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1. Functions 1
1 FUNCTIONS
1.0 RELATIONS
Notes :
(i) Four types of relations : one-to-one, many-to-one, one-to-many and many-to-many.
(ii) Three ways to represent relations : arrowed diagram, set of ordered pairs and graph.
(iii) The arrow diagram representation is of the form below :
In the above relation,
b is the imej of a ;
a is the object of b.
(iv) Set of ordered pairs of a relation takes the form of { (a , b), (c , d), ….. }
EXAMPLE
QUESTION
1.
A relation from P into Q is defined by the set of
ordered pairs
{ (1, 4), (1, 6), (2, 6), (3, 7) }.
State
(a) the image of 1, Ans: 4 and 6
(b) the object of 4, Ans: 1
(c) the domain, Ans: { 1, 2, 3 }
(d) the codomain, Ans: {2,4,6,7,10}
(e) the range, Ans: { 4, 6, 7 }
(f) the type of relation Ans: many-to-many
1.
A relation from P into Q is defined by the set of ordered
pairs
{ (2, 2), (4, 2), (6, 7) }.
State
(a) the image of 4, Ans:
(b) the objects of 2, Ans:
(c) the domain, Ans:
(d) the codomain, Ans:
(e) the range, Ans:
(f) the type of relation Ans:
a is the
object
b is the
image
P
Q
a
b
Q : Codomain
A
B
6
3
4
1
The relation from A into B
above can also be written as set
of ordered pairs
{(1, 4), (3, 6) }
4 is the image of 1 ;
The Object of 6 is 3.
_
5
_
_
_
A : Domain
B : Codomain
Set { 4, 6 } is the RANGE
for the mapping from A to B
P = { 1, 2, 3}
Q = {2, 4, 6, 7, 10}
P = { 1, 2, 3}
Q = {2, 4, 6, 7, 10}
P = { 2, 4, 6}
Q = {2, 3, 6, 7, 10}
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff
pf12
pf13
pf14
pf15
pf16
pf17
pf18
pf19
pf1a

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1 FUNCTIONS

1.0 RELATIONS

Notes :

(i) Four types of relations : one-to-one, many-to-one, one-to-many and many-to-many. (ii) Three ways to represent relations : arrowed diagram, set of ordered pairs and graph. (iii) The arrow diagram representation is of the form below :

In the above relation, b is the imej of a ; a is the object of b.

(iv) Set of ordered pairs of a relation takes the form of { (a , b), (c , d), ….. }

EXAMPLE QUESTION

A relation from P into Q is defined by the set of

ordered pairs

{ (1, 4), (1, 6), (2, 6), ( 3 , 7) }.

State

(a) the image of 1, Ans: 4 and 6 (b) the object of 4, Ans: 1 (c) the domain, Ans: { 1, 2, 3 } (d) the codomain, Ans: {2,4,6,7,10} (e) the range, Ans: { 4, 6, 7 } (f) the type of relation Ans: many-to-many

A relation from P into Q is defined by the set of ordered pairs { (2, 2), (4, 2 ), (6, 7) }.

State (a) the image of 4, Ans: (b) the objects of 2 , Ans: (c) the domain, Ans: (d) the codomain, Ans: (e) the range, Ans: (f) the type of relation Ans:

a is the object

b is the image

P Q

a (^) b

P :Domain

Q : Codomain

A B

The relation from A into B above can also be written as set of ordered pairs {(1, 4), (3, 6) }

4 is the image of 1 ; The Object of 6 is 3.

_

_

_

_

A : Domain (^) B : Codomain

Set { 4, 6 } is the RANGE for the mapping from A to B

P = { 1, 2, 3} Q = {2, 4, 6, 7, 10}

P = { 1, 2, 3} Q = {2, 4, 6, 7, 10}

P = { 2, 4, 6} Q = {2, 3, 6, 7, 10}

1.1 FUNCTION NOTATIONS

Notes:

(i) A function is usually represented using letters in its lower case : f , g , h …..

(ii) Given the function (^) f : x  2 x  1 , we usually write it in the form f( x ) = 2 x + 1 before answering

any question.

