Chepter 1 function 2013

CHAPTER 1 – FUNCTIONS

LEARNING AREA : FUNCTIONS
Learning Objectives : Understand the concept of relations
Learning Outcomes : Student will be able to
1.1 Represent relation using a) arrow diagram , b) ordered pairs, c)graphs
1.2 Identify domain, co domain, object, image and range of a relation.
1.3 Classify a relation shown on the mapped diagram as : one to one many to one ,
one to many or many to many

1.1a Representing a relation between two sets by using an arrow diagram
Example 1
Suppose we have set A = { 0,1,2,3 } and set B = { 0,1,2,3,4,5,6} .Let us examine the relations “is one more than “
from set A to Set B . One way to show the relations is to draw an arrow diagram as shown below . The arrows
relate the elements in A to the elements in B

A

B
is one more than

In the space below give other example to show a relation

 6
5

3

4

2

3

1

2
.

0 .

1
0

FIG 1.1

A relation from set A to set B is an association of elements of A with elements of B .
Exercise 1
1 Two set of numbers, P and Q are shown below
Complete the arrow diagram to show the relations “ is
less than “ from set P to set Q

2. What relation from set S to set T is illustrated in
diagram below .Write your answer in the box below
S

P

is less then

3

8
6

2
.

2

-1 .

1

3. Construct an arrow diagram to show the relation
“ is a factor of “ from set A = {1,4,15,35,40} to set B
={2,5,7,13 }

4

1

3

6

2

 8

10

1 .

T

Q

.  -2

4. Construct an arrow diagram to show the relation
“ is a factor of “ from set A ={1,2,3,4,5,6 } to
set B = {2,3,5}

1

1.1(b) Representing a relation between two sets by using an ordered pairs
The relation can also be shown concisely by ordered pairs (x , y ) . The elements in the pair are in order since the
first element comes from the first set A and the second element comes from the second set B .We have
A = {0,1,2,3} and B = {0,1,2,3,4,5,6, If x  A ( this means that x is a member of A) and y  B, the set of all ordered
pairs (x,y) is { (0,1),(1,2),( 2,3),(3,4)} . This set of ordered pairs defines the relations “ is one less than “ from set A
to set B
Example 2
1. A = { 1,2,3,4,5,6} and B = {2,3,5 }.
Show the relation “ is a factor of” from A to B
as a set of ordered pairs
Solution
The ordered pairs ={ (2,2),(3,3),(4,2),(5,5),(6,2),(6,3) }

2. Write down the relation as a set of ordered pairs
“ is multiple of “from set A ={2,5,7,13 } to set B =
{1,4,15,35,40}
Solution
The ordered pairs = { (2,4) (2,40), (5,15),(5,35),(5,40),(7,35) }

Exercise 1.1 (b)
1. A relation between two sets is defined by the set of ordered
pairs , { (-1,2),(1,4) ,(3,6) (5,8) , (7,10 ) }
Express the relation in function notation..

2. A relation R is defined by { (1,3),(2,6),( 3, 9),(4,12) }
Express the relation in function notation..

3. Given that X ={ 0,1,2,3,4,5,6 } and Y = { 0,1,2,3,4 }
If x  X and y  Y , list the set of ordered pairs in the relation
x “is one less than “ y

4. Given that A = {2,4,6,8} and B = { 1,2,3,4,5 }
If x  A and y  B , list the set of ordered pairs in the relation x is
half of “ y

1.1(c)

Representing a relation between two sets using a graphs

We can plot an ordered pairs in a Cartesian graph of the relations
Example 3
1. We have A = {0,1,2,3} and B = {0,1,2,3,4,5,6, If x  A and
y  B, the set of all ordered pairs ( x,y) is

2. Given that A = B = { 1,2,3,4,5 }
If x  A and y  B , and the relation x “ less than “ y

(0,1),(1,2),(2,3),(3,4)}

Illustrate the relation by means of a Cartesian graph.

we can plot a Cartesian graph to show the relation.

Set B

Set B







Set A
Set A

2

Exercise 1.1(c)
1. Given that A = B = { 1,2,3,4,5,6,}
If x  A and y  B , and the relation x “ is a factor of “ y

Set B

Set B

Illustrate the relation by means of a Cartesian graph

Given that A = {2,4,6 } and B = { 1,2,3,4,5 }
If x  A and y  B , and the relation x “ is double “ y
Illustrate the relation by means of a Cartesian graph

Set A

Set A

Homework : Textbook Exercise 1.1.1 page 3
1.2

Identifying domain, codomain, object, image and range of a relation.

