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0:01
Today is what, Monday, October 10th? Is that right? I don't remember offhand what's the
0:07
date scheduled for example. It's Wednesday of next week, so we've got nine days
0:13
here. So today we're doing Section 4 .2, the Mean Value Theorem. Pretty much the
0:18
entire lecture is going to be the focus and applications of other theorems related to
0:23
the Mean Value Theorem. Wednesday, we'll do a little bit on 4 .3. We're not
0:28
going to do the entire section, but we'll do the first four pages or so,
0:31
and we probably will start Chapter 5, Section 5 .1. Chapter 5
0:37
is our toughest chapter about integration, the theory of integration. It's pretty challenging, especially
0:43
the proofs in the book, to really try to understand those, and to do proofs
0:47
on your arm, too. Wednesday we'll do a little bit of 4 .3, we're not
0:28
going to do that.
0:49
So we'll do Section 5 .1 a little bit on Wednesday, a lot on Friday,
0:55
probably start... Section 5 .2, or at least focus on the main facts from Section
0:59
5 .2 by next Monday or so, and probably your test will
1:05
include certainly definitions and theorems through Section 5 .2, and
1:10
perhaps proofs through Section 5 .1, or maybe an easier type proof from Section 5
1:16
.2 will be what your exam will be through Chapters 3, 4, and the first
1:21
couple sections of chapter 5. So, for the study guide, for chapter 5, I'll have
1:27
to figure that out. So, if you could send me a note out of class
1:30
to remind me, to figure it out. All right, lecture 16, again, our main focus
1:35
is going to be the mean value theorem. We spent time the very first day
1:39
of class talking about the mean value theorem, MBT. right?
1:44
Because you can also call it most valuable theorem, at least for differential But most
1:46
valuable theorem, at least for differential calculus, the book even goes so far to say
1:51
it's sometimes called the fundamental theorem of differential calculus, which is different than the fundamental
1:56
theorem of calculus. Related to derivatives, it's a pretty obvious sounding fact when you look
2:01
at its statement, but the proof, when you sort of deconstruct it out of what
2:06
we'll do today, does ultimately depend on the difference axis. So let's write down
2:18
mean value theorem. It's our
2:24
third value theorem so far, right? We've got the intermediate value theorem, we've got the
2:29
extreme value theorem, and now we have the mean value theorem.
2:35
Suppose you've got a function defined in a closed interval, and here I'm going to
2:40
go We're going to go ahead and use interval notation like the book does. Emphasize
2:44
that we're defining some closed interval. And we're going to assume f is continuous on
2:49
the closed interval.
2:57
And since this is chapter 4, there must be something about derivatives here. And the
3:00
next assumption is that you say, and it is differentiable,
3:06
has a derivative. on the
3:12
open interval. It's sufficient to assume that it's just
3:17
differentiable on the open interval, not the endpoints. Perhaps you've got a function that's got
3:22
a vertical tangent as you approach an endpoint. For example, like the square root function.
3:29
You consider any closed interval with zero as the left endpoint, like zero to five
3:34
or something like 0 to 5 or something like that. It is continuous on the
3:38
entire interval. It fails to be differentiable at zero. It's not even a left or
3:42
not even a right differentiable as a left end point.
3:48
The tangent becomes vertical as you approach that point. That would be one example where
3:53
the mean value theorem would still apply even though the function is not differentiable at
3:57
zero. Another example actually where you could be failed to be differentiable at both endpoints
4:02
would be like the arc sine function. The graph of the arc
4:08
sine function looks like this, and it's got vertical tangents at 1 and
4:14
negative 1. So it fails to be even one
4:20
-sidedly differentiable at those points. I hope you know what the graph of the arc
4:25
sine function looks like. It looks like that. Why don't you just pass it up
4:29
instead of going in front? Why don't you just pass it up? No, I have
4:31
to go in the bathroom. Okay. Sorry.
4:37
Here you've got differentiability not on the closed interval but on the open interval. It's
4:42
continuous on the closed interval. Again, you've got vertical tangents as you approach either endpoint.
