2.2
99
Translations
A translation moves a point or set of points by (a, b), i.e. a units in the x direction and b units in the
y-direction. We can say that the graph of
y − b = (x − a)2
is what we get when every point on the graph of y = x2 is moved by (a, b). In the previous example
we found that y = (x − 2)2 − 1 is just y = x2 shifted by (2, −1), i.e. 2 in the x direction and −1 in
the y-direction.
More generally, we can say that the graph of
y − b = f (x − a)
can be sketched by translating the graph of y = f (x) by (a, b).
Translating by (a, b):
4.2.3
The old x is a less than the new x
The old y is b less than the new y
x→x−a
y →y−b
Applications
We will use translations to help us to identify curves and to write their equations in simpler forms.
Example. What kind of curve is (x − 1)2 + (y + 2)2 = 4?
Since x → x − 1 and y → y + 2, it represents a translation of the circle x2 + y 2 = 4 by (1, −2).
(x − 1)2 + (y + 2)2 = 4
(1, −2)
Example. Give the equation of this translation of the parabola y = x2 whose vertex is at (1, −2).
100
CHAPTER 4. FUNCTIONS 1
(1, −2) vertex
The translation shifts the vertex at (0, 0) by (1, −2), i.e. x → x − 1 and y → y + 2. Therefore the
equation is y + 2 = (x − 1)2 or equivalently y = x2 − 2x − 1.
Many examples involve completing the square.
Example. What kind of curve has the equation x2 + 2x + y 2 − 4y + 1 = 0?
We complete the square separately for x2 + 2x and y 2 − 4y before simplifying:
(x + 1)2 − 1 + (y − 2)2 − 4 + 1 = 0
(x + 1)2 + (y − 2)2 = 4.
The curve is a translation of the circle x2 + y 2 = 4 by (−1, 2).
Example. What kind of curve has the equation 4x2 − 16x + 9y 2 − 18y − 11 = 0?
We first rearrange the equation:
4(x2 − 4x) + 9(y 2 − 2y) − 11 = 0
4([x − 2]2 − 4) + 9([y − 1]2 − 1) − 11 = 0.
Notice how the x terms are separated from the y terms (by the curved brackets) before being factored.
The square is then completed inside each bracket and we fiddle with the result until we get something
familiar:
4(x − 2)2 − 16 + 9(y − 1)2 − 9 − 11 = 0
4(x − 2)2 + 9(y − 1)2 = 36
4(x − 2)2 9(y − 1)2
+
=1
36
36
(x − 2)2 (y − 1)2
+
= 1.
9
4
This shows that the curve is the ellipse
x2
9
+
y2
4
= 1 translated by (2, 1).
4.2. TRANSFORMATIONS (READ ONLY)
4.2.4
101
Stretches
The transformations x → xk and y → ky stretch the curve by a factor of k parallel to the x-axis and
the y-axis, respectively. If k < 0 the curve is also reflected in the other axis. Consider the following
4 curves.
The first is y = x2 . The second (y = ( x2 )2 ) is the first stretched by a factor of 2 parallel to the x-axis:
y = x2
(−2, 4)
y = ( x2 )2
(1, 1)
(−2, 1)
(1, 14 )
(0, 0)
(0, 0)
The third is the first stretched by a factor of 2 parallel to the y-axis: y → y2 . The fourth is the first
y
stretched by a factor of −1 parallel to the y-axis: y → −1
, this reflects the curve in the x-axis.
y
2
= x2 or y = 2x2
(−2, 4)
(0, 0)
(1, −1)
(1, 2)
(−2, −4)
y = −x2
(0, 0)
x
The stretch x → −1
reflects the curve y = x2 in the y-axis, but as it was already its own reflection
x
in this axis, we again have the first curve. The same stretch, x → −1
, applied to y = x2 + 2x + 3
gets y = (−x)2 + 2(−x) + 3 = x2 − 2x + 3. You can easily see below that it is the reflection of
y = x2 + 2x + 3 in the y-axis.
102
CHAPTER 4. FUNCTIONS 1
y = x2 + 2x + 3
y = x2 − 2x + 3
(3, 6)
(−3, 6)
(−0, 3)
(0, 3)
(−1, 2)
(1, 2)
Example. Give the equation of the circle x2 + y 2 = 1 if it is stretched by a factor of 2 parallel to the
y-axis and 3 parallel to the x-axis.
1
1
We apply the transformations x →
x
3
and y →
y
2
to get
x2
9
+
2
3
Example. The graph shown is of the curve y = f (x).
y2
4
= 1, which is an ellipse.
4.2. TRANSFORMATIONS (READ ONLY)
103
Sketch the graph of:
1. y = f (x − 1) + 2,
2. y = 2f (x + 2),
3. y = −3f (x + 2) + 1.
1. Rearrange the equation as y − 2 = f (x − 1), hence the original function f is translated by
(1, 2).
y
2. Rearrange the equation as = f (x + 2), so f is shifted 2 to the left then stretched out by a
2
factor of 2 parallel to the y-axis.
3. Rearrange as follows:
y = −3f (x + 2) + 1
y − 1 = −3f (x + 2)
y−1
= f (x + 2)
−3
The original is stretched out by a factor of 3 parallel to the y-axis, reflected in the x-axis (for
the negative), then shifted left 2 and up 1.
104
CHAPTER 4. FUNCTIONS 1
Practice (Section 4.2).
1. What is the equation of y = 2x2 + 1 when the curve is translated by:
(a) (1, −2),
(b) (3, 2).
2. Identify the following curves:
(a) x2 − 2x + y 2 + 8y − 8 = 0,
(b) x2 + 2x + 4y 2 − 16y + 13 = 0,
(c) x2 − 2x − y 2 + 6y − 9 = 0.
3. Give the equations of:
(a) y = 2x − 1 stretched parallel to the x-axis by a factor of -2,
(b) x2 + y 2 = 4 stretched parallel to both axes by a factor of 2,
(c) y = x2 − 1 stretched parallel to the y-axis by a factor of -2.
4. The following refer to the graph of y = f (x) here shown. The units are all 1.
Sketch:
(a) y = f (x − 4) + 3,
(b) y = −2f (x),
(c) y = 2f (x + 1) − 4.
4.3. ABSOLUTE VALUE (READ ONLY)
4.3
105
Absolute value (read only)
Aim. To learn to understand and use the absolute value function.
Vocabulary and conventions
|x|
⇐⇒
4.3.1
The absolute value (or modulus) of x.
This means “if and only if” and is sometimes written “iff”. That is,
statement a ⇐⇒ statement b means: if statement a is true, then
statement b is true and if statement b is true, then statement a is true.
For example, I am shorter than you ⇐⇒ you are taller than me.
Definition of |x|
We define the absolute value as follows:
(
x, if x ≥ 0
|x| =
−x, if x < 0.
The absolute value effectively makes a negative number positive and leaves a positive number or zero
alone. This is what the definition is saying.
Consider |3|. The argument of the absolute value (the number inside the absolute value brackets) is
3, a positive number. Therefore we must use the top line of the definition above, and write
|3| = 3.
Now consider | − 3|. Since the argument is −3 which is a negative number, we must use the bottom
line of the definition and write
| − 3| = −(−3) = 3.
We can also use the absolute value on expressions with variables. For example, using the definition
with 2x + 1 as the argument, we get
(
2x + 1, if 2x + 1 ≥ 0
|2x + 1| =
−(2x + 1), if 2x + 1 < 0.
This has two inequalities which need to be solved (Section 2.4):
−1
−1
and 2x + 1 < 0 ⇐⇒ x <
.
2
2
The expression above can now be rewritten:
(
2x + 1, if x ≥ −1
2
|2x + 1| =
−(2x + 1), if x < −1
2 .
2x + 1 ≥ 0 ⇐⇒ x ≥
106
4.3.2
CHAPTER 4. FUNCTIONS 1
Working with the absolute value
If the argument can be factored, we can simplify absolute value expressions. This is because
|a|
a
=
b
|b|
|ab| = |a| × |b|,
and
|an | = |a|n .
We can use these to simplify some expressions involving absolute values:
|3x2 + 6x| = |3x(x + 2)| = |3| × |x| × |x + 2| = 3|x||x + 2|,
−3x
| − 3| × |x|
3|x|
=
=
,
y
|y|
|y|
3x3 = 3|x|3 .
Notice that x2 = x2 since a square is positive or zero anyway. See also that
p
√
(−3)2 = 9 = 3 = | − 3|,
so that more generally, we find that
√
a2 = |a|.
We proceed to show that this equation
√ holds true for any x using the same
√ arguments as above. We
2 = x. But now we can consider
have
written
earlier
that
for
x
≥
0,
x
x2 for x < 0. Recall that
√
x2 is the positive square root of x2 . If x is negative,
then −x is positive and so −x must be that
√
positive square root. Hence for x < 0, we have x2 = −x = |x|.
4.3.3
Equalities and inequalities with the absolute value function
Think of the absolute value as a distance. While we can consider one direction to be negative and
the other positive as on a number line, if we have travelled 5 kilometres we don’t think of this as
−5 kilometres even if we travelled it backwards. The absolute value function can be considered as a
distance function, that is, it gives values for the distances between numbers. For example, |3 − 1| = 2.
In words, this says that the distance between 3 and 1 is 2. Equally we have |1 − 3| = 2 because the
distance between 1 and 3 is 2.
The statement |x − 1| = 2 is most easily understood and solved for x if we rewrite it in English as:
‘the distance of x from 1 is 2’. We draw a number line and a picture of this situation below.
x is exactly 2 units away from
1, so it must be −1 on this side
−2
−1
0
x is exactly 2 units away from
1, so it must be 3 on this side
1
2
3
4
This shows that x = 3 or x = −1. Usually we refer to finding a solution in this manner as ‘solving
the equation geometrically’.
4.3. ABSOLUTE VALUE (READ ONLY)
107
We can also solve this problem algebraically. Since
(
x − 1, if x − 1 ≥ 0, i.e. if x ≥ 1
|x − 1| =
−x + 1, if x − 1 < 0, i.e. if x < 1,
we can write two equalities:
when x ≥ 1, |x − 1| = x − 1 = 2
and when x < 1, |x − 1| = −x + 1 = 2
=⇒ x = 3(≥ 1),
=⇒ x = −1(< 1).
Again we have the same result: x = 3 or x = −1.
Example. Solve |x − 2| < 3 for x.
First translate it into English: ‘the distance of x from 2 is less than 3’.
x is less than 3 units away from 2,
so on this side x < 2 but x > −1
−2
−1
0
x is less than 3 units away from 2,
so on this side x > 2 but x < 5
1
2
3
4
5
6
Therefore −1 < x < 5.
The hollow circles indicate that the endpoints of the interval are not solutions for x. Filled circles as
on the previous diagram indicate that the points are solutions for x. The heavy line is used to indicate
places x can be found (solutions for x).
We can also solve this problem algebraically.
For x ≥ 2 :
|x − 2| = x − 2 < 3 =⇒ x < 5,
not forgetting this solution is only valid for x ≥ 2 . So one part of the solution is 2 ≤ x < 5.
For x < 2 :
|x − 2| = −x + 2 < 3 =⇒ x > −1,
and this part of the solution is valid for x < 2. The other part is −1 < x < 2.
Combining these two parts, we have −1 < x < 5.
Example. Solve |x| ≤ 2 for x.
This says that ‘the distance of x from zero is less than or equal to 2’.
x is less than or equal
to 2 units away from
0, so −2 ≤ x ≤ 0 on
this side of 0
−3
Therefore −2 ≤ x ≤ 2.
−2
−1
x is less than or equal
to 2 units away from
0, so 0 ≤ x ≤ 2 on this
side of 0
0
1
2
3
108
CHAPTER 4. FUNCTIONS 1
From these results we can write some general rules:
|x| < a ⇐⇒ −a < x < a
and |x| ≤ a ⇐⇒ −a ≤ x ≤ a,
for a ≥ 0.
See that we can use one of these rules for the previous example:
|x − 2| < 3 ⇐⇒ −2 < x − 2 < 3 ⇐⇒ −1 < x < 5.
Now consider a situation like, solve for x, |x + 1| > 4. There is another rule covering this situation:
|x| > a ⇐⇒ x > a or x < −a,
for a ≥ 0.
Using this rule, we get |x + 1| > 4 ⇐⇒ x + 1 > 4 or x + 1 < −4. Sorting this statement out gives
x > 3 or x < −5.
Alternatively we can just look at the number line: first rewrite |x + 1| > 4 to |x − (−1)| > 4 and
translate:
‘the distance of x from -1 is more than 4’.
x is more than 4 units away from
−1, so x < −5 on this side of −1
−5
x is more than 4 units away from
−1, so x > 3 on this side of −1
−4
−3
−2
−1
0
1
2
3
Then x > 3 or x < −5.
Example. Solve |2x + 3| > 1 for x.
Use the formula: |2x + 3| > 1 ⇐⇒ 2x + 3 > 1 or 2x + 3 < −1. Sorting these out gives 2x > −2
or 2x < −4, that is, x > −1 or x < −2.
Otherwise solve geometrically. This inequality says that the distance of 2x from −3 is greater than 1.
This is not so easy to use because whenever we find a solution, 2x, we will have to halve it to get x.
So we do some work before this. We can divide both sides of the inequality by 2 giving:
|2x + 3|
1
> .
2
2
As
|2x + 3|
|2x + 3|
2x + 3
3
= x+ ,
=
=
2
|2|
2
2
we can replace the original inequality with the statement
x+
That is, the distance of x from
x < −2.
−3
2
3
1
> .
2
2
is more than 12 , and solve as before. Again we have that x > −1 or
4.3. ABSOLUTE VALUE (READ ONLY)
109
We can get some quite complicated problems.
Example. Solve |x + 2| > |x − 1| for x.
We could use all the rules given, but it will get rather complex working out which is true where. Much
the simplest method of solution is the geometric method. Translate the statement to ‘the distance of
x from −2 is greater than the distance of x from 1’.
If x is to be nearer to 1 than to −2, we cannot have x at the midpoint between them which is − 12 , so
we can mark that with a hollow circle (recall that this means it is not a solution). If x is closer to 1
than to −2, the solutions must be on the right hand side of this circle. Choose a few points on this
side – they are all closer to 1 than to −2, so they are all solutions for x. Now look at the left side of
the circle – every point there is closer to −2 than to 1, so they are not solutions for x.
x is closer to 1 than to −2, so every point
on this side of − 12 is a solution for x
−5
−4
−3
−2
x is closer to −2 than to 1, so there
are no solutions on this side of − 12
−1
0
1
2
3
1
We have shown that x > − .
2
4.3.4
The triangle inequality
If we don’t know the value of a, then we do not know the value of |a|. But from the definition of the
absolute value, we know that |a| = a or |a| = −a; that is, we know that
a = |a|
or a = −|a|.
Therefore it is sometimes useful to note that
−|a| ≤ a ≤ |a|.
Now consider |a + b|.
Since −|a| ≤ a ≤ |a| is true, and −|b| ≤ b ≤ |b| is true, we can add these two inequalities and find
that
−|a| − |b| ≤ a + b ≤ |a| + |b|
is true. But it is also true that
a + b = |a + b| or
a + b = −|a + b|.
If we replace a + b in the last inequality with either of its possible values, we get:
either
− |a| − |b| ≤ |a + b| ≤ |a| + |b| or
− |a| − |b| ≤ −|a + b| ≤ |a| + |b|.
110
CHAPTER 4. FUNCTIONS 1
One of these lines must be true, but in fact they both say the same thing because if we multiply the
second line by −1 (and we need then to reverse the inequality) we get the first line, and vice versa.
The fact that |a + b| ≤ |a| + |b| is called the triangle inequality.
Practice (Section 4.3). Solve the following for x:
1. |x − 1| ≤ 4,
3. |2x − 3| > 1,
5. |x| < |x + 2|,
7. 5 − x < |x|,
2. 5 ≤ |x + 13|,
4. |3x + 1| ≤ 2,
6. |x−8| < |x−24 |,
8. 2x − 3 < |x + 1|.
Chapter 5
Trigonometry 1
5.1
Angles
Aim. To learn how the trigonometric functions are defined for all angles.
Vocabulary and conventions
This is a circle with radius 1 and with centre (0, 0).
Often used to refer to angles. Some letters commonly used are: α
(alpha), β (beta), φ (phi) and θ (theta).
The direction of a positive angle is anticlockwise; hence the
direction is a clockwise only if the angle is negative.
unit circle
Greek letters
direction
5.1.1
Angles between 0◦ and 90◦
For the point marked P , at the end of the radius of the unit circle below, we define cos θ (cosine) and
sin θ (sine) by letting (cos θ, sin θ) be the coordinates of P .
P (x, y)
1
θ
1
θ
x
111
y
x = 1 cos θ
y = 1 sin θ
112
CHAPTER 5. TRIGONOMETRY 1
Scaling this triangle up:
h
θ
x = h cos θ
y = h sin θ
y
x
We also define the tangent by
tan θ =
But
sin θ
.
cos θ
sin θ
h sin θ
y
=
=
cos θ
h cos θ
x
from the scaled triangle, so
y
= tan θ =⇒ y = x tan θ.
x
The traditional way of remembering these is to label the sides of the triangle as follows:
hypotenuse = h
opposite = o
θ
adjacent = a
This then gives the traditional relationships
o = h sin θ
a = h sin θ
o = a tan θ
or
sin θ = ho ,
cos θ = ha ,
tan θ = ao .
Example. Find sin θ, cos θ and tan θ for the angle θ shown below.
5
3
θ
4
3
4
3
Using the relations given above, sin θ = , cos θ = and tan θ = .
5
5
4
5.1.2
Angles between 90◦ and 180◦
Extend the definition of cosine and sine to the angle θ shown, by defining them to be the coordinates
of P , the point at the end of the radius. This means that x, and hence cos θ is now negative.
5.1. ANGLES
113
By definition:
x = cos θ
y = sin θ
P (x, y)
θ
1
Relate this new definition back to the basic definitions which use angles less than 90◦ by considering
lengths on the shaded triangle, where φ is the acute angle made with the x-axis, and θ = 180◦ − φ.
This means the coordinates of P are (cos θ, sin θ) = (− cos φ, sin φ).
1
1 sin ϕ
ϕ
1 cos ϕ
Thus
cos θ
sin θ
tan θ
=
=
=
− cos φ
sin φ
− tan φ
− cos(180◦ − θ),
sin(180◦ − θ),
− tan(180◦ − θ).
=
=
=
These equalities follow because 180◦ − θ = φ and tan =
sin
cos .
Example. Find sin θ, cos θ, and tan θ for the angle θ shown below.
√
1
5
ϕ
θ
2
Using the above notation for φ, we have sin φ =
sin θ = sin φ =
5.1.3
√1 ,
5
cos θ = − cos φ =
− √25
√1 ,
5
cos φ =
√2
5
and tan θ = − tan φ =
and tan φ =
1
2.
Therefore
− 12 .
Angles between 180◦ and 270◦
Extend the definition of cosine and sine to the angle θ shown by defining (cos θ, sin θ) to be the
coordinates of P (the point at the end of the radius). This means both x and y and hence cos θ, and
sin θ are both negative.
114
CHAPTER 5. TRIGONOMETRY 1
θ
P (x, y)
By definition:
x = cos θ
y = sin θ
1 cos ϕ
ϕ
1 sin ϕ
1
Relate this new definition back to the basic definitions which use angles less than 90◦ by considering
lengths on the shaded triangle, where θ = 180◦ + φ and φ is the acute (positive) angle made with the
x-axis. This gives P = (− cos φ, − sin φ).
Thus
cos θ
sin θ
tan θ
=
=
=
− cos φ
− sin φ
tan φ
=
=
=
− cos(θ − 180◦ ),
− sin(θ − 180◦ ),
tan(θ − 180◦ ).
These equalities follow because θ − 180◦ = φ and tan =
sin
cos .
Angles between 270◦ and 360◦
5.1.4
θ
P (x, y)
By definition:
x = cos θ
y = sin θ
1 cos ϕ
ϕ
1 sin ϕ
1
cos θ
sin θ
tan θ
=
=
=
cos φ
− sin φ
− tan φ
Here 0◦ ≤ φ ≤ 90◦ . These equalities follow because θ = 360◦ − φ and tan =
sin
cos .
Example. Express sin 130◦ , cos 130◦ and tan 130◦ in terms of acute angles.
1. Draw 130◦ on a set of axes and find the acute angle with the x-axis.
50◦
2. Decide on the sign using the diagram below. This summarises the previous information by
showing where each trigonometric values is positive.
5.1. ANGLES
115
sin
all
tan
cos
The radius for 130◦ falls in the sin positive region, so sin is positive. Similarly, cos and tan are
negative. Therefore
sin 130◦ = sin 50◦ ,
cos 130◦ = − cos 50◦
tan 130◦ = − tan 50◦ .
and
Example. Find the angle θ that the radius passing through (−1, −2) makes with the positive x-axis.
1. Draw the radius on a set of axes and draw the triangle containing the acute angle made with the
x-axis.
1
ϕ
2
(−1, −2)
2. Turn the coordinates into lengths (remember lengths are always positive).
3. Work out the acute angle using the sides of this triangle:
2
1
φ = tan−1 2
tan φ =
(= arctan 2) using the calculator
◦
φ = 63.4 .
4. The final angle is θ = 180◦ + 63.4◦ = 243.4◦ .
Practice (Section 5.1).
1. Express sin, cos and tan of the following angles in terms of the sin,
cos or tan of an acute angle.
(a) 145◦
(b) 310◦
(c) 228◦
(d) 340◦
2. Find the angle the radius passing through each of the following points makes with the positive
x-axis.
(a) (−4, −5)
(b) (−1, 7)
(c) (3, −2)
(d) (1, 7)
116
5.2
CHAPTER 5. TRIGONOMETRY 1
Radians
Aim. To be able to convert between degrees and radians,
To be able to estimate an angle given in radians in degrees,
To know the relationship between simple multiples of π and degrees,
To be able to use the formula for arc length.
Vocabulary and conventions
radian
A measure of angle used especially in Calculus.
subtend
An arc, AB, of a circle subtends an angle, θ, at the centre as shown
below.
A
θ
B
direction
angles
of
Positive angles are measured anticlockwise from the positive
direction of the x-axis, negative clockwise.
135◦
−225◦
d.p.
The number of decimal places used, e.g. 1.231 = 1.23 to 2 d.p.
s.f.
The number of significant figures used, e.g. 34.5637 = 34.6 to 3
s.f.
≈
Approximately equal to e.g. 3.15 ≈ 3.2.
5.2. RADIANS
5.2.1
117
Radians
Consider an equilateral triangle. This is where each angle is 60◦ .
60◦
Now consider the sector of a circle with arc length of one radius, r, as shown.
r
r
1 rad
1 rad
r
We get a figure which is like an equilateral triangle, with all sides having length equal to the radius of
the circle, with one curved side. We call the marked angle a radian:
length of arc created by 1 radian
= 1.
length of radius
We abbreviate radian to rad. In terms of degrees, 1 rad ≈ 60◦ .
The circumference of a circle is 2πr. This is the length of the arc created by a full turn, so that
length of arc created by a full turn
2πr
=
= 2π.
length of radius
r
That is, the radius fits onto the circumference of a circle 2π times so that
2π rad = 360◦ .
Dividing both sides by 2π gives
1 rad =
180◦
.
π
Alternatively, dividing both sides by 360 gives
1◦ =
π
rad.
180
This gives the conversion factors shown in the diagram.
degrees
π
× 180
× 180
π
radians
118
CHAPTER 5. TRIGONOMETRY 1
The only problem with using radians is that there is no simple integer multiple in radians that
corresponds to a full turn (360◦ ) as one full turn is 2π radians. As a consequence, the π is left in
◦
many expressions for angles in radian form, e.g. π2 , π and 3π
2 (respectively a quarter turn (90 ), a half
turn (180◦ ), and a three quarter turn (270◦ ).
In the absence of other information the convention is to specify degrees to 1 dp and radians to 4
significant figures.
Example. Convert:
1. 38◦ to radians,
1. 38◦ = 38 ×
5.2.2
π
180
2. 1.34 radians to degrees.
2. 1.34 rad = 1.34 ×
rad = 0.6632 rad
180◦
π
= 76.8◦ .
To estimate an angle given in radians in degrees
As 1 radian is approximately 60◦ , we can make rough estimations of angles. Each tenth of a radian is
about 6◦ .
Example. Give a rough sketch of an angle of 1.34 rad.
Since this angle is approximately 60◦ +18◦ (ignore the 0.04), i.e. approximately 80◦ , we give a sketch
below.
≈ 80◦
5.2.3
To know the relationship between simple multiples of π and degrees
Memorise the following:
90◦ =
π
2
rad,
180◦ = π rad and 360◦ = 2π rad
and learn to do problems like converting
3π
radians into degrees mentally.
2
Example. Convert mentally:
1.
5π
4
2. 270◦ to radians.
rad into degrees, and
1. Since 90◦ = 90 ×
2. Because 90◦ =
π
2
π
180
rad =
π
2
rad, we have
rad, we have 270◦ =
3π
2
5π
4
rad.
rad = 225◦ .
5.2. RADIANS
5.2.4
119
The formula for arc length
Since 1 radian corresponds to 1 radius arc length, the angle of θ radians correspond to θ radii arc
length. Thus s = rθ, where s is the arc length corresponding to an angle of θ radians.
r
r
s
1 rad
θ rad
r
Remark. If the angle is not already in radians you will need to convert it to use this formula.
Example. Find the arc length subtended by an angle of:
2. 57◦ in a circle of radius 2.1 m.
1. 2.3 rad in a circle of radius 3 cm
1. We use the formula given above: s = rθ = 3 × 2.3 = 6.9 cm.
2. First we convert the angle before finding the arc length: s = rθ ×
2.089 m.
Practice (Section 5.2).
