Now this world is very rapid progress. Technology now can not be separated from the day-to-day activities. One of the very rapid growth in the virtual world we call atu often use the internet with a computer. Through the Internet we can communicate directly with other people even in remote areas. Through the Internet we can add Science's a place we go without that. The one that I know the internet is the search engine data, such as google, yahoo, etc..
And lately I recognize "blog." I know more about the blog from my English lecturer at the university. Now this is my student Yogyakarta State University(UNY. Here I was on campus FMIPA mathematics education department. I know this blog because when the English language teachers to give me the task, the task can be submitted through the blog. For now I know that this blog is about the duty to collect the English language. But I catch my lecturer's explanation, blog is a place to share opinions about a problem. In addition, through this blog can also add a friend ... and experience!
Senin, 12 Januari 2009
English II ???
I'm very happy to get subjects in English II, Yogyakarta State University (UNY). Lecturer in the lead in the English II class, I also very prefesional. Lecturer in English II is Bp. Marsigit M.A.
I know very well from Bp. Marsigit is using the time so well. Bp.Marsigit for almost 1 semester lectures start on time must sometimes enter the lecture room before the English language course that should started. Bp.Marsigit emphasize the use of mathematical vocabulary, as we dive in the world of mathematics. So I also prosecuted for using English not only to the public, but also in mathematics. Sometimes Bp. Marsigit also provide "wejangan" for us to continue in a vibrant learning. Achievements that have been Bp. Marsigit to this is inspiration for us in the struggle to become someone like him. All of this is the first step to a change toward a better tomorrow ... Spirit!
I know very well from Bp. Marsigit is using the time so well. Bp.Marsigit for almost 1 semester lectures start on time must sometimes enter the lecture room before the English language course that should started. Bp.Marsigit emphasize the use of mathematical vocabulary, as we dive in the world of mathematics. So I also prosecuted for using English not only to the public, but also in mathematics. Sometimes Bp. Marsigit also provide "wejangan" for us to continue in a vibrant learning. Achievements that have been Bp. Marsigit to this is inspiration for us in the struggle to become someone like him. All of this is the first step to a change toward a better tomorrow ... Spirit!
Minggu, 11 Januari 2009
Partner
Totologi and withdrawal conclusion
> Every single statement certainly has a value of truth, that is True or False.
> Totologi can be expressed as a function of constant value with the truth contradictions have only one end result.
Some examples totologi
1. Law of Detachement
p & (p => q) => q
2. Mode Tollendo Tollens
-q & (p => q) =>-p
3. Mode Tollendo ponens
(p V q) &-p => q
4. Rules simplification
p & q => p
p & q => q
5. Rules mortgages syllogism
(p => q) & (q => r) => (p => r)
6. Rules contradiction
- (p &-p)
7. Law of Excluded Middle
p V -p
Friday, 9 January 2009 I and my friend, Dindha sit together to carry out a given task. This task given by the lecturers who teach the English language subject. So the task is surely to use the English language. This time I dindha and get a job to explain the subjects of mathematics using the English language. I explain about the withdrawal dindha conclusion in logic. 1 semester because we have only just learn so little, and I explain gave him problems.
Exercise:
- If the glasses have on table my kitchen, and I certainly have seen when a full breakfast
- I read the newspaper in the living room or I read it in the kitchen
- If I read the newspaper in the room, then surely my place the glasses on the table guests
- I do not see my glasses at the time of breakfast
- If I read the newspaper in the kitchen, the glasses in my kitchen table
Specify the location this glasses!
Only take 15 minutes dindha already completed the earlier question and answer correctly.
Answer:
Example:
My glasses on the kitchen table (a)
I have seen when a full breakfast (b)
I read the newspaper in the room (c)
I read it in the kitchen (d)
I put my glasses on the table guests (e)
1. a => b premis
2. c V d premis
3. c => e premis
4. –b premis
5. d => a premis
6. –b => -a (1t)
7. –a (4,6t)
8. –a => -d (5t)
9. –d (7,8t)
10. –c => d (2t)
11. –d => c (2t)
12. c (9,11t)
From the premises to-3 When I read the newspaper in the guest room, then surely my place glasses on the table front. So conclusion glasses are on the table because the room he read the newspaper conclusion on guests room.
When we are probably still the very poor. we are vocabulary still a lot less. But hopefully this is the beginning right to carry the future .. thank guidance on Bp. Marsigit M.A for this ....
Reference: "Logika Dan Himpunan Matematika" By Sukirman, M.Pd.
Partner : Dinda Putra Tanjung
> Every single statement certainly has a value of truth, that is True or False.
