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

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

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### Example Problems III

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
- Example I: Intervals, Local Maxes & Mins, Inflection Points, Intervals of Concavity, and Asymptotes
- Example II: Intervals, Local Maxes & Mins, Inflection Points, Intervals of Concavity, and Asymptotes
- Example III: Cubic Equation f(x) = Ax³ + Bx² + Cx + D
- Example IV: Intervals, Local Maxes & Mins, Inflection Points, Intervals of Concavity, and Asymptotes

- Intro 0:00
- Example I: Intervals, Local Maxes & Mins, Inflection Points, Intervals of Concavity, and Asymptotes 0:11
- Example II: Intervals, Local Maxes & Mins, Inflection Points, Intervals of Concavity, and Asymptotes 21:24
- Example III: Cubic Equation f(x) = Ax³ + Bx² + Cx + D 37:56
- Example IV: Intervals, Local Maxes & Mins, Inflection Points, Intervals of Concavity, and Asymptotes 46:19

### AP Calculus AB Online Prep Course

### Transcription: Example Problems III

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

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

*Let us jump right on in.*0010

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

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

*Use this information to graph the function.*0026

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

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

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

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

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

*Let us see what we can find.*0049

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

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

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

*The first derivative gives us a lot of information.*0063

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

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

*We are going to have x - 4² × the derivative of the top which is going to be 2x - x² × the derivative of the bottom*0072

*which is going to be 2 × x – 4¹ × the derivative of what is inside, 1.*0085

*All of that is going to be / x – 4⁴.*0093

*I’m going to go ahead and pull out the x – 4.*0102

*The numerator, I’m going to write it as 2 x × x - 4 - 2x²/ x – 4⁴.*0106

*I end up getting, this cancels this, turning it into a 3.*0122

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

*I get 2x² - 8x - 2x²/ x - 4³.*0132

*The 2x² cancel leaving me -8x/ x – 4³.*0144

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

*We set that equal to 0 and what we end up getting is -8x = 0.*0157

*Therefore, x is equal to 0.*0164

*0 is our critical point.*0167

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

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

*We have to watch out for those.*0203

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

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

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

*The denominator was x – 4.*0218

*Clearly, we cannot have a denominator equal to 0.*0220

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

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

*4 is also going to be a critical point.*0230

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

*Something we have to elucidate by other means.*0238

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

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

*Here we have a vertical asymptote at x = 4.*0261

*x = 4 is also a critical point.*0278

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

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

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

*We are dealing with f’ and we have our two critical points, we have 0 and we have 4.*0308

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

*F’(x) we said is equal to -8x/ x – 4³.*0327

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

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

*This is a negative number which means decreasing.*0352

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

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

*This is also a negative number.*0370

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

*Positive, we have an increasing.*0379

*It is going to be increasing between 0 and 4.*0382

*Let us go ahead and check a point to the right of 4.*0387

*I’m going to go ahead and try 5.*0390

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

*It is going to be decreasing on that interval.*0398

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

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

*That means that 0 is a local min.*0412

*All of this gives us.*0416

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

*There is an asymptote, a vertical asymptote at x = 4.*0430

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

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

*This is clearly a vertical asymptote.*0452

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

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

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

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

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

*F(0) is equal to 0, the local min, the actual point is, the local min occurs at 0,0.*0485

*We have partial graph.*0501

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

*We know that we have a local min at 0,0.*0508

*We know the graph looks something like this.*0510

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

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

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

*It gives me a lot of information.*0528

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

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

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

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

*F(x) that is equal to x²/ x - 4².*0573

*If I expand the bottom, I get x²/ x² - 8x + 16.*0583

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

*x gets really small, negative infinity.*0599

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

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

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

*What you end up with is x²/ x² which is just 1.*0615

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

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

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

*F(x), we said is equal to x²/ x² - 8x + 16.*0667

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

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

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

*As I take x to + or –infinity, this goes to 0, this term goes to 0, and what I’m left with is 1.*0698

