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 for Max & Min

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: Identify Absolute and Local Max & Min on the Following Graph
- Example II: Sketch the Graph of a Continuous Function
- Example III: Sketch the Following Graphs
- Example IV: Find the Critical Values of f (x) = 3x⁴ - 7x³ + 4x²
- Example V: Find the Critical Values of f(x) = |2x - 5|
- Example VI: Find the Critical Values
- Example VII: Find the Critical Values f(x) = cos²(2x) on [0,2π]
- Example VIII: Find the Absolute Max & Min f(x) = 2sinx + 2cos x on [0,(π/3)]
- Example IX: Find the Absolute Max & Min f(x) = (ln(2x)) / x on [1,3]

- Intro 0:00
- Example I: Identify Absolute and Local Max & Min on the Following Graph 0:11
- Example II: Sketch the Graph of a Continuous Function 3:11
- Example III: Sketch the Following Graphs 4:40
- Example IV: Find the Critical Values of f (x) = 3x⁴ - 7x³ + 4x² 6:13
- Example V: Find the Critical Values of f(x) = |2x - 5| 8:42
- Example VI: Find the Critical Values 11:42
- Example VII: Find the Critical Values f(x) = cos²(2x) on [0,2π] 16:57
- Example VIII: Find the Absolute Max & Min f(x) = 2sinx + 2cos x on [0,(π/3)] 20:08
- Example IX: Find the Absolute Max & Min f(x) = (ln(2x)) / x on [1,3] 24:39

### AP Calculus AB Online Prep Course

### Transcription: Example Problems for Max & Min

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

*Today, we are going to do some example problems for the max and min that we discussed in the last lesson.*0004

*Let us jump right on in.*0010

*Here, identify and estimate the absolute max and min, and the local max and min on the following graph.*0014

*We have got this graph that looks like, in this particular case the domain is constrained.*0021

*We definitely have an endpoint here which belongs to it and an endpoint here which belongs to it.*0029

*This would be our a and this would be our b, something like that.*0037

*In this particular case, the absolute max, yes, definitely here.*0045

*The absolute max is this point right here.*0049

*The absolute max, and it looks like it is equal to about 60, 70, 80.*0054

*It is 80 and looks like it is achieved at x = 16.*0059

*The absolute min is definitely this one over here.*0068

*The absolute min, it is the endpoint, it looks like -0.5, -0.6, -0.7, and it looks like this is -5, 6, 7, 8.*0072

*It looks like absolute min itself is actually equal to -85, 50, 60, 70, 80.*0092

*It looks like it is -85 at x = -8.*0100

*Let me see we have got a local max, we got a local min.*0108

*Then, we have another local min.*0117

*This is actually we can consider that a local max as well.*0123

*That is about it, it just basically identify the points.*0127

*This one, it looks like, we will put it at -4.5 and actual value is 10, 20, 35.*0132

*Let us do this one over here.*0149

*Let us go ahead and put it at 0, not -5, -6, let us go ahead and put it at -6.*0151

*Here we have 5, 6, 7.*0161

*It looks like this point is 7, the value itself was 0.*0163

*This one over here, maybe 1.8.*0169

*The value was at 10, 20, maybe 27, 28, something like that.*0177

*There you go, that is it.*0182

*Absolute, absolute, local, local, local, local, that is all.*0184

*Nothing strange happening here.*0189

*Sketch the graph of a continuous function having the following properties on the closed interval 0,5.*0193

*It has an absolute min at 0, it has an absolute and a local max at 4.*0200

*It has a local max at 2, a local min at 3.5.*0206

*Let us go ahead and take a look.*0213

*I will go ahead and draw myself a little,*0215

*Let us do 1, 2, 3, 4, and 5.*0221

*That is 5 and this is our 0, 1, 2, 3, 4.*0225

*It says we have an absolute min at 0.*0231

*I’m just going to go ahead and put the absolute min right there.*0235

*It says we have a local max at 2, I’m just going to go ahead and put it there.*0238

*We have a local min at 3.5.*0245

*That is 3.5, I will go ahead and just put it there.*0247

*It says we have an absolute max and a local max at 4.*0250

*I will go ahead and put that there.*0254

*There you go.*0257

*That is it, just draw the graph, nice and simple.*0275

*Sketch the following graphs.*0281

*A function having a local min at 3 and differentiable at 3.*0283

*Something that has a local minimum at 3 and is differentiable at 3.*0294

*That is fine, that is just something like that.*0297

*It has a local min and it is differentiable.*0300

*In other words, the derivative is defined there.*0302

*We have to label that a.*0314

*A function having a local min at 3 but not differentiable at 3.*0316

*That is going to look something like this.*0320

*That is one possibility, this is 3.*0325

*Local min, because to the left and to the right of it, function increases, function increases, but is not differentiable there.*0329

