WEBVTT mathematics/ap-calculus-ab/hovasapian
00:00:00.000 --> 00:00:03.900
Hello, welcome back to www.educator.com, and welcome back to AP Calculus.
00:00:03.900 --> 00:00:09.300
Today, we are going to start on the AP practice exam.
00:00:09.300 --> 00:00:15.300
What I would like to do, the practice exam that we are going to use is going to be a 2006 version.
00:00:15.300 --> 00:00:19.200
You can find it on, I will write the link here.
00:00:19.200 --> 00:00:23.700
The link is also going to be down at the bottom of the page in the quick notes.
00:00:23.700 --> 00:00:44.400
www.online.math.uh.edu/apcalculus/exams.
00:00:44.400 --> 00:00:48.000
When you enter that or when you click on the link down below,
00:00:48.000 --> 00:00:51.300
the page will come up and it is going to give you a bunch of different versions.
00:00:51.300 --> 00:00:54.000
The version that we are going to use is going to be version 5.
00:00:54.000 --> 00:01:08.100
Just click on version 5 for part A, the section on part A.
00:01:08.100 --> 00:01:13.000
We are also going to be doing when we get to, it is going to be version 5 for the part B.
00:01:13.000 --> 00:01:19.000
For the written part, the free response questions, we are going to go with version 2.
00:01:19.000 --> 00:01:21.700
You can either pull it up, head on your screen for you.
00:01:21.700 --> 00:01:26.200
If you want, you can print it out, whichever you prefer.
00:01:26.200 --> 00:01:28.600
Let us jump right on in.
00:01:28.600 --> 00:01:36.100
Question number 1, we have got a couple of functions here.
00:01:36.100 --> 00:01:49.600
We have f is equal to -2x + 1 and our g(x), I will write it as g = -x/ x² + 1.
00:01:49.600 --> 00:02:00.100
In this particular case, we are to find f of g(1).
00:02:00.100 --> 00:02:02.800
Nice and straightforward composition of functions.
00:02:02.800 --> 00:02:05.600
First thing we do is we find the g(1).
00:02:05.600 --> 00:02:18.200
g(1) is equal to -1/ 1² + 1 = -½.
00:02:18.200 --> 00:02:25.100
f of g(1) is equal to f(-½).
00:02:25.100 --> 00:02:32.300
We have -2 × -1/2 + 1.
00:02:32.300 --> 00:02:36.500
This is 1 + 1 is equal to 2.
00:02:36.500 --> 00:02:44.600
In this particular case, our answer is c.
00:02:44.600 --> 00:02:50.600
Question number 2, I think I will do this on the next page.
00:02:50.600 --> 00:02:53.600
You know what, I think I’m going to work in blue.
00:02:53.600 --> 00:03:14.100
Question number 2, we have f(x) is equal to -x² – 4√x and we want f’ at 4.
00:03:14.100 --> 00:03:25.500
Take the derivative, plug in 4, and see what it is that you actually get, the slope of the tangent line.
00:03:25.500 --> 00:03:35.100
The slope of the tangent line is the value of the derivative at that point.
00:03:35.100 --> 00:03:42.000
f(x) is that, f’(x), what do we get for f’(x)?
00:03:42.000 --> 00:04:04.800
The derivative of this is -2x and this is going to be -4 × 1/2 × x⁻¹/2 = -2x - 2/ √x.
00:04:04.800 --> 00:04:15.300
f’ at 4 is -2 × 4 - 2/ √4.
00:04:15.300 --> 00:04:21.600
This is equal to -8 – 1, we have -9.
00:04:21.600 --> 00:04:30.600
Therefore, in this particular case, the answer is going to be c.
00:04:30.600 --> 00:04:38.100
Let us see what we have got, let me move this a little bit over here.
00:04:38.100 --> 00:04:46.600
Question number 3, question number 3 is asking us to evaluate,
00:04:46.600 --> 00:05:09.100
they want the limit as x goes to infinity of 2x³ + 4x/ -2x⁵ + x² – 2.
00:05:09.100 --> 00:05:14.500
To evaluate this particular limit, a couple ways that we can do it.
00:05:14.500 --> 00:05:22.000
One of the systematic way, whenever you are dealing with a rational function is to divide the top and bottom,
00:05:22.000 --> 00:05:25.900
the numerator and the denominator by the highest degree in the denominator.
00:05:25.900 --> 00:05:31.300
Basically, divide everything on the top by x⁵ and divide everything in the bottom by x⁵.
00:05:31.300 --> 00:05:33.700
And then, take the limit as x goes to infinity.
00:05:33.700 --> 00:05:36.100
Let us go ahead and do that.
00:05:36.100 --> 00:05:38.200
Let us go ahead and bring this over here.
00:05:38.200 --> 00:05:39.700
This is going to be dealing with a functional.
