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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## Table of Contents

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### Example Problems 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
- Example I: Intervals, Local Maxes & Mins
- Example II: Intervals, Local Maxes & Mins
- Example III: Intervals, Local Maxes & Mins, and Inflection Points
- Example IV: Intervals, Local Maxes & Mins, Inflection Points, and Intervals of Concavity
- Example V: Intervals, Local Maxes & Mins, Inflection Points, and Intervals of Concavity

- Intro 0:00
- Example I: Intervals, Local Maxes & Mins 0:26
- Example II: Intervals, Local Maxes & Mins 5:05
- Example III: Intervals, Local Maxes & Mins, and Inflection Points 13:40
- Example IV: Intervals, Local Maxes & Mins, Inflection Points, and Intervals of Concavity 23:02
- Example V: Intervals, Local Maxes & Mins, Inflection Points, and Intervals of Concavity 34:36

### AP Calculus AB Online Prep Course

### Transcription: Example Problems I

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

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

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

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

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

*Let us jump right on in.*0024

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

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

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

*which graph they actually gave you.*0046

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

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

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

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

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

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

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

*We know that increasing is where the first derivative is greater than 0.*0080

*Decreasing is where the first derivative is less than 0.*0085

*This is the first derivative.*0088

*At this point, between this point and this point, the original function is increasing*0090

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

*To the left of this point which is -1, the first derivative which is this graph is negative, it is below the x axis*0100

*Pass the point 4, it is also negative.*0110

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

*Let me go ahead and work in blue here.*0130

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

*Our intervals of increase, it is when f’ is greater than 0.*0137

*Our interval of increase is from, this is -1.*0148

*I think I read the graph wrong.*0161

*It looks like this is -1 and this looks like this 4, from -1 to 4.*0166

*The interval of decrease is when the first derivative is actually less than 0.*0177

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

*These points, where the first derivative actually equal 0*0211

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

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

*F’(x) = 0 at -1 and 4.*0225

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

*Let us see which one is which.*0244

*At -1, we have decreasing, increasing.*0247

*You have a local min at x = -1.*0253

*Over here, it is increasing/decreasing.*0262

*You have a local max at x = 4.*0267

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

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

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

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

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

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

*This is the process, just be extra careful.*0297

*You have to be very vigilant here.*0300

*Let us do another example, same type.*0305

*The following is a graph of the first derivative.*0307

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

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

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

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

*A lot of information that they require here.*0333

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

*Once again, increasing is where f’ is greater than 0, decreasing is where f’ is less than 0.*0339

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

*Therefore, the function is increasing.*0361

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

*We can read off as, let us just call it 0.9.*0380

*This should be an open.*0385

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

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

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

*Therefore, our interval of decrease, it is negative from -infinity all the way to this -1.8.*0411

*It is also below between this 0.9 and this 2.2.*0422

*These are our intervals of increase and decrease.*0431

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

*Therefore, that makes this point right here.*0445

*I will say the local maxes are going to be this one which is at x = 0.9.*0456

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

*We have -1.8 and our 2.2.*0479

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

*What we had here is our first derivative graph.*0492

*This does not describe the graph.*0497

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

*Increasing/decreasing, positive, negative.*0507

*Let us move on to our next page here.*0512

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

*We have taken care of the increasing/decreasing.*0517

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

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

*We want to talk about some inflection points.*0535

*Inflection points, we said that inflection points are points where f” is equal to 0.*0538

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

*F” is the slope of this graph.*0571

*I hope that make sense.*0584

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

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

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

*Therefore, at x equals -1.25, -0.6, 0, it looks like about 1.6.*0613

*At these points, the second derivative equal 0.*0633

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

*These are points of inflection.*0642

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

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

*To the left or right of those points.*0660

*I have got -1.25, -0.6, 0, and 1.6.*0662

*I need to check where the second derivative.*0674

*We are dealing with the second derivative here.*0677

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

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

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

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

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

*This is where the confusion lies in.*0715

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

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

*Let us write concave up, concave down.*0733

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

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

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

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

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

*We are going to be concave down from -1.25 all the way to -0.6.*0777

*And union where it is negative, it is going to be from 0 all the way to +1.6.*0786

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

*There you go, you got your points of inflection, you have your intervals of concavity*0803

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

*It looks like we have everything.*0815

*Let us try another one.*0820

*The following is a graph of the first derivative.*0823

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

*On which intervals of the function increasing/decreasing?*0832

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

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

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

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

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

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

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

*F’ is negative, it is decreasing.*0874

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

*Trust the math, do not trust your instinct.*0883

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

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

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

*Our intervals of decrease, we have -infinity to 2.8.*0908

*-infinity to 2.8.*0914

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

*Here it is increasing, we are going to have a local min at 2.8.*0925

*That is it, local min at x = 2.8.*0931

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

*In this case, there is no local max.*0947

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

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

*Inflection points, we said inflection points are points where the second derivative is equal to 0.*0961

