## Course diary for Math 135, section F2, summer 2006

Thursday, July 6
Homework problems
Ms. Atkins kindly put her correct solutions of the two homework problems on the board. These are the problems:

Compute the derivatives of the following functions, using only the definition of derivative:
6. g(t)=1/(t2+1)
8. p(t)=sqrt(4t-2).
For example, I think her solution of problem 6 went like this:
• The derivative involves the quotient [f(x+h)-f(x)]/h. So we change f to g and x to t. Plug in.
• Consider [f(x+h)-f(x)]/h=[g(t+h)-g(t)]/h. This is
```     1      1
------ - ---
(t+h)2    t2
----------------
h```
• Combine the fractions. The aim is always to convert this "mess" into an equivalent algebraic form where h does not appear on the bottom, so that we can let h--> by just plugging in h=0 (that is, really, using A1, A2, A3, etc.)
```  t2-(t+h)2            t2-(t2+2th+h2)
----------         -----------------
(t+h)2t2              (t+h)2t2
--------------  =  ---------------------
h                     h```
This whole business is very vulnerable to errors. The fractions on top must be combined, then the (t+h)2 "expanded", and don't lose the darn minus sign! Here you have to hope that the subject makes sense -- you need faith that the computation will come to a good end.
• Now the top cancels and the compound fraction can be restructured as a simple fraction.
```  t2-(t2+2th+h2)      -2th+h2
--------------     ----------
(t+h)2t2         (t+h)2t2             -2th+h2
---------------   = --------------  =  ----------------
h                  h              h((t+h)2t2)```
One of my favorite movies ("Shakespeare in Love") has the following line: "Strangely enough , it all turns out well." Here, too, it all turns out well. The darn h's cancel. (If they didn't cancel, there would be, for most purposes, no calculus!)
• Now the simple fraction, due to -2th+h2=h(-2t+h), becomes
```  -2t+h
----------
(t+h)2t2```
• Finally let's find the limh-->0 by just "plugging in", but now there is no 0 on the bottom. The result is -2t/(t2)2. And this is g´(t).
I think that any sane person would agree that the solutions were tedious. There are details which are easy to get wrong. So what I will do today and Monday is show you ways to get these results with much less effort. I should remark, however, that although the computation of derivatives for functions given by such formulas will turn out to be very routine and, essentially, easy, it is still the definition of derivative and the ideas connected with this definition which make the subject valuable, not the computations with formulas.

Algebra, ugly; pictures, pretty
Well, the title above represents my own "esthetic" but like much opinion, it can be overdone. I wrote it in response to the crawly scrawl which covered the board solving the two homework problems. But in thinking about the homework, I decided that you should have a chance to work with some pictures and maybe learn from them. So I prepared this handout.

Patterns of thought
I asked students to sketch y=f´(x) on page 1 of the handout, and then to sketch y=g(x) on page 2 of the handout. I emphasized that I was interested in the qualitative behavior of the sketches, and the absence of numerical indications on the axes showed that I wasn't giving very definite information, only "relative" information. In particular, I don't think that either request had any particular single sketch as a correct answer.

 Here are the ideas I wanted to discuss when we considered the handout: first, that when the graph of a function goes up (y-values increasing) as we consider "movement" along the graph from left to right (the x-values are increasing), then the slope of tangent lines will be tilted up also: that is, the slopes will be positive. Since one interpretation of the value of the derivative is the slope of the tangent line, I would expect that a "going up" function would have a positive derivative, and a "going down" function would have a negative derivative. We did run into some linguistic problems in our discussion. We needed to talk about lines whose slope was bigger/smaller than other lines. In our discussion, we sometimes confused magnitude with order. So maybe some people might think that -7 and 16 were "larger" than -3 or 5. Other people might state something the other way. But, in fact, -7<-3 and -3<3 and 5<16. This is not "too obvious" because language can be imprecise. Steeper lines have greater magnitude (that is what's left after taking the sign away!). "Shallower" lines have smaller magnitude. So -7 is steeper than -3, but -3>-7: -3 is greater than -7. This can be confusing, so be alert!

Getting a graph of f´(x) from a graph of f(x)
 So here's the initial graph of y=f(x). Let me describe one approach to getting a graph of the derivative. This is certainly not the only approach -- one of the reasons I tried the handout was so that people would get their own ways of describing and understanding what's going on. I think I would first find where the tangent lines are horizontal. In this case, the bumps up and down (we will call them local or relative maxima and minima in more pedantic language) provide points on the graph where the tangent lines are horizontal. Just by accident (nah!) these happen to be where the vertical lines are. The vertical lines go down to the empty axes where we're supposed to draw the graph of y=f´(x). For those values of x, I know that f´(x)=0. So the derivative of f(x) has x-intercepts for those values of x. In order to be able to discuss the graphs more easily, I labeled the four values of x: A, B, C, and D, in increasing order. Now let me concentrate on just two "chunks" of the graph of f(x), one chunk between x=B and x=C, and one chunk between x=C and x=D. The "logic" is sort of the same, just with signs flipped. Let me try. At x=B the tangent line is horizontal. For x's a bit to the right of x=B. the tangent line has positive slope. Move x more to the right, and the slope tilts "up": the line gets steeper. So the slope, a positive number, gets bigger. But as x gets closer to x=C, another value of x where the tangent line is horizontal, the slope decreases. It is still positive, but it gets less steep, and becomes more shallow. Finally the slope returns to 0 at x=C. Please: in pictures, sometimes too much is worse than too little, so I have only drawn two tangent lines between x=b and x=c. The description of the slope transition on the interval from x=C to x=D is similar, only with a sign change. So near and to the left of x=C, the sign of the slope of the tangent line is negative. As x travels more right, the tangent line gets steeper. The magnitude of the slope gets bigger, and the slope itself gets more negative. I am trying to be careful with language here. Even farther to the right, as we get closer to x=D, the tangent line begins to flatten out (?). The slope is still negative, but it is small in magnitude and negative. Finally, the slope comes back to 0 at x=D. Again, I have only drawn two tangent lines between x=c and x=d. The top graph to the right is the graph of f(x). The graph in the middle is a simple "realization" of the graph of f´(x). It has the correct positive and negative portions separating the places where it should be 0. The graph on the bottom is another possible picture of y=f´(x). It has some details noted by Mr. Wientraub. In particular to the right of x=A he noted that the graph of y=f(x) seems to be almost like a straight line. Therefore the graph of the derivative should, as x goes to the left, become almost constant. It should still be positive, because f(x) is still increasing, going up, but the rate of increase, the slope of the tangent line, is just about the same. He observed something similar about the derivative to the right of x=D. The graph of y=f(x) seems to get straighter, almost like a line, so that the derivative again should get more like a constant. Again, the derivative in this region (x>D) is still positive, but the derivative may be sort of constant. (Huh? "sort of"?) Also notice please that maybe something is happening around x=B. The graph of y=f(x) certainly has a horizontal tangent at x=B, but to the left and to the right the slope is horizontalish (not a word to be tried in a Scrabble game!). If we wanted to be fanatical about drawing the derivative maybe I might try to indicate this by flattening the graph of f´(x) near x=B more than at other x-intercepts of the derivative. All this is certainly finer structure than I would expect of a first attempt at drawing the derivative and I congratulate him for his observations.

Getting a graph of g(x) from a graph of g´(x)
Page 2 of the handout had a related question. A graph of a derivative was given. It was labeled a graph of y=g´(x). That is what's shown here. What was requested was a graph of the original function. In addition, a large, fat point was given, with the information that the graph of g(x) should pass through this point.
To the right is the graph. I've labeled the x-intercepts of y=g´(x) with A and B and C and D. (Darn it, I know that there won't always be four such points, but in the two examples considered in class, there were.) Also, I labeled the "other" possibly interesting x value (which connected with the fat point) with x0. I'll ignore x0 temporarily.

 Again, let's start with a small chunk of the graph, between x=B and x=C. We have a bump in g´(x). At x=B, the derivative is 0. This means that the graph of y=g(x) should have a horizontal tangent at x=B. I've tried to draw a short horizontal segment at x=B representing that tangent line. Also the information presented tells us that there should be a horizontal tangent at x=C. Again, a short horizontal segment is presented. I've actually drawn several candidates for that tangent line, at varying heights. Which of these is consistent with the information presented in the graph of y=´(x)? I've also tried to connect each of the candidates with possible graphs of y=g(x). The C3 line segment and connecting graph is lower than the line segment at B. The g(x) graph must move down as x travels from left to right. This seems to mean that the tangent line will slope "down", and g´(x) will be negative. But the derivative graph is positive between x=B and x=C. So the choice of C3 does not seem to be consistent with the information given. Now look at the C2 line segment. That segment is supposed to be at the same level at the B line segment. The graph connecting the two (tangent to horizontal segments at its endpoints) should be either level (so it would alway have derivative 0!) or, if it goes up, then it must go any equal amount down, as x moves from left to right. But the derivative graph is not 0 between B and C, and the derivative graph is always positive between B and C. So choosing the C2 line segment also does not seem to be consistent with the information presented. Finally, let's consider C1. I don't want to just accept it because the other choices seem inconsistent, because maybe my analysis has been wrong. But look at the graph given: the graph moves up, the tangent lines to the graph between x=B and x=C are all tilted up, so their slopes are positive, and this is consistent with the evidence in the graph of y=g´(x).

What I've just described is a major difference from the previous analysis of the transition from the graph of y=f(x) to the graph of y=f´(x). Here, going "backwards" from g´(x) to g(x), the information given only permits us to make a relative placement of the points on the graph of y=g(x). So from x=B to x=C the graph should go "up". But there is not enough information in the graph of y=g´(x) to tell me where it should go up "from"!

On the algebraic side, this is reflected in the fact that all horizontal lines have slope 0. Therefore the lines y=234 and y=-117 both have slope 0. So the functions whose graphs are those lines (say, h(x)=234 and k(x)=117) are both functions so that h´(x)=0 for all x and k´(x)=0 for all x. If you just give me the derivative, you can't tell me where the graph of the function is.

One mathematical model
If the function represents motion along a straight line (driving on a road, say), then the derivative might represent velocity. Telling me that your velocity is 60 mph for 4 hours does not tell me where you are. You need to add one more chunk of information: something about where you were located when you started. Then I can tell where you are after 4 hours.

Another mathematical model
Suppose I look at profit (+) and loss (-) for a business. If I am told that the income for the business has really turned up over the last 2 years, and income has increased, oh, maybe \$100,000 each year, then I know ... well, I don't know if the business is profitable. If before these 2 years have begun, the business is profitable, then, yes, certainly it has continued to be profitable. But if the business started out the period of time with a record of having lost \$300,000 annually, well, over the last 2 years it still hasn't turned profitable. It has just become less of a money sink!

I agree that this is partly linguistic and partly logical, but such precision really might be necessary. Especially if you are thinking of buying the business ... heh, heh, heh.

Back to the graphs
Well, the purpose of the fat point should be clearer now. At x0, the graph goes through the fat point. Now let me try to "draw" a possible graph of y=g(x) consistent with the fat point (sigh: o.k., the mathematic term is "initial value"). I'll start at x0 and move right first.

Moving right from x0
Look at the graph of g´(x). Between x0 and A the graph is positive. Therefore, traveling to the right from the fat point, the graph y=g(x) should go up on its coordinate axes. Then at x=A there should be a horizontal tangent. Between x=A and x=B the graph of the derivative is negative. I think that the graph of the function should decrease, move down, as x goes from A to B. Now we already analyzed what should happen in the interval from x=B to x=C: g(x) increases. So I glue in an increasing chunk of g(x), making sure to have flat places (for the horizontal tangents) at both x=B and x=C. What about the tiny (?) interval between x=C and x=D? Here the derivative is negative, so the original function must be decreasing. So I'll glue in a small interval on the graph of g(x) where the function heads down. Again, remember that there should be horizontal tangents at x=C and x=D. Now for x>D, the derivative is positive, so the function g(x) increases.

