AP Calculus AB Example Problems I

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

In the last couple of lessons, we discussed the first derivative and the second derivative.0005

And how they are used in order to decide what graph of a function looks like.0009

In today's lesson and in the next lesson, we are going to do example problems using these techniques,0015

getting progressively more complex as far as the functions that we are dealing with.0020

Let us jump right on in.0024

The following is the graph of the first derivative of some function,0027

on which the original function increasing or decreasing and are there any local maxes or mins?0032

With these particular problems, you have to be very careful to remember which function you are dealing with,0040

which graph they actually gave you.0046

In this case, they gave you the first derivative.0048

This is not the function itself, this is not the second derivative, it is the first derivative.0051

This is f’(x), we have to remember that.0057

In the process of deciding whether it is increasing/decreasing, concave up, concave down,0060

what is a local max and what is a local min.0067

It is going to start to get confusing because you are going to try to describe the behavior of the function,0069

but what you are given is the first derivative of the function.0075

We know that increasing is where the first derivative is greater than 0.0080

Decreasing is where the first derivative is less than 0.0085

This is the first derivative.0088

At this point, between this point and this point, the original function is increasing0090

because the first derivative is positive, it is above the x axis.0096

To the left of this point which is -1, the first derivative which is this graph is negative, it is below the x axis0100

Pass the point 4, it is also negative.0110

The function itself is decreasing on this interval, increasing on this interval, decreasing on this interval.0117

Let me go ahead and work in blue here.0130

I’m just going to write right on top of the graph.0134

Our intervals of increase, it is when f’ is greater than 0.0137

Our interval of increase is from, this is -1.0148

I think I read the graph wrong.0161

It looks like this is -1 and this looks like this 4, from -1 to 4.0166

The interval of decrease is when the first derivative is actually less than 0.0177

In this case, decrease happens on the interval from -infinity to -1 union 4, all the way to infinity.0184

These points, where the first derivative actually equal 00211

because this is the first derivative of the graph, this one.0214

This -1 and this 4, those are the local extrema.0217

F’(x) = 0 at -1 and 4.0225

These are the local extrema, in other words the local mins and local max.0236

Let us see which one is which.0244

At -1, we have decreasing, increasing.0247

You have a local min at x = -1.0253

Over here, it is increasing/decreasing.0262

You have a local max at x = 4.0267

That is it, that is all that is happening here.0276

Just be very careful which graph they are giving you to read it.0279

In the process, you are going to get confused.0284

That is not a problem, that is the whole process.0287

We have all gone through the process, we have all gotten confused.0290

We have all make some minor mistakes and some gross mistakes, regarding this.0292

This is the process, just be extra careful.0297

You have to be very vigilant here.0300

Let us do another example, same type.0305

The following is a graph of the first derivative.0307

This is f’ of a function, on which intervals is the function, the original function increasing/decreasing.0312

Are there any local maxes or mins, you also want more information,0322

on which interval is the function concave up and concave down?0326

What are the x values of the points of inflection?0330

A lot of information that they require here.0333

Let us go through increasing/decreasing like we did before.0336

Once again, increasing is where f’ is greater than 0, decreasing is where f’ is less than 0.0339

This is the f’ graph, therefore here, between here and here, and from here onward, f’ is positive.0346

Therefore, the function is increasing.0361

Therefore, our increasing interval, I will write it over here, is going to be from, this looks roughly like -1.8 all the way to about here.0363

We can read off as, let us just call it 0.9.0380

This should be an open.0385

This is roughly 2.2, all the way to +infinity.0389

This is where the function, the original function is increasing.0399

It is increasing where the first derivative is positive, positive above the x axis.0404

Therefore, our interval of decrease, it is negative from -infinity all the way to this -1.8.0411

It is also below between this 0.9 and this 2.2.0422

These are our intervals of increase and decrease.0431

What we have here is decreasing on this interval, increasing on this interval, decreasing on this interval, increasing on that interval.0435

