Sign In | Subscribe
INSTRUCTORS Raffi Hovasapian John Zhu
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of AP Calculus AB
  • Discussion

  • Download Lecture Slides

  • Table of Contents

  • Transcription

Bookmark and Share
Lecture Comments (4)

1 answer

Last reply by: Professor Hovasapian
Mon Jul 25, 2016 7:19 PM

Post by Peter Ke on July 23 at 06:42:48 PM

For Problem 5, shouldn't the derivative of-4y^2 be -8y and not -8yy' because you bring down the 2 and multiply it by -4 which is -8 and subtract the exponent by 1?

1 answer

Last reply by: Professor Hovasapian
Wed May 11, 2016 3:15 AM

Post by nathan lau on May 5 at 01:17:18 AM

for question number 9 I keep getting letter b do you know what I could be doing wrong?

AP Practice Exam: Section 1, Part A No Calculator

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
  • Exam Link 0:10
  • Problem #1 1:26
  • Problem #2 2:52
  • Problem #3 4:42
  • Problem #4 7:03
  • Problem #5 10:01
  • Problem #6 13:49
  • Problem #7 15:16
  • Problem #8 19:06
  • Problem #9 23:10
  • Problem #10 28:10
  • Problem #11 31:30
  • Problem #12 33:53
  • Problem #13 37:45
  • Problem #14 41:17

Transcription: AP Practice Exam: Section 1, Part A No Calculator

Hello, welcome back to, and welcome back to AP Calculus.0000

Today, we are going to start on the AP practice exam.0004

What I would like to do, the practice exam that we are going to use is going to be a 2006 version.0009

You can find it on, I will write the link here.0015

The link is also going to be down at the bottom of the page in the quick notes.0019

When you enter that or when you click on the link down below,0044

the page will come up and it is going to give you a bunch of different versions.0048

The version that we are going to use is going to be version 5.0051

Just click on version 5 for part A, the section on part A.0054

We are also going to be doing when we get to, it is going to be version 5 for the part B.0068

For the written part, the free response questions, we are going to go with version 2.0073

You can either pull it up, head on your screen for you.0079

If you want, you can print it out, whichever you prefer.0082

Let us jump right on in.0086

Question number 1, we have got a couple of functions here.0088

We have f is equal to -2x + 1 and our g(x), I will write it as g = -x/ x² + 1.0096

In this particular case, we are to find f of g(1).0109

Nice and straightforward composition of functions.0120

First thing we do is we find the g(1).0123

g(1) is equal to -1/ 1² + 1 = -½.0125

f of g(1) is equal to f(-½).0138

We have -2 × -1/2 + 1.0145

This is 1 + 1 is equal to 2.0152

In this particular case, our answer is c.0156

Question number 2, I think I will do this on the next page.0164

You know what, I think I’m going to work in blue.0170

Question number 2, we have f(x) is equal to -x² – 4√x and we want f’ at 4.0173

Take the derivative, plug in 4, and see what it is that you actually get, the slope of the tangent line.0194

The slope of the tangent line is the value of the derivative at that point.0205

f(x) is that, f’(x), what do we get for f’(x)?0215

The derivative of this is -2x and this is going to be -4 × 1/2 × x⁻¹/2 = -2x - 2/ √x.0222

f’ at 4 is -2 × 4 - 2/ √4.0245

This is equal to -8 – 1, we have -9.0255

Therefore, in this particular case, the answer is going to be c.0261

Let us see what we have got, let me move this a little bit over here.0270

Question number 3, question number 3 is asking us to evaluate,0278

they want the limit as x goes to infinity of 2x³ + 4x/ -2x⁵ + x² – 2.0286

To evaluate this particular limit, a couple ways that we can do it.0309

One of the systematic way, whenever you are dealing with a rational function is to divide the top and bottom,0314

the numerator and the denominator by the highest degree in the denominator.0322

Basically, divide everything on the top by x⁵ and divide everything in the bottom by x⁵.0326

