Lycée Denis-de-Rougemont 2006-2007 TABLE OF CONTENTS · Lycée Denis-de-Rougemont 2006-2007 - 3 - YAy & PRi INTRODUCTION AND HISTORY Mathematics (colloquially, maths or math) is

1MG Level 1 PRECALCULUS Theory Lycée Denis-de-Rougemont 2006-2007

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TABLE OF CONTENTS

§0 VOCABULARY ________________________________________ - 5 -

§1 NUMBERS ___________________________________________ - 7 -

1.1 Sets of numbers ___________________________________________ - 7 -

1.2 Reminder : Exponents and products__________________________ - 9 -

1.3 The real number line______________________________________ - 10 -

1.4 Intervals ________________________________________________ - 10 -

§2 ABSOLUTE VALUE ___________________________________ - 11 -

§3 SYSTÈMS OF ÉQUATIONS ______________________________ - 11 -

3.0 2x2 systems______________________________________________ - 11 -

3.1 3x3 systems______________________________________________ - 13 -

§4 THE STRAIGHT LINE__________________________________ - 14 -

4.0 Generalities _____________________________________________ - 14 -

4.1 Gradient or slope of a line _________________________________ - 14 -

4.2 Vertical shift and intersections with the axes _________________ - 15 -

4.3 Types of lines____________________________________________ - 15 -

4.4 Remarks ________________________________________________ - 15 -

4.5 Equation of a line ________________________________________ - 16 -

§5 QUADRATICS _______________________________________ - 17 -

5.0 Definition and properties__________________________________ - 17 -

5.1 Solving quadratic expressions ______________________________ - 20 -

5.2 Factorisation ____________________________________________ - 21 -

§6 SOLVING EQUATIONS _________________________________ - 22 -

6.0 Equations which reduce to quadratics equations ______________ - 22 -

6.1 Irrational equations_______________________________________ - 23 -

6.2 Rational equations _______________________________________ - 23 -

6.3 Equations involving an absolute value_______________________ - 24 -

6.4 Inequalities______________________________________________ - 25 -

6.5 Systems of inequalities ____________________________________ - 26 -


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§7 APPLICATIONS AND FUNCTIONS_________________________ - 27 -

7.0 Applications_____________________________________________ - 27 -

7.1 Functions _______________________________________________ - 28 -

7.2 Inverse function__________________________________________ - 31 -

7.3 Composite function_______________________________________ - 33 -

§8 POLYNOMIALS ______________________________________ - 34 -

8.0 Reminder _______________________________________________ - 34 -

8.1 Addition and subtraction __________________________________ - 34 -

8.2 Multiplication ___________________________________________ - 35 -

8.3 Long Division ___________________________________________ - 35 -

§9 EXPONENTIAL AND LOGARITHMIC FUNCTIONS _____________ - 38 -

9.0 Definition and properties__________________________________ - 38 -


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INTRODUCTION AND HISTORY

Mathematics (colloquially, maths or math) is the body of knowledge centered on

concepts such as quantity, structure, space, and change. It evolved, through the use of

abstraction and logical reasoning, from counting, calculation, measurement, and the

study of the shapes and motions of physical objects. Mathematicians explore such

concepts, aiming to formulate new conjectures and establish their truth by rigorous

deduction from appropriately chosen axioms and definitions.

Knowledge and use of basic mathematics

have always been an inherent and integral

part of individual and group life.

Refinements of the basic ideas are visible

in ancient mathematical texts originating in

ancient Egypt, Mesopotamia, Ancient

India, and Ancient China with increased

rigour later introduced by the ancient

Greeks. From this point on, the

development continued in short bursts

until the Renaissance period of the 16th

century where mathematical innovations

interacted with new scientific discoveries leading to an acceleration in understanding

that continues to the present day.

Today, mathematics is used throughout the world in many fields, including science,

engineering, medicine and economics. The application of mathematics to such fields,

often dubbed applied mathematics, inspires and makes use of new mathematical

discoveries and has sometimes led to the development of entirely new disciplines.

Mathematicians also engage in pure mathematics for its own sake without having any

practical application in mind, although applications for what begins as pure

mathematics are often discovered later on.

Euclid, Greek mathematician, 3rd century BC, known today as the father of geometry; shown here in a detail of The School of Athens by Raphael.


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The evolution of mathematics might be seen to be an ever-increasing series of

abstractions, or alternatively an expansion of subject matter. The first abstraction was

probably that of numbers. The realization that two apples and two oranges have

something in common was a breakthrough in human thought. In addition to recognizing

how to count physical objects, prehistoric peoples also recognized how to count

abstract quantities, like time days, seasons, years. Arithmetic that is to say addition,

subtraction, multiplication and division, naturally followed. Monolithic monuments testify

to knowledge of geometry.

Further steps need writing or some other system for

recording numbers such as tallies or the knotted strings

called quipu used by the Inca Empire to store numerical

data. Numeral systems have been many and diverse.

From the beginnings of recorded history, the major

disciplines within mathematics arose out of the need to

do calculations relating to taxation and commerce, to

understand the relationships among numbers, to

measure land, and to predict astronomical events.

These needs can be roughly related to the broad

subdivision of mathematics, into the studies of quantity,

structure, space, and change.

Mathematics since has been much extended, and there has been a fruitful interaction

between mathematics and science, to the benefit of both. Mathematical discoveries

have been made throughout history and continue to be made today.

A quipu, a counting device used by the Inca.


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§0 VOCABULARY

French name English name Symbol

Virgule comma ,

Point full stop / period (US) .

Point (produit) dot .

Point décimal point .

Point d’interrogation question mark ?

Point d’exclamation exclamation mark !

