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### Using Derivatives to Graph Functions, Part II

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
• Using Derivatives to Graph Functions, Part II 0:13
• Concave Up & Concave Down
• What Does This Mean in Terms of the Derivative?
• Point of Inflection
• Example I: Graph the Function 13:18
• Example II: Function x⁴ - 5x² 19:03
• Intervals of Increase & Decrease
• Local Maxes and Mins
• Intervals of Concavity & X-Values for the Points of Inflection
• Intervals of Concavity & Y-Values for the Points of Inflection
• Graphing the Function

### Transcription: Using Derivatives to Graph Functions, Part II

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

Today, we are going to continue our discussion of using the derivative to actually graph functions.0005

Let us jump right on in.0011

We have used the first derivative to talk about intervals of increase and decrease.0016

We have also used it to discuss local maxes and mins.0021

We have used the first derivative to find the intervals of increase and decrease,0031

where the actual function itself is increasing and decreasing, as well as local maxes/mins.0047

We know that if a function is from your perspective going from left to right, is decreasing then increasing,0067

we know the function is a local minimum.0074

If the derivative is increasing, it is positive, and then right after a certain point it becomes negative,0079

we know that it hits a local min and local max.0085

Now, let us go ahead and go a little further.0090

Let us look at a couple of segments of a couple of graphs.0094

Segments of the graph, these segments happen to be two increasing segments.0103

Let us look at two increasing segments because we do not know what is happening to the left or right, or what it is that I’m drawing.0116

Let us look at two increasing segments of two functions.0131

One, we have our axis like that and we have something that looks like this.0140

Over here, we have something that looks like this.0148

In both cases, it is increasing, it is increasing.0156

In other words, the derivative is going to be positive, the derivative is going to be positive.0161

But clearly, there is a difference between these two.0166

How do I differentiate, now is where we use the second derivative to differentiate between the two.0168

In both cases, f’(x) is greater than 0, how do we differentiate between the two?0175

One of them, let us call this one number 1 and let us call this one number 2.0203

Number 1 is called concave up, in other words, the concavity is facing upward.0208

Analogously, the other one is concave down.0216

In other words, the inside of the curve, the inside of the concave portion is facing down.0221

A nice definition, clearly, there is a tangent there, there is a tangent there, there is a tangent there.0228

These are all increasing, they are positive tangents.0240

Positive tangent, positive tangent, and positive tangent.0243

A nice definition of concavity is the following.0249

We fill it out here, not that we need it.0259

I know we know what concave up and concave down means.0261

These are pictorial motions, you know what concave is, it is the inside of the curve.0264

The inside of the bowl, as opposed to the outside of the bowl which is called a convex portion of a bowl.0268

A nice definition of concavity is as follows, just for the sake of having a definition.0274

Concave up, I will just use cu, the graph of the function, the curve, the graph was above its tangents.0282

Clearly, as you can see here, the tangents are all below the graph.0302

The graph never falls below the tangents.0306

Concave down is just the opposite.0309

The graph lies below its tangents.0312

In case you are not sure, in case it is really kind of subtle, draw a tangent and see whether the graph is actually below or above the tangent.0320

And that will tell you whether it is concave up or concave down.0328

Here we have a graph, let me go to red.0331

That is your graph, it lies below.0335

Let us go back to blue here.0339

It lies below these tangent lines, it is concave down.0341

Concavity can be very subtle.0348

Clearly, something like this is really obvious.0349

If you have something like this, it is obvious.0351

Something like this, it is obviously concave up.0354

You have something like this, ever so slight a curve, not so obvious.0357

This is where we are going to use our analytical techniques.0363

Now what does this mean in terms of the actual derivative?0369

What does this mean in terms of the derivative?0377

It means the following.0393

If f”(x) is greater than 0 over a given interval, I will call it just a over a given interval a, then f is concave up over that interval.0395

Remember, when we talk about intervals, we are talking about the x axis.0429

Let us say from 1 to 2, every single x value in there, when I put in the second derivative,0433

it happens to be positive then I know that in between 1 and 2, the graph is concave up over that interval.0439

This gives us a numerical way of measuring concavity.0449

In those cases where you have a slight very subtle curvature,0452

where you cannot really decide or you cannot read it off the graph.0456

Everything in math may have a geometric interpretation.0460

We use geometry to help us along with the pictures.0463

I mean, clearly, we are going to use all this to draw the graph of the functions.0466

The graph is important but the algebra, the math, the analytics, that is was paramount.0470

Number 2, if f”(x) is less than 0 over a given interval, then f is concave down over that interval.0478