And then, take the limit as x goes to infinity.0331

Let us go ahead and do that.0334

Let us go ahead and bring this over here.0336

This is going to be dealing with a functional.0338

I do not want to keep writing limit symbol over and over again.0340

When I divide the top, it is going to be 2x³/ x⁵ - 4x/ x⁵.0343

On the bottom, I’m going to get -2x⁵/ x⁵ + x²/ x⁵ - 2/ x⁵.0353

This becomes 2/ x² + 4/ x⁴/ -2 + 1/ x³, I think -2/ x⁵.0365

Now we take the limit as x goes to infinity of this function.0385

As x goes to infinity, this goes to 0 because this becomes really big, therefore, this becomes really small, it goes to 0.0394

This goes to 0, this stays -2, goes to 0, goes to 0.0403

What you are left with is 0/ -2 which is equal to 0.0409

In this particular case, the answer will be a.0414

That is it, nice and straightforward limit.0418

Nothing particularly difficult about it.0421

Question number 4, let us see what question number 4 is asking us.0425

We have a particular function and it is bounded by certain numbers, certain functions.0434

It wants you to express the integral that gives you the area of the particular region that is bounded.0443

We have f(x), in this particular case is equal to x³.0454

Let us go ahead and we know what x³ looks like.0462

x³ looks like that, let us see, from -1 to 0.0470

They say that region 1, x goes from -1 to 0.0480

From -1 to 0, we are bounded by,0490

We have got, this is 1, this is -1.0508

We have this region and we have this region.0518

From -1 to 0, we call this our region 1.0523

From 0 to 1, this is our region 2.0527

Our total area is going to be the area of region 1 + the area of region 2.0534

This is fully symmetric here about the origin.0542

I can basically just take the integral of one of them and multiply it by 2.0545

I can just say it is 2 × the area of region 2.0549

It is equal to 2 × the integral from 0 to 1 of the upper function - the lower function.0555

It is going to be 1 - x³ dx.0565

As far as our choices are concerned, 0 to 1,2, just pull the constant in there, 1 - x³ dx.0573

It looks like our answer is c.0584

They give you a particular graph, they tell you the bounds, and you just have to find a particular area.0589

Let us see what number 5 says.0600

Question number 5 on this particular version.0604

We are given a particular function, we want you to determine the change in y with respect to x.0606

They want us to find the derivative dy dx.0613

We have 3x³ - 4xy – 4y² is equal to 1.0619

We see that in this particular case, our function is given implicitly.0632

We are going to use implicit differentiation.0636

Let us go ahead and do that.0639

The derivative of this with respect to x.0641

We are looking for dy dx.0643

The question asks, determine the change in y with respect to x.0645

Dy dx is what we are looking for.0651

The derivative of this is going to be 9x² – 4.0653

I just tend to pull my constants.0661

It is going to be 4 ×, this is a function of x and this is a function of x.0664

It is going to be this × the derivative of that.0669

x × y’ which is dy dx, just a shortened version, + y × the derivative of that which is 1.0671

- the derivative of this is -8yy’.0681

The derivative of 1 is 0.0686

We have got 9x² - 4xy’ - 4y – 8yy’ is equal to 0.0690

I have got 9x² - 4y, I’m going to put the y’ on one side = 4xy’ + 8yy’.0703

Factor out the y’, I got 9x² – 4y = y’ × 4x + 8y.0720

I’m left with y’ is equal to 9x² – 4y/ 4x + 8y.0733

As far as our choices are concerned, that is equivalent to 9x² - 4y/ - and -4 x is -8y.0756

The answer that you get, depending on how you did it, which you move to the left or the right,0776

maybe slightly different on the choices that you have.0781

What is probably going to be the most tricky part of this is,0784

just because you got something that does not look like any of your choices, it does not mean you are wrong.0788

Just see if you can change what you got and if it is equivalent to one of your choices.0792

