Task 2
Video : Do you believe
Presenting by : “Dalton Sherman”,
I believe in me
Do you believe in me?
Do you believe that I can stand up here fearless and talk over 21 thousand of you?
No matter where we come from, what is sunny salt Dallas, hat is placing brown, what is … or Nort Dallas or West Dallas or whenever, you aren’t better nice give up o us?
No, you are better nice because as you know in some cases you ae all we have got.
Everyone need you, We need you now, wanted ever!
Believe in yourself, finally do you believe that you can do anything, be anything, make anything.
Kamis, 15 Januari 2009
LESSON WITH A FRIEND
MIXTURE PROBLEM
We now consider rate problem involving mixtures. A substances S is allowed fo flow into a certain mixture in a container at a certain rate, and the mixture is kept uniform by stirring. Further. In one such situation, this uniform mixture simultaneously flows out of the container at another (generally different) rate; in another situation this may not be the case. In either case we seek to determine the quantity of the substance S present in the mixture at time t.
Let :
x denote the amount of S present at time t,
dx/dt denotes the rate of changer of x at time t.
INPUT denotes rate S which enter the mixture.
OUTPUT denotes rate S which leaves the mixture.
Then we have at one the basic equation
dx/dt =INPUT-OUTPUT
We have a problem
A tank initially contains 50 gal of pure water. Starting at time t=0, a brine containing 2 lb of dissolved salt per gallon flow into the tank at the rate of 3 gal/min. The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the tank at the same rate. Find the function of salt in the tank!
First we must make the mathematical formula. We apply the basic equation
dx/dt=INPUT-OUTPUT
The brine flows in at the rate of 3 gal/min and each gallon contain 2 lb of salt.
We formulating the INPUT as the weight of salt which flows in tank at time t and OUTPUT as the weight of salt which leaves tank at time t
In Physics, we have formula that
Concentration = weight / volume
Debit =volume/time
Find weight of salt at time t (weight/ time)
Weight/time = (concentration) (debit) = ( weight/ volume ) (volume/time)= weight/time
We apply at the problem to INPUT and OUTPUT
INPUT = (concentration which flows in) (debit which flows in)
= (2 lb/gal) (3 gal/min) = 6 lb/min
To find the rate of salt that leaves the tank at time t, we must find concentration of salt. The tank contains 50 gallon of the mixture at any time t. this 50 gallon contain x lb salt at time t, so the concentration is (x/50) lb/gal and the debit of mixture which leaves tank is 3 gal/min.
OUTPUT = (concentration which leaves tank) (debit which leaves tank)
= (x/50)lb/gal) (3 gal/min) = (3x/50)lb/min
We apply it in basic equation
dx/dt = 6 - (3x/50)
and we have initial condition at t=0 , x=0
x(0)=0
Then we determine this problem.
The equation is both linear and separable.
dx/dt = 6 - (3x/50)
We equaling the denominator, then we have
dx/dt = (300-3x)/50)
We group x with dx and t with dt. We change dt with (100-x) and (100-x) with dt.
dx/(100-x)=3 dt /50
We change this form into
dx/(x-100)= - 3dt/50 with multiply by -1
Then integrating both of side, then we get
ln (x-100) = -3t/50+c1
We use exponential to get the function of x
exp(ln (x-100)) = exp(-3t/50+c1)
Because of exp(ln x) = x, so
x-100=exp(-3t/50+c1)
Then move -100 to other side
x=100+exp(-3t/50+c1)
exp(c1)=c so
x=100+c.exp(-3t/50) ………(1)
We have intial value that x(0)=0 , at t=0 x =0,
then we substitute t=0 and x=0 into (1)
0= 100 + c exp(-3(0)/50)>>> exp(0)=1
0= 100 + c
c= -100
Then c = -100, substitute c into (1), we have
x=100-100.exp(-3t/50), we simply it become
x=100(1-exp(-3t/50)
x=100(1-exp(-3t/50) , is function of how much salt in the tank at t time.
The trouble at I explaining this problem to my friend Bibid are :
1.Sometimes, I forgot some word that I must say to Bibid, like exponent, power, etc.
