INSTRUCTORS Raffi Hovasapian John Zhu

Start learning today, and be successful in your academic & professional career. Start Today!

• ## Transcription

 0 answersPost by R K on April 25 at 09:29:47 PMFor #12, couldn't we use the second derivative test instead of graphing? 0 answersPost by R K on April 25 at 08:42:39 PMFor question number 5, why do you multiply the first integral from 0 to 50 by 1/4?

### AP Practice Exam: Section I, Part B Calculator Allowed

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Problem #1 1:22
• Problem #2 4:55
• Problem #3 10:49
• Problem #4 13:05
• Problem #5 14:54
• Problem #6 17:25
• Problem #7 18:39
• Problem #8 20:27
• Problem #9 26:48
• Problem #10 28:23
• Problem #11 34:03
• Problem #12 36:25
• Problem #13 39:52
• Problem #14 43:12
• Problem #15 47:18
• Problem #16 50:41
• Problem #17 56:38

### Transcription: AP Practice Exam: Section I, Part B Calculator Allowed

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

Today, we are going to continue our practice exam.0005

It is going to be Section 1 Part B, where the calculators are actually allowed.0009

You are going to be using the calculator regularly in this section.0014

Let us go ahead and get started.0018

Let me let you know once again where you can find this, in case you already forgot or did not see it.0020

Of course, there is a link down below in the quick notes.0028

I will go ahead and write it here.0030

You can find this at www.online.math.uh.edu/apcalculus/exams.0032

When you pull out this page, it is going to be section 1 part B.0055

The version that we are going through is version 5.0064

I would recommend just going through all of them as practice.0070

But we are going to be doing version 5 for this.0073

Let us get started, problem number 1.0076

I kind of like this purple, it is nice.0080

That is okay, I will go ahead and work in blue, I think.0082

It is a nice color that we usually work in.0085

Number 1 asks us to give a value of c that satisfies the conclusions of the mean value theorem for a given function.0089

Our function is f(x) = x² – x – 1.0101

The interval that we are concerned with is 1 to 3.0110

The mean value theorem basically says that, if f is continuous on a closed interval, it is differentiable on the open interval.0116

Then, there is some number c in that interval such that f’ at c is going to equal the actual slope from point 1 to point 3.0123

Let us go ahead and run through this.0139

Looking at this, it is definitely continuous.0140

Any polynomial, all polynomials are continuous, or the entire real line.0143

The first hypothesis is satisfied.0148

F is continuous on the closed interval 1,3.0151

F is a definitely differentiable because polynomials are differentiable everywhere.0156

F is differentiable on the open interval 1,3.0163

The conclusion that we can draw, there exists some number c such that c is between 1 and 3,0170

such that f’ at that value of c is equal to f(3) – f(1)/ 3 – 1.0186

That is what the mean value theorem says.0196

Let us go ahead and take the first derivative because we are going to be looking for f’.0199

F’ at x is equal to 2x – 1.0204

F’ at c is nothing more than 2c – 1.0211

We are going to be looking for that value of c.0216

Let us go ahead and calculate this value and set this f’(c) equal to this and solve for c.0219

That is all you have to do.0226

F(3), when I put it in here, I’m going to get a 9 – 3 – 1 is equal to 5.0227

f(1) is equal to 1 - 1 - 1 is equal to -1.0244

f(3) - f(1)/ 3 - 1 is equal to 5 - a -1/ 3 – 1.0255

This is going to be 6/2.0267

It is going to equal 3.0269

Therefore, we have, f’ at c is going to equal 3 or f’ at c is 2c – 1.0271

2c - 1 = 3, then we just solve for c.0283

2c = 4 and c = 2.0287

Our choice is going to be c, for this one.0291

Nice and straightforward, just have to make sure that the hypotheses of the mean value theorem are satisfied.0295

It has to be continuous on the closed interval and it has to be differentiable on the open interval.0300

If either of those two is not satisfied, you cannot use the mean value theorem.0306

