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# module11

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José Guerra Guerra
Precalculus
2017-2018
c José del Cristo Guerra Guerra
Contents
List of Figures
0.1
viii
Mathematic Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
List of Figures and Symbols
I
xiii
xii
Functions
1
1 Functions
2
1.1
Cartesian Product
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.2
Relation and Function . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.2.1
Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.2.2
Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.2.3
Domain of Functions . . . . . . . . . . . . . . . . . . . . . . . . .
9
1.2.4
Quadratic Functions . . . . . . . . . . . . . . . . . . . . . . . . .
15
1.2.5
Graph of a Radical Function . . . . . . . . . . . . . . . . . . . . .
19
1.2.6
Graph of a Piecewise Function . . . . . . . . . . . . . . . . . . . .
20
i
1.3
II
Applied Functions-Variation . . . . . . . . . . . . . . . . . . . . . . . . .
21
1.3.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22
1.3.2
Operations with Functions . . . . . . . . . . . . . . . . . . . . . .
23
Polynomial and Rational Functions
2 Polynomial and Rational Functions
2.1
2.2
2.3
2.4
29
30
Polynomial Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30
2.1.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31
2.1.2
The Intermediate Value Theorem . . . . . . . . . . . . . . . . . .
31
2.1.3
Dividing Polynomial Functions . . . . . . . . . . . . . . . . . . .
33
Real Zeros of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . .
37
2.2.1
Descartes’ Rule of Sign . . . . . . . . . . . . . . . . . . . . . . . .
38
2.2.2
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39
End Behavior of Polynomial Functions . . . . . . . . . . . . . . . . . . .
39
2.3.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
42
2.3.2
Real Zeros of Polynomial Functions . . . . . . . . . . . . . . . . .
42
Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
44
2.4.1
Domain of Rational Functions . . . . . . . . . . . . . . . . . . . .
44
2.4.2
Vertical, Horizontal, and Slant Asymptotes . . . . . . . . . . . . .
48
2.4.3
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50
ii
III
Exponential and Logarithmic Functions
3 Exponential and Logarithmic Functions
51
52
3.1
Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
52
3.2
Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
54
3.2.1
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
56
3.2.2
Graph Logarithmic Functions . . . . . . . . . . . . . . . . . . . .
57
3.2.3
Finding the Domain of Logarithmic Functions . . . . . . . . . . .
57
3.2.4
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
58
3.2.5
Properties of Logarithmic Functions . . . . . . . . . . . . . . . . .
58
3.2.6
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
60
3.3
Exponential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . .
61
3.4
Logarithmic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . .
63
IV
Series
66
4 Series
67
4.1
Arithmetic Progressions . . . . . . . . . . . . . . . . . . . . . . . . . . .
67
4.1.1
Formula to find the nth term of an Arithmetic Progressions . . .
68
4.1.2
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
70
4.1.3
Formula to find the first term, ratio and the number of terms . . .
70
4.1.4
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
73
4.1.5
Formula to find the sum of all terms of Arithmetic Progressions .
73
iii
4.1.6
4.2
V
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
75
Geometric Progressions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
75
4.2.1
Formula to find the nth term of Geometric Progressions . . . . . .
76
4.2.2
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
78
4.2.3
Formula to find the first term and ratio-Geometric Progressions .
78
4.2.4
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
80
4.2.5
Formula to find the sum of all terms of Geometric Progressions
.
81
4.2.6
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83
Systems of Linear Equations and Matrices
5 Systems of Linear Equations
5.1
5.2
84
85
System of Linear Equations 3 × 3 . . . . . . . . . . . . . . . . . . . . . .
85
5.1.1
Equality Method . . . . . . . . . . . . . . . . . . . . . . . . . . .
86
5.1.2
Substitution Method . . . . . . . . . . . . . . . . . . . . . . . . .
87
5.1.3
Elimination Method . . . . . . . . . . . . . . . . . . . . . . . . .
88
5.1.4
Graph Method . . . . . . . . . . . . . . . . . . . . . . . . . . . .
89
5.1.5
Determinants Method . . . . . . . . . . . . . . . . . . . . . . . .
90
5.1.6
Solve Systems of Linear Equations 3 × 3 Using Determinants . . .
94
5.1.7
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
97
Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
99
5.2.1
Operations with Matrices . . . . . . . . . . . . . . . . . . . . . . 101
iv
VI
5.2.2
Zero Matrix, Identity Matrix and Inverse Matrix
5.2.3
Matrices-Elementary Row Operations . . . . . . . . . . . . . . . . 106
5.2.4
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
5.2.5
Matrices and Linear Systems . . . . . . . . . . . . . . . . . . . . . 109
Conics
114
6 Conics
6.1
. . . . . . . . . 105
115
Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
6.1.1
Equation of the Circle-Standard Form . . . . . . . . . . . . . . . 115
6.1.2
Equation of the Circle-General Form
x2 + y 2 + Ax + By + C = 0 . . . . . . . . . . . . . . . . . . . . . 116
6.2
Parabola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
6.3
Equation of the Parabola-Standard Form . . . . . . . . . . . . . . . . . . 117
6.4
Ellipse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
6.5
Equation of the Ellipse-Standard Form . . . . . . . . . . . . . . . . . . . 119
6.6
Hyperbola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
6.6.1
Equation of the Hyperbola-Standard Form . . . . . . . . . . . . . 120
6.6.2
Equation of the Hyperbola-General Form
Ax2 − By 2 + Cx + Dy + E = 0 . . . . . . . . . . . . . . . . . . . 121
6.7
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
v
VII
Trigonometry
124
7 Trigonometry
7.1
7.2
Basic Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
7.1.1
Conversion of degrees to radians and viceversa . . . . . . . . . . . 131
7.1.2
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
Trigonometry of Right Triangles . . . . . . . . . . . . . . . . . . . . . . . 135
7.2.1
Trigonometric Ratios . . . . . . . . . . . . . . . . . . . . . . . . . 135
7.2.2
Unitary Circle or Goniometric Circle . . . . . . . . . . . . . . . . 137
7.2.3
Special Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
7.2.4
Evaluation of the Six Trigonometric Ratios-30o = π6 rad and 60o =
π
rad.
3
7.3
7.4
125
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
7.2.5
Evaluation of the Six Trigonometric Ratios-45o = π4 rad. . . . . . . 140
7.2.6
Evaluation of the Six Trigonometric Ratios-0o = 0rad. . . . . . . . 141
7.2.7
Evaluation of the Six Trigonometric Ratios-90o = π2 rad. . . . . . . 142
7.2.8
Evaluation of the Six Trigonometric Ratios-180o = πrad. . . . . . 143
7.2.9
Evaluation of the Six Trigonometric Ratios-270o =
3π
rad.
2
. . . . . 144
Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
7.3.1
Graph of Trigonometric Functions . . . . . . . . . . . . . . . . . . 146
7.3.2
Transformed Sine and Cosine Graphs . . . . . . . . . . . . . . . . 149
Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
7.4.1
Basic Trigonometric Identities . . . . . . . . . . . . . . . . . . . . 151
vi
7.5
7.6
7.4.2
Sine, Cosine and Tangent of the Sum of Two Angles . . . . . . . . 153
7.4.3
Sine, Cosine and Tangent of Double Angle . . . . . . . . . . . . . 156
7.4.4
Sine, Cosine and Tangent of Half Angle . . . . . . . . . . . . . . . 157
7.4.5
Product to Sum and Sum to Product . . . . . . . . . . . . . . . . 159
Law of Sines and Law of Cosines . . . . . . . . . . . . . . . . . . . . . . 159
7.5.1
Law of Sines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
7.5.2
Law of Cosines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
Inverse of Trigonometric Functions and Trigonometric Equations . . . . . 164
7.6.1
Inverse Function of Sine . . . . . . . . . . . . . . . . . . . . . . . 164
7.6.2
Inverse Function of Cosine . . . . . . . . . . . . . . . . . . . . . . 165
7.6.3
Inverse Function of Tangent . . . . . . . . . . . . . . . . . . . . . 166
7.6.4
Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . 167
vii
List of Figures
1.1
Sagittal diagram and Cartesian plane . . . . . . . . . . . . . . . . . . . .
4
1.2
Domain, image and range. . . . . . . . . . . . . . . . . . . . . . . . . . .
4
1.3
Relation-Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.4
Vertical line test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
1.6
Positive, Negative, Zero and Undefined Slope . . . . . . . . . . . . . . . .
12
1.8
f (x) = x2 + 6x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16
1.9
f (x) = 2x2 − 6x −
5
2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
1.10 f (x) = x2 + 5x + 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
1.13 Horizontal line test. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26
√
2x − 1 and its inverse. . . . . . . . . . . . .
28
2.1
Graph of P (x) = (x − 1)3 (x + 2)2 . . . . . . . . . . . . . . . . . . . . . . .
32
2.2
End Behavior of Polynomial function: odd degree and positive leading
1.14 Graph of the function y =
coefficient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3
40
End Behavior of Polynomial function: odd degree and negative leading
coefficient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
viii
40
2.4
End Behavior of Polynomial function: even degree and positive leading
coefficient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5
41
End Behavior of Polynomial function: even degree and negative leading
coefficient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
2.6
Graph of f (x) = x3 + 3x2 + 2x and zeros of f (x). . . . . . . . . . . . . .
44
2.7
Rational function: f (x) =
1
x
2.8
Rational function: f (x) =
x2 −4
x2 −2x−3
2.9
Rational function: f (x) =
x3 −4x
x2 −1
6.1
Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
6.2
Parabola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
6.3
Ellipse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
6.4
Hyperbola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
7.1
Angle
7.2
Positive angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
7.3
Negative angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
7.4
Quadrants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
7.5
Angle in standard or normal position. . . . . . . . . . . . . . . . . . . . . 128
7.6
Quadrantal angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
7.7
Central angle, and arc length . . . . . . . . . . . . . . . . . . . . . . . . 129
7.8
Reference angle in quadrant I . . . . . . . . . . . . . . . . . . . . . . . . 130
7.9
Reference angle in quadrant II . . . . . . . . . . . . . . . . . . . . . . . . 130
. . . . . . . . . . . . . . . . . . . . . . . . .
49
. . . . . . . . . . . . . . . . . . . . . .
49
. . . . . . . . . . . . . . . . . . . . . . .
50
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
ix
7.10 Reference angle in quadrant III . . . . . . . . . . . . . . . . . . . . . . . 131
7.11 Reference angle in quadrant IV . . . . . . . . . . . . . . . . . . . . . . . 131
7.12 Some angles in degrees and radians. . . . . . . . . . . . . . . . . . . . . . 133
7.13 Right triangle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
7.14 The six trigonometric ratios. . . . . . . . . . . . . . . . . . . . . . . . . . 136
7.15 Unitary circle or Goniometric . . . . . . . . . . . . . . . . . . . . . . . . 137
7.16 Triangle 30o − 60o − 90o . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
7.17 Triangle 45o − 45o − 90o . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
7.18 The six trigonometric ratios: 30o = π6 rad. . . . . . . . . . . . . . . . . . . 139
7.19 The six trigonometric ratios: 45o . . . . . . . . . . . . . . . . . . . . . . . 140
7.20 The six trigonometric ratios: 0o . . . . . . . . . . . . . . . . . . . . . . . . 141
7.21 The six trigonometric ratios: 90o = π2 rad. . . . . . . . . . . . . . . . . . . 142
7.22 The six trigonometric ratios: 180o = πrad. . . . . . . . . . . . . . . . . . 143
7.23 The six trigonometric ratios: 270o = πrad. . . . . . . . . . . . . . . . . . 144
7.24 Signs of six trigonometric functions. . . . . . . . . . . . . . . . . . . . . . 146
7.25 Graph of f (x) = sin(x), domain, range and period. . . . . . . . . . . . . 146
7.26 Graph of f (x) = cos(x), domain, range and period. . . . . . . . . . . . . 147
7.27 Graph of f (x) = tan(x), domain, range and period. . . . . . . . . . . . . 147
7.28 Graph of f (x) = cot(x), domain, range and period. . . . . . . . . . . . . 147
7.29 Graph of f (x) = sec(x), domain, range and period. . . . . . . . . . . . . 148
7.30 Graph of f (x) = csc(x), domain, range and period. . . . . . . . . . . . . 148
x
7.31 Graph of f (x) = sin(x) and g(x) = 3 sin 2 x + π2 + 1.
. . . . . . . . . 150
7.32 Graph of f (x) = cos(x) and g(x) = 4 cos 4 x − π4 − 1. . . . . . . . . . 150
7.33 Unitary circle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
7.34 Unitary circle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
7.35 Non-right triangle
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160
7.36 Non-right triangle
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
xi
xii
0.1
Mathematic Symbols
¬
negation: read “not, is false that, is not true that”
∨
disjunction: read “ or”
∧ conjunction: read “and”
→ conditional, then
much greater than
↔ biconditional, if and only if
⊥ perpendicular to
0
contradiction
1
tautology
±
plus or minus
k
parallel to
∓
minus or plus
∀ for all
<
less than
∈ including
≤
less or equal to
∈
/ no including
>
greater than
S
union
≥
greater or equal than
T
intersection
6=
different , not equal to
⊂ proper subset
≡
equivalent to
⊆ subset equal
≈
approximately equal to
∞ infinity
∼
=
congruent with, congruent to
∠ angle
∝
proportional to
P
much less than
sum
∃ exist
<
the real numbers
N
Z
the integers numbers
Q the rational numbers
C
the complex numbers
xiii
the natural numbers
x absolute value of x
Part I
Functions
1
Chapter 1
Functions
1.1
Cartesian Product
A new set can be constructed by associating every element of one set with every element
of another set. The Cartesian product of two sets A and B, denoted by A × B is the set
of all ordered pairs (a, b) such that a is a element of A and b is a element of B.
Example 1.1.1.
• A = {1, 2} and B = {3, 5, 6}, then:
A × B = {(1, 3), (1, 5), (1, 6), (2, 3), (2, 5), (2, 6)}
c José del Cristo Guerra Guerra
2
3
1.2
1.2.1
Relation and Function
Relation
A relation is a subset of the Cartesian product A × B, where to each element of the set
A is assigned one or more elements of the set B through an association rule.
Example 1.2.1.
A × B = {(1, 3), (1, 5), (1, 6), (2, 3), (2, 5), (2, 6)}. Cartesian product
R = {x/y = 2x + 1} = {(1, 3), (2, 5)}. The second element y must be equal two times
the first element plus one.
R ⊆ A × B. Relation is a subset of Cartesian product A × B.
R = {(x, y)/x ∈ A ∧ y ∈ B}. The relation R is the set of all ordered pair (x, y), such
that, x is include in A and y is include in B.
Representation of Relation
A relation can be represented by Sagittal diagram and Cartesian plane.
Example 1.2.2. Represent the relation R = {x/y = 2x + 1} = {(1, 3), (2, 5)}
Start Set, Arrival Set, Domain, Image and Range
If A × B = {(1, 3), (1, 5), (1, 6), (2, 3), (2, 5), (2, 6)}, then the start set is formed by all
elements of the first set of A × B, in this case A = {1, 2}, and the arrival set is formed
4
Figure 1.1: Sagittal diagram and Cartesian plane
by all elements of the second set of A × B, in this case B = {3, 5, 6}.
If R = {x/y = 2x + 1} = {(1, 3), (2, 5)}, then the domain is the set formed by the first
elements of the relation, in this case {1, 2}. Each second element of this relation is called
image, and the set of all images is called the range.
Figure 1.2: Domain, image and range.
5
1.2.2
Function
Figure 1.3: Relation-Function
A relation is a function when all elements of the start set are part of the domain, and
each element of this domain has one and only one image in the arrival set. According
that, in the figure 1.3, the first relation (to the left) is not a function, the second relation
is a function. Remarks All function is a relation but no all relation is a function.
Characteristic of a Function From a Set A to Set B
• Each element of A must be matched with an element of B.
• Some elements of B may not be matched with any element of A.
• Two or more elements of A may be matched with the same element of B.
• An element of A (the domain) cannot be matched with two different elements of
B.
6
Testing for Functions Represented Graphically
To determine if the graph of a relation correspond to a function, the Vertical Line Test
is used. The vertical line test states that: a curve in the plane is the graph of a function
if and only if vertical line touches one point on the graph, but if the vertical line touches
two or more points, then the graph does not correspond to a function.
(a) A function
(b) Not a function
Figure 1.4: Vertical line test
Testing for Functions Represented Algebraically
A variable y (dependent variable) is a function of the other x (independent variable)
when for each value of x correspond one and only one value of y.
Which of the equations represent(s) y as a function of x?
• x2 + y = 1.
Solving for y, we obtain y = −x2 + 1.
Each value of x corresponds to exactly one value of y. So, y is a function of x.
• −x + y 2 = 1.
√
Solving for y, we obtain y = ± x + 1.
7
The indicates that for a given value of x there correspond two values of y. For
instance, when x = 3, y = 2 or y = −2. So, y is not a function of x.
Exercises
1. Enter Yes or No in each answer space below to indicate whether the corresponding
equation defines y as a function of x.
1. x2 + 3y = 7
2. 9x = y 2
3. 2|x| + y = 5
4. x2 y + y = 8
2. Does the following relation on x and y make for a function of x?
n
o
R = (8, 10), (−7, 4), (10, 6), (−10, 7), (10, 7)
What is the domain of the relation? What is the range of the relation?
3. The following graphs show two relationships. Decide whether each graph shows a
relationship where y is a function of x.
(a)
(b)
(c)
8
Function Notation
A function f is a rule that assigns to each element x in a set A exactly one element,
called f (x), in a set B. That is, x ∈ A and f (x) ∈ B. Set A is called Domain and set
B is called the range. “f (x)” is read “ f of x” and represents the value of f at x, or
the image of x under f . Since y = f (x) depends of the value x, y = f (x) is called the
dependent variable and x is called the independent variable.
input
output
Equation
x
f (x)
f (x) = x2 − 1
1
f (1)
f (1) = (1)2 − 1 = 1 − 1 = 0
2
f (2)
f (2) = (2)2 − 1 = 4 − 1 = 3
Example 1.2.3.
Exercises
1. Find the values of f defined by the equation f (x) =
(a) f (−2)
(b) f (4)
(c) f (0)
2. If f (x) =
1
, find the following:
x+3
(a) f (−11) =
(b) f (−14) =
(c) f (−13) =
3x2 +x
.
2−x
9
1.2.3
Domain of Functions
Domain of Polynomial Functions
Polynomial Functions: Let n be a non negative integer and let an , an−1 , · · · a1 , a0 be
real numbers and an 6= 0. The function f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 is a
polynomial function of x with degree n. The domain of any polynomial function
is all real numbers or D = {x/x ∈ <} or (−∞, ∞)
Example 1.2.4. Find the domain of the function f (x) = x2 + 2x + 1. This is a polynomial function, then, the domain is D = {x/x ∈ <} or (−∞, ∞).
Find the domain of the function f (x) = 4. This is a polynomial function, then, the
domain is D = {x/x ∈ <} or (−∞, ∞).
Domain of Rational Functions
Rational Functions: A Rational Function can be written in the form f (x) =
N (x)
D(x)
where N (x) and D(x) are polynomial functions and D(x) 6= 0.
In general, the domain of a rational function includes all real numbers except x-values
that make the denominator zero.
Example 1.2.5. Find the domain of the function f (x) =
2x2
.
x2 −1
function, We need find the x-values that make the denominator zero.
This is a rational
10
Set the denominator equal 0, x2 − 1 = 0, factoring (x − 1)(x + 1) = 0, (x − 1) =
0 and (x + 1) = 0, x = 1 and x = −1. The domain is D = {x/x ∈ < and x 6= −1, 1} or
(−∞, −1) ∪ (−1, 1) ∪ (1, ∞).
Domain of Radical Functions
Radical Functions: A Radical Function can be written in the form f (x) =
p
n
P (x)
where P (x) is any polynomial function or rational function.
For n even and P (x) is a polynomial function, f (x) is defined only for P (x) ≥ 0.
For n even and P (x) is a rational function, f (x) is defined for P (x) ≥ 0 except x-values
that make the denominator zero.
For n odd and P (x) is a polynomial function, f (x) is defined for all real numbers.
For n odd and P (x) is a rational function, f (x) is defined for all real numbers except
x-values that make the denominator zero.
Example 1.2.6. Find the domain of the function f (x) =
√
4 − 3x. This is a radical
function. We need find the x-values that 4 − 3x ≥ 0.
4 − 3x ≥ 0
−3x ≥ −4
−
3x
−4
≤
−3
−3
4
4
4
x ≤
The domain is D = {x/x ∈ < ∧ x ≤ } or (−∞, ]
3
3
3
11
Exercises
Find the domain of the function of the following functions.
1. f (x) =
4
x+2
5. f (x) =
2. f (x) =
x−9
x3 −25
6. f (x) =
√
3
x−4
√
x2 − 4
√
3. f (x) = x10 + x8 − 3x − 100
4. f (x) =
√
x−4
7. f (x) =
8. f (x) =
x−4
x−6
q
x2 −16x
x4 +1
Polynomial Functions
Let n be a non negative integer and let an , an−1 , · · · a1 , a0 be real numbers and an 6= 0.
The function f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 is a polynomial function of
x with degree n. The domain of any polynomial function is all real numbers or
D = {x/x ∈ <} or (−∞, ∞).
Example 1.2.7. Find the domain of the function f (x) = x2 + 2x + 1. This is a polynomial function, then, the domain is D = {x/x ∈ <} or (−∞, ∞).
Find the domain of the function f (x) = 4. This is a polynomial function (constant function), then, the domain is D = {x/x ∈ <} or (−∞, ∞).
12
Linear Function
A function defined by f (x) = mx + b is called a linear function because the greatest
exponent of the variables is 1 and the graph of the equation y = mx + b is a straight line
with slope m and y-intercept b. When m = 0, the linear function becomes the constant
function, f (x) = b.
Slope
The slope of a line measures the inclination of this.
Rise means how many units you move up or down from point to point. On the graph
that would be a change in the y values. Run means how far left or right you move from
point to point. On the graph, that would mean a change of x values.
Figure 1.6: Positive, Negative, Zero and Undefined Slope
13
Slope Formula Given Two Points
Given two points (x1 , y1 ) and (x2 , y2 ), m =
y2 −y1
x2 −x1
=
change in y
= raise
run .
change in x
Find the slope of the straight line that passes through (7, 5) and (5, 1) or state that the
slope is undefined. Then indicate if the line through the points rises (left to right), falls
(left to right), is horizontal, or is vertical.
(|{z}
7 , |{z}
5 ) and (|{z}
5 , |{z}
1 )
x1
y1
x2
y2
m=
1−5
−4
y2 − y1
=
=
=2
x2 − x1
5−7
−2
(1.1)
Since the slope is positive, the line would rise (left to right).
Example 1.2.8. Find the slope of the straight line that passes through (1, −4) and
(−1, 3) or state that the slope is undefined. Then indicate if the line through the points
rises (left to right), falls (left to right), is horizontal, or is vertical.
(|{z}
−1 , |{z}
3 ) and (|{z}
1 , |{z}
−4 )
x1
y1
x2
y2
m=
y2 − y1
−4 − 3
−7
7
=
=
=−
x2 − x 1
1 − (−1)
1+1
2
(1.2)
Since the slope is negative, the line would fall (left to right).
Example 1.2.9. Find the slope of the straight line that passes through (4, 1) and (−2, 1)
or state that the slope is undefined. Then indicate if the line through the points rises
(left to right), falls (left to right), is horizontal, or is vertical.
14
1 )
(|{z}
4 , |{z}
1 ) and (|{z}
−2 , |{z}
x1
y1
x2
y2
m=
y2 − y1
1−1
0
=
=
=0
x2 − x1
−2 − 4
−6
(1.3)
Since the slope is zero, the line would be horizontal.
Example 1.2.10. Find the slope of the straight line that passes through (−2, 3) and
(−2, 5) or state that the slope is undefined. Then indicate if the line through the points
rises (left to right), falls (left to right), is horizontal, or is vertical.
(|{z}
−2 , |{z}
3 ) and (|{z}
−2 , |{z}
5 )
x1
y1
x2
m=
y2
5−3
2
2
y2 − y1
=
=
= = undef ined
x2 − x1
−2 − (−2)
−2 + 2
0
(1.4)
Since the slope is undefined, the line would be vertical.
Graph of Linear function
Example 1.2.11. If f (x) = 3x + 1 find the domain and range of f and sketch the graph
of f .
This is a linear function with slope 3 and y-intercept 1. Domain = {x/x ∈ <}
15
(a) Table of Values
1.2.4
(b) y = 3x + 1
Quadratic Functions
A quadratic function is a function of the form f (x) = ax2 + bx + c, where a, b and
c are real numbers, with a 6= 0. It is often easier to graph and work with quadratic
functions when we have completed the perfect square trinomial and expressed f in the
form f (x) = a(x − h)2 + k.
Example 1.2.12.
• Express f in the form f (x) = a(x−h)2 +k and sketch the graph
of f : f (x) = x2 + 6x
f (x) = x2 + 6x
f (x) = x2 + 6x + 9 − 9
f (x) = (x + 3)2 − 9
16
Figure 1.8: f (x) = x2 + 6x
• Express f in the form f (x) = a(x − h)2 + k and sketch the graph of f : f (x) =
2x2 − 6x −
5
2
5
f (x) = 2x2 − 6x −
2
9
9
5
2
f (x) = 2 x − 3x +
− −2
4
2
4
f (x) = 2(x − 3)2 − 7
• When a > 0, the parabola f (x) = a(x − h)2 + k, open up. The lowest point is the
vertex (h, k), so the minimum value of the function occurs when x = h, and the
minimum value is f (h) = k.
• When a < 0, the parabola f (x) = a(x − h)2 + k, open down. The highest point is
the vertex (h, k), so the maximum value of the function occurs when x = h, and
the maximum value is f (h) = k.
17
Figure 1.9: f (x) = 2x2 − 6x −
5
2
Example 1.2.13. Determine the maximum or minimum values of f by completing the
perfect square trinomial. Find the intercepts and the sketch the graph. f (x) = x2 +5x+4
f (x) = x2 + 5x + 4
x2 + 5x + 4
25
25
2
f (x) =
x + 5x +
+4−
4
4
2
5
9
f (x) =
x+
−
2
4
f (x) =
minimum value of the function occurs when x = − 25 , and the minimum value is
f (− 25 ) = − 94 .
18
To find the x-intercepts set f (x) = 0 and solve.
f (x) = x2 + 5x + 4 = 0
f (x) = (x + 1)(x + 4) = 0
(x + 1) = 0
x = −1
(x + 4) = 0
x = −4
x-intercepts: -1 and -4.
To find the y-intercept evaluate f (0), f (0) = (0)2 + 5(0) + 4 = 4.
y-intercepts:f (0) = 4
Figure 1.10: f (x) = x2 + 5x + 4
19
Exercises
Determine the maximum or minimum values of f by completing the perfect square trinomial. Find the intercepts and the sketch the graph.
1. f (x) = x2 − 4x − 2
2. f (x) = 3x2 − 6x + 4.
3. f (x) = −2x2 + 5x.
4. f (x) = −5x2 − 10x + 1.
1.2.5
Graph of a Radical Function
Example 1.2.14. If f (x) =
√
2x + 1 find the domain and range of f and sketch the
graph of f .
This is a radical function with Domain = {x/x ∈ < ∧ x ≥ − 12 } or [− 21 , ∞).
(a) Table of Values
(b) Piecewise Function
20
1.2.6
Graph of a Piecewise Function
A function f is called piecewise function if it is defined differently for distinct intervals
of its domain.
Sketch the graph of the function defined by
(a) Table of Values
(b) Piecewise Function
Exercises
Sketch the graph of f .
1.


