Week 7: 0303 INTERNET

Section 8.1
This section presents the topic of functions which was also partially covered at the end of chapter 3.

When you have two functions: f(x) and g(x) you can perform algebraic operations with them. Getting the algebraic expression is the easy part:
(f + g)(x) = f(x) +  g(x)
(f - g)(x) = f(x) -  g(x)
(fg)(x) = f(x) *  g(x)
(f/g)(x) = f(x)/g(x)
Just plug in the functions and simplify as they do in examples 1 & 2.

In example 1, they perform the arithmetic operation, then they plug in the value given for x. In example 2, they perform the arithmetic operation but do not give a value for x that we can plug in.

Example 1a) f(x) = x² + 1, g(x) = 3x + 5
The general function is: (write the f, write a plus sign, then write the g, then simplify)
(f + g)(x) = (x² + 1) + (3x + 5) = x² + 3x + 6
The example says: (f + g)(1) which means they want you to plug in the number (1) in the above expression wherever there's an x.
so, (f + g)(1)= (1)² + 3(1) + 6 = 10

DIFFERENCE QUOTIENT: This formula is written in the center of page 482. Example 3 demonstrates it's use.

Procedure:  Notice that there is a given function f(x). They first plug (x + h) into this given function, then they plug in (x) and take these two answers and subtract them. Then the bottom value is always the letter "h".  This equation is used to find the slope of a secant line. So it's a variation of the slope equation m=(y2-y1)/(x2-x1).  But here we don't know the numbers that need to be plugged in so instead they plug in variables.  Depending on what was given as the function, you may get a single number as an answer or you might get a polynomial as in example 3b.  Example 3a is only figuring out the top part of the equation.

p. 483 COMPOSITION OF FUNCTIONS
(This was also explained in the chapter 3 notes, here's the same explanation I gave there)
Now they give you two functions f(x) and g(x). They want you to plug one into the other in this way:  
    f[g(x)] = replace the x's in the f(x) equation with the whole function g(x)
    g[f(x)] = plug the f function into the g function (replace the x's)

Example 4: a) Plug the 2 into the g function, then plug that answer into the f function.
c) plug -3 into the f function then plug that answer into the g function.

Ex 5: Is doing the same thing but they are not giving you an "x" to plug in.

Ex 6: Domain of composite functions: Find the domain of the f function and then the g function. Then when you have done the composition, find the domain of the new function.

In example 6: Since f(x) = 1/x, it's domain is restricted because x cannot be equal to zero (you can't have zero on the bottom of a fraction). g(x) = square root of (3 - x) restricts it's domain because x cannot be smaller than 3 (if it is smaller than 3, then 3 - 4, for example, gives you a -1. This produces an imaginary number). So the domain of f is {all real numbers x | x (is not equal to) 0} and the domain of g is (-infinity, 3].

In the composition function: after the f is plugged into the g and it is simplified, you have to set the inside of the square root < or equal to 0 and see which values produce a negative square root. Also, the square root (bottom of p.484) is on the bottom of a fraction. So you have to do two tests, (3-x) must be greater than or equal to zero and sqrt(3-x) must not be equal to zero.  The first test produces x (less than or equal to 3), and the second test produces  x not equal to 3.

Next, you have to combine the information from all three domains: f can't equal zero, g can't be bigger than 3. And the composition must be less than or equal to three, but not equal to 3.  So all-together,
the domain of f o g is all real values of x such that (-infinity, 0)U(0,3). (All real numbers up to but not including 3 and also not including zero)

A similar method is done in part b of example 6.  But be aware that since the final composition is an inequality, (3x-1)/1 >(or equal to)0, you will have to draw a graph and test the signs of each region to see which areas are positive. (As in section 7.5)

Example 7 is well explained in the text.

Graphing Functions:   f(x) = a(x - h) + k is the basic form of all the functions. 

The vertex is (h,k) and the altitude is the a. Notice that the h sits inside the parentheses and the k sits by itself outside the parentheses and the a is always to the left of the (x - h) group. 

 f(x)= 2(x - 1) + 3     the a = 2, h = 1, k= 3

If there are no parentheses in the function that means the h had the value zero, so they didn't bother to write (x-0), they just wrote the x.

 f(x)= 2x + 1, or [2(x-0)+1]     the a = 2, h = 0, k= 1

If the k = 0 then it is just missing altogether.   

 f(x)= 2(x + 1),[or 2(x+1)+0]     the a = 2, h = -1, k = 0

NOTE: The formula has a minus sign in front of the h so it changes the sign of it (i.e. it always looks like the h has the wrong sign). You want to use the original sign of h when graphing the vertex. 
  Ex. in f(x)= (x - 1), the h is a +1, 
        in f(x)= (x + 3), the h is a (-3)  [i.e. (x - (-3)].

The a is the altitude (or amplitude) which determines whether the figure opens "up" or "down"
    The a also controls the "stretch or shrink" .
h is the horizontal shift (picture shifts right or left)
k is the vertical shift (picture shifts up or down)
The powers on the group (x - h) determines the shape of the picture

(x-h) is a line
(x-h)^2 is a parabola
(x-h)^3 is a cube
square-root of (x-h) is a square root
1/(x-h) is a rational function

Note: The book always draws the figure at (0,0) then shifts it. You can do the problems this way or you can draw the graph starting from the final location of the vertex, after it has been shifted. I will count either one correct. 
Your textbook concentrates on parabolas and just touches on the other figures.  The exam will test mainly parabolas, but be aware that in the next course you will be expected to know how to draw all the other figures as well. The graphing procedures are the same for the other figures, you are just drawing an altered parabola. (See my "graphing parabolas" notes for details).

The vertex can be found in the function (the (h,k)). The rest of the points can be found by choosing 2 points to the right and left of the vertex and plugging those values into the original function. This will give you a total of 5 points on the graph.  If the figure is too big to fit on your graph, simply label those points that are off the edges so that I will know that you knew where they should have gone.  You do not need to label the points that fit on the graph. 

Using symmetry: If you will notice, each parabola has the same y-values on the left and right sides of the figure.  You can cut down on the amount of calculations you need to do by plugging in only those two values on the right and then graphing the "mirror image" (symmetry) on the other side of the vertex.