Mathematics for Academic Studies 4 Student's Book Sample.

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N O I T A C U D E Y R A D EC O N

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Building SUMA

Blocks piezaS

THIS IS

YOUR BOOK F EACH UNIT

Reading and listening

THE OPENING PAGES O

5

We read and listen a brief historical introduction of the contents you are going to learn in the unit.

EXERCISES AND PROBLEMS EXERCISES AND

Exercises and problems solved.

Methods, suggestions, tricks and thinking strategies that will come in handy to solve similar problems.

ons

functi a certain road and has to brake, actual ers a hazard on the the When a driver encount between their decision to brake and e, the 3/4 of a second. Therefor ore, amount of time passes (reaction time), about in that time. Furtherm an moment they brake the further it will travel it travels faster the car is going, when the brake is applied; ically automat stop the car does not due to inertia. (braking distance) moment the to additional distance hazard the aware of greater the driver becomes (d ) that increases the From the moment an additional distance the car stops, it travels following values: experiment gives the the velocity (v). An

Evidently useful

ns Basic Functio g

Reading and listenin

cy exactly what were debates about th these 20th centuries there . Over the course of During the 19 and in defining a function Riemann and was and was not essential notably Dirichlet, s mathematicians, a function. discussions, numerou formulate a more precise definition of to which is very similar 1923, in Weierstrass, helped offered g definition was Finally, the followin today. use nds to a to the one we value of x correspo function of x if each the equation y = f (x).

One says that y is a ndence is shown in value of y. This correspo

Seeking accura

speed of the car (km/h)

(based on reaction time in m)

distance

30 40 50 60 70

s were of bizarre function the for accuracy, a series unhappy with the direction However, in this search Poincaré, who was in 1899: invented that irritated had taken. This irritation led him to say s s which definition of function a mass of bizarre function seen have we functions century ‘For half a as possible honest to resemble as little new function was a appear to be forced when , purpose. Formerly end. Today, they are which serve some view of some practical ’s reasoning was at invented, it was in to show that our ancestor invented on purpose

80 90 100

The equations are: = 0.21v d reaction

total stopping distance (m)

1

13.5

7

6.5

20.5

12

8.5

29.5

19

10.5

36

14.5

79

60

19

95

74

21

dbraking = 0.0074v

2

2 0.21v dtotal = 0.0074v +

of them g words and use each for each of the followin , span. irritate, mass, resemble in a sentence: bizarre, be slower if a reaction time may to explain why the attention.

2 Use modal verbs and not fully paying tion conversa a person driving is having

1 Write a synonym

can be fault.’ honest functions, which . going to focus on these ting or disproving concepts In this unit we are more than just construc used for something

Solve

for of the formulas above in to check the validity Use your calculator in each row, keeping 1 (only check a few values them. the values in the table and graph approximate values) mind that they are

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12) is one of Henri Poincaré (1854–19 ticians in history. the greatest mathema of spanned all fields His contributions mathematics.

Solve. You can do these motivating activities to activate your previous knowledge.

Unit

This is a very useful formula. You should learn how to use it!

Based on the presumption that the spring becomes deformed when weights of over 6 kg are hung from it, the domain of definition of this function is [0, 6].

y = mx

X

(x2, y2)

lenght (cm)

y = 30 + 15x

Your turn Express each function as a piecewise-defined function:

y2 – y1 m = –––––– x –x 2

1

b) y = |x + 3| + |x|

y = 2x + 3 x +1 Your turn Graph this function: y = 3x – 1 x+2

Your turn. Exercises and problems to practice the strategies you have just learnt.

1

2

3

4

5

6

The icons included with some activities indicate the keys to the project.

n

X

a) It passes through (–5, 7) and its slope is –3 .

5

100 32

F

ã General expression of linear functions: y = mx + n This is represented on a graph as a straight line with slope m that intersects the Y axis at point (0, n).

The number n is called the ordinate at the origin.

Y F = 32 + 1.8C C 50 100

y = mx + n n X

b) P (2, –1), m = –2 d) A (1, 3), m = – 5 3

x<0

0≤x≤2

x>2

–2x + 4

–2x + 4

2x – 4

|x|

–x

x

x

|2x – 4| + |x|

–3x + 4

–x + 4

3x – 4

Therefore, the function is: Z –3x + 4 if x < 0 ]] y = [–x + 4 if 0 ≤ x ≤ 2 ]3x – 4 if x > 2 \

5

D

4 2 –2

2

D

Find the equation of the straight line that passes through points A and B in each case: a) A (3, 0), B (5, 0)

b) A (–2, – 4), B (2, –3)

c) A (0, –3), B (3, 0)

d) A (0, –5), B (–3, 1)

4

6

X

–2

8

B

9

Find the equation in each case and graph it: a) Line that passes through (2, –3) and is parallel to the one passing through (1, –2) and (– 4, 3). b) Proportionality function that passes through (–4, 2).

10

c) Constant function that passes through (18; –1.5).

6

C

6

A

Calculate the equation of these linear functions:

C

Find the value of the unknown parameters so that the straight lines and points comply with these conditions. Graph them.

Graph the following functions by making a table of values like the one below for each of them: x

–4

–3

–2

–1

0

1

2

3

4

y

…

…

…

…

…

…

…

…

…

a) y = x 2 + 1

b) y = –x 2 + 4

c) y = –3x 2

d) y = 0.4x 2

Graph the following parabolas by finding their vertex, some points close to it and the points where they intersect with the axes: b) y = x 2 – 4x a) y = (x + 2)2 d) y = x 2 – 9 c) y = 1 x 2 + 2x + 1 2 Give the point (abscissa and ordinate) where the vertex of the following parabolas is found. In each case, state whether it is a maximum or a minimum.

Then, graph them. a) y = 8 – x 2 c) y = –x 2 – 2x + 4

a) The line that passes through points (4, 0) and (–2, a) has a slope of –1.

e) y = 15 – 1 x 2 + 1 x 2 4 4

b) The line y = bx + 2 passes through point (–3, 4).

11

c) The lines of the equations y = 3x + c and y = cx + 3 intersect at ordinate 2. Which is the corresponding abscissa?

b) y = 4 + (3 – x)2 d) y = 3x – 1 x 2 + 1 2 f ) y = 1 x 2 + 2x + 3 3

The exercises are divided into topics.

Each one is also marked with its degree of difficulty, from one to three.

Calculate the vertex, the axis of symmetry and the points of intersection with the axes (if any) of each of these parabolas: a) y = 2x2

d) The points (d, –2) and (4, e) belong to the line of the equation y = 1 x – 3 .

2

b) y = 2(x – 5)2

c) y = 2(x – 5)2 + 2

d) y = –x2 + 1

e) y = –(x + 1)2 + 1

f ) y = –3x + 2x2 127

For example, the line F = 32 + 1.8C, represented in the margin, allows us to change from degrees Celsius, C, to degrees Fahrenheit, F.

algebraic fraction

anayaeducacion.es GeoGebra. Calculating the slope of a line.

1 Graph the following functions:

b) y = 2 x 3

c) y = – 1 x d) y = – 7 x 3 4

2 Graph the following:

a) y = 3

b) y = –2

c) y = 0

d) y = –5

3 Graph the following functions:

a) y = 2x – 3

analytic expression of a function

We can simplify it: y = 7 – 3 x – 3 · 5 8 y = 4 – 3 x 5 5 5 5 – 7 = –2 = – 1 3 6 4 – (–2)

b) y = 2 x + 2 3 d) y = –3x – 1

c) y = – 1 x + 5 4 4 At an initial point in time, a moving object is 3 m from its origin and is travelling away from it at a velocity of 2 m/s. Find the equation for its distance from the origin as a function of time and graph it.

5

The price of potatoes at the market is €1/kg, and the price of tomatoes is €2/kg. a) Write the equation for the price of a bag of potatoes as a function of its weight. b) Write the equation for the price of a bag of tomatoes as a function of its weight. c) Graph the functions above.

6 Find the equation of these straight lines using the

given data:

a) It passes through (–3, –5) and its slope is 4 .

9 b) It passes through (0, –3) and its slope is 4. c) It passes through (3, –5) and (–4, 7).

111

110

Think and practice. These are exercises to apply the theory you have learnt.

The rate at which a radioactive substance decays is measured by its half-life, which is the time it takes for half of its original mass to decay. On the right, you can see the half-lives of some radioactive substances.

U

Ac

227.028

t

t/28

where M is the amount of actinium remaining after t years.

1

angle bisector apothem

226.025

81

Glossary. We learn the relevant terms that are underlined in the units with a clear definition.

t 57

= 100 · >c 1 m

1 57

t

Tl

204.383

arithmetic

The branch of mathematics that studies numbers and the 2 operations that2we can do with them.

arithmetic progression

A sequence in which the next term is determined by1.2 adding number means that % willthe havesame decayed. (positive or negative), called the difference, to the previous term. 14

axial symmetry axis of symmetry

P = 100 · c 1 m

In other words, in 1 century there will be 100 · 0.988 = 98.8 % of the C14 left, which discrete quantitative variable

• What percentage of C will a 33 000-year-old fossil have compared to a living plant?

dispersion parameters A symmetry in which the points of one figure coincide withathe points another • How old must fossil be if itofonly has 10 % of the C14 of a living plant? figure when reflected across a line. domain of a function

Substitute P inthing the formula for 10equal and isolate t taking logarithms.

An imaginary line that divides a figure, a shape, or any other into two ellipse and symmetric parts.

bar chart

A graphic representation formed of thin 132 bars, where the height of each bar is proportional to the frequency of that value. They are used to represent tables of discrete quantitative or qualitative variables.

biquadratic equation

Fourth-degree equations that have no terms with odd degrees.

box-and-whisker plot

A graphic representation that visually describes the position parameters (median and quartiles).

case

How many eggs did she have at the beginning?

• A farmer shared a flock of sheep among his children.

— He gave the eldest one sheep plus 1/7 of those remaining.

— He gave the second child two sheep plus 1/7 of those remaining.

— He gave the third child three sheep plus 1/7 of those remaining.

— And so on until he got to the youngest.

In this way, all of the children received the same inheritance and no sheep were left over.

How many children does the farmer have?

How many sheep were there in the flock?

1 –– 4

H ≈ 100 · 0.988t 8 P = 100 · 0.988t, t function in centuries discontinuous

The size of a surface.

b) How many candles are needed for 1 000 hours of light?

• A farmer went to the market to sell a basket of eggs. The first customer bought half her eggs plus half an egg. The second customer bought half of the remaining eggs plus half an egg, and the third one did the same. The seller then had no more eggs left.

1 –– 2

direct motion

area

Thallium 3 minutes

mass (g)

There are three naturally occurring isotopes of carbon: C12, C13 and C14. The first two are stable but the third one is radioactive and has a half-life of 5 700 years. This means that over this period of time the amount of C14 reduces by half. like value the other isotopes of carbon, C14 is present in the atmosphere The absolute value of the difference betweenJust a real and a rounded value. cube (CO root2) and is absorbed by plants (photosynthesis) which incorporate it in a specific proportion. C14 is then incorporated A given quotient of two polynomials. cumulative frequency in the same proportion, via these plants, into other living things.

When a living thing dies and Non-adjacent angle that is outside the parallel lines. fossilised, its C14 continues to decay according to the function on the right. becomes decimal number Non-adjacent angle that is inside the parallelEvery lines. 5 700 years, the amount of C14 left reduces by half. Therefore, by finding out the 14 proportion An equation that algebraically relates the two variables.of C in a fossil and checking it against the initial proportion (that in a living plant), the equation above can be used to find the fossil’s geological age,function in other words, decreasing the time when it was formed.

The locus of the points equidistant from the sides of an angle.

If we express the time, t, in centuries, the equation above can be expressed like this: dependent variable A perpendicular segment from the centre of a triangle to its side.

Radon 1 620 years

Ra

Actinium 28 years

M = c 1 m grams 2 where t is the time elapsed taking one half-life as a unit. If the substance was actinium and we wanted to express the time in years, the formula would be: M = c1m 2

88

years

238.029

89

If we have an initial mass of 1 g, the amount of mass of this substance that will be left after a given time is:

a) How many hours of light will 442 candles give you?

1

alternate interior angle

Equation of the line that passes through (–2, 7) and has a slope of – 1 : 3 y = 7 – 1 (x + 2) 8 y = 19 – 1 x 3 3 3

• A candle lasts an hour. With the leftover wax of 10 candles you can make a new one.

Radioactive substances decay by emitting radiation and transforming into other substances. This process takes place over time and its rate varies greatly from one substance to another.

Uranium 92 radioactive substance 8 radiation + different substance 2 500 million

Carbon-14 dating

alternate exterior angle

b) We begin by finding its slope: m =

PRACTICE MAKES PERFECT!

Radioactive decay

glossary absolute error

Examples and solved problems.

To put into practice the most important methods.

Unit 5

LEARN

In order to work out the equation of a line that passes through two points, we do the following:

Think and practise

a) y = 2x

The line y = 0 coincides with the X axis.

For example, the distance of an artificial satellite from Earth is constant. It does not depend on time, t. The equation for this would be d = 36 000, d: distance, in km; t: time, which does not appear in the equation.

200

a) P (0, 0), m = 1 c) A (–2, 1), m = 1 2

A

4

d) y = x 2 – 6x + 6 Y

B

126

Here, you will find readings, activities, advice, information...

➜ finding straight lines using two points

anayaeducacion.es Review the point-slope equation.

b) It passes through (–2, 7) and (4, 5).

Its slope is 0.

y=0

c) y = x 2 – 3

P

Glossary

a) Equation: y = 7 – 3 (x + 5). This is the equation of the line.

5

The slope of the straight line is the constant of proportionality, m.

It is represented with a straight line parallel to the X axis.

y=n

Using the given slope and point, calculate the equation of each line:

Match each expression to its graph: a) y = x 2 b) y = (x – 3)2

MATHS WORKSHO

• We can find the equation based on the slope and one of the points.

Functions of proportionality are graphed with straight lines that pass through the origin. They describe a ratio between the values of both variables.

ã Constant function: y = n

Equation: y = y0 + m(x – x0)

Focus on English. Do you think Mathematics and English haveanything in common?

Discover how language and mathematics are linked so you can learn both: Mathematics and English.

• If we give x the value x0 8 y = y0 + m (x0 – x0) = y0 + m · 0 = y0. x = x0, then

Problem solved

Find the equation of each of the following straight lines:

s = v · t, where v is the slope of the line that relates s to t.

Y

2

if x ≥ 9

We start by dividing: 2x + 3 x + 1 This function can be expressed as: – 2x – 2 2 2x + 3 = 2 + 1 1 x +1 x +1 It is the function y = 1 shifted 1 unit to the left and x 2 units up.

Express the following function k + b. in the form y = x–a Then, graph it.

Graph the following linear functions: a) y = 2x – 3 b) y = 4 x 7 d) y = 2.5 c) y = –3x + 10 5

X

9

6 Functions related to the function of inverse proportionality

• We can find the slope using the two points.

weight (kg)

0

For example, the space covered at a constant velocity, v, as a function of time is:

anayaeducacion.es GeoGebra. Graphic representation of a linear function.

|2x – 4|

a) y = |x 2 – 4| – 2

y = y0. In other words, it passes through (x0, y0).

(x1, y1) x – x 2 1

ã Functions of proportionality: y = mx Y

1

7

1

3

if x ≤ 1 if 1 < x < 9

Exercises and problems. To put into practice all the knowledge acquired throughout the unit.

Unit 5

Quadratic functions

Linear functions

a) The parabola of y = x 2 – 1 intersects with the X axis at x = –1 and x = 1.

Since its branches point upwards, the piecewise-defined function is: Z Z if x < –1 ]]x 2 – 1 + 1 if x < –1 ]]x 2 y = [–x 2 + 1 + 1 if –1 ≤ x ≤ 1 8 y = [–x 2 + 2 if –1 ≤ x ≤ 1 ]x 2 – 1 + 1 if x > 1 ]x 2 if x > 1 \ \ b) We make a table to study how the expression varies within each absolute value depending on the section. The endpoints of each section are those that make each of the expressions equal to 0:

a) y = |x 2 – 1| + 1

• The coefficient of x is m. Therefore, its slope is m.

100

50

Slope: m

• y = y0 + m (x – x0) is a first-degree expression. Therefore, it is a straight line.

y2 – y1

y = 30 + 15x (y: length in cm; x: weight in kg)

Point: P (x0, y0) explanation

If different weights are hung from a spring, they will stretch to different lengths.

In other words, the length of the spring is a function of the weight hanging from it. It is worth pointing out* that this is a linear function.

More specifically, let’s suppose that the spring is 30 cm long before it is stretched and that is stretches 15 cm per kilogram added to it. The relationship is:

We often need to write down the equation of a straight line of which we only know one point and the slope. It is written as follows:

Remember

Science is full of functions in which variations in the cause have a proportional impact on variations in the effect. All such functions are called linear functions and are graphed using straight lines. Let’s look at an example:

point out: to mention some information you think is important.

5

ã Point-slope equation of a line

LINEAR FUNCTIONS ã Linear functions in our daily lives

F ocus on English

Z ] 1 x2 – 5 x + 9 2 ]4 4 ] y = [– 1 x 2 + 5 x – 9 2 4 ] 4 ]] 1 x 2 – 5 x + 9 2 4 \4

b) y = |2x – 4| + |x|

unit for your

PROBLEMS

MATHS WORKSHOP

CONTENT DEVELOPMENT AND ACTIVITIES stretching a spring

1 x 2 – 5 x + 9 = 0 8 x = 1, x = 9 2 4 4 As the branches of the parabola point upwards, since the coefficient of x 2 is positive, the function is only negative between the points of intersection with the X axis.

y = |x 2 – 2x – 8|

The audios of each unit’s content are available at www.anayaeducacion.es

Each unit is divided into epigraphs and subepigraphs. The most important contents are in bold.

y = 1 x2 – 5 x + 9 2 4 4

Express each of the following functions as a piecewise-defined function:

LA

108

In order to express it as a piecewise-defined function, we first need to see where the function within the absolute value intersects with the X axis. In this way, we can work out in which intervals it is positive and in which it is negative.

5 Functions with absolute values

64

47

17

Express the following function without using the absolute value sign and graph it:

Your turn Express as a piecewisedefined function and graph:

50.5

this resources from

EXERCISES AND

Practise

Therefore, we have three sections: (– ∞, 1] and [9, +∞) where the function is positive or zero, so there is no need to change the sign, and the interval (1, 9), in which the function is negative. In this Y section, we change the sign of the function.

Let’s see what it looks like:

39.5

27

12.5

3 7

3

4

20

ing to Poincaré

distance

(braking distance in m)

2

10

accord Honest functions,

1

to choose Remember portfolio.

ED

PROBLEMS SOLV

4 The absolute value of a function

equalisation method

2

3

4

time in half lives

SELF-ASSESSMENT

anayaeducacion.es Answer key and interactive self-assessment.

1 Graph the piecewise function that has the following

equation:

Z ]] 2x + 6 if x < –2 y = [ x/2 + 3 if –2 ≤ x < 2 ] –x + 6 if x ≥ 2 An amount that can be t multiplied by itself three times to result in\the number.

Is it continuous? up with a function with the 1 m 5 700 of a value with the frequencies The sum the of all theCome previous values, P = of 100 · cfrequency same sections that is not continuous. when the values 2 are ordered from smallest to largest.

The result of a non-exact quotient. It has an integer and of a decimal 2 Findportion the vertex each of portion the following parabolas separated by a decimal point. and graph them: 2 independent variable x A function where the dependent variable y decreases asxthe –2 b) y = x 2 + 4x – 5 a) y = increases.

2

c) y =y variable.

(5 – x)(x + 1) The variable represented on the vertical axis. It is the

3 Express as piecewise-defined functions and graph

A variable that can only take certain numerical values. them:

a) y = |2x + 1|

b) y = 1 – x 4 d) y = |9 – (x – 2)2|

Parameters that tell us how far away from the centre the values in a distribution are.

The set of x values that have y values.

c) y = |–x 2 + 4x – 3|

a) y =

1 x +5

d) y = x + 2

b) y = 3 – 2 x

c) y =

e) y = 2 x – 1

f) y = – x – 3

a) y = 1.2x; y = log1.2 x

An equality of algebraic expressions that is only true for certain values of the letters.

equation

An equality statement that includes a letter called an unknown.

equivalent fractions

Two fractions are equivalent when they have the same numerical value.

Every possible result of a random experiment.

equivalent systems

Systems of equations that have the same solution.

central angle in a circle

An angle with its vertex at the centre of the circumference.

exact root

The square root of a perfect square.

chance

A combination of circumstances that supposedly cause an unpredictable event.

factorising polynomials

circumference

A closed two-dimensional curve in which all points are equidistant from the centre.

A method for decomposing a polynomial into the product of other polynomials with the lowest possible degree.

class midpoint

The central value of each interval.

final value

The product of an original value and a variation index.

compatible determinate system

A system that has one unique solution. Its graphic representation is two lines that intersect at one point.

fraction

A given division. It is the result of dividing an integer into equal parts.

frequency polygon

compatible indeterminate system

A system that has infinite solutions. Its graphic representation is two coinciding lines that share all their points.

A graphic representation formed by connecting the ends of the bars in a bar graph or the midpoints of the rectangles in a histogram. It is used to represent quantitative variables.

compound experiment

An experiment that includes more than one random experiment.

frieze

Longitudinal decoration that contains a pattern that repeats in translations.

compound interest

The profit generated over a certain time by adding interest to the original capital.

function

A relationship between two variables that we usually call x and y.

conic section

The curves resulting from different intersections between a cone and a plane.

general term

An expression that represents any term in a sequence.

continuous function

A function without any discontinuities. Its graph can be drawn without lifting the pencil from the paper.

geographic coordinates

A reference system for specifying the location of any point on the Earth. There are two: latitude and longitude.

continuous quantitative statistical variable

A variable that can take any numerical value in an interval.

geometric progression

A sequence in which the next term is determined by multiplying the previous term by a fixed number, called the ratio.

coordinates

A system of values, references... used to determine the position of a point on a plane.

geometric transformation

corresponding angles

Angles that share one side that intersects two parallel lines formed by their other sides.

A transformation in which every point of one figure corresponds to another point of another figure.

geometry

The study of the properties and measurements of figures in the plane or in space.

306

b) y = 2.5x; y = log2.5 x

With respect to which line are the two functions of each pair symmetrical?

b) Which expression gives us the surface area, S, for any base, b ? Graph it. c) For which value of the base do we obtain the maximum surface area? What is the value of that surface area?

A closed geometric curve with two unequal perpendicular axes, resulting from video for target 6.a. Think of something you can do to contribute cutting the surface of a cone with a plane that is not perpendicularWatch to itsthe axis.

Commitment to achieve that goal. Make a commitment to put your idea into practice.

A procedure for solving a system of equations that involves solving for the same unknown in both equations and then equalising the resulting expressions.

equation

3 +1 x –1

5 Graph these pairs of functions:

make a picture frame. a) If the base of the frame is 0.5 m long, how tall is it? What is its surface area?

2

e) y = 2x also + 4xcalled a shift. f ) y = 9 – (x – 1) A type of motion that maintains the direction of rotation,

g) yin= 2(x – 1)(x + 3) gapsh)in y =the (x + 2)2 – 2x 2 A function where the independent variable moves jumps, creating graph.

domains of definition:

6 Using a 3-metre-long strip of wood, we want to

d) y = –(x – 3)2 – 1

2

4 Graph the following functions and find their

Practice makes perfect! In this section you will have to solve many different types of problems.

133

ment. By Self-assess s, se activitie doing the r u o heck y you can c ding and n ta unders you have how much learnt.

307

KEYS

PROJECT

SDG SDG Commitment Discover the Sustainable Development Goals and be an active part of our commitment to make a more equal and liveable world.

Developing thinking Work on strategies for thinking: reflect on the content you are learning, generate ideas, organise them, debate them, explain them…

Cooperative learning Get involved in your learning and participate in the group’s learning; you will find that cooperating improves performance and harmony in the class.

Emotional education Get to know yourself; identify the situations that bring up complicated emotions and manage them with constructive, self-affirming experiences.

Enterprising culture Trust in your skills and knowledge, develop creativity, adapt to changing situations and have a proactive and responsible attitude.

Academic and professional

ICT

orientation

Assessment

Linguistic Plan

Learn how to obtain information, select it and apply it; to plan, manage and work on projects; to collaborate online in an ethical and safe manner.

Evaluate your personal skills, discover and awaken your calling, train yourself to make decisions and learn to choose between different options.

Discover different strategies to analyse what you have learnt and how you learnt it; train yourself to take responsibility or overcome difficulties.

Use your communication skills in the different types of text that you will see.

Language is always present, communicate!

RESOURCE

BANK

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DIGITAL BOOK RESOURCE BANK

resources SUBJECT KEY CONCEPTS Tutorials Activities with GeoGebra Glossary

A digital version of your book to be used online or offline. It offers access to your digital resources which are grouped by type or linked to unit content.

Learn by playing Self-assessments Language Bank

A space with resources, techniques and activities, designed to strengthen your knowledge.

More about the keys

Resources related to THE PROJECT KEYS

SDG

SDG Commitment with short videos that will help you understand the targets for reaching the Sustainable Development Goals worked on in this project.

Linguistic Plan with infographics that will give you models to work with the four linguistic skills, using different text types (descriptive, narrative, explanatory, etc.).

Cooperative learning Preparing for the task In small groups, of four or five members: 1 All the members of the team will review how the assigned task can be accomplished.

2 To do this, the steps can be shared out to each team member who then in turn will explain how each part of the process can be done to the others. The others listen and participate if they think they can contribute something.

Authorship / adaptation : Variant of the Educational Innovation Laboratory of the colegio Ártica - David and Roger Johnson.

Developing thinking with explanations are included on how to apply the different thinking techniques proposed in the project.

3 Once everyone is in agreement on how to do each part, you will all complete the tasks and, finally, verify, among everyone, that you have solved it correctly.

Cooperative learning which includes the descriptions of the cooperative learning techniques proposed in the project.

Thinking techniques

Emotional education with resources to help you overcome any worries that may arise in different situations at school (beginning of the school year, taking a test, etc.).

Logic Wheel This thinking technique will help you to establish phases when analysing specific content that you have to study.

Identify What is it? What is it like? Are there different types?

By following a logical sequence (the logic wheel), and by asking yourself a series of questions in each phase, you can:

1

• Identify content by asking yourself: What is it? What is it like?

Are there different types?

• Compare the content by formulating questions such as: In what way is it similar to ...? In what way is it different from ...?

• Establish cause-effect relationships by asking yourself questions such as: Why? What impact does it have ...?

• Argue, assess and ask yourself questions such as: What conclusions can be drawn after the analysis? What can be assessed or scored about it?

Doing a data dump of these questions into a graphic organiser will help you.

Compare

Argue, assess What can we conclude?

4

Logic wheel

2

In what way is it similar to ...?

In what way is it different from ...?

3 Establish cause-effect relationships Why? What impact does it have ...?

Authorship: Hernández, P., and García, L. A.; adapted by Escamilla, A.

ICT resources to help you use information and communication technology in a healthy, correct and safe way.

Academic and professional orientation with information on different professions linked to the subject content.

0:40/1:33

Assessment which includes resources for your portfolio, as well as rubrics and targets that will help with your self-assessment.

Resources classified by unit

All the resources are classified by unit so that you can find them more easily.

1

course contents REAL NUMBERS

Page 10

1. Irrational numbers...............................................................

2. Real numbers: the real number line..............................

3. Sections of the real number line: intervals and half-lines.........................................................................

4. Roots and radicals...............................................................

5. Approximate numbers. Errors.........................................

6. Numbers in scientific notation. Error control............

7. Logarithms.............................................................................

Exercises and problems solved..........................................

Exercises and problems.........................................................

Maths workshop........................................................................

Self-assessment........................................................................

2

POLYNOMIALS AND ALGEBRAIC FRACTIONS

3

16 18 22 24 26 29 30 34 35

4

FUNCTIONS. CHARACTERISTICS

Page 88

1. Basic concepts...................................................................... 90 2. How functions are presented.......................................... 91 3. Domain of definition........................................................... 94 4. Continuous functions. Discontinuities......................... 95 5. Increase, maximums and minimums............................ 96 6. Tendency and periodicity................................................. 98 Exercises and problems solved........................................... 100 Exercises and problems.......................................................... 102 Maths workshop......................................................................... 106 Self-assessment........................................................................ 107

38 40 42 44 46 48 50 52 54 60 61

5

1. Equations................................................................................

2. Systems of equations.........................................................

3. Inequations with one unknown......................................

Exercises and problems solved..........................................

Exercises and problems.........................................................

Maths workshop........................................................................

Self-assessment........................................................................

64 70 74 78 79 84 85

BASIC FUNCTIONS

Page 108

1. Linear functions.................................................................... 110 2. The parabola: a very interesting curve........................ 113 3. Quadratic functions............................................................ 114 4. Absolute value functions.................................................. 117 5. Radical functions.................................................................. 118 6. Inversely proportional functions.................................... 120 7. Exponential functions........................................................ 122 8. Logarithmic functions........................................................ 124 Exercises and problems solved........................................... 125 Exercises and problems......................................................... 127 Maths workshop......................................................................... 132 Self-assessment......................................................................... 133

6 Page 62

7

SIMILARITY. APPLICATIONS

Page 136

1. Similarity................................................................................. 138 2. Homothety............................................................................. 140 3. Rectangles with interesting dimensions..................... 142 4. Similarity of triangles......................................................... 144 5. Similarity of right-angled triangles............................... 146 6. Similarity of triangles in geometric solids.................. 148 Exercises and problems solved........................................... 150 Exercises and problems.......................................................... 151 Maths workshop........................................................................ 156 Self-assessment........................................................................ 157

TRIGONOMETRY

Page 158

1. Trigonometric ratios of an acute angle....................... 160 2. Basic trigonometric identities......................................... 162 3. Using a calculator in trigonometry............................... 164 4. Solving right-angled triangles........................................ 166 5. Solving non-right-angled triangles............................... 167 6. Trigonometric ratios of 0° to 360°............................... 168 7. Angles of any measurement. Trigonometric ratios.. 171 8. Trigonometric functions. The radian............................ 172 Exercises and problems solved.......................................... 174 Exercises and problems......................................................... 175 Maths workshop......................................................................... 180 Self-assessment......................................................................... 181

8

Page 36

1. Polynomials. Operations...................................................

2. Ruffini’s rule...........................................................................

3. Root of a polynomial. Finding roots.............................

4. Factorising polynomials....................................................

5. Divisibility of polynomials.................................................

6. Algebraic fractions..............................................................

7. Decomposition of algebraic fractions into partial fractions...................................................................................

Exercises and problems solved..........................................

Exercises and problems.........................................................

Maths workshop........................................................................

Self-assessment........................................................................

EQUATIONS, INEQUATIONS AND SYSTEMS

12 14

ANALYTIC GEOMETRY

Page 182

1. Vectors in the plane.......................................................... 184 2. Operating with vectors.................................................... 185 3. Vectors which represent points................................... 187 4. Midpoint of a segment..................................................... 188 5. Aligned points..................................................................... 189 6. Equations of a straight line............................................ 190 7. Straight lines. Parallelism and perpendicularity..... 192 8. Straight lines parallel to the coordinate axes.......... 194 9. Relative positions of two straight lines..................... 195 10. Distance between two points....................................... 196 11. Equation of a circumference......................................... 197 Exercises and problems solved.......................................... 198 Exercises and problems......................................................... 199 Maths workshop........................................................................ 204 Self-assessment........................................................................ 205

9

STATISTICS

Page 208

1. Statistics and statistical methods.................................. 210 2. Frequency tables................................................................. 212 3. Statistical parameters: x and q....................................... 214 4. Dispersion parameters for isolated data.................... 216 5. Dispersion parameters for grouped data................... 218 6. Box-and-whisker plots....................................................... 220 7. Statistical inference............................................................. 222 8. Statistics in the media........................................................ 225 Exercises and problems solved........................................... 227 Exercises and problems......................................................... . 229 Maths workshop......................................................................... 234 Self-assessment ....................................................................... 235

10

TWO-DIMENSIONAL DISTRIBUTIONS

Page 236

1. Two-dimensional distributions........................................ 238 2. Correlation value.................................................................. 242 3. Using the line of best fit to make estimations.......... 244 4. Analyse: does correlation mean cause-effect?........ 246 5. Two-dimensional distributions using a calculator... 247 Exercises and problems solved.......................................... 248 Exercises and problems......................................................... 249 Maths workshop......................................................................... 252 Self-assessment......................................................................... 253

11

COMBINATORICS

Page 254

1. Strategies based on the product................................... 256 2. Variations and permutations (the order matters)... 262 3. When the order does not matter. Combinators....... 264 4. An interesting arithmetical triangle.............................. 266 Exercises and problems solved........................................... 268 Exercises and problems......................................................... 269 Maths workshop......................................................................... 272 Self-assessment......................................................................... 273

12

CALCULATING PROBABILITY

Page 274

1. Random events..................................................................... 276 2. Probability of events........................................................... 278 3. Probability in single experiments.................................. 280 4. Probability in compound experiments........................ 282 5. Creating individual experiments.................................... 283 6. Creating dependent experiments.................................. 284 7. Contingency tales................................................................ 286 Exercises and problems solved........................................... 288 Exercises and problems......................................................... 290 Maths workshop......................................................................... 294 Self-assessment......................................................................... 295

Annex • Glossary............................................................................................ 296

C I T E M H T I R A A R B E G L A D AN K 1 C O BL

BERS

REAL NUM

IONS T C 1 A R F C BRAI E G L A D N IALS A M O N Y L O MS E T 2 P S Y S D NS AN O I T A U Q E S, IN N O I T A U Q 3 E

Remember that you can find academic and professional guidance related to this content at anayaeducacion.es.

1

REAL NUMBERS

The organisation of different types of numbers The sets of numbers that we know and have a clear structure are: • Natural numbers, N.

Z. • If we include fractional numbers as well, we get the rational numbers, Q.

Reading and listening

• If we include their opposites (negatives), we get the set of integers,

Rational and irrational numbers Since ancient times, natural numbers have been used in all civilisations.

The idea of zero and negative numbers took longer to understand. Integers emerged in the late 7th century in India. From this, 9th century Arabic mathematics merged with the decimal-positional numeral system.

Fractions were not used in the way they are today until around the 14th century.

Irrational numbers were discovered by the Pythagoreans around the 5th century BCE. However, for almost 2 000 years they were treated as geometric magnitudes.

The number π is an irrational number. However, over the centuries it has been given many different rational values. Some of these include: Ancient Egypt (approx. 20th century BCE)

3.16

Ancient Babylon (approx. 20th century BCE)

25/8

Archimedes (3rd century BCE)

22/7

Ptolemy

(2nd

century)

377/120

Liu Hiu

(5th

century)

355/113

• If we also include the irrational numbers, do we get a clearly structured set?

rational numbers

Q

natural numbers 8 0, 7, 15, 33 , 3 32 , …

11 integers N Z negative 24 3 integers 8 –13, – 48, – 6 , –27 , … # fractional numbers 8 8.92; –15.863; 7 ; …

11 (rational non-integers)

irrational numbers 8

2,

5, – 8,

3

4, …

1 Use the passive voice to make sentences about natural, negative, and

irrational numbers. For example: Fractions were used differently in the past.

2 Make comparative sentences to show the differences between the

cultures and years which gave rational values to π.

3 Explain who Cantor was. What makes him important for the way we

think of real numbers today?

Solve 1 a) W rite three natural numbers and three integers

that are not natural numbers.

The set of real numbers The concept of real number has emerged over time through the study of functions. A German mathematician, Cantor, changed the way we thought about rational and irrational numbers in 1871. It was then that we started to think of these numbers as forming a single set, with their own characteristics.

10

b) Write three rational numbers that are not integers and three irrational numbers. c) In your notebook, organise the numbers from your previous answers in a diagram like the one at the beginning of this exercise.

2 Given that the value of π is 3.14159265359… give the margin of error

in each of the approximations from the previous page.

ANK ANK B E G A U LANG LANGUAGE B ANK ANK GE BANK B B E E G G A A U U GUA K LANG ANG N L A L AN GE BANK 11 ANK GE BANK B B E E G G A A U U G A LAN LANG LANGUA LANGU

For example: 377 = 3.1416 66… The margin of error is less than 120 2

3.1415 92… 1 ten thousandth: error < 0.0001

Unit

1

IRRATIONAL NUMBERS

ãã The golden ratio: F = √52+ 1

Pentagram

There are infinite irrational numbers. Let’s take a look at some interesting examples.

— √2

1

1

This shape is formed by the five diagonals of a regular pentagon. It was the symbol of the Pythagoreans.

— As such, it could be expressed as the quotient of two integers: Remember When we break down a perfect square into prime factors, each prime number appears an even number of times. For example: N = 22 · 3 · 53 N 2 = (22 · 3 · 53)2 = 24 · 32 · 56 All the exponents are even.

2= a b

2 — We square the two sides: 2 = a 2 8 a 2 = 2b 2 b Since b 2 is a perfect square, it contains the factor 2 an even number of times.

Therefore, 2b 2 has factor 2 an odd number of times, which is impossible since 2b 2 = a 2 is another perfect square.

In this way, our reasoning is as follows: ‘If we presume that 2 is rational, we reach an absurdity’. Therefore, we have shown, through reduction to absurdity, that 2 is not rational.

L =π 2r

It is an irrational number and, therefore, has infinite non-recurring decimal digits.

Remember Unlike 2, 5, F, and other irrational numbers, π and e cannot be represented exactly on the real number line.

π is the Greek letter corresponding to ‘p’. Why is it called Pi? The Greek word perifereia means ‘circumference’ (the periphery of the circle).

ãã The number e The number e is another fundamental irrational number in mathematics. It is named after Leonhard Euler, one of the most important mathematicians in history.

Its approximate value is 2.7182... and from now on you will come across it in many different situations:

Catenary

For the same reason as 2, if p is not a perfect square, p is irrational.

— It is also used to describe the change in radioactivity of a radioactive substance over time.

n

p is an irrational number.

For example, the following are irrational numbers: 2 + 3, 4 – 5 10 ,

12

As you know, π is the relationship between the length of the circumference of a circle and its diameter. You have known and used this number for many years now.

— When describing the growth of an animal or plant population, we use an exponential function which contains the number e.

y=

The result of performing an operation using a rational number and an irrational one is irrational (except in the case of multiplying by zero).

1

L 2r

For example, 8, 3 9 and 5 10 are irrational numbers.

Think and practise

The name, F (phi, letter ‘F’ in Greek), comes from Phidias, a Greek sculptor from the 5th century BCE who often used this ratio.

ãã Other irrational numbers expressed through radicals If p is not an exact nth power,

anayaeducacion.es Recognising irrational numbers.

Greek artists believed that proportions based on the number F were especially harmonious, which is why they decided to call F the golden number.

ãã The number π

anayaeducacion.es Interesting facts about π and other irrational numbers.

We are going to demonstrate that 2 is irrational, in other words, that it cannot be expressed as the quotient of two integers. We will do this through reduction to absurdity (supposing it is rational and then showing it reaches an absurdity).

— Suppose that 2 is rational.

d =F

d

ãã The diagonal of a square: the number √2 The Pythagorean theorem gives us the value of the diagonal of a square with sides measuring 1: d = 12 + 12 = 2

The diagonal of a pentagon with a side length of 1 unit is ( 5 + 1) : 2 which is obviously irrational.

Furthermore, historically, it is the first irrational number to be discovered.

l

Rational numbers are numbers that can be expressed as the quotient of two integers. Their decimal expression is terminating or recurring.

Irrational numbers are non-rational numbers, in other words, they cannot be expressed as the quotient of two integers. Their decimal expression is infinite and non-recurring. For example, π = 3.14159265359…

1

3

9 : 7.

Let’s show that 4 – 5 10 is irrational based on the fact that 5 10 is.

— We will call N = 4 – 5 10 8

5

10 = 4 – N.

— If N were rational, 4 – N would also be rational. In other words, 5 10 would be rational, which is not true.

Demonstrate that the following numbers are irrational:

a) 3

b) 4 3

c) 5 + 4 3

— If we hang a chain, a cable, a rope, etc. with each end at the same height, the curve it forms (a catenary) is also described using the number e.

e x + e –x 2

Think and practise 2 Explain why these constructions contain a segment that measures the same as F = 5 + 1 = 5 + 1 .

2 2 2 F

— √5 — 2

1/2 1

F

1 — 2

1

3 Demonstrate that the golden number, F, is irrational.

4

1

x

is rectangle has the peculiarity that if Th we cut a square out of it, the rectangle that is left is similar to the initial one.

Demonstrate that its longest side is x = F.

1 — 2

13

Unit

2 Interesting fact You may study other sets of numbers in later years, but Û fills the whole number line and leaves no gaps, as we will see below.

ãã Representing numbers on the real number line

REAL NUMBERS: THE REAL NUMBER LINE

Example

The set formed by rational and irrational numbers is called the set of real numbers and is represented by the letter Á.

In other words, both rational and irrational numbers are real numbers. They form all the real numbers there are. With the Á set, we can complete the table of sets of numbers:

integers

rational Z

Q

natural 8 0, 4, 24 , 6 N

➜➜ using thales’ theorem to represent fractional numbers

Representation of 5 : 6

1

2

3

0

5

4

6

The product of two negative integers is not a negative integer. For example: (–2) · (–3) = 6 And the result of adding two fractional numbers is not necessarily a fraction. For example: 1 + 5 = 6 =2 3 3 3

ãã The real number line Rational numbers, as you know, are grouped densely on the number line, in other words, each section, no matter how small, contains an infinite number of rational numbers. However, as strange as it may seem, there are an infinite number of gaps between them. These gaps are filled with the irrational numbers.

All together they fill the line.

Remember The real number line is complete, in other words, there is a real number for each point on the line and a point on the line for each real number.