Hypothesis is very important.0311

Number 2, it asks us to give us a function.0317

Let us go ahead and write this function down.0321

Our f(x), we have got 4x³ + 2 × e ⁺x.0324

It tells us that this function is invertible, it wants us to find the derivative of the inverse at x = 2.0337

What it is asking us to find is f inverse' at 2, that is what we want.0346

We are going to start off by graphing this.0359

Because again, you have this exponential function, you have your calculator, go ahead and graph it to take a look at it.0361

The graph is going to look something like this.0370

This over here, it is going to go something like that.0375

This is f, this is going to be the original function.0379

Now the inverse of that, we know what the inverse is.0382

The inverse is just a reflection about the line y = x.0385

Let me go back to blue here.0394

The inverse is going to look something like that.0396

This value is 2, f(0) is 2.0404

This graph is f inverse, what they want is the slope.0410

They want the derivative of the tangent line.0417

They want the slope of the curve at that point.0422

f inverse' at 2.0426

We actually know how to do this.0432

It turns out that the derivative of the inverse is nothing more than the reciprocal of the derivative of the original function.0434

f inverse', 1/ f’ at a given point.0445

We just have to be very careful at which point we are doing it.0449

Here is what is going to happen.0452

What we are looking for, f inverse' at 2 is actually going to be the inverse of 1/ f’ at 0, not 1/ f’ at 2.0455

Here is why, this point is 2,0, it is inverse point, if you will, is 0,2.0469

The x value for the inverse function is 2 but this 2, because inverse is switch x and y values,0484

when we go up to 2,0, the x value, this point which is the essentially the inverse point of this, its x value is 0.0495

Essentially, it is just going to be the reciprocal of that slope, right there.0503

That slope is f’ at 0.0509

F’ inverse at 2 is not 1/ f’ at 2.0514

You have to find the f inverse’ at 2.0520

You have to find where, that is going to be the y value of the original function.0528

Up here, it is going to be 1/ f’ at 0.0534

We need to find f’at (0), I hope that make sense.0537

Be very careful, that is why it is a good idea to actually graph this.0539

The argument for what we are looking for is going to have to be the y value of the original function,0545

where the inverse function is this.0550

Let us go ahead and do f’(x).0559

F’(x) is going to equal 12x² +,0563

I’m sorry, my real function was wrong, e ⁺2x.0578

There you go.0581

Yes, is it e ⁺2x, now I’m getting confused, 2e ⁺2x.0590

What function am I looking at here?0595

I think we are good.0599

4x³ + 2e ⁺2x, the derivative is going to be 12x² + 2e ⁺2x × 2, 4e ⁺2x.0600

We are going to find f’(0), it is going to be 12 × 0 which is 0.0612

e⁰ is going to be 1, this is going to be 4.0618

f inverse’ at 2 is equal to 1/ f’ at 0.0625

It is just equal to ¼, it is going to be c.0634

Sorry about the little confusion there.0640

I have written two different things on the page and I got a little confused for a moment.0644

Let us go on to question number 3.0649

What is question number 3 asks us?0653

Question 3 is asking us to give a value.0664

We actually have a graph, let me see it here.0670

They gave us a graph and this graph looks like that as it passes through 0 and it passes through 3.0675

1, 2, and 3, it is a little bit of parabola like that.0685

It actually passes through that point.0690

They tell us that this is actually a graph of f’(x).0694

This is the derivative of f.0701

What they are asking us to do is they want us to give a value of x where f actually achieves a local minimum.0703

A local min, we know that local min or local max, we know that f’(x) has to equal 0.0712

Since this is the f’ graph, it is going to be either this point or this point.0719

If I look at 3, if I look at the point 3, and the other point is point 0.0732

It is either going to be 3 or 0, which one of those is it going to be?0742

To the left of 3, the derivative.0746

This is the derivative graph, it is negative, it is decreasing.0748

To the right of 3, it is positive, it is above the x axis.0753

It is increasing.0757

Therefore, decreasing then increasing, our local min is at x = 3.0760

That is how we do with.0767

0, to the left you have a positive, it is increasing and it is decreasing.0768

Here it is actually going to be a local max there.0771

They want the local min, x = 3.0776

Our choice is a.0778

Number 4, basically, you are just going to be graphing the function that they gave you.0786