Deux points colon :

Point virgule semi colon ;

Paragraphe paragraph §

Parenthèse ouvrante opening bracket / (parenthesis) (

Parenthèse fermante closing bracket / (parenthesis) )

Parenthèses parenthesis, round brackets ( )

Crochet ouvrant opening square bracket [

Crochet fermant closing square bracket ]

Accolade ouvrante opening curly bracket {

Accolade fermante closing curly bracket }

Multiplié par, fois multiplied by, times ·

Divisé par divided by : or ÷ or /

Plus plus +

Moins minus -

Strictement inférieur less than <

Strictement supérieur greater than >

Inférieur ou égal less or equal to ≤

Supérieur ou égal greater or equal to ≥

Appartenir à to belong to ∈

Plus ou moins plus or minus ±

Slash slash /

Anti-slash back slash \

Tel que such as |


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Ensemble set

Résoudre to solve

Idem ditto "

Etoile star *

Droite (straight) line

Pontillé dotted

Axe axis

Axes axes

Echelle (graphe) scale

Premier (nombre) prime (number)

Carré square

Au carré squared

Additionner to add (up)

Noter to note (down) / to write down

Règle (loi) rule

Diviseur factor, divisor

Exposant exponent, index (plural indices)

Pair even

Impair odd

Calcul calculation

Calculer to calculate, to compute

Dessiner (graphe) to draw, to plot

Esquisser to sketch

Chiffre digit


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§1 NUMBERS

1.1 Sets of numbers

When we want to treat a collection of similar but distinct objects as a whole, we use the

idea of a set (ensemble). For example, the set of digits consists of the collection of

numbers 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. If we use the symbol D to denote the set of digits,

then we can write :

{ } { }Read as "set of all such that is a digit"

0, 1, 2, 3, 4, 5, 6, 7, 8, 9 is a digit

x x

D x x= =��

Natural numbers : { }0 ; 1 ; 2 ; 3 ; 4 ; ...=�

Integer numbers : { }... ; 2 ; 1 ; 0 ; 1 ; 2 ; ...= − −�

Rational numbers : ; , , 0p

p q qq

= ∈ ∈ ≠

� � �

Real numbers : �

Complex numbers : 2( 1)i = −

Diagram of Venn

Definition

A positive integer p is a prime number (nombre premier) or prime if p ≠ 1 and if its only

positive divisors are 1 and itself.

� � � �

1

4

1097

12

4

-4

-876

0

2

π−

7

3 8−

1,33

5−

3π

54

2

−

984

13

−

1,764

45

7 1,4−1

2

4

« such that »


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Remark

A set of numbers with a « * » is a set without 0.

Thus, { }* 1 ; 2 ; 3 ; 4 ; ...=� , { }* ... ; 2 ; 1 ; 1 ; 2 ; ...= − −� , …

The empty set is denoted by the symbol { } or ∅ .

Generally, when we consider two sets A and B they are formed of elements of a certain

set U called universal set (univers).

Definitions

1) A set A is included in a set B if every element of A belongs to B. We say that A

is a subset (sous-ensemble) of B. We note it BA ⊂ .

Examples

⊂� �

{ } { }; ; ; ;a e i o u alphabet⊂

2) Intersection : { }A B x x A x B∩ = ∈ ∈and

Examples

{ } { } { }1 ; 2 ; 3 2 ; 0 ; 1 ; 3 ; 5 1 ; 3∩ − =

{ } { } { }; ; ; ;red green blue white green yellow green∩ =

3) Union : { }A B x x A x B∪ = ∈ ∈or

Examples

{ } { } { }1 ; 2 ; 3 0 ; 1 ; 3 ; 5 0 ; 1 ; 2 ; 3 ; 5∪ =

{ } { }

{ }

; ; ;

; ; ;

red green white green yellow

red white green yellow

∪ =

=

4) Difference : A \ B { }x x A x B= ∈ ∉and

Examples

{ } { } { }0 ; 1 ; 3 ; 5 \ 1 ; 2 ; 3 0 ; 5=

{ } { }

{ }

; ; \ ; ;

;

red green blue white green yellow

red blue

=

=

B

A

U

B A

U

B A

U

B A

U


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5) Complement : A

UA C= = U \ A

Example

{ } { }

{ }

0 ; 1 ; 2 ; 3 ; 4 ; 5 ; 6 1 ; 2 ; 3

0 ; 4 ; 5 ; 6

U A

A

= =

=

1.2 Reminder : Exponents and products

Integer exponents provide a shorthand device for representing repeated multiplications of

a real number a :

factors

...n

n

a a a a= ⋅ ⋅ ⋅��

where a is called the base and n the exponent or power. We read an as “a raised to the

power n” or as “a to the nth power” and shortly as “a power n”.

We define that :

if 0a ≠ then 0 1a =

if 0a ≠ and n a positive integer then 1n

na

a

−=

if *n ∈� then 1

nna a=

Rules of indices

m nm n aa a+

⋅ = The multiplication rule

m nm n aa a−

÷ = The division rule

)(m n m naa

⋅= The power-on-power rule

( )m m ma b a b⋅ = ⋅ The factor rule

Example

It is possible to combine all these rules : 7 2 3 21 6 3(5 ) 125a r p a r p=

Some important products

1) ( )a b c ab ac⋅ + = + distributivity of the multiplication over the addition

2) If 0ab = than either 0a = , 0b = or 0a b= =

3) 2 2 2( ) 2a b a ab b+ = + + Example 2 2( 5) 10 25x x x+ = + +

4) 2 2 2( ) 2a b a ab b− = − + Example 2 2( 5) 10 25x x x− = − +

5) 2 2( )( )a b a b a b+ − = − Example 2( 5)( 5) 25x x x+ − = −

A

U


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1.3 The real number line

The real numbers can be represented by points on a line called real number line (axe des

reels). Every real number corresponds to a point of the line and each point of the line has

a unique real number associated with it.

The point of the line which is associated with the real number 0 is called the origin and is

noted O. The point 1 unit to the right of O corresponds to the number 1, etc. Notice that an

arrowhead on the right end of the line indicates the direction in which the numbers

increase.

1.4 Intervals

A set containing all real numbers between two given endpoints is called interval.

If a and b are two real number such that a < b, we note :

[ ] { }; anda b x x a x b= ∈ ≤ ≤� closed interval

] [ { }; anda b x x a x b= ∈ < <� open interval

[ [ { }; anda b x x a x b= ∈ ≤ <�

] ] { }; anda b x x a x b= ∈ < ≤�

a and b are called endpoints (bornes) of the interval.

Example ] ] { }1 ; 5 and 1 5x x x= ∈ < ≤�

It is possible to represent an interval on the real number line. For that we place the two

endpoints and we use bold type if the endpoint belongs to the interval. Finally, we thick the

part of the line that corresponds to the interval we have to draw.

Example

Represent, on the real number line, the interval ] ]1 ; 5 .

