Chapter 1- Functions Add Maths

7/31/2019 Chapter 1- Functions Add Maths

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Additional Mathematics Module Form 4

Chapter 1- Functions SMK Agama Arau, Perlis

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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


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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 + 3If 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 ofp can be

calculated.

Remember the concept:

f : x 3x 2 then

f(x)= 3x 2 and when x= 1,

f(-1) = 3( 1) 2


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Example 2:

The function g is defined bymx

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 ofxif 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 functionfis 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 ofxiff(x) =x.

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

7. Given the function 52: xxn . Find the value ofkat which kkn =)( .

mm ++=

2

)2(2

5

)5(2.3

mg

+=

5

)5(2)5( and

mg

+=

5

)2(2)2( . Replace

g(5) withm+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 ofxto make the function

undefined, the denominator must be equal to

zero.


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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: + xx 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 functionsfand g.

Find

(a) the value ofk,

(b) fg(x).

x1

2

3x3

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 offhas

become the domain of function g

or f(x) becomes the domain of

function g.

This is f(x) which has become the object offunction 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)]. Whenxin 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

ex and the ex ression and becomes 3x + 3.

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

kxx

xg

xxf

,1

4:

32:


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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 2x . If 2=x , the denominator would be 0 and the solution will be undefined

or infinity. Soxcannot equal to 2. For the question above, we know thatxcannot 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 1x . 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, 1x

Example 3:

The following information refers to the functionsfand g.

2,2

3:

34:

xx

xg

xxf

Replace g(x) with1

4

x

f(x)=2x 3 ,

)1

4(

xf = 2 )

1

4(

x- 3


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Find

(a) the value ofgf(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


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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

31311)]([2 +

=

xxxg

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


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1

4)(

=

xxg

Hence 1,1

4:

x

xxg

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

Find functionf.

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 ,

1

4

)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


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=

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. Functionfand gfare given as2

1)(

+=

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

2. Given xxf 35)( = and xxg 4)( = . Find functionfg.

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

(i) functionf

(ii) functionfg

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 ofk

(b) find

(i) wv(x)

(ii) vw(x)

Simplify

It meansy

y 128 divided by

y

4


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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

1y

y

f-1

(4) = 13

1x 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

41

X Y

Inverse function

1f

f

1

Ti s

1f

f

1f

f


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1.4.1 Determining the inverse function

Example 1:

Given 2,2

5: xxxg . 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

+

= x

x

xxg

Example 2:

The functionfis defined by xxf 43: . Findf-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


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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 functionf.

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 function1f

(b) state the value of x such that 1f 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 ofp given that 8)(1

=

ph .

2

image

xxf 43)( =

yf = )5(1

image

y -5


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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 0x 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 x2

Replace f(x) with its image which is x2


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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


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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 644 2 =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 kare constants,

find the value ofh and ofk.

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. Wedo 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.


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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

+

x

x

xkxg

Hence ,

12=h and 2=k

1

2

1 2


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CHAPTER REVIEW EXERCISE

1. Diagram 1 shows the relation between setJ 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 thatf(x)= ax+ b.

Find

(a) the value ofa and ofb,

(b) the value ofm ifm is mapped onto itself under the function f,

(c) the value ofxif f-1

(x) =f2(x).

3. Function h is given as .,14)( kx

xkxxh

= Find the value of k.

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

)(1 kxxg

+=

find the values ofh and k.

5. Given 2,2

3:

x

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

1

2

2

4

6

8

10SetJ

Set K

1 3

11

23

x

y

y= f(x)

O

DIAGRAM 2


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

kxg , where h and k are

constants. It is given thatf1

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

7. Given that 23)( = xxf and 1)(

2

= xxfg , find(i) the function g

(ii) the value ofgf(3)

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

(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 ofp and q

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

11. Function g and composite functionfg are defined by 3: + xxg and 76: 2 ++ xxxfg .

(a) Sketch the graph ofy= ( )g x for the domain 5x 3.

(b) Findf(x).

12. Given quadratic function2( ) 2 3 2g x x x= + . find

(a) g(3),

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

13. Functionsfand g are given as2

: xxf and qpxxg +: , wherep and q are constants.

(a) Given that )1()1( gf = and )5()3( gf = , find the value ofp and ofq.

(b) By using the values ofp 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 ofxif 3)( =xf .

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Chapter 1- Functions Add Maths