INSTRUCTORS Raffi Hovasapian John Zhu

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• ## Transcription

 1 answerLast reply by: Professor HovasapianThu Apr 7, 2016 1:37 AMPost by Acme Wang on April 5, 2016Hi Professor,In example II, I am a bit confused about your solving way. I know -2.33 and other 3 values are critical values. Why can we just simply choose a random number such as -4 or -2.48 to discuss the increasing/decreasing property at the left side of -2.33? Would there be the situation that f'(-4) is positive and f'(-2.48) is negative? Then what's the increasing/decreasing property at the left side of this critical point, -2.33? I don't know whether I have made clear about my problem. Hope you can understand and clear my confusion. Thank you very much!Sincerely,Acme

### Using Derivatives to Graph Functions, Part I

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 I 0:12
• Increasing/ Decreasing Test
• Example I: Find the Intervals Over Which the Function is Increasing & Decreasing 3:26
• Example II: Find the Local Maxima & Minima of the Function 19:18
• Example III: Find the Local Maxima & Minima of the Function 31:39

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

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

Today, we are going to talk about using the derivative to actually help us graph a function.0005

Let us jump right on in.0011

I will go ahead and stick with black here.0017

No, I think I will go back to blue.0020

We can use first and second derivatives along with any other information that we have0024

for a function to complete the graph of a function.0062

You are going to be spending a fair amount of time doing this.0080

We ourselves are going to be spending a fair amount of time doing this.0082

The first thing we are going to talk about is something called the increasing/decreasing test.0087

We have the increasing/decreasing test.0092

If f’(x) is greater than 0 on an interval… and remember, when we speak about interval, we are talking about the x axis, the domain.0105

If f’(x) happens to be greater than 0 on an interval, then f is increasing on that interval.0123

The original function, the function itself is on its way up.0135

Again, we are moving from left to right, from negative values towards the positive direction.0138

It is increasing on that interval.0144

If f’(x) is less than 0 on an interval and f is decreasing on that interval, it is on its way down decreasing on that interval.0153

If you have some function, slope is positive 0, decreasing.0181

Now the slope is negative, it is changing, 0, the slope is positive again.0188

It is increasing, function is increasing, function is decreasing, function is increasing.0194

You know this intuitively, in terms of the derivative, what is happening.0199

Let us do an example, find the intervals over which the function f(x) = 2x⁴ – 4x³ – 10 x² + 5 is increasing and decreasing.0206

Find the intervals of increase and find the intervals of which the function is decreasing.0217

When a function goes from increasing to decreasing, in other words,0241

when it goes from f’ greater than 0 to f’ less than 0, it has to pass through f’ = 0.0262

Or it goes from decreasing to increasing, meaning f’ less than 0 going to f’ greater than 0.0275

Then, f’ passes through 0, in other words, there is some point such that f’ is equal to 0.0292

Let us set f’(x) equal to 0.0310

Let us find those points such that f’ is 0.0316

We are going to check points to the left of that and to the right of that, to see whether the derivative is positive or negative.0322

That is going to tell us whether the function is increasing or decreasing.0329

F’(x) is equal to 8x³ - 12x² - 20x.0337

We are going to set that equal to 0.0355

I’m going to factor out a 4x.0357

We have 4x × 2x² - 3x – 5, equal to 0.0360

Therefore, I have 4x ×, I can factor this one, 2x - 5 × x + 1 is equal to 0.0373

Therefore, I have x = 0, x = 5/2, and I have x = -1.0383

This happened to be the critical points of this function.0395

Remember, back when we talked about critical points, maxima and minima,0397

the critical points are the points where the derivative is equal to 0.0400

We know the derivative = 0, the slope is 0.0404

I’m going to check to see a point to the left of these points.0411

I’m going to check a point to the right of these points, and to see if the derivative actually changes sign.0414

If it goes from negative to positive, or it goes from positive to negative.0419

If that is the case, then it will me whether it is increasing or decreasing to the left or right of these points.0423