Moving left from x0
This can be confusing. The derivative is positive for x<x0, so the function y=g(x) should be increasing. But if you are doing this slowly, step-by-step, like me, your graph will be drawn starting at x0 and moving left. The graph in that direction goes down, so that in the standard left-to-right direction the graph of y=g(x) is increasing.

I'd like to emphasize again that these graphs are only qualitative. I have not inserted any numbers or reasoned using anything except where numbers are zero or positive or negative. This means that there could be many answers to the question -- many different valid sketches of y=g(x). In particular, Ms. Smookler asked the following prescient (prescience means "foreknowledge of events") question: look at the interval between x=A and x=C. The graph of y=g(x) dips down and then up. How do we know (or not know!) that the bump in y=g(x) is above (below?) the x-axis? I remarked that we really don't know. All that we can deduce is the correctness of the relative positions of the heights on the graph of y=g(x). I have no good, reliable feeling about exactly where or how high any specific point on the graph is. Later in the course, I hope to show you how to get such information, but we'll need to develop a bit more "technique" than we have now.

Sine: yet another sketch
I tried to draw an accurate picture of sine and then discussed what properties the derivative of sine would have.

• Where would the derivative of sine be 0? Well, where the tangent lines are horizontal. That should be at the tops and bottoms of the sine curve. These occur at Pi/2, 3Pi/2, -Pi/2, etc.: lots of places because sine repeats every 2Pi.
• Let's look carefully at sine between, say, x=-Pi/2 and x=Pi/2. I hope you "see" that the derivative, the slope of the tangent line, would start out at -Pi/2 at 0, then it would increase (as the tangent line began to tilt up). Then it would tilt up more (the slope would be more positive) until it would begin to tilt "down": here the language gets complicated. I am not asserting that the derivative is negative, but I am merely asserting that the slope, which stays positive, begins to decrease. Eventually the slope ends up, when x=Pi/2, at 0 again.
• What should happen between x=Pi/2 and x=3Pi/2, say? If your geometric "intuition" is superb, you can almost see that the curve I should draw on the derivative side starts down, so it is negative, gets more negative, then goes up, and ends up at 0. If you really see things very well, you may see that the shape of the derivative may exactly reflect the shape in the earlier interval, since the sine curve's shape is a flip of the earlier behavior.
• Etc. This word or abbreviation covers a lot. The derivative should be just as periodic as sine is. And it should geometrically be just as nice as sine.

Everything is arranged. Everything works out. What function has a graph which looks like the one drawn for the derivative of sine? Well, heck, we know such a function: cosine.

The derivative of sine is cosine.

In a calculus textbook, something like the following is done when f(x)=sin(x):

```
sin(x+h)-sin(x)     sin(x)cos(h)+sin(h)cos(x)-sin(x)
----------------- = ---------------------------------- = PIECE #1 + PIECE #2
h                          h```
where
```
sin(h)
PIECE #1 = cos(x) --------
h```
and as h-->0 this --> cos(x)·1 because we arranged it this way when we decided to use radian measure! Also,
```
cos(h)-1
PIECE #2 = sin(x) ----------
h```
If we multiply this top and bottom by cos(h)+1, the result on top is [cos(h)]2-1 which is [sin(h)]2. Then
```
[sin(h)]2             sin(h)               1
PIECE #2 = sin(x) ------------- = sin(x) -------- sin(h) ------------
h [cos(h)+1]             h             [cos(h)+1]```
Now as h-->0, I claim:
```
sin(h)                         1
sin(x)-->sin(x); ------ -->1;  sin(h)-->0; ----------- --> 1/2.
Nothing happens!      h                       [cos(h)+1]
```
So the result is sin(x)·1·0·1/2=0.
Don't worry. I believe this is the only time in the course I'll even write the addition formula for sine and you don't need to remember this. Any of this. It is just a bad dream.

Cosine
You can play the same game with cosine as with sine. Here is an impression of what the picture of the derivative of cosine might look like:

If you look closely at the picture, there is a loose minus sign around. This we can't get rid of: I'm sorry.

The derivative of cosine is -sine.

Some practice
If f(x)=-7x3+23sin(x)+56, then f´(x)=(-7)3x2+23cos(x)+0. And if f(x)=88x2-2x+Pi*cos(x)-56, then f´(x)=(88)2x1-2+Pi*{-sin(x)}+0.

Building a table of Func's and Func´'s
So we have some ways of computing derivatives of functions which are defined by formulas:

FunctionDerivative
xn nxn-1
sin(x) cos(x)
cos(x) -sin(x)
Constant 0
c f(x)
if c is a constant
c f´(x)
f(x)+g(x) f´(x)+g´(x)

One short-term goal is to lengthen this table so that we will be able to compute quite efficiently (and, really, easily) the derivatives of functions defined by the formulas used in most common mathematical models.

Valery and Fred: how they grow
Here are two mythical adolescents. I tell you the following "facts" (this means, "Please make the following assumptions."):
Valery is 14 years old, 60 inches tall, and grows 1/2 inch per year.
Fred is 16 years old, 66 inches tall, and grows 1 inch per year.

Valery stands on top of Fred
Suppose Valery stands on top of Fred. The assembled object seems to be 60+66=126 inches tall. How much does it grow in a year? I think it grows (1/2)+1=3/2 inches a year.

Valery and Fred lay decorously on the lawn
Huh: one definition of decorous begins with the phrase characterized by propriety and dignity and good taste in manners and conduct so you should keep your impure thoughts buried. Anyway, they decide to recline on a lawn. Their feet are in contact and they are perpendicular to each other. Together they form two sides of a rectangle. What is the area of the rectangle?

Well, Arearect is the product of the heights of the two people. So Arearect is 60·66=3,960 square inches (in2, I guess). How much is the area in a year? (These are extremely patient, persistent people, who will do anything to help the progress of mathematics.) Well, let's look at the heights and the product of the heights: (60+{1/2})·(66+1)=60·66+(1/2)(66)+(60)(1)+{1/2}(1). I don't want to "simplify" because in this case any messing around with the numbers might disturb the structure in the answer. How much does the area change in a year? That's certainly 60·66+(1/2)(66)+(60)(1)+{1/2}(1)-60·66=(1/2)(66)+(60)(1)+{1/2}(1). The area change is related to the change of the factors in a complicated way.

I can follow the suggestion of Ms. Smookler and look at the grow rate of the area. Suppose we assume that Valery and Fred grow steadily. Of course this is not correct in any "real" situation but, heck, this whole story is totally silly. Well, if h represents that amount of a year which has occurred, then Valery is 60+{1/2}h inches tall and Fred is 66+1h inches tall. The area increase at time h is 60·66+(1/2)h(66)+(60)(1h)+{1/2}h(1)h-60·66=(1/2)h(66)+(60)(1)h+{1/2}(1)h2. The average rate of increase of the area over the time period, h, is [(1/2)h(66)+(60)(1)h+{1/2}(1)h2]h=(1/2)(66)+(60)(1)+{1/2}(1)h. Of course if we wanted the totally mythical instantaneous rate of increase of the area (with units of in2 per year), we'd let h-->0. The result would be (1/2)(66)+(60)(1). I hope that this discussion is a sufficient justification of the following result.

The Product Rule

FunctionDerivative
f(x)·g(x) f´(x)·g(x)+f(x)·g´(x)

Some examples
The derivative of (5x7-19x3+44)(-8x4+23x2+8x) is ({5}7x6+{-19}3x2+0)(-8x4+23x2+8x)+(5x7-19x3+44)({-8}4x3+{23}2x1+8x0).
Here the template for the product rule has the following entries:
f(x)=5x7-19x3+44
g(x)=-8x4+23x2+8x

Another one:
The derivative of (44sin(x)-66x8) (x100+23cos(x)) is (44cos(x)+{-66}8x7)(x100+23cos(x))+(44sin(x)-66x8)(100x99+23{-sin(x)}).
Here the functions in the product rule are:
f(x)=44sin(x)-66x8
g(x)=x100+23cos(x)

A homework problem again
I then did something tricky which I probably should not have done. I looked at a problem that Ms. Atkins had put on the board earlier.
So let's consider g(t)=1/(t2+1). Then (t2+1)g(t)=1. If I differentiate this equation, the right-hand side has derivative equal to 0 and the left-hand side's derivative can be gotten with the product rule:
(2t)g(t)+(t2+1)g´(t)=0.
Now I rewrite:
(t2+1)g´(t)=-(2t)g(t).
And solve:
g´(t)=-[(2t)g(t)]/(t2+1).
Sigh. If you plug in the original definition of g(t), you'll get g´(t)=-(2t)/(t2+1)2.
This was the solution gotten with lots of algebra earlier. In fact, there's a better way to do this.

The Quotient Rule

FunctionDerivative
f(x)/g(x) [f´(x)·g(x)-f(x)·g´(x)]/(g(x))2

Notice that I did not try to illustrate the quotient rule with Fred and Valery. I did not want to put one on top of the other. Sigh.

Example
x Suppose we need to compute the derivative of the function defined by the formula

```  4x7-5sin(x)
--------------
6x3-19cos(x)```
Here the top, f(x), is given by f(x)=4x7-5sin(x) and the bottom, g(x), in the Quotient Rule, is given by g(x)=6x3-19cos(x). The derivative is then:
```
[(4)7x6-5cos(x)][ 6x3-19cos(x)]-[4x7-5sin(x)][(6)3x2+(-19){-sin(x)}]
------------------------------------------------------------------------
[6x3-19cos(x)]2```
And, yeah, even if you're "in the business", this is all fairly ludicrous. In its favor, though, is that fact that the formulas are specific and the process is "straightforward".