We have a graph right here and let us go ahead and say what this graph is.0791

I will go ahead and do over here.0798

f(x) =, the piece wise graph, we have got - x - 5 whenever x is less than -2.0799

We have x² + 1, when x is greater than or equal to -2, less than or equal to 1.0812

We have 2x³ -1, when x is greater than 1.0820

That is it, 3 piece wise function.0826

Question number 1 is, is f continuous at x = -2.0831

-2, no, it is not continuous there.0837

Question 2 is asking us is f differentiable at x = 1?0841

At 1, we also have a discontinuity here.0847

The left hand limit and the right hand limit are not equal, no it is not differentiable at x = 1.0850

3, is x = 0 a local minimum?0856

Is x = 0 a local min?0859

Yes, it is a local min because to the left and right of that, the values of the function are higher than that.0863

Yes, three is definitely true.0869

Number 4, is x - 2 an absolute max?0873

x - 2 an absolute max, no because there is something higher over here.0878

Only 3 is true, our choice is c.0884

That is it, graph it, read it right off the graph.0888

Let us see what question 5 has for us.0895

Question 5, we are given some integrals and we are asked to determine another definite integral.0903

They give us that the integral from 0 to 50 of 4 f(x) dx is equal to 3.0909

They tell us that the integral from 2 to 50 of f(x) dx is equal to 2.0921

They want us to evaluate the integral from 0 to 2 of f(x) dx.0930

This is just an application of the properties of the integral.0938

We are going to go ahead and just write the integral from 0 to 2 of f(x) dx, which is what we want.0943

It is going to equal 1/4 of 4 f(x) dx which is the integral from 0 to 50 +,0950

this upper is going to come down here, so that they end up canceling, so to speak.0965

50 to 2 of f(x) dx.0970

This is equal to this.0975

Now I just plug the values in.0977

This is equal to ¼, I know that the integral from 0 to 50 is this one, 3.0980

This is -2, if I’m not mistaken.0990

Yes, that is -2.0997

The integral from 2 to 50 is -2.0999

The integral from 50 to 2 is going to be the negative of that.1006

It is going to be - -2.1011

When I add these together, I’m going to end up with 11/4, the choice is d.1015

Just be very careful.1023

Here what I have done is, by flipping the integral, I’m basically taking the negative of this one, - -2.1028

Number 6, what is number 6 asks us?1047

It wants us to find an approximate location of the local maximum for the particular function.1051

The particular function that they gave us is this and I have gone ahead and graphed it.1058

F(x) = 4x³ + 3x² - 2x.1063

whatever your calculator uses or calculate, basically, to find this.1077

They want the value of a local maximum.1082

That is it, use the zoom feature, in this particular case, it turns out to be this point -0.7287, 1.502.1088

In this case, our choice is going to be b.1102

That is it, just graph it and read it right off.1105

Either zoom or calculate it directly, however is it that you want, go ahead and do it.1108

Let us move on to number 7.1119

Number 7 asks us to find the approximate average value of function.1127

The function that we are concerned with is going to be 4x ln 3x.1132

Let me just double check and make sure that it is correct, 4x ln 3x.1139

On the interval, of course, average value is always specified on an interval, on the interval 1 to 4.1145

Very simple, we know what the average value is, it is equal to 1/ b – a1153

where the first number is a and the second number is b × the integral from a to b of the function itself.1159

That is it, it is the same as any other average.1167

When you take an average, let us say of 10 numbers, you add those 10 numbers and you divide by 10.1170

An interval is continuous, it is not 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.1175

You are essentially adding an infinite number of numbers.1179

The sum, if you remember the integral is just the sum.1183

Once you add the numbers, that is the integral.1185

And then, you divide by how many there are.1187

Where, how many there are, is basically the length of the interval.1190

This is the same average that you are used to, always.1192

This is nice and simple.1198

1/ 4 – 1/ the integral 1 to 4, 4x of ln of 3x dx.1199

Again, just go ahead and use your calculator to evaluate the integral numerically.1211

