Math 135 diary, third part, spring 2005

Friday, April 15: this is lecture 24

F(x) is an antiderivative of f(x) if F´(x)=f(x).

If you know one, you know them all ...
If F(x) is one antiderivative of f(x), then any other antiderivative of f(x) is F(x)+C.

How to know one
How would you find antiderivatives?

  1. Guess them. This is essentially the "technology" of Math 135. You should look at some familiar functions and be able to guess the antiderivatives. What functions? Well, certainly things like x43 or 7x-8 or ... and sin(x) and cos(x) and ex. But it turns out that finding the antiderivatives of functions defined by even rather simple formulas can be very difficult.
  2. Ask an electronic pal. I just logged onto eden. I typed the word maple on an eden command line. The response was some stuff which looked like this:
    er3% maple
        |\^/|     Maple 9.5 (SUN SPARC SOLARIS)
    ._|\|   |/|_. Copyright (c) Maplesoft, a division of Waterloo Maple Inc. 2004
     \  MAPLE  /  All rights reserved. Maple is a trademark of
     <____ ____>  Waterloo Maple Inc.
          |       Type ? for help.
    The little > invited me to type something. So I did. The program's answer follows. I did a differentiation, an antidifferentiation, another differentiation and another antidifferentiation. Take a look.
    > diff(x^5,x);
                                          5 x
    > int(x^5,x);
    > diff(sqrt(cos(1+x^6)),x);
                                                6   5
                                     3 sin(1 + x ) x
                                   - ----------------
                                               6 1/2
                                      cos(1 + x )
    > int(x^2*cos(5*x),x);
                   1/5 x  sin(5 x) - 2/125 sin(5 x) + 2/25 x cos(5 x)
    > quit
    You should know that such programs exist. The programs won't think for you, but they certainly will help with some computations!
  3. Call me at home and ask ... No, please don't do this. I am a cranky guy, and also I make mistakes.

Joni throws a ball
Most people who have taken a high school physics course have seen problems similar to what follows:
Joni stands on a hill 15 feet high and throws the ball straight up in the air at 30 ft/sec. Remember that gravity pulls a ball down at 32 ft/sec2. What is the position of the ball t seconds after it is thrown?

Well, suppose s(t) is the height (position) of the ball at time t, with up being positive. And v(t) is the velocity of the ball at time t. Finally, a(t) is the acceleration at time t. How do the statements in the previous paragraph translate to calcspeak?

Joni stands on a hill 15 feet highs(0)=15
throws the ball straight up in the air at 30 ft/sec.v(0)=30
gravity pulls a ball down at 32 ft/sec2.a(t)=-32
(the minus sign
means down)

Of course we need to remember that s´(t)=v(t) and v´(t)=a(t). Since a(t)=-32, we guess an antiderivative: v(t)=-32t+C. Since v(0)=30, we know v(t)=-32t+30. How about s(t)? We guess s(t)=-16t2+30t+C. Since s(0)=15, this C is 15. Therefore s(t)=-16t2+30t+15.

Some standard questions

  1. At what time is the ball highest? What is its height at that time?
    The ball is highest when v(t)=0 (critical number!). Since v(t)=-32t+30, -32t+30=0 implies that t=30/32. To get the height at that time, plug into s(t). The height is s(30/32)= -16(30/32)2+30(30/32)+15.
  2. At what time does the ball hit the ground? What is the velocity of the ball when it hits the ground?
    The ball hits the ground when s(t)=0. So we need -16t2+30t+15=0 and (quadratic formula) t={-30+/-sqrt([30]2-4(-16)(15))}/(2[-16]). The roots turn out to be (approximately) -.41 and 2.29. I think that Joni doesn't throw the ball backwards in time, so the ball hits the ground at (about) t=2.29. Its velocity at that time is v(2.29)=-32(2.29)+15=-58.28. The minus sign means the ball is moving down.

As I mentioned in class, specifications such as s(0)=15 and v(0)=30 are called initial conditions. In practice, initial conditions are used to specify one out of a whole bunch of candidate antiderivatives.

Become an expert witness
A car is speeding along at 60 miles per hour (which I remember from high school is 88 ft/sec). Suppose the car hits a wall and comes to a stop in a tenth of a second. If the deceleration is assumed to be constant, what is the deceleration? If you can do such analysis, maybe you could be an expert witness in car accident cases, and make big bucks. Ooops, sorry: maybe you could help that the correct side wins in the court case.

Here a(t)=K, a constant by assumption. I'll use the letter K for a constant here instead of C, since we use that for the constant which arises in antidifferentiation. So I know that a(t)=K, and then by antidifferentiating, v(t)=Kt+C. But v(0)=88. I'm again using distance measured in feet and time measured in seconds. Thus v(t)=Kt+88. Since v(1/10)=0, we know K(1/10)+88=0, so that K=-880 ft/sec2. This is interesting and horrifying. Notice that 880/32 is 27.5: so this deceleration presses the "contents" of the car (the passengers!) with a force twenty-seven and a half times that of normal gravity. Imagine even briefy your weight multiplying by that factor. The consequences are most unpleasant. A chapter in a book I recently read, Stiff: The Curious Lives of Human Cadavers by Mary Roach, has a chapter discussing this situation. Those of you interested in becoming doctors might want to read this book.

An antiderivative
Here's a problem from a past final exam:
a) Suppose A is a constant. What is the derivative of Aexcos(x)?
b) Suppose B is a constant. What is the derivative of Bexsin(x)?
c) Find one antiderivative of exsin(x).
As I remarked, a) and b) were done successfully by most students. Here are the solutions:
a) The derivative of Aexcos(x) is Aexcos(x)-Aexsin(x).
b) The derivative of Aexsin(x) is Aexsin(x)+Aexcos(x).
c) Correct answers to c) were supplied by only 20 to 25% of the students. Let's see: the goal is to get exsin(x). And we know
Aexcos(x) ---d/dx---> Aexcos(x)-Aexsin(x).
Bexsin(x) ---d/dx---> Bexcos(x)+Bexsin(x).
So if we added things, then we'd get (A+B) multiplying excos(x) and we'd get (A-B) multiplying exsin(x). But we want exsin(x): that is, I want no excos(x) and I want exactly one exsin(x). If you can find A and B so that -A+B=1 and A+B=0 then we'll be done. But, goodness, add the equations to get 2B=1 so B=1/2, and then A=-1/2. I'll bet that an antiderivative of exsin(x) is -(1/2)excos(x)+(1/2)exsin(x).
How can you tell if I am correct? You can check by differentiating. The result should be exsin(x).

An introduction to "How many get sick?"
It is extremely doubtful if any student in the class will need to find the slope of a tangent line after they've passed the Math 135 final exam. The major idea of the course still needs to be told to you, though. And it is this major idea which may "flavor" your approach to a very wide variety of quantitative phenomena in the future. I wanted to give a biological example to show what's going on. I couldn't find simple enough data for a real example, so I needed to invent something. So here it is, along with my apologies and acknowledgement that the specifics are so simple as to be almost silly. But there's a profound idea somewhere in what I'll try to describe. Now let's get down to business.

How many get sick?
The scenario is that there's an epidemic, and we'd like to know how many people get sick. One way we might try to get a total is by calling the (only) local hospital at, say, 7 AM in the morning. You could ask, "How many patients (sick with the disease) are waiting for care in the emergency room?" Now I'll add some further simplifying assumptions. I'll assume that patients in the emergency room are seen and diagnosed and stuffed away within an hour. Suppose we get the following information:

Day 1Patient count at 7 AM12
Day 2Patient count at 7 AM9
Day 3Patient count at 7 AM7

So how many people are getting sick? I'd like to use the very simplest kind of estimates. Well, on day 1, there are 12 people in the emergency room. The hospital handles this crowd in an hour. But I guess that the number of people getting sick is going to be constant (this is simplicity!) until I am forced to change my assumption. So on the first day, the hospital has 12·24 people getting sick. There are 24 hours in a day! But on day 2, I've got to change my analysis. I learn that 9 people are getting sick in an hour. So on day 2, there are 9·24 more ill people, and the information for day 3 tells me that there are 7·24 sick folks.
There's a total of 12·24+9·24+7·24 people who have gotten the disease.

This is really a very rough estimate. How could we get better information? Many of the students in the room who were both alive and awake replied that we could get better information by questioning the hospital more often. I certainly agree. So let me assume that I not only call at 7 AM but I also call at 7 PM. So I might get some numbers like this:

Day 1Patient count at 7 AM12
Patient count at 7 PM23
Day 2Patient count at 7 AM9
Patient count at 7 PM7
Day 3Patient count at 7 AM7
Patient count at 7 PM4

Well, what estimate for the number of ill people can we now get? The numbers should now be multiplied by only 12, because each reported number is assumed to represent the patient "flow" in half a "day" (half of 24 is 12, indeed).
The appropriate estimate is a total of 12·12+23·24+9·12+7·12+7·12+4·12 people who have gotten the disease.
This should be a better estimate, closer to the actual number of people who have been infected and been treated by the hospital.

Well, I'm interested in the "structure" of the answers. The second, (presumably) closer answer has more pieces, more things to add up. The pieces, though, each have the reported patient numbers, but are multiplied by a smaller number. Huh? Maybe you can see what is happening if I "refined" this process and imagine "sampling" the population in the emergency room more often. There would be more terms in the sum. Each term would be multiplied by something smaller. Hey, and maybe you might even see that somehow the number of terms getting larger is balanced out by the multiplier getting smaller. Maybe we could also hope that the approximations get better and better.

If you think in terms of pictures, something like the following might help. This picture shows the results of just the 7 AM reports.