Example
I'll do one last random no example. If F(x)=sin(x)/cos(x). In the template for the Quotient Rule, the top, f(x), is f(x)=sin(x), and the bottom, g(x), is g(x)=cos(x). Then the Quotient Rule reports that the answer is:

```cos(x)cos(x)-[sin(x)][-sin(x)]
------------------------------
[cos(x)]2```
The quotient, F(x), occurs often enough in practice (it is tan(x), notice!) that the result is usually simplified. The top is cos2+sin2, and that's 1. Another entry in the table of derivatives is the result:

FunctionDerivative
tan(x) 1/{cos(x)}2 or {sec(x)}2

Wednesday, July 5
A function f(x) is continuous at w if limx-->wf(x) exists and is equal to f(w). So the limit exists, and we can evaluate the limit by "plugging in". This is a very nice situation. If we're told that a function is continuous, then we don't need to do any (possibly tricky) algebraic manipulation and things like sin(h)/h won't occur. We just "plug in" (we decided that "substitute" is the polite word).

To the right is a graph which accompanied review problem 7. The "standard" notation is used with an empty circle and a full circle. On Monday I tried to illustrate the definition of continuous by looking at the behavior near and at various points on this graph.

• At x=1
The function is certainly defined at 1, and f(1)=2. Both the left- and right-hand limits exist as x-->1, and both of them are equal to 2. So this function f(x) is continuous at 1,
• At x=0
Here the left- and right-hand limits exist and agree, so that limx-->0 exists. The value of this limit is 1, the empty circle at (0,1). But f(0)=0. So, even though the limit exists, the function is not continuous at 0 since the limiting value is not the same as the function's value.
• At x=2
Here the left-hand limit, limx-->2-f(x) exists and seems to be 1. The right-hand limit, limx-->2+f(x) exists and is -1. Since these limits do not agree, the full "two-sided" limit limx-->2f(x) does not exist. So the function can't be continuous at 2. Also, of course, 2 does not seem to be in the domain of f(x), so f(2) is not defined, another reason the function is not continuous.
For f(x) to be continuous at x=c, both "handed" limits, limx-->c-f(x) and limx-->c+f(x) must exist and agree, and also f(c) must be defined and agree with that limiting value.

The most interesting consequence of continuity is a result called the Intermediate Value Theorem. I introduced this result with a call on my friend Francine.

The Garden State Parkway, from Cape May to Montvale and my friend Francine ...
The Garden State Parkway runs most of the length of New Jersey. Mile 0 is at Cape May, while the other end, mile 172, seems to be close to Montvale. Suppose that my friend Francine leaves Cape May at 7 AM one morning, and drives north on the Garden State Parkway. Further, suppose she arrives at mile 172, the northern end, at, say, 10 AM. Must Francine at some time be at mile 135 (fairly near Busch campus)? The parkway seal here was "borrowed" from a State of New Jersey webpage.

We discussed various curves which could represent the position of Francine on the parkway in terms of miles from the start of the parkway at time t, in terms of hours elapsed from 7 AM. I tried to show that our everyday intuition lead to the graph being increasing (as you travel from left to right, the points on the graph go up). The graph can have level spots, where Francine pulls over for a rest stop. Legally Francine isn't supposed to drive backwards, though.

If we believe that motion is continuous (so Francine does not have a Star Trek transporter or other device) then the graph of Francine's position goes from (7 AM, 0 miles) to (10 AM, 172 miles) and therefore the graph must have on it at least one point with coordinate description (*,135). All of this, by the way, rests on some complicated assumptions, some of them philosophical (why should motion be continuous?). Today, though, I believe that motion is continuous, and therefore at sometime Francine must be at Mile 135. By the way, I will retain this information for later, when we analyze the rate of change of position (velocity) so that we can see whether Francine deserves a speeding ticket.

The Intermediate Value Theorem
Suppose that the function f is defined and continuous on the interval [a,b]. Then the equation f(x)=y has at least one solution for every y which is between f(a) and f(b).

One application is root finding. If we were desperate to compute sqrt(2) (that is, really, desperate to approximate sqrt(2)), for example, we could look at f(x)=x2-2 on the interval [0,2]. This f(x) is certainly continuous. (We already observed that we could "plug in" values to evaluate limits for polynomials. I know that f(0)=-2<0 and f(2)=+2>0. Therefore according to the Intermediate Value Theorem there will be at least one x inside the interval [0,2] so that f(x)=0: x2-2=0. This is a positive number whose square is 2, which we call sqrt(2). Now we have "trapped" sqrt(2) inside the interval [0,2]. If we compute f(1)=12-2=-1, we know that a root must be inside [1,2] since the signs of f(x) at the two endpoints differ. We can continue this "game", each time halving the interval, and chosing a half-subinterval so that the signs of f(x) differ on the endpoints. The graph of f(x) on the first few subintervals is shown below.
Interval: [0,2] Interval: [1,2] Interval: [1,1.5] Interval: [1.25,1.5] Interval: [1.375,1.5]
This may not seem too impressive, but if you've got a silicon friend who can do arithmetic and make simple decisions quickly, you will be able to "localize" a guess for sqrt(2) very narrowly and very quickly. The name of this scheme is the Bisection Method and it is used frequently: it is fast and reliable.

Another example
We realized that polynomials may not have real roots. I think one example was x4+78.
I tried to give an example that looked like this:
12-8x+15x3+166x12+800x55

The trick here is to think in terms of pictures, but sort of pictures without details. Weird pictures.

 The picture I'm thinking of is not like this, which is a nice tame picture of part of the graph. I'm interested in what happens when x is really large, both positive and negative. There the "dominant" term in the polynomial will be the highest degree term, 800x55. Since the exponent is an odd integer and the coefficient is a positive number, this term and the whole polynomial will be positive for x large positive (this one term will "overwhelm" the other terms) and this term and the whole polynomial will be negative for x large negative (this one term will "overwhelm" the other terms). So the Intermediate Value Theorem states that the graph must cross the x-axis somewhere.

A computer program found a 20 digit approximate root of this polynomial (-.96533812077786645838) in less than a fifth of a second, mostly using the ideas we just learned.

Here is maybe a simpler example, maybe. Look at f(x)=x3+cos(13x-9x2). This is a continuous function. What can be said about the sign of f(2), say? Since 23=8 and the most/least any value of cosine can be is +1/-1, I bet that f(2) is at least 7. Similarly, f(-2) can differ by at most 1 from (-2)3=-8, so I bet that f(-2) must be at most -7. Hey: on the interval [-2,2] f(x) changes sign. Since it is continuous it must have at least one root.

To the right is a graph of f(x) on the interval [-2,2]. It actually has lots of roots. Well, "lots" and "at least one" are logically consistent.

Major definition A function f(x) is differentiable if the limit limh-->0[f(x+h)-f(x)]/h exists. If this limit exists, it is the derivative of f(x), and written f´(x). This is read as "eff-prime of x".

The absolute value function
I looked at f(x)=|x|. Here a piecewise linear definition of |x| reads as follows:

```     / x if x≥0
|x|=<
\ -x if x<0```
f(x) is continuous.
Well, the most "interesting" point is 0. To check continuity, I should look at the "handed" limits and the value at 0 of f(x).
limx-->0-f(x)
This is limx-->0--x because we use the first alternative since here x is negative. But certainly the limit limx-->0--x exists and is 0.
limx-->0+f(x)
This is limx-->0+x since the second part of the definition must be used. But limx-->0+x exists and is 0.
f(0)=|0|=0.
Since the two limits exist and they agree with the value of f(0), we know that f(x) is continuous at 0.

f(x) is not differentiable at 0.
I am supposed to look at the limit defining f´(0). This is limh-->0[f(0+h)-f(0)]/h. I'll look at the two one-sided limits separately.
limh-->0-[f(0+h)-f(0)]/h
f(0+h)=f(h), and since h<0 (the left-hand limit) f(h)=-h. Also f(0)=|0|=0. Therefore limh-->0-[f(0+h)-f(0)]/h= limh-->0--h/h=-1.
limh-->0+[f(0+h)-f(0)]/h
f(0+h)=f(h), and since h>0 (the right-hand limit) f(h)=h. Also f(0)=|0|=0. Therefore limh-->0-[f(0+h)-f(0)]/h= limh-->0-h/h=1.
Since -1 and 1 aren't the same (!) the derivative of f(x) at 0 does not exist.

I'm more of a picture person than an algebra person, so let me show you on a graph what was computed above. Here's a graph of f(x)=|x|. The graph is unbroken, no jumps or breaks, providing evidence that the function is continuous. Near 0, if we look at the slopes of secant lines from the left and from the right, we get completely different pictures. The secant lines from the left all have slope -1, while those from the right all have slope +1. There's no agreement or "approach" of the slopes as the points on the ends of the line segments travel towards 0. This means there is not a unique tangent line, and there is no good unique candidate for the slope of the tangent line. Since that is supposed to be the derivative, there's no f´(0).

Of course as I mentioned in class you are welcome to look at the more complicated but certainly real-world function, the U.S. individual rate schedule for 2006. As a matter of public policy, the function involved is continuous. Due to the progressive and piecewise linear nature of the function there, the function is not differentiable at 5 income values.

If a function is differentiable, then we know that limh-->0[f(x+h)-f(x)]/h=f´(x). This statement may no longer be precisely correct if "limh-->0" is omitted. I tried to illustrate this by looking at f(x)=x2, so f(x+h)=(x+h)2=x2+2xh+h2. Then the fraction [f(x+h)-f(x)]/h becomes exactly

```(x2+2xh+h2)-x2   2xh+h2
------------- = ------ = 2x+h.
h          h```
But f´(x) is 2x, so there's an "error".

In fact, generally there will be an error. So what we have is
[f(x+h)-f(x)]/h=f´(x)+Err(h)
where Err(h) is some sort of error function which -->0 as h-->0. We could then multiply the equation by h and get
f(x+h)-f(x)=f´(x)h+Err(h)h
and add f(x) to both sides and get
f(x+h)=f(x)+f´(x)h+Err(h)h
Certainly you should see, I hope, I hope, again something which was mentioned repeatedly last week. If we kick or perturb the input to f(x) and if we know that f(x) is differentiable, then the kicked or perturbed output will be the old output plus a term directly proportional to the disturbance (this is what's called in economics the marginal response) plus, well, Err(h)h. Since I think of h as small and Err(h) is also small, well, the product is even smaller still. That's the important idea.

Look at f(x+h)=f(x)+f´(x)h+Err(h)h and make h-->0. On the right-hand side, f(x) doesn't change, but both f´(x)h and Err(h)h go to 0. So the right-hand side -->f(x). Goodness, this means the left-hand side, f(x+h), also -->f(x) as h-->0.

A differentiable function must be continuous. BUT there are continuous functions which are not differentiable (such as |x| and the U.S. tax function).

The converse of a statement is not necessarily true even if the statement is true. All frogs are green, but not all green things are frogs (I rarely see grass hop, for example).

Now we can start a feast of computational methods for derivatives or I can tell you about the ideas. Most books begin by computing everything in sight and so will I.

f(x)=x80
Now [f(x+h)-f(x)]/h==[(x+h)80=x80]/h. Now I ain't gonna multiply out
(x+h)(x+h)(x+h)(x+h)····(x+h)
(there are 80 factors) because, realizing that there are 2 choices to make inside each parentheses pair, there would be a total of 280=12089 25819 61462 91747 06176 terms. This is too many even for maniacs. So let's organize things a bit.
We can see what happens if we always choose x's. There will be 80 x's multiplied together: x80.
We could choose h in the first term and x's in all the others. The result will be hx79 (there always will be 80 "things" multiplied together. Or we could choose an x in the first term, then an h, and then all x's. This gets xhx78 which is the same as hx79. In fact, if you think it through, you can choose one h exactly one time from each factor with all the rest x's. There are 80 factors, so there will be a total of 80 terms with one h and 79 x's: 80hx79.
All of the other terms have at least two h's.

Here is how I want to organize (x+h)80:

```        Terms with   Terms with       Terms with
no h's       one h        at least two h's
(x+h)80  =   1x80   +   80hx79   +    h2(JUNK)
```
If the terms have at least two h's then I can certainly "factor out" h2. I don't really care about the JUNK, and you will see why shortly.

```
f(x+h)-f(x)    x80+80hx79+h2JUNK-x80
----------- = ----------------------
h                 h```
Now the x80's cancel. And we can factor h's out of the top and bottom and cancel, and the result is 80x79+h(JUNK) and as h-->0, the JUNK term drops out.

The function f(x)=x80 is differentiable, and its derivative is f´(x)=80x79.

Well, now let's (metaphorically!) go out into the woods. Please imagine we meet yet another pattern beast (mostly green -- it must be a frog). This beast would eat 79's and 80's and excrete n-1's and n's. If we let the beast munch on what was just written, it would become a verification of the following:

The function f(x)=xn is differentiable, and its derivative is f´(x)=nxn-1.
This is amazing, and gives us a very nice template to compute bunches of derivatives.