*This is my horizontal asymptote.*0707

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

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

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

*We know that we have a 1, 2, 3, 4.*0733

*We have a vertical asymptote at 4.*0735

*We know at 1 we have a horizontal asymptote.*0739

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

*We know that it comes probably like this.*0752

*And this one probably goes like that.*0755

*More than likely that is what is happening.*0759

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

*That is what is happening.*0772

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

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

*We said that f’(x) was equal to -8x/ x – 4³.*0812

*Therefore, f”(x) is equal to x – 4³ × -8 - 8x × 3 × x - 4² × 1/ x – 4⁶.*0821

*I’m going to go ahead and factor out an x - 4².*0853

*I’m going to get -8 × x – 4.*0858

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

*All of that is going to be over x – 4⁶.*0871

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

*I apologize for actually going through this process.*0881

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

*This knocks out that, turning that into 4.*0889

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

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

*If my arithmetic is correct, + 32/ x – 4⁴.*0914

*This is what I’m going to set equal to 0 and this gives me 16x + 32 = 0, 16x = -32.*0921

*I get x = -2.*0934

*This is my inflection point.*0937

*We have already seen the graph of this thing.*0955

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

*This is our vertical asymptote.*0963

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

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

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

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

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

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

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

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

*But, let us run through analytically.*1024

*We went ahead and we found that this point of inflection, one of the points of inflection is -2.*1044

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

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

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

*It is a point that is not differentiable.*1059

*Something happens to the left of that point.*1061

*Something happens to the right of that point.*1063

*I have to check 3 regions.*1065

*I’m checking the second derivative.*1068

*The second derivative is equal to 16x + 32/ x – 4⁴.*1072

*Let us go ahead and find f”(-3).*1084

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

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

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

*This means it is concave up, second derivative.*1117

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

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

*This means concave up.*1133

*My intervals of concavity, concave down, like I said –infinity to -2.*1137

*Concave up from -2 to 4 union.*1144

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

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

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

*I have got 1, 2, 3, 4.*1178

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

*I know my graph goes something like this.*1190

*I know it goes like this.*1196

*Let us find what that point is.*1199

*We said that the point of inflection, we said that x = -2.*1201

*When I take f(-2), I end up with 1/9.*1207

*This point right here is the point -2, 1/9.*1214

*The actual coordinates of that point.*1220

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

*This is what the function actually looks like.*1230

*You can see we have a vertical asymptote at 4.*1232

*We have a horizontal asymptote at 1.*1235

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

*There you go, all of the information.*1244

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

*Notice, the first derivative did not give me a critical value of 4.*1258

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

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

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

*Let us go ahead and try another one here.*1284

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

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

*Let us do the asymptotes first.*1298

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

*We are looking at a logarithm function.*1321

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

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

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

*Let us take a look at the domain.*1342

*Let us look at the domain.*1353

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

*This part, the argument of the logarithm function, this has to be greater than 0.*1364

*Let us set it greater than 0.*1377

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

*I have 2 greater than ln x.*1385

*I'm going to exponentiate both sides.*1389

*I get x is less than e².*1391

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

*The domain is from 0 to e² or 0 to about 7.4.*1408

*This is not 1, this is a comma, 0 to about 7.4.*1418

*Now we have our domain.*1423

*Now that we have the domain, let us ask ourselves what happens as x goes to 0 and as x gets close to 7.4.*1425

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

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

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

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

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

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

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

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

*Ln(2) – ln(x) goes to +infinity.*1564

*This is ln(2) – ln(x).*1571

*I'm asking myself, what happens to the function as x gets close to 0.*1574

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

*The –ln(x) is +infinity.*1583

*2 – ln x goes to +infinity.*1585

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

*The function actually goes towards infinity, as we get close to 0.*1593

*That makes 0 a vertical asymptote.*1597

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

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

*I will draw that in just a minute.*1611

*We are just working out the analytics.*1612

*As x approaches e², ln(x) is going to approach 2.*1614

*2 - ln(x) is going to approach 0.*1632

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

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

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

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

*That is a vertical asymptote.*1679

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

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

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

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

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

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

*Let us see what is next.*1712

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

*F(x) is ln of 2 – ln x is equal to 0.*1733

*I’m going to exponentiate both sides.*1739

*When I do that, I'm left with 2 – ln x e⁰ is equal to 1.*1742

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

*I end up with ln x is equal to 1.*1753

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

*It crosses the axis at x = e.*1764

*The graph crosses at x is equal to e.*1773

*Our graph is starting to come together nicely here.*1784

*This is x = e² which is 7.4.*1792

*E is somewhere around there.*1801

*I know the graph looks something like, this is e, 2.718.*1804

*The graph is going to look something like this.*1813

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

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

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

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

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

*We can go ahead and find that.*1839

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

*Let us confirm this analytically.*1862

*Let us confirm this analytically.*1865

*It is exhausting writing all this out.*1882

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

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

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

*That gives us -1/ x × 2 - ln(x).*1912

*We set that first derivative = 0.*1925

*This means -1 is equal to 0 because the only way this is equal to 0, is if the numerator is 0.*1929

