Topik 1 - Fungsi Dan Graf

7/29/2019 Topik 1 - Fungsi Dan Graf

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TOPIK 1 - FUNGSI DAN GRAF

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Relations and Functions

Definition of relation

- A relation between two sets is the correspondence between elements of the first set, called

domain and elements of the other set, called range.

- A relation can be represented by :

(a) an arrow diagram(b) an ordered pairs

- Example 1:Let A = { 2, 3, 5 } and B = { 6, 9, 10 }. Consider the relation is a factor of

This relation can be displayed using an arrow diagram as follow :

The relation can also be written in the form of ordered pairs as

{(2,6), (2,10), (3,6), (3,9), (5,10)}

- There are four types of relations

i) One to one

each element in set X is connected to an element in set Y

2

3

5

6

9

10

is a factor of

AB

1

2

3

1

4

9

X Y

>

>

>

is the square

of

1


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(iii) One to many

(iv) Many to many

Definition of function

- A function is a special case of a relation which takes every element of one set (domain) andassigns to it one and only one element of a second set (range).

- In other words, a function is

one to one relation or

many to one

0

1

4

-2

-1

1

2

YX is the square root of

>

>

>>

a

b

c

d

e

f

a

b

c

d

1

2

3

4

2

ii) Many to


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- A mapping or function f from a set A to a set B is usually writtten as BA:f

- If an element x, of set A is mapped into an element y in set B we say that y is an image of x.

- The image of x is thus represented by f(x) and we write y = f(x)

Example of relations which are not function

(a) (b)

3

BA

a

b

c

1

2

3

BA

a

b

c

1

2

3

BA

a

b

c

1

2

3

4

BA

a

b

c

1

2

3

One-to-one relation and onto One-to-one relation and not onto

Many to one relation and onto Many to one relation and not onto

BA

a

b

c

1

2

3

BA

a

b

c

1

2

3

d


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________________________________________________________________________Example 1

Let A = {1, 2, 3, 4} and B = { set of integers}. Illustrate the function f : x x + 3.

Solution

(i )Quadratic function

(a) f(x) = x2 (b) f(x) = -x2

(ii) Cubic function

(a) f(x) = x3 (b) f(x) = -x3

(iii) surd function

The graph exist only for x 0

4

BA

1

2

3

4

5

6

4 7

f

f(x)

x0

f(x)

x0

f(x)

x0

f(x)

x0


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(a) xxf =)( (b) xxf =)(

(iv) Reciprocal function

(a) (b)

(v) Absolute value function, |f(x)|

(a) f(x) = |x|

Example

Sketch the graph of the following functions.

(a) f(x) = -5 (b) f(x) = -x + 2

5

f(x)

x0

f(x)

x0

xxf

1)( =

xxf

1)( =

f(x)

x0

f(x)

x0

f(x)

x

0


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________________________________________________________________________(c ) f(x) = x2 x 2 (d) f(x) = x2( 2 x )

(e) f(x) = 5+x (f) f(x) = | x 2 |

(g)

+


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________________________________________________________________________

(e) f(x) = 5+x

(f) f(x) = | x 2 |

g)

+


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________________________________________________________________________

Example :

Consider the graphs shown below and state whether they represent functions.

Domain and Range

- Domain of f(x) is the set of values of x for whichf(x) is defined.- Range of f(x) is the set of values ofy for which elements in the domain mapped.

- We can evaluate the domain and range by :

(i) Graph (ii) Algebraic approach

Vertical Line Test :- To test if a graph displayed is a function.

- The graph is a function if each vertical line drawn through the

domain cuts the graph at only one point.

(i) (ii)

2 2

The line cuts the graph at two points.

Not A Function

The line cuts the graph at only one

point.Function

-1

iii)

8

- Vertical lines are drawn parallel to the y-axis

The line cuts the graph at only one point.Function


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________________________________________________________________________Example 1

Consider the function f : x x + 4 with domain A = {1, 2, 3,4}. Then the range of the function canbe shown by the table.

x 1 2 3 4

f(x) 5 6 7 8

The range of the function is B = {5, 6, 7, 8}.

