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Algebraic Skills
Department of Mathematics and Computer Science
June 22, 2011
Preface
Preface
This text aims at refreshing the algebraic skills required for following courses in basic mathematics at an academic level. Familiarity with many basic concepts is implicitly assumed throughout.
The topics covered in Chapters 1–10 show which algebraic skills freshmen students are assumed
to have mastered already. The topics covered in the remaining Chapters 11–14 are typically beyond what is assumed as prior knowledge upon entering university, but nevertheless basic and
essential to any course in calculus. As the main purpose of this text is to be used for self-study,
answers to all the exercises are provided in the final chapter.
Major deficiencies in algebraic skills and lack of familiarity with formulas can not be remedied
within a few practice sessions. Therefore, to allow for extensive and repetitive practicing, many
exercises of the same type are provided for each topic. To make sure that you master a certain
topic the idea is that you do all the exercises and check that you can do these correctly and quickly!
Note: the goal is that you can do the exercises without the use of a formula sheet or calculator.
You should know the rules and formulas in boxes like this by heart or be able to derive these
quickly, without access to a formula sheet or calculator.
Finally, according to a long and widespread tradition, parentheses around the argument to the
standard functions sin, log, etc. are omitted if no confusion arises. So y sin x = sin(x)y 6=
sin x y = sin(x y), sin x cos x = sin(x) cos(x) 6= sin(x cos x) and sin x + y = sin(x)+ y 6= sin(x + y).
Use of superfluous parentheses, especially when in doubt, is allowed.
Note that these functions are typeset in a regular Roman font.
ii
Preface
ii
1 Sets, Logic, Terms, and Factors
1
1.1
Sets and Logic . . . . . . . . . . . . . . . . . . . . . .
1
1.2
Terms and Factors . . . . . . . . . . . . . . . . . . . . .
2
2 Powers
3
3 Simplifying Expressions
5
3.1
Removing Parentheses
. . . . . . . . . . . . . . . . . . .
5
3.2
Factoring Polynomials
. . . . . . . . . . . . . . . . . . .
7
3.3
Factoring Polynomials, continued . . . . . . . . . . . . . . . .
8
4 Fractions
9
5 Trigonometry
11
6 Trigonometric Identities
15
6.1
Basic Formulas . . . . . . . . . . . . . . . . . . . . . .
16
6.2
Factoring . . . . . . . . . . . . . . . . . . . . . . . .
17
6.3
Proofs . . . . . . . . . . . . . . . . . . . . . . . . .
18
7 Differentiation
19
7.1
Derivatives of Elementary Functions . . . . . . . . . . . . . . .
19
7.2
Differentiation Rules . . . . . . . . . . . . . . . . . . . .
20
8 Antiderivatives
23
9 Graphing Functions
25
10 Equations and Inequalities
29
10.1 Polynomial Equations . . . . . . . . . . . . . . . . . . . .
29
10.2 Polynomial Inequalities . . . . . . . . . . . . . . . . . . .
31
10.3 Equations Involving Fractions . . . . . . . . . . . . . . . . .
34
10.4 Inequalities Involving Fractions
. . . . . . . . . . . . . . . .
36
. . . . . . . . . . . . . . . . . . .
38
10.6 Exponential Inequalities . . . . . . . . . . . . . . . . . . .
40
10.7 Logarithmic Equations
. . . . . . . . . . . . . . . . . . .
42
10.8 Logarithmic Inequalities . . . . . . . . . . . . . . . . . . .
44
10.5 Exponential Equations
iii
Contents
10.9 Trigonometric Equations and Inequalities . . . . . . . . . . . . .
46
10.10 Equations Involving Square Roots
. . . . . . . . . . . . . . .
48
10.11 Inequalities Involving Square Roots . . . . . . . . . . . . . . .
50
11 Rationalizing Denominators (extra)
53
12 Partial Fraction Decomposition A (extra)
55
13 Partial Fraction Decomposition B (extra)
57
14 Inverse Trigonometric Functions (extra)
59
15 Answers to the Exercises
61
iv
Chapter 1
Sets, Logic, Terms, and Factors
1.1
Sets and Logic
A set may be defined by enumerating its elements: V = a, b, c, d .
Example: 1, 4, 6, 7 is the set containing the numbers 1, 4, 6, and 7.
If x is an element of the set V we write x ∈ V .
Common sets of interest are: the set of
Natural numbers N = 1, 2, 3, 4, . . .
Integer numbers
Z = . . . , −3, −2, −1, 0, 1, 2, 3, 4, . . .
Rational numbers Q = all numbers that can be written in the form
Real numbers
p
q
with p ∈ Z and q ∈ N
R = all numbers that can be written as an (infinite) decimal expansion
Complex numbers C = all numbers that can be written in the form a + ib with a, b ∈ R
If W is a subset of a set V , we write W ⊂ V .
Hence, N ⊂ Z ⊂ Q ⊂ R ⊂ C.
If W is defined in terms of a given condition B, we write
W = x ∈ V B(x) .
We say: “W consists of all elements x ∈ V such that B(x)”.
Example: the natural numbers which are a multiple of 5:
The union of sets A and B is defined as
A ∪ B = all elements that are in A or in B .
n ∈ N 51 n ∈ N .
The intersection of sets A and B is defined as
A ∩ B = all elements that are in A and in B .
1
1.2 Terms and Factors
Subsets of R can be represented on the real line. A contiguous subset is called an interval. We
use common notation for intervals. Examples:
−3
5
U = x ∈ R −3 < x 6 5 = (−3, 5]
2
V = x ∈Rx >2
= [2, ∞)
−2
W = x ∈ R x 6 −2
= (−∞, −2]
Some intersections and unions of these intervals are
= (−3, ∞)
U ∪ V = x ∈ R x > −3
= [2, 5]
U ∩ V = x ∈ R 2 6 x 6 5}
V ∪ W = x ∈ R x 6 −2 ∨ x > 2 = (−∞, −2] ∪ [2, ∞)
V ∩ W = x ∈ R x 6 −2 ∧ x > 2 = (−∞, −2] ∩ [2, ∞) = ∅
∅ denotes the empty set (the set without any elements).
∨ denotes the logical or:
a ∨ b is true if a is true or b is true (or both).
∧ denotes the logical and:
a ∧ b is true if a is true and b is true.
Note the parallel between ∨ and ∪ , and between ∧ and ∩ .
1.2
Terms and Factors
We distinguish terms and factors. A term is a part of a sum (or difference); a factor is a part of a
product. For example, 3 − a is a sum of the terms 3 and −a; 3a is a product of the factors 3 and
a.
Examples:
• 2a 2 b − 3c consists of two terms, namely 2a 2 b and −3c.
• 2a 2 b consists of three factors: 2, a 2 , and b.
• −3c consists of two factors: −3 and c.
• 2a(3b − 2cd) consists of three factors: 2, a, and (3b − 2cd).
• 3b − 2cd is a sum of which the first term consists of three factors 3 and b; the second term
consists of −2, c, and d.
It is important to realize whether you are dealing with terms or factors since the rules of arithmetic are different!
2
Chapter 2
Powers
The following rules hold, assuming that a > 0 and b > 0:
a p · a q = a p+q ,
(ab) p = a p b p ,
1
ap
−n
,
=
a
= a p−q , (a p )q = a pq ,
an
aq
√
√
√
q
√ √
n
p
an = a 2 ,
aq = a p ,
ab = a b.
For example, using these rules we have:
1
(3a 2 b) 3
1
2a 4 b− 2
1
=
2
1
33 a 3 b3
1
2a 4 b− 2
1
10
5
= 3 3 · 2−1 a − 3 b 6 .
The expression is reduced to a product of numbers and powers of the form
C · a p b q cr · · ·
If x ∈ R then
√
√
3
x 2 = |x| and
x 3 = x,
where |x| denotes the absolute value of x
(
|x| =
Note:
x
if x > 0,
−x
if x < 0.
√
c2 = |c| but x 2 = c2 ⇔ x = c ∨ x = −c.
Powers (exponentials) and logarithms are inverses of each other. For instance:
x
...
2x
...
−6 −5 −4 −3 −2 −1
1
64
1
32
1
16
1
8
1
4
1
2
5
6
...
0
1
2
3
4
1
2
4
8
16 32 64 . . .
log2 y
y
For more properties of logarithms refer to Section 10.7, page 42.
3
2. Powers
Exercises
Reduce the expressions below to the form C · a n bm co . . ..
All variables are positive numbers.
Series A
Series B
1. ( p 4 q 2 )3 · ( p 2 q 5 )2 =
2.
(−a 5 b2 )4
=
(a 3 b)3
1. ( p 3 q 4 )2 · ( p 3 q 2 )4 =
2.
(−2cd 4 )3
=
2(3c2 d)2
√
4. (−3a b)3 =
√
√
5. 2ab2 · 2 a =
p
2 p 3 q2
6. p
=
4
p3 q
(−2c2 d 4 )4
=
−2(3c2 d)3
√
4. (−2ab b)5 =
√
p
5. 2ab3 · 3 ab =
p
2 p 3 q4
=
6. p
p3 q
7. (a 2 b−3 )2 · −3a −7 b−2 =
7. (a −2 b3 )2 · −3a 3 b−2 =
3.
3.
1
1
8. (3a 2 b)− 4 · (6a 3 b2 ) 2 =
1
9.
3a − 3 b2
1
2a 2 b− 3
2
=
9.
=
10.
(2a)− 4
1
2a − 2
3a − 5 b2
1
2a 3 b− 2
=
1
1
10.
1
8. (3a 2 b)− 3 · (6a 3 b2 ) 4 =
2
4
(−a 3 b2 )4
=
(−a 4 b)3
(2a)− 3
1
2a − 2
=
Chapter 3
Simplifying Expressions
3.1
Removing Parentheses
We have:
(a + b)(c + d) = ac + ad + bc + bd.
Special forms of this rule are:
(a + b)2 = a 2 + 2ab + b2 ,
(a − b)2 = a 2 − 2ab + b2 ,
(a − b)(a + b) = a 2 − b2 .
Important for factoring a quadratic polynomial:
(x + a)(x + b) = x 2 + (a + b)x + ab.
Example:
• (x + 3)(x − 7) = x 2 − 4x − 21, where −4 is obtained as +3 plus −7 , and −21 as +3
times −7 .
√
√
√
• (3a√− 2 b)2 = 9a 2 − 12a b + 4b, where −12a b is twice the product of 3a and
−2 b .
We also need to reduce expressions involving square roots.
Examples:
√
√
√
√
72 = 36 · 2 = 6 2 (factor 72 into a “square” times a number,
√
√
3
3 7
3√
• √ =
7 (multiply both the numerator and denominator with 7).
=
7
7
7
•
5
3.1 Removing Parentheses
Exercises
Reduce the expressions below using the above rules. Make sure to remove all square roots from
the denominators. Reduce square roots as much as possible.
Series A
1. (3a − b)2 =
1. (−3a + 2b)2 =
2. (−2a 2 + 3a)2 =
√
√
3. (3 6 − 6 3)2 =
2. (2a 3 − 3a)2 =
√
√
3. (3 15 − 2 3)2 =
4. (m − 2n)(3m + n) =
√ √
5. (6 − 2 3)( 3 + 2) =
4. (2m − 3n)(3m + 2n) =
√ √
5. (1 − 2 5)( 5 + 2) =
6. (−3a 2 b3 + 31 a 4 b)2 =
√
√
√
√
7. (−2a 3 + 21)(2a 3 + 21) =
6. (−2a 4 b3 + 41 a 2 b)2 =
√
√
√
√
7. (−2a 3 + 30)(2a 3 + 30) =
8. (2a − b + 1)(3a + 2b − 5) =
2
9. √220 + √35 =
8. (2a − 3b + 1)(3a + 2b − 4) =
2
9. √220 + √345 =
10. (3a − 1)(3a + 1)(9a 2 + 1) =
6
Series B
10. (2a − 1)(2a + 1)(4a 2 + 1) =
3.2 Factoring Polynomials
3.2
Factoring Polynomials
When factoring a polynomial the goal is to write the polynomial as a product of as many factors
as possible. To do so, first move as many common factors as possible outside the parentheses,
and then try to factor the remaining part.
Examples:
• x 2 + 7x − 44 = x 2 + 11x − 4x − 4 · 11 = (x + 11)(x − 4). Look for two numbers with sum
+7 and product −44.
• 2x 2 + 4x − 6 = 2(x 2 + 2x − 3) = 2(x 2 + 3x − x − 1 · 3) = 2(x + 3)(x − 1). Look for two
numbers with sum +2 and product −3.
• x 3 − 3x 2 − 28x = x(x 2 − 3x − 28) = x(x + 4)(x − 7)
parentheses.
So, first move x outside the
• x 4 − 16 = (x 2 + 4)(x 2 − 4) = (x 2 + 4)(x + 2)(x − 2) Here we have the form a 2 − b2 twice.
Note: a 2 + b2 cannot be factored any further!
• x(x +2)−3(x +2) = (x +2)(x −3) Here you take (x +2) as factor outside the parentheses.
• −x 2 − x + 2 = −(x 2 + x − 2) = −(x + 2)(x − 1)
parentheses.
Here you first take −1 outside the
Exercises
Factor the following polynomials (using factors with integer coefficients only).
Series A.
Series B.
