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This digital version of the ’DictaatRekenvaardigheden’-’Algebraic Skills’ is for personal usebecause of copyright. c©

1

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 mathemat-ics 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 assumedto have mastered already. The topics covered in the remaining Chapters 11–14 are typically be-yond what is assumed as prior knowledge upon entering university, but nevertheless basic andessential 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 remediedwithin a few practice sessions. Therefore, to allow for extensive and repetitive practicing, manyexercises of the same type are provided for each topic. To make sure that you master a certaintopic 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 thesequickly, without access to a formula sheet or calculator.

Finally, according to a long and widespread tradition, parentheses around the argument to thestandard functions sin, log, etc. are omitted if no confusion arises. So y sin x = sin(x)y 6=sin xy = sin(xy), 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

10.5 Exponential Equations . . . . . . . . . . . . . . . . . . . 38

10.6 Exponential Inequalities . . . . . . . . . . . . . . . . . . . 40

10.7 Logarithmic Equations . . . . . . . . . . . . . . . . . . . 42

10.8 Logarithmic Inequalities . . . . . . . . . . . . . . . . . . . 44

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 pq with p ∈ Z and q ∈ N

Real numbers 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:{n ∈ N

∣∣ 15 n ∈ N

}.

The union of sets A and B is defined as

A ∪ B = {all elements that are in A or in B}.

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. Weuse common notation for intervals. Examples:

U = {x ∈ R∣∣ −3 < x 6 5

} = (−3, 5] −3 5

V = {x ∈ R∣∣ x > 2

} = [2,∞) 2

W = {x ∈ R∣∣ x 6 −2

} = (−∞,−2] −2

Some intersections and unions of these intervals are

U ∪ V = {x ∈ R∣∣ x > −3

} = (−3,∞)U ∩ V = {x ∈ R

∣∣ 2 6 x 6 5} = [2, 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 aproduct. For example, 3 − a is a sum of the terms 3 and −a; 3a is a product of the factors 3 anda.

Examples:

• 2a2b − 3c consists of two terms, namely 2a2b and −3c.

• 2a2b consists of three factors: 2, a2, 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 termconsists of −2, c, and d .

It is important to realize whether you are dealing with terms or factors since the rules of arith-metic are different!

2

Chapter 2

Powers

The following rules hold, assuming that a > 0 and b > 0:

a p · aq = a p+q,1an= a−n,

a p

aq= a p−q, (a p)q = a pq,

(ab)p = a pbp,√

an = an2 ,

p√

aq = aqp ,√

ab = √a√

b.

For example, using these rules we have:

(3a2b)13

2a4b−12= 3

13 a

23 b

13

2a4b−12= 3

13 · 2−1a−

103 b

56 .

The expression is reduced to a product of numbers and powers of the form

C · a pbqcr · · ·

If x ∈ R then√

x2 = |x | and3√

x3 = x,

where |x | denotes the absolute value of x

|x | ={

x if x > 0,

− x if x < 0.

Note:√

c2 = |c| but x2 = c2 ⇔ x = c ∨ x = −c.

Powers (exponentials) and logarithms are inverses of each other. For instance:

x . . . −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 . . . log2 y

2x . . . 164

132

116

18

14

12 1 2 4 8 16 32 64 . . . 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 · anbmco . . ..

All variables are positive numbers.

Series A

1. (p4q2)3 · (p2q5)2 =

2.(−a5b2)4

(a3b)3=

3.(−2cd4)3

2(3c2d)2=

4. (−3a√

b)3 =5.√

2ab2 · 2√a =

6.2p 3√

q2

4√

p3q=

7. (a2b−3)2 · −3a−7b−2 =8. (3a2b)−

14 · (6a3b2)

12 =

9.3a−

23 b2

2a2b−13=

10.(2a)−

14

2a−12=

Series B

1. (p3q4)2 · (p3q2)4 =

2.(−a3b2)4

(−a4b)3=

3.(−2c2d4)4

−2(3c2d)3=

4. (−2ab√

b)5 =5.√

2ab3 · 3√ ab =

6.2p 3√

q4√p3q=

7. (a−2b3)2 · −3a3b−2 =8. (3a2b)−

13 · (6a3b2)

14 =

9.3a−

25 b2

2a3b−12=

10.(2a)−

13

2a−12=

4

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 = a2 + 2ab + b2,

(a − b)2 = a2 − 2ab + b2,

(a − b)(a + b) = a2 − b2.

Important for factoring a quadratic polynomial:

(x + a)(x + b) = x2 + (a + b)x + ab.

Example:

• (x + 3)(x − 7) = x2− 4x − 21, where −4 is obtained as +3 plus −7 , and −21 as +3times −7 .

• (3a − 2√

b)2 = 9a2 − 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√7= 3√

77= 3

7

√7 (multiply both the numerator and denominator with

√7).

5


Exercises

Reduce the expressions below using the above rules. Make sure to remove all square roots fromthe denominators. Reduce square roots as much as possible.

Series A

1. (3a − b)2 =2. (−2a2 + 3a)2 =3. (3√

6− 6√

3)2 =4. (m − 2n)(3m + n) =5. (6− 2

√3)(√

3+ 2) =6. (−3a2b3 + 1

3 a4b)2 =7. (−2a

√3+√21)(2a

√3+√21) =

8. (2a − b + 1)(3a + 2b − 5) =9.( 2√

20+ 3√

5

)2 =

10. (3a − 1)(3a + 1)(9a2 + 1) =

Series B

1. (−3a + 2b)2 =2. (2a3 − 3a)2 =3. (3√

15− 2√

3)2 =4. (2m − 3n)(3m + 2n) =5. (1− 2

√5)(√

5+ 2) =6. (−2a4b3 + 1

4 a2b)2 =7. (−2a

√3+√30)(2a

√3+√30) =

8. (2a − 3b + 1)(3a + 2b − 4) =9.( 2√

20+ 3√

45

)2 =

10. (2a − 1)(2a + 1)(4a2 + 1) =

6

3.2 Factoring Polynomials


When factoring a polynomial the goal is to write the polynomial as a product of as many factorsas possible. To do so, first move as many common factors as possible outside the parentheses,and then try to factor the remaining part.

Examples:

• x2+ 7x − 44 = x2+ 11x − 4x − 4 · 11 = (x + 11)(x − 4). Look for two numbers with sum+7 and product −44.

• 2x2 + 4x − 6 = 2(x2 + 2x − 3) = 2(x2 + 3x − x − 1 · 3) = 2(x + 3)(x − 1). Look for twonumbers with sum +2 and product −3.

• x3 − 3x2 − 28x = x(x2 − 3x − 28) = x(x + 4)(x − 7) So, first move x outside theparentheses.

• x4− 16 = (x2+ 4)(x2− 4) = (x2+ 4)(x + 2)(x − 2) Here we have the form a2− b2 twice.Note: a2 + 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.

• −x2 − x + 2 = −(x2 + x − 2) = −(x + 2)(x − 1) Here you first take −1 outside theparentheses.

Exercises

Factor the following polynomials (using factors with integer coefficients only).

Series A.

1. 16x4 − 81 =2. 3x5 − 12x4 − 63x3 =3. x16 − 1 =4. x4 + x2 − 6 =5. x2 − 19x + 34 =6. x2 − 15x − 34 =7. x(x − 1)− (x − 1) =8. x(x2 − 1)+ (x − 1) =9. (3x − 2)2 − (2x + 3)2 =

10. x6 − 6x3 + 9 =

Series B.

1. 81x4 − 16 =2. 3x4 − 15x3 + 12x2 =3. x12 − 16 =4. x4 − x2 − 20 =5. x2 − 21x + 38 =6. x2 − 17x − 38 =7. 2x(x + 1)+ 2(x + 1) =8. x(x2 − 1)− (x − 1) =9. (5x − 3)2 − (3x + 5)2 =

10. x10 + 8x5 + 16 =

7

3.3 Factoring Polynomials, continued


The exercises below are also about factoring polynomials.

Examples:

• 6x3− 18x2− x + 3 = 6x2(x − 3)− (x − 3) = (6x2− 1)(x − 3). Here, it turns out that if youtake a factor outside parentheses for two terms at a time, these factors happen to coincide,and the polynomial can be split into two factors.

• 2x2 + x − 10. In cases like this, sometimes two number a and b can be found such that2x2+x−10 = (2x+a)(x+b). Here, a times b must be equal to −10 and 2b+a must beequal to +1 . This is satisfied by a = 5 and b = −2, hence: 2x2+x−10 = (2x+5)(x−2).

Exercises

Factor the following polynomials (using factors with integer coefficients only).

Series A.

1. 3x2 − 20x + 12 =2. 2x2 + 7x + 6 =3. 3x4 − 11x2 + 6 =4. −2x2 + 7x + 15 =5. 2x4 − x2 − 3 =6. x3 − 4x2 − x + 4 =7. 2x3 − 6x2 + x − 3 =8. x3 + 5x2 − 4x − 20 =9. −3x3 + 6x2 + 2x − 4 =

10. x7 + 2x4 − 15x =

Series B

1. 3x2 − 14x + 15 =2. 2x2 + 9x + 9 =3. 3x4 − 13x2 + 12 =4. −2x2 + x + 21 =5. 2x4 + 2x2 − 4 =6. x3 − 8x2 − x + 8 =7. 3x3 − 12x2 + 2x − 8 =8. 3x3 + 15x2 − 4x − 20 =9. −2x3 + 4x2 + 3x − 6 =

10. x8 − 4x5 − 12x2 =

8

Chapter 4

Fractions

A fraction, numeratordenominator , does not exist if the denominator is equal to 0, and it is indeterminate

if both the numerator and the denominator are equal to 0. Assume throughout this chapter thatdenominators are 6= 0. We have the following rules:

abac= b

cab+ c

d= ad

bd+ bc

bd= ad + bc

bdab· c

d= ac

bdab: c

d= a

b· d

c= ad

bcSo, to divide by a fraction

cd, multiply by its reciprocal

dc.

Simplify fractions by factoring its numerator and denominator.

Examples:

• a2 + abab + b2

= a(a + b)b(a + b)

= ab

• a2 − b2

a2 + 2ab + b2= (a − b)(a + b)

(a + b)2= a − b

a + b

• b − aa − b

= −(a − b)a − b

= −1

For the exercises below you need to write your answer as a single fraction. This is called “findinga common denominator”.