Example 2

The domain of this function is A = {x : -3 x 3} and the range is B = {x : 0 f(x) 9 }.

Example 3

For what values of x are the following functions defined?

i) f(x) = 2x 5 ii) f(x) = 2

1

x

Solution

i) y = f(x) = 2x 5 is defined for all values of x.

ii) y = f(x) =2

1

xis defined for every value of x except x = 2.

Example 4

Sketch the graph of the following functions. Hence, find its domain and range.

(a) f(x) = -3 (b) f(x) = -2x + 1

(c ) f(x) = x2 x 2 (d) f(x) = x2( 4 x )

(e) f(x) = 7+x (f) f(x) = | x 4 |

(g)

+


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________________________________________________________________________Solution

(a) f(x) = -5

Df =

Rf = {-3}

(b) f(x) = -2x + 1

(c ) f(x) = x2 x 2

(d) f(x) = x2( 4 x )

10

f(x)

x0

f(x)

x0

f(x)

x0

f(x)

x0

-3

1

1/2

-1 2

-2

4

),(

Df = (-, )Rf = (-, )

Df = (-, )

Rf = ),[ 4

9

Df = (-, )Rf = (-, )


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________________________________________________________________________

(e) f(x) = 7+x

(f) f(x) = | x 4 |

g)

+


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________________________________________________________________________

Algebraic Approach

Find the domain and range for the following functions :

(a) f(x) = 3x + 1, x R

Df = (-, )Rf = (-, )

b) f(x) = x2 3x, x R

Df = (-,)

c) f(x) = 5+xx + 5 0

x - 5 Df= [- 5,)

By inspection

when x = -5, f(x) = 0

when x , f(x)

Rf = [0, )

Composite Functions

Definition: Consider two functions f(x) and g(x). We define f g(x) = f [g(x)] meaning that theoutput values of the function g are used as the input values for the function f.

Note that ( f g)(x) f(x) g(x).

Similarly, we define g f(x) = g [f(x)] meaning that the output values of thefunction f are used as the input values for the function g.

Note that (g f)(x) g(x) f(x).

Example 1

If f(x) = 3x + 1 and g(x) = 2 - x, find as a function of x

(a) f g (b) g f

Solution

(a) (f g)(x) = f[g(x)]= f(2 - x)= 3(2 - x) + 1

12

4

9

2

3 2 = )x(

),[Rf =4

9


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________________________________________________________________________= 6 - 3x + 1 = 7 - 3x

(b) (g f)(x) = g[f(x)]= g(3x + 1)= 2 - (3x + 1)

= 1 - 3x

Note that ( f g)(x) ( g f)(x).

Example 2

Given that f(x) = 3x and g(x) = 1 + x, find as a function of x

(a) g f (b) f g

Solution

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

= 1 + 3x.

(b) (f g)(x) = f[g(x)]= f(1 + x)

= 3x + 1.

Example 3

If f(x) = 2x - 1 and g(x) = x 3, find the values of(a) gf(3) (b) fg(3) (c) f 2(3)

Solution

(a) gf(3) = g[f(3)]= g[2(3) - 1]

= g(5)

= 53 = 125

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

= f[33]= f[27]

= 2(27) - 1 = 53

(c) f 2(3) = f[f(3)]

= f(5)= 2(5) - 1 = 9

Note that f2(3) [ f(3) ]2

13


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Example 4

The functions f, g and h are defined by f(x) = 2 x, g(x) =1x

3

+(x -1) and h(x) = 2x 1.

(a) Show that f 2(x) = x.

(b) Find an expression for g2(x), and state for which two values of x it isundefined.

(c) Solve the equation h3(x) = x.

Solution

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

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

(b) g2(x) = g[g(x)]

= g(1x

3

+)

=1

1x

3

3

++

=4x

1)x(3

+

+

(c) h3(x) = h2[h(x)]

= h[h2(x)]

= h[h(2x 1)]= h[2(2x 1) - 1]

= h[4x 3]= 2(4x 3) - 1 = 8x 7

So, when h3(x) = x

8x 7 = x7x = 7, x = 1

Example 5

Consider the function f(x) = x2 for x R and g(x) = 2x + 3 for x R .(a) Find g f and determine its domain.