1.2 FINDING THE VALUE OF FUNCTIONS [Mastery Exercise]

f ( a ) represent (a) the value of f (x) when x = a. (b) the image of f when the object is a

EXAMPLE QUESTION

  1. Given the function f : x  2 x  1 ,

find (i) f (3) (ii) f (-4)

Answer : f(x) = 2 x + 1

(i) f ( 3 ) = 2 ( 3 ) + 1

= 6 + 1

(ii) f ( - 2 ) = 2 (- 2 ) + 1

= - 4 + 1

  1. Given the function f : x  2 x  3 ,

find (i) f (2) (ii) f (-1)

Answer : f(x) = 2 x + 3

(i) f( 3 ) = 2( ) + 3

=

(ii) f( - 1 ) = 2 ( ) + 3

=

  1. Given the function : 3 2 g xx  , find

(i) g(0) (ii) g(4)

Answer : g (x) = x

2

  • 3

(i) g ( 0 ) = 0

2

  • 3

= 0 - 3

(ii) g ( 4 ) = 4

2

  • 3

= 16 – 3

  1. Given the function : 5 2 g xx  , find

(i) g(0) (ii) g(2)

Answer : g (x) = x

2

  • 5

(i) g ( 0 ) =

=

(ii) g ( 2 ) =

=

  1. Given the function 3 4

x

h x , 3

x  ,

find h (2).

Jawapan : 3 4

x

h x

h ( 2 ) = 3 ( 2 ) 4

  1. Given the function 2 3

x

h x , 2

x  ,

find h (3).

Jawapan: h ( x ) 

h ( 3 ) =

3

1.3 PROBLEM OF FUNCTIONS INVOLVING SIMPLE EQUATIONS. [1]

Note : Questions on functions usually involve solving linear / quadratic equations.

EXAMPLE QUESTION

  1. Given that f : x  2 x  1 , find the value of x

if f(x) = 5.

Answer : f (x) = 2 x – 1

2 x – 1 = 5

2 x = 6

 x = 3

  1. Given that f : x  4 x  3 , find the value of x if

f ( x ) = 17.

Answer : f (x) = 4 x – 3

4 x – 3 = 17

4 x =

 x = 5

  1. Given that f : x  5 x  3 , find the value of x

if f ( x ) = -7.

Answer : f (x) = 5 x + 3

5 x + 3 = -

5 x = - 10

 x = - 2

  1. Given that f : x  3 x  7 , find the value of x if

f ( x ) = 4.

Answer :

x = - 1

  1. Given that g : xx  2 , find the value of y

if g(y) = 2y - 3

Answer : g(x) = x + 2

g(y) = y + 2

y + 2 = 2 y – 3

3 + 2 = 2 y - y

y = 5

  1. Given that g : x  3 x  2 , find the value of y if

g(y) = 2y + 4.

Answer :

y = 6

  1. Given that f : x  7 x  3 , g : x  4 x  15 ,

find the value of p if f(p) = g(p).

Answer : f(x) = 7x – 3, g(x) = 4x + 15

f(p) = 7p – 3, g(p) = 4p + 15

f(p) = g(p)

7p – 3 = 4p + 15

3p = 18

p = 6

  1. Given that f : x  2 x  10 , g : x  4 x  2 ,

find the value of x if f(x) = g(x).

x = - 6

PROBLEMS OF FUNCTIONS INVOLVING SIMPLE EQUATIONS. [2]

EXAMPLE QUESTION

  1. Given that f : x  2 x  10 , g : xx  7 ,

find the value of x if f(x) =  2 g(x)

A: f(x) = 2x - 10 , g(x) = x – 7

2x – 10 = - 2 (x – 7)

2x - 10 = - 2x + 14

4x = 24

x = 6

  1. Given f : x  3 x  9 , g : xx  7 , find the

value of x if f(x) =  3 g(x).

x = 5

  1. Given that f : x  3 x  4 , find the values

of x if f (x) = x

2 .

A: f(x) = 3x + 4

x 2 = 3x + 4

x

2

  • 3x – 4 = 0

(x + 1) (x – 4) = 0

x = -1 or x = 4

  1. Given that f : x  2 x  8 , find the values of x if

f (x) = x

2 .

x = 4 , - 2

  1. Given that : 3 6

2 g xx  , find the values of

y if g (y) = 6.