2.I forgot how I explaining integral.
3.Sometimes I must read the source book to make Bibid say “I see”
We now consider rate problem involving mixtures. A substances S is allowed fo flow into a certain mixture in a container at a certain rate, and the mixture is kept uniform by stirring. Further. In one such situation, this uniform mixture simultaneously flows out of the container at another (generally different) rate; in another situation this may not be the case. In either case we seek to determine the quantity of the substance S present in the mixture at time t.
Let :
x denote the amount of S present at time t,
dx/dt denotes the rate of changer of x at time t.
INPUT denotes rate S which enter the mixture.
OUTPUT denotes rate S which leaves the mixture.
Then we have at one the basic equation
dx/dt =INPUT-OUTPUT
We have a problem
A tank initially contains 50 gal of pure water. Starting at time t=0, a brine containing 2 lb of dissolved salt per gallon flow into the tank at the rate of 3 gal/min. The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the tank at the same rate. Find the function of salt in the tank!
First we must make the mathematical formula. We apply the basic equation
dx/dt=INPUT-OUTPUT
The brine flows in at the rate of 3 gal/min and each gallon contain 2 lb of salt.
We formulating the INPUT as the weight of salt which flows in tank at time t and OUTPUT as the weight of salt which leaves tank at time t
In Physics, we have formula that
Concentration = weight / volume
Debit =volume/time
Find weight of salt at time t (weight/ time)
Weight/time = (concentration) (debit) = ( weight/ volume ) (volume/time)= weight/time
We apply at the problem to INPUT and OUTPUT
INPUT = (concentration which flows in) (debit which flows in)
= (2 lb/gal) (3 gal/min) = 6 lb/min
To find the rate of salt that leaves the tank at time t, we must find concentration of salt. The tank contains 50 gallon of the mixture at any time t. this 50 gallon contain x lb salt at time t, so the concentration is (x/50) lb/gal and the debit of mixture which leaves tank is 3 gal/min.
OUTPUT = (concentration which leaves tank) (debit which leaves tank)
= (x/50)lb/gal) (3 gal/min) = (3x/50)lb/min
We apply it in basic equation
dx/dt = 6 - (3x/50)
and we have initial condition at t=0 , x=0
x(0)=0
Then we determine this problem.
The equation is both linear and separable.
dx/dt = 6 - (3x/50)
We equaling the denominator, then we have
dx/dt = (300-3x)/50)
We group x with dx and t with dt. We change dt with (100-x) and (100-x) with dt.
dx/(100-x)=3 dt /50
We change this form into
dx/(x-100)= - 3dt/50 with multiply by -1
Then integrating both of side, then we get
ln (x-100) = -3t/50+c1
We use exponential to get the function of x
exp(ln (x-100)) = exp(-3t/50+c1)
Because of exp(ln x) = x, so
x-100=exp(-3t/50+c1)
Then move -100 to other side
x=100+exp(-3t/50+c1)
exp(c1)=c so
x=100+c.exp(-3t/50) ………(1)
We have intial value that x(0)=0 , at t=0 x =0,
then we substitute t=0 and x=0 into (1)
0= 100 + c exp(-3(0)/50)>>> exp(0)=1
0= 100 + c
c= -100
Then c = -100, substitute c into (1), we have
x=100-100.exp(-3t/50), we simply it become
x=100(1-exp(-3t/50)
x=100(1-exp(-3t/50) , is function of how much salt in the tank at t time.
The trouble at I explaining this problem to my friend Bibid are :
1.Sometimes, I forgot some word that I must say to Bibid, like exponent, power, etc.
2.I forgot how I explaining integral.
3.Sometimes I must read the source book to make Bibid say “I see”
Representing The Video of Learning Mathematics
Video 4
INVERSE FUNCTION
INVERSE FUNTION
Before we start study the Inverse Function, better we review the definition of a function.
We have relation of F(x,y), this represent to function y=f(x) if we can solve for y uniquely in term of x as 1-1 function if also we can solve x uniquely in term of y.
If we see y=x2, it is a function but not 1-1 function, because if we choose a value of y, we have two value of x. 1-1 function only mapping a x into a y.
