chapter1 functions

Additional Mathematics Module Form 4

Chapter 1- Functions SMK Agama Arau, Perlis

Page | 1

CHAPTER 1- FUNCTIONS

1.1 RELATIONS

1.1.1 Representing Relations

1- By Arrow Diagram

One to one relation

x x + 2

Many to one relation

x x2

2- By Ordered Pair

(1, 3), (2, 4), (3, 5)

(1, 4), ( − 2, 4), (3, 9), ( − 3, 9)

3. By Graph

y (Image)

4 X

3 X

0 1 2 x (object)

One to many relation

x x

Many to many relation

360 min

3600 sec

60 sec

0.25 day

6 hours

1 min

2160 sec

1 hour

− 2

2

3

− 3

4

9

1

2

3

3

4

5

2

− 2

3

− 3

4

9



Page | 2

domain

codomain

A B

Set A is called domain of the relation and set B is the codomain.

Each of the elements in the domain (Set A) is an object. So 1, 2 and 3 are called object.

Each of the elements in the codomain (Set B) is an image. So 3, 4, 5 and 6 are called image.

The set that contains all the images that are matched is the range.

A B

Domain={ 1,2,3 } codomain= {3,4,5,6}

Object = 1, 2, 3 image = 3, 4, 5, 6

Range= {3, 4, 5}

Object image

1.2 FUNCTIONS

1. Function is a special relation where every object in a domain has only one image. A function is also

known as a mapping.

2. From the types of relation we have learned, only one to one relation and many to one relation are

function.

Function notation

Small letter: f, g, h or something else…

xxf 2: →

Read as function f map x onto 2x

Example:

f : x 2x

f(x) = 2x

x = 1,

f(1) = 2(1)

= 2

1

2

3

3

4

5

6

object

image

Concept:

xxf 2: →

xxf 2)( =

Look at the difference between

these two.



Page | 3

Example 1:

1. Given function 23: −→ xxf . Find

(a) the image of –1

f(x)= 3x – 2

f(-1) = 3(– 1) – 2

= – 3 – 2

= – 5

(b) object which has the image 4

f(x) = 3x – 2

Given that the image is 4,

So f(x) = 4

Compare and ,

Hence,

3x – 2 = 4

3x = 6

x = 2

2. Given that f(x) = px + 3 and f(4)= 5. Find the value of p.

f(x) = px + 3

If x=4,

f(4)= 4p + 3

f(4)= 5

Compare and ,

Hence,

534 =+p

24 =p

2

1=p

If the given is object, so we are going to

find the value of image. -1 is the object

because we have to find the value of its

image.

If the given is image, so we are going to find

the value of object. 4 is the image because we

have to find the value of its object.

1

2

1 2

1

2

1 2

From the information given, we know

that 4 is the object and 5 is the image.

The both function is compared because

they are under the same function or in

other way there are having the same

object which is 4. So the value of p can be

calculated.

Remember the concept:

f : x 3x – 2 then

f(x)= 3x – 2 and when x= –1,

f(-1) = 3(– 1) – 2



Page | 4

Example 2:

The function g is defined by mx

xxg

+=

2)( If g(5) = 3g(2), find the value of m. Hence, find:

(a) image of 4.

(b) the value of x such that the function g is undefined.

Answer:

g(5) = 3g(2)

60 + 12m = 20 + 10m

2m = - 40

m = -20

(a) 204

)4(2)4(

−=g

= 2

1−

(b) 020 =−x

20=x

EXERCISE 1.2

1. Given function f is defined by 45)( −= xxf . Find the value of x if 14)( −=xf .

2. Given that 32)( −= xxf . Find the image of 5.

3. Given that function f such that4

32)(

−

−=

x

xxf .

(i) The image of 5.

(ii ) the value of x such that the function f is undefined

4. Given the function h such that 132)(2

++= kxxxh and 9)1( =h . Find the value of k.

5. Given the function 13: +→ xxf , find

(a) the image of 2,

(b) the value of x if f(x) = x.

