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PROFESSOR: OK, we're ready to
start the eleventh lecture.
We're still in the middle
of sketching.
And, indeed, one of the reasons
why we did not talk
about hyperbolic functions is
this that we're running just a
little bit behind.
And we'll catch up
a tiny bit today.
And I hope all the way on
Tuesday of next week.
So let me pick up where we
left off, with sketching.
So this is a continuation.
I want to give you one
more example of
how to sketch things.
And then we'll go through
it systematically.
So the second example that
we did as one example
last time, is this.
The function is x
+ + 1 / x + + 2.
And I'm going to save you
the time right now.
This is very typical of me,
especially if you're in a
hurry on an exam, I'll
just tell you what
the derivative is.
So in this case, it's
1 / (x + 2)^2.
Now, the reason why I'm bringing
this example up, even
though it'll turn out to be
a relatively simple one to
sketch, is that it's easy to
fall into a black hole with
this problem.
So let me just show you.
This is not equal to 0.
It's never equal to 0.
So that means there are
no critical points.
At this point, students, many
students who have been trained
like monkeys to do exactly
what they've been told,
suddenly freeze and give up.
Because there's nothing to do.
So this is the one thing that
I have to train out of you.
You can't just give
up at this point.
So what would you suggest?
Can anybody get us
out of this jam?
Yeah.
STUDENT: [INAUDIBLE]
PROFESSOR: Right.
So the suggestion was
to find the x values
where f (x) is undefined.
In fact, so now that's a fairly
sophisticated way of
putting the point that I want to
make, which is that what we
want to do is go back to
our precalculus skills.
And just plot points.
So instead, you go back to
precalculus and you just plot
some points.
It's a perfectly reasonable
thing.
Now, it turns out that the most
important point to plot
is the one that's not there.
Namely, the value of x = - 2.
Which is just what
was suggested.
Namely, we plot the
points where the
function is not defined.
So how do we do that?
Well, you have to think about
it for a second and I'll
introduce some new notation
when I do it.
If I evaluate 2 at this place,
actually I can't do it.
I have to do it from the
left and the right.
So if I plug in - 2 on the
positive side, from the right,
that's going to be equal to - 2
+ 1 / - 2, a little bit more
than - 2, +2.
Which is - 1 divided by - now,
this denominator is - 2, a
little more than that, +2.
So it's a little more than 0.
And that is, well we'll fill
that in in a second.
Everybody's puzzled.
Yes.
STUDENT: [INAUDIBLE]
PROFESSOR: No, that's
the function.
I'm plotting points, I'm
not differentiating.
I've already differentiated
it.
I've already got something
that's a little puzzling.
Now I'm focusing on
the weird spot.
Yes, another question.
STUDENT: Wouldn't it be
a little less than 0?
PROFESSOR: Wouldn't it be
a little less than 0?
OK, that's a very good point
and this is a matter of
notation here.
And a matter of parentheses.
So wouldn't this be a
little less than 2.
Well, if the parentheses were
this way; that is, 2+ , with a
- after I did the 2+ , then
it would be less.
But it's this way.
OK.
So the notation is, you
have a number and you
take the part of it.
That's the part which is a
little bit bigger than it.
And so this is what I mean.
And if you like, here I can put
in those parentheses too.
Yeah, another question.
STUDENT: [INAUDIBLE]
PROFESSOR: Why doesn't the
top one have a plus?
The only reason why the top
one doesn't have a plus is
that I don't need it
to evaluate this.
And when I take the limit, I
can just plug in the value.
Whereas here, I'm
still uncertain.
Because it's going to be 0.
And I want to know which
side of 0 it's on.
Whether it's on the positive
side or the negative side.
So this one, I could have
written here a parentheses 2+,
but then it would have just
simplified to - 1.
In the limit.
So now, I've got a negative
number divided by a tiny
positive number.
And so, somebody want to
tell me what that is?
Negative infinity.
So, we just evaluated this
function from one side.
