Stony Brook MAT 122 Fall 2017
Lecture 26: Midterm review, part 1
October 27, 2017

Start   Let's do some practice, we ready?
Have you practice first just finding some derivatives.
Two sneezes is all you get. I think you're the one that got me sick. Let's see... Of course, I went to the doctor's office yesterday and I felt fine going into the office, but two hours later I started to get that.. you know.
1:39Ok, there's 4 to keep you busy.
Ok, let's go through these.
As I said, these aren't so much difficult, as they're just you have to be very careful that you don't make a little mistake, and they're a little on the tedious side.
These equations would never show up on an exam, its more about just making sure you can handle the various rules.
2:03And god forbid we just give you something easy, because then everyone would get an A and we'd all be happy.
So derivative. This whole thing is raised to the 4th, so first we do the derivative of something to the 4th.
So that's 4 times the inside to the third and you don't change the inside. (4(8x^3+6(x)^(1/2)+1)^3 Then we do times the derivative of the inside, and that would be 24x^2+6(1/(2(x)^(1/2)))
2:38Or, you could have made that 1/2(x)^-.5 [?] Now as I said, don't simplify. So you don't have to clean this up, or distribute the 4 or all those things. Just leave it alone.
If we ask you to find the zeros, it would never be on anything that difficult, because it's too much work.
I wouldn't even ask you to find the second derivative of something like that.
3:00Ok? You were all able to do this first one?
By the way, I got a couple emails about the last [?] , couple of you said even though you got the right answer it marked you wrong and all that.
I haven't figured out how to look at what you're putting vs what the computer thinks is the right answer.
So I was supposed to speak to the publishing person today because this is my first class with it, so he's going to walk me through that.
I think it was the last question that was bothering some people? So if they made an error on their part, sometimes its an input error on your part,
3:36like you have a comma in the wrong place or something silly. If I can find that then I'll put the points back.
Although, I can tell you now, 1 point doesn't affect your grade so don't get excited about that. Um, but when I speak to the publisher later today we'll try to figure it out.
Alright, this one. So this is product rule, and inside the product rule you're gonna need chain rule to do the derivative so,
4:00We take the first function and we leave it alone, and I'm going to multiply it by the derivative of the second function.
The derivative of the second function is 3(x^3-4x-1)^2 times the derivative of the inside, (3x^2-4).
I'm going to run out of blackboard space, I have to do it underneath.
Ok, so you leave the first one alone, you take the derivative of the second function which requires chain rule.
4:313 times the inside squared times the derivative of the inside.
Plus, the reverse. The derivative of the first function is 2(x^2+3x+2)^1*(2x+3)*(x^3-4x-1)^3.
And again, you don't have to simplify that. That would not be fun to clean up.
5:04But if you had to take another derivative of it, first, you'd want to make this as clean as possible before you do a second derivative. Ok?
How are we doing on these so far? Good?Okay, yes. Yes, you do that. Thank you for catching that, because otherwise that would be in the video and 100 years from now people would be confused.
5:38I think the idea is even after I retire and or have died, they'll still be running these videos. Still be me. Making stupid jokes.
Walking around the classroom. Maybe they'll make different videos by then, we'll see.
Alright, this you could also think of as ((1-x^3)/(1+x^3))^.5
6:03I might do some playing around with this before I actually take the derivative.
But, it's a real test of your algebra skill. I don't recommend it.
So let's do the derivative. First you do the derivative of the outer function, so its (1/2)((1-x^3)/(1+x^3))^-.5.
Times the derivative of the inside. Which is quotient rule, ((1+x^3)(-3x^2)-(1-x^3)(3x^2))/(1+x^3)^2
6:50This one, you would definitely want to simplify by a lot before you took the second derivative.
Because, for example, you'd get some cancellations in here. You'd get (-3x^5)+3x^5, so you'd get some cancellations it'd be very nice. Ok?
7:09And when you flip this, you'll get a (1+x^3)^.5 on top, and the (1+x^3)^2 on the bottom so you can cancel there, So this would clean up a lot if you had to find a zero or take a derivative, but as I said, I wouldn't ask you to do this, it's a lot of work.
