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INSTRUCTORS Raffi Hovasapian John Zhu
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The Mean Value Theorem

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Rolle's Theorem 0:08
    • Rolle's Theorem: If & Then
    • Rolle's Theorem: Geometrically
    • There May Be More than 1 c Such That f'( c ) = 0
  • Example I: Rolle's Theorem 4:58
  • The Mean Value Theorem 9:12
    • The Mean Value Theorem: If & Then
    • The Mean Value Theorem: Geometrically
  • Example II: Mean Value Theorem 13:43
  • Example III: Mean Value Theorem 21:19

Transcription: The Mean Value Theorem

Hello, welcome back to www.educator.com, welcome back to AP Calculus.0000

Today, we are going to be discussing the mean value theorem.0004

Let us just jump right on in.0007

Before we do the mean value, we are actually going to do a smaller version of the theorem called Rolle’s theorem.0010

Let us begin with that. 0016

We begin to with Rolle’s theorem.0021

Rolle’s theorem is the following.0031

If my three hypotheses, if this function satisfies these three hypotheses then the conclusion we can draw.0034

The first one is if y = f(x), if the function that we are dealing with is continuous on a closed interval ab.0043

B, if y = f(x) is differentiable, in other words, smooth, no sharp corners.0058

If it is differentiable all the open interval ab.0067

The endpoints do not necessarily need to apply.0071

Last, if f(a) is equal to f(b), in other words if the y value at the left endpoint0075

happens to be the same as the y value at the right endpoint.0082

If these three hypotheses are satisfied, then we can conclude the following.0086

Then, there exists at least one number, we will call it c, between a and b, such that f’ at c is actually equal to 0.0093

Geometrically, this means the following.0128

Geometrically, we have the following situation.0135

We got a little function like that.0142

Let us go ahead and call this a, let us go ahead and call this b.0145

If we have the function that satisfies these three hypotheses is continuous, is differential, 0152

and f(a) happens to equal f(b), it does.0158

We can conclude that at least one number c, such that the derivative of that function at c is equal to 0.0161

In other words, the slope is equal to 0 at the y value of that point.0168

That number c happens to be right there.0175

In this particular case, it is only one value of c.0180

You already know this is true intuitively, but again, with mathematics, we need to be very precise.0186

We need to state the hypotheses.0192

If does hypotheses are satisfied, then we can draw the particular conclusion.0194

If this than this.0198

Let me see here, let me go ahead and move on to the next page.0205

As we said, there maybe more than one c such that f’(c) is equal to 0.0216

Geometrically, that just means the following.0232

Something like that.0240

Let us go ahead and call this a, let us call this b.0241

In this particular case, the function is continuous between a and b.0247

The function is differentiable, the open interval a and b.0251

F(a) is equal to f(b).0256

The number c such that f’ at c is equal to 0, the places where the slope is 0 is 1, 2, 3, 4.0259

We have 5 values of c.0268

This is c1, this is c2, this is c3, this is c4, and this is c5.0272

The number of c actually have to do with the function itself.0280

Rolle’s theorem guarantees the existence of at least one.0285

That is it, that is all that is going on here.0288

Let us go ahead and do an example.0290

For the function f(x) = 1/3 x + 3 sin x on the closed interval -3.55 to 3.55.0295

Decide if Rolle’s interval applies, if so, apply it to find all values of c, such that f’(c) is equal to 0.0302

Let us go ahead and check the hypotheses to see if Rolle’s theorem actually applies.0313

A, the question is, is f continuous on the closed interval -3.55 to 3.55, the answer is yes.0318

Part b, is f differentiable on the open interval -3.55 to 3.55.0336

Once again, the answer is yes.0353

Part c, is f(- 3.55) equal to f(3.55).0356

F(-3.55) and f(3.55) both = 0, yes.0367

The three hypotheses are satisfied.0375

Therefore, Rolle’s theorem does apply.0377

There does exist at least one c, such that f’(c) is equal to 0.0389

Let us find what these values of c are.0402

F’(x) is equal to, I have 1/3 + the derivative of sin is cos, 3 cos x.0407

I’m going to go ahead and set that equal to 0.0420

I have 3 cos x is equal to -1/3.0424

I have cos x is equal to -1/9.0429

Therefore, x is the inv cos(-1/9).0434

The values that I get are x = -1.682 and x = +1.682.0440

These are my values of c.0450

C is equal to -1.682, c is equal to 1.682.0452

Here I have two values of c that satisfy this conclusion of Rolle’s theorem.0457

Let us take a look at what this looks like graphically, geometrically, all that wonderful stuff.0466

