AP Calculus AB Example Problems III

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

Today, we are going to do some more example problems using derivatives to graph functions.0004

Let us jump right on in.0010

Our first examples says, for the function f(x) = x²/ x - 4²,0013

find the intervals of increase and decrease, local maxes and mins, intervals of concavity and inflection points, and asymptotes.0018

Use this information to graph the function.0026

In general problems, we will just say graph the function.0028

In these set of problems, we are laying out specifically what we want.0031

In general, all of these things are things that are going to do anyway.0037

You may do some of them, you may do most of them,0041

but we want to get you into the habit of actually doing a nice systematic procedure.0044

Let us see what we can find.0049

This is a rational function, notice this mentioned asymptote.0051

We may or may not be dealing with asymptotes, whenever we deal with rational functions.0055

Let us start off by doing the first derivative, it is always a great place to start.0059

The first derivative gives us a lot of information.0063

I think I’m going to work in blue today.0065

F’(x), we know this is going to be quotient rule.0068

We are going to have x - 4² × the derivative of the top which is going to be 2x - x² × the derivative of the bottom0072

which is going to be 2 × x – 4¹ × the derivative of what is inside, 1.0085

All of that is going to be / x – 4⁴.0093

I’m going to go ahead and pull out the x – 4.0102

The numerator, I’m going to write it as 2 x × x - 4 - 2x²/ x – 4⁴.0106

I end up getting, this cancels this, turning it into a 3.0122

When I multiply that, I get 2x² - 8x.0128

I get 2x² - 8x - 2x²/ x - 4³.0132

The 2x² cancel leaving me -8x/ x – 4³.0144

This is our first derivative, this is the one we set equal to 0, to find our critical points.0152

We set that equal to 0 and what we end up getting is -8x = 0.0157

Therefore, x is equal to 0.0164

0 is our critical point.0167

Recall that a critical point is not only where the first derivative equal 0,0185

it is also where the function might not be differentiable.0197

We have to watch out for those.0203

This procedure, the first derivative setting equal to 0, it is only going to give us those that are.0205

We still have to look at the function and see if there is some place where the graph is not going to be differentiable.0210

In this particular case, we look at the denominator.0215

The denominator was x – 4.0218

Clearly, we cannot have a denominator equal to 0.0220

Therefore, at the point 4, we are going to see some sort of asymptote.0223

The asymptote, it is not going to be differentiable there.0227

4 is also going to be a critical point.0230

A critical point that this first derivative procedure will not give us.0234

Something we have to elucidate by other means.0238

Let us write that out, recall that the critical point, that a critical point, is also where f’ fails to exist.0241

In other words, does not exist, fails to exist.0254

Here we have a vertical asymptote at x = 4.0261

x = 4 is also a critical point.0278

In other words, it is one of the points to the left of which and to the right of which,0285

we are going to have to check to see what the derivative is equal to, positive or negative.0289

In order to decide whether it is increasing or decreasing to the left or to the right of that.0292

We are dealing with f’ and we have our two critical points, we have 0 and we have 4.0308

We are going to be checking points in this interval, in this interval, in this interval, to check increasing and decreasing behavior of the function.0316

F’(x) we said is equal to -8x/ x – 4³.0327

We are going to take points in these intervals, put them in here to see whether we get something that is positive or negative.0338

Let us try -1, when I put -1 into this, I’m going to get something which is a positive number/ a negative number.0343

This is a negative number which means decreasing.0352

I know that to the left of 0 is actually decreasing.0358

Let us try 1, when I put 1, I’m going to get a negative number/ a negative number.0363

This is also a negative number.0370

Definitely, negative divided by a negative is positive, sorry about that.0375

Positive, we have an increasing.0379

It is going to be increasing between 0 and 4.0382

Let us go ahead and check a point to the right of 4.0387

I’m going to go ahead and try 5.0390

When I do 5, I get a negative number on top and I’m going to get a positive number which is negative.0391

It is going to be decreasing on that interval.0398

We said that 0 was a critical point that we got from the first derivative test.0403

