| Start | When a function is when a function is increasing its going up when's decreasing is going down The graph is going up. The graph is going down to the top it stops So the derivative when the derivatives positive that tells you a curve is going up When the derivative is negative, it tells you the curve is going down And we moving the zero that tells you the function the curve has stopped It's not always a maximum or a minimum, but it's a flat spot. It has ceased to go to increase And now it either decreases or increases some more. So if |
| 0:31 | Where's a piece of chalk?
So if you go for positive to negative Right, you have a positive derivative. That's f'(x) > 0 Here f'(x) is less than zero, and here f'(x) equals zero But you can also have a function that goes like this And right there |
| 1:00 | It's zero. So here
It's going up and it's going up again. So you go positive zero and positive, so you can have a flat on the curve
like, um, x^5,
Ok?
And you can go down So if you're down with you're decreasing and then increasing in zero, or of course you go down and get flat but these all assume a continuous curve if you have discontinuous, then you could just sort of stop and then jump and go somewhere else |
| 1:32 | So this is you sticking to smooth stuff now the polynomial. So let's just do another example
Make sure everybody feels good about this
We'll do it together f(x) = x^3/3 -
I'll fix that in a second, where's my eraser?
|
| 2:01 | f(x) = (x^3/3)-2x^2+4x-1
Okay, so we're gonna find
the relative extrema,
mean, maxima, minima,
extrema extremes.
Or be sure extrema includes both maximums are known. Okay. So what you're going to do is find the first derivative and set it equal to zero. |
| 2:30 | So the first derivative,
Let's see bring the 3 down the threes cancel you get x^2-4x+4
and then we're going to set that equal to zero
I'm already regretting picking this problem, but why not?
That's what happens when you just point to one in the book and go. Okay, so now we need to find when this is zero, so this is quadratic |
| 3:00 | you remember eighth-grade you did quadratics and when you wondered why this is why.
So you only get one zero it's going to zero at x =2. And this is (x-2)^2, right? So something squared is always positive or 0 so you don't really need to sign test this it's always going to be positive |
| 3:35 | But just to be sure you picked a number less than 2 like 0,
You plug it in and you're gonna get 4 so it's gonna be positive, so curves going up
It's flat at 2 because the derivative is 0, remember, the derivative = 0, means you have a horizontal tangent line.
So nothing's happening at that instant when I say nothing is happening this mean you stopped Okay, but really only for an instant see you throw the chalk up, you throw the chalk down. |
| 4:00 | It stops at the top but doesn't stay there. It'd be fun, you know, Harry Potter, you get to stay there
they say something usually but I
Don't I didn't go to that class cause I'm a muggle.
See, we're gonna watch this video of 30 years and go 'what's a muggle'? All right Now I pick a number bigger than 2 like 3, 3-2 is positive 3-2 is a positive So it's positive again, so it's still going up. So this is going to be a flat spot So this is a curve where nothing interesting happens at two it just stops for a second |
| 4:38 | You know you're driving you stop to go to the bathroom and then you drive some more
Okay, that's that easily can be something that this is modeling
then you just need to know what's the y-coordinate that goes with x=2 when you plug in 2.
You get f(2), remember that's how you find the y-coordinate you just plug into the original equation, that's (8/3)-8+8-1 which is 5/3. |
| 5:07 | Because the 8 and the 8 cancel.
Okay, so at (2,(5/3)) if you're graphing this You know, let's see what an x intercept at negative 1 and a 2 comma 5/3 for a moment there. The graph is flat. |
| 5:30 | So it either does something weird like that very unlikely, so don't draw that picture. Or,
More likely, it looks like that. Ok?
So as I said, so it's exactly one of those kinds of graphs. It's going up. It stops it goes up some more and when we do second derivatives. Then you can get a better idea of how to sketch it. |
| 6:04 | So let's do one that's almost identical
Do you guys play with it? How about f(x) = -(x^3/3)+2x^2-5x +1.
|
| 6:55 | Yeah, okay try that for a couple minutes.
Let's do this one. So, |
| 7:01 | take the derivative,
you get f'(x) = -x^2+4x+5 = 0.
So some of you stumble at this point because you're not sure how to factor that. You could factor it in from there, it's not it's not impossible. One of your factors is -x and one of your factors is x, but to make life easy if you just multiply this through by negative one you get this |
| 7:32 | Which is definitely factorable. That now factors into (x-5) and (x+1)
So that means that we have two critical values, so what would a test question say, would say find the critical values,
classify them as a maximum or a minimum, sketch.
Ok? So go to the number line |
| 8:01 | And put all the zeros.
So pick a number less than -1 like -2 This will be a negative quantity. This will be negative, 2 negatives make a positive, curves going up. Right, I'm sorry You need a negative outside outside of that huh, always you had a backwards |
| 8:32 | This is negative.
Ok? Remember you're setting it equal to zero to solve it, but you can't throw the negative out of the derivative the derivative is this Ok? Very important the derivative is still -x^2+4x+5. So if you plug in negative 2 this will come out negative. If you plug in number between -1 and 5 you will get positive, so, |
| 9:00 | It's going up and pick a number bigger than 5, like 6,
It'll be negative it goes down again
The zeros are at -1 and 5,
so you can, right, you can pull the negative out,
to find the zeros, but you can't just throw it away.
