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

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

## Discussion

## Download Lecture Slides

## Table of Contents

## Transcription

### Continuity

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
- Definition of Continuity
- Definition of Continuity
- Example: Not Continuous
- Example: Continuous
- Example: Not Continuous
- Procedure for Finding Continuity
- Law of Continuity
- Example I: Determining Continuity on a Graph
- Example II: Show Continuity & Determine the Interval Over Which the Function is Continuous
- Example III: Is the Following Function Continuous at the Given Point?
- Theorem for Composite Functions
- Example IV: Is cos(x³ + ln x) Continuous at x=π/2?
- Example V: What Value of A Will make the Following Function Continuous at Every Point of Its Domain?
- Types of Discontinuity
- Intermediate Value Theorem
- Example VI: Prove That the Following Function Has at Least One Real Root in the Interval [4,6]

- Intro 0:00
- Definition of Continuity 0:08
- Definition of Continuity
- Example: Not Continuous
- Example: Continuous
- Example: Not Continuous
- Procedure for Finding Continuity
- Law of Continuity 13:44
- Law of Continuity
- Example I: Determining Continuity on a Graph 15:55
- Example II: Show Continuity & Determine the Interval Over Which the Function is Continuous 17:57
- Example III: Is the Following Function Continuous at the Given Point? 22:42
- Theorem for Composite Functions 25:28
- Theorem for Composite Functions
- Example IV: Is cos(x³ + ln x) Continuous at x=π/2? 27:00
- Example V: What Value of A Will make the Following Function Continuous at Every Point of Its Domain? 34:04
- Types of Discontinuity 39:18
- Removable Discontinuity
- Jump Discontinuity
- Infinite Discontinuity
- Intermediate Value Theorem 40:58
- Intermediate Value Theorem: Hypothesis & Conclusion
- Intermediate Value Theorem: Graphically
- Example VI: Prove That the Following Function Has at Least One Real Root in the Interval [4,6] 47:46

### AP Calculus AB Online Prep Course

### Transcription: Continuity

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

*Today, we are going to talk about continuity.*0004

*Let us jump right on in.*0006

*Let me work in blue today.*0014

*We mentioned several times already that the limit as x approaches a of f(x) and the value of the function at a,*0016

*f(a), are independent of each other.*0040

*We saw already that you can have a limit exist, as you approach a certain point.*0044

*But that limit is not necessarily the same as the value of the function, at that point.*0049

*It could be, it could not be.*0055

*They are independent of each other.*0057

*The limit as x approaches a of f(x) might or might not equal f(a).*0075

*When it does, that is very special.*0087

*For that, we say that the function is actually continuous there.*0092

*Continuous means there is no gap.*0094

*It is just if you take a pencil, drop it on a piece of paper.*0097

*With one swing of your pencil, without lifting it up, you have the graph that means every single little point is accounted for.*0099

*When it does, we say the function is continuous at that point.*0108

*The function is continuous at a.*0121

*Our definition, just to be reasonably formal about it.*0129

*I just repeated what I just wrote.*0138

*If the limit as x approaches a of f(x) = f(a), then f(x) is continuous at a.*0141

*If we are talking about a particular interval, say from 0 to 5,*0171

*and it is continuous at every point in that interval, we said it is continuous over that interval.*0175

*That is it, pretty straightforward.*0181

*Let us put some graphs to get a feel.*0187

*Let us look at some graphs to get a feel for this notion of continuity.*0190

*Graphs to get a feel what continuity is.*0201

*Again, it is a very intuitive notion.*0208

*Something that you know implicitly, you know intuitively.*0211

*Let us look at some graphs to get a feel for what continuity is.*0215

*Our first graph, let us go ahead and just do something like that.*0231

*Let us say we have this and it goes like that.*0237

*Let us call this 5 and let us say that this is 2.*0241

*The limit of this function, this is our f(x), the graph.*0246

*The limit as x approaches 2, which means from below and from above of f(x), it = 5.*0253

