Topik 1 - Fungsi Dan Graf
Transcript of 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
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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
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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
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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)
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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.
Hence , f ( x ) is one to one function
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
Hence , f ( x ) is one to one function
Method 2 : Horizontal Line Test
y = 3x - 2y
x
y = k
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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
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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
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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
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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(
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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
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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
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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
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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 )