Name : EVRI KURNIAWATI

NIM : 07305144026

Prodi : Math NR 07

E – mail : ev_nye@yahoo.com

Day, date :Monday, Des 01, 2008

English task on Tuesday, November 25, 2008

Class E room 204

VIDEO 1

PRE CALCULUS

Graph of a rational function

Can have discontinuities has a polynomial in the denominator

Example:

F(x)=x-3/x-2. Is called Off Limit if x=1 subtituted in equations F(x)=x-3/x-2 so denominator is equal to zero.

If equations above substituted to x=0 so get:

F(x)=x-3/x-2

F(x)=0-3/0-2

=3/2

the function above is break function if insert 1.

not all rational function will give 0 in denominator and can be 0

Break 2 ways in rational function:

1) Missing point is a loophole

Example:

Y=x^2-2x+1/x-1 if x=1 subtituted in equations beside so denominator is equal to zero

Y=1^2(x)+1/1-1

=0/0 That’s not allowed, missing removable singularity when x leads to 0/0

Y=x^2-2x+1/x-1

=(x-1)(x-1)/(x-1) (using the way to factor)

=x-1

Subtituted x=1

Y=x-1

=1-1

=0

2) Denominator of rasional functions equal to zero

Example : f(x)=x-3/x-2

Video 2

LIMIT by INSPECTION

1. x goes to positive or negative infinity

2. Limit involved a polynomial devided by a polynomial

Example 1:

Lim x^3+7/x^2-x-6

x--~

(polynominer by polynomial with limit x approach infinity )

To solving limit above :

a. looking the power of x in the numerator and denominator

b. must be dividing f by polynomial if power of x in numerator highest than in denominator, limit can be positive or negative infinity

example above power of x in numerator highest than in denominator.

Example 2:

Lim x+3/x^2+1

x--~

Its mean power of x in denominator highest than in numerator, that limit will equal to 0

Example 3 :

Lim x^2-x/2x^2+1

x--~

=Lim ((x^2-x)*1/x^2)/((2x^2)*1/x^2)

x--~

=Lim 1-(x/x^2)/2+(1/x^2)

x--~

=1/2

We get solutions Lim x^2-x/2x^2+1=1/2

x--~

Video 3

PROBLEM SOLVING ABOUT GRAPH MATH

1. The graph y=g(x), if the function is difined by h(x)=g(2x)+2.What is the value of h(1/2)?

Solution:

The fungtions is h(x)=g(2x)+2. If h(1/2) is mean when x =1/2, we substituted this to the functions h(x)=g(2x)+2

H(1/2)=g(2*1/2)+2

= g (1) + 2

g(1) is mean g when x=-2,if we look the graph of y=g(x) we get g(1)=-2

So

h(1/2)+1=g(1)+2

= -2 + 2

= 0

We get the value of h(1/2) is 0.

2. Let the function f be defined by f(x)=x+4. If f(p)= 16,What is the value of f(2p)?

Solution:

The function is f(x)=x+4

The value of f(2p) is mean f when x=2p,we have 2f(p)=16 because each space between have factors 2

So we can over by 2 and get f(p)=8

its mean f when x=p equal to 8 we can substituted this value to the function f(x)=x+4

we getf(p)=2p+4=8

p = 2

So we can get f(2p) If p=2 so 2p=4 is mean f when x=4 we subtitued this value to the function

f(x)=x+4

f(4)=4+4

= 8

We get the value of f(2p) is 8

3. In the xy coordinate x=y^2-25 intersects line at (p,-5) and (7,t).What is the greatest possible value of the slope of l?

The function x=y^2-25 intersects line at (p,-5) and(7,t). .

The formula of slope is m=y2-y1/x2-x1 the coordinate of l are (p,-5) and (7,t)..

thats mean x1=p,y1=5,x2=7,y2=t, we subtitued thus to the m

m=y2-y1/x2-x1

= t-(-5)/7-p

= t+5/7-p

We get the greatest possible value of the slope l is t+5/7-p

Video 4

INVERS FUNCTION

Notation by : F(x,y) = 0

Function y = f (x) is called VLT and x = g (y) is called HLT

• Function x = g(y) : invertible

Example:

1. Line function y=3x-6and y=x

y=3x-6

y=3x-6

-2x=-6

x=-6/-2

x=3

Subtuted x=3 in y=3x-6

y = 6 - 6

= 0

So, line function y=3x-6 and y = x intersect in (3,0).

y=3x-6

y+6=3x

x= y+6/3

x=1/3+2 and y=1/3+2

in invers function the line be write:

f(x) = 3x-6

g(x) = 1/3x+2

f(g(x)) = 3(1/3x+2)-6

= x+6-6

= x

g(f(x)) = 1/3(3x-6)+2

=x-2+2

= x

From the operation above will get the conclusion :

g = f ^-1

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

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