THE PROBLEM SOLVING

1. Make the graph of y = g(x). If the function h is defined by h(x) = g(2x) + 2, what is the value of h(1)?

Solution:
We are looking for h(1). We can get the first information from the graph of y = g(x) and the function of h(x) = g(2x) + 2. So, what’s the value of h(x) when x=1?
We substitute 1 into that situation
h(1) = g(2) + 2
Based on the graph above we find when x is equal to 2, y is equal to 1. Or we can say when x=2, g(x)=1, so h(1) = 1 + 2 = 3
Thus we get h(1) = 3

2. Let the function f be defined by f(x) = x + 1.
If 2f(p) = 20, what’s the value of f(3p)?

Solution:
We are looking for f(3p), it means that we are looking for the value of f(x) when x= 3p.
The first information is the function of f(x) = x + 1. And the next information is 2f(p) = 20. The question is “what’s f(x) when x = 3p”.
By using the function of 2f(p) = 20, we divide both side by 2, so we get f(p) = 10.
f(p) is just the implication of f(x) where x = p. We write
f(p) = 10
p + 1 = 10
p = 9
We are looking for the value of f(x) when x = 3p, so we can take p = 9 into x, thus x = 27.
We back to the first equation f(x) = x + 1. We substitute x = 27 into the equation, so that f(27) = 27 + 1 = 28.
So the value of f(3p) is 28.

INVERSE FUNCTION

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

We can graph the function of x=g(y) with the domain x is bigger than or equal to zero (0 ≤ x) and then we have the square roots of y=x2

Let’s write y=2x-1 and make the graph of that function. There is a straight line with the point of (0,-1), slope 2
(0,-1)

So, in the x-intersect we find point of (½,0).
And let’s look at the line of y=x and for reason we can substitute y=x into the function y = 2x - 1, so : x = 2x - 1
1 + x = 2x
1 = x
So, 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 this equation is also equal to : 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

Let’s do one more example
Write the function of y=(x-1)/(x+2)

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

THE DEFINITION AND THE SENTENCE OF WORDS

A. dispatch
Definition : dispose of rapidly and without delay and efficiently
Sentence : He dispatched the task he was assigned.

B. engaging
Definition : carrying out / participating in an activity; be involved in
Sentence : The students are engaging the National Seminar of Mathematics.

C. pita = ribbon
Definition : any long object resembling a thin line
Sentence : We describe a geometric interpretation of k-ribbon Fibonacci tableaux and use this interpretation to describe a notion of P equivalence for k-ribbon Fibonacci tableaux.

D. kurs = kurs; the exchange rate
Definition : the price of one country’s currency expressed in another country’s currency
Sentence : The higher the exchange rate for one dollar in terms of rupiah, the lower the relative value of rupiah.