Therefore, that makes this point right here.0445

I will say the local maxes are going to be this one which is at x = 0.9.0456

Our local min, decreasing/increasing, decreasing/increasing.0470

We have -1.8 and our 2.2.0479

Remember, what we are talking about here is the actual original function.0486

What we had here is our first derivative graph.0492

This does not describe the graph.0497

We are using the graph to describe the original function and we are doing it analytically, things that we know.0500

Increasing/decreasing, positive, negative.0507

Let us move on to our next page here.0512

We have taken care of the local maxes and mins.0515

We have taken care of the increasing/decreasing.0517

Let us find some points of inflection and some intervals of concavity here.0519

Once again, let us remind ourselves actually that this is f’ not f”.0526

We want to talk about some inflection points.0535

Inflection points, we said that inflection points are points where f” is equal to 0.0538

F” is equal to 0, since this graph that we are looking at, since this graph is f’, f” is the slope of this graph.0553

F” is the slope of this graph.0571

I hope that make sense.0584

We want the slope of this graph, where is it the slope of this graph equal 0?0588

It equal 0 there, there, there, and there.0593

What are these x values, it is going to be some place like right there, right there, right there, and right there.0602

Therefore, at x equals -1.25, -0.6, 0, it looks like about 1.6.0613

At these points, the second derivative equal 0.0633

The second derivative is the slope of this graph which is the graph of the first derivative.0637

These are points of inflection.0642

We take those points of inflection, we put them on a number line.0644

We evaluate whether the second derivative is positive or negative.0652

To the left or right of those points.0660

I have got -1.25, -0.6, 0, and 1.6.0662

I need to check where the second derivative.0674

We are dealing with the second derivative here.0677

I need to check this region, this region, this interval, this interval, and this interval,0679

to see whether it is concave up or concave down.0687

I look to the left of -1.25, the slope here is positive, it is concave up.0692

From -1.25 to -0.6, slope is negative, it is concave down.0704

Not the graph, what we are talking about here is the original function.0711

This is where the confusion lies in.0715

This is not concave up but because this is the f’ graph, f” of the original function is the slope of this graph.0717

The slope is positive, therefore, the original function is concave up.0727

Let us write concave up, concave down.0733

From 0.6 to 0, the slope is positive, this is concave up.0736

From 0 to about 1.6, the slope is negative, we are looking at concave down.0741

From here onward, the slope of this is positive.0749

F’ is positive, this means it is concave up.0753

Therefore, our intervals of concavity, concave up are, -infinity to -1.25 union -0.6 to 0 union 1.6 to +infinity.0757

We are going to be concave down from -1.25 all the way to -0.6.0777

And union where it is negative, it is going to be from 0 all the way to +1.6.0786

I hope it makes sense what it is that we have down here.0795

There you go, you got your points of inflection, you have your intervals of concavity0803

We found our local maxes and mins, and we found out the intervals of increase and decrease of the actual function.0809

It looks like we have everything.0815

Let us try another one.0820

The following is a graph of the first derivative.0823

Let us remind ourselves, we are looking at f’ of a function.0828

On which intervals of the function increasing/decreasing?0832

Are there local maxes and mins, on which intervals of the function concave up and concave down?0835

What are the x values of the points of inflection?0840

The exact same thing as what we just did, we are going to do it again.0842

Increasing/decreasing is where the first derivative is positive/negative respectively.0850

Right about there, it looks like about 2.8, 2.7, 2.8, something like that.0858

To the right of that, that is where the first root is positive.0864

To the left of that, it is all below the x axis.0868

F’ is negative, it is decreasing.0874

Do not let this fool you, just read right off the graph.0878

Trust the math, do not trust your instinct.0883

Your instinct is going to want you to see this as the function.0887

This is not the function, this is the derivative of the function.0891

Our intervals of increase is 2.8 all the way to +infinity, 2.8 onward, that way.0895

Our intervals of decrease, we have -infinity to 2.8.0908

-infinity to 2.8.0914

This is going to be, it is negative so it is decreasing.0921

Here it is increasing, we are going to have a local min at 2.8.0925

That is it, local min at x = 2.8.0931

There are no local maxes because there is no place where it goes from increasing to decreasing.0942