> Totologi can be expressed as a function of constant value with the truth contradictions have only one end result.
Some examples totologi
1. Law of Detachement
p & (p => q) => q
2. Mode Tollendo Tollens
-q & (p => q) =>-p
3. Mode Tollendo ponens
(p V q) &-p => q
4. Rules simplification
p & q => p
p & q => q
5. Rules mortgages syllogism
(p => q) & (q => r) => (p => r)
6. Rules contradiction
- (p &-p)
7. Law of Excluded Middle
p V -p
Friday, 9 January 2009 I and my friend, Dindha sit together to carry out a given task. This task given by the lecturers who teach the English language subject. So the task is surely to use the English language. This time I dindha and get a job to explain the subjects of mathematics using the English language. I explain about the withdrawal dindha conclusion in logic. 1 semester because we have only just learn so little, and I explain gave him problems.
Exercise:
- If the glasses have on table my kitchen, and I certainly have seen when a full breakfast
- I read the newspaper in the living room or I read it in the kitchen
- If I read the newspaper in the room, then surely my place the glasses on the table guests
- I do not see my glasses at the time of breakfast
- If I read the newspaper in the kitchen, the glasses in my kitchen table
Specify the location this glasses!
Only take 15 minutes dindha already completed the earlier question and answer correctly.
Answer:
Example:
My glasses on the kitchen table (a)
I have seen when a full breakfast (b)
I read the newspaper in the room (c)
I read it in the kitchen (d)
I put my glasses on the table guests (e)
1. a => b premis
2. c V d premis
3. c => e premis
4. –b premis
5. d => a premis
6. –b => -a (1t)
7. –a (4,6t)
8. –a => -d (5t)
9. –d (7,8t)
10. –c => d (2t)
11. –d => c (2t)
12. c (9,11t)
From the premises to-3 When I read the newspaper in the guest room, then surely my place glasses on the table front. So conclusion glasses are on the table because the room he read the newspaper conclusion on guests room.
When we are probably still the very poor. we are vocabulary still a lot less. But hopefully this is the beginning right to carry the future .. thank guidance on Bp. Marsigit M.A for this ....
Reference: "Logika Dan Himpunan Matematika" By Sukirman, M.Pd.
Partner : Dinda Putra Tanjung
Determining Limits By Inspection
There are 2 condition to determining limits by inspection
1. x goes to positive or negative infinity
2. limit involves a polynomian devided by a polynomial
For example:
Limit (x^3 - 4) / (x^2 + x + 1) when value of x go to positive infinity
This problem on two condition:
1. polynomial over polynomial
2. x approaches infinity
The key to determining limits by inspection is in looking at power of x in the numerator and the denominator
To apply these rules:
• Must be deviding by polynomials
• X has to be approaching infinity
If the highest power of x is greater in numerator so the limit is positive or negative infinity
Example:
1. Limit (x^3 - 4) / (x^2 + x + 1) when value of x go to positive infinity
- Highest power of x in numerator is 3
- Highest power of x in denominator is 2
Since all the number are positive and x going to positive infinity so value the limit is infinity.
Limit (x^3 - 4) / (x^2 + x + 1) when value of x is going to positive infinity having positive infinity value
If you can’t tell if the answer is positive or negative infinity:
You can substitute a large number for x
See if you end up with a positive or negative number
Whatever sign you get is the sign of infinity for the limit
2. Find solution!
Limit (x^3-4) / (x^4 + 3x +5) when value of x is going to positive infinity
Numerator is x^3 - 4
- Highest power of x in numerator is 3
Dominator is (x^4 + 3x +5)
- Highest power of x in denominator is 4
To know the value of function.
Deviding all of number with x which the power to divided this number is x with highest power of x from numerator and dominator. Because the highest power of x from dominator as high as numerator so we can divide all number in dominator and nominator with x^4.
Attention!