*As we approach 2, the graph, the y value approaches 5.*0264

*As we approach it from above, the y value approaches 5.*0269

*The limit is actually equal to 5.*0274

*Now f(2) is not defined.*0276

*There is not even a value for f(2), that is why there is a hole there.*0282

*This is not continuous.*0286

*That is all continuity means.*0294

*It is not connected.*0296

*Let us do another graph here.*0303

*Now we will just notice nice single sweep of the pencil.*0304

*Let us say we have a point here.*0310

*Let us say the y value of this point is 6.*0312

*Let us say the x value is 5.*0314

*Here the limit as x approaches 5, again when we do not specify, it is from both ends.*0317

*The limit of f(x) as x approaches 5 is equal to 6.*0324

*f(5), f at 5, is actually equal to 6.*0330

*This is continuous because the left hand limit, the right hand limit, and the value of the function at the point equal each other.*0336

*It is that simple.*0345

*Let us look at another graph.*0351

*Let us say we have got this function right here.*0356

*Let us say we have got something that looks like that.*0360

*Let us say this is the point 3, let us say this is 4.*0364

*Let us say the y value of this one is 1.*0367

*Now we have the limit as x approaches 3 from below, we are approaching it this way, of f(x), that is equal to 4.*0371

*The limit as x approaches 3 from above, the value of the function, the limit is 1.*0388

*f(3) = 4, that is the solid dot, right there.*0399

*In this case, even though the value of the function, f(3) happens to equal one of the limits,*0407

*in this case, the left hand limit, they are not all equal.*0413

*The left hand limit has to equal the right hand limit, has to equal the value of the function.*0416

*This is not continuous, we do say that it is left continuous.*0420

*If the function were defined down here at 1, we would say it is right continuous.*0424

*We do differentiate, just like we have left hand derivative, right hand derivative, left hand limit, right hand limit.*0431

*We have left continuity and right continuity, but this function is not continuous.*0437

*Here left hand limit does not equal the right hand limit.*0445

*We know that the limit does not exist.*0462

*I do not need to write all these, it is not a problem.*0470

*You know what, I do not even need to write any of this.*0483

*These do not agree.*0487

*It is that simple, we do not need to label the point with silly formalities.*0488

*Once again, the left hand limit does equal f,*0502

*Let me write this out.*0512

*The limit as x approaches 3 from below of f(x) does equal f(3).*0517

*That is it, we just call this right, it is left continuous.*0526

*f(x) is left continuous.*0532

*That is it, it is continuous from the left, that is all that means.*0537

*Over all, it is not continuous at 3.*0546

*You can see this is not continuous.*0547

*I have to do this then I have to stop.*0549

*I have to pick my pen up, start again over here.*0550

*That is the intuitive notion of continuity, one nice notion.*0555

*It does not need to be a smooth curve motion.*0561

*It can be something like this.*0565

*This is still continuous because I have not lifted my pencil off the paper but these are sharp points.*0572

*The graphs confirm your intuition or should I say your intuitive notion of continuity.*0587

*In other words, without breaks in the graph.*0612

*That is it, that is all it means, without breaks in the graph.*0615

*In other words, when you put your pencil to paper to draw the graph of a continuous function,*0623

*you can draw it without ever stopping and lifting your pencil.*0663

*I should not say stopping because we can stop, without ever lifting your pencil.*0677

*That is all continuity is.*0694

*Once again, the limit as x approaches a of f(x) is equal to f(a).*0706

*One, you find the limit if it exists.*0718

*Two, the second thing you do is you find f(a).*0732

*The third thing that you check is do they equal each other.*0740

*If they equal each other, the function is continuous at that point.*0752

*If they do not equal each other, the function is not continuous.*0754

*You are doing two things.*0757

*You are taking the limit, you are taking the two, and you are checking to see if they are equal.*0758

*That is the question mark.*0763

*This is the definition.*0765

*When they give you a definition, when you see a definition that involves inequality,*0766

*what that means is essentially what you have to do is you have to check the left side, check the right side.*0772

*If the equality is satisfied, then the definition is satisfied.*0777

*Then it is that thing, whatever the definition says.*0781

*Those of you that go on in mathematics, when you take linear algebra,*0785

*you are going to be talking about something called a linear function, which is not exactly what you think it is.*0788

*There is a certain mathematical definition to it.*0794

*A linear function is defined by equality.*0797

*There is something equal to something else.*0800

*What you are going to have to do is you are going to check the left hand side of the equality,*0802

*check the right hand side of the equality, and then you have to confirm that they are equal.*0805