Now we move on to reports at both 7 AM and 7 PM.

Finally, there's this mysterious curve, in magenta. What might this curve represent? It should represent the number of people getting sick (although the number of people is an integer, well, one could imagine that the population was very large, and maybe the curve represents an integer number of people but the scale makes it look smooth). And the total number of people who get sick during all three days is ... the area under the curve.

So this is the big idea?
The big idea is that the total number of people who get sick, which you might somehow think of as a "chunk", a number just sitting somewhere in our heads, is really related to the rate of people getting sick. How? Well, the rate of people getting sick is the derivative of the people who are ill. And you go backwards from the rate of people getting sick to the total number of people who get sick, by taking an antiderivative.
You can measure the total number of people getting sick by questioning the whole population after the epidemic is over. Or: you can look at the process, and see the rate of people getting sick, take its antiderivative, and try to analyze things that way. This is two different points of view, and the two different points of view give a terrific intellectual leverage in understanding the computation. Sometimes one way is more convenient, and sometimes the other one is.

Please begin reading chapter 5. Please.

Tuesday, April 12: this is lecture 23

The fences of a farmer
A farmer wishes to enclose a rectangular area of 1200 square feet with a fence. Additionally, the inside will be divided by two fences which are both parallel to one of the sides of the rectangle. Suppose the outside fence costs $2 per foot and the inside fence costs a half dollar per foot. What dimensions will make the fence cost least?
I tried to go through this problem very slowly, with some special mention of what the inside of my head is doing during the stages.

Step 1: translation
Translating the paragraph into algebra: I would first draw a picture and label it. This is not always necessary, but my approach to these problems frequently needs visual reinforcement. Then I go through the problem sentence by sentence, or phrase by phrase, or even, if I am very confused, word by word. I'll call the two sides of the rectangle L and W. Here I see the constraint LW=1200. And actually the "reality" of the problem imposes L>0 and W>0 also. What about the cost of the fence? The length of the outer fence is 2L+2W (I imagine the units are ), and the cost per foot is $2. Therefore the cost of the outer part of the fence is 2(2L+2W). The inside fence seems to be 2W, with a cost of $1/2 per foot. So its cost is 2W(1/2). The total cost is 2(2L+2W)+2W(1/2), and this is our objective function.

I usually try to check my translation (from language to algebra) at this stage. One way I do this is by looking at extreme or exaggerated situations, and seeing if I get anything that makes sense. In the two cases shown, I think that the area is constant, and I can make L or W very very big, and then the other one is very very small. The cost of the fences then will become large (if L or W is really big, then 2(2L+2W)+2W(1/2) will be large -- remember that both L and W are positive). So, internally, I think, aha, somehow the collection of fence "scenarios" will yield costs, and the costs will be large at the exaggerated cases, and somewhere there should be a minimum cost. As I remarked in class, I might not write this down explicitly if I were doing the problem myself, but I really assure you that I think this way in every problem of this type. If I can't make sense of the problem at this stage, maybe my translation was incorrect.

Step 2: calculus
Minimizing 2(2L+2W)+2W(1/2) while LW=1200 isn't a one variable calculus problem. So I use the constraint to help me: L=1200/W. Then the cost is 2(2L+2W)+2W(1/2)=4L+4W+W=4L+5W=4(1200/W)+5W=[4800/W]+5W. Now I need to minimze C=[4800/W]+5W. Here the domain is W>0. As W-->infinity, the cost-->infinity also. As W-->0+, the term 4800/W-->infinity. I'll bet that there is a minimum somewhere "inside" the domain. An approximate graph of C as a function of W is shown to the right. This is a more precise picture than the preliminary version shown above.

We computed dC/dW and got -[4800/W2]+5. The critical numbers will occur when this is 0, so we must solve -[4800/W2]+5=0. So 5=4800/W2 and 5W2=4800. Next, W2=960, and the solutions of this algebraic equation are +/-sqrt(960). The only solution inside our domain is W=sqrt(960), and this must correspond to the minimum cost.

Step 3: going back to the problem
Why should this be a minimum, what are the dimensions of the resulting rectangle, and what is the cost?
Mimimum? Well, if we have already checked that C-->infinity as W-->0+ or W-->infinity, then I know this value of W will correspond to a minimum.
One could check the first derivative: if 0<W<sqrt(96), then dC/dW is negative and if sqrt(960)<W<infinity, dC/dW is positive. Or we could look at the second derivative. Realize that for this purpose all we need is the sign of the second derivative -- we don't need the exact value. The derivative of -[4800/W2]+5 is 2[4800]/W3. When W=sqrt(960), this is clearly positive so (concave up!) we have found a minimum.
Since LW=1200, at the mimimum cost L=1200/W=1200/sqrt(960). And what is the mimimum cost? Since C=[4800/W]+5W, the cost must be [4800/sqrt(960)]+5(sqrt(960).

Certainly this problem is silly, but I really did it to illustrate the process, which is good even in many more complicated problems: translate, check the translation, get a calculus problem, solve it, go back to the original setting and look at the candidate solution.

One of my WeBWorK problems
Here's a problem about driving a truck in Mexico. The setting is something like this:
The truck costs 51+.28v cents per mile to drive at v miles per hour. (Well, yeah, one could imagine that driving a truck faster might cause parts of the truck to wear out more. So maybe something like this is true in real life.) Also, the cost of a driver is $12 per hour. At what velocity v should one drive the truck on a trip of 2,000 miles in order to spend the least amount of money? Also, we're told that the velocity, v, the "speed", should be at most 100 mph.

The cost is the sum of the cost of the truck plus the cost of the driver. If the truck is driven 2,000 miles, then the truck costs .51+.0028v dollars per mile (thanks, Ms. Dike, for the 00 in the cents) and therefore the cost of the truck is 2000(.51+.0028v) dollars for the trip.

What about the cost of the driver? If the truck moves steadily at v miles per hour, then (dist=rate·time) the time needed to drive the truck 2,000 miles is 2000/v hours. And the driver gets $12/hour, so the resulting driver's cost is 12(2000/v).

The total cost is therefore the sum of the truck cost plus the driver cost, which is 2000(.51+.0028v)+12(2000/v). This function is similar to the one we analyzed in the fencing problem. Oh well: it was silly for me to present both of these problems in class. Again as v gets large, the cost gets large. And as v-->0+, the cost also gets large.

What is the derivative of 2000(.51+.0028v)+12(2000/v)? It is 2000(.0028)-12(2000/v2). This is 5.6-(24000/v2) and we need to know when 5.6-(24000/v2)=0. Again, we can solve this just as before, so v=sqrt(24000/5.6) which is about 65 mph. And, just as before, we can justify our statement that this is a minimum cost.

Section 4.7 has a significant amount of applications in economics (average and marginal cost, marginal profit, etc.). I strongly urge interested students to read this material. I also urge all students to do the next two WeBWorK assignments, which will practice your translation skills on various optimization (max/min) problems.
To the right is a picture of several skuas which is "any large predatory sea bird of the family Stercorariidae".

Moving ahead
We have only a few weeks left in the course. Most people who know and use calculus would agree that the material still to follow is probably the most important and most beautiful in the course. I think there is barely enough instructional time, so I want to begin chapter 5 slightly early.

We'll begin with the principal definition of section 5.1.
F(x) is an antiderivative of f(x) if F´(x)=f(x). I gave several examples of antiderivatives. They were something like what follows.
Uhhh, ________ is an antiderivative of 5x2-7x+3. What is ________ ?
There turn out to be many answers to this. By guessing and then verifying, you can see that (5/3)x3-(7/2)x2+3x+104.3 is an antiderivative of 5x2-7x+3. But you can also see that (5/3)x3-(7/2)x2+3x-Pi/2 is an antiderivative of 5x2-7x+3. We did a few other examples, but what's going on?

The MVT steps in
Here is another consequence of the MVT:
If the derivative of a function is 0, then the function is constant. Why is that? Well, one side of the MVT equation "compares" f(b) and f(a). It declares that f(b)-f(a)=f´(c)(b-a). If the derivative is always 0, then f(b)-f(a)=0 for any choices of b and a, so that f's values stay the same.

If f(t) represents position at time t, then f´(t) is the instantaneous change of position, or velocity. The math statement inthe previous paragraph translates, "If velocity is always 0, then positiion stays constant." Or, more plainly, "If you don't move, you stay in one place." Ain't math profound?

All antiderivatives
Let's ask a more precise question. What are all antiderivatives of 5x2-7x+3? Well, if I know one antiderivative, the previous result states that I can get any other valid antiderivative by adding/subtracting any constant from one known antiderivative, which(at this stage) I can get by guessing and verifying. One antiderivative of the function displayed is (5/3)x3-(7/2)x2+3x+104.3, so that any antiderivative is (5/3)x3-(7/2)x2+3x+104.3+a constant. This is more conventionally written as (5/3)x3-(7/2)x2+3x+C. The constants are pushed together, and the "+C" is a conventional abbreviation.

Exam return
The material in this course is important and I think it is interesting. I tried to grade the exam carefully, and the letter grade equivalents are given in the context of all students taking Math 135. Please evaluate carefully how much time you can spend and will need for success in this course. You should not cheat or restrict your future self: you may well value understanding of this material for applications in both business and the biological sciences.

Although suitable use of "technology" (such as calculators and other silicon pals) is very important, I very much believe that students should be able to write explanations of their work, and that, in a calculus course, students should be able to use calculus to explain their conclusions. Such efforts are important and transferable to other fields. They are therefore suitable examination requests in a college calculus course.