1. 16x 4 − 81 =
1. 81x 4 − 16 =
2. 3x 5 − 12x 4 − 63x 3 =
2. 3x 4 − 15x 3 + 12x 2 =
3. x 16 − 1 =
3. x 12 − 16 =
4. x 4 + x 2 − 6 =
4. x 4 − x 2 − 20 =
5. x 2 − 19x + 34 =
5. x 2 − 21x + 38 =
6. x 2 − 15x − 34 =
6. x 2 − 17x − 38 =
7. x(x − 1) − (x − 1) =
7. 2x(x + 1) + 2(x + 1) =
8. x(x 2 − 1) + (x − 1) =
8. x(x 2 − 1) − (x − 1) =
9. (3x − 2)2 − (2x + 3)2 =
9. (5x − 3)2 − (3x + 5)2 =
10. x 6 − 6x 3 + 9 =
10. x 10 + 8x 5 + 16 =
7
3.3 Factoring Polynomials, continued
3.3
Factoring Polynomials, continued
The exercises below are also about factoring polynomials.
Examples:
• 6x 3 − 18x 2 − x + 3 = 6x 2 (x − 3) − (x − 3) = (6x 2 − 1)(x − 3). Here, it turns out that if you
take a factor outside parentheses for two terms at a time, these factors happen to coincide,
and the polynomial can be split into two factors.
• 2x 2 + x − 10. In cases like this, sometimes two number a and b can be found such that
2x 2 +x −10 = (2x +a)(x +b). Here, a times b must be equal to −10 and 2b+a must be
equal to +1 . This is satisfied by a = 5 and b = −2, hence: 2x 2 + x −10 = (2x +5)(x −2).
Exercises
Factor the following polynomials (using factors with integer coefficients only).
Series A.
1. 3x 2 − 20x + 12 =
1. 3x 2 − 14x + 15 =
2. 2x 2 + 7x + 6 =
2. 2x 2 + 9x + 9 =
3. 3x 4 − 11x 2 + 6 =
3. 3x 4 − 13x 2 + 12 =
4. −2x 2 + 7x + 15 =
4. −2x 2 + x + 21 =
5. 2x 4 − x 2 − 3 =
5. 2x 4 + 2x 2 − 4 =
6. x 3 − 4x 2 − x + 4 =
6. x 3 − 8x 2 − x + 8 =
7. 2x 3 − 6x 2 + x − 3 =
7. 3x 3 − 12x 2 + 2x − 8 =
8. x 3 + 5x 2 − 4x − 20 =
8. 3x 3 + 15x 2 − 4x − 20 =
9. −3x 3 + 6x 2 + 2x − 4 =
9. −2x 3 + 4x 2 + 3x − 6 =
10. x 7 + 2x 4 − 15x =
8
Series B
10. x 8 − 4x 5 − 12x 2 =
Chapter 4
Fractions
numerator
, does not exist if the denominator is equal to 0, and it is indeterminate
denominator
if both the numerator and the denominator are equal to 0. Assume throughout this chapter that
denominators are 6= 0. We have the following rules:
A fraction,
ab
b
=
ac
c
a
c
ad
bc
ad + bc
+ =
+
=
b d
bd
bd
bd
a c
ac
· =
b d
bd
a d
ad
c
d
a c
: = · =
So, to divide by a fraction , multiply by its reciprocal .
b d
b c
bc
d
c
Simplify fractions by factoring its numerator and denominator.
Examples:
•
•
•
a 2 + ab
ab + b2
=
b−a
a−b
=
a(a + b)
b(a + b)
=
−(a − b)
a−b
= −1
a
b
a 2 − b2
(a − b)(a + b)
a−b
=
=
2
2
2
a + 2ab + b
(a + b)
a+b
For the exercises below you need to write your answer as a single fraction. This is called “finding
a common denominator”.
Examples:
•
1
2
2x − 3
2(x − 1)
2x − 3 − 2(x − 1)
−
=
−
=
=
x − 1 2x − 3
(x − 1)(2x − 3) (x − 1)(2x − 3)
(x − 1)(2x − 3)
1
−
(x − 1)(2x − 3)
9
4. Fractions
•
1
3
x
3(x − 2)2
x − 3(x 2 − 4x + 4)
−
=
−
=
=
(x − 2)3
x(x − 2)
x(x − 2)3
x(x − 2)3
x(x − 2)3
−3x 2 + 13x − 12
x − 3x 2 + 12x − 12
=
x(x − 2)3
x(x − 2)3
Remarks on the above example:
- Use the least common denominator, as this simplifies calculations considerably. Compare
1
8
1
17
this to 24
+ 12
= 24
+ 16
= 24
, where using 24 · 12 as denominator instead of 24 would also
24
complicate calculations.
- Make sure to carefully multiply −3 with all terms of x 2 − 4x + 4 between the parentheses.
- Usually, parentheses do not need to be removed from the denominator. Sometimes, the
resulting fraction can be reduced further by also factoring the numerator.
Exercises
Series A.
1.
3
x
+
=
2x − 1 x + 1
2. √
3.
1
1
+√
=
x −1
x +1
3
1
+
=
x −1 x +3
1.
2
2x
−
=
2x − 1 x + 1
1
1
+√
=
2. √
2x − 3
2x + 3
3.
x2
x
2
+
=
− 3 2x + 3
4.
−3
6x
−
=
1−x
x +3
4.
−3
6x
+
=
1−x
x +2
5.
13
5
−
=
2x + 3 x + 1
5.
7
5
− 2
=
2x + 3 x + 1
6.
x
3
−
=
x − 2 2x − 4
6.
x
3
−
=
x + 3 2x + 6
7. −
x2
x
2
+
=
− 3 2x + 1
7. −
x
2x
+
=
x − 3 2x + 1
8.
1
2x
−
x + 1 (x + 1)3
8.
1
2x
+
x − 1 (x − 1)3
9.
1
1
1
+
=
+
2
x − 1 (x − 1)
x +1
9.
1
1
1
+ 2
+
=
x − 1 (x − 1) x + 1
10. 1 −
10
Series B.
1
3
+
=
2x
x +3
10. 1 −
2
3
+
=
x
x −3
Chapter 5
Trigonometry
The trigonometric functions sine, cosine, and tangent are introduced in terms of ratios of sides
in right-angled triangles. This covers the case of angles between 0◦ and 90◦ . We have: tangent(x)
= sine(x) / cosine(x). A natural generalization to arbitrary angles is obtained by using the unit
circle.
We define:
A point P on the unit circle has x-coordinate cos(α) and y-coordinate sin(α), so P =
(cos(α), sin(α)), or in a slightly different notation P = (cos α, sin α).
y
1
P : Hcos Α, sin ΑL
Α
O
1
x
We then have in general:
tan α =
sin α
cos α
and
sin2 α + cos2 α = 1.
Note that the second identity corresponds to Pythagoras’ theorem.
The angle α is commonly measured in radians. In this case α is equal to the length of the
arc on the unit circle corresponding to the angle α. For instance, 2π rad corresponds to 360◦ .
Throughout this syllabus, we measure angles in radians.
11
5. Trigonometry
Exact values of the sine, cosine, and tangent can be given for some special angles. These values
can be found using the triangles depicted below. The table list these values for angles between 0
and 12 π . It is strongly recommended that you memorize this table.
1
π
4
1
π
6
2
√
3
1
2
1
π
4
1
π
2
1
π
2
1
π
3
√
x
0
1
π
6
sin x
0
1
2
cos x
1
tan x
0
√
1
3
2
√
1
3
3
1
π
4
√
2
√
1
2
2
1
2
1
1
π
3
1
2
√
3
1
2
√
3
1
π
2
1
0
und.
1
1
Note that tan x is undefined (und.) for x = 21 π.
For angles exceeding 12 π the corresponding trigonometric function values can be derived directly
from the definitions.
Example:
• The coordinates of point Q which corresponds to an angle 76 π can be derived from the
coordinates of point P corresponding to an angle of 16 π.
1
P
7Π
€€€€€€€€€€
6
Π
€€€€€
6
O
Q
Therefore
sin 67 π = − sin 16 π = − 21
√
cos 76 π = − cos 16 π = − 21 3
tan 67 π =
sin 76 π
cos 76 π
=
√
− 12
1
=
3
√
3
− 12 3
We call this “reducing an angle to the first quadrant”.
12
1
5. Trigonometry
Another type of question goes like this:
√
Given: sin α = − 12 3. How large is α, under the restriction that 0 6 α < 2π?
1
0
1
1 !!!
- €€€€€ 3
2
Use the unit circle
By √
definition, sin α is equal to the y-coordinate of a point on the unit circle. Since
− 21 3 is negative, we know that the y-coordinate is negative. The y-coordinate is
negative in the third and fourth quadrant, so that is where angle α is to be found.
Recall that:
√
sin 31 π = 21 3.
√
From the figure and the above table one can readily see that sin α = − 12 3 corresponds to angles
π + 13 π and 2π − 13 π.
Thus the answer is: α = 43 π ∨ α = 53 π .
13
5. Trigonometry
Exercises
Find the values of the following trigonometric expressions. Reduce angles to the first quadrant
and use the above table.
Series A
Series B
1. sin( 43 π ) =
1. cos( 43 π ) =
2. tan(− 14 π ) =
2. sin(− 41 π ) =
3. cos( 56 π ) =
3. tan( 56 π ) =
4. sin( 23 π ) =
4. cos( 32 π ) =
π) =
5. tan( 85
4
π) =
5. sin(− 85
4
6. sin( 32 π ) =
6. tan( 32 π ) =
7. cos(− 74 π ) =
7. sin(− 47 π ) =
8. tan( 65 π ) =
8. cos(− 56 π ) =
9. cos( 23 π ) =
9. tan( 32 π ) =
π) =
10. sin( 37
4
π) =
10. sin( 34
3
14
Chapter 6
Trigonometric Identities
The following identities can easily be found using the definition in terms of the unit circle:
sin2 x + cos2 x = 1,
sin(−x)
= − sin x,
sin(π − x) = sin x,
sin x
=
cos( 12 π
cos(−x)
− x),
= cos x,
tan(−x)
cos(π − x) = − cos x,
cos x
=
sin( 12 π
− x)
= − tan x
tan(π − x) = − tan x
The following formulas are given without proof. These formulas are strongly connected. For
instance, if you start from one formula you can find the other ones by using the simple identities
above:
sin(x + y) = sin x cos y + cos x sin y
sin(x − y) = sin x cos y − cos x sin y
cos(x + y) = cos x cos y − sin x sin y
cos(x − y) = cos x cos y + sin x sin y
tan x + tan y
tan(x + y) =
1 − tan x tan y
tan x − tan y
tan(x − y) =
1 + tan x tan y
sin 2x = 2 sin x cos x
cos 2x = cos2 x − sin2 x = 2 cos2 x − 1 = 1 − 2 sin2 x
2 tan x
tan 2x =
1 − tan2 x
15
6.1 Basic Formulas
6.1
Basic Formulas
Exercises
1. Derive the formula for sin(x + y) from the formulas for sin(x − y), cos(x + y), and cos(x − y).
2. Derive:
tan(x + y) =
tan x + tan y
,
1 − tan x tan y
tan(x − y) =
tan x − tan y
.
1 + tan x tan y
and
3. Derive from the formulas for sin(x + y), cos(x + y), and tan(x + y):
• sin 2x = 2 sin x cos x
• cos 2x = cos2 x − sin2 x = 2 cos2 x − 1 = 1 − 2 sin2 x
2 tan x
• tan 2x =
1 − tan2 x
4. For an angle x ∈ [0, 12 π] we have:
cos x =
1
2
√
2.
Find cos(x − 61 π ).
5. For an angle x ∈ [ 12 π, π] we have:
cos 2x = 13 .
Find sin x.
6. For an angle x ∈ [0, 12 π] we have:
cos x = 43 .
Find tan 2x.
7. Simplify as much as possible
sin2 x cos2 x + cos2 x + sin4 x.
8. Simplify as much as possible
2(cos2 x − sin2 x)2 tan(2x).
16
6.2 Factoring
6.2
Factoring
Exercises
Use the formulas at the start of this chapter:
Series A
Series B
1. Factor: sin x + sin 2x =
1. Factor: 2 sin2 x − sin 2x =
2. Factor:
2. Factor: cos 2x − cos2 x =
sin(x + y) + sin(x − y) =
3. Factor: 1 − sin 2x − sin2 x =
4. Factor: cos 2x + 7 =
3. Factor: cos 2x + sin2 x =
4. Factor: 1 + sin 2x − cos2 x =
5. Factor: sin2 x − sin x − 6 =
5. Factor: cos 2x − 1 =
6. Factor: cos2 x + 3 sin x + 9 =
r
6. Factor: sin2 x − 5 sin x + 4 =
7. Eliminate the square root:
7. Factor: sin2 x + 5 cos x + 5 =
8. Find the exact value of sin 18 π.
8. Find the exact value of cos 18 π .
9. Simplify
2
9. Simplify
2
cos x
cos x
−
.
1 − sin x
1 + sin x
10. f (x) = 1 − cos 2x − sin2 x can be written
as f (x) = a + b cos(cx).
Determine a, b en c.
1 − cos x
=
1 + cos x
cos4 x − 2 cos2 x sin2 x + sin4 x.
10. f (x) = 3 cos2 x − sin2 x can be written as
f (x) = a + b cos(cx).
Determine a, b en c.
17
6.3 Proofs
6.3
Proofs
Exercises
Use the formulas at the start of this chapter:
Series A
Series B
1. Show that cos2 12 x =
1
2
+ 12 cos x.
2. Show that cos4 x − sin4 x = cos 2x.
3. Show that cos4 x + 12 sin2 2x + sin4 x = 1.
4. Show that cos2 x(1 + tan2 x) = 1.
sin 2x
5. Show that tan x =
.
1 + cos 2x
6. Show that
tan x + tan y
sin(x + y)
=
.
tan x − tan y
sin(x − y)
7. Show that
tan2 x − sin2 x = tan2 x · sin2 x.
