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PROF.
JERISON: So, we're ready to
begin Lecture 10, and what I'm
going to begin with is
by finishing up some
things from last time.
We'll talk about approximations,
and I want to
fill in a number of comments and
get you a little bit more
oriented in the point of view
that I'm trying to express
about approximations.
So, first of all, I want to
remind you of the actual
applied example that I
wrote down last time.
So that was this
business here.
There was something from
special relativity.
And the approximation that
we used was the linear
approximation, with a - 1/2
power that comes out to be t(
1 1/2 v^2 / C^2).
I want to reiterate why
this is a useful way
of thinking of things.
And why this is that this
comes up in real life.
Why this is maybe more important
than everything that
I've taught you about
technically so far.
So, first of all, what this is
telling us is the change in t
/ t, if you do the arithmetic
here and subtract t that's
using the change in
t is t' - t here.
If you work that out, this is
approximately the same as 1/2
(v^2 / C^2).
So what is this saying?
This is saying that if you have
this satellite, which is
going at speed v, and little c
is the speed of light, then
the change in the watch down
here on earth, relative to the
time on the satellite, is going
to be proportional to
this ratio here.
So, physically, this
makes sense.
This is time divided by time.
And this is velocity squared
divided by velocity squared.
So, in each case, the
units divide out.
So this is a dimensionless
quantity.
And this is a dimensionless
quantity.
And the only point here that
we're trying to make is just
this notion of proportionality.
So I want to write this down.
Just, in summary.
So the error fraction, if you
like, which is sort of the
number of significant digits
that we have in our
measurement, is proportional,
in this
case, to this quantity.
It happens to be proportional
to this quantity here.
And the factor is, happens
to be, 1/2.
So these proportionality
factors are what
we're looking for.
Their rates of change.
Their rates of change of
something with respect to
something else.
Now, on your homework,
you have something
rather similar to this.
So in Problem, on Part 2b, Part
II, Problem 1, there's
the speed of a pitch, right?
And the speed of the pitch is
changing depending on how high
the mound is.
And the point here is that
that's approximately
proportional to the change in
the height of the mound.
In that problem, we had this
delta h, that was the x
variable in that problem.
And what you're trying to
figure out is what the
constant of proportionality
is.
That's what you're aiming
for in this problem.
So there's a linear
relationship, approximately,
to all intents and purposes
this is an equality.
Because the lower
order terms are
unimportant for the problem.
Just as over here, this function
is a little bit
complicated.
This function is a little
more simple.
For the purposes of this
problem, they are the same.
Because the errors are
negligible for the particular
problem that we're working on.
So we might as well work with
the simpler relationship.
And similarly, over here, so
you could do this with, in
this case with square roots,
it's not so hard here with
reciprocals of square roots.
It's also not terribly hard
to do it numerically.
And the reason why we're not
doing it numerically is that,
as I say, this is something
that happens all across
engineering.
People are looking for these
linear relationships between
the change in some input and
the change in the output.
And if you don't make these
simplifications, then when you
get, say, a dozen of them
together, you can't figure out
what's going on.
In this case the design of the
satellite, it's very important.
The speed actually isn't
just one speed.
Because it's the relative speed
of u to the satellite.
And you might be, it depends
on your angle of sight with
the satellite what
the speed is.
So it varies quite a bit.
So you really need this
rule of thumb.
Then there are all
kinds of other
considerations in this question.
Like, for example, there's the
fact that we're sitting on
Earth and so we're rotating
around on what's called a
non-inertial frame.
So there's the question
of that acceleration.
There's the question that the
gravity that I experience here
on Earth is not the same
as up at the satellite.
And that also creates a
difference in time, as a
result of general relativity.
So all of these considerations
come down to formulas which
are this complicated or
maybe a tiny bit more.
Not really that much.
And then people simplify them
enormously to these very
simple-minded rules.
And they don't keep track
of what's going on.