Write an equation for a line tangent to y=x4 when x=2
I can write such an equation if I know a point the line goes through and if I know the slope of such a line. Well, if x=2, then y will be 24=16. And now you should remember that one "interpretation" of f´(x) is as the slope of the tangent line to y=f(x) at x. If f(x)=x4, then f´(x)=4x3. When x=2, this is 4·23=32. So an equation for the tangent line is y-16=32(x-2). If you want the equation in slope-intercept form, I guess the answer becomes y=32x-48.

To the right is a picture of part of the curve y=x4 and part of the line y=32x-48. The line certainly seems to be tangent to the curve.

We further deduced two more rules for differentiation. Here they are:

Multiplication by a constant
If c is a constant and f(x) is a differentiable function and its derivative is f´(x), then the function F(x)=cf(x) (this is f(x) multiplied by the constant c) is differentiable, and F´(x)=cf´(x).

Sum of differentiable functions
If f(x) is a differentiable function and its derivative is f´(x), and if g(x) is a differentiable function and its derivative is g´(x), then F(x)=f(x)+g(x), the function defined by taking the sum of f(x) and g(x), is also a differentiable function, and F´(x)=f´(x)+g´(x)

These "rules" allow us to find the derivatives of any polynomial. For example, the derivative of 12-8x+15x3+166x12+800x55 (a polynomial we met earlier today) is 0-8x0+15(3x2+166(12x11)+800(55x54).
I inserted a "zinger" here: the derivative of 12 is 0. In fact, the derivative of any constant is 0. The graph is a horizontal line, and horizontal lines have slope 0, and slopes are values of derivatives. Also I should remark that I am lazy and I wouldn't compute 166·12 or 800·55 unless I needed to.

One last "reward"
To the right is a graph of y=x2-5x3 for x between -.15 and .25. There is a quite delicate little bump. Where is the delicate bump? Well, the delicate bump (a local maximum, not an absolute maximum because there are points on the curve which have larger y values) occurs where the derivative is 0: that's where the tangent line is horizontal. The derivative of x2-5x3 is 2x-15x2=x(2-15x). One root is x=0 and the other comes from 2-15x=0 or 2=15x or 2/[15]=x. And so the delicate bump is at (2/[15],(2/[15])2-5(2/[15])3). I can honestly declare that I don't know any faster or simpler way to locate this point.

Monday, July 3
I defined continuous. We went over the review problems. The semiexam was given.

 Most Math 135 students probably will regard what's done today as somewhat ugly and maybe even distasteful. I apologize, but the intricacy and stringency (which I mostly will dilute!) is part of the culture of math. If I don't display this, my epaulettes may be removed.
Thursday, June 29
 I feel that I have still not convincingly shown students in this class why the language and methods of calculus can be truly relevant and useful. Yes, some aspects of the intellectual game and on disply, but not (yet!) the relevance.

Limits: our obligation to math culture
I would like to display part of the culture of mathematics: the deduction of examples for collections of results. The structure of deductions is an important ingredient of mathematics (not the only one, certainly, but rather important). This is what makes the transition from heuristic to accepted knowledge valid.

In any case, part of taking a math course is to learn how a mathematician "thinks", just as in a history course students should expect some exposure to certain methods of historians. So let's go.
Logic? We don't need no steenkin' badges logic ... Let's just guess!

Algebraic rules
Here are some statements about how limits interact with algebra:

A1 Suppose that limx-->cf(x) exists and equals L1 and that limx-->cg(x) exists and equals L2 then limx-->cf(x)+g(x) exists and equals L1+L2.

A2 Suppose that limx-->cf(x) exists and equals L1 and that limx-->cg(x) exists and equals L2 then limx-->cf(x)·g(x) exists and equals L1·L2.

A3 Suppose that limx-->cf(x) exists and equals L1 and that limx-->cg(x) exists and equals L2 . Also suppose L2 IS NOT 0 then limx-->cf(x)/g(x) exists and equals L1/L2.

These results are true. They can be given detailed proofs using a carefully formulated definition of limit. I don't believe that such detailed proofs are a proper part of Math 135, but there are a number of math courses which do have such discussions. I hope that you are willing to believe if two functions have limits as x approaches c, so they each get close to their limits, then their sum gets close to the limits of the sums. The restriction on L2 in A3 is needed because I don't want to worry about division by 0!

"Doing" the homework problem
One of our homework problems (#2 from chapter 3) asks for the limit of x2+4 divided by x+8 as x-->2. If I wanted to be rather formal I could begin with the observations that limx-->2x=2 (as x gets close to 2, then x gets close to 2 [!]) and limx-->24=4 (the function whose output is always 4 is always close to 4 [11]). Then:
By A2, limx-->2x2=(limx-->2x)·(limx-->2x)=2·2=4.
By A1, limx-->2x2+4=(limx-->2x2)+(limx-->24)=4+4=8.
Again by A1, limx-->2x+8=(limx-->2x)+(limx-->28)=2+8=10.
Finally, by A3, limx-->2(x2+4)/(x+8=(limx-->2x2+4)/(limx-->2x+8)=8/(10). The citation of A3 is valid since 10 is not 0.
This sort of approach is maybe interesting, but (to me) it is hardly interesting more than once. Heh, heh.

Polynomials and "plugging in"
Well, suppose p(x) is a polynomial function of x. What is this? This means that p(x) is a (finite) sum of (integer) powers of x possibly multiplied by real coefficients. Then repeated use of A1 and A2 implies limx-->wp(x)=p(w). That is, a limit of a polynomial can always be obtained by just "plugging in" the target value of the limit.
Example If p(x)=33x47-18x12+13.046, then limx-->79p(x)=33(7947)-18(7912)+13.046. (I don't want to "simplify", please!)

Rational functions and "plugging in"
A rational function is a quotient of polynomials. So a rational function, r(x), is p(x)/q(x) where p(x) and q(x) are polynomials. Suppose that w is a number so that q(w) is not 0. Then limx-->wr(x)=p(w)/q(w). Of course we need the restriction that q(w) not be 0 in order to use A3.
Example If r(x)=(3x4-5x+777)/(44x12+88). I chose the bottom because 12 is even and 88 is positive, so there is no way that the bottom can be 0 for any value of x. Then limx-->-211r(x)=(3({-211}4)-5{-211}+777)/(44({-211}12)+88).

A first strategy: "plugging in"
To me this implies a first strategy for limits. I look at limx-->wf(x), for almost any kind of x, and I try to see what happens if I "plug in" w for x in f(x). What is f(w)? Does it make sense (in math language, is w in the domain of f(x)?)? Then I'd try to convince myself that f(w) is "obviously" ("clearly"?) somehow connected with the f(x) values for x close to w. All this doesn't necessarily need to be true, but it is one strategy, a first strategy.

Let's look at limh-->0{sqrt(5+h)-sqrt(5)}/h. If I just try to generalize the plug in strategy recommended previously, well, then I'll get 0/0, and we'd better not divide by 0.

An apparently intractable problem is tamed
So look at {sqrt(5+h)-sqrt(5)}/h. We want to see what happens when h gets small. Maybe some algebraic "stuff" can help. You may vaguely remember can candidates for action here. The top is the difference of square roots, and maybe the sum of square roots can help. That is, consider

```(A-B)   (A-B)   (A+B)    A2-B2
----- = ----- · ----- = ------
C       C     (A+B)   C(A+B)
```
If we apply this "template" to our situation, here is what happens:
```{sqrt(5+h)-sqrt(5)}   {sqrt(5+h)-sqrt(5)}   {sqrt(5+h)+sqrt(5)}  (sqrt(5+h))2-(sqrt(5))2
------------------- = ------------------- · ------------------ = -----------------------
h                     h             {sqrt(5+h)+sqrt(5)}   h{sqrt(5+h)+sqrt(5)}

```
The square roots and the squares cancel, so the result is
```     (5+h)-5                    h                       1
--------------------- = -------------------- = -------------------
-h{sqrt(5+h)+sqrt(5)}   h{sqrt(5+h)+sqrt(5)}   {sqrt(5+h)+sqrt(5)}```
and the horrible h in the bottom cancels with the h remaining on the top.

What happens to 1/{sqrt(5+h)+sqrt(5)} as h-->0? Well, here I think that algebraic results like A1, A2, etc. help. The result is that we can just plug in, and the limit is 1/{sqrt(5)+sqrt(5)}=1/{2sqrt{5}}.

Limit strategies, part II
First try to "plug in". If that doesn't work, try to use algebraic methods of rewriting things in order to get some equivalent form which allows the limit to be recognized easily (by "plugging in"!).

So, what about the patterns (another part of math culture)
So far what you may have seen in Math 135 about the culture of mathematics includes logical deduction and drawing pictures (I am a picture person, actually). But a good deal of mathematics consists of computing examples, sometimes in an almost drearily repetitive way, and then looking at the results and trying to discern patterns.

Here we know that

```      sqrt(5+h)-sqrt(5)      1
lim   ----------------- = --------
h-->0         h           2sqrt(5)```
What if we wanted to know the limit of {sqrt(3+h)-sqrt(3)}/h as h-->0?
We could imagine an animal roaming the woods. It might eat 5's which are inside square roots, and then excrete (I'm being polite, graphic, but polite!) 3's inside the same square roots as the 5's. So any algebraic mess which had a 5 inside the square root becomes an algebraic mess with the 5 changed to a 3.

This is a metaphor, darn it! Actually the large animals roaming the woods on the Busch campus are mostly deer, too many deer.

Now suppose we allow this animal to browse on our previous collection of formulas. This is what it would eat:

```{sqrt(5+h)-sqrt(5)}   {sqrt(5+h)-sqrt(5)}   {sqrt(5+h)+sqrt(5)}  (sqrt(5+h))2-(sqrt(5))2
------------------- = ------------------- · ------------------ = -----------------------
h                     h             {sqrt(5+h)+sqrt(5)}   h{sqrt(5+h)+sqrt(5)}

(5+h)-5                    h                       1
--------------------- = -------------------- = -------------------
-h{sqrt(5+h)+sqrt(5)}   h{sqrt(5+h)+sqrt(5)}   {sqrt(5+h)+sqrt(5)}```
```      sqrt(5+h)-sqrt(5)      1
lim   ----------------- = --------
h-->0         h           2sqrt(5)```
And this is what it would excrete:
```{sqrt(3+h)-sqrt(3)}   {sqrt(3+h)-sqrt(3)}   {sqrt(3+h)+sqrt(3)}  (sqrt(3+h))2-(sqrt(3))2
------------------- = ------------------- · ------------------ = -----------------------
h                     h             {sqrt(3+h)+sqrt(3)}   h{sqrt(3+h)+sqrt(3)}

(3+h)-3                    h                       1
--------------------- = -------------------- = -------------------
-h{sqrt(3+h)+sqrt(3)}   h{sqrt(3+h)+sqrt(3)}   {sqrt(3+h)+sqrt(3)}```
```      sqrt(3+h)-sqrt(3)      1
lim   ----------------- = --------
h-->0         h           2sqrt(3)```

Another animal
We could imagine another animal eating the same formulas and changing the 5's to 11's. Etc.