14

0

1

0

1

43 2+— 5

The following procedure* allows us to calculate

F ocu s on Eng lish procedure: an accepted way of doing something.

n for any n é N: For example: 3 = ` 2j + 1 2 2

1

0

1

— — √2 √3

2

— √5

—

3 √ 10

10 = 3 2 + 1 2

However, most real numbers cannot be represented exactly using this type of procedure. For example, how would you represent 5 842 ? We normally use an approximate representation. ➜➜ approximate representation of real numbers

We can represent a real number given through its decimal expression with as close an approximation as we like. For example: 5 842 = 3.8464…: 3.8 3.9

0

1

2

Notice how each expansion involves splitting the previous subinterval into ten parts and then taking one of those parts.

Basically, we can get as close to the number we want as we like.

anayaeducacion.es GeoGebra. Representing numbers on the real number line.

3

4 3.84

3.85

3.8

3.9 3.846 3.847

3.84

3.85

Depending on the type, real numbers can be represented on the real number line either exactly or with as close an approximation as we like.

Think and practise 1 a) Justify that the point represented is 21.

If we place an initial point (0, zero) on a line and mark the length of one unit, each point on the line will correspond to a rational or irrational number. In other words, each point on the line corresponds to a real number. This is why we call this number line the real number line.

Once all the real numbers have been put in place, we can be sure that between any two numbers, no matter how close together they are, there are infinite rational and irrational numbers.

2

➜➜ using the pythagorean theorem to represent radicals

We can also extract roots of any index (except roots with even indices of negative numbers) and the result is still a real number. This does not happen with rational numbers.

We can see that the N, Z, Q and, now, Á sets are closed when working with addition and product operations; in other words, both the addition and product of two elements of one of these sets is also part of that set. However, this does not happen with negative integers, fractional and irrational numbers: the sum of two irrationals can be rational (for example: (1 + 2) + (3 – 2) = 4) and the product of two irrational numbers can also be rational (for example: 2 · 8 = 4).

14 = 2 + 4 5 5

With real numbers, we can perform the same operations as with rational numbers: addition, subtraction, multiplication and division (except for zero), and the same properties are maintained.

Observe

14 Have a look at how we use Thales’ theorem to place the number 5 on the real number line:

5 1 — 6

121

negative 8 –11, – 27 , 3 –8 integers 3 Á #

fractional 8 5.84; 1 ; 5.83 ; – 3 10 2 2+ 3

irrational 8 2, 3, F, π, – 5 + 2, 5 real

1

2

What number is the arrow in the following diagram pointing at?

0

0

1

1

2

— √21

b) Represent 27 (27 = 36 – 9) and 40 (40 = 36 + 4).

Represent 2.716 in the same way.

15

Unit

3

SECTIONS OF THE REAL NUMBER LINE: INTERVALS AND HALF-LINES

ãã Half-lines and the real number line Half-lines

(–∞, a) are the numbers less than a: {x / x < a}.

(– ∞, a) = {x / x < a}

In the world of science, we often need to determine the scope of a certain variable.

For example, ‘the period of time between 3 s and 11 s’. Therefore, we need to learn the right names for certain sections of the real number line.

Open interval

The open interval (a, b) is the set of all the numbers between a and b, but not including a or b : { x / a < x < b }.

(a, b) = {x / a < x < b}

It is represented as follows:

a b We read the above expression this way: set of

{ x

/ a <

x <

x in such they are numbers a way greater that than a

b } but less than b

a b

For example, the interval (–2, 1) is formed by the real numbers between –2 and 1, but not including –2 or 1: { x / –2 < x < 1}.

Another example: to make a box using a sheet of card measuring 10 cm × 15 cm, you need to cut four squares out at its corners and then fold it. The sides of the squares, in this case, have to be less than 5 cm 8 (0, 5).

[a, b] = {x / a ≤ x ≤ b}

The closed interval [a, b] is the set of all the numbers between a and b, including both of them: { x / a ≤ x ≤ b }.

It is represented as follows:

a b

a b

For example, the interval [–2, 1] is formed by the real numbers between –2 and 1, including –2 and 1: {x / –2 ≤ x ≤ 1}.

Another example: the parcels we deliver must weigh 2 kg or above, but no more than 5 kg 8 [2, 5].

(a, b] = {x / a < x ≤ b} a b [a, b) = {x / a ≤ x < b} a b

a

[a, +∞) are the numbers greater than a, and a itself: {x / x ≥ a}.

• (– ∞, 2) is the set {x / x < 2} 8 • [2, +∞) is the set {x / x ≥ 2} 8

Observe The union of two intervals or halflines is represented by á: (–∞, 2) á (0, 5] = (–∞, 5] The intersection of two intervals or half-lines is represented by Ü: (–∞, 2) Ü (0, 5] = (0, 2)

2

2 • In order to be able to vote, you need to be 18 8 [18, +∞). Of course, in this context, +∞ has to be put into perspective, because nobody lives forever.

The real number line itself is represented as an interval in this way: Á = (–∞, +∞).

When there is no number that satisfies a specific condition, it is represented by the empty set, whose symbol is Ö. For example, the values of x for which x 2 < 0 correspond to the empty set: {x é Û / x 2 < 0} = Ö.

Problems solved

the following in interval form and represent them:

a) Half-open interval (2, 3]

a) 2 < x ≤ 3

c) Half-line (0, +∞)

b) x ≤ 1

c) x > 0 2 Write

the following as inequalities and represent them:

a) [–2, 0]

b) [–1, +∞)

c) (0, 1)

(x + 2) (x – 3)

2 3

b) Half-line (–∞, 1]

a) {x / –2 ≤ x ≤ 0} b) {x / x ≥ –1} c) {x / 0 < x < 1}

1 0 –2 0 –1 0 1

The square root can be calculated when the radicand is zero or positive, and this happens when one of the factors is zero, both are negative or both are positive. In other words, if x ≤ –2 or if x ≥ 3.

(–∞, –2] ∪ [3, +∞)

–3 –2 –1 0 1 2 3 4

a b

• The interval [a, b) is the set of all the numbers between a and b, including

a but not b : {x / a ≤ x < b }.

a b

For example, the interval (3, 4] is formed by all the real numbers between 3 and 4, including 4 but not 3: {x / 3 < x ≤ 4}.

Another example: a nursery school admits children that have turned 1 but are not yet 4 years old 8 [1, 4).

16

(a, +∞) are the numbers greater than a: {x / x > a}.

a

b but not a: {x / a < x ≤ b}.

It is represented as follows:

a

following expression true?

• The interval (a, b] is the set of all the numbers between a and b, including

It is represented as follows:

(–∞, a] are the numbers less than a, and a itself: {x / x ≤ a}.

3 For which values of x is the

ãã Half-open interval Half-open interval

(a, +∞) = {x / x > a}

a

1 Write

ãã Closed interval Closed interval

(– ∞, a] = {x / x ≤ a}

[a, +∞) = {x / x ≥ a}

ãã Open interval

1

Think and practise 1 Write the following sets in interval form and repre-

sent the numbers that meet the conditions given in each case: a) Numbers between 5 and 6, both included. b) Greater than 7. c) Less than or equal to –5.

anayaeducacion.es GeoGebra. Intervals on the real number line.

2 Write the following in interval form and represent them:

a) {x / 3 ≤ x < 5}

b) {x / x ≥ 0}

c) {x / –3 < x < 1}

d) {x / x < 8}

3 Write as inequalities and represent them.

a) (–1, 4]

b) [0, 6]

c) (–∞, – 4)

d) [9, +∞)

17

Unit

4

c) 4

n n

b) k –243 = –3

k 1 024 = 2 k = 2 d) 3

5 a) 3 – 8 b) 32

c)

80 –32 d)

f ) 3 125

e) 4 81

Radicals have a series of properties that you need to know and be able to use easily. You will find them listed in the margin of this page and the next. All of them are the direct consequences of known properties of powers.

a ) is the name given to a number, b,

We will pay special attention to the operations that they can be used for.

a = b if b n = a

➜➜ simplification of radicals

a is called radical; a, radicand; and n, index of the root.

When working on expressions like this one, you will sometimes need to calculate the numerical value. To do so, you need to bear in mind the definition, like in the exercises in the margin of this page, or use a calculator. However, in other cases, you will need to maintain the radical, simplify it, perform operations with other radicals, etc.

2. Calculate the following roots: 5

n

The nth root of a number, a, (written that meets the following condition:

1. Indicate the value of k in each case: a) 3 k = 2

ãã Operations with radicals

ROOTS AND RADICALS

Mental arithmetic

Property 1 np np

n

a p = a p/np = a 1/n = n a

We have applied property 1 (see margin). ➜➜ reduction of radicals to a common index

It is not always easy to compare two radicals with different indices. If we express them with the same index, it is much easier. Actually, it simply consists of reducing them to a common denominator.

a exists no matter what the value of n is.

For example, to compare 3 586 to 70,

• Although 4 has two square roots, when we write 4 we refer to the positive root: 4 = 2.

In general, a positive number, a, has two square roots: a and – a .

ãã Radicals in exponential form n n

a=

m

am = a n

1 an,

1 (a n )n

n

a=

n

a m = a n , since

since

m

n

=

n an

Property 2

= a1 = a 1

a m = (am) n = a

`6 27j = `6 3 3j = (33/6)2 = 36/6 = 3

3

2

1 m· n

m

=an

2

`3 x 2j a) 5 x b) 5

c)

e) 6

18

a 13 a 6 d) a6 x3 x2

a · b = n a · n b , since:

n

a · b = (a · b) 1/n =

= a 1/n · b 1/n =

=n a·n b

Property 3

form:

4

f ) >e 5 x o H 2 x4

3

586 > 70

In order to simplify some radicals, and to add and subtract them, sometimes we need to take out factors from a root. Let’s look at some examples:

18 = 3 2 · 2 = 3 2 · 2 = 3 2

720 = 2 4 · 3 2 · 5 = 2 4 · 3 2 · 5 = 2 2 · 3 · 5 = 12 5

We have applied property 2 (see margin). ➜➜ product and quotient of radicals with the same index

Think and practise

15

n

64 = 3 2 6 = 26/3 = 22 = 4

1 Express each of the following roots in exponential

586 = 586 1/3 = 586 2/6 = 6 586 2 = 6 343 396 4 8 70 = 70 1/2 = 70 3/6 = 6 70 3 = 6 343 000

➜➜ taking out factors from roots

For example: anayaeducacion.es Activities for reviewing the properties of powers.

3

We have applied property 1 again (see margin).

Radicals can be expressed as powers: 1 an

9 = 4 3 2 = 3 2/4 = 3 1/2 = 3

4

• If a < 0, only exist its roots of odd indices.

Watch out

When expressing the radicals in power form, we will notice that they can sometimes be simplified. For example:

a p = n a , since:

ãã Some peculiarities of roots • If a ≥ 0,

1

2 Calculate.

a) 41/2

b) 1251/3

c) 6251/4

d) 82/3

e) 645/6

f ) 363/2

n

n

a = n a , since: b nb a = a 1/n = a 1/n = n a b l b b b 1/n n b

b) (m 5 · n 5)1/3

c) a 1/2 · b 1/3

d) [(x 2)1/3]1/5

e) [(x 1/2)5]1/3

f ) (y 3 · z 2)2/3

15 = 20

15 = 20

3 (property 3, see margin) 4

➜➜ simplification of products and quotients of radicals

For example: 3 · 3 2 = 6 3 3 · 6 2 2 = 6 3 3 · 2 2 = 6 108

3 Express the following in radical form.

a) x 7/9

For example: 15 · 20 = 15 · 20 = 300 (property 2, see margin)

anayaeducacion.es Practise operations with radicals.

3

16 = 6 32

6

16 2 6 (2 4) 2 6 2 8 6 3 = = = 2 = 2 6 32 25 25

We have applied properties 1, 2 and 3 (see margin).

19

4 ROOTS AND RADICALS

Unit

ãã Rationalising the denominator

➜➜ power of a radical

`n aj = n a p , since: p

`n

aj

p

Observe

For example:

Property 4

= (a 1/n) p = a p/n = n

ap

`

4 2 3j

`5 2j = 5 2 3 = 5 8 3

It is more difficult to calculate: 1.00000000 0100600 016060 01918

➜➜ root of a radical

For example:

mn

a = m · n a , since:

mn

a = (a 1/n) 1/m = a 1/m · n = m · n a

3

2= 2

43

5 = 12 5

than calculating: 1.4142…

014 02 0

6

We have applied property 5 (see margin).

We cannot add two different radicals together unless we find their approximate decimal expressions. We can only add identical radicals. For example:

Only identical radicals can be added.

3+ 2 4 We can only solve them approximately, or 7 – 3 7 leave them indicated.

If you perform the operation both ways, you will see that it is much easier if you remove the radical from the denominator (see margin).

2 0.7071…

Although nowadays we do not need to do so, thanks to the powerful calculation tools available, the final results of problems are still usually given with numerical expressions that do not have radicals in the denominator.

The process we use to remove the radicals from the denominator is called the rationalisation of the denominator.

Remember

7 5 + 11 5 – 5 = 17 5

Sometimes, the possibility of simplifying an addition of radicals is hidden.

First, we need to take out the factors that we can from the roots or simplify them.

F ocu s on Eng lish

anayaeducacion.es Activities to consolidate your knowledge of radicals.

2nd case: other roots. For example:

32 + 5 18 – 50 = 2 5 + 5 3 2 · 2 – 5 2 · 2 =

= 4 2 + 15 2 – 5 2 = 14 2 • (a + b ) · (a – b ) = a 2 – b 2 • We call the expression a – b the

f ) 8 81

5 Which of the two is greater in each case?

a) 4 31 and 3 13

a) 2 ·

20

5

8 Simplify.

5 4 3 5 16 9 a b c a) 3 b) c) 2 3 ab 3 c 3 6

b) 51 and 132 650 3

3 5 a) 3 32x 4 b) 64 81a 3b 5 c c)

` xj · `3 xj d) `3 a 2j e)

9

6 Reduce.

conjugate of

7 Take out all possible factors from the radicals.

12 8 5 10 a) 12 x 9 b) x c) y

3

f ) a

2k

8

9 Calculate.

10 4 6 2 b) a b 6 · 6 3 c) 3

a) 18 + 50 – 2 – 8

2 · (3 – 2) 6–2 2 6–2 2 6–2 2 2 = = 2 = = 2 9–2 7 3 + 2 (3 + 2) · (3 – 2) 3 – ( 2)

Remember

8 + 4 4 = 23 + 4 22 = 2 2 + 2 = 3 2

4 Simplify.

3

1 · ( 5 + 3) 5+ 3 5+ 3 1 = = = 2 2 2 5 – 3 ( 5 – 3) · ( 5 + 3) ( 5) – ( 3)

Think and practise

9 64 d) 6 8 e)

For each case, we should ask ourselves this question: Which expression do I need to multiply the denominator by for the product not to have radicals? Once we have found the expression, we will also multiply the numerator so that the final result does not vary*.

2· 3 2 3 1st case: square roots. For example: 2 = = 3 3 3· 3 5 5 5 1 = 1 · 73 = 73 = 73 5 2 5 2 5 3 5 5 7 7 7 · 7 7 3rd case: addition and subtraction of roots. For example:

vary: change.

For example:

1 = 1· 2 = 2 2 2 2· 2

Rationalising is making something rational that was not rational to begin with.

We can, however, simplify the following expression:

1.4142 0.7071…

and we get the same result.

➜➜ addition and subtraction of radicals Remember

In Antiquity, when the instruments used for calculation did not exist, people had to try and find methods to make operations easier. For example, calculating 1 2 by hand can be done directly (calculating a few figures of 2 and then dividing the result by 1). However, we can simplify the calculations much more if we remember that:

2 = 1.4142…

= 2 12 = 212/2 = 26 = 64

We have applied property 4 (see margin).

Property 5

1

b) 75 + 2 27 – 48

a + b.

And, the other way round, is the conjugate of

a+ b

Some operations are easier to solve if first we rationalise them. For example: 2 3+2 2 2 2 3 2 + 1 – 2 = + – = 1 2 3 3– 2 2 3

a – b.

Think and practise

=

12 3 + 12 2 3 2 4 3 8 3 + 15 2 – + = 6 6 6 6

anayaeducacion.es Consolidate your knowledge with some new rationalisation exercises.

10 Rationalise the following denominators:

5 2 e) 1 d) 4 a) 5 b) c) 3 5 2 2 7 3+ 2 2 3

f )

3 2– 3

21

Unit

5

APPROXIMATE NUMBERS. ERRORS

2 Give a limit of absolute

error and of relative error for each of the estimations given for the amounts in the previous exercise.

ãã Approximations and errors Observe a) 34 m has 2 significant figures. b) 0.0863 hm3 has 3 significant figures. c) It is possible that 53 000 L only has 2 significant figures if the zeros are just used to designate the number. In this case, it would be better to say 53 thousand litres.

Observe a) *

Measurement: 34 m Absolute error < 0.5 m

Z ]] Measurement: 0.0863 hm 3 b) [ Absolute error < 0.00005 hm 3 ] In other words, abs. error < 50 m 3 \ Measurement: 53 thousand L c) * Absolute error < 500 L Observe The relative errors of the previous measurements are: a) R.e. < 0.5 < 0.015 = 1.5 % 34 0 b) R.e. < .00005 < 0.0006 = 0.06 % 0.0863 c) R.e. < 500 < 0.0095 < 0.01 = 1 % 53 000

For practical applications, we generally work with approximate numbers. Let’s review some of the concepts and procedures to manage their use.

Absolute error < 50 000 Relative error < 50 000 = 0.25 = 25 % 200 000 c) Estimation: 3 hundreds of millions = 300 million

value – Approximate

Absolute error < 0.5 tens of millions = 5 million

value|

The real value is generally unknown. Therefore, the absolute error is also unknown. The most important thing is to be able to put limits on it: the absolute error is less than… We get the limit of the absolute error from the last significant figure used. m3),

In the previous example (capacity of the swimming pool: 719 the last significant figure (9) gives units of m3. The absolute error is less than half a cubic metre (error < 0.5 m3).

The relative error is the quotient of the absolute error and the real value. It is therefore lower the more significant figures we use. The relative error is usually also expressed as a percentage (%).

In the example, the relative error is less than: 0.5 < 0.0007 = 0.07 %.

719

1 Use a reasonable number of

significant figures to express the following quantities:

a) It is quite possible that the accuracy of this number is reasonable, as the visitors to a museum pay for a ticket, which is recorded and counted. Let’s suppose that this number, 183 594, is the number of tickets sold.

a) Visitors to an art gallery in one year: 183 594.

Nevertheless, the figure can be simplified: ‘almost two hundred thousand’ or ‘more than one hundred eighty thousand’ are valid estimations.

b) People attending a demonstration: 234 590.

b) It is impossible to count the demonstrators so accurately. Even if this figure has not been ‘inflated’ or ‘played down’ for political reasons, it is impossible to get so close to the actual figure. For example, it would be reasonable to say ‘more than two hundred thousand’, or rather ‘between 200 000 and 250 000’.

22

5 000 < 0.028 < 0.03 8 R.e. < 0.03 = 3 % 180 000 b) Estimation: 200 000

Relative error <

The absolute error of an approximate measurement is the difference between the real value and the approximate value.

Problems solved

c) Number of bacteria in 1 dm3 of a preparation: 302 593 847.

Absolute error < 5 000

For example, if the capacity of a swimming pool is 718 900 L, it would be more reasonable to say it is 719 m3, using only three significant figures. However, if the measurement was not very accurate, or we do not want to give such an exact measure, it would be correct to say 720 m3 or, even better, 72 tens of m3.

Absolute error =

c) One or, at most, two significant figures is the most accurate approximation possible with this type of quantity: 3 hundreds of millions of bacteria, or 30 tens of millions of bacteria.

a) If we say the number of visitors is 180 thousand (or, even better, 18 tens of thousands), we have an absolute error of 183 594 – 180 000 = 3 594 people.

We know this with such accuracy because we know the exact amount. However, whoever we give this information to (18 tens of thousands) will need to understand that there may be an error of up to 5 units from the first unused figure: 5 000 people. To summarise: Estimation: 180 thousand

We call the figures used to express an approximate number significant figures. We only need to use those we are sure of, and in such a way that they are relevant for the information we want to transmit.

|Real

1

Relative error < 5 < 0.017 < 0.02 8 R.e. < 0.02 = 2 % 300

Think and practise 1

anayaeducacion.es Practise calculating errors.

True or false? Explain your answers. a) Saying that a swimming pool holds 147 253 892 thousand drops of water is correct if the measurements are very accurate. b) Saying that a swimming pool holds 147 253 892 thousand drops of water is not reasonable, as it is impossible to achieve such accuracy in measurements. It would be much more sensible to say that the swimming pool holds 15 ten thousands of millions of drops. c) If we correctly estimate that the number of drops of water that a swimming pool holds is 15 ten thousands of millions, the absolute error is less than half of ten thousand million drops; in other words, absolute error < 5 000 000 000 drops. d) If the relative error for a certain measurement is less than 0.019, we can say that it is less than 19 %. e) If the relative error for a certain measurement is less than 0.019, we can say that it is less than 2 %. f ) The calculator tells us that π = 3.14159265. If we take π = 3.14, we can say that the absolute error is less than 0.00159266, but it is more reasonable to say that the absolute error < 0.0016 or, even, that the absolute error < 0.002.

2 Explain why it is not reasonable to say that in a sack there are 11 892 583 grains of rice.

Express it in a suitable way and give a limit of absolute error and relative error for the expression.

3 Give a limit of absolute error and relative error for when you approximate π as 3.1416.

23

Unit

6

NUMBERS IN SCIENTIFIC NOTATION. ERROR CONTROL The numbers 3.845 · 1015 and 9.8 · 10–11 are written in scientific notation because:

anayaeducacion.es Operations with numbers in scientific notation.

➜➜ addition and subtraction

We have to prepare the addends so that they have the same powers with a base of 10. That way we can use it as a common factor. Then, once the addition has been performed, the result is readjusted. For example:

Test yourself

— They are made up of two factors: a decimal number and a power of 10.

3.7 · 1011 + 5.83 · 108 – 4 · 109 = 3 700 · 108 + 5.83 · 108 – 40 · 108 =

Express the following numbers in scientific notation: a) 340 000 b) 0.00000319 c) 25 · 106 d) 0.04 · 109 e) 480 · 10– 8 f ) 0.05 · 10– 8

— The decimal number is greater than or equal to 1 and less than 10.

— The power of 10 is an integer exponent.

Or alternatively:

The first, 3.845 · 1015 = 3 845 000 000 000 000, is a ‘large’ number.

3.7 · 1011 + 5.83 · 108 – 4 · 109 = 3.7 · 1011 + 0.00583 · 1011 – 0.04 · 1011 =

The second, 9.8 · 10–11 = 0.000000000098, is a ‘small’ number. advantages of this notation

As its name suggests, this form of notation is intended to be used in contexts that require precision, not in everyday language. Can you imagine saying something like this?

— How many children do you have? 8 5 · 100 Another example 7.6 · 108 and 7.60 · 108, although apparently equal, are not, as the second is more accurate and is given with an additional significant figure.

— How many students are there in your school? 8 6.74 · 102 — Do you have any pins with a width of 2.5 · 10–4 m?

Obviously not. Clearly, this form of notation is not for use in everyday life.

However, this form of expression is very useful for handling very large or very small approximate quantities, since: • You can see the ‘size’ of the number at a glance. It can be seen in the second factor and it is given by the exponent of 10.

• We can see how accurate the quantity is. The more significant figures given in the first factor, the more accurate the number is.

Interesting fact You might come across an explanation like this: If an integer ends in one or more zeros, you can easily determine the number of significant figures it has by expressing it in scientific notation. However, another way to indicate that these zeros are significant figures is to put a decimal point at the end of the number. Like this: 3 200 would have 2 significant figures.

3 200, would have 4 significant figures.

This rule is not approved by the scientific community (and we will not use it in this book), but if the teacher thinks it might help to clarify things, why not use it?

24

For example, we can see that 7.6 · 108 and 7.603 · 108 are approximately equal (‘very similar sizes’). However, the second one is more accurate, as it has four significant figures, whereas the first one has only two.

ãã Operations with numbers given in scientific notation

We are told that China has 1 400 million inhabitants. a) Express this quantity in scientific notation. b) Is it an exact or approximate amount? c) Give a limit of absolute error bearing in mind how the figure is given. d) Give a limit of relative error.

➜➜ product and quotient

(3.25 · 105) · (4.6 · 1011) = (3.25 · 4.6) · (105 · 1011) = 14.95 · 1016 = 1.495 · 1017 (3.25 · 105) : (4.6 · 1011) = (3.25 : 4.6) · (105 : 1011) = 0.7065 · 10–6 = 7.065 · 10–7 Note that in the quotient 3.25 : 4.6 we have used four significant figures. If the context of the problem does not suggest the right number of significant figures to use, we have to make the decision subjectively.

= 3 665.83 · 108 = 3.66583 · 1011

= 3.66583 · 1011

ãã Controlling error in a number in scientific notation If we are told that ‘there are 2 500 bags of flour in this warehouse’, this could be an approximate quantity. If so, the statement might only mean that there are approximately 25 hundreds of bags, with an error of less than 50 bags. Or it might mean there are 250 tens, with an error of less than 5 bags.

If we use scientific notation, the expression is unambiguous: 2.5 · 102 means that there are only two significant figures. And if there are three, we write 2.50 · 102.

Problem solved

We can easily do operations in scientific notation with a calculator. Remember that to write a number in scientific notation you use the � key. For example, 7.6 · 108 8 7.6 � 8; 2.5 · 10–4 8 2.5 �f 4. But if we wanted to do this ‘by hand’, we have to take great care.

We do the operations involving the decimal components separately from the powers of 10. Then, we adjust the result so that it is expressed correctly using scientific notation. For example:

Attention The information given by the following numbers is different: 2.5 · 102; 2.50 · 102; 2.500 · 102 The zeros added to the end of the decimal number are used to indicate the number of digits that are accurate.

1

a) 1 400 million = 1.4 · 109 inhabitants b) Obviously, this is an approximate number, as it is impossible to calculate such a large, disperse and changing amount with accuracy. c) and d) When we are told ‘1 400 million inhabitants’, we expect the first two figures to be accurate. However, it is possible that one of the zeros that comes after them is also accurate.

— If only the two first figures in the measurement are accurate: Measurement: 14 hundreds of millions of people.

Absolute error < 0.5 hundreds of millions = 50 000 000 Relative error < 0.5/14 < 0.036 = 3.6 % — If the first zero in the measurement is accurate:

Measurement: 140 tens of millions of people. In this case, we can express it using scientific notification as follows: 1.40 · 109. Notice how the presence of 0 behind the decimal point means that this figure is controlled.

Absolute error < 0.5 tens of millions = 5 000 000 Relative error < 0.5/140 < 0.0036 = 0.36 %

Think and practise 1 Calculate and check your answers with a calculator.

a) (6.4 · 105) · (5.2 · 10– 6) b) (2.52 · 104) : (4 · 10– 6) c) 7.92 · 106 + 3.58 · 107 d) 6.43 · 1010 + 8.113 · 1012 – 8 · 1011

2 The distance from the Earth to the Sun is 149 000 000 km.

a) Express it using scientific notation. b) Express it in centimetres to two significant figures. c) Express it in centimetres to four significant figures. d) Give a limit of the absolute and relative errors for the three cases mentioned above.

25

Unit

7

ãã Decimal logarithms

LOGARITHMS

Logarithms with a base of 10 are called decimal logarithms. For a long time, they were the most widely used logarithms. This is why we just write log, without the base.

The equality 23 = 8 can also be written: log2 8 = 3.

log2 8 is read ‘base-2 logarithm of 8’.

Similarly, we can say:

log5 125 = 3 because 53 = 125

log5 5 = 1 because 51/2 = 5 2 log10 1 000 000 = 6 because 106 = 1 000 000

log10 0.0001 = – 4 because 10– 4 = 1/104 = 0.0001

We call the exponent that base a has to be raised to in order to obtain P (a > 0 and a ≠ 1) base-a logarithm of P. It is written loga P. loga P = x ï ax = P

ãã Properties of logarithms 1. Two simple logarithms log a a = 1

The logarithm of the base is 1. The logarithm of 1 is 0 in any base.

3

8 4 = 2

2 3 · 2 2/3

log a P = log a P – log a Q Q

2 1/2

log 2 8 = log 2

= 2 3 + 2/3 – 1/2 = 2 19/6

23 = 3

log 2 3 4 = log 2 2 2/3 = 2 3 log 2 2 = log 2 2 1/2 = 1 2

_ b bb ` 8 b b a

8 3 + 2 – 1 = 19 3 2 6

The logarithm of a product is the sum of the logarithms of the factors.

The logarithm of a quotient is the difference of the logarithms of the divi dend and the divisor.

For example: log2

83 4 = log 2 8 + log 2 3 4 – log 2 2 2

3. Power and root log a

P k

= k log a P

log a

n

P = 1 log a P n

The logarithm of a power (P k or n P = P 1/n ) is equal to the exponent multiplied by the logarithm of the base of the power.

For example: log5 125 = log 5

(125) 1/2

= 1 log 5 125 = 1 log 5 5 3 = 1 · 3 = 3 2 2 2 2

4. Base change log b P =

26

And log 587 = 2. ... since 587 is greater than 100 but less than 1 000.

There is a specific button for these logarithms on calculators. But in modern calculators, you have to press the SHIFT button to access this function: log 200 8 s 200 = 2.301029996

Interesting fact At this level: log means decimal logarithm (base 10) and ln means Napierian logarithm (base e).

However, in more advanced mathematics books log is used to refer to Napierian logarithms, since at these levels this type of logarithm is used almost exclusively.

log a P log a b

If we know how to calculate base-a logarithms, we can use that formula to calculate logarithms with any base, b.

ãã Napierian logarithms Remember the number e? Its value is 2.71828… and at the beginning of this unit we explained that it is used to describe growth in plant or animal populations, radioactive decay, and catenaries.

Logarithms with base e are called Napierian logarithms and they are written ln (in other words, loge x = ln x).

On calculators there is a button,

, for accessing these logarithms.

Problems solved 1 Give the value of these

log a (P · Q ) = log a P + log a Q

Let’s check

For example, log 10 = 1, log 100 = 2, log 1 000 = 3, log 0.001 = – 4.

log a 1 = 0

2. Product and quotient

1

logarithms, putting the numbers in the form of powers:

a) log6 1 296

a) 1 296 = 64. Therefore, log6 1 296 = 4. b) 0.125 = 125 = 1 = 13 = 2–3 1000 8 2 Therefore, log2 0.125 = –3.

b) log2 0.125 2 With the calculator, find

log 5, log 50, log 500 and log 5 000.

All of them have the same decimal part. Why?

log 5 = 0.69897…

log 50 = 1.69897…

log 500 = 2.69897…

log 5 000 = 3.69897…

T hey have the same decimal part because all of them are the following type: log10 (5 · 10n) = log10 5 + n log10 10 = n + log10 5 log10 5 is the decimal part of all of them. Plus, an integer, n, is added in each case.

Think and practise

anayaeducacion.es Practise calculating logarithms.

1 Use the definition above to find these logarithms:

a) log5 125

b) log5 0.04

c) log2 128

d) log2 0.0625

e) log a 1

f ) log10 0.0001

g) log2 `1/ 2j h) log3 (1/3)

i) log3 5 9

2 Work out the base of the following logarithms:

a) log a 10 000 = 2

b) log b 216 = 3

c) log c 125 = 3

d) log d 3 = 1 2

3 Use the calculator to find log 7 and log 70. Explain

why both have the same decimal part.

27

7 LOGARITHMS

BLEMS O R P D N A S E IS EXERC

ãã A bit of history The importance of decimal logarithms Look at these two numbers: A = 6 748 B = 67.48 = A/100 log B = log (A/100) = log A – log 100 = = log A – 2 This means that both logarithms have the same decimal part.

In logarithm tables we would look for the decimal part of the logarithm of 6 748 and then add the integer part as appropriate, depending on whether it was 6 748; 67.48 or 67 480 000.

Several centuries before the emergence of calculators, logarithms were invented to cope with the enormous operations that had to be done by hand. This was done by converting products and quotients into additions and subtractions (which are operations that are much easier to work with than the others). To do this, they had to use enormous tables (in very thick books), which contained the exact or approximate values of the decimal logarithms of the factors.

And what about Napierian logarithms? What role did they play? The number e arose naturally in the laborious process used to obtain the decimal logarithms for many numbers by hand, and therefore, so did Napierian logarithms. In other words, Napierian logarithms were the instrument used to obtain the values of decimal logarithms, which were the ones needed in practice. They were called natural logarithms, while decimal logarithms were known as common logarithms.

Now that we have calculators, why do we need logarithms? Well, it is mostly a question of general culture, but also because we will encounter them in algebraic simplifications (for example, to solve some types of equations) and functional expressions in the world of science and technology.

ãã Logarithms on a calculator As you already know, the calculators we use today have three keys for calculating logarithms: , j and . The first two are easy to find. The third (decimal logarithm), however, is more complicated. It is the second function of the button, so to use it, we have to press s .

With the calculator, find the following logarithms both ways: using the j key and by changing the base. a) log2 1024

b) log5 300

3

6 33 6 –3 2 b) Simplify this expression: 54 3 (3 3 + 6 ) 8 – + 4 2

c) Check that by reducing this expression you get 2 : 3 6 +2 2

2+3 3 Your turn a) Express as a power: 3x : 1 9 3 b) Simplify:

4

4 162 – 2 2 +1

b) What interest applied?

(2 · 3 · 3 3/2) 1/3 2 1/3 · 3 1/3 · 3 1/2 = = 25/6 · 35/6 = 65/6 2 –1/2 2 –3/6 54 3 + b) We perform the multiplication: 3 24 + 48 – 4 2 3 3· 2 6 We rationalise the denominator: = = 2 2 2· 2 We take out all possible factors from the roots: 33 · 2 6 3 6 6 3 23 · 3 + 24 · 3 – + =3·2 6 +4 3 – + = 2 2 4 4 = c6 – 3 + 1 m 6 + 4 3 = 23 6 + 4 3 4 2 4

has

been

a) Abs. e. < 0.0005 · 107 = 5 · 103 yen; Rel. e. < 5 · 103/2.387 · 107 = 0.0002

b) We know that CF = C b1 + r l 8 2.387 · 10 7 = 2 · 10 7 b1 + r l 8 100 100 n

5

5 5 7 r l 8 1.1935 = b1 + r l b 8 2.387 · 10 = + 1 100 100 2 · 10 7 We extract the fifth root from both sides of the equality to clear r :

1.1935 = 5 b1 + r l 8 1.036011 = 1 + r 8 100 100 r

8 0.036011 = 8 r = 3.6 % 100 Your turn How much would this capital turn into at the same annual interest rate over five years with a monthly compounding period?

5

5

3 Logarithms

Find the value of x in each case:

We apply the definition of logarithm in each case:

a) 3 = 5 + log x

a) 3 – 5 = log x 8 x = 10–2

b) logx 36 = 2

b) x 2 = 36 8 x = 6 (x = – 6 does not work) c) We apply the properties of logarithms:

c) log x + 2log 5 = 2

log x + log 52 = 2 8 log (x · 52) = 2 8 25x = 102 8 x = 4

4 Use the j and

keys of your calculator to calculate these logarithms in the two ways we have seen in the previous problem solved: a) log2 740 b) log3 100 c) log5 0.533 d) log8 0.004 e) log

a) We express the roots as powers with fractional exponents. Then, we operate applying the properties of powers:

2 Compound interest and errors

a) Give a limit of absolute error and of relative error for this approximation.

a) log2 1 024 8 j 2 ”1 024 = 10 ln 1024 log2 1 024 = 8 l 1 024 )/l 2 = 10 ln 2 b) log5 300 8 j 5 ”300 = 3.543959311 ln 300 log5 300 = 8 l 300 )/l 5 = 3.543959311 ln 5

1

c) We multiply the numerator and denominator by the conjugated binomial of the denominator: (3 6 + 2 2) (2 – 3 3) 6 6 – 9 18 + 4 2 – 6 6 =

= 4 – 9·3 (2 + 3 3) (2 – 3 3) –9 · 3 2 + 4 2 –23 2 =

= = 2 –23 –23

• log 2 5 8 jí 2 ””5 = 4.64385619

Think and practise

28

a) Express as a power:

A capital of 20 million yen was placed in a bank for 5 years with an annual compounding period, and it has become 2.387 · 107 yen, approximately.

Problem solved

Unit

1 Powers and radicals

To find the logarithm with a base other than 10 and e, newer calculators also include the j button. However, sometimes it may be preferable to use property 4, base change (introduced in page 26), rather than this button. To do this, it is better to use the ln key since that way you do not need to use the shift button.

Let’s look at an example of the two ways of doing it: • log 2 5 = ln 5 8 l 5)/lí 2 = 4.64385619 ln 2

SOLVED

Your turn Calculate x in each case: 3

350

a) log3 x = 1 2

b) 2log x – log 4 = –2 29

you ‘Portfolio’, urce bank on.es reso . ci lio ca fo rt du po ae ur In the anay to create yo ce in how find guidan

will

Unit

1

ROBLEMS

P EXERCISES AND

Practise

Intervals and half-lines

Powers, roots and radicals

Rational and irrational numbers

7

14

1

2

Look at the following numbers: ! !

–2; 1.7; 3; 4.2 ; – 3.75 ; 3π; –2 5; 5e a) Find the rational numbers and express them as a quotient of two integers. b) Which of them are irrational? Which of them can be represented on the real number line exactly and which cannot? a) Classify into rational and irrational numbers. !

3 ; 0.87; – 4; – 7 ; 1 ; 2π; e 3 2 3 2 b) Put them in order from smallest to largest. c) Which are real numbers?

3.5; 11 ; 1 ; 9 4 6; π ; –104 4 5

b) Between –1 and 3, both included.

15

d) Less than 10.

8

Represent each of the following intervals and half-lines on the real number line: A = [–2, 4] B = (1, 6) C = [–7, –3) D = (0, 5]

9

E = (– ∞, 1]

e) 4 < x < 4.1

f ) –3 ≤ x

10

Express each of the following sets of numbers as an interval or half-line and as an inequality: a) b)

–1 0 3

c)

–2

11

0

1 5

d)

(5, 7)

3

12

b) Represent 8 and 11.

6

Which numbers are represented by points A, B, C and D?

c) 0 ≤ x – 5 < 2 13

2 1

Write in interval form the numbers that verify the following inequalities: a) –3 < x + 1 < 3 b) –1 ≤ x – 4 ≤ 7

3 6

0

30

1

2 ABC 3 D

–1 0

e) (a 1/2)1/3

3 f ) (a –1)3/5

Express as powers and solve. a) 2 · 3 4 b) 3 · 4 9

c) 3 3 9

3 d) 5 : 4 5 e) 16 : 3 4

f ) 3 25 : 5

b) a 3 a 4 4 a

24

c) a –1 3 a –3 6 a 5

28

9 4

b) `2 5 – 3j

2

2

d) `3 2 + 5 8j

2

Rationalise and simplify where possible.

29

2 3 2+ 2 c) 3 2 10 e) 33

f ) 5 2 2 81 2 6 b)

Rationalise and simplify where possible. a)

a 1+ 2 3 b) c) a +1 1+ 3 1– 2

d)

4a 11 e) 4 2 2+3 2a 3 b 2

30

f )

5– 3 5+ 3

Simplify.

3

2 6 2 a) 6 + – 5 – 4 9 b) 1 + 6 8 – e o 2 4 2 3 2 3 c)

3 ( 3 – 1)2 8a + 3 a 4 24 – 6 d) 3 – 3 –1 2 a

Approximate numbers. Scientific notation

31

32

4 7

3

2 3 81 2a 3 2a 2 8a 3 4a 2 b) c) 4 18 4 4a 3 (4 2a 5)2

a) 3 3 d) 32 5

Reduce to a common index and solve.

3 4: 2 a) 5 6 · 3 b) ` 2 · 3 3j : `3 2 · 3j c) 6 20 : 4 10 d) Solve.

f ) 2 3

Simplify. a)

Take out all possible factors from the radical.

5 a) 3 28 – 7 – 63 b) 96 + 5 3 32 3 3 72 81 3 + 375 – 3 d) 7 + 7 + c) 2 64 4 3

a) ` 5 – 2 3j` 5 + 2 3j

c) ` 3 2 – 4 – 3 2 + 4 j

f ) 3 a 6 b 9

4 8a 5 a) 3 16a 3 b) 81a 5 b 3 c) 162 d) 3 244 e) f ) 5 9 32 75 a 22 Reduce to a common index and order from smallest to largest.

7, 3 30, 4 40, 6 81

23

Solve.

27

Multiply and simplify. a) 2 3 6

d) 5 < 2x – 1 ≤ 9

Express each of the sets of numbers below as a union of intervals, using the symbol á: a) b) –4 0 0 3

c) d)

d) (a 3)1/4

26

5 15 12 8 a a) 4 3 2 b) a c)

21

[5, 7]

1/3 c4m b) (–3)2/3 c)

3

b) Which of these intervals represents the numbers included in A and B ?

[2, 7)

a) 51/2

34 8 a d) 8 a 2 b 4 e)

18 3 b) c) 23 7 3 5 4

1 12 d) 2 4 5 e) 12 2

x2 1 a) 5 a 2 · a b) c) 4 x a3 18 Express as powers and calculate x in each case, equalling the exponents of the two sides: ( 3)–x 2x 2 a) 3 x + 1 = 1 b) = 1 c) =2 3 81 27 4 19 Simplify.

20

(–3, 5)

Introduce inside the root and simplify.

a) 5

Express the following radicals as powers with fractional exponents and simplify:

0 4

c) Express A á B and A Ü B as intervals and as inequalities.

B 2 C D

17

a) Indicate whether the following numbers are included in A = [–3, 7) or in B = (5, +∞): !

–3; 10; 0.5; 7; – 4; 5; 6.3 ; π; 27 ; 48; 1 – 2 5

A

16

F = (–1, +∞)

Graphically represent and express these inequalities as intervals or half-lines on the real number line: a) –3 ≤ x ≤ 2 b) 5 < x c) x ≥ –2 d) –2 ≤ x < 3/2

1

1

Write as roots.

c) Greater than or equal to 5.

a) Which irrational numbers are represented by points A, B, C and D ?