If number which divide x^4 don’t consist of variable x so the value is zero
If number which divide x^4 consist of variable x but the power variable of x in numerator more than less of x^4 so dividing this variable is zero value
If number which divide x^4 consist of x^4 so the value is a coefficient of x^4 from the numerator
Numerator = (x^3 / x^4) - (4 / x^4)
= 0 + 0 = 0
Dominator = (x^4 / x^4) + (3 / x^4) + (5 / x^4)
= 1 + 0 + 0
= 1
Limit (x^3-4) / (x^4 + 3x +5) when value of x is going to positive infinity has a zero value
3. Limit (3x^3 - 4) / (5x^3 + 3x + 5) when value of x is going to positive infinity
A LITTLE BIT TRICKER
Used when:
Highest power of x in numerator is SAME as highest power of x in denominator
Naminator : 3x^3 - 4
Dominator : 5x^3 + 3x + 5
Limit x going to infinity (positive or negative)
The quotient of the coefficients of two highest power
Numerator = (3x^3 / x^3) - (4 / x^3)
= 3 + 0 = 0
Dominator = (5^3 / x^3) + (3x/ x^3) + (5 / x^3)
= 5 + 0 + 0
= 5
Limit (3x^3 - 4) / (5x^3 + 3x + 5) when value of x is going to positive infinity has positive value. And the value is 3/5
1. x goes to positive or negative infinity
2. limit involves a polynomian devided by a polynomial
For example:
Limit (x^3 - 4) / (x^2 + x + 1) when value of x go to positive infinity
This problem on two condition:
1. polynomial over polynomial
2. x approaches infinity
The key to determining limits by inspection is in looking at power of x in the numerator and the denominator
To apply these rules:
• Must be deviding by polynomials
• X has to be approaching infinity
If the highest power of x is greater in numerator so the limit is positive or negative infinity
Example:
1. Limit (x^3 - 4) / (x^2 + x + 1) when value of x go to positive infinity
- Highest power of x in numerator is 3
- Highest power of x in denominator is 2
Since all the number are positive and x going to positive infinity so value the limit is infinity.
Limit (x^3 - 4) / (x^2 + x + 1) when value of x is going to positive infinity having positive infinity value
If you can’t tell if the answer is positive or negative infinity:
You can substitute a large number for x
See if you end up with a positive or negative number
Whatever sign you get is the sign of infinity for the limit
2. Find solution!
Limit (x^3-4) / (x^4 + 3x +5) when value of x is going to positive infinity
Numerator is x^3 - 4
- Highest power of x in numerator is 3
Dominator is (x^4 + 3x +5)
- Highest power of x in denominator is 4
To know the value of function.
Deviding all of number with x which the power to divided this number is x with highest power of x from numerator and dominator. Because the highest power of x from dominator as high as numerator so we can divide all number in dominator and nominator with x^4.
Attention!
If number which divide x^4 don’t consist of variable x so the value is zero
If number which divide x^4 consist of variable x but the power variable of x in numerator more than less of x^4 so dividing this variable is zero value
If number which divide x^4 consist of x^4 so the value is a coefficient of x^4 from the numerator
Numerator = (x^3 / x^4) - (4 / x^4)
= 0 + 0 = 0
Dominator = (x^4 / x^4) + (3 / x^4) + (5 / x^4)
= 1 + 0 + 0
= 1
Limit (x^3-4) / (x^4 + 3x +5) when value of x is going to positive infinity has a zero value
3. Limit (3x^3 - 4) / (5x^3 + 3x + 5) when value of x is going to positive infinity
A LITTLE BIT TRICKER
Used when:
Highest power of x in numerator is SAME as highest power of x in denominator
Naminator : 3x^3 - 4
Dominator : 5x^3 + 3x + 5
Limit x going to infinity (positive or negative)
The quotient of the coefficients of two highest power
Numerator = (3x^3 / x^3) - (4 / x^3)
= 3 + 0 = 0
Dominator = (5^3 / x^3) + (3x/ x^3) + (5 / x^3)
= 5 + 0 + 0
= 5
Limit (3x^3 - 4) / (5x^3 + 3x + 5) when value of x is going to positive infinity has positive value. And the value is 3/5
SAT Math Lesson 9 - Functions & Graphs! p3_4
13. The figure above shows the graph of y = g(x).
If the function his defined by h(x) = g(2x) + 2. What is the value of h(1)?
Solution:
h(x) = g(2x) + 2
because we need get the value of h(1), so we can substitute 1 for x
h(1) = g(2*1) + 2
h(1) = g(2) + 2 number 2 on g(2) is absis.
g(2) is ordinate when the absis=2.
(2 , g(2)) on h(x)= g(2x) +2
We find ordinate when x=2 which (2,y) on h(x)
From the picture we can see y=1
(2, g(2)) = (2,y)
g(2) = y
g(2) = 1
h(1) = g(2) +2
h(1) = 1 + 2
h(1) = 3
17. In the xy –coordinate plane, the graph of x = y^2 -4 intersects line A at (0,p) and (5,t). What is the greatest possible value of the slope of A?
Solution:
greatest m x = y^2 -4
If we subs x =0 so y = p
If we subs y =t so x = 5
Line A: m=(y2 - y1) / (x2 - x1)
m =(t – p) / (5 – 0)
m =(t – p) / 5
If the function his defined by h(x) = g(2x) + 2. What is the value of h(1)?