We also have a decreasing to the left and an increasing to right of it.0408

That means that 0 is a local min.0412

All of this gives us.0416

There is a local min at x = 0, we have that information.0419

There is an asymptote, a vertical asymptote at x = 4.0430

Even though we have an increasing and a decreasing, this is not a local max.0443

At 4, the function is not differentiable and we see that we have a rational function.0448

This is clearly a vertical asymptote.0452

Basically what is happening is it is going that way and it is coming back down that way.0454

It is going to the same place and we knew that from pre-calculus because it was the original function had a squared term.0460

Whenever it is an even multiplicity on that, it is going to go to the same place.0468

When it is odd multiplicity, it is going to go to different places, the opposite ends.0474

Let us go ahead and see what f(0) is.0481

F(0) is equal to 0, the local min, the actual point is, the local min occurs at 0,0.0485

We have partial graph.0501

You are welcome to graph it as you go along, even if you do not have all the information.0503

We know that we have a local min at 0,0.0508

We know the graph looks something like this.0510

We also know that the value 4, you know we have a vertical asymptote.0514

We know we are looking at something like this and we know we are working at something like that.0520

That is what this first derivative tells me, really nice.0524

It gives me a lot of information.0528

Not done yet but it is a good place to start.0530

Before we take the second derivative to check points of concavity and inflection points,0536

let us see if there are any horizontal asymptotes here.0542

Before we take f”(x), let us see at that horizontal asymptotes.0545

F(x) that is equal to x²/ x - 4².0573

If I expand the bottom, I get x²/ x² - 8x + 16.0583

We caught an asymptote behavior means what happens when x really big, positive infinity.0593

x gets really small, negative infinity.0599

That is it, that is all that it is saying.0602

In this particular case, we notice from pre-calculus, if we remember, the degrees are the same.0604

When x gets really big, these other terms do not even matter.0610

What you end up with is x²/ x² which is just 1.0615

Let me write this down because I have got a polynomial written down here, we might as well be complete.0624

Recall that asymptotic behavior means what happens to f(x), when x goes to + or – infinity, it is really huge.0630

In this particular case, I’m going to run through the process, that is fine.0662

F(x), we said is equal to x²/ x² - 8x + 16.0667

I’m going to multiply the top and bottom by 1/ x².0676

In other words, I’m going to divide the bottom and the top by the highest degree that occurs.0680

I’m going to end up with 1/1 - 8/x + 16/ x².0685

As I take x to + or –infinity, this goes to 0, this term goes to 0, and what I’m left with is 1.0698

This is my horizontal asymptote.0707

You remember this from previous lessons, when we were talking about limits earlier in the course.0711

Horizontal asymptote at 1, at y = 1, horizontal asymptote.0716

Let us go ahead and draw out real quickly what is it we are looking at here.0728

We know that we have a 1, 2, 3, 4.0733

We have a vertical asymptote at 4.0735

We know at 1 we have a horizontal asymptote.0739

What is probably happening is this, we know that we have that.0745

We know that it comes probably like this.0752

And this one probably goes like that.0755

More than likely that is what is happening.0759

Again, the function is getting close to 1, when x gets big or when x gets big in a negative direction.0760

That is what is happening.0772

Let us go ahead and find f” and see what sort of information that gives us.0777

Let us find f”(x) and determine the points of inflection and the intervals of concavity.0789

We said that f’(x) was equal to -8x/ x – 4³.0812

Therefore, f”(x) is equal to x – 4³ × -8 - 8x × 3 × x - 4² × 1/ x – 4⁶.0821

I’m going to go ahead and factor out an x - 4².0853

I’m going to get -8 × x – 4.0858

When I take all of these out, I’m going to get +24x.0867

All of that is going to be over x – 4⁶.0871

I’m sorry, I know that this is a lengthy procedure.0875

I apologize for actually going through this process.0881

I just figured it might as well be as complete as possible.0884

This knocks out that, turning that into 4.0889

It leaves me with -8x + 32 + 24x/ x – 4⁴,0893

which gives me -8x + 24 is going to be 16x.0908

If my arithmetic is correct, + 32/ x – 4⁴.0914

This is what I’m going to set equal to 0 and this gives me 16x + 32 = 0, 16x = -32.0921