6. Given the function 34: +→ xxw , find the value of p such that 23)( =pw .

7. Given the function 52: −→ xxn . Find the value of k at which kkn =)( .

mm ++=

2

)2(2

5

)5(2.3

mg

+=

5

)5(2)5( and

mg

+=

5

)2(2)2( . Replace

g(5) with m+5

)5(2and g(2) with

m+5

)2(2.

Any number that is divided by zero will result

undefined or infinity. The value of denominator

cannot equal to zero because it would cause

the solution becomes undefined or infinity. To

find the value of x to make the function

undefined, the denominator must be equal to

zero.



Page | 5

1.3 COMPOSITE FUNCTIONS

A f:x 3x B g:x x2

C

gf(x) = g[f(x)] gf

= g(3x)

= (3x)2

Example 1:

Given function 1: +→ xxf and xxg 3: → . Find the composite function gf.

f(x) = x + 1

g(x) = 3x

gf(x) = g[f(x)]

= g(x +1)

= 3(x +1)

Hence, gf: x 3(x +1)

Example 2:

The following information refers to the functions f and g.

Find

(a) the value of k,

(b) fg(x).

x

1

2

3x

3

6

(3x)2

9

36

The figure shows Set A, B and C.

The combined effect of two

functions can be represented by

the function gf(x) and not fg(x).

This is because the range of f has

become the domain of function g

or f(x) becomes the domain of

function g.

This is f(x) which has become the object of

function g

The concept is:

f(x) = 3x

f(k) = 3(k)

f(3) = 3(3)

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

(x +1) has become the object of function g which is

replacing x. [g(x)=3x and g(x+1)=3(x+1)]. When x in the

object is replaced by (x+1) so x in the image is also

replaced by (x+1) and become 3(x+1). We can also

expand the expression and becomes 3x + 3.

Replace f(x) with its image which is (x +1)

kxx

xg

xxf

≠−

→

−→

,1

4:

32:



Page | 6

Solution:

In Form One we have learned that any number that divided by zero will result infinity or undefined.

For part (a), if there is given an expression which is a fraction there must be given an information that

the value of unknown in the denominator that it cannot be. For example, 2,2

: ≠−

→ xx

hxg . The

part of denominator is 2−x . If 2=x , the denominator would be 0 and the solution will be undefined

or infinity. So x cannot equal to 2. For the question above, we know that x cannot equal to 1 because if

1=x ,the denominator would be 0 and the solution will be undefined or infinity. It is given that kx ≠ .

We already know that 1≠x . So compare these two equations and it would result 1=k . So 1 is the

answer for part (a)

For part (b),

f(x)=2x – 3 and) 1,1

4)( ≠

−= x

xxg

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

= )1

4(

−xf

= 2 )1

4(

−x- 3

= 1

8

−x- 3

=1

)1(3

1

8

−

−−

− x

x

x

= 1

)1(38

−

−−

x

x

=1

338

−

+−

x

x

=1

311

−

−

x

x, 1≠x

Example 3:

The following information refers to the functions f and g.

2,2

3:

34:

≠−

→

−→

xx

xg

xxf

Replace g(x) with 1

4

−x

f(x)=2x – 3 ,

)1

4(

−xf = 2 )

1

4(

−x- 3



Page | 7

Find

(a) the value of gf(1)

(b) f2

Solution:

For part (a), at first we have to find gf(x).

Given xxf 34)( −= and 2,2

3)( ≠

−= x

xxg

gf(x)= g[f(x)]

= g(4 – 3x)

= 2)34(

3

−− x

= x32

3

−

gf(x)= 3

2,

32

3≠

−x

x

gf(1)= )1(32

3

−

= 32

3

−

= 1

3

−

= 3−

For part (b), f2 means ff.