*If they are equal, the function is linear.*0809

*If they are not equal, the function is not linear.*0811

*That is what equations in definitions mean.*0814

*The laws of continuity correspond to the laws of limit.*0820

*We will write them down.*0826

*The laws of continuity correspond to the limit laws.*0828

*They are entirely obvious.*0845

*We will say let f(x) and g(x) both be continuous.*0854

*They can be continuous over an interval or continuous at a given point.*0868

*Then, the sum of the two functions is continuous.*0872

*The product of two continuous functions is continuous.*0881

*The quotient of functions is continuous, provided, of course, g does not equal 0.*0887

*Any constant times a continuous function is continuous.*0894

*Most functions you deal with are going to be continuous.*0901

*If they have a discontinuity, it is going to be a very few places like for example,*0903

*a place where the function is not defined.*0909

*In other words, a vertical asymptote.*0910

*It is going to be at a place of discontinuity.*0912

*Those are really the only two cases where you are going to deal with something that is discontinuous.*0916

*Most functions you will deal with are continuous.*0920

*Functions are great, they behave very nicely.*0936

*Let us do a few examples.*0940

*We have an example, it says it is going to be example 1.*0955

*We are going to draw a graph and we are going to ask you to tell us where the function is discontinuous.*0964

*Example 1, give the points on the following graph. I decided to draw the graph by hand, where f fails to be continuous.*0969

*There is your graph.*1033

*Here is a vertical asymptote.*1036

*The function is going off to infinity there.*1040

*Where are the points of discontinuity, perfectly obvious.*1043

*It is discontinuous here.*1047

*At x = 1, it is discontinuous here.*1052

*Clearly, at x = 2, it is continuous here.*1053

*Even though it is a shell point, it is still continues.*1057

*I did not have to lift up my pencil to continue again.*1059

*It is discontinuous here.*1062

*This goes off to infinity, never touches this, and never touches this.*1065

*They are not connected.*1068

*Those are your three points of discontinuity, 1, 2, and 6.*1070

*Very intuitive, very obvious.*1074

*Example 2, show that the following function is continuous at a given point.*1081

*Also give the interval over which the function is continuous.*1085

*Now we do not have a graph to help us out.*1090

*If you have a graphing utility, if the particular test you are taking, you are allowed to use a calculator to,*1095

*or graphing calculator or utility, to help you out, by all means use it.*1101

*But again, the whole idea of the calculus is to learn to do things analytically so that we do not have to rely on the graph.*1106

*If we have a graph to help us out, that is fine.*1112

*But there are times you are not going to have a graph.*1114

*We want to develop these notions that are always true.*1117

*Not have to rely on geometric notion of a graph.*1121

*Now we have to do this analytically, show that it is continuous at x = 2.*1123

*We know what the definition of continuity is.*1128

*The definition says the limit as x approaches in this case 2, of this f(x), has to equal f(2).*1130

*That is what we are going to do.*1141

*First, we are going to find the limit as x approaches 2 of this function.*1142

*And then, we are going to evaluate the function.*1146

*We are going to see if they are equal to each other.*1147

*If they are, we are done, it is very simple.*1149

*First, let us evaluate the limit.*1153

*The limit as x approaches 2 of the function x³ - 14 - 4x.*1157

*Again, do not let the symbolism intimidate you.*1167

*It is just a basic limit.*1168

*What you do with a basic limit, what you do to all limit, you put the value in first to see what happens.*1172

*If you get an actual finite number, you can stop.*1177

*You are done, that is your limit.*1180

*If you get something that does not make sense, you try to manipulate the expression and take the limit again.*1181

*That is all we have been doing, that is all you have to do.*1189

*We put 2 in, this is going to be 2³ - √14 - 4 × 2.*1192

*2³ is equal to 8 - the square root of,*1203

*What is 14 – 8, it is 6.*1211

*8 – √6, is 8 - √6 a finite real number?*1213

*Of course it is, yes.*1216

*That is our limit, very simple.*1219

*Now we evaluate the function at 2.*1220

*f(2) = same thing, 2³ - 14 - 4 × 2 = 8 -√6.*1227

*They are equal to each other.*1242

*Yes, this function is continuous.*1243

*You might be saying to yourself, wait a minute, we just did the same thing twice.*1252