And if you follow through the
other side, so this one here,
you get something very similar,
except that this
should be -- whoops, what
did I do wrong?
I meant this.
I want it -2 the same base
point, but I want
to go from the left.
So that's going to be -
2 + 1, same numerator.
And then this - 2 on the left +
2, and that's going to come
out to be - 1 / 0 -, which
is plus infinity.
Or just plain infinity,
we don't have to
put the plus sign.
So this is the first part
of the problem.
And the second piece, to get
ourselves started, you could
evaluate this function
at any point.
This is just the most
interesting point, alright?
This is just the most
interesting
place to evaluate it.
Now, the next thing that I'd
like to do is to pay attention
to the ends.
And I haven't really said
what the ends are.
So the ends are just all the
way to the left and all the
way to the right.
So that means x going to
plus or minus infinity.
So that's the second thing I
want to pay attention to.
Again, this is a little bit
like a video screen here.
And we're about to discover
something that's really off
the screen, in both cases.
We're taking care of what's
happening way to the left, way
to the right, here.
And up above, we just took
care what happens
way up and way down.
So on these ends, I need to
do some more analysis.
Which is related to a
precalculus skill which is
evaluating limits.
And here, the way to do it is
to divide by x the numerator
and denominator.
Write it as (1 + 1
/ x)/ (1 + 2 /x).
And then you can see what
happens as x goes to plus or
minus infinity.
It just goes to 1.
So, no matter whether x is
positive or negative.
When it gets huge, these two
extra numbers here go to 0.
And so, this tends to 1.
So if you like, you could
abbreviate this as f (+ or -
infinity) = 1.
So now, I get to draw this.
And we draw this using
asymptotes.
So there's a level
which is y = 1.
And then there's another
line to draw.
Which is x = - 2.
And now, what information
do I have so far?
Well, the information that I
have so far is that when we're
coming in from the right, that's
to - 2, it plunges down
to minus infinity.
So that's down like this.
And I also know that it goes
up to infinity on the other
side of the asymptote.
And over here, I know it's
going out to the level 1.
And here it's also going
to the level 1.
Now, there's an issue.
I can almost finish
this graph now.
I almost have enough information
to finish it.
But there's one thing
which is making me
hesitate a little bit.
And that is, I don't know, for
instance, over here, whether
it's going to maybe dip below
and come back up.
Or not.
So what does it do here?
Can anybody see?
Yeah.
STUDENT: [INAUDIBLE]
PROFESSOR: It can't dip
below because there
are no critical points.
What a precisely
correct answer.
So that's exactly right.
The point here is that because
f' is not 0, it can't double
back on itself.
Because there can't be any of
these horizontal tangents.
It can't double back, so
it can't backtrack.
So sorry, if f' is not
0, f can't backtrack.
And so that means that it
doesn't look like this.
It just goes like this.
So that's basically it.
And it's practically the
end of the problem.
Goes like this.
Now you can decorate
your thing, right?
You may notice that maybe it
crosses here, the axes, you
can actually evaluate
these places.
And so forth.
We're looking right now for
qualitative behavior.
In fact, you can see where
these places hit.
And it's actually a little
higher up than I drew.
Maybe I'll draw it accurately.
As we'll see in a second.
So that's what happens
to this function.
Now, let's just take a look in
a little bit more detail, by
double checking.
So we're just going to double
check what happens to the sign
of the derivative.
And in the meantime, I'm going
to explain to you what the
derivative is and also talk
about the second derivative.
So first of all, the trick for
evaluating the derivative is
an algebraic one.
I mean, obviously you can do
this by the quotient rule.
But I just point out that this
is the same thing as this.
And now it has, whoops, that
should be a 2 in the
denominator.
And so, now this has the
form 1 - (1 / x + 2).
So this makes it easy to see
what the derivative is.
Because the derivative of
a constant is 0, right?
So this is, derivative,
is just going to
be, switch the sign.