Given xxf 34)( −=

ff(x)= f[f(x)]

= f ]34[ x−

= )34(34 x−−

= x9124 +−

= 89 −x

89)(2

−= xxf

Replace f(x) with x34 −

2

3)(

−=

xxg

2)34(

3)34(

−−=−

xxg

Replace f(x) with x34 −

xxf 34)( −= ,

f ]34[ x− = )34(34 x−−



Page | 8

1.3.1 Determining one of the function in a given composite function

(1)The following information refers to the functions f and fg.

Find function g.

Solution:

Given 32)( −= xxf and 1,1

311)( ≠

−

−= x

x

xxfg

If 32)( −= xxf ,

If 32)( −= yyf

So 3)]([2)]([ −= xgxgf

3)]([2)( −= xgxfg

Given 1,1

311)( ≠

−

−= x

x

xxfg .

Compare the two equations.

Hence,

1

3113)]([2

−

−=−

x

xxg

31

311)]([2 +

−

−=

x

xxg

1

)1(3

1

311)]([2

−

−+

−

−=

x

x

x

xxg

1

)1(3311)]([2

−

−+−=

x

xxxg

1

33311)]([2

−

−+−=

x

xxxg

1

8)]([2

−=

xxg

)1(2

8)]([

−=

xxg

1,1

311:

32:

≠−

−→

−→

xx

xxfg

xxf

Remember the concept…

f(x) = 2x – 3

f(k) = 2k – 3

f(3) = 2(3) – 3

f[g(x)]= 2g(x) – 3

Simplify 8 and 2



Page | 9

1

4)(

−=

xxg

Hence 1,1

4: ≠

−→ x

xxg

(2) The following information refers to the functions f and fg.

Find function f.

Solution:

1

311

1

4

1

311)]([

−

−=

−

−

−=

x

x

xf

x

xxgf

Let

yx

=−1

4

yxy −=4

yxy += 4

y

yx

+=

4

Substitute and into ,

14

)4

(311

)(

−+

+−

=

y

y

y

y

yf

=

y

y

y

y

y

y

−+

+−

4

)312

(11

1,1

311:

1

4:

≠−

−→

−→

xx

xxfg

xxg

1,1

311:

1

4:

≠−

−→

−→

xx

xxfg

xxg

2

3

1

2 3 1

Convert 1 to a fraction



Page | 10

=

y

yy

y

y

y

y

−+

−−

4

31211

=

y

y

yy

4

31211 −−

=

y

y

y

4

128 −

=4

128 y

y

y×

−

= 32 −y

32)( −= yyf

32)( −= xxf

EXERCISE 1.3

1. Function f and gf are given as2

1)(

+=

xxf and 23)( += xxgf respectively. Find function g.

2. Given xxf 35)( −= and xxg 4)( = . Find function fg.

3. Given 6)( −= xxgf and 34)( −= xxg . Find

(i) function f

(ii) function fg

4. Given the function xxw 23)( −= and 1)(2

−= xxv , find the functions

(a) wv

(b) vw

5. Given the function 2: +→ xxw and kxx

xxv ≠

−

−→ ,

1

12)( ,

(a) state the value of k

(b) find

(i) wv(x)

(ii) vw(x)

Simplify

It means y

y 128 −divided by

y

4



Page | 11

1.4 INVERSE FUNCTIONS

If yxf =)( then xyf =−

)(1

A B

Example 1 x 3x + 1

Given 13: +→ xxf

If x=1,

1)1(3)1( +=f

= 13 +

= 4

Hence,

4)1( =f x 3x +1

So, 3

1−y y

f-1(4) = 1

3

1−x x

let

yx =+13

13 −= yx

3

1−=

yx

f-1

(y) 3

1−=

y

f-1

(x) 3

1−=

x

f-1

(4) 3

14 −=

= 1

4

1

X

Y

Inverse function

1−f

f

1≠

Tips…

1−f

f

1−f

f



Page | 12

1.4.1 Determining the inverse function

Example 1:

Given 2,2

5: ≠

−→ x

xxg . Determine the inverse function

1−g .

