AP Calculus AB 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