*The derivative does not exist there, it is a cusp.*0338

*That takes care of that.*0343

*Of course, our last one, a function having a local min at 3 but not continuous at 3.*0345

*I’m going to go ahead and do this.*0352

*Here is our 3, it is a local min.*0359

*In other words, to the left of 3 and to the right of 3, the function is above 3.*0361

*This is definitely a local min but it is not continuous there.*0367

*Very simple, very straightforward.*0373

*Now the good stuff, find the critical values of 3x⁴ – 7x³ + 4x².*0376

*Critical values, we say that we take the derivative and we set it equal to 0.*0386

*We set it equal to 0, we solve for the x values.*0396

*And then, we make sure that there is no point where it is not differentiable, just watching out for that.*0399

*F’(x) = 12x³ - 21x² + 8x.*0405

*We are going to go ahead and set that to 0.*0421

*I’m going to factor an x and this is going to be 12x² - 21x + 8 is equal to 0.*0423

*One of our solutions to this is x = 0.*0435

*When I do this, either quadratic formula or calculator, however you want to do it,*0438

*you are going to get x = 0.559 and x is equal to 1.190.*0443

*There you go, those are our three critical values.*0453

*This is differentiable everywhere.*0457

*I do not have to worry about a critical value showing up at a point that is not differentiable.*0459

*These are the only critical values for this particular function.*0464

*Let us take a look at what it looks like.*0469

*0.559 and 1.190, there you go, this is the function.*0472

*We see it is continuous and differentiable everywhere.*0475

*The critical values that we have, this is the original function, this is f(x).*0479

*This is not the derivative of the function.*0485

*We had 0, we had 0.559, and we had 1.190.*0487

*These are critical values, they are the values in the domain along the x axis.*0500

*Such that, the derivative of the function at that point is equal to 0.*0506

*We can see that we have slope of 0, slope of 0, slope of 0.*0513

*That is all that is going on here.*0519

*Find the critical values of f(x) = absolute value of 2x – 5.*0524

*The easiest way to do this is probably graphically.*0529

*But, let us do it algebraically anyway because we are going to have to.*0531

*We want to know first of all, where this is going to equal 0, where it is going to bounce.*0538

*2x - 5 = 0 which implies that x = 5/2 or 2 ½.*0547

*At 2 ½, we got a 1, 2, 3, at 2 ½, that is where it is going to bounce.*0558

*When x is bigger than 5/2, f(x) is going to equal 2x – 5.*0573

*They are all positive.*0586

*In other words, when x is greater than 5/2, what is in here is a positive number.*0588

*Therefore, I can drop the absolute value signs.*0593

*That means f’(x) is equal to 2.*0596

*2 does not equal 0 anywhere.*0599

*To the right of 5/2, there are no critical values.*0603

*When x is less than 5/2, what is inside here is less than 0 which means that f(x) is actually equal to negative of what is inside.*0612

*-2x-5 which equals 5 - 2x, that is the definition of the absolute value.*0629

*Therefore, f'(x) is equal to -2.*0638

*-2 does not equal 0 anywhere.*0645

*To the left of 5/2, there are no critical values.*0647

*However, we know that at 5/2, where it bounces, it is not differentiable there.*0651

*That is the only critical value, not differentiable at 5/2, at x = 5/2.*0657

*X = 5/2 is the only critical value.*0674

*The graph for this sure enough looks like that.*0687

*There is no place where the derivative is 0, no place where the derivative is 0.*0691

*But here it is not differentiable, so 2.5.*0695

*5/2 is the critical value for that function.*0698

*Find the critical values of f(x) = 3√2x² – x.*0704

*We have f(x) is equal to this.*0710

*Let us rewrite it in a way that it makes it a little bit more tractable for us.*0712

*It is going to be 2x² - x¹/3.*0716

*Therefore, f’(x) is going to equal 1/3 × 2x² – x⁻²/3 × the derivative of what is inside, which is going to be 4x – 1.*0722

*We are going to go ahead and set that equal to 0.*0740

*This is the same as, this is negative so I will bring it down below.*0743

*This is 4x - 1 divided by 3 × 2x² – x²/3 = 0.*0750

*This is equal to 0, when the numerator is equal to 0.*0761

*Despite the fact that it looks complicated, it is only the numerator that I have to deal with.*0764