√



−5 − 3x, if x ≤ −2





f (x) = √5 + 2x,
if −2 < x < 2








−x + 8,
if x ≥ 2
21
1.3
Applied Functions-Variation
The quantities in a mathematical expression can be constants or variables. If the quantity
have a fixed value, the quantity is called a constant, and if the quantity can take many
values, then is called variable. A variable y is a function of the other x when for each
value of x correspond one and only one value of y.
In the direct variation model, y is directly proportional to x if x and y are related
by the equation y = kx for some constant k 6= 0. The constant k is called the constant
of proportionality.
Example 1.3.1.
• Federal excise tax on gasoline is 13 6 c per gallon. Write an
equation that directly relates the tax as a function of gallons. Let g =gallons of
gas and let T =the tax. Federal excise tax is 13 6 c per gallon. Since tax is 13 6 c
per gallon we have T = 0.13g. Here 0.13 is the constant of proportionality.
• The volume of liquid in a soda can is directly related to the height of the liquid in
the can.
• If the can is 12.5cm tall and contains 355ml when full, determine the constant of
proportionality and write the equation for the variation.
Let h =height of the can, V =volume of the can, and k is the constant of proportionality. V = kh, when h = 12.5, V = 355. Substituting, we have 355 = k12.5 ⇔
k=
355
12.5
= 28.4, thus the model is V = 28.4h.
22
2 How much is left in the can when the liquid is 3cm deep? Setting h=3 in the
equation yields V = 28.4h = 28.4(3) = 85.2ml.
3 How deep is the liquid when there is 300ml left? Setting V = 300 and solving for
h yields V = 28.4h ⇒ 300 = 28.4h = 85.2 ⇔ h =
300
28.4
= 10.56cm
In the inversely variation model, y is inversely proportional to x if x and y are
related by the equation y = k x1 =
k
x
for some constant k 6= 0. The constant k is called
the constant of proportionality.
Example 1.3.2.
• Write an equation that relate the width to the length of all rect-
angles whose area is 4 square units.
Let w =width and let l =length. Since A = wl we have 4 = wl ⇔ w =
4
l
thus
width is inversely proportional to length.
1.3.1
Exercises
1. If a 6 pack of books cost \$ 2.82, what is the price per book?
2. If r is directly proportional to the half of s, and if r = 2 when s = 5, find r when
s = 2.5.
3. If s is directly proportional to 3a + 3, and if s = 8 when a = 3, find a when s = 24.
4. If x is inversely proportional to y, and if x = 2 when y = 5, find the formula that
related x and y.
23
5. If A is directly proportional to B, and inversely proportional to C, and if B = 24
when C = 4 and A = 3, find the formula that express A as function of B and C.
1.3.2
Operations with Functions
Sum and Difference of Functions
Let f (x) be a function with domain A and g(x) be a function with domain B. Then
the sum of the functions f (x) and g(x) is f (x) ± g(x) = (f ± g)(x) where the domain is
A ∩ B.
Example 1.3.3. Let f (x) = 8x + 3 and g(x) = x2 − 4. Find f (x) + g(x) and f (x) − g(x)
(f + g)(x) = (8x + 3) + (x2 − 4) = x2 + 8x − 1
(f − g)(x) = (8x + 3) − (x2 − 4) = 8x + 3 − x2 + 4
= −x2 + 8x + 7
Since the domain for both f (x) and g(x) is all real numbers, the domain of (f + g)(x) is
all real numbers.
Product of Functions
Let f (x) be a function with domain A and g(x) be a function with domain B. Then the
product of the functions f (x) and g(x) is f (x)g(x) = (f g)(x) where the domain is A ∩ B.
24
Example 1.3.4. Let f (x) = 8x + 3 and g(x) = x2 − 4. Find f (x)g(x)
(f g)(x) = (8x + 3)(x2 − 4) = 8x3 − 32x + 3x2 − 12
= 8x3 + 3x2 − 32x − 12
Since the domain for both f (x) and g(x) is all real numbers, the domain of (f g)(x) is all
real numbers.
Quotient of Functions
Let f (x) be a function with domain A and g(x) be a function with domain B. Then
(x)
= fg (x) where the domain is
the quotient of the functions f (x) and g(x) is fg(x)
{A ∩ B/g(x) 6= 0}.
Example 1.3.5. Let f (x) = 8x + 3 and g(x) = x2 − 4. Find
f (x)
g(x)
f
(8x + 3)
(x) =
g
(x2 − 4)
Since the domain for both f (x) and g(x) is all real numbers, the domain of
{x/x ∈ < ∧ x 6= −2, 2}.
f (x)
g(x)
is
25
Composition Function
Let f (x) and g(x) be functions. Then the composite function of f (x) and g(x) is
(f og) (x) = f (g(x)). x is the input to g(x) and g(x) is the input to (f og) (x) = f (g(x)).
The domain of (f og) (x) = f (g(x)) is that subset of the domain of g(x) where g(x) is in
the domain of f (x).
Example 1.3.6. Let f (x) = 8x + 3 and g(x) = x2 − 4.
Find (f og) (x) = f (g(x)) and (gof ) (x) = g (f (x))
(f og) (x) = f (g(x)) = 8(g(x)) + 3 = 8(x2 − 4) + 3
= 8x2 − 32 + 3 = 8x2 − 29
(gof ) (x) = g (f (x)) = (f (x))2 − 4 = (8x + 3)2 − 4
= 64x2 + 48x + 9 − 4
= 64x2 + 48x + 5
Since the domain for both f (x) and g(x) is all real numbers, the domain of (f og) (x) =
f (g(x)) and (gof ) (x) = g (f (x)) is all real numbers.
One to One Function and Inverse Function
Definition 1.3.1. A function f (x) with domain A is called a one to one function if
no two elements of A have the same image; that is f (x1 ) 6= f (x2 ) whenever x1 6= x2 .
One test for determining when a function is one to one is called the horizontal line
26
test which states that a function is one to one if and only if no horizontal line intersects
its graph more than once.
Example 1.3.7. Show that f (x) = x2 + 2x − 3 is not one to one function by finding
two points x1 6= x2 where f (x1 ) = f (x2 ).
f (x) = x2 + 2x − 3 = 0
(x − 1)(x + 3) = 0
x−1=0
x=1
x+3=0
x = −3
So, f (1) = f (−3) = 0, thus f (x) is not one to one.
Figure 1.13: Horizontal line test.
27
Definition 1.3.2. Let f (x) be a one to one function with domain A and range B.
Then its inverse function f −1 (x) has domain B and range A and is defined by
f −1 (y) = x ⇔ f (x) = y for any y in B.
The graph of a function and its inverse are symmetric with respect to the line y = x.
Procedure to find the inverse of a function:
1. Change f (x) by y.
2. Interchange x and y.
3. Solve this equation for y in terms of x.
4. Change y by f −1 (x).
Example 1.3.8. Find the inverse of the function f (x) =
1. Change f (x) by y. ⇒ y =
√
2. Interchange x and y. ⇒ x =
2x − 1
√
2y − 1
3. Solve this equation for y in terms of x. ⇒y =
4. Change y by f −1 (x). ⇒ f −1 (x) =
x2 +1
2
x2 +1
2
√
2x − 1. Figure 1.14.
28
Figure 1.14: Graph of the function y =
√
2x − 1 and its inverse.
Exercises
1. Let f (x) =
√
√
x + 3 and g(x) = 9 − x2 . Find (f + g) (x) and find its domain.
2. Let f (x) =
2
x−3
and g(x) =
1
.
x+5
3. Let f (x) = x + 2 and g(x) =
Find (f − g) (x) and find its domain.
2
x+1
− 4. Find (f g) (x) and find its domain.
√
2
4. Let f (x) = x − 1 and g(x) = x − 4. Find fg (x) and find its domain.
5. Let f (x) =
1
x
− 4 and g(x) =
1
.
x+4
Find (f og) (x) = f (g(x)) and (gof ) (x) =
g (f (x)) and find its domain.
6. Find the inverse f (x) = 3x + 4 and find its domain.
7. Find the inverse f (x) = −x − 6 and find its domain.
8. Find the inverse f (x) =
3
2+x
9. Find the inverse f (x) =
3x−1
2+x
and find its domain.
and find its domain.
Part II
Polynomial and Rational Functions
29
Chapter 2
Polynomial and Rational Functions
2.1
Polynomial Functions
Let n be a non negative integer and let an , an−1 , · · · a1 , a0 be real numbers and an 6= 0.
The function f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 is a polynomial function of
x with degree n. The numbers an (leading coefficient), an−1 , · · · , a1 , and a0 (Constant
term) are called the coefficients of the Polynomial Functions. The domain of any
polynomial function is all real numbers or D = {x/x ∈ <} or (−∞, ∞).
When in a polynomial function the terms are written from the highest to lowest degree,
it is said that it is in Standard Form.
c José del Cristo Guerra Guerra
30
31
Example 2.1.1. Find the domain of the function f (x) = x2 +2x+1. This is a polynomial
function in standard form, because the first term has degree 2, the second term has degree
1 and the third term has degree 0. The domain is D = {x/x ∈ <} or (−∞, ∞).
Find the domain of the function f (x) = 4. This is a polynomial function, then, the
domain is D = {x/x ∈ <} or (−∞, ∞).
2.1.1
Exercises
Determine the degree, the leading coefficient, and the constant term of the following
polynomial functions:
1. P (x) = 4x5 − 5x3 + 2x2 − x + 6
2. P (x) = −8x2 − 5x3 + 2x
3. P (x) = 9x8 − 7x3 + 8x2 − 9x − 10
4. P (x) = 15x − 5 + 2x2 − x3
2.1.2
The Intermediate Value Theorem
If P (x) is a polynomial and P (a) and P (b) have opposite signs, then there exists at least
one value c between a and b such that P (c) = 0. To sketch the graph of a polynomial
function P (x), first find all zeros of P (x), then choose points between two consecutive
zeros to determine the sign of P (x). If P (x) is negative, the graph of P (x) is below the
x axis, and if P (x) is positive, the graph of P (x) is above the x axis.
32
If c is a zero of P (x), and (x − c)m is a factor of P (x) which occurs exactly m times in
the factorization of P (x).
m=




odd number,
then the graph cross the x axis at c



even number, then the graph does not cross the x axis at c
Example 2.1.2. Sketch the graph of the polynomial function:
P (x) = (x − 1)3 (x + 2)2 The x intercepts are x = 1 and x = −2. Since 1 occurs three
Figure 2.1: Graph of P (x) = (x − 1)3 (x + 2)2 .
times, the graph cross the x axis at 1. Since −2 occurs two times, the graph does not
cross the x axis at −2.
The y-intercept occurs when x = 0, then P (0) = (0 − 1)3 (0 + 2)2 = (−1)(4) = −4.
33
2.1.3
Dividing Polynomial Functions
Long Division
Definition 2.1.1 (Dividend). Dividend means the number that is to be divided.
Definition 2.1.2 (Divisor). The amount that you want to divide up.
Definition 2.1.3 (Quotient). A result obtained by dividing one quantity (Dividend) by
another (Divisor).
Definition 2.1.4 (Remainder). The number left over is called the remainder.
If D(x) and d(x) are polynomials, with d(x) 6= 0, there exist unique polynomials q(x)
and r(x) such that:
D(x) = d(x)q(x) + r(x),
where r(x) = 0 or the degree of r(x) is less than the degree of d(x).
D(x) is the dividend, d(x) is the divisor, q(x) is the quotient, and r(x) is the remainder.
Remark. The important thing to remember when dividing is to insert 0xj for missing
terms.
34
Example 2.1.3. Let D(x) = x4 + x3 + 4 and d(x) = x2 − 3. Find the polynomials q(x)
and r(x) such that:
D(x) = d(x)q(x) + r(x),
Synthetic Division
The synthetic division is a procedure used in factoring, in finding solutions to P (x) = 0,
and in evaluating polynomials, that is, finding the value of P (c).
To use synthetic division, follow these steps:
1. The quotient is a polynomial in x whose degree is 1 less than the degree of the
35
dividend.
2. The coefficient of the first term of the quotient is equal to the coefficient of the first
term of the dividend.
3. The coefficient of any term of the quotient is obtained by multiplying the coefficient
of the previous term by the second term of the binomial divisor with opposite sign
and adding this product with the coefficient of the term that occupies the same
place in the dividend.
4. The remainder is obtained by multiplying the coefficient of the last term of the
quotient by the second term of the divisor with opposite sign and adding this
product to the independent term (constant term) of the dividend.
(x − 3) divisor
+3 constant term
with change sign
x3
+1
−5x2
-5
(+1)(+3)=3
+1
-2
+3x
+3
(-2)(+3)=-6
-3
+14
Dividend
+14 Coefficients
(-3)(+3)=-9
5
The quotient is x2 − 2x − 3 and the remainder is 5. When the divisor d(x) is linear the
division algorithm becomes the Remainder Theorem:
Definition 2.1.5 (Remainder Theorem). When d(x) = x − c, then
D(x) = (x − c)q(x) + r,
where r = D(c). This means that the remainder of D(x) when divided by (x − c) is the
same as the functional value at x = c.
36
Definition 2.1.6 (Factor Theorem). If the remainder of D(x) when divided by (x − c)
is 0, D(c) = 0, and then (x − c) is a factor of D(x).
Remark.
D(c) = 0 ↔ x − c is a factor of D(x)
Example 2.1.4. Verify the Remainder Theorem by dividing:
P (x) = x3 − 5x2 + 3x + 14
(x − 3) divisor
+3 constant term
with change sign
x3
+1
−5x2
-5
(+1)(+3)=3
+1
-2
by
x − 3. Then find P (3).
+3x
+3
(-2)(+3)=-6
-3
+14
Dividend
+14 Coefficients
(-3)(+3)=-9
5
P (3) = (3)3 − 5(3)2 + 3(3) + 14
= 27 − 5(9) + 9 + 14
= 27 − 45 + 9 + 14
= 5
37
2.2
Real Zeros of Polynomials
Let f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 a polynomial function of x with degree
n. The numbers an 6= 0 (leading coefficient), an−1 , · · · , a1 , and a0 6= 0(Constant term)
are called the coefficients of the Polynomial Functions. The possible rational zeros
(rational roots) of this polynomial function are pq , where p is a factor of constant term
and q is a factor of the leading coefficient, in other words:
p
factors of a0
= ±
q
factors of an
Example 2.2.1. What numbers could be the rational zeros of the polynomial:
Q(x) = x7 − 8x4 + 9x2 − 5x + 16:
The factors of constant term 16 are:
1, 2, 4, 8, 16.
The factors of leading coefficient is:
1
The possibles rational zeros are:
1 2 4 8 16
± ,± ,± ,± ,±
1 1 1 1
1
±1, ±2, ±4, ±8, ±16.
38
2.2.1
Descartes’ Rule of Sign
Rules to determine the amount of zeros of a polynomial function P (x):
1. The number of positive real solutions of P (x) is at most equal to the number of
variations in sign in P (x) or is less than that by an even number.
2. The number of negative real solutions of P (x) is at most equal to the number of
variations in sign in P (−x) or is less than that by an even number.
Example 2.2.2. Use Descartes’ Rule of sign to help find all rational zeros of the polynomial.
P (x) = x4 − 2x3 − x2 − 4x − 6, then factor it completely.
First find P (−x):
P (−x) = x4 + 2x3 − x2 + 4x − 6.
P (x) has one variation in sign so there is one positive real zero.
P (−x) has three variation in sign so there are three or one negative real zeros.
The possible rational zeros are: ±1, ±2, ±3, ±6. Since the result x2 + 2 does no factor
-1
+1
-2
-1
(+1)(-1)=-1 (-3)(-1)=3
+1
-3
+2
-4
-6
(+2)(-1)=-2 (-6)(-1)=+6
-6
0
39
3
+1
-3
(+1)(3)=3
+1
0
+2
-6
(0)(3)=0 (2)(3)=+6
+2
0
over the real numbers (only in complex numbers), then
P (x) = x4 − 2x3 − x2 − 4x − 6 = (x + 1)(x − 3)(x2 + 2)
2.2.2
Exercises
Find all rational zeros and factor completely:
1. P (x) = x3 − 3x2 − 4x + 12
4. P (x) = x3 + x2 − x − 1
2. P (x) = x4 − 11x2 − 18x − 8
5. P (x) = 6x3 + 23x2 + 9x − 18
3. P (x) = −x4 − 7x3 − 7x2 + 22x + 24
2.3
End Behavior of Polynomial Functions
End behavior graph of polynomial function is determined by its degree and leading
coefficient. Let n and an be the degree and leading coefficient respectively of a polynomial
40
function.
If n is odd and =





an > 0 when x → ∞,then f (x) → ∞










and when x → −∞, then f (x) → −∞




an < 0 when x → ∞, then f (x) → −∞










and when x → −∞, then f (x) → ∞
Figure 2.2: End Behavior of Polynomial function: odd degree and positive leading coefficient
Figure 2.3: End Behavior of Polynomial function: odd degree and negative leading coefficient
41
n is even and =