So lets start at y = 2x-1
Let’s look at the graph function
y intercepts -1 , x intersects ½, the slope =m= 2
We have a point (1/2,0) and (0, -1) . To draw this line, find (1/2,0) and (0, -1) first, then draw a line that across the points.
Then attend it
y = x intersects y = 2x-1
x = 2x-1 (add 1 in both of side)
1+x = 2x (move the x into right side)
1 = x
Substitute x = 1 in the equation y = x and y = 2x-1, we get y =1 (x =1 satisfying the both equations because x is intersect of y= x and y = 2x-1
Of course, the point of intersection of y= x and y = 2x-1 is (1,1)
We want to solve for x in term of y
2x-1 = y
2x = y+1
x = ½ (y+1)
x = ½ y + ½
Then in this equation, I want to give exchange rule of x and y, so I’m going to do now is rewrite x into y, and y into x,
We get y = ½ x + ½ ……………..(**)
The x intercepts in -1, and y intercepts in ½
We have point (0,1/2) and (-1,0)
So the inverse function to given function passes through that point (1,1)
We have f(x) = 2x-1
We have g(x) = ½ x + ½
Then take the value of g and plug in for f
We want to compute f(g(x))
f(g(x)) = 2 […] – 1 , then fill the bracket with ½ x + ½
= 2 [½ x + ½] – 1= x + 1 -1= x
in the other hand, we take g(f(x))
g(f(x)) = ½ […..] + ½ = ½ [2x-1] +½ = x-½ +½ = x
Then we write
g = f -1
f(g(x)) = f(f -1 (x)) = x and
g(f(x)) = f -1 (f(x)) = x
Other example. Let’s take the function y =(x-1)/(x+2). This function has a vertical asymptote as line x = -2 and also has horizontal asymptote as a line y = 1
x intercept is approximately 1and y intercept is approximately -½
The function hyperbole
Then let’s take the equation y =(x-1)/(x+2) and solve for x in term of y
y(x+2) = (x-1)
yx-2y = x-1 (take 2y into the right side, move x to the left side)
yx-x =-1-2y
(y-1)x =-1-2y
X = (-1-2y)/(y-1)
we are going to exchange the rule of y and x
y= (-1-2x)/(x-1)
at x = 0 y = -1
at y = 0 -1-2x = 0
-1 = 2x or x =-½
y= (-1-2x)/(x-1) has vertical asymptote as a line x = 1 and also horizontal asymptote as a line y = -2
We have
f(x) = (x-1)/(x+2)
f -1(x) = (-1-2x)/(x-1)
Then
f(f -1 (x)) = (((-1-2x)/(x-1))-1)/(( (-1-2x)/(x-1))+2)
= ((((-1-2x)-(x-1))/(x-1)))/((((-1-2x)+2(x-1)/(x-1)))
= (-1-2x-x+1)/(-1-2x+2x-2)
= -3x/-3
= x
Then find f -1(f(x))
f -1(f(x)) =((-1-2((x-1)/(x+2)))/((x-1)/(x+2) – 1))
= (-1(x+2)-2((x-1)/(x+2))/((x-1)- 1(x+2)/(x+2))
= ((-1x-2-2x+2)/(x+2))/((x-1- x-2)/(x+2)))
= (-1x-2-2x+2)/(x-1- x-2)
= -3x/-3
= x
We find that
f(g(x)) = x
g(f(x)) = x
Range of g is a domain of f, and the range of f is a domain of g
The two function are inverses function.
we just write g = f -1
INVERSE FUNCTION
INVERSE FUNTION
Before we start study the Inverse Function, better we review the definition of a function.
We have relation of F(x,y), this represent to function y=f(x) if we can solve for y uniquely in term of x as 1-1 function if also we can solve x uniquely in term of y.
If we see y=x2, it is a function but not 1-1 function, because if we choose a value of y, we have two value of x. 1-1 function only mapping a x into a y.
So lets start at y = 2x-1
Let’s look at the graph function
y intercepts -1 , x intersects ½, the slope =m= 2
We have a point (1/2,0) and (0, -1) . To draw this line, find (1/2,0) and (0, -1) first, then draw a line that across the points.