Examples:

• 1x − 1

− 22x − 3

= 2x − 3(x − 1)(2x − 3)

− 2(x − 1)(x − 1)(2x − 3)

= 2x − 3− 2(x − 1)(x − 1)(2x − 3)

=

− 1(x − 1)(2x − 3)

9

4. Fractions

• 1(x − 2)3

− 3x(x − 2)

= xx(x − 2)3

− 3(x − 2)2

x(x − 2)3= x − 3(x2 − 4x + 4)

x(x − 2)3=

x − 3x2 + 12x − 12x(x − 2)3

= −3x2 + 13x − 12x(x − 2)3

Remarks on the above example:

- Use the least common denominator, as this simplifies calculations considerably. Comparethis to 1

24 + 812 = 1

24 + 1624 = 17

24 , where using 24 · 12 as denominator instead of 24 would alsocomplicate calculations.

- Make sure to carefully multiply −3 with all terms of x2 − 4x + 4 between the parentheses.

- Usually, parentheses do not need to be removed from the denominator. Sometimes, theresulting fraction can be reduced further by also factoring the numerator.

Exercises

Series A.

1.3

2x − 1+ x

x + 1=

2.1√

x − 1+ 1√

x + 1=

3.1

x − 1+ 3

x + 3=

4.−3

1− x− 6x

x + 3=

5.13

2x + 3− 5

x + 1=

6.x

x − 2− 3

2x − 4=

7. − xx2 − 3

+ 22x + 1

=

8.1

x + 1− 2x(x + 1)3

9.1

x − 1+ 1(x − 1)2

+ 1x + 1

=

10. 1− 12x+ 3

x + 3=

Series B.

1.2

2x − 1− 2x

x + 1=

2.1√

2x − 3+ 1√

2x + 3=

3.x

x2 − 3+ 2

2x + 3=

4.−3

1− x+ 6x

x + 2=

5.7

2x + 3− 5

x2 + 1=

6.x

x + 3− 3

2x + 6=

7. − xx − 3

+ 2x2x + 1

=

8.1

x − 1+ 2x(x − 1)3

9.1

x − 1+ 1(x2 − 1)

+ 1x + 1

=

10. 1− 2x+ 3

x − 3=

10

Chapter 5

Trigonometry

The trigonometric functions sine, cosine, and tangent are introduced in terms of ratios of sidesin 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 unitcircle.

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 = (cosα, sinα).

x

y

O

P : Hcos Α, sin ΑL

Α

1

1

We then have in general:

tanα = sinαcosα

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 thearc 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 valuescan be found using the triangles depicted below. The table list these values for angles between 0and 1

2π . It is strongly recommended that you memorize this table.

13π

16π

12π

1

2 √3

14π

14π

12π

1

1√

2

x 0 16π

14π

13π

12π

sin x 0 12

12

√2 1

2

√3 1

cos x 1 12

√3 1

2

√2 1

2 0

tan x 0 13

√3 1

√3 und.

Note that tan x is undefined (und.) for x = 12π .

For angles exceeding 12π the corresponding trigonometric function values can be derived directly

from the definitions.

Example:

• The coordinates of point Q which corresponds to an angle 76π can be derived from the

coordinates of point P corresponding to an angle of 16π .

O

P

Q

Π��6

7 Π��6

1

1

Therefore

sin 76π = − sin 1

6π = − 12

cos 76π = − cos 1

6π = − 12

√3

tan 76π =

sin 76π

cos 76π

= − 12

− 12

√3= 1

3

√3

We call this “reducing an angle to the first quadrant”.

12

5. Trigonometry

Another type of question goes like this:

Given: sinα = − 12

√3. How large is α, under the restriction that 0 6 α < 2π?

0

1

-1��2�!!!

3

1

Use the unit circle

By definition, sinα is equal to the y-coordinate of a point on the unit circle. Since− 1

2

√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 13π = 1

2

√3.

From the figure and the above table one can readily see that sinα = − 12

√3 corre-

sponds to angles

π + 13π and 2π − 1

3π.

Thus the answer is: α = 43π ∨ α = 5

3π .

13

5. Trigonometry

Exercises

Find the values of the following trigonometric expressions. Reduce angles to the first quadrantand use the above table.

Series A

1. sin( 43π) =

2. tan(− 14π) =

3. cos( 56π) =

4. sin( 23π) =

5. tan( 854 π) =

6. sin( 32π) =

7. cos(− 74π) =

8. tan( 56π) =

9. cos( 23π) =

10. sin( 374 π) =

Series B

1. cos( 43π) =

2. sin(− 14π) =

3. tan( 56π) =

4. cos( 23π) =

5. sin(− 854 π) =

6. tan( 32π) =

7. sin(− 74π) =

8. cos(− 56π) =

9. tan( 23π) =

10. sin( 343 π) =

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, cos(−x) = cos x, tan(−x) = − tan x

sin(π − x) = sin x, cos(π − x) = − cos x, tan(π − x) = − tan x

sin x = cos( 12π − x), cos x = sin( 1

2π − x)

The following formulas are given without proof. These formulas are strongly connected. Forinstance, if you start from one formula you can find the other ones by using the simple identitiesabove:

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 + y) = tan x + tan y1− tan x tan y

tan(x − y) = tan x − tan y1+ 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

tan 2x = 2 tan x1− 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 y1− tan x tan y

,

and

tan(x − y) = tan x − tan y1+ tan x tan y

.

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

• tan 2x = 2 tan x1− tan2 x

4. For an angle x ∈ [0, 12π ] we have:

cos x = 12

√2.

Find cos(x − 16π).

5. For an angle x ∈ [ 12π, π] we have:

cos 2x = 13 .

Find sin x .

6. For an angle x ∈ [0, 12π ] we have:

cos x = 34 .

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

1. Factor: sin x + sin 2x =2. Factor:

sin(x + y)+ sin(x − y) =

3. Factor: cos 2x + sin2 x =4. Factor: 1+ sin 2x − cos2 x =5. Factor: cos 2x − 1 =6. Factor: sin2 x − 5 sin x + 4 =7. Factor: sin2 x + 5 cos x + 5 =8. Find the exact value of cos 1

8π .

9. Simplifycos2 x

1− sin x− cos2 x

1+ sin x.

10. f (x) = 1− cos 2x − sin2 x can be writtenas f (x) = a + b cos(cx).Determine a, b en c.

Series B

1. Factor: 2 sin2 x − sin 2x =2. Factor: cos 2x − cos2 x =3. Factor: 1− sin 2x − sin2 x =4. Factor: cos 2x + 7 =5. Factor: sin2 x − sin x − 6 =6. Factor: cos2 x + 3 sin x + 9 =

7. Eliminate the square root:

√1− cos x1+ cos x

=

8. Find the exact value of sin 18π .

9. Simplify

cos4 x − 2 cos2 x sin2 x + sin4 x .

10. f (x) = 3 cos2 x − sin2 x can be written asf (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

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.

5. Show that tan x = sin 2x1+ cos 2x

.

6. Show that

tan x + tan ytan x − tan y

= sin(x + y)sin(x − y)

.

7. Show that

tan2 x − sin2 x = tan2 x · sin2 x .

8. Show that

tan2 x − 1tan2 x + 1

= sin2 x − cos2 x .

9. Show that

tan( 14π + x) = 1+ tan x

1− tan x.

10. Show that 2 cos2 x − cos 2x = 1.

Series B

1. Show that sin2 12 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 2xsin 2x

.

6. Show that

1+ tan x tan y1− tan x tan y

= cos(x − y)cos(x + y)

.

7. Show that sin 2x − tan x = tan x cos 2x .

8. Show that

sin 2x1+ cos 2x

= 1− cos 2xsin 2x

.

9. Show that

tan( 14π+x)+tan( 1

4π−x) = 2cos2 x − sin2 x

.

10. Show that

cos x − sin xcos x + sin x

+ cos x + sin xcos x − sin x

= 2cos 2x

.

18

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 fis 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 useof terminology.

Under certain conditions we can take the derivative of a function f (x). In terms of the graphof f , the derivative f ′(x) is equal to the slope of the tangent at x to the graph y = f (x). Thederivative of y = f (x) with respect to x is denoted as

f ′(x) orddx

f (x) ord f (x)

dxor y′ or

dydx.

7.1 Derivatives of Elementary Functions

ddx

xn = nxn−1,

ddx

ex = ex ,

ddx

ln x = 1x,

ddx

ax = ax ln a,

ddx

sin x = cos x,

ddx

cos x = − sin x,

ddx

tan x = 1cos2 x

= 1+ tan2 x,

19

7.2 Differentiation Rules


We use the following rules to calculate derivatives:

f (x) = u(x)+ v(x) ⇒ f ′ = u′ + v′ (Sum Rule)

f (x) = cu(x) ⇒ f ′ = cu′ (Rule for Constant Multiples)

f (x) = u(x)v(x) ⇒ f ′ = u′v + v′u (Product Rule)

f (x) = t (x)n(x)

⇒ f ′ = nt ′ − tn′

n2(Quotient Rule)

Given functions f (u) and u(x), the derivative with respect to x of the composite functiony = f (u(x)) is equal to the derivative of f with respect to u multiplied by the derivative of fwith respect to u:

dydx= dy

du· du

dx(Chain Rule)

Examples:

• Calculate the derivative of the function y = 6(3x2 − 2)2.

Define: u(x) = 3x2 − 2. Then we need to differentiate y = 6u2 with respect to u andmultiply the result by the derivative of u = 3x2 − 2 with respect to x .

We have: dydu = 12u and du

dx = 6x . So

dydx= dy

du· du

dx= 12u · 6x = 12(3x2 − 2) · 6x .

Hence: y′ = 72x(3x2 − 2)

• y = 3(x2 − x)5 ⇒ y′ = 15(x2 − x)4(2x − 1).

