| Start | So you're going to have some more work. Brace yourselves.
Midterm is in a couple weeks. Same room as last time. Everybody's doing pretty well so I'm not very worried about everybody. Somebody did not do a good job cleaning this board. So we were talking about graphing. And we said well really when you're doing graphing you don't really- it's not the graph itself that's so important. Of course you should be able to draw it. Because that's just sort of a pictorial idea of what's going on. What's really important is to figure out the maximums and the minimums. |
| 0:31 | So as I said last time you know now how to figure out when a curve is going up and when a curve is going down.
But then a curve is going up, as a function is going up, it could be increasing this way, it could be increasing this way, or that way. So how do you tell them apart? Well if a function is just going straight up then it has a particular derivative that tells you what it's slope is. |
| 1:00 | And that derivative is not changing.
So the first derivative is a constant. That's a line, okay? And you know that because you can take the equation y=mx+b, the first derivative is m, the slope. And that doesn't change. Now what if the curve is doing this? Well then the first derivative is increasing. Because the first derivative gets steeper. So the second derivative will tell us how the first derivative is changing. And similarly, there the first derivative is getting closer to zero. |
| 1:39 | So we can use the second derivative to figure out what's called concavity.
And concavity is the curvature. So there's two types of concavity that we care about. There's concave up like that and concave down like that. |
| 2:00 | Okay? So concave up the second derivative is positive.
And concave down the second derivative is negative. When the second derivative is 0 then you have a straight line. Well if could be a flat spot but it's straight. And where the second derivative is 0 is called the point of inflection. A point of inflection is where the curve does a change in concavity. It goes either from up to down or from down to up. |
| 2:31 | So let's take an example. Let's do one of these cubics again.
So f(x)= x^3 -24x +10. |
| 3:05 | How about 18? Never mind, make that 18x.
It's hard to change. If you do it in pencil you just erase. Alright we keep learning that lesson the hard way. Let's take the first derivative. So the first derivative is 3x^2 -18. Can we do what we did before? Set that equal to 0. I should've stuck with 24. |
| 3:32 | So you get x^2-6=0.
x= + - √6. So if we went to the number line we can test the regions and figure out where the curve is going up and where the curve is going down. So we did this a couple of times already. You pick a number less than -√6 like say -10. This will come out positive. |
| 4:05 | You take a number between those two like 0. This will come out negative and you test and you test again it's positive.
So you know it's going up, then down, then up. So it's just like what we've had like several of these now. Now let's look at the second derivative. So the second derivative is 6x. |
| 4:31 | Where is that 0? That's 0 at x=0.
Now test a value at either side. So take a number less than 0, say -1. The second derivative comes out negative. So that tells you that the curve is concave down there. Now pick a number bigger than 0. Whatever you'd like. 1. This comes out positive so the curve is up there. |
| 5:04 | Which kind of makes sense when you have these arrows but the concavity is not always as obvious on other types of curves.
So now if you want to graph it |
| 5:42 | We now know also that at 0 it switches from concave down to concave up.
So let's find our y coordinates. f(0) is just 10. The √6 you get 6√6 -18√6 so you get 10-12√6. We don't care what that is it's just a negative number. |
| 6:13 | And now you have -6√6 +18√6 so now you get 10+12√6.
|
| 6:31 | √6 is about 2.5, 2.45.
So this is 35-ish. This is -25-ish. Or -15-ish. There you go so let's see we have (0, 10). And we've got √6 it's a negative number. So as I said this now we know that the concavity is concaved down all the way to this spot. And then it's concaved up the rest of the way. |
| 7:04 | So it looks just like the other cubics we've drawn.
So concavity will tell you, will help you figure out like I said, what the curvature is. It's very useful for finding maximums and minimums. In addition to the first derivative and sometimes you'll have an equation where it's difficult to solve, to plug in values to figure out maximums and minimums. Because it's messy so you could use a second derivative to test for maximums and minimums. |
| 7:33 | So how can we do that?