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

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

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

*No solution, no local max or min.*1950

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

*Let us go ahead and find the second derivative.*1971

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

*This is going to be product rule.*2007

*It is going to be this × the derivative of this, which is going to be -1/ x × the derivative of that*2009

*+ 2 - ln x × the derivative of that which is 1.*2028

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

*I’m going to let you work out.*2050

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

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

*We set that equal to 0.*2088

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

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

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

*Now we have our final graph.*2130

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

*1 to 2.718, this is there.*2148

*Our graph is going to look something like this.*2152

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

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

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

*Final graph looks something like this.*2178

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

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

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

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

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

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

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

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

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

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

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

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

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

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

*that achieves a local max at -3, 4 and a local min at 4, -2.*2283

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

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

*I already know what f(-3) is, it is 4.*2303

*I already know what f(4) is, it is -2.*2307

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

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

*I know that equals 4.*2330

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

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

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

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

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

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

*This is our first equation.*2371

*I have a second equation, I know what f(4) is, it is equal to -2.*2374

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

*I know that = -2.*2392

*When I work this out, I get 64a + 16b + 4c + d = -2.*2394

*This is my second equation.*2405

*We also know something else.*2410

*It tells me that at these points, it achieves a local max and a local min which means the derivative of this, at these points, is equal to 0.*2411

*Local max/local min, the derivative there at those points is equal to 0.*2423

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

*I know that f’ at -3 which is going to equal 3a × -3² + 2b × -3 + c, I know that it equal 0.*2441

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

*Its derivative is also equal to 0.*2462

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

*I do the same thing for f’ at 4, that is going to equal 3a × 4² + 2b × 4 + c, I know that equal 0.*2478

*Here I get 48a + 8b + c, that is equal to 0.*2491

*I have my 4 equations and 4 unknowns.*2502

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

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

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

*Do you remember back in pre-calculus, when you guys were doing row reduction*2526

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

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

*I will show you in just a minute.*2550

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

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

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

*Here is what it actually looks like.*2568

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

*Our equations, we just put the coefficients in.*2604

*We end up with -27, 9, -3, 1, is equal to 4.*2606

*We have 64a, 16b, 4c, + d, = -2.*2614

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

*And of course we have 48 and 8 and 1 and 0 and 0.*2632

*This matrix, the first column is the a.*2637

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

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

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

*You get the answer but is not unique.*2661

*Reduced row-echelon is always unique.*2663

*There is only one place it always ends up.*2665

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

*1, 0, 0, 12/343, 0, 1, 0, 0, -18/343, 0, 0, 1, 0, -432/343, and 0, 0, 0, 1, 562/343.*2673

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

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

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

*We get f(x) is equal to 12/ 343 x³ - 18/ 343 x² – 432/ 343 x + 562/343.*2725

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

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

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

*Let us do one more function here.*2771

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

*Use it to graph the function.*2790

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

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

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

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

*which gives 2/ x¹/3 = 1, which gives us x¹/3 is equal to 2.*2836

*Therefore, when I cube both sides, I get x is equal to 8.*2852

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

*That is where the derivative is equal to 0.*2873

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

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

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

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

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

*Notice that f is defined at 0.*2916

*Notice that f(x) is defined at 0, the domain is all real numbers including 0.*2919