00:05:39.700 --> 00:05:42.700
I do not want to keep writing limit symbol over and over again.
00:05:42.700 --> 00:05:53.000
When I divide the top, it is going to be 2x³/ x⁵ - 4x/ x⁵.
00:05:53.000 --> 00:06:05.600
On the bottom, I’m going to get -2x⁵/ x⁵ + x²/ x⁵ - 2/ x⁵.
00:06:05.600 --> 00:06:25.400
This becomes 2/ x² + 4/ x⁴/ -2 + 1/ x³, I think -2/ x⁵.
00:06:25.400 --> 00:06:33.800
Now we take the limit as x goes to infinity of this function.
00:06:33.800 --> 00:06:43.100
As x goes to infinity, this goes to 0 because this becomes really big, therefore, this becomes really small, it goes to 0.
00:06:43.100 --> 00:06:48.800
This goes to 0, this stays -2, goes to 0, goes to 0.
00:06:48.800 --> 00:06:54.500
What you are left with is 0/ -2 which is equal to 0.
00:06:54.500 --> 00:06:58.100
In this particular case, the answer will be a.
00:06:58.100 --> 00:07:00.800
That is it, nice and straightforward limit.
00:07:00.800 --> 00:07:04.700
Nothing particularly difficult about it.
00:07:04.700 --> 00:07:14.100
Question number 4, let us see what question number 4 is asking us.
00:07:14.100 --> 00:07:22.800
We have a particular function and it is bounded by certain numbers, certain functions.
00:07:22.800 --> 00:07:33.900
It wants you to express the integral that gives you the area of the particular region that is bounded.
00:07:33.900 --> 00:07:42.000
We have f(x), in this particular case is equal to x³.
00:07:42.000 --> 00:07:50.400
Let us go ahead and we know what x³ looks like.
00:07:50.400 --> 00:07:59.700
x³ looks like that, let us see, from -1 to 0.
00:07:59.700 --> 00:08:10.500
They say that region 1, x goes from -1 to 0.
00:08:10.500 --> 00:08:28.000
From -1 to 0, we are bounded by,
00:08:28.000 --> 00:08:37.700
We have got, this is 1, this is -1.
00:08:37.700 --> 00:08:42.800
We have this region and we have this region.
00:08:42.800 --> 00:08:47.000
From -1 to 0, we call this our region 1.
00:08:47.000 --> 00:08:53.900
From 0 to 1, this is our region 2.
00:08:53.900 --> 00:09:02.300
Our total area is going to be the area of region 1 + the area of region 2.
00:09:02.300 --> 00:09:05.600
This is fully symmetric here about the origin.
00:09:05.600 --> 00:09:09.500
I can basically just take the integral of one of them and multiply it by 2.
00:09:09.500 --> 00:09:14.900
I can just say it is 2 × the area of region 2.
00:09:14.900 --> 00:09:25.300
It is equal to 2 × the integral from 0 to 1 of the upper function - the lower function.
00:09:25.300 --> 00:09:32.800
It is going to be 1 - x³ dx.
00:09:32.800 --> 00:09:43.900
As far as our choices are concerned, 0 to 1,2, just pull the constant in there, 1 - x³ dx.
00:09:43.900 --> 00:09:49.100
It looks like our answer is c.
00:09:49.100 --> 00:10:00.500
They give you a particular graph, they tell you the bounds, and you just have to find a particular area.
00:10:00.500 --> 00:10:03.800
Let us see what number 5 says.
00:10:03.800 --> 00:10:06.200
Question number 5 on this particular version.
00:10:06.200 --> 00:10:12.800
We are given a particular function, we want you to determine the change in y with respect to x.
00:10:12.800 --> 00:10:18.800
They want us to find the derivative dy dx.
00:10:18.800 --> 00:10:32.600
We have 3x³ - 4xy – 4y² is equal to 1.
00:10:32.600 --> 00:10:35.900
We see that in this particular case, our function is given implicitly.
00:10:35.900 --> 00:10:39.600
We are going to use implicit differentiation.
00:10:39.600 --> 00:10:40.800
Let us go ahead and do that.
00:10:40.800 --> 00:10:42.900
The derivative of this with respect to x.
00:10:42.900 --> 00:10:45.600
We are looking for dy dx.
00:10:45.600 --> 00:10:50.700
The question asks, determine the change in y with respect to x.
00:10:50.700 --> 00:10:53.100
Dy dx is what we are looking for.
00:10:53.100 --> 00:11:01.200
The derivative of this is going to be 9x² – 4.
00:11:01.200 --> 00:11:04.500
I just tend to pull my constants.
00:11:04.500 --> 00:11:09.000
It is going to be 4 ×, this is a function of x and this is a function of x.
00:11:09.000 --> 00:11:11.500
It is going to be this × the derivative of that.