Well let's see you have two options. You take the first derivative and you have some point, c. f(x) is negative when x is less than c. And f(x) is positive when x is greater than c. And that's a minimum. |
| 8:21 | Okay so what we've been doing is you look at the first derivative.
If it's negative to the left of a point, |
| 8:36 | If it's negative to the left of the point and positive to the right of the point and 0 then that tells you you have a minimum.
And if it is positive to the left of the point and negative to the right of the point then you have a maximum. But there's another way you could check if something is a maximum or a minimum. |
| 9:01 | If the second derivative, we'll call it 'a', is 0- sorry ignore that.
If the first derivative at a is 0 and the second derivative at a could be positive or negative, Well if it's positive then you're concaved up. So if it's positive that means that this is a minimum point. And if it's negative it's a maximum point. |
| 9:32 | So this is called the second derivative test and by the way if it's 0 it's neither.
So again if you take the first derivative and you get a 0 like here we got say √6 and now you take √6 and you plug it into the second derivative. So you plug it in you get 6√6 that's positive. So that tells you that's a minimum point. If you take the -√6 that gives you 0 for the first derivative. |
| 10:00 | You plug in a second derivative you get a negative value. That tells you that your curve is concave down.
And therefore that's a maximum. So why would you use the second derivative test vs. the first? We want to kind of be able to do either way. Because sometimes as I said with a problem it's messy to plug in numbers. You might have some complicated radical quotient kind of thing. And finding the second derivative's really impossible so you just plug in the first derivative. Sometimes the second derivative is trivial. Like this. Say the derivative is easy then you know whether something's a maximum or a minimum. |
| 10:33 | You don't have to do the effect or plugging in numbers.
Okay so let's give you guys an example. How are we doing so far? No one is speaking which means you either know it or you're asleep. It's good because then the audience who listens to these videos later won't have to worry about noise. I mean it is a Friday. Yesterday was a Thursday. You know what Thursday is all about. |
| 11:03 | Let's say we give you something really fun.
I'll give you something fairly hard. |
| 11:35 | That's fun.
Let's see if you can find where it's increasing, where it's decreasing, the concavity, and the maximums and the minimums. That's a nice nasty question. Of course if you can handle it that's a good sign. |
| 12:18 | I recommend distributing before you take the derivative.
|
| 15:52 | x * x^1/3 is x^4/3.
|
| 16:23 | Alright a lot of you seem to be waiting for me so let's do this as a team.
First, let's distribute that x. |
| 16:35 | How do we know it's 4/3?
x^1, 1+1/3 is 4/3. So first derivative 1 - 4/3x^1/3. And the second derivative is -4/9x^-2/3. |
| 17:02 | So these are slightly annoying.
Let's set this equal to 0. And you get- I'm sorry equal to 0. You get 1= 4/3x^1/3. Remember 1/3 is cube root. So put the fraction over to the other side you get 3/x = x^1/3. And if you cube that you get 27/64 for x. |
| 17:35 | It's an annoying number.
I wouldn't worry too much about exact values. It's important just to figure out positives and negatives and about how big things are. It's a little less than half. And here this is the same as -4/9 3√x ^2. That's what x^-2/3 means. That's never 0. |
| 18:00 | Why is that never 0 because the numerator can't be 0.
However, the denominator still has a problem when x is 0. So it's possible concavity will change. So if you have to do a hard graph like this. So now you say to yourself so I want to test the numbers from the first derivative? Or do I want to test the numbers in the second derivative? Because do I really want to plug 27/64^2 and put it in that? |
| 18:35 | Not particularly.
So what do we do? Well we use the first derivative. Make a number line. 27/64. Pick a number less than that like 0. Cube root of 0 is just 0 so this is positive. The curve is going up. |
| 19:00 | Pick a number bigger than 27/64 like 1.
1-4/3 is negative so it's now going down so this is a maximum. The y coordinate will be messy. Not really. Alright the second derivative we'll do the second derivative test. We don't have a place where the numerator is 0 so we're not going to have a point of inflection. |
| 19:32 | Something interesting goes on at the second derivative when you plug in 0 it's undefined.