0

Express the following in exponential form: 3 4 a) 5 x 2 b) 2 c) 10 6 d) 20 2 3 15 5 `5 x –2j h) a e) 5 (–3) 3 f ) 4 a g)

a) Greater than 2 and less than 7.

3

Classify the following numbers, indicating which of the N, Z, Q or Á sets they belong to: !

1+ 3 – 4; 13 ; 5; 2.7 ; 152; π; ; e 2 6 4 Place the following numbers on a diagram like the one on the right: # 1; 7.23 ; 1 – 2;

Write the following sets of numbers in interval or half-line form:

25

Give a limit of absolute error and of relative error for these approximations of budgets for sports teams: a) 128 thousand Euros

b) 25 million Euros

c) 6 485 hundred Euros

d) 32 thousand Euros

e) €648 500

f ) €32 000

Give a limit of absolute error for the following approximations and compare their relative errors: a) 8 · 105 b) 5.23 · 106 c) 1.372 · 107 d) 2.5 · 10– 4

e) 1.70 · 10–6

f ) 4.00 · 10–5 31

Unit

EXERCISES AND PROBLEMS 33

Mentally calculate. a) (1.5 · 107) · (2 · 105)

41

b) (3 · 106) : (2 · 1011)

4 10 8 c) (4 · 10–7) : (2 · 10–12) d) · 34

Use scientific notation to perform these calculations and give a limit of absolute error: a) (3.5 · 107) · (4 · 108) b) (5 · 10–8) · (2.5 · 105) c) (1.2 · 107) : (5 · 10–6) e) 5.3 · g) 6 ·

1012

10–9

–3·

–5

1011

· 10–8

d) (6 · 10–7)2 f ) 3 ·

10–5

h) 7.2 ·

+ 8.2 ·

108

10–6

+ 1.5 ·

1010

Logarithms

35

Apply the definition of logarithm and calculate. a) log2 64 b) log2 16 c) log2 1 4 d) log2 2 e) log3 243 f ) log3 1 27 3 g) log3 9 h) log 0.001 i) log5 0.2

36

Calculate the base of the following logarithms: a) logb 10 000 = 2 b) logb 125 = 3

c) logb 1 = –1 d) logb 2 2 = 1 2 4 37 Apply the definition of logarithm and calculate. log4 163 + log4 2 + log 0.0001 + log 38

3

10 100

Use a calculator to find: a) log2 23.4 b) log3 543

c) log5 0.06

d) log6 20.8

f ) log2 0.872

e) log5 123

Problem solving 39

Calculate the perimeter of triangles ABC, DEF and GHI. Use radicals to express the results.

4u

A

D

C

G

The open interval whose midpoint is M and whose endpoints are M – r and M + r is called a neighbourhood with midpoint M and radius r, and it is represented by N(M, r).

For example: N(0, 2) = (–2, 2) and N(2, 3) = (–1, 5) Express the following neighbourhoods as intervals: a) N(0, 1) b) N(0, 3) c) N(3, 5) d) N(–2; 1.5) e) N(–3; 0.3) f ) N(2.1; 3) g) N(–0.2; 5.3)

43

Express each of the following intervals as neighbourhoods of the type N(M, r): a) (–3, 3) b) (2, 4) c) (0, 6) d) (–1, 4) e) (–3, 2) f ) (0; 7.5) g) (–5; –2.2) h) (1.2; 4.7)

44

Is any of the numbers 1 – 3 or 3 + 2 a solution to the equation x 2 – 6x + 7 = 0? Check it without solving the equation.

F

E

Write in interval form the numbers that verify each of the following inequalities: a) | x | < 3 b) | x – 1| ≤ 5 c) | x + 3| < 4 How would you express the numbers that verify the inequalities opposite to the previous ones? | x | ≥ 3

| x – 1| > 5

| x + 3| ≥ 4

Calculate the height of a regular tetrahedron, like the one on the right, with 8 cm edges. Use radicals to express the results.

51

52

53

49

Reference ranges (3.5-11) (4.3-5.9) (1.50-4.50) (0.7-1.3)

Units Ò 103 µL Ò 106 µL Ò 105 µL Ò 105 mg/dL

A limestone rock weighs 830 g. The mass of its molecules is, approximately, 1.66 · 10–22 g. Due to erosion, the stone loses 1013 molecules per second.

If the erosion remains constant, how long will it take for the stone to disappear completely? Give a limit of absolute error.

h/3

6

2h/3

g) Irrational numbers have infinite decimal figures. h) The sum of two irrational numbers is always an irrational number.

h

56

Make up two intervals, B and C, so that: a) B á C = (–2, 7] b) B Ü C = (–1, 0]

57

In an equilateral triangle with 10 cm sides, we cut equilateral triangles with sides x from each corner to obtain an hexagon. Calculate the value of x so that the area of this hexagon is 10 3 cm2.

C This is the logo of a sports B club. It will be reproduced in different sizes. a) Find the radius of each arc in a square with 2 metre sides.

A D b) Check that the relationship between the radii of the arcs is 2 – 1. c) Find the perimeter and the area of the blue part of a square with 2 metre sides.

The diameter of the Milky Way is 105 700 light years, and a light year is 1.461 · 1012 km. a) What is the diameter of our galaxy in kilometres? b) How many millennia would it take for a spacecraft to cross it if travelling at 2 000 km/s? c) Given that the diameter of an electron is 4 · 10–15 m, how many electrons would it take to form a line all the way around the Milky Way? (Assume it is a circumference).

State which of these blood test results are outside the reference ranges: Results 3.16 5.87 1.9 0.68

f ) Some integers are irrational.

h

54

48

White blood cells Red blood cells Platelets Creatinine

Calculate the volume of a regular octahedron with edges measuring 6 cm. Use radicals to express the results.

e) All irrational numbers are real numbers.

6 2 2

b) log M = log (x + 1) – log y + log 3

H

40

50

45

If log x = 1.3 and log y = 0.8, calculate: y a) log (x · y) b) log (x y ) c) log 2 d) log x y x 46 Transform these expressions into other equivalent expressions using logarithms, like in the example: • A = 7x 2 y 8 ln A = ln (7x 2 y ) = ln 7 + 2ln x + 1 ln y 2 3y z a) M = 10xy 3 b) N = 2 c) P = x 2 yz x 47 Express M without logarithms in each case: a) log M = log (x – 3) + 2log x

d) The number 0.83 · 109 is not expressed in scientific notation.

Advanced problem solving

42

I B

32

Find out for which values of x can the following roots be calculated: x2 + 1 a) x – 7 b) 5 – x c) –x d)

True or false? Explain your answers and give examples. a) All decimal numbers are rational. b) There are an infinite number of irrational numbers between two rational numbers. c) The inverse of a recurring decimal number can be a terminating decimal.

Look at this way of representing m on the real number line. Explain it.

m m

0

1

m

Indicate which number is represented by A in this diagram:

–8

58

0

2

A

If x is a number of the interval [–1, 3) and y is a number of the interval (0, 4], explain in which interval you might find x + y.

59

Are these equalities true or false? Why? 3 a) a · 3 b = 6 a · b b) a + b = 3 a + 3 b 4 12 c) a 3 b 2 = 3 (a · b) 2 d) a · b2 = a3 b

60

Are these equalities true or false? Explain your answers. a) log (a · b) = log a · log b

Remember the theory 55

1

b) log b a l = log a – log b b c) log 3 a = 1 log a 3 2 d) log (a · b ) = 2(log a + log b ) 61

Check that it is not possible to use a calculator to obtain 5129 · 463 because the number is too large.

Use the properties of powers to express it in scientific notation.

33

Unit

OP

MATHS WORKSH READ AND LEARN

PRACTICE MAKES PERFECT!

Golden rectangles A rectangle is called a golden rectangle when the ratio between its sides coincides with the golden number. In other words, if we take the shorter side as one unit, the longer side will measure the golden 5 +1 number, F = = 1.618…

2

• Although it may seem strange, the number K that you see here is an integer: F = F = 5 +1 1 2

1 Φ

These rectangles have a curious property: if you put a square on the long side, you get another golden rectangle. Try it:

U U+1 U

F +1 =1+ 1 =1+ 2 =…

F F 5 +1

1

• If you continue adding ever-bigger squares, you will get a series of golden rectangles that you can use to construct a pretty spiral made up of arcs.

2900

899

16 75

150

25

32120

Finally, compare 7300 to 2900 and 25150.

SELF-ASSESSMENT

anayaeducacion.es Answer key and interactive self-assessment.

2900, 899, 1675, 32120 Then, compare the power 25150 to 2900 and 32120.

2.03333…;

81;

3

5 4; ; – 13 ; – 8 9 3

b) Which are irrational?

This spiral is very well known and widely studied in mathematics (the equiangular or logarithmic spiral). But the most surprising thing about it is that it appears naturally in many plant and animal species (flowers, fruit, mollusc shells, etc.).

• Now construct the series of successive Leonardo of Pisa (1170-1250) radii of the spiral, which are the same Also called Fibonacci (which means ‘son of a good man’). length as the sides of the squares that fit together to make it: His father was a merchant and consul of Pisa in the city R1 = F of Béjaïa, in present-day Algeria.

R2 = F + 1 This allowed him to learn Arabic mathematics, especially the decimal numeral system, which he helped to introR3 = 2F + 1 duce to Europe.

R4 = 3F + 2 He was the first to describe the famous sequence: R5 = 5F + 3 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, … … in which each term is obtained by adding the previous two. This Fibonacci Find some relationship between this sequence is closely tied to the golden number, F. sequence and the Fibonacci's?

diagonal of the cube: d = 1 2 + ( 2) 2 = 3

numbers, integers, rational numbers and real numbers: # 6 –4 3 ; 2π; log 2 0.5 ; 3.47 ;

Nautilus shell.

1

d

k = 12 + 12 = 2

Work out whether the distances m and n indicated in this second cube – whose side measures 1 – are rational or irrational.

U

U+1

34

7 300

diagonal of one face:

k

Start by ordering the powers of 2:

2U+1

U

• In a cube with edges measuring 1, the diagonal of one face and the diagonal of the cube are irrational numbers:

63 63 K= 4+ 2 + 4– 2 Can you work out which number it is?

• Order from smallest to largest:

1 a) C lassify the following numbers into natural

3U+2

1

1

c) Put them in order from smallest to largest.

2 a) W rite the following number sets in interval form

and represent them on a number line:

i) {x / –2 ≤ x < 7} ii) {x / x > –1} iii) | x – 3| < 1 b) Write the following intervals as inequalities:

A = [–3, 4)

B = (–∞, 3)

c) Express A á B and A Ü B as intervals and as inequalities.

3 Express in scientific notation and use a calculator to

help you perform the operation. Write the result to three significant figures.

1500 000 · 25 · 10 17 0.00007 · (2000) 4 Then, give a limit of absolute error and of relative error for the approximate value obtained.

Commitment

m n

1

4 Take out all possible factors from the radical.

3

81a 2b 5 16z 4

5 Operate and simplify.

(3 2 + 3 ) 2 b) 54 – 2 6 + 150 3 2 10 c) 5 – d) 2 2 3– 2 50 a)

6 Apply the definition of logarithm, or use a calculator,

to operate: a) log3

3

1 b) log2 c4 1 2m 32 9

7 Express log

4 6 according to log 2 and log 3.

9

8 We cut a right-angled isosceles

triangle from each corner of a square with 10 cm sides to obtain a regular octagon.

x x

s s

a) Find the exact measurement of the side of the octagon. b) Calculate its area. c) Calculate its radius.

Watch the video for target 4.6. Think of something you can do to contribute to achieve that goal. Make a commitment to put your idea into practice.

35

2

POLYNOMIALS AND ALGEBRAIC FRACTIONS Reading and listening

Main phases in the history of algebraic language Today’s algebraic language is simple and easy to use. However, it took a long time to reach this point. We can divide its history into three main phases.

• rhetorical algebra. It was written in everyday language. It was used in Babylon, Egypt, and Greece. In the 9th century, it was used in Arabic mathematics.

• syncopated algebra. Diophantus (3rd c.) pioneered this type of algebra. He made the processes easier by using abbreviations. During the Renaissance (15th and 16th centuries), it was improved through the incorporation of new symbols: operations, coefficients, powers, etc.

• symbolic algebra. This is the predecessor of the algebraic language we know today and consists of a full set of symbols. It was created by Viète in the 16th century. It was then perfected by Descartes in the 17th century.

François Viète (1540-1603). French mathematician who published In Artem Analyticam Isagoge (1591). In the book, he introduced the use of letters in mathematical formulas as common practice.

Let’s look at an example of how Diophantus described polynomials:

Using geometric algebra With our modern form of algebraic notation, it is very easy to prove that: 2

But rhetorical algebra had no way of performing this operation, so mathematicians had to find another way to prove the expression.

Look how Pythagoras did it: If a and b are segments, ab is the area of a rectangle. a b

ss7 c2 x5 M s4 u6

s means ‘squared’ (therefore, ss means ‘fourth power’); c, ‘cubed’; x, ‘the unknown’; u, ‘number’. M means ‘minus’ and precedes all monomials with negative coefficients. Looks odd, doesn’t it?

36

b

a·b

a–b

a–b 2 — 2

( )

a–b — 2

a–b — 2 a+b

a+b — 2 a·b

a–b — 2

b

b

a–b — 2

a+b — 2

— _ 2 Blue area: a · b b 2 b Red area: c a – b m b Therefore, 2 ` 2b 2 2 Total area: c a + b m bb c a + b m – c a – b m = ab 2 2 2 a

1 a) How did Diophantus make algebra easier to use?

b) Write some examples of when we use abbreviations and how they help us in daily life.

2 Form sentences using the comparative form (-er, more than, less than)

to compare our modern algebra to Diophantus’ way.

3 Write the definations of rhetorical, syncopated and symbolic. Then,

describe the difference between syncopated and symbolic algebra.

modern notation 8 7x 4 + 2x 3 – 4x 2 + 5x – 6 diophantus’ notation 8

2

c a + b m – c a – b m = ab 2 2

Solve 1 Use modern algebraic notation to express the following polynomial,

given using Diophantus’ notation: ss3 s5 M c8 x9 u1

Geometric algebra

2 Express the following polynomial using Diophantus’ notation:

The lack of a functional form of algebra meant that mathematicians had to come up with ingenious ways to find or demonstrate algebraic relationships. In many cultures they used geometric shapes, giving rise to geometric algebra.

3

–2x 4 + 5x 3 – 3x 2 – 6x + 8.

ANK ANK B E G A U LANG LANGUAGE B ANK ANK GE BANK B B E E G G A A U U GUA K LANG ANG N L A L AN GE BANK 37 ANK GE BANK B B E E G G A A U U G A LAN LANG LANGUA LANGU

Use Pythagoras’ reasoning to demonstrate the expression above, taking a = 7 and b = 2.

Unit

1 The order of monomials The monomials that make up a polynomial can appear in any order.

However, they are most commonly ordered by degree, from largest to smallest.

ãã Division of polynomials

POLYNOMIALS. OPERATIONS ãã Basic terminology

The result of adding, subtracting or multiplying two polynomials is another polynomial. However, this is not true for division: in general, the quotient of two polynomials is not a polynomial.

As you already know, the following expression is a polynomial:

Let’s take a look:

2x 5 + 3 x 3 – 7

Remember how to divide two polynomials, and how we interpret the result. Let’s divide P (x) = 5x 5 – 6x 3 + 4x 2 – 2x + 1 by Q (x) = x 2 + 2x – 1.

3x 2 + 2.7

It is made up of four monomials: 2x 5; 3/7x 3; – 3x 2 and 2.7, the degrees of which are 5, 3, 2 and 0, respectively (2.7 has a degree of 0, since 2.7 = 2.7 · x0).

The degree of the polynomial is 5 (that of the monomial with the highest degree).

The variable, x, is also called the indeterminate.

The numbers 2, 3/7, – 3 and 2.7 are the coefficients. As there is no thirddegree monomial, we can say that its coefficient is 0.

Remember Degree of the dividend, m

m≥n Degree of the divisor, n Degree of the quotient, m – n Degree of the remainder < n

In a polynomial, the coefficients are any real numbers.

Subtracting polynomials Subtracting is a special type of addition; therefore, the result of subtracting two polynomials is another polynomial.

P – Q = P + (–Q ) The polynomial –Q is obtained by changing the sign for all the Q monomials.

ãã Operations with polynomials. Sum and product

Integer division in Z

The result of adding or multiplying two polynomials is another polynomial.

These operations have certain properties that allow us to perform operations easily and smoothly*. Let’s take a look:

89 5 39 17 4

properties

of the sum

of the product

Associative

(P + Q ) + R = P + (Q + R )

(P · Q ) · R = P · (Q · R )

Commutative

P+Q=Q+P

P·Q=Q·P

Distributive

smoothly: without difficulties.

problems

or

P · (Q + R ) = P · Q + P · R

Curiously, the behaviour of polynomials with regard to addition and multiplication is very similar to that of integers. Let’s look at the above properties in relation to integers: • The associative property of the sum, 2 + [7 + (– 4)] = (2 + 7) + (– 4), allows us to work without the brackets in these cases and write 2 + 7 + (– 4).

The same happens with the product: [3 · (–5)] · 8 = 3 · [(–5) · 8] = 3 · (–5) · 8.

• The commutative property allows us to simply change the order of the factors or the summands.

• The distributive property allows us to extract a common factor: 120 + 220 = 6 · 20 + 11 · 20 = (6 + 11) · 20 = 17 · 20

anayaeducacion.es Practise adding, subtracting and multiplying polynomials.

38

89 = 5 · 17 + 4, or 89 = 17 + 4 5 5

The result of the division is not an integer.

These are all advantages of polynomials.

There are actually many more similarities between integers and polynomials, so we will continue to highlight them throughout the unit.

5x 5 – 6x 3 + 4x 2 – 2x + 1 5 4 –5x – 10x + 5x 3 – 10x 4 – x 3 + 4x 2 – 2x + 1 10x 4 + 20x 3 – 10x 2 19x 3 – 6x 2 – 2x + 1 – 19x 3 – 38x 2 + 19x – 44x 2 + 17x + 1 44x 2 + 88x – 44 105x – 43

x 2 + 2x – 1 5x 3 – 10x 2 + 19x – 44 Quotient: C (x) = 5x 3 – 10x 2 + 19x – 44 Remainder: R (x) = 105x – 43

The result can be written in one of the following two ways: P (x) = Q (x) · C (x) + R (x),

or

P (x) R (x) = C (x) + Q (x) Q (x)

This way of dividing polynomials, where there is a remainder as well as a quotient different from zero, is called integer division.

When the remainder is zero, we say that it is an exact division.

For example, let’s divide the polynomial P (x) = x 4 + x 3 – 10x 2 – 4x + 24 by Q (x) = x 2 + x – 6.

of the product with regard to the sum

ãã Comparing polynomials to integers, Z

F ocu s on Eng lish

2

anayaeducacion.es Practise dividing polynomials.

Breaking down into factors P (x) = x 4 + x 3 – 10x 2 – 4x + 24 has been broken down into a product of two factors: P (x) = (x 2 – 4)(x 2 + x – 6).

And each of these factors can also be broken down: x 2 – 4 = (x – 2)(x + 2) x 2 + x – 6 = (x – 2)(x + 3) Therefore: P (x) = (x – 2)(x + 2)(x – 2)(x + 3) = = (x – 2)2 (x + 2)(x + 3) We need to try and break down all polynomials using this method.

x 4 + x 3 – 10x 2 – 4x + 24 –x 4 – x 3 + 6x 2 – 4x 2 – 4x + 24 4x 2 + 4x – 24 0

x 2 + x – 6 x 2 – 4 The remainder is 0.

It is an exact division.

As it is an exact division, the dividend can be given as the product of two factors: x 4 + x 3 – 10x 2 – 4x + 24 = (x 2 – 4)(x 2 + x – 6) Polynomials, like integers, can be broken down into products of factors. When we try doing this, we face the following problem: Given a polynomial P (x), how can we work out which other polynomial to divide it by for the division to be exact?

When we faced this same question when working with integers, it was answered with the divisibility criteria. However, for polynomials, it is something remotely similar.

In our search, we will need to perform many divisions, most often using expressions of the type x – a. The next section will prepare you to do this.

39

Unit

2

ãã Value of a polynomial for x = a

RUFFINI’S RULE

The numeric value of a polynomial, P(x), for x = a, is the number we get when we replace x with a and perform the operations. This number is called P(a).

ãã Division of a polynomial by x – a

For example, if P(x) = 5x 3 – 3x 2 + 12x – 8, for x = 2 we get:

Let’s use some examples to review how to divide a polynomial by x – a using Ruffini’s rule.

P(2) = 5 · 23 – 3 · 22 + 12 · 2 – 8 = 44 This value, P(2) = 44, coincides with the remainder when we divide P(x) by x – 2. This is not just by chance, it is a general rule.

• We divide P (x) = 3x 4 – 2x 3 – 10x + 7 by x – 2: ‘Ruffini’s’ rule?

Paolo Ruffini was an Italian mathematician who lived in the 18th and 19th centuries. This rule bears his name because he used it to prove an important mathematical property.

However, the same rule had already appeared in a book about algebra written by Pietro Paoli 25 years before.

2

3 –2 6 3 4

7 12 19

It is an integer division.

Quotient: 3x 3 + 4x 2 + 8x + 6

Or:

–

– 10x + 7 = 3x 3 + 4x 2 + 8x + 6 + 19 x–2 x–2

• Another example. We divide P(x) = 2x3 – 5x2 + 4x – 6 by x + 3: – 3

–5 – 6 2 –11

4 33 37

–6 –111 –117

It is an integer division.

Quotient:

2x 2

– 11x + 37

Remainder: –117

Therefore, 2x3 – 5x2 + 4x – 6 = (x + 3)(2x2 – 11x + 37) – 117.

Or:

2x 3

–

5x 2

+ 4x – 6 = 2x 2 – 11x + 37 – 117 x +3 x +3

• We can use Ruffini’s rule to divide by 2x + 8. Let’s see how: Dividing P(x) by (mx + n) n mx + n = m bx + l m Therefore: Do the division P(x) : (x + n/m) (Ruffini’s rule, where a = –n/m).

Quotient: C(x); Remainder: R. So the result of the division P(x) : (mx + n) is: 1 Quotient: C(x); : R m

Since 2x + 8 = 2(x + 4), we divide by x + 4 and ‘work it out’.

For example, we divide x 3 + 2x + 2 by 2x + 8:

1 – 4

0 – 4 1 – 4

2 16 18

2 –72 –70

Quotient: x 2 – 4x + 18 Remainder: –70

P (x) = (x – a) C (x) + r If we make x = a: P (a) = (a – a) C (a) + r = r anayaeducacion.es Applications of Ruffini’s rule with spreadsheets.

Observe The operation performed above is a demonstration. The calculation on the left is a simple check. Nevertheless, you might find this easier to understand than the demonstration.

Take a good look.

divide + a) x – 1 d) x – 4

–

3x 2

+ 3x – 4 by: b) x + 1

c) x – 2

e) x + 4

f ) x – 3

For each of them, say whether the division is an integer or exact division.

40

Therefore, the theorem proves that P(a) = r.

According to this result, we can use Ruffini’s rule to calculate P(a). did you know? If we apply Ruffini’s rule leaving the operations indicated, the remainder takes the form of P(a).

We take P(x) = 5x 3 – 3x 2 + 12x – 8 and a = 2.

5

for x = 3 and for x = –1.27.

5

e) 2x + 3

f ) x – 2

5·2

(5 · 2 – 3) · 2

5·2–3

5 · 22 – 3 · 2 + 12

–8 (5 ·

22

– 3 · 2 + 12) · 2

5 · 23 – 3 · 22 + 12 · 2 – 8

The value of the polynomial for x = 3 7 –42 0 190 0 –13 coincides with the remainder of dividing 3 21 –63 –189 3 9 by x – 3. Therefore, P (3) = –4.

–4 7 –21 –63 1 3

For x = –1.27, it is a good idea to use a calculator:

* 7 =- 42 =* + 0 =* -13 =161.0633922

+190 =*

+0 =

The result is 161.0633922…

2 Divide P (x) = 4x 3 + 12x 2 + 5x – 6 by each of the

d) 2x + 4

12

The remainder is the result of replacing the x in the polynomial with 2.

–1.27

following polynomials and express the result as follows: dividend = quotient + remainder . divisor divisor a) x – 1 b) 2x – 1 c) x + 2

–3

2

*

1 Calculate the quotient and the remainder when you

3x 3

— If in P(x) we replace x with a and perform the operations, we get a number which we call P(a) (value of the polynomial for x = a).

Problem solved

Think and practise

x 4

In other words:

— If we divide P(x) by x – a, the remainder is a number r.

7x 5 – 42x 4 + 190x2 – 13

2 = x – 2x + 9 – 70 2 2x + 8

The value given to a polynomial, P(x), when we perform x = a, coincides with the remainder of the division P(x) : (x – a). That is, P(a) = r.

P (x) x – a r C (x)

Calculate the value of this polynomial:

x 3 + 2x + 2 = 1 · x 3 + 2x + 2 = 1 · cx 2 – 4x + 18 – 70 m = 2x + 8 2 2 x+4 x+4

Polynomial remainder theorem

Demonstration

Remainder: 19

2x 3

2 anayaeducacion.es Activities for consolidating the application of Ruffini’s rule.

0 –10 8 16 8 6

Therefore, 3x 4 – 2x 3 – 10x + 7 = (x – 2)(3x 3 + 4x 2 + 8x + 6) + 19.

3x 4

2

Think and practise

anayaeducacion.es Practise dividing using Ruffini’s rule.

3 Use Ruffini’s rule to find P (a) in the following cases:

a) P (x) = 7x 4 – 5x 2 + 2x – 24, a = 2, a = –5, a = 10

b) P (x) = 3x 3 – 8x 2 + 3x, a = –3, a = 1, a = 8

41

Unit

3

ROOT OF A POLYNOMIAL. FINDING ROOTS

1 2

A number, a, is called a root of a P(x) polynomial if P(a) = 0.

The roots of a polynomial are the solutions of the equation P(x) = 0.

Therefore, if a is the root of P(x), then P(x) = (x – a) Q(x).

For example, 5 is the root of x 3 – 8x 2 + 17x – 10 because: 1 5

–8

5 – 15 1

–3

2

10 0

The division (3x 3 – 7x 2 – 11x + 15) : (x – 3) is exact. For this to be so, in the division process using Ruffini’s rule, the last product 3 · (–5) has to be the same as the independent term, 15, but with the 3 – 7 –11 15 sign changed. In other words, the indepen3 9 6 3 · (–5) dent term must be a multiple of 3.

0 3 2 –5

Key point In order for the division to be exact, the independent term, 15, has to be a multiple of 3 8 15 = 3 · 5

Remember This criterion is very useful for limiting the search for the divisors of a polynomial. It is only valid for polynomials with integer coefficients.

And it can only be used to locate the integer values of a.

+ 17x

6 3

7 4

8 5

If a polynomial has integer coefficients, in order for it to be divisible by x – a, its independent term needs to be a multiple of a (integer a).

9 6

Therefore, to find x – a expressions that are divisors of a polynomial, we try with the integer values of a (positive and negative) that are divisors of the independent term.

• P1(x) is not divisible by x – 5 because its independent term, 18, is not a

multiple of 5.

• P2(x) could be divisible by x – 5 because its independent term, –10, is a multiple

of 5. Let’s see if it is:

5

2 –10 10 2 0

–2 12 0 –10 –2 2

We do not get a remainder of 0 with the numbers 1, –1, 2 and –2.

P (x) = x 3 + x 2 – 7x + 20

1 4 1 1 – 4 1

1 4 5 1 – 4 – 3

–7 20 13 –7 12 5

20 52 72 20 –20 0

We now try with a = 4. The remainder is not 0. Therefore, P (x) is not divisible by x – 4 either.

However, with a = –4 the division is exact, since the remainder is 0.

P (x) divided by (x + 4) is

x 2

– 3x + 5.

careful! It is possible that P (x) was not divisible by x – a for any a divisor of the remainder. In other words, for no a integer.

P2(x) is divisible by x – 5: P2(x) = (x – 5)(2x 3 – 2x + 2)

2x 3 + 40

• P3(x) could be divisible by x – 5 because its independent term, 40, is a multiple

If so, check that it really is.

2 –17 23 40 The division is not exact. Therefore, P3(x) is

5 10 –35 – 60 not a multiple of x – 5.

2 –7 –12 –20

P3(x) 17x 2 +

= 23x

of 5. Let’s see if it really is:

I nvent a 3rd-degree polynomial that has roots of 0, 1 and 2.

This is true for P (x) = x (x – 1)(x – 2).

I nvent a 2nd-degree polynomial that has no roots.

Invent a 4th-degree

polynomial that has no roots.

Any second-degree expression without real roots.

I nvent a 3rd-degree polynomial that has no roots.

Impossible! An odd-degree polynomial, P(x), is bound to have at least one root.

Generally, this is true of any polynomial of this type: Kx (x – 1)(x – 2), with K ≠ 0 For example: P (x) = x 2 + 5x + 10 We find two second-degree polynomials without roots and multiply them.

For example: P (x) = (x 2 + 1)2 = x 4 + 2x 2 + 1

It is not easy to show, so we will have to know it is true without having proved it.

Think and practise 1 Without performing the operations, say whether x = –3

In order for P (x) to be divisible by x – a, a needs to be a divisor of 20. We try by giving a the values ±1, ±2, ±4, ±5, ±10.

–10 10 0

Therefore, x = 5 is only the root of P2(x).

By using this reasoning, we obtain the following rule, which is very useful for locating the integer roots of a polynomial.

Problems solved

42

P2(x) = 2x 4 – 10x 3 – 2x 2 + + 12x – 10

5

ãã Criterion for finding the roots of a polynomial

x – a divisor (where a is an integer) of the polynomial:

3

–

Ruffini’s rule helps us break down a polynomial into factors. However, to do this we need to learn how to find its roots. The following criterion helps us with this search.

x 3 – 8x 2 + 17x – 10 = (x – 5)(x 2 – 3x + 2)

1 Find an

11x 3

P1(x) =

+ 18

–

We can check it using Ruffini’s rule (see margin).

x 5

2

4

53 – 8 · 52 + 17 · 5 – 10 = 0

17 –10

ay whether x = 5 S can be a root of these polynomials:

2

can be the root of each of these polynomials: a) P (x) = x 2 – x – 12 b) P (x) = x 4 + 2x 2 – x + 8 c) P (x) = x 3 + 3x 2 – 5x – 27 d) P (x) = x 3 + 3x 2 + x + 3 If you think it can be, prove whether it is the root or not.

2 Indicate the possible integer roots of each of the

polynomials in the previous exercise. Work out which of them are actual roots.

3 The polynomial x 4 + 3x 3 – 2x 2 – 10x – 12 is divisible

by x – a for two integer values of a. Find them and give the quotient in both cases.

anayaeducacion.es Practise finding the roots of a polynomial.

4 Check that the polynomial x 4 + x 3 + 7x 2 + 2x + 10 is

not divisible by x – a for any integer value of a.

5 Invent a third-degree polynomial that has roots of 3,

–2 and –1.

6 Invent a fourth-degree polynomial that has no roots.

It must be different from the one in problem solved 5.

7 Invent a fourth-degree polynomial that has only two

roots: x = 2 and x = –3.

8 Invent a second-degree polynomial that has x = –3 as

its double root.

9

Is there any first-degree polynomial that has no roots? Explain your answer and give an example.

43

Unit

4

FACTORISING POLYNOMIALS

Problems solved 1 Factorise and identify the

To factorise a polynomial means to break it down into the product of polynomials (factors) of the lowest possible degree.

Observe Ruffini’s rule allows us to find integer roots. However, it is very unlikely that we will be able to factorise a polynomial with a degree greater than two that does not have integer roots. For example, the roots of the polynomial 12x 3 + 4x 2 – 3x – 1 are 1/2, –1/2 and –1/3, but finding them is not easy.

— First-degree expressions of the type x, x – a, x + a.

(*) – 1

–2

— Second-degree expressions without real roots.

Q (x) = 4x 2 – 8x + 3

–9

–9

37 – 35

9 –37

35 – 32

9

Now, we find the roots of P1(x) =

–1 is not a root.

9

18 –28

(**) 1 9

–2

3

9

27

–1

–3

27

–1

–3

0

(***) – 3 9

R (x) = x 3 – x + 6

+

– x – 3:

We have discarded the value –1. We try again with 1 but it is not (in other words, 1 is a simple root). Next, we try with –3 (***), and we see that this time it is a root of P1(x) and, therefore, of P (x).

–3

–27

0

3

Example 2

0

–1

0

We factorise Q (x) = x 5 + 5x 4 – 14x 3.

–3 is a root.

1

–2 1

4 Factorise.

S (x) = 10x 4 – 3x 3 – 41x 2 + + 12x + 4

Now, we solve the equation x 2 + 5x – 14 = 0, obtaining x1 = 2 and x2 = – 7.

Therefore, Q (x) = x 3(x – 2)(x + 7).

10

–3 – 41 12 2 20 34 –14 10 17 –7 –2 –2 –20 6 2 0 10 –3 –1

For example: anayaeducacion.es Factorising polynomials using Ruffini’s rule.

44

+

–

2x 2

– 10x – 12 = (x – 2)(x +

3)(x 2

+ 2x + 2)

• If a polynomial has more than two non-integer roots, although it can be

factorised, we cannot do it easily.

For example: 18x 3 + 9x 2 – 2x – 1 = 18(x – 1/3)(x + 1/3)(x + 1/2)

We find roots of x 2 – 2x + 3: x 2 – 2x + 3 = 0 has no solution.

4 – 4 0

6 2 is a root of S (x).

6 –2 is a root of S (x).

S (x) = 10(x – 2)(x + 2)cx – 1 mcx + 1 m 2 5

• If we reach a second-degree polynomial without roots, that polynomial remains

3x 3

–2 is a root of R (x).

6 – 6 0

We find the integer roots among the divisors of 4:

important

x 4

–1 4 3

As we cannot find any more integer roots, we try to solve the equation: 10x 2 – 3x – 1 = 0 8 x = 1 , x = – 1 2 5 So: 10x 2 – 3x – 1 = 10 cx – 1 mcx + 1 m = (2x – 1)(5x + 1) 2 5 Therefore: S(x) = (x – 2)(x + 2)(2x – 1)(5x + 1), or rather:

We start by taking out x 3 as a common factor: Q (x) = x 3 (x 2 + 5x – 14)

as a single factor (it cannot be broken down into two factors).

0 –2 –2

So: R(x) = (x + 2)(x 2 – 2x + 3)

P (x) = (x – 1)(x + 3)(3x + 1)(3x – 1) = 9(x – 1)(x + 3)(x + 1/3)(x – 1/3)

–1

Use Ruffini’s rule to find a root among the divisors of 6:

We have reached a second-degree polynomial that has no roots.

P1(x) = (x + 3)(9x 2 – 1)

27

4x 2 – 8x + 3 = 0 8 x = 1 ; x = 3 2 2

We see that 9x 2 – 1 = (3x + 1)(3x – 1). Therefore, the final result is:

1 is a root.

9

3 Factorise.

Given that 1 is a root, we can write P (x) = (x – 1)(9x 3 + 27x 2 – x – 3).

27x 2

We look for the roots equalising to 0 and solving:

Q(x) = 2 cx – 1 m 2 cx – 3 m = (2x – 1)(2x – 3) 2 2

In order to locate the roots of P (x), we try with the divisors of 3 (both positive and negative). Let’s begin with –1 (*) and 1 (**) (see margin).

9x 3

All summands have the factor x3. The coefficients 12, –36 and 27 are multiples of 3.

Therefore, we can extract 3x3 as the common factor.

P(x) = 3x 3 (4x 2 – 12x + 9) We see that 4x2 – 12x + 9 is equal to (2x – 3)2.

P(x) = 3x 3 (2x – 3)2 We obtain the roots by equalising each factor to 0.

The roots of P(x) are 0 (triple root) and 3/2 (double root).

Therefore: Q(x) = 4 cx – 1 mcx – 3 m, or rather: 2 2

ãã Procedure for factorising a polynomial

We factorise P (x) = 9x 4 + 18x 3 – 28x 2 – 2x + 3:

3

P (x) = 12x 5 – 36x 4 + 27x 3

2 Factorise.

Example 1

18 –28

roots.

Factorising polynomials is, therefore, a similar process to breaking down an integer into prime factors. The ‘polynomials of the lowest possible degree’ that we will use this year and that play the same role as prime numbers are:

If possible, we start by taking out a common factor. In some cases, we will be able to identify notable products. We use Ruffini’s rule to look for integer roots of the polynomial. We can also find roots of second-degree polynomials by solving the equations. Let’s look at some examples:

9

2

Think and practise

anayaeducacion.es Factorising polynomials.

1 Factorise the following polynomials:

a) 3x 2 + 2x – 8

b) 3x 5 – 48x

c) 2x 3 + x 2 – 5x – 10

d) x 3 – 7x 2 + 8x + 16

e) x 4 + 2x 3 – 23x 2 – 60x

f) 9x 4 – 36x 3 + 26x 2 + 4x – 3 45

Unit

5

ãã Greatest common divisor and lowest common

DIVISIBILITY OF POLYNOMIALS

multiple

In previous sections of this unit we looked at aspects related to the divisibility of polynomials: exact division, breaking them down into factors, etc. Now we are going to systematise this information and complete it by creating a parallel with the divisibility of integers.

We say that D(x) is the greatest common divisor of two polynomials, P(x) and Q(x):

GCD (18, 30) = 6, because 6 is a divisor of 18 and 30, and there is no other number higher than 6 that is a divisor of both.

GCD [P (x), Q (x)] = D (x) if it is the divisor of both and there is no other common divisor polynomial that has a greater degree.

ãã Multiples and divisors A polynomial, D(x), is a divisor of another, P(x), if the division P(x) : D(x) is exact. In this case, P(x) is a multiple of D(x), since P(x) = D(x) · C(x).

For example, x 2 + x is a divisor of x 3 – x because (x 3 – x) : (x 2 + x) = x – 1 is exact. Therefore, x 3 – x is a multiple of x 2 + x, since x 3 – x = (x 2 + x) · (x – 1).

Even though 6x + 3 = 3(2x + 1), 6x + 3 is not really considered to be a multiple of 2x + 1, because the factor 3 is a number and not a polynomial with a degree greater than or equal to 1.

ãã Irreducible polynomials

Reminder: GCD (1 320, 2 100): 1 320 = 23 · 3 · 5 · 11 2 100 = 22 · 3 · 52 · 7 GCD (1 320, 2 100) = = 22 · 3 · 5 = 60

In practice, to obtain this, we do the same as with numbers: we start by breaking down the polynomials into factors, and we take the common factors with the smallest exponent with which they appear. For example: P (x) = (x + 2) (x – 3) 3 (x 2 + 2x + 3) 8 GCD [P (x), Q (x)] = (x + 2) (x – 3)2 4 Q (x) = (x + 2)2 (x – 3)2 (x – 7) We say that M(x) is the lowest common multiple of two polynomials, P(x) and Q(x):

LCM (18, 30) = 90, because 90 is a multiple of 18 and 30, and there is no other number lower than 90 that is a multiple of both.

LCM [P (x), Q (x)] = M (x) if it is a multiple of both and there is no other common multiple polynomial that has a smaller degree.

A polynomial is said to be irreducible if it has no divisors with a degree lower than its own.

Relatively prime polynomials Two polynomials are relatively prime when there is no polynomial divisor of both. For example: • 2x 2 – 2 and 6x – 4 are relatively prime, even though they can both be divided by 2.

• x 2

x2

– 1 and – x are not relatively prime because they can both be divided by x – 1.

For example, x, x – 3, x 2 + 1 and x 2 – 3x + 3 are irreducible polynomials.

6x + 3 is also irreducible, even though it is divisible by 3. x 2 + 8x + 15 is not irreducible because it is equal to (x + 3)(x + 5).

Irreducible polynomials can only be first- or second-degree. Even if we are unable to factorise a fourth-degree polynomial without roots, we know that it is the product of two second-degree polynomials.

Irreducible polynomials play the same role as prime numbers in numerical divisibility.

ãã Breaking down into factors (factorising) Factorising polynomials as a product of irreducible polynomials is similar to breaking down an integer into prime factors. In the process of finding roots, we can show the results as we did with numbers. For example: x 5

–

x 4 x 4

7x 3

7x 2

– + – 36 3 2 – 3x – x + 9x – 18 x 3 – 5x 2 + 9x – 9 x 2 – 2x + 3 1

x+2 x+2 x–3 x 2 – 2x + 3 7

x 5 46

–

x 4

–

7x 3

+

7x 2

– 36 = (x +

2)2

(x – 3)

(x 2

1 –1 –2 1 –3 –2 –2 1 –5 3 3 1 –2

–2

It has no roots.

Therefore, it is irreducible.

2

– 2x + 3)

– 7 7 0 –36 6 2 –18 36 0 – 1 9 –18 10 –18 18 0 9 –9 –6 9 0 3

In practice, to obtain this, we do the same as with numbers: we start by breaking down the polynomials into factors, and we take the common and non-common factors with the largest exponent with which they appear. For example:

Reminder: LCM (1 320, 2 100): 1 320 = 23 · 3 · 5 · 11 2 100 = 22 · 3 · 52 · 7 LCM (1 320, 2 100) = = 23 · 3 · 52 · 7 · 11 = 46 200

P (x) = (x + 2) (x – 3)3 (x 2 + 2x + 3) 8 L CM [P (x), Q (x)] = 4 Q (x) = (x + 2)2 (x – 3)2 (x – 7) = (x + 2)2 (x – 3)3(x – 7)(x 2 + 2x + 3)

Think and practise

anayaeducacion.es Consolidation of the GCD and LCM of polynomials.