Useful to memorize:

y = un ⇒ y′ = nun−1 · u′

y = 1u⇒ y′ = − 1

u2· u′

y = √u ⇒ y′ = 12√

u· u′

y = ln u ⇒ y′ = 1u· u′

20


Examples:

• y = 4(3x2 − 1)2 ⇒ y′ = 8(3x2 − 1) · 6x = 48x(3x2 − 1)

• y = √6x − 1 ⇒ y′ = 1

2√

6x − 1· 6 = 3√

6x − 1

• y = 32x3 − 1

⇒ y′ = − 3(2x3 − 1)2

· 6x2 = − 18x2

(2x3 − 1)2

• y = ln(1− x) ⇒ y′ = 11− x

· −1 = − 11− x

= 1x − 1

• y = 3x2−5 ⇒ y′ = 3x2−5 · ln 3 · 2x = 2x · ln 3 · 3x2−5

• y = (2 sin2 x − 1)3 ⇒ y′ = 3(2 sin2 x − 1)2 · 4 sin x · cos x = 12 sin x cos x · (2 sin2 x − 1)2

21


Exercises

Find the derivatives of:

Series A

1. y = −3(1− 2x)5

2. y = x√x2 − 1

3. y = 35− x2

4. y = 3(3x − 1)3

5. y = 6 · 32x−1

6. y = 2 ex2−1

7. y = x ln(3x + 4)

8. y = ln( x

x + 1

)9. y = x3 · 22x

10. y = 3x√

3x − 1

11. y = ln x1− ln x

12. y = (x2 − 3x) · ex

13. y = tan3 x

14. y = sin x − cos xsin x + cos x

15. y = sin xcos2 x − 1

Series B

1. y = 2x ln 3x

2. y = ln xx

3. y = √sin 2x

4. y = 1√3x2 + 1

5. y = x2 − 7x − 8x2 − 1

6. y = sin2(2x − 16π)

7. y = ln2 xsin x

8. y = x√x2 + 1

9. y = ln(√

3x − 1)

10. y = e2 sin2 x−1

11. y = ex +1ex

12. y = ln4 x

13. y = sin2 x · cos x

14. y = sin x1− cos x

15. y = cos xcos2 x − 1

22

Chapter 8

Antiderivatives

Antiderivatives play an essential role in integral calculus. Taking antiderivatives is the reverseoperation of taking derivatives.

Knowledge of differentiation is therefore required.

Elementary formulas (with a 6= 0, n 6= −1, andc a constant of integration):∫

axn dx = an + 1

xn+1 + c∫ax−1 dx = a ln |x | + c∫eax dx = 1

aeax +c∫

sin(ax) dx = −1a

cos(ax)+ c∫cos(ax) dx = 1

asin(ax)+ c∫

(ax + b)n dx = 1a· 1

n + 1(ax + b)n+1 + c∫

1ax + b

dx = 1a

ln |ax + b| + c∫1

cos2 xdx = tan x + c

Always make sure to check the antiderivative by differentiating it.

23

8. Antiderivatives

Exercises

Find the antiderivatives of the following functions.

Series A

1. (2x − 1)3

2. (5− x)2

3.√

3x − 4

4.1

(2x + 3)4

5. sin 2(x − 16π)

6.1√

x + 3

7.x2 + 1

x

8. sin x + e3x

9. tan2 x

10.2

3− 2x

Series B

1. (3x + 2)3

2. (8− 2x)2

3.√

2x − 3

4.1

(2x − 1)5

5. cos 12(x − 1

3π)

6.2√

x − 5

7.x3 − 1

x

8. cos 2x + e2x

9. tan2 x + 2

10.3

2− 3x

24

Chapter 9

Graphing Functions

The goal of the exercises below is to sketch the graphs of the given functions without using anyelectronic 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 aspoints 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 calculatedeasily.

From the graph of y = f (x) one can obtain the graphs of related functions by scaling the axesand 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 af (x/b) „ multiplying the x -value by bf (x − c) „ shift it c units to the rightf (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

b

)+ d

from the graph of y = f (x), then we have apply this order:

first scale, then shift.

25

9. Graphing Functions

Elementary Functions

−2 −1 0 1 2

0

1

2

3

4

y = x2

0 1 2 3 4 5

0

1

2

y = √x

−2 −1 0 1 2−0.5

0

0.5

1

1.5

2

y = |x|

−2 0 2

0

1

2

3

4

5

y = exp(x)

0 1 2 3 4 5−2

−1

0

1

2

y = ln(x)

−2 −1 0 1 2−2

−1

0

1

2

y = x3

0 0.5π π 1.5π 2π 2.5π

−1

0

1

y = sin(x)

0 0.5π π 1.5π 2π 2.5π

−1

0

1

y = cos(x )

−2 0 2−3

−2

−1

0

1

2

3

y = 1/x

−π 0.5π 0 0.5π π 1.5π 2π−5

0

5

y = tan(x)

26


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 = 12π + kπ , k ∈ Z.

Example of scaling and shifting:

• The graph of y = tan 2(x − 14π) 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 12π ) and shifting it by 1

4π to the right.

− 34π − 1

2π − 14π 3

4π12π1

4π0

0

1

2

−1

−2

27


Exercises

Series A

1. f (x) = (x + 2)4

2. f (x) = (x − 2)2 + 3

3. f (x) = −x3 + 8

4. f (x) = x2 + 6x + 9

5. f (x) = 6(x + 1)5

Series B

1. f (x) = 1x

2. f (x) = 4x − 2

3. f (x) = 3+ 1x − 3

4. f (x) = 3x2

5. f (x) = 4(2− x)2

Series C

1. f (x) = 2x

2. f (x) = ( 12)

x + 3

3. f (x) = 2x−2 + 1

4. f (x) = 3−x − 1

5. f (x) = ex−1

Series D

1. f (x) = log2 x

2. f (x) = log2(x + 2)

3. f (x) = log x2

4. f (x) = 2 log x

5. f (x) = ln(x − e)

Series E

1. f (x) = √x + 2

2. f (x) = 4− 2√

x

3. f (x) = 2+√x + 4

4. f (x) = 3√

x

5. f (x) = 6√

x2 − 1

Series F

1. f (x) = sin 3x

2. f (x) = 2 cosπx

3. f (x) = 12+ 8 sin 13π(x − 1)

4. f (x) = sin2 x

5. f (x) = tan 12(x + 1

4π)

28

Chapter 10

Equations and Inequalities

10.1 Polynomial Equations

The following rules apply when solving equations:

• ax + b = 0 ⇔ ax = −b ⇔ x = −ba

• A · B · C · .... = 0 ⇔ A = 0 ∨ B = 0 ∨ C = 0 ∨ ....

• A2 = B2 ⇔ A = B ∨ A = −B

• A · B = A · C ⇔ A = 0 ∨ B = C

• Quadratic equation: ax2+ bx + c = 0 ⇔ x = −b ±√b2 − 4ac2a

(quadratic formula)

We do not need to use the quadratic formula in simple cases such as

ax2 + c = 0, ax2 + bx = 0, x2 + 2bx + b2 = 0,

or when the expression can easily be factored:

2x2+4x−6 = 2(x2+2x−3) = 2(x2−x+3x−3·1) = 2(x−1)(x+3) = 0 ⇔ x = −3 ∨ x = 1.

Also completing the square1 may be a useful option:

x2+ 2x − 4 = (x2+ 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):

anxn + an−1xn−1 + an−2xn−2 + . . .+ a1x + a0 = 0.

Solutions may sometimes be found by factoring the polynomial.

1 Note: the quadratic formula can be derived by completing the square:

ax2 + bx + c = a(x2 + b

a x + ca) = a

(x2 + 2 b

2a x + ( b2a )

2 − ( b2a)2 + c

a) = a

[(x + b

2a)2 − b2−4ac

(2a)2

]= 0.

29


Examples:

• Solve: x5 − 4x3 − 27x2 + 108 = 0

x3(x2 − 4)− 27(x2 − 4) = 0

(x3 − 27)(x2 − 4) = 0

(x3 − 27)(x − 2)(x + 2) = 0

x = 3 ∨ x = 2 ∨ x = −2

• Solve: (x2 − 14)(x + 4) = 5x(x + 4):

(x2 − 14)(x + 4)− 5x(x + 4) = 0

(x2 − 14− 5x)(x + 4) = 0

(x + 2)(x − 7)(x − 4) = 0

x = −2 ∨ x = 7 ∨ x = −4

• Solve: (x − 4)(x2 + 8x + 12) = (2x + 3)(x2 + 8x + 12):

x2 + 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. 2x2 + 7x − 4 = 0

2. x4 + 6 = 7x2

3. x3 + 6x = 7x2

4. x4 − 42 = x2

5. x4 − 39x2 = 10x3

6. x3 − 3x2 = (x − 3)(x + 20)

7. (x − 2)3 = x − 2

8. 3x3 − x2 − 12x + 4 = 0

9. (x2 − 4)(x + 3) = (x − 2)(4− x2)

10. x6 − 4x4 = 4x2 − 16

Series B

1. 3x2 + 7x − 6 = 0

2. x4 = 2x2 + 24

3. x4 − 24x2 = 10x3

4. x4 − 12 = x2

5. x4 − 33x2 = 8x3

6. x3 + x2 = (x + 1)(x + 2)

7. 3x2 + 4x − 4 = 0

8. x3 − 3x2 = (x − 3)(x + 12)

9. (x2 − 4)(x − 3) = (x + 2)(x − 3)

10. 3(x − 1)2(x + 1) = (x + 1)2

30

10.2 Polynomial Inequalities


Basic rules (assume a > 0)

• A 6 B ⇔ A − C 6 B − C

⇔ a A 6 aB ⇔ −a A > −aB (inequality is reversed)

• A2 6 a ⇔ −√a 6 A 6√

a

A2 > a ⇔ A 6 −√a ∨ A >√

a

Analogously for > and <.

Examples:

−3x + 8 < 4 ⇔ − 3x < −4 ⇔ x >43

as dividing or multiplying by a negative number reverses the inequality. Note:

0 < x < y ⇔ 0 < 1y <

1x

x < y < 0 ⇔ 1y <

1x < 0

x < 0 < y ⇔ 1x < 0 < 1

y

In all these cases we multiply by (xy)−1. In the first two cases this is positive, while in the lastcase 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 thesolution 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 thereal line, we also indicate where the function is equal to zero. Sign changes only take place at thezeros on the real line.

Two example sign charts:

• f (x) = (x + 3)(x − 1)(x − 2) has sign chart:

0 0 0

−3 1 2

− + − +

Evaluate the value of f (x) at simple values for x (other than the zeros). If the function valueis positive (or negative) then the same holds everywhere between the neighboring zeros. Ateach zero on the real line, the sign changes.

31


• g(x) = (x + 3)3(x − 1)(x − 2)2 has sign chart:

000 0 00

−3 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 alsomeans that the sign changes three times, since the sign changes once for each zero. So, ifg(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). Thereforethe 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 signchanges if n is odd.

Examples

• Solve: x3 + 8x 6 6x2.