Domain is the set of x-coordinates of the set of points on a graph or the set of x-coordinates of a given set of
ordered pairs. The value that is the input in a function or relation
Range is the y-coordinates of the set of points on a graph, or the y-coordinates of a given set of ordered pairs.
The range is the output in a function or a relation. A graph of a relation (a set of ordered pairs) is given below.
(Fig 1.2)
Note that the five points on the graph have the ordered pairs
{(-2,1),(3,2),(4,0),(2,-3),(-2,-3)}.
The domain of this relation is the set {-2,2,3,4}. Notice that although 2 is an x-coordinate twice, we need only list it in the domain once.
The range of this relation is the set {-3,0,1,2}. Notice that although -3
is a y-coordinate twice, we need only list it in the range once.
For the ordered pair (  2 , 1 ) the object is  2 and the image is 1
so the objects are -2,2,3,4 and the images are -3,0,1,2
FIG 1.2(a)

Refer to figure 1.2 (b)
The codomain is the set { 2,4,6,8,10 }
The domain is the set { 1,3,5,7,9,}
The range is the set { 2,4,6,8,}
The images are 2,4,6,8
The object s are 1,3,7,9

2

1
3
5
7
9

4
6
8
FIG 1.2(b)

10

3

Exercise 1.2

Complete the table below based on the diagram given
1

2.

Domain

Codomain

Ali

Object

Image

84

Bakar

0
4
8
10
16

-2
0
2
4

90

Samy

Range

Domain

Codomain

Object

Image

Range

Homework : Textbook Exercise 1.1..2 page 4

1.1.3 Classifying a relation
Example 3
Observe the table below and list the characteristic of each relation .
Relations
Arrow diagrams
Ordered pairs

One to one

2
4
6

3
5
7

Graph
y
7

{(2,3),(4,5), (6,7)}

5
3







x

2 4 6
y
One to many

1
3

4
5
6

Many to one

{(1,4),(3,5), (3,6)}



4

1

{(7,6),(9,6), (11,14)}

7
9
11

6
5

6
10
14



3



14
10

 

6

7 9 11

{(a,p),(b,p),(b,q),(c,s),(c,t)}
Many to many

a
b
c

p
q
r
s
t




t
s

r
q

p  
a b c

4

Exercise 1.2
Complete the table below and determine the type of relations
Bil
1

Arrow diagram

-3
1
2

Ordered Pairs

Graf

Relation

9
4
1

4
5
6

One to one





1 2

-3
2
{(-2,1),(-2,3) (0,3)(1,4),(3,6)}

-2
0
1
3

6
4
3
1

1
3
4
6

-2
3

3

1

{(a,2),(b,3) (c,3),(d,5)}

a
b
c
d



5

2
3
5

 

3



2

a

b c d

4

1
2



a
b
c
d






2

1
5
{(4,2),(25,5)}

4
25
36

2
5



5
2


4 25 36

Homework : Textbook Exercise 1.1..3 page 5 and Skill Practice 1.2 page 6
2.0 Learning Objective : Understand the concept of functions
Learning outcomes : Student will be able to
2.1 Recognise functions as a special relation
2.2 Express functions using functions notation
2.3 Determine domain,object,image and range of a function
2.4 Determine the image of a function given the object and vice versa.

5

2.1 Recognising functions as a special relation
Note: A special type of relations between two set A and B can exist when each and every members of A is related
to one and only one member of B, although some of the member of B may not related to any member of A
This type of relation is known as a function. Itis important to relise that a function is a relation but a relation may or
may not be a function . In other words, a function is a special type of relation
Example 4
Determine which of these relations is a functions
a)
b)
A
B
A
B

-2
0
1
3

7
9
11

1
3
4
6

Not a function – 2 has two
images

c)

d)
A

a
b
c
d

6
10
14

A function because each
member in A is related to
member in B

B

A

7
9
11

2
3
5

B

6
10
14

Not a function
No image for d

A function although 9 and
11 in A are both linked to
10

c)

d)

Exersice 2.1
1) Determine which of this relations is a functions
a)

b)

1
3

4
5
6

7
9
11

6
10
14

a
b
c

p
q
r
s

a
b
c
d

2
3
5

Let’s think
Which type of relations ; one to one , one to many, many to one and many to many is a a function.
Discuss it.