Then attend it
y = x intersects y = 2x-1
x = 2x-1 (add 1 in both of side)
1+x = 2x (move the x into right side)
1 = x
Substitute x = 1 in the equation y = x and y = 2x-1, we get y =1 (x =1 satisfying the both equations because x is intersect of y= x and y = 2x-1
Of course, the point of intersection of y= x and y = 2x-1 is (1,1)
We want to solve for x in term of y
2x-1 = y
2x = y+1
x = ½ (y+1)
x = ½ y + ½
Then in this equation, I want to give exchange rule of x and y, so I’m going to do now is rewrite x into y, and y into x,
We get y = ½ x + ½ ……………..(**)
The x intercepts in -1, and y intercepts in ½
We have point (0,1/2) and (-1,0)
So the inverse function to given function passes through that point (1,1)
We have f(x) = 2x-1
We have g(x) = ½ x + ½
Then take the value of g and plug in for f
We want to compute f(g(x))
f(g(x)) = 2 […] – 1 , then fill the bracket with ½ x + ½
= 2 [½ x + ½] – 1= x + 1 -1= x
in the other hand, we take g(f(x))
g(f(x)) = ½ […..] + ½ = ½ [2x-1] +½ = x-½ +½ = x
Then we write
g = f -1
f(g(x)) = f(f -1 (x)) = x and
g(f(x)) = f -1 (f(x)) = x
Other example. Let’s take the function y =(x-1)/(x+2). This function has a vertical asymptote as line x = -2 and also has horizontal asymptote as a line y = 1
x intercept is approximately 1and y intercept is approximately -½
The function hyperbole
Then let’s take the equation y =(x-1)/(x+2) and solve for x in term of y
y(x+2) = (x-1)
yx-2y = x-1 (take 2y into the right side, move x to the left side)
yx-x =-1-2y
(y-1)x =-1-2y
X = (-1-2y)/(y-1)
we are going to exchange the rule of y and x
y= (-1-2x)/(x-1)
at x = 0 y = -1
at y = 0 -1-2x = 0
-1 = 2x or x =-½
y= (-1-2x)/(x-1) has vertical asymptote as a line x = 1 and also horizontal asymptote as a line y = -2
We have
f(x) = (x-1)/(x+2)
f -1(x) = (-1-2x)/(x-1)
Then
f(f -1 (x)) = (((-1-2x)/(x-1))-1)/(( (-1-2x)/(x-1))+2)
= ((((-1-2x)-(x-1))/(x-1)))/((((-1-2x)+2(x-1)/(x-1)))
= (-1-2x-x+1)/(-1-2x+2x-2)
= -3x/-3
= x
Then find f -1(f(x))
f -1(f(x)) =((-1-2((x-1)/(x+2)))/((x-1)/(x+2) – 1))
= (-1(x+2)-2((x-1)/(x+2))/((x-1)- 1(x+2)/(x+2))
= ((-1x-2-2x+2)/(x+2))/((x-1- x-2)/(x+2)))
= (-1x-2-2x+2)/(x-1- x-2)
= -3x/-3
= x
We find that
f(g(x)) = x
g(f(x)) = x
Range of g is a domain of f, and the range of f is a domain of g
The two function are inverses function.
we just write g = f -1
Minggu, 21 Desember 2008
Representing the video of learning mathematics.
Representing the video of learning mathematics
VIDEO I
A. Solving the Problem
We have problem :
1. Find the value of h(1), if the function h is defined by h(x)=g(2x)+2 and
y=g(x), and if x≤0 function is y=x-3, if 0≤x≤3 function is y=-x+3 and if x≥3 the function is y=x+3.
Solution :
We have information that h(x)=g(x)+2. We looking for h(1).
Substitute x=1 to the h(x)=g(x)+2. We get h(1)=g(2)+2 (if x=1 then 2x is equal 2).
From the function of y=g(x), x=2 in the interval 0≤x≤3, substitute x=2 to y =-x+3, and we get y=g(2)=1.
Back to function h(1)=g(2)+2, because g(2) equals 1, h(1)=1+2=3.
