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HERBERT GROSS: Hi, our lecture today, if we're looking at
this from an analytical point view, should be called 'Maxima
and Minima'.
And if we're looking at it from a geometric point of
view, 'High Points and Low Points'.
And actually, whichever point of view we're looking at from,
it's a very nice application off much of the theory that we
have learned up until now.
So without further ado, let's talk a little bit about what
we mean by high points, low points, maxima or minima.
Which is, as I say, I've called the lecture today.
Now, as is always the case, we usually have to have some sort
of a fundamental result from which all of our
other results follow.
And in this context what I call the fundamental theorem
for a study of maxima minima is the following.
Suppose that 'f of c' is at least as great as 'f of x' for
all 'x' in a delta neighborhood of 'c'.
In other words, we have some open interval with delta
surrounding 'c'.
And then for all 'x' in that neighborhood, 'f of c' is at
least as great as 'f of x'.
Or equivalently, it might be that 'f of c' is less than or
equal to 'f of x' for all 'x' in this neighborhood.
And suppose also, that 'f prime of c' exists.
Then the theorem says, 'f prime of c' in this
case, must be 0.
And whereas this can be proven analytically, again the
analytical proof is motivated by what's happening
geometrically.
And the geometric demonstration is particularly
easy to visualize in this particular case.
Let me run the risk of drawing this fairly freehand here.
See what we're saying is this.
Suppose you have at the point (c, 'f of c') on this
particular curve.
Suppose that this particular curve the derivative exists
and is positive.
See that's the way I've drawn this.
In other words, notice that in the neighborhood of the point
'c' here, for example, the derivative here is positive.
The curve is always rising.
Now, what we're saying over here is that if you look at
this particular picture, observe that 'f of c' will not
exceed 'f of x' for all 'x' in this neighborhood.
In fact, I think you can see just by looking at this
picture that as 'x' increases, 'f of x' increases.
In other words, in terms of this picture, if 'x' is less
than 'c', 'f of x' is less than 'f of c'.
And if 'x' is greater than 'c', 'f of x' is
greater than 'f of c'.
In other words, what we're saying is if the derivative is
positive, it means that the curve is always rising.
And hence, the point in the middle of that neighborhood
cannot be the highest point in that neighborhood.
Nor, for example, can it be the lowest point.
In fact, a similar argument holds if we want to illustrate
that 'f of c' is less than or equal to 'f of x' in this
particular neighborhood.
And by the way, again, observe what we're saying.
This is rather crucial over here.
First of all, we're talking about a sufficiently small
neighborhood of 'c'.
What I mean by that is something like this.
And by the way, this could cause a little bit of a
problem if you're not careful with your language.
If we drew a picture like this and you say to somebody, where
are the high-low points on this?
I think it's quite natural that the person would say,
well, this is the high point.
I'll call that 'H'.
And this is the low point.
I'll call that 'L'.
And he might tend to forget about this point here because
even though it's fairly high, it's not nearly as high as
this point.
What I'd like you to see however is that our definition
talks about what?
In a neighborhood of the given point.
You see what we're saying here is that if we pick a
particular neighborhood, an appropriately chosen
neighborhood surrounding 'c', then what is true is that at
the value of 'x' corresponding to 'c', 'f of c' is the
highest point in a suitable neighborhood of 'c'.
In other words, if you knew that for some reason or other,
you had to be working in this neighborhood here, you could
say, well, in the domain in which I'm interested in, this
is a high point.
And this is why we talk about 'local' or 'relative' maxima
or minima in addition to 'absolute' maxima and minima.
In fact, you see, this happens quite frequently in practice.
Suppose you were doing an experiment and you really
wanted to produce the largest possible value of 'y'.
Well, you see, utopianaly you would like to pick 'x' out
here someplace.
But suppose because of some constraint in the laboratory,
the largest value of 'x' that you could choose
might be over here.
And then you see what the equivalent problem would be.
And this is where the domain of the function has such a
powerful meaning in terms of practical applications.
What you're saying is look-it , if the domain of my function
is limited to this interval over here, then this
particular point as far as I'm concerned
is the highest point.
In other words then, notice that the labelling 'N sub
'delta of c'' indicates that you're talking locally rather
than globally in a neighborhood of 'c'.
And this is the important issue here.
Now, the hardest part about this particular result as I
see it, is not understanding the result as much as it is of
reading more into the result than what's really there.