Let us go ahead and do that.0430

I’m going to go ahead and draw the number line.0435

I have my points, -1, 0, and 5/2.0439

Great, I’m going to check a point here, I’m going to check a point here, check a point here,0449

and I'm going to check a point over there.0454

Let us rewrite f’(x), we said that f’.0458

Let me do it over here actually.0465

We said that f’(x) is equal to 4x × 2x - 5 × x + 1.0468

I’m going to check a point to the left of -1.0480

It is going to pick -2.0483

When I do f’ at -2, in other words I’m just putting -2 in for the x values here.0485

I do not have to worry about what the values are.0492

I just need to know whether it is positive or negative.0494

4 × -2 is negative, 2 × -2 - 5 is also negative, -2 + 1 is negative.0498

Negative × a negative × a negative, it is going to be negative, which means that f’ is less than 0.0508

Anything to the left of -1, the function is decreasing.0522

The function is going down, it is decreasing.0526

I’m going to pick a point, I’m going to check this region right here between -1 and 0.0530

I’m going to pick -0.5.0535

F’(-0.5), this is going to be a negative number, this is going to be a negative number, this is going to be a positive number.0539

Negative × negative × positive is a positive number.0549

In other words, f’ is greater than 0, the function is increasing.0554

On this interval, from -1 to 0, the function is increasing.0559

From 0 to 5/2, I’m going to go ahead and do f’(1).0565

I’m going to just choose the point 1.0569

This is positive, this is negative, this is positive, this is negative.0574

Positive × negative × positive is a negative number.0581

Therefore, f’ is less than 0, decreasing.0585

On this interval, between 0, the function is decreasing.0590

I have to pick a point that is in this region right here, greater than 5/2.0595

I’m just going to go ahead and pick 3, f’ at 3.0601

The first one is positive, positive, positive.0606

I have positive, f’ is greater than 0, it is increasing.0611

Therefore, the intervals of increase are, we have -1 to 0 union 5/2 to +infinity.0620

And the intervals over which the function is decreasing is -infinity to -1 union 0 to 5/2.0639

It is at these points -1, 0, 5/2, f’ is equal to 0.0656

The slope is horizontal, decreasing, increasing.0663

Slope is 0 here, slope is 0 here, the slope is 0 here.0669

Just by looking at this, I can tell that I have a minimum, maximum, minimum.0673

Local min, local max, local min.0678

That is our next discussion.0682

Let us go ahead and take a look real quickly what this function actually looks like.0685

This is the function that we had.0689

Again, at -1, it hits a local min, 0 hit a max, and 5/2 it hits a local min.0691

The function is decreasing, it is increasing, it is decreasing, and it is increasing.0698

-infinity to -1, the function is decreasing.0706

-1 to 0, the function is increasing.0709

From 0 to 5/2, the function is decreasing.0711

From 5/2 to infinity, the function is increasing.0716

That is all that is going on here.0721

We are moving in this direction.0721

The graph is going like that because we always move from lower to higher numbers.0724

Let us see here, in this example, we had something that look like this.0736

This procedure that we just followed, finding the critical values,0762

checking points to the left or right of it to see whether increasing or decreasing, plugging values of the first derivative.0768

That is what we are checking.0773

This procedure also tells us, based on what we saw from the example.0775

Also tells us which critical values these are local maxes and local mins.0779

If you go from decreasing to increasing, you are going to hit a local min.0804

If you go from increasing to decreasing, you are going to hit a local max.0808

Increasing to increasing, local min.0811

This is the basis of the first derivative test.0816

It seems like it is a little pointless to actually, like formalize a state, the first derivative test, increasing/decreasing test.0821

These are things that we just sort of layoff for the sake of that being there.0830

The ideas to understand what is happening.0833

Understanding what is happening, you do not to call it a first derivative test, an increasing/decreasing test.0835

This theorem, that theorem, the names are irrelevant.0841

What is important is the concept.0846

Historically, these things evolve and given names, first derivative test.0850

It is like this, it is exactly what we just said.0855

If f’ goes from positive to negative at c, in other words, as you pass through c to the left of c, to the right of c.0859

If you are going from positive to negative, from your perspective positive to negative, then c is a local max.0875

That is here, it is going from positive to negative, the derivative, the slope.0890

The slope is positive, slope is negative, and it passes through c which is that point where the derivative equal 0.0896