So in order to design the
system, you must make these
simplifications, otherwise you
can't even think about
what's going on.
This comes up in everything.
In weather forecasting,
economic forecasting.
Figuring out whether there's
going to be an asteroid that's
going to hit the Earth.
Every single one of these things
involves dozens of
these considerations.
OK, there was a question
that I saw, here.
Yes.
STUDENT: [INAUDIBLE]
PROF.
JERISON: Yeah.
Basically, any problem where
you have a derivative, the
rate of change also depends upon
what the base point is.
That's the question.
You're saying, doesn't this
delta v also depend, I had a
base point in that problem.
I happened to decide that
pitchers pitch on average
about 90 miles an hour.
Whereas, in fact, some pitchers
pitch at 100 miles an
hour, some pitch at 80 miles
an hour, and of course they
vary the speed of the pitch.
And so, this varies
a little bit.
In fact, that's sort of
a second order effect.
It does change the constant
of proportionality.
It's a rate of change at
a different base point.
Which we're considering fixed.
In fact, that's sort of
a second order effect.
When you actually do the
computations, what you
discover is that it doesn't
make that much difference.
To the a.
And that's something that
you get from experience.
That it turns out, which
things matter and
which things don't.
And yet again, that's exactly
the same sort of consideration
but at the next order of what
I'm talking about here is.
You have to have enough
experience with numbers to
know that if you take, if you
vary something a little bit
it's not going to change
the answer that you're
looking for very much.
And that's exactly the point
that I'm making.
So I can't make them all at
once, all such points.
So that's my pitch for
understanding things from this
point of view.
Now, we're going to go
on, now, to quadratic
approximations, which are a
little more complicated.
So, we talked a little bit about
this last time but I
didn't finish.
So I want to finish this up.
And the first thing that I
should say is that you use the
when the linear approximation
is not enough.
OK, so, that's something that
you really need to get a
little experience with.
In economics, I told you they
use logarithms. So sometimes
they use log linear functions.
Sometimes they use log quadratic
functions when the
log linear ones don't work.
So most modeling in
economics is with
log quadratic functions.
And if you've made it
any more complicated
than that, it's useless.
And it's a mess.
And people don't do it.
So they stick with the quadratic
ones, typically.
So the basic formula here, and
I'm going to take the base
point to be 0, is that f (
x ) is approximately f
( 0 ) f' ( 0 )x.
That's the linear part.
Plus this extra term.
Which is f'' ( 0 ) / 2x^2.
And this is supposed to
work for x near 0.
So it shows in the base point
as simply as possible.
So here's more or less where
we left off last time.
And one thing that I said I was
going to explain, which I
will now, is why it's
(1/2) f'' ( 0 ).
So we need to know that.
So let's work that out
here first of all.
So I'm just going to
do it by example.
So if you like, the answer is
just, well, what happens when
you have a parabola?
A parabola's a quadratic.
It had better, its quadratic
approximation
had better be itself.
It's got to be the best one.
So it's got to be itself.
So this formula, if it's
going to work, has
to work on the nose.
For quadratic functions.
So, let's take a look.
If I differentiate,
I get b 2cx.
If I differentiate a second
time, I get 2c.
And now let's plug it in.
Well, we can recover, what is
it that we want to recover?
We want to recover these numbers
a, b and c using the
derivatives evaluated at 0.
So let's see.
It's pretty easy, actually.
f ( 0 ) = a.
That's on the nose.
If you plug in x = 0 here,
these terms drop
out and you get a.
And now, f' ( 0 ), whoops
that was wrong.
So I wrote f' but what
I meant was f.
So f ( 0 ) is a.
Let's back up. f ( 0 ) is a, so
if I plug in x = 0 I get a.
Now, f' ( 0 ), that's this next
formula here, f' ( 0 ), I
plug in 0 here, and I get b.
That's also good.
And that's exactly what the
linear approximation is
supposed to be.
But now you notice, f'' is 2c.