an > 0 when x → ∞, then f (x) → ∞








and when x → −∞, then f (x) → ∞




an < 0 when x → ∞, then f (x) → −∞










and when x → −∞, then f (x) → −∞
Figure 2.4: End Behavior of Polynomial function: even degree and positive leading
coefficient
Figure 2.5: End Behavior of Polynomial function: even degree and negative leading
coefficient
42
2.3.1
Exercises
Describe the end behavior of the following polynomial functions:
1. f (x) = 5x3 − 2x2 + 4x − 5
2. g(x) = x18
3. h(x) = −4x3 + 3
4. f (x) = 3x − 8x2 + 4
5. g(x) = x3
6. h(x) = −3x3 + 7x4 − 2x2 + 5
2.3.2
Real Zeros of Polynomial Functions
If f (x) is a polynomial function and a ∈ <, the following statements are equivalents:
• x = a is a real zero of a polynomial function.
• x = a is a zero of the equation f (x) = 0.
• (x − a) is a factor of polynomial function f (x).
• (a, 0) is a x intercept of the graph of f (x).
43
Guideline to find the real zeros of a polynomial function f (x):
1. f (x) must be factorized completely.
2. f (x) must be equal to zero.
3. Each factor must be equal to zero.
4. Solve the remainder equations.
Example 2.3.1. Find the real zeros of the polynomial function:
f (x) = x3 + 3x2 + 2x.
1. Factoring f (x) = x3 + 3x2 + 2x = x(x2 + 3x + 2) = x(x + 1)(x + 2)
2. Making f (x) = 0 = x(x + 1)(x + 2).
3. Each factor must be equal to zero: x = 0, x + 1 = 0 and x + 2 = 0.
4. Solve the remainder equations: x = 0, x = −1 and x = −2.
• x = 0, x = −1 and x = −2 are real zeros of a polynomial function f (x).
• x = 0, x = −1 and x = −2 are zeros of the equation f (x) = 0.
• x, (x + 1), (x + 2) are factors of polynomial function f (x).
• (−2, 0), (−1, 0), and (0, 0) are x intercepts of the graph of f (x).
44
Figure 2.6: Graph of f (x) = x3 + 3x2 + 2x and zeros of f (x).
2.4
Rational Functions
A rational function is a function of the form
R(x) =
P (x)
Q(x)
where P (x) and Q(x) are polynomials, and Q(x) 6= 0.
2.4.1
Domain of Rational Functions
The domain of rational functions is restricted. The domain of a rational function R(x)
is the set of real numbers where Q(x) 6= 0. The x intercepts are the real numbers where
P (x) = 0. The y-intercept is R(0), provided 0 is in the domain of R(x).
45
Procedure to find the domain of a rational function:
1. The denominator.
2. Set the denominator equal to zero.
3. Set each factor equal to zero.
4. Solve each equation.
Procedure to find the x-intercepts of a rational function:
1. Factor the numerator.
2. Set the numerator equal to zero.
3. Set each factor equal to zero.
4. Solve each equation.
Example 2.4.1. Find the domain, the x-intercepts, and y intercept of the function:
R(x) =
x2 − 4
x2 − 2x − 3
1. Factor the numerator and the denominator.
R(x) =
x2 − 4
(x − 2)(x + 2)
=
2
x − 2x − 3
(x − 3)(x + 1)
46
2. Set the denominator equal to zero.
(x − 3)(x + 1) = 0
3. Set each factor to zero.
(x − 3) = 0
and
(x + 1) = 0
4. Solve each equation.
x=3
and
x = −1
The domain is:
D = {x/x ∈ R, ∧x 6= −1, 3}
or
D = (−∞, −1)
[
(−1, 3)
x-intercepts:
1. Factor the numerator.
R(x) =
(x − 2)(x + 2)
x2 − 4
=
x2 − 2x − 3
(x − 3)(x + 1)
[
(3, ∞)
47
2. Set the numerator equal to zero.
(x − 2)(x + 2) = 0
3. Set each factor equal to zero.
(x − 2) = 0
and
(x + 2) = 0
4. Solve each equation.
x=2
and
x = −2
y − intercept = R(0):
(0 − 2)(0 + 2)
(0 − 3)(0 + 1)
(−2)(+2)
=
(−3)(+1)
−4
=
−3
4
=
3
R(0) =
48
2.4.2
Vertical, Horizontal, and Slant Asymptotes
Definition 2.4.1 (Asymptote). It is a straight line to which, the graph of a function is
approaching and no matter how close it gets, it never touches it.
Definition 2.4.2 (Vertical Asymptote). Vertical asymptotes occurs where the denominator of a rational functions equals zero, in other words, where the rational function is
undefined.
Definition 2.4.3 (Horizontal Asymptote). Horizontal asymptotes occur at the end behavior of a rational function. There are three cases: Let
R(x) =
an xn + an−1 xn−1 + · · · + a1 x + a0
bm xm + bm−1 xm−1 + · · · + b1 x + b0
1. If n < m: y = 0 is a horizontal asymptote.
2. If n = m: y =
an
bn
is a horizontal asymptote.
3. If n > m: no horizontal asymptotes (slant asymptote is possible).
Definition 2.4.4 (Slant Asymptote). Let
N (x)
an xn + an−1 xn−1 + · · · + a1 x + a0
R(x) =
=
D(x)
bm xm + bm−1 xm−1 + · · · + b1 x + b0
49
Slant asymptotes occurs when the degree of the numerator is greater by 1 than the degree
of the denominator (n = m + 1).
The long division is important to write R(x) = q(x) +
r(x)
,
d(x)
where the degree of r(x) <
the degree of d(x).
Example 2.4.2. Graph the rational function f (x) = x1 :
Figure 2.7: Rational function: f (x) =
Example 2.4.3. Graph the rational function f (x) =
1
x
x2 −4
:
x2 −2x−3
Figure 2.8: Rational function: f (x) =
x2 −4
x2 −2x−3
50
Example 2.4.4. Graph the rational function f (x) =
x3 −4x
:
x2 −1
Figure 2.9: Rational function: f (x) =
2.4.3
x3 −4x
x2 −1
Exercises
1. Find the zeros, x-intercepts, y-intercepts of the following polynomial functions:
(a)
i. f (x) = x3 − 5x2 + 2x + 8
ii. h(x) = x2 − 32 x − 1
iii. f (x) = −5x2 + 4 + x4
iv. f (x) = (x − 3)2 (x − 2)2 (x − 1)
2. To the following rational functions, find the horizontal, vertical, and slant asymptotes if any:
(a) f (x) =
x3 −1
x2 −1
(b) f (x) =
x2 −1
x2 +2x−24
(c) f (x) =
x+1
x2 +1
(d) f (x) =
2x3 −1
8x3 −1
Part III
Exponential and Logarithmic
Functions
51
Chapter 3
Exponential and Logarithmic
Functions
3.1
Exponential Functions
Definition 3.1.1. For a > 0, and a 6= 1, the exponential function with base a is
defined f (x) = ax for every real number x.
The value of the base determines the general shape of the graph of f (x) = ax . For
0 < a < 1, the graph of f (x) = ax decreases rapidly. When a = 1, f (x) = 1 is a constant
function. And for a > 1, the function f (x) = ax increases rapidly.
Remark. a−1 = a1 , so a−x =
1
,
ax
a0 = 1, and a positive number to any power is always
c José del Cristo Guerra Guerra
52
53
positive.
Example 3.1.1. Sketch the graphs of the functions f (x) = 2x and g(x) =
1 x
2
on the
same coordinates system.
Since
1
2
= 2−1 , g(x) =
1 x
2
x
= (2−1 ) = 2−x
Example 3.1.2. Sketch the graphs of the functions f (x) = 3x , g(x) = 3x − 2, and
g(x) = −3x + 4 on the same coordinates system.
Definition 3.1.2. As n become larger, the values of 1 +
number called e, e ≈ 2.71828.
1 n
n
approach an irrational
54
The function f (x) = ex is called the natural exponential function. Its domain is all
real numbers (−∞, ∞) and its range is the positive real numbers (0, ∞).
3.2
Logarithmic Functions
Definition 3.2.1. For a > 0, and a 6= 1, the inverse of the exponential function f (x) = ax
is a logarithmic function, written as g(x) = loga (x). The function g(x) = loga (x) is
read as “ g of x is equal to logarithm of base a of x”. So loga (x) is the exponent to which
the base a must be raised to give x.
Logarithmic F orm
Exponential F orm
loga (x) = y ⇔ ay = x
55
In both forms, a is called the base and y is called the exponent.
Definition 3.2.2. The logarithm with base e is called natural logarithmic function
and is given a special name and notation, loge (x) = ln(x). The logarithm with base 10
is called the common logarithm and given abbreviated notation, log10 (x) = log(x).
N atural Logarithmic
Common Logarithmic
F unction
ln(x) = y ⇔ ey = x
Example 3.2.1.
F unction
and
log(x) = y ⇔ 10y = x
• Write the expression log4 (64) = 3 in exponential form
Logarithmic F orm
Exponential F orm
log4 (64) = 3 ⇔ 43 = 64
• Express the expression 63 = 216 in logarithmic form
Exponential F orm
Logarithmic F orm
63 = 216 ⇔ log6 (216) = 3
Example 3.2.2.
• Evaluate the expression log5 (25)
log5 (25) = 2
56
• In the equation log3 (x) = 4, use the definition of logarithmic function to find x:
log3 (x) = 4
x = 34
x = 81
3.2.1
Exercises
1. Write the following expressions in exponential form
(a) log5 (0.04) = −2
(b) log9 (243) =
5
2
2. Express the following expressions in logarithmic form
5
(a) 8− 3 =
(b)
1 −6
2
1
32
= 64
3. Evaluate the expression
(a) log2
1
8
(b) log 1 (27)
3
4. In the following equations, use the definition of logarithmic function to find x:
(a) logx (121) = 2
(b) log 1 (25) = x
5
57
3.2.2
Graph Logarithmic Functions
3.2.3
Finding the Domain of Logarithmic Functions
The argument must be greater than zero (> 0)
Example 3.2.3.
• Find the domain of the function f (x) = log3 (x − 2)
f (x) = log3 (x − 2)
x−2 > 0
x > 2
• Find the domain of the function f (x) = log5 (x2 + 1):
f (x) = log5 (x2 + 1)
58
x2 + 1 > 0 for all x, the domain is (−∞, ∞)
3.2.4
Exercises
1. Sketch the graph of
(a) f (x) = log0.5 (x)
(b) f (x) = − log0.5 (x)
(c) f (x) = log2 (x)
(d) f (x) = − log2 (x − 1)
(e) f (x) = − log0.5 (x − 1)
2. Find the domain of the logarithmic function
(a) f (x) = log(x + 1)
(b) f (x) = log3 (3x + 6)
3.2.5
Properties of Logarithmic Functions
Suppose that x > 0, y > 0, and r is any real number. Then, the laws of logarithms are:
Property
ln(1) = 0
ln(e) = 1
ln(ex ) = x
eln(x) = x
Reason
We must raise e to the power 0 to get 1
We must raise e to the power 1 to get e
We must raise e to the power x to get ex
ln(x) is the power to which e must be raised
to get x
59
Property
loga (xy) = loga (x) + loga (y)
loga ( xy ) = loga (x) − loga (y)
loga (xr ) = r loga (x)
√
loga ( n x) =
loga (x)
n
aloga (x) = x
(loga (x))−1 = logx (a)
loga (x) =
logb (x)
logb (a)
Reason
The logarithm of a product
is the sum of the logarithms.
The logarithm of a quotient is the
difference of the logarithms.
The logarithm of a power is the exponent
times the logarithm of the number.
The logarithm of a root is the logarithms
of the number divided by the index of the root.
is the power to which a must be
raised to get x
The logarithm to power −1 is equal
to the logarithm to power 1 interchange the
base and argument.
The logarithm of a number is equal to the
quotient of the logarithm of the argument and the logarithm
of the base in any base different to the original.
• Use the Laws of Logarithms to rewrite the following expression in
a form that contains no logarithms of products, quotients, powers or root. log5 2x
y
Example 3.2.4.
) = log5 (2x) − log5 (y)
log5 ( 2x
y
The logarithm of a quotient
is the difference of
the logarithms.
2x
log5 ( y ) = log5 (2) + log5 (x) − log5 (y) The logarithm of a product
is the sum of the
logarithms.
Example 3.2.5.
• Use the Laws of Logarithms to rewrite the following expression
as a single logarithm 4 log3 (x) − 37 log3 (y)
Example 3.2.6.
• Evaluate log6 (7) to six decimal places.
60
3
= log3 (x)4 − log3 (y) 7
The logarithm of a power is the
exponent times the logarithm
of the number
p
7
4
3
= log3 (x) − log3 ( y ) Rational exponents
and radicals
x4
√
= log3 7 3
The logarithm of a quotient
y
is the difference of
the logarithms.
log(7)
log(6)
Change of base
0.84509804
≈ 1.086033
0.778151250
ln(7)
log6 (7) =
ln(6)
1.945910149
≈ 1.086033
≈
1.791759469
Change of base
log6 (7) =
≈
3.2.6
Exercises
1. Use the Laws of Logarithms to rewrite the following expression in a form that