2.2 Expressing function using function notation
We have learnt that relations is also a function or mapping as each and every member of set A is linked to one and
only one member of set B . Symbolically , we write f : x  x  2 which means that ” the function f maps x onto ” x
+ 2 ” . In functional notation, we write f ( x)  x  2 . f ( x) is read as ’ function of x ’.
The set of elements in A for the mapping is known as the domain while the set of images in B is known as the
range . In functional notation, if f ( x)  x  2 and x =3 , then f (3)  3  2  5 ( 3 is an object while 5 is the image)
Homework : Textbook Exercise 1.2..2 page 8
2.3 Determining domain,object,image and range of a function
Example 5

Object
x
2
3
-1
a
a+1
2
x
3
x
2
x -1

” times 2 ”
1.

2

4

4

8

x

2x

f:x
2x
f(x) = 2x

6

Images
f(x)
f(2) = 2(2) = 4
f(3) = 2(3) = 6
f(-1) = 2(-1)= - 2
f(a)= 2(a) = 2a
f(a+1)= 2(a+1) = 2a + 2
2
2
f(x ) = 2x
3
3
f(x ) = 2x
2
2
2
f(x – 1) = 2(x -1) = 2x -2

Example 6
Bil

Functions

Functions
notations
f : x  x2

1

-2
-1
0
1
2

0
1
4

f(x) = x

f : x  4x

2
{ (1,4), (2,8), (3,12)}
3

2

7
5

f(-2) = 4
f(2) = 4

3

f(0) = 0




1


p

Exerise 2.2
Complete the folowing table
x value
-2
0
1

Functions
a) f : x  2x + 1

f(p) = 3
f(q) = 1
f(r) = 5

q

r

Calculations image
f(-2) = 2(-2)+1= -3
f(0) = 2(0)+1 = 1

2
b) g; x 

2x  3
7

-3
5

c) h : x  4x - x

-4

2

4
-2
2.4 Determining the image of a function given the object and vice versa
A Find the image for each of the following functions.
2
a) Given that f(x) = 2x - 4x + 5 , find the image
For each of the following .
2

2

2

b) Given that f(x) = (x - 2) (x + 1 )-10 , find the
image for each of the following.

(i)f(-1)=2(-1) – 4(-1) + 5
= 2+4+5
= 11

ii) f(2)

i) f(0)

ii)f(2)

2
iii) f( )
3

iv) f(-4)

v) f(-1)

vi f (-3)

B . Find the object for each of the following functions
c) Given that f(x) = 2x + 6 , find the object when
the image is,

d) Given that f(x) =

2x  8
, find the object when
5

the image is,
(i) 2

ii) - 5

i) 5

7

ii) - 3

iii)

7
3

iv)



3
5

v)

8
9



3
2

C.

a) Given h : x  3x-12 , Find

b) Given that f (x) = 4 - 2x, find the value m
If

i)
the object when the image is 6
ii)
the object which mapped to it’s self
Solution :
(i)
(ii)

f(m) = 10


D. (i) The diagrams below shows part of the mapping
Find
(i)
the values of a and b
(ii)
the image of 3 under f
(iii)
the object whose image is -4

f ( x) : ax  b

Calculations

a)

x  ax+b
4

7
1
8

b)

8


the object whose image is -4

D (ii)
a)

b)

The diagram above shows part of the mapping

The diagram above shows part of the mapping
72
f:x 
. Find ,
ax  b
(i)
(ii)
(iii)
the object of 4

f ( x) :  ax  bx  c . Find
(a) f(x) .
2

(b) the image of - 4 under f

Example 6
12
b
,x  
ax  b
a
Given that f(4) = -3 and f (10) = 6, Calculate
a) the value of a and b
b) the value of x for which f(x) = -x

a) A function f is defined by f : x 

b) A function f is defined by

f: x 

Given that f(2) and f(5) = -1
a) state the value of k
b) find the value a and b
c) find f(4)

Homework : Textbook skill Practice 1.2 page 10

9

a
 b, x  k
x

Absolute Valued Function

Note :
f(x) = x is called absolute valued function. The absolute value of a number is the distance the
number is from zero on the number line. We write the absolute value of -2 as | -2 |.
The absolute value of a number is found by determining how many units the number is from zero. Since
distance is always thought of as positive, the absolute value of any number is positive. For example, -2
is 2 units from zero, so | -2 | = 2. In the picture below, you can see that | 4 | = 4 and | -4 | = 4, because both
4 and -4 are 4 units from zero on the number line.