I get x = -2.0934

This is my inflection point.0937

We have already seen the graph of this thing.0955

We have already seen that it looks, this is our horizontal asymptote.0956

This is our vertical asymptote.0963

We know we are dealing with the function that looks like this.0966

Clearly, this point right here at -2 is the inflection point.0973

From the graph, we already know the intervals of concavity.0977

-infinity to -2, it is going to be concave down.0980

From -2 all the way to 4, it is going to be concave up.0984

From 4 onward, it is also going to be concave up.0988

We know these already, but let us go through the process analytically for the second derivative.0992

From the graph, we already know the intervals of concavity.1002

But, let us run through analytically.1024

We went ahead and we found that this point of inflection, one of the points of inflection is -2.1044

We also have to include the 4, it is an asymptote.1050

There is something going on to the left and right of that.1053

Even though, again, it does not show up, we have to include it.1056

It is a point that is not differentiable.1059

Something happens to the left of that point.1061

Something happens to the right of that point.1063

I have to check 3 regions.1065

I’m checking the second derivative.1068

The second derivative is equal to 16x + 32/ x – 4⁴.1072

Let us go ahead and find f”(-3).1084

When I put that in, I’m going to get a negative number/ a positive number.1090

This is negative which means this is concave down, which I already knew.1095

When I do f” of something between -2 and 4, let us say -1, I end up with a positive/ a positive number.1104

This means it is concave up, second derivative.1117

And then, when I checked f” of let us say 5, something over here to the right of 4,1122

I’m going to end up with a positive number/ a positive number.1129

This means concave up.1133

My intervals of concavity, concave down, like I said –infinity to -2.1137

Concave up from -2 to 4 union.1144

I do not just say -2 to infinity because the 4 separates those two.1152

I do have to break it up into regions, -4 to +infinity, that take care of that.1155

Let us go ahead and find, now I have got my graph.1169

I have got 1, 2, 3, 4.1178

I have my vertical asymptote, I have my horizontal asymptote.1182

I know my graph goes something like this.1190

I know it goes like this.1196

Let us find what that point is.1199

We said that the point of inflection, we said that x = -2.1201

When I take f(-2), I end up with 1/9.1207

This point right here is the point -2, 1/9.1214

The actual coordinates of that point.1220

When I put it all together, let graphing function actually take care of this.1223

This is what the function actually looks like.1230

You can see we have a vertical asymptote at 4.1232

We have a horizontal asymptote at 1.1235

We have our point of inflection from concave down to concave up, still concave up.1239

There you go, all of the information.1244

You are using what you already know, in addition to the tools that you are learning from calculus.1252

Notice, the first derivative did not give me a critical value of 4.1258

But I included it because there is a vertical asymptote there.1262

All of the bits of information that I have at my disposal, I want to be able to bring to there to be able to graph the function.1267

That is my goal to be able to graph the function and to find important points of reference for the graph.1274

Let us go ahead and try another one here.1284

For the function f(x) = ln, 2 - ln x, find the intervals of the normal stuff,1286

increase/decrease, maxes and mins, concavity, inflection points, asymptotes, graph the function.1292

Let us do the asymptotes first.1298

In this case, the first thing we are going to look at, first let us look at the domain.1315

We are looking at a logarithm function.1321

A logarithm function is going to have a restricted domain.1323

We want to see what that domain is first, and more than likely some thing is going to be happening at the endpoints of that particular interval.1327

Asymptotic behavior, domain, they are going to be mixed up here.1335

Let us take a look at the domain.1342

Let us look at the domain.1353

We have the function ln of 2 – ln x.1358

This part, the argument of the logarithm function, this has to be greater than 0.1364

Let us set it greater than 0.1377

We have 2 – ln x greater than 0, let us solve for x.1379

I have 2 greater than ln x.1385

I'm going to exponentiate both sides.1389

I get x is less than e².1391

Or if I like to write it this way, x is less than e².1400

The domain is from 0 to e² or 0 to about 7.4.1408

This is not 1, this is a comma, 0 to about 7.4.1418

Now we have our domain.1423

Now that we have the domain, let us ask ourselves what happens as x goes to 0 and as x gets close to 7.4.1425

