I’ve gotten several requests to explain or teach the mean value theorem. So let’s do that in this video.
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Mean value theorem Derivative applications Differential Calculus
So this is the mean value theorem. And I have mixed feelings about the mean value theorem. It’s kind of neat, but what you’ll see is, it might not be obvious to prove, but the intuition behind it’s pretty obvious. And the reason I have mixed feelings about it, is that even though, as you’ll hopefully see, the intuition is pretty obvious, but they stick it in the math books, and people are just trying to learn calculus, and get to what matters, and then they put the mean value theorem in there, and they have all of this function notation, and they have all of these words, and it just confuses people.
So hopefully this video will clarify that little bit, and I’m curious to see what you think of it. So let’s see. What does the mean value theorem say? Let me draw some axes.
I’ll do a visual explanation first. I think this calls for magenta. So that’s my x-axis.
This is my y-axis. And let’s say I have some function f of x. So let me draw my f of x.
That’s as good as any. And this is some function f of x, and I’m going to put a few conditions on f of x. f of x has to be continuous and differentiable.
And I know a lot of you probably get intimidated when you hear these words. It sounds like what a mathematician would say, and it sounds very abstract. All continuous means is that the curve is connected to itself as you go along it. And here, the conditions are over a closed interval.
This is another very mathy term you’ll see. So you’ll often say, on a closed interval from a to b. All that means is an interval, let’s say a is the low point, let’s say this is a, we don’t know what number that is.
That could be minus 5, or who knows. And let’s say this is b, right here, let me b right here. Let’s say that’s be.
So when people talk about a closed interval, defined on a closed interval, that means that the function needs to be defined at every number between a and b, and the function needs to be defined at a and at b. If they said over an open interval between a and b, that means that it’s only defined at every value between a and b, but not necessarily at a and b. So it has to be continuous, differentiable, and let’s say it’s defined over the closed interval, and this is just the notation for it, a b. So that means, it has to be defined at all of the x values from a to b, including a and b.
If it was an open interval, you would write it like this. You’d write a and b. That means an interval for all the numbers between a and b, but not including those. So let’s ignore that for now.
So back to the mean value theorem. So you know, hopefully, what continuous means. Let me draw here a function that’s not continuous.
So a function that is not continuous would look like this. It would go like this, and it would start up here, go like that. Right? So this would be an example of a function, let’s say, same axes, let me draw it in a different color. If that was our y– no, that’s not a good.
If that was our y-axis, and that was our x-axis, just to give you the reference for what I drew. So if the function is continuous, continuous, continuous, and then it jumps, that disconnect, that would make this function discontinuous, or it would not be a continuous function. So a function just has to be continuous. And now what does differentiable mean? Differentiable means that at every point over the interval that we care about, you have to be able to find the derivative.
That means you can take the derivative of it. It’s a differentiable. And what else does that mean? Well, that means that if you were to graph the derivative of this function, that it is also continuous. And I’ll let you think about that for a second.
And actually, in this video I’m going to show you an example of a function that is continuous, but not differentiable and because of that, the mean value theorem breaks down. But anyway, let’s get back to the mean value theorem. Most of the functions we deal with satisfy all three of these things.
Unless, you know, you’re doing limit problems, and they try to make these things break down. Anyway, back to the function. So this function meets all of these requirements.
So all it says is, if I were take the average slope between point a and point b. So what is the slope, the average slope between point a and point b? Well, slope is just rise over run, right? So what is it? Let me see if I can draw the average slope. So the run would be this distance. That’d be the run, right, and this would be the rise.
So this is the point, right here, that’s the point a, f of a. Over here, this is the point b, f of b. So what’s the average slope between a and b? Well, it’s rise over run. So what’s the rise? What’s this distance? How much have we gone up from f of a to f of b? Well, the rise will just be f of b, this height, minus f of a.
f of b minus f of a. And what’s the run, what’s this distance? Well, it’s just b minus a. And if I were to draw a line that has that average slope, it would look something like this.
We could make it go through those two points, but it really doesn’t have to. Let me do it in a blue. So that’s the average slope between those two points, right? So what does the mean value theorem tell us? It says, if f of x is defined over this closed interval from a to b, and f of x is continuous, and it’s differentiable, that you could take the derivative at any point, that there must be some points c f prime of c is equal to this thing.
So is equal to f prime of c. I shouldn’t have written it here. So what is that telling us? So all that’s telling us, is if we’re continuous, differentiable, defined over the closed interval, that there’s some point c, oh, and c has to be between a and b, there’s some point between a and b, and it could be at one of the points, but there’s some point c where the derivative at c, or the slope at c, the instantaneous slope at c, is exactly equal to the average slope over that interval. So what does that mean? So we can look at it visually.
Is there any point along this curve where the slope looks very similar to this average slope that we calculated? Well, sure, let’s see. It looks like, maybe, this point, right here? Just the way I drew it. This is very inexact. But that point looks like the slope, you know, I could say the slope is something like that, right there.
So we don’t know what, analytically, this function is, but visually, you could see that at this point c, the derivative, so I just picked that point. So this could be our point c. And how do we just say that? Well, because f prime of c is this slope, and it’s equal to the average slope.
So f prime of c is this thing, and it’s going to be equal to the average slope over the whole thing. And this curve actually probably has another point where the slope is equal to the average slope. Let’s see.
This one looks, like, right around there. Just the way I drew it. Looks like the slope there could look something like, could be parallel as well.
