14 - haese mathematics · chapter14 contents: a relations and functions ... f5, 8, 12, 16, 20g. y x...

Relations, functionsand sequences

14Chapter

Contents: A Relations and functionsB FunctionsC Function notationD Composite functions

E Transforming y = f(x)

F Inverse functionsG The modulus functionH Where functions meetI Number sequences

J Recurrence relationships

IB10 plus 2nd edmagentacyan yellow black

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Y:\HAESE\IB_10_PLUS-2ed\IB_10P-2ed_14\325IB_10P-2_14.CDR Wednesday, 13 February 2008 10:35:43 AM PETERDELL

OPENING PROBLEM

326 RELATIONS, FUNCTIONS AND SEQUENCES (Chapter 14)

A piece of paper measures 30 cm by 21 cm.

If squares of equal size are cut from its

corners, the shape remaining can be formed

into an open box.

Things to think about:

1 If 3 cm squares are removed, what is the box’s

a height b length c width d capacity?

2 If 5 cm squares are removed, what is the box’s

a height b length c width d capacity?

3 Does the capacity of the box depend on the size of the squares removed?

4 Is there a formula which allows us to connect the capacity of the box with the side

length x cm of the cut out squares?

5 If such a formula exists, how can it be used to answer questions like:

“What size squares should be removed to create the box of maximum capacity?”

Sue-Ellen wants to send a parcel to a friend. The cost of posting the parcel a fixed distance

will depend on its weight. So, cost is the dependent variable and weight is the independent

variable.

Postal Charges

Weight w kg Cost $C

1 kg 6 w < 2 kg $5:00

2 kg 6 w < 5 kg $8:00

5 kg 6 w < 10 kg $12:00

10 kg 6 w < 15 kg $16:00

15 kg 6 w 6 20 kg $20:00

The table alongside shows the costs of posting parcels

of various weights from Sue-Ellen to her friend.

For example, the table shows that it will cost $8:00 to

post a parcel weighing at least 2 kg but less than 5 kg.

It will therefore cost $8:00 to post a parcel weighing

2 kg or 3:6 kg or 4:955 kg.

We can illustrate the postal charges on a graph.

RELATIONS AND FUNCTIONSA

30 cm

21 cmopen

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25

0 5 10 15 20 25

cost ($)C

weight (kg)

An end point that is included has a filled in circle.

An end point that is not included has an open circle.

We can say that there is a relationship between the

variables weight and cost, so the table of costs is an

example of a relation.

A relation may be a finite number of ordered pairs, for example f(2, 8), (3, 8), (4, 8),

(5, 12)g, or an infinite number of ordered pairs such as the relation between the variables

weight and cost in the postal charges example above.


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RELATIONS, FUNCTIONS AND SEQUENCES (Chapter 14) 327

The table of postal charges shows weights of parcels of 1 kg up to and including 20 kg, so

we can write 1 6 w 6 20. This set of possible values of the independent variable is called

the domain of the relation, so in this case the domain is fw j 1 6 w 6 20g.

We will now look at relations and functions more formally.

RELATIONS

A relation is any set of points on the Cartesian plane.

The following are examples of relations:

(1) The set of 8 points represented by the dots is a

relation.

There is no equation connecting the variables xand y in this case.

(2) The set of all points on and within the illustrated

square is a relation.

It could be specified by

f(x, y) j ¡1 6 x 6 0, 0 6 y 6 1g

(3) The set of all points on this parabola is a relation.

It can be specified by an equation connecting all

points (x, y) lying on the curve

y = ¡x2 + 6x¡ 5.

DOMAIN AND RANGE

The domain of a relation is the set of possible values that x may have.

The range of a relation is the set of possible values that y may have.

The domain and range of a relation are often described using interval notation.

For example, consider the line segment from (1, 3) to (5, ¡2):

The domain consists of all real x such

that 1 6 x 6 5 and we write this as

fx j 1 6 x 6 5g.

The range is fy j ¡2 6 y 6 3g.

y

x�

�

y

x

� �

�

( )��,

y

x

( )��,

(5 -2),domain

range

the set of all such that

The costs for posting the parcels are $5, $8, $12, $16 and $20. This set of possible values

of the dependent variable is called the range of the relation, so in this case the range is

f5, 8, 12, 16, 20g.

y

x

�

�

�

�


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Y:\HAESE\IB_10_PLUS-2ed\IB_10P-2ed_14\327IB_10P-2_14.CDR Thursday, 7 February 2008 10:02:30 AM PETERDELL


In (1) on the previous page, the domain is f¡3, ¡2, ¡1, 1, 2gand the range is f¡2, ¡1, 0, 1, 2g.

In (2), the domain is fx j ¡1 6 x 6 0g and the range is fy j 0 6 y 6 1g.

In (3), the domain is fx j x 2 R g as x can take any real value, and

the range is fy j y 6 4g as the greatest value of y is 4 and yvalues could be very large and negative.

Further examples are:

(4) All values of x < 2 are possible.

) the domain is fx j x < 2g.

All values of y > ¡1 are possible.

) the range is fy j y > ¡1g.

(5) x can take any value.

) the domain is fx j x 2 R g.

y cannot be < ¡2) range is fy j y > ¡2g.

(6) x can take all values except x = 0:

) the domain is fx j x 6= 0g.

y can take all values except y = 0:

) the range is fy j y 6= 0g.

For each of the following graphs, state the domain and range:

a b

a Domain is fx j x 2 R g.

Range is fy j y 6 4g.

b Domain is fx j x > ¡4g.

Range is fy j y > ¡4g.

Example 1 Self Tutor

a bx

a bx

a bx

y

x

( )��

y

x

( )��y

x

( )��,

( )��,

y

x

xy 3

�

y

x

�

� �

xy 221 2��

a bx

For numbers between a and b we write a < x < b.

For numbers ‘outside’ a and b we write x < a or x > b.

would be written as a 6 x < b.

A filled in circle indicates the inclusion of the end point.

An open circle indicates the non-inclusion of that point.


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Y:\HAESE\IB_10_PLUS-2ed\IB_10P-2ed_14\328IB_10P-2_14.CDR Thursday, 14 February 2008 12:32:46 PM PETERDELL


EXERCISE 14A

1 State the domain and range of these relations:

a f(¡1, 5), (¡2, 3), (0, 4), (¡3, 8), (6, ¡1), (¡2, 3)gb f(5, 4), (¡3, 4), (4, 3), (2, 4), (¡1, 3), (0, 3), (7, 4)g.