You are going to get an approximate value of 20.77.1216

Your choice is going to be d.1222

Number 8, let us see what number 8 has for us.1227

We have a region that is enclosed by the graphs of a couple of functions and1235

we rotate it around the y axis to generate a solid.1239

We want to find the volume of this solid.1244

The two functions are y = x³ - 1 and y = x – 1.1248

You can go ahead and graph it or if you know what it looks like,1256

you can graph it by hand and save the calculator for the actual evaluation of the integral.1259

The region is going to look like this.1263

1, this is going to be the x³ - 1 graph.1269

The x - 1 is going to be the line going this way, something like that.1278

Essentially, we are going to be rotating this around the y axis.1287

We are going to end up getting that and we are going to end up getting that.1295

We want the volume of that solid.1305

We can do this with washers, we can do this with shells.1308

I think we can just go ahead and do it with shells.1312

We are basically going to take a volume element.1317

And then, we are going to integrate that volume element.1320

Let me go ahead and work in red for this one.1326

A volume element, I decided I think I'm going to go ahead and go shells.1329

I'm going to take that and that.1335

Basically, a little bit of a cylindrical shell.1341

And then, I’m going to integrate, add them up, all over from here to here, from 0 to 1.1344

That is how I’m going to go ahead and do this.1357

Basically, we are going to take the area of a cylindrical shell.1360

The area of the cylindrical shell is going to be 2π x × the height.1366

This cylindrical shell, if you will, something like that.1375

We are just going to add them all up.1386

This is going to be a cylindrical shell.1387

This thickness is going to be dx.1390

What we want is, we want the surface area then we are going to add them up.1392

That is going to be our dx.1398

2π x × h.1399

X is going to be the radius, then 2π r is going to be the circumference.1404

And that multiply by h which is that, that is going to give me the area of the outside of the circular shell.1410

The dx is going to give me a volume.1417

The volume element is 2π x h dx.1419

H itself, how do we find this height?1426

The height is going to be the difference between that and that.1430

The height is going to be the function x – 1.1436

For any value x, I have this and I have this.1439

I’m going to do x - 1 - x³ – 1.1445

It is going to be this function - this function.1454

That is going to give me the difference, the height.1457

I hope that makes sense.1460

If I have a particular x, it is going to be this height - this height.1462

It is this height, they give me essentially just the difference between the two functions.1471

I get x = x - 1 – x³ + 11478

The 1 cancel, I get a height equal to x - x³.1486

Therefore, my volume element is going to be 2π x.1492

My h his x - x³ and it is going to be × 2 × dx.1500

Now that takes care of this.1509

I have to multiply this by 2 because now I have the bottom part.1513

Therefore, I have a dv element of 4π × x² – x⁴ dx.1525

Now I can integrate.1547

Now the volume is going to be the integral from 0 to 1.1548

I’m going to add up all the cylindrical shells.1552

That is fine, I will just leave it in here, it is not a problem.1560

X² - x⁴ dx and when I evaluate that, I get a volume equal to 1.676 cubic units.1563

The choice is actually d.1574

You could have done this with washers.1578

We take that region and we draw it like that.1582

You could have done it this way but then and you have to integrate along y.1587

You are going to have to change this, which is why I went with shells.1592

It was given in terms of x and y, as a function of x.1596

I integrated along x.1599

Along x, I decided to just use the shells.1601

Number 9, let us see, instantaneous rate of change.1609

They want to know the approximate instantaneous rate of change of a given function.1618

This function f(t) is actually expressed as an integral.1622

0 to 4t of cos(x) dx.1628

Fundamental theorem of calculus tell us that the instantaneous rate of change,1640

we know that an instantaneous rate of change is the derivative.1644

The derivative f’ t is going to be, when I have the function expressed as an integral,1646