In this case, it is equal to one of the choices.0798

It is going to be d.0801

Do not freak out, if what you got is going to be different.0805

Clearly, you figured out by now, having gone through calculus that three of you can do the problem and all get it right.0808

And three of you have different answers, that is not a problem because there is a lot of symbolism going on.0815

The symbolism is going to take different shapes, depending on the particular approach that was used to solve the problem.0820

That was number 5, let us take a look at number 6.0828

We have got f(x) = 4 × sec(x) - 3 × csc(x).0837

We are asked to find the derivative of this.0848

Very straightforward, as long as you know your derivative formulas, = 4.0851

The derivative of sec x is sec tan.0858

4 sec x tan x -, let us do it this way,0861

Let me put -3.0881

The derivative of csc x is –csc cot.0883

-csc x cot x, you end up with 4 sec x tan(x) + 3 csc x cot x.0888

In this particular case, our answer is b.0905

Just a straight application, just have to watch the sign, that is about it.0908

That was number 6, let us go ahead and see what number 7 has to offer.0913

We are asked to compute an integral here.0922

We have the integral from 0 to 1/4 of 16/ 1 + 16t² dt.0926

You remember some of your integral formulas, there was an integral formula0941

where the integral of 1/ 1 + x² dx was equal to the inv tan(x) + c.0945

This thing, then I'm going to rewrite, I’m going to pull the 16 out.0961

16 × the integral from 0 to 1/4 of 1/,0967

I’m going to write this as 1 + 4t².0984

I just change the 16t², I wrote it as (4t)² dt.0994

I’m just going to do a little bit of u substitution here.0999

I’m going to call u for t, therefore du = 4 dt.1003

Therefore, dt is equal to du/ 4.1011

What we get is 16.1019

This integral with the u substitution, we get 16, 0 to ¼, 1/ 1 + u² × dt is du/ 4.1023

I go ahead and pull the 4 out.1041

I can write this as 16/ 4 × the integral from 0 to 1/ 4 of + 1/ 1 + u² du, 1043

which is equal to 4 × the inv tan of u which is equal to 4 × the inv tan or u is 4t.1057

From 0 to 1/4 which is equal to, I will go to the next page, not a problem.1077

Which is equal to 4 × the inv tan of 1 – inv tan of 0.1088

We get = 4 × the inv tan of 1 is π/4 -,1110

The inv tan of 0 is 0 = π.1121

Our answer is d, that is it.1130

A little bit of a recognition of what the integral formula is for 1/ 1 + x².1132

In this case, 1 + 1/ u², slight manipulation.1138

And then, use the u substitution to go ahead and take care of that.1141

Let us move on to number 8.1147

What is question number 8 asking us to do.1156

We have to determine the derivative of this particular expression.1159

ddx, 2x⁴ – 2x/ 2x⁴ + 2x.1168

What is the best way to approach this?1184

What is the best way to do this?1197

Okay, that is not a problem.1204

Before we actually start with the quotient rule, more than likely we are going to be using the quotient rule here.1205

Before we do that, let us see if there is something that we can actually do this function to make it a little easier on us.1210

Especially, when you take a look at the choices.1219

The choices have 2x³ + 2 in the denominator, except one of them which is 1 + x³.1222

Before we actually start taking the derivative, let us see if there is something we can do here.1230

This 2x⁴ - 2x/ 2x⁴ + 2x.1234

Let me factor out a 2x and I end up with x³ - 1, if I’m not mistaken.1247

I got a 2x here and I have got an x³ + 1 over here.1254

2x actually vanishes.1259

What I'm left with is x³ - 1/ x³ + 1.1261

Our f is the x³ - 1 and our g is our x³ + 1.1269

Now f’(x), the quotient rule is gf’ – fg’/ g².1274

We just have to work it out.1285

This is going to equal x³ + 1 × 3x² – fg’ - x³ - 1 × 3x²/ x³ + 1².1288

That is going to equal 3x⁵ + 3x² – 3x⁵ + 3x²/ x³ + 1².1310

3x⁵ and 3x⁵ goes away, that leaves us with a final answer of 6x²/ x³ + 1².1327

This happens to coercive with answer d.1339

That is about it, pretty straightforward.1342

If you just gone ahead and started doing that,1345

you are going to end up with a pretty complicated expression to have to simplify algebraically.1347