A fairly general animal and a result
The pattern recognition is most interesting. What about the animal that would take in formulas with 5's and excrete formulas with x's, say? It would prove (oops!) the following result:
If x is a positive number, then

```      sqrt(x+h)-sqrt(x)      1
lim   ----------------- = --------
h-->0         h           2sqrt(x)```
and we would have verified that if f(x)=sqrt(x), then f´(x)=1/{2sqrt(x)}. Later, later, later.

Order rules and limits
In this case the word "order" doesn't refer to "command" but to "less than ... greater than" type of order. Here is one generally recognized limit rule:
L Suppose that f(x)<g(x) for all x's near (but not necessarily equal to) c, and that limx-->cf(x) exists and equals L1 and that limx-->cg(x) exists and equals L2. Then L1<L2.

People actually use this sort of result frequently in daily reasoning. Rates (as derivatives) are certain kinds of limits. If one faucet and hose arrangement takes 10 minutes to fill a container, and another takes 10 minutes, then I bet the rate of water flow through the first is less than the flow rate through the second. I admit that this observation isn't profound, but ... it is used. Somewhere in your brain there is a mental arrangement, a framework, comparing rates, and this is essentially L. We will use L at various times in the course to get some approximate information about rates.

Example Suppose I know that the values of f(x) are all between -17 and 504. Suppose I also know that, for example, limx-->32f(x) exists. Then that limit must be between -17 and 504.
Please notice that there are functions whose values are between -17 and 504 for which the limit does not exist, such as the piecewise defined function

```       / -5 for x<32
f(x)= <  12 for x=32
\ 22 for x>33```
So you can't conclude anything quantitative about limx-->32f(x) using L because one of the hypotheses of L is not ture or valid or fulfilled: the limit doesn't exist.

The squeeze theorem, a very special result
The squeeze theorem is a very special result which is vital to the success (!) of calculus. It will look quite weird if you have not seen it before. The hypotheses require information about three functions.
The Squeeze Theorem
Hypotheses
The limit of the function f(x) as x-->a exists and is the number L.
The limit of the function h(x) as x-->a exists and is the number L.
These two limits agree.
For x's which are not equal to a, we know that f(x)g(x)h(x).
Conclusion
The limit of the function g(x) as x-->a also exists, and is also equal to L.

You'll have to see how this is used in several important applications. Here, though, is a rather contrived
Example Suppose I know that the function Q(x) is always between -5 and +8. What can I say about the limit as x-->4 of Q(x)? What about the limit as x-->4 of 33+(x-4)Q(x)?
Well, there is not much which can be said, based on the information given, about the limit of Q(x) as x-->4. For example, to the left of 4, Q(x) could be -2 and to the right of 4, it could be +2 (since both -2 and +2 are between -5 and +8, this is a possible example). In this case, the limit of Q(x) as x-->4 certainly does not exist.
What happens to 33+(x-4)Q(x)? Well the largest it could be is 33+|x-4|8, and the smallest it could be is 33+|x-4|(-5). Those are the top and bottom possibilities for the values of Q(x). But examine the top, 33+|x-4|8, as x-->4. This must go to 33. The bottom, 33+|x-4|(-5), as x-->4, also approaches 33. So 33+(x-4)Q(x) is squeezed or sandwiched between two functions which both approach 33 as x-->4. Therefore according to the squeeze theorem, the limit of 33+(x-4)Q(x) as x-->4 must also exist, and it must be 33.
The factor of (x-4) as x-->4 goes to 0. When it multiplies Q(x), since we know Q(x) doesn't get very big either positive or negative, the result also goes to 33. So the 33 remains alone.

Sine
We will eventually want to find the derivative of sin(x). In order to do this, we'll need to handle the limit as h-->0 of {sin(x+h)-sin(x)}/h. Here is, I think, just about the only time in the course where we need the addition formula for sine. It turns out that sin(x+h)=sin(x)cos(h)+sin(h)cos(x). Therefore

```sin(x+h)-sin(x)   sin(x)cos(h)+sin(h)cos(x)-sin(x)
--------------- = --------------------------------
h                         h```
The top of the fraction is sin(x)cos(h)+sin(h)cos(x)-sin(x) which is supposed to help you "organize" the fraction. The fraction is equal to
```      ( cos(h)-1)         (sin(h))
sin(x)(---------) + cos(x)(------)
(    h    )         (   h  )```
Honestly it isn't supposed to be "clear" why the fraction is rewritten or reorganized like this. One advantage, easy to say, is that writing the fraction this way sort of "separates" (decouples!) the h and x parts of the fraction. We can study them separately.

I'll concentrate right now on the behavior of sin(h)/h as h-->0. I had student volunteers show some graphs produced by their graphing calculators of sin(h)/h for h between -5 and 5. Here are the graphs similar to what they got:
sin(h)/h in radians, -5≤h≤5 sin(h)/h in degrees, -5≤h≤5
If you use radians to measure angles, the behavior of sin(h)/h as h-->0 is very neat: it seems to approach 1. If you use degrees to measure angles, the behavior of sin(h)/h as h-->0 is ... well, the number it seems to get close to is something like .01745. If you want to look at the rate of change of trig functions, then degrees will give you formulas with lots of .01745's. Radians will give you formulas with 1's (and therefore, since the 1's will turn out to be multiplicative factors, they won't even be seen!).

A sine limit
It is true that limh-->0 sin(h)/h exists and equals 1 if angles are measured in radians. The numerical simplicity of this result is the chief reason that everyone who studies how things change measures angles in radians rather than in degrees, wombats, or glinkas. (The last two are angle measures for extinct civilizations.) I did not go further in class, but here is further discussion of the limit, including geometric reasoning supporting the statement.

A vote on angle measurement
We voted, 14 to 1, to use radian measure for angles in this course. Sigh: there's alway someone.

Some more limits

• limh-->0sin(h)/(5h)
```sin(h)   ( 1 )   (sin(h))
------ = (---) · -------- --> (1/5)(1)
5h     ( 5 )      h```
as h-->0. So this limit exists and is 1/5.
• limh-->0sin(7h)/h
Here the 7 is inside the function sine, and I can't just pull it out. Why? Because. If you don't believe me, my calculator says that sin(14)=.99061 and sin(2)=.90930 and 7sin(2)=6.36508. So sin(7·2) and 7sin(2) are very different. sin(7h) and 7sin(h) are not necessarily the same!

But if h is very small, 7h is also very small. So I am willing to believe that the ratio sin(7h)/(7h) is close to 1. Therefore I write the following:

```sin(7h)   ( ?  )   (sin(7h))   ( 7 )  (sin(7h))
------- = (----) · --------- = (---) ·--------- --> (7/1)(1)
h      ( ?? )     (7h)      ( 1 )      7h

```
as h-->0. So the limit exists and is 7: limh-->0(sin(7h)/h=7.
• limh-->0sin(33h)/(56h)
Here the limit exists and is (33)/(56): you've got to combine the previous two ideas.
• limh-->Pi/2sin(h)/h
This is a trick question. You are supposed to believe that this is the limit as h-->0 and it is not. In this case, since the bottom (denominator) goes to Pi/2, which is not 0, we can just "plug in". Remember, if you can, that sin(Pi/2)=1, so that this limit is sin(Pi/2)/[Pi/2]=1/[Pi/2]=2/Pi.

Left- and right-hand limits
Real functions in real life have different behavior on the two sides of a point. Look at the income tax function. To the left of \$15,1000, the tax rate is 10% and to the right it is 15%. There is mathematical language to describe this sort of behavior. The language is left- and right- hand limits.

Each of these has the geometric and algebraic definition of limit but restricted to one side or the other. The left-hand limit of f(x) as x approaches c is usually denoted limx-->c-f(x). For consideration of this limit, we are supposed to restrict our attention to x's on the left (negative) side of c . Dually, the right-hand limit of f(x) as x approaches c is usually denoted limx-->c+f(x). For consideration of this limit, we are supposed to restrict our attention to x's on the right (positive) side of c.

A graphical example
Let's look at f(x), a function whose graph is shown to the right. The graphical "notation" which is commonly used is an empty circle where there is no point of the graph (but where, perhaps, your intuition (?) might want there to be one. A solid circle (maybe a thickened-up point) shows a value on the graph. I'm particularly interested in describing what happens at and nearby to -2, 1, 2, and 3. Some of what is describe is "delicate" behavior. In the table below, the entry DNE abbreviates "Does Not Exist." So either the function value doesn't exist (the domain does not include that value of x) or the limit doesn't exist.

c= f(c) Left-hand limit
limx-->c-f(x)
Right-hand limit
limx-->c+f(x)
Two-sided limit
limx-->cf(x)
-22111The graph approaches the point (-2,1) from both sides, but the function value is in the "wrong" place!
1DNE222There doesn't seem to be a value for the function at 1, but the limits are go to 2 there.
2-113DNEFrom the left the limit is 1 and from the right the limit is 3. The two values don't agree, so the two-sided limit which we originally considered doesn't exist.
3220DNEThe two-sided limit doesn't exist since the two one-sided limits disagree.

Prescribing behavior
We could also go the other way. That is, we could "prescribe" various one- and two- sided limits and ask for graphs.

Wednesday, June 28
Some discussion of the homework

Another example
We could imagine that s(t) is the position on the x-axis at time time. Wait, better: we drive along the highway, and s(t) is the distance we've driven on the highway at time t. If we assume we don't drive backwards, then the distance we've driven is actually equal to our position (displacement) on the highway.

Average rate of change to instantaneous rate of change
The average rate of change between the times t and t+h is {s(t+h)-s(t)}/h. Then as h-->0, this approaches the instantaneous rate of change.

Average speed/velocity to instantaneous velocity
The quantity {s(t+h)-s(t)}/h is called, in this physical context, the average speed (or, more correctly, velocity) over time interval [t,t+h]. The limit as h-->0 is called the instantaneous velocity at time t. Interestingly, the speedometer in cars measures average velocity for very small h's, using either a mechanical or an electronic device. This measurement is supposed to approximate the ideal quantity, instantaneous velocity.

Definition of derivative
The derivative of a function, f(x), is another function, f´(x), which is defined by the equation
f´(x)=limh-->0{f(x+h)-f(x)}/h.
This is our first view of one of the major "objects" (ideas?) of the course, and I believe some familiarity is needed to internalize the definition and really use it.

Discussion
Let's see: there are three parts.
f´(x) This is just more mysterious mathematical notation. It is read as "eff prime of ecks". It is intended to represent what is on the other side of the equation.
{f(x+h)-f(x)}/h This is what I called "the average rate of change of f(x) over the interval whose end points are x and x+h". It can represent all sorts of real quantities of interest. It is my job to try to convince you of this with more examples.
limh-->0 Well, this represents ... this represents ... uh, the "instantaneous rate of change of f(x) at x": that just names the thing. It doesn't really tell me what is going on. But maybe I should first discuss an example of the derivative in a bit more detail. And I also would like to discuss it from several different points of view.

f(x)=1/x; f´(3) algebraically
According to the definition (can't argue with a definition, except to suggest that it be modified or replaced, and after about three centuries and millions and millions of people using the definition, replacement seems unlikely!)
f´(3)=limh-->0{f(3+h)-f(3)}/h.
In order to analyze this further, we write f(3+h)=1/(3+h), and then f(3+h)-f(3)={1/(3+h)-1/3}. I can combine fractions and get f(3+h)-f(3)={[3-(3+h)]/[(3+h)3]}. Please: it is very easy to mess up signs in this computation. I like to use lots of parentheses. There is some simplification which can be done:
{[3-(3+h)]/[(3+h)3]}={-h/[(3+h)3]} (the 3's cancel in the top).
Now we will need to handle a compound fraction. The scheme looks like this:

```   A
---      A     1     A
B    = --- · --- = ----
-------    B     C     BC
C```
So
```                   -h
--------
f(x+h)-f(x)      (3+h)3          -h
----------- = ------------ = ------------
h              h          h·(3+h)3
```
Now the h's top and bottom cancel, and we know
{f(3+h)-f(3)}/h=-1/[(3+h)3].
As h-->0, the only thing that changes is on the bottom. I bet that 3+h-->3. Therefore, as h-->0, {f(3+h)-f(3)}/h-->-1/(3·3)=-1/9. The derivative of f(x) at x=3 is -1/9 (for this function).