I go ahead and just drop the integral sign, when I take the derivative.1659

It is going to be cos, I put that in here, of 4t.1663

But because this is not t, it is a function of t, I have to multiply by the derivative of that.1668

They want us to evaluate the instantaneous rate of change at a certain point.1677

What they want is f’ at π/4.1682

f’ at π/4, when I put π/4 into this expression, I get 1.662.1688

That is it, nice and simple, just a straight application of the fundamental theorem of calculus I’m plugging in.1698

Number 10, we have a particular integral and they want to know what the area is1705

when this integral is evaluated by a trapezoidal rule for n = 3.1717

The integral that they gave us is 0 to 1 of sin of π(x) dx.1724

Essentially, what we are going to do is we are going to calculate this integral with our calculator.1732

We are going to do a trapezoidal rule approximation.1735

We are going to subtract 1 for the other and that is going to be our area.1739

In this particular case, n = 3.1742

When n = 3, the integral from a to b of f(x) dx, the trapezoidal approximation is going to b Δ x/ 2.1746

This is just a formula that you have to know.1761

It is going to be f(x) is 0 + 2 f(x1) + 2 × f(x2) + f at x3.1764

n = 3, you are going to have four terms.1779

X0 and x3 and the n, and 2 × down below or Δ x is equal to be - a/ n.1782

In this case, b - a/ n.1792

b - a is equal to 1 - 0/ 3.1799

Δ x = 1/3.1805

We just need to find f(x) 0, f(x1), f(x2), f(x3).1808

When I plug those in, f(x) is 0, it is going to be this is f right here, sin(π) x.1815

It is going to be sin(π) × 0 which is equal to 0.1828

Let me write out everything explicitly here, and a little more clear.1841

We have x(0) = 0, that is going to give us, f(x0), π × 0 = 0.1849

x sub 1 is equal to 1/3 because Δ x is 1/3, it is going to be 0, 1/3, 2/3, 1.1871

We are breaking up this interval into that many parts.1880

x1 is equal to 1/3.1884

Therefore, f of 1/3 is going to equal the sin(π/3) which is equal to 0.86603.1886

x2 is equal to 2/3, therefore, the f(2/3) is going to equal the sin of 2π/ 3.1905

It also equals 0.86603.1917

x of 3 is equal to 1, f(1) is equal to the sin(π).1923

The sin(π) is equal to 0.1931

We have the trapezoidal approximation is equal to, we said Δ x/ 2.1937

It is going to be 1/3/ 2.1944

I will do it in this way.1948

I will actually write everything out.1949

0 + 2 × 0.86603 + 2 × 0.86603 + 0.1953

That is going to give some answer of 0.577353.1966

The actual value of the integral itself, when I evaluate this integral, it is going to equal 0.63662.1975

I take that number - that number.1991

I get an area = 0.63662 - 0.577353.1995

It is going to give me an area of 0.05927.2010

Our choice of is d, that is it, just got to know this formula.2016

That is it, Δ x/ 2, and then, however many you have.2021

If n is 6, you are going to have 7 terms.2024

If n is 19, you have 20 terms.2026

Always n + 1 terms in a trapezoidal approximation.2029

The 1 and n stay, f(x) and f(x) ⁺n, the ones in between all are multiplied by a factor of 2.2032

What does number 11 ask us?2047

They are telling us that the amount of money in a bank is increasing at a certain rate.2050

The rate is going to be dollars per year.2055

They give us a time, a year where t = 0.2057

In this case, it turns out to be 2005.2061

They want to know, what is the approximate total amount of increase from 2005 to 2007?2064

The rate at which the money is increasing is equal to 10,000 × e⁰.06 t.2071

It is going to be that many dollars per year.2086

t is in years.2093

They tell us that t = 0, corresponds to the year 2005.2099

They want to know the increase, this is the rate of increase,2105

they want to know what the increase is from the year 2005 to 2007.2113

If t(0) is 2005, then t = 2 is 2007.2120

What we are going to do, the increase from 2005 to 2007.2127

If the rate of increases this amount per year, I multiply by the number of years.2141

