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Current time:0:00Total duration:16:48

I've gotten several requests
to explain or teach the mean value theorem. So let's do that in this video. 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.