00:11:11.500 --> 00:11:21.400
x × y’ which is dy dx, just a shortened version, + y × the derivative of that which is 1.
00:11:21.400 --> 00:11:26.200
- the derivative of this is -8yy’.
00:11:26.200 --> 00:11:30.100
The derivative of 1 is 0.
00:11:30.100 --> 00:11:43.300
We have got 9x² - 4xy’ - 4y – 8yy’ is equal to 0.
00:11:43.300 --> 00:11:59.800
I have got 9x² - 4y, I’m going to put the y’ on one side = 4xy’ + 8yy’.
00:11:59.800 --> 00:12:13.000
Factor out the y’, I got 9x² – 4y = y’ × 4x + 8y.
00:12:13.000 --> 00:12:35.800
I’m left with y’ is equal to 9x² – 4y/ 4x + 8y.
00:12:35.800 --> 00:12:56.500
As far as our choices are concerned, that is equivalent to 9x² - 4y/ - and -4 x is -8y.
00:12:56.500 --> 00:13:01.600
The answer that you get, depending on how you did it, which you move to the left or the right,
00:13:01.600 --> 00:13:04.100
maybe slightly different on the choices that you have.
00:13:04.100 --> 00:13:07.700
What is probably going to be the most tricky part of this is,
00:13:07.700 --> 00:13:11.900
just because you got something that does not look like any of your choices, it does not mean you are wrong.
00:13:11.900 --> 00:13:18.500
Just see if you can change what you got and if it is equivalent to one of your choices.
00:13:18.500 --> 00:13:21.500
In this case, it is equal to one of the choices.
00:13:21.500 --> 00:13:25.100
It is going to be d.
00:13:25.100 --> 00:13:28.400
Do not freak out, if what you got is going to be different.
00:13:28.400 --> 00:13:35.600
Clearly, you figured out by now, having gone through calculus that three of you can do the problem and all get it right.
00:13:35.600 --> 00:13:40.100
And three of you have different answers, that is not a problem because there is a lot of symbolism going on.
00:13:40.100 --> 00:13:48.600
The symbolism is going to take different shapes, depending on the particular approach that was used to solve the problem.
00:13:48.600 --> 00:13:56.700
That was number 5, let us take a look at number 6.
00:13:56.700 --> 00:14:08.100
We have got f(x) = 4 × sec(x) - 3 × csc(x).
00:14:08.100 --> 00:14:10.800
We are asked to find the derivative of this.
00:14:10.800 --> 00:14:18.000
Very straightforward, as long as you know your derivative formulas, = 4.
00:14:18.000 --> 00:14:20.700
The derivative of sec x is sec tan.
00:14:20.700 --> 00:14:41.400
4 sec x tan x -, let us do it this way,
00:14:41.400 --> 00:14:42.700
Let me put -3.
00:14:42.700 --> 00:14:48.400
The derivative of csc x is –csc cot.
00:14:48.400 --> 00:15:05.200
-csc x cot x, you end up with 4 sec x tan(x) + 3 csc x cot x.
00:15:05.200 --> 00:15:07.900
In this particular case, our answer is b.
00:15:07.900 --> 00:15:13.600
Just a straight application, just have to watch the sign, that is about it.
00:15:13.600 --> 00:15:22.100
That was number 6, let us go ahead and see what number 7 has to offer.
00:15:22.100 --> 00:15:26.600
We are asked to compute an integral here.
00:15:26.600 --> 00:15:41.600
We have the integral from 0 to 1/4 of 16/ 1 + 16t² dt.
00:15:41.600 --> 00:15:44.900
You remember some of your integral formulas, there was an integral formula
00:15:44.900 --> 00:16:00.800
where the integral of 1/ 1 + x² dx was equal to the inv tan(x) + c.
00:16:00.800 --> 00:16:06.800
This thing, then I'm going to rewrite, I’m going to pull the 16 out.
00:16:06.800 --> 00:16:24.600
16 × the integral from 0 to 1/4 of 1/,
00:16:24.600 --> 00:16:34.200
I’m going to write this as 1 + 4t².
00:16:34.200 --> 00:16:39.600
I just change the 16t², I wrote it as (4t)² dt.
00:16:39.600 --> 00:16:42.900
I’m just going to do a little bit of u substitution here.
00:16:42.900 --> 00:16:51.000
I’m going to call u for t, therefore du = 4 dt.
00:16:51.000 --> 00:16:58.800
Therefore, dt is equal to du/ 4.
00:16:58.800 --> 00:17:03.000
What we get is 16.
00:17:03.000 --> 00:17:21.000
This integral with the u substitution, we get 16, 0 to ¼, 1/ 1 + u² × dt is du/ 4.
00:17:21.000 --> 00:17:23.100
I go ahead and pull the 4 out.