Solution:
h(x) = g(2x) + 2
because we need get the value of h(1), so we can substitute 1 for x
h(1) = g(2*1) + 2
h(1) = g(2) + 2 number 2 on g(2) is absis.
g(2) is ordinate when the absis=2.
(2 , g(2)) on h(x)= g(2x) +2
We find ordinate when x=2 which (2,y) on h(x)
From the picture we can see y=1
(2, g(2)) = (2,y)
g(2) = y
g(2) = 1
h(1) = g(2) +2
h(1) = 1 + 2
h(1) = 3
17. In the xy –coordinate plane, the graph of x = y^2 -4 intersects line A at (0,p) and (5,t). What is the greatest possible value of the slope of A?
Solution:
greatest m x = y^2 -4
If we subs x =0 so y = p
If we subs y =t so x = 5
Line A: m=(y2 - y1) / (x2 - x1)
m =(t – p) / (5 – 0)
m =(t – p) / 5
Pre-Calculus
GRAPH OF A RASIONAL FUNCTION
A. Rasional number is number able to be expressed a/b which b=0
For example (discontinuity)
a) f(x) = (x + 2) / (x – 1)
If substitution of x=1 so f(1) = ((1) + 2) / ((1) – 1)
f(1) = i
i = imajiner
f(x) = imajiner when x=1
If substitution of x=2 so f(2) = ((2) + 2) / ((2) – 1)
f(2) = 0/1
f(x) = y, ih f(2) so x=2 and y=0
The co-ordinate is (2,0)
f(x) = (x + 2) / (x – 1) having value If x 1,
• f(x) = (x + 2) / (x – 1) discontinuity
Hp:{x| x < 1 and x > 1, x subset riil number} or can we white (negative infinity,1) and (1,positive invinity)
Example ( continuity )
b) g(x) = 1 / (x^2 +1)
The function of g(x) always having value because the dominator don’t consist of zero for x R.
If x = 1, g(1) = ½
x = 2, g(2) = 1/5
• g(x) = 1 / (x^2 + 1) continuity
Hp={x | x subset R} or can we write (positive infinity, 1 ]
Polynomial
There are two ways to get the value from polynomial function
1. Subtitute of number for the variable.
Example:
g(x) = (x^2 + x – 6) / (x – 3)
x=3, baaad! (not allowed)
This method used direct by including value x
g(x) discontinuity if x=3 because g(x) = 0/0
2. Factor of the polynomial function
Example:
g(x) = (x^2 + x – 6) / (x – 3)
We can formulate (x^2 + x – 6) to (x-3)(x+2), so
g(x) = (x^2 + x – 6) / (x – 3)
= (x - 3).(x + 2) / (x – 3)
= (x + 2)
= x + 2
x=3, no problem!!
Domain g(x) = x subset riil number
A. Rasional number is number able to be expressed a/b which b=0
For example (discontinuity)
a) f(x) = (x + 2) / (x – 1)
If substitution of x=1 so f(1) = ((1) + 2) / ((1) – 1)
f(1) = i
i = imajiner
f(x) = imajiner when x=1
If substitution of x=2 so f(2) = ((2) + 2) / ((2) – 1)
f(2) = 0/1
f(x) = y, ih f(2) so x=2 and y=0
The co-ordinate is (2,0)
f(x) = (x + 2) / (x – 1) having value If x 1,
• f(x) = (x + 2) / (x – 1) discontinuity
Hp:{x| x < 1 and x > 1, x subset riil number} or can we white (negative infinity,1) and (1,positive invinity)
Example ( continuity )
b) g(x) = 1 / (x^2 +1)
The function of g(x) always having value because the dominator don’t consist of zero for x R.
If x = 1, g(1) = ½
x = 2, g(2) = 1/5
• g(x) = 1 / (x^2 + 1) continuity
Hp={x | x subset R} or can we write (positive infinity, 1 ]
Polynomial
There are two ways to get the value from polynomial function
1. Subtitute of number for the variable.
Example:
g(x) = (x^2 + x – 6) / (x – 3)
x=3, baaad! (not allowed)
This method used direct by including value x
g(x) discontinuity if x=3 because g(x) = 0/0
2. Factor of the polynomial function
Example:
g(x) = (x^2 + x – 6) / (x – 3)
We can formulate (x^2 + x – 6) to (x-3)(x+2), so
g(x) = (x^2 + x – 6) / (x – 3)
= (x - 3).(x + 2) / (x – 3)
= (x + 2)
= x + 2
x=3, no problem!!