So to recover c, I better
take half of it.
And that's it.
That's the reason.
There's no chance that any
other formula could work.
And this one does.
So that's the explanation
for the formula.
And now I remind you that I
had a collection of basic
formulas written on the board.
And I want to just make sure
we know all of them again.
So, first of all, there was
sine x is approximately x.
Cosine x is approximately
1 - 1/2 x^2.
And e ^ x is approximately
1 x 1/2 x^2.
So those were three that
I mentioned last time.
And, again, this is
all for x near 0.
All for x near 0 only.
These are wildly wrong, Far
away, but near 0 they're nice,
good, quadratic approximations.
Now, the other two
approximations that I want to
mention are the logarithm and we
use the base point shifted.
So we can put it at x near 0.
And this one - sorry,
this is an
approximately equals sign there.
Turns out to be x - 1/2 x^2.
And the last one is one (1 x)
^ r, which turns out to be 1
rx r ( r - 1) / 2x^2.
Now, eventually, your mind will
converge on all of these
and you'll find them relatively
easy to memorize.
But it'll take some
getting used to.
And I'm not claiming that you
should recognize them and
understand them all now.
But I'm going to put a giant
box around this.
STUDENT: [INAUDIBLE]
PROF.
JERISON: Yes.
So the question was, you get all
of these if you use that
equation there.
That's exactly what are you
going to do. so I already did
it actually for these
three, last time.
But I didn't do it yet
for these two.
But I will do it in
about two minutes.
Well, maybe five minutes.
But first I want to explain just
a few things about these.
They all follow from
the basic formula.
In fact, that one deserves a
pink box too, doesn't it.
That one's pretty important.
Alright.
Yeah.
Maybe even some little
sparkles.
Alright.
OK.
So that's pretty important.
Almost as important as
the more basic one
without this term here.
So now, let me just tell you
a little bit more about the
significance.
Again, this is just to reinforce
something that we've
already done.
But it's closely related to
what you're doing on your
problem set.
So it's worth your while
to recall this.
So, there's this expression
that we were dealing with.
And we talked about
it in lecture.
And we showed that this tends
to e as k goes to infinity.
So that's what we showed
in lecture.
And the way that we did that
was, we took the logarithm and
we wrote it as k times, sorry,
the ln of 1 (1 / k).
And then we evaluated
the limit of this.
And I want to do this limit
again, using linear
approximation.
To show you how easy it is.
If you just remember the
linear approximation.
And then we'll explain
where the quadratic
approximation comes in.
So I claim that this
is approximately
equal to k ( 1 / k).
Now, why is that?
Well, that's just this
linear approximation.
So what did I use here?
I used ln of 1 x is
approximately x.
For x = 1 / k.
That's what I used in this
approximation here.
And that's the linear
approximation
to the natural logarithm.
And this number is relatively
easy to evaluate.
I know how to do it.
It's equal to 1.
That's the same, well, so
where does this work?
This works where this
thing is near 0.
Which is when k is going
to infinity.
This thing is working only when
k is going to infinity.
So what it's really saying, this
approximation formula,
it's really saying that as we
go to infinity, in k, this
thing is going to 1.
As k goes to infinity.
So that's what it's saying.
That's the substance there.
And that's how we want to use
it, in many instances.
Just to evaluate limits.
We also want to realize that
it's nearby when k is pretty
large, like 100 or something
like that.
Now, so that's the idea of
the linear approximation.
Now, if you want to get the rate
of convergence here, so
the rate of what's called
convergence.
So convergence means how fast
this is going towards that.
What I have to do is take
the difference.
I have to take ln ak, and I have
to subtract 1 from it.
And I know that this is going to
0, and the question is how
big is this.
We want it to be very small.
And the answer we're going to
get, so the answer just uses
the quadratic approximation.
So if I just have a little bit
more detail, then this
expression here, in other words,
I have the next higher
order term.
This is like 1 / k, this
is like 1 / k^2.