1. Convert:
(c) 1.78 rad into degrees
2. Sketch an angle of:
(a) 73◦ into radians
π
180
= 2.1 × 57 ×
(b)359◦ into radians
(d) 5.3621 rad into degrees
(a) 0.623 rad
(b) 2.54 rad
3. Mentally convert the following:
(a) 90◦ , 180◦ , 270◦ , 360◦ , 450◦ , 30◦ , 60◦ , 45◦ into radians
(b)
π 3π 5π
5π π 2π
2 , 2 , 2 , 2π, 4 , 6 , 3
into degrees
4. Find the arc length on a circle of the given radius subtended by the given angle:
(a) radius 3.4 m and angle 3.25 rad
(b) radius 2.75 m and angle 135.4◦ .
π
180
=
120
5.3
CHAPTER 5. TRIGONOMETRY 1
Special angles
Aim. To know the trigonometrical ratios corresponding to angles of:
45◦ =
π
,
4
30◦ =
π
6
and
60◦ =
π
.
3
Vocabulary and conventions
Pythagoras
Pythagoras’s theorem states that h2 = a2 + b2
h
a
b
altitude
The altitude of a triangle is the perpendicular from the vertex to the
base.
altitude
5.3.1
Special triangles
The trigonometrical ratios of the special angles are read from two special triangles which you need to
learn to draw:
√
• The 1 : 1 : 2 triangle
Draw the right √
angled isosceles triangle with equal sides 1. Pythagoras gives the length of the
hypotenuse as 2. The angles inside a triangle must add up to π rad. Also, the angles which
are opposite the sides of equal length on the isosceles triangle must be equal. This means that
the acute angles are both π4 rad (= 45◦ ).
√
45◦
1
2
√
π
4
1
π
4
45◦
1
2
1
5.3. SPECIAL ANGLES
121
sin 45◦ = sin
• The 1 : 2 :
√
π
1
π
= √ = cos = cos 45◦
4
4
2
and so
tan 45◦ = tan
π
=1
4
3 triangle
Draw the equilateral triangle with equal sides of length 2. If all sides have the same length,
then all the angles must be equal. This means than each angle is π3 (= 60◦ ). Divide the triangle
in two as shown below. Now we consider only the left √
half of the triangle (the last√
triangle at
right). Pythagoras gives the length of the altitude to be 3. The sides are 1, 2 and 3. These
correspond to the opposite angles π6 , π2 and π3 , respectively.
30◦ =
2
π
6
2
60◦ =
2
π
3
π
3
2
1
√
1
3
◦
◦
cos 60 = sin 30 =
sin 60 = cos 30 =
2
2
◦
π
6
◦
π
π
1
cos = sin =
3
6
2
√
3
π
π
sin = cos =
3
6
2
tan 60◦ =
tan
√
π √
= 3
3
3
√
3
1
tan 30◦ = √
3
tan
π
1
=√
6
3
Remark. From now, all angles will be given solely in radians. Make sure that you are comfortable
with this measure of angles. Further to this, any angle given without units (i.e. degrees or radians) are
assumed to be in radians.
122
CHAPTER 5. TRIGONOMETRY 1
5.4
Graphs of trigonometric functions
Aim. To be able to draw the graphs of sin x, cos x, tan x and related graphs.
Vocabulary and conventions
domain
The set of x-values the function takes, the input to the function.
range
The set of y-values the function gives, the output from the function.
amplitude
Half the “height” of the range, i.e. half the difference between the
maximum and minimum range value.
roots or
intercepts
x-
Trigonometric graphs generally repeat themselves along the x-axis.
The period is the distance between repetitions.
period
5.4.1
The x-values at which the function value is 0.
Drawing trigonometric graphs
The graphs are usually drawn using radians as the scale on the x-axis. They are drawn so that the
scale is the same on each axis.
• The graph y = sin x has range [−1, 1] and period 2π.
1
amplitude = 1
−4π
−3π
−2π
π
−π
−1
period = 2π
2π
3π
4π
5.4. GRAPHS OF TRIGONOMETRIC FUNCTIONS
123
• The graph y = cos x has range [−1, 1] and period 2π.
1
−4π
−3π
−2π
π
−π
2π
3π
4π
−1
• The graph y = tan x has range (−∞, ∞) and period π. It has asymptotes midway between
asymptote
infinite amplitude
−2π
−π
π
period = π
−π π 3π 5π
each multiple of π, i.e. at x = . . . , −3π
2 , 2 , 2, 2 , 2 ,...
2π
124
5.4.2
CHAPTER 5. TRIGONOMETRY 1
Drawing trigonometric graphs
This is best shown by example.
Example. Draw the graph of y = 2 cos x.
The 2 in the front of 2 cos x doubles the amplitude, making the graph range from −2 to 2 instead of
from −1 to 1. This is a stretch in the y-direction: y = 2 cos x is cos x stretched vertically by a factor
of 2. Rewrite the equation as y2 = cos x to make this easier to see (Section 4.2).
2
−4π
−3π
−2π
π
−π
2π
3π
4π
−2
Example. Draw the graph of y = cos x + 1.
The 1 added on moves the whole graph up 1. It is a translation of cos x by (0, 1). Rewrite it y − 1 =
cos x to see this.
2
1
−4π
−3π
−2π
−π
π
2π
3π
Example. Draw the graph of y = sin(3x).
When the argument is not simply x, we find the basic cycle of the graph. A basic cycle is an interval
on the x-axis of length one period. sin x has a basic cycle of 0 ≤ x ≤ 2π (the length of one period).
Replacing the argument in 0 ≤ x ≤ 2π from x to 3x, we get 0 ≤ 3x ≤ 2π. Solving for x, we find
2π
that 0 ≤ x ≤ 2π
3 . So sin 3x does a complete cycle between 0 and 3 , that is, it has the basic sin shape,
2π
but one cycle begins at (0, 0) and finishes at ( 3 , 0).
4π
5.4. GRAPHS OF TRIGONOMETRIC FUNCTIONS
There are 3 stages in sketching this:
(2) Draw a complete cycle between 0 and
125
(1) Mark out lengths of
2π
3 ,
2π
3
on the x-axis,
(3) Copy this cycle as required.
− 8π
3
−2π
− 4π
3
− 2π
3
2π
3
4π
3
2π
8π
3
− 8π
3
−2π
− 4π
3
− 2π
3
2π
3
4π
3
2π
8π
3
− 8π
3
−2π
− 4π
3
− 2π
3
2π
3
4π
3
2π
8π
3
The graph is sin x stretched in the x-direction by 13 . See this by rewriting y = sin(3x) as y = sin( 1x/3 ).
Remark. We have seen that the basic sin shape looks like this:
Similarly, the basic cos shape looks like this:
one cycle starts at (0, 1) and finishes at (2π, 1).
126
CHAPTER 5. TRIGONOMETRY 1
The basic tan shape looks like this:
here we display tan on the interval (− π4 , π4 ).
Example. Draw the graph of y = sin(x − π4 ).
The basic cycle for sin x is 0 ≤ x ≤ 2π, hence sin(x − π4 ) has 0 ≤ x − π4 ≤ 2π as its basic cycle.
This is more easily understood if we rearrange the inequalities as π4 ≤ x ≤ 9π
4 . Therefore one of the
π
π
cycles of sin(x − 4 ) starts at x = 4 .
−4π
−3π
−2π
π
−π
2π
3π
It still has period of 2π (no stretch in the x-direction) and amplitude 2 (no stretch in the y-direction),
so just copy the basic sin shape but starting one cycle at ( π4 , 0) rather than (0, 0).
Practice (Section 5.4). Sketch the following:
1. 3 sin x,
2. sin x − 1,
3. cos 2x,
4. cos(x − π2 ).
4π
5.5. TRIGONOMETRIC EQUATIONS
5.5
127
Trigonometric equations
Aim. To learn how to solve trigonometric equations.
5.5.1
Inverse trigonometric functions
You are already familiar with the inverse trigonometric functions on your calculators. They are
what you use when you are finding an angle in a right angled triangle after you have calculated a
trigonometric ratio of the sides.
There are two common ways of writing the inverse trigonometric functions: using the sin−1 , cos−1
and tan−1 notation, or using the arcsin, arccos and arctan notation.
Remark. There may be some confusion in the −1 notation. In general,
sinn θ = (sin θ)n ,
cosn θ = (cos θ)n
tann θ = (tan θ)n
However for the case n = −1,
except where n = −1.
arcsin = sin−1
is the inverse function of sin, that is, the function which undoes the work of sin. Therefore sin−1 x is
not the same as
1
(sin x)−1 =
,
sin x
which is the reciprocal of sin x. Similarly for cos−1 and tan−1 .
Remark. The argument of an inverse trigonometric function is a ratio and not an angle.
Example. Find the following and give exact values wherever possible:
1. arcsin(0.32),
2. tan−1 (−3.14),
√
3. arccos(
3
2 ),
4. cos−1 ( √12 ).
Using the calculator for the first two and special triangles for the latter two, we find these values to
be:
1. 0.325,
5.5.2
2. −1.262,
3.
π
6,
4.
π
4.
Basic sin equation (+)
√
We outline the steps to solve the equation sin x =
3
2 ,
0 ≤ x ≤ 2π.
Remark. As the interval containing x is given in radians, we must work in radians.
√
1. First draw the unit circle. As this is a problem involving sin, draw the line y = 23 ≈ 0.866
and mark the intersections of this line with the circle. Draw the lines from the origin to these
intersections and mark angle formed by going in the anticlockwise direction from the positive
x-axis.
128
CHAPTER 5. TRIGONOMETRY 1
β
α
√
2. Find the value of the acute angle by finding the inverse sin of 23 . Call this angle α. In this
case we have a special angle (see triangle below), and so α = π3 .
π
6
2
√
3
π
3
1
3. The obtuse angle β is then α short of π (a half turn) and so β = π − α = π −
√
Thus the solution to sin x =
3
2
π
3
=
2π
3 .
on the interval 0 ≤ x ≤ 2π is x = π3 , 2π
3 .
Remark. Many angles have the same trigonometric ratio, but using your calculator only gives you
one of these angles.
5.5.3
Basic sin equation (−)
Solve the equation sin x = −0.9,
0 ≤ x ≤ 2π.
1. First draw the unit circle. As this is a problem involving sin, draw the line y = −0.9 and mark
the intersections of this line with the circle. Draw the lines from the origin to these intersections
and mark angle formed by going in the anticlockwise direction from the positive x-axis. Notice
this time that the angles are both reflexive (i.e. greater than π).
5.5. TRIGONOMETRIC EQUATIONS
129
β
γ
2. Find the value of the acute angle by finding the inverse sin of 0.9 (i.e. omit the negative sign
in the problem) and call this angle α. In this case we need to use a calculator to find that
α = 1.120 (4s.f.).
3. The reflex angles are then α more than a half turn, i.e. β = π + α = π + 1.120 = 4.262 and α
less than a full turn, i.e. γ = 2π − α = 2π − 1.120 = 5.163.
Thus the solution to sin x = −0.9 on the interval 0 ≤ x ≤ 2π is x = 4.262, 5.163.
5.5.4
Basic cos equation (+)
Solve the equation cos x = 12 ,
0 ≤ x ≤ 2π.
1. Draw the unit circle. As this is a problem involving cos, draw the line x = 12 = 0.5. (It may
confuse you that we have an angle called x as well as an x-axis. These are unrelated. If you feel
more comfortable renaming the angle t or θ, do so.) Mark the intersections of the line with the
circle and draw lines from the origin to these points. Now put on the angles formed by going
in the anticlockwise direction from the positive x-axis.
α
β
2. Find the value of the acute angle by finding the inverse cos of 21 , call this angle α. In this case,
we have a special angle, and so α = π3 .
130
CHAPTER 5. TRIGONOMETRY 1
π
6
2
√
3
π
3
1
3. The reflex angle is then α less than 2π (a full turn), and so it is β = 2π − α = 2π −
Thus the solution to cos x =
5.5.5
1
2
π
3
=
5π
3 .
on the interval 0 ≤ x ≤ 2π is x = π3 , 5π
3 .
Basic cos equation (−)
Solve the equation cos x = −0.4,
0 ≤ x ≤ 2π.
1. Draw the unit circle. As this is a problem involving cos, draw the line x = −0.4. Mark the
intersections of the line with the circle and draw lines from the origin to these points. Now put
on the angles formed by going in the anticlockwise direction from the positive x-axis.
β
γ
2. Find the value of the acute angle made with the x-axis by finding the inverse cos of 0.4 (i.e.
omit the negative sign in the problem) and call this angle α. In this case, we need to use a
calculator to find that α = 1.159.
3. The obtuse angle is then α less than π (a half turn), and so it is β = π −α = π −1.159 = 1.982.
The reflex angle is then α more than π, and so it is γ = π + α = π + 1.159 = 4.301.
Thus the solution to cos x = −0.4 on the interval 0 ≤ x ≤ 2π is x = 1.982, 4.301.
5.5. TRIGONOMETRIC EQUATIONS
5.5.6
131
Basic tan equation (+)
Solve the equation tan x = 0.4,
0 ≤ x ≤ 2π.
1. Draw the unit circle. As this is a problem involving tan, and tan gives a slope, draw the line
y = 0.4x. That is, the line through the origin with gradient 0.4. Mark the intersections of the
line with the circle. As this line goes through the origin, we draw no more lines. Put the angles
formed by going in the anticlockwise direction from the positive x-axis.
α
β
2. Find the value of the acute angle made with the x-axis by finding the inverse tan of 0.4, call
this angle α. In this case we need to use a calculator, so α = 0.381.
3. The reflex angle is then α more than π (a half turn), and so it is β = π+α = π+0.381 = 3.522.
Thus the solution to tan x = 0.4 on the interval 0 ≤ x ≤ 2π is x = 0.381, 3.522.
5.5.7
Basic tan equation (−)
Solve the equation tan x = −1,
0 ≤ x ≤ 2π.
1. Draw the unit circle. As this is a problem involving tan, and tan gives a slope, draw the line
y = −x. That is, the line through the origin with gradient −1. Mark the intersections of the
line with the circle. As this line goes through the origin, we draw no more lines. Put the angles
formed by going in the anticlockwise direction from the positive x-axis.
132
CHAPTER 5. TRIGONOMETRY 1
β
γ
2. Find the value of the acute angle α made with the x-axis by finding the inverse tan of 1 (i.e.
omit the negative sign in the problem). We have α = π4 .
√
π
4
1
2
π
4
1
3. The obtuse angle is then α less than π (a half turn) and so it is β = π − α = π −
π
4
=
The reflex angle is then α less than 2π (a whole turn) and so it is γ = 2π − α = 2π −
Thus the solution to tan x = −1 on the interval 0 ≤ x ≤ 2π is x =
3π 7π
4 , 4 .
Practice (Section 5.5). Solve the following equations on the interval 0 ≤ x ≤ 2π:
1. sin x =
√1 ,
2
2. cos x = 0.32,
3. tan x = −1.51,
√
4. cos x = −
3
2 ,
5. tan x =
√1 ,
3
6. sin x = −0.35.
3π
4 .
π
7π
4 = 4 .
5.6. HARDER EQUATIONS
5.6
133
Harder equations
Aim. To learn how to solve harder trigonometric equations.
5.6.1
Domain not 0 ≤ x ≤ 2π
The first problem solved in Section 5.5 was sin x =
√
3
2 ,
0 ≤ x ≤ 2π, with solution x = π3 , 2π
3 .
Even if the domain is different from 0 ≤ x ≤ 2π we always solve by finding values in this interval
first and then proceeding as follows:
• If the domain had been 0 ≤ x ≤ 4π instead, we would have two turns around the circle. That
is, our solutions would have been
2π
π 2π π
, , + 2π,
+ 2π.
3 3 3
3
We can simplify these solutions as
π 2π 7π 8π
, , , .
3 3 3 3
In other words, for every solution in [0, 2π], there is another solution x + 2π in [2π, 4π].
• If the domain had been − π2 ≤ x ≤ π2 , we would proceed in 3 steps:
1. Find a second set of solutions in [−2π, 0] by subtracting 2π from x:
π
2π
− 2π,
− 2π.
3
3
Similarly, simplify these as
−5π −4π
,
.
3
3
2. Add these to our original solutions in 0 ≤ x ≤ 2π, to get all solutions in [−2π, 2π]:
π 2π −5π −4π
, ,
,
.
3 3
3
3
3. Now remove any solutions outside the given domain − π2 ≤ x ≤ π2 .
This leaves
π
3
as the only solution on the interval − π2 ≤ x ≤ π2 .
• If the domain had been −2π ≤ x ≤ 6π, we would proceed in 2 steps:
1. Get a second set of solutions by going around the circle 2 more times:
2π
π
2π
π
+ 2π,
+ 2π, + 4π,
+ 4π.
3
3
3
3
2. Get a third set of solutions by going backwards around the circle by subtracting 2π:
π
2π
− 2π,
− 2π.
3
3
Our solution set is then
5π 4π π 2π 7π 8π 13π 14π
x = − ,− , , , , ,
,
.
3
3 3 3 3 3
3
3
134
5.6.2
CHAPTER 5. TRIGONOMETRY 1
Multiple angles
√
Consider the problem sin(2x) =
First put y = 2x to give sin y =
3
,
√2
3
2 ,
0 ≤ x ≤ 2π.
0≤
y
2
≤ 2π, i.e. sin y =
√
3
2 ,
0 ≤ y ≤ 4π.
Solving exactly as before this gives us the solution π3 , 2π
3 in the interval 0 ≤ y ≤ 2π and hence we
have the solution
π 2π 7π 8π
, , , ,
3 3 3 3
on the interval 0 ≤ y ≤ 4π. However y = 2x, so x = y2 . Now by substituting each of the solutions
into this equation, we have the solution to the original problem in x:
π π 7π 4π
, , , .
6 3 6 3
5.6.3
Composite angles
√
3
,
√ 2
3
2 on
Consider the problem sin(x + π4 ) =
0 ≤ x ≤ 2π.
the interval 0 ≤ y − π4 ≤ 2π. We rearrange this interval as
First let y = x + to get sin y =
9π
π 2π
π
4 ≤ y ≤ 4 . Solving exactly as before, we get the solutions y = 3 , 3 on the interval 0 ≤ y ≤ 2π.
π 9π
The easiest way to solve for y in [ 4 , 4 ] is to get at second set of solutions by adding on 2π:
π
4
2π
π 2π π
, , + 2π,
+ 2π.
3 3 3
3
Simplify this solution as
π 2π 7π 8π
, , , .
3 3 3 3
Now we remove any that lie outside the required interval for y. Because π4 < π3 , the solution π3 is not
7π
7π
8π
excluded. On the other hand, 9π
4 < 3 , so 3 and 3 are both excluded. Therefore the solution is
π
y = π3 , 2π
3 . But recall that y = x + 4 , so the solution to the original question on x is
π 5π
, .
12 12
Practice (Section 5.6). Solve the following trigonometric equations on the given intervals:
1. cos x =
√1 ,
2
0 ≤ x ≤ 4π,
2. tan x = −1.12,
√
3. cos(2x) =
3
2 ,
4. tan(2x) = −0.32,
0 ≤ x ≤ 2π,
−π ≤ x ≤ π,
5. tan(x + π3 ) = − 12 ,
0 ≤ x ≤ 2π,
0 ≤ x ≤ 2π,
6. cos(x + 1) = −0.66,
0 ≤ x ≤ 2π.
Chapter 6
Differentiation 1
6.1
Slopes and simple derivatives
Aim. To learn how to find slopes of functions.
Vocabulary and conventions
coordinates
The position of a point on the (x, y) grid. Saying “the point P =
P (2, −3)” means the the x-value of P is 2 and the y-value is −3.
P (x, y) is a general point. We write this when we want to discuss
a general point, in contrast to a particular point. P (x, f (x)) is a
general point on the graph of a curve y = f (x).
negative slope
positive slope
rise (negative)
rise (positive)
run
run
infinite or vertical
slope (no run)
zero or horizontal
slope (no rise)
slope
The slope of a straight line is given by
slope = rise
run ,
where run is the increase in distance, going from left to right, and
rise is the increase (or decrease) in height.
135
136
CHAPTER 6. DIFFERENTIATION 1
The tangent to a curve at a point is the straight line through P whose
slope is the slope of the curve at P .
tangent
curve
P
tangent at P
Magnifying a very small part of the curve near P shows how very
close the tangent is to the line of the curve near P .
tangent at P
6.1.1
Slopes of functions
Consider the graph of y = 2x, at left, and y = −x, at right:
y
y
y = 2x
Q
P
x
x
Q
P
y = −x
Noting the slope between P and Q, we can see that
rise
increase (or decrease) in y value
=
.
run
increase in x value
6.1. SLOPES AND SIMPLE DERIVATIVES
137
Starting at any point on y = 2x, say, at P = P (−1, −2), and stopping at Q = Q(1, 2), we get:
rise = (y − value at Q) − (y − value at P ) = 2 − (−2) = 4,
run = (x − value at Q) − (x − value at P ) = 1 − (−1) = 2.
Hence, the slope is
4
2
= 2. A similar calculation for y = −x gives a slope of −1.
Remark. For a straight line, the slope is constant. In other words, we get the same value independent
of the points that we use to find the slope.
6.1.2
Slope using function notation
In function notation, we can express the line y = 2x by y = f (x). That is,
f (x) = 2x.
Consider the points P and Q on the line. Let P = P (x, f (x)). Further, let h be the run. In other
words, let h be the increase in x between P and Q. Then we have
Q = Q(x + h, f (x + h)).
f (x) = 2x
Q
f (x + h)
f (x + h) − f (x)
f (x) P
h
x
x+h
Finally, the slope is given by
slope =
f (x + h) − f (x)
2(x + h) − 2x
2h
=
=
= 2.
h
h
h
138
CHAPTER 6. DIFFERENTIATION 1
For a curve, slope changes as x changes. Consider, for example, the following curve:
y
x
tangents
Tangents at different points have different slopes. Finding the exact slope at any point is not as
straightforward as for straight lines. Consider any curve y = f (x) and any point P (x, f (x)) on the
curve. Further, consider a succession of points Q1 , Q2 , . . ., that gets closer and closer to the point P .
We draw lines between each pair of points P and Qi , to show their slopes, for each i = 1, 2, . . .. As
Qi gets closer and closer to the point P , it can be seen that the slope of the line between P and Qi
gets closer and closer to the slope of the tangent line of the curve at the point P .
Q1
Q2
Q3
.
Q4
Q5
Q6
..
P
Let us express this in a different manner. Let Q = Q(x + h, f (x + h)) and let h → 0 (read “h
approaches 0”). As h gets smaller and smaller, Q gets closer and closer to P . In particular, the slope
of the line between them gets closer and closer to the slope of the tangent to the curve at P . The
procedure of letting h → 0 is called finding the limit. In this case, we find the value that the slope is
approaching to the gradient of the tangent of the function f (x) at P .
6.1. SLOPES AND SIMPLE DERIVATIVES
139
Since the slope between P and Q is
f (x + h) − f (x)
,
h
Q(x + h, f (x + h))
f (x + h) − f (x)
P (x, f (x))
h
we express the limit of this quotient as Q approaches P by writing
f (x + h) − f (x)
.
h→0
h
f 0 (x) = lim
We call f 0 (x) the derivative of f , with respect to x. This function which gives the slope of f in
the direction of increasing x, at any value of x. The above formula is the formal definition of the
derivative of f with respect to x.
By the derivative of f with respect to x, we simply mean the slope of the function in the positive
x-direction. Apart form x, we can use other variables, e.g. t or θ, and we can find derivatives with
respect to t or θ, etc.
6.1.3
Easier methods of finding derivatives
The formal definition is often quite difficult and time-consuming to use for finding derivatives. There
are a number of formulae for differentiating different types of functions which are much easier to use.
The problem is to recognise which type of function you have and therefore which formula you need,
and in particular, to use the formula correctly.
Remark. Differentiation is the process of finding a derivative, and a derivative is a function (i.e. a
formula) for finding slope of a curve.
6.1.4
First rules of derivatives
1. Let f (x) = a, where a is a constant. Then f 0 (x) = 0. To see this, note that f (x) = a is a
horizontal line, so its slope is always zero.
f (x)
a
x
140
CHAPTER 6. DIFFERENTIATION 1
Example.
(a) If f (x) = 3, then f 0 (x) = 0. (b) If g(x) = −2, then g 0 (x) = 0 also.
2. Let f (x) = xn , for some constant n. Then the derivative is given by
f 0 (x) = nxn−1 , n 6= 0.
In particular, we have:
Multiply
by the power
xn
nxn−1
and, subtract
1 from the power
Example.
(a) If f (x) = x = x1 , then f 0 (x) = 1 × x1−1 = x0 = 1;
(b) If f (x) = x5 , then f 0 (x) = 5 × x5−1 = 5x4 ;
1
(c) If f (x) = x 2 , then f 0 (x) =
1
2
1
1
× x 2 −1 = 12 x− 2 ;
(d) If f (x) = x−2 , then f 0 (x) = −2 × x−2−1 = −2x−3 .
3. Let f (x) = ag(x) for some constant a and g(x) is some function. Then the derivative of f (x)
is given by
f 0 (x) = ag 0 (x).
In particular, we have
the constant
stays
ag 0 (x)
ag(x)
and, find
the derivative of the function g(x)
Example.
(a) If f (x) = 3x5 , then f 0 (x) = 3 × 5x4 = 15x4 ;
(b) If f (x) = 2x−2 then f 0 (x) = 2 × (−2x−3 ) = −4x−3 ;
1
1
1
(c) If f (x) = −3x 2 , then f 0 (x) = −3 × 12 x− 2 = − 32 x− 2 .
4. Let f (x) = g(x) + h(x). The derivative is given by
f 0 (x) = g 0 (x) + h0 (x).
In other words, we need to find the derivatives of each term.