*When I put 0 in here, I get f = 0.*2929

*It is defined at 0, in other words f(0) = 0.*2933

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

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

*The value of the function does.*2974

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

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

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

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

*When we check our number line, we have to include 0 and 8.*3018

*Let us go ahead and do that.*3023

*0 and 8 are critical points.*3026

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

*F’(x) is equal to 2/ x¹/3 – 1.*3046

*When I check a point to the left here, I’m going to go ahead and check -1.*3053

*When I check -1, I'm going to get 2/-1 -1 which is equal to -3.*3058

*It is going to be decreasing.*3072

*It is going to be decreasing there.*3073

*I’m going to go ahead and check 1.*3076

*When I check 1, I get 2/1 - 1 is equal to 1.*3080

*It is increasing here.*3084

*However, we just said it is actually not differentiable at 0.*3089

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

*It is actually it cusp but not differentiable at 0.*3096

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

*The function is decreasing to the left of 0.*3127

*It is increasing but it is not differentiable there.*3134

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

*That is what the graph looks like there.*3142

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

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

*I’m going to pick the number 27.*3166

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

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

*Therefore, it is actually decreasing there.*3181

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

*It is what we found by setting the derivative, first derivative equal to 0.*3189

*This is definitely a local max.*3193

*We have a local max at x = 8.*3199

*That is a pretty wacky looking 8.*3209

*We have got 8 there.*3213

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

*The function is increasing from 0 to 8.*3231

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

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

*We said that f’(x) is equal to 2x⁻¹/3 – 1.*3257

*We want to take the second derivative.*3266

*F”(x) is equal to -2/3 x ^- 4/3 -1.*3268

*We are going to set that equal to 0.*3281

*We are going to get -2/ 3x⁴/3 - 1 is equal to 0.*3285

*We are going to get -2/3 x⁻⁴/3 = 1.*3295

*Rearranging, we are going to end up with -2 is equal to 3x⁴/3 - 2/3 = x⁴/3.*3304

*This is -2/3 = x¹/3⁴.*3321

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

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

*This is no solution.*3339

*No solution so there are no points of inflection.*3343

*Let us see what we have got here.*3358

*F” at x is equal to -2/ 3x⁴/3 – 1.*3362

*This is actually less than 0 for all x.*3375

*The graph is concave down everywhere.*3381

*F(x) is concave down everywhere.*3386

*Let us see, where else f(x) actually equal 0.*3400

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

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

*Let us see if it actually hits someplace else.*3409

*Let us see where else f(x) = 0, beside x = 0.*3415

*F(x) = 3x²/3 - x is equal to 0.*3431

*I’m going to go ahead and factor out the x²/3.*3444

*I get x²/3 × 3 – x¹/3, that is equal to 0.*3447

*I'm going to get x²/3 = 0 that gives me x = 0.*3456

*I already know that one.*3461

*I'm going to get 3 - x¹/3 = 0, x¹/3 = 3.*3463

*Therefore, when I cube both sides, I get x = 27.*3471

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

*I have all of the information that I need.*3483

*My graph is concave down everywhere.*3489

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

*At 8, it hits a local max of 4.*3503

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

*That is my graph.*3515

*Let us see what it looks like.*3517

*Yes, that is exactly right.*3519

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

*There is a cusp there.*3524

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

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

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

*We will see you next time, bye.*3540

1 answer

Last reply by: Professor Hovasapian

Mon Apr 23, 2018 3:30 AM

Post by Patricia Xiang on April 19 at 07:33:17 PM

Hello professor,

In example III why it is not a local min where there’s a cusp? F(0) is the smallest in the immediate neighborhood. Is it because f’(x) is not differentiable there? If so, why f(x) can has a local min/max where there’s a jump discontinuity as showed in previous videos?

1 answer

Last reply by: Professor Hovasapian

Mon Apr 23, 2018 3:21 AM

Post by Patricia Xiang on April 19 at 07:09:31 PM

Hello professor,

Is Reduced Echelon form required in AP calculus? Or it is just helpful solving the problems? And if it helps a lot, should I acquire how to do it by hands or merely how to do it on the calculator?

Have a great day.

2 answers

Last reply by: Richard Kennesson

Tue Jul 26, 2016 1:21 AM

Post by Richard Kennesson on July 24, 2016

Hello Professor,

I think you may have made an error at 54:40 or maybe I didn't get what you did there.

f(x) = 2x^(-1/3) - 1

f'' is the derivative f' and you just dropped down the 1. I thought that the derivative of a constant is 0.

If I'm right you'd have

f''(x) = -2/3x^(-4/3) = 0

then I multiply by the the denominator 3x^(-4/3) and then get, -2 = 0

Either way, it all worked out in the end as there is no solution.

Have a great day.