(b) Find f g and determine its domain.

Solution

(a) ( g f)(x) = g(x2)= 2x2 + 3

The domain of g f is the domain of f, that is x R.

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

14


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The domain of f g is the domain of g, that is x R.Example 6

The functions f and g are defined by f(x) = 1 + 2x for x 0 and g(x) =1-x

1for x > 1

(a) Find an expression for fg(x) and determine its domain.

(b) Find an expression for gf(x) and determine its domain.

Solution

(a) fg(x) = f(1-x

1)

= 1 + 2(1-x

1) =

1x

1x

+

The domain of fg(x) is the domain of g(x), that is {x R : x > 1}.

(b) gf(x) = g(1 + 2x)

=1)x21(

1+

=x2

1

The domain of gf(x) is the domain of f(x), that is x 0. But then, when x = 0, gf(x) =x2

1

is undefined. So, the domain of gf(x) is {x R: x > 0 }.

Example 7

Given that f(x) = 2x + 3 and fg(x) = 10x - 9, find g(x).

Solution f(x) = 2x + 3

fg(x) = f[g(x)]= 2g(x) + 3

But, fg(x) = 10x - 9

Therefore, 2g(x) + 3 = 10x - 9

2g(x) = 10x - 12g(x) = 5x - 6.

Example 8

If g(x) = 2x + 3 and fg(x) = 10x - 9, find f(x).

Solutionfg(x) = 10x - 9

f(2x + 3) = 10x - 9

Let u = 2x + 3u - 3 = 2x

2

3u = x

15


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________________________________________________________________________

Therefore, f(u) = 10(2

3u ) - 9

f(u) = 5u - 15 - 9f(u) = 5u - 24

By substituting u with x, f(x) = 5x - 24.

Even and Odd Function ?

Inverse Functions

The inverse function f -1 exists only if f is one-to-one.

- Method to test wether a function is 1-1 :

i) Algebraic approach.

If f ( x1 ) = f ( x2 ) , then x1 = x2

ii) Horizontal line test ( graphical approach)

If the horizontal line intersects the graph of the function only once , thenthe function is one to- one.

Example :

Example 2

Show that the following functions are one to - one functions.

a) f ( x ) = 3x 2 , x Rb) g ( x ) = x3 + 7 , x R

one-to-oneNot one-to-one

b)

16

a)


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________________________________________________________________________Solution :

a) f ( x ) = 3x 2 , x R

Method 1 : Use f ( x1 ) = f ( x2 ) to show x1 = x2

f ( x1 ) = f ( x2 )Therefore, 3x1 - 2 = 3x2 - 2

3x1 = 3x2x1 = x2

Hence , f ( x ) is one to one function

Method 2 : Horizontal Line Test

The horizontal line intersects the graph y = 3x - 2 at one point only.


b) g ( x ) = x3 + 7 , x R

Method 1 : Use f ( x1 ) = f ( x2 ) to show x1 = x2

Thus , x13 + 7 = x2

3 + 7

x13 = x2

3

x1 = x2


Method 2 : Horizontal Line Test

y = 3x - 2y

x

y = k

17


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________________________________________________________________________

The horizontal line intersects the graph y = x3 + 7 at one point only.

Hence , f (x) is one to one function

Example 3

Find the inverse function to f(x)=2

3x.

Solution

The function f maps x onto y where y =2

3x.

Make x the subject of this equation,

2

3x= y

x - 3 = 2y

x = 2y + 3, Hence f-1 (x) = 2x + 3.

Suppose - 4 was a value in the original domain . Then f will map this onto 32

1.

f-1 will now map this value onto 2(-32

1) +3 = -4, which is the original value.