And so I have a few cautions for you.
The three commandments they turn out to be.
The first is, beware of false converses.
And before you can be beware of false converses, you have
to know what a converse is.
Roughly speaking, a converse applies only to an if-then
type of statement.
And you get the converse of a given statement by
interchanging the clauses that follow the if and the then.
And the reason you have to beware is that a true
statement can have a false converse.
For example, consider the following true statement.
If a person is listening to me lecture now, then he is alive.
Hopefully, a true statement.
If we now interchange the clauses to form the converse
it says if a person is alive, then he is
listening to me lecture.
A tragically false statement.
But notice the difference between inverting the clauses
of an if-then statement.
And the idea is this.
Notice that our theorem says if 'f prime' exists, then-- or
if there is a local maximum or a local minimum, 'f
prime of c' is 0.
It does not say that if 'f prime of c' is 0 we have a
local maximum or a local minimum.
Perhaps the easiest way to see that is
in terms of an example.
That's the nicest thing to show how a converse is false.
To prove that something is true, you can't do it just by
showing its true in certain cases.
But to show that something is not true, all you have to do
is exhibit one example in which the result is false.
That's enough to prove that it can't always be true.
For example, in this particular diagram, notice
that in the curve 'y' equals 'f of x', which I've drawn
here, the curve at the point 'c' comma 'f of c' has a
horizontal tangent.
In other words, 'f prime of c' is 0 here.
But notice that in any neighborhood that surrounds
'c', in any open interval that's around 'c', notice that
if 'x' is less than 'c', 'f of x' will be less than 'f of c'.
And if 'x' is greater than 'c', 'f of x' will be greater
than 'f of c'.
In other words, notice that except for the stationary
value at which we have a horizontal tangent, the graph
is always rising in any neighborhood of 'c'.
That's the first point I want you to see.
The second point says beware if 'f prime
of c' doesn't exist.
See all our theorem said is, look it.
If you have a relative high point, a relative low point,
relative max, or a relative min at 'x' equal 'c', and if
'f prime of c' exists, then 'f prime of c' is 0.
But who said that 'f prime of c' has to exist?
Again, let's look in terms of an example.
If we let 'f of x' equal the absolute value of 'x', recall
from our previous assignments and the like that the
derivative of 'f of x', in this case, does not
exist when 'x' is 0.
In other words, 'f prime of 0' doesn't exist.
And why is that?
Well, notice that the graph of 'y' equals the absolute value
of 'x' is the straight line 'y' equals 'x' if 'x' is
non-negative and the straight line 'y' equals 'minus x' if
'x' is negative.
In other words, 'f prime of x' is 1 if 'x' is positive.
It's minus 1 if 'x' is negative.
And hence, a jump discontinuity in the
derivative at 0.
But you see, the point is this.
If you look at our graph, do we have a low point
at 'x' equals 0?
In other words, is this the lowest point in the
neighborhood of 0?
And the answer is yes.
In fact, it's an 'absolute' low point.
Meaning that no matter where you go, no point on our graph
can be less or lower than this particular point here.
But that's irrelevant.
The point is what?
If we looked for a place where 'f prime of c' was 0 in this
example, we wouldn't find one.
That would not mean that there wasn't a low point.
What happened was what?
The low point snuck in at a place where the
derivative did not exist.
And again, you have to be careful.
All I said was what?
Beware of points for which 'f prime of c' does not exist.
It does not mean that at each point where 'f prime of c'
does not exist that you're going to have a
high or a low point.
For example, remember that the derivative not existing
loosely speaking means what?
That there's a sharp corner to the curve.
What I'm thinking of is something like this.
Let's take a curve that's always rising, say
something like this.
Now, let's suppose we put a sharp corner in here, but in
such a way that the curve still
continues to always rise.
Say like this.
Now you see, at this particular value of 'c', 'f
prime of c' doesn't exist.
I think you can see intuitively
what's happening here.
The slope approaches one value as you approach 'c' from the
left and another value if you approach 'c' from the right.
Now even know 'f prime of c' doesn't exist at this point,
notice that this point is neither a
high nor a low point.
Meaning that all points what?
To the left of this point are below it and all points to the
right of this point are above it.
So again, what?
Beware when 'f prime of c' equals 0, but don't jump to
any false conclusions.
And the third caution is an extremely important one.