6.1. SLOPES AND SIMPLE DERIVATIVES
141
The derivative of the sum, is the sum of the derivatives:
Example.
g(x)
g 0 (x)
+
+
h(x)
h0 (x)
(a) If f (x) = 3x5 + 2x−2 , then f 0 (x) = 15x4 − 4x−3 ;
1
1
(b) If f (x) = −3x 2 + 4x, then f 0 (x) = − 32 x− 2 + 4;
(c) If f (x) = 3x
−1
3
1
4
+ 4, then f 0 (x) = 3 × − 31 x− 3 −1 + 0 = −x− 3 .
We now give some further examples illustrating these rules used concurrently.
Example. Let f (x) = x2 . We differentiate f with respect to x and we obtain
f 0 (x) = 2 × x2−1 = 2x.
The derivative of x2 with respect to x is 2x. What does that mean to us? It means that we have found
a function, f 0 (x) = 2x, which gives us the slope of the tangent to the curve f (x) = x2 at any value
of x. For example, choose x = 2. Then the slope of f at x = 2 is f 0 (2) = 2 × 2 = 4.
Similarly, we have the following:
• f 0 (1) = 2 × 1 = 2. This means that 2 is the slope of f at x = 1,
• f 0 (0) = 2 × 0 = 0. This means that 0 is the slope of f at x = 0,
• f 0 (−2) = 2 × (−2) = −4. This means that −4 is the slope of f at x = −2.
Example. Let f (x) = x−2 . Find the derivative of f at the points x = −1, x = 1, and x = 2.
Generally, the derivative of f is given by
f 0 (x) = −2x−2−1 = −2x−3 .
Hence,
• f 0 (1) = −2 × 11 − 3 = −2. This means that −2 is the slope of f at x = 1,
• f 0 (−1) = −2 × (−1)−3 = 2. This means that 2 is the slope of f at x = −1,
• f 0 (2) = −2 × 2−3 = − 41 . This means that − 14 is the slope of f at x = 2.
142
6.1.5
CHAPTER 6. DIFFERENTIATION 1
Alternative notation for differentiation
dy
Apart from f 0 (x), we can also use dx
as an alternative notation. In particular, dy expresses the
instantaneous change in y and dx expresses the instantaneous change in x. As Q approaches P ,
slope =
dy
increase (or decrease) in y value
→
= slope at P .
increase in x value
dx
Q
change in y → dy
P
change in x → dx
Given a function y = f (x), we have:
f 0 (x) =
dy
d
df
d[f (x)]
d
=
[y] =
=
=
[f (x)] = (f (x))0 = y 0 .
dx
dx
dx
dx
dx
Indeed when y = f (x), all the expressions in the above chain of equalities represents the same thing.
The reason for the number of different notations with identical meanings is both partly historical and
partly ease of use in different circumstances.
Example. Differentiate f (x) = 3x4 .
f 0 (x) =
d
[3x4 ] = 3 × 4x4−1 = 12x3
dx
1
Example. Differentiate u = 5x−1 + 7x 2 .
−1
1
1
du
d
d
d
7x 2
=
[5x−1 + 7x 2 ] =
[5x−1 ] +
[7x 2 ] = −5x−2 +
dx
dx
dx
dx
2
Example. Differentiate 3 + 2x + −3x−1 as a function of x.
[3 + 2x + −3x−1 ]0 = [3]0 + [2x]0 − [3x−1 ]0 = 2 + 3x−2
6.1.6
Functions involving x in the denominator or in roots
Consider the function given by y = x53 . Although this may look difficult to differentiate, the function
can be rewritten as y = 5x−3 (see Section 1.6). Now, it is easy to take the derivative using the ‘power
rule’:
dy
d 5
d
−15
=
=
[5x−3 ] = 5 × (−3)x−3−1 = −15x−4 = 4 .
3
dx
dx x
dx
x
Now, consider the function
f (x) =
√
3
x2 −
√
x3 .
6.1. SLOPES AND SIMPLE DERIVATIVES
143
Again, simply by rewriting the function, we have
2
3
f (x) = x 3 − x 2 .
Applying the ‘power rule’ will give us:
√
2
3 x
2 2 −1 3 3 −1 2 −1 3 1
3
2
3
2
√ −
− x
= x − x =
.
f (x) = x
3
2
3
2
2
3× 3x
0
Practice (Section 6.1). For each of the following functions, find the slope of the function at the given
value:
1. f (x) = x5 at x = 2,
4. ν = 32 x7 −
2. y = 3x2 at x = 0,
5. h(x) = x
3. g(x) = 6x2 + 5x−4 at x = −2,
6. u =
3x2
2
3
x2
−3
2
−
3
x
+
√
4 x
5
at x = 1,
− 2x−4 + 3x − 3 at x = 1,
+ 1 at x = 2.
144
6.2
CHAPTER 6. DIFFERENTIATION 1
Points and intervals of interest for a function
Aim. To understand information from the derivative of a function.
Vocabulary and conventions
asymptote of a
function
A straight line that the function gets closer and closer to but never
actually joins. Asymptotes may be vertical, horizontal or sloping.
The dashed lines represent the asymptotes and the solid lines
represent the functions.
6.2.1
second derivative
This is the derivative of the derivative. For example, the first
derivative of f (x) = x2 is f 0 (x) = 2x. The second derivative
of f is the derivative of f 0 and is denoted by f 00 . In particular, if
f (x) = x2 then f 00 (x) = 2. If y = f (x), then all of the following
are alternative notations:
d df
d2 f
d dy
d2 y
00
00
f (x) =
=
=
y
=
=
.
dx dx
dx2
dx dx
dx2
sign
The sign of a function at a point is positive or negative when the
function is positive (above the x-axis) or negative (below the xaxis) at that point. Points where functions or their derivatives change
from positive to negative or vice versa are important, and we say a
function changes sign at those points.
Does the first derivative of a function always exist?
First, let us start by rephrasing the title of this section. The title asks whether a function has a finite
slope at every value of x (recall that a vertical line has an infinite slope). In fact, there are a number
6.2. POINTS AND INTERVALS OF INTEREST FOR A FUNCTION
145
of conditions required for the derivative of a function to exist at a value of x. Here, we will mention
some of these requirements.
1. The function must be defined at x
If the function is not defined at a value, we cannot find the slope for it at that value. Sometimes
a function may head towards positive or negative infinity on one or both sides of a vertical line,
i.e. getting closer to the line but never actually touching it. For example, consider the function
f (x) = x2 :
f (x) = x2 , x > 0
f (x) = x2 , x < 0
This function is not defined at x = 0. We can see that for values just bigger than zero, f (x)
is large and positive, and for values of x just smaller than zero, the function f is large and
negative. In fact, we have
f (x) → ∞ as x → 0+
and
f (x) → −∞ as x → 0− .
If a function tends to ∞ or −∞ (often we write x → ±∞) as x approaches some finite value
a, we say that the function has a vertical asymptote at x = a. It is not difficult to see that the
slope of the function is tending towards vertical at such an a. Hence, the derivative does not
exist.
2. The function must be continuous
For the derivative of a function to exist, we need to have no breaks in the graph of a function.
In particular, the derivative does not exist at a point where such a break occurs.
gap
function is defined (solid
line), but there is a break
3. The function must be smooth
The derivative of a function does not exist at points where the graph of a function is sharp. The
reason for this is the existence of ‘more than one tangent line’ at these sharp points.
146
6.2.2
CHAPTER 6. DIFFERENTIATION 1
Points and intervals of interest for a function
The graph of an unknown function y = f (x) is given below.
y
1
a
b
c
d
j
e
i
x
We first abstract some information from the figure:
1. The function is not defined (everywhere)
At x = a, it can be seen that the function should take an infinite value, so that is not defined at
x = a. The slope gets closer and closer to vertical as x → a+ (the notation x → a+ means
moving along the x-axis towards a from right hand side or larger side of a). The function has a
vertical asymptote at x = a and the derivative does not exist (because it is infinite). Moreover,
the function is not defined for x < a and x > i. In particular, we cannot define the derivatives
at these intervals.
2. The function is not continuous
At x = b, the function has a break; so it is not continuous. Although, the function is defined at
b, and the finite value f (b) is given by the height of the solid dot, meanwhile, we cannot find
a tangent for a lone point. Similarly, the function ends at the point (i, f (i)), so the derivative
does not exist at (i, f (i)) either.
3. The function not smooth
At (c, f (c)), the function has a sudden change of direction. The slope of f as x → c− (going
from left hand side of c) is positive, and as x → c+ , we get a negative slope (but on neither side
does the gradient approach zero). Hence, the derivative does not exist at x = c.
4. Intervals with negative derivative (or slope)
For values of x between a and b (i.e., x ∈ (a, b)), the slope is negative. Note that when the
derivative is negative, the function is decreasing. In other words, when the derivative is negative
the values of function is reducing on that interval. It can be readily seen that the function is also
decreasing for x ∈ (c, d).
5. Intervals with positive derivative (or slope)
The slope is positive for values x ∈ (b, c), x ∈ (d, e), and x ∈ (e, i). In these intervals, the
function is increasing (i.e., the value of function is getting bigger on those intervals). Hence, if
the derivative of a function on an interval is positive, the function is increasing on that interval.
6.2. POINTS AND INTERVALS OF INTEREST FOR A FUNCTION
147
6. Zero derivative
At x = d and x = e, we can see that the function in neither increasing nor decreasing. In
particular, the slope of the function at these points is zero. Sometimes, at the points where
the slope is zero, the function turns from an increasing (or decreasing) one to an decreasing or
(increasing) one. These points are known as turning points. A turning point will be either a
relative maximum or a relative minimum. In other words, the point (d, f (d)) is the lowest in
comparison to the points close to it.
relative maximum and turning point
lope
ive s
posit
nega
tive
slope
relative minimum and turning point
nega
tive
slope
ope
ve sl
i
posit
Further, note that a function might have a zero derivative at a point without having relative
minimum or maximum at that point. For example, for the point (e, f (e)), one can see that the
value of the function increases for points smaller and close to e and in addition, the value of
function increases again for the values bigger than e and close to e (while the slope exactly at e
is zero).
zero slope
ive
posit
slope
lope
ive s
posit
zero slope
nega
tive
nega
slope
tive
slope
7. Concavity and points of inflection
Suppose we have found the major points of interest in a graph. How do we join these points
together? In other words, do we join them using a straight line or a curve? If using a curve,
what does our curve look like? We describe curvature using the words concave up and concave
down.
concave down: the curve has a
downward bend
concave up: the curve has an
upward bend
By looking the above graphs, one can see that for the case of concave up, the graph of a function
starts decreasing (hence the slope is negative) and ends with being increasing (hence the slope
148
CHAPTER 6. DIFFERENTIATION 1
is positive). For the case of concave down, it can be seen that the graph of a function starts with
being increasing (hence the slope is positive) and ends with being decreasing (hence the slope
is negative).
Just as we find that the function is increasing when the derivative is positive, we can find
that the derivative is increasing by finding out whether the derivative of the derivative is
positive. Likewise, as the function is decreasing where the derivative is negative, the derivative
is decreasing where the second derivative is negative. The curve is concave down exactly when
the slope is decreasing.
concave up: slope is increasing
concave up: slope is increasing
A point where the second derivative is zero and its sign is changing, from positive to negative
(or negative to positive) is called a point of inflection. The concavity changes at such a point.
The slope at a point of inflection point may or may not be zero.
point of inflection: the curve
changes its convexity here
The graph has points of inflection at (e, f (e)) and also at (j, f (j)). The graph of function is
concave up on (a, b), (c, j) and (e, i). Further, it is concave down on (j, e). Finally, it has zero
concavity on (b, c), because the slope is constant there.
Using the above, we can obtain derivative tests:
1. Second derivative test
At a relative maximum, the curve is concave down and at a relative minimum, the curve is
concave up. To summarise we have the following classification:
• Derivative zero and second derivative negative → relative maximum;
• Derivative zero and second derivative positive → relative minimum;
• Second derivative changes sign → point of inflection.
6.2. POINTS AND INTERVALS OF INTEREST FOR A FUNCTION
149
To use the second derivative test on the point where the first derivative is zero, we find the value
of the second derivative at that point and we see whether the value is positive, negative, or zero.
In the case of zero, we should see whether it changes sign.
2. First derivative test
In some cases the second derive is tedious to find. In these case, it may be faster to look only
at the first derivative in order to find whether the point is relative minimum or maximum. This
method is known as the first derivative test. The following gives us the method for classifying
the points:
• Derivative changes sign from positive to negative → relative maximum;
• Derivative changes sign from negative to positive → relative minimum;
• Derivative zero and the same sign before and after the point → neither a relative minimum
nor a relative maximum.
In particular, to use the above first derivative test at the points where the first derivative is zero,
we find the value of the derivative at two points to the left and right of the point to see the
behaviour of the derivative.
6.2.3
Critical points: derivative is zero or does not exist
In order to obtain information regarding the behaviour of a function (for example, in order to draw
its graph), we need to look for the points known as critical points. These are the points where the
derivative is either zero or does not exist. At these points maxima, minima or points of inflection may
exist, breaks or corners may occur, or vertical asymptotes may be found.
Once more, let us look at the following graph:
y
1
a
b
c
d
j
e
i
x
The critical points of the graph are at x = a, b, c, d, e, and i. In particular, the points (b, f (b)),
(c, f (c)), (d, f (d)), (e, f (e)), (i, f (i)) and the asymptote at x = a give us the major features of the
graph.
In order to decide about the concavity of the function at these points, we can use the second derivative
test. In the above graph, the point (j, f (j)) gives us a point of inflection.
The final job to do when graphing from this information is to connect all the points found with curves
of the correct concavity and/or slope.
150
CHAPTER 6. DIFFERENTIATION 1
Practice (Section 6.2). Use the following figure to decide which points and intervals fit the descriptions
below:
y
a
b
c
d
e
i
j
k
l
1. the function is not defined,
8. there is a relative maximum,
2. the derivative is zero,
9. there are turning points,
3. the derivative is undefined,
10. there is a point of inflection,
4. there are critical points,
11. the function is increasing,
5. the function is concave up,
12. the function is decreasing,
6. there is a relative minimum,
13. the function is positive,
7. the function is concave down,
14. the function is negative.
x
6.3. GRAPHING A FUNCTION
6.3
151
Graphing a function
Aim. To use the information gained from a function in order to graph it.
6.3.1
Graphing functions
To graph a function we need to find and examine a number of points and the intervals between them.
The following is a list of most such points and intervals. As you get more practised at plotting
functions, you will find that you can choose with more accuracy what information from the following
list is essential to graph a particular function:
1. f is zero (x-intercepts),
2. x = 0 (y-intercept),
3. f is positive,
4. f is negative,
5. f is undefined,
6. f 0 is zero (critical point),
7. f 0 is positive,
8. f 0 is negative,
9.
f0
is undefined (critical point),
10. f → ±∞ at a finite value (vertical
asymptote),
11. f 00 is zero,
12. f 00 is positive (concave up),
13. f 00 is negative (concave down),
14. f 00 changes sign (point of inflection),
15. f 0 is zero and f 00 is positive (relative
minimum),
16. f 0 is zero and f 00 is negative (relative
maximum).
Remark. All of the above may not give us enough points to draw a graph. We may still need more
extra points and possibly the slope of the function at these points.
Example. Graph the function f (x) = x2 using the steps in the above list.
1. We need to solve the equation for x. In particular, we get f (x) = x2 = 0. Therefore x = 0.
Hence, (0, f (0)) = (0, 0) is the only x-intercept.
2. We have f (0) = 02 = 0 and so (0, 0) is the only y-intercept..
y
x
(0, 0) is the x and y-intercept
3. We need to solve
f (x) = x2 > 0.
152
CHAPTER 6. DIFFERENTIATION 1
This results in x > 0 or x < 0. Therefore, f is positive for all value except x = 0. This means
that the whole is on or above the x-axis. So we do not need to consider item 4 of the list.
5. The function f (x) = x2 is defined for all (real) values of x (since x2 is a polynomial).
6. The first derivative is
f 0 (x) = 2x.
Now, by solving f 0 (x) = 0, we obtain x = 0. Hence, the derivative is zero only at x = 0. This
gives the critical point (0, f (0)) = (0, 0).
y
no negative y-values
x
(0, 0) is a critical point
7. We know that f 0 (x) > 0 when x > 0. Hence, the function is increasing on the interval
(0, +∞). Note that as x gets large and positive, the slope (2x) will be steep and positive, while
for values near zero the slope will be small.
8. We know f 0 (x) < 0 when x < 0. The function is decreasing on the interval (−∞, 0). As x
gets large and negative, the slope 2x will be steep and negative, while for values to the left and
near zero the slope will be small and negative.
9. As f 0 (x) = 2x is a function that exists for every real value of x (since it is a polynomial), the
derivative is defined everywhere.
10. Using parts 5 and 9 we see there is no such x.
11. The second derivative is given by f 00 (x) = 2. This equation shows that the second derivative is
2 for every value of x. In other words, it is always 2 (and hence, positive), independent of the
choice of x. This gives us items 12, 13 and 14. That is, f is concave up everywhere, there are
points of inflection and no saddle points.
15. We know that f 0 is zero at (0, 0). Further, we know that f 00 is everywhere positive. Hence, we
have a relative minimum at (0, 0).
16. We know that f 0 is zero at (0, 0). Further, we know that f 00 is everywhere positive. Hence, we
have no relative maximum.
We see that the only point distinguished by the items above is (0, 0). As a result, we need to consider
more points. Choose values on either side of x = 0, say, x = 1 and −1. We evaluate f (x) at these
values:
f (1) = 12 = 1,
f (−1) = (−1)2 = 1.
6.3. GRAPHING A FUNCTION
153
Therefore (1, 1) and (−1, 1) are points on the function. Further,
f 0 (1) = 2 × 1 = 2,
f 0 (−1) = 2 × (−1) = −2.
This shows that the slope at (1, 1) is 2 and the slope at (−1, 1) is −2.
f (x) = x2
f is concave up everywhere and is positive everywhere except at (0, 0)
(−1, 1):
here the
slope is −2
(1, 1):
here the
slope is 2
f increasing on (0, ∞)
x
f decreasing on (−∞, 0)
(0, 0) is a relative
minimum and has
zero slope
√
Example. Graph the function f (x) = x.
√
1. If f (x) = x = 0, then x = 0. Therefore the y-intercept is (0, 0). Further, f does not exist for
x < 0 since f is undefined for those values of x. Therefore f is positive for x > 0 and is never
√
negative (as x is the positive square root).
y
no negative x or y values
(0, 0)
x
1
2. Note that we can write f as x 2 . Therefore
1 1
1
f 0 (x) = x− 2 = √ .
2
2 x
This cannot be zero for any real value of x, so there cannot be any relative minimum or
maximum. The derivative is undefined for x = 0. However the function is defined and finite at
x = 0 (namely f (0) = 0), so it does not have a vertical asymptote there. However, the slope
approaches vertical as x → 0+ (i.e., approaches 0 from the right hand side). For x > 0, the
derivative is positive, so f is increasing for x > 0.
154
CHAPTER 6. DIFFERENTIATION 1
3. We find the second derivative:
−1 1 − 3
−1
× x 2 = √ .
2
2
4 x3
For x > 0, the second derivative is always negative, which implies that f is concave down for
x > 0. The second derivative is never zero, so there is no point of inflection.
f 00 (x) =
4. We have only one critical point, which is (0, 0). We need a few more points to make a graph
with. We will use x = 1 and 4, as they are squares (and recall that the function is not defined
for negative numbers).
1
f (1) = 1, f 0 (1) = ,
2
1
0
f (4) = 2, f (4) = .
4
Our new points are (1, 1) and (4, 2), also the slope at these points is positive and quite small.
f (x) =
√
x
f is positive, increasing,
concave down
(4, 2)
(1, 1)
x
(0, 0)
Example. Graph the function f (x) = x2 .
1. Since f (x) → 0 as x → ±∞ (but never actually gets there), f (x) = x2 is never zero. This
means that there is no x-intercept. Also since f is not defined at x = 0, there is no y-intercept
either. The function is positive when x > 0 and is negative when x < 0. Notice that as x → 0+ ,
we have f (x) → ∞. On the other hand, as x → 0− , we have f (x) → −∞.
y
f → −∞ as x → 0−
f → 0 as x → −∞
f → 0 as x → ∞
x
f → ∞ as x → 0+
6.3. GRAPHING A FUNCTION
155
2. Since f (x) = 2x−1 , we have f 0 (x) = −2x−2 = − x22 . Hence f 0 is never zero, it is not defined
at x = 0 and everywhere else it is negative (since x2 is positive when x 6= 0). Hence f has
no relative maxima or minima. In addition, f is decreasing everywhere except x = 0. We saw
above that f (x) → ∞ as x → 0+ (similarly f (x) → −∞ as x → 0− ), so there must be a
vertical asymptote at x = 0.
3. The second derivative of f is given by
f 00 (x) = 4x−3 =
4
.
x3
It can be seen that f 00 is never zero. Further, it is not defined at x = 0, it is positive when x > 0
and negative when x < 0. Therefore f is concave up on x > 0 and concave down on x < 0.
Notice that f has no point of inflection.
Again we need to calculate more points. We have f (1) = 2, f (−1) = −2, therefore our points are
(1, 2) and (−1, −2). Further, we have f 0 (1) = −2 and f 0 (−1) = −2.
In addition to the vertical asymptote at x = 0, we also have a horizontal one: f (x) → 0 as x → ±∞.
The function f is concave up on x > 0 and is concave down on x < 0. The function is decreasing on
x > 0 and x < 0. The fact that that it has a vertical asymptote at x = 0 is now important if we are to
make sense of these statements, so you may wish to mark that in the graph.
f (x) =
2
x
f → ∞ as x → 0+
f is positive, decreasing, concave up
(1, 2)
f → 0 as x → −∞
x
f → 0 as x → ∞
(−1, −2)
f is negative, decreasing, concave down
vertical aymptote at x = 0
f → −∞ as x → 0−
6.3.2
Must you do all this for every graph?
The answer to this question is no. On the other hand, until you have some experience, you may not
recognise the difference between what you need to find and what is useful but not essential.
You have done some graphing in the Basic Algebra chapters without any derivatives. For graphs
whose basic shapes were described there, you should find that graphing is faster with those techniques.
Recall the techniques described for transformation. For example, having sketched f (x) = x2 , it should
be straightforward to sketch the following:
• f1 (x) =
2
x
+ c shifts the graph of f up or down c units (depending on the sign of c),
156
CHAPTER 6. DIFFERENTIATION 1
• f2 (x) = 2k
x stretches or shortens the graph of f by a factor of k units vertically (depending on
the sign of k),
• f3 (x) =
2
x−a
• f5 (x) =
2
(−x)
shifts the graph of f a units to the left or right (depending on the sign of a),
• f4 (x) = − x2 reflects the graph of f in the x-axis,
= − x2 reflects the graph of f in the y-axis.
For most graphs, at least some of the information found from the derivatives will be useful and often
will be necessary. The first derivative is usually very useful. If the second derivative is difficult to find,
use the first derivative test to find if a critical point is a relative maximum or minimum. If the second
derivative is relatively easy to find, use the second derivative test because it includes information
about concavity and points of inflection.
If asked to just sketch a graph, then finding any vertical asymptote and turning points, and joining
these and the y-intercept and x-intercept (if they are easy to find), plus some indication of the direction
of the graph to right and left may be all you need to do. When asked to carefully sketch, then as much
information as possible is expected (like the above examples).
Practice (Section 6.3). Sketch the following:
1. f (x) = x3 + 2x2 − 4x + 1,
2. h(x) = 2x − x4 ,
3. y =
√1 .
x
6.4. RATES OF CHANGE (READ ONLY)
6.4
157
Rates of change (read only)
Aim. To understand that derivatives are rates of change.
Vocabulary and conventions
If y is proportional to x, we can write this relationship
mathematically as
y = kx,
proportional to
where k is a constant. For example, “your height is twice mine”,
can be written as
your height = 2× my height.
The relationships “this length is 5 times that length” and “she is half
my age” are also expressions of proportionality.
6.4.1
Rate of change
Example. Consider a graph giving the height of a person between birth and old age:
h
2
height in
metres
1.5
1
0.5
0
0
10
20
30
40
50
60
70
80
t
age in years
When can we say the person is growing, i.e., her/his height is increasing? When the slope of the graph
is positive, which is when t < 20.
When is the person’s height constant? The graph has zero slope for 20 < t < 50 i.e., there is no
change in height on that interval.
When is the person shrinking? When t > 50 i.e., the slope of the graph is negative showing that
height is decreasing.
Note that the slope of the graph gives us information about the rate of change of height with respect
to time, i.e., the growth. The slope of the graph at any time t0 is the same as the derivative of height
158
CHAPTER 6. DIFFERENTIATION 1
with respect to time at t0 , and the slope gives a rate of change. Thus we can say a derivative is a rate
of change.
Example. Consider a car travelling on the motorway and suppose its speed is 100 kilometres per
hour. This is in fact a rate of change of distance. If we let s represent distance in kilometres, and t
represent time in hours, we can write this fact mathematically as
ds
= 100,
dt
the rate of change in distance with respect to time 100.
In fact, speed is not constant, the driver may go mostly at this rate on the motorway, if they are lucky.
Off the motorway the driver will probably have to go at 50 kilometres per hour or less. If we suppose
they travel for 1 hour on the motorway and for 30 minutes off it, we can draw a graph by using slope
alone:
s
200
distance in km
100
distance travelled after 1 hour 30 minutes is 125 km
slope is 50 km per hour
slope is 100 km per hour
1
time in hours
2
t
and find the distance travelled is 125 kilometres. Because we used an approximation (rough guess) for
speed at various times, this graph will not be very accurate. However it gives a reasonable estimate.
6.4.2
Finding formula for rate of change
Very often we are unable to find a formula for some quantity that we are interested in, like size,
weight, distance or temperature. But we can often find a formula for a rate of change of the quantity
with respect to some variable like time, or distance.
Example.
1. The expression “distance makes the heart grow fonder” suggests that if distance is
s and the quantity of fondness is f , we could write
df
> 0.
ds
More precisely, fondness increases with respect to distance. Without further information we
cannot improve this formula, but it is a start.