*Yes, we do the same thing twice but under two different auspices.*1256

*One, we are evaluating the function at.*1261

*Here we are evaluating the limit but we know from our previous work with limits,*1264

*that the way you evaluate a limit, in the case of a function like this, is to actually plug it in.*1268

*Even though you are doing the same thing, you are doing it for different reasons.*1275

*I know it seems a little weird but that is what is going on.*1281

*It is actually two different processes.*1284

*√14 - 4x, we see that this, our domain is going to be,*1291

*Now I give the interval over which it is continuous.*1298

*We know it is continuous at 2.*1301

*Over which numbers is it actually continuous, besides 2?*1304

*This implies that 14 - 4x has to be greater than or equal to 0, because you cannot take a square root of a negative number.*1308

*Our domain is restricted.*1318

*Therefore, 14 has to be greater than or equal to 4x.*1319

*Therefore, 14/4 has to be greater than or equal to x.*1324

*The domain of definition is, the domain of f is negative infinity to 14/4 exclusive.*1333

*It is continuous over this entire interval.*1346

*There is no point in this interval where it is not continuous.*1351

*The function is defined and it is perfectly valid there.*1355

*Is the following function continuous at a given point.*1364

*f(x) = e ⁺2x, for x less than or equal to 0.*1366

*It is equal to x², for x greater than 0.*1372

*Is it continuous at x = 0?*1375

*Here we have a piece wise continuous function.*1380

*It is one function to the left of, 0 it is another function to the right of 0.*1383

*We need the left hand limit to equal the right hand limit.*1388

*We need it, in order for them to be continuity, we need it.*1397

*First, the limit as x approaches 0 from below is this one.*1400

*We use that function of e ⁺2x.*1415

*As x goes to 0, 2 × 0 is going to be 0.*1421

*It is going to be e⁰ and this can equal 1.*1429

*Now the right hand limit.*1434

*Now it is going to be the limit as x approaches 0 from above of x².*1439

*When I put that in, that is just going to be 0.*1445

*In this particular case, we can if we want to.*1451

*Let us just do it for the heck of it.*1456

*Let us just do f(0).*1458

*f(0), here it is less than or equal to 0.*1462

*We use this function.*1465

*This is going to be e⁰ = 1.*1468

*We see that the left hand limit and the value of the function are equal.*1473

*The function is left continuous, but these three are not equal.*1477

*It is not continuous at 0, it is not continuous at x = 0.*1481

*In this particular case, you do not even have to do this.*1493

*Basically, the left hand limit and the right hand limit are not the same, the function is discontinuous at the point.*1497

*You did not really need to check the value of the function at the given point, at a.*1503

*But in this case, we did, it is not a problem.*1509

*It turns out to be left continuous.*1512

*But overall, not continuous.*1513

*Let us go ahead and do a theorem for composite functions.*1525

*Again, this is mostly just a formality.*1538

*Let us recall what we mean when we say composite function.*1545

*If we are given f(x) and if we are given g(x), two functions of x.*1551

*When we form f(g), that is equal to f of g(x).*1556

*When you form the composite, that is what you are doing.*1561

*If g(x) is continuous at point a and f(x) is continuous at g(a),*1566

*then f(g) is continuous at a.*1597

*Let us do an example.*1615

*Let me go to red.*1619

*I think this is example 4, if I’m not mistaken.*1623

*The question here is, the cos of x³ + natlog of x is continuous at,*1627

*I’m not going to write it all out, I’m just going to say cont.*1643

*Is it continuous at x = π/2, that is our question.*1647

*Using this theorem that we just did, here our f is equal to cos(x).*1658

*Our g is equal to x³ + ln x.*1666

*That is the composite function, it is going to be f(g).*1673

*Cos of x³ + ln x.*1676

*f(g) is equal to the cos of x³ + ln x.*1682

*Let us see what we can do here.*1692

*If g(x) is continuous at a and f is continuous at g(a), then f(g) is continuous at a.*1697

*We ask ourselves, is g(x) continuous at π/2.*1706

*The limit as x approaches π/2 of g(x) is equal to the limit as x approaches π/2 of x³ + ln of x = π/2³ + ln of π/2.*1719