This is what I wrote before.
And that explains it.
But incidentally, it
also shows you that
that this is a hyperbola.
These are just two curves
of a hyperbola.
So now, let's check the sign.
It's already totally obvious
to us that this is just a
double check.
We didn't actually even have to
pay any attention to this.
It had better be true.
This is just going to check
our arithmetic.
Namely, it's increasing here.
It's increasing there.
That's got to be true.
And, sure enough, this is
positive, as you can see it's
1 over a square.
So it is increasing.
So we checked it.
But now, there's one more thing
that I want to just have
you watch out about.
So this means that
f is increasing.
On the interval minus infinity
< x < - 2.
And also from - 2 all the
way out to infinity.
So I just want to warn you, you
cannot say, don't say f is
increasing on minus infinity
< infinity, for all x.
OK, this is just not true.
I've written it on the board,
but it's wrong.
I'd better get rid of it.
There it is.
Get rid of it.
And the reason is, so first of
all it's totally obvious.
It's going up here.
But then it went zooming
back down there.
And here this was true, but
only if x is not - 2.
So there's a break.
And you've got to pay attention
to the break.
So basically, the moral here
is that if you ignore this
place, it's like ignoring
Mount Everest,
or the Grand Canyon.
You're ignoring the most
important feature of this
function here.
If you're going to be figuring
out where things are going up
and down, which is basically all
we're doing, you'd better
pay attention to these
kinds of places.
So don't ignore them.
So that's the first remark.
And now there's just a little
bit of decoration as well.
Which is the role of the
second derivative.
So we've written down the
first derivative here.
The second derivative is now
- 2 / (x + 2)^3, right?
So I get that from
differentiating this formula
up here for the first
derivative.
And now, of course, that's also,
only works for x not
equal to - 2.
And now, we can see that this is
going to be negative, let's
see, where is it negative?
When this is a positive
quantity, so when -2 < x
< infinity.
It's negative.
And this is where this thing
is concave. Let's see.
Did I say that right?
Negative, right?
This is concave down.
Right.
And similarly, if I look at this
expression, the numerator
is always negative but the
denominator becomes negative
as well when x < -2.
So this becomes positive.
So this case, it was negative
over positive.
In this case it was negative
divided by negative.
So here, this is in the range
- infinity < x < -2.
And here it's concave up.
Now, again, this is just
consistent with what we're
already guessing.
Of course we already know it in
this case if we know that
this is a hyperbola.
That it's going to be concave
down to the right of the
vertical line, dotted
vertical line.
And concave up to the left.
So what extra piece of
information is it that
this is giving us?
Did I say this backwards?
No.
That's OK.
So what extra piece of
information is this giving us?
It looks like it's giving
us hardly anything.
And it really is giving
us hardly anything.
But it is giving us something
that's a little aesthetic.
It's ruling out the possibility
of a wiggle.
There isn't anything like
that in the curve.
It can't shift from curving this
way to curving that way
to curving this way.
That doesn't happen.
So these properties say
there's no wiggle in
the graph of that.
Alright.
So.
Question.
STUDENT: Do we define the
increasing and decreasing base
purely on the derivative, or
the sort of more general
definition of picking any
two points and seeing.
Because sometimes there can be
an inconsistency between the
two definitions.
PROFESSOR: OK, so the question
is, in this course are we
going to define positive
derivative as being the same
thing as increasing.
And the answer is no.
We'll try to use these
terms separately.
What's always true is that if
f' is positive, then f is
increasing.
But the reverse is not
necessarily true.
It could be very flat, the
derivative can be 0 and still
the function can
be increasing.
OK, the derivative can
be 0 at a few places.
For instance, like
some cubics.
Other questions?
So that's as much as I need
to say in general.
I mean, in a specific case.
But I want to get you a general
scheme and I want to
go through a more complicated
example that gets all the
features of this
kind of thing.
So let's talk about a general
strategy for sketching.