It is a local max.0903

B, if f’ goes from negative to positive at c, then, c is a local min.0906

Negative slope, positive slope.0924

If it passes from negative to positive, it is a local min.0928

And part c, if f’ does not change sign as it passes from the left of c to the right of c,0934

if it goes from negative to negative, -0 negative, positive 0 positive.0946

If it does not change sign, if f’ does not change sign at c, in other words as it passes through c,0950

then c is neither a local max nor a local min.0968

Part c is a reminder that just because c is a critical point,0992

in other words a place where the derivative = 0 or the derivative fails to exist, in this particular case, if the derivative equal 0.1014

Just because c is a critical number, a critical value of the function, it does not mean that it is a local max or min.1022

Just because something is a critical value, it does not mean that it is a local max or min.1048

If it is a local max or min, it is a critical value but not the other way around.1053

You can have a critical value which is not a local max or min.1059

For example, let us take the function f(x) is equal to x³.1064

F’(x) is equal to 3x², when we set that derivative equal to 0, we see that x = 0 is a critical value.1076

If x is a critical value, let us go ahead and check.1088

To the left of it, to the right of it, to see what happens.1092

0, let us just take the point -1.1096

When I put -1 to the derivative, -1² is 1, 3 × 1 is 3.1099

This is positive, the function is increasing.1104

At 0, the slope is horizontal.1109

Let us check 1, after 1 in there, we get 3, it is positive.1114

The function is increasing again.1119

I know I already know what this graph looks like, it looks like this.1122

It looks like that so can have an increasing positive slope.1128

Horizontal slope and then positive.1132

From positive to positive, there is no local max or min here.1134

It is neither local max nor local min, it does not change sign, that is what part c says.1138

Just because you have a critical value, it does not mean that it is a local max or min.1143

Let us go ahead and do an example here.1153

Find the local maxima and minima of the function f(x) = x + 9 sin 2x/ the interval –π to π.1160

Now, we want to find the local maxes and mins.1170

We are going to find the critical values, we are going to check points to the left and right of them1175

to see what they are and to see whether things change sign, as you pass through those critical points.1180

Let us go ahead and do that.1188

First thing we want to do is find the critical values.1190

Let us go ahead and find f’(x).1194

F’(x) is equal to 1 + 18 × cos(2x), we are going to set that equal to 0.1198

We have 18 × cos(2x) = -1.1214

We have cos(2x) = -1/18.1220

I’m going to set 2x equal to a.1225

What I actually have is the cos(a) is equal to -1/18.1230

A is equal to the inv cos(-1/18).1245

The inv cos(1/18), just the numerical part is going to equal 1.515 rad, it is an angle.1256

However, it is negative.1273

Because it is negative, we are dealing with an angle in the 3rd quadrant and the 4th quadrant.1277

Therefore, what I’m actually going to get is a, the angle a that I’m looking for which is this angle and that angle, is 1.515 rad.1287

That is actually this reference angle right here.1299

A itself, this angle, the one from the positive x axis, it is actually equal to π - 1.515.1302

That is equal to 1.627.1312

A is also equal to this angle which is π + 1.515 which is equal to 4.657.1316

Those are my two values of a.1328

In other words, when I take the inv cos of that, I’m getting this and this.1330

I have just gone through a more basic process and I found the actual reference angle.1337

And then, because I know that it has to be negative, I know that is going to be the 3rd and 4th quadrant.1345

I have just taken the 180 - the reference angle 180 + the reference angle to get my values of a.1349

Because we are talking about a trigonometric function,1359

a is actually equal to 1.627 + multiples of 2 π and a is equal to 4.657 + multiples of 2 π.1362

But a is equal to 2x, and it is x that we are interested in.1378

Therefore, I have got 2x is equal to 1.627 + n 2 π.1385

I have 2x is equal to 4.657 + n 2 π.1401

This is going through these, I mean all of these should be familiar to you from trigonometry, from pre-calculus.1407