Intellectually curious people can ask many questions about this algebraic feast (too much: a mental stomachache!). Why do it? (Answer: yes, there are easier "ways" but more sophistication is needed.) A more interesting question, to me, is why is the answer -1/9? Here there are a variety of answers. One answer is, well, dummy, I just showed you. I transformed the original definition using various algebraic "tricks" until I could get a formula where the limit was more easily recognized. That's a sort of ground floor answer. But what is the significance of the minus sign (if any!) and what about the size (1/9). Well, here are some pictures.

f(x)=1/x; f´(3) geometrically
Here is part of the graph of y=1/x. Suppose P is the point (3,1/3), and Q is the point (3=h,1/(3+h)). Both P and Q are on the graph. I want to think of h as small and positive, maybe h very small (!). The straight line through P and Q is called a secant line of this graph. I'll call its slope msec. I can write this slope in terms of the coordinates of P and Q: it is the difference in the second coordinates divided by the difference in the first coordinates. This turns out to be (no: there aren't any accidents here -- several centuries have gone by with clever people working to make sure that seeming coincidences really happen)

```    1        1
-----  -  ---
3+h       3
-----------------
(3+h) - 3```
This fraction is the same as the average rate of f(x) over the interval [3,3+h]. Now look closely at the picture, because to me, what I am about to describe is difficult to see. When h "gets" smaller and smaller (positive, say) then the point Q "travels" on the curve y=1/x closer and closer to the point P. The QP line seems to more and more look like the line tangent to y=1/x at P. If this is so, then the slope of the secant line, msec, should get closer and closer to the slope of the tangent line, mtan. The limit of msec should be mtan. What we did before should convince you that mtan is actually -1/9.
We can actually write the equation of the line tangent to y=1/x at the point P. We just need some "data": say, a point on the line, and the slope of the line. We've got this:
Point: (3,1/3). Slope: -1/9.
(x,y) will be on the line when [y-(1/3)]/[x-3]=-1/9. so that y-1/3=-1/9(x-3) or y-1/3=-(1/9)x+1/3 or y=-(1/9)x+2/3.

The graph to the right was done by Maple. It shows a portion of y=1/x and the line y=-(1/9)x+2/3. They are not intended to "prove" anything, but rather to help support the preceding discussion. I hope the line does look tangent to the curve at the point where x=3, though. The minus sign in -1/9 occurs because the curve is decreasing as x is increasing. The magnitude of -1/9, that is, 1/9, is not so big: the curve is fairly "flat" around (3,1/3).

f(x)=1/x; f´(3) numerically
Some people like pictures (me!), and some might not like algebra (also me!), but some people do like lots and lots of numbers. So let me try to tell you a numerical interpretation of f´(3).

I tried to convince you that linear functions (functions whose graphs are straight lines) take input differences and multiply them by a fixed number (which happens to be m, the slope). NONlinear functions aren't as nice. The "local multiplier" is {output difference} divided by {input difference}. If h is small, the input difference is x+h minus x, or h. The output difference is f(x+h)-f(x). And, goodness!, the local multiplier is the quotient {f(x+h)-f(x)}/h (this should look familiar!). If h is small, if f(x) is 1/x, if x is 3, then this local multiplier is pretty darn near -1/9. This means f(3+h) should be "close" to (1/3)+(local multiplier)(h), which is (1/3)+(-1/9)h.

Here I show the function machine for f(x)=1/x. (The machine has gears inside, of course! A few of them are showing.) As an example, I computed f(3+.00007). This is 0.33332 55557 to 10 decimal place accuracy. (The purpose of the space is just to help you "see" the darn numbers better.) And (1/3)+(-1/9).00007 is 0.33332 55556. So there is agreement up to one digit in the tenth decimal place (a discrepancy of 10-10). Whether this is interesting or not to you is certainly a matter of personal taste. But it should indicate that maybe for more complicated functions we will be able to compute small perturbations "easily" if we understand f´(x).

This is not enough!
Although these three explanations (?) of f´(x) are a beginning, I still do not believe that I've even begun to approach some sort of realistic payoff for Math 135. I still need to do this.

What's a limit?
The part of the definition of derivative which Here is the definition written by Gootman.

The expression limx-->cf(x)=L means that as x gets closer and closer to c, through values both smaller and larger than c, but not equal to c, then the values of f(x) get closer and closer to the real number L.
This is a version of the definition which took more than 100 years to be generally accepted. I would like to investigate the meaning of the definition by considering a numbers of examples. By the way, the phrase " limx-->cf(x)=L" is usually read as "The limit of f(x) as x approaches c is L."

Easy
I bet limx-->3x2=9. I think this should be emotionally (?) correct. If x is very close to 3 (though not exactly 3) then x2 will be very close to 9. The "=9" in the preceding line of algebra really doesn't just represent equality: it really represents a more complicated logical situation, that the assertion in the definition is fulfilled. You get the range variable, x2, close to 9, by taking x close to 3. Any output tolerance can be obtained by selecting the input close enough.

Another step
In fact, I bet that limx-->6x2=36 and limx-->-5x2=25 and ... This function is maybe too easy.

Change it a bit
Let me define f(x) in a piecewise fashion. Here f(x)=x2 if x is not equal to 1, and f(1)=-2. (Don't complain -- the income tax function is much more complicated!) A chunk of the graph of f(x) is shown to the right. It is a parabola with a hole in it, and the value for x=1 moved to (1,-2). The conventional way to show a "missing value" (hole?) is to draw a circle (an empty circle?) at the point. The conventional way to show a point that is on the graph and otherwise might be missed is to draw a sort of fat point (a filled circle?).

Well, what is limx-->3f(x)? Let's look at values of x close to 3. The definition declares we should examine x's which are "closer and closer to c [c is 3 here], through values both smaller and larger than c [c is 3 here] ..." For close x's, f(x) is actually x2. That means we could just consider x2 in the limit statement, and then the limit statement (as x-->3, x2-->9) is correct if we write limx-->3f(x)=9. O.k.: that seems good. I hope.

What is limx-->1f(x)? Here we have the whole, and the "moved" value of x2. But the definition declares we should look at " values both smaller and larger than c [c is 1 here], but not equal to c [c is 1 here] ..." In this limit statement, we should not consider f(1). Therefore f(x) will be x2! And our limx-->1f(x) is logically the same as limx-->1x2. That value is 1.

The aliens attack ...
The aliens attack (Babylonians?) and they strike at one of the great prizes of the human race, the squaring function: f(x)=x2. Here is what they do: they take 318 different values of x, and for those values of x (as inputs to their version of f(x)) they change the outputs to -9. I will call this alien version of the squaring function, A(x). Notice that I haven't told you which values of x the aliens have changed. Weird, huh? Not very much information. The graph shown maybe gives you some idea, but I have intentionally removed horizontal axis scaling, because we don't know where the holes are.

What can we say about limx-->3A(x)? Well, let me give two arguments.

• Suppose 3 is one of the input values whose output value has been altered by the aliens. But for the purposes of computing limits, I don't care about what happens at 3 ("but not equal to c"). Now close to 3, if we are close enough, close enough, then (since there are only 317 other alien-touched numbers) I can get close enough to exclude these alien-contaminated numbers. So close enough, and not at 3, A(x)=x2. Hey, then we need to "compute" (really better written, we need to think through) limx-->3A(x)=limx-->3x2. This is 9.
• Suppose 3 is not one of the input values whose output value has been altered by the aliens. But who cares (from the point of view of computing limits). If I am interested in limx-->3A(x) I just should pay attention to an interval around 3 which excludes all (318) of the numbers that the evil (did I mention they were evil, and also slimey?) aliens fooled with. So I look close enough to 3, and again, there I find A(x)=x2. Therefore, limx-->3A(x)=limx-->3x2 and this is again 9.
In any case, A(x)=9. In fact, we can easily repair (?) the damage that the aliens did to the squaring function. I claim that for any real number, w, limx-->wA(x)=w2.
This still seems slightly confusing to me, and maybe also to you. Please think about it a bit.

Limits may not "exist"
 Of course graphs may be more complicated. For example, the limiting behavior on two sides may not agree. This can be illustrated with something as simple as the piecewise defined function f(x) which is 0 for x<0 and 1 for x>0. Then the two pieces won't meet up. But in fact situations without limits may be considerably more bizarre. The function pictured is a piecewise linear function which jumps infinitely often between 1 and 0. I don't think I want to describe this with too much precision, but you should appreciate that such things "exist". We won't meet functions with infinitely oscillatory behavior on finite intervals in Math 135, but they do occur in certain applications.

Tuesday, June 27
Analytic geometry
Analytic geometry, that is, geometry combined with algebra, is a very useful tool to use while learning calculus. So let me very rapidly review what we should know.

Distance
If (x1,y1) and (x2,y2) are the coordinates for two points in the plane and D is the distance between the points, then (Pythagoras again):
D2=|x2-x1|2+|y2-y1|2
because D is the hypotenuse of the right triangle shown. If we take square roots, and realize that D, a geometric distance, should be non-negative, and also realize that the darn absolute values are unneeded because of the squaring, we get a formula:
D=sqrt((x2-x1)2+(y2-y1)2)
I remarked in class, and I repeat, that using lots of parentheses to avoid confusion is frequently useful!

Example
The distance between (2,5) and (-2,-7) is sqrt((2-{-2})2+(5-{-7})2)=sqrt(42+122)=sqrt(160).

Circles
Suppose the distance from (x,y) to (2,5) is sqrt(160). This translates geometrically to the requirement
sqrt((x-2)2+(y-5)2)=sqrt(160)
If we square this, we get (x-2)2+(y-5)2=160
The points whose coordinates (x,y) satisfy this equation form a circle, with center (2,5) and radius sqrt(160).

Circles and functions
No circle is the graph of a function, because there are vertical lines which intersect twice (so there isn't a unique output for some of the inputs). But we may want to work with circles in this course, and the methods we'll develop will mostly work with functions. So what can we do, if we must consider circles? Well, look at x2+y2=1 as an example. This is the unit circle, a circle whose center is at (0,0) and whose radius is 1. If I want to work with it, I could solve for y: y2=1-x2 and then y=+/-sqrt(1-x2). There are two choices for most x's (except for x=+/-1). We can dissect (?) the circle into the graph of two functions. One is y=Top(x) defined by Top(x)=sqrt(1-x2), and gives the upper (y>0) semicircle. The other, y=Bottom(x) defined by Bottom(x)=-sqrt(1-x2), is the lower half of the unit circle.

Experimentation
If you weren't there, you should have been. The instructor using extremely complicated and delicate apparatus showed that Hooke's law is almost sort of approximately valid. That is, a rubber band will distort in a way which is proportionate to the load on it. Uhhh ... this is probably only valid for rather small loads (a rubber band supporting an automobile -- probably not). The stretching of the rubber band is linear.

I also discussed (or tried to!) how the relief of muscular ache/pain by a dose of medicine was also (for low doses!!) approximately linear. One might hope or expect that if one pill gives some relief, then two would give maybe twice as much, three ... but biology is more complicated. The result of 30 pills might be very different than 30 times as much relief! And even baby chemistry is "locally" linear: if you dose a cup of coffee or tea with a sugar cube, maybe it would be twice as sweet with two. And even three ... and ... and ... If you played with this, you would see that after a while the sugar would no longer go into solution. It would in fact precipitate into a sludge at the bottom of the cup. So dissolving sugar to get concentrations of sweetness in the fluid is locally linear but the simple changes stop eventually.