Would you have come up with the same answer, honestly,1354

I did not actually carry out that particular algebraic manipulation.1356

I took a look at the choices and I notice the x³.1362

I looked back at the function and thought can I factor it.1366

It turned out that I can.1369

Again, this is calculus we are dealing with.1372

There is a million ways to do something.1377

That is question number 8.1380

Let us go ahead and move on to question number 9 here.1382

How long is 9, that is not a problem.1389

Question number 9, in this particular case, we are given a function and 1392

we are asked to find the equation of the normal line to the graph at a given point.1399

Our f(x) is equal to 3x × √x² + 6 – 3.1408

It is going to be at the point 0 – 3.1423

What we are going to do is we are going to find the slope of the tangent line which is just a derivative,1428

evaluated at a certain point.1439

The normal line, we know that the normal line is perpendicular to the tangent line.1441

All we are going to do is, the slope that we get for the tangent line, we are going to take the negative reciprocal of it.1455

That will give us the slope of the normal line, the slope of the normal line.1460

Because it is still passing through 0 and -3, we have the slope, we have a point.1468

We do y - y1 = m × x - x1.1472

We see the normal line is perpendicular the tangent line.1477

We take the negative reciprocal.1484

Let us go ahead and find f’(x) first.1490

F’(x), we got 3, we got a function of x × a function of x.1494

It is going to be a little long but not too big of a deal.1498

Again, I tend to take that out.1501

It is going to be this × the derivative of that + that × the derivative of this.1503

We are going to get x × 1/2 x² + 6⁻¹/2 × 2x + x² + 6 ^½ × 1.1509

That is going to equal 3 × x²/ √x² + 6 + √x² + 6.1528

We want to find f’ at 0, it is a function of x.1545

We are going to be using the 0 value.1549

That is going to equal 3 × 0/ √0 + 6 + √6.1551

We are going to end up with 3√6.1563

This is the slope of the tangent line, we want the negative reciprocal.1571

The normal line slope = -1/ 3√6.1576

Therefore, our line is equal to y - y1 - 3 = m which is -1/ 3√6 × x – 0.1587

Let us see if I can turn the page here.1605

I have a little difficulty getting to the next page.1607

We end up with y + 3 = -1/ 3√6 × x.1611

When you rearrange this, just multiply by 3√6.1625

Rearrange it to make it correspond with one of the choices.1628

Again, the answer that you got is not going to match one of the choices, but it is the same object.1631

You just have to rearrange it.1637

Rearranging to match one of the choices.1639

We get x – 3√6, y = 9√6 which is choice a.1641

It is going to happen quite a lot.1656

Your answer is going to be slightly different.1658

When you do the rearrangement, make sure you go very slowly.1660

In the choices that they give you, the differences are going to be very subtle.1663

There are going to be + and -, all the symbols are going to be there.1667

9, √6, 3 √6, just be very careful with your choices.1673

Make sure you look at all of your choices to make sure that you have an excluded one.1678

Just because you think you found the right choice, there may be something else going on.1682

Make sure you look at all of your choices.1685

Question number 10, let us see what we got for question number 10.1690

Here we want to find the concavity of a particular graph.1696

In this particular case, our function f(x) is equal to 2 sin(x) + 3 × cos² x.1702

We want to find the concavity at a given point.1716

Let us find and we know that concavity has to do with the second derivative.1722

Let us find f"(x) and evaluate f” at the point π, to see what the concavity of π is.1728

F’x, the derivative of this is going to be 2 × cos(x) + cos(x) – 6 sin x cos x.1747