A: g(x) = 3x 2

  • 6

g(y) = 3y

2

  • 6

3y

2

  • 6 = 6

3y 2

  • 12 = 0

(÷3) : y 2

  • 4 = 0

(y + 2) (y – 2) = 0

y = -2 or y = 2

  1. Given that : 2 5

2 g xx  , find the values of y if

g (y) = 45.

x = - 5 , 5

  1. Given that : 3 6 2 g xx  , find the values of

p if g (p) = 7p.

A: g(x) = 3x 2

  • 6

g(p) = 3p 2

  • 6

3p

2

  • 6 = 7p

3p

2

  • 7p - 6 = 0

(3p+2) (p – 3) = 0

p = 3

 atau p = 3.

  1. Given that : 2 3 2 g xx  , find the values of p if

g( p ) = - 5p.

p = 1/2 , - 3

1.4 FINDING THE VALUE OF COMPOSITE FUNCTIONS [Mastery Exercises]

Notes : To find the value of fg(a) given the functions f and g, we can first find fg(x) and then

substitute x = a. We can also follow the EXAMPLES below.

EXAMPLE 1 :

Given that f : x  3 x  4 and g : x  2 x ,

find fg(3).

Answer : f(x) = 3x - 4 , g(x) = 2x

g(3) = 2(3)

fg(3) = f [ g(3) ]

= f ( 6)

= 3 (6) - 4

EXAMPLE 2 :

Given that f : x  3  2 x and 2 g : xx , find

gf(4).

Answer : f(x) = 3 – 2x , g(x) = x 2 .

f(4) = 3 – 2(4)

gf(4) = g (-5)

2

EXERCISES

  1. Given that f : x  2 x  1 and g : x  3 x ,

find f g(1).

  1. Given that f : x  2 x  9 and g : x  1  3 x ,

find gf(3).

  1. Given the functions f : xx  3 and

g : x  4 x  1 , find

(a) f g(1) (b) gf(1)

  1. Given the functions f : x  3 x  7 and

g : x  4  2 x , find

(a) f g(0) (b) gf(0)

FINDING THE VALUE OF COMPOSITE FUNCTIONS [Reinforcement Exercises]

  1. Given that f : x  2 x  3 and g : x  4 x ,

find fg(2).

  1. Given that f : x  2 x  5 and g : x  5 x , find

gf(7).

  1. Given that f : xx  4 and g : x  2 x  1 ,

find

(a) fg(1) (b) gf(1)

  1. Given that f : x  3 x  4 and g : x  5  2 x ,

find

(a) fg(0) (b) gf(0)

5 Given the functions : 3 1 2 f xx  and

g : x  2  x , find

(a) fg(-1) (b) gf(-1)

6 Given that f : x  3 x and 2 g : x  2  x , find

(a) fg(-2) (b) gf(-2)

  1. Given the functions f : xx  2 and

2 g : x  2  3 xx , find

(a) fg(1) (b) gf(1)

  1. Given the functions f : x  2  x and

2 g : x  1  4 x  3 x , find

(a) fg(-1) (b) gf(-1)

1.6 BASIC EXERCISES BEFORE FINDING THE COMPOSITE FUNCTIONS [ 1 ]

EXAMPLE 1 :

Given f : x  3 x  1 ,

f(x) = 3x + 1

thus (a) f(2) = 3(2) + 1

(b) f(a) = 3 a + 1

(c) f(p) = 3 p + 1

(d) f(2k) = 3 (2k) + 1 = 6k + 1

(e) f(2x) = 3 (2x) + 1 = 6x + 1

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

  • 1

EXAMPLE 2 :

Given g : x  5  4 x ,.

g(x) = 5 – 4x

thus (a) g(2) = 5 – 4(2) = 5 – 8 = -

(b) g(a) = 5 – 4a

(c) g(p) = 5 – 4p

(d) g(3k) = 5 – 4(3k) = 5 – 12 k

(e) g(x

2 ) = 5 – 4x

2

(f) g (3+2x) = 5 – 4 (3+2x)