So the first part of
this strategy, if
you like, let's see.
I have it all plotted
out here.
So I'm going to make sure I
get it exactly the way I
wanted you to see.
So I have, its plotting.
The plot thickens.
Here we go.
So plot, what is it that
you should plot first?
Before you even think about
derivatives, you should plot
discontinuities.
Especially the infinite ones.
That's the first thing
you should do.
And then, you should plot
end points, for ends.
For x going to plus or minus
infinity if there don't happen
to be any finite ends
to the problem.
And the third thing you can do
is plot any easy points.
This is optional.
At your discretion.
You might, for instance, on this
example, plot the places
where the graph crosses
the axis.
If you want to.
So that's the first part.
And again, this is
all precalculus.
So now, in the second part
we're going to solve this
equation and we're
going to plot the
critical points and values.
In the problem which we just
discussed, there weren't any.
So this part was empty.
So the third step is to decide
whether f', sorry, whether, f'
is positive or negative
on each interval.
Between critical points,
discontinuities.
The direction of the sign, in
this case it doesn't change.
It goes up here and it
also goes up here.
But it could go up here and
then come back down.
So the direction can change
at every critical point.
It can change at every
discontinuity.
And you don't know.
However, this particular step
has to be consistent with 1
and 2, with steps 1 and 2.
In fact, it will never, if you
can succeed in doing steps 1
and 2, you'll never
need step 3.
All it's doing is
double-checking.
So if you made an arithmetic
mistake somewhere, you'll be
able to see it.
So that's maybe the most
important thing.
And it's actually the most
frustrating thing for me when
I see people working on
problems, is they start step
3, they get it wrong, and then
they start trying to draw the
graph and it doesn't work.
Because it's inconsistent.
And the reason is some
arithmetic error with the
derivative or something like
that or some other
misinterpretation.
And then there's a total mess.
If you start with these two
steps, then you're going to
know when you get to this step
that you're making mistakes.
People don't generally make
as many mistakes in
the first two steps.
Anyway, in fact you can skip
this step if you want.
But that's at risk of not
double-checking your work.
So what's the fourth step?
Well, we take a look
at whether f''
is positive or negative.
And so we're deciding on things
like whether it's
concave up or down.
And we have these points, f''
( x ) = 0, which are called
inflection points.
And the last step is just
to combine everything.
So this is this the scheme,
the general scheme.
And let's just carry it out
in a particular case.
So here's the function
that I'm going
to use as an example.
I'll use f ( x ) = x / ln x.
And because the logarithm
- yeah, question.
Yeah.
STUDENT: [INAUDIBLE]
PROFESSOR: The question
is, is this optional.
So that's a good question.
Is this optional.
STUDENT: [INAUDIBLE]
PROFESSOR: OK, the question
is is this optional;
this kind of question.
And the answer is, it's more
than just -- so, in many
instances, I'm not going
to ask you to.
I strongly recommend that if I
don't ask you to do it, that
you not try.
Because it's usually awful to
find the second derivative.
Any time you can get away
without computing a second
derivative, you're better off.
So in many, many instances.
On the other hand, if I ask you
to do it it's because I
want you to have the,
work to do it.
But basically, if nobody forces
you to, I would say
never do step 4.
Other questions.
Alright.
So we're going to force
ourselves to do step 4,
however, in this instance.
But maybe this will be
one of the few times.
So here we go, just for
illustrative purposes.
OK, now.
So here's the function that
I want to discuss.
And the range has to be x
positive, because the
logarithm is not defined
for negative values.
So the first thing that I'm
going to do is, I'd like to
follow the scheme here.
Because if I don't follow the
scheme, I'm going to get a
little mixed up.
So the first part is to find
the singularities.
That is, the places where
f is infinite.
And that's when the logarithm,
the denominator, vanishes.
So that's f ( 1 +),
if you like.
So that's 1 / ln 1 +, which is
1 / 0, with a little bit of
positiveness to it.