1 Is there any divisibility relationship between the

following pairs of polynomials? Explain your answer.

4 Mentally calculate the GCD and the LCM of the

a) P(x) = x 3 – 7x 2 and Q(x) = x 3 – 7x

following pairs of polynomials: a) x 2 – 1 and (x + 1)2 b) x 2 + x and x 2 – x

b) P(x) = x 3 – 7x 2 and Q(x) = x 2 – 7x

c) x 3 – x and x 2 – 1

c) P(x) = x 4 – 3x – 10 and Q (x) = x – 2 2 Find two third-degree polynomials that are divisible

by x – 5 and x. Calculate their GCD and LCM.

3 Say which of the following polynomials are irreducible

and factorise those that are not. a) x 2

– 3x + 2

b) x 2

– 5x + 6

d) x 2 + 1 and x 2

5 Find the GCD and LCM of P and Q in each case:

a) P(x) = x 2 – 9, Q (x) = x 2 – 6x + 9 b) P(x) = x 3 – 7x 2 + 12x, Q(x) = x 4 – 3x 3 – 4x 2 c) P(x) = x(x – 3)2(x + 5), Q (x) = x 3(x – 3)(x 2 + x + 2) 6 P (x) = (x – 2)2 x 2. Find a third-degree polynomial,

c) 3x 2 + 5x

d) 3x 2 – 5x – 2

Q (x), which is true for the following: a) GCD [P(x), Q(x)] = x 2 – 2x

e) 3x 2 – 5x + 3

f ) 3x 3 – 5x 2 + 3x

b) LCM [P(x), Q (x)] = (x – 2)2 x 2 (x + 5) anayaeducacion.es Calculate the GCD and LCM of polynomials.

47

Unit

6

ãã Reduction to a common denominator

ALGEBRAIC FRACTIONS

Mental arithmetic

The fractions of polynomials also behave in a similar way to numerical fractions.

Check their similarity in the following definitions and procedures.

We call the quotient of two polynomials an algebraic fraction:

Remember A polynomial can be considered as an algebraic fraction with denominator 1.

P (x) Q (x)

For example, the following are algebraic fractions: 3x + 18 2x 2 – 5x + 1

x3

1 x+7

3x 2

– –2 2 x +1

4x 2 + 1 7

ãã Simplification

Observe An algebraic fraction is irreducible if its numerator and denominator are relatively prime.

If the numerator and the denominator of an algebraic fraction can be divided by the same polynomial (of a degree greater than or equal to 1), the fraction is simplified when we do so.

If we divide the numerator and the denominator of an algebraic fraction by their GCD, we get an irreducible fraction.

Mental arithmetic 1. Simplify these fractions: x +1 a) 22x b) x +x (x + 1) 2 x 2 – 6x + 9 c) x2 + 1 d) x –3 x –1 e)

x2

– 2x x 2 – 3x

f )

x3

3 2 (x 2 – 3x + 4) (x – 2) x 2 – 3x + 4 = For example: x –25x + 10x – 8 = x+7 (x + 7) (x – 2) x + 5x – 14

• One of them is obtained by simplifying the other.

• Both result in the same fraction when simplified.

x 2 + 3x = x (x + 3) = x x 2 + 5x + 6 (x + 2) (x + 3) x + 2 3x 2 – x = x (3x – 1) = x 3x 2 + 5x – 2 (x + 2) (3x – 1) x + 2

anayaeducacion.es Simplify algebraic fractions.

3) 2

2 (x – x (x + 3) a) 2x 3 – 6x b) 2 4x – 2x (x – 3 ) x (x + 2 ) 3 2 x 3 – 5x 2 + 6x c) x + 33x + x2 + 3 d) x 3 – x 2 – 14x + 24 x + 3x

48

Therefore, if we have several algebraic fractions, we can obtain others that, being respectively equivalent to the first ones, have the same denominator. When we do this, we say that we have reduced them to a common denominator.

a) 3x +2 1 and 3 x x x b) 5 and x –1 (x + 1) (x – 1) c) 3 and 22 x +1 x –1

ãã Operations: addition, subtraction, multiplication and division To add algebraic fractions, we reduce them to a common denominator and add the numerators together. Subtracting is a special type of addition.

2. Calculate. a) 3x +2 1 – 3 x x b) 3 + 22 x +1 x – 1

The product of two algebraic fractions is equal to the product of their numerators divided by the product of their denominators.

2 Inverse fractions. The inverse of x – 5x is 32x – 1 , as their product is a 3x – 1 x – 5x fraction equivalent to 1.

2 c) 2x · x – 4 x +2 x

d)

x2

x2 : x – 25 x – 5

The quotient of two algebraic fractions is equal to the product of the first multiplied by the inverse of the second.

When multiplying and dividing algebraic fractions, sometimes it is easier to simplify them first. We do this by factorising the numerator and denominator. For example:

anayaeducacion.es Operations with algebraic fractions.

2 Are the fractions in each pair equivalent? Check it.

3 a) x3 – x2 and 3x – 3 3x x +x 2 (x + 5) b) 3 and x – 3 2 x + 10x 2 + 25x 3x – x

x 2 – 1 · x + 2 = (x + 1) (x – 1) (x + 2) = x + 2 x +1 + 2x + 1 x – 1 (x + 1) 2 (x – 1)

Problems solved 1 Calculate.

x + 7 – x 2 – 2 + 2x – 1 x +1 x x (x + 1)

For example, the following fractions are equivalent because both result in the same fraction when simplified:

Think and practise 1 Simplify the following fractions:

1. Reduce the following to a common denominator.

x2

Two algebraic fractions are equivalent if:

x3

2. Say whether the fractions in each pair are equivalent or not: a) x2 – 3 and x2 x – 3x x b) x and x – 1 x –1 x c) 1 and x2 + 1 x –1 x –1

When we multiply the numerator and the denominator of an algebraic fraction by the same polynomial, we get an equivalent fraction.

ãã Equivalent fractions

– 4x 2

2

We have to reduce the three fractions to a common denominator. The common denominator is x(x + 1). Therefore: 2 (x + 7) (x + 1) (2x – 1) x – x –2 + = x (x + 1) x ( x + 1) x (x + 1) =

2 Calculate.

e 3x + 5 · 82 o : 2x2 – 1 x–3 x x +x

(x 2 + 8x + 7) – (x 2 – 2) + (2x 2 – x) 2x 2 + 7x + 9 = x (x + 1) x2 + x

(3x + 5) · 8 x 2 + x · = e 3x + 5 · 82 o : 2x2 – 1 = x–3 x x +x (x – 3) x 2 2x – 1 =

(3x + 5) · 8 · (x 2 + x) 24x 3 + 64x 2 + 40x 24x 2 + 64x + 40 = = (x – 3) x 2 · (2x – 1) 2x 3 – 7x 2 + 3x 2x 4 – 7x 3 + 3x 2

Think and practise 3 Perform the operations and simplify the result.

2 5x – 10 · x 2 – 9 3 e x – x 2 o c) a) 2x + 1 – x2 + 5 b) x +3 x +3 x–2 x x + 1 x2 – 1 x + 3x 2 2x + 1 : x 2 d) 3x – 1 – 2x + 3 + 2x + 5 e) f ) x : c 1 – 1 m 2x – 1 4x – 2 x –1 x x –1 x x–2 x – 2x

49

7

DECOMPOSITION OF ALGEBRAIC FRACTIONS INTO PARTIAL FRACTIONS A partial fraction is an algebraic fraction whose numerator is a number and whose denominator is a first-degree polynomial of the type x – a.

3 5 = 5/4 –2 For example: 3 x –5 x x – 1/3 4x + 1 x + 1/4 Thus*:

F ocu s on Eng lish

An algebraic fraction whose denominator only has simple roots can be decomposed into the sum of a polynomial and partial fractions.

thus: in this or that way.

Let’s look at some examples to see how this is done: 1 4 1

–5

3

4

–4

–1

–1

Quotient: x – 1. Remainder: –1.

Unit 2

• 3 6x 2+ 6 e roots of the denominator are 0, –1 and 2. Therefore, we try Th x – x – 2x to express the fraction as a sum of the following type:

Observe We can use this interesting procedure to calculate the value of A, B and C by giving different values to x in the following expression: 6x2 + 6 = A(x + 1)(x – 2) + + Bx(x – 2) + Cx(x + 1)

A+ B + C x x +1 x – 2 A + B + C = A (x + 1) (x – 2) + Bx (x – 2) + Cx (x + 1) = x x +1 x – 2 x (x + 1) (x – 2) 2 x ( A + B + C ) + x (– A – 2B + C ) – 2A

= x 3 – x 2 – 2x 2 x 2 ( A + B + C ) + x (– A – 2B + C ) – 2A This must be true: 3 6x 2+ 6 = x – x – 2x x 3 – x 2 – 2x

• If x = –1 8 12 = 3B 8 B = 4 • If x = 0 8 6 = –2A 8 A = –3 • If x = 2 8 30 = 6C 8 C = 5

° A + B + C = 6 (coefficient of x2) ° § § Therefore: ¢ –A – 2B + C = 0 (coefficient of x) ¢ § § –2A = 6 (independent term) £ £

2 • x – 5x + 3 Since the numerator is of a higher degree than the denominator, x–4 we do the division (see margin). We get:

• 27x – 11 The roots of the denominator are 3 and –2. Therefore: x –x–6

x2 – x – 6 = (x – 3)(x + 2) A + B . x –3 x+2 decomposition to be valid, we have to calculate A and B:

➜➜ denominator with double roots 2 • 7x +3 17x +2 10

Roots of the denominator: x = 0 (double); x = –5 x + 5x Since x = 0 is a double root, we decompose it like this:

For the

A + B = Ax + 2A + Bx – 3B = (A + B) x + (2A – 3B) x –3 x+2 (x – 3 ) (x + 2 ) x2 – x – 6 The following must be true: A+B =7 48 A=7 – B 2A – 3B = – 11 2(7 – B) – 3B = –11 8 8 –5B + 14 = –11 8 25 8B= =58 A=7–5=2 5

7x – 11 = (A + B) x + (2A – 3B) 8 * A + B = 7 (coefficient of x) 2A – 3B = –11 (independent term) x2 – x – 6 x2 – x – 6 We solve the system (see margin) and get: A = 2, B = 5 Therefore, we get the following decomposition: 7x – 11 = 2 + 5 x2 – x – 6 x – 3 x + 2 3 2 • 3x – 32x – 11x – 11 Now we start dividing: x –x–6 3x 3 – 3x 2 – 11x – 11 – 3x 3 + 3x 2 + 18x 7x – 11

x 2 – x – 6 3x

It can therefore be expressed like this:

3x 3 – 3x 2 – 11x – 11 = 3x + 7x – 11 x2 – x – 6 x2 – x – 6 The fraction is the same as in the previous example. Therefore: 3x 3 – 3x 2 – 11x – 11 = 3x + 2 + 5 x –3 x+2 x2 – x – 6 We have expressed the initial fraction as the sum of a polynomial, 3x, and two partial fractions.

50

8

° A = –3 § ¢B=4 § £ C = 5

This is how we get the following decomposition: 6x 2 + 6 = –3 + 4 + 5 x x +1 x – 2 x 3 – x 2 – 2x

x 2 – 5x + 3 = (x – 1) + –1 x–4 x–4 Now it is expressed as the sum of a polynomial and a partial fraction.

The fraction can be decomposed like this:

2

anayaeducacion.es Decomposition of algebraic fractions into partial fractions.

7x 2 + 17x + 10 = A + B + C = Ax (x + 5) + B (x + 5) + Cx 2 = x x2 x + 5 x 3 + 5x 2 x 2 (x + 5) x 2 (A + C ) + x (5A + B) + 5B

= x 3 + 5x 2 _ A + B +C = 7 b b We solve this system: 5A + B + C = 17` 8 Solution: A = 3, B = 2, C = 4 5B + C = 10b a 2 The decomposition is: 7x +3 17x +2 10 = 3 + 22 + 4 x x x +5 x + 5x Careful: the second addend is not a partial fraction because x is raised to the second power.

Think and practise 1 Decompose these algebraic fractions:

3x 2

– 7x + 4 a) 3x + 4 b) x +3 2x + 3 2 3x 2 – 5x + 1 c) 3x – 5x + 1 d) 2x + 1 x–4 2 Decompose these algebraic fractions into a sum of partial fractions:

5x – 3 c) 1 a) x2 – 2 b) x +x x2 + x – 6 x3 – x

3 Decompose these algebraic fractions. Use the result

from section a) to find the result of section b): 2 x 3 + 4x 2 – 10x + 7 a) 4x3 – 3x + 13 b) x 3 – 7x + 6 x – 7x + 6

4 Decompose (denominators with double roots).

2 2x – 4 a) x3 + 4x2 + 4 b) x – 2x + x x 3 + x 2 – 5x + 3 3 2 x + 8 d) x 3 – 4x 2 + 4x c) x + 224x – 12 x – 4x 2 x 4 – 2x 3 – 4x 2 + 8x

51

EXERCISES AND

Unit 2

VED PROBLEMS SOL

1 Polynomial remainder theorem

Find the value of a and b so that the polynomial P (x) =

2x 3

+

ax 2

+ bx – 18

is divisible by x + 2 and x + 3.

Your turn Calculate the value of k so that this division is exact: (2x 4

–

5x 3

+

kx 2

– 12) : (x + 2)

For P (x) to be divisible by (x + 2) and by (x + 3), the remainders of the divisions P (x) : (x + 2) and P (x) : (x + 3) have to be 0.

Remember the polynomial remainder theorem:

x–a P (x) r = P (a) C (x)

Therefore, we have to make P (–2) = 0 and P (–3) = 0:

*

P (–2) = –16 + 4a – 2b – 18 = 0 2a – b = 17 8 ) 8 a = 7; b = –3 P (–3) = –54 + 9a – 3b – 18 = 0 3a – b = 24

4 Decompose these algebraic fractions

Decompose these algebraic fractions into a sum of fractions: 210 a) (x – 1) (x + 2) (x – 3 ) (x + 4 )

a) 3x(x – 3) – (x + 1)(x – 3) b) ax 2 – ay + bx 2 – by c) x 3 – a 3

2 c) x – 2x +3 6 (x – 1)

a) We extract the common factor (x – 3): 3x (x – 3) – (x + 1) (x – 3) = (x – 3) [3x – (x + 1)] = (x – 3) (2x – 1) b) In the first two summands, a is the common factor; in the third and the fourth summands, the common factor is b.

Now the common factor is

Your turn Factorise. a) x 2m + x 2n – ym – yn b)

x 3

+

a 3

ax 2

– ay +

bx 2

– by =

a (x 2

a 3.

Let’s try with a:

e 3x 2 – 3 o : 1 x–2 x–2 (x – 2) 52

• If x = 3 ò 210 = 70C 8 C = 3

• If x = –2 ò 210 = 30B 8 B = 7

• If x = –4 ò 210 = –70D 8 D = –3

210 = –7 + 7 + 3 – 3 (x – 1)(x + 2)(x – 3)(x + 4) x – 1 x + 2 x – 3 x + 4

• If x = 1 ò 6 = 6B 8 B = 1 • If x = –5 ò 18 = 36C 8 C = 1/2 • If x = 2 ò 11 = 7A + 7B + C 8 7A = 7/2 8 A = 1/2

We multiply and simplify: 2x + y e x – y o (2x + y) · (x – y) (2x + y) · (x – y) p· = 2 = =1 2 2x + y x – xy (x – xy) · (2x + y) x (x – y) · (2x + y) x b) We break down the denominators of the fractions into factors: x 2 – 15 x–3 x+2 –

+ (x + 5) (x + 2) (x – 5) (x – 3) (x + 5) (x – 5) x 2 – 15 1 – 1 + x + 5 x – 5 (x + 5) (x – 5) We reduce to a common denominator and calculate: 2 2 x 2 – 25 = 1 1 – 1 + x – 15 = x – 5 – x – 5 + x – 15 = x + 5 x – 5 (x + 5) (x – 5) (x + 5) (x – 5) (x + 5) (x – 5)

x 2 + 2 x + 3 = 1/ 2 + 1 + 1/ 2 3 2 2 x +5 x + 3x – 9x + 5 x – 1 (x – 1)

c) Because it is a triple root, we decompose it like this:

x– y 3x – 2x – y e 3x – 1 o = = 2x + y 2x + y 2x + y f

A ( x – 1 ) ( x + 5 ) + B ( x + 5) + C ( x – 1 ) 2 x 3 + 3x 2 – 9x + 5

Numerators: x2+ 2x + 3 = A(x – 1)(x + 5) + B(x + 5) +C (x – 1)2

We get:

We simplify: Your turn Calculate and simplify.

• If x = 1 ò 210 = –30A 8 A = –7

=

a) We do the subtraction in brackets:

Numerators: 210 = A(x + 2)(x – 3)(x + 4) + B(x – 1)(x – 3)(x + 4) + + C(x – 1)(x + 2)(x + 4) + + D(x – 1)(x + 2)(x – 3)

x 2 + 2x + 3 = A + B + C = x 3 + 3 x 2 – 9 x + 5 x – 1 ( x – 1) 2 x + 5

We solve the equation x 2 + ax + a 2 = 0: –a ± –3a 2 x= has no solution (–3a 2 < 0 for any value of a ≠ 0) 2 Therefore, x 3 – a 3 = (x – a)(x 2 + ax + a 2).

+ C (x – 1)(x + 2)(x + 4) + D (x – 1)(x + 2)(x – 3) (x – 1)(x + 2)(x – 3)(x + 4)

Since the denominator has a double root, we decompose it like this:

– y ) + b (x 2 – y ) = (x 2 – y)(a + b )

3 Operations with algebraic fractions

Calculate and simplify.

2x + y a) 2 · e 3x – 1 o x – xy 2x + y – 2 x–3 + b) 2 x + 2 x + 7x + 10 x – 8x + 15 2 + x2 – 15 x – 25

A (x + 2)(x – 3)(x + 4) + B (x – 1)(x – 3)(x + 4) + (x – 1)(x + 2)(x – 3)(x + 4)

b) We factorise the denominator: x3 + 3x2 – 9x + 5 = (x – 1)2(x + 5)

– y ), therefore:

c) We find a root of the polynomial among the divisors of 1 0 0 –a 3 a a a 2 a 3 1 a a 2 0

=

Therefore:

ax 2 – ay + bx 2 – by = a(x 2 – y ) + b (x 2 – y) (x 2

210 = A + B + C + D = (x – 1)(x + 2) (x – 3) (x + 4) x – 1 x + 2 x – 3 x + 4

2 b) 3 x +22x + 3 x + 3x – 9x + 5

2 Factorising

Factorise these expressions:

a) We decompose the algebraic fraction like this:

Your turn Decompose these algebraic fractions: a)

6 (x + 1) (x + 2) (x + 3) (x + 4)

2 b) x3 – 3x2 + 5 x – 3x + 4

c) 2x + 53 (x + 3)

x 2 – 2x + 6 = A + B + C = x – 1 (x – 1 ) 2 (x – 1 ) 3 ( x – 1) 3 A (x –1) 2 + B (x – 1) + C Ax 2 + (–2A + B) x + (A – B + C) = = (x – 1) 3 (x – 1) 3 Numerators: x2 – 2x + 6 = Ax2 + (–2A + B)x + (A – B + C) By writing the expression for the quotients, we get and solve this system: _ A =1 b b –2A + B = –2` 8 A = 1; B = 0; C = 5 A – B –C =6 b a 2 5 We get: x – 2x +3 6 = 1 + x – 1 (x – 1) (x – 1 ) 3 53

r to choose Remembe . lio fo rt po

m this unit resources fro

for your

Unit 2

ROBLEMS

P EXERCISES AND

8

Practise

Divide and express these divisions in this way: dividend = quotient + remainder divisor divisor

2

b) (x 4 – 5x 3 + 3x – 2) : (x 2 + 1)

b) (x 4 – 5x 3 + 7x + 3) : (x + 1)

c) (4x 5 + 3x 3 – 2x) : (x 2 – x + 1)

c) (–x 3 + 4x) : (x – 3)

c) P (x) · Q (x) d) Q (x) · R (x)

d) (x 3 – 5x 2 + 3x + 1) : (x 2 – 5x + 1)

d) (x 4 – 3x 3 + 5) : (x + 2)

Work out the following and simplify the result. + y)2

– x(y + 3)

9

Express the following divisions in this way: a) (6x 3

c) (2y + x + 1)(x – 2y) – (x + 2y)(x – 2y)

a) 49x 2 – 16

b) 9x 4 – y 2

c) 81x 4 – 64x 2

d) 25x 2

e) 2x 2

f ) 5x 2

– 100

–2

Copy and complete these expressions in your notebook so they are the product of two binomials: c) 9 x 2 + 4y 2 + (…)

16

b) (…) + 25y 2 + 60xy y2 4 2 – x y d) (…) + 9 3

Take out the common factor and identify the notable products like in the example.

• 2x 4 + 12x 3 + 18x 2 = 2x 2 (x 2 + 6x + 9) = 2x 2 (x + 3)2

7

a) 20x 3 – 60x 2 + 45x

b) 27x 3 – 3xy 2

c) 3x 3 + 6x 2y + 3y 2x

d) 4x 4 – 81x 2y 2

Find the quotient and the remainder of each of these divisions: a) (7x 2 – 5x + 3) : (x 2 – 2x + 1) b) (2x 3 – 7x 2 + 5x – 3) : (x 2 – 2x) c) (x 3

54

10

Perform the following divisions: a) (2x 3 – x 2 + 3x – 1) : (2x 2 + 2x)

–

5x 2

+ 2x

15

Use Ruffini’s rule to calculate P (3), P (–5) and P (7) in the following cases: a) P(x) = 2x 3 – 5x 2 + 7x + 3 b) P(x) = x 4 – 3x 2 + 7 c) P(x) = x 5 – 2x + 1

+ 4) : (x 2

– x + 1)

16

b) (x 4 – x 3 – 3x + 1) : (2x 2 – 1) c) (x 5 – 3x 2 – 2x – 5) : (3x3 + 4x – 1) d) (x4 – x3 + 3x – 1) : (5x2 – 2) e)

(x3

+ 1) :

(6x3

+ x)

f ) c 1 x 4 + x 2 – 7x + 3m : (3x + 2) 2

Express the following as a product of two binomials:

a) 16x 2 + (…) – 8xy

6

D = d · c + r – 9x) : (3x – 2)

Find out if the polynomial P(x) = x 43 – 2x 2 + 3 is divisible by (x + 1).

c) (4x 4 + 2x 3 – 2x 2 + 9x + 5) : (–2x 3 + x – 5)

Multiply each expression by the LCM of the denominators and simplify: 3x (x + 5) (2x + 1)2 (x – 4) (x + 4) + – a) 5 2 4 2 2 2 2 (8x – 1) (x + 2) (3x + 2) (2x + 3) (2x – 3) – + b) 10 15 6 3 3 (x – 1) + 3 x (x + 2) 2 – x c) 8 10 4

– 3

+

5x 2

14

b) (x 4 – 4x 2 + 12x – 9) : (x 2 – 2x + 3)

d) (x + y) (2x – y) (x + 2y)

5

19

a) (3x 5 – 2x 3 + 4x – 1) : (x 3 – 2x + 1)

b) 3x(x + y) – (x – y)2 + (3x + y)y

4

13

Given the polynomials P (x) = x 3 – 5x 2 – 3; Q (x) = – 1 x 2 + 2x – 1 and R (x) = x 3 – 1 x 2, calculate: 2 3 a) P (x) + Q (x) – R (x) b) 2P (x) – 3Q (x)

a) (2y + x)(2y – x) + (x

3

Factorising polynomials

Apply Ruffini’s rule to find the quotient and remainder of the following divisions: a) (5x 3 – 3x 2 + x – 2) : (x – 2)

Polynomials. Operations

1

Ruffini’s rule. Applications

11

17

In each case, find the value of a that makes the following divisions exact: a) (x 4 + x 3 – 12x 2 + ax + 5) : (x 2 – 2x – 1) b) (x 5 – 2x 4 + 4x 3 + ax 2 + 2x + 2) : (x 2 – 2x + 2) c) (x 4 + 3x 3 + 3x 2 + 3x + a) : (x 2 – 2x – 1) d) (x 3 + x 2 + ax – 3) : (x 2 – 2x – 1)

12

Calculate the value of the dividend in each of the following cases: a) Divisor: (x 2 – 2). Quotient: (x + 3). Remainder: 7 b) Divisor: (x 2 + 2x). Quotient: (x2 + x – 2).

Remainder: (–3x + 4) c) Divisor: (x 2 + 1). Quotient: (x – 2).

Remainder: (2x + 10)

18

Work out which of these numbers 1, –1, 2, –2, 3, –3 are roots of the following polynomials: a) P(x) = x 3 – 2x 2 – 5x + 6 b) Q(x) = x 3 – 3x 2 + x – 3 c) R(x) = x5 + 3x 4 – 5x3 – 15x 2 + 4x + 12 Use Ruffini’s rule to find the quotient and remainder of the following divisions: a) (4x 2 – 8x + 3) : (4x – 2) b) (2x 3 – 4x 2 + 3x – 2) : (2x – 3) c) (3x 3 – 2x – 1) : (3x + 1) Look at the following application of Ruffini’s rule.

In it, we see that (x4 – x3 – 2x2 + x + 3) : (x – 2) is not exact. However, we also notice that (x2 – x – 2) : (x – 2) is exact: 1 2 1

(x 4

2x 2

–2 1

e) Divisor: + – 1). Quotient: 3 Remainder: (3x + x 2 – 5x)

b) (x2

– 2).

– 2 2 0

1 0 1

3 2 5

Now it is your turn: find an exact division in each of these. a) 1 3 5 6 1

(x 3

d) Divisor: – 2). Quotient: x.

Remainder: (2x + 1)

–1 2 1

2 –1 2

–2 1

–2 3

–6 0

0 1

–1 – 2 –3

– 3 3 0

2 0 2

3 – 2 1

Take out the common factor and use notable identities to factorise the following polynomials: a) 3x 3 – 12x b) 4x 3 – 24x 2 + 36x c) 45x 2 – 5x 4 d) x 4 + x 2 + 2x 3 e) x 6 – 16x 2 f ) 16x 4 – 9

20

Factorise the following polynomials: a) x 2 + 4x – 5 b) x 2 + 8x + 15 c) 7x 2 – 21x – 280 d) 3x 2 + 9x – 210 e) 2x 2 – 9x – 5 f ) 3x 2 – 2x – 5 g) 4x 2 + 17x + 15 h) –x 2 + 17x – 72

21

Break down the following polynomials into factors: a) (x 2 – 25)(x 2 – 6x + 9) b) (x 2 – 7x )(x 2 – 13x + 40)

22

Break the following polynomials down into factors and say what their roots are. a) x 3 + 2x 2 – x – 2 b) 3x 3 – 15x 2 + 12x c) x 3 – 9x 2 + 15x – 7

23

d) x 4 – 13x 2 + 36

Factorise the following polynomials and say what their roots are. a) x 3 – 2x 2 – 2x – 3 b) 2x 3 – 7x 2 – 19x + 60 c) x 3 – x – 6

d) 4x 4 + 4x 3 – 3x 2 – 4x – 1

e) 6x 3 + 13x 2 – 4

f ) 4x 3 + 12x 2 – 25x – 75

24

Factorise and say what are the roots of the following polynomials: a) x 4 – 2x 2 + 1 b) x 3 – 2x 2 – 9x + 18

25

c) x 4 – x 3 – 7x 2 + x + 6

d) 8x 3 + 6x 2 – 11x – 3

e) 3x 3 + 8x 2 + 3x – 2

f ) x 3 – 2x 2 + 2x – 4

Write a third-degree polynomial that has the given roots in each case: a) 0, 1 and 2 b) –1 and 3 c) 0 and 5

26

For each of the following, write a polynomial that meets the given condition: a) Fourth-degree without roots. b) That has two double roots, 2 and –2. c) Third-degree with a single root. d) Fourth-degree and with three roots.

55

Unit 2

EXERCISES AND PROBLEMS 27

In each case, indicate the initial polynomial and its factorisation. Complete it, where possible, by finding the roots of the second-degree polynomial. a)

1 –1 2 –3

1 1 1

b)

1 1 –1 2

1 1 1

c)

1 –1 2

1 1

2 –1 1 2 3 –3 0

–4 –1 –5 6 1 0 1

–4 5 1 2 3 –3 0

–5 –1 –6 6 0

5 1 6 –1 5 2 7

–5 6 1 –5 –4 14 10

–25 1 –24 4 –20 20 0

4 –24 –20 20 0

–3 –35 –1 4 –4 –31 2 –4 –2 –35

39 31 70 –70 0

70 –70 0

–6 6 0

31

a)

32

20 –20 0 33

35

2 x+2 a) x – 92 b) (x + 3) x2 – 4

f )

x 2 y – 3xy 2 2xy 2

36

3 2 x 2 – x – 42 c) x –2 3x + 2x d) x 2 – 8x + 7 3x – 9x + 6

56

What algebraic fraction should you multiply the result of each part of the previous exercise by to get 1? And to get the polynomial x 2 + 1?

41

a)

37

42

46

Replace the dots with the suitable expressions to make the fractions equivalent.

2 x = x2 a) x 2 – x = … b) 2x + 1 … x – 1 x +1 …

2 = c) x = 2… d) x + 2 x 2 + 4x + 4 x –3 x –9

47

5 4 3 2 g) 6x – 76x – 55 x – 3x – 25x + 2 x – 2x + 2x – x

Problem solving 43

Find the lowest common multiple and the greatest common divisor for each of the following polynomials: a) x 2; x 2 – x; x 2 – 1 b) x – 3; x 2 – 9; x 2 – 6x + 9 c) 2x; 2x + 1; 4x 2 – 1 d) x + 2; 3x + 6; x 2 + x – 2

Calculate the value of m so that the polynomial – 3x2 + 5x + 9m is divisible by x + 2.

mx3

48

Calculate the value of a and b so that the polynomial P (x) = 2x 3 + 7x 2 + ax + b is divisible by x – 1 and by x + 2.

49

Calculate the value of m and n so that the polynomial P(x) = x 3 – m x 2 + n x + 4 is divisible by x – 2 and x + 2. What are the roots of P(x)?

50

Calculate the value of a and b so that the polynomial P (x) = x 4 – 2x 3 + ax2 + bx + 15 is divisible by (x + 3) and by (x – 5).

51

Calculate the value of m so that the following divisions have the remainder indicated: a) (x 2 – 5x + m) : (x – 2) Remainder = 0 3 2 b) (x – 2x – x + m) : (x + 1) Remainder = –1 c) (2x 3 – 12x + 2m) : (x – 3) Remainder = –5 d) (x 2 – mx + 3) : (x + 3) Remainder = 0

52

The remainder of this division is –8: (2x 4 + kx 3 – 7x + 6) : (x – 2) What is the value of k?

53

What do the values of a and b have to be to ensure that when we divide the polynomial P (x) = 3x 3 + ax 2 – 5x + b by (x – 1) the remainder is 14, and when we divide it by (x + 3) the remainder is – 2.

Decompose these algebraic fractions: 3 3x 3 – 5x + 1 a) 3x – 53 x + 1 b) (x – 2) 3 x c) 4 x + 22

d) 2 1 2 (x – 1) x – 2x + 1 2 5x 2 e) 32x +2 7x – 1 f ) 3 x – 3x 2 + 3x – 1 x + x – x –1

Calculate and simplify where possible.

x + 1 · x2 – 1 a) c 3 – x m : c 1 + 1 m b) x 3 x 3 x (x – 1) 2

Give three examples of pairs of polynomials from exercise 24 that are relatively prime.

3x 3 2x + 3 b) (x – 2) (x + 5) x2 – 4

x2 + 1 1 d) x2 + x (x 2 – 25) (x – 4) 3x – 2 e) 2 1 f ) x2 – 4 x +x –6 x2 g) 2 4 h) 2 x + 4x + 3 x +x –2 2 – 16 i) 2x2 – 5x + 3 j) 2 x – 3x + 2 x – 2x – 15 x –1 k) 2 x + 2 l) 2x + 3x – 1 4x 2 – 9

+ 2x + 3 – 3 x –1 – 2x + 1

Calculate and simplify.

Break down into the sum of partial functions.

c)

x2

2x : c 2x – 1m 2 m · 2 – x d) 2–x x +1 x +1 x2

45

Decomposing algebraic fractions

Work out the following:

c) c1 –

Calculate and simplify.

2 c 1 – 1 m : 32 a) c1 – x – 1 m x – 1 b) x x +3 x x x +3 c1 + y m : c1 + x m c) 4 – 1 c 2 – 12 m d) 2x – 1 x x x y

40

c2 – 2 m: x – 2 a) c 1 : 1 m · x b) x x +1 2 x x+2 x

Break the dividend and the divisor down into factors, and then simplify.

Calculate de GCD and the LCM of the following polynomials from exercise 24: a) Sections a and c. b) Sections b and c. c) Sections c and e. d) Sections b and f.

e) c 1 + 1 – 3x – 4 m · 6x x 2 2x 6 – 2x

c) 2x2 – 3 – x + 1 – x + 2 x –9 x –3 x +3

30

2 x 2 – 3x – 4 a) 2x – 2x b) x – 5x + 6 x3 + x2

39

Calculate.

x2

44

a) >cx + 1 m : cx – 1 mH · (x – 1) b) 2 · c 1 : 1 m x x x x x –1

Reduce to a common denominator and calculate.

b)

2 x 2 + xy c) x +225 – 10x d) x 2 – 2xy + y 2 x – 25

x–2 2 x +x –6

Reduce to a common denominator and calculate.

1 – 1 +1 a) 1 – 1 + 1 b) 2x 4x x x 2 3x x 2 – x +1 c) x + 3 – 1 d) 3x 2 x x2 e) x – 3 f ) x – 3 – x x +3 x–3 x x +1

a) x + 1 + 3 – x2 – 2 x – 1 x +1 x – 1

Break down into factors and simplify.

e)

Calculate and simplify.

a) x –2 2 + x2 + 2 – 2 1 x x – x x –1 – 5 – x–4 b) 2 2x x + x – 2 x + 2 3x + 6 c) x + 2 – 22 + x +1 2x 2x + 1 4x – 1

Check that these pairs of fractions are equivalent:

29

38

x 4 – 5x 2 + 4 x 3 – 4x b) 2 + x – 2x x4 – 1

a) x – 1 – 2 + 2 x x +3 x –3 x –9 b) 2 – 2x + 1 – 2 3 x – 2 x – 2x x – 4 c) 1 + 23x – 3 – x 2x + 2 x – x – 2 x – 2 34

x 2 + x and x a) x – 4 and 1 b) 2x 2 3x – 12 3 x+y 1 d) x 2 c) 2 and and 2 2 – –2 2 x x y x –x x – y

x3

4 3 2 2x 3 – 5x 2 + 3x c) x 4+ 2x 3– 3x 2 d) 2x – 3x + x 2x 4 + x 3 – 6x 2

Algebraic fractions

28

Simplify the following fractions:

54

Find the parameters a and b to ensure that when we divide P(x) = x 5 + ax 4 + x 2 + bx + 8 by (x + 1) the remainder is 9, and when we divide it by (x – 2) the remainder is 6.

57

Unit 2

EXERCISES AND PROBLEMS 55

If P(x) = 3x3 – 11x2 – 81x + 245, find the values P(8.75), P(10.25) and P(–7) with the help of a calculator. Describe the sequence of buttons used, like on page 41.

We inscribe the rhombus inside of this rectangle with sides x and y. Write the perimeter of the rhombus as a function of the sides of the rectangle.

56

Check whether there is any divisibility relationship between these pairs of polynomials: a) P(x) = x 4 – 4x 2 and Q(x) = x 2 – 2x

Take out the common factor in each expression: a) (x + 2)(x – 3) + 2x(x + 2) b) (x – 2)(2x + 3) – (5 – x)(x – 2)

a) ax – ay + bx – by

b) 2x 2y + y + 2x 2 + 1

c) 3x 2y + xy + 3xy 2 + y 2

d) 2ab 3 – ab + 2b 2 – 1

B

2a 2 bx

– – 4x 2 + 8bx + 2ba – ax

A

61

x + 1 is the height of a lateral face.

72

We cut a piece of wire that is 1 m long into two uneven parts. With one of these parts, we form an equilateral triangle, and with the other, a square.

Write the sum of the areas of both figures.

73

Using a rectangular piece of card that measures 30 cm by 20 cm, we make a box without a lid by cutting a square with side x from each corner. Write the volume of the box as a function of x.

x x+1

b) Write the area of rectangle MNPQ using a polynomial at x.

A tap takes x minutes to fill a tank. Another tap takes 3 minutes less to fill the same tank. Express, as a function of x, the part of the tank that fills in a minute if both taps are open.

63

One leg of a right-angled triangle measures 14 cm. Write the perimeter and area of the triangle as a function of the hypotenuse, x.

58

81

A M

x D

x

d) If P (3) ≠ 0, then the polynomial P (x) is not divisible by x – 3.

What do the values of a and b have to be for the polynomials P (x) and Q (x) to be equal?

75

g) It is not possible to write a fourth-degree polynomial that only has a triple root.

The roots of P (x) are 0, 2 and –3. a) Write three first-degree divisors of P(x).

82

b) Write one second-degree divisor of P(x).

C

We have a rectangle with a perimeter of 20 cm. If the base is reduced by 2 cm and the height by 3 cm, by how much is the area of the rectangle reduced?

Express it as a function of the base.

f ) If P (x) = ax 2 + bx + 2 and P (±2) ≠ 0, then P (x) cannot have integer roots.

Q (x) = (a + 3)x 3 + (a + 2)x 2 – 2x + 5

N

Advanced problem solving

e) If P (–2) = 0, then x + 2 is a factor of P (x).

P (x) = x 3 – (4 + a)x + (1 + b )

76

Q

True or false? Explain your answers and give examples. a) If a polynomial has a degree of 3, and another polynomial has a degree of 2, their product will have a degree of 6.

74

D

D'

B P H

69

Prove that the polynomial x2 + (a + b)x + ab is divisible by x + a and by x + b for any value of a and b. What would they look like broken down into factors?

80

c) If we add two third-degree polynomials, we always get a third-degree polynomial.

a) If the division P(x) : (x – 2) is exact, what can you say about the value P(2)? b) If –5 is a root of the polynomial P(x), what can you say about the division P(x) : (x + 5)? c) On which result did you base your answers to these two questions?

62

We mix x kg of paint that costs €5/kg with y kg of another paint that costs €3/kg. What would 1 kg of the mixture cost? Express it as a function of x and y.

By which fraction do we need to multiply x – 5 x –1 2 to get 2x – 5x ? x + 3x – 4

79

b) If P (0) = 1, then P (x) is divisible by (x – 1).

a) Express MN as a function of x. (Use the similarity of triangles AMN and ABC.)

x+2

b) LCM [P(x), Q(x)] = (x – 1)2(x 2 – 9)

x x

We have this polynomial: P(x) = (x – 1)2(x + 3).

Find a second-degree polynomial, Q(x), that meets the following conditions: a) GCD [P(x), Q (x)] = x – 1

Remember the theory

— — In the triangle below, BC = 10 cm, AH = 4 cm.

We draw a point D on the height line of A, in such — a way that AD = x, and through D we draw line MN parallel to BC. Starting at points M and N we draw lines that are perpendicular to BC.

68

A shop owner sold two bicycles. He made a 20 % profit on one of the bicycles, and a loss of 10 % on the other one. In total, he made a profit of 15 %. Use algebra to express this statement.

78

C

A'

60

Express the total surface area of this truncated pyramid as a function of x :

B'

C'

xy 2

– 3a 2 b 2 – 6ab 3 b) 10x – 5y 3a 3 b – 6a 2 b 2

c)

x

Inside rectangle ABCD with sides AB = 3 cm and BC = 5 cm, we have inscribed the quadrilateral A'B'C'D' making AA' = BB' = CC' = DD' = x.

Write the area of A'B'C'D' as a function of x.

Simplify the following algebraic fractions:

4a 2 b 2

y

67

Factorise the following expressions:

a)

71

Two towns, A and B, are 60 km apart. A car leaves A for B at a velocity of v. At the same time, another car leaves B for A at a velocity of v + 3.

Express the time they take to meet as a function of v.

d) (3 – y)(a + b) – (a – b)(3 – y)

2x 2 y

x

66

c) (x + 5)(2x – 1) + (x – 5)(2x – 1)

59

y

Use algebraic language to express the area of the coloured part of this shape using x and y.

c) P(x) = x 3 + x 2 – 12x and Q(x) = x – 3

58

A triangle has a base of 20 cm, and a height of 15 cm. If the height is increased by x % and the base by (x + 2) %, express the new area of the triangle as a function of x.

65

b) P (x) = x 2 – 10x + 25 and Q(x) = x 2 – 5x 57

70

64

77

Invent two second-degree polynomials that meet the condition given in each case: a) LCM [P(x), Q (x)] = x 2 (x – 3)(x + 2) b) GCD [P(x), Q (x)] = 2x + 1

If a polynomial A has a degree of 4 and another polynomial B has a degree of 3, state the degree of the following polynomials: a) Sum of A and B. b) Subtraction of A from B. c) Product of A and B. d) Quotient of A and B.

83

True or false? a) When three polynomials with different degrees are added together, the resulting polynomial has the same degree as the highest of them. b) The result of adding three third-degree polynomials is a sixth-degree polynomial. c) The result of adding two fifth-degree polynomials is another polynomial with a degree of 5 or higher.

59

Unit 2

OP

MATHS WORKSH

FINDING REGULARITIES AND GENERALISING

PRACTICE MAKES PERFECT!

Triangles and powers

• For the following operations, replace each letter with a figure other than zero.

Observe, check and compare: 1 1

(a + b)1 = 1a + 1b

1 2 1

(a + b)2 = 1a 2 + 2ab + 1b 2

1 3 3 1

(a + b)3 = 1a 3 + 3a 2b + 3ab 2 + 1b 3

1 4 6 4 1

(a + b)4 = 1a 4 + 4a 3b + 6a 2b 2 + 4ab 3 + 1b 4

Can you add another row to this number triangle?