The answer is h(1)=3.
2. Let the function f be defined by f(x)=x+1. If 2f(p)=20, what is the value of f(3p)?
Solution :
We looking for f(3p) , what is f when x=3p?
We have f(x)=x+1 and 2f(p)=20.
Divide all side of 2f(p)=20 by 2, we get f(p)=10, because f(p)=p+1, then
p+1=10 >>> p=10-1=9. We get p=9.
x =3p, We get x=3.9=27.
We can find the value of f(3p) by substituting f(x)=f(27)=27+1=28.
So the value of (3p) is 28.
VIDEO II
GRAPH OF RATIONAL FUNCTION
Graph of rational function if can have discontinuities and has a polynomial in the denominator (can be divided by invaluable quantities).
We should find very dangerous situation in pre calculus that deadly convidence.
It possible to some value of x will meet division by zero. Then the value of function is off limits.
Example :
We have f(x)=(x+2)/(x-1). When x=1, the value function become f(1)=(1+2)/(1-1) which is f(1)=3/0, with zero in denominator. Is not good. Choosing x=1 in this function is bad idea. This is break in function graph.
Let x=0, we have f(0)=(0+2)/(0-1)=2/(-1)=-2.
Let x=1, we have f(1)=(1+2)/(1-1) = 3/0. It’s break in function graph because we can’t compute 3/0. The function f(x)=(x+2)/(x-1) don’t have value for x=1. And we can sketch x=1 as asymtot. The graph is break and discontinuity. The conclusion is rational functions don’t always work this way.
Other example, we have f(x)=1/(x^2+1)
Let x=1, then f(1)=1/(1+1)=1
Let x=0, then f(0)=1/(0+1)=1
The function is not break and contoinuity, because x^2 will not able to -1 that can make denominator 0.
Don’t forget, rational functions, denominator can be zero.
For the polynomial, the functions are smooth and unbroken curve.
For the rational function, some value of x can make zero in the denominator. It impossible to find the value and break in the graph.
Break can show up in :
Missing point on the graph
Example :
We have y=(x^2-x-6)/(x-3). This function break in graph at x=3, because x=3 make the numerator and denominator be zero.
We can conclude that x=3 is the missing point.
When you see 0/0, it possible to factor top and bottom of the rational function and simplyfy. See y=(x^2-x-6)/(x-3). We can factoring (x^2-x-6) to (+
y=(x^2-x-6)/(x-3)=(x-3)(x+2)/(x-3), we can simplify, then y=(x+3), it possible to choice y=3. It’s not break in the graph because x=3, y=5. Hahahaa
We can conclude that to avoid 0/0, factoring and simplyfing make the function not break in the graph.
VIDEO III
ALGEBRAIC LONG DIVISION
Example:
Is x-3 a factor of x^3-7x-6?
We can use algebraic long division. When x^3-7x-6 divide by x-3. We can change x^3-7x-6 to x^3+0x^2-7x-6.
Like other division, we must find a multiplier that multiply x into x^3, the multiplier is x^2. Multiply x-3 with x^3. We get x^3-x^2. Then reduce x^3-+0x^2-7x-6 by x^3-x^2, we have 3x^2-7x-6.
We must find a multiplier that multiply x into 3x^2, the multiplier is 3x. Multiply x-3 with 3x^2-9x. Reduce 3x^2-9x+6 by 3x^2-9x, then we have 2x-6. a multiplier that multiply x into 2x the multiplier is 2. Multiply x-3 with 2. We get 2x-6. Recude 2x-6 by 2x-6, we find 0 as final result.
If the reside is not zero, the divider is not factor of the function.
So x-3 is the factor of x^3-7x-6
adding of multiplier which are x^2+3x+2, this also factor of x^3-7x-6. x^2+3x+2 can factoring to (x+2)(x+1).
So the factor of x^3-7x-6 are (x-3)(x+1)(x+2).
And the roots of x^3-7x-6=0 are:
x-3=0 >>> x=3
x+2=0 >>> x=-2
x+1=0 >>> x=-1
Based on above, we can get 3 roots for 3rd degree equation (as the highest power of x^3-7x-6). We remember that quadratic (2nd degree always have at most 2 roots).