Solution:

let

yx

=− 2

5

52 =− yxy

yxy 25 +=

y

yx

25 +=

y

yyg

25)(

1 +=

−

0,25

)(1

≠+

=−

xx

xxg

Example 2:

The function f is defined by xxf 43: −→ . Find f-1

(-5)

Method 1

Let

yx =− 43

yx −= 34

4

3 yx

−=

4

3)(

1 yyf

−=

−

4

3)(

1 xxf

−=

−

4

)5(3)5(

1 −−=−

−f

4

8=

1



Page | 13

2)5(1

=−−f

Method 2

We know that

And

x 3 - 4x

Hence,

543 −=− x

534 +=x

84 =x

2=x

So,

2)5(1

=−−f

EXERCISE 1.4

1. Given xxf 35)( −= and xxg 4)( = . Find

(i) the value of )4(1

−−f

(ii) the value of )4()(1

−−

gf

2. Given xxg 35)( −= and xxgf 4)(1

=−

. Find function f.

3. Given 6)( −= xxgf and 34)( −= xxg . Find

(i) 1

)(−

gf

(ii ) the value of )2(fg

4. Given 1,1

2: ≠

−→ x

x

xxf ,

(a) find the function 1−

f

(b) state the value of x such that1−

f does not exist.

5. Show that the function 1,1

: ≠−

→ xx

xxf , has an inverse function that maps onto its self.

6. If 2: −→ xxh , find the value of p given that 8)(1

=−

ph .

2

image

xxf 43)( −=

yf =−−

)5(1

image

y

-5



Page | 14

1.5 ABSOLUTE FUNCTION

Absolute function means the value of image is always positive because of the modulus sign.

Given xxg −→ 2: and y= ( )g x .Find the value of y if 3=x

y= ( )g x

= |)3(| g

= |32| −

= |1| −

= 1

1.5.1 Graph of absolute function

Example:

Given xxf −= 2)( and |)(| xfy = . Sketch the graph of |)(| xfy = for the domain 0≤ x ≤ 6. Hence

state the corresponding range of y.

|)(| xfy =

|2| x−=

Minimum value of y=0

When 2-x = 0 (2, 0)

X=2

i) x = 0

y |02| −=

|2|= (0, 2)

2=

ii) ) x = 6

y |62| −=

|4| −= (6 , 4)

4=

Then, sketch the graph using the points.

This is called modulus. Any value in the

modulus would be positive. If the number

is positive, it remains positive but it is

negative it would change to positive

when remove the modulus sign.

Minimum value of y must be zero

because of absolute value cannot be

negative so the minimum it could be is

zero.

Replace g(x) with is x−2

Replace f(x) with its image which is x−2



Page | 15

Y

5

4

3 |)(| xfy =

2

1

0 x

1 2 3 4 5 6

Hence, the corresponding range of y is 0≤ y ≤ 4

:: Some important information to be understood ::

1. Concept Of Solving Quadratic Equation by factorization

0432

=−− xx

0)4)(1( =−+ xx

01 =+x or 04 =−x

1−=x or 4=x

Example:

Given xxxf 52)(2

−= . Find the possible values of k such that kf =−

)3(1

x 2x2

– 5x

Solution:

kf =−

)3(1

Hence,

3522

=− xx

03522

=−− xx

0)3)(12( =−+ xx

Range of x

Range of y

3

To make a product equal to zero, one of

them or both must be equal to zero. We

do not know either x-1 is equal to zero or

x-4 equal to zero so that is why we use

the word ‘or’ not ‘and.’

The general form of quadratic equation is

02

=++ cbxax . To factorize and use the

concept, the equation must be equal to zero.

k

image

object



Page | 16

012 =+x or 03 =−x

2

1−=x or 3=x

2

1)3(

1−=

−f or 3)3(1

=−f

Given that kf =−

)3(1

2

1−=k or 3=k

2. Concept of Square root and any root which is multiple of 2

In lower secondary we have learned that if x2

= 4 then x = 2. Actually the right answer for it is not 2 but

2± ( 2± is +2 and -2). This is because if (2)2,

the answer is 4 and if (-2)2, the answer is also 4. So the

square root of 4 is 2± . It is also the same for fourth root, sixth root and so on. But in certain situation,

the answer will be only one.