So it's possible you will change concavity there. Pick a number less than 0 say -1. You get, well this is squared so it's always going to be positive. Cube root is positive. This thing's always going to be negative. So this curve is always concave down. But first it's concave down going up. Then it's concave down going down. |
| 20:08 | So what would that look like?
Well let's see if we can figure out how to graph that. So if we plug in- hang on- well we could do it here. f(27/64) You'll get 27/64(1- the cube root of that. The cube root of that is 3/4). |
| 20:30 | So you get 27/256 which is a very small positive number.
That looks good. When x is 0 you get 0. And we know that the curve is alway going up but it's concave down like that. Then what happens after? Let's see. |
| 21:01 | It's going down and concave down.
It can't be that simple. What else do we know? That's probably about it so yeah it kinda goes like that. I take it back. Easier than it looks. It's not a parabola however. Don't make that mistake. Were any of you able to pull this off? Graph it? Do the curves? No? Too hard? Alright I'll give you an easier one. |
| 21:30 | One person solved it. The rest of you just sort of stared straight ahead.
We'll give you an easier one let's see. |
| 22:28 | Yeah I like that one. Alright how about something that looks like that?
|
| 22:36 | I'll give you a couple minutes.
|
| 30:11 | Okay here's a fun one.
So your attack when you do these problems should always be the same. First, find the derivatives and set them equal to 0. Then sign test them and then do the best you can to graph. |
| 30:32 | Lots of you will have trouble with finding zero's, plugging in those things to find the y coordinates,
and maybe with the graph but you should be able to do the other half effectively.
And if you practice this you'll get good at the first types. So let's take the first derivative. You get f'(x)= 4x^3 -16x. And set it equal to 0. |
| 31:01 | Factor out 4x. 4x(x^2 -4)=0.
Then you get 4x(x+2)(x-2)=0. Were we able to do that? So 3 points. You have 0, 2, and -2. Alright let's do the second derivative while we're here. |
| 31:33 | f''(x)= 12x^2 -16.
By the way usually if the first derivative is easy numbers the second derivative is easy numbers or the other way around. It's very hard to write a problem where they come out integers twice in a row other than a trivial graph. So here you get 12x^2 =16. |
| 32:01 | So that means x^2 =4/3.
x= +-√4/3 So the number line would the be -√4/3, √4/3. 4/3 is a little bit bigger than 1 so the √4/3 will be 1-ish. How'd we do on that? |
| 32:38 | That's an equals sign.
I could. You mean factor a 4 out of here? You'll end up at the same spot. And in fact √4/3 is 2/√3. So some of you may reduce that one more step. Doesn't really matter much either way. Alright I'm going to cover this for a second. |
| 33:01 | So I pick a number less than -2 like -3.
And I plug in. I get a negative value, negative value, negative value. 3 negatives makes a negative value so the curve is going down. I plug in a number between 0 and -2 like -1. Negative, positive because -1+2= -1, negatives. That's 2 negatives, 2 negatives a positive. I mean a negative times a negative makes a positive. |
| 33:31 | You only have to count the number of negative signs by the way. You never care about the positive signs.
The number of negatives multiplied together tells you whether something is positive or negative. If it's an odd number of negatives you get negative. If it's an even number of negatives you get a positive. Pick a number between 0 and 2 like 1. Positive, positive, negative. We only have one negative now. Going down. So this tells us right the minimum, maximum, minimum. |
| 34:06 | And I pick a number bigger than 2 like 3.
And when you plug in everything's positive so it's going up. So if this were a test question or a homework question because you're going to have homework coming, we might ask where is this increasing and where is it decreasing? So you'd say it's increasing when x is between -2 and 0, |
| 34:35 | or when x > 2.
And you could do that in interval notation if you want. Notice you never use an equals. Because at 0, at -2, it's not increasing or decreasing. So these will always be just less than and greater than. Okay there won't be less than or equal to or greater than or equal to. Decreasing, well it's decreasing when x < -2 or when x is between 0 and 2. |
| 35:16 | So far so good?