2 For each of the following graphs, find the domain and range:

a b c

d e f

g h i

3 State the domain and range of:

a b c

A function is a relation in which no two different ordered pairs have the same first member.

So, f(¡1, 3), (2, 2), (¡1, ¡2), (3, 2), (4, ¡1)g is not a function.

FUNCTIONSB

y

x2

y

x

3

3�

�

y

x

( )��,

y

x

y

x

�

y

x

( )��,

y

x

( )��,y

x�

y

x1

y

x

( )��,

(2 ),�

( )��,

( )��,(- )��,

(boundaries not included)

y

x

y

x� �

�

�

different ordered pairswith same first member.

y

x

� �

�


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GEOMETRIC TEST FOR FUNCTIONS: ‘‘VERTICAL LINE TEST”

If we draw all possible vertical lines on the graph of a relation:

² the relation is a function if each line cuts the graph no more than once

² the relation is not a function if any line cuts the graph more than once.

Which of these relations are functions?

a b c

a

) we have a function.

b

) we have a function.

c

EXERCISE 14B

1 Which of the following sets of ordered pairs are functions? Give reasons for your

answers.

a (1, 1), (2, 2), (3, 3), (4, 4) b (¡1, 2), (¡3, 2), (3, 2), (1, 2)

c (2, 5), (¡1, 4), (¡3, 7), (2, ¡3) d (3, ¡2), (3, 0), (3, 2), (3, 4)

e (¡7, 0), (¡5, 0), (¡3, 0), (¡1, 0) f (0, 5), (0, 1), (2, 1), (2, ¡5)

2 Use the vertical line test to determine which of the following relations are functions:

a b c

d e f


y

x

y

x

y

x

y

x

This verticalline cuts thegraph twice.So, the relationis not afunction.

Every vertical line we coulddraw cuts the graph only once.

Every vertical line we coulddraw cuts the graph at mostonce.

y

x

y

x

y

x

y

x

y

x

y

x


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Y:\HAESE\IB_10_PLUS-2ed\IB_10P-2ed_14\330IB_10P-2_14.CDR Tuesday, 5 February 2008 11:53:27 AM PETERDELL


g h i

3 Will the graph of a straight line always be a function? Give evidence.

We sometimes use a ‘function machine’ to illustrate how functions behave.

For example:

So, if 3 is fed into the machine,

2(3) ¡ 1 = 5 comes out.

The above ‘machine’ has been programmed to perform a particular function.

If f is used to represent that particular function we can write:

f is the function that will convert x into 2x¡ 1.

So, f would convert 2 into 2(2) ¡ 1 = 3 and

¡4 into 2(¡4) ¡ 1 = ¡9:

This function can be written as: f : x 7! 2x¡ 1

function f such that x is converted into 2x¡ 1

Another way to write this function is: f(x) = 2x¡ 1

If f(x) is the value of y for a given value of x, then y = f(x).

Notice that for f(x) = 2x¡ 1, f(2) = 2(2) ¡ 1 = 3 and

f(¡4) = 2(¡4) ¡ 1 = ¡9:

Consequently, f(2) = 3 indicates that the point (2, 3)lies on the graph of the function.

Likewise, f(¡4) = ¡9 indicates that the point

(¡4, ¡9) also lies on the graph.

FUNCTION NOTATIONC

y

x

y

x

x

� ��x

I doublethe inputand thensubtract 1

y

x

( )��,

( )��

ƒ( )x x��

y

x89°


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Note: ² f(x) is read as “f of x” and is the value of the function at any value

of x.

² f is the function which converts x into f(x), i.e., f : x 7! f(x).

² f(x) is sometimes called the image of x.

If f : x 7! 3x2 ¡ 4x, find the value of: a f(2) b f(¡5)

f(x) = 3x2 ¡ 4x

a f(2) = 3(2)2 ¡ 4(2)

= 3 £ 4 ¡ 8

= 4

freplacing x by (2)g

b f(¡5) = 3(¡5)2 ¡ 4(¡5)

= 3(25) + 20

= 95

freplacing x by (¡5)g

If f(x) = 4 ¡ 3x¡ x2, find in simplest form: a f(¡x) b f(x + 2)

a f(¡x) = 4 ¡ 3(¡x) ¡ (¡x)2

= 4 + 3x¡ x2freplacing x by (¡x)g

b f(x + 2) = 4 ¡ 3(x + 2) ¡ (x + 2)2

= 4 ¡ 3x¡ 6 ¡ [x2 + 4x + 4]

= ¡2 ¡ 3x¡ x2 ¡ 4x¡ 4

= ¡x2 ¡ 7x¡ 6

freplacing x by (x + 2)g

EXERCISE 14C.1

1 If f : x 7! 2x + 3, find the value of:

a f(0) b f(2) c f(¡1) d f(¡5) e f(¡12)

2 If g : x 7! x +2

x, find the value of:

a g(1) b g(4) c g(¡1) d g(¡4) e g(¡12)

3 If f : x 7! 2x2 ¡ 3x + 2, find the value of:

a f(0) b f(3) c f(¡3) d f(¡7) e f(12)

4 If f(x) = 5 ¡ 2x, find in simplest form:

a f(a) b f(¡a) c f(a + 1) d f(x¡ 3) e f(2x)




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Y:\HAESE\IB_10_PLUS-2ed\IB_10P-2ed_14\332IB_10P-2_14.CDR Tuesday, 5 February 2008 11:58:54 AM PETERDELL

INVESTIGATION 1 FLUID FILLING FUNCTIONS


5 If P (x) = x2 + 4x¡ 3, find in simplest form:

a P (x + 2) b P (1 ¡ x) c P (¡x) d P (x2) e P (x2 + 1)

6 If R(x) =2x¡ 3

x + 2: a evaluate i R(0) ii R(1) iii R(¡1

2)

b find a value of x where R(x) does not exist

c find R(x¡ 2) in simplest form

d find x if R(x) = ¡5:

7 If the value of a car t years after purchase is given

by V (t) = 28 000 ¡ 4000t dollars:

a find V (4) and state what this value means

b find t when V (t) = 8000 and explain what

this represents

c find the original purchase price of the car.

THE DOMAIN OF A FUNCTION

To find the domain of a function, we need to consider what values of the variable make the

function undefined.

For example, notice that for:

² f(x) =px, the domain is fx j x > 0, x 2 R g since

px has meaning only when x > 0.