O

1 5

O 2 3 -1 1 -2 -3 -4 4

0,5 2 -0,5 π �

… …


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§2 ABSOLUTE VALUE

The absolute value or modulus of x, denoted by the symbol x , is defined by :

if 0

if 0

x xx

x x

≥=

− <

Examples 5 5= 5 ( 5) 5− = − − =

§3 SYSTEMS OF EQUATIONS

3.0 2x2 systems

When we solve such a system, we look for x and y that satisfy both equations. Actually we

search the intersection point of two lines. For that we have 3 different methods : addition,

substitution and graphic. We are going to see these 3 methods with an example.

We consider the following system : 5 2 5 (1)

3 2 (2)

x y

x y

− =

− =

1) ADDITION

The goal is to have the same quantity of x or y in both equations but with the

opposite sign. For that, we multiply the equation (2) by –2 and we obtain :

5x – 2y = 5

–6x + 2y = –4 ⊕

–x = 1 ⇒ x = –1

We replace x in the equation (2) to find y : 3·(–1) – y = 2 ⇒ y = – 5


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2) SUBSTITUTION

The purpose is to express x in terms of y or y in terms of x.

We modify the equation (2) to have : y = 3x – 2 (3)

Then we replace y in the equation (1) by the expression (3) to obtain :

5x – 2(3x – 2) = 5

5x – 6x + 4 = 5

–x = 1 ⇒ x = –1

We replace x by –1 in the equation (3) to find y : y = 3·(–1) – 2 ⇒ y = – 5

3) GRAPHIC

The objective is to draw the two lines and the intersection gives us the

solution. The first thing to do is to express y in terms of x and then to draw the

lines :

(1) becomes y = 2

55 −x and (2) becomes y = 3x – 2

This method is not often used. Actually it is relatively long and not really

precise because if the coordinates of the intersection point are not integer

numbers, it is very difficult and even impossible to determine them precisely.

Remarks

1) Two equations multiple of each other are said to be equivalent. Consequently, the

system has an infinite number of solutions.

2) If the system has no solutions (the resolution gives us an impossible result) then the

equations are said to be incompatible.

Example Solve

=−

−=+−−

432

23)12(2

1

yx

yx

( 2) 1 ( 1)12 1 6 4 2 6 5(2 1) 3 2

22 3 4 2 3 4

2 3 4

13 1

3

1 3We replace in the second equation to find : 2 3 4

3 2

andx y x yx y

x y x yx y

y y

y x x x

⋅ − + ⋅ −− − − = − + = −− + = − ⇒ ⇒ ⊕ − = − = − =

= − ⇒ = −

− ⋅ − = ⇒ =


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3.1 3x3 systems

To solve a 3x3 system, we eliminate one of the variables using twice the addition or

substitution method. In other words, we eliminate the variable we choose thanks to two

equations of the system. Then we repeat the operation using the last equation we haven’t

used until now. Thus we obtain a 2x2 system that we solve as in 3.0. Let’s illustrate this

with an example. We consider the following system :

−=+−

−=−−−

=−−

)3(20435

)2(12

)1(722

zyx

zyx

zyx

We add the equations (1) and (2) to eliminate x :

(1) 2 2 7

(2) 2 1

(4) 2 3 6

x y z

x y z

y z

− − =⊕

− − − = −

− − =

We eliminate one more time x but with equations (2) and (3) :

2 1(2) 5

5 3 4 20(3) 2

10 5 5 5

10 6 8 40

(5 ) 11 3 45

x y z

x y z

x y z

x y z

y z

− − − = − ⋅

− + = − ⋅

− − − = −⊕

− + = −

− + = −

We have now a 2x2 system, formed of equations (4) and (5), to solve :

2 3 6(4 )

11 3 45(5)

13 39 3

y z

y z

y y

− − =⊕

− + = −

− = − ⇒ =

We find z by replacing 3y = in the equation (5) : 11 3 3 45 4z z− ⋅ + = − ⇒ = −

Finally, we find the value of x by replacing 3y = and 4z = − in one of the equations (1),

(2), or (3) : 2 3 2 ( 4) 7 1x x− − ⋅ − = ⇒ = .


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line

§4 THE STRAIGHT LINE

4.0 Generalities

The degree of a function f is one if it can be written in the form : y = ax + b ( ,a b ∈ � )

The graph of such a function is a straight line (droite). The coordinates of its points are :

(x ; y). It is very important to notice that for a chosen x the y-coordinate is given by the

function evaluated at x, consequently y = f(x) and the coordinates of the points become :

(x ; f(x)).

The coefficients a and b represent : a : the gradient or slope (pente) of the line

b : the y-intercept of the line

4.1 Gradient or slope of a line

By definition, the gradient is the ratio y

ax

∆=

∆,

where x∆ is the increase in the x-coordinate and

y∆ is the corresponding increase in the y-coordinate.

To find the gradient, you choose any two points lying on the line : );( AA yxA and );( BB yxB .

Then you compute x∆ = the x-step = AB xx − and y∆ = the y-step = AB yy − . Finally, the

gradient is :

B A

B A

y yya

x x x

−∆= =

∆ −

This result does not depend on the point A and B you have selected.

Example

Calculate the gradient a of the line joining (6 ; 9) and ( – 4 ; 12 ).

12 9 3

0,34 6 10

a−

= = = −− − −

x∆

y∆


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4.2 Vertical shift and intersections with the axes

The constant b of the equation of a line represents the vertical shift (décalage vertical) of

the line. It corresponds to the y-coordinate of the intersection point between the line

and the y-axis. Consequently, to find the coordinates of this point we have to replace x by

zero in the equation of the line. We obtain y = b and the coordinates of this point are (0 ; b).

The intersection point between the line and the x-axis is given by the resolution of the

equation : 0 = ax + b because every point on the x-axis has a y-coordinate equals to zero.

The coordinates of this point are (x ; 0).

Vocabulary

The x-coordinate of a point is also called abscissa and the y-coordinate ordinate.

4.3 Types of lines

Depending on the value of the coefficients a and b, the line is :

Linear if 0 and 0a b≠ = : passes through the origin O(0 ; 0).

Constant if 0 anda b= ∈� : parallel to the x-axis.

Affine if , 0a b ≠ : not horizontal, shifted vertically of b over the origin.

Examples

63)(1 −= xxf is affine, xxf 6,4)(2 −= is linear, 8,12)(3 =xf is constant.

4.4 Remarks

1) The sign of the gradient determines if the line increases or decreases (rises or falls).

2) Two parallel lines have the same gradient.