*Inflection points, there are places where f” is actually equal to 0.*0974

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

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

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

*That right there and about right there.*1015

*Points of inflection are going to be x = -0.9 and x = roughly 1.5.*1020

*Once again, we have our -0.9, we have our 1.5.*1045

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

*We have -0.9 and we have our 1.5.*1058

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

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

*Therefore, the slope is concave up.*1077

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

*The slope, remember this is f’.*1093

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

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

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

*We have concave down from -0.9 all the way to 1.5.*1126

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

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

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

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

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

*Let us double check.*1178

*Let us go back to blue here.*1180

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

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

*Yes, decrease, decrease.*1215

*Yes, from 2.8 onward.*1217

*That checks out, very nice.*1221

*We said that it decreased from -infinity all the way to this 2.8.*1225

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

*the function is decreasing, decreasing, decreasing, decreasing, until it hits 2.8.*1235

*Yes, that checks out.*1239

*We said that we had a local min at 2.8.*1242

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

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

*Inflection points, we said that we have an inflection point at x = -0.9.*1261

*We also said we had one at x = 1.5.*1267

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

*Yes, there it is, our blue.*1279

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

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

*There we go at 1.5, at this point.*1289

*It goes from concave down to concave up.*1291

*Yes, these two check out.*1295

*And then, we had our intervals of concavity.*1298

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

*Let us double check.*1311

*Concave up, concave up from -infinity to -0.9.*1312

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

*Yes, that checks out.*1321

*The last thing we want to double check is our concave down, from negative 0.9 all the way to 1.5.*1323

*Yes, from -0.9 all the way to about 1.5.*1332

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

*Here, we see them all together.*1342

*The graph they gave us was this one.*1344

*This is the graph they gave us.*1350

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

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

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

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

*You have to keep track of every little thing.*1376

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

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

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

*Use this information to actually draw the graph.*1400

*Let us go ahead and do it.*1402

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

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

*I got x² + 5 × 20x - 10x².*1420

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

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

*All of that is equal to, when I multiply it out, I get 20x³ + 100x - 20x³/ x² + 5².*1439

*Those cancel, I'm left with 100x/ x² + 5².*1456

*This is my first derivative, I’m going to set that first derivative equal to 0.*1470

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

*It is only 0 when the numerator is 0.*1478

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

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

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

*Let us just go ahead and check -1.*1515

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

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

*This is increasing.*1537

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

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

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

*We have our decreasing interval from -infinity to 0.*1559

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

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

*We know that there is a local min at x = 0.*1576

*This critical value is a local min.*1584

*Let us go ahead and find f”(x).*1588

*Let us go back to blue.*1590

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

*We are left with this thing.*1603

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

*We get x² + 5² × 100 - 100x × 2 × x² + 5 × 2x/ x² + 5⁴.*1609

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

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

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

*I have factored this out as here.*1659

*I get x² + 5⁴, this cancels with one of these, leaving 3.*1661

*I need that equal to 0.*1670

*We have f”(x) is equal to 100x² + 500 – 400x²/ x² + 5³.*1677

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

*I have got 300x² is equal to 500.*1712

*I get x² is equal to 5/3 which gives me the x is equal to + or - 5/3, which is approximately equal to + or -1.3.*1717

*I’m going to set up my -1.3, +1.3.*1731

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

*Let us go ahead and actually do this one.*1747

*F”(x)is equal to -300x² + 500/ x² + 5³.*1750

*I’m going to check the point -2.*1766

*When I check the point -2, I'm going to get a negative number/ a positive number*1769