² f(x) =1px¡ 1

, the domain is fx j x > 1, x 2 R g since, when x ¡ 1 = 0 we are

‘dividing by zero’, and when x ¡ 1 < 0,px¡ 1 is undefined as we can’t find the

square root of a negative in the real number system.

EXERCISE 14C.2

1 Find the domain of the following:

a f(x) =px¡ 2 b f(x) =

p3 ¡ x c f(x) =

px +

p2 ¡ x

d f(x) =1px

e f(x) =1px

+1px + 2

f f(x) =1

xp

4 ¡ x

2 Check each of your answers to 1 using the graphing package.

When water is added to a container, the depth of water is given

by a function of time. If the water is added at a constant rate,

the volume of water added is directly proportional to the time

taken to add it.

So, for a container with a uniform cross-section, the graph of depth against time is a

straight line, or linear. We can see this in the following depth-time graph:

DEMO

GRAPHING

PACKAGE


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The question arises:

‘How does the shape of the container affect

the appearance of the graph?’

For example, consider the graph shown for

a vase of conical shape.

What to do:

1 For each of the following containers, draw a depth-time graph as water is added at

a constant rate.

a b c d

e f g h

2 Use the water filling demonstration to check your answers to question 1.

Sometimes functions are built up in two or more stages.

For example, consider the function F (x) =p

2x¡ 3:

If we let g(x) = 2x¡ 3 then F (x) =pg(x).

If we then let f(x) =px then F (x) = f(g(x)):

So, F (x) =p

2x¡ 3 is composed of f(x) =px and g(x) = 2x¡ 3.

Notice that g(f(x)) = g(px)

= 2px¡ 3, so in general f(g(x)) 6= g(f(x)):

Given f : x 7! f(x) and g : x 7! g(x), the composite function of f and g will

convert x into f(g(x)):

COMPOSITE FUNCTIONSD

water

depth

time

depth

time

depth


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To find welook at the function,

and whenever we seewe replace it by

within brackets.

f g x

f

x

g x

( ( ))

( )


If f(x) = 3x + 2 and g(x) = x2 + 4, find in simplest form:

a f(g(x)) b g(f(x)) c f(f(x))

a f(g(x))

= f(x2 + 4)

= 3(x2 + 4) + 2

= 3x2 + 12 + 2

= 3x2 + 14

b g(f(x))

= g(3x + 2)

= (3x + 2)2 + 4

= 9x2 + 12x + 4 + 4

= 9x2 + 12x + 8

c f(f(x))

= f(3x + 2)

= 3(3x + 2) + 2

= 9x + 6 + 2

= 9x + 8

EXERCISE 14D

1 If f(x) = 3x¡ 4 and g(x) = 2 ¡ x, find in simplest form:

a f(g(x)) b g(f(x)) c f(f(x)) d g(g(x))

2 If f(x) =px and g(x) = 4x¡ 3, find in simplest form:

a f(g(x)) b g(f(x)) c f(f(x)) d g(g(x))

3 Find an f and a g function such that:

a f(g(x)) =px¡ 3 b f(g(x)) = (x + 5)3 c f(g(x)) =

5

x + 7

d g(f(x)) =1p

3 ¡ 4xe g(f(x)) = 3x

2

f g(f(x)) =

µx + 1

x¡ 1

¶2

4 If f(x) = 3x + 1 and g(x) = x2 + 2x, find x when f(g(x)) = 10:

5 If f(x) = 2x + 1 and g(f(x)) = 4x2 + 4x + 3, find g(x):

6 If g(x) = 1 ¡ 3x and f(g(x)) = 9x2 ¡ 6x¡ 2, find f(x).

In Chapter 7 on transformation geometry we observed that under a translation with vectorµhk

¶the equation of an image is found by replacing x by (x¡ h) and y by (y ¡ k).

For example, y = 2x2 under

µhk

¶becomes y ¡ k = 2(x¡ h)2.

TRANSFORMING y = f(x)E



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In general, y = f(x) under

µhk

¶becomes y ¡ k = f(x¡ h):

The following table shows the effect of different transformations on y = f(x):

TransformationTransformation

equations equationsy = f(x) becomes

Translation

µhk

¶ ½x0 = x + h

y0 = y + k

½x = x0 ¡ h

y = y0 ¡ ky ¡ k = f(x¡ h)

² the x-axis

½x0 = x

y0 = ¡y

½x = x0

y = ¡y0¡y = f(x) or y = ¡f(x)

² the y-axis

½x0 = ¡x

y0 = y

½x = ¡x0

y = y0y = f(¡x)

² the line y = x

½x0 = y

y0 = x

½x = y0

y = x0x = f(y)

Dilation:

² factor k

½x0 = kx

y0 = ky

8><>:x =

x0

k

y =y0

k

y

k= f

³xk

´or y = kf

³xk

´

² vertical,

factor k

½x0 = x

y0 = ky

8<:x = x

y =y0

k

y

k= f (x) or y = kf(x)

² horizontal,

factor k

½x0 = kx

y0 = y

8<: x =x0

ky = y0

y = f³xk

´

We see that if y can be written as a function of x then the image of y = f(x) can be

found using the listed transformations.

Find the image equation of:

a y = x3 under a reflection in the line y = x

b y = 4x¡ 1 under a horizontal dilation with factor 2.

a y = f(x) = x3

under a reflection in y = xbecomes

x = f(y) = y3

) y = 3px

b y = f(x) = 4x¡ 1under a horizontal dilation with

factor 2 becomes

y = f¡x2

¢= 4

¡x2

¢¡ 1

) y = 2x¡ 1


Reflection in:

Rearranged

0


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INVESTIGATION 2 INVERSE FUNCTIONS


EXERCISE 14E

1 Find the image equation of:

a y = ¡3x2 under a translation of

µ5¡2

¶b y =

px under a reflection in the x-axis

c y = 4x + 2 under a vertical dilation with factor 12

d y = 3 ¡ 5x under a reflection in the line y = x

e y = x2 ¡ 3x + 1 under a horizontal dilation with factor 12

f y = x2 + x under a reflection in the y-axis.

2 If f(x) = x2, find and graph:

a b c d

In each case state the transformation that has occurred from y = f(x).

3 The graph of y = f(x) is given below. Without finding image equations as in question

1, sketch each of the following on the same set of axes:

a y = f(x) and y = f(x¡ 1) ¡ 2

b y = f(x) and y = ¡f(x)

c y = f(x) and y = f(¡x)

d y = f(x) and y = 2f(x)

e y = f(x) and y = f(2x):

In this investigation we explore the concept of an inverse function. We

will see how it relates to composite functions and also to one of the

transformations studied in the previous section.