Which is infinity.
And second, we do it
the other way.
And not surprisingly,
this comes out
to be negative infinity.
Now, the next thing I want
to do is the ends.
So I call these the ends.
And there are two of them.
One of them is f ( 0 ) from
the right. f ( 0+).
So that is 0 + / ln 0 +, which
is 0 plus divided by, well, ln
0 + is actually minus
infinity.
That's what happens
to the logarithm,
goes to minus infinity.
So this is 0 over infinity,
which is definitely 0, there's
no problem. about
what happens.
The other side, so this is the
end, this is the first end.
The range is this.
And I just did the
left endpoint.
And so now I have to do the
right endpoint, I have to let
x go to infinity.
So if I let x go to infinity,
I'm just going to have to
think about it a little
bit by plugging in
a very large number.
I'll plug in 10 ^ 10, just
to see what happens.
So if I plug in 10 ^ 10
into x ln x, I get
10 ^ 10 / ln 10^10.
Which is 10 ^ 10
/ 10 ( log 10).
So the denominator, this
number here, is about
2.something.
2.3 or something.
So this is maybe 230 in the
denominator, and this is a
number with ten 0's after it.
So it's very, very large.
I claim it's big.
And that gives us the clue that
what's happening is that
this thing is infinite.
So, in other words, our
conclusion is that f of
infinity is infinity.
So what do we have so far
for our function?
We're just trying to build the
scaffolding of the function.
And we're doing it by taking
the most important points.
And from a mathematician's
point of view, the most
important points are the
ones which are sort
of infinitely obvious.
For the ends of the problem.
So that's where we're heading.
We have a vertical asymptote,
which is at x = 1.
So this gives us x = 1.
And we have a value which
is that it's 0 here.
And we also know that when we
come in from the - sorry, so
we come in from the left, that's
f, the one from the
left, we get negative
infinity.
So it's diving down.
It's going down like this.
And, furthermore, on the other
side we know it's climbing up.
So it's going up like this.
Just start a little higher.
Right, so.
So far, this is what we know.
Oh, and there's one other
thing that we know.
When we go to plus infinity,
it's going back up.
So, so far we have this.
Now, already it should be pretty
obvious what's going to
happen to this function.
So there shouldn't be
many surprises.
It's going to come
down like this.
Go like this, it's going to turn
around and go back up.
That's what we expect.
So we don't know that yet,
but we're pretty sure.
So at this point, we can
start looking at
the critical points.
We can do our step 2 here -- we
need a little bit more room
here -- and see what's happening
with this function.
So I have to differentiate it.
And it's, this is the
quotient rule.
So remember the function
is up here, x / ln x.
So I have a ln x^2 in
the denominator.
And I get here the derivative of
x is 1, so we get 1 ( ln x)
- x ( the derivative of
ln x, which is 1 /x).
So all told, that's (ln
x - 1) / ln x^2.
So here's our derivative.
And now, if I set this equal
to 0, at least in the
numerator, the numerator
is 0 when x = e.
The ln e = 1.
So here's our critical point.
And we have a critical
value, which is f(e).
And that's going
to be e / ln e.
Which is e, again.
Because ln e = 1.
So now I can also plot
the critical point,
which is down here.
And there's only one of them,
and it's at (e e).
That's kind of not to scale
here, because my blackboard
isn't quite tall enough.
It should be over here and
then, it's slope 1.
But I dipped it down.
So this is not to scale, and
indeed that's one of the
things that we're not going to
attempt to do with these
pictures, is to make
them to scale.
So the scale's a little
squashed.
So, so far I have this
critical point.
And, in fact, I'm going
to label it with a c.
Whenever I have a critical point
I'll just make sure that
I remember that that's
what it is.
And since there's only one, the
rest of this picture is
now correct.
That's the same mechanism that
we used for the hyperbola.
Namely, we know there's only
one place where the
derivative is 0.
So that means there no more
horizontals, so there's no
more backtracking.