In fact, for the first time from a practical point of
view, we are going to probably see analytically what is the
difference between an open interval
and a closed interval.
Suppose I have the function 'f of x' equals 'x squared', but
the domain of 'f' is now the open interval from 2 to 3.
In other words, the inputs of the 'f' machine are restricted
to all those numbers which are greater than 2,
but less than 3.
Now, let's take a look here.
Let's first of all, see where 'f prime of x' is 0.
First of all, 'f prime of x' is '2x' and that equals 0 if
and only if 'x' equals 0.
Now here's where the domain is very important.
Is 'x' equal 0 in our domain?
The answer is no.
'x' equals 0 is not in our domain.
Our domain is restricted to be the open interval from 2 to 3.
Therefore, as far as our function 'f' is concerned--
and remember, way back in one of our early, I don't even
think it was a lecture.
It was in our supplementary notes.
We pointed out that when you define a function, you need
not only the rule, but you must specify the domain.
Remember, two functions were equal not only if they were
the same rule, but they had to be defined on the same domain.
So f here is defined on a domain from 2 to 3.
And on that particular domain of definition there is no
place where the derivative is 0.
By the same token, since the derivative is a polynomial and
all polynomials are differentiable, and all
differentiable functions are continuous, there certainly
will be no places where 'f prime of x' doesn't exist.
In other words, 'f prime' exists for all 'x' in the
domain of 'f'.
In other words, here is a particular example in which a
particular function on its domain of definition does not
have any high or low points on it.
And I'll illustrate that graphically in a few moments.
I simply wanted to put this on the board first to put it into
sharp contrast with what we're going to say next.
If we're not careful, in fact, the next problem looks exactly
the same as the problem that we just solved.
Namely, what we want to do now is investigate the function 'f
of x' equals 'x squared'.
But we want the domain of 'f' now to be what?
The closed interval from 2 to 3.
In other words, the only difference between this
problem and the problem that we just solved is that now we
want the endpoints included.
Now, there's no sense repeating the
part that we did before.
First of all, will 'f prime of x' equal 0 in the domain of
definition?
As we saw before, no.
Is 'f prime' nonexistent any place in the interval?
No, it's differentiable.
It's a smooth polynomial curve.
Answer is no.
Now, here's where we come to two very important points that
hopefully, will clarify certain conventions that were
made in the textbook.
In our section on continuity there was a little result that
may have seemed a little bit obscure.
It said what?
That a continuous function defined on a closed interval
must take on its maximum and minimum values someplace on
that closed interval.
You see, all we've proven over here is what?
That if 'f' has a max or min, what we've proven is what?
It does not occur in the open interval from 'a' to 'b'.
Well, look it.
If the high and low points have to occur some place on
the closed interval and they can't appear, as we've seen,
in the open interval, where must they occur?
Well, if they can't be inside and they have to be on the
interval, it must be that they take place at the endpoints.
Let's go back without referring
back to another board.
Remember we wrote down when we're talking about high and
low points that we talked about 'x' being in a delta
neighborhood of 'c'.
It meant what?
That you could surround 'c' by some bandwidth 'delta'.
And notice that our definition in the textbook of a
neighborhood was always an open interval.
And the reason that the definition is given to be open
is that notice that what the definition now says is what?
You know what's going on on either side of 'c'.
We know what's going on either side of 'c'.
Notice that in the case where you have a closed interval, by
definition of a closed interval notice that what?
We know what's happening as we come in on 'a' from the right.
And we know what's happening to 'b' as we come in on it
from the left.
But allegedly, meaning that since the function is only
defined on the closed interval from 'a' to 'b', we don't know
what's happening before 'a', and we don't know what's
happening before 'b'.
In other words, this is why the test for 'f prime of c'
equaling 0 applies only to 'c' being in the interior.
In other words, in an open interval.
So in other words then, you see if a function happens to
be continuous on the closed interval and it doesn't have
any high-low points in the interior of the interval, then
it must have its high-low points where?
It must be at the endpoints.
And to show you quite simply what was going on in this
particular problem, notice that if we graph 'f of x'
equals 'x squared' and look at this say, first of all, on the
closed interval from 2 to 3 what we're saying is look it,
any point in which we look at a neighborhood that isn't 2 or
3, the curve is rising on one side of the point.
In other words, the curve is rising on both sides of the
point, so that if you are to the left of the point, the
height will be less than the point you're interested in.
