The problem is question number 13 on page 411. The question is:
The figure shows the graph of y=g(x), in the function h is defined by h(x)=g(2x)+2, what is the value of h(1)?
Answer: We are looking for h(1). The first information is the graph. The next piece of the information is that h(x)=g(2x)+2. What is h equal when x is equal to 1? So let’s substitute 1 into h(x)=g(2x)+2 à h(1)=g(2)+2. Now, how we figure out g(2) or what is g when x is equal to 2? We have the graph above. When x is equal to 2, y is equal to 1. Then g is equal to 1 when x is equal to 2. So, h(1)=g(2)+2 become h(1)=1+2 and we got h(1)=3.
The next problem is another question 13 on page 534:
Let the function f be defined by f(x)=x+1, if 2f(p)=20, what is the value of f(3p)?
Answer: We are looking for f(3p) or in another word what is f when x is equal to 3p? The first piece of information is equation of f(x)=x+1 and the next information is 2f(p)=20. To figure out what is f when x=3p, we need to figure out what p is. Let’s start with the equation of 2f(p)=20. Divide both side by 2, we get f(p)=10. Then, f(p) is just what is function f(x) when x is equal to p. Write that equation. Plug in p in the equation f(x)=x+1à f(p)=p+1=10àp=9. That is not the answer because we are looking for f when x=3p. So if we take p=9 to x=3p, we get x=27. We have the equation for function f which is x+1. We know that x=27, so f(27)=27+1=28.
Let’ move to the next question. Question 17 on page 412:
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?
Answer: We are looking for the greatest slope (m). In the xy-coordinate plane, the graph of x=y^2-4 is:
x=y^2-4 intersect line l at two points (0,p) and (5,t). It does not define exactly where those points are. We have the general idea of the line l on this graph.
Rewrite those two points and put it in a little table to help us.
What is the greatest slope? How do we know about the slope. We know that the slope at the line is going to be m=(y2-y1)/(x2-x1). We can put the values of x and y from the coordinate table that apply to line l. So the slope is going to be m=(t-p)/(5-0)=(t-p)/5. If we want to maximize the slope, we need to maximize the numerator (t-p). How we figure out (t-p)? We have the equation x=y^2-4 and the coordinates (0,p) and (5,t) applied to the equation. The points are the intersection of the graph and the line. So we can plug in the coordinates into the equation of x=y^2-4.
2. Factoring Polynomials
One way to define factor of the polynomials is the rule of algebraic long division that looks like the long division you learn, only harder. For an example, let’s try to see:
Is x-3 a factor of x^3-7x-6?
Answer: When dividing x-3 into x^3-7x-6, first setup the problem like a long division problem from elementary school. That is, you dividing x-3 into x^3-7x-6. Now the zero is in there because there is no second degree term. Now you must ask yourself, what times x gives you x^3? Of course it is x^2. So you write x^2 as the part of the quotient and then multiply x-3 by x^2 which gives you x^3-7x, which you subtract from x^3-0x^2 to get 3x^2. Bringing down the next term -7x, you have 3x^2-7x.
Now we begin again dividing x-3 into 3x^2-7x. Just looking at the first term x goes into 3x^2, 3x. The next part of the answer is 3x. Multiply x-3 by 3x for a product of 3x^2-9x. Subtracting with 3x^2-7x you are left with 2x-6. Now we see that x-3 divides evenly into 2x-6 which equal 2, with no remainder. So the solution to the long division problem x^3-7x-6 divided by x-3 is x^2+3x+2.
Since x-3 divided into x^3-7x-6 evenly with no remainder then x-3 is a factor of x^3-7x-6. The quotient which is x^2+3x+2 is also a factor of x^3-7x-6. We know that x^3-7x-6=(x-3)(x^2+3x+2). The quadratic expression (x^2+3x+2) can be factored into (x+1)(x+2). So x^3-7x-6=(x-3)(x+1)(x+2). Setting the factored form of the equation x^3-7x-6 to zero, we get 0=(x-3)(x+1)(x+2). Those either x-3=0, or x+1=0, or x+2=0. Solving all of this equation for x, we get x=3, x= -1, and x= -2. The roots to x^3-7x-6 are 3, -1, and -2.
Now there are three roots for the third degree equation of x^3-7x-6. On the quadratic or second degree equation we look that always have at most two roots. A fourth degree equation would have four or fewer roots and so on. The degree of a polynomial equation always limits the number of roots.
Let’s summarize the long division process for third order polynomial:
1. Find a partial quotient of x^2, by dividing x into the first term x^3 to get x^2.
2. 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. Pre-Calculus Graph
Let’s begin by discussing the graph of a rational function which can have discontinuities. A rational function has a polynomial in the denominator which means you are dividing by something that is valuable quantity.
It’s possible that some value of x will meet to division be zero. Example:
If f(x)=(x+2)/(x-1), when x=1, the function value become f(x)=(1+2)/(1-1)=3/0 with 0 in the denominator. For this function, choosing x=1 is a bad idea.
