Interpretation of videos

INVERSE FUNCTION

Before we know about the definition of inverse function,we will suppose that a function F(x,y)=0 where function y=f(x) as the Vertical Line T (VLT) and also x=g(y) as the Horizontal Line T (HLT) are have the relation.

And we can solve the function by using function y=f(x) where x=g(x). If we looking for the function of y=x² and then x=g(y) is a horizontal line which intersects the graph in two points.

Let write y= 2x-1

and look at the graph of that function

So, in the x-intersect we find point of (½,0).

look at the line of y=x and we can substitute y=x into the function y = 2x - 1, so : x = 2x - 1

1 + x = 2x

1 = x

And , we have intersection point between line y = 2x – 1 and y = x, of course in point of (1,1).

Now, from these relation we want to solve the equation

2x – 1 = y

2x = y + 1

x = ½(y + 1)

x = ½y + ½

then exchange x into y,and y into x ,so obtain :

y = ½x + ½

We looking back on the graph then we get the other line. Let we graph a line containing of point (1,1) and (0,-1).

We have f(x) = 2x – 1 and g(x) = ½x + ½, so

f(g(x)) = 2(g(x)) – 1

= 2 (½x + ½) – 1

= x + 1 – 1

= x

On the other hand

g(f(x)) = ½(f(x)) + ½

= ½(2x – 1) + ½

= x - ½ + ½

= x

So the important problem of two functions is g = f^-1

f(g(x)) = f(f^-1(x))

= x

g(f(x)) = f^-1(f(x))

= x

The example :

Write the function of y = (x-1 )/(x+2)f

The x-intercept is gonna be equal to 1, and y-intercept is gonna be -1.

The solution is :

y = (x-1 )/(x+2)

y ( x + 2 ) = x - 1

yx + 2y = x – 1

yx – x = -1 – 2y

(y - 1) x = -1 – 2y

x^-1 = (-1-2y) / (y-1) ;

Then exchange x into y,and y into x;so we get:

y^-1 = (-1-2x) / (x-1)

Let; x = 0, y = -1

y = 0, x = -½

There are vertical asymtot x = 1 and horizontal asymtot y = -2

We can see from the figure of function y = (x-1 )/(x+1) and y^-1 = (-1-2x) / (x-1)that the two functions are reflected each other. So, the favorite function to look for the inverses are two function .

Video 1.” Solving the problem”

Question (1)

Figure above shows the graph of y=g(x).If the function h is define by h(x)=g(2x)+2, what is the value of h(1)??

Solution:

First,make the graph,then write the function h(x)=g(2x)+2. To get the value of of g(2),we can look the graph from absis=2,we make an straight line that intersect the function y=g(x),so from here we get g(2)=1.

S0,h(1)=1+2=3;

The solution is 3.

Question (2):

Let function f be define by f(x)=x+1, if 2f(p)=20, what is the value of f(3p)?

Solution:

Looking for f(3p) what is f when x=3p.

Write function f(x)=x+1,

Then write 2 f(p)=20; dividing by 2

f(p)=10

f(p)=f(x)

10=x+1

X=9

After get x=9,substitute to x=3p, x=3.9=27

Then f(3p)=x+1

=27+1=28.

So, value of f(3p) is 28.

Video 2. “Finding the roots of third equation by algebraic long division”.

Example:

Let x-3 a factor of x³-7x-6,

So we can write (x³-7x-6)/ (x-3)

We can find the other factor from that equation by using long division. We still using the method like in the elementary school.

Method:

find a partial quotient of x² by dividing x into x³ to get x²,then multiply x² by the divisor and sustract the product from the dividend. And repeat the process until you either “clear it out” or reach a remainder.

After do this method,we get no remainder. And the solution of (x³-7x-6)/ (x-3) is x²+3x +2.

Since (x³-7x-6)/ (x-3) is no remainder and x-3 is a factor of x³-7x-6, so, x²+3x +2 also a factor of x³-7x-6. Factors of x²+3x +2 are (x+2) and (x+1), and can be written:

x³-7x-6 =(x-3)(x+2)(x+1); setting the factor x³-7x-6=0 ; so

0=(x-3)(x+2)(x+1)

Thus x-3=0 ; x=3

X+2=0 ; x=-2

X+1=0 ; x=-1

So,roots are 3,-2,-1.