*That is the limit.*1753

*Now let us check to see the actual value.*1755

*g(π/2) = π/ 2³ + ln of π/2.*1759

*This and this are equal.*1771

*It is continuous.*1772

*Yes, g(x) is continuous at our point of interest which is π/2.*1775

*That takes care of the first one.*1783

*The second part is, is f(x) continuous at g(a).*1785

*Here is f(x) continuous at g(π/2).*1791

*We said that g(π/2) is equal to this thing.*1811

*Is f(x) continuous at g(π/2)?*1824

*g(π/2) is equal to π/2³ + natlog of π/2.*1828

*Therefore, what we want to check, for continuity we are checking to see that the limit = the value.*1841

*We are going to calculate the limit as x approaches g.*1849

*What we are saying is f(x) continuous at g(π/2).*1855

*It is going to be x approaching g(π/2) of f(x) which = the limit as x approaches the value π/2³ + ln of π/2 of the cos(x).*1860

*That is f, what we separated, that is why we wrote cos(x), instead of cos(x³) + ln x.*1893

*We ask ourselves, is g(x) continuous? Yes.*1902

*Now, is f(x) continuous at g(π/2)?*1906

*We are going to do this.*1909

*This limit =, we just put this here.*1911

*It equals the cos of π/2³ + ln of π/2.*1916

*This is a perfectly valid number.*1925

*We also evaluate f(x) which is f of π/2³ + natlog of π/2*1927

*= the cos of π/2³ + natlog of π/2.*1951

*This and this are equal.*1962

*Yes, that is true.*1964

*Therefore, now we can conclude that fg which is equal to cos(x)³ + ln x is continuous at π/2.*1969

*Those are long process.*1988

*The basic idea is that a composite function is generally going to be continuous.*1990

*If you have f of g(x), if g(x) is continuous and f is continuous at g(x), then your f(g) is going to be continuous.*1998

*You can treat it individually like that or you can basically take a look at the whole function, f(g).*2011

*You can form the function f(g) as a single function of x and check the continuity of that function.*2016

*You can do it either way, it is not a problem.*2022

*A lot of these things pretty straightforward, they are pretty intuitive.*2026

*Sometimes you just break them down so much, that it tends to complicate them more than necessary.*2029

*That is exactly what this problem was.*2036

*It was just an excessive complication of something that is reasonably straightforward.*2037

*Let us see, what value of a will make the following function continuous at every point in its domain?*2049

*Here we have ax² + 7x, 4x less than 4, and x³ - ax for x greater than or equal to 4.*2064

*4 is the dividing point.*2075

*4 is our dividing point.*2079

*We are basically going to be taking limits of the function as x approaches 4, from below and from above.*2085

*In order for this to be continuous, for continuity,*2094

*we need the limit as x approaches 4 from below of f(x) to equal the limit as x approaches 4 from above f(x).*2105

*We need that equal to f(4).*2122

*We need all three things to be equal.*2127

*Let us calculate.*2130

*The limit as x approaches 4 from below, that is this one.*2134

*We are going to use this function of ax² + 7x =, we put x in for there.*2140

*We get a × 4² + 7 × 4.*2148

*You end up with 16a + 28.*2155

*That is our left limit.*2161

*Our right hand limit.*2164

*What is the limit as x approaches 4 from above?*2167

*From above, we are going to use that function.*2170

*It is going to be x³ - ax = 4³ - a × 4.*2174

*This one is going to give us 64 - 4a.*2185

*f(4) less than or equal to, we are going to use that right there.*2195

*That = x³ – ax, it is going to be 4³ – a × 4 = 64 – 4a.*2202

*We see that this, that this is the same as this.*2215

*It is right continuous.*2219

*We want, again, we said that we want the left hand limit to equal the right hand limit to equal f(4).*2222

*It already turns out that those two equal each other.*2231

*We want this equal to that.*2234

*We want this equal to these.*2237

*We are going to set 16a + 28 is equal to 64 - 4a.*2246

*Let us rewrite 16a + 28 = 64 - 4a.*2266

*You are going to get 20a = 36.*2275

*a is equal to 9/5.*2282

*That is it, it is that simple.*2288

*Whenever you need to set some piece wise continuous function,*2292

*you need there to be continuity at the particular point where they are piece wise separate.*2295