This is how you do it, you have a function, you take the first derivative that tells you whether the graph is increasing or decreasing.0514

You take the second derivative and you check points to see where the second derivative is negative,0521

where the second derivative is positive.0527

Let us define something called a point of inflection.0530

It is any point where f is, first of all continuous there and the graph changes from concave up to concave down.0540

In other words, the second derivative goes from negative to positive or positive to negative.0568

The second derivative passes through 0.0574

Setting the second derivative equals 0 gives us an analytic way of actually finding these values of x0576

where there is a change from concave up to concave down.0583

In any point where f is continuous there, and the graph changes from concave up to concave down or concave down to concave up,0587

either way, where it makes the transition.0604

For a point of inflection, f” passes through 0, very important.0611

In other words, f” at a point of inflection is going to equal 0.0632

Let us see what is next here.0642

Let us go ahead and draw a version of this.0646

Here we have something like that and let us say we have a graph that looks like this.0661

This right here, there is a point right here.0669

That right there, from this point to this point over this interval, the graph is concave up.0673

The second derivative is going to be positive.0682

Over this interval, it is concave up which means that f” is greater than 0.0686

From here onward, it is concave down.0692

F”, if I take any x value in here, it is going to be less than 0.0696

At this point, which is a point of inflection, it changes from concave up to concave down.0701

In this particular case, this is a point of inflection.0705

This point happens to be a point of inflection, the x value.0710

If I want the y value, I put it into the original function to find the y value of where it actually flips.0714

At this point, f”(x) is going to equal 0.0722

This is a point of inflection, that is it.0727

That is your geometric explanation of what a point of inflection is.0729

Once again, over a certain interval, if the second derivative is greater than 0, the graph is concave up.0734

It could be very clear concave up or it can be very subtle.0742

If the second derivative over a certain interval is less than 0, it is concave down.0747

At a point of inflection, it is going to equal 0, provided it is actually continuous there.0752

Most of the functions that we are going to be dealing with are going to be continuous0756

but there are going to be some instances where we have to be a little extra careful.0759

We do not just take these rules and just apply them blindly.0763

We always want to step back a little bit to use other resources at our disposal,0768

to make sure we know exactly what is going on with the graph.0772

That is really what mathematics is all about, is every class that you take teaches you a little bit more of something,0776

a little bit more of something, gives you something else to put in your toolbox.0781

You are going to apply some or all of these tools to a particular situation, depending on how complex or how easy the situation is.0784

Let us do an example.0796

Here says sketch the graph of a function that satisfies the following conditions.0801

It is telling me that the f’(0) is equal to the f’(3), is equal to the f’(5), which is equal to 0.0806

The first derivative of these points is equal to 0.0814

That tells me that I’m hitting either a local max or a local min, or it is going to be not differentiable there.0817

It tells me that the first derivative is greater than 0, when x is less than 0.0829

Which means that when x is less than 0, the graph is actually rising, it is increasing.0834

It is also rising between 3 and 5, which means between 0 and 3, it is actually falling.0841

After 5, it is also falling.0850

It tells me that f’(x) is less than 0, the first derivative is less than 0 between 0 and 3.0855

It is decreasing from 0 and 3.0864

It is decreasing when it is greater than 5.0866

It tells me that the second derivative f” is positive between 2 and 4.0870

It is negative to the left of 0 or when x is less than 2, or x is greater than 4.0881

Let us see here, I think there might be a mistake in something here.0896

But let us go ahead and see if we can work this out.0906

Let us go ahead and draw it.0908

I think I have might mis-worded something over here, but let us see.0912

F’ at 0 and 3 and 5, I have 1, 2, 3, 4, and 5.0918

It is a 5, 4, 3, 2, 1, and this is our origin 0.0926

I’m going to hit a local max or local min at this point, this point, and this point.0937

But I do not know whether it is going to be local max or local min yet, so I have to look at these.0945

F’(x) is greater than 0 when x is less than 0.0952

F’(0) is less than 0 between 0 and 3.0956

The function is rising to the left of 0 and it is falling between 0 and 3.0962

Therefore, I actually hit a local max at 0.0974

It is rising over here and it is falling over here.0981

F’(x) is greater than 0, the function is rising between 3 and 5.0992

That means here, it actually hits a minimum.0996

It is rising here, that means this comes this way and that is confirmed with this right here.1001

The second derivative between 2 and 4, it is positive between 2 and 4, yes it is concave up.1009

And then, it is less than 0 when x is bigger than 4, it is concave down.1020

At 5, it looks like we are probably going to f’(x) is a greater than 0, when x is between 3 and 5.1026