4:48
But the mean value theorem would still apply to it. So by only assuming differentiability
4:52
on the open interval, you get a stronger theorem that does apply to these examples
4:56
too. The conclusion is then,
5:00
there exists a point C in the open interval
5:05
such that the derivative at C, the derivative of
5:11
F at C, which will be the slope of the tangent line, of course, right?
5:14
Don't forget that. is equal to the slope of the secant line
5:19
between the points A comma F of A and B comma F of B.
5:26
Drawing a generic picture makes it easy to convince yourself that this is
5:31
true on an intuitive level. If this is A and this is B,
5:38
the secant line between the points A comma F of A
5:43
and B comma F of B looks like that. The mean
5:49
value theorem says there's some number C between A and B strictly... between A and
5:53
B. I even made it look like almost a vertical tangent here today. Where the
5:57
slope of the tangent is the same as the slope of the secant. So it
6:00
would be parallel. Perhaps right about there
6:04
is a point where the tangent has a slope that's the same as the slope
6:09
of the secant. So right about there would be the C that is guaranteed to
6:14
exist in the conclusion of the mean value theorem. There can be more than one
6:18
C. Graph looks like that. There's
6:24
your secant line. There's more than one spot where
6:30
the slope of the tangent is the same as the slope of the secant. So
6:34
there can be more than one C. But there's guaranteed to exist at least one
6:37
C. So that's the statement. There's a visual
6:43
emphasizing the geometric meaning. Here's a visual emphasizing the geometric meaning. Here are a couple
6:46
examples emphasizing that you only need differentiability on the open interval. You should also know,
6:52
as a real basic thing, that if your function fails to be differential at even
6:55
one point, the conclusion might not hold. And the simplest example is
7:01
the absolute value function. Take it to be on any interval where a
7:07
is negative and b is positive. For example, the interval is negative 2 to 7.
7:09
And call the interval from negative 2 to 7. f of
7:15
7 minus f of negative 2 over 7 minus negative 2.
7:21
That'll be 7 minus 2 over 9.
7:27
5 ninths is the slope of the secant line. The graph looks
7:33
like v. Slope of that line. It is supposed to be 5
7:39
ninths here. There is no point where the slope of that line is supposed to
7:39
be 5 ninths here. derivative equals 5
7:44
ninths. The derivative equals negative 1 over here when x is negative and positive 1
7:50
when x is positive and it does not exist at 0. f prime of 0
7:54
does not exist. Never does the derivative equal 5 ninths. If you fail to be
7:59
differentiable at even one point it is possible that the conclusion will not hold.
8:06
On the other hand, you should also be aware that the hypotheses might not be
8:11
hold for a certain situation, but the conclusion could hold. That can happen.
8:17
For example, take the qVert function.
8:26
On any interval containing zero in the interior, where the left end point Where the
8:30
left endpoint is negative and the right endpoint is positive, and I think the is
8:31
negative and the right end point is positive. conclusion will hold, and certainly think about
8:37
particular examples here, like consider the interval from negative 8 to positive 27,
8:42
f of 27 minus f of 8
8:47
divided by 20, or minus f of negative 8, excuse me, divided by f 27
8:53
minus negative 8, cube root of 27 is 3, cube Q root of 27 is
8:58
3. Q root of negative 8 is negative 2. 27 minus negative
9:04
8 will be plus 35. root of negative 8 is negative 2. 5 30 fifths
9:10
or 1 7th is the slope of the secant line. Is there a place where
9:15
the derivative has a value equal to 1 7th where the slope of the
9:21
tangent to the original function does have that same
9:27
slope? I believe it will be true, but there will be a place where that
9:32
is true, and you can graph it to see that visually. In fact, there might
9:36
need to be two places. There's the derivative.
9:42
Set it equal to one -seventh and see if you can solve for x.
