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I've gotten several requests to explain or teach the

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mean value theorem.

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So let's do that in this video.

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So this is the mean value theorem.

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And I have mixed feelings about the mean value theorem.

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It's kind of neat, but what you'll see is, it might not be

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obvious to prove, but the intuition behind it's

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pretty obvious.

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And the reason I have mixed feelings about it, is that even

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though, as you'll hopefully see, the intuition is pretty

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obvious, but they stick it in the math books, and people are

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just trying to learn calculus, and get to what matters, and

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then they put the mean value theorem in there, and they have

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all of this function notation, and they have all of these

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words, and it just confuses people.

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So hopefully this video will clarify that little bit,

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and I'm curious to see what you think of it.

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So let's see.

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What does the mean value theorem say?

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Let me draw some axes.

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I'll do a visual explanation first.

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I think this calls for magenta.

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So that's my x-axis.

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This is my y-axis.

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And let's say I have some function f of x.

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So let me draw my f of x.

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That's as good as any.

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And this is some function f of x, and I'm going to put

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a few conditions on f of x.

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f of x has to be continuous and differentiable.

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And I know a lot of you probably get intimidated

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when you hear these words.

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It sounds like what a mathematician would say, and

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it sounds very abstract.

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All continuous means is that the curve is connected to

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itself as you go along it.

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And here, the conditions are over a closed interval.

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This is another very mathy term you'll see.

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So you'll often say, on a closed interval from a to b.

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All that means is an interval, let's say a is the low point,

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let's say this is a, we don't know what number that is.

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That could be minus 5, or who knows.

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And let's say this is b, right here, let me b right here.

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Let's say that's be.

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So when people talk about a closed interval, defined on a

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closed interval, that means that the function needs to be

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defined at every number between a and b, and the function needs

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to be defined at a and at b.

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If they said over an open interval between a and b, that

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means that it's only defined at every value between a and b,

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but not necessarily at a and b.

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So it has to be continuous, differentiable, and let's say

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it's defined over the closed interval, and this is just

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the notation for it, a b.

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So that means, it has to be defined at all of the x values

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from a to b, including a and b.

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If it was an open interval, you would write it like this.

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You'd write a and b.

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That means an interval for all the numbers between a and

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b, but not including those.

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So let's ignore that for now.

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So back to the mean value theorem.

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So you know, hopefully, what continuous means.

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Let me draw here a function that's not continuous.

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So a function that is not continuous would

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look like this.

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It would go like this, and it would start up

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here, go like that.

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Right?

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So this would be an example of a function, let's say,

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same axes, let me draw it in a different color.

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If that was our y-- no, that's not a good.

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If that was our y-axis, and that was our x-axis, just to

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give you the reference for what I drew.

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So if the function is continuous, continuous,

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continuous, and then it jumps, that disconnect, that

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would make this function discontinuous, or it would not

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be a continuous function.

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So a function just has to be continuous.

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And now what does differentiable mean?

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Differentiable means that at every point over the interval

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that we care about, you have to be able to find the derivative.

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That means you can take the derivative of it.

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It's a differentiable.

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And what else does that mean?

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Well, that means that if you were to graph the derivative

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of this function, that it is also continuous.

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And I'll let you think about that for a second.

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And actually, in this video I'm going to show you an example of

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a function that is continuous, but not differentiable and

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because of that, the mean value theorem breaks down.

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But anyway, let's get back to the mean value theorem.

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Most of the functions we deal with satisfy all

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three of these things.

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Unless, you know, you're doing limit problems, and they try to

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make these things break down.

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Anyway, back to the function.

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So this function meets all of these requirements.

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So all it says is, if I were take the average slope

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between point a and point b.

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So what is the slope, the average slope between

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point a and point b?

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Well, slope is just rise over run, right?

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So what is it?

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Let me see if I can draw the average slope.

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So the run would be this distance.

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That'd be the run, right, and this would be the rise.

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So this is the point, right here, that's

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the point a, f of a.

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Over here, this is the point b, f of b.

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So what's the average slope between a and b?

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Well, it's rise over run.

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So what's the rise?

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What's this distance?

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How much have we gone up from f of a to f of b?

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Well, the rise will just be f of b, this

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height, minus f of a.

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f of b minus f of a.

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And what's the run, what's this distance?

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Well, it's just b minus a.

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And if I were to draw a line that has that average slope, it

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would look something like this.

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We could make it go through those two points, but it

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really doesn't have to.

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Let me do it in a blue.

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So that's the average slope between those

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two points, right?

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So what does the mean value theorem tell us?

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It says, if f of x is defined over this closed interval from

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a to b, and f of x is continuous, and it's

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differentiable, that you could take the derivative at any

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point, that there must be some points c f prime of c is

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equal to this thing.