(It is called Pascal’s triangle).

yz yz yz yz + yz xyz

ab x c de + fg hi

Can you develop the polynomial for (a + b )5 without multiplying the binomial (a + b ) by itself five times?

READ AND LEARN Cutting up Cutting this trapezium into two or three equal pieces is very simple, but cutting it into four is much harder.

Try it.

a a

THINK AND EXPRESS YOURSELF

a

2a

Make the largest number you can with them... x y z

The largest number ........ 853

Make the smallest number .............................. z y x

The smallest number ...... 358

Subtract one from the other ........................... x y z – z y x

The difference ................ 853 – 358 = 495

Using algebraic language, show that the observation above is true for any trio of figures, x, y, z, as long as at least two of them are different.

Tip:

x y z = 100x + 10y + z z y x = 100z + 10y + x 60

1 Multiply by the LCM of the denominators and

7 Decompose the following fractions into a sum of

simplify.

(x – 2) (x + 1) (3x – 1) 2 (2x – 3) (2x + 3) – + 3 8 12

3 Calculate the value of parameter m so that the

For example, 5, 8 and 3.

•

anayaeducacion.es Answer key and interactive self-assessment.

(3x 4 – 5x 3 + 4x 2 – 1) : (x 2 + 2)

Think of any three numbers that are not all the same

• Check that the difference is always a multiple of 9 and 11.

SELF-ASSESSMENT

2 Find the quotient and remainder for this division:

Interesting finding

• Solve the following problems without using algebra: a) A pond is fed by two water channels. If we open the gate of the first of these channels, the pond fills in 8 hours. If we open both, it fills in 3 hours. How long would it take to fill the pond if we only open the gate of the second channel? b) A reservoir has a tap and a drain. When the drain is open, the reservoir drains in 2 hours.

One day, without realising it, and when the reservoir was full, we opened the drain, but left the tap on. It took the reservoir 5 hours to drain.

How long does it take the tap to fill the reservoir?

polynomial P(x) = 7x3 – mx2 + 3x – 2 is divisible by x + 1.

4 Break down the following polynomials into factors:

a) x 4 – 12x 3 + 36x 2

b) 2x 3 + 5x 2 – 4x – 3

5 Simplify the following algebraic fractions:

2x 2 y – xy 2 3a 2 b 2 – 6ab 3 a) b) 10x – 5y 3a 3 b – 6a 2 b 2

partial fractions: a)

–9x + 6 (x – 1) (x + 2) (x – 2)

3 2 b) x +310x 2+ 26x + 23 x + 4x + x – 6

8 Given the polynomial P(x) = (x – 1)2(x – 3):

a) Invent a second-degree polynomial Q(x) such that P(x) and Q(x) are relatively prime. b) Find a polynomial R(x) for which this is true: GCD [P(x), R(x)] = (x – 1)(x – 3) LCM [P(x), R(x)] = (x – 1)2(x – 3)2(x + 2) 9 Find the value of a and b so that when we divide

x 3 + ax 2 + bx – 4 by x + 1 the remainder is –10, and when we divide it by x – 2 the remainder is 2.

10 A plot has sides

2 x2 – 6 – x – 3 a) 2x : 3 8 2 b) x – 3 x – 3x (x – 2) 2 x – 2

x and y, and we build a house on it in the coloured zone indicated on the plan.

c) 1 – 2 a + 23a + 1 a a –1 a – a

Express the area of the undeveloped plot as a function of x and y.

6 Work out the following and simplify where possible:

Commitment

Watch the video for target 3.9. Think of something you can do to contribute to achieve that goal. Make a commitment to put your idea into practice.

50 m

y 30 m x

61

3

EQUATIONS, INEQUATIONS AND SYSTEMS

... and systems of equations Systems of linear equations had already been solved by trial and error 4 000 years ago, in ancient Babylon. In China by the 2nd century BCE they were already solving them using a system similar to the one we use today.

Two Chinese problems

Reading and listening

Writings found on strips of bamboo dating back more than 2 200 years: • Calculate the depth of a circular pond

Milestones in the history of equations...

• the pioneer: Diophantus (3rd century) came up with and solved

complex algebraic problems using some very interesting and original methods. However, he did so without any defined methodology, so his work had little value for teaching others.

• systematisation: Al-Khwarizmi (9th century) was the first to carry out

with a surface area of 10π square feet if a cane growing in the centre, which stands one foot out of the water, reaches exactly to the surface when it is tilted against the edge of the pond.

a systematic process for solving first- and second-degree equations. His book, al-Jabr wa’l-Muqabala, was comprehensive and groundbreaking, and was translated into many languages.

10 foot

• the 16th century Italians: In the 16th century, several Italian algebraists

• A 10-foot-tall bamboo cane has

snapped and now the top is resting on the ground 3 feet away from its base. How far up did the cane snap?

(Tartaglia, Cardano, etc.) had lively and productive debates bringing up ideas and solving different types of cubic (third-degree) equations. This brought about a great leap forward in solving higher-degree equations.

A Babylonian problem A Babylonian tablet dating back more than 4 000 years poses the following problem:

3 foot 1 What do these phrasal verbs mean? Make a sentence with each one.

come up with - carry out - bring up - bring about

What is the length of the side of a square if we know that its area minus the length of its side equals

?

The symbol isequalto10and equals1.Therefore, the first set of symbols has a total value of 10 + 1 + 1 + 1 + 1 = 14, and the second has a value of 10 + 10 + 10 = = 30. The Babylonians used a positional numeral system with 60 as its base. Therefore, the number above was 14 · 60 + 30 = 870.

Caution! The area is a surface area and the side is a length. We cannot subtract one from the other. Therefore, for the statement to be correct, we would have to say, ‘the number of u2 of the surface area minus the number of u of the length of the side is 870’.

2 Make sentences using the third conditional to imagine different ways

the ancient people could have solved their problems.

Solve 1 Find the length of the side of a square if the square metres of the surface

area minus the metres of the length of its side is 870. Do not use the formula for a second-degree equation to solve it. Remember that x 2 – x = = x (x – 1) is the product of two consecutive numbers. (Factorise 870.)

2 Find the depth of the pond in the first Chinese problem.

ANK ANK B E G A U LANG LANGUAGE B ANK ANK GE BANK B B E E G G A A U U GUA K LANG ANG N L A L AN GE BANK 63 ANK GE BANK B B E E G G A A U U G A LAN LANG LANGUA LANGU

3 In the second Chinese problem, how far up did the cane snap?

62

Unit

1

EQUATIONS

ax 4

ãã Second-degree equations Second-degree equations take this form: no solution

ax 2 + bx + c = 0, with a ≠ 0

two solutions

• Complete equations. When b ≠ 0 and c ≠ 0, we say that the equation is complete and we can solve it by applying the following formula: Z ]] if b 2 – 4ac > 0, there are two solutions.

2 –b ± b – 4ac 8 [ if b 2 – 4ac = 0, there is one solution. x= 2a ] if b 2 – 4ac < 0, there is no solution. \ • Incomplete equations. If b = 0 or c = 0, we say that the equation is incomplete and can be solved very simply without the need to apply the formula mentioned above:

one solution

anayaeducacion.es •H ow to find the formula for solving second-degree equations.

• S econd-degree equations and problems to consolidate solving mechanisms.

Problems solved 1 Solve

these equations:

incomplete

a) 5x 2 – 45 = 0 b) 5x 2

c) 3x 2 – 21x = 0

these equations:

a) 5x 2 – 45 = 0 8 x 2 = 45 = 9 8 x = ± 9 = ±3 5 Solutions: x1 = 3, x2 = –3 b) 5x 2 + 45 = 0 8 x 2 = – 45 = –9 No solution.

5 x =0 c) 3x 2 – 21x = 0 8 x (3x – 21) = 0 8 * 3x – 21 = 0 8 x = 7 Solutions: x1 = 0, x2 = 7

+ 45 = 0

2 Solve

b = 0 8 ax 2 + c = 0 8 x 2 = – c 8 x 1 = – c , x 2 = – – c a a a c = 0 8 ax 2 + bx = 0 8 x (ax + b ) = 0 8 x1 = 0, x2 = – b a

complete

a) x 2 + x – 6 = 0 b) 9x 2 + 6x + 1 = 0 c) 5x 2 – 7x + 3 = 0

–1 ± 1 + 24 –1 ± 5 = +x–6=0 8 x= 2 2 Solutions: x1 = 2, x2 = –3 – 6 ± 36 – 36 – 6 ± 0 b) 9x 2 + 6x + 1 = 0 8 x = = =– 6 =– 1 18 18 3 2·9 Single solution: x = – 1 3 7 ± 49 – 60 7 ± –11 c) 5x 2 – 7x + 3 = 0 8 x = = No solution.

10 10 a) x 2

Think and practise

anayaeducacion.es Solving second-degree equations.

1 Solve.

a) 2x 2

3 The long side of a right-angled triangle is

– 50 = 0

b) 3x 2

+ 5 = 0

c) 7x 2 + 5x = 0

2 Solve the following: a) 10x 2 – 3x – 1 = 0

b) x 2 – 20x + 100 = 0

64

c) 3x 2 + 5x + 11 = 0

3 cm longer than its medium side, while its medium side is 3 cm longer than its short side.

What is the length of each side?

+

bx 2 x2

x4

az 2

3

ãã Biquadratic equations: ax 4 + bx 2 + c = 0

+c=0

Biquadratic equations are fourth-degree equations with no odd-degree terms. To solve them, we change the variable x2 = z and, therefore, x4 = z2. This gives us a second-degree equation with the unknown z:

=z = z2

az 2 + bz + c = 0

+ bz + c = 0

Once it has been solved, it gives us the values corresponding to x. For each positive value of z there will be two values of x, since x 2 = z 8 x = ± z .

x=± z

A biquadratic equation can have one, two, three or four solutions, or none.

Problem solved

Solve this equation: x 4 – 13x 2 + 36 = 0

x 4 – 13x 2 + 36 = 0

x2 = z

z 2 – 13z + 36 = 0

z =9 8 x =±3 13 ± 169 – 144 13 ± 5 = 8 * 2 2 z =4 8 x =±2 Solutions: x1 = 3, x2 = –3, x3 = 2, x4 = –2

anayaeducacion.es Consolidate solving equations with x in the denominator.

z=

ãã Equations with x in the denominator

Remember

Like numerical denominators, algebraic denominators are eliminated by multiplying by the product of all of them or, rather, by the lowest common multiple. This can give us an equation we know how to solve.

In order to eliminate denominators in equations, we multiply both sides by the lowest common multiple of the denominators.

Remember that you should check all the solutions when you have finished.

False solutions can appear during the process of multiplying by polynomial expressions. Therefore, every time you do so you need to make sure you check all the solutions.

Equations with the x in the denominator are also called rational equations.

Problem solved

Multiply both sides by 10x (x + 3).

Solve 1 – 1 = 3 . x x + 3 10

10(x + 3) – 10x = 3x (x + 3) 8 10x + 30 – 10x = 3x 2 + 9x 8 –3 ± 9 + 40 –3 ± 7 = 2 2 There are two solutions: x1 = 2, x2 = –5. Checking they fit the initial equation, we see that both solutions are valid: 1 – 1 =– 1 + 1 = 3 1 – 1 = 5–2 = 3 –5 –2 5 2 10 2 5 10 10 8 3x 2 + 9x – 30 = 0 8 x 2 + 3x – 10 = 0 8 x =

anayaeducacion.es Consolidate solving equations with x in the denominator.

Think and practise

anayaeducacion.es Solving biquadratic equations.

4 Solve.

a) 3x 4

5 Solve the following equations:

–

12x 2

= 0

b) 3x 4

+

75x 2

=0

c) 7x 4 – 112 = 0

d) x 4 – 9x 2 + 20 = 0

e) 4x 4 + 19x 2 – 5 = 0

f ) x 4 + 9x 2 + 18 = 0

5 + x =3 a) x + 2x = 3 b) x +2 x +3 2 x – 1 x +1 x +1 + 1 – x = 5 c) 1 + 12 = 3 d) x x x +5 x – 4 2 4

65

1 EQUATIONS

Unit

Remember You must check all possible solutions.

ãã Equations with radicals

ãã Exponential equations

We sometimes find equations where the x is under a square root. In order to solve this type of equation, it is generally a good idea to eliminate the root by isolating it first on one side and then squaring both sides. Be careful, though! In this squaring process, although all the solutions remain, new solutions can appear that you obviously need to reject. This is why you must check all the solutions in this type of equation.

Exponential equations are equations where the unknown is in the exponent.

Let's look at some exponential equations that require different solving processes: Problem solved

Equations where the unknown is inside a radical are also called irrational equations.

b) 5 x + 1 = 183

The equation looks like this: 5x

c) 3 x + 3 x + 2 = 90

The solutions to the equation x 2 – 4x + 3 = 0 are x1 = 1 and x2 = 3.

Problems solved 1 Solve this equation:

x 2 + 7 + 2 = 2x 8

a) The left side is a power of base 5, and so is the right side:

Solve the following equations: 2 a) 5 x – 4x = 1 125

1 = 1 = 5 –3 125 5 3 2

4x 2

3x 2

(x + 1) ln 5 = ln 183 8 x = ln 183 – 1 = 2.2368 ln 5

• 3 2 + 7 + 2 = 16 + 2 = 6 ; 2 · 3 = 6. The first one, x1 = 3, is a solution.

2

• c– 1 m + 7 + 2 =

3 2 c– 1 m = – 2 3 3

_ 64 + 2 = 8 + 2 = 14 b 9 3 3 b` The second one, – 1 , is not a solution.

3 bb

c) Bear in mind that 3x + 2 = 3x · 32 = 9 · 3x.

Therefore: 3x + 3x + 2 = 3x + 9 · 3x = 10 · 3x.

anayaeducacion.es Solving exponential equations.

The equation looks like this: 10 · 3x = 90 8 3x = 9 8 x = 2

ãã Logarithmic equations

a

Therefore, the solution is x = 3.

2 Solve this equation:

Move

x+4 – 6 – x=2

Logarithmic equations are equations where the unknown is in an expression affected by a logarithm. We solve them by looking at the properties of logarithms.

6 – x to the right side:

x + 4 = 4 + (6 – x) + 4 6 – x 8 x + 4 = 10 – x + 4 6 – x 8 8 2x – 6 = 4 6 – x 8 x – 3 = 2 6 – x

Square again: x 2 – 6x + 9 = 4(6 – x) 8 x 2 – 6x + 9 = 24 – 4x 8 x 2 – 2x – 15 = 0 8

8 x=

2 ± 4 + 60 2 ± 8 = 2 2

x1 = 5 x 2 = –3

Check both possible solutions: anayaeducacion.es Consolidate solving equations with radicals.

a) log3 (5x – 7) = 2

a) Given the definition of logarithm: 32 = 5x – 7 5x – 7 = 9 8 5x = 16 8 x = 16 5

b) log5 (x 2 – 8) = 0

b) 50 = x 2 – 8. Since 50 = 1, the equation looks as follows:

Solve these equations:

x 2 – 8 = 1 8 x 2 = 9 8 x1 = 3, x2 = –3

c) logx 16 = 2

c) x2 = 16 8 x1 = – 4, x2 = 4. The solution x2 = – 4 is not valid, because the base of the logarithm cannot be negative. The only solution is x = 4.

• –3 + 4 – 6 – (–3) = 1 – 9 = 1 – 3 = –2 8 –3 is not a solution.

Think and practise

Therefore, the solution is x = 5.

anayaeducacion.es Solving logarithmic equations.

7 Solve the following exponential equations:

6 Solve the following equations:

66

Problem solved

• 5 + 4 – 6 – 5 = 9 – 1 = 3 – 1 = 2 8 x1 = 5 is a solution.

Think and practise

a) x – 2x – 3 = 1

In this type of equations you must also check all the solutions. Remember that only positive numbers have logarithms and that the base of the logarithm must be positive and a number other than 1.

x + 4 = 2 + 6 – x . Square both sides:

= 5–3. Therefore: x 2 – 4x = –3.

ln 5x + 1 = (x + 1) ln 5 ( the logarithm of a power is equal to the exponent multiplied by the logarithm of the base).

8 ± 64 + 36 8 ± 10 +7= – 8x + 4 8 – 8x – 3 = 0 8 x = = 6 6 There are two possible solutions: x1 = 3, x2 = – 1 . Check them: 3 x 2

– 4x

b) Since the right side cannot be given as a power of base 5 like the left side, we cannot apply the previous method. We apply logarithms on both sides:

x 2 + 7 = 2x – 2

Square both sides:

x 2 + 7 + 2 = 2x

3

b) x + 4 – 6 – x =

x 2 + 2x –2 c)

+ 9 – 7 = 2x d) 20 – x = x – 8

2 a) 3x – 5

b) 2x + 1

c) 4x

d) 2x

= 81 + = 272 4x + 2

e) 5x = 193

= 4 + 2x + 3 = 36

x 2

f ) 2

3

–2

= 835

8 Apply the definition of logarithm to calculate:

a) log2 (2x – 1) = 3

b) log2 (x + 3) = –1

c) log 4x = 2

d) log (x – 2) = 2.5

e) log (3x + 1) = –1

f ) log2 (x 2 – 8) = 0 67

1 EQUATIONS

Unit

ãã Equations of the type (…) · (…) · (…) = 0

Think and practise

In order for a product to equal 0, we just need one of its factors to be 0. Therefore, we can solve an equation of this type as long as each of the brackets presents an equation we know how to solve. For example: anayaeducacion.es •C onsolidate using factorisation to solve equations.

x 2(x 2 – 36) · (3 2x – 1 – 243) = 0. We make each of the brackets equal zero:

x 2 = 0 8 x = 0 (double root)

•C onsolidate solving problems with equations.

x 2 – 36 = 0 8 x 2 = 36 8 x = ±6

3 2x – 1 – 243 = 0 8 3 2x – 1= 35 8 2x – 1 = 5 8 x = 3

(x 2 – 2x – 15)(3 x – 1 – 27) = 0

x 2

– 2x – 15 = 0 8 x1 = –3, x2 = 5

In the previous unit we looked at the factorisation of polynomials, which allows us to express polynomials as a product of factors. We can then solve them directly.

For example, we factorised the polynomial P(x) = 10x4 – 3x3 – 41x2 + 12x + 4 in the previous unit: P(x)= 10x 4 – 3x 3 – 41x 2 + 12x + 4 = 10(x – 2)(x + 2)(x – 1/2)(x + 1/5) Now, the solution for P(x) = 0 is obvious: x1 = 2, x2 = –2, x3 = 1/2, x4 = –1/5

In the last school year you learnt that to solve a polynomial equation with a calculator, we have to press � and select A:Equation/Function. Then, when we choose 2: Polynomial, we are asked for the degree. We write the degree and we introduce the coefficients of the equation. Finally, we press = several times to obtain the solutions. ax4 +bx3 +cx2 +dx+e 3x3 10x4 + 12x + 4

10 =- 3 =- 41 =12 = 4 =

Every time we press = we get one solution.

b) (x 4 – 13x 2 + 36) c 1 + 12 – 10 m = 0 x x 9

68

a) 4x + 5 = x + 2

d) x 4 + 5x 2 + 4 = 0

41x2

10

10 Solve.

a) x 4 – 10x 3 + 5x 2 + 40x – 36 = 0 b) 9x4 + 18x3 – 28x2 – 2x + 3 = 0 Check the results with a calculator.

x 2 4 x – 5 2 x + 1 = 0 8 (2 ) – 5 2 x + 1 = 0 4 4 4 4 x x 2 We make 2 = t and, therefore, (2 ) = t 2. We get: t 2 – 5 t + 1 = 0 8 t 2 – 5t + 4 = 0 8 t = 4 and t = 1 4 4 x We put 2 back in: 2x = 4 8 x1 = 2; 2x = 1 8 x2 = 0

a) 23 + 2x – 3 · 2x + 1 + 1 = 0 b) 4x + 1 = 2 – 7 · 2x c) 3x + 1 + 6 · 3x = 3 b) x + 2 = x

15 Solve.

a) 6 – x = 3

b) 2x – 4 – 2 x – 3 = 0

c) 2x – 1 = 5

d) 2 x – 1 – 6 – x = 0

16 Solve.

17 Solve.

x –1 = 1 x –1 = 1 b) 2 4 x + 6x x 2 + 6x 2x 1– x = x +3 c) 3 x –1 +2= 5 3 d) 10 + x x –1 18 Solve.

2 a) 4 x – 2x – 8 = 1 b) 32x – 1 = 27 1024 a)

c) 2x + 1 + 2x + 3 = 320

Solve this equation: 4 x – 1 – 5 · 2 x – 2 + 1 = 0 We express it as a function of 2x. To do this, we have to keep in mind that 4x = (22)x = (2x)2.

21 Solve these equations:

c) ` x – x + 2j` x – 3j` x + 3j = 0

19 Solve.

Think and practise

a) ` x – x + 2j` x – 3j` x + 3j = 0

f ) 7x 2 – 3x + 4 = 0

c) x – 2 – 12 – x = 2 d) x – 5 + x =5

a calculator

9 Solve the following equations:

e) 10x 2 + 9x = 5.2

13 – x 2 + x = 5 a) x + 4 + 7 = 2x b)

ãã Solving equations of up to fourth degree with

For example, let’s enter the polynomial from the section above:

d) 6x 2 – x – 1 = 0

14 Solve.

ãã Polynomial equations with a degree higher than 2

When solving a polynomial equation with a calculator, you may get solutions which contain the letter i.

These solutions are not valid.

For example, when we solve the equation x3 – 3x2 + x – 3 = 0, we get x1 = 3, x2 = i, x3 = –i. The equation only has one solution, x = 3.

c) 5x 2 – 7x = 0

2 a) x + 7 + x 2 – 3x + 6 = 1 x + 3 x + 2x – 3 b) 2x + 1 + x – 1 = 2 x x – 2x

Solutions: x1 = –3, x2 = 5, x3 = 4

Invalid solutions

b) 3x 2 + 48 = 0

13 Solve.

3x – 1 – 27 = 0 8 3x – 1 = 33 8 x – 1 = 3 8 x = 4

When using a calculator to solve polynomial equations, it is important to select the mathematical input and output setting (1:MathI/MathO), so that your results are expressed as an exact decimal, not an approximation.

a) 3x 2 – 48 = 0

c) 4x 4 – 5x 2 + 1 = 0

When we make each of the brackets equal zero, we get two equations:

20 Problem solved

a) 7x 4 = 63x 2 b) x 4 – 10x 2 + 9 = 0

Problem solved

Attention

11 Solve these equations:

12 Solve the following equations:

Solutions: x1 = 0 (double), x2 = 6, x3 = –6, x4 = 3 Solve the following equation:

3

d) 2.5x = 49

22 Solve.

a) log7 (5x + 6) = 2

c) log ` x – 3j = –1

b) log3 (2 – 3x) = 0 d) log2 (x 2 – 3x) = 2

23 Solve.

a) log3 x + log3 (x + 2) = 1 b) log2 (x – 3) + log2 (x – 4) = 1 24 Problem solved

Solve the equation log9 4 + log3 (x + 4) = 2log3 x. log 3 4 We express log9 4 as , which gives us: log 3 9 log 3 4 + log3 (x + 4) = log3 x2 8 log 3 9 log 3 2 2 + log3 (x + 4) = log3 x2 8 2 8 log3 2 + log3 (x + 4) = log3 x2 8 8

8 log3 2(x + 4) = log3 x2 8 2(x + 4) = x2 8 8 x 2 – 2x – 8 = 0 8 x1 = 4, x2 = –2 x = –2 is not a valid solution because log3 x does not exist (in the initial equation) if x is negative. Solution: x = 4

a) 5x2 – 1 = 1

25 Solve the equation log5 x – log25 (x + 6) = 0.

b) 3x + 2 = 1

26 Solve, keeping in mind that logx 5 =

c) 92x + 1 – 27x = 0 d) 5x – 1 – 25x + 1 = 0

a) log5 x = 4logx 5 + 3 b) log3 x = 9logx 3

log 5 5 = 1 . log 5 x log 5 x

c) log3 x + 2 = 3logx 3

69

Unit

2

SYSTEMS OF EQUATIONS

ãã Systems of non-linear equations

ãã Systems of linear equations

The methods we know for solving systems of linear equations, as well as what we know about solving non-linear equations, allow us to solve many different types of systems almost without any additional effort.

Let’s review systems of equations and how to solve them.

anayaeducacion.es Review what you already knew about the number of solutions of a system of equations.

When we have to eliminate a square root (by squaring) or some denominators (by multiplying by their lowest common multiple) while solving a system, a false solution may appear. Therefore, in such cases, we need to check that all the solutions fit the initial system of equations.

Two equations form a system of equations when our aim* is to find their common solution.

If both equations are linear, it is called a linear system.

F ocu s on Eng lish

*

aim: the result you want to achieve by doing something.

ax + by = c a'x + b'y = c'

Problems solved 1 Use the most suitable method

There are three methods for solving systems of equations: substitution, equalisation and reduction. Let’s review some of them.

to solve the following.

*

y – x =1 x2 + y2 = 5

Problems solved 1 Solve this system:

3x + 2y = 7 * 5x – y = 16

2 Solve.

anayaeducacion.es x – ya5ysabías = –3sobre núRepasa lo6que * mero de soluciones 3x + 2y =de12un sistema de ecuaciones.

3

*

y – x =1 We use the substitution method. + y2 = 5

x2

Isolate y in the first equation: y = x + 1 Substitute it in the second equation: x 2 + (x + 1)2 = 5

Isolate the y in the second equation: y = 5x – 16

Develop the brackets: x 2 + x 2 + 2x + 1 – 5 = 0

Substitute it in the first equation: 3x + 2(5x – 16) = 7 (substitution method) Substitute in the isolated y equation: y = 5 · 3 – 16 = 15 – 16 = –1

Group: 2x 2 + 2x – 4 = 0 –2 ± 4 + 32 –2 ± 6 = Solve: x = 8 x1 = 1, x2 = –2 4 4 If x = 1, then y = 1 + 1 = 2.

Solution: x = 3, y = –1. Or (3, –1).

If x = –2, then y = –2 + 1 = –1.

Multiply the 2nd equation by 2 to equal the coefficients of x :

Solutions: *

Solve the equation: 3x + 10x – 32 = 7 8 13x = 39 8 x = 3

6x – 5y = –3 4 Subtract, 2nd – 1st: 9y = 27 8 y = 3 (reduction method) 6x + 4y = 24

2 Solve the following system:

*

Substitute in one of the two initial equations: 6x – 5 · 3 = –3 8 x = 2 Solution: x = 2, y = 3. Or (2, 3).

x 2 + y 2 = 58 x 2 – y 2 = 40

*

x 1 = 1, y 1 = 2 x 2 = –2, y 2 = –1

x 2 + y 2 = 58 Use the reduction method. x 2 – y 2 = 40

Adding both equations, we get: 2x 2 = 98 8 x 2 = 49 8 x = 7, x = –7 If x = 7, then y 2 = 58 – 7 2 = 9 8 y = 3, y = – 3

ãã Solving linear systems with a calculator anayaeducacion.es Consolidate solving systems of equations.

If x = –7, then y 2 = 58 – (–7)2 = 9 8 y = 3, y = – 3 Z ] x 1 = 7, y 1 = 3 ]] x = 7, y = –3 2 2 Solutions: [ ] x 3 = –7, y 3 = 3 ] x 4 = –7, y 4 = –3 \

To solve a system of linear equations using a calculator press � and select A: Equation/Function. Then, choose 1:Simul equation and select the number of unknowns. Finally, enter the coefficients and the independent term of each equation and press = twice, first to get the – – value of x, and then to get the value of y. { 6x3x + 5y2y == 123 12

Think and practise

Think and practise

1 Solve the following systems:

x + 5y = 7 a) * 3x – 5y = 11

70

5x + y = 8 b) * 3x – y = 11

anayaeducacion.es Solving systems of non-linear equations.

2 Solve the following systems:

22x + 17y = 49 3x + 10y = 6 x + 5y = 7 3x – 5y = –26 c) * d) * e) * f) * 31x – 26y = 119 x + 2y = 1 3x – 5y = 11 4x + 10y = 32

a) *

x – y = 15 x 2 + y 2 = 41 x 2 + xy + y 2 = 21 b) * 2

c) * xy = 100 x – y2 = 9 x + y =1

71

2 SYSTEMS OF EQUATIONS Unit

Problems solved 1 Solve the following system:

Use the substitution method, replacing the x in the second equation with the value of x in the first equation:

x = 2y + 1 * x+y– x – y=2

3 Solve the following system:

2x – 2 – 3 y –1 = 5 * x +1 2 – 3 y + 2 = –17

(2y + 1) + y – (2y + 1) – y = 2 Simplify:

3y + 1 – y + 1 = 2

Isolate a root:

3y + 1 = 2 + y + 1

Square: 3y + 1 = 4 + ( y + 1) + 4 y + 1 Simplify: y – 2 = 2 y + 1

x 2x = z; therefore, 2x – 2 = 2 2 = 2 y 3y = t; therefore, 3y – 1 = 3 = 3 The system becomes:

z ; 2x + 1= 2 · 2x = 2z 4 t ; 3 y + 2 = 32 · 3 y = 9 · 3 y = 9t 3

Square: y 2 – 4y + 4 = 4( y + 1)

We isolate z = 77 – 5t and substitute it into the second equation:

Simplify: y 2 – 8y = 0 8 y1 = 0, y2 = 8

2(77 – 5t) – 9t = –17 8 154 – 10t – 9t = –17 8 19t = 171 8 t = 9

Find the corresponding values for x :

z = 77 – 5t = 77 – 45 = 32

If y1 = 0, 8 then x1 = 2 · 0 + 1 = 1

Now we reverse the change of variables:

If y2 = 8, 8 then x2 = 2 · 8 + 1 = 17 x = 1, y = 0 Possible solutions: x = 17, y = 8 Now, check that these possible solutions fit the initial system of equations: — x = 1, y = 0 is true for the 1st equation, but not for the 2nd. It is not a solution.

— x = 17, y = 8 is true for both equations. It is the solution.

z = 32 8 2 x = 32 = 2 5 8 x = 5 4 8 Solution: x = 5, y = 2 t = 9 8 3 y = 9 = 32 8 y = 2 4 Solve this system:

*

log 5 x 5 – log 5 y 4 = –2 log 5

(xy 2) = 8

Solution: The only solution for the system is x = 17, y = 8.

We apply the properties of logarithms: 5 log 5 x – 4 log 5 y = –2 4 We change the variables: log5 x = z; log5 y = t log 5 x + 2 log 5 y = 8 The system becomes: We have multiplied the second equation 5z – 4t = –2 5z – 4t = –2 4 8 4 by 2 so that we can apply the reduction z + 2t = 8 2z + 4t = 16 method.

Begin by simplifying the first equation: y + x xy – 1 = 8 y + x = xy – 1 xy xy

Z ]] 1 + 1 = 1 – 1 xy [x y ] xy = 6 \

We change the variables:

z – t =5 3z – 4t = 60 4 8 We subtract: z + 5t = 77 4 3 4 8 2z – 9t = –17 2z – 9t = –17

Simplify and isolate the root: 2y – 4 = 4 y + 1

2 Solve.

3

We add them together: 7z = 14 8 z = 2; and substitute 2t = 8 – 2 8 t = 3 Finally, we reverse the change of variables and get the solution:

y + x = xy – 1 The system then looks like this: * xy = 6

z = 2 8 log 5 x = 2 8 x = 5 2 = 25 t = 3 8 log 5 y = 3 8 y = 5 3 = 125

Substitute the value of xy of the 2nd equation in the 1st equation:

4 8 Solution: x = 25, y = 125

y+x=6–1 8 y=5–x You can now substitute this value into the second equation: x1 = 3 x (5 – x) = 6 8 –x 2 + 5x = 6 8 x 2 – 5x + 6 = 0 x2 = 2

anayaeducacion.es Consolidate solving systems of non-linear equations.

If x1 = 3 8 y1 = 5 – 3 = 2 8 Solution: x1 = 3, y1 = 2 If x2 = 2 8 y2 = 5 – 2 = 3 8 Solution: x2 = 2, y2 = 3 Check that both solutions are valid.

72

*

2x – y – 1 = 0 x2 – 7 = y + 2

4 Solve the following systems of equations:

Z ]] log 2 x 2 – log 2 y = 2 x + 1 y – 2 3 +4 =5 a) * x + 2 b) [ x3 3 + 4 y –1 = 19 ] log 2 e y o = 6 \ x + 1 y – 2 2 +5 =5 log x 2 + 2 log y = 2 c) * x d) * 4 – 5 y – 1 = –1 x – y =3

5 Solve.

Think and practise

3 Solve: a)

Think and practise

b) *

x + y = 18 xy = y + 6x + 4

c) *

y + 8 = x2 y – 2x = 0

Z ]] 2 + 3 – 6 = 1 d) [ x y xy ]x + y = 5 \

e) *

x – y – x =2 5x = 4y

Z ] x + y – 2 (x – y) = ] 3 5 [ ]] 3x + y + 2 (x – y) = 2 \

11 5 –9 2

6 Solve these systems of non-linear equations:

a) *

x2 + y2 = 5

x2 – y2 = 3

b) *

x + y=2 x 2 + 2y = 7

c) *

2x + y = 3

x2 – y2 = 0

d) *

3x + y = 1 xy = –2

e) *

3x 2 – 5y 2 = 30

x 2 – 2y 2 = 7

f) *

Z ] 1 + 1 = 17 ] x2 y2 g) [

]] e 1 + 1 o = 9 x y \

x + y =5 x – y =1

Z ] 2 + 3 =1 ] x +1 y +1 h) [ ]] 4 – 1 = 1 x –1 y–3 \ 73

Unit

3

INEQUATIONS WITH ONE UNKNOWN

ãã Solving inequations by graphing Let’s see how we can use graphs to solve the three inequations, a), b) and c), on the previous page.

Sometimes, the statements that result in an algebraic expression do not say ‘is equal to’, but ‘is greater than’, ‘is less than’, ‘is greater than or equal to’ or ‘is less than or equal to’. For example:

a) 2x + 4 > 0

For which values of x is 2x + 4 greater than 0?

y = 2x + 4

The length of a table is 141 cm. If I measure it with my hand, 6 hand spans fall short of the end of the table. What does that tell me about my hand span?

In other words: For which values of x does line y = 2x + 4 lie above the X axis?

2x ++ 44 yy == 2x

Let’s translate that statement into algebraic language:

My hand span is the unknown. We will call it x.

Six times my hand span does not reach 141 cm 5 6x < 141

3

x > –2

If we look at the graphed line on the margin, the answer is clear: x > –2.

–2 xx >> –2

This is, any number greater than –2 is the solution.

The set of solutions is therefore (–2, +@).

y = –2x + 7

This expression is called an inequation.

b) –2x + 7 ≤ x – 3 2 Line y = –2x + 7 lies below or on line y = x – 3 for values of x greater than 2 4 and for 4 itself.

yy == –2x –2x ++ 77 x –3 y=— 2 xx – 3 — –3 yy == — x 22Ó 4

According to this, an inequation is a statement in which an inequality appears: For which values of x is it true that one thing is less (or greater) than another?

The answers to this question are the solutions of the inequation.

The solution to this inequation is any number x ≥ 4.

xx Ó Ó 44

Inequations generally have infinite solutions (there is only one number equal to another, but there are infinite numbers that are less than the mentioned number).

The solutions, in our case, are obtained in this way: x < 141 = 23.5 5 My hand span can be any length less than 23.5 cm.

6

The set of solutions is the interval [4, +@).

y = 2x – 3

c) –x 2 + 4x > 2x – 3

yy == 2x 2x –– 33

The parabola y = –x 2 + 4x lies above line y = 2x – 3 for values of x between –1 and 3.

y = –x 2 + 4x

The set of solutions for this inequation is the interval (–1, 3).

2 yy == –x –x 2 ++ 4x 4x

An inequation is an algebraic inequality. It has two sides with one of the following symbols between them: <, ≤, >, ≥.

–1 < x < 3

To use graphs to solve an inequation with a single unknown, f (x) ≤ g (x) or f (x) ≥ g (x):

–1 –1 << xx << 33

For example, the following expressions are inequations: a) 2x + 4 > 0 b) –2x + 7 ≤ x – 3 c) –x 2 + 4x > 2x – 3 2

1. Draw graphs y = f (x) and y = g (x), clearly marking the points of intersection.

2. Observe in which intervals the desired inequality is true.

Any value of the unknown that makes the inequality true is called a solution.

Solving an inequation is giving all of its solutions.

For example, x = 5 is the solution to inequation a), because 2 · 5 + 4 = 14 is greater than 0. As you can see, x = 5 is also a solution to inequation b), but not of c).

Think and practise

1 Give two integer solutions for each of the following

inequations: a) 3x < 50

b) 2x + 5 ≥ 25

c) 7x + 4 < 19

d) x 2 + x < 50

c) 1.1

d) 2

e) 5 f ) 3.2 g) 5.3 h) 10 2

74

Translate these statements into algebraic language:

b) The total number of students in my class is less than 35. c) If my money tripled and I also won another €20, I would have at least €110. d) I still have 20 monthly payments left on my mortgage. In other words, at least €6 000.

anayaeducacion.es GeoGebra Solving inequations by graphing.

4 Use graphs to solve the following inequations:

a) The triple of a number plus eight units is less than 20.

2 Which of the following values are solutions to the

inequation x 2 – 8x < 12? a) –5 b) 0

3

Think and practise

5

a) 3x > 9

b) 3x ≥ 9

c) 3x + 2 < 11

d) 3x + 2 ≥ 11

e) 2x – 3 < 5

f ) 2x – 3 ≤ 5

6 Solve the following inequations by graphing in your

notebook, bear in mind the representation of the function y = x 2 – 5x + 4: a) x + 4 ≤ x 2 – 5x + 4 b) –x + 4 < x 2 – 5x + 4

Look at the following dialogue: — How often have you been to a football match?

c) x 2 – 5x + 4 < x – 1

— Triple the number of times plus 2 does not reach 10.

d) x 2 – 5x + 4 ≥ x + 1

Express the answer in algebraic language, solve it and then, provide the solutions bearing in mind that they have to be non-negative integers.

e) x 2 – 5x + 4 < – 6 + 2x

f ) x 2 – 5x + 4 ≤ –2

75

3 INEQUATIONS WITH ONE UNKNOWN

Unit

ãã Using algebra to solve an inequation

ãã Systems of inequations

We are going to use algebra to solve the three inequations, a), b) and c), which we solved by graphing on the previous page.

Re-read the problem with which we started this section (page 74):

The length of a table is 141 cm. If I measure it with my hand, 6 hand spans fall short of the end of the table. What does that tell me about my hand span?

subtract 4 divide by 2 a) 2x + 4 > 0 ÄÄÄÄÄ8 2x > – 4 ÄÄÄÄÄÄÄ8 x > –2

Solutions: x > –2. Interval (–2, +@).

If we call my hand span x, the above statement is expressed algebraically as follows: 6x < 141. Its solution is x < 23.5.

As you can see, we performed the same operations that we perform when solving equations.

: ax + bx + c>0 : ax + bx + c<0 : ax + bx + c>0 – : ax + bx + c<0 – 2 2

For example, let’s solve the inequation from section c) with a calculator. x2 – 2x – 3 < 0 1 =-2 =-3 = –1 < x < 3

again, we get

The solutions of a system of inequations are the solutions that are common to all the inequations that form the system.

Solve the following system of inequations:

By studying the sign of x 2 – 2x – 3 in a value of each section, we can see whether or not the inequality in it is true.

02 – 2 · 0 – 3 = –3

8 Yes

42 – 2 · 4 – 3 = 5

8 No

No

Yes –2 –1

0

1

2

3

d) –2x + 7 ≥ x – 3 2 f ) –x 2 + 4x < 2x – 3

8 Solve the following using algebra:

a) 3x – 5 ≥ 13

b) 5x + 1 < x + 9

c) 3 – 2x > x + 5

d) 7 – 11x + 2 ≤ 23 + 4x

e) x 2 9

x –3 2

2nd inequation: –2x + 7 ≥ x – 3 8 – 4x + 14 ≥ x – 6 8 –5x ≥ –20 8 2

8 x ≤ –20 = 4. Solution: (–@, 4] –5 Solutions to the 1st inequation –2

4

anayaeducacion.es Reinforce your knowledge on how to solve inequations.

Notice how they are very similar to those we have solved above: a) 2x + 4 ≥ 0 b) 2x + 4 < 0

2x + 4 > 0

*–2x + 7 ≥

1st inequation: 2x + 4 > 0 8 2x > – 4 8 x > –2. Solution: (–2, +@)

No

We conclude that the solutions form the interval (–1, 3).

7 Solve the following inequations in algebraic form.

76

Conclusion: my hand span measures between 20.1 cm and 23.5 cm.

Problem solved

c) –x 2 + 4x > 2x – 3 8 –x 2 + 2x + 3 > 0 8 x 2 – 2x – 3 < 0

(–2)2 – 2(–2) – 3 = 5 8 No

6x < 141 x < 23.5 its solution is * .

7x > 141 x > 20.1

In other words, x é (20.1; 23.5).

In order to solve a first-degree inequation, we proceed as if it were an equation, but with the following difference: if we multiply or divide by a negative number, the sign in the inequality changes.

Think and practise

c) –2x + 7 > x – 3 2 e) –x 2 + 4x ≥ 2x – 3

*

very important! When multiplying or dividing both sides of an inequality by a negative number, the sign in the inequality changes.

We define three sections: (– @, –1), (–1, 3) and (3, + @).

2

=

Of the length, x, of a hand span, we can say that:

divide by –5 8 –5x ≤ –20 ÄÄÄÄÄÄÄÄÄ8 x ≥ –20 = 4 . Interval [4, +@). on both sides –5

We find the roots of the second-degree expression: x 1 = –1, x2 = 3.