Roots of equation:

· 3 roots for this third degree equation.

· Quadratic equation always have at most 2 roots.

· A fourth degree equation

Video 3

“PRECALCULUS”.

The graph of rational function can have discontinuities,because has a polynomial in the denominator.

Example:

f(x)=(x+2) /(x-1) ; when x=1 we will get 3/0, so it will break in graph.

Let, insert x=0 ; f(0)=-2

Insert x=1; f(1)=???(it’s imposible)

But, the rational function don’t always work this way:

Let f(i)=1 / (i +1), this function never 0 because of the denominator is i+1.

But, rational function, the denominator can be zero and polynomial also have smooth / broken rational function.

For example:

Y= (x² -x-6) / (x-3) ; when x=3 so we’ll find 0/0????

So,to get the value of function y we can simplify the numerator.

My Experience..

Tell my friend about Limit Function

On 9th january 2009, I try to explain the material of mathematic to my friend. My friend is Alin. She is my classmate. I tell her about Limit. In senior high school ,we have got this material. So ,she know less about limit. First, I tell her the definition limit function. I tell,” Function f is defined at opened interval that having a, maybe at a have no value definition. Limit f(x) is L for x approach a,can be written:

Lim f(x) =L “.

x-->a

then I tell her about the postulates of limit, I just tell her a part of them. It’s just a introduction to learn about limit. So I don’t explain her of all the limit. In process I explain the limit, we always laughing together. And I try to be serious. I try to think that my friend is the students, and I am the teacher. I can explain easily, because she can catch my explain fast. After I tell her, I give her some exercises. I give her five exercises, and she can do three well, but not of two numbers. Then I give the explaining for her to get the correct answer. So, she can catch my explaining clearly now. I try not to be the real teacher for her, I just try to explain her, and we enjoy it. We learn together. Learn about limit.
From this moment, I can do the activity that can make my friend be clearly in limit. From she know about limit, now she more be professional (hehehe..just kidding..). And this is my experience. Try to be the great teacher (without angry…hehehe..).
This is my experience, maybe I’m not a perfect person..sorry I mean that I’m not the great teacher. And thanks for viewing my blog or thanks for reading.

The concept of limit:
Limit concept of limit is fundamental in understanding the differential calculus and integral calculus,that one of kinds of mathematics.
Definition
Function f is defined at opened interval that having a, maybe at a have no value definition. Limit f(x) is L for x approach a,can be written:
Lim f(x) =L
x-->a
If for every positive number ε, however so tidy will be got positive number δ, so
|f(x) – L| < ε is followed by 0 < |x – a| < δ

The Postulates
Postulate 1

If m and b are constanta, so

Lim (mx+b) = ma +b
x-->a

Postulate 2
If c is constanta, so for every arbitrary number a
Lim c =c
x-->a

Postulate 2
Lim x = a
x-->a

Postulate 4
If lim f(x) =L and lim g(x)= M, so
x-->a x-->a
lim [f(x) + g(x)] = L+M
x-->a

Postulate 5
If lim f(x) =L and lim g(x)= M, so
x-->a
lim [f(x) . g(x)] = L.M
x-->a

Postulate 6
If lim f(x) =L and n is arbitrary positive number, so
x-->a
lim [f(x)]^n = L^N

Postulate 7
If lim f(x) = L and lim g(x) = M, and M is not 0, so
x-->a x-->a
lim [f(x) / g(x)]= L/M
x-->a

The Right Limit and Left Limit

Definition
Function f is defined for every number at interval (a,c). So limit f(x) for x approach a from the right is R. can be write:

Lim f(x) = R
x-->a+

if for every positive number ε, however so tidy will be got positive number δ . so ;
|f(x) – R| < ε is followed by 0 < |x – a| < δ

Definition
Function f is defined for every number at interval (d,a). So limit f(x) for x approach a from the left is L. can be write:

Lim f(x) = L
x-->a-

if for every positive number ε, however so tidy will be got positive number δ . so ;
|f(x) – L| < ε is followed by 0 < |a-x| < δ

By the definition of left and right limit, so, all of postulates that have learn before is used for right limit and left limit.

Source: “Kalkulus I.by Drs.Soemoenar”