Example:

Solve the equation 6442

=x .

Solution:

4

16

644

2

2

±=

=

=

x

x

x

3. Comparison method

Example:

Given the functions 2,2

: ≠−

→ xx

hxg and 0,

12:

1≠

+→

− xx

kxxg where h and k are constants,

find the value of h and of k.

Solution:

From the information given, we know that the method that we have to use to solve the question is

comparison method.

To make a product equal to zero, one of

them or both must be equal to zero. We

do not know either 2x+1 is equal to zero

or x-3 equal to zero so that is why we use

the word ‘or’ not ‘and.’



Page | 17

First, we have to find the inverse function of g.

Let

yx

h=

− 2

hyyx =− 2

yhyx 2+=

y

yhx

2+=

y

yhyg

2:

1 +→

−

0,2

:1

≠+

→− x

x

xhxg

Given that

0,12

:1

≠+

→− x

x

kxxg

Both of them are inverse function of g.

So compare and ,

0,)(

)2()(:

1≠

+→

− xx

xhxg

0,)(

)()12(:

1≠

+→

− xx

xkxg

Hence ,

12=h and 2=k

1

2

1 2



Page | 18

CHAPTER REVIEW EXERCISE

1. Diagram 1 shows the relation between set J and set K.

DIAGRAM 1

State

(a) the codomain of the relation,

(b) the type of the relation.

2. Diagram 2 shows the graph of the function y = f(x).

It is given that f(x)= ax + b.

Find

(a) the value of a and of b,

(b) the value of m if m is mapped onto itself under the function f,

(c) the value of x if f -1

(x) = f 2

(x).

3. Function h is given as .,1

4)( kx

x

kxxh ≠

−

−= Find the value of k.

4. Given function 5)( += hxxg and ,2

)(1 kx

xg+

=−

find the values of h and k.

5. Given 2,2

3: ≠

−→ x

xxg . Find the values of x if xxg =)( .

1•

2•

•2

•4

•6

•8

•10 Set J

Set K

1 3

11

23

x

y

y= f(x)

O

DIAGRAM 2



Page | 19

6. Functions f and g are defined by hxxf −→ 3: and 0,: ≠→ xx

kxg , where h and k are

constants. It is given that f −1

(10) = 4 and fg(2) = 16. Find the value of h and of k.

7. Given that 23)( −= xxf and 1)(2

−= xxfg , find

(i) the function g

(ii) the value of gf(3)

8 Given 22)( += xxf and |)(| xfy = . Sketch the graph of |)(| xfy = for the domain 0≤ x ≤ 4. Hence

state the corresponding range of y.

9. Given that function h is defined as 12: +→ pxxh . Given that the value of 3 maps onto itself under

the function h, find

(a) the value of p

(b) the value of )2(1

−− hh

10. Given that the functions 12)( −= xxh and qpxxhh +=)( , where p and q are constants, calculate

(a) the value of p and q

(b) the value of m for which mh 2)1( =−

11. Function g and composite function fg are defined by 3: +→ xxg and 76:2

++→ xxxfg .

(a) Sketch the graph of y = ( )g x for the domain −5≤ x ≤ 3.

(b) Find f(x).

12. Given quadratic function 2

( ) 2 3 2g x x x= − + . find

(a) g(3),

(b) the possible values of x if 22)( =xg

13. Functions f and g are given as 2

: xxf → and qpxxg +→: , where p and q are constants.

(a) Given that )1()1( gf = and )5()3( gf = , find the value of p and of q.

(b) By using the values of p and q obtained in (a), find the functions

(i) gg

(ii) 1−

g

14. If 63)( −= xxw and 16)( −= xxv , find

(a) )(xwv

(b) the value of x such that xxwv =− )2( .

15. Given 63)( −= xxf . Find the values of x if 3)( =xf .

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