9:50
Multiply both sides by three to get x to the negative two -thirds is three
9:54
-sevenths. Take the reciprocal of both sides, x to the positive two -thirds is seven
10:00
-thirds. Raise both
10:05
sides to the three -halves power and take plus or minus square
10:11
roots, really. Plus or minus seven -thirds to the three -halves power. Looks like it
10:17
works. That would be the same as plus or minus
10:22
seven -thirds times the square root of seven -thirds. That would be the same as
10:21
plus or minus... Do you want to
10:27
double check if those are both between negative eight and 27? I'm pretty sure they
10:31
are. Seven -thirds, yeah, it's about two, a little bit bigger than two, take the
10:34
square root. Yeah, definitely both of those are in the interval. If you draw a
10:39
graph, it seems pretty clear that it's true.
10:47
Remember, the cube root function I said last time fails to be differentiable at zero.
10:52
We have an infinite tangent there. I should say a vertical tangent. Undefined
10:58
slope, really. It fails to be differentiable at zero, but there are points where
11:04
the slope of the tangent equals the slope of the secant. Right about here
11:09
and right about there. I'm not drawing very well.
11:16
Those points actually should be equal distance from x equals 0, but I didn't draw
11:20
it that way. So you get the main points here.
11:26
This is a theorem. We are going to prove it. Here's the intuition by why
11:32
it seems plausible, but not a proof. It's good to consider
11:37
even the simplest example, the absolute value function, fails to be differentiable at one point
11:42
on any interval where a is differentiable at one point. negative and b is positive,
11:45
the conclusion of the mean value theorem will not hold. But there are examples like
11:50
this one where even though the hypotheses do not hold, the conclusion still does. That
11:55
can happen for a specific example, some guarantee from the theorem.
12:04
How do you prove the theorem? Well, again, we're going to deconstruct it here. We're
12:08
going to go backwards, so to speak. Well again, we're going to deconstruct it here.
12:08
We're going to go backwards, so to speak. Although the first step in this backwards
12:13
kind of approach to the proof,
12:19
I don't know if it's the best way to describe it. It might seem a
12:21
little funny or artificial or something like that. We're going to use something called Rolle's
12:27
Theorem to prove it. And Rolle's Theorem, it's kind of interesting, is actually a special
12:32
case of the mean value theorem.
12:42
And so then the crux of the matter will then be proving Rohl's theorem.
12:48
says suppose we've got a function in the same kind of situation
12:54
that is continuous
13:03
on the open interval, and
13:08
suppose f of a
13:14
equals f of b, the conclusion is
13:20
then there exists a c between a and b. So what happens? Well,
13:25
if the function values are the same, that means the slope of the secant must
13:28
be zero. There must be a c where f prime of c is zero,
13:34
where the tangent has a slope of zero.
13:40
A simpler looking conclusion, but it does follow from the mean value theorem. We can
13:46
easily prove Rolle's theorem from the mean value theorem. What's not so obvious and
13:52
what's pretty interesting is you can actually take this special case of the mean value
13:55
theorem and prove the more general case. Once you see the
14:01
trick, it's actually pretty easy. Though it does involve, Charles, a trick.
14:08
You love tricks? Yeah. All right. These are all good tricks to know and
14:14
to learn about and to try to remember. But then the harder work will be
14:20
proving Rolle's theorem. So we are, in
14:26
a sense, by using Roll's Theorem to prove the mean value theorem, we're avoiding the
14:31
hard work for the moment.
14:38
Proof sketch. I'm just going to do sketches of
14:44
the mean value theorem. Proof sketch of the pf there. This sketch is a PF
14:49
there. For the sake of time, we've got lots to talk about today.
14:56
It's a little trick, yeah. Well, maybe you'd call it a big trick. Actually, maybe
15:01
you should have the picture back up here.
15:07
It's kind of similar to the trick that you used to prove that a function
15:10
that maps an interval into itself and is continuous has a fixed point. You took
15:15
the function minus x to create a new function. that you apply the intermediate value
15:20
theorem to. It's a similar trick here. If this is the graph of f, we've
15:25
got a secant line here between the endpoints. That's the graph of some
15:31
function, a linear function. Maybe, well, if I subtract that linear function from f,
15:37
I get a function whose values at a and he would be the same, because
15:42
these two graphs come together at the endpoints. Because these two graphs come together at
15:44
the end points. And if the values of that new function of A and B
15:49
are the same, I could apply Rohl's theorem to it. And maybe that application of
15:53
Rohl's theorem to that new function would get us what we want, maybe. It's not
15:58
obvious that it will, but maybe if we're lucky here. Let g of x
16:04
be f of x minus l of x, where l of x is a
16:10
function whose graph is this secant line.
