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
^^ :
- 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 + 36as 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 + 8How 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 ^^ :
- Function Compositions is NOT commutative
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
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
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)
------------------------------ -
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
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
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