So we can conclude that n-th degree equation always have at most n roots.
Long division for a 3rd order polynomial :
1.Fing a partial quotient of x^2, by dividing into x^3 to get x^2.
2.Multiply x^2 by the divisor & substract the product from the dividend.
3.Repeat the process until you either “clear it out: or reach a remainder.
VIDEO I
A. Solving the Problem
We have problem :
1. Find the value of h(1), if the function h is defined by h(x)=g(2x)+2 and
y=g(x), and if x≤0 function is y=x-3, if 0≤x≤3 function is y=-x+3 and if x≥3 the function is y=x+3.
Solution :
We have information that h(x)=g(x)+2. We looking for h(1).
Substitute x=1 to the h(x)=g(x)+2. We get h(1)=g(2)+2 (if x=1 then 2x is equal 2).
From the function of y=g(x), x=2 in the interval 0≤x≤3, substitute x=2 to y =-x+3, and we get y=g(2)=1.
Back to function h(1)=g(2)+2, because g(2) equals 1, h(1)=1+2=3.
The answer is h(1)=3.
2. Let the function f be defined by f(x)=x+1. If 2f(p)=20, what is the value of f(3p)?
Solution :
We looking for f(3p) , what is f when x=3p?
We have f(x)=x+1 and 2f(p)=20.
Divide all side of 2f(p)=20 by 2, we get f(p)=10, because f(p)=p+1, then
p+1=10 >>> p=10-1=9. We get p=9.
x =3p, We get x=3.9=27.
We can find the value of f(3p) by substituting f(x)=f(27)=27+1=28.
So the value of (3p) is 28.
VIDEO II
GRAPH OF RATIONAL FUNCTION
Graph of rational function if can have discontinuities and has a polynomial in the denominator (can be divided by invaluable quantities).
We should find very dangerous situation in pre calculus that deadly convidence.
It possible to some value of x will meet division by zero. Then the value of function is off limits.
Example :
We have f(x)=(x+2)/(x-1). When x=1, the value function become f(1)=(1+2)/(1-1) which is f(1)=3/0, with zero in denominator. Is not good. Choosing x=1 in this function is bad idea. This is break in function graph.
Let x=0, we have f(0)=(0+2)/(0-1)=2/(-1)=-2.
Let x=1, we have f(1)=(1+2)/(1-1) = 3/0. It’s break in function graph because we can’t compute 3/0. The function f(x)=(x+2)/(x-1) don’t have value for x=1. And we can sketch x=1 as asymtot. The graph is break and discontinuity. The conclusion is rational functions don’t always work this way.
Other example, we have f(x)=1/(x^2+1)
Let x=1, then f(1)=1/(1+1)=1
Let x=0, then f(0)=1/(0+1)=1
The function is not break and contoinuity, because x^2 will not able to -1 that can make denominator 0.
Don’t forget, rational functions, denominator can be zero.
For the polynomial, the functions are smooth and unbroken curve.
For the rational function, some value of x can make zero in the denominator. It impossible to find the value and break in the graph.
Break can show up in :
Missing point on the graph
Example :
We have y=(x^2-x-6)/(x-3). This function break in graph at x=3, because x=3 make the numerator and denominator be zero.
We can conclude that x=3 is the missing point.
When you see 0/0, it possible to factor top and bottom of the rational function and simplyfy. See y=(x^2-x-6)/(x-3). We can factoring (x^2-x-6) to (+
y=(x^2-x-6)/(x-3)=(x-3)(x+2)/(x-3), we can simplify, then y=(x+3), it possible to choice y=3. It’s not break in the graph because x=3, y=5. Hahahaa
We can conclude that to avoid 0/0, factoring and simplyfing make the function not break in the graph.
VIDEO III
ALGEBRAIC LONG DIVISION
Example:
Is x-3 a factor of x^3-7x-6?
We can use algebraic long division. When x^3-7x-6 divide by x-3. We can change x^3-7x-6 to x^3+0x^2-7x-6.