If you're to the right of it, the height will exceed the
point that you're interested in.
But notice that at the end points themselves, you do have
what in this case?
Not a relative high or low, but actually an
absolute high or low.
In other words, the point '2 comma 4'.
See, what is the endpoint here?
One endpoint is 4, the other endpoint going up on the
y-direction here is 9.
What you're saying is what?
4 is less than 9.
Every value of 'x squared' between 2 and 3 falls
between 4 and 9.
And what you're saying is what?
That the lowest point occurs when 'x' is 2, the highest
point occurs when 'y' is 3.
The interesting point to note is that if you now look at the
open interval and exclude the endpoints, you cannot get a
lowest point or a highest point.
Namely, notice that if you allow yourself to get as close
to 2 as you want without ever getting there, it means you
could have done what?
Moved closer to 2.
In other words, if 'x' is greater than 2, you can pick
another value that's what?
Between 'x' and 2.
There's a space there.
In other words, what you saying is that no matter how
low you get here, as long as you're not exactly at 2, you
can find the point that's lower.
And in the same way as you move out this way, as you move
closer and closer to this point, if you exclude this
point itself, no matter where you stop, you could have
always found the point that was a little bit higher.
And again, you have to be where?
When I say check the endpoints of a closed interval, it does
not mean that the endpoints will give
you high or low points.
For example, look at the following curve.
It looks something like this.
It's continuous.
It's defined on the closed interval from 'a' to 'b'.
Notice that at a we do not get an absolute maximum.
In fact, what?
All of these points on the curve are higher than what's
happening here.
Same thing happens, what?
All of these points are lower than what the height is
corresponding to 'x' equals 'a'.
And in the similar way, this is what happens at 'b'.
You see, with the endpoints, obviously since you can't see
what's happening before, this will be either the highest
point or the lowest point near here depending on how the
curve is sloped.
But all we're saying is what?
In terms of absolute high values and absolute low
values, meaning the highest possible points and the lowest
possible points, we must always check the endpoints.
But we can't be positive that the endpoints
are going to be chosen.
In fact, let's summarize.
And I'm going to summarize again at
the end of the lecture.
But the idea is this.
If 'f of x' is continuous on the closed interval from 'a'
to 'b'-- and here's the key word.
To find candidates--
you know the old cliche about many are
called, but few are chosen.
In this case, few are called and even fewer are chosen.
Namely, what we're saying is look it.
When we're looking for high-low points, the
candidates are what?
Those points for which the derivative is 0.
We can check those out because those are possibilities.
Those points for which the derivative fails to exist.
We can check those out because they're possibilities.
And the endpoints.
Namely, if the function is continuous,
we check the endpoints.
By the way, if the function is not continuous, then there is
no need to check-- well, let's put it this way.
If you're on an open interval there's no need to check the
endpoints because there aren't any.
In other words, notice that I'm talking about what?
That the function 'f' is not only continuous, but on a
closed interval.
That's all there is to this.
In other words, these are all the possible candidates.
Now you see the bigger question is,
how do you use this?
And I thought that I would make up a makeshift exercise,
one that's rather easy to do at the blackboard.
For deeper exercises, for more quantitative results, we have
several exercises in the learning exercises.
Several exercises worked out
illustratively in the textbook.
But let's pick a particularly straightforward example.
Let's suppose that what I want to do is construct a cylinder.
This is a cross sectional view of a cylinder.
'x' represents the radius of the base and 'y' represents
the height.
This is a right circular cylinder.
I'm given a constraint, namely I'm told that for some reason
or other, and I don't know why anybody would ever want to
impose this condition other than the fact that we want
some condition imposed here to see what's happening.
I'm told that I want the sum of the radius of the base and
the altitude to be exactly 30.
In other words, if the radius of the base is going to be 6
inches, I want the altitude to be 24 inches.
That's a constraint.
And by the way, I'll come back to this later also.
Notice how you're almost begging a related rates
relationship here.
Or an implicit relationship that 'x' and 'y' are not
independent.
But I've now put a constraint on here.
At any rate the question is, how shall I use up my 30
inches, say, if I want to make the volume of the resulting
cylinder as large as possible?
And notice how we work this thing.
We say, well, the volume is equal to 'pi 'x squared' y'.
In this particular case, it's easy to see explicitly that
'y' is equal to '30 - x'.