I find that it is nice to go through the process itself.1414

This was the calculus part, this is just the trigonometry part.1418

Therefore, x is equal to 0. 814 + n π and x is equal to 2.329 + n π.1423

The possible values, let us list the first couple of possible values.1439

If n is 0, n is 1, n is 2, do it that way.1444

This is going to be + or -.1451

We have 0.814, we have 3.96, we have -2.33.1454

Here we have x = 2.33, - 0.814, 5.47.1465

This is etc, etc.1482

Let me rewrite these values for the x that I got.1491

X is equal to 0. 814, 2.33.1494

Let us do +, it is going to be 3.96, -2.33, etc.1505

X = 2.33, - 0.814, 5.47, etc.1514

The only values that are in –π to π are that one, that one, and that one.1526

X = -2.33, -0.814, + 0.814, and +2.33.1551

These are my critical values, this is what I have found here.1574

I have got -2.33, -0.814, + 0.814, 2.33.1578

-2.33, -0.814, 0.814, and 2.33.1587

I have to check this interval, this interval, this interval, and this interval.1597

I have to find values that I put into the derivative f’ to check to see if they are positive or negative.1604

To see if they are increasing or decreasing.1611

Let us go ahead and do that.1615

Let us go to the next page here and write.1620

We have f’(x), we set it is equal to 1 + 18 × cos(2x).1627

We have this number line that we were going to check.1638

The one point, one point, one point, one point.1641

This was our -2.33, -0.814, 0.814, and 2.33.1645

I need to check a point over to the left of this.1657

I’m going to check f’(-3).1660

When I do that and I put it in, I end up with 18.28.1665

This is greater than 0, it is increasing.1677

It is increasing on that interval.1681

I’m going to check something here.1685

I’m going to use -1.1688

F’(-1), it ends up being -6.49, it is less than 0.1690

It is decreasing, decreasing on that interval.1696

That makes -2.33 a local max.1702

I’m going to check a point in between the -0.814 and 0.814.1708

I guess we can actually use 0, it is probably the easiest.1727

I, actually, turns out I did not use 0 but we can go ahead and do it here.1731

F’(0), the cos(2) × 0 is 0, the cos(0) is 1.1735

18 × 1 is 18, 18 + 1 is 19.1743

It is 19, it is greater than 0, it is increasing on this interval.1746

That makes -0.814 a local min.1751

I go ahead and check another point.1759

Now between here and here, I’m going to use the point 1.1760

F’(1) is going to equal -6.49, less than 0, decreasing.1765

It is decreasing on that interval which makes 0.814 a local max.1773

If I check f(3), it is going to end up being 18.28 which is greater than 0, which means that it is increasing.1778

That makes 2.33 a local min.1788

There you go, -2.33 is a local max, -0.814 local min, using this increasing/decreasing, instead of positive negative.1793

I think it is better because you can actually see pictorially going up, coming down.1813

You can see the max and min.1818

That is a local min, and then 0.814 is another local max.1819

2.33 happens to be another local min.1828

Take the function, find the critical values, examine points to the left and right of those critical values1836

to see if your derivative is positive or negative.1843

If it is positive, the function is increasing.1846

If it is negative, the function is decreasing.1848

If the function passes from positive to negative, from your perspective, it is a local max,1851

If it passes from negative to positive, it is a local min.1858

Let us go ahead and take a look at what it looks like.1865

π is somewhere around like there, π is somewhere around here.1871

Here we have our -2.33 local max, -0.814 local min, 0.814 local max, 2.33 local min.1877

There we go, that is it, nice and straightforward, nothing strange about it.1890

Let us go ahead and do another example.1899

Find the local maxima and minima of the function f(x) = x³ + 4x²/ x² + 2, rational function.1901

Let us go ahead and find the derivative of this.1910

F’(x), it is going to be this × the derivative of that - that × the derivative this/ this².1913

We are going to get x² + 1.1922

I think I used slightly different set of numbers here.1930

Let me just double check to make that I’m not going to end up with some values that are different than normal.1944

This 3x², 3x².1953

You know what, I actually ended up doing my arithmetic wrong on this one.1955

That is not a problem, we will just go ahead and do it.1959

We will finish it off based on what is written here.1962

It is going to be x² + 2 × the derivative of this which is going to be 3x² + 8x – the numerator × the derivative of the denominator.1965

