Representing the video of learning mathematics.

Representing the video of learning mathematics
VIDEO I
A. Solving the Problem
We have problem :
1. Find the value of h(1), if the function h is defined by h(x)=g(2x)+2 and
y=g(x), and if x≤0 function is y=x-3, if 0≤x≤3 function is y=-x+3 and if x≥3 the function is y=x+3.
Solution :
We have information that h(x)=g(x)+2. We looking for h(1).
Substitute x=1 to the h(x)=g(x)+2. We get h(1)=g(2)+2 (if x=1 then 2x is equal 2).
From the function of y=g(x), x=2 in the interval 0≤x≤3, substitute x=2 to y =-x+3, and we get y=g(2)=1.
Back to function h(1)=g(2)+2, because g(2) equals 1, h(1)=1+2=3.
The answer is h(1)=3.

2. Let the function f be defined by f(x)=x+1. If 2f(p)=20, what is the value of f(3p)?
Solution :
We looking for f(3p) , what is f when x=3p?
We have f(x)=x+1 and 2f(p)=20.
Divide all side of 2f(p)=20 by 2, we get f(p)=10, because f(p)=p+1, then
p+1=10 >>> p=10-1=9. We get p=9.
x =3p, We get x=3.9=27.
We can find the value of f(3p) by substituting f(x)=f(27)=27+1=28.
So the value of (3p) is 28.



VIDEO II
GRAPH OF RATIONAL FUNCTION
Graph of rational function if can have discontinuities and has a polynomial in the denominator (can be divided by invaluable quantities).

We should find very dangerous situation in pre calculus that deadly convidence.
It possible to some value of x will meet division by zero. Then the value of function is off limits.
Example :
We have f(x)=(x+2)/(x-1). When x=1, the value function become f(1)=(1+2)/(1-1) which is f(1)=3/0, with zero in denominator. Is not good. Choosing x=1 in this function is bad idea. This is break in function graph.

Let x=0, we have f(0)=(0+2)/(0-1)=2/(-1)=-2.
Let x=1, we have f(1)=(1+2)/(1-1) = 3/0. It’s break in function graph because we can’t compute 3/0. The function f(x)=(x+2)/(x-1) don’t have value for x=1. And we can sketch x=1 as asymtot. The graph is break and discontinuity. The conclusion is rational functions don’t always work this way.

Other example, we have f(x)=1/(x^2+1)
Let x=1, then f(1)=1/(1+1)=1
Let x=0, then f(0)=1/(0+1)=1
The function is not break and contoinuity, because x^2 will not able to -1 that can make denominator 0.
Don’t forget, rational functions, denominator can be zero.
For the polynomial, the functions are smooth and unbroken curve.
For the rational function, some value of x can make zero in the denominator. It impossible to find the value and break in the graph.

Break can show up in :
Missing point on the graph
Example :
We have y=(x^2-x-6)/(x-3). This function break in graph at x=3, because x=3 make the numerator and denominator be zero.
We can conclude that x=3 is the missing point.
When you see 0/0, it possible to factor top and bottom of the rational function and simplyfy. See y=(x^2-x-6)/(x-3). We can factoring (x^2-x-6) to (+
y=(x^2-x-6)/(x-3)=(x-3)(x+2)/(x-3), we can simplify, then y=(x+3), it possible to choice y=3. It’s not break in the graph because x=3, y=5. Hahahaa
We can conclude that to avoid 0/0, factoring and simplyfing make the function not break in the graph.


VIDEO III
ALGEBRAIC LONG DIVISION
Example:
Is x-3 a factor of x^3-7x-6?
We can use algebraic long division. When x^3-7x-6 divide by x-3. We can change x^3-7x-6 to x^3+0x^2-7x-6.
Like other division, we must find a multiplier that multiply x into x^3, the multiplier is x^2. Multiply x-3 with x^3. We get x^3-x^2. Then reduce x^3-+0x^2-7x-6 by x^3-x^2, we have 3x^2-7x-6.
We must find a multiplier that multiply x into 3x^2, the multiplier is 3x. Multiply x-3 with 3x^2-9x. Reduce 3x^2-9x+6 by 3x^2-9x, then we have 2x-6. a multiplier that multiply x into 2x the multiplier is 2. Multiply x-3 with 2. We get 2x-6. Recude 2x-6 by 2x-6, we find 0 as final result.

If the reside is not zero, the divider is not factor of the function.
So x-3 is the factor of x^3-7x-6
adding of multiplier which are x^2+3x+2, this also factor of x^3-7x-6. x^2+3x+2 can factoring to (x+2)(x+1).

So the factor of x^3-7x-6 are (x-3)(x+1)(x+2).
And the roots of x^3-7x-6=0 are:
x-3=0 >>> x=3
x+2=0 >>> x=-2
x+1=0 >>> x=-1

Based on above, we can get 3 roots for 3rd degree equation (as the highest power of x^3-7x-6). We remember that quadratic (2nd degree always have at most 2 roots).
So we can conclude that n-th degree equation always have at most n roots.

Long division for a 3rd order polynomial :
1.Fing a partial quotient of x^2, by dividing into x^3 to get x^2.
2.Multiply x^2 by the divisor & substract the product from the dividend.
3.Repeat the process until you either “clear it out: or reach a remainder.

Assignment I

SUMMARY
1.Function and Exponent Function Graph
a.Exponent Function is functions that mapping each real number of x to real number ax, with a > 0 and a ≠ 1.
b.Exponent Function Graph
i.Definite if x elemen of R and the graph always on the top of x axis.
ii.Intersect with y axis at (0,1).
iii.Decline monotonous curve for 0 < a < 1.
iv.The horizontal asymptote is y = 0 or x axis

2.Exponent Equation
a.Exponent Equation is equations that obtain exponent with form of main number or exponent is function of variable.
b.Determined exponent equation; a > 0, a ≠ 1, b > 0, b ≠ 1
i.If af(x) = 1, then f(x) = 0
ii.If af(x) = ab, then f(x) = b
iii.If af(x) = ag(x), then g(x) = 0
iv.If af(x) = bf(x), then f(x) = 0
v.af(x) = bf(x) can be determined with logarithm help’s.
vi.If f(x)g(x) = f(x)h(x )there is 4 possible solution
1.g(x) = h(x)
2.f(x) = 1
3.f(x) = -1 if g(x) and h(x) is odd or both of them is even
4.f(x) = 0 if g(x) and h(x) is positive
vii. If f(x)h(x) = g(x)h(x) there is 2 possible solution
1.h(x) = 0, if f(x), g(x) ≠0
2.f(x)= g(x)
viii.A[af(x)]2 + B[af(x)] +C = 0 determine this in quadrate equation


3.Inequation Exponent
a.Inequation exponent is inequation which that exponent has form as variable obtain inequation sign ( >, < , ≥, ≤)

b.Determined inequation exponent
i.if af(x) < ag(x), then f(x) < g(x), for a>1
ii.if af(x) > ag(x), then f(x) > g(x), for a>1
iii.if af(x) < ag(x), then f(x) > g(x), for 0iv.if af(x) > ag(x), then f(x) < g(x), for 0