contains no logarithms of products, quotients, powers or root.
(a) log2 (9x)
(b) log3 (x4 y 3 )
5
16w
(c) log4 √
3
x2 y
61
2. Use the Laws of Logarithms to rewrite the following expression as a single logarithm.
(a) 2 log(x) + 3 log(y) − 21 log(8)
(b) 4 log5 (x) − log5 (y) − 2 log5 (w)
3. If log2 (5) = x, express log2
q
4
25
4
in terms of x.
4. Simplify (log3 (8))(log8 (7))
3.3
Exponential Equations
Guidelines for solving exponential equations:
1. Isolate the exponential expression on one side of the equation.
2. Take the logarithm of each side, then use the Laws of Logarithms to “bring down
the exponent.”
3. Solve for the variable.
62
Example 3.3.1. Solve for x: 23x+1 = 32
23x+1 = 32
log2 (23x+1 ) = log2 (32)
(3x + 1) log2 (2) = 5
| {z }
1
3x + 1 = 5
3x = 5 − 1
3x = 4
x =
4
3
Example 3.3.2. Solve for x: x5 ex + 3x4 ex = 4x3 ex
x5 ex + 3x4 ex − 4x3 ex = 0
x3 ex (x2 + 3x − 4) = 0
x3 ex (x + 4)(x − 1) = 0
x3 = 0, (x + 4) = 0
x = 0, x = −4
ex 6= 0, (x − 1) = 0
x=1
63
3.4
Logarithmic Equations
Guidelines for solving logarithmic equations:
1. Isolate the logarithmic term on one side of the equation; you may need to first
combine the logarithmic terms.
2. Write the equation in exponential form (or raise the base to each side of equation).
3. Solve for the variable.
Example 3.4.1. Solve for x: log5 (62 − x) = 3
Check
log5 (62 − x) = 3
(62 − x) = 53
62 − x = 125
x = 62 − 125
x = −63
log5 (62 − x) = 3
log5 (62 − (−63)) = 3
log5 (62 + 63) = 3
log5 (125) = 3
3 = 3
64
Example 3.4.2.
Solve for x: log3 (x) + log9 (x) = 6
log3 (x)
= 6
log3 (9)
log3 (x)
log3 (x) +
= 6
2
3
(log3 (x)) = 6
2
log3 (x) +
3 log3 (x) = 12
log3 (x) = 4
x = 34 ⇒ x = 81
Example 3.4.3. Check
log3 (x) + log9 (x) = 6
log3 (81) + log9 (81) = 6
4+2 = 6
6 = 6
65
Exercises
1. Solve for x
2. 3x = 243
3. e2x − 4ex + 3 = 0
4. 4e2x−1 = 7
5. Solve for x
6. log8 (x + 2) + log8 (x + 5) =
2
3
7. log2 (x) − log2 (1 − x) = 5
8. What is the value of b if the succession {logb (1), logb (2), logb (4), logb (8), · · · } is
equivalent to the succession {0, 1, 2, 3, 4, · · · }?
9. Solve for x: log
√
5x + 5 + 21 log(2x + 1) − log(15) = 0
10. Solve for x: (log(x))ln(log(x)) = 1
11. If log(3) = L and log(5) = M , express log
81×50
√
15
in terms of L and M .
Part IV
Series
66
Chapter 4
Series
Definition 4.0.1. Series is a succession of terms which are formed according to one rule.
Example 4.0.1. 1, 3, 5, 7, . . . is a series whose rule is that of each term is obtained add
2 to the before term.
1, 2, 4, 8, . . . is a series whose rule is that of each term is obtained multiplying 2 to the
before term.
The series which we study are the progressions.
4.1
Arithmetic Progressions
Definition 4.1.1. Arithmetic Progressions are sequences of numbers such that the difference between the consecutive terms is constant. For instance, the sequence 1, 3, 5, 7,
c José del Cristo Guerra Guerra
67
68
. . . is an arithmetic progression with a common difference of 2. This common difference
o constant is called ratio.
4.1.1
Formula to find the nth term of an Arithmetic Progressions
Let the Arithmetic Progressions: a1 , a2 , a3 , a4 , a5 , . . ., an , where an is the nth term and
whose ratio is r. Then:
a2 = a1 + r
a3 = a2 + r = a1 + r + r = a1 + 2r
a4 = a3 + r = a1 + 2r + r = a1 + 3r
a5 = a4 + r = a1 + 3r + r = a1 + 4r
..
.
..
.
an = a1 + (n − 1)r
an = a1 + (n − 1)rnth term
Example 4.1.1. Find the 15th term of the arithmetic progression 4, 7, 10. . .
69
Here a1 = 4, n = 15 and r = 7 − 4 = 10 − 7 = 3, then:
an = a1 + (n − 1)r
a15 = 4 + (15 − 1)3
a15 = 4 + (14)3
a15 = 4 + 42
a15 = 46
Example 4.1.2. Find the 42th term of the arithmetic progression −2, −1 25 , − 54 . . .
Here a1 = −2, n = 42 and r = −1 52 − (−2) = − 75 − 2 = 53 , then:
an = a1 + (n − 1)r
a42 = −2 + (42 − 1)
a42 = −2 + (41)
a42 = −2 +
113
5
3
= 22
5
a42 =
a42
123
5
3
5
3
5
70
4.1.2
Exercises
Find the
1. 9th term of progression 7, 10,13. . .
2. 12th term of progression 5, 10,15. . .
3. 19th term of progression -4, − 32 ,. . .
4. 13th term of progression 3, -1, -5. . .
5. 12th term of progression 21 , 34 , 1. . .
4.1.3
Formula to find the first term, ratio and the number of
terms
an = a1 + (n − 1)r Solving for a1 ⇒ a1 = an − (n − 1)r
an = a1 + (n − 1)r Solving for r ⇒ r =
an = a1 + (n − 1)r Solving for n ⇒ n =
an − a1
(n − 1)
an − a1 + r
r
71
a1 = an − (n − 1)r
r =
n =
an − a1
(n − 1)
an − a1 + r
r
Example 4.1.3. Find the first term of the arithmetic progression such that a11 = 10
and the ratio is 21 .
Here an = a11 = 10 and r = 21 , then:
a1 = 10 − (11 − 1)
a1 = 10 − (10)
a1 = 10 − 5
a1 = 5
1
2
1
2
72
Example 4.1.4. Find the ratio of the arithmetic progression whose a1 = − 43 and an =
a8 = 3 81
an − a1
n−1 3 18 − − 34
r =
8−1
25
3
+
4
r = 8
7
r =
r =
31
8
7
31
r =
56
Example 4.1.5. How many terms does the progression 2, 1 23 . . . − 4 13 have?
Here, r = 1 32 − 2 = − 31 , then:
an − a1 + r
r
1
−4 3 − 2 + − 13
n =
− 31
n =
−2−
− 13
3
n =
− 13
− 20
3
n =
− 13
n = 20
1
3
73
4.1.4
Exercises
1. The 15th term of an arithmetic progression is 20, and the ratio is 27 . Find the first
term.
2. Find the ratio of 3, . . . , 8 where 8 is the 6th term.
3. How many terms it has the progression 4, 6, . . . , 30?
4. How many terms it has the progression 5, 5 31 , . . . , 18?
5. The 5th term of an arithmetic progression is 7, and the 7th term is 8 13 . Find the
first term.
4.1.5
Formula to find the sum of all terms of Arithmetic Progressions
Let the progression a1 , a2 , a3 , a4 , . . ., an−3 , an−2 , an−1 , an , which it has n terms. Then
the sum S of all terms is: S = a1 + a2 + a3 + a4 + . . . + an−3 + an−2 + an−1 + an and
S = an + an−1 + an−2 + an−3 . . . + a4 + a3 + a2 + a1 .
Add the two equality,
2S = (a1 + an ) + (a2 + an−1 ) + (a3 + an−2 ) + (a4 + an−3 ) + . . . (an−3 + a4 ) + (an−2 + a3 ) +
(an−1 + a2 ) + (an + a1 )
All binomial are equal to (a1 + an ), then, 2S = (a1 + an )n, here
74
S =
(a1 + an )n
2
Example 4.1.6. Find the sum of all twelve terms of the progression 7, 13, 19 . . .
We need to know the a12 term, using the formula an = a1 + (n − 1)r, then:
a12 = 7 + (12 − 1)6
a12 = 7 + (11)6
a12 = 7 + 66
a12 = 73 then,
(a1 + an )n
2
(80)12
(7 + 73)12
=
S =
2
2
960
S =
2
S =
S = 480
Example 4.1.7. Find the sum of all thirteenth terms of the progression:
5 1
, ,
6 12
...
The ratio r is r =
1
12
−
5
6
= − 43 . We need to find the a13 term, using the formula
75
an = a1 + (n − 1)r, then
5
3
5
3
5
a13 =
+ (13 − 1) −
= + (12) −
= −9
6
4
6
4
6
49
a13 = −
then,
6
5
5
+ − 49
13
− 49
13
(a1 + an )n
6
6
6
6
S =
=
=
2
2
2
44
22
286
− 6 13
− 3 13
− 3
−286
2
=
=
=
= −47
S =
2
2
2
6
3
4.1.6
Exercises
Find the sum of the
1. 8 first terms of the 15, 19, 23, . . .
2. 19 first terms of the 31, 38, 45, . . .
3. 80 first terms of the −10, −6, −2, . . .
4. 17 first terms of the −2, 41 , . . .
5. 12 first terms of the −5, −4 85 , . . .
4.2
Geometric Progressions
Definition 4.2.1. Geometric Progressions are sequences of numbers where each term
after the first is found by multiplying the previous one by a fixed, non-zero number
76
called the common ratio. For instance, the sequence 5, 10, 20, 40, . . . is a geometric
progression whose ratio is 2. Then, 5 × 2 = 10, 10 × 2 = 20, 20 × 2 = 40, 5 . . ..
To find the ratio r in all geometric progressions divide one term by the precedent term
to it.
4.2.1
Formula to find the nth term of Geometric Progressions
Let the Geometric Progressions: a1 , a2 , a3 , a4 , a5 , . . ., an , where an is the nth term and
whose ratio is r. Then:
a2 = a1 r
a3 = a2 r = (a1 r)r = a1 r2
a4 = a3 r = (a1 r2 )r = a1 r3
a5 = a4 r = (a1 r3 )r = a1 r4
..
.
..
.
an = a1 rn−1
77
Example 4.2.1. Find the 5th term of the geometric progression 2, 6, 18. . .
Here a1 = 2, n = 5 and r = 6 ÷ 2 = 3, then:
an = a1 rn−1
a5 = 2(3)5−1
a5 = 2(3)4
a5 = 2(81)
a5 = 162
Example 4.2.2. Find the 8th term of the geometric progression 6, 4, . . .
Here a1 = 6, n = 8 and r = 4 ÷ 6 = 23 , then:
an = a1 rn−1
8−1
2
a8 = 6
3
7
2
a8 = 6
3
128
a8 = 6
2187
256
a8 =
729
78
4.2.2
Exercises
Find the
1. 7th term of the progression 3, 6,12. . .
2. 8th term of the progression 8, 4, 2. . .
3. 9th term of the progression 3, −1, 31 , . . .
1
, ...
4. 10th term of the progression − 34 , − 14 , − 12
5. 5th term of the progression 13 , 1, 3 . . .
4.2.3
Formula to find the first term and ratio-Geometric Progressions
an = a1 rn−1
Solving for a1
a1 =
an
rn−1
79
an = a1 rn−1
Solving for r
an
a1
r
rn−1 =
r =
n−1
an
a1
Example 4.2.3. The 6th term of the geometric progression is a6 =
1
.
2
1
16
and the ratio is
Find the first term.
Here an = a6 =
1
,
16
r = 12 , and n = 6. Then:
a1 =
a1 =
a1 =
a1 =
an
rn−1
1
16
1 6−1
2
1
16
1
32
=
1
16
1 5
2
32
16
a1 = 2
Example 4.2.4. The 1st term of the geometric progression is a1 = 3 and the 6th is
a6 = −729 . Find the ratio.
Here a1 = 3, a6 = −729, and n = 6. Then:
80
r
r =
n−1
r
r =
r =
r =
6−1
√
5
√
5
an
a1
−729
3
−243
−35
r = −3
4.2.4
Exercises
1. The ratio of a geometric progression is 12 , and the 7th term is
1
.
64
Find the first
term.
2. Find the ratio of 2, . . . , 64 of six terms.
3. The 9th of a geometric progression is
64
2187
and the ratio is 23 . Find the firs term.
1
4. Find the ratio of 8, . . . , 512
of seven terms.
2
5. The 5th term of a geometric progression is − 81
, and the first term is
ratio.
27
.
64
Find the
81
4.2.5
Formula to find the sum of all terms of Geometric Progressions
Let the geometric progression a1 , a2 , a3 , . . ., an−2 , an−1 , an , which has n terms, and
whose ratio is r. Then the sum S of all terms is:
S = a1 + a2 + a3 + + . . . + an−2 + an−1 + an
Multiplying both sides equation 4.1 by the ratio r.
Sr = a1 r + a2 r + a3 r + . . . + an−2 r + an−1 r + an r
Subtract equation 4.1 of 4.1, we obtain:
Sr = a1 r + a2 r + a3 r + . . . + an−2 r + an−1 r + an r
−S = a1 − a2 − a3 − . . . − an−2 − an−1 − an
Sr − S = an r − a1
S(r − 1) = an r − a1 Using common factor
S =
an r − a1
r−1
82
S =
an r − a1
r−1
Example 4.2.5. Find the sum of the first six terms of the progression 4, 2, 1 . . .. We
need to know the a6 term, using the formulas
an r − a1
r− 1
1
1
−4
S = 8 1 2
−1
2
1
−4
S = 16 1
−2
S =
an = a1 rn−1
6−1
1
a6 = 4
2
5
1
1
a6 = 4
=4
2
32
4
1
a6 =
=
32
8
− 63
63
16
S =
1 =
8
−2
7
S = 7
8
Example 4.2.6. Find the sum of the first eight terms of the progression 9, −3, 1 . . ..
Here the ratio is −3 ÷ 9 = − 13 . Find the a8 using.
an = a1 rn−1
8−1
1
a8 = 9 −
3
7
1
1
a8 = 9 −
=9 −
3
2187
9
1
a8 = −
=−
2187
243
83
an r − a1
r − 1
1
− 13 − 9
− 243
S =
− 13 − 1
1
−9
729
S =
4
−3
S =
− 6560
1640
729
=
4
243
−3
182
S = 6
243
S =
4.2.6
Exercises
Find the sum of the
1. 5 first terms of the 6, 3, 1 12 , . . .
2. 6 first terms of the 4, −8, 16, . . .
3. 10 first terms of the 14 , 12 , 1, . . .
4. 6 first terms of the 9, −3, 1, . . .
5. 10 first terms of the −6, −3, −1 12 , . . .
Part V
Systems of Linear Equations and
Matrices
84
Chapter 5
Systems of Linear Equations
5.1
System of Linear Equations 3 × 3