What to do:

1 Consider f(x) = 3x + 2 which has graph y = 3x + 2:

a Interchange x and y and then make y the subject of this new equation.

b Let g(x) be the new function in a. Hence show that f(3) = 11 and g(11) = 3.

c From b notice that g(11) = g(f(3)) = 3: Show that f(g(3)) = 3 also.

d Prove that f(g(x)) = x and g(f(x)) = x for f and g above.

e f(x) and g(x) are said to be inverse functions, and g(x), the inverse function

of f(x), is often written as f¡1(x).

Find f¡1(x) for f(x) = 3 ¡ 4x using the above method, and check that

f(f¡1(x)) = f¡1(f(x)) = x.

INVERSE FUNCTIONSF

y

x

y x�� ( )

( )��, ( )��,

y = f(¡x) y = ¡f(x) y = f(3x) y = 2f(x).


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2 a Draw the graph of f(x) and f¡1(x) for f(x) = 3x + 2 on the same

set of axes and also draw the graph of y = x.

b Draw the graph of f(x) and f¡1(x) for f(x) = 3 ¡ 4x on the same

set of axes and also draw the graph of y = x.

c What do you notice from a and b?

INVERSE FUNCTIONS

The inverse function f¡1(x) of the function y = f(x)

is the reflection of y = f(x) in the line y = x.

f¡1(x) has the property that f(f¡1(x)) = f¡1(f(x)) = x:

f¡1(x) can be found algebraically by interchanging x and y and then making y the subject

of the resulting formula. The new y is f¡1(x):

Consider f(x) = 12x¡ 1:

a Find f¡1(x): b Check that f(f¡1(x)) = f¡1(f(x)) = x:

c Sketch y = f(x), y = f¡1(x) and y = x on the same axes.

a y = 12x¡ 1 has inverse function x = 1

2y ¡ 1 finterchanging x and yg) 2x = y ¡ 2

) y = 2x + 2

) f¡1(x) = 2x + 2

b f(f¡1(x)) = f(2x + 2)

= 12(2x + 2) ¡ 1

= x + 1 ¡ 1

= x

f¡1(f(x)) = 2f(x) + 2

= 2(12x¡ 1) + 2

= x¡ 2 + 2

= x

c


y = f¡1(x) is a

reflection of

y = f(x) in the liney = x.

y

x

2

2

-1

-1

y x��ƒ�( ) y x��ƒ( )

y x��


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EXERCISE 14F

1 For each of the following functions:

i find f¡1(x) ii sketch y = f(x), y = f¡1(x) and y = x on the same axes.

a f(x) = 2x + 5 b f(x) =3 ¡ 2x

4c f(x) = x + 3

2 Copy the following graphs and draw the graph of each inverse function:

a b c

3 If f(x) = 2x + 7, find:

a f¡1(x) b f(f¡1(x)) c f¡1(f(x))

4 If f(x) =2x + 1

x + 3, find:

a f¡1(x) b f(f¡1(x)) c f¡1(f(x))

5 a Sketch the graph of y = x2 and reflect it in the line y = x.

b Does f(x) = x2 have an inverse function?

c Does f(x) = x2, x > 0 have an inverse function?

6 The horizontal line test says that ‘for a function to have an inverse function, no

horizontal line can cut it more than once’.

a Explain why this is a valid test for the existence of an inverse function.

b Which of the following have an inverse function?

i ii iii iv

�

�

y

x

y

x

�

�

y

x

1

y

x-3

3

y

x

3

y

x

y

x

2

�

7 a Explain why y = x2¡2x+5 is a function but does not have an inverse function.

b Explain why y = x2 ¡ 2x + 5, x > 1 has an inverse function.

c Show that the inverse function of b is f¡1(x) = 1 +px¡ 4:

Hint: Swap x and y then use the quadratic formula to solve for y in terms of x.


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INVESTIGATION 3 THE MODULUS FUNCTION


What to do:

1 Suppose y = x if x > 0 and y = ¡x if x < 0.

Find y for x = 5, 7, 12 , 0, ¡2, ¡8, and ¡10:

2 Draw the graph of y =

½x if x > 0

¡x if x < 0using the results of 1.

3 Consider M(x) =px2 where x2 must be found before finding the square root.

Find the values of M(5), M(7), M(12), M(0), M(¡2), M(¡8) and M(¡10).

4 What conclusions can be made from 1 and 3?

MODULUS

The modulus or absolute value of a real number is its size, ignoring its sign.

We denote the modulus of x by jxj.

For example, the modulus or absolute value of 7 is 7, and

the modulus or absolute value of ¡7 is also 7.

GEOMETRIC DEFINITION OF MODULUS

jxj is the distance of x from 0 on the number line.

Because the modulus is a distance, it cannot be negative.

If x > 0 : If x < 0 :

For example:

ALGEBRAIC DEFINITION OF MODULUS

From Investigation 3, jxj =½

x if x > 0

¡x if x < 0or jxj =

px2

y = jxj has graph:

THE MODULUS FUNCTIONG

7 7

-7 �

| |x

0 x

| |x

0x

y

x

This branchis , 0.y x x�� "

This branch

is , 0.y x x� >


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Y:\HAESE\IB_10_PLUS-2ed\IB_10P-2ed_14\340IB_10P-2_14.CDR Tuesday, 5 February 2008 1:54:47 PM PETERDELL


If a = ¡7 and b = 3 find: a ja + bj b jabj

a ja + bj= j¡7 + 3j= j¡4j= 4

b jabj= j¡7 £ 3j= j¡21j= 21

Draw the graph of f(x) = x + 2 jxj :

If x > 0,

f(x) = x + 2(x) = 3x

If x < 0,

f(x) = x + 2(¡x) = ¡x

EXERCISE 14G.1

1 If x = ¡4, find the value of:

a jx + 6j b jx¡ 6j c j2x + 3j d j7 ¡ xj

e jx¡ 7j f¯̄x2 ¡ 6x

¯̄g¯̄6x¡ x2

¯̄h

jxjx + 2

2 a Find the value of a2 and jaj2 if a is:

i 3 ii 0 iii ¡2 iv 9 v ¡9 vi ¡20

b What do you conclude from the results in a?

a b jabj jaj jbj¯̄̄ab

¯̄̄ jajjbj

12 3

12 ¡3

¡12 3

¡12 ¡3

3 a Copy and complete:

b

4 In Investigation 3 we discovered that jxj =px2. Using this we find jxj2 = x2.

Consequently jx¡ 2j2 = (x¡ 2)2.