6.4. RATES OF CHANGE (READ ONLY)
159
2. If we have a population P of rabbits (with plenty of food and space available), we can expect
that a certain (constant) proportion of rabbits will be breeding at any time t. Another reasonably
constant proportion of rabbits will die of old age or disease. Gathering these constants together,
we can say that the change in P with respect to time is proportional to P . Hence,
dP
= kP,
dt
where k is the resulting constant gained from birth and death rates as a fraction (we will assume
that it is positive). Given a starting value for P , that is some value for P at, say, t = 0, and
a value for k, we could construct a reasonable approximation of the graph of P . Notice that
its slope, which is kP , will get steeper as P gets bigger. In other words, slope is increasing so
the graph will be concave up. In Chapter 12, we learn that the graph of P is an exponential (or
power) graph.
This section is read only. However, understanding change from a graph or being able to find a rate of
change and to make predictions from it are very useful skills.
160
CHAPTER 6. DIFFERENTIATION 1
Chapter 7
Trigonometry 2
7.1
Addition formulae
Aim. To learn how to manipulate the addition formulae.
7.1.1
Major values
Using the unit circle, we can quickly find the following:
sin 0 = sin 2π = 0
cos 0 = cos 2π = 1
π
=1
2
π
cos = 0
2
sin
sin π = 0
3π
= −1,
2
3π
cos
= 0.
2
sin
cos π = −1
1
(x, y) = (cos θ, sin θ)
−1
θ
1
−1
7.1.2
Addition formulae
We will not derive the following formulae, but they are worth noting:
sin(θ + φ) = sin θ cos φ + cos θ sin φ,
sin(θ − φ) = sin θ cos φ − cos θ sin φ,
cos(θ + φ) = cos θ cos φ − sin θ sin φ,
cos(θ − φ) = cos θ cos φ + sin θ sin φ.
161
162
CHAPTER 7. TRIGONOMETRY 2
Using the previous formulae, we can obtain the following:
tan(θ + φ) =
tan θ + tan φ
1 − tan θ tan φ
and
tan(θ − φ) =
tan θ − tan φ
.
1 + tan θ tan φ
In particular, these results demonstrate that
sin(θ + φ) 6= sin θ + sin φ.
Otherwise, we get the following nonsense result:
π π π
π
+
= sin + sin = 1 + 1 = 2.
0 = sin π = sin
2
2
2
2
3π
Example. Expand sin(θ − 3π
2 ) and cos(θ − 2 ).
3π
3π
• sin θ − 3π
2 = sin θ cos 2 − cos θ sin 2 = sin θ × 0 − cos θ × (−1) = cos θ,
3π
3π
• cos θ − 3π
2 = cos θ cos 2 + sin θ sin 2 = cos θ × 0 + sin θ × (−1) = − sin θ.
3π
Note that the values of cos 3π
2 and sin 2 could be obtained from the graphs or from inspecting the
unit circle.
Example. Expand tan θ − 3π
2 .
Because tan 3π
2 is not defined, we first need to rewrite the formula in the following way:
sin θ − 3π
3π
2 .
tan θ −
=
2
cos θ − 3π
2
Now by expanding the numerator and the denominator using the equations in the above example, we
get
3π
cos θ
tan θ −
=
= − cot θ.
2
− sin θ
Example. Simplify sin 2θ cos θ − cos 2θ sin θ.
Notice the pattern
sin cos
− cos
sin
.
This is the same as in the right hand side of the expansion
sin(θ − φ) = sin θ cos φ − cos θ sin θ.
Using this formula backwards, we have
sin 2θ cos θ − cos 2θ sin θ = sin(2θ − θ) = sin θ.
7.1. ADDITION FORMULAE
163
Practice (Section 7.1).
1. Expand the following formulae:
(a)
(b)
(c)
(d)
sin(X − 2Y ),
tan(3A − B),
cos π2 − 2θ ,
tan θ − π2 .
2. Simplify the following formulae:
(a) cos A sin B − cos B sin A,
(b) cos 3ζ cos 2ζ − sin 3ζ sin 2ζ.
164
7.2
CHAPTER 7. TRIGONOMETRY 2
Multiple angles
Aim. To learn how to use the multiple angle formulae.
Vocabulary and Conventions
identity
7.2.1
An expression which relates one mathematical expression to another
and is always true. For example, 2x = x + x, for every x.
Double angle formulae
By letting θ = φ in the addition formulae, we get the following:
sin 2θ = sin(θ + θ) = sin θ cos θ + cos θ sin θ = 2 sin θ cos θ,
cos 2θ = cos(θ + θ) = cos θ cos θ − sin θ sin θ = cos2 θ − sin2 θ.
As a result, we have
sin 2θ = 2 sin θ cos θ,
cos 2θ = cos2 θ − sin2 θ.
Recall Pythagoras’s theorem, sin2 θ + cos2 θ = 1.
sin θ
1
θ
cos θ
By rearranging the identity, we have the following formulae:
cos2 θ = 1 − sin2 θ,
sin2 θ = 1 − cos2 θ.
We now use the above and the formula for cos 2θ that we obtained earlier to obtain the following:
cos 2θ = 2 cos2 θ − 1 = 1 − 2 sin2 θ,
tan θ + tan θ
2 tan θ
tan 2θ =
=
.
1 − tan θ tan θ
1 − tan2 θ
7.2. MULTIPLE ANGLES
7.2.2
165
Half angle formulae
As sin θ = sin( 2θ + 2θ ) and cos θ = cos( 2θ + 2θ ), we replace θ with
in Section 7.2.1 to get:
θ
2
in the formulae that we obtained
sin θ = 2 sin 2θ cos 2θ ,
cos θ = 2 cos2
θ
2
− 1 = 1 − 2 sin2
tan θ =
2 tan( θ2 )
1−tan2 ( θ2 )
θ
2
,
.
Practice (Section 7.2). Simplify the following formulae:
1. 2 sin
π
π
cos ,
12
12
2π
,
2. 1 − 2 sin
9
2
3. 2 cos2 x − 1,
5. cos2 4x − sin2 4x,
α
4. 1 − 2 sin ,
2
6.
2
2 tan π8
.
1 − tan2 π8
166
7.3
CHAPTER 7. TRIGONOMETRY 2
Sums and products
Aim. To learn how to manipulate between sums and products.
7.3.1
Replacing a product with a sum
Adding the formulae for sin(θ + φ) and sin(θ − φ) gives us,
sin(θ + φ) + sin(θ − φ) = sin θ cos φ + cos θ sin φ + sin θ cos φ − cos θ sin φ = 2 sin θ cos φ.
Hence, we have
1
sin θ cos φ = (sin(θ + φ) + sin(θ − φ)).
2
Similarly, by subtracting the formulae for sin(θ + φ) and sin(θ − φ), we have,
1
cos θ sin φ = (sin(θ + φ) − sin(θ − φ)).
2
By adding and subtracting the formulae for cos(θ + φ) and cos(θ − φ) we have the following:
1
cos θ cos φ = (cos(θ + φ) + cos(θ − φ)),
2
1
sin θ sin φ = − (cos(θ + φ) − cos(θ − φ)).
2
In particular, the above formulae allow us to replace a product of two trigonometric functions by a
sum.
7.3.2
Replacing a sum by a product
Let x = θ + φ and y = θ − φ. Notice that their sum and difference are particularly nice:
x + y = 2θ
and
x − y = 2φ.
Substituting these results into sin θ cos φ = 12 (sin(θ + φ) + sin(θ − φ)), we will get the following
formula:
x+y
x−y
1
sin
cos
= (sin x + sin y).
2
2
2
Rearranging and replacing x and y by θ and φ, we obtain the following:
θ+φ
θ−φ
cos
,
2
2
θ+φ
θ−φ
sin θ − sin φ = 2 cos
sin
,
2
2
θ+φ
θ−φ
cos θ + cos φ = 2 cos
cos
,
2
2
θ+φ
θ−φ
cos θ − cos φ = −2 sin
sin
.
2
2
sin θ + sin φ = 2 sin
7.3. SUMS AND PRODUCTS
167
To summarise, we have the following:
sin θ cos φ =
1
2
(sin(θ + φ) + sin(θ − φ)),
cos θ sin φ =
1
2
(sin(θ + φ) − sin(θ − φ)),
cos θ cos φ =
1
2
(cos(θ + φ) + cos(θ − φ)),
sin θ sin φ = − 12 (cos(θ + φ) − cos(θ − φ)),
and
θ−φ
sin θ + sin φ = 2 sin θ+φ
2 cos 2 ,
θ−φ
sin θ − sin φ = 2 cos θ+φ
2 sin 2 ,
θ−φ
cos θ + cos φ = 2 cos θ+φ
2 cos 2 ,
θ−φ
cos θ − cos φ = −2 sin θ+φ
2 sin 2 .
Example. Write:
1. sin 2A cos 3A as a sum,
2. sin 2A + sin 3A as a product.
Note that sin(−A) = − sin A and cos(−A) = cos A. You may check this by drawing angles of A
and −A on the unit circle then considering the definition of sin and cos.
1.
1
1
1
[sin(2A + 3A) + sin(2A − 3A)] = [sin 5A + sin(−A)] = [sin 5A − sin A],
2
2
2
2. 2 sin
2A − 3A
5A
−A
5A
A
2A + 3A
cos
= 2 sin
cos
= 2 sin
cos .
2
2
2
2
2
2
Practice (Section 7.3).
1. Write the following as sums:
(a) sin 3θ cos θ,
(b) sin 5θ sin 3θ.
2. Write the following as products:
(a) sin 3θ − sin θ,
(b) cos 4θ − cos 6θ,
(c) sin[(n + 1)θ] + sin[(n − 1)θ].
168
7.4
CHAPTER 7. TRIGONOMETRY 2
Basic identities (read only)
Aim. To learn how to manipulate basic trigonometric identities.
7.4.1
Reciprocal functions
We define three new trigonometric functions from old:
sec x =
7.4.2
1
,
cos x
csc x =
1
,
sin x
and
cot x =
1
cos x
=
.
tan x
sin x
Pythagoras
Recall that sin2 x+cos2 x = 1. There are two other forms of this identity. Dividing sin2 θ+cos2 θ = 1
through by cos2 θ, we have
sin2 θ
cos2 θ
1
+
=
,
2
2
cos θ cos θ
cos2 θ
which gives
tan2 θ + 1 = sec2 θ.
Similarly by dividing sin2 θ + cos2 θ = 1 through by sin2 θ, we get
1 + cot2 θ = csc2 θ.
Remark. Do not memorise the above formulae. Instead, work them out as needed by writing out
sin2 θ + cos2 θ = 1 and dividing through by the appropriate function.
Example. Show that 3 cos2 θ − 2 sin2 θ = 5 cos2 θ − 2.
3 cos2 θ − 2 sin2 θ = 3 cos2 θ − 2(1 − cos2 θ) = 3 cos2 θ − 2 + 2 cos2 θ = 5 cos2 θ − 2.
Example. Prove that csc θ sec θ = 2 csc 2θ.
csc θ sec θ =
1
1
1
2
×
=
=
= 2 csc 2θ.
sin θ cos θ
sin θ cos θ
2 sin θ cos θ
Practice (Section 7.4). Prove the following formulae:
1
= sin θ,
cot θ sec θ
1
2. cot θ ×
= cos θ,
csc θ
3. csc θ ÷ cot θ = sec θ,
1.
4. sin2 θ + 3 cos2 θ = 1 + 2 cos2 θ,
5. sin2 θ + 3 cos2 θ = 3 − 2 sin2 θ,
6. sec4 θ − tan4 θ = sec2 θ + tan2 θ.
7.5. IDENTITIES (READ ONLY)
7.5
169
Identities (read only)
Aim. To learn how to prove trigonometric identities.
Example. Prove that tan(θ + φ) =
tan θ + tan φ
.
1 − tan θ tan φ
sin(θ + φ)
cos(θ + φ)
sin θ + cos φ + cos θ sin φ
=
cos θ cos φ − sin θ sin φ
tan(θ + φ) =
7.5.1
=
sin θ cos φ
cos θ cos φ
cos θ cos φ
cos θ cos φ
=
tan θ + tan φ
1 − tan θ tan φ
+
−
cos θ sin φ
cos θ cos φ
sin θ sin φ
cos θ cos φ
expanding top and bottom
divide every term by cos θ cos φ
What to do if you are stuck
Look at the result you wish to get. What kind of arguments do the trigonometric functions have? In
the above problem, the resulting arguments were θ and φ, while those of the starting statements were
θ + φ. The addition formulae
sin(θ + φ) = sin θ cos φ + cos θ sin θ,
cos(θ + φ) = cos θ cos φ − sin θ sin φ
make these conversions, so use them. The following problem is similar in that we need to convert
from 2θ arguments to a θ argument. So look at:
sin(θ + θ) = 2 sin θ cos θ,
cos(θ + θ) = cos2 θ − sin2 θ = 1 − 2 sin2 θ = 2 cos2 θ − 1,
and observe which of the formulae at right will be useful for the problem.
Example. Prove that
sin 2θ
1−cos 2θ
= cot θ.
sin 2θ
2 sin θ cos θ
2 sin θ cos θ
cos θ
=
=
=
= cot θ
2
2
1 − cos 2θ
sin θ
1 − (1 − 2 sin θ)
2 sin θ
sin θ+sin φ
θ−φ
θ+φ
sin(θ+φ) = cos 2 sec 2 .
θ−φ
with θ+φ
2 and 2 . Therefore it
Example. Show that
seems a good idea to look at at formulae which
We want to end up
convert to these arguments. You also need to recognise that sec x = cos1 x .
θ−φ
2 sin θ+φ
cos θ−φ
sin θ + sin φ
sin θ + sin φ
θ−φ
θ+φ
2 cos 2
2
=
=
=
= cos
sec
θ+φ
θ+φ
θ+φ
θ+φ
sin(θ + φ)
2
2
2 sin 2 cos 2
cos 2
sin 2 2
170
CHAPTER 7. TRIGONOMETRY 2
Practice (Section 7.5). Prove the following:
1.
2.
tan θ − tan φ
,
1 + tan θ tan φ
(b) cos(θ − φ) − cos(θ + φ) = 2 sin θ sin φ,
cos θ + sin θ
π
+θ =
(Note that tan
(c) tan
4
cos θ − sin θ
(a) tan(θ − φ) =
(a)
(b)
(c)
3.
(a)
(b)
π
4
= 1).
sin 2θ
= tan θ,
1 + cot 2θ
1 − cos 2θ
= tan2 θ,
1 + cos 2θ
tan θ cot θ = 2 csc 2θ
Hint: Rewrite tan and cot in terms of sin and cos and then give the expression a common
denominator.
π
1
π
+ θ sin
− θ = cos 2θ,
sin
4
4
2
sin 8θ cos θ − sin 6θ cos 3θ
= tan 2θ
cos 2θ cos θ − sin 4θ sin 3θ
Hint: Use products to sums and then sums to products.
Chapter 8
Differentiation 2
Aim. To learn how to find derivatives of simple trigonometric functions.
8.1
Derivatives of trigonometric functions
1
−2π
− 3π
2
−π
− π2
π
2
π
3π
2
2π
−1
Consider the function f (x) = sin x, and its slope at various values for x. Observe that the slope of f
at x = π2 is zero, at x = π4 is a bit less than 1, and at x = 0 it appears to be about 1. These values are
in fact those of cos x. That is, cos π2 = 0, cos π4 = √12 and cos 0 = 1.
For all values of x,
d
dx [sin x]
= cos x,
d
dx [cos x]
= − sin x
and
d
dx [tan x]
= sec2 x =
1
.
(cos x)2
−π
The points where cos x = 0 (x = π2 , 3π
2 , 2 , . . . ) give the maxima and minima of sin x, the points
where − sin x = 0 (x = 0, π, 2π, −π, etc.) give the maxima and minima of cos x.
π
d
1
2
Notice that for x = π2 , 3π
2 , − 2 , . . . , dx [tan x] = sec x = (cos2 x) does not exist. This is because
cos x = 0 at these points. These are exactly the points where the graph of tan x has vertical
asymptotes. Since sec2 x > 0, for all x, we know that tan x has no relative maxima or minima,
and the slope of its graph is always positive.
Example. Differentiate f (θ) = sin θ and evaluate the slope of f at θ = − π2 .
We have f 0 (θ) = cos θ, and so f 0 (− π2 ) = cos − π2 = 0. The slope of f at θ = − π2 is zero.
Notice that we differentiated the function with respect to θ, since f was a function of θ, and θ was
the variable we were interested in. The name of the variable makes no difference to the process of
differentiation.
171
172
CHAPTER 8. DIFFERENTIATION 2
Remark. You are really going to need to know the major values of sin, cos and tan. Recall the unit
circle and the two special triangles.
1
(x, y) = (cos θ, sin θ)
sin θ
−1
θ
cos θ
1
−1
π
4
π
6
2
√
√
3
π
3
2
1
π
4
1
1
1
In particular, we needed the above to evaluate cos(− π2 ) in the previous example. First do a quick
sketch of the unit circle, along with the point at the angle θ = − π2 from the positive x-axis.
1
−1
1
sin(− π2 )
− π2
P (cos(− π2 ), sin(− π2 )) −1
At θ = − π2 , the value of the x-coordinate of P is zero. So cos(− π2 ) = 0. Notice also that sin(− π2 ) =
−1 because it is the y-coordinate of P .
8.1. DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
173
Example. Differentiate f (t) = 2 tan t and evaluate the slope at t = π.
Recall that
d
[ag(x)] = ag 0 (x).
dx
Therefore,
f 0 (t) = 2
d
2
[tan t] = 2 sec2 t =
.
dx
cos2 t
This means that we need cos π to find f 0 (π).
1
−1
P (cos π, sin π)
π
1
cos π
−1
The value of the x-coordinate at the point on the unit circle with angle π from the positive x-axis is
−1. Therefore cos π = −1, and so
f 0 (π) =
Example. Differentiate f (θ) =
3 sin θ
2
2
2
=
= 2.
cos2 π
−1
+ 5 cos θ and evaluate the slope of f at θ = π3 .
3
d
d
× [sin θ] + 5 × [cos θ]
2 dθ
dθ
3
= cos θ + 5 × (− sin θ)
2
3 cos θ
=
− 5 sin θ.
2
f 0 (θ) =
What is f 0 ( π3 )?
174
CHAPTER 8. DIFFERENTIATION 2
1
P (cos π3 , sin π3 )
sin π3
π
3
−1
cos
1
π
3
−1
√
Using the special triangle, we have cos π3 = 12 and sin π3 = 23 , hence
√
√
3 1
3
3 − 10 3
0 π
f
=
×
−5×
=
.
3
2 2
2
4
√
Example. Differentiate f (t) = 3 cos t and evaluate the slope at 3π
4 .
√
√
d
3π
0
0
Firstly, f (t) = 3 × dt [cos t] = − 3 sin t. But what is f ( 4 )?
1
(cos π4 , sin π4 )
3π
(cos 3π
4 , sin 4 )
π
4
−1
3π
4
1
−1
Draw in P as at above, at an angle of 3π
4 with the positive x-axis. P makes the acute angle
x-axis. Check the first quadrant triangle whose angle is π4 , namely sin π4 = √12 .
π
But sin 3π
4 (i.e. the y-coordinate of P ) has the same length and sign as sin 4 , therefore
sin
3π
π
1
= sin = √ .
4
4
2
We can now answer the question:
r
√
3π
3
0 3π
f
= − 3 sin
=−
.
4
4
2
π
4
with the
8.1. DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
175
Example. Differentiate f (θ) = 2 sin θ − 3 tan θ and evaluate the slope at θ =
We know
3
.
f 0 (θ) = 2 cos θ − 3 sec2 θ = 2 cos θ −
(cos θ)2
What is f 0 7π
6 ?
7π
6 .
1
(cos π6 , sin π6 )
7π
6
−1
1
π
6
7π
(cos 7π
6 , sin 6 )
−1
Draw in P as at above, at an angle of
7π
6
π
√6
3
2 .
with the positive x-axis. P makes the acute angle
with the x-axis. Check the first quadrant triangle whose angle is π6 , namely we know cos π6 =
π
Now cos 7π
6 (i.e. the x-coordinate of P ) has the same length as cos 6 but has a negative value. So
π
cos 7π
6 = − cos 6 =
f0
8.1.1
7π
6
√
− 3
2 ,
and
= 2 cos
7π
6
−
3
cos
7π 2
6
=2
√ !
− 3
−
2
3
−
√ 2
− 3
2
√
= − 3 − 4.
Finding the critical points of a trigonometric function
Recall that the critical points of any function occur where the slope is zero or not defined. So to graph
a function that includes trigonometric expressions, we may need to solve trigonometric equations in
order to find critical points. Note that another method of graphing trigonometric functions is given in
Section 5.4, however functions which are more complicated need the approach given here.
Example. Classify the critical points of f (θ) = sin θ + cos θ + 1, for θ ∈ [0, 2π].
We need to find the values for θ such that f 0 (θ) = cos θ − sin θ is zero or undefined. Since cos θ and
sin θ are defined for all θ, the derivative f 0 is always defined. First, we find the values of θ ∈ [0, 2π]
such that f 0 (θ) is zero so we solve
cos θ − sin θ = 0,
176
CHAPTER 8. DIFFERENTIATION 2
or equivalently
cos θ = sin θ.
If we divide both sides of the second equation by cos θ, we get tan θ = 1. If you don’t already see
the answer, don’t reach straight for the calculator but instead draw the unit circle (for harder problems
revise Chapter 7).
For tan θ = 1, the line through the origin with slope 1 cuts the unit circle in two places, giving us two
angles. Look at the angle in the first quadrant. Since any triangle where the two sides √
next to the right
angle are equal in length (so that the gradient is 1) has the same angles as the 1 : 1 : 2 triangle, the
first quadrant angle must be θ = π4 .
The third quadrant triangle is identical, so the third quadrant angle must be θ = π + π4 = 5π
4 .
1
(cos π4 , sin π4 )
5π
4
−1
π
4
1
5π
(cos 5π
4 , sin 4 )
−1
Therefore f 0 (θ) = 0 at θ = π4 and
√
√
( π4 , 1 + 2) and ( 5π
4 , 1 − 2).
5π
4 ;
we have found that the critical points (between 0 and 2π) are
To see whether these points are minima or maxima, we check the second derivative:
f 00 (θ) = − sin θ − cos θ.
√
Because f 00 π4 = − 2 < 0, we have a relative maximum at θ = π4 .
√
Similarly f 00 5π
2 > 0, so we have a relative minimum at θ = 5π
4 =
4 .
3π 7π
00
Noticethat f (θ) changes
sign
at
θ
=
,
(check
the
circle)
so
that
we have points of inflection at
4
4
3π
7π
4 , 1 and 4 , 1 . You might like to try to sketch one period of the graph of f .
Example. Find the critical points of f (θ) = sin θ + tan θ, for θ ∈ [0, 2π].
We know
1
cos3 θ + 1
f 0 (θ) = cos θ + sec2 θ = cos θ +
=
.
cos2 θ
cos2 θ
Note that it is much easier to find critical points when working with a single term, which is why the
derivative has been rewritten as a quotient.
When is the derivative undefined?
8.1. DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
177
Since it is a quotient, it must be undefined where the denominator is zero. This happens when cos θ =
π 3π
3π
0, which is true for θ = π2 and θ = 3π
2 . Since f is also undefined at θ = 2 , 2 (because tan 2 and
3π
tan 2 are undefined), f has vertical asymptotes at these values.
When is the derivative zero?
Since it is a quotient, the derivative is zero when the numerator (cos3 θ + 1) is zero. That is, when
cos3 θ = −1. Equivalently, when cos θ = −1. Check the unit circle: cos θ = −1 only at θ = π.
Note that f (π) = sin π + tan π = 0, therefore (π, 0) is a critical point. There are vertical asymptotes
at θ = π2 and 3π
2 . To see whether (π, 0) is a maximum or minimum, we will use the first derivative
test by choosing two points either side of this critical point (taking care there are no further critical
4π
points between them and (π, 0)), then evaluating f 0 at these points. We choose θ = 2π
3 and θ = 3 :
f
0
f
0
2π
3
4π
3
=
3
( −1
−4
7
2 ) +1
=
+4= ,
−1 2
8
2
( 2 )
=
3
( −1
−4
7
2 ) +1
=
+4= .
−1 2
8
2
( 2 )
This tells us that the slope is positive to the left of (π, 0) and positive to the right. In other words,
(π, 0) is not a turning point. In fact it is a point of inflection (i.e. a point where concavity changes
sign), which can be seen by examining the second derivative.
Practice (Section 8.1). Differentiate the following functions and find their slopes at the given values.
Also, find the critical points of the functions on [0, 2π] for questions 1, 3, 4 and 5.
1. 3 sin θ,
θ = π4 ,
2. 2 tan t − cos t,
3. 3 cos θ + sin θ,
t=
5π
3 ,
4. 2 sin t − 3 tan t,
θ=
5π
4 ,
5. cos θ − sin θ,
t = −π, 6. tan t + 4 cos t,
θ=
7π
4 ,
t=
11π
6 .
178
CHAPTER 8. DIFFERENTIATION 2
8.2
Chain rule 1
Aim. To learn how to differentiate composite functions.
8.2.1
Functions of functions (composite functions)
We have differentiated a range of functions so far:
• powers of a variable, e.g. 5x2 , 3x−1/2 , 4x + 1,
• trigonometric functions, e.g. sin x, sec x, 3 cos x − tan x.
Recall that a function of a function (i.e. a composite function) is one in which the variable in one
function (the outer function) is replaced by another function (the inner function). Let
f (x) = x2 ,
g(x) = 3x + 1
and
h(x) = sin x.
If we replace the variable x in f (x) with the function g(x), we get
(f ◦ g)(x) = f (g(x)) = f (3x + 1) = (3x + 1)2 .