Domain g(x) = x subset riil number
Inverse Functions part I
F(x,y)=0
Function y = f(x) : Vertical Line
1.1. Function x = g(y) : Horizontal Line
For example:
1. y= x^2
domain : (y) subset riil number
range : (y) always on positive value
x=g (y) : Horizontal Line : Invertible
2. y = 2x - 1
If we subs zero to x, so y = -1
If we subs zero to y, so x = 1/2
Function y = 2x – 1 is a line which (0, -1) and (1/2, 0) is intersecting graph on axis
y = x other line on the picture
- substitute y = x to y = 2x-1
x =2x-1 making a move internode (from right to left)
x-2x = -1
- x = -1 multiple with -1
x = 1 …………………………….( i )
Substitute ( i / x = 1) to y = x (or y = 2x – 1)
y = 1
intersection two line y = x and y = 2x – 1 on (1,1)
From y = 2x -1, we will change to become function of x, so
y = 2x -1 making a move internode (from right to left)
y – 1 = 2x multiple with ½
½ y + ½ = x
½(y + 1) = x
Because y = x and x = ½(x +1)
y = ½(x +1)
We find new function!
from all that we calculate above, we get:
f( x) = 2x – 1
g(x) = = ½(x +1)
f( g(x) ) we substitute g(x) to variable on the function f.
g( f(x) ) we substitute f(x) to variable on the function g.
f ( g(x) ) = 2 ( g(x) ) - 1
f ( g(x) ) = 2( ½(x+1) - 1
f ( g(x) ) = x + 1 – 1
f ( g(x) ) = x
g ( f(x) ) = ½( (2x – 1) +1)
g ( f(x) ) = ½(2x )
g ( f(x) ) = x
g = inverse f
f ( g(x) ) = f (f inverse (x) ) = x
g( f(x) ) = f (f inverse (x) ) = x
3. y = (x - 1) / (x + 2)
Method 2 find y inverse
y = (x - 1) / (x + 2) multiple with x+2
y ( x+2 ) = x - 1
xy + 2y = x – 1
xy – x = -1 -2y
x ( y – 1 ) = -1 -2y
x = (-1 – 2y) / (y - 1)
So y inverse = (-1 – 2y) / (y - 1)
When x = 0 y = -1
When y = 0 -1 – 2x = 0
-2x = 1
x = ½
Function y = f(x) : Vertical Line
1.1. Function x = g(y) : Horizontal Line
For example:
1. y= x^2
domain : (y) subset riil number
range : (y) always on positive value
x=g (y) : Horizontal Line : Invertible
2. y = 2x - 1
If we subs zero to x, so y = -1
If we subs zero to y, so x = 1/2
Function y = 2x – 1 is a line which (0, -1) and (1/2, 0) is intersecting graph on axis
y = x other line on the picture
- substitute y = x to y = 2x-1
x =2x-1 making a move internode (from right to left)
x-2x = -1
- x = -1 multiple with -1
x = 1 …………………………….( i )
Substitute ( i / x = 1) to y = x (or y = 2x – 1)
y = 1
intersection two line y = x and y = 2x – 1 on (1,1)
From y = 2x -1, we will change to become function of x, so
y = 2x -1 making a move internode (from right to left)
y – 1 = 2x multiple with ½
½ y + ½ = x
½(y + 1) = x
Because y = x and x = ½(x +1)
y = ½(x +1)
We find new function!
from all that we calculate above, we get:
f( x) = 2x – 1
g(x) = = ½(x +1)
f( g(x) ) we substitute g(x) to variable on the function f.
g( f(x) ) we substitute f(x) to variable on the function g.
f ( g(x) ) = 2 ( g(x) ) - 1
f ( g(x) ) = 2( ½(x+1) - 1
f ( g(x) ) = x + 1 – 1
f ( g(x) ) = x
g ( f(x) ) = ½( (2x – 1) +1)
g ( f(x) ) = ½(2x )
g ( f(x) ) = x
g = inverse f
f ( g(x) ) = f (f inverse (x) ) = x
g( f(x) ) = f (f inverse (x) ) = x
3. y = (x - 1) / (x + 2)
Method 2 find y inverse
y = (x - 1) / (x + 2) multiple with x+2
y ( x+2 ) = x - 1
xy + 2y = x – 1
xy – x = -1 -2y
x ( y – 1 ) = -1 -2y
x = (-1 – 2y) / (y - 1)
So y inverse = (-1 – 2y) / (y - 1)
When x = 0 y = -1
When y = 0 -1 – 2x = 0
-2x = 1
x = ½
Langganan:
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