Errors? Please ...
In a class of this size, I almost certainly have made errors in grading. These errors may range from the purely arithmetic ("2+7+3= ... 10 (?) ...") to more subtle ("You ignored my correct answer on the back of the page."). I try to grade correctly. Please look at the answer sheet supplied and also look at the description of partial credit assignment. If you believe I have made errors in grading your exam, please write a statement supporting your belief and give me the statement together with your exam.

Please read sections 4.3 and 4.4 and hand in these problems on Thursday:
4.6: 11, 17, 24
4.7: 5, 7, 43

Friday, April 8: this is lecture 22

The second exam.

Tuesday, April 5: this is lecture 21

Third max/min problem
Suppose a line passes through the point (2,3) and cuts off a triangle in the first quadrant. What is the area of the triangle which has the smallest area?

The setup here was more complicated. I drew a picture, something like what's shown here. and then ... got a triangle. I'll call the base, x, and the height, y, so that the area is (1/2)xy. We need to minimize this. But this is two variables. We need a constraint, one or more restrictions on the variables, perhaps relating them, so that the function we want to minimize will have only one variable, and the techniques of Math 135 can be used.

I find the picture useful here. What restrictions on x and y seem to "follow" (clearly?) from the picture. Certainly x>2 and y>3, because otherwise a line passing through (2,3) will not cut both the x and y axes in the first quadrant. There's a more precise restriction on x and y, caused by the fact that (x,0), (2,3), and (0,y) must all be on the same straight line. One way of getting an algebraic condition is to follow a suggestion of a student: these points are on the same straight line exactly when the slope of the line segment connecting (x,0) and (2,3) equals the slope of the line segment connecting (2,3) and (0,y). This means:
0-3 3-y ----- = ----- x-2 2-0 which (cross-multiplying) gives (-3)(2)=(3-y)(x-2). This means 3-y=-6/(x-2) so -y=[-6/(x-2)]-3 so y=[6/(x-2)]+3. I need to minimize (1/2)xy, so now replacing y by [6/(x-2)]+3 the resulting function is f(x)=(1/2)x{[6/(x-2)]+3}. The domain of this function is 3<x<infinity.

Does this function make sense? Well, if you look at the picture and you let x-->2+ from the right, then you should see that you'll get a triangle whose base, x, is very near 2, but whose height, y, is quite large. Hey: the area-->infinity. On the algebraic side, consider f(x)=(1/2)x{[6/(x-2)]+3}. If x is near 2 but slightly larger than 2, the x-2 is a small positive number. But it is in the bottom of a fraction whose top is 6. I bet that fraction is very large positive: the function reflects the growth of the area as x-->2+.

On the other hand, if x-->infinity, the geometry as shown gives me a very wide triangle, whose height gets near 3. But the area will also be large. And the algebra agrees, since f(x)=(1/2)x{[6/(x-2)]+3}, as x-->infinity becomes (1/2)[large{[6/large-2]+3}, and this is (1/2)large{small+3} so it is guaranteed to be large also.

If you think about it, we have just verified that if there is only only critical point "inside" the interval, then that critical point corresponds to a minimum which is an absolute minimum. So we need to search for critical points, and hope (since this a problem in a calculus class rather than a real problem) there will be only one, and we will just about be done.

The vicissitudes of finding f´(x) were many. By the way, "vicissitude" means, according to one dictionary, " One of the sudden or unexpected changes or shifts often encountered in one's life ..." The derivative is:
Never rely on student answers!
and this is the same as (3/2)[x(x-4)]/(x-2)2. The only way this can be 0 is if the top is 0. But the top is x(x-4). The only root in the domain of the f(x) in this problem is x=4. So I bet that f(4)=12 is the minimum area.

A graph of f(x)is displayed to the right. I hope you can "see" the absolute minimum at the point (4,12).

If you look very carefully at the structure of the last two max/min problems, you will see a weird correspondance between the objective function and the constraint. They interchange in the two problems. This weirdness has a name in mathematical economics: the problems are called dual to one another. This pairing is used to analyze more complicated problems.

Fourth max/min problem
My darling ... is in trouble. She is a half-mile offshore, 4 miles down from me along a straight beach. Suppose that I can run (hah, if only I could!) at 10 mph, and swim at 4 mph. What is the best strategy for me to get to her in the least amount of time?

This is a more complicated problem, and we needed to discuss it quite a while. There are a variety of strategies for "Me" to get to "My darling". The strategies can be analyzed by assuming that they all consist of running along the beach for a distance of x miles in a straight line, followed by swimming along a straight line to the darling. One extreme "pure" strategy is gotten by taking x=0. This is shown in the yellow line straight from me to darling. I might want to follow that strategy if I were a huge turtle. My running speed would then be rather low, but my swimming speed would be relatively high. I hope you can see that the "all swimming" strategy then would be a winner, the least time path. On the other hand, if I were, uhhhh ...., something which could run very fast but swim really slowly (a cheetah? Animal science majors: can cheetahs swim?) the best strategy would likely be to run as much as possible and swim as little as possible. This would correspond to x=4, shown in red on the picture.

Analysis of a general mixed strategy, the dashed green path
If I run x miles, then (at 10 miles per hour, with distance=rate·time) the running time would be x/10. How long a distance will I need to swim? Well, the tilted part of the dashed green path is the hypoteneuse of a right triangle. The remaining beach distance is one leg of the right triangle, and is 4-x miles, while the other leg is 1/2, the miles the darling is offshore. Then Pythagoras gives the swimming distance as sqrt{(x-4)2+(1/2)2}, so that (rate/time/distance again) the swimming time is sqrt{(x-4)2+(1/2)2}/4. Hey, this is some function.

Darling time
So we need to find the minimum of f(x)=(x/10)+sqrt{(x-4)2+(1/2)2}/4 with domain 0<-x<=4. Well, we can graph the darn thing, and see that there is actually an interior minimum (inside the intervalP, fairly close to 4. I gave up in class, but (sigh!) you can actually differentiate this and then look for that critical number, and then plug in. It turns out that the critical number is 4-(21)-1/2 (this comes from solving a quadratic equation) and this is approximately 3.781782110 miles and the minimum time needed is (sigh!) about .5145643941 hours and this is definitely not clear at all. Here is my secret: I admit that I did this all with the help of a program called Maple. This program is available on eden and many other Rutgers computer systems. Maple can do many of the routine computations of calculus and can be used as if it were a very smart graphing calculator with algebraic manipulation capabilities. You should know that such programs exist. You might need one of them some time.

Outline of things you should know
in preparation for the exam on Friday

I will not be able to have a review session before this exam. I hope that Mr. Lin will help you during your recitations. You should attend the recitations with questions. Specific questions are better than general inquiries such as, "Do you think this will be on the exam?" When I run review sessions, the majority of the preparation I do is writing an outline of the material on the exam. Such an outline is below. Please let me know if you think I have omitted anything important.

Students should be prepared for questions on major topics. They should realize that there are usually several ways to ask about a topic. This is perhaps most easily illustrated by the direct applications of the MVT. How could I ask about this? I can give you a formula and ask you to draw a picture. I could give you numerical properties of a function, or I could ask about such properties. I could give you a graph of a function and then ask you about the graph of the derivative or properties of the derivative. Or I could give the graph of a derivative and ask about the graph of the function or properties of the derivative.

There will be a formula sheet. The formula sheet has information about almost every topic below. I have observed that a need to consult formula sheets frequently during an exam is usually correlated with poor performance. Please: you should be familiar with the information on the formula sheet.

Apology Should I be writing l'Hopital as l'Hôpital? If so, I apologize to those of French ancestry whom I may have offended.

Friday, April 1: this is lecture 20

I discussed certain WeBWork problems which can be handled by l'Hopital's Rule. The due time for this WeBWork assignment has been changed to 9 PM, Tuesday, April 5.

L'Hopital's Rule
This is a computational method which can be applied to certain recalcitrant limits. I don't believe that many Math 135 students will need to use l'Hop after the course is over, but it will appear on certain exams. l'Hop is on the formula sheet.

I began with an example, something like this:
What is limx-->0[sqrt(1+3x)-1]/sin(5x)?
Here if we try to plug in (the first thing I'd do if I had a limit problem involving familiar functions), we get 0/0. This is called an indeterminate form because you can't conclude anything from just that much evidence.
Why indeterminate?
Look at these examples which all have the plug in appearance of 0/0):

I hope the examples persuade you that you can't predict anything from just knowing that the situation is 0/0.

But what if we want limx-->0f(x)/g(x) where we know that f(0)=0 and g(0)=0. The following is not a real proof, but is intended to be an argument which might help you agree with l'Hop.
Since f(0)=0 and g(0), I know that f(x)/g(x)={f(x)-f(0)}/{g(x)-g(0)} but we can go on a bit, and divide the top and the bottom both by x-0 to get a complicated quotient which looks like this:
If you are clever now, you can see that the top alone would have limit f´(0) as x-->0 and the bottom similarly would be g´(0). This is not a real proof, because we used "the limit of the quotients is the quotient of the limits" and that is really generally valid only when we know all the limits exist, and l'Hop actually applies where we don't know the original limit exists. Oh well. Here is a statement for us to use:

L'Hop Rule
Suppose we want to compute limx-->af(x)/g(x) where f(x) and g(x) are differentiable functions.
Eligibility criterion Suppose we know that f(0)=0 and g(0)=0.
If limx-->0f´(x)/g´(x) exists, then limx-->0f(x)/g(x) exists and is equal to limx-->0f´(x)/g´(x)
This is really what you wanted the quotient rule to be, isn't it? Admit it!

