TEST 9_MATHEMATIC NR 07

Nama : Erlina Kurniasih Widyaningrum
NIM : 07305144040
Prodi : Matematika NR 2007
Mata Kuliah : Bahasa Inggris II
Hari, tgl, ruang : Selasa, 18 November 2008, 204
No HP : 081567741616
Email : erlina_dear@yahoo.co.id

VIDEO I
PRE CALCULUS: Graphs of a rational function
• Can have discontinuities has a polynomial in denominator.
• For example:
We have a function f(x) = (x + 2) / (x – 1)
v If value x = 1, we must substitute x = 1 at a function f (x) = (x + 2) / (x – 1) f (1) = (1 +2) / (1 – 1), because one plus two equal three {(1 + 2 = 3)} and one minus one equal zero {(1 – 1) = 0)}, so that the value of three over zero equal infinity {(3 / 0 = ~)}. After value x =1 substitute at rational function will give 0 (zero) in denominator. Further, value x = 1 used bad.
v If value x = 0, we must substitute x = 0 at a function f(x) = (x + 2) / (x – 1)
F (0) = (0 +2) / (0 – 1), because zero plus two equal two {(0 + 2 = 2)} and zero minus one equal negative one {(0 – 1 = 1)}, so that the value of two over negative one equal negative two {(2 / -1= -2)}.
• Not all rational function will give 0 in denominator and can be 0 in denominator.
• Break 2 ways:

v Missing point is a loophole
For example: we have a function y =(x2 – x – 6) / (x – 3).
If value x = 3, we must substitute x = 3 at a function
y = (x2 – x – 6) / (x – 3). We get y = (32 – 3 – 6) / (3 -3). Because three squared minus three minus six equal zero {(32 – 3 – 6 = 0)} and three minus three equal zero {(3 – 3 = 0)}, so that zero over zero not allowed.
We have to factor of x2 – x – 6 become (x – 3) (x + 2). At numerator and denominator there are (x – 3), so that it’s a factor top and bottom. Further, we get value (x – 2) at a function y = (x2 – x – 6) / (x – 3).
v X → 0 in the denominator
v Removable singularity
When x leads 0 / 0, we can do with:
Ø factoring at function
Ø multiplying with the conjugate number

VIDEO II
LIMIT BY INSPECTION
1. x goes to positive / negative infinity
2. Limit involve a polynomial divide by a polynomial
For example:
1. Lim (x3 + 4) / (x2 + x +1)
x→~
Solution:
Easy solution is numerator divide denominator with highest to the power of variable. We look the power of x in numerator and denominator.
At a function highest power of x in numerator is three (3) and highest power of x in denominator is two (2). Because to the power of numerator greater from to the power of denominator, so that the value of Lim (x3 + 4) / (x2 + x +1) = ~
x→~

2. Lim (x2 + 4) / (x3 + 1)
x→~
Solution:
Easy solution is numerator divide denominator with highest to the power of variable. We look the power of x in numerator and denominator.
At a function highest power of x in numerator is three (2).
At a function highest power of x in denominator is two (3).
Because to the power of numerator smaller from the power of denominator, so that the value of Lim (x2 + 4) / (x3 + 1) = 0
x→~
3. Lim (4x3 + x2 + 1) / (3x3 +4)
x→~
Solution:
Easy solution is numerator divide denominator with highest to the power of variable. We look the power of x in numerator and denominator.
At a function highest power of x in numerator is three (3).
At a function highest power of x in denominator is two (3).
Because to the power of numerator equal from the power of denominator, so that the value of Lim (4x3 + x2 + 1) / (3x3 +4) = 4/3
x→~

VIDEO III
EXERCISE
1. If the function h is defined by h(x) = g (2x) + 2. What is the value of h (1)?
Answer:
We will look for value h (1) with h (x) = g (2x) + 2
Because h (1), so that x = 1. We can substitute x = 1 at h (x) = g (2x) + 2
Further, we get h (1) = g (2) + 2
To look for value g (2), we make graph y = g (x). At a graph value from g (2) is 1, so that value from h (1) is three because one plus two equal three.

2. Let the function f be defined by f (x) = x +1. If 2 f (p) = 20
What is the value of f (3p)?
Answer:
What is the f when x = 3p?
2 f (p) = 20, so that twenty divided two equal ten. Further, f (p) = 10
We have a function f (x) = x + 1
For example x = p
We substitute x = p at a function f (x) = x +1, so that it’s become f (p) = p + 1. Because the value of f (p) = 10, so that 10 = p + 1. And the value of p is nine (9). After we get p = 9, we have to substitute p = 9 at x = 3p. The value of x after three times nine is twenty seven.
Because x = 27 and the function f (x) = x + 1 so that, we must substitute x = 27 at a function f (x) = x +1. And the value of f (27) is 28.

3. In the x-y coordinate plane, the graph of x = y2 – 4 intersects line l at (0, p) and (5, t). What is the greater possible value of the slope of l?
Answer:
The graph of x = y2 – 4 intersects line l at (0, p) and (5, t).
For example (0, p) = (x1, y1) and (5, t) = (x2, y2).
The formula of a slope is m = (y2 – y1) / (x2 – x1)
Substitute x1 = 0, x2 = 5, y1 = p and y2 = t at the formula of a slope.
m = (t – p) / (5 – 0)
So that, the greater possible value of the slope of l is m = (t –p) / 5

VIDEO IV
INVERSE FUNCTION
F(x, y) = 0
There are a function y = f(x) (VLT) and x = g(y) (HLT)
For example y = 2x-1 and y = x
§ We get meeting between y = 2x – 1 and y = x
Substitute y =x at y =2x – 1 so that, we get x = 2x-1. After in formulate, we get the value of x is one (1). Because y = x, the value of y also 1 so that, meeting y = 2x-1 with y = x is (1, 1).
§ We get a function y = f(x) and x = g(y)
We have a function y =2x – 1. Because y = f(x) so that, we get the value of f(x) = 2x – 1. We looking for the value of x = g(y), this is the same looking for the inverse of f (x). After y = 2x – 1 formulated, we get x = 1/2 y +1/2. Further, we change the value of y become x and x become y. so that, we get the function of g(x) =1/2x + 1/2.
§ We looking for the composition of a function f g(x) and g f(x)
Firstly, we looking for the composition of a function from f g(x) with substitute g(x) = 1/2x + 1/2 at a 2x – 1 so that, we get f g(x) = 2 (1/2x -1/2) + 1 and if simplified become f g(x) = x.
Second, we looking for the composition of a function from g f(x) with substitute f(x) = 2x – 1 at a 1/2x +1/2 so that, we get g f(x) = 1/2 (2x – 1) + 1/2 and if simplified become g f(x) = x.
Because g = f-1 so that, f g(x) = f (f-1 (x)) = x and also g(x) = f-1 (f(x)) = x given by a function y = (x – 1)/ (x +2). What is the inverse of y(y-1)?
Solution:
Firstly, to looking for the inverse of y = (x – 1) / (x + 2), we must multiplied y and (x + 2) so that, we get y(x + 2) = (x – 1). After that, we formulate the function and we get a function x = (-1 – 2y) / (y – 1). Further, we change the variable of y become x and also x become y. so that, we get the function of y-1 = (-1 – 2x)/(x-1). If y-1 = 0 so that, the function become (-1 – 2x) / (x – 1) = 0. Further, multiplied (x – 1) with 0 so that, we get x = - 1/2.