If we replace the variable x in f (x) with the function h(x), we get
(f ◦ h)(x) = f (h(x)) = f (sin x) = (sin x)2 = sin2 x.
If we replace the variable x in h(x) with the function g(x), we get
(h ◦ g)(x) = h(g(x)) = h(3x + 1) = sin(3x + 1).
If we replace the variable x in g(x) with the function h(x), we get
(g ◦ h)(x) = g(h(x)) = g(sin x) = 3 sin x + 1.
We have differentiated functions like 3 sin x + 1 before without having to remark that they can be
called composite functions, and by expanding we could differentiate (3x + 1)2 . But when we have
functions like sin(3x + 1) and (3x + 1)10 , it begins to be necessary to see them as functions of
functions when we want to differentiate them.
We are going to provide two approaches to the chain rule. The first is a purely practical method which
can be used with some basic functions.
The second (in the following section) gives two versions of a theoretical approach which is more
generally useful.
8.2. CHAIN RULE 1
8.2.2
179
Chain rule
The following is a list of practical rules:
1. Powers of a function
subtract 1 from the power
d
([g(x)]n ) = n × g 0 (x) × [g(x)]n−1 , n 6= 0
dx
multiply by the power
multiply by the derivative of the inner function g(x)
•
d
10
9
dx [(3x + 1) ] = 10 × 3 × (3x + 1) = 30(3x +
d
d
2
2
2
dx tan x = dx [(tan x) ] = 2 sec x × (tan x)
•
d
dx
•
d
dx
•
h
1
x2 +2x
h√
i
=
d
2
dx [(x
i
3x2 + 1 =
2. sin of a function
1)9
+ 2x)−1 ] = −1 × (2x + 2) × (x2 + 2x)−2 =
d
2
dx [(3x
1
+ 1) 2 ] = 12 (6x)(3x2 + 1)
−1
2
=
−2(x+1)
(x2 +2x)2
√ 3x
3x2 +1
replace the outer function sin x with its derivative cos x
d
[sin(g(x))]
dx
= g 0 (x) × cos(g(x))
multiply by the derivative of the inner function
•
•
d
4
3
4
dx [sin(3x + 4x + 2)] = (12x + 4) cos(3x + 4x + 2)
d
dθ [3 sin(πθ)] = π × 3 cos(πθ) = 3π cos(πθ)
3. cos of a function
= 4(3x3 + 1) cos(3x4 + 4x + 2)
replace the outer function cos x with its derivative sin x
d
[cos(g(x))]
dx
= −g 0 (x) × sin(g(x))
multiply by the derivative of the inner function
•
d
dx
π cos
x
2
=
1
2
× −π sin
x
2
=
−π
2
sin
x
2
180
CHAPTER 8. DIFFERENTIATION 2
4. tan of a function
replace the outer function tan x with its derivative
d
[tan(g(x))]
dx
= g 0 (x) × sec2 (g(x))
multiply by the derivative of the inner function
•
d
dθ [5 tan(4θ)]
= 4 × 5 sec2 (4θ) = 20 sec2 (4θ)
You are probably beginning to see a pattern, and this is what we will discuss in the next section.
Sometimes the inner function may itself be a composite function. In this case it is a good idea to write
down what you need to find out and have a working column to calculate this before replacing it in
your answer.
Example. Differentiate sin2 (3x).
need to find the derivative of the inner function sin 3x
d
2
dx [sin (3x)]
=
d
dx
h
i
(sin(3x))2 = 2 ×
d
dx [sin(3x)]
× [sin(3x)]2−1
multiply by the power
subtract 1 from the power
We have
d
dx [sin 3x]
= 3 cos 3x and so,
d (sin 3x)2 = 2 × 3 cos 3x × (sin 3x) = 6 cos 3x sin 3x = 3 sin 6x.
dx
The last simplification comes from sin(A + A) = 2 sin A cos A and substituting 3x for A. The earlier
answer is corect, although the final expression is more useful if we have further work to do with the
result.
√
Example. Differentiate tan( x2 + 1).
Firstly,
p
i
i
p
d h
d hp 2
tan
x2 + 1 =
x + 1 × sec2
x2 + 1 .
dx
dx
h
i
1/2
d
Since dx
x2 + 1
= 21 × 2x × (x2 + 1)−1/2 = √xx2 +1 , we have
p
i
p
d h
x
tan
x2 + 1 = √
sec2
x2 + 1 .
dx
x2 + 1
8.2. CHAIN RULE 1
181
Remark. To use this method, we must ensure that the function considered is indeed a composite
function (i.e. function of a function). In particular, you should figure out what the inner and outer
functions are. If the composition is of the form ‘trigonometric function of another function’, then the
outer function is the trigonometric function. Similarly if we have a power of another function, then
the outer function is the power function (xn , for some n).
Practice (Section 8.2). Differentiate the following functions using the methods given above. Make
sure you have figured what the inner and outer functions are before proceeding. Simplify your results
where possible.
1. f (θ) = 3 sin2 θ,
6. g(x) = 2(4x2 + x − 5)−2 ,
2. g(x) = 2(5x3 + 4x2 − 3x − 10)5 ,
7. f (θ) = cot(θ),
3. y = 3 tan(3θ),
8. h(x) = tan2 (x2 + 4x),
√
9. y = 3x2 + 1,
4. h(x) = cos(x2 + 4x),
5. f (θ) = 3 cos2 (3θ),
10. f (x) =
4
.
5x2 +6x−2
182
8.3
CHAPTER 8. DIFFERENTIATION 2
Chain rule 2
Aim. To learn to use the general formula for the chain rule.
8.3.1
Methods in our madness
You will have noticed there is a pattern in the formulae given for differentiating composite functions.
In fact, there is a general rule which can be applied to any composite function. There are two main
formulae for the chain rule, which of them you choose to use depends on which you find easier to
apply.
8.3.2
Chain rule Formula 1
Suppose we want to differentiate the composite function f (g(x)). Here f is the outer function and g
is the inner function. Then the derivative of this composition is given by
d
[f (g(x))] = f 0 (g(x)) × g 0 (x).
dx
(1)
The notation f 0 (g(x)) is intended to mean the derivative of f with respect to g, or df
dg . To find this,
0
we first find f (x), i.e. the derivative of f where g(x) is replaced with x, then substitute g(x) for x
back into f 0 (x) to obtain f 0 (g(x)). The product of f 0 (g(x)) and g 0 (x) is the derivative required. This
is what we have already been doing in Section 8.2.
Example. Find the derivative of h(x) = (3x + 1)10 using Formula 1.
What is the inner function, and what is the outer function? The brackets are a good clue: choose
3x + 1 inside the brackets as the inner function, that is g(x) = 3x + 1. That leaves us ( )10 not yet
accounted for, so we will use that as the outer function. The outer function is not h given above,
strictly speaking it is a different function (it has a different operation) and we should call it a different
name, something like f (x) = x10 , so that now we have
h(x) = f (g(x)) = (g(x))10 = (3x + 1)10 .
Then apply the formula:
h0 (x) =
d
[f (g(x))] = f 0 (g(x)) × g 0 (x).
dx
The outer function f is ( )10 (i.e. x10 ):
d 10
[x ] = 10x9 −→ 10( )9 .
dx
The inner function g is 3x + 1:
g 0 (x) = 3.
Hence we have
h0 (x) =
d
[f (g(x))] = f 0 (g(x)) × g 0 (x) = 10(3x + 1)9 × 3.
dx
Finally, simplifying this we have h0 (x) = 30(3x + 1)9 .
8.3. CHAIN RULE 2
8.3.3
183
Chain rule Formula 2
We wish to differentiate f (g(x)) with respect to x. First we rename the inner function g as u to
emphasise that it is just a variable. Now f (g(x)) can be written f (u), this latter form is a function of
the variable u. The second chain rule formula is given by:
df
df
du
=
×
dx
du dx
(2)
We now motivate how this formula arises. We wish to find df
dx , but f has been rephrased as a function
df
of u. We can however work out what du is. Thinking of derivatives as fractions, we can say
df
df
=
× something,
dx
du
and that something must ‘cancel out’ the du by replacing it with dx. We can do this by letting that
something be du
dx , because then
df
df
du
=
×
.
dx
du dx
Indeed, ‘cancelling the du’s’ (as functions) gives equality. If we can find the two derivatives on the
right hand side, then their product is the derivative we wanted in the first place.
When in use, we tend not to bother with a name change: if the original function is h(x) = f (g(x)) =
f (u), then we usually write the chain rule as
dh
dh du
=
×
,
dx
du dx
similarly to what we have above.
Example. Differentiate h(x) = (3x + 1)10 using Formula 2.
You do not need to remember the chain rule in this version, you can invent it every time by saying: I
10 = u10 is a function
wish to find dh
dx , but since I renamed the inner function u, now h(x) = (3x + 1)
dh
dh
dh
of u. However, we can find du . Starting with dx = du × something, we find that the something is
d
u
du
dh
dh
×
dx , in which case dx = d
dx balances correctly.
u
h(x) = (3x + 1)10
dh
dh du
=
×
dx
du dx
= 10u9 × 3
outer function h(u) = u10 , inner function u = 3x + 1
dh
d 10
du
=
[u ] = 10u9 ,
=3
du
du
dx
replace u with 3x + 1
= 10(3x + 1)9 × 3
= 30(3x + 1)9
Follow the examples, find which method suits you best. Now practice using it. Then practice a bit
more. In case we haven’t said this before, practising maths is like practising the piano: you cannot
expect to play the piano like a maestro without many hours of practice, and nor can you master
mathematics without many hours of practice.
184
CHAPTER 8. DIFFERENTIATION 2
Example. Differentiate f (x) = sin 4x using both methods.
Method 1
f (x) = sin 4x
outer function h = sin( )
d
[h(g(x))] = h0 (g(x)) × g 0 (x)
dx
= cos 4x × 4
inner function g = 4x
h0 = cos( ),
g0 = 4
= 4 cos 4x
Method 2
f (x) = sin 4x
df
df
du
=
×
dx
du dx
= cos u × 4
du
= 4,
dx
u = 4x, f (u) = sin u
df
d
=
[sin u] = cos u
du
du
= 4 cos 4x
replace u with 4x
√
Example. Differentiate f (x) = 3 x2 + 4x + 2 using both methods.
Method 1
p
f (x) = 3 x2 + 4x + 2
d
[h(g(x))] = h0 (g(x)) × g 0 (x)
dx
3
= √
× (2x + 4)
2
2 x + 4x + 2
3(2x + 4)
= √
2 x2 + 4x + 2
outer function h = 3( )1/2
inner function g = x2 + 4x + 2
3
h0 = ( )−1/2 ,
2
g 0 = x2 + 4
Method 2
p
f (x) = 3 x2 + 4x + 2
df
df
du
=
×
dx
du dx
3
= √ × (2x + 4)
2 u
3(2x + 4)
= √
2 x2 + 4x + 2
u = x2 + 4x + 2,
f (u) = 3u1/2
du
= 2x + 4
dx
df
d
3
=
[3u1/2 ] = u−1/2
du
du
2
replace u with x2 + 4x + 2
8.3. CHAIN RULE 2
185
Example. Differentiate h(x) = sin2 (3x) using both methods.
First we rewrite h as (sin 3x)2 .
Method 1
2
outer function k = 3( )2
h(x) = (sin 3x)
d
[k(g(x))] = k 0 (g(x)) × g 0 (x)
dx
d
= 2(sin 3x) ×
[sin 3x]
dx
inner function g = sin 3x
outer function of sin 3x is sin( ), with derivative cos( )
and inner function is 3x, with derivative 3, so g 0 = 3 cos 3x
= 2(sin 3x) × 3 cos 3x
= 6 sin 3x cos 3x
sin 2x = 2 sin x cos x
= 3 sin 6x
Method 2
h(x) = (sin 3x)2
dh du
dh
=
×
dx
du dx
dh du dv
=
×
×
du dv
dx
= 2u × cos v × 3
u(x) = sin 3x,
v(x) = 3x,
h(u) = u2
u(v) = sin v
dh
du
dv
= 2u,
= cos v,
=3
du
dv
dx
replace u with sin 3x and v with 3x
= 2 sin 3x × cos 3x × 3
= 3 sin 6x
Whichever method you choose to use, practice and familiarity will make it possible to use shortcuts,
and you will see that the practical methods given first are in fact shortcuts. However, you also need
to be familiar with at least one of the formulae given, in order to be able apply the chain rule with
functions other than the basic ones we have so far discussed. Using a formula has the merit of being
able to keep track of all the derivatives that have to be found. This means that if you get lost, there is
a record of where you were going.
8.3.4
Other trigonometric derivatives (read only)
With the chain rule it is quite easy to find the derivatives of csc, csc and cot:
1 d
d
d
−1 = − cos x(sin x)−2 = − cot x csc x,
• dx
csc x = dx
sin x = dx (sin x)
1 d
d
d
−1 = −(− sin x)(cos x)−2 = tan x sec x,
sec x = dx
• dx
cos x = dx (cos x)
1 d
d
d
−1 = − sec2 x(tan x)−2 = − csc2 x.
• dx
cot = dx
tan x = dx (tan x)
In practice, these functions do not appear so often as sin, cos and tan. You can choose to memorise
these derivatives, or be prepared to rewrite csc, sec and cot in terms of sin, cos and tan and differentiate
them in that form. However it is not recommended that these be memorised.
186
CHAPTER 8. DIFFERENTIATION 2
Practice (Section 8.3). Differentiate the following using whichever method you prefer:
1. f (x) =
3
,
x2 +2x+1
6. f (x) = 2 cot(πx),
3. g(θ) = 2 tan(θ2 ),
7. y = cos3 (3θ),
√
8. g(x) = 4 3 x2 − 4x + 7,
4. h(x) = (x4 + 3x2 + 3)6 ,
9. y =
2. y =
sin(2x+1)
,
2
5. ν = 5 tan3 θ,
√ 2
,
x4 +1
10. h(x) =
4
.
(x2 +x+1)3
8.4. PRODUCT RULE
8.4
187
Product rule
Aim. To learn how to find the derivative of the product of two or more functions.
8.4.1
Which rule to use when differentiating
The basic functions that we have so far discussed are:
• powers of a variable,
• trigonometric Functions of a Variable.
In Sections 8.2 and 8.3, we investigated basic functions used to create other functions, called composite
functions. Most functions of a variable are made from basic functions, i.e. they are sums of functions,
functions of functions, and also product of functions. We will discuss a few other basic functions
before the end of the course.
To be able to differentiate any particular one of a large range, we need to recognise how the function
is made from the basic functions. A check list may be helpful; is the function:
• a sum of basic functions,
• a function of basic functions (function composition),
• a product of basic functions, or
• a sum, product or function of any of the above?
If it does not look like one of these, can you rewrite it so it does?
8.4.2
Differentiating products of functions: product rule
We have already dealt briefly with products of basic functions when we differentiated a constant
multiplied by a function. Recall that
d
[af (x)] = af 0 (x),
dx
where a is a constant. When we find that a function is a product of two functions, we can use the
product rule to differentiate it:
d
[f (x)g(x)] = f (x)g 0 (x) + f 0 (x)g(x),
dx
or more briefly:
(f g)0 = f g 0 + f 0 g.
188
CHAPTER 8. DIFFERENTIATION 2
Observe that if we used the rule for the product af (x) = a × f (x), we would get
d
d
[af (x)] = af 0 (x) +
[a] × f (x) = af 0 (x) + (0 × f (x)) = af 0 (x),
dx
dx
since the derivative of a constant (function) is zero.
Example. Differentiate h(x) = 3x(x2 + 4)5 .
The first step is to recognise that this is a product of 3x and (x2 + 4)5 , then use the product rule to
make a statement of the derivatives needed:
h0 (x) =
d
d
d
[3x(x2 + 4)5 ] = 3x [(x2 + 4)5 ] + (x2 + 4)5 ×
[3x]
dx
dx
dx
= 3x × 5(x2 + 4)4 × 2x + (x2 + 4)5 × 3
= 30x2 (x2 + 4)4 + 3(x2 + 4)5
= 3(x2 + 4)4 (11x2 + 4).
It does not matter in the least which function you differentiate first you second, only that you get the
combination right: one function times the derivative of the other plus the other function times the
derivative of the first. You may need a working column to find some of the more difficult derivatives.
Try to simplify the result were possible, especially if you need to continue working with the resulting
derivative.
Example. Find the derivative of f (x) = 3x2 tan x.
View f as a product of 3x2 and tan x, then we have f 0 (x) = 6x tan x + 3x2 sec2 x.
√
Example. Find the derivative of g(x) = 2x2 5x + 1.
First, we need to use the product rule
g 0 (x) =
√
d
d√
[2x2 ] × 5x + 1 + 2x2 ×
5x + 1.
dx
dx
We know
and
Hence,
d
[2x2 ] = 4x
dx
d √
d 1
1
5
du
[ 5x + 1] =
[u 2 ] ×
= u−1/2 × 5 = √
.
dx
du
dx
2
2 5x + 1
√
√
5
5x2
g 0 (x) = 4x 5x + 1 + 2x2 × √
= 4x 5x + 1 + √
.
2 5x + 1
5x + 1
Example. Differentiate f (x) = 5x4 (x3 + 4x)6 .
Note that this is basic function times a composite function. Using the product rule,
f 0 (x) = 5x4 ×
d 3
d
(x + 4x)6 +
[5x4 ] × (x3 + 4x)6 .
dx
dx
8.4. PRODUCT RULE
Further, we have
189
d 3
(x + 4x)6 = 6(3x2 + 4)(x3 + 4x)5 ,
dx
and
d
[5x4 ] = 20x3 .
dx
Hence,
f 0 (x) = 5x4 × 6(3x2 + 4)(x3 + 4x)5 + 20x3 × (x3 + 4x)6
= 30x4 (3x2 + 4)(x3 + 4x)5 + 20x3 (x3 + 4x)6
= (x3 + 4x)5 30x4 (3x2 + 4) + 20x3 (x3 + 4x)
4
3
9
2
5
2
= 10x (x + 4x) (11x + 20)
5
factoring
simplify and factoring
2
= 10x (x + 4) (11x + 20).
Example. Differentiate f (θ) = sin θ cos θ + tan2 θ.
This is the sum of two functions, namely f 0 (θ) is be the sum of the derivatives of sin θ cos θ and
tan2 θ. Note that sin θ cos θ is a product. Further, observe that tan2 θ can be seen as the product
tan θ tan θ, or as the composite function (tan θ)2 which we can differentiate with the chain rule. We
have
d
d
[sin θ cos θ] + [tan2 θ].
f 0 (θ) =
dθ
dθ
As
d
d
d
[sin θ cos θ] =
[sin θ] × cos θ + sin θ × [cos θ] = cos2 θ − sin2 θ,
dθ
dθ
dθ
and
d
[tan θ tan θ] = tan θ × sec2 θ + sec2 θ × tan θ = 2 tan θ sec2 θ,
dθ
using the trigonometric identity cos 2A = cos2 A − sin2 A, we have
f 0 (θ) = cos2 θ − sin2 θ + 2 tan θ sec2 θ = cos 2θ + 2 tan θ sec2 θ.
Example. Differentiate f (x) =
First, rewrite this as
3x+2
.
x2 +4
f (x) = (3x + 2) × (x2 + 4)−1 .
This is the product of two functions, of which one is composite function.
d
d 2
[3x + 2] × (x2 + 4)−1 + (3x + 2) ×
(x + 4)−1
dx
dx
= 3 × (x2 + 4)−1 + (3x + 2) (−1)(x2 + 4)−2 × 2x .
f 0 (x) =
Now we simplify:
f 0 (x) =
2x(3x + 2)
3(x2 + 4) − 2x(3x + 2)
12 − 4x − 3x2
3
−
=
=
.
x2 + 4
(x2 + 4)
(x2 + 4)2
(x2 + 4)
190
CHAPTER 8. DIFFERENTIATION 2
Example. Differentiate f (x) = x2 sin x tan x.
d
d
First note that dx
[x2 sin x] = 2x sin x + x2 cos x and dx
[sin x tan x] = tan x sec x + sin x. This is a
product of three function. We can differentiate in three different ways:
• f 0 (x) = x2 ×
d
dx [sin x tan x]
+
d
2
dx [x ]
× sin x tan x,
• f 0 (x) = x2 sin x ×
d
dx [tan x]
+
d
2
dx [x sin x]
× tan x,
• f 0 (x) = x2 sin x ×
d
dx [tan x]
+
d
2
dx [x sin x]
× tan x,
using any of the above, we get f 0 (x) = x2 tan x sec x + 2x tan x sin x + x2 sin x.
We also remark that the function can be viewed as a product of four functions by separating the x2 ,
however this is not recommended because x2 is already very easy to work with.
Practice (Section 8.4). Differentiate and simplify where possible:
1. g(x) = x2 (3x3 + 1),
4. y = (x + 1)3 sin x,
2. f (θ) = 3θ sin θ,
5. h(x) = 3x2 cos(πx),
3. f (x) =
x
x+1 ,
6. ν = x4 [sin2 x − cos x],
7. g(x) = x(2x + 1)3 sin x,
√
8. f (x) = (x3 − 2x + 1)3 x2 + 1,
9. y = 3x(x2 + 1)10 .
Chapter 9
Functions 2
Aim. To learn how to work with logarithmic functions.
9.1
Logarithms
The graph below shows the function f (x) = 2x .
f (x) = 2x
y
y=x
1
x
Its inverse, f −1 (x), is the reflection of f about the line y = x, and is given by
f −1 (x) = log2 x,
which is read “log base 2 of x”.
191
192
CHAPTER 9. FUNCTIONS 2
We graph this inverse function by reflecting along the line y = x:
y
y=x
f −1 (x) = log2 (x)
x
1
Note that log2 x is not defined for x ≤ 0. This is because f (x) = 2x does not cross the x-axis,
therefore its reflection does not cross the y-axis. Note that log2 x is negative for 0 < x < 1, and is
positive for x > 1. Its slope is always positive. By considering other exponential functions and their
inverses, we can create any number of logarithmic functions. For example, h(x) = 3x is a bit steeper
than f (x) = 2x , and consequently its inverse, h−1 (x) = log3 x is a bit flatter than log2 x (for x > 1).
y
g −1 (x) = log1.5 (x)
1
h−1 (x) = log3 (x)
g(x) = 1.5x
h(x) = 3x
1
x
y=x
For any positive number a, f (x) = ax is an exponential function. For a > 1, the graph of ax has the
same general shape as that of 2x . Similarly, the inverse function f −1 (x) = loga x is a logarithmic
function, and for a > 1, the graph of loga x has the same genral shape as that of log2 x.
9.1. LOGARITHMS
193
y
y=x
ax
(b, c)
loga (x)
1
(c, b)
1
x
If we have a point (b, c) = (b, ab ) on the graph f (x) = ax , observe that as the reflection of the point
(c, b) = (c, loga c) must be on the graph f −1 (x) = loga x.
This leads us to see that if c = ab then b = loga c.
Similarly if b = loga c then c = ab .
We now give a list of properties of logarithms.
1. From the above argument, we have that for all a > 0,
y = ax ⇐⇒ x = loga y.
This rule shows how an exponential function is related to its logarithmic function, and therefore
how the logarithmic function works, and we can use it to write a number of basic log rules.
2. Substituting x = 0 in Property 1 (or looking at the graph) gives us a0 = 1 ⇐⇒ 0 = loga 1.
In other words, for all a > 0, we have
loga 1 = 0.
3. Instead substitute x = 1 into Property 1, then we get a1 = a ⇐⇒ 1 = loga a. Hence for all
a > 0,
loga a = 1.
Example. log4 1 = 0 and log2 1 = 0;
log2 2 = 1 and log4 4 = 1.
4. Because the inverse of the function undoes the work of the original function, i.e. f (f −1 (x)) =
x with f (x) = ax , then we find
aloga x = x.
194
CHAPTER 9. FUNCTIONS 2
5. If instead we let f (x) = loga x and compose with its inverse (ax ), we get
loga (ax ) = x.
Example. 2log2 3 = 3, 2log2 5 = 5, 3log3 5 = 5;
log10 (104 ) = 4, log2 (27 ) = 7, log3 (37 ) = 7.
6. From Property 5, we have loga (ab ) = b, loga (ac ) = c and loga (ab+c ) = b + c. But also
ab+c = ab × ac , therefore
loga (ab × ac ) = loga (ab+c ) = b + c = loga (ab ) + loga (ac ).
Substituting x = ab and y = ac , we have
loga (xy) = loga x + loga y.
7. Using a similar argument to Property 6 but noting that ab−c =
ab
ac ,
we have
x
= loga x − loga y.
loga
y
Example.
6
3
(a) log2 6 = log2 (3 × 2) = log2 3 + log2 2 = log2 3 + 1,
2,3
7
(b) log3 ( 31 ) = log3 1 − log3 3 = 0 − 1 = −1,
6
5
(c) log10 300 = log10 (3 × 100) = log10 3 + log10 100 = log10 3 + log10 102 = log10 3 + 2,
6
5
(d) log2 20 + log2 ( 58 ) = log2 (20 × 85 ) = log2 32 = log2 (25 ) = 5,
6
6
(e) log10 1000 = log10 (10 × (10 × 10)) = log10 10 + log10 (10 × 10) = log10 10 + log10 10 +
3
log10 10 = 1 + 1 + 1 = 3.
From these examples, you may see a pattern for logs of a power. Indeed, the next property will
describe this property.
8. We can use Property 6 (n − 1)-times on loga bn to get
loga bn = loga (b × b × · · · × b) = loga b + loga b + · · · + loga b = n loga b.
Therefore
loga bn = n loga b.
Example.
(a) log10 1000 = log10 103 = 3 log10 10 = 3,
(b) log2 x4 = 4 log2 x,
√
(c) log3 3 = log3 31/2 =
1
2
log3 3 = 21 .
9.1. LOGARITHMS
9.1.1
195
Applications to solving equations
We now have the ability to take an equation involving exponentials and transform the problem that it
represents to one involving logarithms, or vice-versa. We now give examples of such equations.