Example 4

Find the inverse of f(x) = 3 x

Solution

f maps x onto y where y = 3 - x,so x = 3 - y

f-1 (x) = 3 - x

y = k

y = x3 + 7y

x

18


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Example 5 :

Find the inverse for each of the following function :

Rx,x)x(f)a 12 =

Rx,x)x(f)b 12 3 +=Rx,x)x(f)c

3

23+=

Rx,x)x(f)a 12 =

Solution :

x)]x(f[f =1

x)x(f = 12 1

2

11 += x)x(f

Rx,x)x(f)b 12 3 +=

x))x(f( =+ 12 31

31

2

1=

x)x(f

Rx,x)x(f)c

3

23+=

y)x(fLet =1

x)y(f =

x)]x(f[f =1

xy =+3

23

3

3

2= xy

31

3

2=

x)x(f

19


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________________________________________________________________________Example

Given the function f(x) =3

+x

px, (x 3) , where p is a constant,

(a) Find the value of p if f(5) = 12

1,

(b) Find f

-1

in a similar form.(c) State the value of x for which f-1 is undefined.

Solution

(a) f(5) =35

5

+ p

= 12

1

5 + p = 3

p = -2

(b) From (a), f (x) =3

2

x

x

f[f-1(x) = x

x)x(f

)x(f=

3

21

1

x)x(xf)x(f 32 11 =

x)x)(x(f 3211 +=

11

321 +

= x,x

x)x(f

(c) f-1 is undefined for x =1 (This means that there is no value of x in the original domain which

had an image of 1. So, 1 does not exist in the range and therefore cannot be used.

Example 7

Given thatf(x) = 3x + 5. Find

(a) (f1 )2 (b) (f2)-1

Solution

(a) (f1 )2

Let y = 3x + 5

x =

f-1(x) =

(f-1 )2 (x) =f-1[f-1(x)]

20

3

5y

3

5x

= 3

51 xf3

53

5

=

x


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________________________________________________________________________(b) (f2)-1

f2(x) = f[f(x)]=f(3x + 5)

= 3(3x + 5) +5

= 9x + 20

Let w = 9x + 20

x =

Note that :

Example 8

The functions f and g are defined by f : x 2x + 3 and g : x x - 1. Find

(a) f -1 and g-1

(b) g f-1 and f g-1

(c) (f g)-1

(d) f -1 g-1

Solution

(a) x)]x(f[f =1

x)x(f =+ 32 1

2

31 = x)x(f

x)]x(g[g =1

x)x(f = 11

11 += x)x(f

(b) (g f-1)(x) = g[f -1(x)]

= g(2

3x )

=2

3x - 1 =

2

5x

(f g-1)(x) = f[g-1(x)]= f(x + 1)

= 2(x + 1) + 3

= 2x + 5

21

9

20=x

9

20w ( ) ( )9

2012 =

xxf

( ) ( ) 1221 = ff


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(c) (f g)(x) = f[g(x)]= f(x - 1)

= 2(x 1) + 3= 2x + 1

Let a)x()gf( =1

12 +==

ax

)a)(gf(x

2

1=

xa (f g)-1(x) =

2

1x

(d) f -1 g-1 = f-1(x + 1)

=2

3)1x( +=

2

2x

Note that (f g)-1(x) f-1 g-1(x)

Example 9

Given that f(x) = 1 - x and g(x) =2x

1

+, x -2.

Find (f g)-1 and g-1 f-1. What conclusion can you draw?

Solution

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

2x

1

+)

= 1 -2x

1

+=

2x

1x

+

+

Let

2a

1

-1

g)(a)(fx

1

+=

==

a)x()gf(

xa

=+

12

1

x

a

=+1

12 , 2

1

1

=

xa

x

x)x()gf(

=

1

121

22


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________________________________________________________________________

b)x(f

Let

=11-g

2b

1

1

+=

= )b(g)x(f

+=

2

1

bfx

2

11

+=

b

xb

=+

12

1

xb

=+

1

12 , 2

1

1

=

xb ,

x

x)x(fg

=

1

1211

Therefore, (f g)-1

= g-1

f-1

Example 10

Given that f(x) =x1

x1

+and g(x) =

x

1are defined for all x R (excluding 1, 0 and 1).

Show that (f g)-1 = f g.

Solution

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

1

)

=

x

11

x

11

+

=1x

1x

+

a)x(Let =-1g)(f x)a)(gf( =

x

a

a=

+

1

1

a + 1 = ax x

a(1 x) = -(1+x)

1

1

+

=x

xa

1

11

+=

x

x)x()gf(

23


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________________________________________________________________________

Graphical Illustration of an inverse Function.