00:17:23.100 --> 00:17:37.000
I can write this as 16/ 4 × the integral from 0 to 1/ 4 of + 1/ 1 + u² du,
00:17:37.000 --> 00:17:56.700
which is equal to 4 × the inv tan of u which is equal to 4 × the inv tan or u is 4t.
00:17:56.700 --> 00:18:08.400
From 0 to 1/4 which is equal to, I will go to the next page, not a problem.
00:18:08.400 --> 00:18:30.300
Which is equal to 4 × the inv tan of 1 – inv tan of 0.
00:18:30.300 --> 00:18:41.100
We get = 4 × the inv tan of 1 is π/4 -,
00:18:41.100 --> 00:18:49.800
The inv tan of 0 is 0 = π.
00:18:49.800 --> 00:18:52.300
Our answer is d, that is it.
00:18:52.300 --> 00:18:57.700
A little bit of a recognition of what the integral formula is for 1/ 1 + x².
00:18:57.700 --> 00:19:00.700
In this case, 1 + 1/ u², slight manipulation.
00:19:00.700 --> 00:19:06.700
And then, use the u substitution to go ahead and take care of that.
00:19:06.700 --> 00:19:15.700
Let us move on to number 8.
00:19:15.700 --> 00:19:19.600
What is question number 8 asking us to do.
00:19:19.600 --> 00:19:28.000
We have to determine the derivative of this particular expression.
00:19:28.000 --> 00:19:44.200
ddx, 2x⁴ – 2x/ 2x⁴ + 2x.
00:19:44.200 --> 00:19:57.300
What is the best way to approach this?
00:19:57.300 --> 00:20:03.900
What is the best way to do this?
00:20:03.900 --> 00:20:05.100
Okay, that is not a problem.
00:20:05.100 --> 00:20:10.200
Before we actually start with the quotient rule, more than likely we are going to be using the quotient rule here.
00:20:10.200 --> 00:20:19.500
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.
00:20:19.500 --> 00:20:22.200
Especially, when you take a look at the choices.
00:20:22.200 --> 00:20:29.800
The choices have 2x³ + 2 in the denominator, except one of them which is 1 + x³.
00:20:29.800 --> 00:20:34.300
Before we actually start taking the derivative, let us see if there is something we can do here.
00:20:34.300 --> 00:20:47.100
This 2x⁴ - 2x/ 2x⁴ + 2x.
00:20:47.100 --> 00:20:53.700
Let me factor out a 2x and I end up with x³ - 1, if I’m not mistaken.
00:20:53.700 --> 00:20:59.100
I got a 2x here and I have got an x³ + 1 over here.
00:20:59.100 --> 00:21:01.500
2x actually vanishes.
00:21:01.500 --> 00:21:09.300
What I'm left with is x³ - 1/ x³ + 1.
00:21:09.300 --> 00:21:14.100
Our f is the x³ - 1 and our g is our x³ + 1.
00:21:14.100 --> 00:21:24.900
Now f’(x), the quotient rule is gf’ – fg’/ g².
00:21:24.900 --> 00:21:27.900
We just have to work it out.
00:21:27.900 --> 00:21:49.700
This is going to equal x³ + 1 × 3x² – fg’ - x³ - 1 × 3x²/ x³ + 1².
00:21:49.700 --> 00:22:07.100
That is going to equal 3x⁵ + 3x² – 3x⁵ + 3x²/ x³ + 1².
00:22:07.100 --> 00:22:18.800
3x⁵ and 3x⁵ goes away, that leaves us with a final answer of 6x²/ x³ + 1².
00:22:18.800 --> 00:22:22.200
This happens to coercive with answer d.
00:22:22.200 --> 00:22:25.200
That is about it, pretty straightforward.
00:22:25.200 --> 00:22:27.300
If you just gone ahead and started doing that,
00:22:27.300 --> 00:22:33.900
you are going to end up with a pretty complicated expression to have to simplify algebraically.
00:22:33.900 --> 00:22:36.600
Would you have come up with the same answer, honestly,
00:22:36.600 --> 00:22:42.300
I did not actually carry out that particular algebraic manipulation.
00:22:42.300 --> 00:22:46.000
I took a look at the choices and I notice the x³.
00:22:46.000 --> 00:22:48.700
I looked back at the function and thought can I factor it.
00:22:48.700 --> 00:22:52.000
It turned out that I can.
00:22:52.000 --> 00:22:57.100
Again, this is calculus we are dealing with.
00:22:57.100 --> 00:23:00.400
There is a million ways to do something.
00:23:00.400 --> 00:23:02.200
That is question number 8.
00:23:02.200 --> 00:23:08.900
Let us go ahead and move on to question number 9 here.
00:23:08.900 --> 00:23:11.900
How long is 9, that is not a problem.