We ask ourselves, now we work out the asymptotic behavior.1437

We ask ourselves, one, what happens to f(x) as x approaches 0 from the left.1445

Because there is nothing less than 0, we are approaching 0 from the right, from the positive side.1462

Two, what happens to f(x), as x approaches e² from the left.1470

Number 1, as x gets close to 0 from the right, the ln(x), it is going to go to –infinity.1492

Negative of the ln(x) is going to go to +infinity.1509

2 – ln(x) is going to go to +infinity so that makes 0 a vertical asymptote.1516

Remember, my function here, my f(x) is equal to, 2 – ln(x).1546

Ln(2) – ln(x) goes to +infinity.1564

This is ln(2) – ln(x).1571

I'm asking myself, what happens to the function as x gets close to 0.1574

As x gets close to 0, the ln(x) part, that goes to –infinity.1577

The –ln(x) is +infinity.1583

2 – ln x goes to +infinity.1585

As this goes to positive infinity, the ln goes to infinity.1590

The function actually goes towards infinity, as we get close to 0.1593

That makes 0 a vertical asymptote.1597

In other words, from your perspective 0 is here, e² is here.1600

As we get close here, the function is going to fly up vertically towards infinity.1605

I will draw that in just a minute.1611

We are just working out the analytics.1612

As x approaches e², ln(x) is going to approach 2.1614

2 - ln(x) is going to approach 0.1632

Ln of 2 - ln x is going to approach negative infinity.1639

e² is also a vertical asymptote, when the function goes down.1649

From your perspective, 0, e², as I get close to e², my function is going to go down to negative infinity.1658

What I have is this, I will just go ahead and put e² there.1671

That is a vertical asymptote.1679

I know that as I get close to 0, my function is going to fly up.1681

As I get close to this, my function is going to fly down.1687

It is going to cross somewhere there, we will deal with that later.1694

At least I have my end behavior of my graph.1697

Nothing happens over here, nothing happens over here, nothing happens over here.1700

Because again, this is our domain in the center.1706

Let us see what is next.1712

It crosses the x axis when f(x) is equal to 0, the root.1716

F(x) is ln of 2 – ln x is equal to 0.1733

I’m going to exponentiate both sides.1739

When I do that, I'm left with 2 – ln x e⁰ is equal to 1.1742

I’m going to move the ln x over to this side, bring the 1 over this side.1751

I end up with ln x is equal to 1.1753

I’m going to exponentiate both sides, I get x is equal to e.1758

It crosses the axis at x = e.1764

The graph crosses at x is equal to e.1773

Our graph is starting to come together nicely here.1784

This is x = e² which is 7.4.1792

E is somewhere around there.1801

I know the graph looks something like, this is e, 2.718.1804

The graph is going to look something like this.1813

Let us go ahead and see what we can do.1817

Clearly, there is no local max or min here.1822

We already know what the graph looks like, just from this basic analysis which did not involve any derivatives.1824

We already know that there is no local max, local min.1830

But there is a concavity and there is a point of inflection.1835

We can go ahead and find that.1839

Clearly, there is no, I should say local max, min.1842

Let us confirm this analytically.1862

Let us confirm this analytically.1865

It is exhausting writing all this out.1882

We have f(x) is equal to ln(2) - ln(x).1885

Our f’(x) is going to equal 1/ the argument which is 2 – ln x × the derivative of what is inside.1893

The derivative is going to be -1/ x.1904

That gives us -1/ x × 2 - ln(x).1912

We set that first derivative = 0.1925

This means -1 is equal to 0 because the only way this is equal to 0, is if the numerator is 0.1929

We get -1 is equal to 0, there is no solution.1936

This confirms the fact that there is no critical point.1939

There is no possibility of a maximum or a minimum, local max or min, high point or low point.1942

No solution, no local max or min.1950

We have f’(x), we said it is equal to -1/ x × 2 - ln(x).1961

Let us go ahead and find the second derivative.1971

F”(x) equals this × the derivative of that is 0 - that × the derivative of what is down below, which is going to be,1976