Prove the following properties of modulus for all real numbers a and b:

a jabj = jaj jbj b ja¡ bj = jb¡ aj c¯̄̄ab

¯̄̄=

jajjbj provided b 6= 0.

Hint: In a, (LHS)2 = jabj2 = a2b2. In b, ja¡ bj = j¡(b¡ a)j :



Perform all operationsinside the modulus signsbefore actually finding

the modulus.

What can you conclude fromthe results in ?a

y

x

y x��y x��


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a b ja + bj jaj + jbj ja¡ bj jaj ¡ jbj2 5

2 ¡5

¡2 5

¡2 ¡5

5 a Copy and complete:

b

6 On the same set of axes, graph the following pairs of modulus functions. Clearly state

the transformation which maps the first function on to the second.

a y = jxj and y = jx¡ 2j b y = jxj and y = jx + 2jc y = jxj and y = 2 jxj d y = jxj and y = ¡ jxje y = jxj and y = jxj + 2 f y = jxj and y = jx¡ 1j + 3

7 By replacing jxj with x for x > 0 and (¡x) for x < 0 write the following functions

without the modulus sign and hence graph each function:

a y = ¡ jxj b y = jxj + x c y =jxjx

d y = x¡ 2 jxj8 Use technology to graph:

a y = j(x¡ 2)(x¡ 4)j b y = jx(x¡ 3)jHow are these functions related to y = (x¡ 2)(x¡ 4)

and y = x(x¡ 3)?

SUMMARY OF MODULUS PROPERTIES

² jxj is never negative, i.e., jxj > 0 for all x

² jxj2 = x2 for all x

² jxyj = jxj jyj for all x and y

²¯̄̄̄x

y

¯̄̄̄=

jxjjyj for all x and y, y 6= 0

² ja¡ bj = jb¡ aj for all a and b.

MODULUS EQUATIONS

It is clear that jxj = 2 has two solutions x = 2 and x = ¡2.

In general, if jxj = a where a > 0, then x = §a.

Proof: If jxj = a

then jxj2 = a2

) x2 = a2

) x2 ¡ a2 = 0

) (x + a)(x¡ a) = 0

) x = §a

We use this rule to solve modulus equations.

What can you concludefrom the results in ?a

GRAPHING

PACKAGE

GRAPHING

PACKAGE


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Solve for x: a j2x + 5j = 1 b j2x + 5j = ¡2

a j2x + 5j = 1

) 2x + 5 = §1

) 2x = 1 ¡ 5 or ¡1 ¡ 5

) 2x = ¡4 or ¡ 6

) x = ¡2 or ¡ 3

b j2x + 5j = ¡2has no solution as the LHS

is never negative.

EXERCISE 14G.2

1 Solve for x:

a jxj = 5 b jxj = ¡6 c jxj = 0

d jx + 1j = 4 e j4 ¡ xj = 5 f jx + 6j = ¡3

g j3x¡ 5j = 2 h j3 ¡ 2xj = 4 i j12 ¡ 7xj = 5

j

¯̄̄̄2x

x + 1

¯̄̄̄= 5 k

¯̄̄̄2x¡ 1

x + 3

¯̄̄̄= 2 l

¯̄̄̄x + 4

1 ¡ 2x

¯̄̄̄= 3

4

2 Use technology to graph:

a y = j3 ¡ 2xj and y = 4

b y =

¯̄̄̄2x¡ 1

x + 3

¯̄̄̄and y = 2

Consider the graphs of a quadratic function and a linear function on the same set of axes.

Notice that we could have:

If the graphs meet, the coordinates of the point(s) of intersection can be found by solving the

two equations simultaneously.

WHERE FUNCTIONS MEETH


GRAPHING

PACKAGE

cutting

(2 points of intersection)

touching

(1 point of intersection)

missing

(no points of intersection)

and hence check your answer to 1h

and hence check your answer to .1k


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Find the coordinates of the points of intersection of the graphs with equations

y = x2 ¡ x + 3 and y = 2x + 7:

The graphs meet when x2 ¡ x + 3 = 2x + 7

) x2 ¡ 3x¡ 4 = 0

) (x + 1)(x¡ 4) = 0

) x = ¡1 or 4

Substituting into y = 2x + 7, when x = ¡1, y = 2(¡1) + 7 = 5

when x = 4, y = 2(4) + 7 = 15

) the graphs meet at (¡1, 5) and (4, 15).

EXERCISE 14H

1 Find the coordinates of the point(s) of intersection of the graphs with equations:

a y = x2 + 2x¡ 1 and y = x + 5 b y =2

xand y = x¡ 1

c y = 3x2 + 4x¡ 1 and y = x2 ¡ 3x¡ 4 d y =1

xand y = 5x¡ 4

2 Use a graphing package or a graphics calculator to find the coordinates of the points

of intersection (to two decimal places) of the graphs with equations:

a y = x2 + 3x + 1 and y = 2x + 2

b y = x2 ¡ 5x + 2 and y =3

x

c y = ¡x2 ¡ 2x + 5 and y = x2 + 7

d y = x2 ¡ 1 and y = x3:

NUMBER SEQUENCESI


GRAPHING

PACKAGE

Consider the illustrated pattern of circles:

The first layer has just one blue ball.

The second layer has three pink balls.

The third layer has five black balls.

The fourth layer has seven green balls.

If we let un represent the number of balls in the nth layer then u1 = 1, u2 = 3,

u3 = 5, and u4 = 7.

The pattern could be continued forever, generating the sequence of numbers:

1, 3, 5, 7, 9, 11, 13, 15, 17, ......

The string of dots indicates that the pattern continues forever.


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A sequence is a function whose domain is the set of all positive integers

1, 2, 3, 4, 5, ......

The sequence for the pattern of balls can be specified:

² using words “The set of all odd numbers starting with 1.”

² using an explicit formula un = 2n + 1 generates all terms.

² using a recursive formula u1 = 1 and un+1 = un + 2 for all n > 1

Check: u1 = 1

u2 = u1 + 2 = 1 + 2 = 3 X

u3 = u2 + 2 = 3 + 2 = 5 X

ARITHMETIC SEQUENCES

An arithmetic sequence is a sequence in which each term differs from the previous one

by the same fixed number. We call this number the common difference.

For example: 1, 5, 9, 13, 17, .... is arithmetic as 5 ¡ 1 = 9 ¡ 5 = 13 ¡ 9 = 17 ¡ 13, etc.

