Pembahasan Soal Fungsi Matematika

1.      G(x) = x² – x + 3

      Fog)(x) = 3x² – 3x + 4

      F(x-2) = ....

    

Function Composition

WELCOME EVERYBODY! ^^ TODAY I'LL POST ABOUT FUNCTION COMPOSITION, HIGHSCHOOL GRADE. THIS MATERIAL IS SO CHALLENGING. BUT YOU CAN GET USED TO IT! ^^
 

The Basic Characteristic Of Function Compositions

       I.          Composition Function is :

( f o g ) (x) =  f( g(x) )  = f composition g for x

Example ^^ :

1.  f(x) = 5x – 4 and g(x) = 2x + 8

( f o g )? = f( g(x) ) 

f( 2x + 8 )  ----->  5( 2x + 8 ) – 4 ---> 10x + 40 – 4 ----> 10x + 36

as you can see above ...

Basically change the X value of F(x) into g(x) so become f( g(x) ) = f( 2x + 8 )   and because the F(x) = 5x – 4 logically F( 2x + 8) = 5( 2x + 8) – 4     ^^   ----->   10x + 8

How about ( g o f )?

Value still the same : f(x) = 5x – 4 and g(x) = 2x + 8

( g o f ) =  g( f(x) )

g(5x -4) ------>  2(5x – 4) + 8   ----->  10x – 8 + 8 = 10x    ^^

The answer is different so this is mean ( f o g )  IS NOT THE SAME AS ( g o f )

2.      F(x) = x² + 2x – 3 and g(x) = 5x -1

( f o g )? = f( g(x) ) 

f(5x – 1) ------>  (5x – 1)² + 2(5x – 1) – 3

               ------->  25x – 10x + 1 + 10x – 2 - 3

               ------->  (25x – 4) the answer!

How about ( g o f )(-1) ?

Value still the same :  F(x) = x² + 2x – 3 and g(x) = 5x - 1

( g o f ) =  g( f(x) ) ---> ( g o f )(-1) = g( f(-1) )

5(x² + 2x – 3) – 1 ----->  What was asked is ( g o f )(-1) . so change the x from ( f(x) )into value -1 FIRST like this  -----> 5((-1)² + 2(-1) – 3) – 1 

---> 5(1 -2 – 3) – 1

---> 5(-4) – 1 = -21 the answer!

Here some other Characteristic ^^   :

1.  Function Compositions is NOT commutative 

           ( f o g )  IS NOT THE SAME AS ( g o f )

     2.  Function Compositions IS Associative

            ( ( f o g ) o h)(x)  =  ( f o (g o h ) )(x)

     3.  There is Identity Function I(x) = x

           so ---> ( f o I )(x) = ( I o f )(x) = f(x)

How To Determine Circle Equations

II.        LOOK FOR CIRCLE EQUATIONS ^^

         Now the reverse, how to find the equation of the circle if known its center and radius.

     Determine the equation of the circle below :

1.     Center (0,0) r = 3

2.     Center (-3, -1) r = 6
3.   Point (6, 8) is on a Circle A the center is (0.0). The circle equations?
            Answer 

  1. For the center (0,0) The circle equation  : x² + y² = r²     

o   So, the equation circle = x² + y² = 3

---

2. For the center not (0,0) r The circle equation  : (x-a)² + (y-b)² =r²

o   So, the equation circle = (  x-(-3) )² + ( y – (-8) )² = 6²

o   The equation circle = (x+3)² + (y+8)² = 6²

---

3. For the Center (0,0) and point (6.8) are in the circle, 
because passed on the circle. (6.8) as (x, y)
• Equation circle when center (0,0) = x² + y² = r²   (x,y) = (6,8)
Because radius unknown we seek it first:
x² + y² = r²
6² + 8² = r²
r² = 100
then the equation circle = x² + y² = 10²

---

So now, how to determine Circle Equations if the circle passed on some 3 points  that located in the circle itself ^^

Example :

 A Circle x² + y² + Ax + By + C = 0 passed on 3 points (2,0) , (0, -2) , (4, -2) The Circle Equations is?

