- [Instructor] So we've been given the value of h of x
at a few values of x, and then we're told,
James said that since h of seven minus h of three
over seven minus three is equal to one.
So this is really the average rate of change
between x is equal to three and x is equal to seven,
between that point and that point right over there.
So since that is equal to one,
there must be a number c in the closed interval
from three to seven for which the derivative
at that value c is equal to one.
So what James is trying to do is apply
the Mean Value Theorem,
which tells us if the conditions apply
for the Mean Value Theorem,
it tells us that if I'm going through two values,
so let's say this is a, let's say this is b,
and let's say the function does something like this.
If we meet the conditions for the Mean Value Theorem,
it tells us that there's some c in the closed interval
from a to b where the derivative of c is equal to
the average rate of change between a and b.
So the average rate of change between a and b
would be the slope of the secant line right over there,
and then we could just think about well,
looks like there are some points.
The way I've drawn it,
that point right over there seems to have the same slope.
This point right over there seems to have the same slope.
And so that's all what the Mean Value Theorem is claiming,
that there's going to be at least one c,
if we meet the conditions for the Mean Value Theorem,
where the derivative at that point is the same
as the average rate of change from the first endpoint
to the second.
Now what are the conditions for the Mean Value Theorem
to apply, and we've reviewed this in multiple videos.
One way to think about it,
if we're talking about the closed interval
from three to seven,
one condition is that you have to be differentiable,
differentiable, over the open interval
from three to seven.
So that's the interval but not including the endpoints.
And you have to be continuous
continuous over the entire closed interval
so including the endpoints.
And one interesting thing that we've mentioned before
is that differentiability implies continuity.
So if something is differentiable over the open interval,
it's also going to be continuous over this open interval,
and so the second condition would just say well then we also
have to be continuous at the endpoints.
Now let's look at the answers.
So it says which condition makes James's claim true?
So we have to feel good about these two things
right over here in order to make James's claim.
Choice A, h is continuous over the closed interval
from three to seven.
So that does meet this second condition,
but continuity does not imply differentiability.
So that doesn't give us the confidence
that we're differentiable over the open interval.
If you're differentiable you're continuous,
but if you're continuous
you're not necessarily differentiable.
Classic example of that is if we have a sharp turn,
something like that,
we wouldn't be differentiable at that point
even though we are continuous there.
So let me rule that one out.
I'm gonna rule that one out.
Choice B, h is continuous and decreasing
over the closed interval from three to seven.
Now that doesn't help us either
because it still doesn't mean you're differentiable.
You could be continuous and then decreasing
and still have one of these sharp turns
where you're not differentiable.
So we will rule this one out.
H is differentiable over the closed interval
from three to seven.
This one feels good
because if you're differentiable over the closed interval,
you're definitely gonna be differentiable
over the open interval that does not include the endpoints.
This is a subset of this right over here.
And if you're differentiable over a closed interval,
you're gonna be continuous over it.
Differentiability implies continuity.
So I like this choice right over here.
Now this last choice says the limit as x approaches five
of h prime of x is equal to one.
So they're saying the limit of our derivative
as we approach five is equal to one.
Now that doesn't, we don't know for sure,
this limit might be true,
but that still does not necessarily imply
that h prime of five is equal to one.
We sill don't know that.
And you know five is in this interval,
but we still don't know just from this statement alone
that there is definitely some c in the interval
whose derivative is the same as the average rate of change
over the interval.
So I would rule this one out as well.