In equalities of a bsolute Valued Function
(i)
x <k  -k<x<k
(ii)

ax  b < k  - k < ax + b < k

(iii)

ax  b > k  ax + b < -k or ax + b > k

Example 10
a) Given that f(x) = 3x  2 , find the value of
i) f(2)

ii) f (

1
)
2

Exercise 2.4
3
a) Given that f (x) =  x – 4  , find
i) f( 3)
ii) f(-2 )

c) Given the function f :x = 2x - 5 ,find the values
of x such that f(x) = 7

b) Given that f : x  3x  5 . Find the domain of
the following functions
i) f(x)  4

ii) f(x) > 3

b) Given that f :x = 2x - 8  . Find the domain of the
following functions
i) f(x)  2
ii) f(x) > 4

d) Given the function f :x = 2x - 5 ,find the range of
values of x such that f(x) < 7

10

Example 11:
Sketch the following function . Hence, state the range that match with the domain given .
a) f : x  x  3 ,
- 4  x  2 b) f: x  x - 3,
-1  x  3 c) f: x  x  2 + 1

-2  x  3

2.5 Exercise
Sketch the following function . Hence, state the range that match with the domain given .
a) f : x  x  2 for

-1  x  3

b) g : x  2 x  1 - 2 for -1  x  2

c) h : x  4  x 2 for

-1  x  3

Homework

1.

Sketch the graph of function f:x   2x-3 
for 0  x  4. Hence state the range that
match with the domain given

f(x) =  7 – 2x  for 0  x
range that match to the domain given.

3. Given that



8 . find the

2. . Sketch the graph of function   2x-5 
for 0  x  6 . Find the value of x if f(x)  4.

4. A function f is defined by f : x  3x-5 
a) find f(4) , f(10) and f( -5)
b) if f (a) = 26, find the possible value of a

11

Learning Objective : Understand the concept of composite functions.
Learning Outcomes : Student will be able to
3.1 Determine composition of two functions.
3.2 Determine the image of composite functions given the object and vice versa.
3.3 Determine one of the functions when the composite function and the other
function are given.

3.0 Understanding the concept of composite functions.
Note : 1. Composite functions is a function composed of two or more algebraic functions.
2. Functions can be combined to give a composite function. If a function f is followed by a function g ,
we obtain the composite function g f . In general gf  fg .

f followed by g followed by h is donated by hgf . The order is relevant and important.
3.1 Determine composition of two functions.
Example 3. 1 Determining the composite functions.
a) Two functions f and g are defined by
2
f : x  x - 2 and g : x  x – 1. Obtained
expressions for
(i) fg ,
(ii) gf

Exercise 3.1
a) Two functions

f and g are defined by

f : x  x + 1 and g: x 
2

i)

gf

2

and g

4
. Find
x
ii)

b) Two functions f and g are defined by f : x  5x - 2
2
and g :x  x – x Find,
i)

g2

ii)

f 2 g2

b) Two functions f and g are defined by
2
f : x  x + 6 and g : x  3x + 4.
Find the values of x if gf (x) = fg (x)

fg and f 2


12

3.2 Determining the image of composite functions given the object and vice versa.
Example 3.2
Two functions f and

g are defined by

2

f : x  x + 1 and g: x 
(i)

gf (2)

2

and g (4)

4
. Find
x
2
ii) fg (-4) and f ( 3)

Exercise 3.2
c) Two functions f and g are defined by
2
f : x  x - 2 and g : x  x – 1. Obtained
expressions for

fg (2) , gf (-2) , f 2 (2) , g 2 (8)

Note : It is common to express

Example 3.2(ii) The function g is defined by g:x 
3

d) Two functions f and g are defined by f : x  a x - b
and g :x  2x - 5 . Given that f g (1) = 10 and
g f (2) = 8 . Calculate the values of a and b

ff (x) (x) as f2 (x) , fff (x) as f3(x) etc.


following functions
2
a) g (x)

i) Two functions f and g are defined by
f : x  2x + 6 and g : x  a x + b
Given that f g (1) = 10 and g f (2) = 8 . Calculate the
values of a and b

b) g (x)

1 x
, x  1 . Express in their simplest forms each of the
1 x
4

c) g (x)

13

16

c) g (x)

Exercise 3.2(ii) The function g is defined by f: x 

5
, x  0 Express in their simplest forms each of the
x

following functions

a) f 2(x)

b) f3(x)

c) f4(x)

d) f31 (x)