2

When we press –1 < x < 3

In other words, 7x > 141. The solution of this inequation is x > 20.1.

move x to the left side –4x + 14 ≤ x – 6 ÄÄÄÄÄÄÄÄÄÄÄÄ8 –4x – x ≤ –6 – 14 8 and 14 to the right side

Many calculators can help us solve polynomial inequations with a degree of up to 4. To access this function, we press �, select B:Inequality and select the degree. A screen then appears where we have to select the type of inequation. When doing so, it is important to keep in mind that the inequation should be ‘ordered’, in other words, one of its sides should be equal to zero.

1 2 3 4

However, if we look more closely, we see that, although not explicitly mentioned, we can presume that 7 hand spans will be longer than the table.

b) –2x + 7 ≤ x – 3 2 Multiply the whole inequation by 2 to remove the denominator:

Solving inequations with a calculator

3

– 3x + 2 ≤ 4x – 8

Solve the inequations a), c) and d) from exercise 3 on page 74 and interpret the solution.

0

4

Solutions to the 2nd inequation anayaeducacion.es Reinforce your knowledge on solving systems of inequations.

Solutions to the system

Solution: (–2, +@) Ü (– @, 4] = (–2, 4]

Think and practise

anayaeducacion.es GeoGebra Solving systems of inequations.

10 Solve the following systems of inequations:

a) *

x +3<7 2x – 3 < 3x + 5 2x – 3 < 3x + 5 b) c) * * x – 1≥ 0 7x + 1 ≥ 13 + 4x 7x + 1 ≥ 13 + 4x

d) *

x 2 – 7x + 6 ≤ 0 x 2 – 7x + 6 ≤ 0 e)

* 3x + 2 > 17 2x + 5 < 7

f ) *

x 2 – 7x + 6 ≤ 0 2x + 5 ≤ 7

77

EXERCISES AND

VED PROBLEMS SOL

r to choose Remembe . lio fo rt po

Your turn The base of a rectangle is 10 cm longer than its height. If the base is increased by 20 % and the height by 30 %, the perimeter increases by 24 %. Find the dimensions of the rectangle.

Your turn On a 450-km journey, the velocity of the outward journey was 15 km/h less than the return journey and it took an hour longer. Find the velocity of each journey and the time they took.

If the base, b, is increased by x %, it will be multiplied by b1 + x l.

100 If the height, a, is increased by 2x %, it will be multiplied by c1 + 2x m.

100 If the area, b · a, increases by 32 %, it will be multiplied by c1 + 32 m.

100 Therefore: b b1 + x l · a c1 + 2x m = b · ac1 + 32 m 100 100 100 We divide both sides of the equation by b · a and we obtain:

2x + 3 ≥ 1 x–2

Your turn Solve this inequation: x – 4 ≥1 x–2

78

7

b1 + x l c1 + 2x m = c1 + 32 m 8 (100 + x)(100 + 2x) = 13 200 100 100 100 x = 10 10 000 + 300x + 2x 2 = 13 200 8 x 2 + 150x – 1 600 = 0 x = –160 Doesn’t work.

Therefore, the solution we seek is x = 10.

Equations

1

(x – 1) (x + 2) x – 3 (x + 1) (x – 2) – =1+ 12 3 6 2 2 2 2 b) (x + 1) – (x – 2) = (x + 3) + x – 20 c) 2

The distance travelled is the same in both cases, therefore:

3

vt = 300 3 v = 10t – 10 8 (10t – 10) t = 300 8 10t 2 – 10t – 300 = 0 8 10t – v = 10 t = 6 8 v = 300 : 6 = 50 1 ± 1 + 120 1 ± 11 = 8 t 2 – t – 30 = 0 8 t = 2 2 t = –5 Doesn’t work.

(–5, 2)

(2, +∞)

+ –

+ +

–

+

x

+5

x

–2

– –

x+5 x–2

+

Furthermore, since x – 2 is in the denominator, x = 2 is not part of the solution, but x = –5 is. Therefore, the solution is (–∞, –5] á (2, +∞).

8

a) x 4 – 4x 2 + 3 = 0

b) x 4 – 16 = 0

c) x 4 – 25x 2 = 0

d) x 4 – 18x 2 + 81 = 0

4

c) 2x + 5x – 6 = 4 d) 5x + 1 – x + 1 = 2 9

5

2

c) 101 – x = 0.001 10

3x 2 – 2 = x + 2 x e) 23x = 1 + f ) x 2 (x – 3) x –9 2x 2 – x 6

Solve. a) x + 25 – x 2 = 2x + 1 b) 3x + 6x + 10 = 35 4x 2 + 7x – 2 = x + 2 c) x + 1 – 5x + 1 = 0 d)

x

d) 81 c 1 m = 3 x + 2 3

Solve. a) 3 · 5x + 5x + 1 = 200 b) 7 · 2x – 1 – 5 · 2x = – 3 4 x + 1 x – 1 c) 2 · 3 +3 – 5 · 3x = 108 d) 2x – 1 + 2x – 2 + 2x – 3 = 224

11

Solve by applying the definition of logarithm. a) log5 (2x – 3) = 1 b) log 4c x + 1 m = –2 2 x c) log 2 ( x – 1) = 3 d) log (2 – 15) = 0

12

Apply the properties of logarithms to solve the following equations: a) log3 x2 – log3 4 = 4 b) log2 x – log2 3 = 2 c) log2 (x – 3) + log2 x = 2 d) log (x – 9) – log x = 1

13

Break down into factors and solve. a) x 3 – 4x = 0 b) x 3 + x 2 – 6x = 0 c) x 3 + 2x 2 – x – 2 = 0

Solve the following equations: 3x + 1 – 1 = 3 a) x + 1 – 3 = 2 – x b) x –1 x 4x + 3 x 1 – 2 = 2 – 5x c) 3x + 4 – 1 = x + 19 d) x + 3 x x 2 + 3x x +3 2 4x + 6

Solve the following exponential equations: 3 x = 17 a) 2x + 1 = 8 b)

Solve. x – 4 – x – 1 = –3x a) x + 2 + 3x = 5x + 6 b) x 2 x 4x x – x – 1 = 3x – 1 c) x – 3 + x +2 3 = 2 d) x 3 x +1 2 x

Solve. a) x + 7 – 3x = –1 b) x + 3x – 2 = 2

Solve.

e) (2x 2 + 1)2 – 5 = (x 2 + 2)(x 2 – 2)

x+5 2x + 3 2x + 3 2x + 3 – x + 2 ≥1 8 – 1≥ 0 8 ≥0 8 ≥0 x –2 x –2 x –2 x –2

(–∞, –5)

c) 5x – 7 – 1 – x = 0 d) 2 5 – 4x + 4x = 5

2

d) 2cx + 1 m + 25x = c 1 – x m(7x + 1) – 4 2 2 2

vt = 300 We develop the second equation and simplify by 3 (v + 10) (t – 1) = 300 replacing vt with 300:

We analyse the sign of the numerator and the denominator. The numerator changes to x = –5, and the denominator changes to x = 2. The solutions are the values of x for which both have the same sign.

x2 + 3 – 3 – x = 0 a) x – 17 = 169 – x 2 b)

Solve the following equations: (x – 3) 2 (2x – 1) 2 35 – = 16 16 4 2 (x + 1) (x – 1) 2 2 + x b) – 1+ x = – 2 16 16 4 c) (x + 1)2 = x (5x + 6) – (2x 2 + 1) 2

On the return journey, the velocity is v + 10 and the time is t – 1.

He travels at 50 km/h and takes 6 hours on the outward journey. He travels at 60 km/h and takes 5 hours on the return journey.

x ( x – 2) x + 1 x – 3 x – 4 – = – 2 3 6 4

a)

On the outward journey, the velocity is v and the time is t.

vt + 10t – v – 10 = 300 8 10t – v – 10 = 300 – vt 8 10t – v – 10 = 0

Solve the following equations. Solve those that are incomplete second-degree equations without applying the general formula.

Two of the following equations have no solution.

Work out which they are and solve the others.

a)

3 Solving a rational inequation

Solve the following inequation:

Unit 3

ROBLEMS

Practise

2 Studying and solving a system

A lorry driver travels to a city 300 km away. On his return journey, his average velocity is 10 km/h above his velocity on the outward journey, and it takes him one hour less. Calculate the velocity of each journey and the time each one took.

for your

P EXERCISES AND

1 Studying and solving an equation

If the height of a rectangle is increased by x % and the base by 2x %, its area increases by 32 %. Find the value of x.

m this unit resources fro

14

d) x 3 – x 2 – 5x – 3 = 0

Solve the following equations: a) (x – 2)(x 2 – 2x – 3) = 0 b) x (x 2 + 3x + 2) = 0 c) (x 2 – 3x)(2x + 1 – 1) = 0 d) (x + 5) log2 (x – 3) = 0 e) (x 4 – 5x 2 + 4)(5x – 10) = 0 f ) (x 2 + 5) ( x – 3) = 0

15

Isolate the unknown and solve. a) x 3 – 64 = 0 b) 625 – x 3 = 0 x x – 2 =0 c) 3x + 162 = 0 d) 8 4 9x 81x 3 79

Unit 3

EXERCISES AND PROBLEMS Systems of equations

Inequations and systems of inequations

Problem solving

16

23

31

Solve the following systems by applying the reduction method twice: a) *

13x – 12y = 127 8.6x + 5.4y = 11 b) * 21x + 17y = 96 25x – 12y = –245

17

Solve the following systems of equations: Z Z ] x +1 + y = 5 ] x + 2 – 3y – 1 = –3 ] 8 ] 5 10 10 4 8 a) [ b) [ ] 3x – 1 + y = 1 ] 2x + 3 + y + 7 = 19 ] 12 ] 8 8 2 6 4 \ \ 18 Solve. x – y +3=0 x + y =1 a) * 2 b) * x + y2 = 5 xy + 2y = 2 c) * 19

2x + y = 3 3x + 2y = 0 d) * 2 xy – y = 0 x (x – y) = 2y 2 – 8

Solve the following systems of equations: x 2 + y 2 = 41 3x 2 + 2y 2 = 35 a) * 2 b) * x – y2 = 9 x 2 – 2y 2 = 1 + + x + y = 32 c) * 2 x – y 2 + x – y = 28 x2

20

y2

24

25

log x – log y = 1 2x + 1 – y = 0 c) * x d) * x + y = 22 2 + y = 12 80

b) x 2 – 3x – 10 ≤ 0

c) x 2 – 4x – 5 < 0

d) 2x 2 + 9x – 5 ≥ 0

27

Solve. b) –x 2 + 2x + 3 ≤ 0

Solve the following inequations: a) 3x(x + 4) – x(x – 1) < 15 b) 2x(x + 3) – 2(3x + 5) + x > 0 2 2 c) x – 9 – x – 4 < 1 – 2x 15 3 5

28

Solve these systems of inequations: 2 – x >0 5x – 3 ≤ x + 1 b) * 2+x >0 2x + 6 ≥ x + 2 Z Z ] x + 13 < 39 – 2x ] 2x + 5 < x – 1 ] 3 ] 6 18 c) [ d) [ ]] x – 1 < 2x – 1 ]] 3x – 5 < –1 3 5 \ \ 4 a) *

29

Translate into algebraic language and solve: a) Half of a number minus 10 is less than 7. b) If I subtract 2 from three quarters of a number, I get more than if I add 5 to half of that number. c) The sum of two consecutive numbers does not exceed 8. d) The perimeter of a rectangle whose base measures 3 cm more than its height is less than 50 m long.

Solve the following systems of inequations: Z Z ] x – 1 + 2x + 2 > 3x – 7 ] x+2 < x –3 ] 4 ] 2 3 2 6 a) [ b) [ ]] 8 – x < 1 + x – 1 ]] 2x – 1 + 2x < 2x – 9 3 2 4 \ \ 4 30 Check that these two systems of inequations have no solution: 3x + 5 < 2x – 3 8x + 7 < 16 – x a) * b) * x +3 <x –3 –3x + 5 < 2x 7

Solve these systems of inequations: Z Z ]x – y = 0 ]x – z = 4 ] ] a) [ x – 2z = 6 b) [ 2x + y = 7 ]] y + z = 3 ]] x + y = 2z \ \

33

Problem solved

Solve the equation x6 + 7x3 – 8 = 0 The equation only has terms with the degrees 6, 3, and 0. We can do the change of variable t = x3, like we did for biquadratic equations. t=1 t2 + 7t – 8 = 0 8 t = –7 ± 9 t = –8 2 We reverse the change of variable: t = 1 8 x3 = 1 8 x = 3 1 8 x = 1 t = –8 8 x3 = –8 8 x = 3 –8 8 x = –2 Solutions: x1 = 1, x2 = –2 34

Solve the following equations: a) x6 – 2x3 + 1 = 0 b) x6 + 19x3 – 216 = 0 c) x8 – 2x4 + 1 = 0 d) x12 + x6 – 2 = 0

35

ax2

Given that x = 2 is a solution for the equation – 5x – 6 = 0, find the other solution.

36

Solve the following equations: a) 4x + 1 + 2x + 3 = 320 b) 4x – 8 = 2x + 1 c) 23 – x = 5 – 2x – 1 d) 3 · 4x + 9 · 2x – 30 = 0 e) 9x + 1 = 243 + 234 · 3x

37

Solve and check the solutions. a) log (x – 2) + log (x – 3) = 1 – log 5 b) 1 log (3x + 5) + 1 log x = 1 2 2 c) 2log x – 3log 2 = log (x + 6)

Problem solved

Solve the equation 2logx 8 – logx 4 = 2.

82 = 2 8 logx 16 = 2 8 4

8 x2 = 16 8 x = ±4 We can discard the solution x = –4 because the base of a logarithm cannot be negative. Therefore, x = 4. logx 82 – logx 4 = 2 8 logx

39

Solve. a) logx (3 – x) + logx 4 = 2 c) 3logx 2 + logx 18 = 2

32

c) x 2 + 7 < 5x d) x 2 + 4x + 4 > 0

21

Solve. xy = 15 xy = 12 a) * 2 b) * 2 2 x + y = 34 x – 5y 2 = 16 Z ]] x 2 + y 2 = 82 xy = 4 c) * d) 9 [ (x + y) 2 = 25 ] xy = –1 \ 22 Solve. x – y =1 log x + log y = 1 a) * x b) * y 2 –2 =4 x – y =9

a) x 2 + 2x – 3 > 0

c) x 2 – 2x – 7 > 5 – x d) x2 < x + 7 6 26 Some inequations have no solution, while others have any number as a solution. Which of the following inequations are of each type? a) x 2 + 4 > 3 b) x 2 + x + 2 < 0

+ 2y 2

Solve and check the solutions.

Z Z ]] 1 + 1 = 1 ]] x + y = 2 a) [ 1 1 b) [ x y 20 2 ] x + 2y = 3 ]x + y =– 3 \ \ 2 2 x +1 = y +1 y – 2y + 1 = x c) * d) * 2x – 3y = 1 x + y =5

Solve the following inequations:

a) –x 2 + 3x – 2 ≥ 0

+ x +1= 0 d) * 2 x – 2y 2 + 3x + 1 = 0 x2

Solve. x + 4 + 3 x + 10 a) 7 – 3x < x + 1 b) ≥ 2 3 6 c) 2x – 2(3x – 5) < x d) x – 1 – x –1 <0 2

38

b) logx 25 + logx 5 = 3 d) logx 54 + logx 4 = 3

40

Solve the equation 5x – 5x – 2 = 120 √5.

41

Problem solved

Solve this system: *

(x – 2) (y – 1) = 0 ( x + 3) ( y – 5 ) = 0

Z ]] x = 2(*) 8 5 · (y – 5) = 0 8 y = 5 (x – 2)(y – 1) = 0 [ y = 1(*) 8 (x + 3) · (–4) = 0 8 ] 8 x = –3 \

(*) We

substitute the value into the other equation.

Solutions: x1 = 2, y1 = 5; x2 = –3, y2 = 1 42

Solve these systems of equations: Z ]] (x – 3) c y – 1 m = 0 x (x + 2) = 0 2 a) [ b) * 2 x + y=4 ] (x – 1) (y + 2) = 0 \ 43 Solve the following inequations: a)

3x – 3 ≥1 2x + 1

b)

2–x <1 3x – 4

c)

x 2 + x – 6 ≥0 3 – 2x

44

An integer multiplied by another integer two units higher, is less than 8.

What can the number be?

45

If we subtract three times a number from its square, we get more than 4. What can we say about that number?

46

Three friends earn €756 altogether for a job. The first of them worked for 12 hours; and the third, who worked twice as many hours as the second, earned €360. How many hours did each one work and how much did each one earn?

81

Unit 3

EXERCISES AND PROBLEMS The total area of a cylinder is 112π cm2, and its radius plus its height equals 14 cm. Find its volume.

47

48

The average mark in the maths exam taken by class 4 C was 5.4, and the average mark of class 4 B was 6.4. How many students are there in each class if there are 50 in total and their average mark is 5.88?

58

A farm has sheep and chickens. The total number of legs on the farm is 20 plus twice the number of heads. How many sheep are there?

59

We have a rectangular plot. If its base is reduced by 80 m, and its height is increased by 40 m, it becomes a square. If its base is reduced by 60 m, and its height is increased by 20 m, then its area is reduced by 400 m2. What are its dimensions?

49

The perimeter of a right-angled triangle is 36 cm, and one of its legs measures 3 cm less than the other.

Find the length of the sides of the triangle.

x – 80

67

x2

What values can a have in order for the equation + 2ax + 7a – 10 = 0 to have only one real root?

We know that x2 = 8x + y and that y2 = 8y with x ≠ y. Find the value of x2 + y2.

68

70

An athlete is at point A, in the sea, 120 m from the beach, BD, which is 1 510 m long.

52

53

An antiques dealer sold two pocket watches for €210. He made 10 % profit on one and lost 10 % on the other. In total he made 5 % profit. What was the purchase price of each of the watches?

Yago bought some books, each for the same price, and he paid €90. However, because he was a good customer, Sara, the bookshop owner, gave him 3 more for free, which meant that each book cost him €5 less. How many books did he take home and how much did he pay for each one?

Calculate the time it will take for a capital of €10 000 deposited in a bank to increase by 50 % in the following cases: a) At an annual interest rate of 4 %. b) At a monthly interest rate of 3.6 %.

54

A group of friends orders one drink each and they have to pay a total of €9. As there are two people who can only put €1 in, the others have to increase their contribution by €0.25 each. How many friends are there?

55

In order to fill a 36 m3 tank, we open a tap, A, for 2 hours and another tap, B, for 10 hours. If we want to fill just 28 m3 with these taps, we open A for 3 hours and B for 5 hours. How many litres per hour come out of each tap?

56

The side of a rhombus measures 5 cm, and its area is 24 cm2. Calculate the length of its diagonals.

57

82

The sum of the two figures of a number is 8. If we add 18 units to the number, the resulting number is formed by the same figures in the reverse order.

What is this number?

y + 40

A y

c) x 4 – 16 = 0

d) x 4 + x 2 = 0

e) x 4 + 3x 2 + 2 = 0

f ) x 4 – 4x 2 + 4 = 0

76

Look at the graphic representation of the lines y = 2 – x and y = 2x – 3: 2 x y=2–— 2

In order to reach point D, he swims to C at 40 m/min and walks from C to D at a speed of 90 m/min.

120 m

51

How many solutions can a biquadratic equation have? Check your answer by solving the following equations: a) x 4 – 10x 2 + 9 = 0 b) x 4 – 4x 2 = 0

Solve: (x2 – x)2 = 18(x2 – x) – 72

69

50

One person takes 3 hours longer than another to do the same job. If they do it together, they take 2 hours. How long does each one take separately?

75

Advanced problem solving

2 –2

2 y = 2x – 3

x

60

The hypotenuse of a right-angled triangle is 10 m, and its area is 24 m2. How long are its legs?

61

There is one unit of difference between the two figures of a number. If we divide this number by the number resulting from inverting the order of its figures, the quotient is 1.2. What is the number?

62

Find the radius and the generatrix of a cone that is 15 cm high and has a lateral area of 136π cm2.

B

65

A box contains black and white balls. If we add one white ball, they will represent 25 % of the contents of the box. If we remove one white ball, the white balls that remain will represent 20 % of the contents of the box. How many balls of each colour are there in the box?

66

We know that two fractions have the same numerator, their denominators are consecutive numbers, and the sum of both fractions is equal to 27/20. We also know that the sum of the numerator and the denominator of the smaller of the two fractions is 8. What are these fractions?

1 510 – x

D

77

Look at the representation of line y = –x – 1 and of the parabola y = x 2 – 2x – 3.

71

I climb a mountain at a speed of 4 km/h and want to ensure my average speed of the ascent and descent is 6 km/h. At what speed do I need to descend?

72

In an exam with 40 questions, you get two points for each correct answer and lose 0.5 points for each mistake. How many questions do you have to answer correctly to get a minimum of 40 points if it is mandatory to answer all of them?

How many kilograms of paint worth €3.50/kg do we need to mix with 6 kg of another worth €5/kg for the price of the mixture to be less than €4/kg?

C

Calculate the distances he travelled, both swimming and walking, if the total time it took was 20 minutes.

63

64

x

Is it possible to plant 275 trees on a rectangular plot of 72 m × 30 m in a way that they form a regular grid as shown in the figure? x

x

If so, work out the distance there needs to be between two trees in a row.

Answer without performing the operations: for which values of x is 2x – 3 ≥ 2 – x ?

2

y = x 2 – 2x – 3

Answer without performing the operations: for which values of x is x 2 – 2x – 3 < –x – 1?

y = –x – 1

78

Solve without performing any operation. x 2 – 4x + 3 ≤ x – 1 * 2 –x + 4x > 8 – 2x y = 8 – 2x

73

The surface areas of the faces of a cuboid are 35 cm2, 60 cm2 and 84 cm2. What is its volume?

Calculate the length of its edges.

y=x –1 1 1

Remember the theory 74

True or false? Explain why and give examples. a) If b 2 – 4ac = 0, the equation ax 2 + bx + c = 0 has no solution. b) If k < 1, the equation 9x 2 – 6x + k = 0 has two solutions. c) (x 2 + 5)(2x – 5) = 0 has two solutions. d) (x – 1)2 + (x + 1)2 – 2(x 2 + 1) = 0 has infinite solutions. e) Some systems of inequations have no solution. f ) An inequation always has infinite solutions.

y = x – 4x + 3 2

1 1 y = –x 2 + 4x

79

Write second-degree equations for which the solutions are: a) 2 and –3 b) 4 and 5 c) –2 and – 8 d) 2 and 1 3 Look at your equations and find a relationship between the coefficients a, b and c of each equation and the sum and product of its solutions.

80

Show that if x1 and x2 are the solutions of then x1 + x2 = – b and x1 · x2 = c . a a

ax 2 + bx + c = 0,

83

Unit 3

OP

MATHS WORKSH INVESTIGATE

PRACTICE MAKES PERFECT!

Diophantine problems

Did you know…?

Below are two problems that can be solved using Diophantine equations.

Diophantine equations are characterised by having natural numbers (and sometimes integers) as their solutions. They are named after Diophantus of Alexandria, a mathematician who lived in the 3rd century and who is considered to be the first algebraist.

These types of problems generally have several solutions.

If there is more than one solution, you have to find all of them. anayaeducacion.es How to solve Diophantine problems involving an equation with two first-degree unknowns. anayaeducacion.es Some other Diophantine problems.

Problem 1

Problem 2

A 4-cm leg of a sideboard has broken off.

On a test with 20 questions, you get 5 points for each correct answer, and lose 3 points for each incorrect answer and 2 points for each unanswered question.

We have several wooden disks we can use to balance the sideboard provisionally, some of them are 5 mm thick and some others are 3 mm thick. How many disks of each type will we need?

What should happen for you to get 0 points? And to get 50?

Try solving the problems below without using algebra.

• A driver leaves home at five in the afternoon to go to an appointment. He realises that if he travels at 60 km/h he will arrive a quarter of an hour late, but if he travels at 100 km/h he will get there a quarter of an hour before the appointment. What time is the appointment? What is the distance to his destination?

• In a training session, a tennis coach gives each player three balls and has 11 remaining. The next day, he takes 20 more balls to the session, meaning each player gets five and there is one remaining. How many players are there?

• A train is travelling at 300 km/h on a straight piece of track. A car is travelling at 120 km/h in the same direction on a parallel road.

• You and your friends are going to put in €6 each to buy a basketball for your friend Jordi. However, Iván and Julia cannot pay, so you will each need to put in €10.

How many of you are there in your group of friends?

UNANS

WERED

–2

CORRECT

+5

ECT

INCORR

–3

If it takes 4 s to fully pass the car, how long is the train?

USING ALGEBRAIC LANGUAGE

USE YOUR INTELLECT

Equalizing

Fruit

The eldest of three brothers wins the lottery. He is generous, so he decides to double the money both his younger brothers have. When he has done this, they realise that the richest of the three is now the middle brother. He is also generous and doubles the money of the other two.

What fruits can we swap for two pears?

SELF-ASSESSMENT

anayaeducacion.es Answer key and interactive self-assessment.

1 Solve the following equations:

5 A shopkeeper wants to sell the computers in the

1 – x +1 + 5 = 0 a) x + 1 – x = x – 7 b) x x –1 2 4 2 Solve. a) *

x =4– y xy = 15 b) * y2 = 4 + x 4x 2 – y 2 = 11

3 Solve.

a) 102x – 1 = 0.001

b) 25x = 500

c) 2x – 1 + 2x + 3 = 17 8 1 d) log 2 (3x + 3) – 1 log 2 (2x – 3) = log 2 2 2 2 Now it turns out that the richest one is the youngest, so he doubles the money of the older two. Finally! Now they are all equal, and each of them has €400.

How much did each of them have at the beginning?

84

4 Solve.

a) 3x 2

– 5x – 2 ≤ 0

Commitment

2x – 3 < 4 b) * 4 – x ≥ –1

warehouse for €60 000. However, two get broken and he has to increase the price of the others by €50 each to earn the same amount of money.

How many computers were there and how much were they sold for?

6 The diagonals of a rhombus added together measure

42 m and its area is 216 m2. What is the perimeter of the rhombus?

7 In a class there are 5 more boys than girls. You know

that there are more than 20 students, but there are not as many as 25.

What might the composition of the class be?

8 How many litres of wine at €5/L have to be mixed

into 20 L of another at €3.50/L in order for the price of the mixture to be less than €4/L?

Watch the video for target 5.2. Think of something you can do to contribute to achieve that goal. Make a commitment to put your idea into practice.

85

C N FU

S N O I T

2 K C O BL

4 5

CS

RISTI E T C A R A . CH

S

FUNCTION

ONS

NCTI U F Y R A T ELEMEN

Remember that you can find academic and professional guidance related to this content at anayaeducacion.es.

4

Contributions by great mathematicians

FUNCTIONS. CHARACTERISTICS

• Nicolas Oresme (14th century) described the laws of nature as relationships between dependent variables. Galileo (16th century) was the first to use experimentation to establish these numerical relationships.

• Descartes (17th century) used algebra to represent geometry, making it possible to graph functions.

Reading and listening

• Leibniz (17th century) used the word function to express these relationships for the first time in 1673.

Evolution of the concept of function

• Euler (18th century) made the concept more precise. His definition of function is similar to what we use today. He introduced the notation f (x).

The concept of function has evolved, giving rise to these characteristics: • A function links two variables.

• Dirichlet (19th century) accepted that a relationship between two variables can be a function even if there is no analytical expression to describe it.

• Functions describe natural phenomena.

• Functional relationships can be described using formulas (algebraic relationships).

At what speed do objects fall?

• Functions can be represented on graphs.

Aristotle’s beliefs (4th century BCE) were rarelyquestiones in 16th century Europe. According to him, if two bodies of different weight were dropped, the heavier one would reach the ground first. It is said that Galileo used a public experiment to disprove this: he dropped two metal objects of very different weights from the Tower of Pisa. They fell at the same speed. He succeeded in demonstrating his thesis, but he was expelled from the university.

An example of a function Sundials are imprecise because the Earth does not travel at a constant speed as it revolves around the Sun.

We don’t know if this story is true, but Galileo did experiments on bodies falling along an inclined plane, controlling the distances and times.

distance (cm)

This graph shows how many minutes fast or slow a sundial is over the course of a year.

250

We can use the graph to answer different questions:

200

15

MINUTES

10

• On what date is the time shown by the sundial furthest ahead of the actual time?

3N

5

15 M 16 A

14 J

25 D

J F M A M J J A S O N D

5 –10 –15 11 F

88 88

27 J

1 Answer the questions in the section ‘An example of a function’ using the

future tenses to make sentences.

2 In your opinion, which of these mathematicians are the most influential

to our modern day mathematics.

Solve

150

1 A ball rolls down a slightly inclined rail and the distance it travels in

• On what date is it furthest behind?

• On what dates is the time shown most accurate?

• What will happen the following year? And the year after that?

This behaviour is repeated every year.

Functions that repeat themselves in this way are called periodic functions.

different lengths of time is measured:

100

time in

s (t )

distance in

50

1

2

3

4

5

time (s)

cm (e)

0 0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

0 2.5 10 22 40 63 90 123 160 202 250

a) Plot the data above on a grid like the one on the left. Use it to find the corresponding curve. b) Check that the resulting values correspond (with a good degree of approximation) to the following relationship: e = 10t 2

ANK ANK B E G A U LANG LANGUAGE B ANK ANK GE BANK B B E E G G A A U U GUA K LANG ANG N L A L AN GE BANK 89 ANK GE BANK B B E E G G A A U U G A LAN LANG LANGUA LANGU

Unit

1

2

BASIC CONCEPTS y is the dependent variable.

via: by way of.

ãã As graphs

y = f (x ) In order to visualise the behaviour of a function, we graph it.

Axis of ordinates

Y

Observe

Coordinates of the point

Ordinate of the point y

The graph of a function lets us see its overall behaviour at a glance.

(x, y)

Y

Range of f

Functions appear in many different forms: as graphs, as tables of values, as formulas or as verbal descriptions (statements).

The relationship between the variables via* the f function gives each value of x a single value of y. It is expressed in the following way:

F ocu s on Eng lish

X x Abscissa of the point Axis of abscissas f

X

Domain of f

The first one is the independent variable.

The second one is the dependent variable.

At each moment, the phone has a certain percentage of battery. So for each value of t there is a unique value of p.

property price index

110% 105% 100% 95% 90% 85% 9 10 111213141516 1718192021

year

When a function appears as a statement or a description, it often lacks quantitative accuracy. However, if the statement is accompanied by numerical data, the function can be perfectly well determined. Let’s look at two examples that relate the height above sea level to the time passed:

Problem solved

Charge your phone to 100 % and make a note of the battery percentage every 15 min until it runs out.

The best way to see the overall behaviour of a function is by graphing it.

115%

ãã As statements

The set of output values is called the range of f. In other words, the set of values of y for which there is an x value in such way that f (x) = y.

Two variables are linked: the time passed, t, measured in minutes, and the battery percentage, p.

The function on the right is presented as a graph. It describes the evolution of property prices in a specific region in recent years.

Therefore, whenever we want to analyse a function, we try by graphing it, whatever its original form.

The set of input values of x is called the domain of definition of the function, f, and is expressed using the notation Dom f.

Explain why the following relationship is a function:

HOW FUNCTIONS ARE PRESENTED We frequently come across functions in the study of mathematics and other sciences, and even in daily life.

A function links two numerical variables which are generally named x and y: x is the independent variable.

4

• Félix left his house in the country in the morning and hiked up a path to the

top of a mountain. He had lunch at the top and returned home just before nightfall.

anayaeducacion.es Work on interpreting graphs.

• María left her house by the beach at 9 in the morning. She walked for

45 minutes to the top of a hill that is 250 metres above sea level and stopped for 10 minutes to admire the view. It took her 30 minutes to return home.

Therefore, p is a function of t: p = f (t).

Think and practise

anayaeducacion.es GeoGebra. The concept of function.

1 This graph describes the temperature at which the water

comes out of a tap that has been left on for a while:

60 50 40 30 20 10

temperature (ºC)

time (min)

b)

Y

1

3

c)

Y

Graph

X

Y 1

2

1

b) Explain why it is a function.

1 2 3 4 5 6

90

a) What are the two variables?

c) What are the domain of definition and the range of the function?

1

2 What are the domain and range of these functions?

a)

1X

10

Think and practise

X

a function for which the domain and range are [–2, 5] and [2, 7] respectively. Make up another one with a domain of [0, 5] and range of {1}.

Let’s analyse the graph above: a) Between what years does it show the evolution of property prices? b) Why does the graph begin at 100 %? Do you think this makes sense? c) The maximum reached was 115 %. What was the minimum? When was each of these reached? d) At which two moments is the price index 105 %?

What was the price index in the first quarter of 2016?

2 Look at the functions height above sea level – time

passed described above in reference to Félix and María’s trips. a) Draw a graph for Félix. b) Draw a graph for María. c) If you compared your graphs with those of your classmates, which would be the most similar, Félix’s graphs or María’s?

Explain why this is.

91

2 HOW FUNCTIONS ARE PRESENTED

Exemple Someone who earns €54 000: • Corresponds to the third row of the table.

• For the first €45 000 they have to pay €7 250, and for the rest (€54 000 – €45 000 = €9 000) they have to pay 35 %.

35 % of €9 000 = €3 150 • Therefore, they have to pay: €7 250 + €3 150 = €10 400

Unit

ãã As tables of values

ãã As analytical expressions or formulas

The values of a function are often shown in a table that directly provides the data we need. However, sometimes we have to perform complex calculations to obtain the information needed.

Analytical expressions are the most accurate* and operational way of expressing a function. However, in order to visualise it, we then need to study it in detail.

Let’s look at some examples:

Exemple: the table showing income tax payable

Example 1

Look at this table that shows what each person (taxpayer) has to pay as income tax (gross tax payable) based on what they earned the previous year, minus any legally deductible items (taxable income). taxable income (€)

gross tax (thousand €)

gross tax payable (€)

remaining taxable income (€)

The space covered, s, in centimetres, as a function of time, t, in seconds, is given by the formula s = 10t 2.

rate applied (%)

10 000

0

up to 15 000

15

25 000

2 250

up to 20 000

25

45 000

7 250

up to 25 000

35

70 000

16 000

and upwards

45

There is an example on the margin, however, this is how the table is used: • Earnings of 0 to €10 000 are not taxed (0 %).

10

• Earnings between €10 000 and €25 000 are taxed 15 %.

t (s)

Example 2

F ocu s on Eng lish accurate: correct, exact.

The tax payable by somebody who earns €54 000 is calculated as follows:

Example 4

54 000 = 10 000 + 15 000 + 20 000 + 9 000

The magnification, M, of an object viewed through a magnifying glass is M = 2 .

2–d d : distance between the magnifying glass and the object in cm.

0 % of 10 000 15 % of 15 000 25 % of 20 000 35 % of 9 000

9↓ 9↓ 9 9 €0 + €2 250 + €5 000 + €3 150 = €10 400 Problem solved

Find the gross tax payable for each of the following taxable incomes:

a) This amount corresponds to the first row, they do not have to pay anything.

a) €9 500

c) This corresponds to the third row. For the first €45 000 they have to pay €7 250.

For the remaining €5 000, they have to pay 35 %, which is €1 750. Therefore, they will have to pay €7 250 + €1 750 = €9 000.

b) €25 000 c) €50 000 d) €85 000

Think and practise

taxable incomes: a) €12 000 b) €20 000 c) €45 000 d) €100 000

92

4 We can calculate the values in the second column of

the table using the data from the other columns.

Explain how.

M M

d (cm)

Think and practise 5 In example 1, calculate the distance the ball travels in

1, 2, 3, 4 and 5 seconds. What time corresponds to a distance of 2 m?

In example 2, find the volume of a sphere with a radius of 5 cm and the radius of a sphere with a volume of 800 cm3.

anayaeducacion.es GeoGebra. Functions defined using tables of values.

3 Find the gross tax payable for each of the following

l (m)

M: magnification (number by which the real size is multiplied).

b) The tax corresponding to this amount is already given (no calculations needed) in the second row. 8 Gross tax payable: €2 250.

d) This corresponds to the last row. For the first €70 000 they have to pay €16 000.

For the remaining €15 000, they have to pay 45 %, which is €6 750. Therefore, in total they have to pay €16 000 + €6 750 = €22 750.

Example 3

The period is the time, in seconds, of one oscillation from side to side.

• Earnings over €70 000 are taxed 45 %.

25 45 70 taxable income (thousand €)

r (cm) T (s)

The period, T, of a pendulum is given as a function of its length, l (in m), by the formula T = 2 l .

• Earnings between €45 000 and €70 000 are taxed 35 %.

10

V (cm3)

The volume of a sphere as a function of its radius is:

V = 4 πr 3 (r in cm, V in cm3) 3

• Earnings between €25 000 and €45 000 are taxed 25 %.

5

s (cm)

A ball rolling down a slightly inclined plane has an acceleration of 20 cm/s2.

note: the table used by the tax agency is simiar to this one but a lot more complex.

15

4

800 cm3 5 cm

6 In example 3, find the period of a pendulum which

is 1 m long. What is the length of a pendulum with a period of 6 seconds?

7 Calculate the apparent size, M, of an object

(example 4) for the following values of d : 0; 0.5; 1; 1.5; 1.9; 1.99

If d = 4, we get that M = –1. This means that the object appears as the same size, but inverted. Interpret the values of M for d : 10; 5; 2.5; 2.1; 2.01

93

Unit

3 Y

4

DOMAIN OF DEFINITION In the function y = x 2, we can give x any value and we will get the corresponding value of y. We say that this function is defined in all Á or, rather, that its domain of definition is Á or (–∞, +∞).

y=x2

CONTINUOUS FUNCTIONS. DISCONTINUITIES The function shown in the margin is continuous in all its domain of definition.

However, the three functions below are discontinuous: a)

However, in the function y = x , we cannot give negative values to x . Its domain of definition is [0, +∞).

y=√x

b)

ãã Why is the domain of a function restricted?

X

f (x) = x 2 8 Dom f = (–@, +@) g (x) = x 8 Dom g = [0, +@)

a) There is a jump at point a on the X axis. b) Its branches tend to infinite at point a. In other words, the values of the function increase indefinitely when the x approaches a.

• Denominators. The values that make zero a denominator are not in the

• Square roots. Values that make the expression under the root negative are

not in the domain of definition.

Observe Time lines, like temperature lines, are continuous, but the function of house prices is made up of points joined by a line drawn to see the evolution.

For example, for f (x) = x – 2 , the values x < 2 are not in the domain of definition. Therefore, Dom f = [2, +∞).

— The real context the function has been taken from.

For example, if A = s 2 represents the area of a square as a function of its side, the domain is (0, +∞), as the length of the side must be a positive number.

If Dom f is the set of all the real numbers except x = –3, we can express it by joining the intervals, (– ∞, –3) « (–3, +∞), or as follows: Dom f = Á – {–3}

— The will of the person proposing the function.

We can talk about the function y = 2x defined in (0, 4] simply because we want to.

Unless otherwise stated, the domain of definition is as broad as the operations comprising the analytical expression of the function allow it to be.

b) y = x + 5

a) x 2 – 2x – 8 = 0 8 x =

2 ± 4 + 32 2 ± 6 = = 2 2

4 –2

The values x = –2 and x = 4 cancel out the denominator, which is why they do not belong to the domain of definition.

Therefore, Dom f = (–@, –2) « (–2, 4) « (4, +@) = Á – {–2, 4}

b) x + 5 ≥ 0 8 x ≥ –5. The domain of definition is Dom f = [–5, +@).

Think and practise

anayaeducacion.es Work on calculating domains.

1 Find the domain of definition of the following functions:

a) y =

94

x2

1 1 b) y = x – 5 c) y = 1 d) y= 2 + 2x – 8 x –5 x – 2x – 8

c) There is a point missing. In other words, it is not defined in x = a.

A function is continuous when it has no discontinuities of any kind.

A function is continuous in an interval [a, b] if there are no discontinuities in it.

Examples

Important!

In a continuous function, ‘small’ variations in x correspond to equally ‘small’ variations in y. However, at points of discontinuity (with a jump) a small variation of x (one more minute in the car park) can result in a large variation (€2) in y.

Many car parks still charge ‘by the hour’. This means that you have to pay for one hour just by entering. If you stay 1 h and 10 min you pay for 2 h. The first of the two graphs below describes this form of payment. It is a function with several points of discontinuity shown as steps.

10 cost (€) 8 6 4 2

10 cost (€) 8 6 4 2

1

1 2 3 4 5 time (h)

2

1 2 3 4 5 time (h)

Users prefer prices to be in line with the function shown in graph 2 . This one, obviously, is continuous. The following graphs show examples of discontinuity:

Problem solved

Find the domain of definition of the following functions given as analytical expressions: a) y = 2 1 x – 2x – 8

a

— The impossibility of performing an operation. This happens if in f (x) there are:

1 , the domain is the set of all the real numbers x +3 except x = –3. In other words, Dom f = (–∞, –3) « (–3, +∞).

Notation

a

Why are they discontinuous?

For example, for f (x) =

anayaeducacion.es Review calculating domains.

a

c)

Dom f can be restricted by any of the following causes:

domain of definition.

4

Observe 2 x (x – 2) y = x – 2x = =x x –2 x –2 In other words, y = x if x ≠ 2, because we cannot divide by zero. For this reason, we leave a gap at this point.

This one has branches tending to infinity.

And this one is missing a point.

3

4

2

Think and practise 1

x 2 – 2x y=— x–2

1 y=— (x – 2)2 2

anayaeducacion.es GeoGebra Continuous and discontinuous functions.

Construct a function similar to 1 , but where each half hour costs €1. Which of the payment options do you think is the fairest?

2 Analyse function 3 for values ‘close to 2’. Check that

when x equals 1.9; 1.99; 1.999; 2.01; 2.001, the values of y become ‘very large’.