Likewise, 42, 37, 32, 27, ....

Notice in the sequence 1, 5, 9, 13, 17, .... that u1 = 1

u2 = 1 + 4

u3 = 1 + 4 + 4

u4 = 1 + 4 + 4 + 4

= 1 + 1(4)

= 1 + 2(4)

= 1 + 3(4), etc.

This suggests that: If un is arithmetic then the nth term is un = u1 + (n ¡ 1)d

where u1 is the first term and d is the constant common difference.

Consider the sequence 3, 9, 15, 21, 27, .....

a Show that the sequence is arithmetic.

b Find the formula for the general term un.

c Find the 100th term of the sequence.

d Is i 489 ii 1592 a member of the sequence?

a 9 ¡ 3 = 6, 15 ¡ 9 = 6, 21 ¡ 15 = 6, 27 ¡ 21 = 6So, assuming that the pattern continues, consecutive terms differ by 6:) the sequence is arithmetic with u1 = 3 and d = 6.

b un = u1 + (n¡ 1)d ) un = 3 + 6(n¡ 1) i.e., un = 6n¡ 3

c If n = 100, u100 = 6(100)¡ 3 = 597:


un is called the nth term or the general term.

is arithmetic as 37 ¡ 42 = 32 ¡ 37 = 27 ¡ 32, etc.


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d i Let un = 489

) 6n¡ 3 = 489

) 6n = 492

) n = 82

) 489 is a term of the sequence.

In fact it is the 82nd term.

ii Let un = 1592

) 6n¡ 3 = 1592

) 6n = 1595

) n = 26556

which is not possible as n is an

integer.

) 1592 cannot be a term.

Find k given that k + 5, ¡1 and 2k¡ 1 are consecutive terms of an arithmetic

sequence, and hence find the terms.

Since the terms are consecutive,

¡1 ¡ (k + 5) = (2k ¡ 1) ¡ (¡1) fequating common differencesg) ¡1 ¡ k ¡ 5 = 2k ¡ 1 + 1

) ¡k ¡ 6 = 2k

) ¡6 = 3k

) k = ¡2

) the terms are 3, ¡1, ¡5:

Find the general term un for an arithmetic sequence given that

u3 = 4 and u7 = ¡24:

u7 ¡ u3 = (u1 + 6d) ¡ (u1 + 2d)

= u1 + 6d¡ u1 ¡ 2d

= 4d

But u7 ¡ u3 = ¡24 ¡ 4 = ¡28

) 4d = ¡28

) d = ¡7

Now un = u1 + (n¡ 1)d

) u3 = u1 + (2)(¡7)

) 4 = u1 ¡ 14

) u1 = 18

) un = 18 + (n¡ 1)(¡7)

) un = 18 ¡ 7n + 7

) un = 25 ¡ 7n

Check:

u3 = 25 ¡ 7(3)

= 25 ¡ 21

= 4 X

u7 = 25 ¡ 7(7)

= 25 ¡ 49

= ¡24 X




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EXERCISE 14I.1

1 Consider the sequence 4, 11, 18, 25, 32, .....

a Show that the sequence is arithmetic. b Find the formula for its general term.

c Find its 30th term. d Is 340 a member?

e Is 738 a member?

2 Consider the sequence 67, 63, 59, 55, .....

a Show that the sequence is arithmetic. b Find the formula for its general term.

c Find its 60th term. d Is ¡143 a member?

e Is 85 a member?

3 An arithmetic sequence is defined by un = 11n¡ 7:

a Find u1 and d. b Find the 37th term.

c What is the least term of the sequence which is greater than 250?

4 A sequence is defined by un =21 ¡ 4n

2:

a Prove that the sequence is arthmetic. b Find u1 and d. c Find u55:

d For what values of n are the terms of the sequence less than ¡300?

5 Find k given the consecutive arithmetic terms:

a 31, k, 13 b k, 8, k + 11 c k + 2, 2k + 3, 17

6 Find the general term un for an arithmetic sequence given that:

a u4 = 37 and u10 = 67 b u5 = ¡10 and u12 = ¡38

c the fourth term is ¡4 and the fifteenth term is 29

d the tenth and sixth terms are ¡16 and ¡13 respectively.

7 Consider the finite arithmetic sequence 3, 212 , 2, ...., ¡6.

a Find u1 and d. b How many terms does the sequence have?

8 An arithmetic sequence starts 17, 24, 31, 38, ..... What is the first term of the sequence

to exceed 40 000?

GEOMETRIC SEQUENCES

A sequence is geometric if each term can be obtained from the previous one by multiplying

by the same non-zero constant.

For example: 2, 6, 18, 54, .... is a geometric sequence as

2 £ 3 = 6 and 6 £ 3 = 18 and 18 £ 3 = 54.

Notice that u2 = u1 £ 3

and u3 = u2 £ 3 = u1 £ 3 £ 3 = u1 £ 32,

u4 = u3 £ 3 = u1 £ 32 £ 3 = u1 £ 33,

u5 = u4 £ 3 = u1 £ 33 £ 3 = u1 £ 34.


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This suggests the following algebraic definition:

If un is geometric then un = u1 £ rn¡1 for all positive integers n.

u1 is the first term and r is a constant called the common ratio.

Notice: ² r is called the common ratio becauseun+1

un= r for all n:

² 2, 6, 18, 54, .... is geometric with r = 3.

² 2, ¡6, 18, ¡54, .... is geometric with r = ¡3.

For the sequence 16, 8, 4, 2, 1, ......

a Show that the sequence is geometric. b Find the general term un.

c Hence, find the 10th term as a fraction.

a 816 = 1

2 , 48 = 1

2 , 24 = 1

2 , 12 = 1

2

So, assuming the pattern continues, consecutive terms have a common ratio of 12 :

) the sequence is geometric with u1 = 16 and r = 12 :

b un = u1rn¡1 ) un = 16 £ ¡12¢n¡1

or un = 24 £ (2¡1)n¡1

= 24 £ 2¡n+1

= 24+(¡n+1)

= 25¡n

c u10 = 16 £ (12)9 =24

29=

1

25= 1

32

k ¡ 1, k + 2 and 3k are consecutive terms of a geometric sequence. Find k.