It has to come down to here.
Get to there.
This is the only place
it can turn around.
Goes back up.
It has to start here and it
has to go down to there.
It can't go above 0.
Do not pass go, do
not get positive.
It has to head down here.
So that's great.
That means that this
picture is almost
completely correct now.
And the rest is more
or less decoration.
We're pretty much done with
the way it looks, at least
schematically.
However, I am going to punish
you, because I warned you.
We are going to go over here and
do this step 4 and fix up
the concavity.
And we're also going to do
a little bit of that
double-checking.
So now, let's again, just,
I want to emphasize.
We're going to do
a double-check.
This is part 3.
But in advance, I already have,
based on this picture I
already know what
has to be true.
That f is decreasing on
0 to 1. f is also
decreasing on 1 to e.
And f is increasing
on e to infinity.
So, already, because we've plot
a bunch of points and we
know that there aren't any
places where the derivative
vanishes, we already know
it goes down, down, up.
That's what it's got to do.
Now, we'll just make sure that
we didn't make any arithmetic
mistakes, now.
By actually computing
the derivative, or
staring at it, anyway.
And making sure that
it's correct.
So first of all, we just take
a look at the numerator.
So f,' remember, was (ln
x - 1) / ln x^2,
the quantity squared.
So the denominator
is positive.
So let's just take a look
at the three ranges.
So we have 0 < x < 1.
And on that range, the logarithm
is negative, so this
is negative divided by positive,
which is negative.
That's decreasing,
that's good.
And in fact, that also works
on the next range.
1 l< x < e, it's negative
divided by positive.
And the only reason why we
skipped 1, again, is that it's
undefined there.
And there's something dramatic
happening there.
And then, at the last range,
when x is bigger than e, that
means the logarithm is already
bigger than 1.
So the numerator is now
positive, and the
denominator's still positive,
so it's increasing.
So we've just double-checked
something
that we already knew.
Alright, so that's pretty
much all there is to
say about step 3.
So this is checking the
positivity and negativity.
And now, step 4.
There is one small point
which I want to make
before we go on.
Which is that sometimes, you
can't evaluate the function or
its derivative particularly
well.
So sometimes you can't plot
the points very well.
And if you can't plot the points
very well, then you
might have to do 3 first, to
figure out what's going on a
little bit.
You might have to skip.
So now we're going to go on
the second derivative.
But first, I want to use an
algebraic trick to rearrange
the terms. And I want to notice
one more little point.
Which I, as I say, this is
decoration for the graph.
So I want to rewrite
the formula.
Maybe I'll do it right
over here.
Another way of writing this is
(1 / ln x) - (1 / (ln x)^2).
So that's another way of
writing the derivative.
And that allows me to
notice something
that I missed, before.
When I solved the equation ln
x - 1 - this is equal to 0
here, this equation here.
I missed a possibility.
I missed the possibility
that the
denominator could be infinity.
So actually, if the
denominator's infinity, as you
can see from the other
expression there, it actually
is true that the derivative
is 0.
So also when x = 0 +, the
slope is going to be 0.
Let me just emphasize
that again.
If you evaluate using this other
formula over here, this
is (1 / ln 0+) - (1
/ (ln 0+)^2).
That's 1 / - infinity
- 1 / infinity,
if you like, squared.
Anyway, it's 0.
So this is 0.
The slope is 0 there.
That is a little piece of
decoration on our graph.
It's telling us, going back to
our graph here, it's telling
us this is coming in with
slope horizontal.
So we're starting
out this way.
That's just a little start
here to the graph.
It's a horizontal slope.
So there really were two places
where the slope was
horizontal.
Now, with the help of this
second formula I can also
differentiate a second time.
So it's a little bit easier to
do that if I differentiate 1 /
ln, that's -( ln x) ^ - 2 ( 1
/ x) + 2 (ln x) ^ -3 (1/x).
And that, if I put it over a
common denominator, is x ln
x^3 times, let's see here.