3) Two lines with gradient m1 and m2 are perpendicular if m1·m2 = –1

4) To draw a line :

a) Place a first point thanks to the y-intercept (0 ; b). Then, from that point, draw

the gradient (move of x-step to the right and move up or down of y-step) or…

b) Place a first point thanks to the y-intercept (0 ; b). Then find a second point

thanks to the equation of the line.


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4.5 Equation of a line

To determine the equation of a line means to determine the value of the constants a and b.

For that we have two different ways to proceed :

a) Use two points on the line to calculate the gradient. Then find the y-intercept by

substituting the coordinates of one of your two points in the equation y ax b= + , of

which only b is now unknown. Solve this equation to obtain b or…

b) Use two points of the line and replace x and y of the line by the coordinates x and

y of one of your two points. Ditto with the second point. Now you have to solve a

2x2 system.

Examples

1) The two following lines are parallel : f(x) = 3x + 4 and f(x) = 3x – 5

2) The two following lines are perpendicular : f(x) = 2x – 1 and f(x) = x2

1− + 3

3) Find the equation of the line that passes through A(–2 ; 3) and B(4 ; –5) :

a) The general equation of a line is y = ax + b and we have to find a and b. The

gradient is 5 3 8 4

4 ( 2) 6 3

ya

x

∆ − − − −= = = =

∆ − −, so

4

3y x b

−= + . To determine b we

substitute x and y of the line by the coordinates x and y of the point A :

4

3 ( 2)3

b−

= ⋅ − + ⇒ 1

3b = and finally,

4 1

3 3y x

−= + .

b) We substitute x and y of the line by the coordinates x and y of the points A

and B :

3 2 4 1

and 3 35 4

a ba b

a b

= − + −⇒ = =

− = + so y =

3

4−x +

3

1.


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−3 −2 −1 1 2 3

2

4

6

8

10

−2 −1 1 2 3 4

2

4

6

8

10

−2 −1 1 2 3 4

−2

2

4

6

8

−5 −4 −3 −2 −1 1

−6

−4

−2

2

4

§5 Quadratics

5.0 Definition and properties

A quadratic function is s function defined by a second-degree polynomial in one variable.

Thus it is a function of the form :

2( )f x ax bx c= + +

where a, b, c are real numbers and a ≠ 0. The graph of such a function is a parabola.

Examples

y = x2 y = (x – 1)2

y = (x – 1)2 – 2 y = – (x + 2)2 + 3


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We can prove that :

2 22 4

2 4

b b acax bx c a x

a a

− + + = + −

. Consequently, any quadratic

function can be written as 2( ) ( )f x a x h k= − + where 2

bh

a= − and

2 4

4

b ack

a

−= − . Its graph

is a parabola whose vertex is V(h ; k). As ( )y f x= we have that ( )k f h= and V(h ; f(h)). The

axis of symmetry is the vertical line with equation b

x2a

= −= −= −= − .

Remarks

1. If a > 0 then the parabola opens up (smile).

If a < 0 then the parabola opens down (grimace).

2. The process of writing 2( )f x ax bx c= + + as 2( ) ( )f x a x h k= − + is called

« completing the square » and the form 2( ) ( )f x a x h k= − + is called

« completed square form » (« forme du sommet »).

3. The point where a parabola meets its axis of symmetry is called the vertex

(sommet).

4. If a is positive, the vertex is at the lowest point of the graph (minimum); if a is

negative the vertex is at the highest point (maximum).

5. Suppose that you have two functions y = f(x) and y = g(x). The coordinates (xi , yi)

of the intersection point satisfy both graphs. Therefore, to find the intersection

point of two functions we have to solve the equation f(x) = g(x). After having found

x, don’t forget to calculate y.

2( ) with y a x h k= − + a < 0 2( ) with y a x h k= − + a > 0

k

h

V

k

h

V


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Examples

1. Find the completed square form of the parabola 2( ) 10 32f x x x= + + .

METHOD 1

First of all we consider only the terms with x and we want to form (x – h)2.

In this example we need to write x2 + 10x as (x – h)2. If we develop (x – h)2 we

obtain x2 – 2hx + h2. As we have 2h = 10 and we want h, we have to divide the

coefficient in front of x, which is 10, by 2 so we have (x + 5)2. But if we develop it,

we obtain :

(x + 5)2 = x2 + 10x + 25

that is not exactly x2 + 10x. Therefore, we have to subtract 25 from the

parenthesis and we obtain (x + 5)2 – 25 = x2 + 10x. Finally we have to add 32 on

both sides to have the equation of our example which is x2 + 10x + 32 :

x2 + 10x = (x + 5)2 – 25

x2 + 10x +32 = (x + 5)2 – 25 + 32

x2 + 10x + 32 = (x + 5)2 + 7

METHOD 2

We use the formula 10

52 2

bh

a= − = − = − . We replace this value in the parentesis

so (x – (-5))2 = (x + 5)2. As in method 1 we don’t exactly have x2 + 10x, but

x2 + 10x + 25 so we have to subtract 25 to the parenthesis and then add 32 to

have the given quadratic.

2. Find the vertex and the axis of symmetry of the parabola 2 2 15y x x= + − .

METHODE 1 2

Axis of symmetry : 1 12 2

: It is on the axis of symmetry therefore the -coordinate is 1.

To find the -coordinate, we replace 1 in the equation of the parabola :

Vertex

bx x

a

x

y x

− −= = = − ⇒ = −

−

= −

1 2 15 16 and ( 1; 16 )y V= − − = − − −

METHODE 2

2 2

2 2

We write the equation into ( ) .

( 1) 1 15 ( 1) 16 thus the vertex is ( 1 ; 16 )

y ax bx c y a x h k

y x x V

= + + = − +

= + − − = + − − −

METHODE 3 Cf exercise 25


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5.1 Solving quadratic expressions

When we search the intersection of f(x) = ax2 + bx + c with the x-axis we have to solve the

equation ax2 + bx + c = 0, where a, b and c are constants. The solution(s) of ax2+ bx + c = 0,

where 0a ≠ , is (are) given by the fomula :

2

1,2

4

2

b b acx

a

− ± −= Viète

Remark

The solutions of ax2+ bx + c = 0 are called zeros or roots.