Here is our function y = 1/3 x +, let me go ahead and write that down.0474

Y is equal to 1/3 x + 3 sin x, this is our original function f.0481

From -3.55 to +3.55, that is here and here.0493

We notice that f(a) = f(b), in this particular case, it happens to equal 0.0499

We know that it is continuous, it is differentiable.0506

There are two places where the slope is 0 here and here.0508

C value that is one of them and that is two of them.0515

This was our -1.682, that is our c1, and this is our +1.682.0521

These was our others c value.0538

That is it, that is all that is going on here.0540

You are finding the values of c, inside the domain that satisfy the conclusions of Rolle’s theorem which is of f’(c) is equal to 0.0542

Let us go ahead and state the mean value theorem.0555

The mean value theorem, the theorem itself geometrically, it makes total sense.0559

It is completely intuitive but we need to state it explicitly and precisely.0568

The mean value theorem is very important in calculus for theoretical reasons more than anything else.0575

The mean value theorem says the following.0582

My hypotheses are as follows, if a, my hypotheses are the same as Rolle’s theorem except for the third.0584

It does away with that particular hypothesis.0592

The first one is y = f(x) is continuous on the closed interval ab.0596

If it is continuous and if y = f(x) is differentiable on the open interval ab.0609

Then, I can conclude the following.0625

Now, I'm no longer restricted to f(a) equaling f(b).0628

Then, there is at least one number c between a and b, such that, the following is true.0634

F’ at c is equal to f(b) - f(a)/ b – a.0658

Geometrically, it means the following.0670

I have got some function here.0677

Let us go ahead and take a here, let us take b here, randomly.0682

This is f(a) and this is f(b).0687

I’m going to draw the cord, the line that connects them.0692

If this continuous on the closed interval, if it is differentiable on the open interval, there is some number c between them.0701

Such that f’ at that c, in other words, the slope where it actually touches the graph is equal to f(b) - f(a)/ b – a.0710

F(b) – f(a)/ b – a is the slope of this cord.0723

Because this point is a, f(a).0728

This point is b, f(b).0734

F(b) – f(a)/ b - a is just the change in y/ the change in x, that is the slope here.0738

They are telling me that there is some number c such that it actually has, the f' at that c has the same slope as this.0743

They end up being parallel, makes total sense.0757

The slope has to pass through, there is going to be a point where the slope along the curve is going to equal the slope between the 2 points, at the endpoints.0760

That is all this is saying.0774

This slope, this slope, it has to pass through this, in order for it to go that down.0775

This is the mean value theorem.0782

Very important theorem.0785

Sometimes, you will see it stated this way.0789

F’(c) × b - a is equal to f(b) – f(a).0793

Sometimes, they write it this way.0804

I actually prefer this way.0806

The expression for the slope is equal to the expression of a slope, the derivative at some point.0808

That is the mean value theorem, nothing particularly strange.0814

It is nice that we do not have to worry about that third hypothesis.0818

Let us go ahead and do an example.0820

For the function f(x) = 4 - x²/ the interval – 2 to 0, decide if the mean value theorem applies.0826

If so, apply it to find all values of c such that f’(c) = f(b) – f(a)/ b – a, which is the conclusion of the mean value theorem.0834

Let us go ahead and do this.0844

Part a, we need to check, is f continuous on the closed interval from -2 to 0?0846

The answer is yes, it is continuous.0865

In case you are wondering, you are more that welcome to use a graph to decide whether it is continuous, differentiable.0867

You want to use all the resources at your disposal.0875

This is just the equation y = √4 - x², which is nothing more than x² + y² = 4, except for -2 to 0.0879

This is nothing more than, in the equation of the circle of radius 2 going from here to here, including the endpoints.0892

Here is -2 and here is 0, that is all.0901

This graph, that is all this is.0905

Certainly, it is continuous and it is also differentiable.0909

Is f differentiable, it is differentiable on the open interval from -2 to 0.0914

Yes, it is perfectly smooth.0923

Was it differentiable, yes.0924

Since that is the case, therefore, there exists some c such that f’(c) is actually equal to f(b) – f(a)/b – a.0930