It's also easy to see physically that 'x' must be
more than 0.
You can't have a negative radius of the base.
It must be less than 30 because if you used up more
than 30 inches in the radius of your base, how can the
radius of the base plus the altitude add up to exactly 30?
Because physically, the constraint is that the
altitude can't be negative.
It's certainly a positive value.
So our analytic relationship is what?
That 'v' equals 'pi 'x squared' y', which can be
written as 'pi 'x squared' times '30 - x''.
Which in turn, can be written as the polynomial '30 pi 'x
squared'' minus 'pi 'x cubed'', where 'x' is the open
interval or defined on the open interval from 0 to 30.
Now, how do I proceed?
What was my test for membership?
I have three ways of checking out where
high-low points will occur.
Or max-min points in this case, since notice that this
function does not require a graph to understand it.
I first check out to see where the derivative is 0.
Well, the derivative is simply '60 pi x'
minus '3 pi 'x squared''.
If I set this thing equal to 0, I find upon factoring that
either 'x' must be 0 or 'x' must be 20.
I can immediately exclude 'x' equals 0 because notice that
my function 'v' had its domain of definition on the open
interval from 0 to 30.
'x' equals 0 is not in the open interval from 0 to 30.
Consequently, we can rule this thing out.
And what we discover is that 'x' equals 20.
So 'x' equals 20 is the only possible
candidate that we have.
And by the way, since the sum of 'x' and 'y' must be 20, if
'x' equals 20, 'y' must be 10.
So the only candidate that we have by setting the derivative
equal to 0 is that 'x' should be 20 and 'y' should be 10.
We do not get any candidates in the sense of where the
derivative doesn't exist.
Because if you look at '60 pi x' minus '3 pi 'x squared'',
this certainly exists for all values of 'x'.
And finally, we have no endpoints to check.
Because again, our function 'v' is
defined on an open interval.
By the way, if I tried to draw this somewhat to scale, a very
interesting result turns up, which may show the power of
analytical methods versus more intuitive types of methods.
You see, what we showed here was what?
Without going into the details, that to get the
largest volume cylinder, 'x' should be 20.
In other words, the radius of your base should be 20 and the
altitude should be 10.
Let's take a look at that drawn roughly to scale.
The radius of the base here is 20 and the altitude is 10.
By the way, that means that the cross
section will be what?
A rectangle whose height is 10 and whose base is 40.
Now intuitively, I think it's easy to see that for a
rectangle, for a given perimeter the largest possible
area rectangle is the one which is a square.
Well, without even trying to prove that, let's go to this
thing here.
Instead of using up the 20 and the 10, let's draw a second
rectangle, or a second cross section of a cylinder where
the radius is 15 and the height is 15.
Notice that this still satisfies the fact that the
sum of the radius and the altitude are 30.
Now, look at this.
If we compute the area of this particular rectangle, 40 times
10, it's 400.
On the other hand, the volume is what?
Pi times 20 squared.
The radius of the base squared.
Times the height, which is 10.
And that yields the result of 4,000 pi.
On the other hand, if we look at our second rectangle, its
area is 450.
But its volume is what?
It's pi times 15 squared times 15, which is 3,375 pi.
The interesting result here is what?
The area of this rectangle is greater than the
area of this rectangle.
In fact, the area of the second rectangle is 450.
The area of the first rectangle is only 400.
But notice that when you revolve this thing to form the
cylinder, the smaller cross sectional area generates the
larger volume.
And the reason, of course, for that is that the relationship
in our variables was not linear.
In other words notice that when 'x' is large, a
relatively small change in 'x' produces a large
change in 'x squared'.
In other words, a relatively small value in 'x' can offset
a relatively large value or increase in 'y'.
And this is kind of interesting because try to
figure out intuitively how you would figure out where these
stop compensating for one another?
Where does it turn out that finally you've taken so much
away from 'x' that even though you square it, it can't
compensate the change in 'y'?
How would you intuitively pick off where the high-low points
occur in a problem like this?
And all I want you to see is again, the beautiful gentle
balance between intuitive calculus
and rigorous calculus.
That we don't throw away our intuition.
But notice how, in many cases, where our intuition fails us,
the analytic recipes come to our rescue.
But enough said about that, let me now again, highlight
the difference or the relationship between functions
and graphs.