00:23:11.900 --> 00:23:19.100
Question number 9, in this particular case, we are given a function and
00:23:19.100 --> 00:23:28.400
we are asked to find the equation of the normal line to the graph at a given point.
00:23:28.400 --> 00:23:42.900
Our f(x) is equal to 3x × √x² + 6 – 3.
00:23:42.900 --> 00:23:48.600
It is going to be at the point 0 – 3.
00:23:48.600 --> 00:23:59.100
What we are going to do is we are going to find the slope of the tangent line which is just a derivative,
00:23:59.100 --> 00:24:01.200
evaluated at a certain point.
00:24:01.200 --> 00:24:15.000
The normal line, we know that the normal line is perpendicular to the tangent line.
00:24:15.000 --> 00:24:20.100
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.
00:24:20.100 --> 00:24:28.200
That will give us the slope of the normal line, the slope of the normal line.
00:24:28.200 --> 00:24:32.400
Because it is still passing through 0 and -3, we have the slope, we have a point.
00:24:32.400 --> 00:24:37.200
We do y - y1 = m × x - x1.
00:24:37.200 --> 00:24:43.800
We see the normal line is perpendicular the tangent line.
00:24:43.800 --> 00:24:50.100
We take the negative reciprocal.
00:24:50.100 --> 00:24:54.300
Let us go ahead and find f’(x) first.
00:24:54.300 --> 00:24:58.300
F’(x), we got 3, we got a function of x × a function of x.
00:24:58.300 --> 00:25:01.000
It is going to be a little long but not too big of a deal.
00:25:01.000 --> 00:25:02.800
Again, I tend to take that out.
00:25:02.800 --> 00:25:08.800
It is going to be this × the derivative of that + that × the derivative of this.
00:25:08.800 --> 00:25:28.600
We are going to get x × 1/2 x² + 6⁻¹/2 × 2x + x² + 6 ^½ × 1.
00:25:28.600 --> 00:25:45.100
That is going to equal 3 × x²/ √x² + 6 + √x² + 6.
00:25:45.100 --> 00:25:49.000
We want to find f’ at 0, it is a function of x.
00:25:49.000 --> 00:25:51.400
We are going to be using the 0 value.
00:25:51.400 --> 00:26:03.100
That is going to equal 3 × 0/ √0 + 6 + √6.
00:26:03.100 --> 00:26:11.500
We are going to end up with 3√6.
00:26:11.500 --> 00:26:16.000
This is the slope of the tangent line, we want the negative reciprocal.
00:26:16.000 --> 00:26:27.300
The normal line slope = -1/ 3√6.
00:26:27.300 --> 00:26:45.300
Therefore, our line is equal to y - y1 - 3 = m which is -1/ 3√6 × x – 0.
00:26:45.300 --> 00:26:46.800
Let us see if I can turn the page here.
00:26:46.800 --> 00:26:51.300
I have a little difficulty getting to the next page.
00:26:51.300 --> 00:27:05.100
We end up with y + 3 = -1/ 3√6 × x.
00:27:05.100 --> 00:27:08.500
When you rearrange this, just multiply by 3√6.
00:27:08.500 --> 00:27:11.300
Rearrange it to make it correspond with one of the choices.
00:27:11.300 --> 00:27:16.800
Again, the answer that you got is not going to match one of the choices, but it is the same object.
00:27:16.800 --> 00:27:18.900
You just have to rearrange it.
00:27:18.900 --> 00:27:21.600
Rearranging to match one of the choices.
00:27:21.600 --> 00:27:36.300
We get x – 3√6, y = 9√6 which is choice a.
00:27:36.300 --> 00:27:37.800
It is going to happen quite a lot.
00:27:37.800 --> 00:27:40.200
Your answer is going to be slightly different.
00:27:40.200 --> 00:27:43.200
When you do the rearrangement, make sure you go very slowly.
00:27:43.200 --> 00:27:47.200
In the choices that they give you, the differences are going to be very subtle.
00:27:47.200 --> 00:27:52.900
There are going to be + and -, all the symbols are going to be there.
00:27:52.900 --> 00:27:58.400
9, √6, 3 √6, just be very careful with your choices.
00:27:58.400 --> 00:28:01.700
Make sure you look at all of your choices to make sure that you have an excluded one.
00:28:01.700 --> 00:28:05.300
Just because you think you found the right choice, there may be something else going on.
00:28:05.300 --> 00:28:10.100
Make sure you look at all of your choices.
00:28:10.100 --> 00:28:16.400
Question number 10, let us see what we got for question number 10.
00:28:16.400 --> 00:28:22.100
Here we want to find the concavity of a particular graph.
00:28:22.100 --> 00:28:35.700
In this particular case, our function f(x) is equal to 2 sin(x) + 3 × cos² x.
00:28:35.700 --> 00:28:42.000
We want to find the concavity at a given point.