This is going to be product rule.2007

It is going to be this × the derivative of this, which is going to be -1/ x × the derivative of that2009

+ 2 - ln x × the derivative of that which is 1.2028

It is going to be all of that /2x – x ln x².2038

I’m going to let you work out.2050

I write this on this paper, as opposed to here, what we end up with is 2 - 1 - ln x/ 2x – x ln x².2053

We end up with f”(x) is equal to 1 - ln x/ 2x – x ln x².2074

We set that equal to 0.2088

We have 1 - ln x is equal to 0, we have 1 = ln x.2090

We exponentiate both side, we get x is equal to e.2101

Therefore, at x = e which happened to be also be the root, there is a point of inflection.2107

Now we have our final graph.2130

Let us go ahead and, 1, 2, 3, 4, 5, 6, 7, 8, 7.4, this is that.2138

1 to 2.718, this is there.2148

Our graph is going to look something like this.2152

We are going to be concave up from 0 to e.2159

We are going to be concave down from e to e².2165

I think that is it, we have everything that we need.2175

Final graph looks something like this.2178

There you go, 0, notice it is steep rise, this is our e.2179

It also happens to be our inflection point, in addition to being our root.2187

Over here 7.4 that is our vertical asymptote, that is a graph.2191

Make sure you examine the graph, examine the function.2208

Take a good look at the function before you just jump right on in and start taking derivatives.2215

Make sure you examine the function before you just jump in and start differentiating.2220

It is very tempting to want to just run through the process, and hopefully that algorithmic process will give you what you need.2238

It is not always true, it was true in simple math but this is reasonably sophisticated math.2247

We have to bring all of all our resources to bear.2253

Examine the function, check the domain, check the asymptotic behavior.2257

If you take the first derivative, is it going to give you all the critical points.2262

We have to places where the function is not differentiable.2265

There is a lot going on and that is the nature of more complex material, more complex functions.2267

Let us take a look at this one, find the cubic equation ax³ + bx² + cx + d,2277

that achieves a local max at -3, 4 and a local min at 4, -2.2283

They are telling me that, basically, I need to find a, I need to find b,2291

I need to find c, and I need to find d, that satisfies these conditions.2297

I already know what f(-3) is, it is 4.2303

I already know what f(4) is, it is -2.2307

I can plug that in here to get two equations.2309

F(-3) = 8 × -3³ + b × -3² + c × - 3 + d.2315

I know that equals 4.2330

When I solve this, I get -27a + 9b, I hope to God that I have done my arithmetic correct.2335

-3c + d, that is equal to 4, this is the first of my four equations.2345

I have 4 variables, I’m going to need 4 equations and 4 unknowns, it is that simple.2352

That is the only tool that we have at our disposal, at this point.2357

We have dealt with linear systems, four equations and four unknowns.2363

This is a linear system now because everything is to the first power.2367

This is our first equation.2371

I have a second equation, I know what f(4) is, it is equal to -2.2374

F(4) which = a × 4³ + b × 4² + c × 4 + d.2378

I know that = -2.2392

When I work this out, I get 64a + 16b + 4c + d = -2.2394

This is my second equation.2405

We also know something else.2410

It tells me that at these points, it achieves a local max and a local min which means the derivative of this, at these points, is equal to 0.2411

Local max/local min, the derivative there at those points is equal to 0.2423

I find the derivative f’(x) is equal to 3ax² + 2bx + c.2428

I know that f’ at -3 which is going to equal 3a × -3² + 2b × -3 + c, I know that it equal 0.2441

F’ at that point = 0 because it is a local max and f’ at this point is a local min.2456

Its derivative is also equal to 0.2462

I get 27 a - 6 b + c = 0, this is the third of my four equations.2465

I do the same thing for f’ at 4, that is going to equal 3a × 4² + 2b × 4 + c, I know that equal 0.2478

Here I get 48a + 8b + c, that is equal to 0.2491

I have my 4 equations and 4 unknowns.2502

This is my 1st, this is my 2nd, this is my 3rd, this is my 4th.2505

I’m going to solve this by putting it in matrix form and I’m going to convert that matrix to something called reduced row-echelon form.2514