95

Unit

5

ãã Average rate of change (ARC)

INCREASE, MAXIMUMS AND MINIMUMS Function f is increasing in this interval because:

In contrast, function f is decreasing in this interval because:

if x1 < x2, then f (x1) < f (x2)

if x1 < x2, then f (x1) > f (x2)

y = f (x )

f (x1)

x1

y = f (x )

f (x2)

In order to measure the speed of change (increase or decrease) of a function in an interval, we use the average rate of change or ARC. The average rate of change of function f in B interval [a, b ] is the quotient between the f (b) change in the function and the length of the f (b) – f (a) interval.

A b–a f (b) – f (a) f (a)

ARC of f in [a, b ] = b–a

anayaeducacion.es GeoGebra. Average rate of change.

a

f (x1)

x1

b

Note how the ARC of f in [a, b ] is the slope of segment AB.

f (x2) x2

4

x2

An interesting case of ARC is average velocity:

A function can be increasing in some intervals and decreasing in others.

space covered . time elapsed

Problems solved 1 Calculate the average rate

A function has a relative maximum at a point when the value of that point is higher than the values of all the points around it. In such cases, the function is increasing up to the maximum, and decreasing after it.

A function can have other points with higher values than a relative maximum, or lower values than a relative minimum.

of change for the function graphed on the right for intervals [1, 5] and [5, 8].

In contrast, if f has a relative minimum at one point, it is decreasing before that point and increasing after it.

1

2 Find

the ARC of the function y = x 2 – 4x + 5 for intervals [2, 4] and [0, 3].

Problem solved

In which intervals is the function shown on this graph increasing, and in which interval is it decreasing?

The function is defined by the interval [–7, 11].

–7

What are the relative maximums and minimums?

11

It is increasing in intervals (–7, –3) and (1, 11).

It is decreasing in interval (–3, 1).

It has a relative maximum at point –3 on the X axis. Its value is 2.

It has a relative minimum at point 1 on the X axis. Its value is –5.

There are points where the function has values that are lower than the relative minimum. For example, for x = –7, the function has the value –6.

5

8

ARC of f in [1, 5] =

f (5) – f (1) 9 – 6 3 = = 5–1 4 4

ARC of f in [5, 8] =

f (8) – f (5) 3 – 9 – 6 = = = –2 8–5 3 3

f (4) – f (2) 5 – 1 4 = = =2 4–2 4–2 2 f (3) – f (0) 2 – 5 –3 = = = –1 ARC in [0, 3] = 3–0 3–0 3 ARC in [2, 4] =

5

1234

3 The altitude a stone reaches

when thrown into the air is given in the equation a = 40t – 5t 2 (a, in m; t, in s). Find the average velocity in intervals [0, 2] and [4, 6].

ARC in [0, 2] =

a (2) – a (0) 60 – 0 = 30 m/s = 2–0 2

a (6) – a (4) 60 – 80 = –10 m/s = 2 6–4 Velocity is considered to be positive when the stone is going up and negative when it is coming down.

ARC in [4, 6] =

2 Find the average rate of change

1 Look at the function on the right and answer:

of the function f graphed on the left in intervals [1, 3], [3, 6], [6, 8], [8, 9] and [3, 9].

f

a) In which intervals is it increasing and in which is it decreasing?

96

f (1) = 6, f (5) = 9, f (8) = 3. Therefore:

80

a

60 40 20 2

4

6

8

t

Think and practise

Think and practise

b)What are its relative maximums and minimums?

In the figure, we can see that:

f

1

3

6

89

y = x 2 – 4x + 5 (problem solved 2) in [0, 2], [1, 3] and [1, 4].

3 Find the ARC of the function

4 Find the average velocity of the stone thrown into the

air in problem solved 3 for intervals [0, 1], [0, 3], [3, 4] and [4, 8].

97

Unit

6

4

ãã Periodicity

TENDENCY AND PERIODICITY A parachutist jumps out of a plane at a certain height. His velocity increases rapidly at the beginning but, with time, it tends to stabilise since the friction force of the air is equal to gravity. This is the graph of the function that relates the velocity to time:

height (m)

40

40

On the margin, you can see a graph representing the change in height of a basket on a ferris wheel as it turns. It takes half a minute (30 seconds) to go up, reach its highest point, come down and reach the ground. This movement is repeated over and over again. On a graph it looks like this:

velocity (m/s)

10

60 50

5

15

30

time (s)

40 30

30

10

We can see that, after a specific point in time, the velocity stabilises at a certain value. We can say that: With time, the velocity tends towards 55 m/s (198 km/h) With some functions, even if we only know a piece of them, we can predict how they behave far outside the interval studied, because they have branches with a very clear tendency.

Problems solved 1 After

cleaning out the freezer, you have a piece of ice in a glass. Graph the change in temperature of the water, knowing that the ice comes out of the freezer at –10 °C, takes half an hour to reach 0 °C and takes two more hours to fully melt. The room temperature is 20 °C.

2 What does the volume of a

cube tend towards when the length of its edges increases?

Problem solved

This graph represents the beginning of a periodic function with a period of 7.

Work out the values of this function at the following points on the X axis: a = 10; b = 19; c = 418.5 and d = 1 778.

The ice increases in temperature until it reaches 0 °C. Then, it begins to melt little by little, remaining at 0 °C until it has melted completely. The temperature of the water then increases and tends to become equal to the room temperature.

5 4 3 2 1

temperature (ºC)

20 10 0

a = 10 8 f (10) = f (3) = 3 (since 10 = 7 · 1 + 3, and every 7 units the value of the function is repeated) b = 19 8 f (19) = f (5) = 4 (since 19 = 7 · 2 + 5) 5 4 3 2 1 1 2 3 4 5 6 7 8 910 12 14 16 18 20 22

c = 418.5 8 f (418.5) = f (5.5) = 3 (since 418.5 = 7 · 59 + 5.5) 1234567

d = 1 778 8 f (1 778) = f (0) = 0 (since 1 778 = 7 · 254)

Think and practise 1

2

3

4

1 The amount of radioactivity that a substance has is

time (h)

–10

The volume of a cube as a function of its edges is V = s 3. The longer the edges, the greater the volume.

In other words, the volume increases indefinitely. We express this in the following way: When the edges increase indefinitely, the volume tends to infinity.

reduced by half each year. The graph below represents the quantity of radioactivity that there is in a portion of this substance over time.

volume (cm3)

8 7 6 5 4 3 2 1

2

The water tank in some public toilets fills and empties automatically every two minutes following the pattern on this graph: 30

volume (L)

a) Draw the graph that corresponds to a 10-minute span.

radioactivity

What is the tendency of the amount of radioactivity over time?

1 1

2

3

4

edge (cm)

98

120

A periodic function is one where the behaviour is repeated every time the independent variable passes a certain interval. The length of this interval is called a period.

anayaeducacion.es GeoGebra. Periodic functions.

time (s)

10

5

These tendencies are called limits of functions. You will learn how to recognise and calculate them in bachillerato.

90

In this function, what takes place in interval [0, 30] is repeated over and over again. It is a periodic function with a period of 30.

20

Limit of a function

60

12

time (years)

20

b) How much water will there be at these points in time?

10 time (min)

1

2

I) 17 min II) 40 min 30 s III) 1 h 9 min 30 s

99

EXERCISES AND

Unit 4

VED PROBLEMS SOL

1 Domain of definition

Find the domain of definition of this function:

We find the values of x that make + 6x – 9 either positive or zero. We 2 solve 3x + 6x – 9 = 0 8 x1 = –3, x2 = 1. We graph it: • If x ≤ –3 8 3x 2 + 6x – 9 ≥ 0

3x 2 + 6x – 9 > 0

3x 2 + 6x – 9 < 0

• If –3 < x < 1 8 3x 2 + 6x – 9 < 0 • If x ≥ 1 8

y = –2x 2 + 14x – 20

–3

+ 6x – 9 ≥ 0

Therefore, we can give x any value less than or equal to –3 or any value grater than or equal to 1.

2

Your turn Calculate the domain of definition of the following function:

3x 2

1

Therefore, the domain of this function, f, is: Dom f = (– @, –3] « [1, +@)

We have included the graph so you can clearly see what we are doing.

1 x–3

b) g(x) =

x +1 4 – x2

a) I n f (x) there are two restrictions: a root, whose radicand cannot be negative, and a denominator, which cannot be 0.

For x the restriction is x é[0, +@); and for x – 3 ≠ 0, it must be the case that: x ≠ 9. Therefore, Dom f = [0, 9) « (9, +@). b) This works in the same way as section a), but now there are 3 restrictions: • x + 1 8 x é[–1, +∞) • 4 – x 2 8 x é[–2, 2]

c) h(x) = f(x) + g(x)

• 4 – x 2 ≠ 0 8 x ≠ ±2

° ¢ £

a) f(x) =

–1 –2

x é(–2, 2)

Therefore, Dom g = [–1, +∞) » (–2, 2) = [–1, 2) c) Dom h = Dom f » Dom g = = ([0, 9) « (9, +∞)) » [–1, 2) = [0, 2)

2 –1

2

9 2

0

2

3 The cyclist

Pablo goes on a bike ride. The graph represents his velocity during the whole journey. a) For the first 15 minutes he cycles on flat terrain. At what velocity does he travel? What distance does he travel? b) Between minutes 18 and 27, is he travelling uphill or downhill? At what velocity? c) Give a 5-minute interval when he is travelling downhill. What’s his velocity?

100

40

a) For the first 15 min he cycles at a velocity of 25 km/h.

velocity (km/h)

35 30

25 20

15 min = 1 h = 0.25 h 4

In this time, he has travelled:

15

10 5

0.25 · 25 = 6.25 km

time (min) 10

20

30

40

50

60

x 40 – 2x

a) Calculate the function that relates the volume of the box to the length of the sides of the squares we cut out.

Therefore, the volume as a function of x is:

x

b) Calculate the domain of definition of the function.

V = (40 – 2x) · (30 – 2x) · x = = 4x 3 – 140x 2 + 1 200 x

b) In order for there to be a box, the x cannot be negative or 0, and it must also be less than 15 cm. Therefore, the domain of the function is (0, 15).

Your turn Repeat this activity for a piece of cardboard measuring 1 m × 70 cm.

Given the graph on the right: a) Find its domain. b) Is the function increasing or decreasing at the interval [0, 1]? Find the ARC of the function for this interval. c) Indicate the points of discontinuity of the function.

What type of discontinuity is there at each one?

a) A t first glance, the domain appears to be [–4, 5), but if you look carefully, the function is not defined for x = –1, therefore: Dom =[–4, –1) « (–1, 5) b) At the interval [0, 1] the function is decreasing.

ARC [0, 1] = –1 – 2 = –3 1– 0 c) Th ere is a discontinuity at x = –1 because the function’s branches tend to infinity, and at x = 3 because there is a jump.

6 Average velocity

0 –1

x

a) If we cut a square out of each corner with side x, the rectangle we get that forms the base of the box measures 40 – 2x long by 30 – 2x wide. The height of the box will be x.

5 Discontinuous graph

2 More domains of definition

Find the domains of definition of the following functions:

40 cm

30 cm

3x 2 + 6x – 9 > 0

We want to make a box from a piece of cardboard that measures 40 cm × 30 cm. To do this, we cut a square with side x from each corner.

30 – 2x

y = 3x 2 + 6x – 9

4 Building a box

3x 2

70

b) Between minutes 18 and 27, he is travelling uphill since his velocity has reduced by 9 km/h. In other words, he has gone from 25 km/h to 16 km/h. c) For example, between minutes 40 and 45 he is travelling downhill at 38 km/h.

The function represented in the graph on the right relates the distance travelled by a motorbike to time. a) What is the distance to the rest stop? b) Find the ARC of the function in intervals [0, 9], [0, 5] and [6, 9]. c) What is the average velocity in the first leg of the journey before the rest stop? And in the second? What is the relationship between the average velocities and the ARCs in section b)? d) Find the average velocity of the journey.

a) The distance to the rest stop is 400 km, which is where the horizontal part begins, showing that 700 the motorbike stopped for 1 h.

500 300 b) ARC [0, 9] = 700 – 0 = 700 = 77.78 9–0 9 100

ARC [0, 5] = 400 – 0 = 400 = 80 1 2 3 4 5 6 7 8 9 5–0 5 time (h)

ARC [6, 9] = 700 – 400 = 300 = 100 3 9–6 c) The average velocities in the first and second legs of the journey coincide with the average rates of change of the function in intervals [0, 5] and [6, 9], respectively.

In functions that relate distance travelled to time, when we find the average rate of change in an interval, what we are really doing is calculating the average velocity in this interval of time. d) The average velocity for the journey is the average rate of change of the function in interval [0, 9], which was already calculated in section b), 77.78 km/h.

Your turn Find the average velocity in the first hour, in the first two hours, in the first three hours… and so on. distance covered (km)

m this resources fro r to choose Remembe portfolio.

unit for your

Unit 4

ROBLEMS

P EXERCISES AND

Practise

4

Interpreting graphs

1

This graph shows the temperature of a glass of water taken out of a fridge over time: 22

temperature (ºC)

30

40

50

(hm3) 2019

252

203

385

412

372

328

(hm3) 2020

327

341

393

378

342

307

j) y = x + 7 k) y = 1 – x l) y = 3x – 9

307

310

371

360

316

286

m) y = –x n) y = 3 3x – 4 o) y = 1 – 5 2x + 2

20

40

2021

60

a) Draw the graphs corresponding to each year for the function week n.º - amount of water on the same axes. Describe them. b) According to the data, in what month and year was there the most water in the reservoir? When was there the least? c) What is the domain? d) Can we use this data to find the range of each function?

a) What is the temperature inside and outside the fridge? b) You take a glass of water out of the microwave at a temperature of 98 °C. Draw a graph showing the temperature of the water over time.

The average atmospheric pressure at sea level is the pressure exerted by a column of mercury of 760 mm. This is usually expressed as 760 mm of Hg. The graph shows how it changes with altitude.

800 pressure (mmHg) 700

a) What does the pressure tend to as the altitude increases?

600

b) What pressure is exerted on the outside of a plane flying at an altitude of 10 km?

5

500 400 300

6

200 100

altitude (km)

10 20 30

This graph shows the progress of the first three competitors to reach the finish line in a 1 000 m race:

The sound of a speaker becomes less intense as we move away from it. The intensity also decreases more slowly as we are farther from it. a) Show the intensity of the sound as a function of the distance to the speaker. b) What is the tendency?

The volume, in cm3, of a cylinder with a 1 cm radius as a function of its height, h, given in cm, is V = πh. The volume of another cylinder which is 1 cm tall as a function of the radius, r, in cm, of its base is V = πr 2. h 1 cm

800 600 400 200 50

100

150

200

250

time (s)

a) How long did each of them take to run the race? b) When did they overtake each other? c) What was the average speed of each runner in the first half of the race? What about in the second half?

r

1 cm

distance (m)

1000

102

20

amount (hm3)

time (min)

Find the domain of definition in each case: 2 a) y = x2 – 1 b) y = x + 2x c) y = 2 x3 + 7x – 2 x 5 d) y = 1 e) y = –3x f ) y = 2x2 – 1 x–3 2x + 10 x +1 x – 1 2 g) y = h) y= 2 i ) y = 2 1 –x x +x –6 x –x

10

amount

2

7

Characteristics of functions

0

amount

8

3

This table shows the amount of water in a reservoir over three years. We have specific data for five points over each year: week of the year

16

2

Domain of definition

Statements, formulas and tables

a) Calculate the volume of a cylinder with a 1 cm radius for heights of 1, 2, 3, 4 and 5 cm. Graph the function. b) Find the volume of a 1 cm high cylinder for radii of 1, 2, 3, 4 and 5 cm. Graph the function. c) How high is a cylinder with a 1 cm radius and a volume of 37.68 cm3? d) What is the radius of a 1 cm high cylinder with a volume of 803.84 cm3?

8

h(x) = x 2

10

4 2 –6

X

X c)

Y c)

X

X

6

Find the ARC of y = 3x 3 + 9x 2 – 3x – 9 in intervals [–2, 0], [–1, 0], [–3, –1] and [0, 1].

Given the following periodic function: Y

1 1

X

2

3

4

5

6

7

8

9

X

Find its period and the values of the function at x = 1; x = 3; x = –1,5; x = –9; x = 20; x = 23 and x = 42.

Y X

Y d)

d)

Y X

Y b)

b)

4

2

• The domain of definition of the function y = 1 f (x) is Á – {–2, 0, 4}.

• The domain of definition of the function y = f (x) is [–2, 0] « [4, +∞).

Following the above example, give the domain of definition of y = 1 and y = f (x) , for the f (x) following graphed functions: Y

13

Y

Bearing this in mind, we can say that:

Y a)

2

–4

12

Given the functions from the previous exercise, find the following domains of definition: j (x) 1 a) Dom[f (x) + g(x)] b) Dom c) Dom g (x) h (x)

a)

–2

Copy the graph in your notebook and draw the segment you are finding the slope for in each case.

j(x) = 4 – x 2

Look at the graph of the function y = f (x). It crosses the X axis at x = –2, x = 0 and x = 4. Its values are positive in intervals (–2, 0) and (4, +∞), and negative in (–∞, –2) and (0, 4).

–4

–2

g (x) = x 2 + 6x – 7

i(x) = –x 2

Look at this function and find its ARC in intervals [0, 4], [0, 5], [5, 7], [0, 7], [– 4, 0] and [– 4, –2].

Y

Find the domain of definition of: f (x) = x 2 – 9

9

11

X

Look at these discontinuous graphs and answer: Y 2

I

–2

–4

–2 Y

III

2

2 –2 –2

4

X

–4 –2

IV

h 2

Y 2

II

f

4

X

–4 –2

g 2

–2

4

X

Y 2 –2

i 2

4

X

a) Which are the points of discontinuity? Explain why there is discontinuity at each point. b) What is the domain of definition?

Y X

14

c) Indicate the relative maximums and minimums.

X

d) In which intervals is it increasing? And decreasing? e) Find f (2), g (2), h (2) and i (0).

103

Unit 4

EXERCISES AND PROBLEMS

15

Look at the following graphs of functions: 23

a) Create a table of values and use it to write the function that relates the value of y to that of x.

temperature (ºC)

2

An isosceles triangle has a perimeter of 20 cm.

Call x the unequal side, and y the equal sides.

time

–12

18

a) Relate each curve to these statements about the temperature of a glass of water: II. It is taken out of the fridge and left on the table.

III. It is moved from the table to the freezer.

165 160 155 150 145 140 135 130 125 120 115 110 105 100 95 90 85 80 75 70

25

The analytical expression of a function takes the form y = ax 3 + bx 2 + c. If we know that the points A (0, –2), B (1, 5) and C (–2, –22) belong to the graph, what are the values of a, b and c?

26

Calculate the value of a, b and c so that the points A (–12, a), B (3/4, b ) and C (0, c) belong to the graph of the function y = 3x 2 – x + 3.

27

a) Calculate the ARC of the function y = 2x – 3 in intervals [0, 1], [5, 6], [1, 5] and [0, 7]. b) Note how the value obtained is the same in all intervals. With which characteristic element of the straight line does this value coincide? c) Make a generalisation by completing this sentence in your notebook: ‘In linear functions, the ARC in any interval is equal to …’.

28

Explain why these statements are true or false: a) If a function is discontinuous at a point, that point does not belong to the domain of definition. b) If a point does not belong to the domain of definition of a function, it cannot be continuous at this point. c) We can be sure that periodic functions are continuous. d) The slope of a line is the ARC of any of its intervals. e) The ARC of a periodic function in any interval that is the same length as the period is 0. f ) If in a parabola the ARC of an interval is 0, the vertex is at the mid-point of that interval. g) All non-linear functions have at least one relative maximum or minimum. h) If a periodic function is decreasing throughout its domain, it is not continuous. i) The domain of a periodic function is Á. j) The domain of definition of a polynomial function is Á. k) The domain of the height - time function of a stone that is thrown 10 m away is [0, 10]. l) The domain of the function length of base - area of a square is R, since it is a polynomial function.

The orbit of Halley’s comet is an eccentric ellipse.

One of its foci is the Sun. This function relates the distance between the comet and the Sun to time:

x

(years)

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

x

x

a 20 cm x

x 50 cm

b

a

30

b) Give values to x and graph the function V (x).

20

c) What is the domain of V (x)?

21 1832

1909

1986 year

a) Is it a periodic function? What is its period? b) When will the comet come close to the Sun again?

The graph below shows the evolution of the value of a company since it was founded:

Draw a square ABCD with 7-cm sides. On side AB mark a point, P, which is at a distance x from A, then draw a new square PQRS inscribed in the first one.

A

x

P

B Q

S D

R

C

a) Note how if x = 3 cm, then AS = 7 – 3 = 4 cm.

How long is PS ? What is the area of the new square?

value (million Euros)

b) Graph the function that relates x to the area of the inscribed square. What is its domain?

2

1

age

a

a) Find the expression for the volume.

22

4

8

12

16

20

24

In a square ABCD with a side of 1 dm, draw the diagonal AC. For each point, P, on 1 dm this diagonal, we get a rectangle D C like the one in the figure. a) Find its area when P is at 1/4 dm, 1/2 dm and 3/4 dm from AB.

time

28 (months)

a) What was its value when it first opened?

P

A

B

As you can see, Ana, a 15-year-old girl who is 170 cm tall, is in the 90th percentile. This means that she is taller than 90 % of the population and shorter than 10 %.

b) What was its value after 4 months?

b)Graph the function that relates the distance from P to AB to the area of the rectangle.

c) What is the ARC in interval [4, 12]? Give the result in thousands of Euros per month.

c) Indicate the domain of definition of the function.

a) Estimate the percentile of the following girls:

d) What is the ARC in [12, 14] and in [14, 20]?

• Esther: 13 years; 160 cm • Érica: 11 years; 135 cm

104

Find the equation of the quadratic function whose graph, a parabola, passes through (0, 0), (1, –3) and (5, 5).

b

1755

19

170

24

We want to make a box with a lid out of a piece of card measuring 50 cm × 20 cm. x

10

Look at the graph below. It shows what approximate height percentile girls are in based on their age and height: 99 90 75 50 25 10 3

20

b) What is its domain of definition?

b) What is the temperature of the house? And of the freezer? And of the fridge?

190 height (cm) 185 180 175

Remember the theory

distance to the Sun (UA)

I. It is moved from the table to the fridge.

16

Advanced problem solving

1 dm

17

Problem solving

• María: 8 years; 113 cm • Marta: 12 years; 150 cm

e) This function has a relative maximum and a relative minimum. Describe them.

b) If Olivia is 13 years old and in the 75th percentile, how tall is she?

f ) What does the tendency seem to be for this function for the following months?

c) How old is Leonor if she is 105 cm tall and in the 25th percentile?

g) Write a comprehensive description of the value of this company in its first three years.

23

A function, f (x), is periodic with a period of 5, and its ARC in [1, 3] is 1. a) What can we say about the ARC of the function in interval [6, 8]? And in interval [11, 13]? b) What ARC does the function have in [3, 6]? c) What is the average rate of change of the function in interval [4, 9]? And in [8, 43]?

105

In your resource bank you have:

•

OP

MATHS WORKSH

Unit 4

different worksheets to deal with your emotions

and evaluate them.

•

FIND IRREGULARITIES AND GENERALISE

worksheets to improve your digital citizenship.

PRACTICE MAKES PERFECT!

Find the dependence relationship and write the equation of the function in each case:

•

x

f(x)

x

g(x)

x

h(x)

x

t(x)

1 2 3 5 10

3 5 7 11 21

1 2 3 5 10

2 5 8 14 29

1 2 3 5 10

2 5 10 26 101

1 2 3 5 10

4 10 18 40 130

help: To solve the last one, relate t(x) to g(x) and h(x).

A

a) T ransform figure A into just three squares by removing three toothpicks. b) Transform figure A into just three squares by removing two toothpicks. c) Cut figure B into three pieces along two straight lines.

Use them to make a square.

INVESTIGATE

• a) T urn figure C into four squares by changing the position of two toothpicks.

C D

B

C

b) Cut figure D into four identical pieces.

D

d) Cut figure B into four identical pieces.

Linked functions

Imagine you build a software application program for the function f1(x) = 1 + x , 1– x so that when you write a number, x, and press ‘enter’, it is transformed into f1(x).

If you press ‘enter’ again, it repeats the transformation using the previous result, and so on as many times as you press the button.

• Calculate the successive results using a starting point of x = 3 and pressing ‘enter’ several times.

3

f

1

–2

f1

–1/3

f1

f1

Let’s now call the function that transforms x into the result obtained from pressing ‘enter’ twice f2(x); the result of pressing three times f3(x); …; the result of pressing n times fn(x).

• Find f12(3), f13(3), f14(3) and f1 f1 f1 f1 x f15(3). f2 f3 • Generalise: what is the value of fn(3)?

with a calculator

If you want, try doing these operations with a calculator by pressing the following sequence of buttons: 3 =(1 +q)/(1Each time you press =, you get the next result.

Crossroads Imagine that there are several houses around a field and that all of them are joined by separate paths. A B A

B

A

C

Three points can be joined by roads that do not cross. f (3) = 0

D

C

This is the same for four points. f (4) = 0

E

106

crossroad

D

Five points cannot be joined unless there is one crossroad. f (5) = 1

What is the value of f (6)? What about f (7)? You are better off not trying the second one, no one has ever been able to solve it!

1 This curve shows the television audience on a

weekday.

40

4 Have a look at this periodic function:

30 20

Y 2

10 2

the

4

6

8

10 12 14 16 18 20 22 24 time (h)

a) Describe it. Note the most significant times. b) What is its domain of definition? And its range? c) In your notebook, graph what you think a Sunday would look like. d) In your notebook, draw the curve for the 31st of December.

f –4 –2

3 Find the domain of definition of these functions:

a) f (x) = x2 – 16 b) f (x) = 4x + 8 c) g (x) = 1 x–7 f (x) 1 2 d) h(x) = x + 2x – 15 e) f)

h (x) f (x)

audience (%)

2 Look at the graph and find the following:

B

C

anayaeducacion.es Answer key and interactive self-assessment.

50

q)=

anayaeducacion.es Generalisation.

Obtaining functions fn(x).

SELF-ASSESSMENT

Y 2 –2 –4

2

4 X

a) Domain and range. b) Maximums and minimums. c) Increasing and decreasing intervals. d) Where it is continuous and the points of discontinuity.

e) Calculate f (–4), f (–2), f (1), f (3) and f (5).

Commitment

2

4

6

8

X

a) What is its period? b) Find the values of the function at the following points on the X axis: x = 0; x = –6; x = –3; x = 2; x = 4; x = 40; x = –40 and x = 42.

5 Graph the function y = –x 3 + 9x 2 – 15x + 26,

defined in [0, 5], giving x integer values.

Let y be the stockmarket value, in millions of Euros, of a company that has just come under new management. Let x be the number of months since the change of management.

Describe its evolution over five months, showing increases, decreases, maximums and minimums.

6 Calculate the average rate of change of the function

equation y = x 2 + 4x – 5 in intervals [–5, 2], [–2, 1] and [1, 2].

Watch the video for target 7.2. Think of something you can do to contribute to achieve that goal. Make a commitment to put your idea into practice.

107

5

Basic Functions Reading and listening

Seeking accuracy During the 19th and 20th centuries there were debates about exactly what was and was not essential in defining a function. Over the course of these discussions, numerous mathematicians, notably Dirichlet, Riemann and Weierstrass, helped to formulate a more precise definition of a function.

Finally, the following definition was offered in 1923, which is very similar to the one we use today.

Evidently useful functions When a driver encounters a hazard on the road and has to brake, a certain amount of time passes between their decision to brake and the actual moment they brake (reaction time), about 3/4 of a second. Therefore, the faster the car is going, the further it will travel in that time. Furthermore, the car does not stop automatically when the brake is applied; it travels an additional distance (braking distance) due to inertia.

From the moment the driver becomes aware of the hazard to the moment the car stops, it travels an additional distance (d ) that increases the greater the velocity (v). An experiment gives the following values: speed of the car (km/h)

One says that y is a function of x if each value of x corresponds to a value of y. This correspondence is shown in the equation y = f (x).

Honest functions, according to Poincaré However, in this search for accuracy, a series of bizarre functions were invented that irritated Poincaré, who was unhappy with the direction the definition of functions had taken. This irritation led him to say in 1899: ‘For half a century we have seen a mass of bizarre functions which appear to be forced to resemble as little as possible honest functions which serve some purpose. Formerly, when a new function was invented, it was in view of some practical end. Today, they are invented on purpose to show that our ancestor’s reasoning was at fault.’

In this unit we are going to focus on these honest functions, which can be used for something more than just constructing or disproving concepts.

(based on reaction time in m)

distance

distance

(braking distance in m)

total stopping distance (m)

10

2

1

3

20

4

3

7

30

6.5

7

13.5

40

8.5

12

20.5

50

10.5

19

29.5

60

12.5

27

39.5

70

14.5

36

50.5

80

17

47

64

90

19

60

79

100

21

74

95

The equations are: dreaction = 0.21v

dbraking = 0.0074v 2

dtotal = 0.0074v 2 + 0.21v

1 Write a synonym for each of the following words and use each of them

in a sentence: bizarre, irritate, mass, resemble, span.

2 Use modal verbs to explain why the reaction time may be slower if a

person driving is having a conversation and not fully paying attention.

Henri Poincaré (1854–1912) is one of the greatest mathematicians in history.

His contributions spanned all fields of mathematics.

108

Solve 1

Use your calculator to check the validity of the formulas above for the values in the table (only check a few values in each row, keeping in mind that they are approximate values) and graph them.

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Unit

1

ãã Point-slope equation of a line

LINEAR FUNCTIONS Science is full of functions in which variations in the cause have a proportional impact on variations in the effect. All such functions are called linear functions and are graphed using straight lines. Let’s look at an example: If different weights are hung from a spring, they will stretch to different lengths.

In other words, the length of the spring is a function of the weight hanging from it. It is worth pointing out* that this is a linear function.

100

y = 30 + 15x (y: length in cm; x: weight in kg)

50

F ocu s on Eng lish

lenght (cm)

Y

y = mx

X

• If we give x the value x0 8 y = y0 + m (x0 – x0) = y0 + m · 0 = y0. x = x0, then

(x2, y2)

y = y0. In other words, it passes through (x0, y0).

2

2

3

4

5

6

• We can find the slope using the two points.

• We can find the equation based on the slope and one of the points.

Problem solved

anayaeducacion.es Review the point-slope equation.

Find the equation of each of the following straight lines:

a) Equation: y = 7 – 3 (x + 5). This is the equation of the line.

5

Functions of proportionality are graphed with straight lines that pass through the origin. They describe a ratio between the values of both variables.

a) I t passes through (–5, 7) and its slope is –3 .

5

The slope of the straight line is the constant of proportionality, m.

b) I t passes through (–2, 7) and (4, 5).

We can simplify it: y = 7 – 3 x – 3 · 5 8 y = 4 – 3 x 5 5 5 b) We begin by finding its slope: m = 5 – 7 = –2 = – 1 3 6 4 – (–2)

ãã Constant function: y = n n

It is represented with a straight line parallel to the X axis.

y=n

Its slope is 0.

y=0

X

The line y = 0 coincides with the X axis.

For example, the distance of an artificial satellite from Earth is constant. It does not depend on time, t. The equation for this would be d = 36 000, d: distance, in km; t: time, which does not appear in the equation.

200 100 32

F

ãã General expression of linear functions: y = mx + n This is represented on a graph as a straight line with slope m that intersects the Y axis at point (0, n).

The number n is called the ordinate at the origin.

Y F = 32 + 1.8C C 50 100

y = mx + n n X

110

In order to work out the equation of a line that passes through two points, we do the following:

1

s = v · t, where v is the slope of the line that relates s to t.

Y

➜➜ finding straight lines using two points

y 2 – y1 m = –––––– x –x

(x1, y1) x – x 2 1

1

Equation: y = y0 + m(x – x0)

• The coefficient of x is m. Therefore, its slope is m.

weight (kg)

0

Slope: m

• y = y0 + m (x – x0) is a first-degree expression. Therefore, it is a straight line.

For example, the space covered at a constant velocity, v, as a function of time is:

anayaeducacion.es GeoGebra. Graphic representation of a linear function.

Point: P(x0, y0) explanation

y = 30 + 15x

ãã Functions of proportionality: y = mx

point out: to mention some information you think is important.

This is a very useful formula. You should learn how to use it!

y2 – y1

More specifically, let’s suppose that the spring is 30 cm long before it is stretched and that is stretches 15 cm per kilogram added to it. The relationship is: Based on the presumption that the spring becomes deformed when weights of over 6 kg are hung from it, the domain of definition of this function is [0, 6].

We often need to write down the equation of a straight line of which we only know one point and the slope. It is written as follows:

Remember

ãã Linear functions in our daily lives

stretching a spring

5

For example, the line F = 32 + 1.8C, represented in the margin, allows us to change from degrees Celsius, C, to degrees Fahrenheit, F.

Equation of the line that passes through (–2, 7) and has a slope of – 1 : 3 y = 7 – 1 (x + 2) 8 y = 19 – 1 x 3 3 3

Think and practise

anayaeducacion.es GeoGebra. Calculating the slope of a line.

1 Graph the following functions:

a) y = 2x b) y = 2 x c) y = – 1 x d) y = – 7 x 3 3 4

2 Graph the following:

a) y = 3

b) y = –2

c) y = 0

d) y = –5

3 Graph the following functions:

a) y = 2x – 3

b) y = 2 x + 2 3 d) y = –3x – 1

c) y = – 1 x + 5 4 4 At an initial point in time, a moving object is 3 m from its origin and is travelling away from it at a velocity of 2 m/s. Find the equation for its distance from the origin as a function of time and graph it.

5

The price of potatoes at the market is €1/kg, and the price of tomatoes is €2/kg. a) Write the equation for the price of a bag of potatoes as a function of its weight. b) Write the equation for the price of a bag of tomatoes as a function of its weight. c) Graph the functions above.

6 Find the equation of these straight lines using the

given data:

a) It passes through (–3, –5) and its slope is 4 .

9 b) It passes through (0, –3) and its slope is 4. c) It passes through (3, –5) and (–4, 7).

111

1 LINEAR FUNCTIONS

Unit

2 y=x+4

y = –2x + 13

y=1

(3, 7)

ãã Piecewise linear functions

The tennis player on the left takes advantage of the fact that his opponent is close to the net to beat him with a lob. The curve described by the ball is a parabola.

It is common to see functions whose graphs are made up of different pieces of straight lines. Look at the function represented in the graph on the margin: 4. Its equation is: y = x + 4.

• The second piece, 3 < x < 6, is a line that passes through (3, 7) with a slope

3<x<6

We encounter all sorts of parabolas in everyday life: jets of water, the flights of balls and other objects, sections of parabolic antennas and car headlights, etc. All of them are described using quadratic functions.

of –2. Therefore, in the point-slope form its equation is y = 7 – 2(x – 3) 8 y = –2x + 13.

5 x≤3

x≥6

• The third piece, x ≥ 6, is the constant function y = 1.

➜➜ quadratic functions

Z if x ≤ 3 ]]x + 4 The function is described as follows: f (x) = [–2x + 13 if 3 < x < 6 ]1 if x ≥ 6 \ anayaeducacion.es Exercises to consolidate functions defined by two or three pieces.

• The function y = x 2 gives us a typical parabola, which serves as a point of reference for the others.

• The equation of the parabola described by the tennis player’s ball in the image 2 above is y = 5 – x , where x is its horizontal path, in metres, and y is its 20 height, also in metres. The origin of coordinates is at the player standing at the net.

The analytical description of a piecewise graph made up of straight lines gives the equation of the different pieces in order from left to right. It indicates the x values for which the function is defined on each of the pieces.

• The area of a square as a function of the length of a side (A = s 2), or of a circle as a function of its radius (A = πr 2).

Problem solved

Graph the function that has the following analytical expression: Z if x < –1 ]4 ]]0.5 (x – 1) + 5 if –1 ≤ x ≤ 3 y =[ if 3 < x < 5 ]–2x + 12 ]–1 if x ≥ 5 \

We find the endpoints of each of the pieces, starting, for example, at x = –5 and ending at x = 11(we could also take –4 and 8): 1st 2nd 3rd 4th (–5, 4) 0.5(–1 – 1) + 5 = 4 8 (–1, 4) –2 · 3 + 12 = 6 8 (3, 6) (5, –1) (–1, 4) 0.5(3 – 1) + 5 = 6 8 (3, 6)

We can see that the function is continuous at x = –1 because at that point the two sections have the same value, 4. This is also true at x = 3, but not at x = 5, where the function is discontinuous.

(–1, 4) (5, 2) (5, –1)

Think and practise 7 Graph the function that has the following analytical

expression: Z if x < 0 ]]–3 y = [x – 3 if 0 ≤ x ≤ 5 ]2 if x > 5 \ Find the slope of each of the pieces that form the function. Is it a continuous function?

• The height of a stone thrown up in the air, as a function of time (height = 20t – 5t 2). y = x2

–2 · 5 + 12 = 2 8 (5, 2) (11, –1)

We plot the four pieces (taking into account if the endpoints are included in the mentioned piece or not) to obtain the graph of the function.

(3, 6)

112

THE PARABOLA: A VERY INTERESTING CURVE

• The first piece, x ≤ 3, is a line with a slope of 1 and its ordinate at the origin is

5

anayaeducacion.es GeoGebra. Graphing piecewise linear functions.

8 Write the equation that corresponds to the following

graph:

5

x2 y = –– 4

➜➜ are all parabolas the same?

Yes, or to be more precise, they are all similar. Some parabolas may seem much wider than others, but that is because we are only looking at the ‘end’, the part near to the vertex. If we extended them, they would be the same shape as other seemingly narrower parabolas. ➜➜ do the parabolas of these functions fit their definitions as loci?

anayaeducacion.es • All parabolas are similar.

• Parabolas can be expressed using quadratic equations. They have a focus and a directrix.

Last year you learnt that a parabola is defined by a point, F (focus) and a line, d (directrix). The points P of the parabola are equidistant from F and d: 2 PF = dist(P, d). So, do the parabolas y = x 2 or y = x fit this definition? The 4 answer is, of course, yes. For y = x 2, the focus is F(0, 1/4), and the directrix 2 is d: y = –1 . For y = x , the focus is F(0, 1), and the directrix is d: y = –1.

4 4 You can check it yourself. ➜➜ what is special about parabolic antennas and car headlights?

5

A ray parallel to the axis of the parabola will be reflected by it so that it passes through the focus, which concentrates all of the rays at this one point.

5

10

Similarly, rays emitted from the focus are reflected in the parabola and are projected outwards as a beam parallel to the axis. anayaeducacion.es Property of parabolas which justify parabolic antennas, solar furnaces, car headlights, etc.

113

Unit

3

QUADRATIC FUNCTIONS. PARABOLAS

ãã Graphing quadratic functions

Tables of values on a calculator

ãã Quadratic functions Last year you learnt some things about parabolas. Let’s review what you know and learn a bit more.

Look at the following parabolas and their respective equations:

In order to graph a quadratic function given as an equation, we only need to find a few of its points. We begin by calculating the vertex of the parabola in order to then find some of the points around it. These are the steps you should follow: • Find the abscissa of the vertex of the parabola y = ax 2 + bx + c 8 p = – b 2a • Then, calculate the value of the function at a few values close to the vertex.

We can use a calculator to prepare a table of values quickly and efficiently corresponding to any function, for the interval we want and with the desired increments. Look at how we do this for the function y = –x 2 + 3x + 4.

• The points of intersection with the axes can be helpful for graphing it:

We select 9:Table from �. f (x) appears and we enter the expression.

Remember that to write x we have to press

-

y y==x x2y 2= x 2

1— 2+ 21 x2x –+42x4 + 4 y1x=x2 — ––2x y y==— 22 2

1— 21 x 2 y1x=x2 — y y==— 22 2

y 2=2 3x 2 y y==3x3x 2 –+18x y 2=2–3x y y==3x3x –18x 18x +2424+ 24

x )

x )

.

x+3

x )

— With the X axis, we can solve the equation: ax 2 + bx + c = 0.

— With the Y axis, the (0, c ).

Problem solved

+4=

Graph the parabola for the equation y = –x 2 + 3x + 4.

Then, this screen appears, where we have to indicate the first and last values of x and the size of the steps:

First, we find the vertex: Abscissa: p = – 3 = 1.5 8 Ordinate: f (1.5) = 6.25 8 Vertex: (1.5; 6.25) –2 Then, we find points close to the vertex:

2+ x6x –+66x6 + 6 y y==x x2y 2–=–6x

2– +–66x6 – 6 y 2=2+–x +6x6x y y==–x–x

y 2=2 –x 2 y y==–x–x

2 +–18x 2++18x y =2 –3x 18x –2424– y y==–3x –3x

2 2 y =2 –3x y y==–3x –3x

In our case, since the vertex is at x = 1.5, we give the values from –2 to 5 and choose an increment of 1 step.

24 12 x 2 1=— 1x–x2 — y y==––y— 22 2

To see the values that are not shown on the screen, you have to scroll down with the cursor.

The functions y = ax 2 + bx + c, with a ≠ 0 are called quadratic functions. They are all graphed using parabolas and are continuous in all Á. anayaeducacion.es Further theory and practice on the translation of parabolas.

eje axis

Their shape depends on a, the coefficient of following way:

1

2

3

4

5

y

– 6

0

4

6

6

4

0

– 6

Match each of the coefficients of x 2 to its corresponding parabola: C

• a = – 1 3 1 • a = 2

• If a > 0, their branches point upwards, and if a < 0, they

114

1

D

• a = 2

x 2, their corresponding parabolas are identical, although they can be located in different positions.

• The larger the value of |a |, the narrower the parabola.

0

anayaeducacion.es GeoGebra. Graphing quadratic functions.

• a = –1

• If two quadratic functions have the same coefficient of

point downwards.

–1

We see that – x 2 + 3x + 4 = 0 has two solutions, x = –1 and x = 4; and that f (0) = 4. Nevertheless, these points of intersection with the axes already appear in the table.

A

in the

–2

Think and practise

Each of these parabolas has an axis parallel to the Y axis. x 2,

x

Note how the points at the same distance from the vertex coincide in the value of their ordinate.