Then I can understand how big
the difference is between the
expression that I've
got and its limit.
And so that's what's
on your homework.
This is on your problem set.
OK, so that is more or less an
explanation for one of the
things that quadratic
approximations are good for.
And I'm going to give you
one more illustration.
One more illustration.
And then we'll actually
check these formulas.
Yeah, another question.
STUDENT: [INAUDIBLE]
PROF.
JERISON: That's a very
good question here.
When they, which in this case
means maybe, me, when I give
you a question, does one specify
whether you want to
use a linear or a quadratic
approximation.
The answer is, in real life
when you're faced with a
problem like this, where some
satellite is orbiting and you
want to know the effects of
gravity or something like
that, nobody is going to
tell you anything.
They're not even going to tell
you whether a linear
approximation is relevant, or
a quadratic or anything.
So you're on your own.
When I give you a question, at
least for right now, I'm
always going to tell you.
But as time goes on I'd like you
to get used to when it's
enough to get away with a
linear approximation.
And you should only use a
quadratic approximation if
somebody forces you to.
You should always start trying
with a linear one.
Because the quadratic ones are
much more complicated as
you'll see in this
next example.
OK, so the example that I want
to use is, you're going to be
stuck with it because I'm asking
for the quadratic.
So we're going to find the
quadratic approximation near,
for x near 0.
To what?
Well, this is the same function
that we used in the
last lecture.
I think this was it. e
^ - 3x (1 x) ^ - 1/2.
OK.
So, unfortunately, I stuck it in
the wrong place to be able
to fit this very long
formula here.
So I'm going to switch it.
I'm just going to
write it here.
And we're going to just
do the approximation.
So we're going to say quadratic,
in parentheses.
And we'll say x near 0.
So that's what I want.
So now, here's what
I have to do.
Well, I have to write in the
quadratic approximation for e
^ - 3x, and I'm going to use
this formula right here.
And so that's (1 (-
3x) (- 3x)^2 / 2).
And the other factor, I'm going
to have to use this
formula down here. because
r is - 1/2.
And so that's (1 - 1/2 x 1/2
( - 1/2)( - 3/2)x^2).
So this is the r term, and
this is the r - 1 term.
And now I'm going to do
something which is the only
good thing about quadratic
approximations.
They're messy, they're long,
there's nothing particularly
good about them.
But there is one good
thing about them.
Which is that you always get
to ignore the higher order
terms. So even though this
looks like a very ugly
multiplication, I can
do it in my head.
Just watching it.
Because I get a 1 * 1, I'm
forced with that term here.
And then I get the cross terms
which are linear, which is -
3x - 1/2 x.
We already did that when we
calculated the linear
approximation, so that's
this times the 1 and
this times that 1.
And now I have three
cross-terms which are quadratic.
So one of them is these two
linear terms are multiplying.
So that's plus 3/2 x^2.
That's (- 3)( - 1/2).
And then there's this term,
multiplying the 1,
that's plus 9/2 x^2.
And then there's one
last term, which is
this monster here.
Multiplying 1, and
that is - 3/8.
So the great thing is, we drop
x^3, x ^ 4, etc., terms. Yeah?
STUDENT: [INAUDIBLE]
PROF.
JERISON: OK, well
so copy it down.
And you work it out as
I'm doing it now.
So what I did is, I
multiplied 1 by 1.
I'm using the distributive
law here.
That was this one.
I multiplied this 3x by this
one, that was that term.
I multiplied this by this,
that's that term.
And then I multiplied
this by this.
In other words, 2 x terms
that gave me an x^2
and a (- 3)( - 1/2).
And I'm going to stop
at that point.
Because the point is it's
just all the rest of the
terms that come up.
Now, the reason, the only reason
why it's easy is that I
only have to go up to
x squared term.
I don't have to do
the higher ones.
Another question,
way back here.
Yeah, right there.
STUDENT: [INAUDIBLE]
PROF.