Some of you have actually seen reduced row-echelon form.2524

Do you remember back in pre-calculus, when you guys were doing row reduction2526

and solving simultaneous systems, 3 equations, 4 equations, row reduction is what you guys did.2531

Reduced row reduction is taking it a step further and making it so all of the coefficients in that matrix end up just equaling 1 along the diagonal.2538

I will show you in just a minute.2550

I’m not going to run through the process, there is plenty of mathematical software available online.2551

Just do a Google search for reduced row-echelon calculator and a whole number of things will come up.2557

You basically plug these numbers in and it will give you the answers.2563

Here is what it actually looks like.2568

Solve by converting the augmented matrix of equations to reduced row-echelon form.2570

Our equations, we just put the coefficients in.2604

We end up with -27, 9, -3, 1, is equal to 4.2606

We have 64a, 16b, 4c, + d, = -2.2614

27a - 6b + c, 0, 0, there is no d.2622

And of course we have 48 and 8 and 1 and 0 and 0.2632

This matrix, the first column is the a.2637

The 2nd column is the variable b, 3rd column is the variable c, 4th column is the variable d.2641

When I subject this to reduced row-echelon form, a reduced row-echelon form of a matrix is unique.2647

Row-echelon form, Gaussian elimination is not a unique matrix.2657

You get the answer but is not unique.2661

Reduced row-echelon is always unique.2663

There is only one place it always ends up.2665

In this particular case, what you end up with is the following, when you have your computer do it.2667

1, 0, 0, 12/343, 0, 1, 0, 0, -18/343, 0, 0, 1, 0, -432/343, and 0, 0, 0, 1, 562/343.2673

There you go, a is equal to this, b is equal to this, c is equal to this, d is equal to this.2699

That is it, that simple, that is what beautiful about reduced row-echelon form.2707

This is a, this is b, this is c, and this is d.2712

We get f(x) is equal to 12/ 343 x³ - 18/ 343 x² – 432/ 343 x + 562/343.2725

Again, you can use whatever method you want to solve the system.2756

I think reduced row-echelon is just about fastest and best because it just gives you your answer, your final matrix.2758

You just read it off, it is the last column, a, b, c, and d.2763

Let us do one more function here.2771

For the function f(x) = 3x²/3 – x, find the intervals of increase/decrease, local maxes and mins, intervals of concavity, normal stuff.2780

Use it to graph the function.2790

The domain here is all real numbers so we can just start with f’(x).2793

You know what, I will stick with blue, sorry about that.2805

I will just use red if I absolutely need to.2811

We have f’(x) is equal to 2x⁻¹/3 - 1 is equal to 0, which gives us 2/ x¹/3 - 1 is equal to 0,2815

which gives 2/ x¹/3 = 1, which gives us x¹/3 is equal to 2.2836

Therefore, when I cube both sides, I get x is equal to 8.2852

I have 8, that is one of the critical points.2869

That is where the derivative is equal to 0.2873

Let us find other critical points, if they exist.2876

In this particular case, places where f’(x) does not exist.2886

We just have to be careful because when we see this function,2898

we are not dealing with some ordinary polynomial where you have integral powers.2903

This right here, leads me to believe that there might be places where it is not differentiable.2908

Notice that f is defined at 0.2916

Notice that f(x) is defined at 0, the domain is all real numbers including 0.2919

When I put 0 in here, I get f = 0.2929

It is defined at 0, in other words f(0) = 0.2933

But f’(0) which = 2 × 0⁻¹/3 - 1 which = 2/ 0¹/3 – 1, does not exist.2940

2/0 is undefined, so the derivative at 0 does not exist.2967

The value of the function does.2974

The graph passes through 0, hits 0, but it is not differentiable there.2976

F’ at 0 is undefined, in other words, f, the original function is not differentiable there.2985

0 is another critical point, not differentiable at x = 0, even though it is defined there.3004

It is defined but it is not differentiable which means that it is another critical point.3014

When we check our number line, we have to include 0 and 8.3018

Let us go ahead and do that.3023

0 and 8 are critical points.3026

0, 8, I’m going to check here, I’m going to check here, and I’m going to check there.3039