16:16
Stylistically, I should really say what L of X is first, but oh well. What
16:22
is L of X? How would you make a line whose graph is this? You've
16:27
got to get the slope right, and you've got to get a point right. You
16:31
know the slope, you know the point, you can get the formula right. But we
16:33
are doing it in an abstract form. We don't have specific numbers here. The slope
16:38
of that line will be that.
16:45
That's the slope of the secant line. The slope of that line will be... It
16:48
has to get multiplied by x in the equation for this linear function, so
16:54
that it is the slope of that linear function. You can also multiply it by
16:58
x minus a, and it would still have the right slope. The benefit of
17:04
writing x minus a, though, is that it helps me guarantee that I get the
17:08
right point, this point on the graph of L of x. If I just
17:14
now add f of a to this, that will do it.
17:22
This quantity, this quotient there, that's the slope of a secant line. It's an average
17:26
rate of change. It's being multiplied by x, so it is the slope of this
17:31
linear function. And if you plug So it is the slope of this linear function.
17:32
And if you plug in the point of whose x -coordinate is a into here,
17:37
you get 0 plus f of a. does give you f of a.
17:44
That does go through the right point. It's certainly differentiable. It's a linear
17:49
function. We know we can differentiate it. We know, well, okay, let's check. Does g
17:55
of a equal g of b? We're going to apply Rohl's theorem to this new
18:00
function, g, which I was trying to say We're going to apply Rolle's theorem to
17:59
this new function g. before that is going to be continuous
18:05
on the closed interval and differential and open interval because f is and certainly because
18:10
l is. What is g of a?
18:18
Well, I already talked about what you give when you plug in a into l.
18:21
You get zero there plus f of a. You get f of a, we're subtracting
18:27
it, g of a is zero.
18:32
What about g of b? It's less clear that g of b is going to
18:35
be zero, but it will be. You'll get f of b minus
18:41
in parentheses here, f of b minus f of a divided by b minus
18:47
a
18:53
The b minus a is cancelled there. Inside the parentheses here, the
18:59
f of a is cancelled. You ultimately get f of b
19:05
minus f of b. You get zero.
19:16
Got that? All right. Just replacing x first with a to get that. then with
19:21
b to get this. Divide those out. In the parentheses here, cancel the f of
19:27
a. We've got a minus f of a there and a plus f of a.
19:30
Then ultimately, we just have f of b minus f of b, which is still
19:34
zero. So for g, there must be some c between a and b where g
19:39
prime of c equals zero. The only thing left you can figure out, then, is
19:44
does that mean this is true? How do you relate that to f? Well, you
19:49
can relate the derivative of g to the derivative of f. g prime
19:55
of x will be f prime of x minus l prime of x. But
20:01
l prime of x is easy. It's a linear function. Its derivative isn't slow.
20:07
It's a constant function f of b minus
20:13
f of a over b minus a. So by saying G prime of C equals
20:18
0, from Rohl's theorem,
20:24
G prime of C will be 0. That means at C, this will be 0.
20:31
And I hope you can see it in your mind's eye without me writing it
20:34
down. That would imply that this is true. At that value of C. Now you
20:38
can see.
20:44
Should I go through it quick again? Okay. It's a trick.
20:51
It's an ingenious trick to use the special case of the theorem to prove the
20:54
theorem itself. Create a new function. That's the original function minus
21:00
a function that's linear where the graph of that linear function is the secant line.
21:06
Actually, I could get away with using any other parallel line, too. Using
21:12
the seeking line. ultimately allowed me to say g of a and g of b
21:15
were zero. What is the formula for that linear function? That does work.
21:21
It's got the right slope. It goes to the right point.