It is –x³ + 4x² × 2x, all divided by x + 2², the denominator².1982

We are going to set that equal to 0.2001

I'm going to go ahead and multiply it out.2005

X² and 3x², this × that is going to be 3x⁴.2010

X² × 8x is going to be + 8x³.2020

2 × 3x² is going to be 6x².2028

2 × 8x is going to be 16x – x³ × 2x.2036

It is going to be -2x⁴.2045

4x² × 2x is going to be -8x³/ x + 2² equal to 0.2050

The 8x³ and 8x³ go away.2068

I have 3x⁴ – 2x⁴, that takes care of the x⁴.2071

I have a 6x², that takes care of that.2078

I have + 16x/ x + 2² = 0.2090

The denominator goes away, it is just x⁴ + 6x² + 16x is equal to 0.2105

When I factor out the x, I’m going to get x × x³ + 6x + 16 is equal to 0.2116

I know that one of the values is equal to 0.2129

Again, because I did my arithmetic on my piece of paper here incorrectly,2131

I used an x² + 1 instead of x² + 2, basically, we just have to find the other root of whatever it happens to be.2139

Let us say it happens to be, my guess is only one root here.2151

Let us just call it x = a.2156

Let us just call it a.2161

When we have that, we are going to end up doing this.2164

We have x = 0 and x = a are our two critical points.2170

We are going to end up with 0 and we are going to end up with a.2174

We are going to have to check a value in this interval, a value in this interval,2178

and a value in this interval and plug in the derivative to see what they are.2183

We can go ahead and do this interval right here, it is not a problem.2194

Let us check, I’m not even sure whether a is positive or negative.2197

You know what, let us take that back and let us do it this way.2204

We said that, we have x = 0 and we have x is equal to some number a.2212

A itself can be a positive or negative root.2221

I’m not exactly sure which it is.2223

If a is a positive root, what I have is 0 and I have a over here to the right of 0.2227

I have to check a value there, check a value there, and check value there,2232

to see whether the function is increasing, decreasing, or otherwise.2236

If it is negative number, and of course 0 is here, and is going to be to the left of 0.2243

I have to check a point here, check a point here, and check a point here.2249

Since I do not know what a is because of my arithmetic, I cannot go ahead and at the very least,2255

check what is to the right and left of 0.2262

Let me go ahead and do that at least, that should not be a problem.2265

Let me go ahead and take a negative one and a positive one.2270

If I do f’(-1), what did we say what f’ was, let us make sure we have that right.2274

Let us go ahead and write what f’ is.2288

We have an f’(x) is equal to x⁴, we said we have x⁴ + 3x² + 8x/ x + 2².2290

This was our f’.2313

Again, I do not know what a is because of the arithmetic issue but I can at least check to the left of 0 and to the right of 0.2317

Let us take f’(-1).2323

When I put -1 into the f’, I’m going to end up with, this is going to be 1,2329

this is going to be +3, -8, this is always going to be + because it is square.2336

1 + 3 is 4, 4 – a, I’m going to end up with which is negative number/ a positive number which is a negative number.2352

F’ at -1 is going to be less than 0, this is going to be decreasing.2360

To the left, it is going to be decreasing.2367

Let us go ahead and check f’ at 1.2370

It is going to be 1, this is positive, this is positive, this is positive.2374

This is going to be 0, it is going to be increasing.2377

At the very least, I can tell you that my 0 is going to be a local minimum.2381

Because I do not what a is, I'm not exactly sure.2393

But again, if you go back and find the root of that particular function that we had,2395

you will find what a is an it will tell you whether it is either positive or negative.2400

And then, you can check points to the left or right of that.2405

My guess is that no matter what a is positive or negative, it is going to end up being actually a local max.2408

It will probably be increasing here and decreasing here.2415

Or it is going to be increasing to the left and decreasing.2419

It is going to make a local max.2423

My guess is that it actually going to be a local max.2425

The only other option is that it ends up not changing sign at all.2428

It ends up going from negative to negative or positive to positive.2431

In which case, it is neither a local max nor a local min.2435

But I can say with certainty to 0 is a local min.2438

There you have it and I apologize for the arithmetic error.2442

Let us go ahead and take a look at the graph as is, to confirm whether it is local max.2449

Yes, it is going to be a local max.2454

It is increasing, it looks like the root was somewhere around -2.2456

Therefore, we know 0 was the local min and sure enough this is the actual function, the correct original function.2465

Yes, it looks like our root is going to be here, the derivative is going to be 0 at, it looks like somewhere like -1., something like that.2476

There you have it, local max local min.2487

Thank you so much for joining us here at www.educator.com.2491

We will see you next time, bye.2493