8. Show that
tan x − 1
= sin2 x − cos2 x.
tan2 x + 1
2
9. Show that
tan( 14 π + x) =
1 + tan x
.
1 − tan x
10. Show that 2 cos2 x − cos 2x = 1.
18
1. Show that sin2 21 x =
1
2
− 12 cos x.
2. Show that cos4 x(1 − tan4 x) = cos 2x.
3. Show that 4 sin2 x − 4 sin4 x = sin2 2x.
4. Show that
cos4 x(1 + tan4 x) = 1 − 2 sin2 x cos2 x.
5. Show that tan x =
1 − cos 2x
.
sin 2x
6. Show that
1 + tan x tan y
cos(x − y)
=
.
1 − tan x tan y
cos(x + y)
7. Show that sin 2x − tan x = tan x cos 2x.
8. Show that
sin 2x
1 − cos 2x
=
.
1 + cos 2x
sin 2x
9. Show that
tan( 14 π +x)+tan( 41 π −x) =
cos2
2
.
x − sin2 x
10. Show that
2
cos x − sin x cos x + sin x
+
=
.
cos x + sin x cos x − sin x
cos 2x
Chapter 7
Differentiation
If a function f assigns the value f (x) to x, we also call f (x) the mapping rule. The graph of f
is the curve in the (x, y)-plane given by the equation y = f (x). We will often use a function f ,
a mapping rule f (x), and the equation y = f (x) interchangeably, even though this is sloppy use
of terminology.
Under certain conditions we can take the derivative of a function f (x). In terms of the graph
of f , the derivative f 0 (x) is equal to the slope of the tangent at x to the graph y = f (x). The
derivative of y = f (x) with respect to x is denoted as
f 0 (x) or
7.1
d
f (x) or
dx
d f (x)
dx
or
y0
or
dy
.
dx
Derivatives of Elementary Functions
d n
x
dx
d x
e
dx
d
ln x
dx
d x
a
dx
d
sin x
dx
d
cos x
dx
d
tan x
dx
= nx n−1 ,
= ex ,
=
1
,
x
= a x ln a,
= cos x,
= − sin x,
=
1
= 1 + tan2 x,
cos2 x
19
7.2 Differentiation Rules
7.2
Differentiation Rules
We use the following rules to calculate derivatives:
f (x) = u(x) + v(x)
f (x) = cu(x)
f (x) = u(x)v(x)
t (x)
f (x) =
n(x)
⇒ f 0 = u 0 + v0
0
⇒ f = cu
0
0
(Sum Rule)
0
(Rule for Constant Multiples)
0
⇒ f =uv+vu
nt 0 − tn 0
⇒ f0 =
n2
(Product Rule)
(Quotient Rule)
Given functions f (u) and u(x), the derivative with respect to x of the composite function
y = f (u(x)) is equal to the derivative of f with respect to u multiplied by the derivative of f
with respect to u:
dy
dy du
=
·
dx
du dx
(Chain Rule)
Examples:
• Calculate the derivative of the function y = 6(3x 2 − 2)2 .
Define: u(x) = 3x 2 − 2. Then we need to differentiate y = 6u 2 with respect to u and
multiply the result by the derivative of u = 3x 2 − 2 with respect to x.
We have:
dy
du
= 12u and
du
dx
= 6x. So
dy
dy du
=
·
= 12u · 6x = 12(3x 2 − 2) · 6x.
dx
du dx
Hence: y 0 = 72x(3x 2 − 2)
• y = 3(x 2 − x)5 ⇒ y 0 = 15(x 2 − x)4 (2x − 1).
Useful to memorize:
y = u n ⇒ y 0 = nu n−1 · u 0
1
1
y=
⇒ y0 = − 2 · u0
u
u
√
1
0
y = u ⇒ y = √ · u0
2 u
1
y = ln u ⇒ y 0 = · u 0
u
20
7.2 Differentiation Rules
Examples:
• y = 4(3x 2 − 1)2 ⇒ y 0 = 8(3x 2 − 1) · 6x = 48x(3x 2 − 1)
• y=
√
1
3
6x − 1 ⇒ y 0 = √
·6= √
2 6x − 1
6x − 1
• y=
3
3
18x 2
0
2
⇒
y
=
−
·
6x
=
−
2x 3 − 1
(2x 3 − 1)2
(2x 3 − 1)2
• y = ln(1 − x) ⇒ y 0 =
• y = 3x
2 −5
⇒ y 0 = 3x
1
1
1
· −1 = −
=
1−x
1−x
x −1
2 −5
· ln 3 · 2x = 2x · ln 3 · 3x
2 −5
• y = (2 sin2 x − 1)3 ⇒ y 0 = 3(2 sin2 x − 1)2 · 4 sin x · cos x = 12 sin x cos x · (2 sin2 x − 1)2
21
7.2 Differentiation Rules
Exercises
Find the derivatives of:
Series A
Series B
1. y = −3(1 − 2x)5
2. y = √
x
x2
−1
3. y =
3
5 − x2
4. y =
3
(3x − 1)3
5. y = 6 · 32x−1
6. y = 2 ex
2 −1
7. y = x ln(3x + 4)
x 8. y = ln
x +1
9. y = x 3 · 22x
√
10. y = 3x 3x − 1
11. y =
ln x
1 − ln x
12. y = (x 2 − 3x) · ex
13. y = tan3 x
14. y =
sin x − cos x
sin x + cos x
15. y =
sin x
cos2 x − 1
22
1. y = 2x ln 3x
ln x
x
√
3. y = sin 2x
2. y =
1
4. y = √
3x 2 + 1
5. y =
x 2 − 7x − 8
x2 − 1
6. y = sin2 (2x − 61 π )
ln2 x
sin x
x
8. y = √
2
x +1
√
9. y = ln( 3x − 1)
7. y =
2
10. y = e2 sin
11. y =
x−1
ex +1
ex
12. y = ln4 x
13. y = sin2 x · cos x
sin x
1 − cos x
cos x
15. y =
cos2 x − 1
14. y =
Chapter 8
Antiderivatives
Antiderivatives play an essential role in integral calculus. Taking antiderivatives is the reverse
operation of taking derivatives.
Knowledge of differentiation is therefore required.
Elementary formulas (with a 6= 0, n 6= −1, and
c a constant of integration):
Z
a
ax n dx
=
x n+1 + c
n+1
Z
ax −1 dx
= a ln |x| + c
Z
1
eax dx
= eax +c
a
Z
1
sin(ax) dx = − cos(ax) + c
a
Z
1
cos(ax) dx = sin(ax) + c
a
Z
1
1
(ax + b)n dx = ·
(ax + b)n+1 + c
a n+1
Z
1
1
dx = ln |ax + b| + c
ax + b
a
Z
1
dx
= tan x + c
cos2 x
Always make sure to check the antiderivative by differentiating it.
23
8. Antiderivatives
Exercises
Find the antiderivatives of the following functions.
Series A
Series B
1. (2x − 1)3
1. (3x + 2)3
2. (5 − x)2
√
3. 3x − 4
2. (8 − 2x)2
√
3. 2x − 3
4.
1
(2x + 3)4
5. sin 2(x − 16 π )
6. √
7.
x +3
x2 + 1
x
1
(2x − 1)5
5. cos 12 (x − 31 π )
6. √
7.
2
x −5
x3 − 1
x
8. sin x + e3x
8. cos 2x + e2x
9. tan2 x
9. tan2 x + 2
10.
24
1
4.
2
3 − 2x
10.
3
2 − 3x
Chapter 9
Graphing Functions
The goal of the exercises below is to sketch the graphs of the given functions without using any
electronic tools.
The scales used for the x-axis and the y-axis do not have to be the same.
Each time choose the domain such that the properties of the graph are clearly visible, such as
points of intersection with the axes, asymptotes, periods, and so on.
Also put values next to the points of intersection and asymptotes if these values can be calculated
easily.
From the graph of y = f (x) one can obtain the graphs of related functions by scaling the axes
and by horizontal or vertical shifts.
the graph of:
a · f (x) is obtained from the graph of f (x) by multiplying the y-value by a
f (x/b)
„
multiplying the x-value by b
f (x − c)
„
shift it c units to the right
f (x) + d
„
shift it d units upward
Note the order in which these steps are applied:
If we want to construct the graph of
y = a· f
x − c
+d
b
from the graph of y = f (x), then we have apply this order:
first scale, then shift.
25
9. Graphing Functions
Elementary Functions
4
3
2
2
2
y=x
1
1
0
0
−2
−1
0
1
y = |x|
1.5
2
1
y = √x
0.5
0
0
2
1
2
3
4
5
−0.5
−2
−1
0
1
2
5
2
4
2
y = exp(x)
3
1
1
2
0
0
1
y = x3
−1
y = ln(x)
−1
0
−2
0
−2
2
0
1
2
3
4
5
−2
−2
y = sin(x)
−1
0
1
y = cos(x )
1
1
0
0
−1
−1
0
0.5π
π
1.5π
2π
2.5π
3
0
0.5π
1
π
1.5π
2π
5
2
y = tan(x)
y = 1/x
0
0
−1
−2
−3
26
2
−2
0
2
−5
−π 0.5π
0 0.5π π 1.5π 2π
2.5π
9. Graphing Functions
The graph of y = tan(x) (see previous page) has the following properties:
• periodic with period π;
• zeros at x = kπ,
k ∈ Z;
• asymptotes at x = 12 π + kπ,
k ∈ Z.
Example of scaling and shifting:
• The graph of y = tan 2(x − 14 π ) is obtained from the graph of y = tan x by shrinking the
x-value by a factor of 2 (the period changes from π to 21 π) and shifting it by 14 π to the right.
2
1
0
−1
−2
− 34 π
− 21 π
− 41 π
0
1
π
4
1
π
2
3
π
4
27
9. Graphing Functions
Exercises
Series A
Series B
1. f (x) = (x + 2)4
1. f (x) =
1
x
2. f (x) = (x − 2)2 + 3
2. f (x) =
4
x −2
3. f (x) = −x 3 + 8
3. f (x) = 3 +
4. f (x) = x 2 + 6x + 9
5. f (x) = 6(x + 1)5
Series C
4. f (x) =
3
x2
5. f (x) =
4
(2 − x)2
Series D
1. f (x) = 2x
1. f (x) = log2 x
2. f (x) = ( 12 )x + 3
2. f (x) = log2 (x + 2)
3. f (x) = 2x−2 + 1
3. f (x) = log x 2
4. f (x) = 3−x − 1
4. f (x) = 2 log x
5. f (x) = ex−1
5. f (x) = ln(x − e)
Series F
Series E
2.
3.
4.
5.
√
x +2
√
f (x) = 4 − 2 x
√
f (x) = 2 + x + 4
√
f (x) = 3 x
√
6
f (x) = x 2 − 1
1. f (x) =
28
1
x −3
1. f (x) = sin 3x
2. f (x) = 2 cos π x
3. f (x) = 12 + 8 sin 31 π(x − 1)
4. f (x) = sin2 x
5. f (x) = tan 12 (x + 14 π )
Chapter 10
Equations and Inequalities
10.1
Polynomial Equations
The following rules apply when solving equations:
⇔
⇔
ax = −b
• A · B · C · .... = 0
⇔
A = 0 ∨ B = 0 ∨ C = 0 ∨ ....
• A2 = B 2
⇔
A = B ∨ A = −B
• A· B = A·C
⇔
A=0 ∨ B=C
• Quadratic equation: ax + bx + c = 0 ⇔ x =
2
x =−
b
a
• ax + b = 0
−b ±
√
b2 − 4ac
(quadratic formula)
2a
We do not need to use the quadratic formula in simple cases such as
ax 2 + c = 0,
ax 2 + bx = 0,
x 2 + 2bx + b2 = 0,
or when the expression can easily be factored:
2x 2 +4x −6 = 2(x 2 +2x −3) = 2(x 2 −x +3x −3·1) = 2(x −1)(x +3) = 0 ⇔ x = −3 ∨ x = 1.
Also completing the square1 may be a useful option:
√
√
x 2 + 2x − 4 = (x 2 + 2x + 1) − 5 = (x + 1)2 − 5 = 0 ⇔ x + 1 = ± 5 ⇔ x = −1 ± 5.
In general, polynomial equations can be written in the form (with n a positive integer):
an x n + an−1 x n−1 + an−2 x n−2 + . . . + a1 x + a0 = 0.
Solutions may sometimes be found by factoring the polynomial.