00:28:42.000 --> 00:28:48.300
Let us find and we know that concavity has to do with the second derivative.
00:28:48.300 --> 00:29:06.900
Let us find f"(x) and evaluate f” at the point π, to see what the concavity of π is.
00:29:06.900 --> 00:29:40.200
F’x, the derivative of this is going to be 2 × cos(x) + cos(x) – 6 sin x cos x.
00:29:40.200 --> 00:29:44.400
That is one of our f’.
00:29:44.400 --> 00:29:47.100
We want to do our f”.
00:29:47.100 --> 00:29:51.200
F”(x), I will just take the derivative of what it is that we just got.
00:29:51.200 --> 00:30:11.900
We are going to end up with -2 × sin(x) - 6 × sin x × -sin x + cos x cos x,
00:30:11.900 --> 00:30:26.300
which is equal to -2 × sin(x) + 6 sin² x - 6 cos² x.
00:30:26.300 --> 00:30:30.500
Now we go ahead and evaluate f” at π.
00:30:30.500 --> 00:30:51.500
When I put π in for this, -2 × sin(π) + 6 × sin² of π - 6 × cos² of π.
00:30:51.500 --> 00:30:59.000
sin(π) is 0, sin(π)² is 0, cos(π) is -1.
00:30:59.000 --> 00:31:09.500
-1² is 1, we get -6.
00:31:09.500 --> 00:31:15.800
The answer is d because -6 is less than 0.
00:31:15.800 --> 00:31:17.900
We are concave down.
00:31:17.900 --> 00:31:20.000
That is it, nice, straight application.
00:31:20.000 --> 00:31:23.600
Second derivative is positive, you are concave up.
00:31:23.600 --> 00:31:29.600
Second derivative is negative, you are concave down.
00:31:29.600 --> 00:31:33.600
Let us go to question number 11.
00:31:33.600 --> 00:31:37.800
Here it looks like we are computing a derivative.
00:31:37.800 --> 00:31:44.700
Number 11, our particular derivative, we are evaluating at an indefinite integral.
00:31:44.700 --> 00:31:46.500
My apologies.
00:31:46.500 --> 00:31:56.700
The integral of -3x² × √x³ + 3 dx.
00:31:56.700 --> 00:32:01.500
I think it is just going to be the straight u substitution.
00:32:01.500 --> 00:32:04.500
I noticed x³ and I noticed an x².
00:32:04.500 --> 00:32:19.500
I’m going to try u is equal to x³ + 3, du = 3x² dx.
00:32:19.500 --> 00:32:34.800
Our integral, -3x² × the integral of x³ + 3 dx, is actually going to equal -the integral,
00:32:34.800 --> 00:32:45.000
3x² x is du, this is going to be e ^½.
00:32:45.000 --> 00:32:47.700
We are accustomed to seeing du on the right.
00:32:47.700 --> 00:32:51.500
It is not a big deal, the order does not matter because you are just multiplying two things.
00:32:51.500 --> 00:32:54.800
The multiplication is commutative, it does not matter the order.
00:32:54.800 --> 00:33:02.900
You are used to seeing the du or the dx, or the dy, on the right hand side in the integrand.
00:33:02.900 --> 00:33:05.000
Let us do it that way.
00:33:05.000 --> 00:33:25.100
- the integral of u ^½ du which is equal to -u³/2/ 3/2 + c which is equal to -2/3 u³/2 + c.
00:33:25.100 --> 00:33:29.300
Of course, we plug the u back in, x³ + 3.
00:33:29.300 --> 00:33:39.800
Our final answer is -2/3 × x³ + 3³/2 + c.
00:33:39.800 --> 00:33:44.400
Our final answer, the choice is going to be e.
00:33:44.400 --> 00:33:50.600
I hope that was reasonably straightforward.
00:33:50.600 --> 00:33:59.000
Let us take a look at the problem number 12.
00:33:59.000 --> 00:34:01.800
Here we have a particular function.
00:34:01.800 --> 00:34:06.600
We want to give the value of x where the function has a local extrema.
00:34:06.600 --> 00:34:10.200
In this particular case, a local maximum.
00:34:10.200 --> 00:34:17.700
Local maximum means you take the derivative, you set the derivative equal to 0.
00:34:17.700 --> 00:34:23.700
You draw yourself a number line and you check points to the left of the critical values,
00:34:23.700 --> 00:34:27.900
to the right of the critical values, to see whether the derivative is increasing or decreasing.
00:34:27.900 --> 00:34:32.400
You decide which is local min and local max.
00:34:32.400 --> 00:34:44.100
f(x) is equal to x³ + 6x² + 9x + 4.
00:34:44.100 --> 00:34:47.100
f’(x), very simple.
00:34:47.100 --> 00:34:52.500
We have 3x² + 12x + 9.