 

Insert the points to the x² + y² + Ax + By + C = 0 as (x,y)

(2,0) ---> 2² + 0² + A(2) + B.0 + C = 0

                 4 + 2A + 0 + C = 0 ........................ (1)

(0, -2) --> 0² + (-2)² + A.0 + B(-2) + C = 0

                             4 – 0 - 2B + C = 0 .......................... (2)

(4, -2) ---> 4² + (-2)² + A(4) + B(-2) + C = 0

                  16 + 4 + 4A - 2B +  C = 0

                  20 + 4A - 2B +  C = 0 ....................(3)

 
there are 3 quations and let's solve the equations!

Eliminate (1) dan (2)

4 + 2A + 0 + C = 0 .... (1) 

4 – 0 - 2B + C = 0 ..... (2)        

--------------------------------   -

0 + 2A + 2B + 0 = 0 .... (4)  

 

Eliminate (3) dan (2)

20 + 4A - 2B +  C = 0 ...(3)

4 – 0 - 2B + C = 0 ..... (2)    

------------------------------    -

 16 + 4A = 0

 

A = -4 -------> Substitute A as -4 to .... (4)

2A + 2B  = 0 ---> -8 + 2B = 0 --> B =4 

Substitute A and B as its value to equations number (1) / (2) / (3) 

4 + 2A + 0 + C = 0 .... (1) 
 4 – 8 + 0 + C = 0 

C = 4

after we found all the value A, B, C insert them to : x² + y² + Ax + By + C = 0 

So the Circle Equations  x² + y² - 4x + 4y + 4 = 0 ^^ 

Cara Menentukan Persamaan Lingkaran

Menentukan persamaan lingkaran jika diketahui pusat dan jari-jari.

     Tentukan persamaan lingkaran berikut :


1. Pusat (0,0) Jari-jari = 3

2. Pusat (-3, -1) Jari-jari = 6

3. Titik (6, 8 ) melewati lingkaran A yang pusatnya (0,0). Persamaan Lingkaran A?

To Know kinds of Circle Equations, Look For Center and Radius

    THERE ARE 3 TYPES OF CIRCLE EQUATIONS:

1.    The center (0,0) equation : x² + y² = r²     

2.    The center (a, b) equation 1 : (x-a)² + (y-b)² =r²

3.     The center (a, b) equation 2 :  x² + y² + Ax + By + C = 0       atau

                                           x² + y² - 2ax – 2by + a²  + b² - r² = 0     

Description :   -    a,b  =          center of circle

-        x,y =          a point on the circle

-        r     =          radius

-        A    =          - ½ a

-        B    =          - ½ b

-        C =             a²  + b² - r²     

---

I.                HOW TO FIND RADIUS AND CENTER 

 if equation of circle is known

Specify the center and radius of the following circle equation:

1.     Center (0,0) equation : x² + y² = r²   example :

·       x² + y² = 100  -->  center (0,0) radius = 10 because100 = r²  not r ^^

·       x² + y² + 19 = 100 -->  center  (0,0) radius = √100-19 = √81 = 9

·        

2.     Center (a,b) equation 1 : (x-a)² + (y-b)² =r² example :

·       (x+3)² + (y+8)² = 6² --> center (3,8) ; r = 6

·       (x+9)² + (y+1)² = 100 --> center (9,1) ; r = 10

3.    Center (a,b) equation 2 :  x² + y² + Ax + By + C = 0 

·       2x² + 2y² - 8x + 5 = 0 

A = 2 -----> - ½ a = -2 ----> a = 4

B = 0 -----> -  ½ b = 0 ----> b = 0

C = a²  + b² - r² 

5 =  (4)²  + 0² - r²     

r² =  16 – 5 = 9 à r = 3

center = (4,0) r = 3

---

Mengetahui Jenis-jenis Persamaan Lingkaran, Mencari Pusat dan Jari - Jarinya

    ADA 3 JENIS PERSAMAAN LINGKARAN :

1.     Pusat (0,0) persamaan : x² + y² = r²     

2.     Pusat (a,b) persamaan 1 : (x-a)² + (y-b)² =r²

3.     Pusat (a,b) persamaan 2 :  x² + y² + Ax + By + C = 0       atau

                                           x² + y² - 2ax – 2by + a²  + b² - r² = 0     

Keterangan :   -    a,b =          pusat lingkaran

-        x,y =          suatu titik di lingkaran

-        r     =          jari-jari

-        A    =          - ½ a

-        B    =          - ½ b

-        C =             a²  + b² - r²     

---