In this case, there is no local max.0947

Let us go ahead and talk about some inflection points.0954

I think I can go ahead and do this one in red.0957

Inflection points, we said inflection points are points where the second derivative is equal to 0.0961

Inflection points, there are places where f” is actually equal to 0.0974

F” is f’ of f’, it is going to be the derivative of this graph.0985

It is going to be the slope of that graph.0997

F” which is the slope of this graph is equal to 0 at 2 points.1004

That right there and about right there.1015

Points of inflection are going to be x = -0.9 and x = roughly 1.5.1020

Once again, we have our -0.9, we have our 1.5.1045

Let us go ahead and do our f’ check.1050

We have -0.9 and we have our 1.5.1058

I need to check this region, this region, and this region.1063

To the left of 0.9, the slope of this graph is positive that means the f” is positive.1069

Therefore, the slope is concave up.1077

From -0.9 to 1.5, from here to here, the slope is negative, this means it is concave down.1082

The slope, remember this is f’.1093

The slope is f” of the original function, concave down.1098

From this point onward, we have a positive slope, positive slope, it is concave up.1103

Therefore, our intervals of concavity are concave up from -infinity to -0.9 union at 1.5, all the way to +infinity.1111

We have concave down from -0.9 all the way to 1.5.1126

Again, be very careful, when you know you are dealing with f’.1138

Now I’m going to show you the image of all three graphs right on top of each other, to see what is going on.1143

This was the graph that we were given, this is f’.1155

This is the graph of f”, in other words, it is the slope of this purple.1162

This is actually f(x), this is the one whose behavior we listed and elucidated.1170

Let us double check.1178

Let us go back to blue here.1180

We said the following, we said our interval of increase is 2.8 to +infinity.1182

We look at our original function, it is this one the blue.1211

Yes, decrease, decrease.1215

Yes, from 2.8 onward.1217

That checks out, very nice.1221

We said that it decreased from -infinity all the way to this 2.8.1225

Sure enough, the function from -infinity as we move from left to right,1230

the function is decreasing, decreasing, decreasing, decreasing, until it hits 2.8.1235

Yes, that checks out.1239

We said that we had a local min at 2.8.1242

Yes, there is our local min at 2.8, right there, that checks.1248

We said we had inflection points, let us see what we have got.1255

Inflection points, we said that we have an inflection point at x = -0.9.1261

We also said we had one at x = 1.5.1267

Let us go to -0.9, roughly right about there.1275

Yes, there it is, our blue.1279

There is our inflection point, it changes from concave up to concave down.1280

And then, roughly around 1.5, right about there.1285

There we go at 1.5, at this point.1289

It goes from concave down to concave up.1291

Yes, these two check out.1295

And then, we had our intervals of concavity.1298

We said that we have an interval of concavity from -infinity to -0.9 union at 1.5, all the way to +infinity.1301

Let us double check.1311

Concave up, concave up from -infinity to -0.9.1312

From 1.5 all the way to +infinity, 1.5 to +infinity.1317

Yes, that checks out.1321

The last thing we want to double check is our concave down, from negative 0.9 all the way to 1.5.1323

Yes, from -0.9 all the way to about 1.5.1332

Yes, the graph is concave down, there we go.1337

Here, we see them all together.1342

The graph they gave us was this one.1344

This is the graph they gave us.1350

From that, I was able to elucidate all of this information that corroborated the actual function, which is this.1352

Be very careful with this, that is the take home lesson, just be vigilant.1361

As in all things with, when it comes down to higher math and higher science.1369

You just have to be extra vigilant, there is a lot happening.1373

You have to keep track of every little thing.1376

Let us go ahead and do some analytical work here.1380

For the function 10x²/ x² + 5, find the intervals of increase and decrease,1385

the local maxes and mins, the points of inflection, and the intervals of concavity.1395

Use this information to actually draw the graph.1400

Let us go ahead and do it.1402

I’m going to go back to blue here.1406

I have got f’(x) is equal to, I have this × the derivative of that - that × the derivative of this/ that².1409