I guess I'll have to
take the 2 - ln x.
So I've now rewritten the
formula for the second
derivative as a ratio.
Now, to decide the sign, you see
there are two places where
the sign flips.
The numerator crosses when the
logarithm is 2, that's going
to be when x = e ^2.
And the denominator flips when
x = 1, that's when the log
flips from positive
to negative.
So we have a couple
of ranges here.
So, first of all, we have
the range from 0 to 1.
And then we have the range
from 1 to e^2.
And then we have the
range from e ^2 all
the way out to infinity.
So between 0 and 1, the
numerator is, well this is a
negative number in this, so
minus a negative number is
positive, so the numerator
is positive.
And the denominator is negative,
because the ln is
negative it's taken to
the third power.
So this is a negative numbers,
so it's positive divided by a
negative number, which
is less than 0.
That means it's concave down.
So this is concave down plot.
And that's a good thing, because
over here this is
concave down.
So there are no wiggles.
It goes straight down,
like this.
And then the other two pieces
are f'' is equal to, well it's
going to switch here.
The denominator becomes
positive.
So it's positive
over positive.
So this is concave up.
And that's going over here.
But notice that it's not the
bottom where it turns around,
it's somewhere else.
So there's another
transition here.
This is e ^2.
This is e.
So what happens at the end is,
again, the sign flips again.
Because the numerator,
now, when x >
e ^2, becomes negative.
And this is negative divided by
positive, which is negative.
And part is concave down.
And so we didn't quite
draw the graph right.
There's an inflection point
right here, which
I'll label with i.
Makes a turn the other
way at that point.
So there was a wiggle.
There's the wiggle.
Still going up, still
going to infinity.
But kind of the slope of
the mountain, right?
It's going the other way.
This point happens to
be (e^2, e ^2 / 2).
So that's as detailed
as we'll ever get.
And indeed, the next game is
going to be avoid being, is to
avoid being this detailed.
So let me introduce
the next subject.
Which is maxima and minima.
OK, now, maxima and minima,
maximum and minimum problems
can be described graphically
in the following ways.
Suppose you have a function,
right, here it is.
OK?
Now, find the maximum.
And find the minimum.
OK.
So this problem is done.
The point being, that it is easy
to find max and the min
with the sketch.
It's very easy.
The goal, the problem, is that
the sketch is a lot of work.
We just spent 20 minutes
sketching something.
We would not like to spend all
that time every single time we
want to find a maximum
and minimum.
So the goal is to do
it with, so our
goal is to use shortcuts.
And, indeed, as I said earlier,
we certainly never
want to use the second
derivative if we can avoid it.
And we don't want to decorate
the graph and do all of these
elaborate, subtle, things which
make the graph look
nicer and really, or
aesthetically appropriate.
But are totally unnecessary
to see whether the
graph is up or down.
So essentially, this whole
business is out, which is a
good thing.
And, unfortunately, those early
parts are the parts that
people tend to ignore.
Which are typically, often,
very important.
So let me first tell you
the main point here.
So the key idea.
Key to finding maximum.
So the key point is,
we only need to
look at critical points.
Well, that's actually what
it seems like to, in many
calculus classes.
But that's not true.
This is not the end
of the sentence.
And, end points, and points
of discontinuity.
So you must watch
out for those.
If you look at the example that
I just drew here, which
is the one that I carried out,
you can see that there are
actually five extreme points
on this picture.
So let's switch.
So we'll take a look.
There are five places where the
max or the min might be.
There's this important point.
This is, as I say, the
scaffolding of the function.
There's this point, there
down at minus infinity.
There's this, there's this,
and there's this.
Only one out of five is
a critical point.
So there's more that
you have to pay
attention to on the function.
And you always have to keep the
schema, the picture of the
function, in the back
of your head.
Even though this may be the most
interesting point, and
the one that you're going
to be looking at.
So we'll do a few examples
of that next time.