Namely, in all of this discussion that we've done on
this particular board, we're talking about what?
'v' being a function of 'x'.
We do not have to visualize this thing pictorially.
But if we wish, what we can say is let's graph 'v' as a
function of 'x'.
And you see, going back to the material of last time.
And notice, you see how interrelated these things are.
Notice how curve plotting ties in very nicely with
derivatives and the like.
All we're saying is what?
Given this relationship, which is how 'v' is related to 'x',
we can form the first derivative, we can form the
second derivative, we can look to see where the first
derivative is 0, we can look to see where the second
derivative is 0.
I leave these details to you because after the homework
assignment you did last time these should be fairly trivial
exercises to do.
But the idea is if you now utilize all of this
information, you find that if you plot 'v' verses 'x', you
get a graph something like this.
By the way, again notice that we were not talking about just
'v' being a function of 'x'.
The domain of 'v' was restricted to be the open
interval from 0 to 30, and this is a very crucial thing
to keep in mind.
You can get into a whole bunch of trouble if you start
looking to see what happens out here.
For example, you say, hey, this curve is always going to
keep going up.
Won't this be greater than this maximum value over here
eventually?
The answer is yes, it will be.
But what does it mean to say that 'x' is negative?
'x' was the radius of our base.
So in other words, what we should
really say here is what?
That the function that we're talking about is not this
whole curve.
Yikes, that wasn't a very good job of drawing.
But rather what?
Just this portion of the curve defined on the open interval
from 0 to 30.
By the way, let me point out something else.
You recall in an earlier lecture we talked about
related rates.
We had an assignment with other lecture where you've
solved some problems using related rates.
Let me show you how related rates play a very important
computational role in dealing with max-min problems.
In this particular problem, we had that 'v' equals 'pi 'x
squared' y', where 'y' happened to be a particular
function of 'x'.
In fact, implicitly it was given by the fact that 'x + y'
happened to equal 30.
Now let's keep track of something here.
In this particular problem, notice that it was very easy
to change this implicit relationship
to an explicit one.
It was also easy once you expressed 'y' explicitly in
terms of 'x' that when you wanted to substitute into
here, it was a very easy computational job to carry out
the operations.
But suppose there happened to be cube roots in here, or all
sorts of nasty things, whatever they might be.
And suppose instead of 'x + y' equals 30, you had our old
friend something like 'x to the eighth' plus ''x to the
sixth' 'y squared'' plus 'y to the sixth' equals 3.
How would you solve for 'y' explicitly in
terms of 'x' there?
And the point that I'd like you to see is that we can
solve this problem very nicely without having to resort to
explicitly replacing 'y' as a function of 'x'.
Namely, implicitly assuming that 'y' is a differentiable
function of 'x', we can differentiate
this thing as a product.
'dv dx' will be what?
The derivative of the first factor, which is '2 pi x',
times the second.
Plus the first factor, which is 'pi 'x squared', times the
derivative of 'y' with respect to 'x'.
So that's 'dv dx'.
On the other hand, from this relationship here, we can
conclude by differentiating implicitly that '1
+ 'dy dx'' is 0.
And therefore, 'dy dx' is minus 1.
And putting that value for 'dy dx' in here, we wind up with
this explicit relationship.
And we can now see that 'dv dx' is 0 if and only if just
by solving this thing, setting it equal to
0, 'x' equals '2y'.
By the way, that's exactly what happened in our problem.
You'll notice that 'x' turned out to be 20 and 'y' turned
out to be 10.
You see again, 'x' and 'y' are related.
You can say, gee, couldn't 'x' be 60 and 'y' be 30?
Answer, no.
Because the constraint is that 'x + y' must be 30.
Well, look it.
I don't want us to get too wrapped up on the idea of
computational differences now.
What I do want to do is review our basic result.
And in fact, let me come over here.
I hope this doesn't spoil you.
I'd like to come back to the board.
And actually, since I've got this all written down, let's
close on this particular result again.
To find the high-low points of a function 'f of x', we first
of all, check out when 'f prime of c' is 0 for
candidates.
We check out where 'f prime of c' fails to exist.
And if it's a closed interval, we check the endpoints.
This is the mechanism behind what we're doing from that
point on, as the cliche goes, it's all engineering's baby.
It's all computational know how.
At any rate, I think this is enough in terms of emphasizing
the points that we wanted to make.
And so until next time, goodbye.
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