It is rising between 3 and 5, and is less than 0 when x is greater than 5.1040

It hits another max over here and that is less than 0, for x greater than 5.1044

It drops off like that.1054

We have a local max, a local min, local max, and these corroborates.1057

Let us make sure what we have here.1063

F’(x) is greater than 0, when x is greater than 2 and less than 4.1065

Greater than 2 and less than 4, yes, concave up.1070

F’(x) is less than 0, when x is less than 2, yes, concave down.1074

That is our concave down right here.1079

This is concave down right here, or x is greater than 4.1082

Yes, greater than 4, there was no mistake, it was exactly as written.1089

Concave down, concave up.1093

It certainly satisfies these conditions.1101

You are just using the information given.1105

Once again, when f’ = 0, it is a critical point.1108

It may or may not be a local max or min.1114

In this case, workout to be local maxes and mins.1116

When f’, the first derivative is greater than 0, the function is increasing.1120

When the first derivative is less than 0, the function is decreasing, the original function.1123

When f” is greater than 0 over that interval, the graph is concave up.1128

When f’ is less than 0, the graph is concave down, the original function.1134

Let us do another example.1143

Now we can bring all of our analytics to there.1145

For the function x⁴ – 5x², discuss the intervals of increase and decrease,1148

local maxes and mins, points of inflection, the intervals of concavity.1153

And then, use this information to graph the function.1157

There is no law that says you have to do anything at a given order.1161

Generally, you are going to take the first derivative first, and then take the second derivative afterward.1165

You do not have to wait until the end to actually draw your graph.1170

You can draw it any time you want.1172

If you have enough information and you feel comfortable enough1174

to actually conclude the things that you can conclude from the mathematics given.1176

You may use all the information, you may use only some of the information.1181

You may use other information that you have, asymptotes, roots,1184

whatever it is that you need, whatever things that you have in your toolbox.1189

This first derivative and second derivative stuff, these are just two more tools in your toolbox.1193

Let us go back to blue here.1201

F’(x) that is going to equal of 4x³ - 10x.1206

We want to go ahead and set that equal to 0.1216

I’m going to factor out the x.1218

X × 4x² - 10 that is equal to 0.1220

I get x is equal to 0 and I get 4x² - 10 = 0.1227

That is going to be one of my possible points, one of my critical values.1239

I’m going to have 4x² that is equal to 10, x² = 10/4 which is equal to 5/2.1246

Therefore, x is equal to + or – √5/2 which is approximately equal to 1.58.1257

I need to check my intervals of increase and decrease.1268

Now that I find my critical values, 0, positive, this is positive or negative, +1.58 and -1.58.1271

I have to draw my number line and I have to check points in those intervals to see where I have increase or decrease.1282

Let us go ahead and do that.1287

I have 0, I have +1.58 and I have -1.58.1290

We are checking f’, I have to check points in this interval, this interval, this interval, this interval, to see,1298

I’m going to put them back in the first derivative to see whether the first derivative is positive or negative, increasing or decreasing.1306

F’(x), we said is equal to x × 4x² – 10.1314

I’m going to try some points, I’m going to try a point here, here, here, and here.1322

Over here, I’m going to try -2.1325

When I put -2 into the first derivative, I do not really need to calculate the value.1329

I just need to know whether it is positive or negative.1337

-2 that means the x is negative.1340

When I put -2 in here, I’m going to end up with a positive number.1343

A positive × a negative is a negative.1349

F’ is negative which means that the function is decreasing on that interval.1356

I’m going to try the value here, I’m going to try -1.1362

When I put -1 in for x into the first derivative, this becomes negative, this becomes negative, it is positive.1367

The first derivative is positive in this interval which means the function is actually increasing.1377

I'm going to try one over here, I get negative.1382

1 is a positive number.1391

If I put 1 in here, I get a negative number.1393

I get a negative which means the function is decreasing on that interval.1395

I have 2 which is going to be positive.1402

If I put that there, 2 × 2 is 4, 4 - 10 is also positive.1405

It is positive which means it is going to be increasing along that interval.1410

The derivative = 0 at this point, this point, and this point.1416

Those are our critical points.1419

Our intervals of increase are this interval right here, let me go to red.1423

This interval right here and this interval right here.1429

Our intervals of increase are -1.58 to 0 union +1.58 to +infinity.1433

Our intervals of decrease are going to be that and that.1447

Our intervals of decrease are -infinity to -1.58 union 0 to 158, 0 to 1.58.1458

We have also taken care of local maxes and mins here.1471

We have decreasing/increasing, there is a local min right here.1475

Increasing/decreasing, there is a local max, here, somewhere.1482

There is a local min for x = 1.58.1487

I have taken care of that as well.1496

Let us go ahead and deal with those.1496

We also know that -1.58 is a local min.1501

We know that 0 is a local max and we know that +1.58 is a local min, as well.1517