= 5 – 12 – 8x

= - 7 – 8x

EXERCISES

  1. Given f : x  2 x  3 ,

f(x) = 2x + 3

thus (a) f( 2 ) = 2(2) + 3 =

(b) f(a) =

(c) f(p) =

(d) f(2k) =

(e) f(x

2 ) =

(f) f(-x

2 ) =

(g) f( - 3x) =

2 Given g : x  2  4 x.

g(x) =

thus (a) g (2) =

(b) g(a) =

(c) g(s) =

(d) g(3x) =

(e) g(x

2 ) =

(f) g (3+2x) =

(g) g(2 – 4x) =

BASIC EXERCISES BEFORE FINDING THE COMPOSITE FUNCTIONS [ 2 ]

  1. Given f : x  4  2 x ,

f(x) = 4 – 2x

thus (a) f( 3 ) = 4 – 2 ( 3 ) =

(b) f(-x) =

(c) f(2+x) =

(d) f(3 - x) =

(e) f(x 2 ) =

(f) f(-x

2 +2) =

2 Given g : x   2 x  1.

g(x) =

thus (a) g(-1) =

(b) g(2x) =

(c) g(x-2) =

(d) g(-3x) =

(e) g(x 2 ) =

(f) g (1-2x) =

  1. Given f : x  1  x ,

f(x) =

thus (a) f(10) =

(b) f(3x) =

(c) f(2 - x) =

(d) f(4+x) =

(e) f(2x- 3) =

(f) f(x

2 ) =

4 Given g : x  2 ( x  2 ),.

g(x) =

thus (a) g(-1) =

(b) g(2x) =

(c) g(x-1) =

(d) g(-x) =

(e) g(x 2 ) =

(f) g (1+2x) =

REINFORCEMENT EXERCISES FOR FINDING COMPOSITE FUNCTIONS

  1. Given the functions f : x  3 x  2 and

g : x  2  2 x , find

(a) fg(x) (b) gf(x)

  1. Given that f : x  3 x and 2 g : x  2  x , find

the composite functions

(a) fg (b) gf

  1. Given the functions f : xx  4 and

g : x  1  3 x , find

(a) fg(x) (b) gf(x)

  1. Given that f : x  3  x and g : x   4 x  3 ,

find the composite functions

(a) f g (b) gf

  1. Given the functions f : xx  2 and

g : x  5  2 x , find

(a) fg(x) (b) gf(x)

  1. Given that f : x  2  5 x and 2 g : x  1  x ,

find the composite function gubahan

(a) fg (b) gf

REINFORCEMENT EXERCISES FOR FINDING COMPOSITE FUNCTIONS [ f 2 and g 2 ]

  1. Given that f : x  3 x  2 , find f

2 ( x ).

f(x) = 3x + 2

f 2 (x) = f [f(x)]

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

= 3(3x+2) + 2

=

  1. Given that f : x  3 x , find the function f

2 .

  1. Given that g : x  1  3 x , find gg ( x ) 4. Given that g : x   4 x  3 , find the function

g 2 .

  1. Given that f : xx  2 , find ff ( x ). 6. Given that f : x  2  5 x , find the function f 2 .
  2. Given that f : x  3  4 x , find f

2 ( x ). 8. Given that f : x  5 x , find the function f

2 .

  1. Given that g x x 2

:  , find g 2 ( x ). 10. Given that 2

x

h x , find the function f 2 .

DRILLING PRACTICES ON CHANGING SUBJECT OF A FORMULA [ 2 ]

  1. If y = 4

2 x  5 , then x =

2 x  5 = y

2x – 5 = 4y

2x =

x =

  1. If y = 5

3 x  2 , then x =

  1. If y = 5 + 6x , then x =
    1. If y = 4 – 2

x , then x =

  1. If y = 6 + 2

x , then x =

  1. If y = 6x – 9 , then x =
  2. If y = 3 2

x

, then x = 8. If y = 4  x

, then x =

  1. If y = 4 2

x

x , then x = 10. If y = x

x

3 

, then x =

DETERMINING THE INVERSE FUNCTION FROM A FUNCTION OF THE TYPE

ONE-TO-ONE CORRESPONDENCE

EXAMPLE 1 :

Given that f(x) = 4x – 6 , find f

  • 1 (x).

EXAMPLE 2 : Given that f : xx

x

, x 2,

find f

  • 1 ( x ).