3.3 Determining one of the functions when the composite function and the other function are given .
Example 3.3
b) If f (x) = x +1 find the function g such that
1
2
( x  1) find the function f such that
a) If g(x) =
g f (x) = x + 2x + 12
g f (x) =

2
2x 1
,
3

Exercise 3.3
a) If f (x) = x + 4 find the function g such that
f g(x) =

2( x  1)
,x  2
x2

[g(x)

2(3  x)
,x  2 ]
x2

c) Given that hg(x) = 2x + 1 and h(x) =

x
. Find the
4

2

4

2

b) if f(x) =x + 1 and g f (x) = x + 2x + 9. Find the
2
function g . [ Ans g(x)=x + 8 ]

d) Given that fg(x) = 5 – x and f(x) = 8x +3. Find the
function g

function g

14

Learning objective 4.0 Understand the concept of inverse functions.
Learning outcomes : Student will be able to:
4.1 Find the object by inverse mapping given its image and function.
4.2 Determine inverse functions using algebra.
4.3 Determine and state the condition for existence of an inverse function.

Homework : Textbook Exercise 1.3.3 page 15 and Skill Practice 1.3 page 16.

4.0 Understanding the concept of inverse functions.
Suppose f is a function “multiply by 3” .If. this function is applied to x, then the images of x is 3x

What function should we

apply to 3x in order to get back to x ?.This function which maps the image back to its initial value is known as the inverse
function of f, and it is denoted by

f 1 . Symbolically we write f ( x)  3x

and

f 1 ( x) 

3
x

If f is a many – to – one function, then its inverse is not a function but a one- to- many relation. Only one – to – one functions will
give one – to – one inverse functions.

4.1 Finding the object by inverse mapping given its image and function.

f = x–1

f -1

3

2

2

3

4

3

3

4

5

4

4

5

6

5

5

6

-1

f(x) = x – 1
f(3) = 2
f(4) = 3
f(5) = 4

f ( 2) = 3
-1
f ( 3) = 4
-1
f ( 4) = 5
-1
f ( 5) = 6

Example 4.1
Find the value of a and b in the following diagram by the inverse mapping

a)

f x :

x  4  3x

3

1

15

17

1

1

13

b

8

-3

4

13

b

x  2x  5

a

-5

a

b) f x :

7

Exercise 4.1 Find the value of a and b in the following diagram by the inverse mapping

a)

b)

x  2 x

f x :

f x :

x  2x  1

a

-3

1

4

9

1

b

17

8

1

3

-

3
a
b
1

1

4.2 Determining inverse functions using algebra
Example 4. 2 (i)
a) Find the inverse function for the function
f : x  3x + 4 then find f

1

(2)

3x  1
, x  2. Find f -1 (x) ,
x2
1
f 1 (4) , f 1 (2) and f 1 ( )
3

b) Given f : x 

Exercise 4.2
a) Given

h:x 

3 x
–1
, find h (x) , h 1 (3)
2

1
and h ( )
4
1

b) Given that f :x 
:x 


16

hx  k
, x  2 and its inverse function f -1
x2

2x  9
, x  4 . Find the value of h and k.
x4

Example 4.2(i) :

a) A function f is defined by f : x  x + 3. Find
1

(i) ff
2
(ii) the function g such that gf : x  x + 6x + 2.
(i) f f

-1

b) A function f is defined by f : x  x + 2. Find
the function g such that gf: x  x + 9

ii) function g

A function f is defined by f: x  2x-1. . Find
i) f g : x  7 -6x

the function g such that
5
ii) g f: x 
2x

4.3 Determining and stating the condition for existence of an inverse function.
.

(b)

(a)
2

Diagram (a) shows function g(x) = x , which is not a one to one function. Reversing the arrows, you do not have a
one to one function.. In general, for a function f to have an inverse function, f must be a one to one function.
Note : If f is a many – to – one function, then its inverse is not a function but a one – to – many relation.
Only one – to – one function will give one- to-one functions
Remember : If a function f is one to one, then the inverse function

17

f 1 does exist !!!

Example 4.3
Determine whether the inverse of the following function is a function or not .
2
a) f(x) = (x+1)
1
b) g ( x)   4 , x0

x

Homework : Textbook Exercise 1.4.3 page 20 and Skill Practice 1.4

FUNCTIONS – SPM QUESTIONS
2004 – P1
1. Diagram 1 shows the relation between set P and
set Q.