*Therefore, 4x - 1 = 0.*0768

*Therefore, x = ¼ or 0.*0772

*I have taken care of the critical value and so far as the derivative setting it equal to 0.*0781

*At 1/4, the slope is going to be 0.*0787

*In other words, there is going to be a local max and a local min.*0792

*I have to account for the possible places where this thing is not differentiable.*0796

*Are there places where f(x) is defined.*0803

*It is very important, a function has to be defined there, in order for it to have a critical value there.*0817

*It was defined but are there places where f(x) is defined but f’(x) does not exist?*0822

*Yes, f’(x) is this thing right here which is this thing right here.*0838

*It fails to exist, it is not defined but the derivative fails to exist when that is equal to 0.*0847

*This happens and the answer is yes.*0859

*The denominator is 0 when x is equal to 0.*0869

*In other words, if you put 0 into here, you get 0 – 0, the denominator is 0, x = 0, and x = 0.5.*0877

*When you put 0.5 into here, you also get 0 in the denominator.*0886

*At 0 and 0.5, the derivative does not exist.*0891

*At these two points, the function exists.*0897

*If I put them into the original function, I get f(0) = 0 and f(0.5) = 0.*0901

*The function is defined but the derivative is not defined.*0915

*The derivative does not exist.*0919

*Do you want to call it a critical value?*0925

*Basically, what happens is the slope actually ends up going to infinity.*0929

*Is it a differentiable there?*0933

*No, technically, these are critical values.*0934

*I’m just going to say, so I suppose.*0937

*It is not the end of the world if you do not call it critical values.*0939

*I suppose, we can say that x = 0 and x = 0.5 are also critical values.*0941

*There are places where their function is defined but the derivative does not exist are also critical values.*0954

*Again, it is really not the end of the world.*0967

*I myself, personally, I do not consider these critical values.*0969

*Do they fit the definition?*0974

*Yes, they do but for the most part they are irrelevant and insignificant.*0975

*Let us take a look at what the graph looks like.*0983

*It looks like this, we know that at 0.25, ¼, yes, it is a place where the derivative = 0, that is a critical value.*0987

*Over here at 0 and.5, the function exists.*0998

*At 0, it passes through there, the derivative does not exist.*1004

*It is going to be a fully vertical slope.*1007

*Technically, we consider them critical values.*1010

*Let us see what we have got.*1017

*Find the critical values of f(x) = cos² 2x on 0 to 2π.*1018

*Let us go ahead and do f’(x) and set it equal to 0.*1027

*F’(x) is equal to 2 × cos(2x) × -sin(2x) × 2.*1031

*That is equal to -4, I’m going to put the sin 2x first, cos 2x.*1048

*I’m going to rewrite that as -2 × 2 sin 2x cos 2x.*1057

*Hopefully, you recall an identity, that says the sin(2θ) = 2 sin θ cos θ.*1065

*We have 2 sin θ cos θ.*1076

*Therefore, it is equal to sin 2 θ.*1080

*This becomes -2 × sin(4x).*1082

*Therefore, we take -2 × sin (4x).*1090

*We set the derivative equal to 0, we have sin(4x) is equal to 0.*1098

*We have 4x is equal to 0 + n π.*1106

*Sin(0) = 0, sin(π) = 0, 2π, 3π, 4π.*1111

*Therefore, x is equal to 0 + n × π/4.*1117

*I just take n, 1, 2, 3, 4, 5, on this particular interval, here is what I have.*1127

*On the interval from 0 to 2π, x is equal to 0, π/4, 2π/4, 3π/4, 4π/4, 5π/4, 6π/4, 7π/4, and 8π/4,which puts us at 2π.*1134

*Of course, it is differentiable everywhere so I do not have to worry about points where the derivative does not exist.*1166

*These are my critical values.*1172

*There are places where the derivative = 0.*1174

*Let us take a look at what it looks like.*1177

*There you go, 2π is 6.28.*1180

*It is going to be somewhere around there.*1185

*Sorry, we are doing critical values.*1194

*There, there, there, there, and there.*1197

*There you go, that is it.*1204

*Let us take a look at example 8.*1209

*It says find the absolute max and min of this function on the closed interval 0 to π/3.*1211

*We have a function defined at a closed interval.*1219

*We know by the extreme value theorem that the absolute max and the absolutely min do exist.*1222

*We have a procedure for finding them.*1227

*We take the derivative, we find the critical values.*1229

*We evaluate those critical values in the original equation f(x).*1232

*And then, we evaluate the endpoints, in this case 0 to π/3.*1238

*We have a bunch of numbers, we take the biggest of those numbers, that is the absolute max.*1242