00:34:52.500 --> 00:34:54.900
We want to set the derivative equal to 0.
00:34:54.900 --> 00:34:56.100
We can factor out the 3.
00:34:56.100 --> 00:34:57.600
We are going to get the same roots.
00:34:57.600 --> 00:35:03.900
We have got x² + 4x + 3 is equal to 0.
00:35:03.900 --> 00:35:06.700
We can actually factor this one.
00:35:06.700 --> 00:35:10.900
We get x + 3, we get x + 1.
00:35:10.900 --> 00:35:18.700
Therefore, we have x is equal to -3 and we have x is equal to -1.
00:35:18.700 --> 00:35:22.600
Go ahead and draw myself a little number line here.
00:35:22.600 --> 00:35:24.700
I have got -3, I will put it over here.
00:35:24.700 --> 00:35:26.800
I have got -1, I will put it over here.
00:35:26.800 --> 00:35:32.300
I’m going to check a point here, check a point here, and check a point here, in those intervals.
00:35:32.300 --> 00:35:33.700
I’m going to put them into the derivative.
00:35:33.700 --> 00:35:40.000
To see if the derivative is less than 0, decreasing, or greater than 0, increasing.
00:35:40.000 --> 00:35:42.700
That is all I'm doing.
00:35:42.700 --> 00:35:55.600
Let me go ahead and rewrite f’(x), I’m going to use this version right here.
00:35:55.600 --> 00:35:57.100
I should actually use the original version.
00:35:57.100 --> 00:36:02.500
I should have actually written this as 3 × that.
00:36:02.500 --> 00:36:10.900
This is 3 × x + 3 × x + 1.
00:36:10.900 --> 00:36:17.000
To the left of -3, when I check something in this interval, over here, let us try -4.
00:36:17.000 --> 00:36:25.400
When I plug in -4, I'm going to get, for x, I'm going to get 3 × -4 + 3 is a negative number.
00:36:25.400 --> 00:36:29.300
-4 + 1 is a negative number.
00:36:29.300 --> 00:36:34.100
3 × a negative × a negative is definitely a positive number.
00:36:34.100 --> 00:36:37.100
Here the slope is increasing.
00:36:37.100 --> 00:36:39.300
Or if you want I can put a positive sign.
00:36:39.300 --> 00:36:43.100
Some of you use positive, some of you use increasing/decreasing with an arrow.
00:36:43.100 --> 00:36:45.300
It is that way.
00:36:45.300 --> 00:36:47.400
Let us try a point in between here.
00:36:47.400 --> 00:36:49.900
Let us try -2.
00:36:49.900 --> 00:36:57.700
When I plug -2 in to the derivative, I get 3, -2 + 3 is a positive number.
00:36:57.700 --> 00:37:00.400
-2 + 1 is a negative number.
00:37:00.400 --> 00:37:05.700
3 × a positive × a negative gives me a number that is a negative.
00:37:05.700 --> 00:37:09.200
It is less than 0, it is decreasing there or negative slope.
00:37:09.200 --> 00:37:11.900
There you go, that pretty much takes care of it.
00:37:11.900 --> 00:37:14.900
Because it is increasing, the graph is increasing.
00:37:14.900 --> 00:37:21.800
Then, the graph is decreasing, that is a local max.
00:37:21.800 --> 00:37:24.800
To the left it is increasing, to the right it is decreasing.
00:37:24.800 --> 00:37:31.100
That means it is hitting a maximum point.
00:37:31.100 --> 00:37:40.500
There is a local max at x = -3.
00:37:40.500 --> 00:37:46.500
I think the particular choice was choice e.
00:37:46.500 --> 00:37:48.900
Nothing too crazy so far.
00:37:48.900 --> 00:37:57.000
Here we have number 13 and let us see what number 13 is asking us.
00:37:57.000 --> 00:38:02.700
The slope of the tangent line of the graph that will give the value of c.
00:38:02.700 --> 00:38:04.200
We are give a particular function.
00:38:04.200 --> 00:38:18.000
We have 4x² + cx + 2, e ⁺y is equal to 2.
00:38:18.000 --> 00:38:40.500
They are telling us that the slope of the tangent line of this graph, at x = 0 is 4.
00:38:40.500 --> 00:38:46.500
They are asking us to find c.
00:38:46.500 --> 00:38:51.600
The slope of the tangent line, for this graph, at x = 0 is equal to 4.
00:38:51.600 --> 00:38:57.700
They want us to find the value of c.
00:38:57.700 --> 00:39:01.900
Let us see what we can do.
00:39:01.900 --> 00:39:04.300
Let us go ahead and take the derivative.
00:39:04.300 --> 00:39:08.200
In this particular case, let us go ahead and take the derivative with respect to,
00:39:08.200 --> 00:39:15.700
8x + c +, this is the function of y, we are going to do implicit differentiation here.
