| Start | Some of the function stuff from the very beginning were not going to test on the final. Like the domain thing and stuff like that.
It's ok, I mean part of the course is a pre-calculus review, but since none of you are going on to do more math from here, at least I hope none of you are, if you are, you're in the wrong course. This is a terminal course, as they say, so you'd have to go back and take other things. You know, domain is important, you want to know what you can and can't use, like negative dollars, that doesn't make sense in a lot of problems. |
| 0:34 | But um, generally, you know you can't take the square root of a negative number, so we're not going to get too hung up on some of those things.
Generally, the exam will cover limits, derivatives, some integration, you should be able to find when things are going up and down, increasing, decreasing. You should be able to find concavity, So I'll repeat. Limits, derivatives, increasing/decreasing, so when a function is going up and when it's going down, concavity, when it's concave up and concave down, |
| 1:09 | maximum/minimum, what else, with integration you should be able to do regular basic integrals, integrals with substitution, integral approximations with rectangles,
show you use the integral to find the total of something.
|
| 1:30 | You should be able to take derivatives of a bunch of things, like the product rule, quotient rule, chain rule.
Eh, I think that's about it. Didn't put any continuity on the exam. Again, it's important, but actually mini business functions are discontinuous. There's an abrupt jump, so if you look at say, amazon prices. One day it's $379, the next day it's $479. So there's no smooth transition from one to the other. |
| 2:00 | Um, I think that's about it, at the moment the exam is 9 questions with sub-parts.
So, as I said, I figure if you really know what you're doing you'll be out of there in an hour, hour-hour and a half for most of you. Some of you will sit there for the entire 2 hours and 15 minutes, driving yourselves and the proctors crazy, while you hope that some last answer strikes you from above. But you never know, it might. Ok? Don't be too anxious about this exam. The exam will be in Javits 101, again Javits 101. I will put this on blackboard. |
| 2:36 | I think tomorrow. Blackboard is misbehaving right now, so I can't get to it to do some stuff.
They'll have it fixed in a couple hours. Right now when I go into blackboard it shows me last semester's courses. Which either, this isn't December, it's may, or they have some work to do. Is this May? It's really cold for May. And dark. I'm going to go with they're wrong. Um, and there you go. Alright, so let's do some practice stuff. So I'll put a few problems up for everybody, and then we'll go over them. |
| 3:09 | And as I said, I'll make a list of everything for you guys later.
Why don;'t you practice |
| 4:35 | Alright, let's warm up with some of those.
Let's see how well everyone remember's their limits. Ok, let's start going over these. So we look at the first one. So remember when you get the limit, you're trying to see what happens when x gets really close to 3. A little bit less or a little bit more. So the first thing we do is always just plug in 3 and see what happens. |
| 5:01 | So we plug in 3 here and we get 3^2-25/(3-5)
Which is -16/-2, which is 8. So nothing special happens.
Ok? There's no issues at x=3. So if you look at the curve, it basically goes through the point (3,8). So you don't have a problem at 3, however you do have a problem at 5. Cause here if you plug in 5, |
| 5:34 | You get 25-25=0 on top and 5-5=0 on the bottom, so you get 0/0, which makes you nervous.
But before you decide you can't take the limit, you see if you can rewrite the problem. So you notice that this factors into (x+5)(x-5)/(x-5) |
| 6:04 | So the problem is right when you get close to 5, you're getting the factor on top is making the 0, and the factor on the bottom is making a 0.
Remember you're doing the limit as x approaches 5. Not actually at 5. You're at 4.999 and 5.0001, you're very close to 5, but you're not actually at 5. So that means that these are the same, they're not 0, so you can cancel them. |
| 6:33 | Now when you plug in 5, you get 10.
So if you graphed this, there would be a hole at (5,10) so if you did the graph, right at 5, there would be no value. But if there were a value, it would be y=10. But at 4.999, you're right around 10, I forget if you're bigger or smaller, but it doesn't really matter. you're at 9.999, ok? |
| 7:02 | Now what about when x goes to infinity?
Well we know that this thing is like x+5 anyway, cause we just did that. And as you go to infinity, the top power is squared. And the bottom power is just linear. So the top function will affect what is happening much more than the bottom function, the point of the top function was to dominate the whole thing. So if you plugged in 1000, you'd have 1,000,000-25/1000-5 |
| 7:33 | So the million would be significantly more, have significantly more impact than 1000. If you put in 1,000,000, you have a trillion-25/million-5 so it starts irrelevant what goes on on the bottom.
So this is just going to go to infinity. So far so good? There will be some limits on the final. And you know, some of them you have to show some work, some of them you don't have to show some work. Ok? |
| 8:04 | Even if you look at it and you go this is obvious, show a little work so we don't think you're just cheating ok? Protect yourself.
You can determine what's a lot of work and what's not. Alright, limit as x approaches 2-. Well 2- is a little bit less than 2. if you square a number a little less than 2, you get a number a little less than 4. Since that's less than 4, this will be taking the square root of a negative number, so the limit does not exist. |
| 8:41 | I'm not sure what work you can show on that.
However, when x is just a teeny bit bigger than 2, When you square it you'll get 4, 4-4=0, so this will be 0. |
| 9:01 | So then if you wanted to find it exactly at 2, you always test whether the left side agrees with the right side. Since the left side does not exist, this limit does not exist.
The only has a right hand limit, or a right side limit. It doesn't have a left side limit. So you can't say what the limit is at 2 because at the limit you have to squeeze in from both sides. And when you squeeze in from both sides, you only an answer when you go from the one side, not the other side. |
| 9:35 | Alright, let's look at the next one.