Ok, another one. Quotient rule.
7:36Bottom function times the derivative of the top, and the derivative of the top would be the chain rule.
So that's 4(8x^2-10x+2)^3*(16x-10) Minus, the other way around. Top function times the derivative of the bottom which is 2(x^5-x^3)^1*(5x^4-3x^2).
8:23All over the bottom squared.
8:31And notice, this has an (x^5-x^3) term, this has an (x^5-x^3), and this, so'd you'd be able to cancel 1 from each part of the fraction.
Theres 1 place you could simplify, so again, this can simplify a lot if you wanted it to. How'd we do on these? Feel good about these?
You guys were pretty fast, which is a good sign. [?] Alright? Next one.
10:05Yup. Find the equation of the tangent line to the curve y = (x^3+1)^(.5) at x=2.
Alright.
One thing about the cold is my voice gets deeper.
Ok, we want to do the derivative, something you need for the tangent line, and what do we need?
You need y coordinates and you need a slope.
10:32Sometimes they give you the y coordinates, sometimes they don't. And then you can plug it into his equation.
Ok, so let's find y. When x =2, y= (2^3+1)^.5, or y=3.
Your equation would be y-3 = m(x-2)
11:03Now we need to find the slope. Take the derivative, as I always say, this is calculus class, you only have 1 trick which is take the derivative.
This is to the 1/2, dy/dx = (1/2)(x^3+1)^(-1/2)*(3x^2)
11:34And don't simplify that at all, immediately plug in.
(dy/dx)(2) = (1/2)(9)^(-1/2)*(12) What is 9^(-1/2), well 9^(1/2) is 3 because it's the square root.
12:05So this is 1/3. So we get (1/2)*(1/3)*(12), that comes out to 2.
You see what I did? So if that's 2, our equation is y-3 = 2(x-2).
12:33And we can leave it like that. We don't have to make it 2x - 1.
Good? How'd we do on that? Easy? Eh? Alright, I'll give you some more practice.
15:11Ok, given y = 2x^3 -3x^2 -12x + 8, find the integrals where the function is increasing or decreasing, find the coordinates of the maximums or minimums, Find where it is concave up or down, find coordinates of points of inflection and graph.
15:33Alright that's enough time, lets get started on it.
So problems like this, theres lots to do, so it'll be very sequential.
First we said well where is it increasing or decreasing? So a function is increasing where the derivative is positive, and a function is decreasing when the derivative is negative.
16:00So let's take the derivative. The derivative is dy/dx or y' = 6x^2 - 6x -12.
If you want to find whether it is increasing or decreasing now, we'll set it equal to zero.
And that factors into (x-2)(x+1)= 0.
16:31When I give you these on an exam, they'll be relatively easy to factor and you'll get relatively easy numbers, you may get a square roots, but nothing terribly messy.
So now you take the 0s of this, which are x =2 and x = -1.
And those are potential maximums and minimums. I'll go to a number line,
17:01And test whats going on with the derivative by plugging in numbers.
So if we take a number less than -1, like -2, this becomes negative, and this becomes negative, so 6*negative*negative is a positive. So the curve is going up.
Pick a number between -1 and 2, like 0. Plug it into the derivative, by the way you can plug it into any of these forms, it's just usually easiest when you plug it into the factored form.
17:35You get 6(-2)1 = -12, thats negative, so the function is now going down.
Then pick a number bigger than 2, like 3, or a trillion. Then plug in 3 you get 6(1)(4), so the curve is going up again.
Now you know you have a maximum at -1 and a minimum at 2.
18:05Ok? Now let's go work on where it's concave up and concave down, and later we're gonna need those y coordinates.
So we take the second derivative.
18:35And you get 12x-6.
Set that equal to 0, you get x=1/2.
So 1/2 is a potential point of inflection. Yes?
Correct. You can plug into any of these forms, whatever is easiest for you. Generally it's easiest to plug in here, because you have to square a number when you plug in there.
19:08So again a number line, pick a number less than 1/2, like 0, when you plug it in the second derivative you get -6. It's negative so the function is concave down.