Example.
1. Solve 2x+1 = 72 for x.
If the above equation is true, then it is also true that
log2 (2x+1 ) = log2 72.
(This process is called “taking logs of both sides” of an equation. This is a good idea when x is
a power of a number.)
First we note that
log2 (2x+1 ) = x + 1.
This can be seen either by Properties 8 and 3 (as shown below) or using Property 5:
log2 2x+1 = (x + 1) log2 2 = (x + 1) × 1 = x + 1.
Therefore,
x + 1 = log2 72.
Now we rearrange and simplify:
x = log2 72 − 1
= log2 (32 × 23 ) − 1
factoring 72
= log2 32 + log2 23 − 1
Property 6
= 2 log2 3 + 2.
Property 3
= 2 log2 3 + 3 log2 2 − 1
Property 8
In fact, we need not have taken log base 2 of each side of the original equation, even though a
power of 2 was involved. Instead we could have taken, say, log base 10:
log10 (2x+1 ) = log10 72.
Simplifying the second equation with Property 8 gives
(x + 1) log10 2 = log10 72,
and after some algebra,
x=
log10 72
− 1.
log10 2
Remark. Here be warned of a common error:
loga x
6= loga (x − y).
loga y
196
CHAPTER 9. FUNCTIONS 2
Check Property 7, it is the one that is wrongly remembered, and instead it says
x
= loga x − loga y.
loga
y
Therefore if we cannot simplify a term like
loga c
loga b
by the rules given previously, we can do nothing except look for a calculator and evaluate the
problem with that, if possible. Likewise loga (c − b) is difficult to simplify.
2. Solve log10 x = 3 for x.
Since x is in the argument of log10 in the equation, we will “take powers both sides”, namely
powers of 10. Since the equation above is true, then it must also be true that
10log10 x = 103 .
Use Property 4 to simplify the left hand side to get
x = 103 .
3. Solve 4x+2 = 32 for x.
Since 4 and 32 are both powers of 2, rewrite 4x+2 as a power of 2:
x+2
4x+2 = 22
= 22x+4 .
This implies
2x + 4 = log2 22x+4 = log2 4x+2 = log2 32 = log2 25 = 5 log2 2 = 5.
Then 2x + 4 = 5, and so x = 21 .
4. Solve log3 x−1 = 4 for x.
Using Property 8, we have − log3 x = 4. Otherwise this can also be seen using Properties 7
and 2:
1
−1
log3 x = log3
= log3 1 − log3 x = − log3 x.
x
Now log3 x = −4, so x = 3−4 =
5. Simplify
We have
1
81 .
log10 36
log10 6 .
log10 36
log10 (62 )
2 log10 6
=
=
= 2.
log10 6
log10 6
log10 6
There are often several ways of solving these kinds of equations, so that your calculations may not
follow the same order as those above. However you should still arrive at the correct answer.
9.1. LOGARITHMS
197
Practice (Section 9.1).
1. Simplify:
(a) log2 6 + log2 8 − log2 3,
log2 18 − log2 3
(b)
,
log2 2 + log2 3
(c) log5 (x + 2)3 − log5 (x + 2).
2. Solve:
(a) log2 x = 4,
(b) 22x+1 = 8,
(c) log10 (x−5)= log10 (7x) − log10 (x + 4).
198
9.2
CHAPTER 9. FUNCTIONS 2
The exponential function and the natural logarithmic function
Aim. To understand the exponential function and the natural logarithmic function.
9.2.1
The exponential function
There is one exponential function whose slope at every point on its graph has the same value as the
function itself. This means that there is a number a such that where f (x) = ax ,
f 0 (x) = f (x).
This number has been given the name e. It is an irrational number (like π, it cannot be expressed as
a quotient of two whole numbers) and its value is approximately 2.718281823. The function ex is
known as the exponential function, and is often written exp(x).
We give here some properties of the exponential function:
1. on x > 0, the graph climbs very steeply,
2. it has y-intercept 1, i.e. it passes the point (0, 1),
3. its value is always positive,
4. its slope is always positive,
5. ex → 0 as x → −∞,
6. ex → ∞ as x → ∞,
7. it ‘grows faster’ than any positive power of x,
8. it has the same shape as every other exponential function ax , where a > 1.
y
ex
x
Here are some simple transformations of the exponential function:
1. e−x is a reflection of ex in the y-axis, therefore its slope is the negative of its own value, for
every x,
2. ex+1 is ex shifted to the left by one unit,
9.2. THE EXPONENTIAL FUNCTION AND THE NATURAL LOGARITHMIC FUNCTION 199
3. e2x is the square of ex and so its slope is twice its own value at every point. It can also be seen
as a stretch of ex by a factor of half in the direction parallel to the x-axis, i.e. it is squeezed into
half the space horizontally.
Knowing the y-intercept and the slope of ex makes it possible to quickly sketch these variations and
many other.
e2x
ex+1
y
ex
x
The function ex works like any other power, we still have the properties:
ex+y = ex × ey ,
9.2.2
ex−y = ex × e−y =
ex
ey
(ex )y = exy .
and
The natural logarithm function
Reflecting ex in the line y = x gives a logarithmic function, the natural logarithm function, which is
denoted loge x or ln x.
ex
y
ln x
1
1
x
Notice how the properties of the natural logarithm contrast those of the exponential function:
1. ln x becomes increasingly steep as x → 0+ ,
200
CHAPTER 9. FUNCTIONS 2
2. it becomes decreasingly steep as x → ∞,
3. it is negative on 0 < x < 1 and is positive on x > 1, with x-intercept 1,
4. ln x → −∞ as x → 0+ ,
5. ln x → ∞ as x → ∞,
6. it ‘grows slower’ than any positive power of x,
7. it has the same shape as every other logarithm function loga x, where a > 1.
The function ln x is the inverse function of ex . These two functions prove to be so useful that
we hardly ever need any other exponential or logarithmic functions. All the rules established for
logarithms in the previous section hold for natural logarithm:
7. ln( xy ) = ln x − ln y,
1. y = ex ⇐⇒ ln y = x,
4. eln x = x,
2. ln 1 = 0,
5. ln(ex ) = x,
3. ln e = 1,
6. ln(xy) = ln x + ln y,
9.2.3
8. ln(xn ) = n ln x.
Converting powers of a to powers of e
All the exponential functions can be written in terms of ex = exp(x):
ax = ex ln a = exp(x ln a).
To see this, let y = ax and take natural logs on both sides of this equation to get
ln y = ln ax = x ln a.
Now taking powers of e both sides give us
eln y = ex ln a .
But as eln y = y and y = ax , we have ax = ex ln a as required.
9.2.4
Converting loga to loge = ln
All the other logarithmic functions can be written in terms of ln x:
loga x =
ln x
.
ln a
This follows first by taking power of a on both sides of y = loga x to get ay = x. Then taking the
natural log on both sides, we have y ln a = ln x. This can then be algebraically manipulated to the
boxed equation above.
9.2. THE EXPONENTIAL FUNCTION AND THE NATURAL LOGARITHMIC FUNCTION 201
9.2.5
Working with these function
Example. Solve the following for x:
1. ex = 2,
2. ex
2 −4
= 1,
3. e2x = 24,
5. ln x2 = 2,
4. ln(x + 2) = 2,
6. ln 3 + ln x = 2,
7. ln(2x) − ln 2 = 1.
1. Take logs on both sides (the natural log since we have a power of e) to get ln ex = ln 2. Then
simplify the left hand side to x = ln ex = ln 2.
2. Take logs on both sides to get x2 − 4 = ln 1 = 0. We have used here Properties 8, 3 and 2.
Now, we have a quadratic equation which factors to (x + 2)(x − 2) = 0, therefore x = −2 or
x = 2.
3. Take logs to get 2x = ln 24 = ln(23 × 3) = 3 ln 2 + ln 3. Therefore x = 12 (3 ln 2 + ln 3).
4. Note that eln(x+2) = x + 2, so we take powers on both sides (powers of e since there is a natural
log) to get x + 2 = e2 . Therefore x = e2 − 2.
5. ln x2 = 2 ln x = 2, therefore ln x = 1 and so x = e.
6. Using Property 6, we get ln 3 + ln x = ln(3x) = 2, so 3x = e2 , hence x = 13 e2 .
7. ln(2x) − ln 2 = ln( 2x
2 ) = ln x = 1, therefore x = e.
Practice (Section 9.2). Solve the following for x, leaving solutions as ln a or eb as necessary:
1. ex−2 = 84,
3. ex e3 = 1,
5. e7x = e3x+1 ,
7. 5 ln(x − 3) = 1,
2. ln(x) = 3,
4. ln x + ln 5 = ln 6,
6. ln(e2x ) = 6,
8. 2e−x = 1.
202
9.3
CHAPTER 9. FUNCTIONS 2
Differentiating exp and natural log
Aim. To learn how differentiate the exponential function and the natural logarithm function.
9.3.1
Differentiating the exponential function
Since the slope of the graph of ex has the same value as ex , for every x, we have
d x
[e ] = ex .
dx
We add ex to our list of basic functions. To differentiate composite functions, like e2x , e2x+3 , etc.,
we will need to use the chain rule shown below. In fact, the following is a consequence of the chain
rule in Section 8.3:
d f (x)
[e
] = f 0 (x)ef (x) .
dx
To see this, let f (x) = u so that ef (x) = eu and
du
dx
= f 0 (x). Then
d[eu ]
d[eu ] du
d[ef (x) ]
=
=
×
= eu × f 0 (x) = f 0 (x)ef (x) .
dx
dx
du
dx
Example. Differentiate the following:
2
1. g(x) = e3x+2 ,
3. f (x) = esin x ,
5. 3t2 et ,
7. 2x .
2. h(θ) = 5eπθ ,
4. ex sin(x),
6. xe2x − 4e−x ,
1. Using the specialised chain rule given above, immediately we have g 0 (x) = 3e3x+2 .
Alternatively, let u = 3x + 2. Then g = eu and so
du
dx
= 3. Also
d u
dg
=
[e ] = eu = e3x+2 ,
du
du
hence
g 0 (x) =
dg
dg
du
=
×
= 3e3x+2 .
dx
du dx
2. h0 (θ) = 5πeπθ
3. f 0 (x) = cos x × esin x = esin x cos x
4. We use the product rule:
d x
dx [e sin(x)]
= ex ×
d
dx [sin x]
+
d x
dx [e ]
d x
[e sin(x)] = ex cos x + ex sin x.
dx
× sin x. Therefore
9.3. DIFFERENTIATING EXP AND NATURAL LOG
203
d
5. Note that dt
[3t2 ] = 6t and using the chain rule we have
product rule,
d t2
dt [e ]
2
= 2tet . Therefore by the
d
d 2
d 2 t2
2
2
2
2
2
2
[3t e ] = [3t2 ]×et +3t2 × [et ] = 6tet +3t2 ×2tet = 6tet +6t3 et = 6t(1+t2 )et .
dt
dt
dt
6. We have a sum of two functions. Note that
d
d
d 2x
[xe2x ] =
[x] × e2x + x ×
[e ] = e2x + 2xe2x
dx
dx
dx
and
d
[−4e−x ] = −4 × (−1) × e−x = 4e−x ,
dx
hence
d
[xe2x − 4e−x ] = e2x + 2xe2x + 4e−x .
dx
7. First recall that 2x = ex ln 2 , but because ln 2 is a constant, we have
d x ln 2
d x
[2 ] =
[e
] = ln 2 × ex ln 2 = 2x ln 2.
dx
dx
9.3.2
Differentiating the natural log function
You may have noticed that no function had a derivative involving x1 in all the earlier basic rules for
differentiation. It turns out that the slope of ln at x is x1 , for every value of x,
y
slope at x = 2 is
slope at x = 1 is 1
slope at x =
1
3
ln x
1
2
1
x
is 3
y=x
therefore
d
1
[ln x] = ,
dx
x
x > 0.
We can in fact give a more general rule. The following is the graph of ln |x|, the natural logarithm of
the absolute value of x.
204
CHAPTER 9. FUNCTIONS 2
y
−1
We have
1
x
(
ln |x| =
ln x, if x > 0
ln(−x), if x < 0.
This function is not defined at x = 0. For negative values of x, the graph of ln |x| is the reflection of
the graph for positive values of x in the y-axis. You can see that at x = −1, the slope must be −1,
and at = −2 the slope must be − 21 , etc.
Therefore the derivative of ln |x| is
1
d
[ln |x|] = ,
dx
x
x 6= 0,
and the specialised chain rule for ln |x| is
f 0 (x)
d
[ln |f (x)|] =
,
dx
f (x)
f (x) 6= 0.
We will show this using the usual chain rule. Write u = f (x) so that ln |f (x)| = ln |u|. Then
d
d
d
du
1
f 0 (x)
[ln |f (x)|] =
[ln |u|] =
[ln |u|] ×
= × f 0 (x) =
.
dx
dx
du
dx
u
f (x)
Example. Differentiate the following on the domain for which the argument of ln is positive:
1. ln(7x),
3. ln x2 ,
5. t2 ln(2t + 1),
2. ln(7(x − 2)),
4. log2 x,
6. ln(e2x + 3x2 ).
1. Somewhat surprisingly,
d
[ln(7x)] =
dx
The following is the graph of ln |7x|.
d
dx [7x]
7x
=
7
1
= .
7x
x
9.3. DIFFERENTIATING EXP AND NATURAL LOG
205
y
− 17
1
7
x
Compare it with the graph of ln |x| and see that everything that happened between say x = 0
and x = 7 in ln |x| occurs between x = 0 and x = 1. The slope of ln |x| at x = 7 is 17 , but
because the graph of ln |7x| is squeezed into a seventh of the horizontal space, the slope of
ln |7x| at x = 1 is much steeper, in fact it is 1.
2.
d
dx [ln 7(x
− 2)] =
7
7(x−2)
=
1
x−2
3. This can be tackled in two different ways. See that the function can be simplified using the log
rules:
d
d
2
ln x2 = 2 ln x and
[2 ln x] = 2 ×
[ln x] = .
dx
dx
x
Alternatively, using the chain rule, we get
d
[ln x2 ] =
dx
4. Rewriting log2 x as
ln x
ln 2 ,
d
2
dx [x ]
x2
=
2x
2
= .
2
x
x
we have
d
1 d
1
[log2 x] =
[ln x] =
.
dx
ln 2 dx
x ln 2
5. Using the product and chain rules:
d
d
d 2
[t ln(2t + 1)] = [t2 ] × ln(2t + 1) + t2 [ln(2t + 1)]
dt
dt
dt
2
2
= 2t ln(2t + 1) + t ×
2t + 1
2
2t
= 2t ln(2t + 1) +
.
2t + 1
6. Using the fact that
d 2x
[e + 3x2 ] = 2e2x + 6x,
dx
we have
2e2x + 6x
d
[ln(e2x + 3x2 )] = 2x
.
dx
e + 3x2
206
CHAPTER 9. FUNCTIONS 2
Practice 9.3
Differentiate the following:
1. f (x) = ln(x + 2),
7. f (θ) = 3 sin θ + 2θ ln(θ + π),
2. f (t) = e3t ,
8. g(x) = x2 − ln(3 − 2x),
3. g(x) = 2e2x + ln(2x),
9. h(x) = e3x cos x,
4. h(x) = x ln x,
10. f (t) = 2et ln(3t),
5. y = (x2 + 2) ln(3x),
11. h(x) = x ln(ex+1 ),
6. g(x) = ex+2 ln x,
12. g(t) = ln(t3 ).
Chapter 10
Differentiation 3
Aim. To learn how to differentiate quotients of functions.
10.1
Quotients of functions
We have already met quotients of functions. Some examples are:
f (x) =
3x
,
ex
g(θ) =
sin θ
πθ
and h(t) =
4t
.
(3t2 + 1)2
These functions are fractions and they can be differentiated using the product rule (and potentially
chain rule) by rephrasing them in the following way:
f (x) = 3xe−x ,
g(θ) = (πθ)−1 sin θ
and
h(t) = 4t(3t2 + 1)−2 .
However you will have noticed that when using the product rule, you often end up with some nasty
fractions to add. The quotient rule generally comes up with a simpler result because the fractions are
already considered.
(x)
Writing h(x) = fg(x)
= f (x)(g(x))−1 , and differentiating we get
h0 (x) = f 0 (x)(g(x))−1 − f (x)g 0 (x)(g(x))−2
f 0 (x) f (x)g 0 (x)
−
g(x)
(g(x))2
f 0 (x)g(x) − f (x)g 0 (x)
=
.
(g(x))2
=
Usually we phrase the quotient rule as follows:
d f (x)
f 0 (x)g(x) − f (x)g 0 (x)
=
,
dx g(x)
(g(x))2
or more briefly,
0
f 0g − f g0
f
=
.
g
g2
207
208
CHAPTER 10. DIFFERENTIATION 3
Example. If g(t) =
et
4t ,
then
g 0 (t) =
et (t − 1)
et × 4t − et × 4
=
.
(4t)2
4t2
Remark. The major problem with the quotient rule is forgetting which term in the final numerator is
subtracted. The way to remember this is to look at the following derivative:
0
1
= [(g)−1 ]0 = −g −2 × g 0 .
g
The negative sign here arises from the fact that we are differentiating the reciprocal of g, i.e. something
to the −1 power. Using the chain rule, this −1 comes down.
Generally, the quotient rule is easier than the product rule when the denominator is a polynomial.
For other types of quotients it may be no faster than the product rule, and may even be harder to
simplify when a logarithmic function or surd is involved. It is a good idea to make a statement first,
and calculate the derivatives next until you are comfortable with the rule, or when the derivatives are
complicated.
Example. Differentiate:
1. f (x) =
x
,
2x + 1
3. f (x) =
2. h(t) =
4t
,
(3t2 + 1)2
4. f (θ) = tan θ,
3x
,
ex
5. h(x) =
1+x
,
x4
6. g(t) =
2t
.
ln(t + 1)
1. By the quotient rule,
f 0 (x) =
d
dx [x]
× (2x + 1) − x ×
(2x + 1)2
d
dx [2x
+ 1]
=
1 × (2x + 1) − x × 2
1
=
.
2
(2x + 1)
(2x + 1)2
2. We use the quotient rule:
d
dt [4t]
d
× (3t2 + 1)2 − 4t × dt
[(3t2 + 1)2 ]
(3t2 + 1)4
4(3t2 + 1)2 − 4t × 2 × 6t(3t2 + 1)
=
(3t2 + 1)4
4(3t2 + 1) − 48t2
=
(3t2 + 1)3
4 − 36t2
=
.
(3t2 + 1)3
h0 (t) =
3t2 + 1 is a common factor
3. We give two ways to differentiate this function. By the quotient rule, we get
f 0 (x) =
3ex − 3xex
3(1 − x)
=
.
e2x
ex
10.1. QUOTIENTS OF FUNCTIONS
209
Writing the function as f (x) = 3xe−x , we can also differentiate using the product rule:
f 0 (x) = 3ex − 3xe−x =
3(1 − x)
.
ex
4. Recall the trigonometric identity sin2 θ + cos2 θ = 1:
f 0 (θ) =
cos θ × cos θ − sin θ × (− sin θ)
cos2 θ + sin2 θ
1
=
=
= sec2 θ.
2
2
cos θ
cos θ
cos2 θ
5. Using the quotient rule,
h0 (x) =
1 × x4 − (1 + x)4x3
x − (1 + x)4
−4 − 3x
=
=
.
8
5
x
x
x5
6. Using the quotient rule, we have
d
dt [2t]
× ln(t + 1) − 2t ×
g (t) =
(ln(t + 1))2
1
2 ln(t + 1) − 2t × t+1
=
(ln(t + 1))2
2(t + 1) ln(t + 1) − 2t
.
=
(t + 1)(ln(t + 1))2
0
d
dt [ln(t
+ 1)]
Otherwise using the product rule with g(t) = 2t(ln(t + 1))−1 ,
d
d
[2t] × (ln(t + 1))−1 + 2t × [(ln(t + 1))−1 ]
dt
dt
1
= 2(ln(t + 1))−1 + 2t ×
× (−1)(ln(t + 1))−2
t+1
2
2t
=
−
.
ln(t + 1) (t + 1)(ln(t + 1))2
g 0 (t) =
Practice (Section 10.1). Differentiate the following using the quotient rule:
1. f (x) =
2x + 1
,
x+3
5. g(x) =
2. g(t) =
t2
,
2t + 1
6. y =
3. h(θ) =
sin 2θ
,
3θ + 4
7. h(θ) =
t+1
4. f (t) = √
,
2t + 1
8. y =
3x2 + 4x + 1
,
e2x−1
x
,
ln x
3 tan2 θ
,
θ
2 4x − 1
−
.
x x2 + 1
210
CHAPTER 10. DIFFERENTIATION 3
10.2
Summary of differentiation rules
10.2.1
Sums, products, quotients and compositions of functions
In this section, we summarise the techniques covered for differentiation.
If asked to differentiate, first ask whether the function is:
• a sum of functions (which ones?)
• a product of functions (which ones?)
• a quotient of functions (what is the numerator and denominator?)
• a function composition (what are the inner and outer functions?)
Once the above have been decided, use the below list to find the correct rule to use to differentiate the
function. If the function is complicated, you may need to apply these rules again to find the derivatives
that help make up the derivative of the original function:
• a sum of functions:
d
d
d
[f (x) + g(x)] =
[f (x)] +
[g(x)]
dx
dx
dx
or
(f + g)0 = f 0 + g 0 ,
• a product of functions:
d
d
d
[f (x)g(x)] =
[f (x)] × g(x) + f (x) ×
[g(x)]
dx
dx
dx
or
(f g)0 = f 0 g + f g 0 ,
• a quotient of functions:
d f (x)
=
dx g(x)
or
× g(x) − f (x) ×
(g(x))2
d
dx [g(x)]
0
f 0g − f g0
f
=
,
g
g2
• a composite function:
or
d
dx [f (x)]
d
[f (g(x))] = f 0 (g(x))g 0 (x)
dx
d
d[f (u)]
df
du
[f (g(x))] =
=
×
,
dx
dx
du dx
where u = g(x).
10.2. SUMMARY OF DIFFERENTIATION RULES
10.2.2
211
Summary of differentiation rules for basic functions
Knowing the derivatives of simple functions is as important as knowing how to reduce the derivative
of a complicated function to a combination of simple ones.
We now give a list of derivatives of some simple functions:
•
d
[a] = 0, where a is a constant,
dx
•
d
[tan x] = sec2 x,
dx
•
d n
[x ] = nxn−1 ,
dx
•
d x
[e ] = ex ,
dx
•
d
[sin x] = cos x,
dx
•
1
d
[ln |x|] = , when x 6= 0,
dx
x
•
d
[cos x] = − sin x,
dx
•
d
1
[ln x] = , when x > 0.
dx
x
10.2.3
Shortcuts for the chain rule
Finally, here are some shortcuts when differentiating function compositions with the chain rule:
•
d
[f (x)]n = nf 0 (x)(f (x))n−1 ,
dx
•
d
[sin(f (x))] = f 0 (x) cos(f (x))
dx
•
d
[cos(f (x))] = −f 0 (x) sin(f (x))
dx
(cos f )0 = −f 0 sin f ,
•
d
[tan(f (x))] = f 0 (x) sec2 (f (x))
dx
(tan f )0 = f 0 sec2 f ,
•
d f (x)
[e
] = f 0 (x)ef (x)
dx
•
d
f 0 (x)
[ln |f (x)|] =
,
dx
f (x)
(f n )0 = nf 0 f n−1 ,
n 6= 0
(sin f )0 = f 0 cos f ,
(ef )0 = f 0 ef ,
(ln |f |)0 =
f (x) 6= 0
f0
,
f
Practice (Section 10.2). Differentiate the following potpourri of functions:
5. −4(7x3 − 6x + 5)3 + 4x ln(4x + 1),
1. ln(x2 ) + 4ex ,
2. 4x2 sin(3x),
3.
7x3 + 2x
,
ex
4. (x2 + 1)e
5x3 −1
,
n 6= 0,
6.
xe2x
,
x2 + 1
7.
tan3 (x2 )
+ 2017.
x
xex − ln |x2 sin( x−cos
x )|
f 6= 0.
212
CHAPTER 10. DIFFERENTIATION 3
Chapter 11
Summation
11.1
Sigma notation
Aim. To learn how to use sigma notation.
Vocabulary and conventions
P
11.1.1
Sigma, the letter for capital S in the Greek alphabet. It is used in
mathematics to mean sum of.
Sequences – functions on the set of integers
We have been working with functions whose domain is the set of all real numbers. There are also
functions whose domain is the set of integers Z (i.e. . . . , −3, −2, −1, 0, 1, 2, 3, . . . ), or sometimes
just the set of natural numbers N (i.e. 1, 2, 3, . . . ), and these functions have a different notation.
Where a function on the real numbers would be written, for example,
f (x) = 2x,
x ∈ R,
a function on N will look like this:
ai = 2i,
i∈N
(or i = 1, 2, 3, . . . ).
This means that for every i ∈ N, there is a corresponding number ai whose value is 2i, that is
a1 = 2 × 1 = 2,
a2 = 2 × 2 = 4,
and so on.
213
a3 = 2 × 3 = 6,
214
CHAPTER 11. SUMMATION
Another example of this type of function is:
bn = n2 − 2,
n ∈ Z,
that is,
. . . b−2 = (−2)2 − 2 = 2,
b−1 = (−1)2 − 2 = −1,
b0 = 02 − 2 = −2,
b1 = 12 − 2 = −1, . . . ,
and so on.
The numbers written here as i or n are called indices (singular is index), and the letters most commonly
used as indices are i, j, k, m and n. A list of numbers given by a function on the set of integers, like
2, 4, 6, 8, . . .
or
. . . , 2, −1, −2, −1, 2, 7, 14, . . .
is called a sequence. The dots to left or right say that the sequence continues in the given pattern. The
most basic sequence is the set of natural counting numbers N:
1, 2, 3, 4, 5, 6, . . .
and we could write this in function notation by writing
ai = i,
11.1.2
i ∈ N.