I got x² + 5 × 20x - 10x².1420

I really hope to God that I did my arithmetic correctly.1427

All over x² + 5, I’m going to rely on you to double check that for me.1431

All of that is equal to, when I multiply it out, I get 20x³ + 100x - 20x³/ x² + 5².1439

Those cancel, I'm left with 100x/ x² + 5².1456

This is my first derivative, I’m going to set that first derivative equal to 0.1470

When I set it equal to 0, the denominator was irrelevant.1474

It is only 0 when the numerator is 0.1478

100x = 0, which means that x = 0, that is my critical value.1481

I need to check values to the left of 0, values to the right of 0, to see whether I have a local max or min, and increasing/decreasing.1492

Let us see here, I'm going to go ahead and check.1506

Let us just go ahead and check -1.1515

When we put -1 into the first derivative, you are going to end up with a negative number on top.1516

It is going to be positive, this is going to be negative.1532

This is increasing.1537

If I check the number 1, which is to the right of 0, you are going to end up with a positive number on top.1539

Positive, positive, this is going to be increasing.1544

Therefore, we have our increasing interval from 0 to +infinity.1548

We have our decreasing interval from -infinity to 0.1559

That takes care of our increasing/decreasing of the actual function.1565

We also know that this is decreasing, this is increasing.1572

We know that there is a local min at x = 0.1576

This critical value is a local min.1584

Let us go ahead and find f”(x).1588

Let us go back to blue.1590

F”(x), we are looking for inflection points, in order to check intervals of concavity.1594

We are left with this thing.1603

It is going to be this × the derivative of that - that × the derivative of this/ this².1604

We get x² + 5² × 100 - 100x × 2 × x² + 5 × 2x/ x² + 5⁴.1609

I’m going to factor out an x² + 5 here, which leaves me with x² + 5 × 100 - 400x²/,1633

I multiply this, this, this, to get 400²/,1649

I’m sorry, I multiply this, this, and this.1656

I have factored this out as here.1659

I get x² + 5⁴, this cancels with one of these, leaving 3.1661

I need that equal to 0.1670

We have f”(x) is equal to 100x² + 500 – 400x²/ x² + 5³.1677

Set that equal to 0 and I’m going to get -300x² + 500 = 0 because it is only the numerator that matters.1699

I have got 300x² is equal to 500.1712

I get x² is equal to 5/3 which gives me the x is equal to + or - 5/3, which is approximately equal to + or -1.3.1717

I’m going to set up my -1.3, +1.3.1731

I’m doing f”, I need to check points there, points there, and points there.1739

Let us go ahead and actually do this one.1747

F”(x)is equal to -300x² + 500/ x² + 5³.1750

I’m going to check the point -2.1766

When I check the point -2, I'm going to get a negative number/ a positive number1769

which is a negative number, which is concave down.1779

When I check 0, I’m going to end up with a positive number on top of a positive number.1783

When I put 0 into the second derivative, which means positive, which means concave up.1789

When I check 2, I'm going to get a negative/ a positive number that is going to be negative, this is going to be concave down.1796

Let us move on to the next page here.1811

I have got concave down from -infinity to -1.3 union at 1.3 to +infinity.1814

I have got concave up from -1.3 all the way to 1.3.1826

Let us see what happens.1835

We are dealing with a rational function.1839

We have to check the asymptotic behavior.1841

We have to check to see what happens when x gets really big, in a positive or negative direction.1843

Let us see what happens when x goes to + or -infinity.1850

F(x) is equal to 10x²/ x² + 5.1866

As x goes to infinity, this drops out.1874

The x² cancel and you are left with f(x) ends up approaching 10.1878

10 is the horizontal asymptote of this function.1884

What do we have, let us go back and put it all together.1890

We have a local min at 0.1898

When we check f (0), it equal 0.1905

We are looking at the point 0,0.1908

We know that the function is increasing from 0 to +infinity.1912

We know the function is actually decreasing from -infinity to 0.1918

We know that we have points of inflection at x = -1.3.1926

When I check the y value, I get 2.53.1941

Therefore, at -1.3 and 2.53 is my actual point of inflection, both xy coordinate.1944

