Representing The Video of Learning Mathematics

1. Video 1

SOLVING PROBLEM IN GRAPH

Problem 1:

The figure shows 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)?

We are looking for h(1).

We can see in the graph. The next pieces information is h(x) = g(2x)+2. we can say this equation with equation 2.

We are looking for h(1), so we can substitute to the equation 2, and we get

h(x) = g(2x)+2

h(1) = g(2)+2

Then, we will looking for g(2) in the graph. There are 2 in axis x, so we can see the value in axis y, that is g(2) is 1.

Then we change g(2) with 1, so h(1) = 1+2 = 3

So we get h(1) = 3

Problem 2:

Let the function be defined by f(x) = x+1.If 2f(p) = 20. What is the value of f(3p)?

We are looking for f(3p). it means what is f when x = 3p.

We have already known that f(x) = x+1 and 2f(p) = 20.

To find f(3p), first we must find the value of p.

2f(p) = 20, so f(p) = 10

f(p) is just same with f(x) since p equal to x. So let plug-in p as x. Then we can write f(x) = x+1 equal with f(p) = p+1.

In above shows that f(p) = 10 , then f(p) = p+1 = 10.And we get p = 9

We are looking for f(3p), so take p = 9 to x = 3p.

And we get x = 3.9 = 27.

Then we substitute x = 27 to the equation

f(x) = x+1

f(27) = 27+1 = 28

And the answer of f(3p) is 28.

Problem 3:

In the xy-coordinate plane, the graph of x = y^2-4 intersect line l at (0,p) and (5,t). What is the greatest possible value of the slope of l?

We are looking for the greatest slope (m) of l. In the equation shows that x = y^2-4. So we can draw the graph. There are two points that intersect the axis of y, then the graph intersect line l in coordinate (0,p) and(5,t). we rewrite the value of x and y.

If x = 0 y = p

If x = 5 y = t

We are looking for the slope, we know that the slope of a line is the different of y coordinate divided by the different of x coordinate.

m = (y2-y1)/(x2-x1)

We can substitute to the equation of slope

m = (t-p)/5

and what is the maximize, so we must maximize the nominator (t-p). We figure t-p with equation x = y^2-4, so we plug-in the pair of coordinate to the x = y^2-4.

2. Video 2.

FACTORING POLYNOMIAL

One way to find factoring polynomial is with Algebraic long division. For example: Is x-3 a factor of x^3-7x-6.

First we can use method in elementary school by dividing x^2-7x-6 by x-3. Because there is no second degree term, then x^3-0x^2-7x-6 divide by x-3.

Now, what times x to be x^3, of course is x^2 and then multiply with x-3. There is x^3-3x^2. Then subtract the x^3-0x^2-7x-6 with x^3-3x^2, we get 3x^2-7x-6. And we must divide that by x-3, that must 3x. Multiply 3x with x-3, that is 3x^2-9x. Subtracting 3x^2-7x-6 with 3x^2-9x. So it must be 2x-6, then 2x-6 divided by x-3, that is 2. And multiply 2 with x-3, it is 2x-6. then subtract 2x-6 with 2x-6 the answer is 0. So there is no remainder. So the solution of long division problem x^3-7x-6 divide by x-3 is x^2+3x+2.

Since x-3 division to x^3-7x-6, evenly with no remainder the x-3, is a factor of x^3-7x-6. The solution of that problem is x^2+3x+2 is also a factor of x^3-7x-6.

We now know that x^3-7x-6 = (x-3)( x^2+3x+2). The quadratic can be factor into (x+1)(x+2), so x^3-7x-6 = (x-3) (x+1)(x+2).

Since x^3-7x-6 = 0 , we get 0 = (x-3) (x+1)(x+2). Thus x-3 = 0 or x+1 = 0 or x+2 = 0 solve the equation we get x = 3, x = -1 and x = -2.

There are 3 roots for 3^rd degree equation and quadratic (2^nd degree) equation always have at most 2 roots. A 4^th degree equation would have 4 or fewer roots and so on. The degree of a polynomial equation always limits the number of roots.

Let the summary the long division for a 3^rd order polynomial:

1. Find a partial quotient of x^3 by dividing x into x^3 to get x^2.

2. Then multiply x^2 by the divisor and subtract the product from the dividend

3. Repeat the process until you either “clear it out” or reach a remainder.

3. Video 3

GRAPH OF RATIONAL FUNCTION

Graph of a rational function can have discontinuities because a rational function has a polynomial in the denominator.

Example if f(x) = (x+2)/(x-1), when x = 1 the equation become (1+2)/(1-1) which is 3/0, with 0 in the denominator , so that is bad idea. It is bad choice to chose x = 1, it will break in function graph. Example f(x) = (x+2)/(x-1), try to inserting 0 for x, we get -2. So we put point down on the graph at (0,-2). Now try x = 1, equal 3/0. It is impossible, it means that the graph not have any point at x = 1. Rational functions don’t always work this way, not all rational function will give zero in denominator.

Another example, f(x) = 1/(x^2+1). No matter what we chose for x, the denominator will never to be zero. But don’t forget that rational function, denominator can be zero. For polynomial, the graph is smooth and unbroken curve. But for rational function, somehow for x it can be zero in denominator. That is impossible situation, because there is no value for the function and break in the graph.

The break graph can show up in two ways. It is missing point on the graph. Example y = (x^2-x-6)/(x-3), at point x = 3 the graph is break, because if we substitute the x-3 for the equation, the result is 0/0, it also tell we that it should be possible to factor the top and the bottom on rational function and simplify. Other example, y = (x^2-x-6)/(x-3) we can factor, so y = (x-3)(x+2)/(x-3), and simplify y = x+2 .

If y = x +2 it is no problem if substitute x=3. It is the one of the key idea on calculus. Removable singularity is missing point on the graph, then we factor and simplify rational function, division by zero can be avoided.

4. Video 4

INVERS FUNCTION

Let, f(x, y) = 0

y = f(x)

function x = g(y) invertible

Example:

y=2x+1

the invers of y=2x-1 is
2x-1=y
2x =y+1
x =1/2(y+1)
x =(1/2)y+(1/2)

so y=(1/2)x+(1/2)
Then f(x)=2x-1
g(x)=1/2x+1/2
substitute that x equal to g(x)
f(g(x)) =2(1/2x+1/2)-1
=x+1-1
=x
substitute that x equal to f(x)
g(f(x)) =1/2(2x-1)+1/2
=x-1/2+1/2
=x
Then g=f ^-1
f(g(x))=f( f ^-1 (x))
g(f(x))= f ^-1 (f(x))=x

Next Example:

y=(x-1)/(x+2)
We are looking for the invers of 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)

so y=(-1-2x)/(x-1)
when x=0 then y=-1
y=0 then -1-2x =0
-2x =1
so we get x =-(1/2)