*You need to set the left hand limit and the right hand limit equal to each other,*2301

*and it has to equal the value of the function there.*2304

*Just use the definition of continuity which says that the limit as x approaches a of f(x) must equal f(a).*2308

*This left hand limit must equal the right hand limit.*2335

*A couple of nomenclature issues, just words that seem to come up.*2359

*I will just say, by the way.*2364

*Discontinuities are given different names.*2368

*I do not why, but okay.*2371

*That is just the way it is.*2376

*What if we have a situation like this.*2377

*We call this removable discontinuity.*2380

*The reason I call it removable discontinuity is because basically at a point where it is discontinuous,*2383

*whatever that is, let us say a, you are going to define it anyway you want.*2392

*You can just say f(a) = either that point or another point.*2396

*You can give any definition.*2400

*It is a discontinuity that is removable, we can remove it.*2402

*Here this type of discontinuity, we call this a jump discontinuity.*2407

*Again, these names are completely ridiculous.*2416

*They are completely meaningless.*2421

*People use them but for all practical purposes, it is continuity and discontinuity.*2423

*We do not need to know what type of continuity it is or discontinuity it is.*2428

*Of course, there is this one, you have some sort of an asymptote and you have a function going this way and a function going this way.*2432

*This is called an infinite discontinuity.*2440

*Just to let you know that.*2444

*You will see these terms, they do not mean anything.*2453

*The most important and practical theorem regarding continuity,*2459

*there are several of them but for our purposes,*2479

*is something called the intermediate value theorem.*2487

*The intermediate value theorem.*2492

*Here are the hypotheses, the if section.*2503

*If then, hypothesis, or hypotheses, conclusion.*2508

*The hypotheses, there are few of them.*2513

*Hypotheses are if a is less than b on the interval, if f(a) is different than f(b),*2517

*if they are not equal, if f(x) is continuous on the closed interval ab.*2530

*If all of these three are satisfied then the conclusion, the then statement, the conclusion is there exists,*2543

*I will write it out, there exists, I will put it in parentheses.*2556

*The symbol for there exists, the logical symbol is a reverse e.*2563

*There exists a number c somewhere between a and b, such that f(c) is equal to some number a,*2567

*or a is between f(a) and f(b).*2600

*This is a formal statement of the theorem.*2607

*We have the hypothesis which is your if clause and we have your conclusion which is your then clause.*2609

*Let us go ahead do what this actually means.*2616

*Pictorially, it looks like this.*2619

*I have got something that looks.*2628

*We have our axis.*2631

*We have a nice continuous function.*2636

*We have our point, let us say a is here that means f(a) is there.*2640

*Let us say that b is over here.*2649

*This is f(b) is over there.*2653

*It says that this is some number c and this is the number a.*2661

*Basically, what this is saying is that if a is less than b on the x axis, here we have a definitely less than b.*2674

*If f(a) and f(b) are different, here f(a) and f(b) are definitely different.*2682

*And if the function itself is continuous.*2689

*This function f(x), yes, we can see that it is nice and continuous.*2691

*If that is the case, then the conclusion I can draw is that there is some number, at least one, there may be more than one,*2696

*that some number c between a and b such that the f of that number is actually going to fall between f(a) and f(b).*2701

*That is what the intermediate value theorem says.*2714

*Is that, if the three hypotheses are satisfied then there is some number between the two,*2717

*such that when I take the f of that number, the y value is going to fall between f(a) and f(b).*2722

*All three hypotheses have to be satisfied.*2729

*a less than b, f(a) not equal to f(b), and it has to be a continuous function.*2731

*That is the real important one.*2737

*It has to be continuous.*2739

*If it is not continuous then there is no guarantee that a value of a actually even exist at all.*2740

*That is the whole idea because if we were to break the graph like that, if there is a discontinuity in the graph,*2744

*you might have a number but it is not defined there.*2753

*There is no guarantee that some number a in between f(a) and f(b) has some c value.*2758

*Continuity of the graph is profoundly important.*2768

*Let us write, since f is continuous as x goes from a to b, as x moves along from a towards b, f(x) hits every value.*2774

*In other words, as we move from here to here, f actually hits every single value between here and here.*2806