Lines, again
Non-vertical lines also can be written in slope-intercept form: y=mx+b. How can we get m and b if we're given two points on the line? Well, suppose we want m and b for a line containing the points (6,9) and (2,4)? Here is one way to find them. We know that:
9=m(6)+b
4=m(2)+b
by just "plugging in" the values of the coordinates of the points. We can subtract these equations. The b's will cancel, and we get:
9-4=m(6-2), so m=5/4. Once we know m we can get b by using either equation. So 9=(5/4)(6)+b and b=9-(5/4)(6). The geometric meaning of b is that (0,b) is the y-intercept of the line (the coordinates of the point the line intersects with the y-axis).

Multipliers ...
What about m? If x1 and x2 are inputs to the function f(x)=mx+b, then the corresponding outputs will be mx1+b and mx2+b. The difference in inputs is x2-x1. The difference in outputs is (mx2+b)-(mx1+b). The output difference can be "simplified". The b's cancel, and the m can be factored out, so the output difference is m(x2-x1). For a linear function, output differences are proportional to input differences. m is the constant of proportionality.

Notation is the key to secrecy in calculus. So maybe I could write x2-x1 as x and then write (mx2+b)-(mx1+b) as y. The is the Greek capital letter delta, and it abbreviates the difference of the inputs and the outputs. Lines and linear functions are so "neat" because there is a magnification or multiplier factor changing input differences to output differences. The ratio, y/y, is m.

Positive and negative slopes
If m is positive, then the input/output changes have the same sign. If m is negative, the signs of the changes are different. Geometrically, the lines slope up when m>0 (moving left to right) and the lines slope down when m<0 (moving left to right). m also has another interpretation, which can be read from the triangles shown several times above. m is the tangent of the angle that a tilted line makes with the positive x-axis. This angle is usually called (the small Greek letter theta).

Exercises with lines
Find a line parallel to y=3x+7 which goes through (2,3).
Parallel lines have equal 's.

Trig review, more or less
Here is sort of the minimal "stuff" you need to know about trig for this course. I hope.

Height of a tree
Well, suppose I am standing near a tree on a sunny day. Assume I am 6 feet tall, standing perpendicular to the (flat) earth, and that my shadow is 2 feet long. The tree is also perpendicular to the earth, and its shadow, formed by sunlight assumed parallel to the sunlight on me, is 5 feet long. How tall is the tree?

Well the picture sort of shows the situation. The yellow sunlight (?) forms the hypotenuse of two right triangles (the "right angles" are signaled in red). The right triangles have one acute angle equal, and since the sum of the angles of each triangle is always the same ("180o") the third angle is the same. So the sides are proportional. Therefore
Heighttree/Shadowtree=Heightme/Shadowme and ?/5=6/2, so that ?=3·5=15. The tree must be 15 feet tall.

Similar triangles
Two triangles are similar if their corresponding angles are equal. To the right is supposed to be a picture of a pair of such triangles. The colors and patterns are supposed to identify the equal corresponding angles. Then people in Egypt and Greece noticed (and verified!) a long time ago that the ratios of corresponding sides in the two triangles must be equal. This allows a lot of indirect measurement problems to be solved, such as the tree height problem.

Right triangles
Two right triangles which have one acute angle the same must also have the other acute angle the same. This is because in any triangle, the sum of the three angles is 180o, and if the triangles are "right" one angle is 90o. Therefore the sum of the two acute angles must be 90o so if one pair is equal so is the other. In the picture the two red (right) angles are equal. If one pair of acute angles (green) are equal, then the other acute angles are equal, and the right triangles are similar.
This is neat because you only need to check one pair of angles to insure similarity of triangles and, consequently, that the ratios of the lengths of corresponding pairs of triangles will be equal!

Named ratios of side of right triangles
The ratios of sides of right triangles have special names. If is the acute angle, then
Here HYP means hypotenuse, OPP means opposite side, and ADJ means adjacent side (as shown). There are three more ratios (sec, csc, cot) but I think that the three I just named are enough for this course.

Sine and cosine equations useful in this course
We will sometimes need to know sin2+cos2=1 (another Pythagorean equation). And maybe we will once need to know sin(1+2)=sin(1)cos(2)+cos(1)sin(2). You will see how this is used, I promise.

Measuring angles
Two systems are in common use: degrees and radians.

Degrees
There are three hundred and sixty degrees (360o) in a circle. This is not because Babylonians came from outer space and had 360 fingers on each hand. Then a quarter circle (a right angle) has 90o.

The radian measure of an angle is done this way: take the vertex of the angle (the corner) and put it at the center of a circle of radius 1. The length of the arc between the two sides of the angle is called the radian measure of the angle. Since the total circumference in a circle of radius 1 is 2Pi, a right angle in radian measurement is Pi/2,a quarter circle.

We will see why (for calculus and lots of other purposes)
radian measurement is better than degree measurement.
Besides, since you are neither Babylonian nor from outer space, 360 doesn't really matter that much.

Back to rate of change, multiplier, etc.
The "response" of linear functions to changes in inputs are fairly easy to understand. If f(x)=mx+b, and if we change the input x to x+h, then the output changes by mh. There is a fixed multiplier, m, which sort of "transmits" the effect of the input change, h, to the output.
If a function is not linear, such simple behavior won't usually happen when the input value is changed. First I'll give one piece of vocabulary: the average rate of change of f(x) on the interval [x1,x2] is defined to be the quotient {f(x2)-f(x1)}/(x2-x1). There's no immediate conceptual value to this -- it is just the definition of a phrase.

A function that is not linear
The text (see sections 2.1 and 2.2) has several examples of the analysis which is done here. Let's look at the function f(x)=x3. I would like to compute the average rate of change of f(x) on the interval [2,5]. The definition above states that we should compute {f(x2)-f(x1)}/(x2-x1) with x1=2 and x2=5. Here is some arithmetic:
{53-23}/(5-2)=(125-8)/3=117/3=39.
So my "guess" for a multiplier, m, for f(x) using data from the interval [2,5] is 39. The input 2 is kicked by a change of 3, and the output response is changed by 39·3=117.

Now consider the average rate of change on [2,10] of f(x)=x3. Since f(10)=103=1000, this average rate of change is {1000-8}/(10-2)=992/8=124. Here the input 2 is kicked by 8 and the output changes by 992. The estimate for the multiplier m must therefore be 124.

Which is correct for m, 39 or 124? Well, neither and both.
Both are correct because for the designated intervals they are the arithmetically correct multipliers, transmitting input changes to output changes. Hey, that is exactly what the arithmetic describing the average value computes.
Neither is correct because if there were a multiplier, m, then that multiplier would be the same. It wouldn't change.

The geometry of what this is about is shown in the picture here. The curve is supposed to be y=x3. The point A is (2,8), the point B is (5,125) and the point C is (10,1000). Please: this picture is mostly "poetic". It is certainly not quantitatively correct, otherwise the base would be about, oh 15 units wide and the height of the graph would be more than 1000 units. I don't want to draw something that tall and narrow! The line AB has slope 39 and the line AC has slope 124. The lines (almost naturally in the geometric context!) get steeper as x2 increases and x1=2 steadily. The corresponding multipliers are the slopes of the lines. We are not doing something right if we want our estimates for m, the multiplier, to be more stable. Instead of making x2 grow larger than x1=2, we should try to make x2 closer to 2.

The real-life examples imply that we can get behavior which is almost linear only "locally" -- near a fixed number. So we should consider small h's, not large h's.

Various h's
I'll look at {f(2+h)-f(2)}/h for h's which are small positive numbers. This is the average value of f(x)=x2 on the interval [2,2+h]. What's below are 20 digit accuracy (twenty! computations of h's, f(2+h), and the average value. It is probably worth examining the numbers a bit. The h's are successively smaller powers of 10: that is, 1/10, then 1/100, then 1/1,000, etc. (No, I didn't do this myself: I had a silicon friend do it.)

hf(2+h)
f(2+h)-f(2)
h
.1 9.261 12.61
.01 8.120601 12.0601
.001 8.012006001 12.006001
.0001 8.00120006000112.00060001
.00001 8.000120000600001 12.0000600001
.000001 8.000012000006000001 12.000006000001
.0000001 8.0000012000000612.0000006
.00000001 8.0000001200000006 12.00000006
.000000001 8.000000012000000006 12.000000006
.0000000001 8.0000000012000000001 12.0000000006

These computations were done to 30 digit accuracy and then truncated to 20 digit accuracy. If all computations were done with "only" 20 digit accuracy, the last entry would have been 12.0000000001, not 12.0000000006, because even 20 digit accuracy errors accumulate. Amusing, but not very amusing.

Average and instantaneous
So what do we learn? Well, the second column tells me that f(2+h)=(2+h)3 is darn close to f(2)=23=8 when h is small. This maybe is not too surprising, although we'll talk a bit more about it in a day or so. But it is the changes that are more interesting.

We seem to see that the average rate of change of f(x)=x3 in the interval [2,2+h] gets really close to 12. So for this f(x) and for this value of x (that is, x=2), we seem to see that f(2+h) is approximately 8+12h for h very small.

The ugly algebra
Let me show another way to look at what we're doing. The average rate of change of f(x)=x3 on the interval [2,2+h] is [(2+h)3-23]/h. I will actually compute out (2+h)3=(2+h)(2+h)(2+h) in order to really, really see what is happening. Let's begin.

I compute (2+h)(2+h)=22+2·2·h+h2=4+4h+h2. This maybe is not too bad. Now I need to multiple the result by 2+h. So I first multiply by 2:
2·(4+4h+h2)=8+8h+2h2.
Now multiply by h:
h·(4+4h+h2)=4h+4h2+h3.
Now add these results, making sure to keep track of all the powers of h:
(2+h)3=8+12h+6h2+h3.
Finally, the algebraic form of the average rate of change:
[(2+h)3-23]/h=(8+12h+6h2+h3-8)/h.
Stay calm. Stay very, very calm. Notice that the 8's cancel on top:
(12h+6h2+h3)/h.
Now the h on the bottom lowers the degree of the h's in all the terms on the top:
12+6h+h2
The result is that the average rate of change of f(x)=x3 on the interval [2,2+h] is 12+6h+h2. So therefore (!!!) when h is very very small, this is close to 12. Hey, our observation with the numerical data is confirmed.

Pictures?
Here are some pictures of y=x3. The pictures need to be considered carefully. The software which produces these pictures (a program called Maple) automatically tries to fill up a square box. This means that the vertical or horizontal scale is adjusted (squeezed or expanded) to show the graph without "wasting" any space. For example, in the first picture, the width of the base is 3, but the height goes from 03=0 to 33=27. The "square" should actually be almost a 1-to-10 thin rectangle.
In each graph, the green dot is supposed to be the point (2,8).
x3 in [0,3] x3 in [1.5,2.5] x3 in [1.95,2.05]
What you should see in these pictures is that as we get closer and closer to (2,8) the curve seems to resemble more and more a straight line. This is the geometric version of the numerical computations above: smaller and smaller changes in the input will get more and more like the output responses to a suitable linear function. In this case, the "linear function" is 8+12h. The 8 comes from 23, and is shown at (2,8), the green dot. Where the heck is the the 12? Well, if I force my last picture above (the third, on the interval for x's in [1.95,2.05]) to be displayed with correct, real proportion, the result is to the right. Hey: if we could measure this graph, an almost line-looking graph, you would see that its slope is 12!
That's where the 12 is hiding.

Monday, June 26
I discussed the course and this specific section of the course. What I said is here along with other information. I handed out student information forms.

