For more information, please see full course syllabus of AP Calculus AB

For more information, please see full course syllabus of AP Calculus AB

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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

### AP Calculus AB Online Prep Course

### 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 endpoint*0075

*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

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