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Lecture Comments (6)

2 answers

Last reply by: Professor Hovasapian
Fri Mar 25, 2016 8:30 PM

Post by Acme Wang on March 12 at 08:42:52 PM

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 at 07:56:48 AM

Professor Raffi Hovasapian,
Why f(a) cannot equal to f(b) in intermediate value theorem?

Thanks for the answer!

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 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

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