JERISON: OK.
So somebody can check
my arithmetic, too.
Good.
STUDENT: [INAUDIBLE]
PROF.
JERISON: Why do I get to drop
all the higher-order terms.
So, that's because the situation
where I'm going to
apply this is the situation
in which x is, say, 1/100.
So if here's about 1/100.
Here's something which is
on the order of 100.
This is on the order
of 1/100^2.
1/100^2, all of these terms.
Now, these cubic and quartic
terms are of the order
of 1/100^3.
That's 10 ^ - 6.
And the point is that I'm
not claiming that I
have an exact answer.
And I'm going to drop things
of that order of magnitude.
So I'm saving everything
up to 4 decimal places.
I'm throwing away things which
are 6 decimal places out.
Does that answer
your question?
STUDENT: [INAUDIBLE]
PROF.
JERISON: So.
That's the situation, and now
you can combine the terms. I
mean, it's not very
impressive here.
This is equal to 1 - 7/2
x, maybe 51/8 x^2.
If I've made that, if those
minus signs hadn't canceled, I
would have gotten the
wrong answer here.
Anyway.
So, this is a 2 here, sorry.
7/2.
This is the linear approximation
we got last time
and here's the extra information
that we got from
this calculation.
Which is this 51/8 term.
Right, you have to accept that
there's a certain degree of
complexity to this problem and
the answer is sufficiently
complicated so it can't be less
arithmetic because we get
this peculiar 51/8
there, right.
So one of the things to realize
is that these kinds of
problems, because they involve
many, many terms are always
going to involve a little bit
of complicated arithmetic.
Last little bit, I did promise
you that I was going to derive
these two relations,
as I said.
Did the ones in the
left column.
So let's carry that out.
And as someone just pointed out,
it all comes from this
formula here.
So let's just check it.
So we'll start with
the ln function.
This is the function, f,
and then f' is 1/1 x.
And f'', so this is f', this
is f'', is - 1/1 x^2.
And now I have to
plug in x = 0.
So at x = 0 this is
ln 1, which is 0.
So this is at x = 0.
I'm getting 0 here, I plug
in 0 and I get 1.
And here, I plug in
0 and I get - 1.
So now I go and I look up at
that formula, which is way in
that upper corner there.
And I see that the coefficient
on the constant is 0.
The coefficient on x is 1.
And then the other coefficient,
the very last
one, is - 1/2.
So this is the - 1 here.
And then in the formula,
there's a 2 in the
denominator.
So it's half of whatever
I get for this second
derivative, at 0.
So this is the approximation
formula, which is way up in
that corner there.
Similarly, if I do it for
(1 x) ^ r, I have to
differentiate that I get r( 1
x) ^ r - 1, and then r ( r -
1( x 1) ^ r - 2.
So here are the derivatives.
And so if I evaluate them
at x = 0, I get 1.
That's 1 ^ r = 1.
And here I get r.
(1 ^ r - 1)r.
And here, I plug in x = 0
and I get r ( r - 1).
So again, the pattern is
right above it here.
The 1 is there, the
r is there.
And then instead of r ( r
- 1), I have half that.
For the coefficient.
So these are just examples.
Obviously if we had a more
complicated functional, we
might carry this out.
But as a practical matter, we
try to stick with the ones in
the pink box and just use
algebra to get other formulas.
So I want to shift gears now and
treat the subject that was
supposed to be this lecture.
And we're not quite caught up,
but we will try to do our best
to do as much as we can today.
So the next topic is
curve sketching.
And so let's get started
with that.
So now, happily in this subject,
there are more
pictures and it's a little
bit more geometric.
And there's relatively
little computation.
So let's hope we can do this.
So I want to, so here
we go, we'll
start with curve sketching.
And the goal here
STUDENT: [INAUDIBLE]
PROF.
JERISON: So that's like,
liner, the last time.
That's kind of sketchy
spelling, isn't it?