Painful? It's a lot. Okay second derivative test. Pick a number less than -√4/3 like -2. |
| 35:30 | This will come out positive, bigger than 16 so it'd be positive.
So it's concave up. Which kind of agrees with that shape. Between -√4/3 and √4/3 like 0. This is negative because you get -16. So it's negative so it's concave down, And bigger than √4/3 like 2. Positive because 48 > 16. |
| 36:02 | So it is concave up when x < -√4/3 or when x > √4/3.
And it is concave down when -√4/3 < x < √4/3. |
| 36:36 | Okay now let's graph it.
|
| 37:09 | Okay so we need some y coordinates.
We've got f(0), f(2), f(-2), f(√4/3), and f(-√4/3). Okay f(0) is easy that's 10. |
| 37:32 | f(2)= -6.
f(-2)= -6. √4/3 let's see. It's -22/9 I think. |
| 38:00 | And so is that one.
I think that's right. Doesn't matter. Okay let's graph it. This curve is symmetric. Symmetric because these are matching, these are matching. So let's see. We have (0, 10). (2, -6) (-2, -6) Then a couple of values like that. |
| 38:30 | Okay this is a maximum.
This one's a minimum. And then we switch right there from concave up to concave down. We switch right there from concave down to concave up. So these are called the points of inflection. So you have a point of inflection at (+-√4/3, -22/9). |
| 39:10 | How'd we do on this?
Eh? Well fourth degrees are hard. I can give you a 5th you know. Alright let's just do a little more. Make sure everybody gets the concept. |
| 39:33 | And as I said notice I'm sticking to polynomials. That won't last forever.
So let's do another where I'm not going to graph them I'm just going to analyze it. |
| 40:18 | That says x^5 -80x +1.
The 80 looks scary but it should make it easier. |
| 40:30 | So we're just going to find where is this increasing, decreasing, where is the concave up, where is the concave down?
I won't worry about the picture. So first derivative you get 5x^4 -80. Set that equal to 0 so let's see. 5x^4 =80. x^4= 16. x= +-2. That's not so bad. |
| 41:21 | Let's take the second derivative.
It's just 20x^3. So that equals 0 when x equals 0. |
| 41:38 | Okay now let's do a little analysis.
Don't get your finger caught in there. I know from experience. Okay so number line. |
| 42:00 | So our number line for the first derivative.
We've got -2 and 2. We're missing a couple by the way because notice it's x^4 =16. We're missing the two imaginary complex roots. But that's okay don't worry about that. That never effects this class. So at -3. -3^4 is 81 times 5 is going to come out bigger than 80. So you get a positive first derivative there so it's increasing. |
| 42:34 | So plug in 0. 5(0)^4 is 0 so it's negative. And then this will go positive again.
These alternate a lot but not always. And we do the second derivative. We just have 0. And since it's 20x^3 and a negative number will make that negative and a positive number will make tat positive. So the curve is going up, down, up. Concave down, concave up. |
| 43:06 | So if I say on what intervals is this increasing it is increasing when x < -2 or x > 2.
And then it's decreasing between them. |
| 43:32 | It has a maximum at x= -2.
I won't worry about the y coordinate for now. It's findable. And it has a minimum at x=2. As I said you can use interval notation it doesn't matter. The TAs probably like it but it's not necessary. |
| 44:04 | Certainly won't be necessary on the midterm.
It's concave up when x>0 and concave down when x<0. And there's a point of inflection, well we could find the coordinate (0, 1). Point of inflection- the concavity changes goes from negative to positive and has a zero there. |
| 44:40 | How do we feel about that?
So you'll have some MyMathLab stuff of this to practice. Plus on the paper homework there's two of these. An easy one and a not so easy one. Remember you could use your calculator so feel free to graph it in your calculator for your homework. You just can't use it on the exam. But you can feel free to put it in your calculator so you can get a picture to know what the graph looks like. |
| 45:02 | And then start to interpret it from there.
After all if I told you couldn't you could anyway. I can't enforce it. Alright see everybody on Monday. |