Plug in a number bigger than 1/2, like 1, 12-6 is positive, its concave up.
19:31So far so good? Ok, now we need some y coordinates.
And uh, let's see. We have -1, (1/2) and 2, oh 1/2 is slightly annoying but not too bad.
So you always find y coordinates by plugging the values into the original equation, So you plug -1 into the original equation, you get 2(-1)^3-3(-1)^2+12(-1)-8 = 15.
20:05You plug in 2 and you get 2(2)^3-3(2)^2+12(2)-8 = -12.
And you plug in 1/2, 2(.5)^3-3(.5)^2+12(.5)-8 = 3/2.
Do you agree with me on this? I don't need a calculator yet. It's my test. Every year. If it's wrong, put me in a box, send me to Florida.
20:38Now we just plot. So let's see, we got (-1,15) is a maximum, so we have the curve doing something like that and its concave down.
And that at ((1/2),(3/2)), it's going to shift to be concave up.
And then at (2,-12), that looks like (2,-12), it goes up again. Ok?
21:07These cubics all have the same general shape.
How'd we do on this? Alright, I wrote one more, we could do together, or you guys could work on it for like 4 minutes. Whatever you prefer.
21:30Together after 1 vote. Can one vote carry?Ok, I guess, if that's what you want.
x^4-6x^2+4.
Try to get 1 step in front of me, while I walk here and walk back, and erase.
22:15Ok, so first derivative. First derivative is 4x^3-12x.
And that's not so bad. I want you to find where that's 0.
22:30Pull out 4x and you have 4x(x^2-3), and that's 4x(x+(3)^.5)(x-(3)^.5).
So then, theres a zero at 0, -(sqrt)3 and (sqrt)3.
23:05Ok, pick a number less than -(sqrt)3, like -2.
That will be negative, that will be negative, that will be negative. So it's going down.
Pick a number between -(sqrt)3 and 0, like -1.
This will be negative, this will be negative, -1+(sqrt)3 is positive, the (sqrt)3 is bigger than 1. So it is going up.
23:33Now pick a number between 0 and the (sqrt)3, like 1.
And you get 4 times a negative times a positive, is a negative, so it's going down again.
Pick a number bigger than the (sqrt)3, like 2, and everything is positive, so it is going up. By the way, with these polynomials, they usually alternate. Ok?
24:00So we've got a minimum, a maximum and a minimum.
And you end up with that sort of W shape.
Ok, second derivative.
Second derivative is 12x^2-12, set that equal to 0, factor out 12, you get 12(x-1)(x+1).
24:41So that has zeros at 1 and -1.
Ok, go to the number line, pick a number less than -1, like -2,
25:03This comes out to a negative and a negative, which makes a positive. So concave up here.
And then a number between -1 and 1, like 0, so that is a negative, so it's concave down.
Bigger than 1 is a positive, so it goes up again.
25:31Ok, now we just need some y coordinates and we can graph this.
Oh, let's see. I've got -(sqrt)3, 0, (sqrt)3, -1 and 1.
Let's do the easy ones, this always goes into the original equation. So when x=0, y=4.
26:03When I plug in x=1, I get -1, So when I plug in 1 and when I plug in -1, x^4 and x^2 are the same, so we get the same answer when we plug in positive and negative 1.
It's x^4 +6x^2 so everything's an even power.
When I plug in (sqrt)3, I get 9-18+4 = -5, so they would both be -5.
26:37You can do it more slowly, I'm just saying.
Alright, let's graph it. This function has a maximum at (0,4), has minimums at ((sqrt)3,-5) and (-(sqrt)3,-5),
27:00And it changes its concavity at -1.
So around there, it switches from being concave down to concave up.
Ok, and when you draw a picture, make sure you label these points or you do something like this.
Be organized so TA's don't have to figure out what you're up to.
27:37Ok, you could do that, you could do this, next to your graph.
Ok? Try to remember when you take the exam you don't want to confuse the TA's.
They're grading 100 of you, so we want things to be relatively easy to understand.
Alright, we'll do some more on Monday.