Task English II
Tuesday, November 25, 2008
(My Created from listening video mathematics in English language)

Video 1
PRE CALCULUS
A. Graph of a rational function
Can have discontinuities has a polynomial in the denominator
Example:
f(x) = (x + 2)/(x - 1)
Is called Off Limit, when x = 1 because if x =1 substituted to function f the denominator can be 0.
f(1) = (1 + 2) /(1 - 1)
f(1) = 3 / 0
The function above is break function if insert 1.
If the function insert 0
f(0) = (0 + 2)/(0 – 1)
f(0) = 2 /(-1)
f(0) = -2
and not all rational function will give 0 in denominator and can be 0.

B. Break 2 ways in rational function:
1. Missing point is a loophole
Example:
y = (x^2 – x – 6)/(x – 3)
if x =1 substituted to function f not allowed because
y = (3^2 – 3 - 6)/(3 - 3)
= 0
Good choice from the function above
y = (x - 3)(x + 2)/(x - 3) (cancelled the factor)
= x + 2
2. Zero in denominator
Example: in function f(x) = (x + 2)/(x - 1) and insert 1.

Video 2
LIMIT by INSPECTION
1. x goes to positive or negative infinity
2. Limit involved a polynomial divided by a polynomial

Example 1:
lim (x^3 + 4) / (x^2 + x + 1)
x--~
(Polynomial over by polynomial with limit x approach infinity)
To solving limit above attention this:
- looking the power of x in the numerator and denominator
- must be dividing f polynomial if power of numerator highest than denominator, limit can be + positive or negative infinity
example above power of x in numerator highest than in denominator.
Example 2:
lim (x^2 + 3) / (x^3 + 1)
x--~
Its mean power of x in denominator highest than in numerator
Example 3:
lim (4x^3 + x^2 + 1) / (3x^3 + 4)
x--~
Limit above power of x in denominator = in numerator, so the solution of the limit is same of the coefficient from the highest power of x in denominator and in numerator it self. Solution of limit:
lim (x^3 + 4) / (x^2 + x + 1) = 3/4
x--~

Video 3
PROBLEM SOLVING ABOUT GRAPH MATH
1. The graph y = g(x) if the function is defined by h(x) = g(2x) + 2. What the value of h(0)?
Solution:
The functions is h(x) = g(2x) + 2. If h(0) is mean when h = 1, we substituted this to the functions
h(x) = g(2x) + 2
h(0) = g(2.0) + 2
= g(0) + 2
g(0) is mean g when x = 0, if we look the graph of y = g(x), we get g(0) = 3 because x = 0
and y = 3
So h(1)= g(0)+2
= 3 + 2
= 5
We get the value of h(0) is 5.
2. Let the function f be defined by f(x) = x + 3. If 3f(p) = 15, What is the value of f(2p)?
Solution:
The function is f(x) = x + 3
The value of f(2p) is mean f when x = 2p, we have 3f(p) = 15 because each edge have factors 3. So we can over by 3 and we get f(p) = 5
f(p) = 5 is mean f when x = p is 5, we can substituted this value to the function f(x) = x + 3 we get f(p) = p + 3 = 5
p = 5 - 3
p = 2
With move 1 to the left space between become we get the value of p.
So we can get f(2p) If p = 2 so 2p = 4 is mean f when 2p = 4 we substituted this value to the function f(x) = x + 3
f(4) = 4 + 1
= 5
We get the value of f(2p) is 5

3. In the xy coordinate x = y^2 – 9 intersects line at (p,-3) and (7,t).What is the greatest possible value of the slope of g?
The function x = y^2 – 9 intersects line at (p,-3) and (7,t). The formula of slope is m = (y2 – y1)/(x2-x1) the coordinate of g are (p,-3) and (7,t). That’s mean x1 = p, y2 = -3, x2 = 7, y2 = t, we substituted thus to the m
m = (y2 – y1)/(x2-x1)
= {t – (-3)} /(7 – p)
= (t +3)/(7 – p)
We get the greatest possible value of the slope g is the (t + 3) / (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

A. Function x = g(y) : invertible
Example:
1. Line function y = 3x - 2 and y = x
y = 3x - 1
x = 3x – 1
1 + x = 3x
1 = 2x
x = 1/2
Substituted x = 1/2 in y = 3x - 2
y = 3.1/2 - 1
= 1/2
So, line function y = 3x - 2 and y = x intersect in (1/2,1/2)

3x – 2 = y
3x = y + 2
x = 1/3 (y + 2)
x = 1/3y + 2/3 and y = 1/3x + 2/3


in inverse function the line be write:
f(x) = 5x -2
g(x) = 1/5x + 2/5

f(g(x)) = 5(1/5x + 2/5) - 2
= x + 2 – 2
= x
g(f(x)) = 1/5(5x – 2) + 2/5
= x – 2/5+ 2/5
= x
From operation above can be conclusion:
g = f^(-1)
f(g(x)) = g(f^(-1) (x)) = x
g(x) = f^(-1) (f(x)) = x

2. Line function y = (x - 3)/(x + 5)
y(x + 5) =
yx – x = (- 3) – 5y
(y – 1)x = (- 3) – 5y
x = {(- 3) - 5y}/(y-1) and
y = {(- 3) – 5x}/(x-1)