My other one is at +1.3 and the y value is 2.53.1955

1.3 and 2.53 is my other point of inflection.1963

I am concave down from -infinity to -1.3 union 1.3 to +infinity.1969

I am concave up from -1.3 all the way to +1.3.1981

I have a horizontal asymptote at y is equal to 10.1990

When I put all of this together, what I end up getting is the following graph.2007

Here is my local min, here is one of my points of inflection.2017

Here is my other point of inflection.2024

It is decreasing all the way to 0, increasing past 0.2028

It is concave down all the way to -1.3, concave down up to this point.2035

It is concave up from this point to this point.2044

It is concave down again, notice that it approaches 10.2047

That is it, first derivative, second derivative, local maxes and mins, points of inflection, intervals of concavity.2055

I need horizontal asymptotes, vertical asymptotes, everything that I need in order to graph this function.2064

Let us go ahead and try another example here.2073

For the function x² ln 1/3 x, find the intervals of increase/decrease,2077

local maxes and mins, points of inflection, intervals of concavity.2081

Use this information to draw the graph.2085

Let us go ahead and do it.2088

F’(x) is equal to x, this is a product function.2091

This × the derivative of that + that × the derivative of this.2097

X² × 1/ 1/3 x × 1/3 + ln of 1/3 x × 2x.2101

1/3, 1/3, x, x, what I should end up with is x + 2x ln 1/3 (x).2116

We want to set that equal to 0.2129

I'm going to go ahead and factor out an x.2131

I get x × 1 + 2 ln of 1/3 x that is equal to 0, that gives me x = 0.2134

It gives me 1 + 2 ln of 1/3 x is equal to 0.2153

0 is one of my critical points.2160

When I solve this one, I will go ahead and solve it up here.2165

I get 2 × ln of 1/3 x is equal to -1.2169

Ln of 1/3 x is equal to -1/2.2177

I exponentiate both sides, I get 1/3 x = e ^-½.2181

I get x is equal to 3 × e⁻¹/2.2190

X is approximately equal to 1.8.2194

Let us go ahead and do it.2202

These give me my critical points.2204

I have 0 and I have got 1.8.2206

I need to check a point here, check a point here, check a point here.2210

Put them into my first derivative to see whether I get a positive or negative value.2213

Let me go ahead and write it out.2223

F’(x), I’m going to write out the multiplied form, = x × 1 + 2 ln of 1/3 x, that is my f’(x).2225

I’m going to pick a point again here, here, here.2243

Put it into this to see what I get.2246

When I check the point, this here is not the domain.2247

I do not have to check a point there.2257

The reason is I cannot take the log of a negative number.2258

That is not a problem, I do not have to check a point here and here.2262

I'm going to go ahead and check 1.2265

For 1, when I put 1 into here, this is going to be positive, this is going to be negative.2268

Therefore, it is going to be decreasing.2274

We are going to be decreasing on that interval.2277

When I check the number 2, I get positive and positive which means it is increasing on that interval.2279

I’m decreasing from 0 to 1.8, I’m increasing from 1.8 to +infinity.2294

I have a local from decreasing to increasing.2300

I have a local min at 1.8.2302

That is what this information tells me.2305

Let us write that down.2307

I’m decreasing from 0 to 1.8, I am increasing from 1.8 to +infinity.2309

I have a local min at x = 1.8.2324

The y value at x = 1.8, that is just the original function.2335

F(1.8), it gives me -1.7.2345

My local min is going to be the point 1.8, -1.7.2353

Notice that we have x = 0, what is the other critical point.2364

But I cannot say that there is actually a local max there.2368

The reason I cannot say that is because there is nothing to the left of 0, the domain.2374

In order to have a local max or a local min, I have to have the point2378

defined to the left on to the right of that particular critical point.2381

Here it is only defined to the right.2385

The positive and negatives do not count.2389

Let me write that out.2395

We cannot say that there is a local max.2400

In some sense there is, but not by definition, that there is a local max at x = 0 because f is not defined for x less than 0.2410

In other words, our function, we know there is a local min at 1.8 and -1.7.2435

We know it is here.2442

We know it is going to be something like this.2444

We cannot necessarily say that this is a local max because it is not defined over to the left of it.2446