Yeah, there are certain kinds of
things which I can't spell.
But, alright.
Sketching.
Alright.
So here's our goal.
Our goal is to draw the graph
of f, using f' and f''.
Whether they're positive
or negative.
So that's it.
This is the goal here.
However, there is a
big warning that I
want to give you.
And this is one that
unfortunately I now have to
make you unlearn, especially
those that you that have
actually had a little bit of
calculus before, I want to
make you unlearn some of your
instincts that you developed.
So this will be harder for those
of you who have actually
done this before.
But for the rest of you, it
will be relatively easy.
Which is, don't abandon your
precalculus skills.
And common sense.
So there's a great deal of
common sense in this.
And it actually trumps
some of the calculus.
The calculus just fills
in what you
didn't quite know yet.
So I will try to illustrate
this.
And because we're running a bit
late, I won't get to the
some of the main punchlines
until next lecture.
But I want you to do it.
So for now, I'm just going to
tell you about the general
principles.
And in the process I'm going to
introduce the terminology.
Just, the words that we need
to use to describe what is
that we're doing.
And there's also a certain
amount of carelessness with
that in many of the treatments
that you'll see.
And a lot of hastiness.
So just be a little patient
and we will do this.
So, the first principle
is the following.
If f' is positive, then
f is increasing.
That's a straightforward idea,
and it's closely related to
this tangent line approximation
or the linear
approximation that I just did.
You can just imagine.
Here's the tangent line,
here's the function.
And if the tangent line is
pointing up, then the function
is also going up, too.
So that's all that's
going on here.
Similarly, if f' is negative,
then f is decreasing.
And that's the basic idea.
Now, the second step is also
fairly straightforward.
It's just a second order effect
of the same type.
If you have f'' as positive,
then that means that f' is
increasing.
That's the same principle
applied one step up.
Right?
Because if f'' is positive,
that means it's the
derivative of f'.
So it's the same principle
just repeated.
And now I just want to draw
a picture of this.
Here's a picture
of it, I claim.
And it looks like something's
going down.
And I did that on purpose.
But there is something that's
increasing here.
Which is, the slope is very
steep negative here.
And it's less steep negative
over here.
So we have the slope which
is some negative
number, say, - 4.
And here it's - 3.
So it's increasing.
It's getting less negative,
and maybe eventually it'll
curve up the other way.
And this is a picture of what
I'm talking about here.
That's what it means to say
that f' is increasing.
The slope is getting larger.
And the way to describe a curve
like this is that it's
concave. So f is concave up.
And similarly, f'' negative is
going to be the same thing as
f concave, or implies
f concave down.
So those are the ways in which
derivatives will help us
qualitatively to draw graphs.
But as I said before, we still
have to use a little bit of
common sense when we
draw the graphs.
These are just the additional
bits of help that
we have from calculus.
In drawing pictures.
So I'm going to go through
one example to
introduce all the notations.
And then eventually, so probably
at the beginning of
next time, I'll give you a
systematic strategy that's
going to work when what I'm
describing now goes wrong, or
a little bit wrong.
So let's begin with a
straightforward example.
So, the first example that I'll
give you is the function
f (x) = 3x - x^3.
Just, as I said, to be able to
introduce all the notations.
Now, if you differentiate
it, you get 3 - 3x^2.
And I can factor that.
This is 3 ( 1 - x)( 1 x).
OK?
And so, I can decide whether
the derivative
is positive or negative.
Easily enough.
Namely, just staring at this,
I can see that when -1 < x <
1, in that range there, both
these numbers, both these
factors, are positive.
1 - x is a positive number and
1 x is a positive number.
So, in this range, f'
( x ) is positive.
So this thing is, so
f is increasing.
And similarly, in the other
ranges, if x is very, very
large, this becomes, if it
crosses 1, in fact, this
becomes, this factor becomes
negative and
this one stays positive.
So when x > 1, we have that
f ' (x) is negative.