2x + 3y = 13 This is a system of two linear equations with two



4x − y = 5
variables whose solution is x = 2 and y = 3.
To solve systems of two linear equations use the following method: Equality, Substitution,
Elimination, Graph and Determinants.
c José del Cristo Guerra Guerra
85
86
5.1.1
Equality Method
Example 5.1.1. Solve the system




7x + 4y = 13 eq.(1)



5x − 2y = 19 eq.(2)
1. In both equations, solve for the same variable.
Solving for x eq.(1),
7x = −4y + 13 ⇒ x =
−4y+13
7
eq.(3)
Solving for x eq.(2)
5x = 2y + 19 ⇒ x =
2y+19
eq.(4)
5
2. From eq.(3) and eq.(4)
−4y+13
7
=
2y+19
5
⇒ 5(−4y + 13) = 7(2y + 19)
⇒ −20y + 65 = 14y + 133
⇒ −20y − 14y = 133 − 65
⇒ −34y = 68
⇒
−34y
−34
=
68
−34
⇒ y = −2
3. Substituting y = −2 into eq.(3) or eq.(4), x =
⇒ x = 3.
The solutions is x = 3 and y = −2 or (3, −2).
−4(−2)+13
7
=
8+13
7
=
21
7
87
5.1.2
Substitution Method
Example 5.1.2. Solve the system




7x + 4y = 13
1



5x − 2y = 19
2
1. Choose one of the two equations and solve for one of two variables.
Solving for x eq. 1 ,
7x = −4y + 13 ⇒ x =
−4y+13
7
3
2. Substitute the result of the step 1 in the eq. that was not used in step 1.
5x − 2y = 19 ⇒ 5
−4y+13
7
− 2y = 19
3. Solve the resulting linear equation in one variable. Solve for y in the last equation
5
−4y+13
7
−20y+65
7
− 2y = 19
− 2y = 19
(6 7) −20y+65
− 7(2y) = 7(19)
67
−20y + 65 − 14y = 133
−34y = 133 − 65
−34y = 68 ⇒ y = −2
4. Substitute the result of the step 2 into the expression obtained in step 1 to find the
value of the second variable. Put y = −2 into eq. 3
x=
−4y+13
7
⇒x=
−4(−2)+13
7
88
8+13
7
x=
=
21
7
⇒x=3
Solution: x = 3 and y = −2 or (3, −2).
5.1.3
Elimination Method
Example 5.1.3. Solve the system




7x + 4y = 13
1



5x − 2y = 19
2
1. Is necessary to have the same coefficients
with opposite signs in one of the two




7x + 4y = 13 1 × 1
variables to eliminate its variable.




5x − 2y = 19 2 × 2




1
7x + 4y = 13



10x − 4y = 38
2
2. Add the last two equations
7x
+
4y
=
13
10x
-
4y
=
38
=
51
=
3
17x
x
3. Substituting x = 3 back into either of the two original equations to solve for y. We
choose eq. 1
7x + 4y = 13 ⇒ 4y = −7x + 13 ⇒ y =
⇒y=
−7(3)+13
4
=
−21+13
4
=
−8
4
−7x+13
4
⇒ y = −2
Solution: x = 3 and y = −2 or (3, −2)
89
5.1.4
Graph Method
Example 5.1.4. Solve the system




7x + 4y = 13
1



5x − 2y = 19
2
1. Solve for y in both equations.
y=
−7x+13
4
and y =
5x−19
2
2. For each line give two x values to obtain two y values
−7x+13
4
x
-1
5
1
-7
3
-2
3
-2
x
y=
y=
5x−19
2
3. Graph each line using the values of x and y
90
5.1.5
Determinants Method
a b
= ad − bc
c d
a and c are in the column 1.
b and d are in the column 2.
a and b are in the row 1.
c and d are in the row 2.
a and d are extremes.
b and c are means.
To solve a determinant the second order (2 row and 2 columns) to the product of the
extremes subtract the products of the means.
3 2
Example 5.1.5.
= (3)(4) − (2)(5) = 12 − 10 = 2
1.
5 4
3 −5
= (3)(−2) − (−5)(1) = −6 + 5 = −1
2.
1 −2
−2 −5
= (−2)(−9) − (−5)(−3) = 18 − 15 = 3
3.
−3 −9
91
Solve Systems of Linear Equations by Determinants
Example 5.1.6.
 Solve the following system of linear equations by the determinant



6x − 5y = −9 1
method each.