The example
What is limx-->0[sqrt(1+3x)-1]/sin(5x)?
is exactly suited for l'Hop. We have already checked (with our plug in attempt) that the eligibility criterion is satisfied (0/0). So we compute the derivative of the top: (1/2)(1+3x)-1/2(3) and the derivative of the bottom: cos(5x)5. Now plug in and the quotient f´(0)/g´(0) is (1/2)3/5, or 3/10.
Note L'Hop can also be used with limmits with x-->infinity. And it can also be used for indeterminate forms of the type infinity/infinity.

Student question
"Hey, why didn't you teach us this nifty way of computing limits back when we started doing limits?" My answer was that we needed to get the technology of derivatives before we could even state and use l'Hop.

I reminded people that the eligibility criterion must be checked, or else you run into something like limx-->0(5+6x)/(7+2x) which is certainly not the same value as limx-->0(6)/(2) which could be gotten by applying L'Hop without thought.

Some WeBWorK problems
These are from my own (personal!) set of problems.
8. (First part)
What is limx-->0[e-15x-1]/[sin(4x)]?
Plug in x=0, and get [1-1]/0, certainly 0/0. Now take the derivative of the top divided by the derivative of the bottom. The result is e-15x(-15)/cos(4x)4. Evaluating at x=0 (pugging in) gets us -15/4.

(Second part)
Consider limx-->infinity[-15sin(x)+4cos(x)]/x. Why can't we use l'Hop? Then compute this limit with the Squeeze Technique.
The top function has no limit as x-->infinity, since it wiggles. What do I know about the top? since the values of sine and cosine are always between -1 and +1, I know that the top is between -15-4=-19 and +15+4=19. Therefore we can sandwich or squeeze when x>0:
But +/-19/x-->0 as x-->infinity, so the original fraction must also -->0 as x-->infinity.

9. (First part)
Compute limx-->0[5x-2x]/x.
When x=0, the fraction becomes [1-1]/0, another 0/0. Remember the derivative of ax is axln(a). Then the fraction of derivatives becomes [5xln(5)-2xln(2)]/1. As x-->0, this -->ln(5)-ln(2), which is the value of the original limit by l'Hop.

The second part of problem 9 has a limit which reeds two uses of l'Hop, and the third part suggests a dubious use of l'Hop.

10. (Just part 1)
Evaluate limx-->0+-2x13ln(x).
Several clever ideas are needed here. First, if we plug in, we get 0·(-infinity), which isn't a l'Hop-type fraction. But we can change this by algebra:
We changed a multiplication into a simple fraction, a very weird thing to do. In this case, however, if we evaluate (plug in) now, we get 0/0. Let's try l'Hop: the derviative of the top is -2/x and the derivative of the bottom is -13x-14. Now what? We've gotten rid of ln(x) and are left with powers of x: [-2/x]/[-13x-14]. But please realize that if we write this as one power of x, the result is (2/13)x13, and, as x-->0, this -->0. Whew.
So there are several tricks. One is to convert a multiplication into a fraction. Then use l'Hop, and then convert back. If you don't believe the result, I suggest that you ask your graphing calculator for a picture of -2x13ln(x) for x positive and near 0. I hope the result will give support to the fact that as x-->0+, -2x13ln(x)-->0.

I then went on and did some max/min problems, as discussed in section 4.6, the last section of material which will be eligible for next week's exam.

First max/min problem
Two non-negative numbers have sum 20. How should we choose the numbers so that the product of the cube of one of them plus the square of the other is a maximum? There didn't seem to be a useful picture to draw. So we decided that x+y=20 with x>=0 and y>=0. We needed to minimize x3·y2. This we reduced to a calculus problem: find the minimum of f(x)=x3(20-x)2 with 0<=x<=20.

f(0)=0 and f(20)=0. To be sure of finding the minimum, we need to look at critical numbers. f´(x)=3x2(20-x)2+x32(20-x)1(-1). When is this equal to 0? The easiest way is to factor the mess, since a product is zero exactly when at least one factor is zero. So:
f´(x)=x2(20-x)(3(20-x)+2x(-1)). This is zero when x=0 and x=20 (whose values we previously checked) and for x=12 also. If we cube 12 and multiply by the square of 8, we will get the max. value requested. (110,592).

In some economics applications, "x+y=20 with x>=0 and y>=0" would be called the constraint and "x3·y2" would be called the objective function.

Second max/min problem
What is the rectangle of largest area which has two sides on the positive x- and y-axes and one corner on the line 2x+3y=6?

First we drew some rectangles, just to get a feeling for the problem. Then I asked how to make the geometry more algebraic: how can we introduce variables and try to get a function and turn this into a more routine calculus question. We decided that the corner of the rectangle on the tilted line could be called (x,y) and then the area of the rectangle would be xy, so we needed to maximize A=xy. A constraint relating the two variables x and y was given by the line 2x+3y=6, so that y=2-(2/3)x, and thus A=x(2-(2/3)x). Which x's would be allowed, I asked? Could we have x=10,000 or x=-10,000? We would be guided in this by the original problem, and by the diagram which we had sketched. It showed that candidate x's could vary from 0 to 3. Now we had a "pure" calculus problem.

What is the maximum of f(x)=x(2-(2/3)x) when 0<=x<=3? Since f(0)=0 and f(3)=0, we just need to find the critical numbers of f(x). I multiplied out and got f(x)=2x-(2/3)x2, so that f´(x)=2-(4/3)x. Since f´(x) exists everywhere in the domain, the only critical numbers occur when f´(x)=0. The only critical number of this function was x=6/4=3/2. Then we can find y, etc. So I declared that we had found the rectangle whose area was maximum.

Review for the exam!

Tuesday, March 29: this is lecture 19

I very vaguely discussed the First and Second Derivative Tests, and asked people to read the presentation and problem solutions which I wrote in the last diary entry.

Which grows bigger faster?
A casual look at the graphs of x3 and ex shows that when x gets large positive, the values of these functions are also large positive numbers. The question "Which gets larger faster?" sounds funny but really represents key knowledge in many applications, where two "processes" are occurring and we'd like to know, in terms of our mathematical model, what happens for large values of x. In typical applications, x represents time.

One way to analyze the situation, to see how the growth compares, is to consider the quotient: x3/e3. This function is easy enough to graph but I would like to use the methods of calculus to be sure we understand what happens.

Step 1
This is artificial and a trick, but I hope that you'll see the need for the trick and where the trick comes from after we're done. So:

 x3      x3          1
---- = -------- · ------
 ex     e(1/2)x    e(1/2)x
What happens on the bottom is a use of eA+B=eAeB with A=B=(1/2)x. Now I just want to analyze the first factor, which I will call f(x). So f(x)=x3/e(1/2)x.

Step 2
If x>0, then f(x) is positive since both x3 and ex are positive. Now differentiate: f´(x)=[3x2e(1/2)x-x3e(1/2)x(1/2)]/[e(1/2)x]2. Here I used the quotient rule and then the chain rule applied to the exponential part. The bottom is positive. What about the top? We can factor it:
I hope that what you have learned in the last few lectures is that the sign of f´(x) is important. Because the bottom is always positive, we need to know the sign of the top. Well, if x>0, the first factor, x2 is certainly positive, and the second factor, e(1/2)x is always positive. What about (3-(1/2)x)? If x>6, 3-(1/2)x) is negative. Therefore f´(x) is negative for x>6.

Step 3
Since f´(x) is negative for x>6, f(x) is decreasing there, so f(x)<f(6) when x>6. But f(x)<f(6) exactly means
x3/e(1/2)x<63/e(1/2)6. I will give a name to the number 63/e(1/2)6 since I will get tired typing it. I will call this number TOAD.
So we know that if x>6, f(x)<TOAD.

Step 4
Now we reassemble the original fraction, and make some conclusions. We started with the positive function x3/ex. We split it up in step 1, and then in step 3 concluded something. The specific application we can now make is: if x>6, then

 x3      x3          1               1
---- = -------- · ------ < TOAD · ------
 ex     e(1/2)x     e(1/2)x           e(1/2)x
Hey: TOAD is a constant, and as x-->infiinity, 1/e(1/2)x=e-(1/2)x decays (exponentially) to 0.

Now I can answer the original question. As x-->infinity, x3/e3-->0. So ex grows bigger faster than x3. The picture should show you that things work o.k.

Huh? What happens when we deal with ... more numbers
I wanted to illustrate that the qualitative behavior is much that same, so I asked, "Which of x12,304 and e.00027x grows bigger faster?" Here the numbers are chosen so that the functions aren't very easy to graph and aren't even very easy to compute. When, say, x=2, well 212,304 is about 103703 and e(.00027)2 is about 1.00054.

The general shape of x12,304 for large positive x is "clear": it goes up.

What about e.00027x? Maybe the weird number here confuses you. Well, this is an expoential so that its values are always positive. And the value when x=0 is 1. What could I tell you about other values for positive x's? The derivative of e.00027x is e.00027x(.00027). This is a positive number. So e.00027x is increasing. Let's try a specific number. How big is e.00027(555)? According to the MVT,

 e.00027(555)- e.00027(0)
--------------------- =  e.00027(c).00027
The stuff on the right-hand side is the derivative at c. The value of the right-hand side is at least .00027 since the exponential is at least 1. If you untwist this fraction, you can see that e.00027(555) is at least 1+(.00027)(555). Hey, the bigger the number is, the bigger this underestimate is.
Any exponential of the form ekx with k positive is increasing, and eventually grows beyond any bound.

Now back to business:
Which of x12,304 and e.00027x grows bigger faster? I need to analyze x12,304/e.00027x as x-->infinity.
I will write everything in a fashion similar to the analysis above.