1 Note: the quadratic formula can be derived by completing the square:
h
i
b x + ( b )2 − b 2 + c = a x + b 2 − b2 −4ac = 0.
ax 2 + bx + c = a x 2 + ab x + ac = a x 2 + 2 2a
a
2a
2a
2a
(2a)2
29
10.1 Polynomial Equations
Examples:
• Solve: x 5 − 4x 3 − 27x 2 + 108 = 0
x 3 (x 2 − 4) − 27(x 2 − 4) = 0
(x 3 − 27)(x 2 − 4) = 0
(x 3 − 27)(x − 2)(x + 2) = 0
x = 3 ∨ x = 2 ∨ x = −2
• Solve: (x 2 − 14)(x + 4) = 5x(x + 4):
(x 2 − 14)(x + 4) − 5x(x + 4) = 0
(x 2 − 14 − 5x)(x + 4) = 0
(x + 2)(x − 7)(x − 4) = 0
x = −2 ∨ x = 7 ∨ x = −4
• Solve: (x − 4)(x 2 + 8x + 12) = (2x + 3)(x 2 + 8x + 12):
x 2 + 8x + 12 = 0 ∨ x − 4 = 2x + 3
(x + 6)(x + 2) = 0 ∨ − x = 7
x = −6 ∨ x = −2 ∨ x = −7
Exercises
Solve the following equations:
Series A
1. 2x 2 + 7x − 4 = 0
1. 3x 2 + 7x − 6 = 0
2. x 4 + 6 = 7x 2
2. x 4 = 2x 2 + 24
3. x 3 + 6x = 7x 2
3. x 4 − 24x 2 = 10x 3
4. x 4 − 42 = x 2
4. x 4 − 12 = x 2
5. x 4 − 39x 2 = 10x 3
5. x 4 − 33x 2 = 8x 3
6. x 3 − 3x 2 = (x − 3)(x + 20)
6. x 3 + x 2 = (x + 1)(x + 2)
7. (x − 2)3 = x − 2
7. 3x 2 + 4x − 4 = 0
8. 3x 3 − x 2 − 12x + 4 = 0
8. x 3 − 3x 2 = (x − 3)(x + 12)
9. (x 2 − 4)(x + 3) = (x − 2)(4 − x 2 )
9. (x 2 − 4)(x − 3) = (x + 2)(x − 3)
10. x 6 − 4x 4 = 4x 2 − 16
30
Series B
10. 3(x − 1)2 (x + 1) = (x + 1)2
10.2 Polynomial Inequalities
10.2
Polynomial Inequalities
Basic rules (assume a > 0)
• A6B
• A2 6 a
A2 > a
⇔
A−C 6 B −C
⇔
aA 6 aB
⇔
√
√
− a6A6 a
√
√
A6− a ∨ A> a
⇔
⇔
−a A > −a B
(inequality is reversed)
Analogously for > and <.
Examples:
4
3
as dividing or multiplying by a negative number reverses the inequality. Note:
−3x + 8 < 4
0<x <y
x <y<0
x <0<y
⇔
⇔
⇔
⇔
− 3x < −4
0<
1
y
1
x
<
1
y
1
x
<
⇔
x>
1
x
<0
<0<
1
y
In all these cases we multiply by (x y)−1 . In the first two cases this is positive, while in the last
case it is negative hence the inequality is reversed.
For a quadratic inequality the solution set may consist of one or two intervals:
√
√
x2 6 5 ⇔ − 5 6 x 6 5
√
√
x2 > 5 ⇔ x < − 5 ∨ x > 5
If the graph can be sketched quickly, one can also first solve the equation and then read out the
solution to the inequality from the graph. Also take the domain into account.
In more advanced cases of polynomial inequalities a sign chart is a useful tool. By means of +
and − signs on the real line, sign charts show where a function is positive or negative. On the
real line, we also indicate where the function is equal to zero. Sign changes only take place at the
zeros on the real line.
Two example sign charts:
• f (x) = (x + 3)(x − 1)(x − 2) has sign chart:
−
0
−3
+
0
1
− +
0
2
Evaluate the value of f (x) at simple values for x (other than the zeros). If the function value
is positive (or negative) then the same holds everywhere between the neighboring zeros. At
each zero on the real line, the sign changes.
31
10.2 Polynomial Inequalities
• g(x) = (x + 3)3 (x − 1)(x − 2)2 has sign chart:
+
−
000
0
−3
+ +
00
1
2
At −3 on the real line 0 is written three times to represent the solution to
(x + 3)3 = 0, that is (x + 3)(x + 3)(x + 3) = 0. We call this a triple zero (or root). This also
means that the sign changes three times, since the sign changes once for each zero. So, if
g(x) is positive for x < 3 then g(x) will be negative for x > −3.
At 2 on the real line 0 is written twice as (x − 2)2 can be written as (x − 2)(x − 2). Therefore
the sign changes twice, which amounts to no change in sign at 2.
Hence, at a zero of multiplicity n there is no change in sign if n is even, and the sign
changes if n is odd.
Examples
• Solve: x 3 + 8x 6 6x 2 .
First rewrite as: x 3 − 6x 2 + 8x 6 0. Next factor the left-hand side:
x(x 2 − 6x + 8) = 0
x(x − 2)(x − 4) = 0
−
0
Sign chart:
+
0
0
−
2
0
+
4
Hence the solution is: x 6 0 ∨ 2 6 x 6 4
or, in terms of intervals: (−∞, 0] ∪ [2, 4]
• Solve x(x + 3)3 (x − 2)2 > (x + 3)3 (x − 2)2
x(x + 3)3 (x − 2)2 − 1(x + 3)3 (x − 2)2 > 0
(x − 1)(x + 3)3 (x − 2)2 = 0
+
Sign chart:
Hence the solution is: x 6 −3 ∨ x > 1
000
−3
or, in terms of intervals: (−∞, −3] ∪ [1, ∞)
32
−
0
1
+ +
00
2
10.2 Polynomial Inequalities
Exercises
Solve the following inequalities:
Series A
Series B
1. x 2 6 25
1. x 2 > 16
2. (x − 3)2 > 1
2. 9x 2 6 16
3. (x − 2)2 6
3. (x − 2)2 >
1
4
1
9
4. 8 < x 2 + 2x
4. 15 < x 2 + 2x
5. 3(x + 1) − 2(2x + 3) > −(x − 2)
5. 3(x − 1) − 2(2x + 3) > 5(x − 2)
6. 35 < x 2 + 2x
6. x 2 − 1 > 9 − 3x
7. (x 2 − 4)(x − 4)2 6 0
7. (x 2 − 9)(x − 3)2 6 0
8. 3(x + 1) − 2(2x + 3) > 5(x − 2)
8. x 2 (3x − 5) − (2x + 3)(3x − 5) > 0
9. (x 2 − 7x + 12)(x 2 + 2x − 24) 6 0
9. (x − 1)(x − 2)2 (x − 3)3 > 0
10. x 6 − 9x 3 + 8 6 0
10. 3x 3 − 12x > x 4 − 4x 2
33
10.3 Equations Involving Fractions
10.3
Equations Involving Fractions
The following rules apply when solving equations:
•
A
=0
B
⇔
A = 0 ∧ B 6= 0.
•
A
C
=
B
D
⇔
A · D = B · C ∧ B 6= 0 ∧ D 6= 0
(cross-multiplication)
Cross-multiplication and finding common denominators are important techniques when solving
equations involving fractions.
Make sure to check the solutions found, as the denominators should all be different from 0!
Examples:
•
2x + 8
x +5
=
3x − 6
5
Cross-multiplication yields:
(3x − 6)(x + 5) = 5(2x + 8)
3x 2 + 9x − 30 = 10x + 40
3x 2 − x − 70 = 0
3x(x − 5) + 14(x − 5) = 0
(3x + 14)(x − 5) = 0
∨ x = 5 (all solutions are valid)
x = − 14
3
•
x 2 − 3x + 2
=5
x −2
We obtain:
x 2 − 3x + 2 = 5x − 10
x 2 − 8x + 12 = 0
(x − 2)(x − 6) = 0
x = 2 ∨ x = 6 (solution x = 2 is invalid)
x =6
34
10.3 Equations Involving Fractions
•
6
5
+
=3
x −1 x +1
We obtain:
6(x + 1) + 5(x − 1)
=3
(x − 1)(x + 1)
11x + 1
=3
x2 − 1
11x + 1 = 3x 2 − 3
3x 2 − 11x − 4 = 0
3x 2 − 12x + x − 4 = 0
3x(x − 4) + 1(x − 4) = 0
(x − 4)(3x + 1) = 0
x = − 13 ∨ x = 4 (all solutions are valid)
Exercises
Solve the following equations:
Series A
1.
2.
3
2
=
x +1
x +2
x −3
x
−3=
x −1
x +2
Series B
1.
2.
x 2 + 3x − 2
=4
x +1
3x 2 + 5x − 2
=2
x 2 + 3x + 2
3.
x −3
x −1
−2=
x −1
x −3
3.
4.
x 3 − 4x 2
=3
x −4
4.
5.
1
1
−2=
x2
x
3x 3 + 6x
=2
x2 + 2
5.
4
4
+ =3
x2
x
6.
3
+x =4
2x − 1
6.
3x − 1
−3= x
x −5
x
x
7
−
=−
x −4 x +3
2
3x 2 + 6x + 1
=2
x2 + 2
7.
2x + 3 x + 1
−
=7
x
x −2
7.
8.
2x − 1 x 2 − 3
+
=2
5
2x
8.
9.
1
x +8
= 3
(x + 1)2
x +2
x2 − 4
1
−
=1
2
x +4 x −3
9.
x
2x
+
=3
x −1 x +2
10.
3x − 4
x2
+
=2
2
x +2
10.
1
1
1
+ 2 − 3 =1
x
x
x
35
10.4 Inequalities Involving Fractions
10.4
Inequalities Involving Fractions
Fractions may change in sign not only at zeros of the numerator, but also at zeros of the denominator. Sign charts therefore indicate both the zeros of the numerator (0) and the zeros of the
denominator (×). The function is undefined for the zeros of the denominator.
Example
•
2
5
−
61
x −1 x +3
2(x + 3)
5(x − 1)
(x − 1)(x + 3)
−
−
60
(x − 1)(x + 3) (x − 1)(x + 3) (x − 1)(x + 3)
−x 2 − 5x + 14
60
(x − 1)(x + 3)
(x + 7)(x − 2)
>0
(x − 1)(x + 3)
Sign chart:
+
0
−
−7
so x 6 −7 ∨ − 3 < x < 1 ∨ x > 2 .
X
−3
+
X
1
− +
0
2
Again: if a zero of the numerator (0) or the denominator (×) occurs an even number of
times, the sign of the function does not change.
36
10.4 Inequalities Involving Fractions
Exercises
Solve the following inequalities involving fractions.
Series A
Series B
x −3
<0
− 3x − 28
1.
x 2 − 2x − 15
60
x 2 + 4x + 3
1.
2.
1
>0
x 2 − 3x − 28
2.
3.
5
2
−
63
2x + 1 x − 3
3.
1
1
1
− 2 − > −1
3
x
x
x
4.
x2
2x 2
−
> −3
2x + 3 x + 2
4.
x
6
+
62
2x − 5 x + 1
5.
(x 2 − 1)(x + 3)
60
(x + 1)(x − 3)2
5.
6.
3x
2x − 5
>
2x − 1
x −2
x −2
x −1
6
+1
x
x −3
x2
10x
>1
+x
x3
6.
x
x −3
−16
x +4
3
7.
3x − x 2
61
2x − 2
7.
8.
1
1
1
−
<1
+
2
x − 1 (x − 1)
(x − 1)3
x 2 − 4x + 3
5
>
2
x − x − 12
6
8.
9.
x 2 − 2x − 15
−1>0
x 2 − 4x + 3
3
x
+
>1
3x − 1 2x + 1
9.
10.
2x + 3
3x − 8
−2>
x
x −2
10.
x3
x
1
>
− 2x
2
24
3
7
−
<
x +3 x −4
6
37
10.5 Exponential Equations
10.5
Exponential Equations
For the exercises below you are required to reduce the given equation to an exponential equation,
that is, an equation of the form
a expression in x = a number
with a > 0.
Since a x is an increasing function (if a > 1) or a decreasing function (if a < 1), one can simply
equate the exponents. Subsequently, solve the resulting equation.
Use the rules for exponents, for example 4x = (22 )x = (2x )2 = 22x .
Examples
• ex+1 =
1 x−6
e
ex+1 = e−1·(x−6)
x + 1 = −x + 6
2x = 5
x = 2 12
• 2x+3 − 3 · 2x = 80
23 · 2x − 3 · 2x = 80
8 · 2x − 3 · 2x = 80
5 · 2x = 80
2x = 16 = 24
x =4
• 2x + 23−x = 6
2x + 8 · 2−x = 6.
Suppose 2x = a, then
a + 8 · a −1 = 6
a 2 − 6a + 8 = 0
(a − 2)(a − 4) = 0
a=2 ∨ a=4
2x = 2 ∨ 2x = 4
x =1 ∨ x =2
38
10.5 Exponential Equations
Exercises
Solve the following exponential equations.
Series A
Series B
2. ex+1 = ( e1 )x−2
1. 3x+3 = 6 + 3x+2
√
2. 9x = 13 3
3. 3x+3 = 2160 + 3x−1
3. 3x+2 + 3x−1 =
4. e2x −12 · ex +25 = 0
4. 4 · 3x+1 − 32x = 27
5. 2x+3 = 60 + 2x−1
5.
1. 2x+1 + 2x+3 = 320
6. 163x+3 = 8x
2 +4
1
2
28
27
ex = 5 − e2x
6. 4 · 32x+1 − 33x = 3x+3
8. 33x − 2 · 32x+1 = 3x+3
7. 2log3 x = 41
√
8. ( 13 3)x = 9
9. 2x + 23−x = 6
9. 6 · ( 14 )x = 23 ( 12 )x
7. 22x + 64 = 2x+4
10. ex = 2 · e−x +1
10. 600 · (0.4)x = 150 · (0.8)x
39
10.6 Exponential Inequalities
10.6
Exponential Inequalities
For exponential inequalities you must be careful with bases smaller than 1.
For instance, 2x > 24 ⇔ x > 4 but ( 21 )x > ( 21 )4 ⇔ x < 4.
You may reach the same conclusion as follows:
(2−1 )x > (2−1 )4 ⇔ 2−x > 2−4 ⇔ − x > −4 ⇔ x < 4.
Examples
√
• Solve: 5x−1 + 5x−2 > 6 5
√
1
5x−1 + 5x−2 > 6 5 ⇔ 5x · 5−1 + 5x · 5−2 > 6 · 5 2 .
Multiply both sides by 52 :
5
5
5
5 · 5x + 5x > 6 · 5 2 ⇔ 6 · 5x > 6 · 5 2 ⇔ x > .