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So is equal to f prime of c.

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I shouldn't have written it here.

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So what is that telling us?

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So all that's telling us, is if we're continuous,

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differentiable, defined over the closed interval, that

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there's some point c, oh, and c has to be between a and b,

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there's some point between a and b, and it could be at one

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of the points, but there's some point c where the derivative at

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c, or the slope at c, the instantaneous slope at c, is

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exactly equal to the average slope over that interval.

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So what does that mean?

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So we can look at it visually.

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Is there any point along this curve where the slope looks

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very similar to this average slope that we calculated?

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Well, sure, let's see.

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It looks like, maybe, this point, right here?

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Just the way I drew it.

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This is very inexact.

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But that point looks like the slope, you know, I could

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say the slope is something like that, right there.

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So we don't know what, analytically, this function is,

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but visually, you could see that at this point c, the

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derivative, so I just picked that point.

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So this could be our point c.

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And how do we just say that?

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Well, because f prime of c is this slope, and it's equal

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to the average slope.

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So f prime of c is this thing, and it's going to be equal to

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the average slope over the whole thing.

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And this curve actually probably has another point

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where the slope is equal to the average slope.

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Let's see.

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This one looks, like, right around there.

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Just the way I drew it.

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Looks like the slope there could look something like,

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could be parallel as well.

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These lines should be parallel.

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The tangent lines should be parallel.

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So hopefully that makes a little sense to you.

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Another way to think about it is that your average, actually,

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let me draw a graph just to make sure that we

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hit the point home.

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Let's draw my position as a function of time.

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So this is something, this'll make it applicable

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to the real world.

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So that's my x-axis, or the time axis, that's

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my position axis.

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This is going back to our original intuition of what

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even a derivative is.

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So this is time, and I call this position, or distance,

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or it doesn't matter.

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Position.

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And if I was moving at a constant velocity, my position

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as a function of time would just be a straight line, right?

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And the velocity is actually your slope.

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But let's say I had a varying velocity.

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And in reality, if you're driving a car, you are always

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at a variable velocity.

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So let's say I start at a standstill at time t equals 0,

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and then I accelerate, then I decelerate a little bit,

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decelerate a little bit, I keep decelerating, and then I come

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to a standstill, so my position stays still.

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Then I accelerate again, decelerate,

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accelerate, et cetera.

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Right?

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So this could be, you know, I have a variable velocity, and

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this could be my position as a function of time.

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So all this says, let's say that after, this

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is time 0, position 0.

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Let's say after 1 hour, let's say that is 1 hour, this time

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equals 1 hour, let's say I have gone 60 miles.

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So what can you say?

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You could say that my average velocity equals just change in

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distance divided by change in times.

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It equals 60 miles per hour.

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So what the mean values theorem says, is OK.

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Your average velocity, so you could almost view it as the

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average slope between this point and this point with 60,

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if your average velocity was 60 miles per hour, there was some

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point in time, maybe more, but there was at least one point in

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time, where you were going exactly sixty miles per hour.

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That make sense, right?

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If you average 60 miles per hour, maybe you're going 40

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miles per hour some of the point, but at some point you

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went 80, and in between you had to be going 60 miles per hour.

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So let me see if I can draw that graphically.

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So this slope is my average velocity, and the way I drew

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it, there's probably two points, let's see, probably

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right around here, I was probably going 60 miles per

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hour, the slope is probably 60 there, the instantaneous

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velocity probably there, as well.

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So before I leave, let's do this analytically, just

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to work with numbers.

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And the reason why I have mixed feelings about the mean value

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theorem, it's useful later on, probably if you become a math

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major you'll maybe use it to prove some theorems, or maybe

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you'll prove it, itself.

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But if you're just applying calculus for the most part,

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you're not going to be using the mean value

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theorem too much.

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But anyway, if you've got to know it, you've got to know it,

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and it tells you something else about the world, so it's

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interesting that way.

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So let's say we have the function f of x is equal to x

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squared minus 4x, and the interval that I care about

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here is between, is a closed interval, so I'm including

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2, from 2 to 4.

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Now, the mean value theorem tells us that if this

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function is defined on this interval, and it is, right?

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We could put any number.

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The domain of this is actually all real numbers, I could put

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any number here, so obviously it's going to be defined

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over this interval.

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But so it's defined over the interval, this is continuous,

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this is differentiable.

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You could take the derivative, and the derivative

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is continuous.

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So the mean value theorem should apply here.

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So let's see what value of c is equal to the average

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slope between 2 and 4.

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So what's the average slope between 2 and 4?