That is one of our f’.1780

We want to do our f”.1784

F”(x), I will just take the derivative of what it is that we just got.1787

We are going to end up with -2 × sin(x) - 6 × sin x × -sin x + cos x cos x, 1791

which is equal to -2 × sin(x) + 6 sin² x - 6 cos² x.1812

Now we go ahead and evaluate f” at π.1826

When I put π in for this, -2 × sin(π) + 6 × sin² of π - 6 × cos² of π.1830

sin(π) is 0, sin(π)² is 0, cos(π) is -1.1851

-1² is 1, we get -6.1859

The answer is d because -6 is less than 0.1869

We are concave down.1876

That is it, nice, straight application.1878

Second derivative is positive, you are concave up.1880

Second derivative is negative, you are concave down.1883

Let us go to question number 11.1889

Here it looks like we are computing a derivative.1893

Number 11, our particular derivative, we are evaluating at an indefinite integral.1898

My apologies.1905

The integral of -3x² × √x³ + 3 dx.1906

I think it is just going to be the straight u substitution.1917

I noticed x³ and I noticed an x².1921

I’m going to try u is equal to x³ + 3, du = 3x² dx.1924

Our integral, -3x² × the integral of x³ + 3 dx, is actually going to equal -the integral,1939

3x² x is du, this is going to be e ^½.1955

We are accustomed to seeing du on the right.1965

It is not a big deal, the order does not matter because you are just multiplying two things.1968

The multiplication is commutative, it does not matter the order.1971

You are used to seeing the du or the dx, or the dy, on the right hand side in the integrand.1975

Let us do it that way.1983

- the integral of u ^½ du which is equal to -u³/2/ 3/2 + c which is equal to -2/3 u³/2 + c.1985

Of course, we plug the u back in, x³ + 3.2005

Our final answer is -2/3 × x³ + 3³/2 + c.2009

Our final answer, the choice is going to be e.2020

I hope that was reasonably straightforward.2024

Let us take a look at the problem number 12. 2030

Here we have a particular function.2039

We want to give the value of x where the function has a local extrema.2042

In this particular case, a local maximum.2046

Local maximum means you take the derivative, you set the derivative equal to 0.2050

You draw yourself a number line and you check points to the left of the critical values,2058

to the right of the critical values, to see whether the derivative is increasing or decreasing.2064

You decide which is local min and local max.2068

f(x) is equal to x³ + 6x² + 9x + 4.2072

f’(x), very simple.2084

We have 3x² + 12x + 9.2087

We want to set the derivative equal to 0.2092

We can factor out the 3.2095

We are going to get the same roots.2096

We have got x² + 4x + 3 is equal to 0.2097

We can actually factor this one.2104

We get x + 3, we get x + 1.2107

Therefore, we have x is equal to -3 and we have x is equal to -1.2111

Go ahead and draw myself a little number line here.2119

I have got -3, I will put it over here.2122

I have got -1, I will put it over here.2125

I’m going to check a point here, check a point here, and check a point here, in those intervals.2127

I’m going to put them into the derivative.2132

To see if the derivative is less than 0, decreasing, or greater than 0, increasing.2134

That is all I'm doing.2140

Let me go ahead and rewrite f’(x), I’m going to use this version right here.2143

I should actually use the original version.2155

I should have actually written this as 3 × that.2157

This is 3 × x + 3 × x + 1.2162

To the left of -3, when I check something in this interval, over here, let us try -4.2171

When I plug in -4, I'm going to get, for x, I'm going to get 3 × -4 + 3 is a negative number.2177

-4 + 1 is a negative number.2185

3 × a negative × a negative is definitely a positive number.2189

Here the slope is increasing.2194

Or if you want I can put a positive sign.2197

Some of you use positive, some of you use increasing/decreasing with an arrow.2199