F’(x) is equal to 2/ x¹/3 – 1.3046

When I check a point to the left here, I’m going to go ahead and check -1.3053

When I check -1, I'm going to get 2/-1 -1 which is equal to -3.3058

It is going to be decreasing.3072

It is going to be decreasing there.3073

I’m going to go ahead and check 1.3076

When I check 1, I get 2/1 - 1 is equal to 1.3080

It is increasing here.3084

However, we just said it is actually not differentiable at 0.3089

Even though it is decreasing and increasing, this is not a local min.3092

It is actually it cusp but not differentiable at 0.3096

This is not a local min, what you have is a cusp here.3116

The function is decreasing to the left of 0.3127

It is increasing but it is not differentiable there.3134

The graph actually looks like this, there is a cusp there.3137

That is what the graph looks like there.3142

Let us go ahead and try, we have done these two, we still have try a region and number over there.3149

I’m going to pick a number that is actually convenient to work with.3162

I’m going to pick the number 27.3166

When I try the number 27 because the 3√27 is really easy to take,3167

it is going to give me 2/3 – 1, which is a negative number.3173

Therefore, it is actually decreasing there.3181

Increasing/decreasing, 8 was a normal critical point.3185

It is what we found by setting the derivative, first derivative equal to 0.3189

This is definitely a local max.3193

We have a local max at x = 8.3199

That is a pretty wacky looking 8.3209

We have got 8 there.3213

F(8) is equal to 4, the point 8, 4 is our local max.3216

The function is increasing from 0 to 8.3231

Our function is decreasing from negative infinity to 0 union 8 to +infinitive.3240

Let us go ahead and deal with some other aspects, the second derivative.3253

We said that f’(x) is equal to 2x⁻¹/3 – 1.3257

We want to take the second derivative.3266

F”(x) is equal to -2/3 x ^- 4/3 -1.3268

We are going to set that equal to 0.3281

We are going to get -2/ 3x⁴/3 - 1 is equal to 0.3285

We are going to get -2/3 x⁻⁴/3 = 1.3295

Rearranging, we are going to end up with -2 is equal to 3x⁴/3 - 2/3 = x⁴/3.3304

This is -2/3 = x¹/3⁴.3321

I separated this 4/3 that way, this is even.3329

Even, something when you have an even exponent, you are never going to get a negative number.3335

This is no solution.3339

No solution so there are no points of inflection.3343

Let us see what we have got here.3358

F” at x is equal to -2/ 3x⁴/3 – 1.3362

This is actually less than 0 for all x.3375

The graph is concave down everywhere.3381

F(x) is concave down everywhere.3386

Let us see, where else f(x) actually equal 0.3400

In other words, let us see if it actually hits the x axis, someplace else.3403

We know it hits the x axis at 0, that cusp.3406

Let us see if it actually hits someplace else.3409

Let us see where else f(x) = 0, beside x = 0.3415

F(x) = 3x²/3 - x is equal to 0.3431

I’m going to go ahead and factor out the x²/3.3444

I get x²/3 × 3 – x¹/3, that is equal to 0.3447

I'm going to get x²/3 = 0 that gives me x = 0.3456

I already know that one.3461

I'm going to get 3 - x¹/3 = 0, x¹/3 = 3.3463

Therefore, when I cube both sides, I get x = 27.3471

27 is the other root, it is the other place where it actually hits the x axis.3477

I have all of the information that I need.3483

My graph is concave down everywhere.3489

It hits at 0 and it hits at 27, there is a cusp here and there is a cusp here.3493

At 8, it hits a local max of 4.3503

When it comes back down, passes through 27 again, and goes down that way and goes up that way.3508

That is my graph.3515

Let us see what it looks like.3517

Yes, that is exactly right.3519

Here is my point where f is defined but it is not differentiable.3521

There is a cusp there.3524

Concave down, concave down, there is no concave up.3525

Here is my max at 8, 4, and here is where I’m at 27, 0, it is my other root.3530

There we go, thank you so much for joining us here at www.educator.com.3537

We will see you next time, bye.3540