Now we need the Y coordinates, so let's see f(-1) = -(1/3)+2-5+1. |
| 9:36 | which is -5/3.
Think that's right Ok, get that one right? And f(5) is really annoying, f(5) = -125/3+50+25+1 |
| 10:02 | so 76/3,
228/3, so f(5) = 103/3.
Did I get that one right? Yeah. All right, I'll trust you. So it says that we have (-1, -5/3) and that is a minimum. So it's going down and up and And, we have (5, 103/3), as I said, on the exam I wouldn't give you one that's quite this messy. |
| 10:41 | And if you were to graph it,
you'd have some value about there, some value about there, so your graph would look something like that.
Okay, how'd we do? |
| 11:01 | Well, you don't need the negative if you're just thinking about where this is zero.
Ok? Because you would just get negative zero which is a silly thing, but, You still need to keep it in the derivative in general to figure out where the derivative is positive or where the derivative is negative or you'll get it backwards. Ok? Alright so this one confused everybody, but this is good practice, we'll do one more. We'll do one that is slightly different. So not simple polynomial. |
| 11:41 | Right, so what are you plugging into? You can plug it into so if you have say -2,
you can either plug it into this,
which is circled, or this in factored form.
It's easier in factored form because remember I said you only care about the sign So if you plug in -2 you say this is -7 that's -1, you multiply them you get 7, |
| 12:03 | times the negative makes -7 the number isn't important you just care whether it's positive or negative.
So you usually factored form. It's easier for plugging in. Unless you're plugging in 0 and you just look at the last number But that's only good for polynomials you can do messier problems. Question? Touching your head in despair? |
| 12:30 | Other questions? All right, let's do one that's different, that'll take a little bit more work.
How about, Want to make sure he gets this this is like a crucial thing. Crucial topic before we move on. |
| 13:50 | Okay, let's see if we can do that. One more round of max, min, increasing and decreasing.
|
| 14:00 | Okay, same concept. Take the derivative that's your first step.
This is calculus class, you only have one tool/ Derivative, so if you don't know what to do, take the derivative, you'll get something, you'll get some points. f'(x) = 8x^3-64x. Now you have to set that equal to zero to find the critical values. So, 8x^3-64x=0. |
| 14:30 | You can pull out an 8x,
and you're left with 8x(x^2-8)
You would think you could factor x^2-8, or just take x^2-8 and set it equal to zero.
If you factor it, you get (x+ sqrt(8)) and (x-sqrt(8)). You don't really care what the square root of 8 is, it's somewhere between two and three, right the square root of 9 is 3. |
| 15:10 | So now let's sign test that.
That's zero -(sqrt)8, sqrt(8) so as I said, (sqrt)8 is less than 3, So let's try -3. By plugging -3, -24 is negative, -3+8 will be negative |
| 15:30 | -3 minus sorry -3 - (sqrt)8 will be negative.
-3 minus the (sqrt)8 will be negative, three negatives makes it negative. Grass going down. Okay, again, the (sqrt)8 is 2 point something, it doesn't really matter now what it is, but it's like 2.8 whatever Now pick a number between -(sqrt)8 and 0, like -1. This is still negative, this will now be positive, |
| 16:00 | This is negative, 2 negatives make a positive, so curves going up.
That's a min minimum Now we plug in 0 0 we get 0 I'm sorry, put up a number between (sqrt)8 and 0, so like 1 we get positive positive negative, let's go down again. So this one was a maximum. |
| 16:33 | And now pick a number bigger than the (sqrt)8 like three or a trillion.
It doesn't matter because this will be positive that'll be positive and that would be positive. I mean, if you're not sure you pick a number that you know is bigger than sqrt(8), like 10 ok? All right, so that's a minimum again. So you have sort of an idea what the curse doing? It's gonna be sort of w-shaped |
| 17:00 | Okay, so go down and go up go down and go up
Now we need to find the Y coordinates
so f(-(sqrt)8), f(0), f((sqrt)8),
Well (sqrt)8 ^2 is 8.
So, (sqrt)8 ^4 is 8^2. So 2((sqrt)8^4)-32(sqrt(8))^2+18. |
| 17:33 | So if you have a calculator you can do that. Ok, you get
- 110? Right?
Anybody check this in their calculator? I'm good? I'm gonna just trust you. All right, if I plug in 0 I get 18. Then I plug in (sqrt)8, |
| 18:00 | That means the same thing is when I plug in -(sqrt)8, so you can get -110 again
So therefore I have a minimum at (-(sqrt)8, -110)
maximum at (0,18),
The new minimum we get at ((sqrt)8, -110).
Remember what I told you, for maximum it's a relative maximum, |
| 18:31 | So (0,8), maybe down there, down there,
if you're graphing, it doesn't have to be to scale, just label your points.
Let's label them (0,18) |
| 19:03 | It's when they truly understand what you're doing,
when you label these.