These lines should be parallel. The tangent lines should be parallel. So hopefully that makes a little sense to you.
Another way to think about it is that your average, actually, let me draw a graph just to make sure that we hit the point home. Let’s draw my position as a function of time. So this is something, this’ll make it applicable to the real world.
So that’s my x-axis, or the time axis, that’s my position axis. This is going back to our original intuition of what even a derivative is. So this is time, and I call this position, or distance, or it doesn’t matter. Position.
And if I was moving at a constant velocity, my position as a function of time would just be a straight line, right? And the velocity is actually your slope. But let’s say I had a varying velocity. And in reality, if you’re driving a car, you are always at a variable velocity.
So let’s say I start at a standstill at time t equals 0, and then I accelerate, then I decelerate a little bit, decelerate a little bit, I keep decelerating, and then I come to a standstill, so my position stays still. Then I accelerate again, decelerate, accelerate, et cetera. Right? So this could be, you know, I have a variable velocity, and this could be my position as a function of time. So all this says, let’s say that after, this is time 0, position 0.
Let’s say after 1 hour, let’s say that is 1 hour, this time equals 1 hour, let’s say I have gone 60 miles. So what can you say? You could say that my average velocity equals just change in distance divided by change in times. It equals 60 miles per hour. So what the mean values theorem says, is OK.
Your average velocity, so you could almost view it as the average slope between this point and this point with 60, if your average velocity was 60 miles per hour, there was some point in time, maybe more, but there was at least one point in time, where you were going exactly sixty miles per hour. That make sense, right? If you average 60 miles per hour, maybe you’re going 40 miles per hour some of the point, but at some point you went 80, and in between you had to be going 60 miles per hour. So let me see if I can draw that graphically. So this slope is my average velocity, and the way I drew it, there’s probably two points, let’s see, probably right around here, I was probably going 60 miles per hour, the slope is probably 60 there, the instantaneous velocity probably there, as well.
So before I leave, let’s do this analytically, just to work with numbers. And the reason why I have mixed feelings about the mean value theorem, it’s useful later on, probably if you become a math major you’ll maybe use it to prove some theorems, or maybe you’ll prove it, itself. But if you’re just applying calculus for the most part, you’re not going to be using the mean value theorem too much.
But anyway, if you’ve got to know it, you’ve got to know it, and it tells you something else about the world, so it’s interesting that way. So let’s say we have the function f of x is equal to x squared minus 4x, and the interval that I care about here is between, is a closed interval, so I’m including 2, from 2 to 4. Now, the mean value theorem tells us that if this function is defined on this interval, and it is, right? We could put any number. The domain of this is actually all real numbers, I could put any number here, so obviously it’s going to be defined over this interval.
But so it’s defined over the interval, this is continuous, this is differentiable. You could take the derivative, and the derivative is continuous. So the mean value theorem should apply here.
So let’s see what value of c is equal to the average slope between 2 and 4. So what’s the average slope between 2 and 4? Well, it’s going to be f of 4, so the change in the function, f of 4 minus f of 2 divided by the change in x, so 4 minus 2. So this equal to the average slope. So f of 4 is 16 minus 16, right? So that’s 0.
Let me make sure of that. 4 times 4, 16, minus 4 times 4, 16, right. Minus f of 2.
f of 2 is 2 squared, is 4, right, and then minus 4 times 2. So minus 8. All of that over 2. And so this equals minus 4.
So this equals 4 over 2. So the average slope from x is equal to 2 to x is equal to 4 is 2. And now the mean value theorem tells us, that there must be some point that’s between these two, maybe including one of those, where the slope at that point is exactly equal to 2. Let’s figure out what point that is.
That c. Let’s take the derivative, because the derivative at c is going to be equal to 2. So we just take the derivative.
So let’s say f prime of x is equal to 2x minus four. And we want to figure out, at what x value does this equal 2. So we say, 2x minus 4 is equal to 2.
Where does the slope equal 2? And you get 2x is equal to 6, x is equal to 3. So if x is equal to 3, the derivative is exactly equal to the average slope. But let me see if I can, let me get the graphing calculator here. Let me what I can do.
OK. So here’s the graph of x squared minus 4x. Let me see if I can make it a little bit bigger. The interval that we care about is from here to here.
So the average slope over that interval was 2. So if we were to draw the slope, it was like that, the slope would look like that. And at the point 3, the slope is exactly 2. So let me actually draw that.
This isn’t too hard to draw, for myself. Let me see. So if that’s the x-axis, I’ll want that graph out of the way.
That’s the y-axis. So the graph goes through the point 0, 0 as neatly as possible. Nope, that’s not neat.
So the graph goes something like this, it dips up, then it goes like that, and actually it keeps going straight up, like that, it’s a parabola. So this is point 4. The point 2 is here.
And at 2 we’re at negative 4, so the vertex is at the point 2, minus four. So what we said, the average slope, so the closed interval that we care about, between 2 and 4, it’s from 2 here to 4 here. That’s the interval, 2 to 4.
The average slope is 2. Doesn’t look like it, only because I’ve kind of compressed the y-axis. And we’re saying, at the point x is equal to 3, the slope is equal to exactly that. So at x is equal to three, the slope is equal to the same thing.
That’s all the mean value theorem is. I know sounds complicated. People talk about continuity, and differentiability, and f prime of c, and all this, but all it says is, there’s some point between these two points where the instantaneous slope, or slope exactly at that point, is equal to the slope between these two points. Hope I didn’t confuse you.
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