Demonstration

We have seen that we can write :

2 22 4

2 4

b b acax bx c a x

a a

− + + = + −

We look for the solution(s) of ax2 + bx + c = 0 :

2 2 22 2

2

2 2 22

1,2

22

2

4 4 40 ( ) ( )

2 4 2 24

4 4 4( )

2 4 2 2 2

4( )

2 4

b b ac b b ac b acax bx c a x x

a a a aa

b b ac b b ac b b aca x a x

a a a a a

b b acx

a a

− − −+ + = = + − + = ± = ±

− − − − ± −+ = ÷ = ± =

−+ =

Example

Find the zeros of 2 2 15y x x= − − + .

2

12

1,2

2

zeros 0 2 15 0

2 85

2( 2) ( 2) 4 ( 1) 15 2 64

2 ( 1) 22 8

32

y x x

x

x

x

⇒ = ⇒ − − + =

+= = −

−− − ± − − ⋅ − ⋅ ±

= = =⋅ − −

−= =

−

↗

↘


- 21 - YAy & PRi

Remark

The expression b2 – 4ac is called discriminant. We note it ∆ = b2 – 4ac. It determines

the number of solutions of the equation ax2 + bx + c = 0 :

If 2 4b ac− > 0 the equation ax2+ bx + c = 0 has two distinct solutions.

If 2 4b ac− = 0 the equation ax2+ bx + c = 0 has only one solution or sometimes it

is said two coincident roots or a repeated root.

If 2 4b ac− < 0 the equation ax2+ bx + c = 0 has no real solutions.

Graphically this means :

If ∆ > 0 : 2 solutions If ∆ = 0 : a unique solution If ∆ < 0 : no real solutions

Example

We consider the equation f(x) = 2x2 + 5x – 3. The discriminant ∆ = 52 – 4 · 2 · (-3) = 49 is

positive, therefore the equation has two solutions :

x1 = 34

495−=

−− x2 =

2

1

4

495=

+−

5.2 Factorisation

Given the quadratic f(x) = ax2 + bx + c.

if ∆ > 0

f has two roots / zeros x1 and x2 and f can be written in the form :

f(x) = a(x – x1)(x – x2)

We call it the factor form.

We also have : x1 + x2 = a

b− and x1 · x2 =

a

c

These two relations are called relations of Viète (François Viète 1540 – 1603)

x1 x2 x


- 22 - YAy & PRi

if ∆ = 0

f has a unique zero / root (repeated root) x1 and we can write :

f(x) = a(x – x1) (x – x1) = a(x – x1)2

if ∆ < 0

f has no zeros / roots and it is not possible to write it as a product.

Example

f(x) = 2x2 – 8 has two roots x1 = – 2 and x2 = 2, so its factor form is f(x) = 2(x – 2)(x + 2).

§6 Solving equations

6.0 Equations which reduce to quadratics equations

Some equations are not quadratic equations, but sometimes it can be changed into

quadratic equation, usually by making the right substitution. Let’s consider an example to

illustrate this.

Example

Find the solutions of : x4 – 45x2 – 196 = 0

As 4 2 2( )x x= , we substitute t = x2 and the equation becomes : t2 – 45t – 196 = 0.

It is now possible to use Viète’s formula and we find two values for t : t1 = 49 and t2 = –4.

But we look for the values of x that satisfy x4 – 45x2 – 196 = 0, that is x1,2 and x3,4.

Actually the degree of the equation is four so the equation may have at maximum four

roots. We had substitued x2 = t thus :

1,2 1 49 7x t= ± = ± = ± and 3,4 2x t= ± = ± − ∉��

Therefore there are only two solutions : 1,2 7x = ±


- 23 - YAy & PRi

6.1 Irrational equations

Their name comes from the fact that they contain one or more square roots of the variable.

The strategy is to isolate the square root on one side of the equation and then to square it.

It is very important to check the answers at the end, because when we square an

expression we create, sometimes, alien solutions.

Example

-x + x−9 = 3 +x

x−9 = 3 + x ( )2

9 – x = 9 + 6x + x2 –9 and +x

x2 + 7x = 0 Factorize

x(x + 7) = 0

x1 = 0 and x2 = –7

Check : 0 + 09 − = 3 OK!

– (–7) + )7(9 −− = 3 x2 = –7 is an alien solution!

There is only one solution : x = 0

6.2 Rational equations

We call rational equation an equation that contains fractions with the variable at the

denominator. Before solving such an equation we have to find the excluded values of the

equation, that is the x for which we have a division by zero. Consequently, we have to

find the zeros of the denominator(s). Let’s consider an example.

Example

Solve the equation 52

3

14

2

+

−=

− xx

The exluded values are the zeros of the denominator. We find the by solving :

1) 4x – 1 = 0 x = 4

1

2) 2x + 5 = 0 x = 2

5−


- 24 - YAy & PRi

Let’s solve the equation by eliminating the fractions :

52

3

14

2

+

−=

− xx · (4x – 1)

52

3122

+

+−=

x

x · (2x + 5)

312104 +−=+ xx –10 and +12x

16x = –7 /16

x = -7

16

We check that the answer is not one of the excluded values (it is not the case here).

6.3 Equations involving an absolute value

We have defined the absolute value in §2. To solve equations involving an absolute value,

we use the same method as for irrational equations, that is to say we isolate the absolute

value on one side of the equation. Then we have two equations to solve :

a) One without the modulus.

b) One without the modulus and changing the sign on one side of the equation.

Finally we have to check the obtained solutions.

Example

Solve –5 + 2·| 2x – 2 | = 11

–5 + 2·| 2x – 2 | = 11 +5 and /2

| 2x – 2 | = 8

a) (2x – 2) = 8

2x = 10

x = 5

b) (2x – 2) = – 8

2x = – 6

x = –3

Check : –5 + 2·| 2·5 – 2 | = 11 OK!

–5 + 2·| 2·(–3) – 2 | = 11 OK!


- 25 - YAy & PRi

6.4 Inequalities

An inequality is an equation where the « = » sign has been replaced by one of the

following signs : « ≤ , ≥ , <, > ». The solution is not a number any more but an interval.

Each time you divide or multiply the equation by a negative number you have to

change the direction of the inequality. The symbols « ≤ » and « ≥ » are called weak

inequalities and the symbols « < » and « > » are called strict inequalities.