But I integrate, because it is a function of t.2147

It is the integral from 0 to 2 of r(t) dt.2150

That is it, I just calculate that integral with my calculator and it turns out to be \$21,250.2156

That is it, I hope that makes sense.2163

Dollars per year × year.2167

Years cancels year, you are left with dollars.2171

But because the rate of increase, the dollars per year is not constant.2174

I cannot just multiply them, I have it integrate it.2178

That is it, very simple.2181

Number 12, let us see what is number 12 asking us.2189

A particle is moving with a certain acceleration which they give us here and its initial velocity is 0.2195

For how many values of t does the particle change the direction?2202

A particle changes direction, when velocity goes from positive to negative or negative to positive.2207

When the velocity curve actually crosses the x axis, that is when it changes direction.2213

The acceleration that they give us here,2222

Our acceleration that they give us is 2t² - 4t.2227

They tell us that the initial velocity is 0.2234

Was that correct , yes, the initial velocity is 0.2237

For how many values of t does it change direction?2244

For how many values of t does the velocity curve cross the x axis?2255

We need to find the function for the velocity.2260

We know what the velocity is.2263

The velocity is just the integral of the acceleration.2266

It is going to be integral of 2t² - 4t dt.2269

When we do that, we get 2t³/ 3 - 2t² + c.2279

They tell us that the initial velocity is actually equal to 0.2288

v(0) is nothing more than, I put 0 into this for t.2292

It is going to be 0 + 0 + c which implies that c itself = 0,2299

which means now I can go ahead and put that into here to get my velocity function is equal to 2t³/ 3 – 2t².2305

Now I graph that function and I see where it crosses the x axis.2316

Not hits the x axis only, it have to fully cross.2321

It has to go from positive to negative or negative to positive.2323

Let us see what we do next.2331

This is a graph of the function that we just dealt with.2336

This is the velocity function 2t³/ 3 - 2t².2340

This is the velocity function, if something changes direction, it is what they want.2355

Does the particle change direction?2359

It changes direction when the velocity goes from positive to negative or negative to positive.2360

Notice here, the velocity goes to 0 but it stays negative.2364

It means it is moving to the left.2367

It stops but it keeps moving to the left.2369

Here it only crosses once, there is only one place where it actually changes direction.2373

Our choice is e, hope that makes sense.2380

There is only one place where it actually changes direction.2386

Number 13, let us see what we have got.2395

This is a related rates problem.2401

We have a sphere, they want to know fast the volume of the sphere is changing,2403

when the surface area is 3m² and the radius is increasing at a rate of ¼ m/ min.2413

Let us see what we have got.2426

They tell me that the radius is changing.2428

This is a basic related rates problem.2430

You may be given at least one rate, at least one, you might be given more.2433

You are going to asked to find another rate.2438

Your goal, your task is only to find the relationship between the two variables that you choose.2440

And then, differentiate implicitly and just plug your values in.2446

They tell me that the radius is changing at ¼ m/ min.2451

I’m going to skip the units here.2469

The rate of change of the radius is ¼.2473

It is increasing at ¼ m / min.2476

What they want is the rate of change of volume.2480

They want dv dt, that is what they want.2482

Our task is to find the relationship between that and that.2484

We know what that is, the volume of the sphere is 4/3 π r³.2490

Now we differentiate, dv dt is equal to 4 π r² dr dt.2499

They are telling me, they want to know what this value is when the surface area is equal to 3.2516

We know the formula for surface area, it is 4 π r².2527

They want to know when that is equal to 3.2532

It is 4 π r², the rate of change of the volume, when the surface area is 3,2537

I just put in the 3 where I see 4 π r², and I multiply by dr dt which I have.2545

3/4 m³/ min is my answer, which is going to be choice e because that is in decimal form, 0.75.2552

I hope that makes sense.2564

All related rates problems are the same.2565

They will give you a rate, they will ask for a rate.2568

Your job is to find an explicit expression between this variable and that variable.2571

Using whatever other information is given to you, and then differentiate that expression.2576