And so f is decreasing.
And the same thing goes
for the other side.
When it's less than - 1, that
also works this way.
Because when it's less
than - 1, this
number factors positive.
But the other one is negative.
So in both of these cases, we
get that it's decreasing.
So now, here's the schematic
picture of this function.
So here's - 1, here's 1.
It's going to go
down, up, down.
That's what it's doing.
Maybe I'll just leave
it alone like this.
That's what it looks like.
So, this is the kind of
information we can get right
off the bat.
And you notice immediately that
it's very important, from
the features of the function,
the sort of key features of
the function that we see here,
are these two places.
Maybe I'll even mark them
in a, like this.
And these things are
turning points.
So what are they?
Well, they're just the
points where the
derivative changes sign.
Where it's negative here and
it's positive there, so there
it must be 0.
So we have a definition, and
this is the most important
definition in this subject,
which is that is if f' (x0) =
0, we call x0 a critical
point.
The word 'turning point' is
not used just because, in
fact, it doesn't have to turn
around at those points.
But certainly, if it turns
around then this will happen.
And we also have another
notation, which is the number
y0 which is f ( x0 ) is called
a critical value.
So these are the key numbers
that we're going to have to
work out in order to understand
what the function
looks like.
So what I'm going to do
is just plot them.
We're going to plot
the critical
points and the values.
Well, we found the critical
points relatively easily.
I didn't write it down here
but it's pretty obvious.
If you set f(x) = 0, that
implies that (1 - x)( 1 x) =
0, which implies that
x is or - 1.
So those are known as
the critical points.
And now, in order to get the
critical values here, I have
to plug in f (1), for instance,
the function is 3x -
x^2, so there's this 3 *
1 - 1^3, which is 2.
And f ( - 1), which is 3 ( -
1) - (- 1)^3, which is - 2.
And so I can plot the
function here.
So here's the point - 1 and
here's, up here, is 2.
So this is - whoops,
which one is it?
Yeah.
This is - 1, so it's
down here.
So it's (- 1, - 2).
And then over here, I have
the point (1, 2).
Alright, now, what information
do I get from - so I've now
plotted two, I claim, very
interesting points.
What information do
I get from this?
The answer is, I know something
very nearby.
Because I've already checked
that the thing is coming down
from the left, and
coming back up.
And so it must be shaped
like this.
Over here.
The tangent line is 0, it's
going to be level there.
And similarly over here,
it's going to do that.
So this is what we know
so far, based
on what we've computed.
Question.
STUDENT: [INAUDIBLE]
PROF.
JERISON: The question
is, what happens if
there's a sharp corner.
The answer is, calculus, it's
not called a critical point.
It's a something else.
And it's a very important
point, too.
And we will be discussing
those kinds of points.
There are much more dramatic
instances of that.
That's part of what we're
going to say.
But I just want to save
that, alright.
We will be discussing.
Yeah.
Question.
STUDENT: [INAUDIBLE]
PROF.
JERISON: The question that was
asked was, how did I know at
the critical point that it's
concave down over here and
concave up over here.
The answer is that I actually
did not use the second
derivative yet.
What I used is another
piece of information.
I used the information that
I derived over here.
That f' is positive, where
f' is positive
and where it's negative.
So what I know is that the graph
is going down to the
left of - 1.
It's going up to the
right, here.
It's going up here and it's
going down there.
I did not use the second
derivative.
I used the first derivative.
OK, but I didn't just use the
fact that there was a turning
point here.
So, actually, I was using
the fact that it
was a turning point.
I wasn't using the fact
that it had the
second derivative, though.
OK.
For now.
You can also see it by the
second derivative as well.
So now, the next thing that
I'd like to do, I need to
finish off this graph.
And I just want to do it a
little bit carefully here.
In the order that
is reasonable.
Now, you might happen to notice,
and there's nothing
wrong with this, so let's
even fill in a guess.