It is that way.2203

Let us try a point in between here.2205

Let us try -2.2207

When I plug -2 in to the derivative, I get 3, -2 + 3 is a positive number.2210

-2 + 1 is a negative number.2218

3 × a positive × a negative gives me a number that is a negative.2220

It is less than 0, it is decreasing there or negative slope.2226

There you go, that pretty much takes care of it.2229

Because it is increasing, the graph is increasing.2232

Then, the graph is decreasing, that is a local max.2235

To the left it is increasing, to the right it is decreasing.2242

That means it is hitting a maximum point.2245

There is a local max at x = -3.2251

I think the particular choice was choice e.2260

Nothing too crazy so far.2266

Here we have number 13 and let us see what number 13 is asking us.2269

The slope of the tangent line of the graph that will give the value of c.2277

We are give a particular function.2283

We have 4x² + cx + 2, e ⁺y is equal to 2.2284

They are telling us that the slope of the tangent line of this graph, at x = 0 is 4.2298

They are asking us to find c.2320

The slope of the tangent line, for this graph, at x = 0 is equal to 4.2326

They want us to find the value of c.2331

Let us see what we can do.2338

Let us go ahead and take the derivative.2342

In this particular case, let us go ahead and take the derivative with respect to,2344

8x + c +, this is the function of y, we are going to do implicit differentiation here.2348

This is going to be 2e ⁺yy’ is equal to 0.2356

When I rearrange this, I'm going to get y’ is equal to -8x - c/ 2 × e ⁺y.2363

I will get 2 × e ⁺y.2377

2 × e ⁺y, if I move these two over to the right, 2 × e ⁺y from the original function is equal to 2 - 4x² – cx.2386

Therefore, if I plug in this into here, I get y’ is equal to -8x - c/ 2 – 4x² – c ⁺x.2401

They are telling me that y’ at 0 is equal to 4.2421

y' at 0 is equal to -8.2426

y’ at 0 is equal to 0 - c/ 2 - 0 – 0.2431

They are telling me that this is actually equal to 4.2440

I get - c/ 2 is equal to 4 which implies that c is equal to 8.2445

Our choice is e.2457

Just differentiate, in this particular case, you are going to get something which is 2e ⁺y.2460

You notice the 2e ⁺y is actually separate here.2464

You can move these over, plug in for x, and then solve for c.2467

I hope that made sense.2472

Let us try number 14, see how we are doing here.2477

Number 14, what is number 14 asking us.2483

It looks like it is asking us to evaluate this particular integral.2486

Evaluate the integral of 7 ⁺x - 4e⁷ ln x dx.2492

Let us see what we can do with this.2509

I’m going to separate this out.2511

The integral is linear, the linear of the integral of the sum is the sum of the integrals.2512

I’m going to do this as, this is the integral of 7 ⁺x dx - 4 × the integral of e⁷ ln x dx.2518

Let us go ahead and deal with the first integral.2534

This one right here.2536

The first integral, the integral of 7 ⁺x dx.2538

I’m going to do a u substitution here.2548

I’m going to let u equal to 7 ⁺x.2550

Du is going to be 7 ⁺x ln 7 dx, du/ ln 7 = 7 ⁺x dx.2554

Therefore, this integral actually = the integral of du/ ln 7 = 1/ ln 7 × the integral of du = 1/ ln 7 u + c.2572

I plug u back in which is 7 ⁺x.2601

I get 7 ⁺x/ ln 7 + c, that is our first integral.2604

Let me see here, where am I?2614

The second integral is -4, it is that one, × the integral of e ⁺ln x dx = -4 × the integral of e ⁺ln(x)⁷ dx2619

= 4 × the integral of e ⁺ln is just x⁷ dx = -4x⁸/ 8 + some constant c.2648

The c and c are not the same.2664

If you want you can just c1 and c2, it does not really matter.2666

This is this one, which is –x⁸/ 2 + c.2672

Now we just put them back together.2691

We have the integral that we wanted, it is equal to 7 ⁺x/ ln 7 – x⁸/ 2 + c.2694

It looks like that is option a.2709

That takes care of the first of 14 of the problems.2714

For the next lesson, we are just going to continue on with this particular section and go on with the practice exam.2718

Thank you so much for joining us here at

We will see you next time, bye.2728