Here is the expression, differentiate the expression.2582

This is what we want.2585

We arrange it for whatever they ask, that is all.2586

Let us see what do we got number 14.2597

Rectangle, upper two vertexes, the graph, give the decimal approximation of the maximum possible area.2604

What we have is this situation.2611

We have a function which is e ⁻9x².2619

When I graph this function, I get something like this.2631

It is called a dump function, because it looks like a little dump.2639

They are telling me that there is this rectangle that is in here.2643

They want to know what the maximum area of the rectangle can be.2649

Let us take a look at what we have got.2656

This is going to be x, this is going to be f(x).2659

Therefore, the area of this rectangle is going to be 2x × f(x).2666

2x × f(x) which is going to equal 2x × e ⁻9x².2675

What we are going to do, we are going to graph the area.2686

We are going to graph the derivative because the area is what we want to maximize.2689

When you maximize something, we maximize the function like this one.2696

You take the derivative and you set it equal to 0.2700

Since we have our calculator, we are just going to graph the function, graph the derivative,2703

and we are going to see where the derivative crosses the x axis.2707

We are going to find where the derivative = 0.2710

That is the x value that we are going to use to put in our particular function.2714

Let us go ahead and do that.2721

Once again, maximization means this.2723

Find the x values where a’(x) = 0.2729

This green right here, this is our original function.2750

This is our area function.2754

This is the function that we want to maximize.2755

This purple function, this is a’(x).2758

That is the derivative of that.2763

Our a(x), remember we said this is 2x e ⁻9x².2765

It does not matter what this is.2771

It is the derivative of this.2772

I need to know where the derivative crosses the x axis, where the derivative is equal to 0.2773

It is equal to 0 to places here and here.2778

This is a negative value, I cannot have a negative length because this is a box that we are talking about.2781

A box whose length is this part is x.2787

That is the value that I'm interested in.2792

That value, I read it off the graph, zoom in, use my calculator, however it is that I want to use it.2795

I end up with this value is 0.236.2800

a(0.236), you can see that the area function is maximized here.2808

The x value is here, the derivative confirms that.2815

That is all we are doing.2818

It is a 0.236, the area of 0.236, when I put 0.236 into here, I get an area of 0.2859.2819

My choice is a, I hope that made sense.2831

Number 15, let us see what number 15 is asking us.2839

They are saying that a rough approximation from the natlog of 5 is going to be 1.609.2848

They want us to use this approximation and differentials or linear approximations to approximate.2855

They tell us that the natlog of 5 is equal to 1.609.2864

They want us to find an approximate value for the natlog of 521/ 100.2876

That is just the natlog of 5.21.2884

We are going to use a linear approximations and differentials here.2890

Actually linear approximations not differentials.2892

The linear approximation of a value is equal to that function at that value + the derivative at that value × x - x0.2898

In this particular case, we want to find the value at 5.21.2913

It is going to be the f(5) + f’ at 5 × 5.21 – 5.2920

f(x) is just equal to the natlog of x.2934

f’(x) = 1/ x, f’ at 5 = 1/5.2939

Our answer, the approximation of the natlog of 5.21 is equal to f(5),2949

which is f(5) which is ln of 5 which is 1.609 + f'(5) which is 1/5 × 5.21 – 5.2958

When I calculate this, I end up with 1.651.2976

My answer is c, linear approximation, that is it.2980

That is all you are doing.2986

This is going to be the difference between what you know and what you are looking for.2988

That is all that is happening here.2993

Just remember, linear approximation.2998

Let us say this is a function, the whole idea of the derivative is 2.3001

As long as you are staying pretty close to the curve, the line itself is actually pretty good approximation.3005

This point was our ln 5.3014

This point was going to be our ln 5.21.3019

These are linear approximation, we used the derivative instead of actually using the logarithm function because it is very close to it.3027

That is all we have done here.3036

Let us take a look at number 16, let us see what we have got.3043

What is 16 asking us, it is giving us a function and it is telling us that it is differentiable everywhere.3050