First rewrite as: x3 − 6x2 + 8x 6 0. Next factor the left-hand side:

x(x2 − 6x + 8) = 0

x(x − 2)(x − 4) = 0

Sign chart:0 0 0

0 2 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:000 0 00

−3 1 2

+ − + +

Hence the solution is: x 6 −3 ∨ x > 1

or, in terms of intervals: (−∞,−3] ∪ [1,∞)

32


Exercises

Solve the following inequalities:

Series A

1. x2 6 25

2. (x − 3)2 > 1

3. (x − 2)2 6 14

4. 8 < x2 + 2x

5. 3(x + 1)− 2(2x + 3) > −(x − 2)

6. 35 < x2 + 2x

7. (x2 − 4)(x − 4)2 6 0

8. 3(x + 1)− 2(2x + 3) > 5(x − 2)

9. (x2 − 7x + 12)(x2 + 2x − 24) 6 0

10. x6 − 9x3 + 8 6 0

Series B

1. x2 > 16

2. 9x2 6 16

3. (x − 2)2 > 19

4. 15 < x2 + 2x

5. 3(x − 1)− 2(2x + 3) > 5(x − 2)

6. x2 − 1 > 9− 3x

7. (x2 − 9)(x − 3)2 6 0

8. x2(3x − 5)− (2x + 3)(3x − 5) > 0

9. (x − 1)(x − 2)2(x − 3)3 > 0

10. 3x3 − 12x > x4 − 4x2

33

10.3 Equations Involving Fractions


The following rules apply when solving equations:

• AB= 0 ⇔ A = 0 ∧ B 6= 0.

• AB= C

D⇔ A · D = B · C ∧ B 6= 0 ∧ D 6= 0 (cross-multiplication)

Cross-multiplication and finding common denominators are important techniques when solvingequations involving fractions.

Make sure to check the solutions found, as the denominators should all be different from 0!

Examples:

• 2x + 83x − 6

= x + 55

Cross-multiplication yields:

(3x − 6)(x + 5) = 5(2x + 8)

3x2 + 9x − 30 = 10x + 40

3x2 − x − 70 = 0

3x(x − 5)+ 14(x − 5) = 0

(3x + 14)(x − 5) = 0

x = − 143 ∨ x = 5 (all solutions are valid)

• x2 − 3x + 2x − 2

= 5

We obtain:

x2 − 3x + 2 = 5x − 10

x2 − 8x + 12 = 0

(x − 2)(x − 6) = 0

x = 2 ∨ x = 6 (solution x = 2 is invalid)

x = 6

34


• 6x − 1

+ 5x + 1

= 3

We obtain:

6(x + 1)+ 5(x − 1)(x − 1)(x + 1)

= 3

11x + 1x2 − 1

= 3

11x + 1 = 3x2 − 3

3x2 − 11x − 4 = 0

3x2 − 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

1.3

x + 1= 2

x + 2

2.x − 3x − 1

− 3 = xx + 2

3.x − 3x − 1

− 2 = x − 1x − 3

4.x3 − 4x2

x − 4= 3

5.1x2− 2 = 1

x

6.3

2x − 1+ x = 4

7.2x + 3

x− x + 1

x − 2= 7

8.2x − 1

5+ x2 − 3

2x= 2

9.1

(x + 1)2= x + 8

x3 + 2

10.3x − 4

2+ x2

x + 2= 2

Series B

1.x2 + 3x − 2

x + 1= 4

2.3x2 + 5x − 2x2 + 3x + 2

= 2

3.3x2 + 6x + 1

x2 + 2= 2

4.3x3 + 6x

x2 + 2= 2

5.4x2+ 4

x= 3

6.3x − 1x − 5

− 3 = x

7.x

x − 4− x

x + 3= −7

2

8.x2 − 4x2 + 4

− 1x − 3

= 1

9.x

x − 1+ 2x

x + 2= 3

10.1x+ 1

x2− 1

x3= 1

35

10.4 Inequalities Involving Fractions


Fractions may change in sign not only at zeros of the numerator, but also at zeros of the denom-inator. Sign charts therefore indicate both the zeros of the numerator (0) and the zeros of thedenominator (×). The function is undefined for the zeros of the denominator.

Example

• 2x − 1

− 5x + 3

6 1

2(x + 3)(x − 1)(x + 3)

− 5(x − 1)(x − 1)(x + 3)

− (x − 1)(x + 3)(x − 1)(x + 3)

6 0

−x2 − 5x + 14(x − 1)(x + 3)

6 0

(x + 7)(x − 2)(x − 1)(x + 3)

> 0

Sign chart:

0 0XX

−7 −3 1 2

+ − + − +

so x 6 −7 ∨ − 3 < x < 1 ∨ x > 2 .

Again: if a zero of the numerator (0) or the denominator (×) occurs an even number oftimes, the sign of the function does not change.

36


Exercises

Solve the following inequalities involving fractions.

Series A

1.x2 − 2x − 15x2 + 4x + 3

6 0

2.1

x2 − 3x − 28> 0

3.5

2x + 1− 2

x − 36 3

4.2x2

2x + 3− x2

x + 2> −3

5.3x

2x − 1>

2x − 5x − 2

6.x − 2

x6

x − 1x − 3

+ 1

7.3x − x2

2x − 26 1

8.1

x − 1+ 1(x − 1)2

− 1(x − 1)3

< 1

9.x2 − 2x − 15x2 − 4x + 3

− 1 > 0

10.2x + 3

x− 2 >

3x − 8x − 2

Series B

1.x − 3

x2 − 3x − 28< 0

2.10x

x3 + x> 1

3.1x3− 1

x2− 1

x> −1

4.x

2x − 5+ 6

x + 16 2

5.(x2 − 1)(x + 3)(x + 1)(x − 3)2

6 0

6.x

x + 4− 1 6

x − 33

7.x2 − 4x + 3x2 − x − 12

>56

8.x

3x − 1+ 3

2x + 1> 1

9.x

x3 − 2x>

12

10.24

x + 3− 3

x − 4<

76

37

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

aexpression in x = anumber

with a > 0.

Since ax is an increasing function (if a > 1) or a decreasing function (if a < 1), one can simplyequate the exponents. Subsequently, solve the resulting equation.

Use the rules for exponents, for example 4x = (22)x = (2x)2 = 22x .

Examples

• ex+1 =(1

e

)x−6

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

a2 − 6a + 8 = 0

(a − 2)(a − 4) = 0

a = 2 ∨ a = 4

2x = 2 ∨ 2x = 4

x = 1 ∨ x = 2

38


Exercises

Solve the following exponential equations.

Series A

1. 2x+1 + 2x+3 = 320

2. ex+1 = ( 1e )

x−2

3. 3x+3 = 2160+ 3x−1

4. e2x −12 · ex +25 = 0

5. 2x+3 = 60+ 2x−1

6. 163x+3 = 8x2+4

7. 22x + 64 = 2x+4

8. 33x − 2 · 32x+1 = 3x+3

9. 2x + 23−x = 6

10. ex = 2 · e−x +1

Series B

1. 3x+3 = 6+ 3x+2

2. 9x = 13

√3

3. 3x+2 + 3x−1 = 2827

4. 4 · 3x+1 − 32x = 27

5. 12 ex = 5− e2x

6. 4 · 32x+1 − 33x = 3x+3

7. 2log3 x = 14

8. ( 13

√3)x = 9

9. 6 · ( 14)

x = 32(

12)

x

10. 600 · (0.4)x = 150 · (0.8)x

39

10.6 Exponential Inequalities


For exponential inequalities you must be careful with bases smaller than 1.

For instance, 2x > 24 ⇔ x > 4 but ( 12)

x > ( 12)

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

5x−1 + 5x−2 > 6√

5 ⇔ 5x · 5−1 + 5x · 5−2 > 6 · 5 12 .

Multiply both sides by 52:

5 · 5x + 5x > 6 · 5 52 ⇔ 6 · 5x > 6 · 5 5

2 ⇔ x >52.

• 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 + 273x

6 12.

Assume 3x = a and note that therefore a > 0 !

a + 27a

6 12 ⇔ a2 + 27 6 12a.

This is valid because a > 0.

a2 − 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


Exercises

Solve the following exponential inequalities.

Series A.

1. ( 12)

2x−1 < 8

2. 3+ 2x 6 2x+2 − 3

3. (2x − 4)(2x − 8) > 0

4. 8x−1 > ( 14)

x

5. 2x + 32 · 2−x > 12

6.6− 5x

51−x< 1

7. 4x > 14 · 23x

8. 2x + 8 · 2−x > 6

9. 26x − 4x+1 > 0

10. 33−2x − 4 · 31−x + 3 > 0

Series B

1. 1− ( 12)

2x−2 > 0

2. 9x + 3x+1 > 18

3.22x − 82x − 4

6 0

4. 9x+1 > 127

5. 2x + 8 · 2−x > 9

6. 3x + 33−x < 12

7. 5x − 2 · 5x−1 < 75

8. ( 12)

3x − ( 12)

2x > 0

9. 6 · 5x − 52x < 5

10. ( 13)

2x−3 − 4 · ( 13)

x−1 − 15 < 0

41

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 = 1, loga a2 = 2, (etc.)

loga ax = x

aloga x = x provided x > 0

loga x + loga y = loga xy

loga x − loga y = loga (xy )

loga xr = r loga x

loga x = loga p · logp x

loga x = logp x

logp a= ln x

ln a= log x

log a

Note: y = n loga x ⇒ y = loga xn , but y = loga xn 6⇒ y = n loga x .

Make sure to verify the validity of the solutions found, as the argument of a logarithm must bepositive (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 = 67 (check: the solution is valid).

• Solve: ln2 x − ln x2 = 3.

This is equivalent to (why?): ln2 x − 2 ln x − 3 = 0.

Suppose ln x = a

a2 − 2a − 3 = 0 ⇔ (a + 1)(a − 3) = 0 ⇔ a = −1 ∨ a = 3

⇔ ln x = −1 ∨ ln x = 3 ⇔ x = e−1 ∨ x = e3

42


Exercises

Solve the following equations.

