Chapter 1 Notes for Math 176
Precalculus with Limits, Second Edition

In order to tie comments to specific locations in the book, I have used the available page ruler sheet to identify lines in the text. A copy of that page ruler sheet can be printed from The Index Sheet.

Chapter 1: Functions and Their Graphs
1.1 Rectangular Coordinates
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2	30	It is somewhat unfortunate, but mathematics, which is often praised for being so exact in its meanings, uses the same notation for multiple things. Thus, (4,7) could indeed be the open interval on the number line between 4 and 7. Also, as noted in the text, it could be the point over 4 and up 7 from the origin on the number plane. It could also be a complex number, one that could also have been writeen as 4+7i. The precise meaning has to be understood in the context of the symbols. [Note that the symbols (7,4) could not mean an open interval since we always have the left point of the interval on the left and the right point on the right. However, (7,4) could be either the complex number 7+4i or the coordinate plane point that is over 7 and up 4 from the origin.]
3	34	A scatter plot does give the correct representation of data such as that presented in Example 2. Consider the following reprint of the data, a scatter plot, and a line graph: Data in the spreadsheet Scatter Plot of the data   Line graph of the data The line graph merely connects the points of the scatter plot. And, it might seem reasonable to suppose that as the year 1997 progresses and finally becomes 1998 that the number of subscribers slowly but steadily increases from 55.3 million to 69.2 million. But supposition is not reality. In fact, all that we have is the data points represented in the scatter plot. To connect them with the line graph may make the graph look better, and it may give us a fealing for the growth of subscribers over the years, but the line graph really misrepreents tha known data by implying that we know that growth within each day of each year.
4	16	The book really should have given some explanation of the use of the absolute value in |x2 - x1|, and especially how the absolute value gets dropped in moving from |x2 - x1|2 to (x2 - x1)2. The distance between points on the x-axis is the absolute value of the difference of the x-coordinates. The distance between points on the y-axis is the absolute value of the difference of the y-coordinates. The value of x2 -  x1 could be positive or negative. We take the absolute value of that expression so that we always have a positive value (or rather, a non-negative value) for the distance. That gives the distance to be |x2 -  x1| and if we want to squarwe that we really should write that as ( |x2 -  x1| )2. However, since the square of a number is the same as the square of the negative of the number, we really do not need to take the absolute value before we square the difference, giving rise to (x2 -  x1)2.
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4	16	The Battleship program and problem will be presented in class to reinforce the coordinate plane and the distance formula. There is a special web page devoted to to go through a logical solution to the Battleship problem. You will find that page at battleexplain.htm.
1.2 Graphs of Equations
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13-15	all	The web page Graph, Trace, Window, Plot has a lengthy presentation of using various aspects of the TI-83 to do this material. It is well worth the time to go through that page.
1.3 Linear Equations in Two Variables
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24-29	all	The web page Solving for Linear Equations goes over all of this, but with some added discussion related to using the TI-83 family of calculators with the slope and slope1 programs.
1.4 Functions
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41	24	Function notation is one of the more sloppy parts of our mathematical language. As noted in the book, we start with functions that we describe as y = 1 - x2 This is a formula that allows us to put in any value we wish for x and then, as the result of the expression on the right side, obtain a correponding value for y. Then, we find a variation of this style when we move to f(x) = 1 - x2 This says the same thing, but it has the neatness that we clearly wish to substitute a value for x in order to get a value for the function. Thus, for example, f(3) = 1 - 32 f(3) = 1 - 9 f(3) = -8 On the other hand, none of these reflect the original definition of a function as a set of ordered pairs, (x,y), where no value of x appears more than once. Really, both of the two symbolic versions are shorthand for the more complete statement f = {(x,y) | y = 1 - x2 } This version clearly identifies that the function f is, by definition, a set of ordered pairs. Each ordered pair in that set must follow the restriction that the second coordinate, y must be equal to 1 minus the square of the first coordinate, x. The complete statement is read as "the function f is defined to be the set of all ordered pairs x and y such that y is equal to one minus the square of x.
42	28	The book uses the standard and almost universal style of defining a piecewise function definition. Thus we have a format that looks something like I find this presentation to be less than intuitive because it puts the "if" part of the definition at the end and has no clear "then" portion, an unnatural approach in terms of everyday language. Thus, we read the definition as "the function f of x is defined to be x squared plus one if x is less than or equal to zero; to be x minus 1 if x is greater than 0 and less than or equal to 5; and to be 3 times x minus 11 if x is greater than 5." Although the presentation is almost universally common, our everyday language would have something more like "If x is less than or equal to 0 then f of x is defined as x squared plus one; if x is greater than 0 and less than or equal to 5 then f of x is defined as x minus 1; and if x is greater than 5 then f of x is defined as 3 times x minus 11." This is in a more traditional if then sequence. We could present this as Go to the web page on piecewise functions to see how to define a piecewise function on a TI-83.
44	3	As the book notes, an IMPLIED DOMAIN is the set of all values for which the function definition yields a value. An EXPLICIT DOMAIN needs to be given as part of the function definition. An EXPLICIT DOMAIN is always a part of the IMPLIED DOMAIN. For example, assuming that we are working within the Real number system, then has an IMPLICIT DOMAIN of real numbers greater than or equal to 4. On the other hand, has an EXPLICIT DOMAIN of real numbers between 10 and 24 inclusive.