2
• Solve: 3x + ( 13 )x−3 6 12
3x + ( 13 )x−3 6 12 ⇔ 3x + (3−1 )x−3 6 12 ⇔ 3x + 33−x 6 12
and
3x + 33 · 3−x 6 12 ⇔ 3x +
27
6 12.
3x
Assume 3x = a and note that therefore a > 0 !
a+
27
6 12 ⇔ a 2 + 27 6 12a.
a
This is valid because a > 0.
a 2 − 12a + 27 6 0 ⇔ (a − 3)(a − 9) 6 0 ⇔ 3 6 a 6 9
Hence
31 6 3x 6 32 ⇔ 1 6 x 6 2.
40
10.6 Exponential Inequalities
Exercises
Solve the following exponential inequalities.
Series A.
Series B
1. ( 12 )2x−1 < 8
1. 1 − ( 12 )2x−2 > 0
2. 3 + 2x 6 2x+2 − 3
2. 9x + 3x+1 > 18
3. (2x − 4)(2x − 8) > 0
3.
4. 8x−1 > ( 41 )x
5. 2 + 32 · 2
x
−x
4. 9x+1 >
> 12
6−5
<1
51−x
x
6.
7. 4x >
1
4
22x − 8
60
2x − 4
· 23x
1
27
5. 2x + 8 · 2−x > 9
6. 3x + 33−x < 12
7. 5x − 2 · 5x−1 < 75
8. 2x + 8 · 2−x > 6
8. ( 21 )3x − ( 21 )2x > 0
9. 26x − 4x+1 > 0
9. 6 · 5x − 52x < 5
10. 33−2x − 4 · 31−x + 3 > 0
10. ( 13 )2x−3 − 4 · ( 31 )x−1 − 15 < 0
41
10.7 Logarithmic Equations
10.7
Logarithmic Equations
The logarithm is defined as (with a > 0 and a 6= 1):
aL = x
⇔
L = loga x
log10 x is commonly used in scientific texts, written as log x.
loge x is called the natural logarithm and is usually written as ln x
(in some mathematical texts also as log x !)
Important rules:
loga x exists only if x > 0
loga x is an increasing as a function of x if a > 1 and decreasing if 0 < a < 1
loga 1 = 0,
loga a x = x
loga a = 1,
loga a 2 = 2, (etc.)
a loga x = x provided x > 0
loga x + loga y = loga x y
loga x − loga y = loga ( xy )
loga x r = r loga x
loga x = loga p · log p x
log p x
ln x
log x
loga x =
=
=
log p a
ln a
log a
Note: y = n loga x ⇒ y = loga x n , but y = loga x n 6⇒ y = n loga x.
Make sure to verify the validity of the solutions found, as the argument of a logarithm must be
positive (larger than 0)!
Examples:
• Solve:
log2 (x + 2) = 2 + log2 (2x − 1).
Then we have:
log2 (x + 2) = log2 4 + log2 (2x − 1) = log2 4(2x − 1)
Under the condition that the arguments must be positive this is equivalent to:
x + 2 = 8x − 4 ⇔ 6 = 7x ⇔ x =
• Solve:
6
7
(check: the solution is valid).
ln2 x − ln x 2 = 3.
This is equivalent to (why?): ln2 x − 2 ln x − 3 = 0.
Suppose ln x = a
a 2 − 2a − 3 = 0 ⇔ (a + 1)(a − 3) = 0
⇔
⇔ ln x = −1 ∨ ln x = 3
42
a = −1 ∨ a = 3
⇔
x = e−1 ∨ x = e3
10.7 Logarithmic Equations
Exercises
Solve the following equations.
Series A
Series B
1. logx 16 = 8
1. log2x 27 = 3
2. 3log2 (x−1) = 9
2. 4log4 (8−2x) = 2
3. ln(x 2 − 7x + 7) = 0
3. ln(x + e) − 2 = ln x
4. log2 x − 5 = log 1 (x + 14)
2
5. log2 (x − 1) + log2 (x + 13) = 5
6. log(x 2 − 20x) = 2
7. ln(7 − x) =
1
2
ln(x − 1)
8. ln(x + 1) − 2 · ln 5 = 3
9. ln2 x + 6 = ln x 5
10. ln2 x + 2 ln x − 3 = 0
4. log2 (5 − x) + log2 x = 2
5. 2 + log 1 (2x − 1) = 1
3
6. log x + log(x + 32 ) = 1
7. log2 (x + 1) + log 1
2
1 =5
x −3
8. ln(x + 3) − ln(x + 1) = 1
9. ln2 x − ln x 3 + 2 = 0
1 10. ln(x 2 − 8) = − ln
−2 − x
43
10.8 Logarithmic Inequalities
10.8
Logarithmic Inequalities
An important step in solving logarithmic inequalities is determining the domain. Of course,
solutions need to be in the domain.
Again, be careful:
log2 x < log2 4 ⇒ 0 < x < 4
but
log 1 x < log 1 4 ⇒ x > 4 !
2
2
Just as for exponentials, the logarithm function is decreasing if the base is less than 1, and we
need to reverse the inequality sign.
Examples:
• Solve: log 1 x > 3 + log 1 (x + 3).
2
2
Here: x > 0 ∧ x > −3, so D = (0, ∞).
log 1 x > log 1
2
2
1
1
+ log 1 (x + 3) ⇔ log 1 x > log 1 (x + 3)
2
2
2
8
8
hence
1
x 6 (x + 3) ⇔ 8x 6 x + 3 ⇔ 7x 6 3 ⇔ x 6
8
3
7
Taking into account the domain D, the solution is: (0, 37 ]
• Solve: log3 (2x − 3) < 3 − log3 x.
We have:
(x > 32 ) ∧ (x > 0) ⇒ D = ( 23 , ∞).
On this domain:
log3 (2x − 3) + log3 x < log3 27 ⇔ log3 x(2x − 3) < log3 27
2x 2 − 3x − 27 < 0 ⇔ (2x − 9)(x + 3) < 0 ⇔ − 3 < x <
Taking into account the domain D, the solution is: ( 23 , 92 )
44
9
2
10.8 Logarithmic Inequalities
Exercises
Solve the following logarithmic inequalities.
Series A
Series B
1. log5 (2x + 1) 6 2
1. log2 (x 2 − x) 6 1
2. log4 (x 2 − 3x) > 1
2. log4 (x 2 + 6x) 6 2
3. log2 (x 2 − 4x − 5) 6 4
3. log2 (x 2 − 8x + 7) 6 4
2 − ln x
>0
1 + ln x
2 − ln x
60
2 + ln x
4.
5. 3 − log3 x > log3 (x − 6)
5. log2 (x − 2) < 3 − log2 x
6. log 1 x 2 > −2
6.
7. ln(x − e)2 > 1
7. log3 (22x + 1) 6 2
4.
3
8. x ln x 3 − ln x > 0
9. log3 (x − 1) 6 2 − log3 (x + 7)
10. ln |x| > ln(3 − 12 x)
ln(x − 3) − 1
60
ln x
8. x log(x + 4) + 4 log(x + 4) 6 0
9. log2 (2x − 8) < 3
10.
log(2x + 3)
<2
log x
45
10.9 Trigonometric Equations and Inequalities
10.9
Trigonometric Equations and Inequalities
A few rules:
sin x = sin a
cos x = cos a
tan x = tan a
⇔
x = a + 2kπ ∨ x = π − a + 2kπ
⇔
x = a + 2kπ ∨ x =
⇔
x = a + kπ
− a + 2kπ
Here k ∈ Z, so k = 0, ±1, ±2, . . .
Examples:
• Solve: cos(x − 34 π ) = sin 2x
cos(x − 34 π ) = sin 2x
⇔
cos(x − 34 π ) = cos( 12 π − 2x)
x − 34 π = 12 π − 2x + k · 2π ∨ x − 43 π = − 21 π + 2x + k · 2π
3x = 45 π + k · 2π ∨ − x = 14 π + k · 2π
x=
5
π
12
+ k · 23 π ∨ x = − 14 π + k · 2π
• Solve: 2 sin2 x = 3 cos x
2 sin2 x = 3 cos x
⇔
2(1 − cos2 x) = 3 cos x
⇔
(a + 2)(2a − 1) = 0
⇔
2 cos2 x + 3 cos x − 2 = 0
Suppose cos x = a then
2a 2 + 3a − 2 = 0
⇔
a = −2 ∨ a =
1
2
Note that cos x = −2 has no solution, hence
cos x =
1
2
⇔
x = ± 13 π + k · 2π
Inequalities:
1. Determine where equality holds.
2. Determine where the function is undefined.
3. This yields all points where the inequality possibly reverses. These boundary points define the
intervals where the inequality, or its reverse, holds.
4. Draw a sketch of the function and write down the solution.
Example:
• Solve: tan(2x − 12 π ) > 1.
We have tan(2x − 12 π ) = 1 als 2x − 12 π = 14 π + kπ, dus x = 83 π + 21 kπ .
The function is undefined if 2x − 12 π = 12 π + kπ, so x = 21 kπ .
Consider the graph in Chapter 9 on page 27.
Hence the solution is 38 π + 12 kπ 6 x < 21 π + 12 kπ.
46
10.9 Trigonometric Equations and Inequalities
Exercises
Solve the following trigonometric equations and inequalities. Use R as domain.
Series A.
Series B
1. sin 2x = sin x
1. cos 2x = cos 3x
2. tan 2x + tan x = 0
2. tan x = sin 2x
3. sin(2x + π4 ) + sin(3x − π4 ) = 0
3. cos(2x − 31 π ) + sin(x − 16 π ) = 0
4. cos 2x + cos 3x = 0
4. sin x − sin 3x = 0
5. cos2 x + 3 sin x = 3
5. cos2 x + 2 sin x cos x = sin2 x
6. 2 sin2 x cos x − sin x = 0
6. sin 2x = 2 cos2 x
7. tan 2x = 3 tan x
7. 2 tan x = tan 2x
8. sin 2x =
1
tan x
8. 2 sin 2x =
1
tan x
9. 3 cos2 x − sin2 x = 0
9. cos2 x + cos x = sin2 x
10. 2 sin2 x + sin 2x = 1
10. 2 sin2 2x + 6 sin2 x = 3
11. sin 2x − cos 2x = 1
11. 6 cos2 x + 11 sin x = 10
12. 2 cos x + 3 tan x = 0
√
13. 2 − 2 cos 2x = − sin x
12. sin 2x − cos 2x = 1
14. (tan x − sin x)(tan x + sin x) = cos2 x
14. sin 2x − tan x = sin x
15. sin x · sin 2x = cos x
15. sin x + cos x = 0
Series C.
13. sin2 x + 2 cos2 x = 1 + sin x cos x
Series D
1. 2 sin 12 x > 1
1.
2. tan 12 x > 1
2. tan 2x 6 0
3. sin2 x < cos2 x
3. tan x > sin x
1
2
cos 2x <
1
2
47
10.10 Equations Involving Square Roots
10.10
Equations Involving Square Roots
Equations involving square roots are often solved by squaring both sides of the equation. However, this introduces additional solutions compared to the solutions to the original equation.
Phrased differently:
√
y = x ⇒ y 2 = x,
but, in general,
y 2 = x 6⇒ y =
√
x.
Always remember that the expression under the square root sign cannot be negative, and that
the square root is a non-negative number too. Check the solution found by substituting it in the
original equation.
Examples:
• Solve:
√
2x + 1 = 2x − 5
√
2x + 1 = 2x − 5 ⇒ 2x + 1 = (2x − 5)2
2x + 1 = 4x 2 − 20x + 25 ⇔ 4x 2 − 22x + 24 = 0
2x 2 − 11x + 12 = 0 ⇔ (x − 4)(2x − 3) = 0
3
x =4 ∨ x =
2
Since 2x − 5 < 0 for x = 32 , only x = 4 remains as solution.
√
√
• Solve: x + 2x + 1 = 5
p
√
√
x + 2x + 1 = 5 ⇒ x + 2 2x 2 + x + 2x + 1 = 25
p
2 2x 2 + x = 24 − 3x ⇒ 8x 2 + 4x = 576 − 144x + 9x 2
x 2 − 148x + 576 = 0 ⇔ (x − 4)(x − 144) = 0
x = 4 ∨ x = 144.
The only solution is x = 4.
48
10.10 Equations Involving Square Roots
Exercises
Solve the following equations involving square roots.
Series A
1.
2.
3.
Series B
√
4x + 1 = x − 1
1.
√
2x − 1 = x − 8
√
x −2=
√
2.
3.
2x + 3 − 2
√
4. 2x − 3 14 − x = 8
4.
x +3
5. √
=3
2x + 1
√
√
x +3
6. x x + 3 =
x
√
7. x 2 − 3 13 + x 2 + 3 = 0
8.
p
x−
√
5.
6.
7.
x −1= x
√
√
√
9. 4 4 − p − 13 ( 4 − p)3 − p 4 − p =
10. For which values of p is the graph of
√
f (x) = 3x − 2x + 1
tangent to the graph of
g(x) = 2x + p ?
8.
8
3
√
2x + 1 = x − 7
√
2x + 11 = 12 − x
√
√
x − 1 = 2x + 5 − 2
√
x+ x
√ =2
x− x
√
x + 13 − x
=1
2x − 1
√
√
4x 2 + x
= x
x −1
√
x2
2x 2 + 3
=
√
x2
2x 2 + 3
q
√
3
x − 8−x =2
2
√
√
√
9. 4 2 + x − ( 2 + x)3 + 2x 2 + x = 27
10. For which values of p is the graph of
√
f (x) = 5 − x
tangent to the graph of
g(x) = p − 21 x + 1?