Series A

1. logx 16 = 8

2. 3log2(x−1) = 9

3. ln(x2 − 7x + 7) = 0

4. log2 x − 5 = log 12(x + 14)

5. log2(x − 1)+ log2(x + 13) = 5

6. log(x2 − 20x) = 2

7. ln(7− x) = 12 ln(x − 1)

8. ln(x + 1)− 2 · ln 5 = 3

9. ln2 x + 6 = ln x5

10. ln2 x + 2 ln x − 3 = 0

Series B

1. log2x 27 = 3

2. 4log4(8−2x) = 2

3. ln(x + e)− 2 = ln x

4. log2(5− x)+ log2 x = 2

5. 2+ log 13(2x − 1) = 1

6. log x + log(x + 32) = 1

7. log2(x + 1)+ log 12

( 1x − 3

)= 5

8. ln(x + 3)− ln(x + 1) = 1

9. ln2 x − ln x3 + 2 = 0

10. ln(x2 − 8) = − ln( 1−2− x

)

43

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 12

x < log 12

4 ⇒ x > 4 !

Just as for exponentials, the logarithm function is decreasing if the base is less than 1, and weneed to reverse the inequality sign.

Examples:

• Solve: log 12

x > 3+ log 12(x + 3).

Here: x > 0 ∧ x > −3, so D = (0,∞).

log 12

x > log 12

18+ log 1

2(x + 3) ⇔ log 1

2x > log 1

2

18(x + 3)

hence

x 618(x + 3) ⇔ 8x 6 x + 3 ⇔ 7x 6 3 ⇔ x 6 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 = ( 3

2 ,∞).On this domain:

log3(2x − 3)+ log3 x < log3 27 ⇔ log3 x(2x − 3) < log3 27

2x2 − 3x − 27 < 0 ⇔ (2x − 9)(x + 3) < 0 ⇔ − 3 < x < 92

Taking into account the domain D, the solution is: ( 32 ,

92)

44


Exercises

Solve the following logarithmic inequalities.

Series A

1. log5(2x + 1) 6 2

2. log4(x2 − 3x) > 1

3. log2(x2 − 4x − 5) 6 4

4.2− ln x2+ ln x

6 0

5. 3− log3 x > log3(x − 6)

6. log 13

x2 > −2

7. ln(x − e)2 > 1

8. x ln x3 − ln x > 0

9. log3(x − 1) 6 2− log3(x + 7)

10. ln |x | > ln(3− 12 x)

Series B

1. log2(x2 − x) 6 1

2. log4(x2 + 6x) 6 2

3. log2(x2 − 8x + 7) 6 4

4.2− ln x1+ ln x

> 0

5. log2(x − 2) < 3− log2 x

6.ln(x − 3)− 1

ln x6 0

7. log3(22x + 1) 6 2

8. x log(x + 4)+ 4 log(x + 4) 6 0

9. log2(2x − 8) < 3

10.log(2x + 3)

log x< 2

45

10.9 Trigonometric Equations and Inequalities


A few rules:

sin x = sin a ⇔ x = a + 2kπ ∨ x = π − a + 2kπ

cos x = cos a ⇔ x = a + 2kπ ∨ x = − a + 2kπ

tan x = tan a ⇔ x = a + kπ

Here k ∈ Z, so k = 0,±1,±2, . . .

Examples:

• Solve: cos(x − 34π) = sin 2x

cos(x − 34π) = sin 2x ⇔ cos(x − 3

4π) = cos( 12π − 2x)

x − 34π = 1

2π − 2x + k · 2π ∨ x − 34π = − 1

2π + 2x + k · 2π3x = 5

4π + k · 2π ∨ − x = 14π + k · 2π

x = 512π + k · 2

3π ∨ x = − 14π + k · 2π

• Solve: 2 sin2 x = 3 cos x

2 sin2 x = 3 cos x ⇔ 2(1− cos2 x) = 3 cos x ⇔ 2 cos2 x + 3 cos x − 2 = 0

Suppose cos x = a then

2a2 + 3a − 2 = 0 ⇔ (a + 2)(2a − 1) = 0 ⇔ a = −2 ∨ a = 12

Note that cos x = −2 has no solution, hence

cos x = 12 ⇔ x = ± 1

3π + 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 − 12π) > 1.

We have tan(2x − 12π) = 1 als 2x − 1

2π = 14π + kπ , dus x = 3

8π + 12 kπ .

The function is undefined if 2x − 12π = 1

2π + kπ , so x = 12 kπ .

Consider the graph in Chapter 9 on page 27.Hence the solution is 3

8π + 12 kπ 6 x < 1

2π + 12 kπ .

46


Exercises

Solve the following trigonometric equations and inequalities. Use R as domain.

Series A.

1. sin 2x = sin x

2. tan 2x + tan x = 0

3. sin(2x + π4 )+ sin(3x − π

4 ) = 0

4. cos 2x + cos 3x = 0

5. cos2 x + 3 sin x = 3

6. 2 sin2 x cos x − sin x = 0

7. tan 2x = 3 tan x

8. sin 2x = 1tan x

9. 3 cos2 x − sin2 x = 0

10. 2 sin2 x + sin 2x = 1

11. sin 2x − cos 2x = 1

12. 2 cos x + 3 tan x = 0

13.√

2− 2 cos 2x = − sin x

14. (tan x − sin x)(tan x + sin x) = cos2 x

15. sin x · sin 2x = cos x

Series B

1. cos 2x = cos 3x

2. tan x = sin 2x

3. cos(2x − 13π)+ sin(x − 1

6π) = 0

4. sin x − sin 3x = 0

5. cos2 x + 2 sin x cos x = sin2 x

6. sin 2x = 2 cos2 x

7. 2 tan x = tan 2x

8. 2 sin 2x = 1tan x

9. cos2 x + cos x = sin2 x

10. 2 sin2 2x + 6 sin2 x = 3

11. 6 cos2 x + 11 sin x = 10

12. sin 2x − cos 2x = 1

13. sin2 x + 2 cos2 x = 1+ sin x cos x

14. sin 2x − tan x = sin x

15. sin x + cos x = 0

Series C.

1. 2 sin 12 x > 1

2. tan 12 x > 1

3. sin2 x < cos2 x

Series D

1. 12 cos 2x < 1

2

2. tan 2x 6 0

3. tan x > sin x

47

10.10 Equations Involving Square Roots


Equations involving square roots are often solved by squaring both sides of the equation. How-ever, this introduces additional solutions compared to the solutions to the original equation.Phrased differently:

y = √x ⇒ y2 = x,

but, in general,

y2 = x 6⇒ y = √x .

Always remember that the expression under the square root sign cannot be negative, and thatthe square root is a non-negative number too. Check the solution found by substituting it in theoriginal equation.

Examples:

• Solve:√

2x + 1 = 2x − 5√

2x + 1 = 2x − 5 ⇒ 2x + 1 = (2x − 5)2

2x + 1 = 4x2 − 20x + 25 ⇔ 4x2 − 22x + 24 = 0

2x2 − 11x + 12 = 0 ⇔ (x − 4)(2x − 3) = 0

x = 4 ∨ x = 32

Since 2x − 5 < 0 for x = 32 , only x = 4 remains as solution.

• Solve:√

x +√2x + 1 = 5

√x +√2x + 1 = 5 ⇒ x + 2

√2x2 + x + 2x + 1 = 25

2√

2x2 + x = 24− 3x ⇒ 8x2 + 4x = 576− 144x + 9x2

x2 − 148x + 576 = 0 ⇔ (x − 4)(x − 144) = 0

x = 4 ∨ x = 144.

The only solution is x = 4.

48


Exercises

Solve the following equations involving square roots.

Series A

1.√

4x + 1 = x − 1

2.√

2x − 1 = x − 8

3.√

x − 2 = √2x + 3− 2

4. 2x − 3√

14− x = 8

5.x + 3√2x + 1

= 3

6. x√

x + 3 =√

x + 3x

7. x2 − 3√

13+ x2 + 3 = 0

8.√

x −√x − 1 = x

9. 4√

4− p − 13(√

4− p)3 − p√

4− p = 83

10. For which values of p is the graph of

f (x) = 3x −√2x + 1tangent to the graph of

g(x) = 2x + p ?

Series B

1.√

2x + 1 = x − 7

2.√

2x + 11 = 12− x

3.√

x − 1 = √2x + 5− 2

4.x +√xx −√x

= 2

5.x +√13− x

2x − 1= 1

6.

√4x2 + xx − 1

= √x

7.x2

√2x2 + 3

=√

2x2 + 3x2

8.√

32 x −√8− x = 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− xtangent to the graph of

g(x) = p − 12 x + 1?

49

10.11 Inequalities Involving Square Roots


Sometimes it helps to draw two graphs and compute the points of intersection. But keep thedomain 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 + 4x2

4x2 − 89x + 400 = 0

The quadratic formula yields: x = 254 ∨ x = 16 . Only x = 25

4 satisfies the originalequation.If we now substitute, for example, x = 4 (4 < 25

4 ) in the inequality, we get a value lessthan 20. But if we substitute, for example, x = 9 , we get a value larger than 20.

The solution is: [0, 254 ).

• Solve:

3+√x√x − 5

6 3.

The domain is D = (5,∞). Then the fraction is always positive.

First solve:

3+√x√x − 5

= 3.

We have:

3+√x√x − 5

= 3 ⇔ 3+√x = 3√

x − 5 ⇒ 9+ 6√

x + x = 9(x − 5)

6√

x = 8x − 54 ⇔ 3√

x = 4x − 27 ⇒ 9x = 16x2 − 216x + 729

16x2 − 225x + 729 = 0

The quadratic formula yields: x = 8116 ∨ 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 lessthan 3. But if we substitute, for example, x = 14 , we get a value larger than 3. This meansthat we need to take values larger than 9.

The solution is: [9,∞).

50


Exercises

Solve the following inequalities involving square roots.

Series A

1.√

2x + 7 6 x − 4

2. x − 5√

x + 6 < 0

3.

√5− x√5+ x

> 1

4. x −√2x + 1 6 1

5.√(x − 3)2 > 1

3

6. 7+√3x − 6 > 2x

7.√

x2 + x + 5 6 x + 1

8.√

x + 2 6 |x |

9.x +√xx −√x

< 3

10.√

x + 1 < |x − 5|

Series B

1.√

x − 3 < 12 x − 1

2. x −√x − 3 < 5

3.

√3x + 4√2x − 4

> 2

4. x −√x − 4 6 6

5.√(x − 3)2 > 1

6. 12

√x2 + 3 <

√x

7.(x − 1)

√x − 1√

x + 5> 1

8.√

x >√

2x − 7− 3

9.6√

x2 − 1> 3

10.√

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 frac-tions. This is called rationalizing denominators. In case of just a square root, this is done bymultiplying top and bottom by this square root.