*It hits every single value.*2818

*It might hit it once, we do not know what the graph looks like.*2820

*But we know that it hit it at least once.*2825

*Because f is continuous, as x goes from a to b, f(x) hits every value between f(a) and f(b).*2827

*In another words, because the function itself is continuous, as we move from here to here,*2849

*we move continuously from f of f(a) to f(b).*2854

*There is always going to be some number between them.*2858

*I hope that make sense.*2861

*Again, I think intuitively, it absolutely does make sense.*2864

*Let us do a problem, prove that the following function has at least one real root in the interval for 6.*2867

*Here we have a function and they want us to prove it has at least one real root.*2875

*What that means is that, you do not know what the function looks like.*2881

*We can graph it, if we want to.*2887

*But again, we do not have a graph at our disposal.*2888

*We are going to use this theorem, analytical methods,*2890

*to show that this function actually crosses the x axis at least once, between 4 and 6.*2893

*It is either going to cross like this way or it is going to across this way, going from negative to positive or positive to negative.*2903

*We do not know which one first but we want to show that at least it hits the x axis, at least one time.*2911

*That is what this says, has at least one real root.*2917

*That means it hits the real axis at least one time.*2919

*The y value is going to be 0, that is what the root means, in the interval 4 to 6.*2923

*Let us check our hypotheses.*2928

*I think i will do this in blue actually.*2932

*Our hypotheses, the first one is a less than b?*2942

*Yes, 4 is definitely less than 6.*2950

*Our second hypotheses, our second hypothesis is, is f(4) less than f(6)?*2956

*We have to check that.*2979

*f(4) = 1.643, when I put 4 into there, I get 1.643.*2981

*Definitely, make sure that your calculators are in radian mode not degree mode.*2990

*When you put in 4 in here, all the functions in calculus,*2996

*when you use you calculator to solve trigonometric function, make sure you are in radian mode.*2999

*You will get a number if you are in degree mode, but the numbers are going to be wrong.*3005

*f(4) is that, f(6) = -1.497.*3008

*Yes, this is true.*3014

*f(4) is different than f(6).*3017

*Hypothesis three, is f(x) continuous over the closed interval of 4 to 6.*3021

*Yes, sin(2x) is a completely continuous function.*3034

*Sin and cos are continuous everywhere.*3037

*Cos is continuous and we know that the sum of the difference of a continuous function is continuous.*3039

*Yes, that is continuous, that hypotheses are satisfied.*3044

*Therefore, there exists at least one number between 4 and 6, such that f of this number, some number, let us call it c.*3048

*There exists at least one number c between 4 and 6.*3089

*There is some number here in that interval, such that f(c) is equal to 0, because 0 is between 1.643 and -1.497.*3092

*f(4) is 1.643, let us just put it right there.*3119

*f(6) is -0.497, let us just put it right over there.*3123

*The function is continuous.*3128

*Because it is continuous, when I'm going from here to here, I cannot lift my pencil.*3130

*Therefore, somewhere, it does not matter what trajectory it takes, some path,*3135

*whatever the function looks like, it has to pass through 0.*3145

*That is what we are saying.*3149

*That is what the intermediate value theorems says.*3150

*If a is less than b, if the f values are different from each other and if it is continuous,*3154

*that there is some number between the x values, such that the f of that number is between the f values of the two numbers.*3162

*That is it, thank you so much for joining us here at www.educator.com.*3174

*We will see you next time, bye.*3179

2 answers

Last reply by: Professor Hovasapian

Fri Mar 25, 2016 8:30 PM

Post by Acme Wang on March 12, 2016

Hello Professor Hovasapian,

In example II, I just confused why the function is continuous in the interval (-?, 14/4]. We just got the answer that the function is continuous at x=2 ONLY right?

Also, if f(x) and g(x) are both continuous, is f-g also continuous?

And for example VI, from my point of view, since the f(4) is positive and f(6) is negative, definitely f(x) would have only one real root (f(x) would pass through x-axis only ONCE :). I don't understand why there are more than one root between this interval?

Thank you for your time to answer my question.

Sincerely,

Acme

2 answers

Last reply by: Professor Hovasapian

Fri Mar 25, 2016 8:16 PM

Post by john lee on March 9, 2016

Professor Raffi Hovasapian,

Why f(a) cannot equal to f(b) in intermediate value theorem?

Thanks for the answer!