They want to know what n is.3056

f(x) is equal to, it is a piece wise function here.3060

We have got nx³ - x for x less than or equal to 1.3067

We have got nx² + 5 for x greater than 1.3074

Once again, they tell us that it is differentiable everywhere.3082

They want to know what is n.3091

We want to know what n is.3095

Let us go ahead and work in red here.3098

Differentiable means it is continuous.3101

Because differentiability implies continuity, not the other way around, continuous.3107

We know that this function is continuous everywhere.3115

The differentiability also means that the left hand derivative = the right hand derivative.3122

The left hand derivative, I take the derivative of this, the derivative of that.3139

The left hand derivative, as I approach it from the left.3148

I have got 3nx² - 1 is equal to 2nx.3150

I got myself one equation there.3160

Because I know that the left hand derivative is equal to the right hand derivative.3161

That has to be true.3164

The limit as x approaches 1 from the left of f(x) = the limit as x approaches 1 from the left of nx³ - x is equal to, I plug 1 in for x.3170

It is going to be n - 1 and the limit as x approaches 1 from above, they are differentiable, left hand and right hand derivative are equal.3201

What I’m saying is that it is continuous which means that3216

the left hand limit and the right hand limit are going to be the same because it is continuous everywhere.3219

Therefore, those two things have to be the same.3226

If f(x) = the limit as x approaches 1 from above of that function mx² + 5 = m + 5.3232

These are equal, precisely because f is continuous everywhere.3252

It is continuous everywhere because it is differentiable everywhere.3259

Now I have got, we have got n - 1 is equal to m + 5.3263

Let us go ahead and write this as n - 6 is equal to m.3274

I’m going to go ahead and plug n – 6, wherever I see that.3283

I got 3nx² - 1 = 2 × n - 6 × x.3290

I get 3nx² – 1 = 2nx - 12x.3303

If I did my math correctly and I end up with x = 1.3311

I know the x = 1 because I'm talking about 1,3324

because 1 is the point at which it is going to be both continuous and differentiable.3329

X = 1, I get 3n.3334

When I plug in x = 1 to there and to there, I get 3n - 1 = 2n – 12.3340

I get n = -11, my choice is e.3351

In this particular case, when they said it is differentiable everywhere, it implies that it is continuous everywhere.3355

Because it is differentiable, the left hand and right hand derivative have to be the same.3360

I take the derivative of both functions, to the left of 1 or the right of 13366

because it is differentiable at 1 and I set them equal to each other, that is one of my equations.3370

Because it is continuous, I know that the left hand limit, the original function = the right hand limit of the original function.3375

I set those two things equal to each other.3381

That gives me another equation, an m and n.3383

Now I just solve the two equations.3387

In this case, they just wanted n.3388

I could have done both, not a problem.3390

Let us see number 17, which I think is our last problem here.3396

They want to know which of the following functions has a vertical asymptote at -1 and a horizontal asymptote at y = 2.3403

They gave us some choices.3413

Vertical asymptote at x = -1 and they want a horizontal asymptote at y = 2.3414

This is what they are asking.3435

Basically, you are just going to check each function, that is it.3439

Denominator, where the denominator is 0, that is going to be your vertical asymptote.3442

Horizontal asymptote, it depends on the nature of the rational function, what the top is and what the bottom is.3450

I notice that the 2x² + 1/ x² – 1, it goes to 2 as x goes to infinity.3457

Horizontal asymptote means taking x to infinity and see what happens.3471

Essentially, as x goes to infinity, these do not really matter.3475

x²/ x², they cancel.3479

You are just left with 2/1 which is 2.3481

Sure enough, y = 2 is a horizontal asymptote and x² – 1, this is x – 1 × x + 1.3484

x cannot be -1 because it is actually going to end up making the denominator 0.3501

Our choice is b, nice and straightforward.3507

That takes care of Section 1, Part B, where the calculator was allowed for the exam.3513

Next time, we are going to start with our free response question.3520

Thank you so much for joining us here at www.educator.com.3524

We will see you next time, bye.3526