Examples

a) Linear inequalities : Solve 4 7 13x− + < −

4 7 13 7

4 20 ( 4)

5

x

x

x

− + < − −

− < − ÷ −

>>>>

Division by a negative number

Finally the answer is : ] [5;x ∈ ∞

b) Quadratic inequalities : Solve 2 3 7 5x x− + ≥

The first step is to write all the x and all the numbers on the same side :

2

2

3 7 5 5

3 2 0

x x

x x

− + ≥ −

− + ≥

As a parabola changes its sign only when it intersects the x-axis we have to find

two solutions of the equation 2 3 2 0x x− + = . For that we use the quadratic

formula :

1

1,2

2

13 9 4 2

22

x

x

x

=± − ⋅

= =

=

↗

↘

Thanks to the two roots we can find the solution by making the sign table for 2 3 2 0x x− + = or sketching its graph. We want the x such that 2 3 2 0x x− + ≥ . It

is the grey cells :

x 1 2 2 3 2y x x= − + + 0 – 0 +

Finally the answer is : ] ] [ [;1 2;x ∈ − ∞ ∪ ∞

x

1 2


- 26 - YAy & PRi

6.5 System of inequalities

A system of inequalities is a group of many inequalities that have to be satisfied at the

same time. We solve such a system by using a graphic method. We start by drawing the

lines that form the system. Then we hatch the area that contains the points (x ; y) that

satisfy the system.

Example

Solve the system

2 32 3

12 4 2

2

y xy x

y x y x

≤ −≤ −

⇔ + > > − +

We draw the two lines of the system

2 3

12

2

y x

y x

= −

= − +

.

Now we hatch the the points (x ; y) that satisfie both inequalities. The couples (x ; y) which

satisfy 2 3y x≤ − are under and on the line ( ≤ ). The couples (x ; y) that satisfy

12

2y x> + are over the line (>). Finally, we obtain the grey area and the line in bold type.

12

2y x= − +

-4 -2 2 4 6 8 10

-10

-5

5

10

15

2 3y x= −


- 27 - YAy & PRi

§7 Applications and functions

7.0 Applications

Definition Application

An application is a relation that associates to each element of a first set A (preimages)

one and only one element of a second set B (images).

We can illustrate this notion with a concrete example. We consider :

a relation : the flights of Swiss airline,

a set A : Madrid, Paris, London and Bern

a set B : The United States.

This relation is an application, if and only if, for each cities of A there is one and only one

flight to a state of The United States.

Examples

a)

Not an application

It is an application

Not an application

X

X

X

X

X

X

X

A B b)

X

X

X

X

X

X

X

A B c)

X

X

X

X

X

X

X

A B

A B


- 28 - YAy & PRi

d) A = { The students of the group 111 }, B = � and R (relation) = “To be born in …”

Yes it is an application because every student has a unique birth date.

e) A = { The students of the group 111 }, B = A and R = “To be taller than …”

No it is not an application because Léon is taller than Roxane and Caroline.

7.1 Functions

Definition Function

A function f is an application where A and B are sets of numbers.

Notation A → B

x � y = f(x)

Examples

a) A = B = �

f : x � y = x + 2 It is a function

f : x � y = x – 1 It is not a function : 0 – 1 = –1∉� so 0 doesn’t have

any image.

b) A = B = �

f : x � y = x – 2 It is a function

f : x � y = 2

2−x It is not a function :

5 2 3

2 2

−= ∉� so 5 doesn’t have

any image.

c) A = B = �

f : x � y = x – 3

2 It is a function

f : x � y = x

1 It is not a function, 0 doesn’t have any image

(division by zero).

d) A = B = �

f : x � y = x It is not a function, negative numbers don’t have any

image.

e) A = B = +�

f : x � y = x It is a function

f


- 29 - YAy & PRi

Definition Domain

The domain D of a function f is the set of x (often x ∈ � ) for which the function f(x)

exists.

Examples

a) f(x) = 2

2−x Df = �

b) f(x) = 4

32 −x

Df = { }\ 2±�

c) f(x) = x−1 Df = ] ] ] ]\ 1 ; ; 1∞ = − ∞�

Definition Range

The range of a function f is the set of all values produced by the function f. Sometimes

we call it the image.

Definition Graph

The graph of a function f : A → B

x � y = f(x)

is the collection of all pairs (x, f(x)) = (x ; y) of the function, where x belongs to A and y

belongs to B. In particular, graph means the graphical representation of this collection, in

the form of a curve, together with axes.

Definition Injective function or injection

An injective function or injection is a function which maps distinct input values to

distinct output values. More formally, a function f : A → B is injective if, for every y in B

there is at most one x in A with f(x) = y.

Examples

Not an injection It is an injection

a)

X

X

X

X

X

X X

A B b)

X

X

X

X

X

X

X

A B


- 30 - YAy & PRi

Definition Surjective function or surjection

A surjective function or surjection is a function with the property that all possible

output values of the function are generated when the input ranges over all the values in

the domain. More formally, a function f : A → B is surjective if, for every y in B, there is at

least one x in A with f(x) = y.

Examples

It is a surjection Not a surjection

Definition Bijective function or bijection

A function f : A → B is a bijection or one-one if it is both injective and surjective. This

means that for each number y in B there is only one number x in A with y = f(x).

Examples

Not a bijection It is a bijection Not a bijection

d) f : →� �

x � y = 0,5x – 2

It is a bijection

e) f : →� �

x � y = x2

Not a bijection, because it is not an injection nor a surjection, but if is

defined like that : f : + +

→� � it is a bijection.

X

X

X

X

X

X X

a) A B

b) X

X

X

X

X

X

X

A B

a)

X

X

X

X

X

X X

B A b)

X

X

X

X

X

X

B A c) X

X

X

X

X

X

X

B A


- 31 - YAy & PRi

7.2 Inverse function

If a function is a bijection, then an inverse function (fonction réciproque) exists and we

note it if. It is defined from B to A. It allows us to “come back” :

f : A → B if : B → A

x � y = f(x) y � x = if(y)

Remark

We know that an inverse function exists only if the function is a bijection. Thus we should

check, before any calculations, that the function is a bijection. However it is not always an

easy exercise. So we are going to simplify this problem by supposing that the functions

we inverse are bijections. If it is not the case we calculate the inverse function and after

that we choose the domain and the range to make a bijection.