*We take the smallest of those numbers, that is the absolute min.*1245

*Let us do it.*1248

*We have f’(x) is equal to 2 cos x – 2 sin x.*1251

*We set that derivative equal to 0, we are trying to find the critical values here.*1263

*I’m going to rewrite this as 2 cos x = 2 sin x which gives me sin x/ cos x is equal to 1.*1267

*That is the same as the tan(x) equaling 1.*1287

*If the interval from 0 to π/3, x is equal to π/4 or 45°.*1295

*That is our critical value.*1302

*We go ahead and evaluate the critical values.*1307

*We have f(π/4), we put it back into the original.*1313

*That is going to equal, we put π/4 back into the original.*1322

*2 × sin(π/4) + 2 × cos(π/4).*1327

*We end up with the answer, it is going to be 4/√2, when you work it out.*1331

*That is equal to 2.83, that is one possible value.*1338

*We evaluated at 0 and π/3.*1346

*When we do f(0), when we put 0 in for here, we are going to end up with 2.*1349

*When we put f(π/3), we end up with √3 + 1 which = 2.732.*1356

*Of these numbers, the biggest number is 2.83.*1371

*Our absolute max = 2.83 and it happens at x = π/4.*1379

*Our minimum value is 2.*1391

*Our absolute min is equal to 2 and it happens at x = 2.*1393

*Let us take a look at what this looks like.*1402

*π/4, it looks like we are going from 0 to π/3.*1416

*I’m sorry, it did not happen at 2, it happened at 0, I apologize.*1427

*Here is your absolute min at 0 and at π/4, just right about there, we achieve our absolute max.*1431

*There you go, that is about it.*1446

*Our domain was 0 to π/3.*1448

*π/3 puts us right about there.*1454

*The graph that we are looking at is just about like that.*1457

*There you go, you see that this is the absolute max because if you go to the right of it, it drops a little.*1466

*You go to the left of it, it drops a little.*1472

*This is the highest point on the domain.*1474

*Find the absolute max and min of ln of 2x/x on the interval 1,3, nice and simple.*1481

*Once again, we form f’(x).*1489

*F’(x) this is a quotient, it is going to be this × the derivative of that - that × the derivative of this/ this².*1495

*This × the derivative of that, x × the derivative of ln 2x is 1/2x × 2 – the ln of 2x × the derivative of x which is × 1/x².*1504

*This is equal to, the 2 cancel, the x’s cancel, w get – ln of 2x/x².*1525

*We set that equal to 0 and that is equal to 0, when the numerator is equal to 0.*1536

*We have 1 – ln of 2x is equal to 0.*1544

*We have ln of 2x is equal to 1, we exponentiate both sides, e to this this or e to this power.*1550

*I'm left with 2x = e, that means x = e/2 which actually equals 1.359.*1557

*This is our one of our critical points.*1570

*Now we are going to evaluate.*1574

*We are going to take f(1.359), we are going to evaluate at the critical points.*1577

*When we do that, it is going to be, we put the 1.359 at the original equation.*1585

*It is going to equal ln(2) × 1.359 divided by 1.359.*1593

*We get the value 0.736.*1602

*That is the only critical value.*1608

*We evaluate at 1 and evaluate the function at 3.*1611

*F(1) is equal to ln(2) × 1/1 = ln(2) and that = 0.693.*1615

*F(3) that is going to equal the nat-log of 2 × 3/3 which is going to be ln(6)/3, that is going to give us 0.597.*1628

*The largest value here is this one, this is our absolute max.*1644

*The smallest value is this one, and this is going to be our absolute min.*1651

*Is that correct? Yes.*1663

*It is still the numbers that I got.*1666

*Let us take a look at what this looks like.*1676

*We are moving from 1 to 3, that is our domain.*1681

*The part of the graph that we are concerned about is this part right here.*1689

*Sure enough, as you can see, at about 1.359 that is where it hits its max.*1694

*Here, notice that it goes this way.*1702

*It drops a little bit, if you look really carefully.*1704

*Here is your absolute max on this domain.*1707

*Here is your absolute min.*1712

*That is all, if you have a function to find on a closed interval, find the critical values, evaluate f(x) at those critical values.*1716

*Evaluate f(x) at the two endpoints.*1724

*The largest number is your absolute max, the smallest number is your absolute min on that domain.*1727

*Thank you so much for joining us here at www.educator.com.*1733

*We will see you next time, bye.*1736

2 answers

Last reply by: Professor Hovasapian

Wed Jan 18, 2017 7:49 PM

Post by Sarmad Khokhar on January 5 at 05:49:37 AM

Professor what is difference between stationary point and critical point .

Thanks