4x + 3y = 13
2
−9 −5
13
3
x =
6 −5
4
3
determinant formed by change coefficients of x by constant terms and coefficient of y
determinant formed by coefficients of the variables
−27 + 65
38
(−9)(3) − (−5)(13)
=
=
=1
=
(6)(3) − (−5)(4)
18 + 20
38
=
6 −9
4 13
y =
6 −5
4
3
determinant formed by coefficient of x and by change coefficients of y by constant terms
determinant formed by coefficients of the variables
(6)(13) − (−9)(4)
78 + 36
114
=
=
=
=3
(6)(3) − (−5)(4)
18 + 20
38
=
Solution: x = 1 and y = 3.
92
Systems of Linear Equations 3 × 3





x





Example 5.1.7. Solve the system
2x








 3x
+
4y
-
z
=
6
1
+
5y
-
7z
=
-9
2
-
2y
+
z
=
2
3
1. Combining eq. 1 and 2 and eliminating the x. Multiplying the eq. 1 by 2 and
subtracting





2x +





-2x 








2 from 1 , we obtain:
8y
-
2z
=
12
1
5y
+
7z
=
9
2
3y
+
5z
=
21
4
2. Combining eq. 3 with any of the others equations 1 or 2 and eliminating the x.
We use the eq. 1 . Multiplying the eq. 1 by 3 and subtracting 3 from 1 , we
obtain:
3x
+
12y
-
3z
=
18
1
-3x
+
2y
-
z
=
-2
3
14y
-
4z
=
16
Divide by 2
7y
-
2z
=
8
5
3. Now, we take equations 4 and 5
3y
+
5z
=
21
4
Eliminating z mul-
7y
-
2z
=
8
5
93
6y
+
10z
=
42
4
35y
-
10z
=
40
5
=
82
y
=
82
41
y
=
2
tiplying eq. 4 by 2 and eq. 5 by 5: 41y
4. Substituying y = 2 into eq. 5 we obtain:
7(2)
-
2z
=
8
14
-
2z
=
8
-
2z
=
8-14
-
2z
=
-6
−2z
−2
=
−6
−2
z=
3
5
5. Substituting y = 2 and z = 3 into any of eq. 1 , 2 or 3 , by example 1 , we
obtain:
x
+
4(2)
-
3
=
6
x
+
8
-
3
=
6
x
+
5
=
6
x
=
6-5
x
=
1
1





x= 1





Solution : y = 2








z = 3
94
Solve Systems of Linear Equations 3 × 3 Using Determi-
5.1.6
nants
a1 b1 c1
A determinant such that a2 b2 c2 is called of third order or 3 × 3
a3 b3 c3
1
Example 5.1.8. Solve the determinant −4
5
−2 −3
2
1
−1
3
Using the Sarrus Rule:
1
1. Below of the third row, put the two first rows
−2 −3
−4
2
1
5
−1
3
1
−2 −3
−4
2
1
2. We draw 3 diagonal line from the right to left and 3 from left to right.
&
1
&
−4
&.
5
.
1
.
−4
−2
.
−3
&.
.
2
1
&.
− 1
&.
− 2
2
&.
3
&
−3
&
1
95
3. Multiplying the three numbers in each diagonal. The product of the numbers in
the diagonals from left to right, stay with the same sign and the product of the
numbers in the diagonals from right to left change the sign.
(1)(2)(3) − (4)(1)(3) − (5)(2)(1) + (3)(2)(5) + (1)(1)(1) − (3)(2)(4) = 6 − 12 − 10 +
30 + 1 − 24 = −9
Solve Systems of Linear Equations by Determinants





x





Example 5.1.9. Solve the system by determinants
2x








 3x
+
4y
-
z
=
6
1
+
5y
-
7z
=
-9
2
-
2y
+
z
=
2
3
−164
−82
=2
6
4
−1
1
−9
5
−7
2 −9 −7
2
−2
1
x=
=
−82
−82
3
=1
6
2
−1
1
y=
=
1
4
−1
1
4
−1
2
5
−7
2
5
−7
3 −2
1
3 −2
1
96
1
4
6
2
5
−9
3 −2
2
z=
=
1
4
−1
2
5
−7
3 −2
1
−246
−82
=3





x= 1





Solution : y = 2








z = 3
97
5.1.7
Exercises
Solve the following systems of linear equations by the four method each.
1.
2.




6x − 5y = −9
1



4x + 3y = 13




7x − 15y = 1



−x − 6y = 8
3.




3x − 4y = 41



11x + 6y = 47




x − 2y = 6
1
2



2x − 4y = 5
2
1




x − 2y = 5
1



2x − 4y = 10
2
4.
5.
2
1
6.
2




x − 5y = 8
1



−7x + 8y = 25
2
Solve the following systems of linear equations by determinants method.
1.
2.
3.




6x − 5y = −9




x − 2y = 6
1
1



4x + 3y = 13
2



2x − 4y = 5
2




7x − 15y = 1
1




x − 2y = 5
1



−x − 6y = 8
2



2x − 4y = 10
2
4.
5.




3x − 4y = 41
1



11x + 6y = 47
2
6.




x − 5y = 8
1



−7x + 8y = 25
2
98
Solve the determinants:
−9
1 2 1
3.
1. 1 3 4
1 0 2
11
−5 7
2. −12
3
8
−13
1
9
Solve the





x





1.
x








 2x





x





2.
2x








 x
3.



















4.
7
3
−4
−5 −3
4
6
1
12
5
10
8 −6
9
7
4
−2
following system by elimination and determinants:
+
y
+
z
=
11
1
-
y
+
3z
=
13
2
+
2y
-
z
=
7
3
+
y
+
z
=
-6
1
+
y
-
z
=
-1
2
-
2y
+
3z
=
-6
3
x
3
-
y
4
+
z
4
=
1
1
x
6
+
y
2
-
z
=
1
2
x
2
-
y
8
-
z
2
=
0
3
99
5.2
Matrices
What is a matrix? A matrix is a rectangular arrangement of numbers composed of row
and columns.


2 3 3




1 0 8
Frequently, the matrices are denoted by parenthesis or brackets.


2 3 3




1 0 8
Each horizontal line of numbers is called a ROW, and each vertical line of numbers is
called a COLUMN.
2 3 3 ⇒ This is a row
1 0 8
2
3
3
1
0
8
⇓
This is a column
What is the order of a matrix? The order of a matrix is determined by the number of
rows and the number of columns. If the matrix has m rows and n columns, the matrix
has an ORDER m × n.
100
What is the order of the following matrix?


2 3 3




1 0 8
What is a square matrix? When in a matrix, if, the number of rows and the number of
columns are equals, the matrix is SQUARE. The order of a square matrix is n × n for
certain natural number n.




2 3 3




1 0 8
2 3




1 0
To indicate the elements of a matrix, is used the rows and columns. The position of each
element is indicated saying the row number first and then the column number. To denote
matrices capital letters are used, and the elements with lowercase letter. Let the matrix


2 3 3

A=


1 0 8
The element in the row 2 and column 1 is: 1
The element in the row 1 and column 3 is: 3
This is not a practical way to describe the matrix elements.
to shortcut these expressions, say A = {aij }, where i: indicates the row and j: indicates
101
the column.


2 3 3

A=


1 0 8
With this representation, the nomenclature of each element can be determined.
a21 = 1 it means: 1 is the element in the second row and the first column. What
represents a13 = 3?
What represents a22 ?
5.2.1
Operations with Matrices
Add and Subtract Matrices
To Add or Subtract Matrices:
1. The matrices must have the same order.
2. To find the result, the correspondent elements are added or subtracted.
3. The resultant matrix must have the same order as the matrices which are being
added or subtracted.
Add A = a11 a12 a13 a21 a22 a33
with 

b11 b12 b13 

B=


b21 b22 b23
102
The procedure is




a11 a12 a13  b11 b12 b13 
+

A+B =

 

a21 a22 a23
b21 b22 b23


a11 + b11 a12 + b12 a13 + b13 

=


a21 + b21 a22 + b22 a23 + b23
Add








2 3 3 22 35 −3  2 + 22 3 + 35 3 + (−3) 24 38 0

=
=
+


 
 
 

2 2 9
1+1 0+2
8+1
1 2
1
1 0 8
103
 To subtract,
  the same principle
  of the addition is used:

2 3 3 22 35 −3  2 − 22 3 − 35 3 − (−3) −20

−
=
=

 
 
 
1 0 8
1 2
1
1−1 0−2
8−1
0




2 3
 22
yB=
Let the matrices A = 



1 0
1

−32
6


−2 7
35 


3
Which is the result of A + B?, Which is the result of A − B?, is A + B = B + A?, and
A − B = B − A?.
Product of Matrices
To multiply matrices:
1. The number of column of the first matrix must be equal to the number of row of
the second matrix.
2. Multiply each row of the first matrix by each column of the second matrix.
3. The resultant matrix must have the same number of rows of the first matrix and
the same number of columns of the second matrix.
104



 a11 a12 . . . a1n   b11 b12 . . . b1n 






a
b

b
.
.
.
b
a
.
.
.
a
 21
22
2n   21
22
2n 



 .
 .

..
.. 
.
.
 ..


.
.
.
.
.  .
.
. 








am1 am2 . . . amn
bm1 bm2 . . . bmn


 a11 b11 + · · · + a1n bm1 a11 b12 + · · · + a1n bm2 a11 b1n + · · · + a1n bmn 




 a b + ··· + a b
a21 b12 + · · · + a2n bm2 a21 b1n + · · · + a2n bmn 
 21 11

2n m1

=


..
..
..


.
.
.






am1 b11 + · · · + amn bm1 am1 b12 + · · · + amn bm2 am1 b1n + · · · + amn bmn
Given the matrices


 −3 5 5 
,
B=


2 5 5


 3 −4 2 
,
C=


1 −3 4
can the operation BC be performed? No, because the first matrix is the order 2 × 3 and
the second matrix is the order 2 × 3. The number of columns of the first matrix is 3 and
the number of row of the second column is 2.
Let


 −1 4 −2 
,
A=


1 −2 3


 −2 3 





B =  −1 −4 
.




2 −3
105
If possible, compute the following.
AB = and BA =
Is AB = BA
Is the matrices’ product commutative?
5.2.2
Zero Matrix, Identity Matrix and Inverse Matrix
Zero Matrix
The zero matrix isa matrix whose all elements are zeros.
0 0 0
.
01×1 = , 02×3 = 


0 0 0
Main Diagonal
The main diagonal of a square matrix is the diagonal that goes from the upper left corner
to the lower right corner.
The Identity Matrix
The identity matrix is the square matrix in which all elements of the
are
 main diagonal



1 0 0


1 0


 , I3 = 0 1 0 · · ·
1 and the other elements are zeros. I1 = , I2 = 






0 1


0 0 1
The product of any matrix by the identity is the same matrix (provided the product is
defined).
106
The Inverse Matrix
If A is a square matrix, the inverse of A is other matrix B such that, AB = BA = I.
(Identity Matrix). Usually, B = A−1 . Here −1 is not an exponent, this is the symbol to
indicate the inverse of A.
5.2.3
Matrices-Elementary Row Operations
One matrix can be transformed into an equivalent matrix by developing the following
elementary row operations:
1. Interchange two rows. Ri ↔ Rj
2. Multiply a row by a nonzero constant. cRi
3. Replace one row by adding c times other row to it. Ri + cRj → Ri
Example 5.2.1. Given the augmented matrix




A=




1 −3 1 


2 −5 6 
,


−3
5 6
perform each row operation in the order specified and write the final result.
1. First: R1 ↔ R2
107




A=



2. Second: 2R1 → R3

2 −5 6 


1 −3 1 
,


−3
5 6




A=




1 −3 1 


2 −5 6 
,


−1 −1 8
3. Third: 4R2 + R3 → R3

 1 −3 1


A=
 2 −5 6


5 −15 30
5.2.4




,



Exercises
1. If A, B, and C are 3 × 2, 2 × 5, and 5 × 6 matrices respectively, determine which of
the following products are defined. For those defined, enter the size of the resulting
matrix.
• BA:
• AB:
• CB:
• A2 :
108




1 −2 
 4
 3 −2 −3






 and B =  1 −3 1
2. If A = 
−2
−3
2









−3 −4 1
0 −3 −4





AB = 






, then















and BA = 






.



3. AB = BA for any two square matrices, are A and B of the same size?
4. Let



2 −5 

A=


0
7
−3 1





C = −4 5





−1 2
,


 8 −5 

B=


−5
4
,

y


 −7 2

D=


3 1
109
Develop the following operations (if is possible):
5.2.5
• A+B
• A×B
• A×C
• A−B
• C ×D
• B×A
• A+C
• D×C
• A+D
• B×C
Matrices and Linear Systems
Let the following linear system




4x + 7y = 7
eq.1



3x + 3y = −3 eq.2,
The array which corresponds to this system is the following:

 


4 7 x  7

  =  

   
y
−3
3 3
A Coefficient is the numerical part of a term (number that accompany a variables), and
a constant term, is the term that only contains numbers.
When a variable does not appear in an equation, the corresponding coefficient is zero.
Steps
1. Take the coefficients of each equation as the row of a matrix. Remember, if the
110
variable

 does not appear in one equation the coefficient is zero.
4 7

 (Coefficients Matrix)


3 3
2. 
Form
 a column matrix with the variables.
x
  .(Column Matrix)
 
y
3. 
Forma column matrix with the constant terms.
 3
  .(Column Matrix)
 
−3
Example 5.2.2. Let




4x + 7y = 2 eq.1



3x + 5y = 1 eq.2
Which
 array
  that corresponds to the last system of linear equations?
is
 the

4 7x 2

 =  .

   
1
3 5 y
Example 5.2.3. Let





x + 2y + 3z = 1 eq.1





2x − y − 3z = 1 eq.2








x + 3y + z = −2 eq.3
Linear systems can be solved using matrix perform row operation, to obtain the ma-
111
trix in row echelon form: (The principal diagonal is 1 and the elements below this are all
zeros)


3
1
1 2




2 −1 −3 1  .






1 3
4 −2


3
1
1 2





R2 − 2R1 → R2 0 −5 −9 −1
.




1 3
4 −2

3
1
1 2





R3 − R1 → R3 
0 −5 −9 −1 .




0 1
1 −3




1 
1 2 3




.
R3 − R2 → R3 
0
1
−3
−19






0 0 −4 −16

1 
1 2 3





6R3 + R2 → R2 
0 1 −3 −19 .