49
10.11 Inequalities Involving Square Roots
10.11
Inequalities Involving Square Roots
Sometimes it helps to draw two graphs and compute the points of intersection. But keep the
domain in mind when determining the solution.
√
For instance, for x − 1 < 2, the solution is [1, 5) because x > 1.
Examples:
√
• Solve: 2x + 3 x < 20
√
√
2x + 3 x < 20 ⇔ 2x + 3 x − 20 < 0
with x > 0. First solve:
√
√
2x + 3 x − 20 = 0 ⇔ 3 x = 20 − 2x ⇒ 9x = 400 − 80x + 4x 2
4x 2 − 89x + 400 = 0
∨ x = 16 . Only x = 25
satisfies the original
The quadratic formula yields: x = 25
4
4
equation.
If we now substitute, for example, x = 4 (4 < 25
) in the inequality, we get a value less
4
than 20. But if we substitute, for example, x = 9 , we get a value larger than 20.
The solution is: [0, 25
).
4
• Solve:
√
3+ x
6 3.
√
x −5
The domain is D = (5, ∞). Then the fraction is always positive.
First solve:
√
3+ x
= 3.
√
x −5
We have:
√
√
√
√
3+ x
= 3 ⇔ 3 + x = 3 x − 5 ⇒ 9 + 6 x + x = 9(x − 5)
√
x −5
√
√
6 x = 8x − 54 ⇔ 3 x = 4x − 27 ⇒ 9x = 16x 2 − 216x + 729
16x 2 − 225x + 729 = 0
81
The quadratic formula yields: x = 16
∨ x = 9 . Only x = 9 satisfies the original equation.
If we now substitute, for example, x = 6 (6 < 9) in the inequality, we get a value less
than 3. But if we substitute, for example, x = 14 , we get a value larger than 3. This means
that we need to take values larger than 9.
The solution is: [9, ∞).
50
10.11 Inequalities Involving Square Roots
Exercises
Solve the following inequalities involving square roots.
Series A
1.
2.
3.
4.
5.
√
2x + 7 6 x − 4
√
x −5 x +6<0
√
5−x
>1
√
5+x
√
x − 2x + 1 6 1
p
(x − 3)2 > 13
√
6. 7 + 3x − 6 > 2x
√
7. x 2 + x + 5 6 x + 1
√
8. x + 2 6 |x|
√
x+ x
9.
√ <3
x− x
√
10. x + 1 < |x − 5|
Series B
1.
2.
3.
4.
5.
6.
7.
8.
√
x − 3 < 12 x − 1
√
x − x −3<5
√
3x + 4
>2
√
2x − 4
√
x − x −466
p
(x − 3)2 > 1
√
√
1
x2 + 3 < x
2
√
(x − 1) x − 1
>1
√
x +5
√
√
x > 2x − 7 − 3
9. √
10.
√
6
x2 − 1
>3
x − 3 < |2x − 9|
51
52
Chapter 11
Rationalizing Denominators (extra)
In the exercises below you need to remove the square roots from the denominators of the fractions. This is called rationalizing denominators. In case of just a square root, this is done by
multiplying top and bottom by this square root.
Example:
√
√
3
2
2
2 3
2√
=
3.
√ =√ ·√ =
3
3
3
3
3
In more complex cases we can apply the rule: (a − b)(a + b) = a 2 − b2 .
Example:
√
√
√
3+ 3
1
3+ 3
3+ 3
1 1√
1
=
= +
3
√ =
√ ·
√ =
√
6
2 6
3− 3
3− 3 3+ 3
32 − ( 3)2
53
11. Rationalizing Denominators (extra)
Exercises
Rationalizing the denominators:
Series A
1
2
1. √ − √
=
5
2−1
2
3
1. √ − √
=
3
5−1
1
2
2. √ − √ =
3
6
1
3
2. √ − √ =
6
2
3
1
√ +√ =
2 3
3
√
2+ 3
4. √
=
3
√
√
3+ 2
5. √
√ =
3− 2
3
1
√ −√ =
2 5
5
√
3+ 6
4. √
=
3
√
√
5− 2
5. √
√ =
5+ 2
3.
1
√
√ =
3 2−2 3
√
2
7. √
√ =
18 − 8
6.
8.
1
=
√
(1 + 2)2
1
=
9. √
2−1
√
3
10.
√ =
2+ 3
54
Series B
3.
2
√
√ =
2 5− 3
√
3
7. √
√ =
48 − 12
6.
8.
1
=
√
(3 − 2)2
1
9. √
=
3−1
√
5
10.
√ =
1− 5
Chapter 12
Partial Fraction Decomposition A (extra)
In the fractions given below, the denominator is a product of two factors or can be written as
such.
The goal is to split the given fraction into a sum of two fractions of which the denominators are
these two factors, respectively. This is called partial fraction decomposition.
Examples:
• We have:
x + 10
3
1
=
−
.
(2x − 1)(x + 3)
2x − 1 x + 3
This can be checked easily afterwards, but how do we actually find the numerators?
A
B
x + 10
by
+
where we need to compute A and B.
We replace
(2x − 1)(x + 3)
2x − 1 x + 3
Write this expression as a single fraction:
A(x + 3) + B(2x − 1)
(A + 2B)x + (3A − B)
=
.
(2x − 1)(x + 3)
(2x − 1)(x + 3)
Then A + 2B = 1 and 3A − B = 10 must hold. If you solve this system of two equations,
you get A = 3 and B = −1.
• Consider:
3
=?
(2x + 3)(x + 4)
We put:
3
A
B
A(x + 4) + B(2x + 3)
(A + 2B)x + (4A + 3B)
=
+
=
=
.
(2x + 3)(x + 4)
2x + 3 x + 4
(2x + 3)(x + 4)
(2x + 3)(x + 4)
Then A + 2B = 0 and 4A + 3B = 3 must hold. Solving this system yields A =
B = − 53 .
Therefore we can replace
6
5
and
3
6
3
by
−
.
(2x + 3)(x + 4)
5(2x + 3) 5(x + 4)
55
12. Partial Fraction Decomposition A (extra)
• Because of the double zero in the denominator, the next fraction is slightly different:
2x + 1
=?
(x − 3)2
We change the numerator into the following form containing x − 3:
2x + 1
2(x − 3) + 7
2
7
=
=
+
.
2
2
(x − 3)
(x − 3)
x − 3 (x − 3)2
The resulting partial fraction decomposition has as denominators all powers of (x − 3) up
to the power (x − 3)2 occurring in the given fraction.
Exercises
Apply partial fraction decomposition to the following fractions:
Series A
Series B
1.
5
=
(2x − 1)(x − 3)
1.
5
=
(2x + 1)(x + 3)
2.
1
=
x2 − 1
2.
1
=
4x 2 − 9
3.
2x
=
(x − 1)(x − 3)
3.
4x
=
(x + 1)(x − 3)
3x
=
x2 − x − 6
4.
5.
6.
7.
x2
x
=
−x −2
4.
x2
2x
=
− 2x
5.
2x
=
(x − 2)2
6.
9x 2
x +4
=
− 6x + 1
7.
3x
=
+ 3x
−x 2
6x
=
(x − 3)2
4x 2
x +4
=
− 4x + 1
8.
x2 − 1
=
(x − 1)2
8.
x2 − 4
=
(x − 2)2
9.
2x
=
(x − 1)2
9.
6x
=
(x − 3)2
10.
x4 − 4
=
(x 2 + 1)2
10.
x4 − 9
=
(x 2 + 3)2
56
Chapter 13
Partial Fraction Decomposition B (extra)
Below a few more exercises are included, where a given fraction is decomposed into several parts.
Examples
• Consider
2
23
=3+ .
7
7
We may obtain this result in two ways:
21 + 2
21 2
2
23
=
=
+ =3+ .
7
7
7
7
7
We also have:
7/23\3
21
2
hence:
23
2
=3+ .
7
7
The is always possible if the numerator exceeds the denominator.
• The same applies to rational functions:
Also:
3x + 1
3x + 9 − 8
8
=
=3−
x +3
x +3
x +3
x + 3/3x + 1\3
3x + 9
−8
hence:
3x + 1
8
=3−
.
x +3
x +3
57
13. Partial Fraction Decomposition B (extra)
• We have:
2x − 1/ 12 x 2 −3x + 1\ 41 x −
1 2
x − 14 x
2
x +1
− 11
4
− 11
x + 11
4
8
− 38
hence:
11
8
− 3x + 1
= 14 x −
2x − 1
1 2
x
2
11
8
−
3
8
2x − 1
.
The method used in the last two examples applies if the degree of the numerator is larger than or
equal to the degree of the denominator.
Exercises
Apply partial fraction decomposition in the exercises below.
Series A
1.
2.
x 2 + 8x
=
2x − 1
1.
2.
4x 2 + x − 5
=
x −1
x 2 + 4x
=
2x − 1
3.
x 2 − 3x + 2
=
x −2
3.
x 2 + 5x − 14
=
x −2
4.
x3
=
x2 − 1
4.
x4
=
x2 + 1
5.
6.
7.
8.
9.
10.
58
6x 2 − 11x + 5
=
x −1
Series B
x4 − 1
=
x −1
x 2 + 2x − 3
=
1−x
2x 2 − 4x + 5
=
x −1
x 3 − 3x 2 + 3x − 1
=
x −1
x 3 − 2x 2 + 3x − 2
=
x2 − 4
x 4 + x 2 − 2x 3 − 2x + 1
=
x2 + 1
5.
6.
7.
8.
9.
10.
x 4 − 16
=
x +2
x 2 + 2x − 3
=
3+x
2x 2 − 4x + 5
=
x −2
x 3 − 6x 2 + 5x + 6
=
x −2
2x 3 − 2x 2 + 3x − 1
=
x2 − 1
x 4 + x 2 + 2x 3 − 2x + 1
=
x2 + 1
Chapter 14
Inverse Trigonometric Functions (extra)
In high school you have already seen inverse functions. For instance, log x and 10x are inverses
√
of each other and so are x and x 2 , for x > 0. On a scientific calculator, these function often
share the same key. So you can tell that also ln x en ex are each other inverses.
A property of inverse functions is that the graphs of a function and its inverse are reflections of
each other in the line y = x. The domain and range of a function and its inverse need not be
identical; you can check this for the above functions. The functions sin x, cos x and tan x have
inverse functions . sin−1 x, cos−1 x and tan−1 x is often used on calculators; we will use arcsin x,
arccos x and arctan x.
We will now check the domain and range of these functions by reflecting their graphs. In the
figure on the left you may recognize the graph of y = sin x reflected in the line y = x.
π
1
π
2
1
π
4
y = arcsin(x)
1
π
2
3
π
4
1
π
4
− 14 π
1
π
4
− 12 π
0
−1 −0.5
0
1
π
2
0
0
0.5
1
− 41 π
y = arccos(x)
−1 −0.5
− 12 π
−5
0
0.5
y = arctan(x)
−4
−3
−2
−1
0
1
2
3
4
5
1
We note that we take only a small piece of the sin-graph, since otherwise the reflection of the
graph would not be a function. If the domain for sin x is equal to D = [− 12 π, 12 π], the inverse is
indeed a function. In choosing the domain we include the origin and ensure that all values for
sin x are reached (from −1 to 1). See the figure above.
The range of y = sin x is R = [−1, 1]. The inverse function of y = sin x is called y = arcsin x. Its
domain is D = [−1, 1], while its range is R = [− 21 π, 21 π].
In this way we can also reflect the graphs of y = cos x and y = tan x in the line y = x.
1. Check that for y = arccos x the domain is D = [−1, 1] choosing as range R = [0, π].
2. Check that for y = arctan x the domain is D = (−∞, ∞) choosing as range R = (− 21 π, 12 π ).
59
14. Inverse Trigonometric Functions (extra)
√
Similar as before when we determined which x satisfy sin x = − 12 3, we can now determine a
√
value for arcsin(− 21 3).
The important difference is that we now get exactly one result, as a function may take on one
value only.
Therefore:
√
arcsin(− 12 3) = − 31 π.
Or, in general:
y = arcsin x ⇒ x = sin y, but x = sin y 6⇒ y = arcsin x,
and similarly for arccos and arctan.
Exercises
Give your answer in radians (in terms of π ).
Series A
Series B
1. arcsin( 12 ) =
√
2. arccos(− 12 2) =
√
3. arctan( 3) =
√
4. arccos( 12 3) =
1. arccos( 12 ) =
√
2. arcsin(− 21 2) =
√
3. arctan(− 3) =
√
4. arcsin( 12 3) =
5. arctan(1) =
5. arccos(1) =
6. arcsin(−1) =
√
7. arccos( 12 2) =
√
8. arctan( 13 3) =
6. arctan(−1) =
√
7. arcsin( 21 2) =
√
8. arccos( 12 3) =
9. arcsin(0) =
9. arctan(0) =
10. arccos(−1) =
10. arcsin(−1) =
60
Chapter 15
Answers to the Exercises
2 Powers
Series B
Series A
1. p 16 q 16
1. p 18 q 16
2. a 11 b5
2. −b5
3. − 49 c−1 d 10
8 2 13
c d
3. − 27
3
4. −27a 3 b 2
√
5. 2 2ab
1
15
4. −32a 5 b 2
√
5. 3 2ab
5
1
5
6. 2 p 4 q 12
6. 2 p − 2 q 6
7. −3a −3 b−8
7. −3a −1 b4
1
1
3
8. 3 4 2 2 ab 4
9.
3 − 83 73
a b
2
5
1
10. 2− 4 a 4
1
1
1
1
8. 3− 12 2 4 a 12 b 6
9.