So these are absolutely most they are the absolute lowest that the curve goes. This is not. This is a relative maximum because the curve goes higher on your end. But in this area locally, it's a maximum. All right. Now, let's try a more complex graph. How about, |
| 19:53 | f(x) = (-8)/(x^2+1).
Which when we take the derivative, you could rewrite it as -8(x^2+1)^-1. |
| 20:09 | If you want to take the derivative of that, you have to do chain rule.
f'(x)= Bring down the -1, and you have 8(x^2_1)^-2 times the derivative the inside |
| 20:36 | Ok, so again, do the derivative of the outside part.
That's the -1(x^2+1)^-2 * the derivative of the inside. So this is a chain rule which we can then rewrite as (16x)/(x^2+1)^2. And now we need to find critical values, like where's that 0. |
| 21:01 | Well, (x^2+1) is always positive,
because x^2 is always either positive or 0.
When x^2 is 0 you get 1 otherwise you get a number bigger than 1. The bottom of this fraction is always positive, so, this will only be 0 at x =0. That's your only point you have to worry about. So when you do the derivative and you end up with a rational expression, |
| 21:30 | you're only going to care where the numerator is 0 to find maxes and mins.
You do care about where the denominator is 0, because you could still change sign there. So you can switch from increasing to decreasing but that won't be a maximum or a minimum. But I don't think we're gonna do any of those in this course. I'm not really too concerned. It's not that kind of a class. All right. So sign test. Take your number less than zero about -1. You plug negative 1 is the numerators negative. The bottom is always positive, |
| 22:01 | so you get negative, curve is going down.
Pick a number bigger than 0, like 1. This is 16 that's positive, top is positive. I mean the bottom is positive. So it's positive. So now it's going up. But it can't be as easy as a parabola. Just something funny has to be going on. |
| 22:39 | Now we can get the y-coordinate so when x=0, f(0) = -8,
so, this has a minimum (0,-8).
So you want to make this into a parabola, but it's not quite got that shape |
| 23:06 | How do I know? Well, what happens when x is a really big number?
Think about the limit for a second, like what happens when X is a billion? The denominator is really big, so it's 0ish. And when X is a negative really big number the denominator would be big, so it'll be 0 again. This graph is actually gonna do something like that and in |
| 23:33 | Friday's class I'll show you how you start to figure that stuff out.
Okay, so I wouldn't worry just yet about how I got that graph. Okay, just know that that's what it looks like. All right. Let's do another one, I really just want to analyze the maximum and minimums. How about, |
| 24:00 | x^(5/3)
f'(x) = 5/3x^(2/3)
Don't you hate those faction powers? They're really annoying.
So the only place this is 0, is at x=0. |
| 24:35 | What does x^2/3 mean? It means take the cube root of x and square it so you could think of this, you don't have to write it this way, that you think of it as, Squared in other words. This is always going to be a positive quantity or a zero. So this derivative is always positive |
| 25:05 | F at zero is zero. So goes through the origin.
And it's flat right here Oops Ignore that picture. |
| 25:34 | It's very hard to do it with the board. It's much easier to do it again on paper.
Turns over. One, one more eraser like that ok. And it's flat here, so I'm not doing a good job of getting the flatness in there. So those are not asymptotes. It's going up this is going up slowly. |
| 26:03 | Do you guys know what the cube root of x looks like?
Cube root of x, looks like that. So cube root of x^5, also looks like that. Ok? It's only a small variation on it, so I'm just trying to draw that right, but it's flattening there in the middle. Ok? And yeah, how can you tell? Well, you can do some limits when x is a very big number, this will be a very big number. If you take a trillionth and you take the 5/3 of it, it's still a big number. |
| 26:33 | It's just not a trillion anymore, in fact it's bigger than a trillion.
And negative it'll go down. So it's gonna go up and down. But it's gonna go slowly, because you take the cube roots. So the only interesting thing is happening is at zero. Ok, how do you make this more annoying? This is already annoying, but you know, let's make it more annoying. Let's do x^(2/3) instead of x^(5/3). |
| 27:09 | Now if you take the derivative,
You get (2/3)x^(-1/3).
Where is that zero? You want to say 0 don't you? It is not because remember this, is 1/(cuberoot of x) so it's actually undefined at 0. |
| 27:38 | What do we mean by undefined? That means we have 0 in the denominator, so it's infinitely some way.
So this is not a maximum or a minimum. So we'll deal with this as I said, when's Friday when instructed to can concavity, but this, is a cusp. It's a point where the Denominator goes to 0 so the slope becomes infinite. In fact, it's even more, |
| 28:07 | you know like that. It's like a ball that's bouncing, so it's very steep and then goes back up. Ok?
However, it still changes from increasing to decreasing at zero. So that's what I said. You have to still check a critical point Because to the left of 0 this quantity is negative that's going down and to the right of 0 it's positive. So it's going up. |
| 28:30 | So even though it's not going to be a maximum or minimum, it still helps you figure out if the graph is increasing or decreasing.
Confused? Probably a little confused, I'd be a little confused. But that's what we have more classes for, so I'll see everybody on Friday and we'll clear it up then. |