Sigma notation
We very often need to add up all the numbers in a sequence. Such a sum could take forever to write,
so we use sigma notation to write it in a compact way. For example, the sum of the squares of all the
natural numbers can be written
∞
X
n2 .
n=1
The notation at the bottom of the sigma sign, “n = 1”, means the first value for for n is 1, so the first
term of the sequence n2 will be 12 . Then the second value of n will be 2, and the second term of the
sequence is 22 , and so on.
The sigma sign means add together the terms of the sequence (in this case squares of each value for n).
At the top, the infinity sign (∞) means we go on adding forever. The numbers 1 and ∞ are the bounds
of the variable n of the sum. In particular, 1 is the lower bound and ∞ is the upper bound. The sum
of a sequence is called a series. We now write the above sum in expanded form:
∞
X
n=1
n2 = 12 + 22 + 32 + 42 + 52 + 62 + 72 + · · · .
The dots indicate that the sum keeps on going in this pattern. We can write series with different
bounds and different sequences very compactly in sigma notation.
11.1. SIGMA NOTATION
Example.
1.
P5
j=2 j
3
215
means sum j 3 from j = 2 until j = 5. That is,
5
X
j 3 = 23 + 33 + 43 + 53 = 8 + 27 + 64 + 125 = 224.
j=2
2.
P6
i=0 ai
means sum ai from i = 0 until i = 6. That is,
6
X
ai = a0 + a1 + a2 + a3 + a4 + a5 + a6 .
i=0
We cannot find a value for this last sum until we know what the value of ai is for each i between
0 and 6.
If the only difference between two series is the name of the index, then the sums must be identical;
observe that
5
5
X
X
j3,
k 3 and 23 + 33 + 43 + 53
j=2
k=2
mean exactly the same thing. For this reason, the index is often called a dummy variable. That is, the
name of the index is unimportant, so we can change its name without affecting the sum at all.
• Property 1 – if the bounds are the same and the sequences being added are the same, then the
sums are the same:
n
X
j=m
11.1.3
aj =
n
X
ak .
k=m
Working with sigma notation
It is useful to be able to simplify problems given in sigma notation. Since, for example,
4
X
i=−1
2i2 = 2 × (−1)2 + 2 × (0)2 + 2 × (1)2 + 2 × (2)2 + 2 × (3)2 + 2 × (4)2
= 2 × [(−1)2 + (0)2 + (1)2 + (2)2 + (3)2 + (4)2 ]
=2×
4
X
i2 .
i=−1
This gives us the next rule.
• Property 2 – we can factor out a constant multiple from all the summands:
n
X
i=m
cai = c
n
X
ai ,
i=m
where c is a constant or does not depend on i (that is, it does not change when i does).
216
CHAPTER 11. SUMMATION
We can also add series with the same bounds:
4
X
4
X
2i +
(3i + 2)
3
i=2
i=2
3
= 2(2) + 2(3)3 + 2(4)3 + [3(2) + 2] + [3(3) + 2] + [3(4) + 2]
= [2(2)3 + 3(2) + 2] + [2(3)3 + 3(3) + 2] + [2(4)3 + 3(4) + 2]
=
4
X
(2i3 + (3i + 2)).
i=2
This idea is generalised below.
• Property 3 – the sum of two summations is the summation of the sums:
n
X
(ai + bi ) =
i=m
n
X
ai +
i=m
n
X
bi .
i=m
That is, we can split up a sum into several parts if we wish.
In a similar fashion, we may combine sums if their bounds are consecutive (i.e. follow each other
without a break).
• Property 4 – consecutive sums can be combined:
n
X
ai =
i=m
Example.
1.
3
P
k
X
n
X
ai +
i=m
ai , where k < n.
i=k+1
i2 = 12 + 22 + 32 = 1 + 4 + 9 = 14.
i=1
2. We can then use the first example together with Property 2 to evaluate
3
P
3i2 :
i=1
3
X
3i2 = 3
i=1
3. Write
1
3
+
1
4
+
1
5
3
X
i=1
i2 = 3 × 14 = 42.
in sigma notation.
There are a variety of possible answers, two of them being:
5
1 1 1 X1
+ + =
3 4 5
n
n=3
3
and
1 1 1 X 1
+ + =
.
3 4 5
n+2
i=1
11.1. SIGMA NOTATION
11.1.4
217
Adding with sigma notation
Notice that
50
X
c = c + c + c + · · · + c + c = 50c,
{z
}
|
i=1
50-times
where c is a constant (i.e. it does not change when i changes). This means that we can have
• Property 5 – summation of constants:
n
X
c = nc,
i=1
where c is either a constant or is independent of the index i (i.e. it is constant with respect to
the index i).
The following are some other useful rules for addition which will not be proved here.
• Property 6 – formula for summation of consecutive natural numbers:
n
X
i=
i=1
n(n + 1)
.
2
• Property 7 – formula for summation of consecutive squares:
n
X
i2 =
i=1
n(n + 1)(2n + 1)
.
6
Example. The following illustrate Properties 6 and 7:
1.
20
X
(3i + 2) =
i=1
2.
7
X
j=1
20
X
3i +
i=1
(5j 2 − 3) = 5
7
X
j=1
20
X
2=3
i=1
j2 −
20
X
i+
i=1
7
X
j=1
3=5×
20
X
2=
i=1
3 × 20(20 + 1)
+ 20 × 2 = 670,
2
7(7 + 1)(2 × 7 + 1)
− 3 × 7 = 679.
6
We now separate out two common types of sequences. The first are arithmetic sequences, where each
term of the sequence is some constant plus the previous term. An example is
2, 5, 8, 11, . . .
where the first term is 2 and the constant difference is 3. If we call the first term in the sequence a,
and this constant difference between the terms d, then we can write the sum of the first n terms of the
arithmetic sequence 2, 5, 8, 11, . . . in sigma notation:
n−1
X
i=0
(a + di) =
n−1
X
i=0
(2 + 3i) = 2 + 5 + 8 + 11 + · · · + [2 + 3(n − 1)].
218
CHAPTER 11. SUMMATION
The sum of the first n terms of an arithmetic sequence with first term a and constant difference d is
• Property 8 – sum of arithmetic series:
n−1
X
(a + di) =
i=0
n(2a + (n − 1)d)
,
2
where a is the first term and d is the difference between consecutive terms.
Property 8 is quite easy to show this by using rules 4, 5 and 6:
n−1
X
n−1
X
i=0
i=1
n−1
X
(a + di) = a + d × 0 +
=a
n−1
X
a+d
i=1
(a + di)
i
i=1
= a + a(n − 1) + d ×
=
(n − 1)n
2
n[2a + (n − 1)d]
.
2
The second common type of sequence is a geometric sequence, where every preceding succeeding
term is a constant multiple of the previous. An example is
3, 6, 12, 24, 48, . . .
where the first term is 3 and the ratio (i.e. constant multiple) is 2. If we call the first term a and the
ratio r, then we can write the sum of the first n terms of a geometric sequence as follows.
• Property 9 – sum of geometric series:
n−1
X
ari =
i=0
a(1 − rn )
.
1−r
If |r| < 1, then we can let n → ∞ to find the sum of an infinite geometric sequence. In particular,
|r| < 1 means that the infinite sum is a finite number.
• Property 10 – infinite sum of geometric series:
∞
X
i=0
ari =
a
,
1−r
when |r| < 1.
11.1. SIGMA NOTATION
219
A useful trick (from Property 4) is to be able to change
bounds of a sum by adding and subtracting
P8 the
1 i
appropriate numbers. Suppose we want to know i=3 ( 4 ) . Because this is a portion of a geometric
sequence, we would want to use Property 9. Here we are missing the i = 0, 1, 2 terms (short of
P
8
1 i
i=0 ( 4 ) ). Therefore we may rearrange this to get
8 i
8 i
2 i
2 i
8 i
2 i
X
X
X
X
X
X
1
1
1
1
1
1
=
+
−
=
−
.
4
4
4
4
4
4
i=3
i=3
i=0
i=0
i=0
i=0
We have a formula for both terms in the final equation, so that gives us a way to find the sum we
wanted initially.
Example.
1. Using Rule 9 with a = 5 and r = 13 , we have
1 4
3
n
5
1
−
X
5
3
1
=
=
5
1
3
1− 3
n=0
2. Find the sum
1
2
+
1
4
+
1
8
80
81
2
3
=
200
.
27
1
+ 16
+ ···.
P∞ 1 n
This sum can be written n=1 2 . However to use the summation formula of Property 10,
we must first include n = 0 (then subtract that term so that we have our original sum back):
∞ n
∞ n
X
X
1
1
1
=
−1=
− 1 = 1.
2
2
1 − 21
n=1
n=0
Here a = 1 and r = 21 .
If you are not sure of the logic of the result of this example, think of walking half way cross a
room, then half of that distance in the same direction, then half of that again, and then again,
and again. If your halves are precise, you won’t get to the other side of the room in a finite
number of movements, but you will get close, and closer, and closer. We say in fact you do
get there if you are allowed to make any number of movements, because no matter how small
a measure you choose to use (can you get closer than a centimetre or a millimetre?), you can
make enough movements to find your distance from the far wall is less than that measure. You
are not, however, going to get past the far wall, no matter how many movements you make.
220
CHAPTER 11. SUMMATION
3. Calculate
10
X
j=2
P10
j=2 (3
+ 4j).
10
1
X
X
(3 + 4j) =
(3 + 4j) −
(3 + 4j)
j=0
=
11−1
X
j=0
j=0
(3 + 4j) −
2−1
X
(3 + 4j)
preparing to use Property 8
j=0
11(2 × 3 + 10 × 4) 2(2 × 3 + 1 × 4)
−
2
2
= 253 − 10 = 243.
a = 3 and d = 4
=
Practice (Section 11.1). Solve:
1.
6
P
i(i + 1),
5.
n
P
n(k + 1)(k + 2),
k=1
3.
(2j + j 2 ),
j=0
i=2
2.
4
P
3
P
6.
k
P
3i,
i=1
23−j ,
i=−2
7.
4
P
cos(kπ),
k=0
4.
100
P
(4i + 1),
i=2
8. 1 +
1
4
+
1
16
+ ··· +
1
.
410
Chapter 12
Integration
12.1
Anti-differentiation
Aim. To learn how to reverse differentiation.
Vocabulary and conventions
anti-differentiation
Rindefinite integration
f (x) dx
12.1.1
All these mean “reverse the process of differentiation”.
What is integration and why integrate?
Differentiation deals with quantities that change and finds how they change. Integration lets us work
back from change to the quantities themselves. If we have, for example, an expression for velocity
(rate of change of distance with respect to time) we can integrate to find distance at any time.
To integrate is to reverse the process of differentiation. Unlike differentiation where there are a few
simple rules, a function written down at random probably cannot be integrated. This makes integration
harder.
R
The easiest way of to think of f (x) dx is:
•
R
means integrate the function at right, in this case f (x). In other words, find the function
whose derivative is f (x).
• dx tells you which variable is under consideration, in this case x.
Thus
R
3x2 dx means find the function whose derivative with respect to x is 3x2 .
221
222
CHAPTER 12. INTEGRATION
12.1.2
Integrating polynomials
Integration is done by reversing the differentiation rules. We must reverse both the individual steps
and their order. The following table summarises the information and will be made clear in the
examples:
Differentiating xn ,
Integrating xn ,
n 6= 0
First multiply by the power (n), then
subtract 1 from the power: xn 7→ nxn−1 .
n 6= −1.
Add 1 to the power, then divide by the new power:
n+1
xn 7→ xn+1 . Then add a constant (explained below).
Remember to add a constant (an arbitrary value c) when integrating as differentiating will have have
wiped out any constants.
Note that integrating constants is a special case of the above rule. This is because any constant a can
be written ax0 , so its integral is 1 × ax1 + c = ax + c.
Example.
1. Calculate
R
x4 dx.
First we add 1 to the power (4 becomes 5), then divide by the new power (x5 becomes
Finally add a constant (add c). Therefore
Z
x4 dx =
x5
5 ).
x5
+ c.
5
Note that this integral can also be written 15 x5 + c.
R
2. Calculate (4x + 2) dx.
We integrate the terms 4x and 2 separately. First add 1 to the power of x in 4x (4x becomes
4x2 ), then divide by the new power (4x2 becomes 42 x2 ). Recall that the constant term 2
integrates to 2x. Add a constant (add c). Therefore
Z
4
(4x + 2) dx = x2 + 2x + c = 2x2 + 2x + c.
2
R
3. Find (4x5 − 3x2 − 4x + 2) dx.
Z
4
3
4
2
(4x5 − 3x2 − 4x + 2) dx = x6 − x3 − x2 + 2x + c = x6 − x3 − 2x2 + 2x + c.
6
3
2
3
12.1. ANTI-DIFFERENTIATION
12.1.3
223
Integrating expressions involving fractional and negative powers
The method is exactly the same but may involve fractions.
R
1. Find x−4 dx.
Z
x−4+1
x−3
1
x−4 dx =
+c=
+ c = − 3 + c.
−4 + 1
−3
3x
2. Solve
R
4x2/3 dx.
Adding 1 to the power and dividing by the new power:
2
2 3
5
+1= + =
3
3 3
3
5
4 3
12
= × = .
3
1 5
5
then 4 ÷
Therefore we have
Z
12.1.4
4x2/3 dx =
2
12
4
x 3 +1 + c = x5/3 + c.
2/3 + 1
5
Turning an expression into an integrable form
At the moment all we can integrate is a string of simple power terms joined together by plus or minus.
We can often rewrite an expression involving such things as surds into an expression we can integrate,
provided we can write everything in terms of powers. It is usual to write the final answer in the form
the question was given.
Example.
R √
1. Compute (3 x3 −
3
) dx.
x3
First we write each term as a power of x. The two terms are 3x3/2 and −3x−3 . To each of these
we add 1 to the power:
3
3 2
5
+1= + =
2
2 2
2
and
− 3 + 1 = −2,
and
3
− 3 ÷ −2 = .
2
then divide by that new power:
3÷
5
3 2
6
= × =
2
1 5
5
Therefore we have
Z √
Z
3
3
3 x − 3 dx = (3x3/2 − 3x−3 ) dx
x
6
3
= x5/2 + x−2 + c
5
2
6√ 5
3
=
x + 2 + c.
5
2x
write as simple factors
integrate
write in original form
224
CHAPTER 12. INTEGRATION
2. Because we cannot yet integrate products and can only integrate a sum of simple power terms,
we need to multiply out the bracket before we can integrate (x + 1)(x − 1):
Z
Z
x3
− x + c.
(x + 1)(x − 1) dx = (x2 + 1) dx =
3
3. Similarly, we cannot integrate polynomial quotients, so we must expand by writing it as a sum
of simple power terms. That is, we need to multiply out the brackets before we can integrate:
Z 2
Z Z 2
x −x
x 1
x
x
dx =
dx
dx =
−
−
2x
2x 2x
2 2
Z
1
1 x2
x2 x
=
(x − 1) dx =
−x +c=
− + c.
2
2 2
4
2
Notice that we have factored out a constant factor of
1
2
from inside the integral sign.
In general:
R
R
• cf (x) dx = c f (x) dx, where c is a constant,
R
R
R
• [f (x) + g(x)] dx = f (x) dx + g(x) dx.
It is sometimes useful to rewrite the integral with these rules before beginning to integrate. For
example,
Z
Z
Z
2
2
[3x + x] dx = 3 x dx + x dx.
Practice (Section 12.1). Solve:
(4x3 + 3x − 1)dx,
1.
R
2.
R
3.
R√
4.
R
2 4
3x
− 2x3 + 2 dx,
x3 dx,
1
2x4
−
1
6x3
dx,
5.
R
1
3x2
−
√
3
x dx,
6.
R
8.
R 3x5 −4x (2x + 1)(2x − 1)dx,
R x2 +3x 7.
dx,
x
2x3
dx.
12.2. SUBSTITUTION
12.2
225
Substitution
Aim. To learn how to integrate expressions using substitution.
12.2.1
Integration by substitution
What we want to do is to simplify the expression in order to get an easier integral. In general, we
begin by looking for the “complicating” part of the expression and rename it.
R
Example.
1. Find I = (x + 1)4 dx.
This is a power of a function, where the function is x+1. It would be easy to integrateRif we had
the power of a variable, integrated with respect to that variable. On the other hand, u4 du is
very easy to integrate. The x+1 is the complication, and so we begin by renaming it u = x+1.
So by substituting u = x + 1, we have
Z
u4
.
Now we need to write the remaining parts of the integrand in terms of u. We must also not
forget dx. So we first find out how u changes when x does, that is, we find du
dx . We then make
dx the subject of the resulting equation. In particular,
du
= 1,
dx
which means
du = dx.
When we substitute u = x + 1 and du = dx, we end up with
Z
I = u4 du
which we are able to integrate:
Z
I=
1
u4 du = u5 + c.
5
We have now integrated the expression but need to get the answer back in terms of x. So we
replace u by x + 1. This is called substituting back:
1
I = (x + 1)5 + c.
5
2. Find I =
R
4(3x − 1)8 dx.
What we want to do is turn into something like the easier integral
extra constant).
R
u8
du (with perhaps an
226
CHAPTER 12. INTEGRATION
We see that the most complicated part is 3x − 1, so we let that be u. Therefore
4(3x − 1)8 = 4u8 .
This would give I =
R
4u8
, but we still need to replace the dx.
Differentiating u = 3x − 1 gives
du
= 3,
dx
which we can rearrange to
dx =
Now substitute u for 3x − 1 and
1
du.
3
1
3
du for dx, then integrate:
Z
4 8
4
I=
u du = u9 + c.
3
27
Finally we substitute back in terms of x:
I=
4
(3x − 1)9 + c.
27
Note that if not all the x terms can be replaced by u terms, you have probably used a substitution
that will not work for the problem. Try a different one. It may happen that all the x terms can
be replaced, but the problem is still too hard. In this case also, try a different substitution.
√
3x2 2x3 − 1 dx.
R
First rewrite this as I = 3x2 (2x3 − 1)1/2 dx. Set u = 2x3 − 1 because this is the most
complicated part, then we have (2x3 − 1)1/2 = u1/2 . This will give us I = u1/2
, but
2
leaves us with 3x dx needing to be written in terms of u. We will fix up the dx first by
differentiating:
du
= 6x2 .
dx
Rearrange this to get
1
x2 dx = du.
6
3. Find I =
R
Notice how this sorts out the 3x2 dx, since
3x2 dx =
1
3
du = du.
6
2
Substituting u = 2x3 − 1 and 3x2 dx = 12 du, integrating, then substituting back 2x3 − 1 for
u, we have
Z
1 1/2
1
1
u du = u3/2 + c = (2x3 − 1)3/2 + c.
I=
2
3
3
12.2. SUBSTITUTION
227
Practice (Section 12.2). Solve:
1.
R
(x − 3)11 dx,
3.
R
(2x + 1)(x2 + x + 1)4 dx,
2.
R
x(3x2 − 1)5 dx,
4.
R
(x − 1)(x2 − 2x + 7)6 dx.
228
12.3
CHAPTER 12. INTEGRATION
Trigonometric functions
Aim. To learn how to integrate expressions using trig functions.
12.3.1
Integrating the basic trigonometric functions
Reversing the results from differentiation gives:
Z
d[sin x]
= cos x =⇒
cos x dx = sin x + c,
dx
Z
d[cos x]
d[− cos x]
= − sin x =⇒
= sin x =⇒
sin x dx = − cos x + c,
dx
dx
Z
d[tan x]
= sec2 x =⇒
sec2 x dx = tan x + c.
dx
Be very careful of the signs in integrating sin and cos:
sin differentiates to cos but sin integrates to − cos
cos differentiates to − sin but cos integrates to sin.
12.3.2
Substitution
The method is the same as in the previous section. We now calculate
Z
I = cos 2x dx.
We
cos x, but the 2x complicates things. So we want to simplify
I =
R
R know how to integrate
R
. Set u = 2x so that cos u = cos 2x. This gives us I = cos u
,
cos 2x dx to I = cos u
but we need to substitute for the dx by differentiating u = 2x. Because
du
= 2,
dx
we rearrange this to get
1
dx = du.
2
Substitute u = 2x and dx = 12 du before integrating:
Z
I=
1
1
cos u du = sin u + c.
2
2
Substituting back gives
I=
1
sin 2x + c.
2
12.3. TRIGONOMETRIC FUNCTIONS
Example.
1. Find I =
R
229
2x2 sec2 (x3 + 1) dx.
R
Here the most unpleasant function is sec2 (x3 + 1), but we know that sec2 x dx = tan x + c,
so it is really the x3 + 1 that is causing problems.
R
, but we still need
Let u = x3 + 1, so that sec2 (x3 + 1) = sec2 u. This gives I = sec2 u
to sort out 2x2 dx.
As earlier, we deal with the dx by differentiating u = x3 + 1:
du
= 3x2 ,
dx
rearrange to get
1
du = x2 dx.
3
So we can rewrite the whole 2x2 dx term as
2x2 dx =
2
du.
3
Substitute u = x3 + 1 and 2x2 dx = 32 du, integrate and back substitute to get
Z
2
2
2
I=
sec2 u du = tan u + c = tan(x3 + 1) + c.
3
3
3
2. Find I =
R
sin3 x cos x dx.
Note that whenever you have a power of a trigonometric function such as sin3 x, write it as
(sin x)3 .
R
Using this trick, the integral now becomes (sin x)3 cos x dx. The most complicated term here
is the (sin x)3 . However, we know how to integrate
R the power of a variable, so let u = sin x,
which gives (sin x)3 = u3 . We now have I = u3
, but cos x dx is still left over. As
before, we worry about the dx by differentiating u = sin x and then rearranging to get all xterms on one side. We have du
dx = cos x, so we can rearrange this equation to du = cos x dx.
we substitute u = sin x and du = cos x dx, integrate and substitute back to get
Z
1
u4
I = u3 du =
+ c = sin4 x + c.
4
4
Practice (Section 12.3). Find the following integrals. For question 5, it may be useful to note that
2 cos2 (4x) = cos(8x) + 1.
Z
Z
Z
1.
sin 2xdx,
3.
x sec2 (x2 + 1)dx,
5.
cos2 (4x)dx.
Z
2.
Z
cos(4x + 1)dx,
4.
cos4 x sin xdx,
230
12.4
CHAPTER 12. INTEGRATION
Exponentials and logarithms
Aim. To learn how to integrate expressions using exponentials and log functions.
12.4.1
Integrating the basic functions
Reversing the results from the differentiation gives:
Z
d[ex ]
x
= e =⇒
ex dx = ex + c,
dx
Z
1
1
d[ln |x|]
=
=⇒
dx = ln |x| + c, x 6= 0.
dx
x
x
We can also write the log result as
Z
x−1 dx = ln |x| + c.
Note that we need to write ln |x| because ln x is only defined for x > 0, whereas ln |x| is defined as
long as x 6= 0.
VERY IMPORTANT
The only power that integrates to give ln is −1 (i.e. a denominator of power 1).
The other powers all obey the usual rule for integration.
12.4.2
Polynomial integration and logs
The method in Section 12.1 can be used, except for the case when we have a power of −1: this will
give a log term in the result.
Example.
Z Z
1
2
1
2
x − 2x + − √ + 2 dx = (x2 − 2x + x−1 − 2x−1/2 + x−2 ) dx
x
x x
x3
=
− x2 + ln |x| − 4x1/2 − x−1 + c
3
√
x3
1
=
− x2 + ln |x| − 4 x − + c.
3
x
12.4.3
Substitution
The method of Section 12.2 carries over exactly.
R
1. Find I = e2x dx.
12.4. EXPONENTIALS AND LOGARITHMS
231
x
It is easy
R uto integrate e , so the problem is the inner function (i.e. 2x). Let u = 2x, so that
I = e
. We also need to deal with dx. Differentiate u = 2x and rearrange to get
dx = 21 du. Substitute u = 2x and dx = 21 dx, integrate, then substitute back to get
Z
1 u
1
1
I=
e du = eu + c = e2x + c.
2
2
2
2. Find
R
cos x
1+sin x
dx.
the most complicated part is the denominator, so let u = 1 + sin x. This gives I =
RHere
1
, leaving behind cos x dx. Differentiate u = 1 + sin x and rearrange to get cos x dx =
u
du. Substitute u = 1 + sin x and cos x dx = du, integrate, then substitute back:
Z
1
I=
du = ln |u| + c = ln |1 + sin x| + c.
u
3. Find I =
R
3
4x−1
dx.
The denominator is the complicated part. Let u = 4x − 1, find du
dx , then rearrange to get
dx = 14 du. This means 3 dx = 43 du. Substitute this and u = 4x − 1, integrate, then substitute
back:
Z
3
3
3
du = ln |u| + c = ln |4x − 1| + c.
I=
4u
4
4
R
3
4. We now illustrate the warning about logs coming from the power −1: evaluate I = √4x−1
dx.
Let u = 4x − 1. Note that in I, the power of u is − 12 and not −1. Proceed as usual to get
R
I = u−1/2
. Now differentiate u = 4x − 1 and rearrange to find dx = 41 du. Substitute
this and u = 4x − 1, integrate, then back substitute:
Z
3 −1/2
3 2
3√
I=
u
du = × u1/2 + c =
4x − 1 + c.
4
4 1
2
12.4.4
A short method
Use the following result only if you are very confident: if the integral is of the form
Z 0
f (x)
dx
I=
f (x)
some function f (x), then I = ln |f (x)| + c. Indeed, in the example above we found that
Rfor cos
x
0
1+sin x dx = ln |1 + sin x| + c without the trick (because [1 + sin x] = cos x).
Practice (Section 12.4). Integrate:
1.
R
e−3x dx,
2.
R
xex
2 −1
dx,
3.
R
4.