Example:

2√3= 2√

3·√

3√3= 2√

33= 2

3

√3.

In more complex cases we can apply the rule: (a − b)(a + b) = a2 − b2.

Example:

1

3−√3= 1

3−√3· 3+√3

3+√3= 3+√3

32 − (√3)2= 3+√3

6= 1

2+ 1

6

√3

53

11. Rationalizing Denominators (extra)

Exercises

Rationalizing the denominators:

Series A

1.1√5− 2√

2− 1=

2.1√3− 2√

6=

3.3

2√

3+ 1√

3=

4.2+√3√

3=

5.

√3+√2√3−√2

=

6.1

3√

2− 2√

3=

7.

√2√

18−√8=

8.1

(1+√2)2=

9.1√

2− 1=

10.

√3

2+√3=

Series B

1.2√3− 3√

5− 1=

2.1√2− 3√

6=

3.3

2√

5− 1√

5=

4.3+√6√

3=

5.

√5−√2√5+√2

=

6.2

2√

5−√3=

7.

√3√

48−√12=

8.1

(3−√2)2=

9.1√

3− 1=

10.

√5

1−√5=

54

Chapter 12

Partial Fraction Decomposition A (extra)

In the fractions given below, the denominator is a product of two factors or can be written assuch.

The goal is to split the given fraction into a sum of two fractions of which the denominators arethese two factors, respectively. This is called partial fraction decomposition.

Examples:

• We have:

x + 10(2x − 1)(x + 3)

= 32x − 1

− 1x + 3

.

This can be checked easily afterwards, but how do we actually find the numerators?

We replacex + 10

(2x − 1)(x + 3)by

A2x − 1

+ Bx + 3

where we need to compute A and B.

Write this expression as a single fraction:

A(x + 3)+ B(2x − 1)(2x − 1)(x + 3)

= (A + 2B)x + (3A − B)(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(2x + 3)(x + 4)

= A2x + 3

+ Bx + 4

= A(x + 4)+ B(2x + 3)(2x + 3)(x + 4)

= (A + 2B)x + (4A + 3B)(2x + 3)(x + 4)

.

Then A + 2B = 0 and 4A + 3B = 3 must hold. Solving this system yields A = 65 and

B = − 35 .

Therefore we can replace3

(2x + 3)(x + 4)by

65(2x + 3)

− 35(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(x − 3)2

= 2(x − 3)+ 7(x − 3)2

= 2x − 3

+ 7(x − 3)2

.

The resulting partial fraction decomposition has as denominators all powers of (x − 3) upto the power (x − 3)2 occurring in the given fraction.

Exercises

Apply partial fraction decomposition to the following fractions:

Series A

1.5

(2x − 1)(x − 3)=

2.1

x2 − 1=

3.2x

(x − 1)(x − 3)=

4.x

x2 − x − 2=

5.2x

x2 − 2x=

6.2x

(x − 2)2=

7.x + 4

9x2 − 6x + 1=

8.x2 − 1(x − 1)2

=

9.2x

(x − 1)2=

10.x4 − 4(x2 + 1)2

=

Series B

1.5

(2x + 1)(x + 3)=

2.1

4x2 − 9=

3.4x

(x + 1)(x − 3)=

4.3x

x2 − x − 6=

5.3x

−x2 + 3x=

6.6x

(x − 3)2=

7.x + 4

4x2 − 4x + 1=

8.x2 − 4(x − 2)2

=

9.6x

(x − 3)2=

10.x4 − 9(x2 + 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

237= 3+ 2

7.

We may obtain this result in two ways:

237= 21+ 2

7= 21

7+ 2

7= 3+ 2

7.

We also have:

7/23\321

2

hence:

237= 3+ 2

7.

The is always possible if the numerator exceeds the denominator.

• The same applies to rational functions:

3x + 1x + 3

= 3x + 9− 8x + 3

= 3− 8x + 3

Also:

x + 3/3x + 1\33x + 9

−8

hence:3x + 1x + 3

= 3− 8x + 3

.

57

13. Partial Fraction Decomposition B (extra)

• We have:

2x − 1/ 12 x2 −3x + 1\ 1

4 x − 118

12 x2 − 1

4 x− 11

4 x +1− 11

4 x + 118

− 38

hence:12 x2 − 3x + 1

2x − 1= 1

4 x − 118 −

38

2x − 1.

The method used in the last two examples applies if the degree of the numerator is larger than orequal to the degree of the denominator.

Exercises

Apply partial fraction decomposition in the exercises below.

Series A

1.6x2 − 11x + 5

x − 1=

2.x2 + 8x2x − 1

=

3.x2 − 3x + 2

x − 2=

4.x3

x2 − 1=

5.x4 − 1x − 1

=

6.x2 + 2x − 3

1− x=

7.2x2 − 4x + 5

x − 1=

8.x3 − 3x2 + 3x − 1

x − 1=

9.x3 − 2x2 + 3x − 2

x2 − 4=

10.x4 + x2 − 2x3 − 2x + 1

x2 + 1=

Series B

1.4x2 + x − 5

x − 1=

2.x2 + 4x2x − 1

=

3.x2 + 5x − 14

x − 2=

4.x4

x2 + 1=

5.x4 − 16x + 2

=

6.x2 + 2x − 3

3+ x=

7.2x2 − 4x + 5

x − 2=

8.x3 − 6x2 + 5x + 6

x − 2=

9.2x3 − 2x2 + 3x − 1

x2 − 1=

10.x4 + x2 + 2x3 − 2x + 1

x2 + 1=

58

Chapter 14

Inverse Trigonometric Functions (extra)

In high school you have already seen inverse functions. For instance, log x and 10x are inversesof each other and so are

√x and x2, 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 ofeach other in the line y = x . The domain and range of a function and its inverse need not beidentical; you can check this for the above functions. The functions sin x , cos x and tan x haveinverse 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 thefigure on the left you may recognize the graph of y = sin x reflected in the line y = x .

y = arcsin(x)14π

12π

− 14π

− 12π

0

0−0.5−1 0.5 1

y = arccos(x)14π

34π

12π

π

0

0−0.5−1 0.5 1

y = arctan(x)− 1

2π

− 14π

14π

12π

0

0 1 2 3 4 5−1−2−3−4−5

We note that we take only a small piece of the sin-graph, since otherwise the reflection of thegraph would not be a function. If the domain for sin x is equal to D = [− 1

2π,12π ], the inverse is

indeed a function. In choosing the domain we include the origin and ensure that all values forsin 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 . Itsdomain is D = [−1, 1], while its range is R = [− 1

2π,12π ].

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 = (− 12π,

12π).

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(− 12

√3).

The important difference is that we now get exactly one result, as a function may take on onevalue only.

Therefore:

arcsin(− 12

√3) = − 1

3π.

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

1. arcsin( 12) =

2. arccos(− 12

√2) =

3. arctan(√

3) =4. arccos( 1

2

√3) =

5. arctan(1) =6. arcsin(−1) =7. arccos( 1

2

√2) =

8. arctan( 13

√3) =

9. arcsin(0) =10. arccos(−1) =

Series B

1. arccos( 12) =

2. arcsin(− 12

√2) =

3. arctan(−√3) =4. arcsin( 1

2

√3) =

5. arccos(1) =6. arctan(−1) =7. arcsin( 1

2

√2) =

8. arccos( 12

√3) =

9. arctan(0) =10. arcsin(−1) =

60

Chapter 15

Answers to the Exercises

2 Powers

Series A

1. p16q16

2. a11b5

3. − 49 c−1d10

4. −27a3b32

5. 2√

2ab

6. 2p14 q

512

7. −3a−3b−8

8. 314 2

12 ab

34

9. 32 a−

83 b

73

10. 2−54 a

14

Series B

1. p18q16

2. −b5

3. − 827 c2d13

4. −32a5b152

5. 3√

2ab

6. 2p−12 q

56

7. −3a−1b4

8. 3−112 2

14 a

112 b

16

9. 32 a−

175 b

52

10. 2−43 a

16

61

15. Answers to the Exercises


Series A

1. 9a2 − 6ab + b2

2. 9a2 − 12a3 + 4a4

3. 162− 108√

2

4. 3m2 − 5mn − 2n2

5. 2√

3+ 6

6. 9a4b6 − 2a6b4 + 19 a8b2

7. 21− 12a2

8. 7b − 7a + ab + 6a2 − 2b2 − 5

9. 165

10. 81a4 − 1

Series B

1. 9a2 − 12ab + 4b2

2. 9a2 − 12a4 + 4a6

3. 147− 36√

5

4. 6m2 − 5mn − 6n2

5. −3√

5− 8

6. 116 a4b2 − a6b4 + 4a8b6

7. 30− 12a2

8. 14b − 5a − 5ab + 6a2 − 6b2 − 4

9. 45

10. 16a4 − 1


Series A

1. (2x − 3)(2x + 3)(4x2 + 9)

2. 3x3(x + 3)(x − 7)

3. (x − 1)(x + 1)(x2 + 1)(x4 + 1)(x8 + 1)

4. (x2 − 2)(x2 + 3)

5. (x − 2)(x − 17)

6. (x + 2)(x − 17)

7. (x − 1)2

8. (x − 1)(x + x2 + 1)

9. (5x + 1)(x − 5)

10. (x3 − 3)2

Series B

1. (3x − 2)(3x + 2)(9x2 + 4)

2. 3x2(x − 1)(x − 4)

3. (x3 − 2)(x3 + 2)(x6 + 4)

4. (x2 + 4)(x2 − 5)

5. (x − 2)(x − 19)

6. (x + 2)(x − 19)

7. 2(x + 1)2

8. (x − 1)(x + x2 − 1)

9. 4(4x + 1)(x − 4)

10. (x5 + 4)2

62


Series A

1. (3x − 2)(x − 6)

2. (x + 2)(2x + 3)

3. (3x2 − 2)(x2 − 3)

4. (2x + 3)(5− x)

5. (2x2 − 3)(x2 + 1)

6. (x − 1)(x − 4)(x + 1)

7. (x − 3)(2x2 + 1)

8. (x + 5)(x − 2)(x + 2)

9. (2− x)(3x2 − 2)

10. x(x3 − 3)(x3 + 5)

Series B

1. (3x − 5)(x − 3)

2. (x + 3)(2x + 3)

3. (3x2 − 4)(x2 − 3)

4. (x + 3)(7− 2x)

5. 2(x2 + 2)(x − 1)(x + 1)