So now we look and the degree of the top is 8x^3, and the degree is the bottom is also 4x^3, so these are moving/growing at the same rate. So if you wanted to do the limit as x goes to infinity, you could ignore these terms. the fact that this is added and this is subtracted doesn't mean anything because they're only x. |
| 10:03 | So the only thing you care about is that this behaves like 8x^3/4x^3, which is 2.
Alright, what about that last one. Well, that last one is saying, let's see. I'm doing x approaches a number, you have to check the left hand and the right hand limits, |
| 10:32 | So let's see the limit as x approaches 8 from the plus side of 9/x-8, that'll give me 9/0. Which is infinite.
So the question is, is it +infinity or -infinity? Well since x is just a little bigger than 8, this is just bigger than 0, so it's positive, so that's positive infinity. |
| 11:01 | And then if we did the other side, the left hand limit,
Now you get a number just smaller than 8. 7.9999.
So this will come out negative, so this will be -infinity. Since these don't agree, the limit does not exist. |
| 11:31 | So remember, when you're doing to a number, if we don't give you a +/-, you have to check both sides.
Ok? How'd we do on these? Are we loving the limits? Good. Alright, now we can give you different types of limits. |
| 12:53 | Ok, use the definition of the derivative to find f'(x) if f(x)=x^2-x. Now you know what the derivative is.
|
| 13:04 | Ok, you can do that in your head. So you know what you're supposed to get in your answer so let's see how you do.
You want to do this with the definition of the derivative. In the book they refer to this as the simplified difference quotient. So if you're looking for problems in the textbook, they call it the simplified difference quotient.I have no idea why. |
| 13:32 | Well, I do, but we can call it the derivative.
So what you want to do is the limit as h goes to 0 of f(x+h)-f(x)/h and we know we have to end up with 2x-1 because we know what the derivative is x^2-x is, right? Ok, so this will be the limit as h goes to 0 of [(x+h)^2-(x+h)]-[x^2-x]/h. |
| 14:16 | And the answer to your question, no you actually don't have to write limit at every step, but you need to write limits at the first step and the last step.
Which should demonstrate that you understand that you are simplifying a limit before you take the limit. |
| 14:33 | Ok? So,
Let's work on the numerator for a minute. That's (x^2+2x+h^2-h-x)-(x^2-x)/h.
So you have x^2-x^2, you have -x-(-x), so you can simplify this to (2xh+h^2-h)/h |
| 15:17 | Remember what I told you guys earlier in the semester. You should have an h in every term. Every term should contain at least 1 h,.
So you can factor it out. So that becomes lim h goes to 0, h(2x+h-1)/h |
| 15:42 | And now you let h be 0. And you get 2x-1.
Alright let's do another one of these, make sure you get it, and then move on to some other stuff. |
| 16:08 | So again, we're going to need to do the limit as h goes to 0, etc, so limit as h goes to 0 of (1/(x+h+3)-(1/x+3))/h
So right now if you plug in 0, (1/x+3)-(1/x+3)/0, so we have some work to do.
|
| 16:35 | So how will we simplify the top? Well we can get a common denominator. Take the left hand fraction and multiply it by (x+3)/(x+3).
Take the right hand and multiply it by (x+h+3)/(x+h+3), in other words, this |
| 17:07 | Ok? So common denominator. We want to combine those into a single fraction.
And now add the 2 fractions together since you have a common denominator, |
| 17:30 | and you get ((x+3-(x+h+3))/(x+3)(x+h+3))/h
Ok, you have x-x, you have 3-3, so this simplifies to the limit as h goes to 0 of -h/(x+h+3)(x+3)*h.
|
| 18:12 | Why can you do this because you can think of this as h/1, so you can flip and multiply.
Ok, now the h's cancel, |
| 18:30 | -1/(x+h+3)(x+3), now when you plug h in as 0,
you get -1 on top, and (x+3)(x+3), or -1/(x+3)^2
Now you could find the derivative of that, with the chain rule, and you should get the same answer. Don't want to get ahead of ourselves.
|
| 19:08 | Ok, let's practice some more stuff.
Let's take some derivatives. |
| 20:02 | So each time you want to find the derivative.
|
| 20:48 | Ok, there's our first set.
Let's do these. So these are all very simple derivatives. The first one, you just use power rule, you get 12x^2-22x+6, you know you should be able to do them as fast as I'm writing them. You should be able to just do them in your head. |
| 21:14 | Ok, the second one, well you could think of this
as x^1/2-2x^(1/3) if you're not good with the radicals.
But remember the derivative of (sqrt) of x is just (1/2(sqrt)x), or you could make it (1/2)x^(-1/2), it doesn't matter. |
| 21:37 | And this would be -2/3x^(-2/3), which is the same as (1/2(sqrt)x)-(2/3(sqrt)x^2)), no you do not have to do that if you don't want to.
You should be able to do that, but you don't need to do that. |
| 22:04 | Same here, you should be able to think of this either way.
The derivative is -2/x^2 +10/x^3 Or you could do, -2x^-2+10x^-3, either answer is acceptable. |
| 22:46 | Ok, product rule.
We're not going to do it here. Ok, first function, times the derivative of the second function + second function times the derivative of the first function. |
| 23:20 | And last, but certainly not least, quotient rule.
Bottom function, (low*d-high - high*d-low)/(low)^2 |
| 23:56 | How did we do so far?
Good sign, ok, Wednesday we'll do some more review, Friday we'll do some more review. |
| 24:04 | I will put up in blackboard at some point today a list of the various ranges of problems for the exam, you guys can use the book to practice those. Ok? Good luck.
|