Verify that the inverse of f(x) = 2x -3 is,

f-1 (x) =2

3+x

-2 -1 1 2 3 4 5

The graph of y = f(x) and f

-1

(x) are reflection of each other in the line y = x.

24

1

2

3

4

5

y = x

A'

y =3

2x +

-1

-2

-3

A

y = 2x - 3

y

x

32

32

f-1

f

y

yy = x

x


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________________________________________________________________________

f(x) =x

1, then f-1 (x) =

x

1

One final point that is worth noting is that

f ( f-1 (x) ) = f-1 ( f (x) ) = x

Df = Rf-1

Rf= Df-1

Example 10

For each the functions below, find the inverse function and state the domain and the range of f-1.Sketch the graph of f and f-1 on the same axes.

a) f(x) = -x2 + 5 , x 5b) f(x) = 2x , x 2

c) f(x) = 33

2>

x,

x

Solution

a) Let f[f-1(x)] = x

[-f-1(x)]2 + 5 = x(f-1(x))2 = x 5

f-1(x) = ,x 5Df = [ 0, + ) = Rf-1Rf = ( -, 5 ] = Df-1

b) f(x) = 2x , x 2Let f[f-1(x)] = x

25

y = x

x

5

5f

f-1 y = x

f(x)

x

y = xf(x)f-1


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________________________________________________________________________

x)x(f = 21

21 2 x)x(f =

f-1(x)= x2 + 2

Dh = [ 2, + ) = Rh-1

Rh = [ 0, + )= Dh-1

c) f(x) = 33

2>

x,

x

Let f[f-1(x)] = x

x)x(f

= 3

21

x

x)x(f

321 +=

Df = ( 3, + ) = Rf-1

Rf = ( 0, + ) = Df-1

Example 11

Find the inverse of f(x)= ( x + 4)(x 1), x 2

3 and stating its domain. Then, on the same axes,

sketch the graph of f and its inverse.

Solution

y = ( x + 4)(x 1) x 23

= x2 + 3x 4

(By completing the square )

26

x

f2

2

y = x

f(x)

x

3

3

f

f-1

4

25

2

3 2 += )x(

4

25

2

32

+=

+ yx

4

25

2

3+=+ yx

4

25

2

3+= yx

4

25

2

31++=

x)x(f

4

25

4

25

2

3

2

3

f-1

f

y = x

) += ,D

f 4

251


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________________________________________________________________________

Example 12

Given that f ( x ) = ( x 9 )2 , x R , x 9.

i) Find the inverse of function f .ii) Sketch the graph of f and f -1 on the same plane.

iii) State the domain and range of f-1.

Solution

a) To find f-1

We may use f [ f-1(x) ] = x

( f-1(x) - 9) 2 = x

f-1(x) - 9 = x ( taking the + sign )

= x + 9

27


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________________________________________________________________________

c) Domain of f-1 : [ 0 , )Range of f-1 : [ 9 , )

Example 13

The function g is defined by

Explain why g has no inverse.

If the domain of g is x R , x 0 , find g -1 and sketch the graph of g and g-1. State the domainand range of g-1.

Solution

By Horizontal Line Test ,

b) Sketch the graphs

y

x

f

9

9

f-1

y =k

y

x

y = ( x + 1)2 - 2

28

Rx,21)(xx:g 2 +


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________________________________________________________________________Graph y = k intersects y = (x +1)2 - 2 at two points. It is seen that g(x) is not one to one. Hence,

g has no inverse.

To find g-1 ,

x + 1 = 2+ y

x = 12 +y

That is g-1 ( y ) = 12 +yLet y = x,

Hence,

If the domain of g is x R , x 0 , then g is one to one - and hence g -1 exists .

Dg-1

= [-1 , ) , Rg-1= [0 , )

y

x

g y = k

-1

g-1

-1

Let y = ( x + 1)2 - 2

y + 2 = ( x + 1)2

121 + x)x(g

( taking the + sign because x 0 )

Topik 1 - Fungsi Dan Graf

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