Step 1
This is artificial and a trick, but I hope that you'll see the need for the trick and where the trick comes from after we're done. So:

 x12,304      x12,304       1
-------- = --------- · -------
 e.00027x     e.0002x     e.0007x
What happens on the bottom is a use of eA+B=eAeB with A=.0002x and B=.00007x. Now I just want to analyze the first factor, which I will call f(x). So f(x)=x12,304/e.0002x.

Step 2
If x>0, then f(x) is positive since both x12,304 and e.0002x are positive. Now differentiate: f´(x)=[12,304x2e.0002x-x12,304e.0002x(.0002)]/[e.0002x]2. Here I used the quotient rule and then the chain rule applied to the exponential part. The bottom is positive. What about the top? We can factor it:
I hope that what you have learned in the last few lectures is that the sign of f´(x) is important. Because the bottom is always positive, we need to know the sign of the top. Well, if x>0, the first factor, x12,303 is certainly positive, and the second factor, e.0002x is always positive. What about (12,304-.0002x)? This factor is 0 if x=12,304/.0002. The name I'll give this number is FROG. If x>FROG, 12,304-.0002x) is negative. Therefore f´(x) is negative for x>FROG.

Step 3
Since f´(x) is negative for x>FROG, f(x) is decreasing there, so f(x)<f(FROG) when x>FROG. But f(x)<f(FROG) exactly means
x12,304/e.0002x<FROG12,304/e(.0002)FROG. I will give a name to the number FROG12,304/e(.0002)FROG since I will get tired typing it. I will call this number TOAD.
So we know that if x>FROG, f(x)<TOAD.

Step 4
Now we reassemble the original fraction, and make some conclusions. We started with the positive function x12,304/ex. We split it up in step 1, and then in step 3 concluded something. The specific application we can now make is: if x>FROG, then

 x12,304      x12,304        1                 1
-------- = --------- · -------- < TOAD · -------
 e.00027x     e.0002x     e.00007x          e.00007x
Hey: TOAD is a constant, and as x-->infinity, 1/e.00007x=e-.00007x decays (expponentially) to 0.

Now I can answer the original question. As x-->infinity, x12,304/e.00027x-->0. So e.00027x grows bigger faster than x12,304. In fact, (1010)12,304 is 10123,040 and e.00027(1010) is larger than 101,172,595. So there!

Poly versus exp
Any sort of polynomial growth, no matter how big the degree is, is eventually slower than any sort of exponential growth, no matter how small the positive exponential constant is.

There is a companion geometric language to the algebraic stuff. We verified that limx-->infinityx12,304/e.00027x=0. This means that the graph of the function gets closer to the x-axis, the line y=0. This horizontal line is called a horizontal asymptote of y=x12,304/e.00027x. There are also vertical asymptotes.

Graphing [(x-3)(x-5)]/(x-1)
Here are some comments and a series of pictures which I hope will show you how relatively straightforward manipulations can give a fairly good picture of a graph.

I tried to consider the graph of y=[(x-3)(x-5)]/(x-1). I remarked that I would like to see how much information I could get about the graph without using a graphing calculator. Certainly I can get the x-intercepts. The only places that y=0 on this graph are at 3 and 5, so that (3,0) and (5,0) are on the graph.

What about when x gets large positive? Well, there are two powers of x on top, and only one power of x on the bottom, so I bet that y gets large also. What is perhaps more interesting is the x-1 on the bottom of the fraction.
1 is not in the domain of [(x-3)(x-5)]/(x-1). What happens as x-->1+, that is, x approaches 1 from the right? The top gets close to (1-3)(1-5) which is 8. The bottom, x-1, is a small positive number. So the quotient is 8/[small +]. This is a large positive number. So as x-->1+, y-->+infinity.
What happens as x-->1-, that is, x approaches 1 from the left? The top gets close to (1-3)(1-5) which is 8 (just the same as before). The bottom, x-1, is now a small negative number. So the quotient is 8/[small -]. This is a large negative number. So as x-->1-, y-->-infinity.
The vertical line x=1 is a vertical asymptote of y=[(x-3)(x-5)]/(x-1). Now as x-->-infinity, the sign of y is negative, and, again because there are two powers of x on the top and only one on the bottom, y-->-infinity.
As I remarked in class, if we sort of interpolate the information we already have in the simplest manner possible, we can guess at what the graph looks like. In particular, the graph seems to have two critical points. I wanted to check that.

Well, y=[(x-3)(x-5)]/(x-1). I'll expand the top to differentiate it more easily: y=[x2-8x+15]/[x-1]. Then
dy/dx=[(2x-8)(x-1)-(x2-8x+15)(1)]/(x-1)2= [2x2-10x+8-x2+8x-15)]/(x-1)2 [x2-2x-7]/(x-1)2. Where is this fraction equal to 0? Well, the roots of the quadratic top are at [+2+/1sqrt{(-2)2-4(-7)}]/2. This is 1+/-sqrt{29}. The roots are at about 3.828 and -1.828, which you can see in the "real" graph I had a computer draw. Thursday:
4.3: 19, 28, 37, 40;
4.4: 9, 11, 15, 23.

Exam warning
The second exam will be on Friday, April 8. It will concentrate on the material covered since the first exam, up to and including section 4.6. Please see the home page for review material, including the formula sheet which will be distributed with the exam.

Friday, March 25: this is lecture 18

The Oxford English Dictionary says that paradigm, meaning "A pattern, exemplar, example" was first used in (written) English in 1483. I graphed y=x3-x and y=3x2-1 and y=6x. I wanted to use the function f(x)=x3-x as a (relatively) simple example of the material in section 4.3. There is a lot of material in section 4.3.

This part of the text discusses the relationships between the graphs of a function and its first and second derivatives. For example, (this was my one moment of theory in the lecture), we know from the MVT that there's a c between a and b so that f´(c)=[f(b)-f(a)]/[b-a]. But look: if b>a and if f´(c) is positive, then f(b)-f(a) must be positive so f(b)>f(a). Therefore:

If f´(x) is positive, then f(x) increases (the graph goes up
as we travel from left to right).


If f´(x) is negative, then f(x) decreases (the graph goes down
as we travel from left to right).

It's usually easier to see if a function is positive or negative, rather than try to deduce directly if another function is increasing and decreasing.

For example, here I look at f´(x)=3x2-1. I hope that you believe the graph and, what's more, you see that the roots of this quadratic are at {+/-}1/sqrt(3). Hey: to the left of -1/sqrt(3), the function is increasing. I don't think the following is obvious: f(-500)=(-500)3-(-500) is less than f(-500)=(-400)3-(-400). And I don't need to compute those numbers, either.

The second derivative shows how the slope of the tangent line changes as we travel from left to right. If the second derivative is positive in an interval, then the first derivative is increasing in that interval, so as we "move" from left to right, the slopes rotate "up" or counterclockwise. The graph of the original function lies above the tangent lines in this case, and the graph bends up: it is concave up. A similar phenomenon occurs when the second derivative is negative. Summarizing:

If f´´(x) is positive, then f(x) lies above its tangent lines:
the graph is concave up.

If f´´(x) is negative, then f(x) lies above its tangent lines:
the graph is concave down.

Points where the graph switches concavity are called inflection points. The situation is very analogous with first derivatives and relative {max|min}s. So:

If a function is differentiable, its relative extrema (maxes
and mins) must occur at critical points where f´(x)=0.

If a function is differentiable, its inflection points (where concavity changes) must occur where f´´(x)=0.

Graph of the function, f(x)=x3-x

Graph of the first derivative, f´(x)=3x2-1

Graph of the second derivative, f´´(x)=6x

The success of this approach is really how well it does in practice. So let me do a sequence of problems, first one from WeBWork and then a bunch of problems from section 4.3.

A WBWorK problem
Suppose f(x)=9e-x2/50. Find the absolute maximum of f(x) and find the x-coordinates of the inflection points of f(x).
I said in class that generally the first thing I'd probably do is use a graphing calculator to get some idea of what the function's graph looks like. But you may not be very familiar with how the calculator works, and many keystrokes can be necessary to enter the function. A more irritating problem is that the window for a suitable graph may not be obvious. Unfortunately, the window problem is real and can't be avoided, except for almost the simplest functions. And, of course, it reflects difficulties in reality: what sizes are useful to deal with particular phenomena (weights in milligrams might be good to handle hummingbird food, but not so useful in discussing the diets of elephants).

A graph is displayed to the right. The window is [-15,15] for x and [0,10] for y. The absolute max seems to occur when x=0. The value of this max is f(0)=9 since e0=1. Near the max, the curve is concave down. Out towards the edges (where the curve is asymptotic to the x-axis) the curve seems to be concave up. An initial guess would be that there are at least two inflection points, one positive and one negative, symmetric with respect to the origin.

Confirming the picture
Here we differentiate and do the algebra. If f(x)=9e-x2/50 then f´(x)=9e-x2/50(-2x/50) using the chain rule. The second derivative needs both the product rule and the chain rule: f´#180;(x)=9e-x2/50(-2x/50)2+9e-x2/50(-2/50).
Now we try to find critical points, and therefore we search for x's satisfying f´(x)=0 which is the same as 9e-x2/50(-2x/50)=0. Notice
The exponential function is always positive and never 0.
Therefore the only way 9e-x2/50(-2x/50) can be 0 is if (2x/50)=0 or x=0. So 0 is the only critical point. Also please notice that if x<0, 9e-x2/50(-2x/50)=0 is positive (because of the minus sign in front of the 2x). Also the first derivative is negative for x>0. This means that f(x) increases to the left of 0 and decreases to the right of 0. So f(0)=9 must be the absolute max of f(x).