44	5	The little TECHNOLOGY box at the left of the page suggests that you you should be able to use your graphing calculator to determine the domains of two functions. To some extent this is true, but it is really not a good thing to do, other than to get a general idea. Here are some screen images to demonstrate this. Figure 1 In Figure 1 we se that we have defined and graphed the function y=sqrt(4 - x2). Furthermore, we have moved the calculator into TRACE mode and we have moved the blinking highlight to the right so that the x value is just less than 2. We see that the function is still defined because we calculatr displays the corresponding value of y. Figure 2 Pressing the key one time brings us to Figure 2. Now the x value is 2.1276596 (approximately?) and the y value is undefined. Therefore, we have left the DOMAIN. The question is, where? Figure 3 For FIgure 3 we have forced the calculator to ignore its usual steps between x values and to go to the point where x = 2. The fact that the corresponding y value is 0 indicates the 2 is in the DOMAIN. That is all well and good, but we do not always have such easy, even numbers. Figure 4 Figure 4 uses a different function. Again we have turned on the TRACE feature. In figure 4 it would seem that the function never gets to zero. The graph stops well above the x-axis. Figure 5 Figure 5 captures the moment that the highlight is on the last, the rightmost, dot of the graph. In this case x=1.2234043. And, since we are on a dot of the graph, it is no surprize to find that the TRACE feature gives us a corresponding value for y. Figure 6 However, if we move one step to the right, as in Figure 6, we have moved into the region where the function is not defined. We must assume that the function stops being defined at some point between the x value in Figure 5 and the x value in Figure 6. But what is the value of that point? Figure 7 We could choose some value between 1.2234043 and 1.2765957. For Figure 7 we have done exactly that, choosing 1.225 as our value for x. The calculator does indicate that the function is defined at 1.225 since there is a y value associated with that value of x. It does seem, because the y value is greater than 0, that there are more points to the right of 1.225 that are also in the DOMAIN of the function. We just do not know what they are yet. In addition, a close examination of Figure 7 raises the question of just why is the highlight on a pixel that was not part of the graph in Figures 4, 5, and 6? The page pixels3 should us some idea of the challenge that the calculator is facing here.
45	6	Example 8 depends upon the statement "the ratio of the height to the radius is 4." Thus, as indicated in the box above the picture of the can, (h/r)=4. We could also write that as h=4r or as r=(h/4). These are the substitutions that are used in the solutions given by the text. It is also somewhat strange that the picture given in Figure 1.49 does not correspond to that restriction. The r given for that picture is 1.25 mm long and the h is 5.20 mm long. Thus the value of (h/r) = 5.20/1.25 = 4.16.
45	24	In order to get the calculator had to have a WINDOW setting .
46	3	I could not let Example 10 go by without offering a solution using tables on the TI-83. Figure 1 Figure 1 shows the Y= screen. Note that there are four defined functions, but that only Y2 and Y3 are active. We can tell that they are active becuase the = sign is highlighted for those two functions whereas the = sign is not highlighted for Y1 and Y4. The GRAPH feature and the TABLE feature will ignore inactive function definitions. Y2 is the first part of our piecewise defined function. Y3 is the second part of the piecewise defined function. Figure 2 In Figure 2 we are looking at the TBL SET screen where we have modified that screen so that TblStart has the value 5. Figure 3 We then move to the TABLE screen in Figure 4. Here we see all of the values that we need for year 5, 6, 7, 8, 9, 10 and 11. Although the calculator computes all of the values, we know that Y2 is only defined for years 5 through 9 and Y3 is defined for years 10 through 16. Therefore, we will read off the Y2 values only for the lines where x is 5, 6, 7, 8, and 9. We will read off the Y3 values for lines 10 and 11. Figure 4 Then, just to get the rest of our Y3 values, we just move down the table to see Figure 16.
46	23	Just a small but important point: the Difference Quotient is extremely important. It will not go away. Do not ignore it!
1.5 Analyzing Graphs of Functions
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		No comments.
1.6 A Library of Parent Functions
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		No comments.
1.7 Transformations of Functions
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73	33	The usual approach is to try to memorize this list of transformations. I suggest that it is far better and in the long run easier to spend that time understanding them so that you can figure them out when the appear.
1.8 The Combinations of Functions: Composite Functions
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2	36	The box showing the sum, difference, product, and quotient of functions represents another instance of mathematical abbreviations. We use (f + g)(x) to mean f(x) + g(x).
1.9 Inverse Functions
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96	9	If the method in the book makes sense to you then there is no need to be confused by reading my version of the process. On the other hand, if you cannot make sense of the book's process, you might want to try doing it this way. I will do Example 6 with my process and my explanations. We are asked to find the inverse of: f(x) = 5 - 3x 2 Of course, the real function definition is f = { (x,y) | y = 5 - 3x } 2 We might as well use the other shorthand y = 5 - 3x 2 Solve for x. First we will multiply both sides of the equation by 2. 2y = 5 - 3x Add -5 to both sides. 2y - 5 = -3x Multiply both sides by (-1/3), The result is tht we have expressed x as a function of y. Drop a value of y into the expression, do the calculations, and the result is the value of x. This is really the inverse function, which we wish to call f-1. 5 - 2y  = x 3 We then have the definition of f-1 given as shown to the right. Notice that this is a set of ordered pairs (y,x). The ordered pair is (y,x) becuase we put the independent variable first and the dependent variable, the one that has its value computed based on the choice of the independent variable, second. f-1 = { (y,x) | x = 5 - 2y } 3 However, in the symbolism that we used in the previous step, x and y are just placeholders. In reality we could rewrite the definition using symbol for x and any symbol for y. Because we are used to seeing functions as sets of ordered pairs (x,y) we will replace the symbol x with y and the symbol y with x to get a rewrite of the inverse function as: f-1 = { (x,y) | y = 5 - 2x } 3 We can use the shorter form to rewrite that as f-1(x) = 5 - 2x 3
1.10 Mathematical Modeling and Variation
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		I do have a separate web page giving another regression example from a computational perspective. Then the web page Linear Regression on a TI-83 walks you through another example, this time using the TI-83 to do all the work.