0 1 1 −3


1 
1 2 3




1
.
− 4 R3 → R3 
0
1
−3
−19






0 0 1
4
112
So, z = 4. Back-substituting into y − 3z = −19,
y − 3(4) = −1
y − 12 = −19
y = −19 + 12
y = −7
Back-substituting into x + 2y + 3z = 1,
x + 2(−7) + 3(4) = 1
x − 14 + 12 = 1
x−2 = 1
x = 3
113
Solution





x=3





y = −7








z = 4
Part VI
Conics
114
Chapter 6
Conics
6.1
6.1.1
Circle
Equation of the Circle-Standard Form
Example 6.1.1. Find the equation of the circle whose center is (2, −3) and radius r = 3
Center: (h, k) = (2, −3)
c José del Cristo Guerra Guerra
115
116
Figure 6.1: Circle
The equation of the circle is:
(x − h)2 + (y − k)2 = r2
(x − 2)2 + (y + 3)2 = 32
(x − 2)2 + (y + 3)2 = 9
6.1.2
Equation of the Circle-General Form
x2 + y 2 + Ax + By + C = 0
Example 6.1.2. x2 + y 2 − 6x + 4y + 4 = 0
We need write the equation in standard form by completing the perfect square trinomial:
1. Group the terms containing x and the terms containing y and pass to the other
117
side the constant term.
(x2 − 6x) + (y 2 + 4y) = −4
2. Complete the perfect square trinomial in each variable, and factor each trinomial
and simplify
(x2 − 6x + 9) + (y 2 + 4y + 4) = −4 + 9 + 4
(x − 3)2 + (y + 2)2 = 9 Center : (3, −2) r = 3
6.2
Parabola
6.3
Equation of the Parabola-Standard Form
Figure 6.2: Parabola
118
Example 6.3.1. Find the equation of the parabola whose vertex is (2, −3), open downward, and the focal distance p = 1
Vertex: (h, k) = (2, −3)
The equation of the parabola is:
1
(x − h)2
4p
1
(y + 3) = − (x − 2)2
4
(y − k) =
Example 6.3.2. Calculate the vertex, focus and directrix of the parabola whose equation
is x2 − 2x − 6y − 5 = 0
The equation of the parabola is x2 − 2x − 6y − 5 = 0, then;
x2 − 2x = 6y + 5
x2 − 2x + 1 = 6y + 5 + 1
(x − 1)2 = 6y + 6
(x − 1)2 = 6(y + 1)
1
(x − 1)2 = (y + 1)
6
3
1
3
1, −1 +
=
1,
V ertex : (1, −1), 4p = 6 ⇒ p = F ocus :
2
2
2
3
5
Directrix : y = k − p ⇒ y = −1 − = −
2
2
119
6.4
Ellipse
6.5
Equation of the Ellipse-Standard Form
Figure 6.3: Ellipse
Example 6.5.1. Write the equation of the ellipse whose major semiaxis a = 4, minor
semiaxis b = 3, and the center is (3, 4)
(x − h)2 (y − k)2
+
= 1
a2
b2
(x − 3)2 (y − 4)2
+
= 1
42
32
(x − 3)2 (y − 4)2
+
= 1
16
9
120
6.6
Hyperbola
Figure 6.4: Hyperbola
6.6.1
Equation of the Hyperbola-Standard Form
Example 6.6.1. Write the equation of the ellipse whose major semiaxis a = 4, minor
semiaxis b = 3, and the center is (3, 4).
(x − h)2 (y − k)2
−
= 1
a2
b2
(x − 3)2 (y − 4)2
−
= 1
42
32
(x − 3)2 (y − 4)2
−
= 1
16
9
b
T he Asymptotes are : y = ± (x − h) + k
a
3
y = ± (x − 3) + 4
4
121
6.6.2
Equation of the Hyperbola-General Form
Ax2 − By 2 + Cx + Dy + E = 0
Example 6.6.2. 5x2 − 9y 2 − 50x + 36y + 116 = 0
1. Group the terms containing x and the terms containing y:
(5x2 − 50x) − (9y 2 − 36y) = −116
2. Complete and factor the perfect square trinomial in each variable:
5(x2 − 10x) − 9(y 2 − 4y) = −116
⇒ 5(x2 − 10x + 25) − 9(y 2 − 4y + 4) = −116 + 125 + 36
(x − 5)2 (y − 2)2
−
= 1
9
5
√
Center : (5, 2), a = 3, b = 5,
5(x − 5)2 − 9(y − 2)2 = 45 ⇒
c 2 = a2 − b 2 = 9 − 5 = 4 ⇒ c = 2
Example 6.6.3. −9x2 + 16y 2 + 54x − 128y + 319 = 0
1. Group the terms containing x and the terms containing y:
(−9x2 + 54x) + (16y 2 − 128y) = −319
122
2. Complete and factor the perfect square trinomial in each variable:
−9(x2 − 6x) + 16(y 2 − 8y) = −319
−9(x2 − 6x + 9) + 16(y 2 − 8y + 16) = −319 − 81 + 256
−9(x − 3)2 + 16(y − 4)2 = −144 ⇒
(x − 3)2 (y − 4)2
−
= 1
16
9
Center : (3, 4), a = 4, b = 3,
c2 = a2 + b2 = 16 + 9 = 25 ⇒ c = 5
6.7
Exercises
Find the Equation in Standard Form and General Form of Each Conic.
1. Ellipse
123
2. Circle
3. Hyperbola
4. Parabola
Part VII
Trigonometry
124
Chapter 7
Trigonometry
7.1
Basic Concepts
Definition 7.1.1 (Trigonometry). Is the branch of mathematics that studies the relationship between the angles and the sides of a triangle.
Definition 7.1.2 (Angle). An angle is the space or aperture that exists between two
rays that has a common point called a vertex.
c José del Cristo Guerra Guerra
125
126
Figure 7.1: Angle
127
Definition 7.1.3 (Positive Angle). An angle is positive if the terminal side moves counterclockwise.
Figure 7.2: Positive angle
Definition 7.1.4 (Negative Angle). An angle is negative if the terminal side moves in a
clockwise direction.
Figure 7.3: Negative angle
Definition 7.1.5 (Quadrants). Four parts of the Cartesian plane between two consecutive semi axis x and y.
Figure 7.4: Quadrants
128
Definition 7.1.6 (Angle in standard or normal position). An angle in normal or standard
position is an angle whose vertex coincides with the origin of the rectangular coordinate
system and the initial side is on the positive semi axis x.
Figure 7.5: Angle in standard or normal position.
Definition 7.1.7 (Quadrantal Angle). Angles in the standard position where the terminal side lies on the x or y axis. For example: 90o , 180o , etc.
Figure 7.6: Quadrantal angle
Definition 7.1.8 (Degree). A complete circle measures 360 degrees. A grade (1o ), is a
1
three hundred and sixtieth part ( 360
) of the complete circle.
Definition 7.1.9 (Radians). A radian is the measure of the central angle of a circle that
intersects an arc of a circle that has the same length as the radius of the circle. The
129
Figure 7.7: Central angle, and arc length
length of the circumference of a circle of radius r is 2πr. The measurement in radians of
a positive angle of a complete rotation is:
θ =
2πr
= 2π
r
Example 7.1.1. What is the radians measure of a central angle θ opposite an arc of 24
meters in a circle of radius 6 meters?
Solution:
θ =
24 meters
= 4 radians
6 meters
Example 7.1.2. What is the radian measure of a central angle θ opposite an arc of 60
feet in a circle of radius 12 feet?
Solution:
θ =
60 feet
= 5 radians
12 feet
130
Definition 7.1.10 (Complementary Angles). Complementary angles are two positive
angles whose sum of their measures gives 90o .
Definition 7.1.11 (Supplementary Angles). Supplementary angles are two positive angles whose sum of their measures gives 180o .
Definition 7.1.12 (Reference Angles). This is the acute angle that is formed with the
terminal side of an angle and the semi-axis of x that is closer to that side.
Figure 7.8: Reference angle in quadrant I
Figure 7.9: Reference angle in quadrant II
131
Figure 7.10: Reference angle in quadrant III
Figure 7.11: Reference angle in quadrant IV
7.1.1
Conversion of degrees to radians and viceversa
• To convert degrees to radians, multiply the degrees by
π
.
180
• To convert radians to degrees, multiply the radians by
180
.
π
132
Example 7.1.3.
1. Convert 30o to radians.
30o = 30o ×
π
180
30π
180
π
=
6
=
2. Convert 60o to radians.
60o = 60o ×
π
180
60π
180
π
=
3
=
Example 7.1.4.
1. Convert
π
4
to degrees.
π
π 180
=
×
4
4
π
180π
=
4π
= 45o
133
2. Convert
3π
2
to degrees.
3π
3π 180
=
×
2
2
π
540π
=
2π
= 270o
Figure 7.12: Some angles in degrees and radians.
7.1.2
Exercises
1. Find the complement and supplement of the following angles:
(a) 70o
(b) 80o
134
(c)
2π
3
(d)
π
3
2. Which is the angle that is equal to half of its complement?
3. Which is the angle that is the double of its complement?
4. Which is the angle that is equal to its complement?
5. Which is the angle that is equal to half of its supplement?
6. Which is the angle that is equal to its supplement?
7. Which is the reference angle of an angle of 25o ?
8. Which is the reference angle of an angle of 125o ?
9. Which is the reference angle of an angle of
10. Which is the reference angle of an angle of
11. Covert to degrees:
(a)
3π
2
(b)
7π
4
12. Covert to radians:
(a) 540o
(b) 390o
5π
?
4
11π
?
6
135
7.2
Trigonometry of Right Triangles
Definition 7.2.1 (Right Triangle). A right triangle is a triangle with an internal 90o
angle.
Figure 7.13: Right triangle.
7.2.1
Trigonometric Ratios
The six trigonometric ratios of a right triangle are:
opposite side
hypotenuse
adjacent side
cosine = cos =
hypotenuse
opposite side
tangent = tan =
adjacent
sine = sin =
hypotenuse
opposite side
hypotenuse
secant = sec =
adjacent side
adjacent side
cotangent = cot =
opposite side
cosecant = csc =
136
Figure 7.14: The six trigonometric ratios.
Example 7.2.1. Find the value of the six trigonometric ratios for the following right
triangle:
12
6
=
10
5
8
4
cos(α) =
=
10
5
12
3
tan(α) =
=
8
2
sin(α) =
10
5
=
12
6
10
5
sec(α) =
=
8
4
8
2
cot(α) =
=
12
3
csc(α) =
137
7.2.2
Unitary Circle or Goniometric Circle
Definition 7.2.2 (Unitary Circle or Goniometric). Unitary circle is a circle with a radius
equal 1.
Figure 7.15: Unitary circle or Goniometric
Remark. According to the unitary circle, y = sin(α), x = cos(α), and tan(α) = xy .
7.2.3
Special Triangles
Definition 7.2.3 (Triangle 30 − 60 − 90). In all triangles 30 − 60 − 90, the length of the
√
shortest side is one half the length of the hypotenuse, and the longest side measures
the length of the hypotenuse or
√
3
2
3 the length of the shortest side. Figure (7.16)
Definition 7.2.4 (Triangle 45 − 45 − 90). In all triangle 45 − 45 − 90, the length of the
hypotenuse is
√
2 the length of one of the other side. Figure (7.17)
138
Figure 7.16: Triangle 30o − 60o − 90o
Figure 7.17: Triangle 45o − 45o − 90o
7.2.4
Evaluation of the Six Trigonometric Ratios-30o = π6 rad and
60o = π3 rad.
In the unitary circle, the hypotenuse length is 1, then the shortest side is
√
longest side measures is
3
.
2
1
2
and the
139
Figure 7.18: The six trigonometric ratios: 30o = π6 rad.
According to the figure 7.18, the six trigonometric ratios for 60o are:
√
3
sin(60o ) =
=
2
1
cos(60o ) = =
2
√
tan(60o ) = 3 =
4
5
4
5
3
2
2
csc(60o ) = √ =
3
2
sec(60o ) = = 2
1
1
cot(60o ) = √ =
3
√
2 3
3
√
3
3
Remark.
√
3
2
1
cos(60o ) = sin(30o ) =
2
√
tan(60o ) = cot(30o ) = 3
o
o
sin(60 ) = cos(30 ) =
csc(60o ) = 2
√
2 3
sec(60 ) = sec(30 ) = csc(30 ) =
3
√
3
cot(60o ) = tan(30o ) =
3
o
o
o
140
This is because the angles of 30o and 60o are complementary angles.
7.2.5
Evaluation of the Six Trigonometric Ratios-45o = π4 rad.
In the unitary circle, the hypotenuse length is 1, then the measures for the remainder
√
sides are:
2
.
2
Figure 7.19: The six trigonometric ratios: 45o .
Remark.
√
sin(45o ) = cos(45o ) =
2
2
tan(45o ) = cot(45o ) = 1
sec(45o ) = csc(45o ) =
√
2
This is because the angles of 45o and 45o are complementary.
141
7.2.6
Evaluation of the Six Trigonometric Ratios-0o = 0rad.
Figure 7.20: The six trigonometric ratios: 0o .
Figure 7.20 show that the coordinate of the terminal point is (1, 0). x = 1 and y = 0.
then:
sin(0o ) = 0
csc(0o ) = undef ined
cos(0o ) = 1
sec(0o ) = 1
tan(0o ) = 0
cot(0o ) = undef ined
Remark. x = cos(0o ), y = sin(0o ) and tan(0o ) = xy .
142
7.2.7
Evaluation of the Six Trigonometric Ratios-90o = π2 rad.
Figure 7.21: The six trigonometric ratios: 90o = π2 rad.
Figure 7.21 show that the coordinate of the terminal point is (0, 1). x = 0 and y = 1.
then:
sin(0o ) = 1
csc(0o ) = 1
cos(0o ) = 0
sec(0o ) = undef ined
tan(0o ) = undef ined
cot(0o ) = 0
Remark. x = cos(90o ), y = sin(90o ) and tan(90o ) = xy .
143
7.2.8
Evaluation of the Six Trigonometric Ratios-180o = πrad.
Figure 7.22: The six trigonometric ratios: 180o = πrad.
Figure 7.22 show that the coordinate of the terminal point is (−1, 0). x = −1 and
y = 0. then:
sin(180o ) = 0
cos(180o ) = −1
tan(180o ) = 0
csc(180o ) = undef ined
sec(180o ) = −1
cot(180o ) = undef ined
Remark. x = cos(180o ), y = sin(180o ) and tan(180o ) = xy .
144
7.2.9
Evaluation of the Six Trigonometric Ratios-270o =
3π
2 rad.
Figure 7.23: The six trigonometric ratios: 270o = πrad.
Figure 7.23 show that the coordinate of the terminal point is (0, −1). x = 0 and
y = −1. then:
sin(270o ) = −1
cos(270o ) = 0
tan(270o ) = undef ined
csc(270o ) = −1
sec(270o ) = undef ined
cot(270o ) = 0
Remark. x = cos(270o ), y = sin(270o ) and tan(270o ) = xy .
145
angle α
Degree-Radians
0o = 0rad
30o = π6 rad
45o = π4 rad
60o = π3 rad
90o = π2 rad
180o = πrad
rad
270o = 3π
2
o
360 = 2πrad
sin(α)
0
cos(α)
tan(α)
cot(α)
1
0
√
3
1
√
undefined
√
1
√2
2
√2
3
2
√
3
√2
2
2
1
2
1
0
-1
0
0
-1
0
1
3
3
3
3
undefined
0
undefined
0
1
√
3
0
undefined
0
undefined
sec(α)
csc(α)
1
undefined
2
√
2
√
√
2 3
√3
2
2
undefined
-1
undefined
1
2 3
3
1
undefined
-1
undefined
Table 7.1: Values of the six trigonometric ratios for: 0o , 30o , 45o , 60o , 90o , 180o , 270o ,
360o .
7.3
Trigonometric Functions
The six trigonometric functions are;
f (x) = sin(x)
odd function
f (x) = cos(x)
even function
f (x) = tan(x)
odd function
f (x) = cot(x)
odd function
f (x) = sec(x)
even function
f (x) = csc(x)
odd function
Remark. If f (x) = f (−x), then f (x) is an even function.
If f (−x) = −f (x), then f (x) is an odd function.
146
The sign of each trigonometric function depends on the quadrant where it is being
evaluated. Figure 7.24 shows each quadrant and the signs of each trigonometric function.
Figure 7.24: Signs of six trigonometric functions.