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Well, it's going to be f of 4, so the change in the function,

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f of 4 minus f of 2 divided by the change in x, so 4 minus 2.

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So this equal to the average slope.

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So f of 4 is 16 minus 16, right?

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So that's 0.

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Let me make sure of that.

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4 times 4, 16, minus 4 times 4, 16, right.

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Minus f of 2.

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f of 2 is 2 squared, is 4, right, and then

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minus 4 times 2.

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So minus 8.

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00:13:23,629 --> 00:13:27,809


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All of that over 2.

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And so this equals minus 4.

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So this equals 4 over 2.

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So the average slope from x is equal to 2 to x

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is equal to 4 is 2.

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And now the mean value theorem tells us, that there must be

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some point that's between these two, maybe including one of

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those, where the slope at that point is exactly equal to 2.

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Let's figure out what point that is.

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That c.

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Let's take the derivative, because the derivative at c

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is going to be equal to 2.

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So we just take the derivative.

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So let's say f prime of x is equal to 2x minus four.

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And we want to figure out, at what x value does this equal 2.

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So we say,

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2x minus 4 is equal to 2.

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Where does the slope equal 2?

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And you get 2x is equal to 6, x is equal to 3.

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So if x is equal to 3, the derivative is exactly equal

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to the average slope.

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00:14:28,629 --> 00:14:30,990
But let me see if I can, let me get the graphing

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calculator here.

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Let me what I can do.

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OK.

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So here's the graph of x squared minus 4x.

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00:14:43,090 --> 00:14:44,600
Let me see if I can make it a little bit bigger.

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00:14:44,600 --> 00:14:47,920


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The interval that we care about is from here to here.

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So the average slope over that interval was 2.

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00:14:55,909 --> 00:14:59,009
So if we were to draw the slope, it was like that, the

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00:14:59,009 --> 00:15:00,899
slope would look like that.

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And at the point 3, the slope is exactly 2.

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So let me actually draw that.

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00:15:07,539 --> 00:15:11,319
This isn't too hard to draw, for myself.

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Let me see.

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00:15:13,539 --> 00:15:21,269
So if that's the x-axis, I'll want that graph out of the way.

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00:15:21,269 --> 00:15:22,159
That's the y-axis.

315
00:15:22,159 --> 00:15:24,959


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00:15:24,960 --> 00:15:28,820
So the graph goes through the point 0, 0 as

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00:15:28,820 --> 00:15:31,010
neatly as possible.

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00:15:31,009 --> 00:15:31,990
Nope, that's not neat.

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00:15:31,990 --> 00:15:35,680


320
00:15:35,679 --> 00:15:40,609
So the graph goes something like this, it dips up, then it

321
00:15:40,610 --> 00:15:43,610
goes like that, and actually it keeps going straight up,

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00:15:43,610 --> 00:15:45,659
like that, it's a parabola.

323
00:15:45,659 --> 00:15:47,620
So this is point 4.

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00:15:47,620 --> 00:15:51,210
The point 2 is here.

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00:15:51,210 --> 00:15:56,150
And at 2 we're at negative 4, so the vertex is at

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00:15:56,149 --> 00:15:57,709
the point 2, minus four.

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00:15:57,710 --> 00:16:01,389
So what we said, the average slope, so the closed interval

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00:16:01,389 --> 00:16:04,279
that we care about, between 2 and 4, it's from 2

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00:16:04,279 --> 00:16:05,959
here to 4 here.

330
00:16:05,960 --> 00:16:09,160
That's the interval, 2 to 4.

331
00:16:09,159 --> 00:16:12,850
The average slope is 2.

332
00:16:12,850 --> 00:16:14,570
Doesn't look like it, only because I've kind of

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00:16:14,570 --> 00:16:17,010
compressed the y-axis.

334
00:16:17,009 --> 00:16:20,090
And we're saying, at the point x is equal to 3, the slope

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00:16:20,090 --> 00:16:22,009
is equal to exactly that.

336
00:16:22,009 --> 00:16:25,179
So at x is equal to three, the slope is equal

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00:16:25,179 --> 00:16:26,859
to the same thing.

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00:16:26,860 --> 00:16:28,620
That's all the mean value theorem is.

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00:16:28,620 --> 00:16:29,840
I know sounds complicated.

340
00:16:29,840 --> 00:16:32,810
People talk about continuity, and differentiability, and f

341
00:16:32,809 --> 00:16:36,519
prime of c, and all this, but all it says is, there's some

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00:16:36,519 --> 00:16:41,019
point between these two points where the instantaneous slope,

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00:16:41,019 --> 00:16:43,689
or slope exactly at that point, is equal to the slope

344
00:16:43,690 --> 00:16:45,510
between these two points.

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Hope I didn't confuse you.