5
3 − 17
a 5 b2
2
4
1
10. 2− 3 a 6
61
15. Answers to the Exercises
3.1 Removing Parentheses
Series A
Series B
1. 9a 2 − 6ab + b2
1. 9a 2 − 12ab + 4b2
2. 9a 2 − 12a 3 + 4a 4
√
3. 162 − 108 2
2. 9a 2 − 12a 4 + 4a 6
√
3. 147 − 36 5
4. 3m 2 − 5mn − 2n 2
√
5. 2 3 + 6
4. 6m 2 − 5mn − 6n 2
√
5. −3 5 − 8
6. 9a 4 b6 − 2a 6 b4 + 19 a 8 b2
6.
7. 21 − 12a 2
7. 30 − 12a 2
8. 7b − 7a + ab + 6a 2 − 2b2 − 5
8. 14b − 5a − 5ab + 6a 2 − 6b2 − 4
9.
16
5
10. 81a 4 − 1
9.
1 4 2
a b
16
− a 6 b4 + 4a 8 b6
4
5
10. 16a 4 − 1
3.2 Factoring Polynomials
Series A
1. (2x − 3)(2x + 3)(4x 2 + 9)
1. (3x − 2)(3x + 2)(9x 2 + 4)
2. 3x 3 (x + 3)(x − 7)
2. 3x 2 (x − 1)(x − 4)
3. (x − 1)(x + 1)(x 2 + 1)(x 4 + 1)(x 8 + 1)
3. (x 3 − 2)(x 3 + 2)(x 6 + 4)
4. (x 2 − 2)(x 2 + 3)
4. (x 2 + 4)(x 2 − 5)
5. (x − 2)(x − 17)
5. (x − 2)(x − 19)
6. (x + 2)(x − 17)
6. (x + 2)(x − 19)
7. (x − 1)2
7. 2(x + 1)2
8. (x − 1)(x + x 2 + 1)
8. (x − 1)(x + x 2 − 1)
9. (5x + 1)(x − 5)
9. 4(4x + 1)(x − 4)
10. (x 3 − 3)2
62
Series B
10. (x 5 + 4)2
3.3 Factoring Polynomials, continued
Series B
Series A
1. (3x − 2)(x − 6)
1. (3x − 5)(x − 3)
2. (x + 2)(2x + 3)
2. (x + 3)(2x + 3)
3. (3x 2 − 2)(x 2 − 3)
3. (3x 2 − 4)(x 2 − 3)
4. (2x + 3)(5 − x)
4. (x + 3)(7 − 2x)
5. (2x 2 − 3)(x 2 + 1)
5. 2(x 2 + 2)(x − 1)(x + 1)
6. (x − 1)(x − 4)(x + 1)
6. (x − 1)(x − 8)(x + 1)
7. (x − 3)(2x 2 + 1)
7. (x − 4)(3x 2 + 2)
8. (x + 5)(x − 2)(x + 2)
8. (x + 5)(3x 2 − 4)
9. (2 − x)(3x 2 − 2)
9. (2 − x)(2x 2 − 3)
10. x(x 3 − 3)(x 3 + 5)
10. x 2 (x 3 + 2)(x 3 − 6)
4 Fractions
Series A
Series B
−2x−1)
1. − 2(2x
(2x−1)(x+1)
2
1.
2x 2 +2x+3
(2x−1)(x+1)
2.
√
2 x
x−1
2.
√
2 2x
2x−9
3.
4x
(x−1)(x+3)
3.
4x 2 +3x−6
(x 2 −3)(2x+3)
−3x−3)
4. − 3(2x
(x−1)(x+3)
4.
3(2x 2 −x+2)
(x−1)(x+2)
2
5.
3x−2
(x+1)(2x+3)
5.
(x−2)(7x+4)
(x 2 +1)(2x+3)
6.
2x−3
2(x−2)
6.
2x−3
2(x+3)
7.
x+6
(3−x 2 )(2x+1)
8.
x 2 +1
(x+1)3
8.
x 2 +1
(x−1)3
9.
2x 2 −x+1
(x−1)2 (x+1)
9.
2x+1
(x−1)(x+1)
10.
2x 2 +11x−3
2x(x+3)
10.
7x
7. − (x−3)(2x+1)
x 2 −2x+6
x(x−3)
63
15. Answers to the Exercises
5 Trigonometry
Series A
√
1. − 12 3
2. −1
√
3. − 21 3
√
4. 21 3
5. 1
6. −1
√
7. 12 2
√
8. − 13 3
9. − 12
√
10. − 12 2
Series B
1. − 12
√
2. − 12 2
√
3. − 13 3
4. − 12
√
5. 12 2
6. undefined.
√
7. 12 2
√
8. − 12 3
√
9. − 3
√
10. − 12 3
6.1 Trigonometric Identities: Basic Formulas
1. We have:
sin(x − y) = sin(x + (−y)) = sin x cos(−y) + cos x sin(−y) = sin x cos y − cos x sin y
cos(x + y) = sin( 12 π − x − y) = sin(( 21 π − x) − y)
= sin( 12 π − x) cos y − cos( 12 π − x) sin y = cos x cos y − sin x sin y
cos(x − y) = cos(x + (−y)) = cos x cos(−y) − sin x sin(−y) = cos x cos y + sin x sin y
2. We have:
sin(x + y)
sin x cos y + cos x sin y
tan x + tan y
=
=
cos(x + y)
cos x cos y − sin x sin y
1 − tan x tan y
tan x + tan(−y)
tan x − tan y
tan(x − y) = tan(x + (−y)) =
=
.
1 − tan x tan(−y)
1 + tan x tan y
tan(x + y) =
3. We have:
sin 2x = sin(x + x) = sin x cos x + cos x sin x = 2 sin x cos x
cos 2x = cos(x + x) = cos x cos x − sin x sin x = cos2 x − sin2 x =
= (1 − sin2 x) − sin2 x = 1 − 2 sin2 x = cos2 x − (1 − cos2 x) = 2 cos2 x − 1
2 tan x
tan x + tan x
tan 2x = tan(x + x) =
=
.
1 − tan x tan x
1 − tan2 x
4. We have: cos(x − 16 π ) = cos x cos 16 π + sin x sin 16 π. Since x lies in the first quadrant also
√
√
√
√
√
√
sin x = 21 2. So cos x cos 16 π + sin x sin 16 π = 12 2 · 12 3 + 21 2 · 21 = 14 6 + 41 2.
√
5. We have: cos 2x = 1 − 2 sin2 x = 13 ⇒ 13 = 1 − 2 sin2 x ⇒ sin2 x = 13 ⇒ sin x = ± 31 3.
√
Since x ∈ [ 21 π, π] we find: sin x = 13 3.
64
9
7
6. sin2 x = 1 − cos2 x = 1 − 16
= 16
, hence sin x =
√
√
1
So tan x = 3 7, and tan 2x = 3 7
q
7
16
because x ∈ [0, 21 π ].
7. sin2 x cos2 x + sin4 x + cos2 x = sin2 x(cos2 x + sin2 x) + cos2 x = sin2 x + cos2 x = 1.
8. 2(cos2 x − sin2 x)2 tan(2x) = 2 cos(2x)2 sin(2x)/ cos(2x) = 2 cos(2x) sin(2x) = sin(4x).
6.2 Trigonometric Identities: Factoring
Series A
Series B
1. (2 cos x + 1) sin x
1. 2 sin x(sin x − cos x)
2. 2 sin x cos y
2. (cos x − 1)(cos x + 1) = − sin2 x
3. (1 − sin x)(1 + sin x) = cos2 x
3. (cos x − 2 sin x) cos x
4. sin x(sin x + 2 cos x)
5. 2(cos x − 1)(cos x + 1) = −2 sin2 x
4. −2(sin x − 2)(sin x + 2)
5. (sin x + 2)(sin x − 3)
7. (cos x + 1)(6 − cos x)
p
√
8. 21 2 + 2
6. (sin x + 2)(5 − sin x)
1 − cos x 1 = tan x 7. 2
sin x p
√
8. 12 2 − 2
9. 2 sin x
9. cos2 2x
6. (sin x − 1)(sin x − 4)
10.
1
, − 12
2
en 2
10. 1, 2, 2
65
15. Answers to the Exercises
7.2 Differentiation
Series A.
Series B
1. 30(1 − 2x)4
2. −
1. 2 ln x + 2 ln 3 + 2
1
(x 2
− 1)
2.
3
2
1 − ln x
x2
1
3. 6x(x 2 − 5)−2
3. cos 2x(sin 2x)− 2
4. −27(3x − 1)−4
4. −3x(3x 2 + 1)− 2
5. 4 ln 3 · 32x
5. 7(x − 1)−2
6. 4x ex
2 −1
7. ln(3x + 4) +
8.
3x
3x + 4
1
x + x2
9. x 2 · 22x (2x ln 2 + 3)
10.
27x − 6
√
2 3x − 1
1
11.
x(ln x − 1)2
12. ex (x 2 − x − 3)
13. 3(tan2 x)(tan2 x + 1)
2
14.
sin 2x + 1
cos x
15. − 2
cos x − 1
66
3
6. 2 sin(4x − 13 π )
7.
2 ln x sin x − x ln2 x cos x
x sin2 x
3
8. (x 2 + 1)− 2
9. 3(6x − 2)−1
10. 2 sin(2x) e− cos 2x
11. − e−x
12.
4 ln3 x
x
13. sin x(2 cos2 x − sin2 x)
14.
1
cos x − 1
15.
1 + cos2 x
sin3 x
8 Antiderivatives
Series A
1.
Series B
1
(2x
8
− 1)4
2. − 13 (5 − x)3
3.
2
(3x
9
3
− 4) 2
1.
1
(3x
12
+ 2)4
2. − 16 (8 − 2x)3
3.
1
(2x
3
3
− 3) 2
4. − 16 (2x + 3)−3
4. − 18 (2x − 1)−4
5. − 12 cos 2(x − 16 π )
√
6. 2 x + 3
5. 2 sin 21 (x − 13 π )
√
6. 4 x − 5
− ln |x|
7.
1 3
x
3
8. − cos x + 13 e3x
8.
1
2
9. tan x − x
9. x + tan x
7.
1 2
x
2
+ ln |x|
10. − ln |3 − 2x|
sin 2x + 12 e2x
10. − ln |2 − 3x|
67
15. Answers to the Exercises
9 Graphing Functions
f HxL
f HxL
f HxL
f HxL
f HxL
9
8
7
3
-1
-2
f HxL = Hx + 2L4
x
O
O
A1
2
f HxL = Hx - 2L2 + 3
f HxL
f HxL
x
f HxL = 8 - x3
A2
f HxL
x
O
O
x
A3
O
f HxL = 6 Hx + 1L5
A4
-2
1
O
4
f HxL = €€€€€€€€€€€€€€€€
x-2
1
f HxL = €€€€€
x
B1
x
3
f HxL
f HxL
f HxL = 2x
B3
f HxL
B4
f HxL = 3 + 2-x
O
C1
f HxL = 1 + 2x-2
x
C2
B5
f HxL
x
f HxL = logHxL
O
f HxL = logHx2 L
2
D1
D2
f HxL
x
1
O
f HxL = 2 logHxL
D3
D4
D5
f HxL
-1
2
x
O
f HxL
E1
4
!!!
f HxL = 4 - 2 x
f HxL
2
x
O
-4
!!!!!!!!!!!
f HxL = x + 4 + 2
E2
E3
O
E4
f HxL
20
E5
f (x)
f (x)
1
4
0.8
f (x) = tan 12 (x + 41 π )
2
0.6
Π
€€€€€
3
2Π
€€€€€€€€€€
3
x
O
1
€€€€€
2
3
€€€€€
2
x
-1
f HxL = sinH3 xL
F1
-2
f HxL = 2 cosHΠ xL
F2
–6
0.4
0.2
4
68
x
1
-1
6
!!!!!
f HxL = x2 - 1
3
!!!
f HxL = x
x
1
O
x
ã+1
4
2
x
ã
f HxL = 2 lnHx - ãL
f HxL
f HxL
4
O
C5
f HxL
x
f HxL = logHx + 2L
2
O
x
O
f HxL = ãx-1
C4
1
-1
1ã
-1
f HxL = -1 + 3-x
f HxL
x
-2
x
C3
f HxL
1
!!!