When there is a bad choice of x when it makes the bottom of the rational function, it shows up a break in a function graph. For example, suppose you to finding the graph f(x)=(x+2)/(x-1). Start with inserting 0 for x. So now we have f(0)=(0+2)/(0-1)=2/(-1)=-1. So, you put off point down on the graph at (0,-2).
Next you try x=1. This time you get f(1)=(1+2)/(1-1)=3/0. That is you know is impossible and it means that you can not compute y-value when x=1.
It also mean that the graph of this function will not have any point for x=1. The graph is separated into two disconnected pieces at x=1.
Take the graph of f(x)=1/(x^2+1). No matter what numbers you choose for x, the denominator will never zero. The graph is smooth and unbroken.
Don’t forget that the general rule on the rational function, you must expect the possibility that the denominator may turn out to be zero.
Review: For polynomials, the graph is a smooth unbroken curve. For a rational function, sometimes the value of x may be zero in the denominator. That is an impossible situation because there is no y for that x. At that point, there is no value for the function and there is a break in the graph.
A break can show up in two ways. The simpler type of a break is just a missing point. This function is an example of this type of break: y=(x^2-x-6)/(x-3).
The gate that appear in the graph is at points where x=3. If you try to substitute 3 for x in the equation, the result is y=(3^2-3-6)/(3-3)=0/0. That is not possible, not feasible, and not allowed. So, there is no y for x=3. That is the typical example of the missing point syndrome and conveniently it always goes with the result like 0/0.
When you see the result of 0/0, it also tells you that it should be possible to factor the top and the bottom of the rational function and simplify. For an example: y=(x^2-x-6)/(x-3)=(x-3)(x-2)/(x-3)=(x+2).
For this kind of the break, the missing point is a loop hole. For the original function without simplify, x=3 is a bad point because it leads to the division be zero, y=0/0. But if you simplify first, then there’s no problem with x=3, y=x+2. It can be one idea or key in calculus.
Removable singularity appears simplest missing points on the graph when x leads to 0/0. For this kind of a break, if you factor and simplify the rational function, the division by zero can be avoided.
4. Inverse Function
We will talk about the inverse function. To talk about it, we need to review the definition of the function. If we have the relation of f(x,y)=o, then function y=f(x). That is 1.1 function if we can represent it in x=g(y). The function y=f(x) is “VLT” (Vertical Line Test). Every vertical line intersect the graph in at most one point. The 1.1 function satisfy “HLT” (Horizontal Line Test).
If we look at function y=x^2, this is not 1.1 function because if we graph any horizontal line on the graph, we get two intersection points.
To make it as a 1.1 function, we need a domain of 0<=x, so the graph will be:
We can figure out the other squared root of the left side so if we have function of 1.1 then the function is “invertible”.
Let’s start out with the function y=2x-1. And let’s look at the graph of that function.
This is a straight line with y intersect (-1) and x intersect (1/2). Look at the line y=x. That line intersect the graph of y=2x-1. So, we get x=2x-1à 1+x=2x à 1=x. So, the intersection of the line y=x and y=2x-1 with x=1 is (1,1).
See this relation for y=x. Write 2x-1=y à 2x=y+1 à x=(y+1)/2 à x=(y/2)+1/2. Changing x to y in the last equation, we get y=(x/2)+1/2. Looking back on the graph, then we get another line. That line contain the point (1,1). The x-intercept is (0,-1) and the y-intercept is (0,1/2). So, the invers line to the given line y=2x-1 passes through the same point.
We have f(x)=2x-1 and on the other hand, we have g(x)=(x/2)+1/2. We want to compute f(g(x)). f(g(x))=2(…)-1=2((x/2)+1/2)-1à x+1-1=x. g(f(x))=1/2(…)+1/2à g(f(x))=x-1/2+1/2=x.
So, the important problem is g=f-1 à f(g(x))=f(f^(-1)(x))=x and g(f(x))=f^(-1)(f(x))=x.
For another example, take y=(x-1)/(x+2). This is the line with a vertical asymptot at the line x=2 and x=1. The x-intercept is equal to 1, so the point is (1,0). The y-intercept is equal to (0,-1/2). And otherwise the function graph is a hyperbola in a second and fourth quadrant and respect to its asymptoth.
Take a look at the equation of y=(x-1)/(x+2). Multiply both side with (x+2), we get y(x+2)=(x-1) à yx+2y=x-1 à yx-x=-1-2y à (y-1)x=-1-2y à x=(-1-2y)/(y-1). Now, interchange the x to y and y to x à y=(-1-2x)/(x-1). When x=0, we get y=-1 and when y=0 we get -1-2x=0 à -2x=-1 à x= -1/2.
With the vertical asymptot at x=1 and horizontal asymptot at y=-2, again we get the hyperbola. If we look carefully, we will see two functions that reflect each other.
The invers function of y=2^x is: 
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