6. (x − 1)(x − 8)(x + 1)

7. (x − 4)(3x2 + 2)

8. (x + 5)(3x2 − 4)

9. (2− x)(2x2 − 3)

10. x2(x3 + 2)(x3 − 6)

4 Fractions

Series A

1. 2x2+2x+3(2x−1)(x+1)

2. 2√

xx−1

3. 4x(x−1)(x+3)

4. − 3(2x2−3x−3)(x−1)(x+3)

5. 3x−2(x+1)(2x+3)

6. 2x−32(x−2)

7. x+6(3−x2)(2x+1)

8. x2+1(x+1)3

9. 2x2−x+1(x−1)2(x+1)

10. 2x2+11x−32x(x+3)

Series B

1. − 2(2x2−2x−1)(2x−1)(x+1)

2. 2√

2x2x−9

3. 4x2+3x−6(x2−3)(2x+3)

4. 3(2x2−x+2)(x−1)(x+2)

5. (x−2)(7x+4)(x2+1)(2x+3)

6. 2x−32(x+3)

7. − 7x(x−3)(2x+1)

8. x2+1(x−1)3

9. 2x+1(x−1)(x+1)

10. x2−2x+6x(x−3)

63


5 Trigonometry

Series A

1. − 12

√3

2. −1

3. − 12

√3

4. 12

√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 ycos(x + y) = sin( 1

2π − x − y) = sin(( 12π − x)− y)

= sin( 12π − x) cos y − cos( 1

2π − x) sin y = cos x cos y − sin x sin ycos(x − y) = cos(x + (−y)) = cos x cos(−y)− sin x sin(−y) = cos x cos y + sin x sin y

2. We have:

tan(x + y) = sin(x + y)cos(x + y)

= sin x cos y + cos x sin ycos x cos y − sin x sin y

= tan x + tan y1− tan x tan y

tan(x − y) = tan(x + (−y)) = tan x + tan(−y)1− tan x tan(−y)

= tan x − tan y1+ tan x tan y

.

3. We have:

sin 2x = sin(x + x) = sin x cos x + cos x sin x = 2 sin x cos xcos 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

tan 2x = tan(x + x) = tan x + tan x1− tan x tan x

= 2 tan x1− tan2 x

.

4. We have: cos(x − 16π) = cos x cos 1

6π + sin x sin 16π . Since x lies in the first quadrant also

sin x = 12

√2. So cos x cos 1

6π + sin x sin 16π = 1

2

√2 · 1

2

√3+ 1

2

√2 · 1

2 = 14

√6+ 1

4

√2.

5. We have: cos 2x = 1− 2 sin2 x = 13 ⇒ 1

3 = 1− 2 sin2 x ⇒ sin2 x = 13 ⇒ sin x = ± 1

3

√3.

Since x ∈ [ 12π, π] we find: sin x = 13

√3.

64

6. sin2 x = 1− cos2 x = 1− 916 = 7

16 , hence sin x =√

716 because x ∈ [0, 1

2π ].So tan x = 1

3

√7, and tan 2x = 3

√7

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

1. (2 cos x + 1) sin x

2. 2 sin x cos y

3. (1− sin x)(1+ sin x) = cos2 x

4. sin x(sin x + 2 cos x)

5. 2(cos x − 1)(cos x + 1) = −2 sin2 x

6. (sin x − 1)(sin x − 4)

7. (cos x + 1)(6− cos x)

8. 12

√2+√2

9. 2 sin x

10. 12 ,− 1

2 en 2

Series B

1. 2 sin x(sin x − cos x)

2. (cos x − 1)(cos x + 1) = − sin2 x

3. (cos x − 2 sin x) cos x

4. −2(sin x − 2)(sin x + 2)

5. (sin x + 2)(sin x − 3)

6. (sin x + 2)(5− sin x)

7.

∣∣∣∣1− cos xsin x

∣∣∣∣ = ∣∣tan 12 x∣∣

8. 12

√2−√2

9. cos2 2x

10. 1, 2, 2

65


7.2 Differentiation

Series A.

1. 30(1− 2x)4

2. − 1

(x2 − 1)32

3. 6x(x2 − 5)−2

4. −27(3x − 1)−4

5. 4 ln 3 · 32x

6. 4x ex2−1

7. ln(3x + 4)+ 3x3x + 4

8.1

x + x2

9. x2 · 22x(2x ln 2+ 3)

10.27x − 6

2√

3x − 1

11.1

x(ln x − 1)2

12. ex(x2 − x − 3)

13. 3(tan2 x)(tan2 x + 1)

14.2

sin 2x + 1

15. − cos xcos2 x − 1

Series B

1. 2 ln x + 2 ln 3+ 2

2.1− ln x

x2

3. cos 2x(sin 2x)−12

4. −3x(3x2 + 1)−32

5. 7(x − 1)−2

6. 2 sin(4x − 13π)

7.2 ln x sin x − x ln2 x cos x

x sin2 x

8. (x2 + 1)−32

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

66

8 Antiderivatives

Series A

1. 18(2x − 1)4

2. − 13(5− x)3

3. 29(3x − 4)

32

4. − 16(2x + 3)−3

5. − 12 cos 2(x − 1

6π)

6. 2√

x + 3

7. 12 x2 + ln |x |

8. − cos x + 13 e3x

9. tan x − x

10. − ln |3− 2x |

Series B

1. 112(3x + 2)4

2. − 16(8− 2x)3

3. 13(2x − 3)

32

4. − 18(2x − 1)−4

5. 2 sin 12(x − 1

3π)

6. 4√

x − 5

7. 13 x3 − ln |x |

8. 12 sin 2x + 1

2 e2x

9. x + tan x

10. − ln |2− 3x |

67


9 Graphing Functions

x

f HxL

O-2

f HxL = Hx + 2L4 x

f HxL

O

3

7

2

f HxL = Hx - 2L2 + 3

x

f HxL

8

2f HxL = 8 - x3

x

f HxL

O-3

9

f HxL = x2 + 6 x + 9

x

f HxL

O-1

f HxL = 6 Hx + 1L5A1 A2 A3 A4 A5

x

f HxL

O

f HxL = 1��x

x

f HxL

O

-2

2

f HxL = 4

��x - 2

x

f HxL

O

3

3

f HxL = 3 +1

��x - 3

x

f HxL

O

f HxL = 3

��x2

x

f HxL

O

1

2 4

f HxL = 4

��H2 - xL2B1 B2 B3 B4 B5

x

f HxL

O

1

f HxL = 2x

x

f HxL

O

3

f HxL = 3 + 2-x

x

f HxL

O

1

f HxL = 1 + 2x-2

x

f HxL

O

-1f HxL = -1 + 3-x

x

f HxL

O

1�ãf HxL = ãx-1

C1 C2 C3 C4 C5

x

f HxLO 1

f HxL = 2logHxL

x

f HxL

O-1-2

f HxL = 2logHx + 2L

x

f HxLO

1

f HxL = logHx2L

x

f HxLO 1

f HxL = 2 logHxL

x

f HxL

O ã ã + 1

f HxL = 2 lnHx - ãLD1 D2 D3 D4 D5

x

f HxL

O

2

f HxL = �!!!

x + 2

x

f HxL

O

4

4

f HxL = 4 - 2�!!!

x

x

f HxL

O

2

4

-4

f HxL = �!!!!!!!!!!!

x + 4 + 2

x

f HxL

O

f HxL = �!!!

x3

x

f HxLO

-1

1-1

f HxL = �!!!!!

x26

- 1

E1 E2 E3 E4 E5

x

f HxLO Π

��3

2 Π��3

1

-1

f HxL = sinH3 xLx

f HxL

O 1��2

3��2

2

-2

f HxL = 2 cosHΠ xL x

f HxL

O 1 7

20

4

f HxL = 8 sinJ 1��3

Π Hx - 1LN + 12

0

0.2

0.4

0.6

0.8

–4 –2 2 4

1

f (x)

f (x) = sin2 x–4

–2

0

2

4

–6 –4 –2 2 4 6

f (x) = tan 12 (x + 1

4π)

f (x)

F1 F2 F3 F4 F5

68


Series A

1. x = −4 or x = 12

2. x = 1 or x = −1 or x = √6or x = −√6

3. x = 0 or x = 1 or x = 6

4. x = √7 or x = −√7

5. x = 0 or x = −3 or x = 13

6. x = 5 or x = −4 or x = 3

7. x = 1 or x = 2 or x = 3

8. x = 2 or x = 13 or x = −2

9. x = 2 or x = − 12 or x = −2

10. x = 2 or x = −2 or x = √2 or

x = −√2

Series B

1. x = −3 or x = 23

2. x = √6 or x = −√6

3. x = 0 or x = −2 or x = 12

4. x = 2 or x = −2

5. x = 0 or x = −3 or x = 11

6. x = 2 or x = −1

7. x = −2 or x = 23

8. x = −3 or x = 3 or x = 4

9. x = −2 or x = 3

10. x = 2 or x = 13 or x = −1

69



Series A

1. [−5, 5]2. (−∞, 2] ∪ [4,∞)3. [ 32 , 5

2 ]4. (2,∞) ∪ (−∞,−4)

5. ∅6. (5,∞) ∪ (−∞,−7)

7. {4} ∪ [−2, 2]8. (−∞, 7

6)

9. {4} ∪ [−6, 3]10. [1, 2]

Series B

1. [4,∞) ∪ (−∞,−4]2. [− 4

3 ,43 ]

3. (−∞, 53 ] ∪ [ 73 ,∞)

4. (3,∞) ∪ [−∞,−5)

5. (−∞, 16)

6. [2,∞) ∪ (−∞,−5]7. [−3, 3]8. (−1, 5

3) ∪ (3,∞)9. {2} ∪ (−∞, 1] ∪ [3,∞)

10. [−2, 0] ∪ [2, 3]


Series A

1. x = −4

2. x = 0 or x = −1

3. x = 1+√2 or x = 1−√2

4. x = √3 or x = −√3

5. x = −1 or x = 12

6. x = 1 or x = 72

7. x = 1

8. x = − 59 or x = 3

9. x = − 65 or x = − 1

2

10. x = − 85 or x = 2

Series B

1. x = −2 or x = 3

2. x = 3

3. x = 2√

3− 3 or x = −2√

3− 3

4. x = 23

5. x = − 23 or x = 2

6. x = −2 or x = 7

7. x = −4 or x = 3

8. x = −10 or x = 2

9. x = 2

10. x = −1 or x = 1

70


Series A

1. (−1, 5]2. (−∞,−4) ∪ (7,∞)3. [ 23 , 2] ∪ (3,∞) ∪ (−∞,− 1

2)

