Video 1.” Solving the problem”

Question (1)

Figure above shows the graph of y=g(x).If the function h is define by h(x)=g(2x)+2, what is the value of h(1)??

Solution:

First,make the graph,then write the function h(x)=g(2x)+2. To get the value of of g(2),we can look the graph from absis=2,we make an straight line that intersect the function y=g(x),so from here we get g(2)=1.

S0,h(1)=1+2=3;

The solution is 3.

Question (2):

Let function f be define by f(x)=x+1, if 2f(p)=20, what is the value of f(3p)?

Solution:

Looking for f(3p) what is f when x=3p.

Write function f(x)=x+1,

Then write 2 f(p)=20; dividing by 2

f(p)=10

f(p)=f(x)

10=x+1

X=9

After get x=9,substitute to x=3p, x=3.9=27

Then f(3p)=x+1

=27+1=28.

So, value of f(3p) is 28.

Video 2. “Finding the roots of third equation by algebraic long division”.

Example:

Let x-3 a factor of x³-7x-6,

So we can write (x³-7x-6)/ (x-3)

We can find the other factor from that equation by using long division. We still using the method like in the elementary school.

Method:

find a partial quotient of x² by dividing x into x³ to get x²,then multiply x² by the divisor and sustract the product from the dividend. And repeat the process until you either “clear it out” or reach a remainder.

After do this method,we get no remainder. And the solution of (x³-7x-6)/ (x-3) is x²+3x +2.

Since (x³-7x-6)/ (x-3) is no remainder and x-3 is a factor of x³-7x-6, so, x²+3x +2 also a factor of x³-7x-6. Factors of x²+3x +2 are (x+2) and (x+1), and can be written:

x³-7x-6 =(x-3)(x+2)(x+1); setting the factor x³-7x-6=0 ; so

0=(x-3)(x+2)(x+1)

Thus x-3=0 ; x=3

X+2=0 ; x=-2

X+1=0 ; x=-1

So,roots are 3,-2,-1.

Roots of equation:

· 3 roots for this third degree equation.

· Quadratic equation always have at most 2 roots.

· A fourth degree equation

Video 3

“PRECALCULUS”.

The graph of rational function can have discontinuities,because has a polynomial in the denominator.

Example:

f(x)=(x+2) /(x-1) ; when x=1 we will get 3/0, so it will break in graph.

Let, insert x=0 ; f(0)=-2

Insert x=1; f(1)=???(it’s imposible)

But, the rational function don’t always work this way:

Let f(i)=1 / (i +1), this function never 0 because of the denominator is i+1.

But, rational function, the denominator can be zero and polynomial also have smooth / broken rational function.

For example:

Y= (x² -x-6) / (x-3) ; when x=3 so we’ll find 0/0????

So,to get the value of function y we can simplify the numerator.