Let us go ahead and find the y values of these things, just for the sake of it.1534

Let us find the y values of these critical points, these local maxes or mins.1540

We just want to know where they are exactly, y values of these points.1549

When I do f(-1.58), I get -6.25.1559

I hope my arithmetic is correct.1566

When I do f(0), that is going to equal 0.1567

When I do f(1.58), I get a -6.25 again.1572

We already have much of our graph.1585

We do not yet know where it passes through the x axis yet.1604

We do not know its roots yet.1615

We can get those just by setting the original function equal to 0 and solving by with whatever means we have our disposal.1616

In general for these graphing ones, whether you are doing them for calculus class or for the AP test itself,1627

you may not necessarily need to know where it passes the x axis.1635

What they want is a good solid graph showing increase/decrease things like that.1639

But not necessarily if the function is too complicated and1645

you do not have a calculator at your disposal or a computer algebra system at your disposal,1648

it might be very hard to find a particular root.1652

That is your secondary concern, where it passes through the axis.1656

Let us go ahead and draw what we have so far.1664

F(0) is 0.1675

Let us do 1, 2, 3, 4, 5, 6, 7.1681

Let us do 1, 2, 3, 1, 2, 3.1686

At -1.58, at -6.25.1694

1.58, -6.25, it is going to put us somewhere around there.1699

It is also the same over here, somewhere around there.1703

I know these are the local maxes and mins.1706

It was a local max here, it is a local min here, it is a local min here.1709

We have our graph, at least some of our graph.1715

Of course we want to finish off with the other information which is points of inflection, intervals of concavity.1722

We are looking for points where the concavity changes from positive to negative.1728

We see here clearly this is concave up.1734

But there are some points around here where the concavity goes from concave up,1736

all of the sudden it is concave down, or from concave down to concave up.1742

It is going to go up and pass through here.1747

Let us go ahead and deal with those next.1750

Intervals of concavity, let us go back to blue.1754

Our points of inflection, we are going to take f” and we are going to set it equal to 0.1769

Solve for x, that is going to give us our points of inflection.1781

We are going to do the number line again.1785

We are going to check points to the left and to the right of these points of inflection,1786

to see where the second derivative is positive or negative, concave up, concave down, respectively.1790

Let us rewrite, we have our original function f(x) is equal to x⁴ – 5x².1799

We have f’(x) which is 4x³ - 10x, that makes our double prime.1807

3 × 4 that is going to be 12x² – 10.1815

We go ahead and we set that equal to 0.1821

We get 12x² is equal to 10.1827

We get x² is equal to 10/12 which equals 5/6.1832

Therefore, x is equal to positive or negative √5/6 which is approximately equal to 0.913, + or -.1838

Let us go ahead and do our little number line.1852

We have +0.913 and we have -0.913.1854

We are going to check points to the left in between and to the right, to see where it is concave up or down.1862

We already know the answer, using the first derivative, you are able to get 90% of our graph.1869

The only thing we do not really know is where it hits the x axis.1875

We know where it is concave up and concave down.1879

This procedure, this analytical procedure, gives us the actual numbers 0.913 to let us decide.1882

We know that we are talking about something which is going to be concave up, concave down, concave up.1889

Based on what we drew on the previous page.1898

We just confirmed that here.1900

This we are working with double prime.1905

The hardest part of these problems is going to be, once you get the values of x and1906

putting it back into the original function, the first derivative, the second derivative.1912

You just have to be very careful which function you put them in.1916

You are going to be dealing with 3 functions.1920

The function of the first and second derivative.1922

You have to make sure you know which one you are putting them in.1924

Here, we are taking x values and we are putting them into the second derivative to see what sign the second derivative is.1927

F”(x) is equal to 12x² – 10.1934

I’m going to pick a point over on this side.1940

I’m going to pick a point -1.1945

When I put -1 into this, I'm going to get a positive.1947

The second derivative, f” is positive here, this is concave up.1954

When I check something in this region, I’m going to go ahead and check 0.1960

I’m going to get -, 0 - 10 is negative.1964

Negative means it is concave down.1968

When I check +1 which is in this region over here, +1, I’m going to get positive which means it is concave up here.1972

Let us go ahead and re-graph our graph.1989

We have this and we have something that looked like this, something like that.2003