This is because the parabola is symmetric with regard to its axis. In other words, as the vertex is at x = 1.5, then f (1) = f (2); f (0) = f (3)…

12++2x 1=— 1x–x2 — 2– x2x +–42x4 – 4 y y==––y— 22 2

By analysing them, we can reach the following conclusions:

vértice vertex

5

2 Graph the following parabolas:

a) y = x 2 – 2x + 2

b) y = –2x 2 – 2x – 3

c) y = 1 x 2 + x – 2 3 e) y = – 1 x 2 + 2 2

d) y = –x 2 + 4 f ) y = 3x 2 + 6x + 4

3 In your notebook, graph the following quadratic

functions:

E

• a = –3 B

a) y = (x – 1) · (x – 3)

b) y = 2(x – 2)2

c) y = 1 (x + 2) · (x – 2) 2

d) y = (x – 1)2 + 5

115

3 QUADRATIC FUNCTIONS. PARABOLAS

Unit

4

ãã Lines and parabolas To solve some problems we have to perform a combined study of a linear function and a quadratic function, or of two quadratic functions. To do this, we sometimes have to solve a second-degree system, whose solutions (two, one or none) will be the points at which the two graphs intersect.

|a | = )

y = |x|

1 Solve

the following system analytically and graphically: y = –2x 2 + 5x – 2 y=x – 2

|x | = )

Analytically:

ãã The absolute value of a function: y = |f (x)| Look at how we find the absolute value of some functions:

y=x –2

Graphically: We graph the parabola y = –2x 2 + 5x – 2 and the straight line y = x – 2.

The points of intersection of the two curves are the solutions of the system: (0, –2) and (2, 0).

2 Graph this function:

y = |2x – 4|

(2, 0)

(0, –2)

defined for x ≤ 1/2. The vertex is at point (0, 1), it intersects the X axis at (–1, 0) [we do not consider the point (1, 0), as it has to be x ≤ 1/2], and passes through points (1/2, 3/4) and (–2, –3).

–1

–1

y = |x2 – 3x – 4|

y = |x 2 – 3x – 4|

2

3

and graphically:

The analytical expression is achieved by changing the sign of the function in the sections where f (x) takes negative values. To do this, we need to know the points where y = f (x) intersects with the X axis.

y = – 6x + 5

y=x –5

b) *

x2

y = – 2x y = –x + 2

Z 2 ]] y = x – 4x + 2 y = 2x + 2 2 c) * d) [ y = 5 – x2 ] y = –x + x + 6 2 \ 116

5 Graph the following functions:

Z 2 ]x + 2 ] a) y = [3 ]] –3x + 15 2 \

if x ≤ 1 if 1 < x < 3 if x ≥ 3

4–x if x ≤ 1 b) y = )–x 2 + 4x if x > 1

Z 2 ]]x – 3x – 4 if x ≤ –1 In other words: y = [–x 2 + 3x + 4 if –1 < x < 4 ]x 2 – 3x – 4 if x ≥ 4 \

Graphing y = |f (x)| from y = f (x) is very simple: we just need to ‘make a negative positive’. In other words, the part of the graph that is under the X axis is replaced by its symmetrical version with respect to this axis.

–2

Think and practise 4 Solve the following systems of equations analytically

if x ≤ –1 or x ≥ 4 x 2 – 3x – 4 y = |x 2 – 3x – 4| = * 2 – (x – 3x – 4) if –1 < x < 4

y = x 2 – 3x – 4

1

– (2x – 4) if x < 2 y = |2x – 4| = ) 2x – 4 if x ≥ 2 –2x + 4 if x < 2 In other words: y = ) 2x – 4 if x ≥ 2

intersects at x = –1 x=4

1

begins at (1/2, 3/4) and ends at (2, 3/4).

point (2, 3/4) and passes through (3, 7/4).

y = |2x – 4|

2

• The second section is a piece of horizontal line that • The last section is a piece of straight line that starts at

intersects at x = 2

y = 2x – 4

y = –2x 2 + 5x – 2

• The first section of the function corresponds to a piece of parabola and is only

Z 2 ]]–x + 1 if x ≤ 1/2 [3/4 if 1/2 < x < 2 ]x – 5/4 if x ≥ 2 \

–x if x < 0 x if x ≥ 0

This is called an absolute value function and it is graphed as shown on the left margin.

y = –2x 2 + 5x – 2 4 –2x 2 + 5x – 2 = x – 2 8 –2x 2 + 4x = 0 8 y=x –2 x = 0 8 y = –2 8 x (–2x + 4) = 0 x =2 8 y =0

There are two solutions. When graphed, we see that the parabola and the straight line intersect at points (0, –2) and (2, 0).

a) *

a if a ≥ 0 –a if a < 0

As a consequence, y = |x | is a piecewise function like the one below:

Problems solved

x2

ABSOLUTE VALUE FUNCTIONS Remember that the absolute value of a number a coincides with a if it is positive or zero, or with its opposite, –a, if it is negative:

In some piecewise functions you will also find straight lines and parabolas together.

*

5

Think and practise 1 In your notebook,

show the absolute value of these functions:

2 Graph these functions and rewrite them in their A

B

analytical expression without the absolute value: a) y = |3 – 2x | b) y = 1 x + 1 2 c) y = |x 2 – 4x |

d) y = |–x 2 + 6x – 5|

117

Unit

5

RADICAL FUNCTIONS

5

Problems solved 1 Graph this function and

indicate its definition:

Let’s look at some examples of radical functions: • The length of the side of a square as a function of its area is:

domain

of

y = –2 + x – 1

l= A

• The x in the radicand means that the curve goes to the right.

• The first value we give x is 1, because it cancels out the radicand.

• We give x the values 1, 2, 5, 10 and 17, which make the root exact. x

1

2

5

10

17

x–1

0

1

4

9

16

0 1 √x – 1 y = –2 + √x – 1 –2 –1

2

3

4

0

1

2

• The radius of a circle as a function of its surface area is: S π • The period of a simple pendulum as a function of its length is: r=

Y

T = 2 l ; l in m, T in s • The speed at which a ball that we drop from a height h reaches the ground is:

5

10

15

X

v = 20 h ; v in m/s, h in m Now we are going to study them theoretically.

Domain of definition: [1, +∞)

ãã Different types of radical functions Remember The function y = x is defined and is continuous in [0, +∞). It is also increasing, although it increases more and more slowly.

On the contrary, y = – x is defined and is continuous in [0, +∞) and decreasing.

2 Graph this function:

The functions y = x and y = – x can be plotted* point by point and result in the graphs below. They are halves of parabolas and together they describe a parabola that is identical to y = x2, but with the X axis as its axis of symmetry.

Y

y = –3 –x + 2

• The coefficient –3 produces negative ordinates. Therefore, the curve is below the X axis.

y = √x X y = –√x

F ocu s on Eng lish plot: to mark the position of a point using coordinates.

the following function and indicate its domain of definition: y= 3x

The following curves belong to the same family: Y

— y = 2√x – 4

— y = 2√–x + 1

— y = –2√x – 5

— y = –2√–x – 3

The functions y = a x + b and y = a –x + b are graphed using half parabolas.

Their domains of definition are, respectively:

118

3

12 (1, –3)

(–2, –6) (–7, –9) (–14, –12)

y = √x

–8

–1

0

1

8

27

–2

–1

0

1

2

3

Y

8

X

[–b, +@) and (– @, b]

x

Y

X

–2

Domain of definition: (–@, 2] 3 Graph

The domain of definition of these functions is [0, +∞).

–7

• The first value we give x is 2, because it cancels out the radicand.

• We give x the values 2, 1, –2, –7 and –14, with which the root is exact.

Y

X

• The –x tells us that the curve goes to the –14 left.

27 X

Domain of definition: (–∞, +∞) Think and practise

anayaeducacion.es GeoGebra. Graphing radical functions.

1 Graph the following functions and find the domain of definition of each one:

a) y = 2 x b) y = –2 x c) y = 2 x + 3 d) y = –2 x + 3 e) y = 4 –2 x + 3 f ) y = 2 –x g) y = –2 –x h) y = 2 –x + 3 i) y = –2 –x + 5 j) y = –3 –2 –x + 5

119

Unit

6

INVERSELY PROPORTIONAL FUNCTIONS

ãã Functions related to y = k/x

5 10 15 20 25 30 V (litres) 6

Inversely proportional relationships are very common in nature, physics, economics, etc. Now we are going to analyse them theoretically.

1 1

P (atm)

6

2 , which related 2–d the magnification, M, produced by a magnifying glass to the distance, d, of the object being magnified. The graph of this function is also a hyperbola. Its asymptotes are the axis of abscissas and the line d = 2.

• In the previous unit we looked at the function M = • We have a rectangle with an area of 100 cm2 and sides of unknown length. We call the sides x and y. We know that xy = 100. We express it like this: y = 100 (For a given area, the sides are inversely proportional). x • There are 6 L of air in a tank at atmospheric pressure. If we compress the air (increase the pressure), the volume decreases. It satisfies the relationship P · V = 6. We express it like this: 6 V = P (The pressure and volume of a given mass of gas are inversely proportional).

30 25 20 15 10 5

• The functions

Problems solved

the following function and indicate its domain of definition:

This is the graph for the function y = 1 : x

y= y

x

y

1 2 3 4 5 …

1 0.5 !

0. 3 0.25 0.20 …

1 0.5 !

0. 3 0.25 0.20 …

1 2 3 4 5 …

1 y=— x

1

• This curve is called a hyperbola. Its

asymptotes are the coordinate axes.

3 … 5

6

7

8 10 …

x – 4 –6 –4 –3 –2 –1 … 1

2

3

4

3

2 1.5 1 …

y

• If x gets close to 0, y takes on

• If x takes ever-increasing values, y

The graph of this function will be like the one of the function y = 6/x, shifted 4 units to the right. Let’s check this using a table of values: x

• It is not defined for x = 0.

gets closer to 0. This is why the X axis is another asymptote.

When the sign of x changes, the sign of y changes as well.

6 x–4

–2

0

1

2

–1 –1.5 –2 –3 –6 … 6

6 …

Its asymptotes are x = 4, y = 0.

ever-larger values. This is why we say that the Y axis is an asymptote.

1

k y= x –a, y= k , y= k +b a–x x–a are generally hyperbolas related to functions of inverse proportionality. Let’s look at some examples.

anayaeducacion.es Extension: translation of hyperbolas.

1 Graph

ãã Studying the function y = k/x

x

5

Its domain of definition is (–∞, 4) « (4, +∞).

2 Graph this function and

indicate its definition: y=

domain

of

6 +2 x–4

The graph will be like the previous one, but shifted 2 units upwards. In other words, it can be obtained from the previous graph by shifting the X axis down two units.

Its domain of definition is (–∞, 4) « (4, +∞).

The following curves are from the same family: 6 y = –– x

1 y = – –– x

4 y = – –– x

Think and practise

anayaeducacion.es GeoGebra. Graph functions of inverse proportionality.

1 Match each of the following graphs with its corresponding function. Indicate the domain of

The functions y = k are called functions of inverse proportionality. x They are represented with hyperbolas whose asymptotes are the coordinate axes.

Their domain of definition is formed by the sections (–@, 0) and (0, +@).

It is expressed like this: Dom = (–@, 0) « (0, +@) 120

each one: a) y = 2 b) y = – x c) y = 2 d) y = x–3 x

2 x 2 + 5 –3

A

B

2 Graph each function and indicate its domain:

a) y = 8 b) y = – 8 c) y = 8 x x x–2

d) y =

C

8 e) y = 8 – 3 2–x x

D

f) y =

8 +3 x–2

121

Unit

7 15 y = 2x

EXPONENTIAL FUNCTIONS

ãã Applications of exponential functions

ãã Increasing exponential functions: y = a x, a > 1

Exponential growth is very common in nature (microorganism cultures, animal and plant populations, etc). They are also used to describe economic phenomena, among other things.

The graph on the right shows the exponential function of base 2: y = 2x.

Let’s look at some examples.

x ≥ 0: 10

2x

–4

anayaeducacion.es

x < 0:

0 1 2 3 4 …

–1 x x

5

1

2

4

8

16 …

2x

2–1

–2

= 0.5

2–2

= 0.25

–3 2–3

= 0.125

When x takes increasingly larger values, 2x tends to infinity. However, when x takes the values –4, –5, –10, …, 2x becomes really small, it tends to 0.

0

4

• They are continuous, their domain is all Á and they pass through (0, 1) and Interesting fact

• If the base is greater than 1 (a > 1), then they are increasing.

• The greater a is, the more quickly they increase.

( )

1 y = –– 2

x

10

The functions of the type

y = a kx = (a k)x

are exponential functions with base

a k.

x

y = ax

0

4

Functions where 0 < a < 1 also pass through (0, 1) and (1, a). They are continuous and their domain is all Á, but they are decreasing. They decrease more quickly the closer a is to 0.

The functions y = ax and y = (1/a)x are symmetric with respect to the Y axis.

Problem solved

Graph this function: y = 10 0.2x Then, plot the following function on the same axes: y=c

1 x m 1.6

100.2x = (100.2)x = 1.6x It passes through points (0, 1) and (1; 1.6).

Since 1.6 > 1, the function is increasing.

We find some more points around the Y axis. x 1 m is symmetrical to The function y = c 1.6 the first one with respect to the Y axis.

Think and practise 1 Graph the following functions in your notebook.

What is the domain of definition of each of them? a) y = 1.25x b) y = 0.8x

122

In the Maths Workshop (p. 132) we study the application of the exponential function to radioactive decay and carbon-14 dating.

ãã Decreasing exponential functions: 0 < a < 1 y = c 1 m is also exponential. Since its base (1/2) is less than 1, the function is 2 decreasing. Its graph is symmetrical to the graph y = 2x with respect to the Y axis.

5

–4

Semi-logarithmic graph paper is very useful for graphing and comparing functions involving rapid growth, such as exponential functions.

Exponential functions are those that take the form y = ax.

(1, a).

15

5

( )

1 y= — 1.6

The number of amoebas that there will be after t hours is N = 2t. Its graph is like the first graph on the previous page, but it is only valid for t ≥ 0. ➜➜ example 2. capital growth

There are €20 000 in a bank with a monthly interest rate of 0.5 %, which means that each month the amount grows by 0.5 %. Therefore, the money in the account at the start of each month is multiplied by 1.005.

C = 20 000 · 1.005T, T ≥ 0

Problems solved 1 We deposit €60 000 in an

account with 5 % annual interest. How much money will we have in 8 years and 3 months?

decreases at a rate of 8 % per year. If it is currently worth €120 000, how much will it be worth in 11 years?

y = 1.6x

X

2 Write these expressions in exponential form:

b) 100.01x

Suppose that the conditions of a culture mean that the number of amoebas approximately doubles every hour, and that there is one amoeba in the beginning.

The expression that gives the money in the account after T months is:

2 The value of a machine

anayaeducacion.es GeoGebra. Graphing exponential functions.

a) 20.4x

Amoebas, as you know, are unicellular beings that reproduce by splitting in two (binary fission). This can happen at different rates.

An annual increase of 5 % means multiplying by 1.05 each year. The money will grow according to the equation: C = 60 000 · 1.05T The money in the account after 8 years and 3 months (8 and a quarter years) is the value of C for T = 8.25: C (8.25) = 60 000 · 1.058.25 = €89 735.23

Y x

➜➜ example 1. growth of a population

x/28

c1m c) 1.0112x d) 2

An annual devaluation of 8 % means multiplying by 0.92 at the end of each year (1 – 0.08 = 0.92). The value of the machine changes as follows: V = 120 000 · 0.92T For T = 11, we get V (11) = 120 000 · 0.9211 = €47 956.49.

Think and practise 3 Write the equation expressing the approximate num-

ber of amoebas that there would be after t hours in a culture like the one in example 1, if there were 200 amoebas at the beginning.

How many amoebas would there be after 8 hours?

anayaeducacion.es More on the applications of exponential functions.

4 €130 000 is deposited in a bank account with an

annual interest rate of 12 %. Express the value of the capital, C, as a function of the time, T, expressed in years, that it is in the bank. How much money will there be after 6 years and 9 months?

123

Unit 5

LOGARITHMIC FUNCTIONS

1 Piecewise continuous function

8

y = log2 x

4 –4

II

4

8

12

16

(1, 0), (2, 1), (4, 2), (8, 3), (16, 4)… are points of II. In general, if point (a, b) belongs to I, then (b, a) belongs to II. This is why we say the function described by II is the inverse of I.

20

–4

• 2–1 = 1 8 log2 1 = –1 2 2

The function described by the red graph is called logarithmic function of base 2, and is written: y = log2 x • The logarithmic function y = loga x with a > 1

is the inverse of the exponential function y = a x.

1 1

a

y = log1.5 x

Find the equation of the parabola whose vertex is (3, –1) if it passes through the point (2, –2).

Graph it on coordinate axes.

(2; 2.25)

(2.25; 2)

(3; 3.38)

(3.38; 3)

(4; 5.06)

(5.06; 4)

(5; 7.59)

(7.59; 5)

(6; 11.39)

(11.39; 6)

Think and practise

The parabola passes through (3, –1) and (2, –2): (*)

–1 = 9a + 3b + c 8 –9a + c = –1

The equation is y = –x2 + 6x – 10.

Your turn Find the equation of the parabola whose vertex is at the point (–2, –9) and that passes through (0, 1)

* y = log1.5 x 1 1

X

anayaeducacion.es GeoGebra. Graphing logarithmic functions.

a) y = ex; y = ln x b) y = 3x; y = log3 x c) y = 10x; y = log x d) y = 2.2x; y = log2.2 x

124

Therefore: –b/2a = 3 8 b = –6a (*)

(*)

Show this system analytically and graphically:

y = 1.5x

1 In your notebook, plot each pair of functions on the same coordinate axes:

Find the domain of definition of each of them.

The abscissa of the vertex is x = 3.

3 Straight lines and parabolas

Y

In order to graph it, we use the inverse, the exponential function y = 1.5x to help us: y = log1.5 x

The equation for a parabola has the form y = ax 2 + bx + c.

–2 = 4a + 2b + c 8 –8a + c = –2 9a – c = 1 ) 8 a = –1; b = 6; c = –10 –8a + c = –2

words, its domain of definition is (0, +@).

• It passes through points (1, 0) and (a, 1).

• It is increasing, but as x increases, the velocity at which it does so drops to very small values.

• It has a branch tending to infinity on the Y axis.

y = 1.5x

2x – 5 if x < 2 The function looks like this: y = ) –2x + 3 if x ≥ 2

2 Finding the equation of a parabola from its vertex and one of its points

• It is defined by values greater than 0. In other

We know that it passes through (1, 0) and (1.5; 1).

2 · 2 – 5 = 2m + 3 8 m = –2

X

Your turn Find the value of k so that this function is continuous: 4x + 2 if x < –1 y=) Graph it for the value of k you get. kx – 3 if x ≥ –1

Problem solved

Graph the following function:

–5

Graph it for the value of m you get.

2

+3

I

12

(0, 1), (1, 2), (2, 4), (3, 8), (4, 16)… are points of I.

Y

–2x

y=x

16

These two curves are related analytically in the following way:

2x – 5 if x < 2 y=* mx + 3 if x ≥ 2

The function is represented with two lines. In order for it to be continuous, we just need these to coincide at x = 2, where it changes from the first to the second. To do this, 2x – 5 has to coincide with mx + 3 at x = 2:

y=

Logarithms Remember that log a P is the exponent to which the base, a, must be raised to get P. In other words: log a P = x ï a x = P For example: • 23 = 8 8 log2 8 = 3

y = 2x

Calculate the value of m so that this function is continuous:

2x

The graph of the function y = 2x is plotted on these coordinate axes in blue, and its symmetrical function with respect to the line y = x, in red.

20

SOLVED

y=

8

BLEMS O R P D N A S E IS C R E EX

e) y = 4x; y = log4 x

y = x2 – 2 y=x

Describe and graph the continuous function that is obtained from both equations and that takes the form lineparabola-line.

We find the points where the parabola and the straight line intersect. In order to do this, we isolate y in each equation and equalise: y = x2 – 2 4 8 x 2 – 2 = x 8 x 2 – x – 2 = 0 8 x = –1, x = 2 y=x They intersect at (–1, –1) and (2, 2).

In the statement, it says that the first is a piece of straight line, then there is a piece of parabola, and then there is another piece of straight line. This is the piecewise-defined function: Z if x < –1 ]]x 2 y = [x – 2 if –1 ≤ x ≤ 2 ]x if x > 2 \ It is represented in the graph on the right. y = 4 – x2 Your turn Now do the same with this system: * y = –x – 2

Y

X

125

EXERCISES AND

m this resources fro r to choose Remembe portfolio.

VED PROBLEMS SOL

ROBLEMS

Practise

Express the following function without using the absolute value sign and graph it:

In order to express it as a piecewise-defined function, we first need to see where the function within the absolute value intersects with the X axis. In this way, we can work out in which intervals it is positive and in which it is negative.

y = 1 x2 – 5 x + 9 2 4 4

1 x 2 – 5 x + 9 = 0 8 x = 1, x = 9 2 4 4 As the branches of the parabola point upwards, since the coefficient of x 2 is positive, the function is only negative between the points of intersection with the X axis.

y = |x 2 – 2x – 8|

a) y = |x 2 – 1| + 1 b) y = |2x – 4| + |x|

Your turn Express each function as a piecewise-defined function: a) y = |x 2 – 4| – 2 b) y = |x + 3| + |x|

Z ] 1 x2 – 5 x + 9 2 4 ]4 ] 1 2 5 y = [– x + x – 9 2 4 ] 4 5 9 1 2 ]] x – x + 2 4 \4

3

if 1 < x < 9

1

9

y = 2x + 3 x +1 Your turn Graph this function: y = 3x – 1 x+2

126

7

Graph the following linear functions: a) y = 2x – 3 b) y = 4 x 7 – 3 x + 10 c) y = d) y = 2.5 5

C

2 –2

x<0

0≤x≤2

x>2

|2x – 4|

–2x + 4

–2x + 4

2x – 4

|x|

–x

x

x

|2x – 4| + |x|

–3x + 4

–x + 4

3x – 4

Therefore, the function is: Z ]]–3x + 4 if x < 0 y = [–x + 4 if 0 ≤ x ≤ 2 ]3x – 4 if x > 2 \

We start by dividing: 2x + 3 x + 1 This function can be expressed as: – 2x – 2 2 2x + 3 = 2 + 1 1 x +1 x +1 It is the function y = 1 shifted 1 unit to the left and x 2 units up.

D

Find the equation of the straight line that passes through points A and B in each case: b) A (–2, – 4), B (2, –3)

c) A (0, –3), B (3, 0)

d) A (0, –5), B (–3, 1)

Find the equation in each case and graph it: a) Line that passes through (2, –3) and is parallel to the one passing through (1, –2) and (– 4, 3). b) Proportionality function that passes through (–4, 2). c) Constant function that passes through (18; –1.5).

6

Find the value of the unknown parameters so that the straight lines and points comply with these conditions. Graph them.

c) The lines of the equations y = 3x + c and y = cx + 3 intersect at ordinate 2. Which is the corresponding abscissa? d) The points (d, –2) and (4, e) belong to the line of the equation y = 1 x – 3 .

2

4

6

X

Graph the following functions by making a table of values like the one below for each of them: x

– 4

–3

–2

–1

0

1

2

3

4

y

…

…

…

…

…

…

…

…

…

a) y = x 2 + 1

b) y = –x 2 + 4

c) y = –3x 2 d) y = 0.4x 2 9

Graph the following parabolas by finding their vertex, some points close to it and the points where they intersect with the axes:

a) y = (x + 2)2 b) y = x 2 – 4x c) y = 1 x 2 + 2x + 1 d) y = x 2 – 9 2 10 Give the point (abscissa and ordinate) where the vertex of the following parabolas is found. In each case, state whether it is a maximum or a minimum.

Then, graph them. a) y = 8 – x 2 b) y = 4 + (3 – x)2 c) y = –x 2 – 2x + 4 d) y = 3x – 1 x 2 + 1 2 e) y = 15 – 1 x 2 + 1 x f ) y = 1 x 2 + 2x + 3 2 3 4 4

a) The line that passes through points (4, 0) and (–2, a) has a slope of –1. b) The line y = bx + 2 passes through point (–3, 4).

2 –2

8

a) A (3, 0), B (5, 0)

D

4

b) P (2, –1), m = –2

B

C

6

Calculate the equation of these linear functions: A

5

Y

A

c) A (–2, 1), m = 1 d) A (1, 3), m = – 5 3 2

4

d) y = x 2 – 6x + 6

B

Using the given slope and point, calculate the equation of each line: a) P (0, 0), m = 1

X

a) T he parabola of y = x 2 – 1 intersects with the X axis at x = –1 and x = 1.

Since its branches point upwards, the piecewise-defined function is: Z 2 Z 2 if x < –1 ]]x – 1 + 1 if x < –1 ]]x 2 2 y = [–x + 1 + 1 if –1 ≤ x ≤ 1 8 y = [–x + 2 if –1 ≤ x ≤ 1 ]x 2 – 1 + 1 if x > 1 ]x 2 if x > 1 \ \ b) W e make a table to study how the expression varies within each absolute value depending on the section. The endpoints of each section are those that make each of the expressions equal to 0:

Match each expression to its graph: a) y = x 2 b) y = (x – 3)2 c) y = x 2 – 3

if x ≥ 9

6 Functions related to the function of inverse proportionality

Express the following function k in the form y = + b. x–a Then, graph it.

1

Therefore, we have three sections: (– ∞, 1] and [9, +∞) where the function is positive or zero, so there is no need to change the sign, and the interval (1, 9), in which the function is negative. In this Y section, we change the sign of the function.

Let’s see what it looks like: if x ≤ 1

Quadratic functions

Linear functions

2

5 Functions with absolute values

Express each of the following functions as a piecewise-defined function:

Unit 5

P EXERCISES AND

4 The absolute value of a function

Your turn Express as a piecewisedefined function and graph:

unit for your

11

Calculate the vertex, the axis of symmetry and the points of intersection with the axes (if any) of each of these parabolas: a) y = 2x2 b) y = 2(x – 5)2 c) y = 2(x – 5)2 + 2

d) y = –x2 + 1

e) y = –(x + 1)2 + 1

f ) y = –3x + 2x2 127

Unit 5

EXERCISES AND PROBLEMS Piecewise-defined functions

Absolute value of a function

Other functions

12

16

20

Graph these functions. Which are continuous?

Z ]] 2x if x ≤ –1 –3 if x < 0 a) y = [ –2 if –1 < x ≤ 3 b) y = ) 2x + 1 if x ≥ 0 ] x – 5 if x > 3 \ Z ]] –x + 3 if x < 1 if 1 ≤ x < 2 c) y = [ 2 ]x if x ≥ 2 \ Z if x < –2 ]] 0 d) y = [ x + 2 if –2 ≤ x ≤ 0 ] 3x – 2 if x > 0 \ 13 Graph the following functions: Z ]] –1 – x if x < –1 a) y = [ 1 – x 2 if –1 ≤ x ≤ 1 ] x – 1 if x > 1 \ x 2 if x < 0 b) y = * 2

–x if x ≥ 0 c) y = * 14

15

a) y = |2x – 2|

Match these functions to their corresponding graph and give the domain of definition of each one: I) y = 1 II) y=3– 1 x–3 2–x III) y = 2 + 2 IV) y = – 1 x x +3 Y Y a) b)

A

4

4

2

2

–4 –2

2

c) 17

a) y =

|x 2

c) y =

|x 2

– 6x + 5|

d) y =

|–x 2

A

C

D

Y

B

4

6

8 X

18

19 X

X

D

Y

X

Express each function as a piecewise-defined function. Remember that to define the intervals you need the points of intersection with the X axis. a) y = 4 – 1 x b) y = |2x + 2| 3

Y

X

C

–4 –2 2 Y

c)

2

c) y = |x 2 – 2x – 3|

d) y = |–x 2 + 2x – 1|

e) y = 1 x 2 – 4x + 7 2 2

f ) y = |9 – x 2|

Write the piecewise-defined function for each of the functions in the graph.

Express them as absolute value functions.

a

–4

–4

a) y = x + 2

4

b)

4

2 X 6 X

e) y = –2 – x f ) y = –2 – 2 –x g) y = 2 + –x h) y = 2 –x + 2

Y 2 2

4

d)

c) y = –x – 1 d) y=2+ x +3

X

6 Y

e) y = 1 + x – 1 f ) y = 2(x – 1)

2

g) y = – – (x + 2) h) y = 1 + 1 – (x + 1) X

27

Match these functions to the graphs: I) y = 3x II) y = 1.5x IV) y = 0.7x III) y = 0.4x

4

2

2

b –4 –2

2

4

8

–4 –2 d

4

4

2

2 4

28

–4 –2

d) y = 0.75–x

Plot each pair of functions on the same coordinate axes. What is the relationship between them? x

a) y = c 1 m ; y = 3x b) y = 0.25x; y = 4x 3

4

8 6

2

x

2

6

Graph the following functions by making a table of values for each of them: a) y = 2–x b) y = 3x + 1 c) y = c 2 m + 3 3

6

4

–4 –2 c

b

6

Work out the domain of definition of the following functions and graph them: a) y = 2 – x b) y = 7 – 2x + 4

–6 –4 –2

a

b) y = 2 – x

c) y = 2 –x d) y = – –x

26

29

2

Say whether they are increasing or decreasing.

128

25

–2

2

–2

22

a) y = – 1 b) y= 2 x x c) y = 1 – 3 d) y = 3 + 2 x x 24 Give the domain of definition and asymptotes of the following functions. Graph them. a) y = 1 b) y = – 3 x +1 x +3 c) y = 1 + 2 d) y = 1 + 2 x –1 1– x

–2

Y 2

4

Graph the following functions:

Graph the following functions and find the domain of definition of each one:

–2

III) y = 3 – –x IV) y = –3x

Y

2

X

23

Match the functions to their corresponding graph and give their domains of definition: I) y = x – 3 II) y= x –3 a)

–6 –4 –2

6 X

4

Y 2

X

– 4x – 3| 21

2

d)

–4 –2

b) y = |–x 2 – 10x – 22.75|

– 4|

4 X

Y 2

Match each function to its graph:

B

Write the analytical expressions of each function:

Y

D

C

B

We know that the equations of the parabolas that appear in the graphs are: y = x 2; y = –x 2 – 4x; y = 4 – x 2; y = x 2 – 6x + 5.

A

b) y = |4 – x |

c) y = 1 x + 2 d) y = 1– 1 x 2 2

x 2 + 4x if x < 0 –x 2 + 4x if x ≥ 0

Write the equation of the piecewise function that corresponds to this graph.

Match each function to its corresponding graph:

When completing the exercises below, we recommend using a calculator to create tables of values.

4

a) Graph the following functions: y = 3x and y = log3 x b) Check whether the following points belong to the graph of y = log3 x : (243, 5)

c 1 , –3m 27

` 3; 0.5j

(–3, –1)

129

Unit 5

EXERCISES AND PROBLEMS 40

Problem solving 30

Solve the following systems of equations analytically and graphically: a) *

31

y = 2x 2 – 5x – 6 y = –x 2 + 5x * b) 2 y = 3x + 4 y = x + 3x – (15/2)

a) Calculate b and c so that the vertex of the parabola y = x 2 + bx + c is at point (3, 1). b) What is its axis of symmetry? c) What are its points of intersection with the axes?

32

The parabola y = + bx + c passes through the origin. What would the value of c be?

If we also know that it passes through points (1, 3) and (4, 6), find a and b and graph the parabola.

33

Calculate a and b so that the function y = a passes through points (2, 2) and (–1, –1). x –b

34

The graph of an exponential function of the type y = ka x passes through points (0, 3) and (1; 3.6). a) Calculate k and a. b) Is it increasing or decreasing? c) Graph the function.

35

36

41

ax 2

42

The exponential function y = ka x passes through points (0, 2) and (2; 1.28). Calculate k and a. Graph it. a) Graph the exponential function y = 1.2x using a table of values to help you. b) What is the inverse function of y = 1.2x? Graph it on the same axes.

43

37

Graph the following functions: Z 2 ]]x if x < 1 x if x < 0 y = [1 a) y = ) b) x if x ≥ 0 ] x if x ≥ 1 \ 38 Calculate the value of the parameter k so that the following function is continuous: Z 2 ]]–x – kx – 5 if x ≤ –2 y = [1 if x > –2 ]2 x + 4 \ 39 Graph these functions: a) y = |x 2 – 1| – 2 b) y = 1 + |x | c) y = 1 – |x 2 – 6x + 5| 130

d) y = |x | – |4 – x |

44

The boiling point of water is 100 ºC at 0 m above sea level. For every 300 m increase in altitude, the boiling point decreases by 1 ºC. In other words, at 3 000 m above sea level the boiling point of water is 90 ºC. a) Write the function that relates the boiling point of water, T, to the altitude, a. b) Graph the function. c) What temperature does water boil at where you live? What about at the top of Mount Everest?

An arrow is shot upwards at a velocity of 40 m/s. Its height, h, at each moment in time, t, is h = 40t – 5t 2. a) Graph the function. b) What it is its domain of definition? c) At what moment in time does it reach its maximum height? What is that height? d) At what moment in time does the arrow hit the ground? e) In what interval of time is the arrow above a height of 35 metres?

Andrea spent €100 on a birthday present for Carlos. The rest of the friends in the group decide to share the cost of the present.

Write a function that shows the amount of money each friend has to contribute depending on the number of friends in the group and graph it.

Does it make sense to join the points? Why? Ana’s basic annual salary is €24 000. In her work contract, its says she will receive an 8 % raise per year. a) How much will she be earning in 10 years time? b) Write the function that relates the salary to time. c) For which values of the variable is it defined?

The intensity of the sound coming out of a speaker is inversely proportional to the square of the distance we are from it. For example: d, distance in metres I = 252 , d I, intensity of the sound Graph it taking 1 square = 1 m on the X axis, and 1 square = 5 u in the Y axis.

At what distance must a person with hearing disorders who can only hear sounds over 100 u be from the speaker?

Advanced problem solving

Remember the theory

45

51

Explain why these expressions cannot be piecewise-defined functions: Z ]]2x + 1 if x ≤ –2 x + 1 if x ≤ 1 a) y = [x 2 – 4 if –2 ≤ x ≤ 1 b) y = ) x – 1 if x > 0 ]2x + 1 if x ≥ 1 \

52

Graph and write the equation for each of the parabolas that meet these conditions: a) Its axis is x = 2, the coefficient of x 2 is –1 and intersects the X axis at a single point. b) It has the vertex at point (3, –2) and is of the type y = x 2. c) It has the vertex at the origin and passes through point (–3, –18).

53

Construct piecewise-defined functions that meet the following conditions and graph them:

Choose a suitable scale and graph the following: 2 a) y = x b) y = –75x 2 + 675 100 c) y = 0.002x 2 – 0.04x d) y = –10x 2 – 100x

46

Express the following functions in the form y = k + b and graph them: x–a a) y = 3x – 1 b) y = x + 1 c) y = 1 – x x –1 x +1 x+2 47 This year, Veronica has managed to harvest 240 kg of avocados that she can sell today at €1.20/kg. From now on, every day that goes by, 4 kg will go off, but the price will increase by €0.10/kg. When should she sell the avocados to get the most profit? What would that profit be?

48

49

50

The manufacturing cost per unit of a type of box decreases according to the number of units manufactured and is given by the function: y = 0.3x + 1000 x a) What values does x take ? b) Calculate the cost per unit and the total cost for manufacturing 10 and 100 000 boxes. c) How much do you think the cost per unit will get close to when the number of boxes gets very large?

What is the analytical expression of each of the parabolas graphed on the right?

a) It is continuous and made up of two pieces of straight lines. It passes through the origin and has a slope of –2 at x = 4. It has a maximum at (3, 7). b) It is continuous and is made up of a piece of parabola and a piece of straight line, in that order.

It has a relative minimum at (0, 0) and a relative maximum at (2, 4).

54

All the exponential functions of the type y = a x pass through the same point. State which point this is and explain why. For what values of a is the function decreasing?

55

True or false? a) The functions y = x and y = –x form a sideways parabola when plotted on the same axes. b) If the axis of a parabola is x = 2, it cannot pass through points (–1, 6) and (5, 8). c) The functions y = 4x and y = – 4x are symmetrical with respect to the Y axis. x d) The functions y = 4x and y = c 1 m are 4 symmetrical with respect to the Y axis. e) The function y = log3 x has two asymptotes, a vertical one and a horizontal one. f ) The functions y = log x and y = 10x are symmetrical with respect to the Y axis. g) The functions y = log x and y = 10 x are symmetrical with respect to the line y = x.

f

g

A pool has a diving board 8 m above the water.

Esther rolls a ball off the diving board and it falls 12 m from the vertical line of the diving board.

Write the equation for the trajectory of the ball from when 8m it leaves the diving board until it touches the water. What is its O 12 m domain of definition?

The trajectory is a parabola y = ax 2 + bx + c with its vertex at the point where it falls. Take O as the origin and bear in mind that the vertex is (0, 8).

131

Unit 5

OP

MATHS WORKSH LEARN

PRACTICE MAKES PERFECT!

Radioactive decay

• A candle lasts an hour. With the leftover wax of 10 candles you can make a new one.

Radioactive substances decay by emitting radiation and transforming into other substances. This process takes place over time and its rate varies greatly from one substance to another.

Uranium 92 radioactive substance 8 radiation + different substance 2 500 million The rate at which a radioactive substance decays is measured by its half-life, which is the time it takes for half of its original mass to decay. On the right, you can see the half-lives of some radioactive substances.

If we have an initial mass of 1 g, the amount of mass of this substance that will be left after a given time is:

U

238.029

89

Ac

227.028

t

M = c 1 m grams 2 where t is the time elapsed taking one half-life as a unit. If the substance was actinium and we wanted to express the time in years, the formula would be: t/28

M = c1m 2

where M is the amount of actinium remaining after t years.

a) How many hours of light will 442 candles give you?

88

Ra

years

226.025

81

Actinium 28 years 1

204.383

1 –– 2 1 –– 4

There are three naturally occurring isotopes of carbon: C12, C13 and C14. The first two are stable but the third one is radioactive and has a half-life of 5 700 years. This means that over this period of time the amount of C14 reduces by half.

Just like the other isotopes of carbon, C14 is present in the atmosphere (CO2) and is absorbed by plants (photosynthesis) which incorporate it in a specific proportion. C14 is then incorporated in the same proportion, via these plants, into other living things. When a living thing dies and becomes fossilised, its C14 continues to decay according to the function on the right.

Every 5 700 years, the amount of C14 left reduces by half. Therefore, by finding out the proportion of C14 in a fossil and checking it against the initial proportion (that in a living plant), the equation above can be used to find the fossil’s geological age, in other words, the time when it was formed.

If we express the time, t, in centuries, the equation above can be expressed like this: P = 100 · c 1 m 2

= 100 · >c 1 m 2

t

H ≈ 100 · 0.988t 8 P = 100 · 0.988t, t in centuries

In other words, in 1 century there will be 100 · 0.988 = 98.8 % of the C14 left, which means that 1.2 % will have decayed.

• What percentage of C14 will a 33 000-year-old fossil have compared to a living plant?

• How old must a fossil be if it only has 10 % of the C14 of a living plant?

Substitute P in the formula for 10 and isolate t taking logarithms.

132

b) How many candles are needed for 1 000 hours of light?

• A farmer went to the market to sell a basket of eggs. The first customer bought half her eggs plus half an egg. The second customer bought half of the remaining eggs plus half an egg, and the third one did the same. The seller then had no more eggs left.

How many eggs did she have at the beginning?

mass (g)

Carbon-14 dating

1 57

Thallium 3 minutes

Tl

1

t 57

Radon 1 620 years

• A farmer shared a flock of sheep among his children.

— He gave the eldest one sheep plus 1/7 of those remaining.

— He gave the second child two sheep plus 1/7 of those remaining.

— He gave the third child three sheep plus 1/7 of those remaining.

— And so on until he got to the youngest.

In this way, all of the children received the same inheritance and no sheep were left over.

How many children does the farmer have?

How many sheep were there in the flock?

2

3

4

5

time in half lives

P = 100 · c 1 m 2

t 5 700

SELF-ASSESSMENT

anayaeducacion.es Answer key and interactive self-assessment.

1 Graph the piecewise function that has the following

equation:

Z ]] 2x + 6 if x < –2 y = [ x/2 + 3 if –2 ≤ x < 2 ] –x + 6 if x ≥ 2 \ Is it continuous? Come up with a function with the same sections that is not continuous.

2 Find the vertex of each of the following parabolas

and graph them:

2 a) y = x – 2 2 c) y = (5 – x)(x + 1)

b) y = x 2 + 4x – 5 d) y = –(x – 3)2 – 1

e) y = 2x 2 + 4x f ) y = 9 – (x – 1)2 g) y = 2(x – 1)(x + 3)

h) y = (x + 2)2 – 2x 2

3 Express as piecewise-defined functions and graph

them:

a) y = |2x + 1| c) y = |–x 2 + 4x – 3|

Commitment

b) y = 1 – x 4 d) y = |9 – (x – 2)2|

4 Graph the following functions and find their

domains of definition: a) y =

1 b) y = 3 – 2 x x +5

d) y = x + 2 e) y = 2 x – 1

c) y =

3 +1 x –1

f ) y = – x – 3

5 Graph these pairs of functions:

a) y = 1.2x; y = log1.2 x b) y = 2.5x; y = log2.5 x With respect to which line are the two functions of each pair symmetrical?

6 Using a 3-metre-long strip of wood, we want to

make a picture frame. a) If the base of the frame is 0.5 m long, how tall is it? What is its surface area?

b) Which expression gives us the surface area, S, for any base, b ? Graph it. c) For which value of the base do we obtain the maximum surface area? What is the value of that surface area?

Watch the video for target 6.a. Think of something you can do to contribute to achieve that goal. Make a commitment to put your idea into practice.

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