For more information, please see full course syllabus of AP Calculus AB

For more information, please see full course syllabus of AP Calculus AB

### 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

### AP Calculus AB Online Prep Course

### 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 have *0024

*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

*You already know this geometrically.*0180

*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 them*1175

*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

*Just put -3 into here, make sure that your calculator is in radian mode because these are all radians, real numbers, radians.*1668

*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 values*1836

*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

1 answer

Last reply by: Professor Hovasapian

Thu Apr 7, 2016 1:37 AM

Post by Acme Wang on April 5 at 10:58:11 PM

Hi 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