Let's look at f´´(x)=0 which is 9e-x2/50(-2x/50)2+9e-x2/50(-2/50)=0. This is a more complicated expression. To see if "something"=0, if I can factor the "something" then I will get A·B=0 which occurs exactly when either A=0 or B=0. The factors here practically yell at us, I think. 9e-x2/50(-2x/50)2+9e-x2/50(-2/50)=0 becomes
[9e-x2/50]·[(-2x/50)2+(-2/50)]=0. Again, the exponential function (the "A") is never 0, so we need to know when (-2x/50)2+(-2/50)=0. This is (square it correctly!) [(4x2)/502]-2/50=0 or [(4x2)/502]=2/50. Multiply by 502 and divide by 4 to get x2=25 so that x=+/-5. Hey: here are the x-coordinates of two inflection points which we saw.

Textbook problem 4.3 #17
Well, I certainly wasted some class time with this problem. Let's start: Here f(x)=(x-1)/(x2+3). The textbook requires (please look at it) all of the information that can be gotten from the first and second derivatives. So we need to find critical numbers, intervals where the function is {in|de}creasing, critical points and relative {max|min}'s, intervals of concavity ({up|down}), and inflection points. To the right is a graph in the window where x is between -10 and 10. I tried several candidates for the y interval to get the good view shown of the curve. There seems to be two critical points, one a max and one a min. Close inspection allows us toe guess that there will be three inflection points. "Far" to the left the curve looks concave down, then at the relative min it is concave up, and then at the relative mx the concavity flips again down, and, finally, away to the right the concavity is up. Please keep this in mind as we do the algebra and the resulting computations.

The first derivative
The quotient rule gives f´(x)=[1·(x2+3)-2x·(x-1)]/(x2+3)2. At least this isn't as horrible as one might think. The bottom is a square, and it is always positive. The sign of the fraction and whether or not the fraction is 0 is totally ruled by the top. The top is 1·(x2+3)-2x·(x-1)=x2+3-2x2+2x=-x2+2x+3. Well, -x2+2x+3=-(x2-2x-3)=-(x-3)(x+1). Should I say "clearly" here since I messed up my first factorization attempt in class. Well, the roots are x=3 and x=-1. I bet these are critical numbers (look at the picture!). What about the critical points? These should be the points on the curve corresponding to the critical numbers. The second coordinates will be gotten using f(x)=(x-1)/(x2+3). So (-1,-1/2) and (3,1/6) are the critical points. The function is increasing on the interval [-1,3] and is decreasing on the two intervals (-infinity,-1] and [3,infinity).

The second derivative
We've got to differentiate (-x2+2x+3)/(x2+3)2 so we will use the quotient rule. Sigh. Therefore f´´(x)=TOP/BOT where
Some thought will save some pain. Since TOP and BOT are part of a fraction, we can factor and then delete any multiples which all terms have. One thing that appears everywhere is x2+3. There are four of these in BOT, and two in the first part of TOP and one in the second part of TOP. Therefore I will drop one power of x2+3, and get
f´´(x)=[ (-2x+2)(x2+3)-(-x2+2x+3)2(2x)]/(x2+3)3. The bottom of this expression is always positive, so I can check for sign and for where f´´(x)=0 just be considering the top. Wow, I guess. Now the top of this expression for f´´(x) is (-2x+2)(x2+3)-(-x2+2x+3)2(2x) which is -2x3-6x+2x2+6+4x3-8x2-12x and this is 2x3-6x2-18x+6

How to waste time pointlessly ...
Well, everyone has their favorites, but one of mine is to try to "solve" an equation like 2x3-6x2-18x+6=0 or (dividing by 2) x3-3x2-9x+3=0. I thought (darn it!) since this was a textbook problem we could maybe get integer or simple fraction solutions. Let me be polite to myself: this turns out not to be the case. The only nice observation I made at this stage was to note that since we've got a degree 3 polynomial, the roots will probably be the three inflection points we guessed when we looked at the graph. Well, after several students pushed and pushed, I yielded: the roots seem to be (approximately!) -2.06417 and .30561 and 4.75877. Yes, these are gotten from silicon-based objects, not carbon-based objects, that is, from calculators/computers, not my feeble brain. Sigh. These are the x-coordinates of the three inflection points. From these, you can state where the graph is concave up and concave down. I am embarrassed.

Textbook problem 4.3 #21
Here f(x)=1+2x+18/x. A quick graph is shown to the right, but even with the relatively simple function, I admit that I played with the window (finally showing you [-12,12] by [-21,21]) in order to get something that I thought showed the curve nicely.

The first derivative
So f´(x)=0+2-18x-2. Hey I thought 18/x=18x-1 so the derivative of that piece is -18x-2. When is 2-18x-2=0? I would multiply by x2 and divide by 2 so that I get x2-9=0. I bet that the critical numbers are +/-3 (look at the picture!). What about the critical points? Well f(3)=1+2(3)+18/3=13 and f(-3)=-11, so that (3,13) is a critical point (and a relative min) and (-3,-13) is a critical point (and a relative max). The intervals of increase and decrease should be apparent to you, I hope. f(x) is increasing on (-infinity,-3] and [3,infinity) and f(x) is decreasing on [-3,0) and (0,3]. Please, there is one "wrinkle" that happens here which has not previously appeared. The domain of f(x) does not include 0, so therefore 0 should not be in any of the intervals we specify ({in|de}crease, concave {up|down}). In fact, you can look at the picture and see that f(x) is not decreasing if you pack the values [-3,0) and (0,3] together. Why? Well, f(-1)=-19 and f(1)=21 so f(-1)<f(1). So there is some care and precision needed when the domain has "holes".

The second derivative
Since f´(x)=0+2-18x-2, f´´(x)=36x-3. Look at the graph, please! When x is negative, the second derivative is negative (-3 is an odd integer and odd powers of negative numbers are odd). Therefore f(x) is concave down in (-infinity,0). Similary, f(x) is concave up in (0,infinity). But here is the wrinkle again: f(x) has no points of inflection, since 0 is not in f(x)'s domain. The official definition of a point of inflection (p.207 of the text) begins, "A point P(c,f(c) on the curve is called an inflection point ..." and there is no point on the curve when c=0. So trick answer: this curve has intervals where it is concave up and concave down, but it has no point of inflection.

Textbook problem 4.3 #26
Here f(t)=(t3+3t2)3. O.k.: I said in class that you should reach for the graphing calculator first. Here's a case where good pictures are hard to get.

too big
Look at the function. What are some values? Well, f(0)=0 (not too bad) but f(1)=43=64 and f(2)=(8+3·4)3=(20)3=8,000. So if you look at this thing in a window like [-5,5] by [-5,5] you will miss lots of stuff because the function is too big.


too small
Actually the situation is even worse for the casual user of an electronic graphing device. Look at the first picture, which I think is in the window [-4.5,2.5] by [-2,100]. Of course we seem to be missing information on the left. Why? Well, for example, f(-4) is the third power of (-4)3+3(-4)2=-16 and this is -4096. O.k., so we lose that. What is worse is the situation around x=-3. Golly, here the function is too small. I don't see anything too interesting in the big picture near -3. The second picture, below the first, is in the ludicrous (at first consideration, anyway) window [-3.1,-2.9] by [-.85,.62]. This window is really focused at (-3,0). I only asked for this picture after I did the work below. In some sense, I cheated. In fact, I got both windows only after I did the algebraic work. In this case, the algebra guided my use of the graphing device. Sometimes things can be very, very strange.

The first derivative
Since f(t)=(t3+3t2)3, f´(t)=3(t3+3t2)2(3t2+6t) (the chain rule). This is already in a nicely factored form. It will be equal to 0 when either t3+3t2=0 or when 3t2+6t=0. So t3+3t2=t2(t+3) so there are critical numbers at 0 and -3. And 3t2+6t=3t(t+2), so there is an additional critical number at -2. What kinds of realtive {max|min} can we expect? Hey, if we already know the pictures, then we know the answers, but, as I wrote above, I got the pictures after I did this. Since f´(t)=3(t3+3t2)2(3t2+6t), I can tell you that (after I think a bit!) f´(t)=3(t3+3t2)2(3t2+6t) is actually 3t3(t+2)(t+3)2. Here is where it gets interesting (or difficult, depending on your attitude). There is a sign change in f´(t) at -2 and 0 since the exponents of the corresponding factors are both odd (1 and 3). But we've got (t+3)2. When you look near the left and right sides of -3, the square doesn't change sign. So -3 won't correspond to either a relative max or a relative min. Sigh. Or look at the pretty pictures, and hope that someone else will always be responsible for the pretty pictures.

The second derivative
Since f´(t)=3(t3+3t2)2(3t2+6t), f´´(t)=6(t3+3t2)1(3t2+6t)2+3(t3+3t2)2(6t+6), using both the product rule and the chain rule. Let me try to factor this mess.


I guess
When it this mess equal to 0? One factor is 6(t3+3t2) and this is 0 when t=0 and t=-3.
The other factor is (3t2+6t)2+3(t3+3t2)2(t+1). Ummmm .... not clearly, this is 18 t4(2t+5)(2t+3)(t+3). I refuse to write out the details. There are exactly three inflection points. 0 is not an inflection point, because it is a relative min, and the curve is concave up in an interval around 0. But the numbers -5/2 and -3/2 and -3 are all inflection points. This computation should not be shown to anyone under the age of 25.