21:27
That's the main trick right there. Then there's this wondering, if we apply Roll's
21:33
Theorem to G, will we get the right conclusion? And fortunately, we do. You've got
21:38
to check that the hypotheses of the Roll's Theorem applies to G. is going to
21:44
be differentiable in the closed interval and continuous, excuse me, continuous in the closed interval
21:48
and differentiable in the open interval because F is assumed to be. And certainly linear
21:52
functions definitely are everywhere, continuous and differentiable. Check
21:58
that G of A and G of E are equal. B and E are equal.
22:00
They are. They're both zero. They are. They're both zero. Therefore, by Rohl's theorem, there
22:03
is a C between A and B, so that G prime of C is zero,
22:08
because I'm applying Rohl's theorem to G. Because G prime and F prime are related,
22:16
that means something about F prime as well. It means F prime of C minus
22:20
this is zero, which means this is true. Okay,
22:26
but this is begging the question, how do you prove Rohl's theorem? This is begging
22:26
the question, how do you prove Roll's Theorem?
22:57
continuing to work backwards, that your book doesn't give a name to.
23:04
I've seen it called Fermat's theorem before.
23:10
That's not the same as Fermat's little theorem from algebraic structures. It's also
23:16
not the same as Fermat's last theorem, proved some
23:21
25 years ago, 21 years ago, I think it's when they finally had an airtight
23:27
proof, which the person who proved it became world famous, okay? There's newspapers
23:33
everywhere. They made a normal video about him. His name is Andrew
23:39
Wiles. Anyway, this
23:44
Fermat's theorem is different than those. It says that
23:48
if you It says that if you've got a function that's got a local max
23:54
or min at an interior point of an interval, and if the function is differentiable
23:59
there, the derivative must be zero at the max or min.
24:09
The derivative must be zero there.
24:16
Let's sketch the proofing. So again, we're trying to prove rules there.
24:23
So we are assuming that a function is continuous in the closed interval and differentiable
24:27
in the open interval. If the function is constant over the interval,
24:32
if f of x equals f of a
24:37
for all x in the interval,
24:44
Including at b where you're already assuming f of a equals f of b. Then
24:48
the conclusion is obvious. The function is a constant function. Its graph is horizontal. Its
24:53
derivative is always here. Including at every interior point
24:59
of the close interval. Then the conclusion
25:18
right because F prime of C will always equal zero for all C between A
25:23
and B I am assuming A is strictly less than B here
25:25
if not
25:30
if this is not true then the function by the extreme value theorem
25:36
over this closed interval will have a global maximum in somewhere,
25:42
a global extreme value, which will also be a local extreme value.
25:50
And by Fermat's theorem, that means where you have the
25:55
extreme values, f prime of c will be zero.
26:02
So actually, that doesn't feel like much of a proof either, but it does. prove
26:06
it.
26:13
There's just a C and AD where F has either
26:18
a global max or global min.
26:30
So it's a local max or a local min as well.
26:43
Our book doesn't really use the word global anywhere. I like using it. They just
26:47
say extreme value, max or min. They avoid the word local, but they use a
26:52
different word. They say relative. The author, I should say, does. I like using the
26:58
words global and local for emphasis. Global is over the entire domain. I like using
26:58
the words global and local for emphasis. Local means nearby
27:04
some point in some neighborhood of the point.
27:11
That's the EVT rule.
27:30
Not Fermat's Little Theorem, not Fermat's Last Theorem. And the book doesn't call
27:35
Fermat's Theorem. I will point it out to you here. It's at the beginning of
27:41
Section 4 .2. It's Theorem 4 .10, page 139.
27:52
That's what I'm calling Fermat's Theorem. Since the function
27:58
is differentiable there That's what I'm calling for miles through. as well, f prime of
28:02
c must be zero. The book states the conclusion of Fermat's Theorem as
28:08
either f is not differentiable at c if you've got a local maximum in there,
28:12
or f prime of c is zero. In either case, you've got a critical point
28:18
for the function at c. Okay, so you've got a critical point for the function
28:19
at C.