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

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

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

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

*Let us move on to the next page here.*1811

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

*I have got concave up from -1.3 all the way to 1.3.*1826

*Let us see what happens.*1835

*We are dealing with a rational function.*1839

*We have to check the asymptotic behavior.*1841

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

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

*F(x) is equal to 10x²/ x² + 5.*1866

*As x goes to infinity, this drops out.*1874

*The x² cancel and you are left with f(x) ends up approaching 10.*1878

*10 is the horizontal asymptote of this function.*1884

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

*We have a local min at 0.*1898

*When we check f (0), it equal 0.*1905

*We are looking at the point 0,0.*1908

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

*We know the function is actually decreasing from -infinity to 0.*1918

*We know that we have points of inflection at x = -1.3.*1926

*When I check the y value, I get 2.53.*1941

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

*My other one is at +1.3 and the y value is 2.53.*1955

*1.3 and 2.53 is my other point of inflection.*1963

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

*I am concave up from -1.3 all the way to +1.3.*1981

*I have a horizontal asymptote at y is equal to 10.*1990

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

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

*Here is my other point of inflection.*2024

*It is decreasing all the way to 0, increasing past 0.*2028

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

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

*It is concave down again, notice that it approaches 10.*2047

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

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

*Let us go ahead and try another example here.*2073

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

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

*Use this information to draw the graph.*2085

*Let us go ahead and do it.*2088

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

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

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

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

*We want to set that equal to 0.*2129

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

*I get x × 1 + 2 ln of 1/3 x that is equal to 0, that gives me x = 0.*2134

*It gives me 1 + 2 ln of 1/3 x is equal to 0.*2153

*0 is one of my critical points.*2160

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

*I get 2 × ln of 1/3 x is equal to -1.*2169

*Ln of 1/3 x is equal to -1/2.*2177

*I exponentiate both sides, I get 1/3 x = e ^-½.*2181

*I get x is equal to 3 × e⁻¹/2.*2190

*X is approximately equal to 1.8.*2194

*Let us go ahead and do it.*2202

*These give me my critical points.*2204

*I have 0 and I have got 1.8.*2206

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

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

*Let me go ahead and write it out.*2223

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

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

*Put it into this to see what I get.*2246

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

*I do not have to check a point there.*2257

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

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

*I'm going to go ahead and check 1.*2265

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

*Therefore, it is going to be decreasing.*2274

*We are going to be decreasing on that interval.*2277

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

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

*I have a local from decreasing to increasing.*2300

*I have a local min at 1.8.*2302

*That is what this information tells me.*2305

*Let us write that down.*2307

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

*I have a local min at x = 1.8.*2324

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

*F(1.8), it gives me -1.7.*2345

*My local min is going to be the point 1.8, -1.7.*2353

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

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

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

*In order to have a local max or a local min, I have to have the point*2378

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

*Here it is only defined to the right.*2385

*The positive and negatives do not count.*2389

*Let me write that out.*2395

*We cannot say that there is a local max.*2400

*In some sense there is, but not by definition, that there is a local max at x = 0 because f is not defined for x less than 0.*2410

*In other words, our function, we know there is a local min at 1.8 and -1.7.*2435

*We know it is here.*2442

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

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

*That is all that is going on here.*2452

*Let us do point of inflection.*2454

*Let me go back to red.*2457

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

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

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

*X × 1/1/3 x × 1/3 + the nat-log of 1/3 x × 1.*2488

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

*1/3, 1/3, x, x, this becomes 1.*2508

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

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

*That is our second derivative.*2535

*We need to set that equal to 0.*2540

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

*We need to set that equal to 0.*2554

*We get 2 ln 1/3 x is equal to -3, ln of 1/3 x = -3/2.*2556

*We exponentiate both sides, we get 1/3 x is equal to e⁻³/2*2566

*which gives us x is equal to 3 × e⁻³/2, which is approximately equal to 0.7.*2581

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

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

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

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

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

*When I check the value, I will just check 0.*2626

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

*I’m not going to check 0 actually.*2649

*I will check my 0.5.*2652

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

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

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

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

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

*That takes care of that.*2692

*Now I have my intervals of concavity.*2706

*It is going to be concave down from 0 all the way to this 0.7.*2708

*Let me double check, yes.*2724

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

*Let us find where f(x) actually equal 0.*2742

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

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

*F(x) is equal to x² ln 1/3 x, we want to set that to equal to 0.*2758

*That gives us x² is equal to 0 and it also gives us ln of 1/3 x is equal to 0.*2767

*This is 0, it attaches at 0,0.*2776

*This one you get 1/3 x e⁰ is 1, that means x = 3.*2779

*It touches the x axis at 0 and 3.*2788

*Let us list what we have got.*2794

*Our root x = 0, x = 3.*2800

*We have a local min at the point .1.8, -1.7.*2809

*We have a point of inflection at 0.7, -0.7.*2819

*We are concave down from 0 to 1.4.*2828

*I’m sorry not 1.4, it is going to be 0.7.*2841

*Concave down from 0 to 0.7.*2844

*We are concave up from 0.7 to +infinity.*2847

*The function is decreasing from 0 to 1.8.*2856

*The function is increasing from 1.8 to +infinity.*2862

*When we put all of that together, local min at 1.8, -1.7.*2875

*1.8, -1.7 probably puts us right there.*2884

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

*Our graph goes something like this.*2894

*It is decreasing from 0 to 1.8, decreasing, concave down from 0 to 0.7.*2896

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

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

*Here is our root, here is our root.*2921

*Points of inflection is somewhere around there.*2925

*Local min somewhere around there.*2931

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

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

*The graph goes, passes through 3.*2941

*There you go, that is it, wonderful.*2946

*Thank you so much for joining us here at www.educator.com*2949

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

*Take care, bye.*2959

1 answer

Last reply by: Professor Hovasapian

Tue Jul 19, 2016 6:07 AM

Post by Peter Ke on July 15, 2016

For example V, how is the point of inflection is (0.7, -0.7)?

2 answers

Last reply by: Professor Hovasapian

Mon Jul 25, 2016 6:49 PM

Post by Acme Wang on April 7, 2016

Hi Professor,

I felt confused in finding the horizontal asymptote of 10x^2/(x^2+5). When x goes to infinity, would f(x) also approaches infinity since 10x^2 goes to infinity and (x^2+5) goes to infinity?

Thank you very much!

Sincerely,

Acme