4. (−∞,−2) ∪ (− 32 ,∞)

5. (2, 5] ∪ ( 12 , 1]

6. (0, 2] ∪ (3,∞) ∪ (−∞,−3]7. [−1, 1) ∪ [2,∞)8. (1, 2) ∪ (−∞, 0) ∪ (2,∞)9. (1, 3) ∪ (9,∞)

10. (2, 3) ∪ (0, 23)

Series B

1. (3, 7) ∪ (−∞,−4)

2. (−3, 0) ∪ (0, 3)

3. (−∞,−1) ∪ (0, 1) ∪ (1,∞)4. (−∞,−1) ∪ [5,∞) ∪ [ 43 , 5

2)

5. [−3,−1) ∪ (−1, 1]6. (−4,−1] ∪ [0,∞)7. (4, 6) ∪ (−∞,−3) ∪ (13,∞)8. ( 1

3 , 2] ∪ (− 12 ,

14 ]

9. (√

2, 2] ∪ [−2,−√2)

10. (4, 6) ∪ (−∞,−3) ∪ (13,∞)


Series A

1. x = 5

2. x = 12

3. x = 4

4. x = ln 4 or x = ln 8

5. x = 3

6. x = 0 or x = 4

7. x = 3

8. x = 2

9. x = 1 or x = 2

10. x = ln 2

Series B

1. x = −1

2. x = − 14

3. x = −2

4. x = 1 or x = 2

5. x = ln 2

6. x = 1 or x = 2

7. x = 19

8. x = −4

9. x = 2

10. x = 2

71



Series A

1. (−1,∞)2. [1,∞)3. (−∞, 2) ∪ (3,∞)4. [ 35 ,∞)5. (−∞, 2] ∪ [3,∞)6. (−∞, 0) ∪ (1,∞)7. (−∞, 2)

8. (−∞, 1] ∪ [2,∞)9. ( 1

2 ,∞)10. (−∞,∞)

Series B

1. [1,∞)2. (1,∞)3. [ 32 , 2)

4. [− 52 ,∞)

5. (−∞, 0] ∪ (3,∞)6. (1, 2)

7. (−∞, 3)

8. (−∞, 0]9. (−∞, 0) ∪ (1,∞)

10. (0,∞)


Series A

1. x = √2

2. x = 5

3. x = 1 or x = 6

4. x = 2

5. x = 3

6. x = 10− 10√

2 or x = 10+ 10√

2

7. x = 5

8. x = 25 e3−1

9. x = e2 or x = e3

10. x = e or x = e−3

Series B

1. x = 32

2. x = 3

3. x = e /(e2−1)

4. x = 1 or x = 4

5. x = 2

6. x = 52

7. x = 7

8. x = e−31−e

9. x = e or x = e2

10. x = −3

72


Series A

1. (− 12 , 12]

2. (−∞,−1) ∪ (4,∞)3. [−3,−1) ∪ (5, 7]4. (0, e−2) ∪ [e2,∞)5. (6, 9]6. (−3, 0) ∪ (0, 3)

7. (−∞, e−√e) ∪ (e+√e,∞)8. (0, 1

3) ∪ (1,∞)9. (1, 2]

10. (−∞,−6] ∪ [2, 6)

Series B

1. [−1, 0) ∪ (1, 2]2. [−8,−6) ∪ (0, 2]3. [−1, 1) ∪ (7, 9]4. (e−1, e2)

5. (2, 4)

6. (3, e+3]7. (−∞, 3

2 ]8. (−4,−3]9. (3, 4)

10. (0, 1) ∪ (3,∞)

73



Series A

1. {2kπ} ∪ { 13π + 23 kπ}

2. { 13 kπ}3. { 25 kπ} ∪ { 32π + 2kπ}4. { 15π + 2

5 kπ}5. { 12π + 2kπ}6. {kπ} ∪ { 14π + kπ}7. {kπ} ∪ { 16π + kπ} ∪ { 56π + kπ}8. { 12π + kπ} ∪ { 14π + 1

2 kπ}9. { 13π + kπ} ∪ { 23π + kπ}

10. { 18π + 12 kπ}

11. { 14π + kπ} ∪ { 12π + kπ}12. { 76π + 2kπ} ∪ { 11

6 kπ + 2kπ}13. {kπ}14. { 14π + 1

2 kπ}15. { 12π + kπ} ∪ { 14π + 1

2 kπ}

Series B

1. {2kπ} ∪ { 25 kπ}2. {kπ} ∪ { 14π + 1

2 kπ}3. { 23 kπ}4. {kπ} ∪ { 14π + 1

2 kπ}5. { 38π + 1

2 kπ}6. { 12π + kπ} ∪ { 14π + kπ}7. {kπ}8. { 16π + 1

3 kπ}9. { 13π + 2

3 kπ}10. { 16π + kπ} ∪ { 56π + kπ}11. { 16π + 2kπ} ∪ { 56π + 2kπ}12. { 14π + kπ} ∪ { 12π + kπ}13. { 14π + kπ} ∪ { 12π + kπ}14. { 23 kπ} ∪ {kπ}15. { 34π + kπ}

Series C

1. ( 13π + 4kπ, 5

3π + 4kπ)

2. [ 12π + 2kπ, π + 2kπ)

3. [− 14π + kπ, 1

4π + kπ ]

Series D

1. (0+ kπ, π + kπ)

2. ( 14π + 1

2 kπ, 12π + 1

2 kπ ]3. [0+ kπ, 1

2π + kπ)

74


Series A

1. x = 6

2. x = 13

3. x = 3 or x = 11

4. x = 314

5. x = 12 or x = 0

6. x = −3 or x = 1 or x = −1

7. x = 2√

3 or x = −2√

3

8. x = 1

9. p = 4− 2 3√

2

10. p = −1

Series B

1. x = 12

2. x = 7

3. x = 10 or x = 2

4. x = 9

5. x = 4

6. x = 0 or x = 6

7. x = √3 or x = −√3

8. x = 4

9. x = 7

10. p = 2


Series A

1. [9,∞)2. [−6, 30)

3. (−5, 0)

4. [− 12 , 4]

5. (−∞, 83) ∪ ( 10

3 ,∞)6. [2, 5)

7. [4,∞)8. [−2,−1] ∪ [2,∞)9. (0, 1) ∪ (4,∞)

10. [−1, 3) ∪ (8,∞)

Series B

1. [3, 4) ∪ (4,∞)2. [3, 7)

3. (2, 4]4. [4, 8]5. (−∞, 2) ∪ (4,∞)6. (1, 3)

7. (3,∞)8. [ 72 , 64]9. (−√5,−1) ∪ (1,√5)

10. [3, 4) ∪ ( 214 ,∞)

75


11 Rationalizing Denominators (extra)

Series A

1. 15

√5− 2

√2− 2

2. 13

√3− 1

3

√6

3. 56

√3

4. 23

√3+ 1

5. 2√

6+ 5

6. 12

√2+ 1

3

√3

7. 1

8. 3− 2√

2

9.√

2+ 1

10. 2√

3− 3

Series B

1. − 14

√5− 5

4

2. 12

√2− 1

2

√6

3. 110

√5

4.√

3+√2

5. 73 − 2

3

√10

6. 217

√3+ 4

17

√5

7. 12

8. 649

√2+ 11

49

9. 12

√3+ 1

2

10. 23

√3− 3

4

√5− 3

4

12 Partial Fraction Decomposition A (extra)

Series A

1. (x − 3)−1 − 2(2x − 1)−1

2. 12(x − 1)−1 − 1

2(x + 1)−1

3. 3(x − 3)−1 − (x − 1)−1

4. 13(x + 1)−1 + 2

3(x − 2)−1

5. 2(x − 2)−1

6. 2(x − 2)−1 + 4(x − 2)−2

7. 13(3x − 1)−1 + 13

3 (3x − 1)−2

8. 2(x − 1)−1 + 1

9. 2(x − 1)−1 + 2(x − 1)−2

10. 1− 3(x2 + 1)−2 − 2(x2 + 1)−1

Series B

1. 2(2x + 1)−1 − (x + 3)−1

2. 16(2x − 3)−1 − 1

6(2x + 3)−1

3. (x + 1)−1 + 3(x − 3)−1

4. 65(x + 2)−1 + 9

5(x − 3)−1

5. 3(3− x)−1

6. 6(x − 3)−1 + 18(x − 3)−2

7. 12(2x − 1)−1 + 9

2(2x − 1)−2

8. 4(x − 2)−1 + 1

9. 6(x − 3)−1 + 18(x − 3)−2

10. 1− 6(x2 + 3)−1

76

13 Partial Fraction Decomposition B (extra)

Series A

1. 6x − 5

2. 12 x + 17

4 + 174 (2x − 1)−1

3. x − 1

4. x + 12(x − 1)−1 + 1

2(x + 1)−1

5. x + x2 + x3 + 1

6. −x − 3

7. 2x − 2+ 3(x − 1)−1

8. x2 − 2x + 1 = (x − 1)2

9. x − 2+ (7x − 10)/(x2 − 4)= x + (x − 2)−1 + 6(x + 2)−1 − 2

10. x2 − 2x + (x2 + 1)−1

Series B

1. 4x + 5

2. 2x + 9+ 9(2x − 1)−1

3. x + 7

4. x2 − 1+ (x2 + 1)−1

5. x3 − 2x2 + 4x − 8

6. x − 1

7. 2x + 5(x − 2)−1

8. x2 − 4x − 3

9. 2x − 2+ (5x − 3)/(x2 − 1)

10. x2 + 2x + (1− 4x)/(x2 + 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 agraph of a function if we take [− 1

2π,12π ] 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 ev-erywhere. So, for the domain [ 12π, 3

2π ] the resulting image should also be the graph of afunction.

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 cannow check the same for y = cos x on [0, π].

2. As above.

77


14 Inverse Trigonometric Functions (extra)

Series A

1. 16π

2. 34π

3. 13π

4. 16π

5. 14π

6. − 12π

7. 14π

8. 16π

9. 0

10. π

Series B

1. 13π

2. − 14π

3. − 13π

4. 13π

5. 0

6. − 14π

7. 14π

8. 16π

9. 0

10. − 12π

78

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