Textbook problem 4.3 #38
The text states: "use the first-derivative test to classify each of the given critical numbers as a relative minimum, a relative maximum, or neither." The function in this problem is f(x)=(x2-4)4(x2-1)3 and the critical numbers given are x=1 and x=2.

The first derivative test for what happens at a critical point

immediately to the left, the first derivative is positive
immediately to the right, the first derivative is negative
there's a relative max.

Immediately to the left, the first derivative is negative
immediately to the right, the first derivative is positive
there's a relative min.

the signs of the derivative are the same on both sides
the critical point is an inflection point.

The picture gives it all away, but I will do the horrible algebra. Since f(x)=(x2-4)4(x2-1)3 we get
using the chain rule and the product rule. I'll factor this mess:
f´(x)=(x2-4)3(2x)(x2-1)2 [4(x2-1)+(x2-4)3]

What happens near x=1?
Near x=1, the factor (x2-4)3 looks to me like (-3)3, negative.
Near x=1, the factor 2x looks like 2, positive.
Near x=1, the factor (x2-1)2 looks, well, the exponent is even so I bet it is positive (except exactly at x=1 where it is 0).
Near x=1, the factor 4(x2-1)+(x2-4)3 looks like ... uhhh ... 0 (from the 4(x2-1) term) and -9 (from the (x2-4)3 term), so it looks negative.
The result (the product of everything) is that the sign of the derivative looks positive on both sides of x=1.

What happens near x=2?
Near x=2, the factor (x2-4)3 does change sign because the power, 3, is odd. On the left, if x is a bit less than 2, x2 is less than 4, so x2-4 is negative, and the cube of this is negative. On the right, if x is a bit more than 2, x2 is larger than 4, so x2-4 is positive and the cube of this is positive.
Near x=2, the factor 2x looks like 4, positive.
Near x=2, the factor (x2-1)2 looks like (22-1)3 which is positive.
Near x=2, the factor 4(x2-1)+(x2-4)3 looks like ... uhhh ... 12 (from the 4(x2-1) term) and 0 (from the (x2-4)3 term), so it looks positive.
If you collect all this sign information, then the sign of the derivative goes from negative to positive at x=2, so this must be a relative minimum.

Hey: we can see this in the darn graph! This is a lot of work. Who cares?

Textbook problem 4.3 #42
Here the text states: "use the second-derivative test to classify each of the given critical numbers as a relative minimum, a relative maximum, or neither." The function in this problem is f(x)=(x2-3x+1)e-x and the critical numbers given are x=1 and x=4.

The second derivative test for what happens at a critical point

the second derivative at the
critical number is positive
there's a relative max.

the second derivative at the
critical number is negative
there's a relative max.

the second derivative at the
critical number is zero
no conclusion can be made.

Again, the picture tells us what's going on, but I'll do the second derivative test. Since f(x)=(x2-3x+1)e-x, f´(x)=(2x-3)e-x+(x2-3x+1)e-x(-1) (product rule and chain rule). Let me rearrange this slightly:
Therefore f´´(x)=(-2x+5)e-x+(-x2+5x-4)e-x(-1)= [(-2x+5)+(-x2+5x-4)(1)]e-x= (x2-7x+9)e-x.

What happens near x=1?
Since f´´(1)=(12-7·1+9)e-1=3e-1 and values of ex are all positive, we see that f´´(1)>0, so by the Second Derivative Test, f(x) has a relative minimum at x=1.

What happens near x=4?

f´´(4)=(42-7·4+9)e-4=-3e-1 and values of ex are all positive, we see that f´´(1)<0, so by the Second Derivative Test, f(x) has a relative maximum at x=4.

Again, why does this matter, since we can see all this in the darn graph! This is a lot of work. Who cares?

I drew a graph of a function on the board. I asked that students do the following:

This exercise is an antidote to all the algebra. It is a qualitative exercise, which means that reasonable people (even the instructor!) can disagree about exact details. For example, I did not mean to have inflection points out on the edges, but I am sure that some students saw such points there.

A solution to the QotD

A simple sketch of the derivative follows mostly from noticing the signs of the the slopes of the tangent lines.

Here is one possible answer:

Why should anyone care?
So there seem to be several ways to characterize maxes and mins. This is good because sometimes what works in one situation doesn't work in another. Also, because in some applications, the first and second derivatives turn out to have important meaning in the subjects. For example, I loathe the second derivative test because computing second derivatives turns out to be very irritating for any moderately complicated function. But in economics, the second derivative test is very useful to describe certain max/min situations. Also, you should realize that the examples we've been looking at usually can be graphed easily (but not even all of them). In practice, the scale of certain processes is not clear. Again, in biochemical reactions, some processes occur much faster than others. Just graphing without any thinking frequently may omit very useful information. The information could be discerned by looked at the first and second derivatives.


Tuesday, March 22: this is lecture 17

Welcome to spring!

Review of the last lecture
Stuff to memorize:

One example
Here is an example of a problem which our "technology" can now handle quite well. Consider the graph of y=1-x2. This is a parabola opening downwards, with its vertex at (0,1) as shown in the picture to the right. There is a region in the plane whose boundary is an interval ([-1,1]) of the x-axis and the "top" arch of the parabola. What is the rectangle inside that region whose area is largest?

I do not believe this is a profound problem, but it is certainly a problem whose exact solution is not obvious. There are many rectangles inside that region of the plane. We can "swell up" a rectangle until it touches the boundary, and then rotate the rectangle. Some thought should convince you that the largest rectangle will sit as shown, with one side along the x-axis, and two corners touching the arch of the parabola. But there are many such rectangles:

How can we find the rectangle which has the most area? Well, let's call the coordinates of the upper-righthand corner of the rectangle (x,y). Then the area of the rectangle is 2xy (the "2" comes from the total length of the horizontal edge). Since (x,y) is on the parabola, y=1-x2, and the area must be 2x(1-x2)=2x-2x3. What x's are "eligible"? Surely not x=500. If you look carefully, you can see that any x in the interval [0,1] describes exactly one of the eligible rectangles. So now we have a calculus problem.

The resulting calculus problem
What is the maximum of the function A(x)=2x-2x3 when the domain of A(x) is [0,1]. Well, this absolute maximum will be attained either at the endpoints or at a critical point. The endpoints are 0 and 1. What are the critical points? A´(x)=2-6x2. This will be 0 when 6x2=2 or x2=1/3. Since x can't be negative in the domain of A(x), the only solution is x=1/sqrt(3).

There are three candidates for values of x at which A(x) can have a maximum value. The simplest way to find the absolute maximum and the value of x at which it occurs is to evaluate A(x) at these numbers. Now remember that A(x)=2x-2x3.
Left endpoint A(0)=0 (the rectangle of the five displayed which is thin but about one unit high has very little area: look at the rectangle on the right).
Right endpoint A(1)=0 (the rectangle of the five displayed which is very short but about two units long has very little area: look at the rectangle on the left).
Critical number A(1/sqrt(3))=2/sqrt(3)-2(1/sqrt(3))3. This works out to 4/(3 sqrt(3)). This is the largest rectangle.

Again, this is not a profound problem, but the method is straightforward and can be applied in many different ways. We will come back to other such applications, but now I need to go on, following the textbook:

Two dots on the graph ...
I drew coordinate axes with two big dots on the x-axis and urged students to come up and draw a graph going through the two dots. I got several graphs similar to what is displayed.

I observed that in each case there were points on the graph where the tangent line was horitizontal. Below are the graphs with those points and parts of the tangent lines displayed. Only, I was rude and after the students' graphs were analyzed, I drew something like the first graph below, with a bad point. People really like to draw s-m-o-o-t-h curves.

The logic of the situation is this: if the function is positive between the two points, then it has a positive relative max inside the interval. That must be a critical point (horizontal tangent or not differentiable). If the function is negative, then ... the same situation below the x-axis. (Hey, yeah, if the function is never positive and never negative, then it has lots of horizontal tangents, because the graph is itself a horizontal line.)

Several versions of Rolle's Theorem
My informal description was: if there are two dots on the x-axis which are part of the graph of a function, then either the function has a bad place in between or has a flat place somewhere in between. Since "bad" and "flat" are not usually considered precise and suitable words in Math 135 (and there are other implicit assumptions such as continuity), I rephrased this statement. Also, the way the result is applied usually assumes that the function is differentiable everywhere, so the "bad" alternative vanishes.

Rolle's Theorem (official version)
Suppose f(x) is a differentiable function, and f(a)=0 and f(b)=0. Then there is at least one number c between a and b so that f´(c)=0.
Tilting Rolle's Theorem
People rarely use Rolle's Theorem directly. The hypotheses are rather strict. Usually the result is tilted, as in the accompanying picture. But the picture is not labeled conventionally, so the second picture is what you see in calculus books most of the time. There are some lines in the second picture which are parallel. The line connecting the points (a,f(a)) and (b,f(b)) has slope [f(b)-f(a)]/[b-a] (difference in second coordinates divided by the difference in the first coordinates). The other lines have slope given by the derivative of f(x). So you get the statement of the Mean Value Theorem, which is below.

The Mean Value Theorem (MVT)
Suppose f(x) is differentiable. Then there is at least one number c between a and b so that f´(c)=[f(b)-f(a)]/[b-a].

This is one of the major structural (?) results of calculus, and I will call it MVT. I asserted in class, and really believe, that the reasoning usind the MVT is often used by everyone. Please keep this in mind as you read these examples.