Examples

a) For easy bijections from →� � such as 3 2, 5 7, ...y x y x= − = − + it is easy to

find an inverse function. For that, we break up the function. Let’s consider, for

example, the function ( ) 4 6f x x= − . It is built like that :

4 64 4 6x x y x⋅ −→ → = −

To find the inverse function, we have to make the inverse operations of « 4⋅ »

and « – 6 » that are « 4÷ » and « + 6 ». We still have to determine in which

order we have to implement them. To answer this question compare this situation

with the fact of getting dressed. Firstly we put our socks ( 4⋅ ) and then our shoes

(– 6). When we get undressed, we take off our shoes (+ 6) and then our socks

( 4÷ ). It is now possible to find the inverse function of ( ) 4 6f x x= − :

6 4 64 6 6 4

4

yy x y x x+ ÷ +

= − → + = → =

We rename this function to obtain : 6

( )4

i xf x y

+= =

f if

x y

A B f

if


- 32 - YAy & PRi

As the lines are bijections from →� � , we don’t have to find the domain and the

range.

b) For rational functions, the method is always the same, we have to express x in

terms of y. Let’s consider an example : 2

2

xy

x

−=

+.

2( 2)

2

( 2) 2 distributivity

2 2 and 2

2 2 factorize

( 1) 2 2 ( 1)

2 2

1

xy x

x

y x x

yx y x x y

yx x y

x y y y

yx

y

+= ⋅ −

−

− = +

− = + − +

− = +

− = + ÷ −

+=

−

We rename it to obtain : 2 2

( )1

i xf x y

x

+= =

−

The domain of ( )i f x is : { }\ 1i fD = �

Finally :

{ } { }: \ 2 \ 1

2

2

f

xx y

x

→

+=

−

� �

� is a bijection.

Its inverse function is :

{ } { }: \ 1 \ 2

2 2

1

i f

xx y

x

→

+=

−

� �

�

c) We proceed exactly as above for irrational functions. Let’s consider an

example : 2 4 1y x= − + .

2

2

2

2

2 4 1 1

1 2 4 ( )

( 1) 2 4 4

( 1) 4 2 2

( 1) 4

2

y x

y x

y x

y x

yx

= − + −

− = −

− = − +

− + = ÷

− +=

We rename it and we obtain : 2( 1) 4

( )2

i xf x y

− += =


- 33 - YAy & PRi

d) The last kind of functions that we are going to consider in order to find their

inverse function is the quadratic functions. We have to start by writing the

quadratic in completed square form. As in the other examples, let’s work with an

example: 2 2y x x= − . The completed square form is 2( 1) 1y x= − − . It is now

possible to find its inverse function :

2

2

( 1) 1 1

1 ( 1)

1 1 1

1 1

y x

y x

y x

x y

= − − +

+ = −

+ = − +

= + +

The function :

[ [ [ [

2

: 1; 1;

2

f

x y x x

∞ → − ∞

= −� is a bijection.

Its inverse function is :

[ [ [ [: 1; 1;

1 1

i f

x y x

− ∞ → ∞

= + +�

Remark

If f is a one–one function or bijection, then the graphs of y = f(x) and y = if(x) are

reflections of each other in the line y = x.

7.3 Composite function

Since secondary school, we know how to compose symmetries, rotations, translations, etc

together. Now we are going to learn how to compose functions. Given two functions f and

g. We define the composite function (fonction composée) of f and g, in this order, like

that :

( )( ) ( ( ))g f x g f x∗ =

Example

Given the functions ( ) 2 4f x x= − and 2( ) 3g x x= . Calculate andg f f g∗ ∗ .

2 2 2( )( ) ( ( )) (2 4) 3(2 4) 3(4 16 16) 12 54 54g f x g f x g x x x x x x∗ = = − = − = − + = − +

2 2 2( )( ) ( ( )) (3 ) 2 3 4 6 4f g x f g x f x x x∗ = = = ⋅ − = −


- 34 - YAy & PRi

Remark

This example illustrates the fact that andg f f g∗ ∗ are usually two different functions.

§8 POLYNOMIALS

As for numbers, it is possible to add, multiply, subtract or divide polynomials.

8.0 Reminder

A polynomial, p(x), is an expression in x of the form 1 ...n nax bx jx k−+ + + + ,

where a, b, c, … , k are real numbers, 0a ≠ and n is a non-negative integer.

The number n is called the degree of the polynomial. The expressions 1, , ... , andn nax bx jx k− are called terms (monômes). The numbers a, b, c, … , k

are called the coefficients and a is the leading coefficient (coefficient dominant).

The coefficient k is the constant term.

When a polynomial is ordered with the terms of highest degree first and the other

terms in descending degree order finishing with the constant term, the terms are

said to be in the descending order (ordre décroissant). If the terms are written in

reverse order, they are said to be in ascending order (ordre croissant).

8.1 Addition and subtraction

To add or subtract two polynomials, you simply add or subtract the coefficients of

corresponding powers. In other words, you collect like terms.

Example

Let’s consider the polynomials 4 2 4 3 2( ) 4 2 6 16 and ( ) 3 8 7p x x x x q x x x x= − + − + = − + − .

4 3 2( ) ( ) 10 6 9p x q x x x x x+ = − − + − +

4 3 2( ) ( ) 7 6 6 23p x q x x x x x− = − + − − +


- 35 - YAy & PRi

8.2 Multiplication

When we multiply polynomials it is fundamental to bear in mind that the multiplication is

distributive over the addition, this means that all the terms of the right parenthesis are

multiplied by all the terms of the left parenthesis. We multiply the coefficients between

them and the rules of indices tell us that we have to add the degrees. Finally we add the

terms in x that have the same degree.

Example

Given two polynomials 4 2 2( ) 4 2 16 and ( ) 2 1p x x x q x x= − + + = − , then

2 4 2 6 4 2 4 2

6 4 2

( ) ( ) (2 1)( 4 2 16) 8 4 32 4 2 16

8 8 30 16

p x q x x x x x x x x x

x x x

⋅ = − − + + = − + + + − − =

= − + + −

8.3 Long Division

The problem we have to solve is to divide 4 3 22 3 5 7x x x x− + + − by 2x + . For that, we

are going to copy the division of two numbers. Let’s divide 67 by 5 :

67 5

(65) 13

2

quotient

remainder

− =

=

We can write : 67 5 13 2= ⋅ +

If we apply this method to polynomials we obtain the following process :

1st step We consider only the terms of highest degree 42x :

42 :x x the result is 32x

So we write :

4 3 22 3 5 7x x x x− + + − 2x +

32x

2nd step We multiply 2+x by 32x the result is 4 32 4x x+

Then we write :