00:39:15.700 --> 00:39:22.900
This is going to be 2e ⁺yy’ is equal to 0.
00:39:22.900 --> 00:39:37.300
When I rearrange this, I'm going to get y’ is equal to -8x - c/ 2 × e ⁺y.
00:39:37.300 --> 00:39:45.700
I will get 2 × e ⁺y.
00:39:45.700 --> 00:40:00.700
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.
00:40:00.700 --> 00:40:20.800
Therefore, if I plug in this into here, I get y’ is equal to -8x - c/ 2 – 4x² – c ⁺x.
00:40:20.800 --> 00:40:26.500
They are telling me that y’ at 0 is equal to 4.
00:40:26.500 --> 00:40:31.600
y' at 0 is equal to -8.
00:40:31.600 --> 00:40:40.300
y’ at 0 is equal to 0 - c/ 2 - 0 – 0.
00:40:40.300 --> 00:40:45.200
They are telling me that this is actually equal to 4.
00:40:45.200 --> 00:40:57.200
I get - c/ 2 is equal to 4 which implies that c is equal to 8.
00:40:57.200 --> 00:41:00.200
Our choice is e.
00:41:00.200 --> 00:41:04.400
Just differentiate, in this particular case, you are going to get something which is 2e ⁺y.
00:41:04.400 --> 00:41:07.100
You notice the 2e ⁺y is actually separate here.
00:41:07.100 --> 00:41:12.500
You can move these over, plug in for x, and then solve for c.
00:41:12.500 --> 00:41:17.000
I hope that made sense.
00:41:17.000 --> 00:41:23.300
Let us try number 14, see how we are doing here.
00:41:23.300 --> 00:41:26.300
Number 14, what is number 14 asking us.
00:41:26.300 --> 00:41:31.800
It looks like it is asking us to evaluate this particular integral.
00:41:31.800 --> 00:41:49.500
Evaluate the integral of 7 ⁺x - 4e⁷ ln x dx.
00:41:49.500 --> 00:41:51.300
Let us see what we can do with this.
00:41:51.300 --> 00:41:52.500
I’m going to separate this out.
00:41:52.500 --> 00:41:58.500
The integral is linear, the linear of the integral of the sum is the sum of the integrals.
00:41:58.500 --> 00:42:13.800
I’m going to do this as, this is the integral of 7 ⁺x dx - 4 × the integral of e⁷ ln x dx.
00:42:13.800 --> 00:42:16.500
Let us go ahead and deal with the first integral.
00:42:16.500 --> 00:42:17.700
This one right here.
00:42:17.700 --> 00:42:28.100
The first integral, the integral of 7 ⁺x dx.
00:42:28.100 --> 00:42:30.200
I’m going to do a u substitution here.
00:42:30.200 --> 00:42:34.500
I’m going to let u equal to 7 ⁺x.
00:42:34.500 --> 00:42:52.500
Du is going to be 7 ⁺x ln 7 dx, du/ ln 7 = 7 ⁺x dx.
00:42:52.500 --> 00:43:21.300
Therefore, this integral actually = the integral of du/ ln 7 = 1/ ln 7 × the integral of du = 1/ ln 7 u + c.
00:43:21.300 --> 00:43:24.300
I plug u back in which is 7 ⁺x.
00:43:24.300 --> 00:43:34.500
I get 7 ⁺x/ ln 7 + c, that is our first integral.
00:43:34.500 --> 00:43:38.700
Let me see here, where am I?
00:43:38.700 --> 00:44:08.200
The second integral is -4, it is that one, × the integral of e ⁺ln x dx = -4 × the integral of e ⁺ln(x)⁷ dx
00:44:08.200 --> 00:44:23.900
= 4 × the integral of e ⁺ln is just x⁷ dx = -4x⁸/ 8 + some constant c.
00:44:23.900 --> 00:44:26.300
The c and c are not the same.
00:44:26.300 --> 00:44:32.000
If you want you can just c1 and c2, it does not really matter.
00:44:32.000 --> 00:44:50.900
This is this one, which is –x⁸/ 2 + c.
00:44:50.900 --> 00:44:54.500
Now we just put them back together.
00:44:54.500 --> 00:45:09.000
We have the integral that we wanted, it is equal to 7 ⁺x/ ln 7 – x⁸/ 2 + c.
00:45:09.000 --> 00:45:14.400
It looks like that is option a.
00:45:14.400 --> 00:45:18.600
That takes care of the first of 14 of the problems.
00:45:18.600 --> 00:45:25.200
For the next lesson, we are just going to continue on with this particular section and go on with the practice exam.
00:45:25.200 --> 00:45:28.200
Thank you so much for joining us here at www.educator.com.
00:45:28.200 --> 00:45:29.000
We will see you next time, bye.