28:24
Yeah, they use the word relative extreme value, local max or local min is another
28:30
way to phrase that. They don't assume F is differentiable at C, so they say
28:35
either F is not differentiable at C or F prime of C is zero. But
28:40
we're in a situation here where F is differentiable on the opening rule and the
28:43
assumption of rolls theorem, and therefore we don't have to say well, maybe f is
28:48
not differential at c. It is. And f prime c must be zero. So
28:54
actually that proof is not too bad once you see that kind of trick. Which
28:59
again brings us back in our deconstruction of the proof to, well, how do you
29:03
prove Fermatis Theorem?
29:11
Sketch or proof of Fermat. I think calling it
29:17
Fermat's Theorem is not completely standard. Maybe that's why our book
29:23
didn't call it that. Some people call it Fermat's Theorem.
29:29
And I'm pretty sure I do remember hearing or reading that Fermat did work on
29:33
differential calculus. He didn't see, he was before Newton and Leibniz and so he didn't
29:38
have the fundamentals to connect And so he didn't have the fundamental calculus to connect
29:40
integrals and derivatives, integrals and . but he did work on slopes of tangents and
29:46
calculating those slopes before Newton and Leibniz came
29:52
up with the full -fledged calculus. Essentially, you've got to think about
29:58
this kind of difference quotient here. Carefully. This
30:04
is a real sketch. okay? Mostly I'm just writing this and I'm going to think
30:08
about it here, okay? Mostly, I'm just writing this and I'm going to think about
30:08
it here. So what are we assuming with Fermat's Theorem? If we assume the
30:14
function is differentiable at C, where you've got a local maximum A, you can look
30:19
at the statement of Fermat's Theorem if it helps, page 139.
30:25
You've got an interval, a function defined in the closed interval, you've got an interior
30:30
point, not an endpoint C, where you've got a local Then point C, where you've
30:35
got a local max or min, high point or low point, not necessarily global, but
30:41
local, then the F is either not differentiable
30:46
there or F prime of C is zero. You can prove that by saying, well,
30:52
let's say F is differentiable at C, because otherwise the conclusion is obvious. It would
30:56
not be differentiable at C. So you can say, okay, assume f is differentiable at
31:02
c and prove f prime of c to 0. So you can say, okay, assume
31:00
F is differentiable at C and prove that. Essentially, it's just thinking about this
31:06
difference quotient as x approaches c from the right or the left and thinking about
31:12
the fact that f has a local max remittance c, say a local max, just
31:16
for sake of completeness here.
31:26
that x minus c is positive,
31:32
greater than zero, and f of x minus f of c, if f of
31:38
c is a local max, that can't be positive. It could be zero.
31:44
That's less than or equal to zero when x is bigger than c. Actually, also
31:50
when x is less than c, but when x is bigger than c, the bottom
31:52
is positive. Making this fraction
31:57
as a whole less than or equal to zero, not only for fixed x
32:03
greater than c, but also as x approaches c in the limit. So the
32:09
right -hand limit as x approaches c from right here
32:13
must be less than or equal to zero
32:16
because when x is bigger than c, the
32:22
bottom is positive. and because you've got a local max at C, the top is
32:27
less than or equal to zero, can't be positive. Makes the whole fraction non
32:32
-positive, and the limit then must be non -positive. On the other hand, if you
32:38
let an X approach C from the left,
32:45
the top is still less than or equal to zero, but the bottom
32:51
is now negative, making a fraction as a whole
32:57
greater than or equal to zero for any fixed x less than c. Less than
33:02
c now. The slope of the secant is positive, or at least not negative,
33:08
and therefore the limit has got to be greater than or equal to zero.
33:16
Non -strict inequalities are preserved by the limits. Just a second.
33:22
If I'm assuming F is differentiable of C, that means these two limits must be
33:26
the same. The one -sided limit must equal the two -sided limit. Since this one's
33:32
less than or equal to zero, this one's greater than or equal to zero, they
33:34
must both be zero. And the two -sided limit must be zero. F times C
33:40
must be zero. Travis? So how would you do this with like a x squared
33:44
function? You mean with a local min instead of a local max?
33:50
Yeah, like where it would be? Instead of this x squared, you know, this
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