Answer :

Given f(x) = 4x – 6

so f (y) = 4y – 6

then f

  • 1 ( 4y – 6 ) = y.

f

  • 1 ( x ) = y when x = 4y – 6

x + 6 = 4y

y = 4

x  6

 f

  • 1 (x) = 4

x  6

Given f( x ) = x

x

so f( y ) = y

y

2

2 1

then f

  • 1 ( y

y

2

2 1 ) = y

f

  • 1 ( x ) = y when x = y

y

2

2 1

x (2 - y ) = 2 y + 1

2 xxy = 2 y + 1

2 x – 1 = 2 y + xy

2 x – 1 = y (2 + x )

y = x

x

2

2 1

 f

  • 1 ( x ) = 2

x

x , x -2.

EXERCISES

  1. Given that f : x  4 + 8x , find f
    • 1 . 2. Given that g : x  10 – 2x , find g - 1 .
  2. Given that f : x  4 – 3x , find f
    • 1 . 4. Given that g : x  15 + 6x , find g - 1 .

FINDING THE VALUE OF f

- 1

(a) WHEN GIVEN f(x)

Example : Given that f : x  2x + 1, find the value of f

  • 1 (7). METHOD 1

[ Finding f

  • 1 first ]

METHOD 2

[ Without finding f

  • 1 (x) ]

f(x) = 2x + 1

f(y) = 2y + 1

f

  • 1 (2y + 1) = y

f

  • 1 (x) = y when x = 2y + 1

x – 1 = 2y

y = 2

x  1

f

  • 1 (x) = 2

x  1

f

  • 1 (7) = 2

f(x) = 2x + 1

Let f

  • 1 (7) = k

f(k) = 7

2k + 1 = 7

2k = 6

k = 3

 f

  • 1 (7) = 3

( If you are not asked to find f

- 1 ( x ) but only want to have the value of f - 1 ( a ), then use Method 2 )

EXERCISES

  1. Given that f : x  x + 2, find the value of

f

  • 1 (3). 2. Given that g : x  7 – x , find the value of

g

  • 1 (2).
  1. Given that f : x  7 – 3x , find the value of

f

  • 1 (-5). 4. Given that g : x  5 + 2x , find the value of

g

  • 1 (5).
  1. Given that f : x  3x - 2, find the value of

f

  • 1 (10). 6. Given that g : x  7 – x , find the value of

g

  • 1 (2).
  1. Given that f : x  4 + 6x , find the value of

f

  • 1 (-2). 8. Given that g : x  2
  • x , find the value of

g

  • 1 (4).
  1. Given that f : x  3

x

, find the value of

f

  • 1 ( - 1 ). 10. Diberi g : x  2  x

, cari nilai g

  • 1 (-1).

Finding the Other Related Component Function when given a Composite Function

and One of its Component Function

TYPE 1 ( Easier Type )

Given the functions f and fg, find the function g.

OR

Given the functions g and gf, find the function f.

TYPE 2 ( More Challenging Type )

Given the functions f and gf , find the function g.

OR

Given the functions g and fg , find the function f.

EXAMPLE :

  1. Given the functions f : x  2 x  3 and

fg : x  6 x  1 , find the function g.

Answer : f(x) = 2x + 3

fg(x) = 6x – 1

Find g(x) from fg(x) = 6x – 1

f [ g(x) ] = 6x – 1

2 g(x) + 3 = 6x – 1

2 g(x) = 6x – 4

g(x) = 3x - 2

  1. Given the functions f : x  2 x  5 and

gf : x  10 x  25 , find the function g.

Answer : f(x) = 2x – 5

gf(x) = 10x – 25

Find g(x) from gf(x) = 10x – 25

g [ f(x) ] = 10x – 25

g ( 2x – 5) = 10x – 25

g ( 2y – 5) = 10y – 25

g(x) = 10y – 25 when x = 2y – 5

x + 5 = 2y

y = 2

x  5

So : g(x) = 10 ( 2

x  5 ) – 25

= 5x + 25 – 25

 g(x) = 5x

EXERCISES

  1. Given the functions f : x  2 x  2 and

fg : x  4  6 x , find the function g.

Answer :

  1. Given the functions f : x  2 x  2 and

g f : x  5  6 x , find the function g.

Answer :