Like other division, we must find a multiplier that multiply x into x^3, the multiplier is x^2. Multiply x-3 with x^3. We get x^3-x^2. Then reduce x^3-+0x^2-7x-6 by x^3-x^2, we have 3x^2-7x-6.
We must find a multiplier that multiply x into 3x^2, the multiplier is 3x. Multiply x-3 with 3x^2-9x. Reduce 3x^2-9x+6 by 3x^2-9x, then we have 2x-6. a multiplier that multiply x into 2x the multiplier is 2. Multiply x-3 with 2. We get 2x-6. Recude 2x-6 by 2x-6, we find 0 as final result.
If the reside is not zero, the divider is not factor of the function.
So x-3 is the factor of x^3-7x-6
adding of multiplier which are x^2+3x+2, this also factor of x^3-7x-6. x^2+3x+2 can factoring to (x+2)(x+1).
So the factor of x^3-7x-6 are (x-3)(x+1)(x+2).
And the roots of x^3-7x-6=0 are:
x-3=0 >>> x=3
x+2=0 >>> x=-2
x+1=0 >>> x=-1
Based on above, we can get 3 roots for 3rd degree equation (as the highest power of x^3-7x-6). We remember that quadratic (2nd degree always have at most 2 roots).
So we can conclude that n-th degree equation always have at most n roots.
Long division for a 3rd order polynomial :
1.Fing a partial quotient of x^2, by dividing into x^3 to get x^2.
2.Multiply x^2 by the divisor & substract the product from the dividend.
3.Repeat the process until you either “clear it out: or reach a remainder.
Senin, 08 Desember 2008
Assignment I
SUMMARY
1.Function and Exponent Function Graph
a.Exponent Function is functions that mapping each real number of x to real number ax, with a > 0 and a ≠ 1.
b.Exponent Function Graph
i.Definite if x elemen of R and the graph always on the top of x axis.
ii.Intersect with y axis at (0,1).
iii.Decline monotonous curve for 0 < a < 1.
iv.The horizontal asymptote is y = 0 or x axis
2.Exponent Equation
a.Exponent Equation is equations that obtain exponent with form of main number or exponent is function of variable.
b.Determined exponent equation; a > 0, a ≠ 1, b > 0, b ≠ 1
i.If af(x) = 1, then f(x) = 0
ii.If af(x) = ab, then f(x) = b
iii.If af(x) = ag(x), then g(x) = 0
iv.If af(x) = bf(x), then f(x) = 0
v.af(x) = bf(x) can be determined with logarithm help’s.
vi.If f(x)g(x) = f(x)h(x )there is 4 possible solution
1.g(x) = h(x)
2.f(x) = 1
3.f(x) = -1 if g(x) and h(x) is odd or both of them is even
4.f(x) = 0 if g(x) and h(x) is positive
vii. If f(x)h(x) = g(x)h(x) there is 2 possible solution
1.h(x) = 0, if f(x), g(x) ≠0
2.f(x)= g(x)
viii.A[af(x)]2 + B[af(x)] +C = 0 determine this in quadrate equation
3.Inequation Exponent
a.Inequation exponent is inequation which that exponent has form as variable obtain inequation sign ( >, < , ≥, ≤)
b.Determined inequation exponent
i.if af(x) < ag(x), then f(x) < g(x), for a>1
ii.if af(x) > ag(x), then f(x) > g(x), for a>1
iii.if af(x) < ag(x), then f(x) > g(x), for 0
1.Function and Exponent Function Graph
a.Exponent Function is functions that mapping each real number of x to real number ax, with a > 0 and a ≠ 1.
b.Exponent Function Graph
i.Definite if x elemen of R and the graph always on the top of x axis.
ii.Intersect with y axis at (0,1).
iii.Decline monotonous curve for 0 < a < 1.
iv.The horizontal asymptote is y = 0 or x axis
2.Exponent Equation
a.Exponent Equation is equations that obtain exponent with form of main number or exponent is function of variable.