Equating common ratios gives3k

k + 2=

k + 2

k ¡ 1

) 3k(k ¡ 1) = (k + 2)2

) 3k2 ¡ 3k = k2 + 4k + 4

) 2k2 ¡ 7k ¡ 4 = 0

) (k ¡ 4)(2k + 1) = 0

) k = 4 or ¡ 12

Check: If k = 4, the terms are: 3, 6, 12. X fr = 2gIf k = ¡1

2 , the terms are: ¡32 , 3

2 , ¡32 . X fr = ¡1g




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A geometric sequence has u2 = ¡5 and u5 = 40. Find its general term.

u2 = u1r = ¡5 ...... (1) fusing un = u1rn¡1 with n = 2g

and u5 = u1r4 = 40 ...... (2) fusing un = u1r

n¡1 with n = 5g

So,u1r

4

u1r=

40

¡5f(2) ¥ (1)g

) r3 = ¡8

) r = 3p¡8

) r = ¡2

and so in (1) u1(¡2) = ¡5

) u1 = 52

Thus un = 52 £ (¡2)n¡1

EXERCISE 14I.2

1 For the geometric sequence with first two terms given, find b and c:

a 3, 6, b, c, .... b 8, 2, b, c, ..... c 15, ¡5, b, c, .....

2 a Show that the sequence 1, 3, 9, 27, ..... is geometric.

b Find un and hence find the 10th term.

3 a Show that the sequence 40, ¡20, 10, ¡5, ..... is geometric.

b Find un and hence find the 12th term as a fraction.

4 Show that the sequence 16, ¡4, 1, ¡0:25, .... is geometric and hence find the 8th

term as a decimal.

5 Find the general term of the geometric sequence: 3, 3p

2, 6, 6p

2, ......

6 Find k given that the following are consecutive terms of a geometric sequence:

a k, 2, 6 b 4, 6, k c k, 2p

2, k2

d 3, k, 27 e k, 3k, 10k + 7 f k, k + 4, 8k + 2

7 Find the general term un of the geometric sequence which has:

a u3 = 16 and u8 = 512 b u3 = 32 and u6 = ¡4

c u7 = 24 and u15 = 384 d u3 = 3 and u9 = 38 :

8 A geometric sequence has general term un with u3 = 12 and u7 = 34 . Find u12.

9 un is the general term of a geometric sequence.

a If u2 = ¡212 and u5 = 5

16 , find u10.

b If u3 = 7 and u8 = ¡7, find u88.

c If u3 = 18 and u5 = 162, find u11.



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Perhaps the most famous example of a recurrence relationship

is that which generates the Fibonacci sequence:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ......

The pattern starts with two 1s, and after this each term is

obtained by adding the two terms preceding it:

1 + 1 = 2

1 + 2 = 3

2 + 3 = 5

3 + 5 = 8...

Leonardo Fibonacci

We can hence write down a recurrence relationship:

u1 = u2 = 1 and un+2 = un+1 + un for n = 1, 2, 3, 4, 5, 6, 7, ......

Check: If n = 1,

If n = 2,

If n = 3,

u3 = u2 + u1 = 1 + 1 = 2 X

u4 = u3 + u2 = 2 + 1 = 3 X

u5 = u4 + u3 = 3 + 2 = 5 X

Leonardo Fibonacci (1170-1250) noticed that the sequence 1, 1, 2, 3, 5, 8, ...... occurred

frequently in the natural world.

For example, he noticed that flowers of particular species have the same number of petals

and that these numbers are all members of the Fibonacci sequence.

3 petals lily, iris

5 petals buttercup

8 petals delphinium

13 petals cineraria

21 petals aster

34 petals pyrethrum

Fibonacci also observed the number sequence in:

² the number of leaves arranged about the stem in plants

² the seed patterns of the sunflower

² the seed pattern on the pine cone.

The recurrence relationship is very useful for dealing with the Fibonacci sequence because

its explicit formula is very complicated. In fact, it is:

un =1p5

"Ã1 +

p5

2

! n

¡Ã

1 ¡p5

2

! n#for n = 1, 2, 3, 4, 5, .....

RECURRENCE RELATIONSHIPSJ


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ARITHMETIC SEQUENCES

The sequence 4, 7, 10, 13, .... can be generated using un = 3n + 1 or by using the

recurrence relationship:

“u1 = 4 and un+1 = un + 3 where n is a positive integer”.

This is easily seen as u1 = 4 and u2 = u1 + 3 = 4 + 3

and u3 = u2 + 3 = 7 + 3

and u4 = u3 + 3 = 10 + 3

= 7

= 10

= 13 etc.

In general, if

then

un = u1 + (n¡ 1)d

un+1 = un + d.

where u1 and d are constants

GEOMETRIC SEQUENCES

The sequence 3, 6, 12, 24, .... can be generated using the general term formula

un = 3 £ 2n¡1 where n is a positive integer.

However, the same sequence can be generated using the recurrence relationship:

“u1 = 3 and un+1 = 2un where n is a positive integer”.

We have u1 = 3

and u2 = 2u1 = 2 £ 3 = 6 fletting n = 1gand u3 = 2u2 = 2 £ 6 = 12 fletting n = 2g

etc.

In general, if

then

un = u1rn¡1

un+1 = run .

where u1 and r are constants

Find the next three members of the sequence given by:

a u1 = 7 and un+1 = un ¡ 2 b u1 = 24 and un+1 = 12un

a If n = 1, u2 = u1 ¡ 2

= 7 ¡ 2

= 5

If n = 2, u3 = u2 ¡ 2

= 5 ¡ 2

= 3

If n = 3, u4 = u3 ¡ 2

= 3 ¡ 2

= 1

b If n = 1, u2 = 12u1

= 12 £ 24

= 12

If n = 2, u3 = 12u2

= 12 £ 12

= 6

If n = 3, u4 = 12u3

= 12 £ 6

= 3



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Find the explicit formula for un if:

a un+1 = un + 7 and u1 = 5 b un+1 = 0:4un and u1 = 100

a We identify this sequence as arithmetic

with d = 7:Now un = u1 + (n¡ 1)d

) un = 5 + (n¡ 1)7

) un = 7n¡ 2

b We identify this sequence as geometric

with r = 0:4 .

Now un = u1rn¡1

) un = 100 £ (0:4)n¡1

Sequences do not have to be arithmetic or geometric.

For example, the Fibonacci sequence is neither arithmetic nor geometric.

Given u1 = 1 and un+1 = 2un + 3 for all positive

integers n, find the next three terms of the sequence.

Letting n = 1, u2 = 2u1 + 3 = 2(1) + 3 = 5

Letting n = 2, u3 = 2u2 + 3 = 2(5) + 3 = 13

Letting n = 3, u4 = 2u3 + 3 = 2(13) + 3 = 29

The Fibonacci sequence is also an example of a recurrence relationship that relates any term

to the previous two terms.