7.3.1
Graph of Trigonometric Functions
Figure 7.25: Graph of f (x) = sin(x), domain, range and period.
147
Figure 7.26: Graph of f (x) = cos(x), domain, range and period.
Figure 7.27: Graph of f (x) = tan(x), domain, range and period.
Figure 7.28: Graph of f (x) = cot(x), domain, range and period.
148
Figure 7.29: Graph of f (x) = sec(x), domain, range and period.
Figure 7.30: Graph of f (x) = csc(x), domain, range and period.
149
7.3.2
Transformed Sine and Cosine Graphs
The graph of functions of the form:
f (x) = A sin[K(x + B)] + C
f (x) = A cos[K(x + B)] + C,
are transformed sine and cosine curves, where:
• |A| is called the amplitude, and is the maximum value that the function can take.
• The number K (K > 0) produces change in the period at
2π
.
K
• The number B is the phase change. This number determines how much the graph
moves horizontally (B > 0, the graph moves to the left, and if B < 0, the graph
moves to the right).
• The number C determines how much the graph moves vertically. If C > 0, the
graph moves C units up, and if C < 0, the graph move C units down.
150
Figure 7.31: Graph of f (x) = sin(x) and g(x) = 3 sin 2 x + π2 + 1.
Figure 7.32: Graph of f (x) = cos(x) and g(x) = 4 cos 4 x − π4 − 1.
7.4
Trigonometric Identities
Definition 7.4.1 (Identity). An identity is an equation that is valid for all real numbers.
In other words, the solutions or root are all real numbers.
151
Figure 7.33: Unitary circle.
7.4.1
Basic Trigonometric Identities
According figure 7.33:
sin(α) = y
cos(α) = x
sin(α)
cos(α)
cos(α)
cot(α) =
sin(α)
1
sec(α) =
cos(α)
1
csc(α) =
sin(α)
tan(α) =
152
Using Pythagorean Theorem, we can obtain the following identities:
sin(α)
=
y
cos(α)
=
x
x2 + y 2
=
1
cos2 (α) + sin2 (α)
=
1
Dividing both sides of the last equation by cos(α) the result is
tan2 (α) + 1
=
sec2 (α)
Dividing both sides of the last equation by sin(α) the result is
cot2 (α) + 1
=
csc2 (α)
The basic trigonometric identities can be used to simplify complex trigonometric expressions.
cos2 (α) + sin2 (α) = 1
Remark.
tan2 (α) + 1 = sec2 (α)
sin(α)
cos(α)
cos(α)
cot(α) =
sin(α)
1
sec(α) =
cos(α)
1
csc(α) =
sin(α)
tan(α) =
cot2 (α) + 1 = csc2 (α)
153
Example 7.4.1. Simplify the trigonometric expression sin(x) sec(x) + cos(x) csc(x) in
terms of sines and cosines.
1
1
+ cos(x)
cos(x)
sin(x)
sin(x) cos(x)
+
cos(x) sin(x)
sin(x) sin(x) cos(x) cos(x)
+
cos(x) sin(x) sin(x) cos(x)
cos2 (x)
sin2 (x)
+
cos(x) sin(x) sin(x) cos(x)
sin2 (x) + cos2 (x)
cos(x) sin(x)
1
cos(x) sin(x)
sin(x) sec(x) + cos(x) csc(x) = sin(x)
=
=
=
=
=
7.4.2
Sine, Cosine and Tangent of the Sum of Two Angles
Figure 7.34: Unitary circle.
154
Acoording figure 7.34:
sin(α + β) =
sin(α) =
cos(α) =
sin(β) =
cos(β) =
BE
= BE
1
DF
DE
EF
DE
DE
= DE
1
AD
= AD
1
but
sin(α + β) = BE
= EF + F B
= DE cos(α) + AD sin(α)
= sin(β) cos(α) + cos(β) sin(α)
In the same way, it’s possible obtain:
cos(α + β) = cos(β) cos(α) − sin(β) sin(α)
The sine is an odd function, sin(−α) = − sin(α) and the cosine is an even function
155
cos(−α) = cos(α)
sin(α + (−β)) = sin(−β) cos(α) + cos(−β) sin(α)
sin(α − β) = − sin(β) cos(α) + cos(β) sin(α)
sin(α − β) = cos(β) sin(α) − sin(β) cos(α)
and
cos(α + (−β)) = cos(−β) cos(α) − sin(−β) sin(α)
cos(α − β) = cos(β) cos(α) + sin(β) sin(α)
sin(α ± β)
cos(α ± β)
sin(β) cos(α) ± cos(β) sin(α)
=
cos(β) cos(α) ∓ sin(β) sin(α)
tan(α ± β) =
Divide each term by cos(β) cos(α):
tan(α ± β) =
tan(β) ± tan(α)
1 ∓ tan(β) tan(α)
156
7.4.3
Sine, Cosine and Tangent of Double Angle
From equations for the sine of sum of two angles making α = β:
sin(α + α) = sin(α) cos(α) + cos(α) sin(α)
sin(2α) = 2 sin(α) cos(α)
From cosine of sum of two angles making α = β:
cos(α + α) = cos(α) cos(α) − sin(α) sin(α)
cos(2α) = cos2 (α) − sin2 (α)
This last equation can be written in other ways using the Pythagorean identity cos2 (α) +
sin2 (α) = 1, where cos2 (α) = 1 − sin2 (α) or sin2 (α) = 1 − cos2 (α):
cos(2α) = 2 cos2 (α) − 1
= 1 − 2 sin2 (α)
and from the tangent of the sum of two angles making α = β:
tan(2α) =
2 tan(α)
1 − tan2 (α)
157
Example 7.4.2. Find the exact value of sin(75o ), cos(75o ), and tan(75o ).
Using the formulas for the sum of two angles: as 75o = 45o + 30o
sin(45o + 30o ) = sin(45o ) cos(30o ) + cos(45o ) sin(30o )
√ √
√ 2 3
2 1
=
+
2 2
2
2
√
√
6
2
=
+
√4
√4
6+ 2
=
4
cos(45o + 30o ) = cos(45o ) cos(30o ) − sin(45o ) sin(30o )
√ √
√ 2 3
2 1
=
−
2 2
2
2
√
√
6
2
=
−
√4
√4
6− 2
=
4√
1 + 23
o
o
√
tan(45 + 30 ) =
1 − (1) 23
=
√
2+ 3
2
√
2− 3
2
√
2+ 3
√
=
2− 3
7.4.4
Sine, Cosine and Tangent of Half Angle
From equations of the cosine and tangent of double angle, changing α by
cos(α) = 2 cos
2
α
2
−1
α
2
158
then
2 cos2
α
2 sin2
α
= cos(α) + 1
2
α
cos(α) + 1
cos2
=
2
r 2
α
cos(α) + 1
cos
=
2
2
= 1 − cos(α)
2
α
1 − cos(α)
sin2
=
2
r 2
α
1 − cos(α)
sin
=
2
2
and for tangent of half angle:
tan
α
2
sin α2
=
cos α2
q
= ±q
s
= ±
1+cos(α)
2
cos(α)+1
2
1 − cos(α)
1 + cos(α)
159
7.4.5
Product to Sum and Sum to Product
From equation of the sum and difference of two angles it’s easy to find product to sum:
cos(α − β) − cos(α + β)
2
cos(α − β) + cos(α + β)
cos(α) cos(β) =
2
sin(α + β) − sin(α − β)
sin(β) cos(α) =
2
sin(α + β) + sin(α − β)
sin(α) cos(β) =
2
sin(α) sin(β) =
and sum to product making a change of variables:
α−β
α+β
cos
2 sin
2
2
α+β
α−β
2 cos
sin
2
2
α−β
α+β
cos
2 cos
2
2
α+β
α−β
−2 sin
sin
2
2
sin(α) + sin(β) =
sin(α) − sin(β) =
cos(α) + cos(β) =
cos(α) − cos(β) =
7.5
7.5.1
Law of Sines and Law of Cosines
Law of Sines
This activity comes from c 2008 National Council of Teachers of Mathematics.
Right triangle trigonometry can be used to solve problems involving right triangles. How-
160
ever, many interesting problems involve non-right triangles. In this lesson, you will use
right triangle trigonometry to develop the Law of Sines. The law of sines is important
because it can be used to solve problems involving non-right triangles as well as right
triangles.
Consider the triangle ABC shown below.
Figure 7.35: Non-right triangle
1. Sketch an altitude from vertex B.
2. Label the altitude h.
3. The altitude creates two right triangles inside triangle ABC. Notice that ∠A is
contained in one of the right triangles, and ∠C is contained in the other. Using right
triangle trigonometry, write two equations, one involving sin A, and one involving
sin C.
4. Notice that each of the equations in Question 3 involves h. Solve each equation for
h.
161
5. Since both equations in Question 4 are equal to h, they can be set equal to each
other. Set the equations equal to each other to form a new equation.
6. Notice that the equation in Question 5 no longer involves h. Write an equation
equivalent to the equation in Question 5, regrouping a with sin A and c with sin C.
7. Now, sketch an altitude from vertex C.
8. Label the altitude h.
9. The altitude creates two right triangles inside triangle ABC. Notice that ∠A is
contained in one of the right triangles, and ∠B is contained in the other. Using right
triangle trigonometry, write two equations, one involving sin A, and one involving
sin B.
10. Notice that each of the equations in Question 9 involves h. Solve each equation for
h.
11. Since both equations in Question 10 are equal to h, they can be set equal to each
other. Set the equations equal to each other to form a new equation.
12. Notice that the equation in Question 11 no longer involves h. Write an equation
equivalent to the equation in Question 11, regrouping a with sin A and b with sin B.
13. Use the equations in Question 6 and Question 12 to write a third equation involving
b, c, sin B and sin C.
162
14. Together, the equations in Questions 6, 12, and 13 form the Law of Sines. The law
of sines is important, because it can be used to solve problems involving both right
and non-right triangles, because it involves only the sides and angles of a triangle.
sin(A)
sin(B)
sin(C)
=
=
a
b
c
7.5.2
Law of Cosines
The law of sines can be used to determine the measures of missing angles and sides of
triangles when the measures of two angles and a side (AAS or ASA) or the measures of
two sides and a non-included angle (SSA) are known. However, the law of sines cannot
be used to determine the measures of missing angles and sides of triangles when the
measures of two sides and an included angle (SAS) or the measures of three sides (SSS)
are known. Since the law of sines can only be used in certain situations, we need to
develop another method to address the other possible cases. This new method is called
the Law of Cosines.
Figure 7.36: Non-right triangle
163
To develop the law of cosines, begin with triangle ABC. From vertex C, altitude h
is drawn and separates side c into segments x and c − x.
1. The altitude separates triangle ABC into two right triangles. Use the Pythagorean
theorem to write two equations, one relating h, b, and c − x, and another relating
a, h, and x.
2. Notice that both equations contain h2 . Solve each equation for h2 .
3. Since both of the equations in Question 2 are equal to h2 , they can be set equal to
each other. Set the equations equal to each other to form a new equation.
4. Notice that the equation in Question 3 involves x. However, x is not a side of
triangle ABC. As a result, we will attempt to rewrite the equation in Question 3
so that it does not include x. Begin by expanding the quantity (c − x)2 .
5. Solve the equation in Question 4 for b2 .
6. The equation in Question 5 still involves x. To eliminate x from the equation,
we will attempt to substitute an equivalent expression for x. Write an equation
involving both cos B and x.
7. Solve the equation from Question 6 for x.
8. Substitute the equivalent expression for x into the equation from Question 5. The
resulting equation contains only sides and angles of triangle ABC. This equation
is called the Law of Cosines.
164
9. Using a similar method, two other forms of this law could be developed for a2 and
c2 . Based on your work for Questions 1 to 8, write the two other forms of the law
of cosines for triangle ABC.
a2 = b2 + c2 − 2bc cos(A)
b2 = a2 + c2 − 2ac cos(B)
c2 = a2 + b2 − 2ab cos(C)
7.6
Inverse of Trigonometric Functions and Trigonometric Equations
If a function is one to one, it has an inverse. The sine, cosine, tangent functions are not
one to one. Since these trigonometric functions are not a one-to-one function, we restrict
the domain.
7.6.1
Inverse Function of Sine
The domain of the sine function can be restricted to [− π2 , π2 ]. The inverse of the sine
function is the function with domain [−1, 1] and range [− π2 , π2 ].
y = arcsin(x)
= sin−1 (x)
165
According to the last equation;
x = sin(y)
√ Example 7.6.1. Find the exact value of arcsin − 23 :
The angle in the interval [− π2 , π2 ] whose sine is negative must be in the IV quadrant and
is − π3 . Then:
√ !
3
π
arcsin −
= −
2
3
7.6.2
Inverse Function of Cosine
The domain of the cosine function can be restricted to [0, π]. The inverse of the cosine
function is the function with domain [−1, 1] and range [0, π].
y = arccos(x)
= cos−1 (x)
According to the last equation;
x = cos(y)
166
√
2
2
Example 7.6.2. Find the exact value of arccos −
:
The angle in the interval [0, π] whose cosine is negative must be in the II quadrant and
is
3π
.
4
Then:
√ !
2
3π
=
arccos −
2
4
7.6.3
Inverse Function of Tangent
The domain of the tangent function can be restricted to [− π2 , π2 ]. The inverse of the
tangent function is the function with domain (−∞, ∞) and range [− π2 , π2 ].
y = arctan(x)
= tan−1 (x)
According to the last equation;
x = tan(y)
Example 7.6.3. Find the exact value of arctan −
√
3
3
:
The angle in the interval [− π2 , π2 ] whose tangent is negative must be in the IV quadrant
167
and is − π6 . Then:
√ !
3
π
arctan −
= −
3
6
7.6.4
Trigonometric Equations
A trigonometric equation is an equation that contains trigonometric functions. If a
trigonometric equation has a solution, then it would have infinite solutions. First we find
the solutions at an appropriate interval (usually a period) and then use the period of the
trigonometric function to determine the other solutions.
Remark. Like polynomial equations, the trigonometric equations are solved by moving
all the terms to one side of the equation and factorizing, then each factor must be equal
to zero. Finally, find all solutions using the period of the trigonometric function.
168
Example 7.6.4. Find the solutions of the equation 2 sin(x) cos(x) = sin(x).
2 sin(x) cos(x)
=
sin(x)
2 sin(x) cos(x) − sin(x)
=
0
sin(x)(2 cos(x) − 1)
=
0
sin(x) = 0 and 2 cos(x) − 1 = 0
x = sin−1 (0)
x = 0, π
x = 0 + k2π
x = π + k2π
k∈Z
k∈Z
2 cos(x) = 1
1
2
1
−1
x = cos
2
π 5π
x= ,
3 3
π
x = + k2π
3
5π
x=
+ k2π
3
cos(x) =
The angle must be in
II and IV quadrant
because the cosine is positive
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