b.Determined exponent equation; a > 0, a ≠ 1, b > 0, b ≠ 1
i.If af(x) = 1, then f(x) = 0
ii.If af(x) = ab, then f(x) = b
iii.If af(x) = ag(x), then g(x) = 0
iv.If af(x) = bf(x), then f(x) = 0
v.af(x) = bf(x) can be determined with logarithm help’s.
vi.If f(x)g(x) = f(x)h(x )there is 4 possible solution
1.g(x) = h(x)
2.f(x) = 1
3.f(x) = -1 if g(x) and h(x) is odd or both of them is even
4.f(x) = 0 if g(x) and h(x) is positive
vii. If f(x)h(x) = g(x)h(x) there is 2 possible solution
1.h(x) = 0, if f(x), g(x) ≠0
2.f(x)= g(x)
viii.A[af(x)]2 + B[af(x)] +C = 0 determine this in quadrate equation
3.Inequation Exponent
a.Inequation exponent is inequation which that exponent has form as variable obtain inequation sign ( >, < , ≥, ≤)
b.Determined inequation exponent
i.if af(x) < ag(x), then f(x) < g(x), for a>1
ii.if af(x) > ag(x), then f(x) > g(x), for a>1
iii.if af(x) < ag(x), then f(x) > g(x), for 0
Minggu, 23 November 2008
Zulistya AS
HOW TO EXPRESS MATHEMATICS
ZULISTYA ADI SAPUTRA
07305141034
Matematika Reguler 2007
a. compulsory
in indonesia mean : wajib
Definisi
compulsory
[kuhm-puhl-suh-ree]
adjective, noun
–adjective (kata sifat)
1. required (diwajibkan); mandatory (wajib); obligatory (wajib)
–noun
1. something that needed or obligated (sesuatu yang dibutuhkan atau diwajibkan)
Contoh kalimat :
I believe that music should be compulsory in Junior High School
b. perceive
in indonesia mean : merasa
Definisi
[per-seev]
–verb (used with object), -ceived, -ceiving.
1. to become aware of (dapat merasa) , know (tahu) , or identify by means of the senses: (identifikasi dari sebuah rasa)
2. to recognize (mengenal), discern( memahami), or understand (memahami):
Contoh kalimat:
I perceived an object looming through the mist.
c. Bangun sisi lengkung
Definisi :
Bangun yang mempunyai sisi yang lengkung
Contoh : tabung, kerucut, bola
English : Curve shapes
Explanation : a shape that contain curve side, like cylinder, cone and sphere
Contoh kalimat :
The examples of curve shape are cylinder, cone and sphere.
d. Bidang sebelah kanan
Definisi :
Bidang yang terletak sebelah kanan pada suatu bangun
English : right side
A sector in the right of a shapes.
Contoh kalimat :
The right side of cube ABCD.EFGH is ADEF.
ZULISTYA ADI SAPUTRA
07305141034
Matematika Reguler 2007
a. compulsory
in indonesia mean : wajib
Definisi
compulsory
[kuhm-puhl-suh-ree]
adjective, noun
–adjective (kata sifat)
1. required (diwajibkan); mandatory (wajib); obligatory (wajib)
–noun
1. something that needed or obligated (sesuatu yang dibutuhkan atau diwajibkan)
Contoh kalimat :
I believe that music should be compulsory in Junior High School
b. perceive
in indonesia mean : merasa
Definisi
[per-seev]
–verb (used with object), -ceived, -ceiving.
1. to become aware of (dapat merasa) , know (tahu) , or identify by means of the senses: (identifikasi dari sebuah rasa)
2. to recognize (mengenal), discern( memahami), or understand (memahami):
Contoh kalimat:
I perceived an object looming through the mist.
c. Bangun sisi lengkung
Definisi :
Bangun yang mempunyai sisi yang lengkung
Contoh : tabung, kerucut, bola
English : Curve shapes
Explanation : a shape that contain curve side, like cylinder, cone and sphere
Contoh kalimat :
The examples of curve shape are cylinder, cone and sphere.
d. Bidang sebelah kanan
Definisi :
Bidang yang terletak sebelah kanan pada suatu bangun
English : right side
A sector in the right of a shapes.
Contoh kalimat :
The right side of cube ABCD.EFGH is ADEF.
Langganan:
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