Given u1 = 1, u2 = 3 and un+2 = un+1¡un for all

positive integers n, find the next three terms of the sequence.

Letting n = 1, u3 = u2 ¡ u1 = 3 ¡ 1 = 2

Letting n = 2, u4 = u3 ¡ u2 = 2 ¡ 3 = ¡1

Letting n = 3, u5 = u4 ¡ u3 = ¡1 ¡ 2 = ¡3:

EXERCISE 14J

1 Find the next three members of the sequence defined by:

a u1 = 2 and un+1 = un + 3 b u1 = 11:7 and un+1 = un ¡ 2:1

c u1 = 1268 and un+1 = un + 23:9 d u1 = 4 and un+1 = 3un

e u1 = 1000 and un+1 = 110un f u1 = 128 and un+1 = ¡1

2un:





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INVESTIGATION 4 SEQUENCES IN FINANCE


2 Find the explicit formula for un if:

a u1 = 41 and un+1 = un + 5 b u1 = 11:8 and un+1 = un ¡ 1:7

c u1 = 12 and un+1 = 13un d u1 = 36 and un+1 = ¡1

2un:

3 Find the next three members of the sequence defined by:

a u1 = 1 and un+1 = 2un ¡ 1 b un+1 = 3 ¡ un and u1 = 2

c un+1 = 13un ¡ 1 and u1 = 3 d un+1 = u 2

n ¡ 1 and u1 = 1:

4 Find the next four terms of the sequence defined by:

a u1 = 1, u2 = 0 and un+2 = un+1 ¡ un

b u1 = u2 = 1 and un+2 = 2un+1 + 3un:

Arithmetic and geometric sequences are observed in many financial

calculations.

What to do:

1 $1000 is invested at a simple interest rate of 7% per year with the interest paid at

the end of each year.

After 1 year, its value is $1000 £ 1:07 fto increase by 7% we multiply by 107%gAfter 2 years, its value is $1000 £ 1:14 fan increase of 14% on the originalg

a Find the value of the investment at the end of:

i 3 years ii 4 years iii 5 years.

b Do the amounts form an arithmetic or geometric sequence or neither of these?

Give reasons for your answer.

c Give an explicit formula for finding any term un of the sequence being generated.

d Give a recursive formula for the sequence.

2

After 1 year, its value is $6000 £ 1:07

After 2 years, its value is ($6000£ 1:07) £ 1:07 = $6000 £ (1:07)2

a Explain why the amount after 3 years is given by $6000 £ (1:07)3.

b Write down, in the same form, the amount after:

i 4 years ii 5 years iii 6 years.

LINKSclick here

FIBONACCI

Areas of interaction:Human ingenuity

$6000 is invested at a fixed rate of 7% p.a. compound interest over a lengthy period.

Interest is paid at the end of each year.


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REVIEW SET 14A


c Do the amounts at the end of each year form an arithmetic or geometric sequence

or neither of these? Give reasons for your answer.

d Give an explicit formula for finding any term un of the sequence being generated.

e Give a recursive formula for the sequence.

3 A photocopier originally cost $12 000 and it depreciates in value (loses value) by

20% each year.

a Find its value at the end of:

i one year ii two years iii three years.

b Do the resulting annual values form an

arithmetic or geometric sequence?

c If the answer to b is yes, find:

i an explicit formula for finding its

value at the end of the nth year

ii a recursive formula for the value.

1 For the following graphs, determine:

i the range and domain ii whether the relation is a function.

a b

c d

2 If f(x) = 3x¡ x2, find: a f(2) b f(¡1) c f(12):

3 If g(x) = x2 ¡ 3x, find in simplest form: a g(¡x) b g(x + 2):

4 If f(x) = 5 ¡ 2x and g(x) = x2 + 2x find:

a f(g(x)) b x when g(x) = 20 c f¡1(x) in the form ax + b.

5 Consider f(x) = j2x¡ 4j :a Sketch the graph of the function f(x).

b Find the solutions of j2x¡ 4j = 5 from the graph in a.

c Check your answer to b by solving the equation algebraically.

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y

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y

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(-2 ), 5y

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REVIEW SET 14B


6 Consider the sequence ¡6, ¡2, 2, 6, ......

a Show that the sequence is arithmetic.

b Find a formula for the general term un.

c Find the 100th term of the sequence.

d Find the largest term that is less than 500.

7 Find the general term un of the geometric sequence with u4 = 24 and u7 = ¡192:

8 Find the next three members of the sequence with u1 = 2 and un+1 = 2un + 4:

9 Find the coordinates of the points of intersection of the graphs with equations

y = x2 + 4x¡ 2 and y = 2x + 1:

10 a Find the image of y = 2x2 under:

i a translation of

µ¡13

¶ii a vertical dilation with factor 3.

b On the same set of axes, draw the graphs of f(x) = 3 ¡ x2, y = f(¡x)

and y = 2f(x).

1 For the following graphs, determine:

i the range and domain ii whether the relation is a function.

a b

c d

2 If f(x) = 5x¡ x2, find in simplest form:

a f(¡3) b f(¡x) c f(x + 1)

3 If g(x) = x2 ¡ 1 and h(x) = 3x + 2 find:

a g(h(x)) b h(g(x)) c h¡1(x) in the form ax + b.

y

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2

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4 Consider f(x) = j3 ¡ 4xj :a Sketch the graph of y = f(x):

b Find the solutions of j3 ¡ 4xj = 7 from the graph in a.

c Check your answer to b algebraically.

5 Find the general term un for an arithmetic sequence given that the third term is 24and the 11th term is ¡36.

6 Show that the sequence 64, ¡32, 16, ¡8, 4, ...... is geometric, and hence find the

16th term as a fraction.

7 2k + 7, 1 ¡ k and k ¡ 7 are consecutive terms of a geometric sequence. Find kand hence find the three terms given.

8 Find an explicit formula for un if:

a un+1 = un ¡ 3 and u1 = 7 b u1 = 24 and un+1 = ¡23un:

9 Find the coordinates of the points of intersection of the graphs with equations y =3

xand y = 2x¡ 1:

10 a Find the image of y =5

xunder:

i a reflection in the x-axis ii a dilation with factor 34 .

b On the same set of axes, graph f(x) = 2x+1, y = 2f(x) and y = f(2x):


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14 - haese mathematics · chapter14 contents: a relations and functions ... f5, 8, 12, 16, 20g. y x...

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