Video 1
Pre-calculus: Graph of A Rational Function
We start by graph of rational function. which can have discontinuities. Why? Because has a polynomial in denominator.
Example:
If f(x) = (x + 2)/(x-2)
When x =1, the function becomes
f(1)=3/0
With zero is denominator, for this function choice x=1 is bad idea. This resulted in the break in function graph. For example,
f(x)= (x + 2)/(x-2)
When insert 0 for x. so,
f(0)=0+2/0-1
f(0)=-2
So, point in the graph is (0,-2). Next we try, insert 1 for x. so,
f(1)=3/0. 0 is denominator.
That is you know, is impossible. So, x value that may be entered when x is equal to zero function.
Pictures on the graph if the function f(x) = (x+2)/(x-1), with x=1 then the graph will not be through the line x=1. So, the rational function don't always work this way. Check the graph. f(x)=1/(x^2+1). When insert -1 for x so,
f(-1)=1/((-1)^2+1)
f(-1)= ½. But the graph is not broken. So, not all rational function give zero in denominator.
Polynomials for graph smooth unbroken curve. There is no value for the function, will lead to break in the function. for example,
y= (x^2-x-6)/(x-3). when insert 3 for x, so y =0/0. That not possible, not feasible and not allowed. Another way is, with factor top and bottom the rational function and simplify. The above examples are, y= (x^2-x-6)/(x-3). With factor polynomial,
y=((x-3)(x+2))/(x-3). Will become more simple so that, y = x +2. When insert 3 for x, is not bad.
Video 2
Limit By Inspection
Determining limit by inspection, there are two conditions. First, x goes to positive or negative infinity. Second, limit involves a polynomial divided by a polynomial. For example,
Limit (x^3+4)/(x^2+x+1) as x approaches infinity.
This problem are two conditions where: first, polynomial over polynomial and second, x approaches infinity. The key to determining limits by inspection is in looking at power of x in the numerator and the denominator. Remember, to apply these rules, must be dividing by polynomials and x has to be approaching infinity. Another first shortcut rule if the highest power of x is greater in numerator, and limit is positive or negative infinity. Last looking the example,
Limit (x^3+4)/(x^2+x+1) as x approaches infinity.
Highest power of x in numerator 3 and greater highest power of x in denominator 2. The limit can be positive or negative infinity since all the numbers are positive and x is going to positive infinity, the limit is positive infinity. If you can’ tell if the answer is positive or negative infinity you can substitution a large number for x and see if you end up with a positive or negative number, whatever sign you get is the sign of infinity for the limit.
Now, the second shortcut rule.
If the highest power of x is in the denominator and the limits is zero.
Example,
Limit (x^2+3)/(x^3+1) as x approaches infinity.
Hence, highest power of x in numerator 2 last highest power of x in denominator 3. Then value the limit equal to zero.
Now, last shortcut rule.
This rule use when highest power of x in numerator is same as highest power of x in denominator. Limit is x positive and negative infinity is just the quotient of the coefficients of the two highest powers. For example,
Limit (4x^3+x^2+1)/(x^3+4) as x approaches infinity.
Then, as highest power in both nominator and denominator is 3. According in this rule then means limit is just the coefficients of x cubic’s over each other. The coefficient x^3 in nominator is 4 and coefficient x^3 in denominator is 3. So, limit =4/3.
Video 3
Solusing Problem Graph Math 1
First problems, the figure above 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)?
Solution:
h(x)=g(2x)+2, when subtitud h(1) for this function then
h(1)=g(2x)+2. we can see the graph, for g(2) then g(2)=1. hence,
h(1)=1+2
h(1)=3
Second problems, let function f be defined by f(x)= x+1. If two f(p) =20, what is the value of f (3p)?
Solution:
When f(3p), what is f when x=3p
When f(x)=x+1 and
2f(p) =20. so,
f(p) =10. when f(x) insert p, then
f(p) =p+1=10
p =9
When x =3p, then insert p=9
x =3*9
x =27
So, x =27. later then, insert x =27 for the function.
f(27) =27+1
F(27)=28
Last problems, in the xy-coordinate plane, the graph x =y^2-4. Intersect line l at (0,p) and (5,t). what is the greathest possible value the slope of l ?
Solution:
Search greatest m, where x =y^2-4 and intersect line l.
x one is 0 and x two is 5. Then y one is p and y two is t.
Then, slope in the graph is line l=m=(y two – y one)/( x two - x one).
So, slope m =(t–p)/5.
Video 4
Invers Function
An inverse function uniquely identifies the input x of another function based only on its output y, for all y is a member of Y. A function is invertible if and only if this rule defines a function. Not all functions have an inverse. For this rule to be applicable, each element y is a member of Y must correspond to exactly one element x is a member of X.
f(x,y)=0 and function y =f(x).
No 1. one function x =g(x) at y =y(x), v intersect t and x =g(y), h intersect t.
For example, line y =2x-1 and line y =x have the point. So, to find the point we are the substitution y =x at function y =2x-1. so established,
x=2x-1
1+x=2
x=1, then
2x-1=y
2x=y+1
x=1/2(y+1)
x=1/2y+1/2
x=1/2y+1/2 we can change to y =1/2x+1/2.
The function f(x) =2 x and g(x) =1/2 x +1/2. So,
f(g(x)) =2 sometimes -1. Sometime is 1/2x-1/2. So,
f(g(x)) = x.
g(f(x)) =1/2 sometime +1/2. Sometime is x -1. So,
g(f(x)) = x.
So, is the conclusion g =f invers.
f(g(x)) =(f invers(x))=x and g(f(x) =f invers(f(x)) =x.
Line y = (x -1)/(x +1), the graphics will not be the same with the 2x and x is the same as the one, but one line will go through point (1, 0) and (0, 1/2). So,
y(x+2) =x-1
xy+2y =x-1
xy-x =-1-2y
(y-1)x =-1-2y
x =-1-2y/y-1
We can be change to y =-1-2x/x-1.
When x =0, y =-1. When y =0, x =-1/2. And when x =1, y =-2.