I asked students to work on some starter problems, and discussed solutions.

1. A sign at the track near the Werblin recreation center declares that 9 laps is 4 miles. What is the length in miles of 7 laps?
If x is the length in miles of 7 laps, we set up a proportion, so 4/9=x/7 and x=(28)/9 or three and one-ninth miles.

I mentioned that people should be comfortable with arithmetic, and mentioned another problem I could have asked:
Suppose that A=1010 and B=10-10 and C=-1010 and D=-10-10. Can we "order" these numbers on the number line?
There was some early misunderstanding. Exponentiation is usually supposed to be done "first" before subtraction, so -1010 is -(1010), and so -1010 is a negative number. It is the number -100000 00000: this is minus ten billion. Well we saw that C and D were negative numbers and A and B were positive numbers. A is ten billion, and B is .00000 00001, one ten-billionth.

A and C were very far away from zero, and D and B were very close to zero.
I do note that it isn't hard to get numbers farther away from 0 than A and C (just 1011 and -1011) or closer to 0 than B and D (10-11 and -10-11). There are many big, big numbers and many small, small numbers. The minus signs in the exponent are very different from minus signs in "front" of the number.

 2. The graph to the right is taken from a webpage of the Financial Forecast Center. It records and predicts the Dow Jones Industrial Average (DJIA) for the period indicated.(The graph was copied on 6/22/2006 from http://www.forecasts.org/djia.htm.) a) What is the (approximate) highest value of the DJIA shown? What date does it occur on? Also answer the corresponding questions for the lowest value. The highest value seems to occur on about the beginning of May 06 and the value is about 11350. The lowest value is near the beginning of March 04 and the value is about 10280. The top is technically called a maximum and the bottom is called a minimum. There is other "local" tops (at about late January 04 and about July-August 05): these are called local maxima and there is a "local" bottom around October 05: this is a local mimimum. We will discuss these terms later. b) What (approximately) is the length of the longest period of time shown when the DJIA seems to be increasing? What are the beginning values (date, DJIA amount) and ending values for this period of time? This seems to be the period from October 05 to May 06. The DJIA increases from about 10350 to 11350. The function shown is increasing. This means, specifically, that for any two time values, say t1 and t2 in this interval, with t1

3. The Body Mass Index (BMI) is computed by taking a person's weight (in kilograms), and dividing by the person's height (in meters) squared.
a) Write a formula for BMI using the variables W for weight and H for height. Your formula should begin BMI(W,H)=....
Here BMI(W,H)=W/H2.

b) Describe how the BMI changes if weight increases and height stays the same. What happens exactly to BMI if the weight is doubled and the height is fixed? Verify your assertion algebraically.
I think that if W increases and H stays the same, then BMI(W,H) will increase. This is because (??) W is on top and H2 is on the bottom. If the weight is doubled while the height is fixed, then the BMI will double. Here is algebraic verification:
BMI(2W,H)=(2W)/H2=2(W/H2)=2BMI(W,H).

c) Describe how the BMI changes if height increases and weight stays the same. What happens exactly to BMI if the height is doubled and the weight is fixed? Verify your assertion algebraically.
If W is fixed and H increases, then the BMI will decrease. Again, this is because the H stuff is on "the bottom" and the W stuff is on "the top". If the height is doubled while the weight is fixed, then the BMI will be divided by four. Here is algebraic verification:
BMI(W,2H)=W/(2H)2=W/(4H2)=(1/4)(W/H2)=(1/4)BMI(W,H).

A big chunk of this course will be devoted to understanding increasing and decreasing, both graphically and algebraically.

Here is my "tribute" to standard geometrical problems.
4. A right triangle and a square share sides, as shown in the diagram to the right. The hypotenuse of the right triangle has length x, and one "leg" of triangle (not shared with the square) has length 3.

a) What is the side length, s, of the square? (Find this by computing the length of the other leg of the triangle.)
In a right triangle, the square of the hypotenuse is the sum of the squares of the two "legs" (Theorem of Pythaogoras) so that x2=s2+32. We solve for s: s2=x2-9, and then ... well, there can be two square roots of numbers (the square of both 5 and -5 is 25). So s is the square root (I'll abbreviate this as sqrt!) of x2-9. But which one? Well, we'd better take the one that is not negative. A triangle whose side length is negative doesn't feel good. So s=sqrt(x2-9).

b) What is the area of the square? What is the area of the triangle?
The area of the square is s2=(sqrt(x2-9))=x2-9. The area of the triangle is half the base multiplied by the height. The base is 3 and the height is s, so the area of the triangle is (3/2)s=(3/2)sqrt(x2-9).

c) Define a function, f(x), to be the sum of the areas of the square and the triangle. Write an algebraic formula for f(x). What is the "natural domain" of this function?
f(x)=x2-9+(3/2)sqrt(x2-9). This function is defined if x2-9 is at least 0 (we can't take square roots of negative numbers in this course). This means that x should be at least 3 (greater than or equal to 3) or (actually!) at most -3 (less than or equal to -3).
Please notice that the geometrically "meaningful" part of f(x)'s domain is x>=3.

d) Find x so that the sums of the areas of the square and rectangle is 7. (One way is to write f(x) in terms of s, find s so that the total area is 7, and then compute x.)
We did not do solve this in class since I wanted to have a break. If we put s=sqrt(x2-9) then f(x)=x2-9+(3/2)sqrt(x2-9)=s2+(3/2)s. To get s2+(3/2)s=7 we multiply by 2: 2s2+3s=14 and then change this to 2s2+3s-14=0. But amazingly (!!) 2s2+3s-14=(2s+7)(s-2). (Not too amazing, this is a problem in a math course. In real life, nothing as nice as this happens!) Now if (2s+7)(s-2)=0 either s=-7/2 or s=2. The side length of the square should be positive, so take s=2. Since s=sqrt(x2-9), we know that 2=sqrt(x2-9) and 4=x2-9. This means x2=13, so that x is either -sqrt(13) or sqrt(13). I think the length of the hypotenuse should not be negative, so x=sqrt(13).

More to come!!!

Function
A function is a rule transforming (real) numbers to (real) numbers. I frequently think of functions in terms of "boxes". This metaphor (?) can help organize my thoughts.
The collection of all inputs is called the domain of the function, and the collection of all possible outputs is called the range of the function.

In the DJIA example above (#2) the domain seems to be dates from from about Dec 04 to Dec 06. The range, the collection of possible values, goes from 10280 to 11350. In interval notation, the range is [10280,11350]. The square brackets indicate that the numbers on the boundary are included. Otherwise, round brackets are used.

Natural domain of a function defined by a formula
Many of the functions we'll consider will be defined by simple algebraic formulas. For example, f(x)=x2. Here the input is a number x, and the output is x2. When we have such a function, the domain will usually be the natural domain implied by the formula. This is all numbers for which the formula gives a real answer. In our course, the obstacles we will encounter to "a real answer" are created by division and square root. We can't divide by 0 and we can't take the square root of a negative number.

Example
If f(x)=(1/{x-2})+sqrt(6-x), what is the natural domain of f(x)? Since we can't divide by 0, the first part of the formula implies that x should not equal 2. But the input to sqrt must be non-negative, so we need 6-x>=0, which means 6>l;=x. Therefore the domain of the function f(x) defined by this formula is all numbers less than or equal to 6 except for 0.

An even sillier example
What if we wanted a function whose domain was the following collection of numbers:
all numbers greater than 5 except for the numbers 7 and 8.
This is the sort of weird question which I believe that none of you will ever see again outside of Math 135. Well, we can get rid of 7 and 8 by somehow making sure that pieces of the function involve dividing by x-7 and also x-8. How about excluding numbers less than 5? Well, a suitably shifted square root can do that: sqrt(x-5) excludes everything to the left of 5. So one possible function whose domain is as requested is f(x)=(1/{x-7})+(1/{x-8})+sqrt(x-8).
Please notice that this is not the only f(x). For example, the even more silly formula f(x)=(1/{x-7}19)+(1/{x-8}84)+sqrt(x-8) has the same domain. Sigh.

Graph of a function
If f(x) is a function, the collection of points in the plane whose coordinates are the order pairs (Input,Output) for all choices of Inputs is called the graph of the function. So if f(x)=x2, then (2,4) and (-2,4) and (3,9) are on the graph of f(x).

Condition for a collection of points in the plane to be a graph (the Vertical Line Test)
A bunch of points in the plane will be the graph of a function exactly when each vertical line hits the points at most one time. This is because inputs are supposed to have only one output. So if the point with coordinates (64,221) is on the graph of a function, the point (64,-117) cannot be on the graph of that function. If the input 64 produces 221 as its output, it can't also produce -117. So the line x=64 only intersects the graph at y=221.
Therefore if we want to "study" circles, for example, we'd better think carefully, because a circle is not the graph of a function.

Some graphs
Student should be familiar with the graph of f(x)=x2. This is a parabola. It touches 0 at 0 (I mean that (0,0) is on the graph and nothing else on the x axis is on the graph.) Also it is symmetric with respect to the y-axis. This is because f(x)=f(-x).

Another useful function is f(x)=1/x. When x is large positive, f(x) is small positive. And when x is small and positive, f(x) is large and positive. Since f(-x)=-f(x), what happens in quadrant I is flipped into quadrant III.

Straight lines
We will study rates of change a great deal, and those functions whose graphs are straight lines have the simplest rates of change to analyze.

Vertical and horizontal lines
x=3 is a vertical line, and it is not the graph of a function (because it hits itself more than once!).
Horizontal lines are the graphs of functions, although as functions they aren't too interesting (the range of a horizontal line function is just one number!). y=2 is a horizontal line,

Lines given in slope-intercept form: these are graphs of functions.
Functions defined by formulas such as f(x)=mx+b have graphs in the xy-plane defined by y=mx=b. This is called the slope-intercept form. The b refers to the point (0,b) on the y-axis where the line intersects the y-axis. The m refers to the amount of tilt in the function.

How functions related to lines change when inputs change
 If f(x)=mx+b, then the point (x,mx+b) is on the line y=mx+b. If we increase x to x+h, the corresponding point is (x+h,f(x+h)). The formula for f(x) makes this equal to (x+h,m(x+h)+b). This is (x+h,mx+b+mh). The change in the height is mh. This means that the change in x (from x to x+h) becomes a change in y of mh: the change in y is directly proportional to the change in x, and the constant of proportionality is m. Here is another way of looking at this computation. Think of the function as a box with various inputs. One input/output pair (x,mx+b). If we "kick" the input by h, then the corresponding pair is (x,m(x+h)+b). The output changes in a particularly simple way.

Let's look at some numbers ...

• One linear function
If f(x)=3x+7, then f(x+2)=3(x+2)+7=3x+6+7=(3x+7)+6. A change of 2 in the input gets "transmitted" as a change of 6=3·2 in the output.
If f(x)=3x+7, then f(x+5)=3(x+5)+7=3x+15+7=(3x+7)+15. A change of 2 in the input gets "transmitted" as a change of 15=3·5 in the output.
• Another linear function
If f(x)=8x-11, then f(x+3)=8(x+3)-11=8x+24-11=(8x-11)+24. A change of 3 in the input gets "transmitted" as a change of 24=8·3 in the output.
If f(x)=8x-11, then f(x-2)=8(x-2)-11=8x-16-11=(8x-11)-16. A change of -2 in the input gets "transmitted" as a change of -16=8·(-2) in the output. (So negative changes also are transmitted.)
• A nonlinear function
If f(x)=x2, things get more complicated.
Then f(x+3)=(x+3)2=x2+6x+9. The change of 3 gets transmitted as a change of 6x+9.
And f(x+5)=(x+5)2=x2+10x+25. The change of 5 gets transmitted as a change of 25x+25.
So changes percolate through the (non)linear function box in a much more complicated way.