4 3 22 3 5 7x x x x− + + − 2x +

4 32 4x x+ 32x


- 36 - YAy & PRi

3rd step SUBTRACT 4 32 4x x+ from 4 3 22 3 5 7x x x x− + + −

We obtain :

4 3 22 3 5 7x x x x− + + − 2x +

4 3(2 4 )x x+−−−− 32x

3 27 5 7x x x− + + −

Next step Repeat those three steps with the new term of highest degree 37x− :

( )3 27 : 7x x x− = − and ( )2 3 27 ( 2) 7 14x x x x− ⋅ + = − −

Thus :

4 3 22 3 5 7x x x x− + + − 2x +

4 3(2 4 )x x+−−−− 3 22 7x x−

3 27 5 7x x x− + + −

3 27 14x x− −

SUBTRACTION : ( ) ( )3 2 3 2 27 5 7 7 14 19 7x x x x x x x− + + − − − − = + −

4 3 22 3 5 7x x x x− + + − 2x +

4 3(2 4 )x x+−−−− 3 22 7x x−

3 27 5 7x x x− + + −

3 2( 7 14 )x x− −−−−−

219 7x x+ −

Repeat these three steps with the new term of highest degree 219x :

( )219 : 19x x x= and 219 ( 2) 19 38x x x x⋅ + = +

We have :

4 3 22 3 5 7x x x x− + + − 2x +

4 3(2 4 )x x+−−−− 3 22 7 19x x x− +

3 27 5 7x x x− + + −

3 2( 7 14 )x x− −−−−−

219 7x x+ −

2(19 38 )x x+−−−−

37 7x− −

NEW


- 37 - YAy & PRi

Repeat those three steps with the new term of highest degree 37x− :

( )37 : 37x x− = − and ( 37) ( 2) 37 74x x− ⋅ + = − −

We obtain :

4 3 22 3 5 7x x x x− + + − 2x +

4 3(2 4 )x x+−−−− 3 22 7 19 37x x x− + −

3 27 5 7x x x− + + −

3 2( 7 14 )x x− −−−−−

219 7x x+ −

2(19 38 )x x+−−−−

37 7x− −

( 37 74)x− −−−−−

67 The division is over because we can’t divide 67 by 2x + .

Conclusion The division 4 3 22 3 5 7x x x x− + + − : 2x + has…

a quotient : 3 22 7 19 37x x x− + −

a remainder (reste): 67

Thus we can write :

4 3 2 3 22 3 5 7 ( 2)(2 7 19 37) 67x x x x x x x x− + + − = + − + − + or

( ) ( )4 3 2 3 2 672 3 5 7 : 2 2 7 19 37

2x x x x x x x x

x− + + − + = − + − +

+

So we can check the result by computing : quotient ⋅ divisor + remainder

and we must obtain the polynomial 4 3 22 3 5 7x x x x− + + − .

Here : ( ) ( )3 2 4 3 22 7 19 37 2 67 2 3 5 7x x x x x x x x− + − ⋅ + + = − + + − OK !

Two important results

When a polynomial p(x) is divided by a non-constant divisor, d(x), the quotient

q(x) and the remainder r(x) are defined by the equality :

( ) ( ) ( ) ( )p x d x q x r x= +

where the degree of the remainder is less than the degree of d(x).


- 38 - YAy & PRi

Let p(x) be a polynomial. Then

If x t− is a factor of p(x), i.e. ( ) ( ) ( )p x x t a x= − ⋅ then p(t) = 0 (t is a zero or

solution of p(x)).

If p(t) = 0, i.e. t is a solution or zero of p(x), then x t− is a factor of p(x) so we

can write : ( ) ( ) ( )p x x t a x= − ⋅ where a(x) is the quotient of the division of p(x)

by x t− . This is the factor theorem.

In other words, if t is a zero or solution of p(x), then p(x) is divisible by x t− .

If p(x) is divisible by x t− then t is a zero or solution of p(x).

§9 EXPONENTIAL AND LOGARITHMIC FUNCTIONS

9.0 Definition and properties

Here are the graphs of 10 , Log( ) andxy y x y x= = =

As these graphs are symmetrical about the line

xy = , we deduce that (cf. 7.2) :

Log( ) and 10xx are inverse functions.

This means : ( )x Log( )Log 10 and 10 xx x= =

A function of the form ( ) where , 0xf x b x b= ∈ >� is called an exponential function.

We call Log logarithm to base 10 (logarithme de base 10). Calculating Log(b) means looking for

the power to which we have to rise 10 to find b. Actually :

3Log(0,001) 3 10 0,001 −= − ⇔ =

It is possible to calculate logarithm to any base, but here we are going to consider only the

base 10. Using Log, it is now possible to solve new equations. For that we use a property

of logarithm :

Log( ) Log( )ba b a= ⋅

-2 2 4

-3

-2

-1

1

2

3

4

5

10xy=

Log( )y x=

y x=


- 39 - YAy & PRi

Let’s check this with 3 and 5a b= = : 5Log(3 ) Log(243) 2,38= ≅ and 5 Log(3) 2,38⋅ ≅

Thanks to the multiplication rule, it is now possible to solve equations like : 114 =x

4 11 Log

Log(4 ) Log(11) rule

Log(4) Log(11) Log(4)

Log(11)1,73

Log(4)

x

x

x

x

=

=

⋅ = ÷

= ≅

Verification : 1,73...4 11 !OK=

As exponential and logarithmic functions are inverse function, it is also possible to solve

for instance Log( ) 2x = :

Log( ) 2

Log( ) 2 10

10 10

100

x

x

x

inverse functions

x

=

=

=

Verification : Log(100) 2 !OK=

Remark

These two new functions are used in domains as chemistry, biology or economy as well.

Actually, the growth of a bacteria population, the decomposition of a radioactive element

or the banking interests can be described by exponential functions.

Another concrete example is the Richter scale which measures the power of an earthquake. It is

a logarithmic scale : the magnitude of Richter corresponds to the logarithm of the amplitude of

the vibrations recorded by a seismograph calibrated according to the distance from the

epicentre. Thus, for example, this means that an earthquake of magnitude 6 is 10 times more

powerful than a seism of magnitude 5.

Documents

Lycée Denis-de-Rougemont 2006-2007 TABLE OF CONTENTS · Lycée Denis-de-Rougemont 2006-2007 - 3 - YAy & PRi INTRODUCTION AND HISTORY Mathematics (colloquially, maths or math) is