f HxL = x + 2
x
4
f HxL
O
x
O
O
O
2
4
f HxL = €€€€€€€€€€€€€€€€€€€€2€€
H2 - xL
3
f HxL = €€€€€2€€€
x
1
x
O
f HxL
O
3
1
O
x
O
1
f HxL = 3 + €€€€€€€€€€€€€€€€
x-3
B2
f HxL
f HxL
A5
f HxL
f HxL
3
x
2
-3
f HxL = x2 + 6 x + 9
x
2
x
O
O 1
7
1
f HxL = 8 sinJ €€€€€ Π Hx - 1LN + 12
3
F3
x
–4
–2
0
–4
–2 0
2
–2
2
4
–4
f (x) = sin2 x
F4
F5
4
6
10.1 Polynomial Equations
Series A
Series B
1. x = −4 or x =
1
2
2. x = 1 √
or x = −1 or
or x = − 6
x =
√
6
1. x = −3 or x =
2. x =
2
3
√
√
6 or x = − 6
3. x = 0 or x = −2 or x = 12
3. x = 0 or x = 1 or x = 6
√
√
4. x = 7 or x = − 7
4. x = 2 or x = −2
5. x = 0 or x = −3 or x = 13
5. x = 0 or x = −3 or x = 11
6. x = 5 or x = −4 or x = 3
7. x = 1 or x = 2 or x = 3
8. x = 2 or x =
1
3
or x = −2
9. x = 2 or x = − 12 or x = −2
√
10. x = 2 or x = −2 or x = 2 or
√
x =− 2
6. x = 2 or x = −1
7. x = −2 or x =
2
3
8. x = −3 or x = 3 or x = 4
9. x = −2 or x = 3
10. x = 2 or x =
1
3
or x = −1
69
15. Answers to the Exercises
10.2 Polynomial Inequalities
Series A
Series B
1. [−5, 5]
1. [4, ∞) ∪ (−∞, −4]
2. (−∞, 2] ∪ [4, ∞)
2. [− 34 , 43 ]
3. [ 23 , 52 ]
3. (−∞, 35 ] ∪ [ 73 , ∞)
4. (2, ∞) ∪ (−∞, −4)
4. (3, ∞) ∪ [−∞, −5)
5. ∅
5. (−∞, 16 )
6. (5, ∞) ∪ (−∞, −7)
6. [2, ∞) ∪ (−∞, −5]
7. {4} ∪ [−2, 2]
7. [−3, 3]
8. (−∞, 67 )
8. (−1, 53 ) ∪ (3, ∞)
9. {4} ∪ [−6, 3]
9. {2} ∪ (−∞, 1] ∪ [3, ∞)
10. [−2, 0] ∪ [2, 3]
10. [1, 2]
10.3 Equations Involving Fractions
Series A
Series B
1. x = −4
1. x = −2 or x = 3
2. x = 0 or x = −1
√
√
3. x = 1 + 2 or x = 1 − 2
√
√
4. x = 3 or x = − 3
2. x = 3
√
√
3. x = 2 3 − 3 or x = −2 3 − 3
4. x =
5. x = −1 or x =
5. x = − 23 or x = 2
6. x = 1 or x =
1
2
7
2
6. x = −2 or x = 7
7. x = 1
7. x = −4 or x = 3
8. x = − 59 or x = 3
8. x = −10 or x = 2
9. x = − 65 or x = − 12
9. x = 2
10. x = − 85 or x = 2
70
2
3
10. x = −1 or x = 1
10.4 Inequalities Involving Fractions
Series A
Series B
1. (−1, 5]
1. (3, 7) ∪ (−∞, −4)
2. (−∞, −4) ∪ (7, ∞)
2. (−3, 0) ∪ (0, 3)
3. [ 32 , 2] ∪ (3, ∞) ∪ (−∞, − 12 )
3. (−∞, −1) ∪ (0, 1) ∪ (1, ∞)
4. (−∞, −2) ∪ (− 32 , ∞)
4. (−∞, −1) ∪ [5, ∞) ∪ [ 43 , 25 )
5. (2, 5] ∪ ( 21 , 1]
5. [−3, −1) ∪ (−1, 1]
6. (0, 2] ∪ (3, ∞) ∪ (−∞, −3]
6. (−4, −1] ∪ [0, ∞)
7. [−1, 1) ∪ [2, ∞)
7. (4, 6) ∪ (−∞, −3) ∪ (13, ∞)
8. (1, 2) ∪ (−∞, 0) ∪ (2, ∞)
8. ( 13 , 2] ∪ (− 12 , 14 ]
√
√
9. ( 2, 2] ∪ [−2, − 2)
9. (1, 3) ∪ (9, ∞)
10. (2, 3) ∪ (0, 23 )
10. (4, 6) ∪ (−∞, −3) ∪ (13, ∞)
10.5 Exponential Equations
Series A
Series B
1. x = 5
1. x = −1
2. x =
1
2
2. x = − 14
3. x = 4
3. x = −2
4. x = ln 4 or x = ln 8
4. x = 1 or x = 2
5. x = 3
5. x = ln 2
6. x = 0 or x = 4
6. x = 1 or x = 2
7. x = 3
7. x =
8. x = 2
8. x = −4
9. x = 1 or x = 2
9. x = 2
10. x = ln 2
1
9
10. x = 2
71
15. Answers to the Exercises
10.6 Exponential Inequalities
Series A
Series B
1. (−1, ∞)
1. [1, ∞)
2. [1, ∞)
2. (1, ∞)
3. (−∞, 2) ∪ (3, ∞)
3. [ 32 , 2)
4. [ 35 , ∞)
4. [− 25 , ∞)
5. (−∞, 2] ∪ [3, ∞)
5. (−∞, 0] ∪ (3, ∞)
6. (−∞, 0) ∪ (1, ∞)
6. (1, 2)
7. (−∞, 2)
7. (−∞, 3)
8. (−∞, 1] ∪ [2, ∞)
8. (−∞, 0]
9. ( 21 , ∞)
9. (−∞, 0) ∪ (1, ∞)
10. (−∞, ∞)
10. (0, ∞)
10.7 Logarithmic Equations
Series A
1. x =
Series B
√
2
3
2
2. x = 5
2. x = 3
3. x = 1 or x = 6
3. x = e /(e2 −1)
4. x = 2
4. x = 1 or x = 4
5. x = 3
5. x = 2
√
√
6. x = 10 − 10 2 or x = 10 + 10 2
6. x =
7. x = 5
7. x = 7
8. x = 25 e3 −1
8. x =
9. x = e2 or x = e3
9. x = e or x = e2
10. x = e or x = e−3
72
1. x =
5
2
e −3
1−e
10. x = −3
10.8 Logarithmic Inequalities
Series A
Series B
1. (− 12 , 12]
1. [−1, 0) ∪ (1, 2]
2. (−∞, −1) ∪ (4, ∞)
2. [−8, −6) ∪ (0, 2]
3. [−3, −1) ∪ (5, 7]
3. [−1, 1) ∪ (7, 9]
4. (0, e−2 ) ∪ [e2 , ∞)
4. (e−1 , e2 )
5. (6, 9]
5. (2, 4)
6. (−3, 0) ∪ (0, 3)
√
√
7. (−∞, e − e) ∪ (e + e, ∞)
6. (3, e +3]
8. (0, 31 ) ∪ (1, ∞)
8. (−4, −3]
9. (1, 2]
9. (3, 4)
10. (−∞, −6] ∪ [2, 6)
7. (−∞, 32 ]
10. (0, 1) ∪ (3, ∞)
73
15. Answers to the Exercises
10.9 Trigonometric Equations and Inequalities
Series A
Series B
1. {2kπ} ∪ { 13 π + 32 kπ }
1. {2kπ} ∪ { 52 kπ}
2. { 13 kπ}
2. {kπ} ∪ { 41 π + 12 kπ}
3. { 52 kπ} ∪ { 32 π + 2kπ }
3. { 32 kπ }
4. { 15 π + 25 kπ}
4. {kπ} ∪ { 14 π + 12 kπ}
5. { 12 π + 2kπ}
5. { 83 π + 12 kπ}
6. {kπ } ∪ { 41 π + kπ}
6. { 21 π + kπ } ∪ { 14 π + kπ}
7. {kπ } ∪ { 16 π + kπ} ∪ { 56 π + kπ }
7. {kπ}
8. { 12 π + kπ} ∪ { 14 π + 12 kπ }
8. { 61 π + 13 kπ}
9. { 13 π + kπ} ∪ { 23 π + kπ }
9. { 31 π + 23 kπ}
10. { 18 π + 12 kπ}
10. { 61 π + kπ } ∪ { 56 π + kπ}
11. { 14 π + kπ} ∪ { 12 π + kπ }
11. { 61 π + 2kπ} ∪ { 56 π + 2kπ}
12. { 76 π + 2kπ} ∪ { 11
kπ + 2kπ}
6
12. { 41 π + kπ } ∪ { 12 π + kπ}
13. {kπ }
13. { 41 π + kπ } ∪ { 12 π + kπ}
14. { 14 π + 12 kπ}
14. { 32 kπ } ∪ {kπ }
15. { 12 π + kπ} ∪ { 14 π + 12 kπ }
15. { 34 π + kπ }
Series C
74
Series D
1. ( 13 π + 4kπ, 53 π + 4kπ )
1. (0 + kπ, π + kπ )
2. [ 12 π + 2kπ, π + 2kπ )
2. ( 41 π + 12 kπ, 21 π + 12 kπ]
3. [− 14 π + kπ, 14 π + kπ]
3. [0 + kπ, 21 π + kπ )
10.10 Equations Involving Square Roots
Series A
Series B
1. x = 6
1. x = 12
2. x = 13
2. x = 7
3. x = 3 or x = 11
3. x = 10 or x = 2
4. x =
4. x = 9
31
4
5. x = 12 or x = 0
5. x = 4
6. x = −3 or x = 1 or x = −1
√
√
7. x = 2 3 or x = −2 3
6. x = 0 or x = 6
√
√
7. x = 3 or x = − 3
8. x = 1
8. x = 4
√
3
9. p = 4 − 2 2
10. p = −1
9. x = 7
10. p = 2
10.11 Inequalities Involving Square Roots
Series A
Series B
1. [9, ∞)
1. [3, 4) ∪ (4, ∞)
2. [−6, 30)
2. [3, 7)
3. (−5, 0)
3. (2, 4]
4. [− 12 , 4]
4. [4, 8]
, ∞)
5. (−∞, 83 ) ∪ ( 10
3
5. (−∞, 2) ∪ (4, ∞)
6. [2, 5)
6. (1, 3)
7. [4, ∞)
7. (3, ∞)
8. [−2, −1] ∪ [2, ∞)
8. [ 72 , 64]
√
√
9. (− 5, −1) ∪ (1, 5)
9. (0, 1) ∪ (4, ∞)
10. [−1, 3) ∪ (8, ∞)
10. [3, 4) ∪ ( 21
, ∞)
4
75
15. Answers to the Exercises
11 Rationalizing Denominators (extra)
Series A
1.
2.
3.
4.
5.
6.
Series B
√
1. − 14 5 − 45
√
√
2. 12 2 − 21 6
√
1
3. 10
5
√
√
4. 3 + 2
√
5. 73 − 23 10
√
√
4
2
3 + 17
5
6. 17
√
√
5−2 2−2
√
√
1
3 − 13 6
3
√
5
3
6
√
2
3+1
3
√
2 6+5
√
√
1
2 + 13 3
2
1
5
7.
7. 1
1
2
√
2 + 11
49
√
9. 12 3 + 21
√
√
10. 23 3 − 43 5 −
√
8. 3 − 2 2
√
9. 2 + 1
√
10. 2 3 − 3
8.
6
49
3
4
12 Partial Fraction Decomposition A (extra)
Series B
Series A
1. (x − 3)−1 − 2(2x − 1)−1
2.
1
(x
2
− 1)−1 − 12 (x + 1)−1
3. 3(x − 3)−1 − (x − 1)−1
4.
1
(x
3
+ 1)−1 + 23 (x − 2)−1
2.
1
(2x
6
− 3)−1 − 61 (2x + 3)−1
3. (x + 1)−1 + 3(x − 3)−1
4.
6
(x
5
+ 2)−1 + 59 (x − 3)−1
5. 2(x − 2)−1
5. 3(3 − x)−1
6. 2(x − 2)−1 + 4(x − 2)−2
6. 6(x − 3)−1 + 18(x − 3)−2
7.
1
(3x
3
− 1)−1 +
13
(3x
3
− 1)−2
7.
1
(2x
2
− 1)−1 + 29 (2x − 1)−2
8. 2(x − 1)−1 + 1
8. 4(x − 2)−1 + 1
9. 2(x − 1)−1 + 2(x − 1)−2
9. 6(x − 3)−1 + 18(x − 3)−2
10. 1 − 3(x 2 + 1)−2 − 2(x 2 + 1)−1
76
1. 2(2x + 1)−1 − (x + 3)−1
10. 1 − 6(x 2 + 3)−1
13 Partial Fraction Decomposition B (extra)
Series B
Series A
1. 6x − 5
2.
1
x
2
+
17
4
1. 4x + 5
+
17
(2x
4
− 1)−1
−1
1
(x
2
2. 2x + 9 + 9(2x − 1)−1
3. x − 1
4. x +
1
(x
2
− 1)
+
−1
+ 1)
5. x + x 2 + x 3 + 1
3. x + 7
4. x 2 − 1 + (x 2 + 1)−1
5. x 3 − 2x 2 + 4x − 8
6. −x − 3
6. x − 1
7. 2x − 2 + 3(x − 1)−1
8. x − 2x + 1 = (x − 1)
2
2
9. x − 2 + (7x − 10)/(x − 4)
= x + (x − 2)−1 + 6(x + 2)−1 − 2
2
10. x 2 − 2x + (x 2 + 1)−1
7. 2x + 5(x − 2)−1
8. x 2 − 4x − 3
9. 2x − 2 + (5x − 3)/(x 2 − 1)
10. x 2 + 2x + (1 − 4x)/(x 2 + 1)
14 Inverse Trigonometric Functions (extra)
1. The graph of y = sin x can be reflected in the line y = x, where the resulting image is a
graph of a function if we take [− 21 π, 12 π ] as domain for x. For this domain we have a graph
which is increasing everywhere.
It should be no problem to see that this also holds for a graph which is decreasing everywhere. So, for the domain [ 21 π, 32 π ] the resulting image should also be the graph of a
function.
This can be checked using a graphing calculator, by entering on the TI-83:
y1=sin(x)
and then using DrawInv y1 on the Draw-menu to inspect the reflected graph. You can
now check the same for y = cos x on [0, π].
2. As above.
77
15. Answers to the Exercises
14 Inverse Trigonometric Functions (extra)
Series A
1.
1
π
6
1.
2.
3
π
4
2. − 14 π
3.
1
π
3
3. − 13 π
4.
1
π
6
4.
5.
1
π
4
5. 0
6. − 12 π
1
π
3
1
π
3
6. − 14 π
7.
1
π
4
7.
1
π
4
8.
1
π
6
8.
1
π
6
9. 0
10. π
78
Series B
9. 0
10. − 12 π