That is all that is going on here.2452

Let us do point of inflection.2454

Let me go back to red.2457

Points of inflection, we have f’(x) is equal to x + 2 ln 1/3 x.2459

Therefore, f”(x), I’m going to take the derivative of this.2471

It is going to be 1 + 2 ×, this is 2x, 1 + 2 × x × the derivative.2476

X × 1/1/3 x × 1/3 + the nat-log of 1/3 x × 1.2488

I just pulled out the 2, just for the hell of it.2505

1/3, 1/3, x, x, this becomes 1.2508

You end up with 1 + 2 × 1 is going to be 2 + 2 × the ln of 1/3 x.2512

We are going to get 3 + 2 × the ln of 1/3 x.2527

That is our second derivative.2535

We need to set that equal to 0.2540

F”(x)is equal to 3 + 2 × the ln of 1/3 x.2545

We need to set that equal to 0.2554

We get 2 ln 1/3 x is equal to -3, ln of 1/3 x = -3/2.2556

We exponentiate both sides, we get 1/3 x is equal to e⁻³/22566

which gives us x is equal to 3 × e⁻³/2, which is approximately equal to 0.7.2581

We have to check, this is a point of inflection.2593

We need to check a point to the left of 0.7 to the right of 0.7, to see whether the second derivative is positive or negative.2596

In other words, concave up or concave down, respectively.2603

We have got 0.7 here, we are checking f”.2609

F” is equal to 3 + 2 × the nat-log of 1/3 x.2615

When I check the value, I will just check 0.2626

I check the value 0, it is going to be concave down.2633

I’m not going to check 0 actually.2649

I will check my 0.5.2652

0.5, you are going to end up getting a negative number which is going to be concave down.2654

And then, I'm going to go ahead and check some other number 1, 2, 3, does not really matter.2666

Let us just check 3, positive, that is going to be concave up.2672

In other words, I’m putting these values into the second derivative to tell me whether something is positive or negative.2677

This is going to be positive which is going to be concave up.2686

That takes care of that.2692

Now I have my intervals of concavity.2706

It is going to be concave down from 0 all the way to this 0.7.2708

Let me double check, yes.2724

And then, concave up from 0.7 to +infinity.2728

Let us find where f(x) actually equal 0.2742

Let us see where it actually crosses the x axis, if in fact it actually does so.2746

Let me go ahead and do that on the next page here.2753

F(x) is equal to x² ln 1/3 x, we want to set that to equal to 0.2758

That gives us x² is equal to 0 and it also gives us ln of 1/3 x is equal to 0.2767

This is 0, it attaches at 0,0.2776

This one you get 1/3 x e⁰ is 1, that means x = 3.2779

It touches the x axis at 0 and 3.2788

Let us list what we have got.2794

Our root x = 0, x = 3.2800

We have a local min at the point .1.8, -1.7.2809

We have a point of inflection at 0.7, -0.7.2819

We are concave down from 0 to 1.4.2828

I’m sorry not 1.4, it is going to be 0.7.2841

Concave down from 0 to 0.7.2844

We are concave up from 0.7 to +infinity.2847

The function is decreasing from 0 to 1.8.2856

The function is increasing from 1.8 to +infinity.2862

When we put all of that together, local min at 1.8, -1.7.2875

1.8, -1.7 probably puts us right there.2884

Point of inflection at 0.7, -0.7, 0.7, -0.7, somewhere around there.2887

Our graph goes something like this.2894

It is decreasing from 0 to 1.8, decreasing, concave down from 0 to 0.7.2896

It is concave down here, concave up from 0.7 to infinity.2907

That is our graph, let us see a better version of it.2915

Here is our root, here is our root.2921

Points of inflection is somewhere around there.2925

Local min somewhere around there.2931

As you can see, we have concave down from here to here.2933

Concave up from here to here, and continuously concave up.2937

The graph goes, passes through 3.2941

There you go, that is it, wonderful.2946

Thank you so much for joining us here at www.educator.com2949

We will see you next time for a continuation of example problems on using the derivative to graph functions.2952

Take care, bye.2959