R
1
3x−1
dx,
5.
R
ex
ex −1
4x
dx,
6.
R
ex
(ex −1)2
x2 −2
dx,
dx.
232
CHAPTER 12. INTEGRATION
Chapter 13
Answers to exercises
Basic Algebra Section 1.1
1. 96
3. 52
2. 3
4. 18
Basic Algebra Section 1.2
1. 2x − 6
3. x2 − 2xy
5. −y 2 − 3y
7. x2 − 2xy + y 2
2. 5a + 10b
4. −x + 1
6. 2x2 − 3xy + y 2
8. −x2 − xy + 2y 2
Basic Algebra Section 1.3
1. 2(2x − 3)
5. x(y − x2 )
2. 5b(a + 4)
6. y(xy − 2 + y)
3. x(x + 2y)
7. a(a − b2 )
4. 2b(2a − 3b)
8. [a(a + 2b) − 1](a + 2b) or (a2 + 2ab − 1)(a + 2b)
Basic Algebra Section 1.4
1.
7x
3y
4.
2.
b
ac
25z
5.
4y
ab(1 + c) b(1 + c)
b(1 + c)2
8. ×
=
a
ac
abc
3.
2c
(3c + 1)
6. 1
9.
18y(x − y)
x
233
7.
2b
c
x/4
3/2x
=
x
3
x 2x
x2
÷
= ×
=
4 2x
4
3
6
234
CHAPTER 13. ANSWERS TO EXERCISES
Basic Algebra Section 1.5
1.
7
10a
5.
5x + 3
3x
2.
ab − c
3a
6.
10b − a
6a2 b2
3.
2y + x
xy
7.
b
a(a − b)
4.
xz(3 + y) − 2z 2 + xy(4 + 3x)
xyz
8.
5xy(x + y) + 2x(x + y) + 3y 2
y(x + y)
Basic Algebra Section 1.6
1. 8x6
64
2. 4
y
3. 6ab3
5. 15x3 y 4
4. −27a6 b3
6. 2xy 2
Basic Algebra Section 2.1
√ √ √
√
√
√
√
√
4+ 2 3+ 2
4+ 2 3+ 2
12 + 2 + 3 2 + 4 2
14 + 7 2
√ ×
√ =
√ √ =
√
√ =
1.
=2+ 2
7
3− 2 3+ 2
3− 2 3+ 2
9−2−3 2+3 2
√
√
√
√
2. 5 2(1 − 10) or 5 2 − 10 5
3. −3x2
√
√
√
√
4. 28 3 + 6 or 3(28 + 2)
√
√
√
10 x
x−5
10x − 50 x
5. √
×√
=
x − 25
x+5
x−5
6.
7.
8.
9.
10.
3x2
2y
p
√
25a2 (a + 4b) = 5a a + 4b
√
x+y
√
√
3x2 x + 6xy 2 or 3x(x x + 2y 2 )
p
p
3
3
b3 (81b2 − 64a3 ) = b × 81b2 − 64a3
235
Basic Algebra Section 2.2
1. 3
5.
77
4
6.
−10
9
7.
5
3
2. 13
3.
4.
−2
−10
=
5(a + 2b − 2)
a + 2b − 2
y2 − 3
3
8. 2a − 2b
Basic Algebra Section 2.3
1. (x − 4)(x + 1) = 0
x = 4 or x = −1
5. (3x + 1)(x + 2) = 0
−1
or x = −2
x=
3
2. (2x + 3)(3x + 4) = 0
−4
−3
or x =
x=
2
3
6. (4y − 1)(y + 2) = 0
1
y = or y = −2
4
3. (2x + 1)(x − 3) = 0
−1
or x = 3
x=
2
7. (2x + 5)(x − 1) = 0
−5
x=
or x = 1
2
4. (3x + 5)(3x − 5) = 0
−5
5
x=
or x =
3
3
8. (3x − 8)(x + 1) = 0
8
x = or x = −1
3
Basic Algebra Section 2.4
6. x ≥ −2
1. 5 < x
2. x > b2 +
3. x <
b
3
7. x >
−10
9
8. −12 ≤ x < 5
9. x < 0 or x >
4. x ≥ 3
10. −1 < x ≤ 0
5. x < 0
Basic Algebra Section 2.5
3 2 25
1. x −
−
=0
2
4
2. (x − 3)2 − 2 = 0
20
27
x = 4 or x = −1
x=3+
√
2 or x = 3 −
√
2
5
3
236
CHAPTER 13. ANSWERS TO EXERCISES
"
3. 2
5
x−
4
2
#
49
=0
−
16
4. 9x2 − 25 = 0
x=
x = 3 or x =
−1
2
5
−5
or x =
3
3
5. (x − 3)2 = 0
x=3
"
#
3 2 31
6. 2 x −
+
=0
4
16
No solution
7. (x − 1)2 − 25 = 0
x = −4 or x = 6
2
5
81
8. x −
−
=0
x = −2 or x = 7
2
4
#
"
8
5 2 121
−
=0
x = −1 or x =
9. 3 x −
6
36
3
Basic Algebra Section 2.6
1. x3 − 3x2 − x + 3 = (x − 1)(x2 − 2x − 3) = (x − 1)(x − 3)(x + 1) = 0. x = 1, 3, −1.
2. x4 − 8x2 + 16 = (x2 − 4)2 = (x − 2)2 (x + 2)2 = 0. x = 2, −2
3. z 3 − z 2 − z − 2 = (z − 2)(z 2 + z + 1) = 0. z = 2
4. 27y 3 + 64 = (3y + 4)(9y 2 − 12y + 16) = 0. y =
−4
3
√
√
√
5. (x2 − 3)2 − 25 = (x2 − 3 − 5)(x2 − 3 + 5) = (x − 2 2)(x + 2 2)(x2 + 2) = 0. x = ±2 2
6. x4 + 10x3 + 35x2 + 50x + 24 = (x + 1)(x3 + 9x2 + 26x + 24) = (x + 1)(x + 2)(x2 + 7x + 12)
= (x + 1)(x + 2)(x + 3)(x + 4). x = −1, −2, −3, −4
7. 2x − 7 +
9
x+1
8. x2 + 2x + 3 +
12x + 5
− 2x − 2
x2
Basic Algebra Section 3.1
7
1. (a) , (b) gradient is undefined since run = 0
2
2. (a) 3x − 2y − 5 = 0, (b) 7x + 3y − 16 = 0
3. (a) y = 3,
(b) x = 1
4. (a) x + 3y − 7 = 0,
(b) 3x − y − 11 = 0
237
Basic Algebra Section 3.2
7 6
22 5
2.
1.
,
,
13 13
5 5
3.
22 13
,−
15 10
4. No solution
Basic Algebra Section 3.3
1. Circle with radius 5
2. Ellipse: x-intercepts ±4, y-intercepts ±3
3
3. Hyperbola: x-intercepts ±4, asymptotes y = ± x
4
4. Ellipse: x-intercepts ±3, y-intercepts ±4
4
5. Hyperbola: x-intercepts ±3, asymptotes y = ± x
3
6. Power function: y-intercept at (0, 3)
7. Power function: y-intercept at (0, 1), but notice this one slopes down from left to right.
Basic Algebra Section 3.4
1. (−1, −3)
(5, −9)
2. (1, 4)
3. (1, 2)
4. No solution
Functions Section 4.1
1. 7
3. −3
2. −1
4.
4−x
x
5. x
7.
1
(x + 1)
4
6. 16x2 − 8x − 3
8.
1
x
9. R, R ie all real numbers
10. For both, all the real numbers except zero. (Refer back to graph of rectangular hyperbola).
11. R, y ≥ −4. Draw graph to see this
x2 − 4
√
2
12. Restrict range to y ≥ 0, giving h−1 (x) =
√
x + 4.
−4
−2
−4
2
x+4
238
CHAPTER 13. ANSWERS TO EXERCISES
Functions Section 4.2
1.
(a) y = 2x2 − 4x + 1
(c) Hyperbola:
(1, 3)
3.
(a) Circle: x2 + y 2 = 25 translated (1, −4)
(a) y = −x − 1
(b) y = 2x2 − 12x + 21
2.
x2
(b) Ellipse:
+ y 2 = 1 translated (−1, 2)
4
4.
x2 − y 2 = 1
translated
(b) x2 + y 2 = 16
(c) y = −2x2 + 2
(a) Translated right 4 and up 3
(b) Stretched parallel to y-axis with factor -2, i.e. reflected in x-axis and stretched by a factor
of 2
(c) Stretched 2 parallel to y-axis then translated (−1, −4)
Functions Section 4.3
1. −3 ≤ x ≤ 5
3. x > 2 or x < 1
2. x ≥ −8 or x ≤ −18
4. −1 ≤ x ≤
1
3
5. Read as this: the distance of x from 0 is less than the distance of x from −2. Using a number
line you can see that we cannot have x = −1, and we can have x > −1. Every other value of
x is closer to −2 than to 0. x > −1.
6. Similarly: the distance of x from 8 is less than the distance of x from 16, giving x < 12.
7. Try using the definition, and so writing:
5
which is OK, as then x ≥ 0.
2
(b) 5 − x < −x, for x < 0. This has solution 5 < 0, which is not true.
(a) 5 − x < x, for x ≥ 0. This has solution x >
5
So the only solution comes from the first line, x > .
2
8. Try the definition:
(a) 2x − 3 < x + 1, if x + 1 ≥ 0. This has solution x < 4, which is OK as long as also
x ≥ −1. First part gives −1 ≤ x < 4.
2
(b) 2x − 3 < −(x + 1), if x + 1 < 0. This has solution x < which is OK as long as also
3
x < −1. Second part gives x < −1.
Total solution is x < 4.
239
3
, the left side is positive or zero and so we can use the rule
2
3
|x| > a ⇐⇒ x > a for a ≥ 0, getting ≤ x < 4 (the second part gives no solutions). If
2
3
however x < , the left side is negative, and since everything in an absolute value bracket is
2
3
either positive or 0, the inequality must be true for every x < .
2
Alternatively: See that if x ≥
Trigonometry Section 5.1
1.
(a) sin 35◦ , − cos 35◦ , − tan 35
2.
(a) 231.3◦
(b) − sin 50◦ , cos 50◦ , − tan 50
(b) 98.1◦
(d) − sin 20◦ , cos 20◦ , − tan 20
(d) 81.9◦
(c) − sin 48◦ , − cos 48◦ , tan 48
(c) 326.3◦
Trigonometry Section 5.2
1.
(a) 1.274 rad
150◦
(b) 6.266 rad
(b)
(c) 102.0◦
3.
(d) 307.2◦
4.
2.
(a)
40◦
3π
5π π π π
π
, π, , 2π, , , ,
2
2
2 6 3 4
(b) 90◦ , 270◦ , 450◦ , 360◦ , 225◦ , 30◦ , 120◦
(a)
(a) 11.05 m
(b) 6.499 m
Trigonometry Section 5.4
1. sin graph between −3 and 3 on y-axis
2. sin graph moved down one unit
3. cos graph, but with a complete cycle between 0 and π
4. Same as sin x
Trigonometry Section 5.5
1.
π 3π
,
4 4
2. 1.244, 5.039
3. 2.156, 5.297
4.
5π 7π
,
6 6
5.
π 7π
,
6 6
6. 3.499, 5.925
240
CHAPTER 13. ANSWERS TO EXERCISES
Trigonometry Section 5.6
π 7π 9π 15π
, , ,
4 4 4
4
2. −0.8419, 2.3
4. 1.4160, 2.9868, 4.5576, 6.1283
1.
3.
5. 1.6307, 4.7723
π 11π 13π 23π
,
,
,
12 12 12 12
6. 1.2916, 2.9916
Differentiation Section 6.1
1. f 0 (x) = 5x4 : slope is 80
2.
4.
dy
= 6x: slope is 0
dx
3. g 0 (x) = 12x − 20x−5 : slope is −23
dv
2
1
14
6
= x6 + 3 + √ : slope is 11
dx
3
x
15
5 x
5. h0 (x) =
3
8
6.
−3 −5
1
x 2 + 8x−5 + 3: slope is 9
2
2
du
3
3
= 3x + 2 : slope is 6
dx
x
4
Differentiation Section 6.2
1. x < a, x > l
8. x = i
2. x = d, i, k
9. x = d, i, k
3. x = a, c, l
10. x = b, e, j
4. x = a, c, d, i, k, l
11. x ∈ (d, i), x ∈ (k, l)
5. x ∈ (a, b), x ∈ (c, e), x ∈ (j, l)
12. x ∈ (a, b), x ∈ (b, c), x ∈ (c, d), x ∈ (i, k)
6. x = d, k
13. x ∈ [a, c]
7. x ∈ (b, c), x ∈ (e, j)
14. x ∈ (c, l)
241
Differentiation Section 6.3
1. f (x) = x3 + 2x2 − 4x + 1
9
1
1
f (0) = 1: y-intercept (0, 1), Notice that f (1) = 0 giving an x-intercept at (1, 0). f 0 (x) =
2 4
3x2 + 4x − 4: use the quadratic formula to find that the derivative is zero when x = − ± ,
3
3
2 −13
giving turning points at (−2, 9) and
,
. f 00 (x) = 6x + 4: second derivative is zero
3 27
−2
at x =
, positive for values to the right of this and negative to the left. Then the graph
3
2 −13
is concave down at (−2, 9), it must be a relative maximum. Similarly,
,
must be a
3 27
relative minimum. A few more points, say f (−4) and f (3) will be sufficient for a good graph.
Sketch the curves in keeping concavity in mind.
2. h(x) = 2x − x4
√
3
2
1
3. y = √
x
1
1
242
CHAPTER 13. ANSWERS TO EXERCISES
Trigonometry Section 7.1
1.
(a) sin X cos 2Y − cos X sin 2Y (c) sin 2θ
tan 3A − tan B
(b)
(d) − cot θ
1 + tan 3A tan B
(a) sin(B − A)
2.
(b) cos 5ζ
Trigonometry Section 7.2
1. sin 30◦ =
1
2
2. cos 80◦
3. cos 2x
5. cos 8x
4. cos α
6. tan
π
=1
4
Trigonometry Section 7.3
1.
1
(sin 4θ + sin 2θ)
2
1
(b) − (cos 8θ − cos 2θ)
2
(a)
2.
(a) 2 cos 2θ sin θ
(b) −2 sin 5θ sin(−θ) = 2 sin 5θ sin θ
(c) 2 sin nθ cos θ
Differentiation Section 8.1
3
π 3π
1. 3 cos θ, √ , θ = ,
2 2
2
√
3
2
2. 2 sec t + sin t, 8 −
2
√
3
1
3. −3 sin θ + cos θ, √ − √ = 2, θ = 0.3218, 3.4634
2
2
4. 2 cos t − 3 sec2 t, −5, t =
π 3π
,
2 2
1
1
3π 7π
5. − sin θ − cos θ, √ − √ = 0, θ =
,
4 4
2
2
6. sec2 t − 4 sin t,
4
10
+2=
3
3
Differentiation Section 8.2
1. 6 sin θ cos θ = 3 sin(2θ)
5. −18 sin(3θ) cos(3θ) = −9 sin(6θ)
2. 10(15x2 + 8x − 3)(5x3 + 4x2 − 3x − 10)4
6. −4(8x + 1)(4x2 + x − 5)−3
3. 9 sec2 (3θ)
7. − sec2 (θ)(tan(θ))−2 = − csc2 (θ)
4. −(2x + 4) sin(x2 + 4x)
8. 2(2x + 4) tan(x2 + 4x) sec2 (x2 + 4x)
243
9.
−1
1
3x
(3x2 + 1) 2 × 6x = √
2
3x2 + 1
10.
−4(10x + 6)
−8(5x + 3)
=
2
2
(5x + 6x − 2)
(5x2 + 6x − 2)2
Differentiation Section 8.3
1. −3(2x + 2)(x2 + 2x + 1)−2 =
−6(x + 1)
(x2 + 2x + 1)2
2. cos(2x + 1)
3. 4θ sec2 (θ2 )
4. 12(2x3 + 3x)(x4 + 3x2 + 3)5
dν
= 15 sec2 (θ) tan2 (θ)
dθ
d 6.
2(tan πx)−1 = −2π sec2 πx × (tan πx)−2 = −2π csc2 πx
dx
5. ν = 5 tan3 (θ),
dy
= −9 sin 3θ cos2 3θ
dx
i 4
−2
1
8(x − 2)
d h 2
p
8.
4(x − 4x + 7) 3 = (2x − 4)(x2 − 4x + 7) 3 =
3
dx
3
3 × (x2 − 4x + 7)2
7. y = cos3 (3θ),
9. y = 2(x4 + 1)
10.
−1
2
,
−3
dy
−4x3
= −(x4 + 1) 2 × 4x3 = p
dx
(x4 + 1)3
d 2
−12(2x + 1)
4(x + x + 1)−3 = −12(x2 + x + 1)−4 × (2x + 1) = 2
dx
(x + x + 1)4
Differentiation Section 8.4
1. 2x(3x2 + 1) + x2 (9x2 ) = 15x4 + 6x3 + 2x
2. 3 sin θ + 3θ cos θ
3.
d 1
x(x + 1)−1 = (x + 1)−1 − x(x + 1)−2 =
dx
(x + 1)2
4.
dy
= 3(x + 1)2 sin(x) + (x + 1)3 cos(x) = (x + 1)2 [3 sin(x) + (x + 1) cos(x)]
dx
5. 6x cos(πx) − 3πx2 sin(πx) = 3x(2 cos πx − πx sin πx)
6.
dν
= 4x3 [sin2 x − cos x] + x4 [2 sin x cos x + sin x]
dθ
= 4x3 [sin2 x − cos x] + x4 [sin 2x + sin x]
= x3 [4 sin2 x − 4 cos x + x sin 2x + x sin x]
244
CHAPTER 13. ANSWERS TO EXERCISES
7. (2x + 1)3 sin x + x 6(2x + 1)2 sin x + (2x + 1)3 cos x
= (2x + 1)2 (8x + 1) sin x + x(2x + 1)3 cos x
= (2x + 1)2 [(8x + 1) sin x + x(2x + 1) cos x]
p
−1
2x 3
8. 3(3x2 − 2)(x3 − 2x + 1)2 x2 + 1 +
(x − 2x + 1)3 (x2 + 1) 2
2
p
x(x3 − 2x + 1)3
2
3
2
2
√
= 3(3x − 2)(x − 2x + 1) x + 1 +
x2 + 1
(x3 − 2x + 1)2 (10x4 + x2 + x − 6)
√
=
x2 + 1
9. 3(x2 + 1)10 + 3x × 10(x2 + 1)9 × 2x = 3(x2 + 1)10 + 60x2 (x2 + 1)9
Functions Section 9.1
1.
(a) log2 (2 × 3) + log2 23 − log2 3 = log2 2 + log2 3 + 3 log2 2 − log2 3 = 4
(b)
log2 32 × 2 − log2 3
2 log2 3 + 1 − log2 3
log2 3 + 1
=
=
=1
1 + log2 3
1 + log2 3
1 + log2 3
(c) log5 (x + 2)3 − log5 (x + 2) = 2 log5 (x + 2)
2.
(a) log2 x = 4 =⇒ x = 24 = 16
(b) 22x+1 = 8 =⇒ 2x + 1 = log2 8 = 3 log2 2 = 3 =⇒ x = 1
7x
(c) log10 (x − 5) = log10
x+4
7x
=⇒ x − 5 =
x+4
(x − 5)(x + 4) = 7x
x2 − 8x − 20 = 0
(x − 10)(x + 2) = 0 =⇒ x = 10, x = −2.
Notice however that the original statements must have a positive number in the arguments
of all the logarithms, that is, we must have x > 5. Therefore x = 10 is the only solution.
Functions Section 9.2
1
4
1. x = ln 7 + ln 3 + 2 ln 2 + 2
5. 7x = 3x + 1 =⇒ x =
2. x = e3
6. x = 3
3. x = −3
7. x = e 5 + 3
1
1
8. −x = ln
=⇒ x = − ln
= ln 2
2
2
4. x =
6
5
1
245
Functions Section 9.3
1.
1
x+2
7. 3 cos θ + 2 ln(θ + π) +
2. 3e3t
3. 4e2x +
8. 2x +
1
x
2
3 − 2x
9. 3e3x cos x − e3x sin x
4. ln x + 1
5. 2x ln(3x) +
2θ
θ+π
x2 + 2
x
6. ex+2 ln(x) +
ex+2
x
10. 2et ln(3t) +
2et
t
11. 2x + 1
12.
3
t
Differentiation Section 10.1
1.
2(x + 3) − (2x + 1)
5
=
2
(x + 3)
(x + 3)2
2.
2t(2t + 1) − t2 × 2
2t(t + 1)
=
2
(2t + 1)
(2t + 1)2
6θ cos 2θ + 8 cos 2θ − 3 sin 2θ
2 cos 2θ × (3θ + 4) − 3 sin 2θ
=
(3θ + 4)2
(3θ + 4)2
√
1
1 × 2t + 1 − (t + 1) × √2t+1
t
2t + 1 − (t + 1)
√
√
4.
=
=
2t + 1
(2t + 1) 2t + 1
(2t + 1) 2t + 1
3.
5.
(6x + 4)e2x−1 − (3x2 + 4x + 1)2e2x−1
(−6x2 − 2x + 2)
=
(e2x−1 )2
e2x−1
6.
ln x − 1
(ln x)2
7.
6 tan θ sec2 θ 3 tan2 θ
−
θ
θ2
8.
0 × x − 2 × 1 4(x2 + 1) − (4x − 1)2x
−2 4 + 2x − 4x2
−
=
−
x2
(x2 + 1)2
x2
(x2 + 1)2
Differentiation Section 10.2
1.
d
d
2x
d
2
(ln(x2 ) + 4ex ) =
(ln(x2 )) +
(4ex ) = 2 + 4ex = + 4ex
dx
dx
dx
x
x
246
CHAPTER 13. ANSWERS TO EXERCISES
d
d
d
(4x2 sin(3x)) =
(4x2 ) × sin(3x) + 4x2 ×
(sin(3x))
dx
dx
dx
= (8x) × sin(3x) + 4x2 × 3 × cos(3x) = 8x sin(3x) + 12x2 cos(3x)
d
d
3
x
3
x
d 7x3 + 2x
dx (7x + 2x) × e − (7x + 2x) × dx (e )
=
3.
dx
ex
(ex )2
2
x
3
x
(21x + 2)e − (7x + 2x)e
21x2 + 2 − 7x3 − 2x
=
=
(ex )2
ex
i
d 2
d 5x3 −1 d h 2
3
3
(x + 1)e5x −1 =
(x + 1) × e5x −1 + (x2 + 1) ×
e
4.
dx
dx
dx
5x3 −1
2 2
5x3 −1
= 2xe
+ 15x (x + 1)e
2.
d
−4(7x3 − 6x + 5)3 + 4x ln(4x + 1)
dx
d
d
= −4
(7x3 − 6x + 5)3 +
(4x ln(4x + 1))
dx
dx
d
d
= −4 × 3(7x3 − 6x + 5)2 × (21x2 − 6) +
(4x) × ln(4x + 1) + +4x ×
(ln(4x + 1))
dx
dx
16x
= −12(7x3 − 6x + 5)2 (21x2 − 6) + 4 ln(4x + 1) +
4x + 1
d
d
2x
2
2x
2
d
xe2x
dx (xe ) × (x + 1) − xe × dx (x + 1)
6.
=
dx x2 + 1
(x2 + 1)2
d
d
(x) × e2x + x × dx
(e2x ) × (x2 + 1) − xe2x × 2x
= dx
(x2 + 1)2
2x
2x
2
e2x [2x3 − x2 + 2x + 1]
[e + 2xe ](x + 1) − 2x2 e2x
=
=
(x2 + 1)2
(x2 + 1)2
5.
7. This function is differentiable because it is a combination of simple functions (albeit a rather
complicated one) which we have summarised in this section. WolframAlpha can differentiate
this for you.
Summation Section 11.1
1.
6
X
2
i +
i=2
2. n
i=2
n
X
k=1
5
6
X
i=
6
X
i +
6
X
i=1
i=1
i − (1 + 1) = 110
n(n + 1)(2n + 1)
n(n + 1)
(k + 3k + 2) = n
+3×
+ 2n
6
2
2
4
0
3. 2 + 2 + · · · + 2 =
4. 4 ×
2
5
X
i=0
2i =
1 − 26
= 63
1−2
100 × 101
+ 100 − 5 = 20295
2
247
5.
1 − 25 4 × 5 × 9
+
= 61
1−2
6
6.
3k(k + 1)
2
7. cos 0 + cos π + cos 2π + cos 3π + cos 4π = 1
11
10 i
4194303
X
1 − 14
1398101
1
4193404
=
=
8.
=
= 1.3333
1
3
4
1048576
1− 4
4
i=0
Integration Section 12.1
3
1. x4 + x2 − x + c
2
2 5 1 4
x − x + 2x + c
2.
15
2
√
2
x5 + c
3.
5
1
1
+
+c
3
6x
12x2
3√
1
3
−
x4 + c
5. −
3x 4
4
6. x3 − x + c
3
4. −
7.
1 2
x + 3x + c
2
8.
1 3 2
x + +c
2
x
Integration Section 12.2
1.
1
(x − 3)12 + c
12
3.
1 2
(x + x + 1)5 + c
5
2.
1
(3x2 − 1)6 + c
36
4.
1 2
(x − 2x + 7)7 + c
14
Integration Section 12.3
1
1. − cos 2x + c
2
1
2. sin(4x + 1) + c
4
1
3. tan(x2 + 1) + c
2
1
4. − cos5 x + c
5
5.
1
1
x+
sin 8x + c
2
16
Integration Section 12.4
1
1. − e−3x + c
3
1 2
2. ex −1 + c
2
1
3. ln |3x − 1| + c
3
4. 2 ln |x2 − 2| + c
5. ln |ex − 1| + c
6. −
ex
1
+c
−1