2. Given the functions h : x
-1

h

d

w

:x

2kx +

4x + m and

5
, where m and k are
8

constants , find the value of m and of k .
[3 marks]

e

x

f

y
z

Set P

Set Q

.
DIAGRAM 1
State

3.

(a) the range of the relation,
(b) the type of the relation
[2 marks]

Given the function h(x) =

6
, x ≠ 0 and
x

composite function hg(x) = 3x,
Find (a) g(x) ,
(b) the value of x when gh(x) = 5 .
[4 marks]

18

2003 – P1
4.

8. The following information refers to the functions h
and g.

h : x  2x  3
g : x  4x  1

P={1,2,3}
Q = { 2 , 4 , 6 , 8 , 10 }
-1

Find gh (x).
Based on the above information, the relation
between P and Q is defined by the
set of ordered pairs { (1,2) , (1,4) , (2,6) , (2,8) } .
State (a) the image of 1 ,
(b) the object of 2 ,
[2 marks]

[3 marks]

5. Given that g : x
5x + 1 and
2
h:x
x – 2x + 3 , find
–1
(a) g (3) ,
(b) hg(x) .

2006 – P1
9. In Diagram 1, set B shows the images of certain
elements of set A.
[4marks]

5
4
-4
-5
2005 – P1
6. In Diagram 1, the function h maps x to y and the
function g maps y to z.

x

h

y

g

25
16
Set B

Set A
Diagram 1
(a)

z
8

(b)

State the type of relation between set A and
set B.
Using the function notation, write a relation
between set A and set B.

5
2
Diagram 1
10. Diagram 2 shows the function
Determine (a)

h 1 (5),
(b) gh(2).

h: x 
x

[2 marks]

mx
, x  0, where m is a constant.
x
mx

x

8

DIAGRAM 2

7. The function w is defined as

5
, x  2.
2 x
1
(a) w ( x),
1
(b) w (4).

w( x) 



1
2

Find the value of m.
[2 marks]
[3 marks]

19

2006 – P2
11. Given that

f : x  3x  2 and g : x 

15. Given the functions f : x
ax + b , a > 0 and
2
f :x
9x – 8 . Find
(a) the value of a and of b
[3 marks]
–1 2
(b) ( f ) (x)
[3 marks]

x
 1,
5

find
(a)

f 1 ( x),
[1 mark]
1

(b)

f g ( x),

(c)

h(x) such that hg ( x)  2 x  6.

[2 marks]
[3 marks]

2000 – P1
16. Given that g

–1

(x) =

5  kx
2
and f(x) = 3 x – 5
3

. Find

2002 – P1

(a) g(x)

12. (a) Given that f : x

3x + 1 , find f

[2 marks]

–1

(5) ,
[2 marks]

2

(b) the value of k such that g(x ) = 2 f (- x)
[3 marks]

(b) Given that f(x) = 5 – 3x and g(x) = 2ax + b
where a and b are constants .
If fg(x) = 8 – 3x , find the value of a and of b
[3 marks]

2007 – P1
17. Diagram 1 shows the linear function h.

x
13. (a) Find the range of value of x if

0
1

2 x  3 for the domain 0  x  4.

Hence, state the corresponding range for the
domain.
[3 marks]

4

5

(b) Sketch the graph of the function

1
2

m

x < 5.
[2 marks]

f :x

h(x)

6
Diagram 1

(a) State the value of m.
(b) Using the function notation, express h in terms
of x.
[2 marks]
14. Given that f(x) = 4x – 2 and g(x) = 5x + 3 . Find
–1
(i) fg (x)
(ii) the value of x such that fg

–1

x
2
 =
.
2

5

[5 marks]

2001 – P1

20

f : x  x  3 , find the

18. Given the function

values of x such that f(x) = 5 .

20. Diagram 1 shows the graph of the function

f ( x)  2 x  1 , for the domain

[2 marks]

21. Given the functions

g : x  5x  2 and

h : x  x  4 x  3 , find
1
(a) g (6)
(b) hg (x)
2

[4 marks]
08P1. 2

19. The following information is about the function h
2
and the composite function h .

h : x  ax  b, where a and b are constants,
a > 0 h 2 : x  36 x  35
22. Given the functions

Find the value of a and of b .

f ( x)  x  1 and

g ( x)  kx  2 , find

[3 marks]

(a) f(5),
(b) the value of k such that gf(5) = 14.
[3 marks]
2008 P1

08P1. 3

y

1
t

5

x

(a) the value of t,
(b) the range of f(x) corresponding to the given domain.

21

Chepter 1 function 2013

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