In order to fill in a guess,
though, and have it be even
vaguely right, I do have to
notice that this thing
crosses, this function
crosses the origin.
The function f(x) = 3x - x^3
happens also have the property
that f ( 0 ) = 0.
Again, common sense.
You're allowed to use
your common sense.
You're allowed to notice
a value of the
function and put it in.
So there's nothing
wrong with that.
If you happen to have
such a value.
So, now we can guess what
our function is
going to look like.
It's going to maybe come
down like this.
Come up like this.
And come down like this.
That could be what
it looks like.
But, you know, another
possibility is it sort of
comes along here and
goes out that way.
Comes along here and goes
out that way, who knows?
It happens, by the way, that
it's an odd function.
Right?
Those are all odd powers.
So, actually, it's symmetric
on the right half
and the left half.
And crosses at 0.
So everything that we do on the
right is going to be the
same as what happens
on the left.
That's another piece
of common sense.
You want to make use of that as
much as possible, whenever
you're drawing anything.
Don't want to throw
out information.
So this function happens
to be odd.
Odd, and f ( 0 ) = 0.
I'm considering those to be
kinds of precalculus skills
that I want you to use
as much as you can.
So now, here's the first feature
which is unfortunately
ignored in most discussions
of functions.
And it's strange, because
nowadays we
have graphing things.
And it's really the only part
of the exercise that you
couldn't do, at least on this
relatively simpleminded level,
with a graphing calculator.
And that is what I would call
the ends of the problem.
So what happens off the screen,
is the question.
And that basically is the
theoretical part of the
problem that you have
to address.
You can program this.
You can draw all the pictures
that you want.
But what you won't see is
what's off the screen.
You need to know something
to figure out
what's off the screen.
So, in this case, I'm talking
about what's off the screen
going to the right, or
going to the left.
So let's check the ends.
So here, let's just
take a look.
We have the function f(x), which
is, sorry, 3x - x^3.
Again this is a precalculus
sort of thing.
And we're imagining now,
let's just do x
goes to plus infinity.
So what happens here.
When x is gigantic, this term
is completely negligible.
And it just behaves like -
x^3, which goes to minus
infinity as x goes
to plus infinity.
And similarly, f (x) goes to
plus infinity if x goes to
minus infinity.
Now let me pull down this
picture again, and show you
what piece of the information
we've got.
We now know that it is
heading up this way.
It doesn't go like this,
it goes up like that.
And I'm going to put an
arrow for it, And it's
going down like this.
Heading down to minus infinity
as x goes out
farther to the right.
And going out to plus infinity
as x goes farther to the left.
So now there's hardly
anything left of
this function to describe.
There's really nothing left
except maybe decoration.
And we kind of like that
decoration, so we will pay
attention to it.
And to do that, we'll have to
check the second derivative.
So if we differentiate a
second time, the first
derivative was, remember,
3 - 3x^2.
So the second derivative
is - 6x.
So now we notice that f'' (x) is
negative if x is positive.
And f'' ( x) is positive
if x is negative.
And so in this part
it's concave down.
And in this part it's
concave up.
And now I'm going to switch
the boards so that
you'll, and draw it.
And you see that it was begging
to be this way.
So we'll fill in the
rest of it here.
Maybe in a nice color here.
So this is the whole graph and
this is the correct graph.
It comes down in one swoop down
here, and comes up here.
And then it changes to concave
down right at the origin.
So this point is of interest,
not only because it's the
place where it crosses the axis,
but it's also what's
called an inflection point.
Inflection point, that's a point
where because f'' at
that place is equal to 0.
So it's a place where the
second derivative is 0.
We also consider those to
be interesting points.
Now, so let me just making
one closing remark here.
Which is that all of this
information fits together.
And we're going to have much,
much harder examples of this
where you'll actually have to
think about what's going on.
But there's a lot of stuff
protecting you.
And functions will behave
themselves and turn around
appropriately.
Anyway, we'll talk about
it next time.