Return to Francine: distance gives information about "speed"
On February 1st we considered my friend Francine. Remember her, please. She was driving on the Garden State Parkway. Here is the key information about her travel:

Time7 AM10 AM
(in miles on GSP)

Let me make some assumptions about the Garden State, which are actually not true, but will allow me to use some calculus. So assume that:
First, access to the GSP is controlled by a ticket system (similar to what the Turnpike uses) and these tickets are dispensed and read by machines at each end of the trip.
Second, that the speed limit on the GSP is 55 miles per hour on its whole length. What happens then?

Francine gets a speeding ticket
Here is the setup: suppose that s(t) describes the distance that Francine travels along the GSP. I will meaaure time in hours, starting the measurement at 7AM. I will measure s(t) in miles from the beginning of the parkway in Cape May. What do I know from the table above? I know that s(0)=0 and s(3)=172. I will also assume that Francine's travel is differentiable, and now look at s´(t). This is officially her velocity but I will call it "speed" here, since that's what this quantity is usually called. The MVT declares
There is at least one number c between 0 and 3 so that s´(c)=[s(3)-s(0)]/[3-0]=[172-0]/[3-0]=172/3. But this number is 57 (and a third) and is greater than 55, the official speed limit. Therefore she must have been speeding at some time during the trip.

Term paper (optional)
Explore whether this sort of thing is actually done in New Jersey or elsewhere, wherever there are roads whose access is controlled in such fashions. Explore that political and legal implications of using these ideas in enforcing speed restrictions.
I am curious. Please tell me if you know any situations like this.

The reverse: "speed" gives information about distance
In Francine's case, information about the function was used to get information about the derivative. The MVT can also be used the other way, and I said that such ideas are used very often in real life. So here is a sample situation.
Suppose you are driving down a road, and at time 0 you are 20 miles from the start of the road: s(0)=20. Also suppose that you are driving so that your speed is always between 30 and 40 miles per hour. What can you say about your distance from the start of the road 3 hours later?

The reaction from the tired students (who had allo just spent a week in a luxury hotel in Cozumel, I think) was not enthusiastic in response. I did hope that people would start giving me answers. But let's use the MVT: I know that s´(c)=[s(3)-s(0)]/[3-0]=[s(3)-20]/3.
But we are told 30<=s´(c)<=40, so that
30<=[s(3)-20]/3<=40. Multiply by 3 and get
90<=s(3)-20<=120. Add 20 and get
Therefore three hours later the traveler will be at least 110 miles from the start of the road, and will also be no farther than 150 miles from the start of the road. So information about the derivative gives you information about the function.

Increasing and decreasing
If the derivative is positive then the function increases (drive forward and increase the distance from the start).
If the derivative is negative then the function decreases (drive backward and decrease the distance from the start).

A specific function
This one is much uglier than what I wrote in class. Please: what I am about to write is, of course, not a "random" function, but an example I have invented to show you a sequence of ideas.
Suppose you want to look at f(x)=445x301+678x501+19x+sin(3sin(2x)). (Is that ugly enough?) Now I'd like to compare f(100) and f(50). Which number is larger? Please realize that part of the consideration in writing this function was to make the exponents so darn large that our simple calculators can't compute f(100) and f(50). So some reasoning is needed.
Step 1 Compute the derivative. By now, I hope you will agree that this is basically easy. Yes, mistakes can be made in these computations but we can probably usually get it correct.
      f´(x)=445·301x300+ 678·501x500 +19+cos(3sin(2x))3·cos(2x)2.
Step 2 Look at the derivative and try to figure out the sign of the derivative. Look at the pieces:      

So we have four pieces. If you consider now the sum of these, I bet that the sum is at least 19-6=13.
Step 3 The derivative is positive. Therefore this function is increasing: bigger values of the inputs to the function result in bigger values of the outputs. So now I know that f(100) is larger than f(50).

Yes, this function is ludicrous, but the ideas are what I want to show you. Here they are:

Step 1: What is the behavior of a function defined by a complicated formula?

Step 2: Differentiate the function and get a mess. Now look for any simple information you can about the sign of the derivative. This search is frequently not difficult.

Step 3: Get conclusions about the {in|de}decreasing behavior of the original function.

Another specific function
I think I looked at f(x)=sqrt(5x500+4). I computed (by hand, a marvel of intelligence!) that f(0)=2 and f(1)=3. I asked what could we conclude about f(x) for x between 0 and 1. I was told that these values would be between 2 and 3.

QotD (Maybe an 8 point question on the next exam) Explain why f(x) is between 2 and 3.
Hint Use the sign of the derivative of f(x).
Hint2 After a few minutes I think I wrote this on the board: (1/2)(5x500+4)-1/2(2500x499).

Please read sections 4.1 and 4.2 and hand in these problems on Thursday:
4.1: 8, 10, 20, 25;
4.2: 5, 13, 16, 30.

Friday, March 11: this is lecture 16

This is the diary entry for the day before vacation.

I did a WeBWorK problem involving error and relative error.

I try to be enthusiastic, and I was there. What did I want to cover? Basically, I wanted to recite five definitions and an inference, and give enough supporting examples so that students would believe the inference. I did discuss adequately three of the fivce definitions and the inference, but I ran out of time before I covered the last two definitions very well. That's sort of annoying, and I apologize, especially to the students who were present (about 60% of the usual crowd).

Definition #1
A function f(x) has a relative maximum at c if

Definition #2
A function f(x) has a relative minimum at c if

Then I sketched many many pictures on the board to illustrate these concepts. The only amusing thing was that I used as examples graphs we had already drawn in past lectures, and maybe that was sort of fun (trees growing, tax revenue, etc.). Here are some examples of the creatures I drew. In each case I labeled any relative minima with rm and I labeled any relative maxima with RM. I have also tried to include some useful comments on the results, that is, the labels or the lack of labels, which sometimes is more surprising.

The top of the graph is a Relative Max and the bottom is a Relative Min. The endpoints are not eligible to be relative extrema because the function is not defined in an interval containing either of them in its inside.

There is no Relative Max (again each endpoint is not eligible since the function is not defined in an interval containing that point). There's a Relative Min where indicated, because nearby points (on both sides) have larger function values.

The bottom of the graph is a Relative Min. There are no points representing a Relative Max.

There are no Relative Max and no Relative Min in this picture. The highest and lowest points are on the ends which aren't eligible.

I actually went through about ten or a dozen such pictures. This was not the best use of instructional time.

Definition #3
Suppose the function f(x) is defined at c. Then f(x) has a critical number at c if

Then (c,f(c)) is called a critical point of f(x).

The tangent line is horizontal at the top and the bottom. At these points, f´(c)=0, so these points are Critical Points.

The function is not continuous at a point so the derivative doesn't exist there. But points where the derivative doesn't exist are still Critical Points.

The function is not differentiable at the corner since the left- and right-hand limits defining the derivative don't agree there. The corner is a Critical Point.

There are two points where the derivative doesn't exist and these are Critical Points.

Here's the inference
These examples and the others I displayed were supposed to help people to believe that if f´(c) exists and wasn't 0, then nearby values of f(x) were higher or lower, so that c was not a relative extremum. The following statement (the "inference" I mentioned) would be true:

Relative extrema occur only at critical points.

Here is an "if ... then" statement of the same inference:

If c is a relative max or min, then (c,f(c)) is a critical point.

One definition of inference is "the forming of a conclusion from premisses." Of course, a definition of premiss is "a previous statement from which another is inferred." Does this help?

Some algebraic examples of critical points

The functionsSketch of the graphs
(A machine helped me with these.)
#1 What are the critical numbers of f(x)=(1/3)x3+(1/2)x2-6x?

So f´(x)=x2+x-6=(x+3)(x-2), and this is 0 when x=-3 or x=2.
The critical numbers are c=-3 and c=2.

#2 What are the critical numbers of f(x)=x2e-3x?

Here f´(x)=2xe-3x+x2(-3e-3x)=(2x-3x2)e-3x. Since the exponential function is never 0, the derivative is 0 only when 2x-3x2=0, which is when x(2-3x)=0.
The critical numbers are c=0 and c=2/3.

#3 What are the critical numbers of f(x)=x1/3(x-2)?

Now f´(x)=(1/3)x-2/3(x-2)+x1/3(1). Stay alert! Algebra coming. I will rewrite this as a product, because when I write it as a product, the result will be equal to zero exactly when either of the factors is 0.
I get f´(x)=x-2/3[(1/3)(x-2)+x]. Now when x=0, the first factor, x-2/3, doesn't make any sensed because the negative exponent means that we would be dividing by 0. So 0 is a critical number of f(x), because the derivative is not defined at 0. The second factor, (1/3)(x-2)+x, is equal to 0 when (1/3)(x-2)+x=0, which happens when x-2+3x=0 or when 4x=2 or when x=1/2. So 1/2 is also a critical number of this function.
The critical numbers are c=0 and c=1/2.

In general, if I give you a "random" function, finding out where the function is "highest" or "lowest" might be very difficult. But the inference allows one to throw out points which can't be highest or lowest. But here are two more definitions:

Definition #4
A function f(x) has an absolute maximum at c in an interval if f(c)>=f(x) for all x in that interval.
Definition #5
A function f(x) has an absolute minimum at c in an interval if f(c)<=f(x) for all x in that interval.

If you look at the examples and at the inference above then you should believe that

Absolute maxes or mins can occur only at critical numbers or endpoints.

QotD I asked people to answer problem #25 in section 4.1: find the absolute maximum and the absolute minimum of s(t)=t cos(t)-sin(t) on the interval [0,2Pi].
This was not a good QotD. I had not adequately explained the ideas and had certainly not discussed the statement in blue written above. I need to spend more time in class on this and I plan to.

Maintained by and last modified 3/22/2005.