Let us see what is that we have got.0953

Let us go ahead and find f’ first.0956

F(x) = this.0962

I will go ahead and draw a little line here.0966

Let me rewrite f(x), let me actually do this in red.0971

I have f(x), I’m going to write it as 4 - x² ^½.0979

It makes differentiation a little bit easier.0987

F'(x) is going to equal ½ × 4 - x²⁻¹/2 × -2x.0989

That is going to equal equaling, this 2 and this 2 cancel.1003

We are going to end up with –x/ √4 – x².1008

This is our f'.1017

We want to set this equal to 0.1019

Sorry, we are not setting it equal to 0.1025

We are setting it equal to f(b) – f(a)/ b – a.1027

Let us go ahead and find f(b).1037

F(b) = the f(0) which = 4 - 0², all under the radical, that is equal to 2.1042

F(a) is equal to f(-2) which is equal to 4, - and -2², under the radical, that = 0.1055

We have f(b) - f(a) is 2 – 0/ b - a which is 0 -2.1067

Remember, this is a and this is b.1082

F(b) is 2, f(a) is 0.1086

B is 0, a is -2.1092

We end up with 2/2 which is equal to 1.1098

We need to solve –x/4 – x² is equal to 1.1104

Is that correct? Yes, that was correct.1119

Let us go ahead and do that.1122

We get - x, I will just go ahead and call it c.1124

Let us go ahead and go √–c/4 – c² = 1.1131

That becomes -c = 4 - c².1141

Square both sides, I end up with c² = 4 - c².1148

Let me go to the next page here.1158

I have got c² = 4 - c².1164

I have got 2 c² = 4, c² = 2, c = + or –√2.1169

I'm going to go ahead and take the -√2 because that is the one that is actually in the domain between -2 and 0, which is equal to -1.414.1177

That is it, we just found it, c is equal to –√2.1189

Let us go ahead and see what that is geometrically.1197

Here we have our function y is equal to 4 - x², all under the radical.1203

Our c was –√2, it is -1.414.1214

We come up there, the slope of this line which is f' at c, f’(-√2) that happens to equal the slope of this line.1220

We found the value of c that satisfies this relationship.1242

The value of c such that when I go up there, the slope of the graph at that c 1246

happens to be the same as the slope between the two endpoints.1255

Here and here.1261

That is it, nice and simple, nothing very strange about this at all.1263

Let us go ahead.1271

Let us try another example here, see how that one goes.1277

For the function f(x) = x²/5 + cos(x) on the interval from -1 to 1, whose graph is shown below.1280

This graph right here.1289

In case it is not too clear, here is the -1, here is the 1.1293

Decide if the mean value theorem applies.1298

If so, apply it to find all the values of c such that f’(c) = that right there.1300

Part a, is f continuous between -1 and 1?1310

Yes, it is continuous.1318

B, is it differentiable on the open interval between -1 and 1?1323

It is not, there is a sharp point there.1331

It is not smooth there, it is not differentiable.1334

No, it is not differentiable.1337

Therefore, the mean value theorem does not apply.1344

I cannot do this.1349

In this particular case, we have the graph to let us know that something is differentiable or not differentiable.1355

We just look and we found that this low cusp.1361

It is not differentiable.1363

What if you did not have the graph?1364

What if you needed to actually decide whether this thing was going to be differentiable or not?1366

Something not being differentiable means the derivative does not exist.1369

Let us go ahead and do this analytically.1373

Let me actually work in blue.1378

If we did not have a graph to decide differentiability, then, we would just do it analytically.1382

Let us go ahead and find f', f’(x).1408

When I take the derivative of this, it is going to be 2/5 × x⁻³/5 – sin(x).1412

I'm going to rewrite this as 2/5, I’m just going to bring this down.1429

X³/5 - sin(x), watch what happens as x gets close to 0.1441

It is not a problem, if x is any number, and this is certainly valid.1454

It is certainly differentiable everywhere.1458

But what happens when x gets close to 0?1461

As x gets close to 0, if this goes to 0, 0.5, 0.4, 0.3, 0.2, 0.1, the denominator is going to go to 0 also.1467

As this goes to 0, the denominator goes to 0.1483

As the denominator goes to 0, the fraction, the first term, 2/0 is going to go to infinity.1487

The first term, this whole thing is going to end up going to positive or negative infinity.1501

Depending on if we approach 0 from the right or from the left.1508

Therefore analytically, f’ at 0 does not exist.1512

We can conclude it is not differentiable.1523

We do not need a graph where we see a cusp, clearly it is not differentiable.1525

Clearly, the mean value theorem does not apply.1530

If we did have a graph, we will just go ahead and take the derivative 1533

and see if there is any point where the derivative does not exist.1536

Sure enough the derivative does not exist at 0.1539

0 happens to be right in the middle of this domain.1541

Therefore, the mean value theorem does not apply.1545

That is it, nice and straightforward.1548

Thank you so much for joining us here at www.educator.com.1551

We will see you next time, bye. 1553