This point was 1.58, that is where it hit the local min.2017

This was -1.58, that hit a local min.2023

Let me draw this a little bit better.2028

It hit the local min, the points of inflection are right here.2033

These x values, that +0.913 and -0.913.2038

That is a point of inflection, that is the point of inflection,2042

it is where it changes from concave up, concave down, back to concave up, over that number line that we did in the previous page.2047

Let us go ahead and find the y values.2058

In other words, we found the x 0.913, -0.913.2060

Let us go ahead and find the y values for these points.2064

Let us find y values for the points of inflection.2069

Remember, very important, if you are looking for y values, these x values of the points of inflection,2092

they go back to the original function.2101

Do not worry, we all make mistake like we all did.2115

That is the process, the original function, in other words f(x).2118

We have f(-0.913) = -3.4 and f(0.913) = -3.4.2133

Our points of inflection are -0.913, -3.4, and 0.913, -3.4.2157

Let us go ahead and write down the intervals of concavity, up and down.2194

Sorry about that, I do not think I actually wrote them down formally.2200

Intervals of concavity, it is going to be concave up from -infinity, we said to -0.913 union 0.913 to +infinity.2203

It is concave down from -0.913 all the way to +0.913.2224

Just to write them down formally.2233

That takes care of that, the last thing we should do is actually find out where it crosses the x axis, just for good measure.2237

The last thing we should do is find where f(x), the original function, crosses the x axis.2246

In other words, the roots, the x axis.2266

There is no guarantee that the function is actually going to cross the x axis.2273

It could not cross at all, in which case you are going to set f(x) equal to 0 and you are going to get no solution.2276

F(x), the original function is equal to x⁴ - 5x², we set that equal to 0.2284

I’m going to go ahead and factor out the x² and I'm going to get x² – 5 and that is equal to 0.2293

I get x² = 0 and I get x² - 5 = 0.2302

Therefore, it crosses x = 0.2308

It is a double root, it actually touches, we know that already from the graph.2312

It touches the graph there, it does not cross.2315

X² = 5, therefore, x = + or -√5 which is equal to + or -, roughly, approximately equal to 2.24.2320

There you go, final graph.2333

I will go ahead and put a point here, put a point there.2340

They cross there, cross there.2346

This is 2.240, this point is -2.240.2359

This point is a point of inflection, this point is of course 0,0.2372

This point is 0.913, -3.4.2379

This point is -0.913, -3.4.2387

This point, the local min, that is -1.58 and -6.25.2393

If I’m not mistaken, if I remember properly.2400

This one is 1.58 and -6.25.2402

Let us see what a better version of the graph looks like.2408

Local max, local min, local min.2416

Where it crosses the x axis, where it crosses the x axis, those are two roots.2421

Points of inflection are here and here.2426

Concave up until this point, this interval all the way to -0.913, it is concave up then it is concave down until it hits +0.913.2431

When it flips over and becomes concave up again, all the way towards infinity.2443

They are you have it.2449

The first derivative will tell us whether we have a local max or min.2455

It does not tell us increasing/decreasing.2480

It also tells us when it passes from increasing to decreasing or decreasing to increasing, we are going to hit a local max or min.2481

The first derivative is all we need for local max or min.2487

There is also a second derivative test, if we want it.2493

There is also a second derivative test for local maxes and mins.2496

It is generally unnecessary but we will go ahead and lay it out.2517

Let f”(x) be continuous near a point a.2524

To the left and right of it, it is going to be continuous.2540

If f’(a) is equal to 0 and f” at a is greater than 0, f’(a) is equal to 0.2543

It is either a local max or min.2565

F” if a is greater than 0, which means it is concave up, it is a local min.2567

Then f has a local min at a, and the opposite.2577

If f’(a) = 0, f” at a is less than 0, which means concave down.2588

Then, f has a local max at the point a.2605

What you are looking at is this.2618

You know this already, that point, that point.2626

Here f’ is equal to 0, f” greater than 0, concave up, local min.2629

Here you have got f’ is equal to 0, f” less than 0.2645

This is concave down, this is your local max.2652

That is it, second derivative test, in case you want to use it.2657

You really do not need it but it appears in the calculus books, we are going to go ahead and give it to you.2659

I’m going to go ahead and stop the lesson here.2669

Do not worry that there is only been a couple of examples in this lesson and in the previous lesson.2672

This was mostly just theory, something to get your feet wet.2677

The next couple of lessons are going to be nothing but example problems, using what we just did in these last two lessons.2680

Again, the next set of lessons is going to be just example problems.2687

We are going to do plenty of these, no worries.2692

Thank you so much for joining us here at www.educator.com.2695

We will see you next time, bye.2697