- [Instructor] This table gives us a few values
of the function g.
So we know what g of x is equal to with these values
right over here.
X is equal to negative two, negative one, zero, and one.
It says Rafael said that since g of one minus g of zero
over one minus zero is equal to negative five, there must be
a number c in the interval in the closed interval from zero
to one for which g prime of c is equal to negative five.
Which condition makes Rafael's claim true and they give us
four choices here.
So I encourage you to pause this video and try to work
through it on your own.
Alright before we even look at these choices let's just
revisit what's going on over here.
So Rafael is saying since this, so he's looking at g of one
minus g of zero, so this is g of one, minus g of zero
that's this right over here, over one minus zero, over one
minus zero.
So what is this?
This right over here is the average rate of change,
rate of change between x equals zero and x is equal to one.
Another way to think about it this is the slope of the line
that connects the point 0,8 to the point 1,3.
And that average rate of change is negative five.
And that he's saying there must be a number c
in the interval, in the closed interval from zero to one
for which the derivative of that x value what x is equal
to c is the same thing as this average rate of change.
So this is what the mean value theorem is all about,
it's this notion that...
We plot it a little bit just as a bit of a review,
so we're going from 0,8 which is maybe right over here
to 1,3 so that's one, 1,3 might look something like this.
So we're seeing the average rate of change, the slope
of the line that connects those two points is like this.
What the mean value theorem tells us if we have
the conditions for the mean value theorem, that let's say
that our function does something like this, as long
as our functions meet the conditions of the mean value
theorem, which essentially has to be differentiable
over the open interval and continuous over the closed
interval, then there has to be a point c where the slope
of the tangent line at point c is equal to the average
rate of change.
So over here the slope of the tangent line looks the same
as our average rate of change, right of there just
like that.
So this is the mean value theorem, but in order to make
this claim that in order to apply the mean value theorem
to say there must be a c in the interval where the slope
of the tangent line at x equals c is the same as the average
rate of change, we have to feel very confident that we are
meeting the conditions for the mean value theorem, namely
that the function is continuous over the closed interval
and differentiable over the open interval from zero to one.
So let's look at the choices now.
G is continuous at x equals zero and x equals one.
Well that by itself isn't sufficient, we also have
to be continuous at all the points in between, so that alone
does not make this claim true.
G is continuous over the closed interval from zero to one.
So that's nice, but that doesn't ensure that we're
differentiable over the open interval from zero to one,
we have to be continuous over the closed interval
and differentiable over the open interval, so we can rule
that one out, that's not enough.
G is differentiable over the open interval from negative one
to one and so that includes this right over here includes
the open interval that we care about.
This includes the open interval from zero to one, that's
what we care about, so that is a check that we're
differentiable over the open interval and continuous
over the closed interval from negative one to one.
Well once again this includes the closed interval that we
care about.
This includes the closed interval from zero to one
and so this is looking good.
What we care about is that we're differentiable over this
interval and if we're differentiable here we're definitely
differentiable here and then we're continuous over
this interval.
So I like this choice.
This right over here says g is differentiable over
the open interval and at x is equal to zero, so this
is getting us close because if you are differentiable
you are continuous.
So one way to interpret this is that we are differentiable
over the interval that is from zero to one that includes
the point zero.
Now this is close if they said that we are differentiable
for over the closed interval from zero to one, that actually
would have been sufficient because if you are differentiable
then you are also continuous but they did not say that.
So we can rule this one out as well, we would have to know
this last choice still does not make us confident that
we are continuous at the point one, that's what being
continuous over the closed interval would have told us.
Let's do one more example that's a little bit different
than this one.
So here we are told that g is a differentiable function
once again they've given us the function sample that some
values of x.
Max was asked whether there's a solution to g prime
of x is equal to two on the closed interval from three
to six.
So we see three here, we see three here.
So we want to care about the closed interval from three
to six.
So the first thing he did, he says okay let's find
the average rate of change from x equals three to x equals
six, so this is our change in our function, so g of six
minus g of three over six minus three, let's see g of six
is equal to one, g of three is equal to negative five,
this is three.
One minus negative five over three, that's six over three,
it is indeed equal to two.
So that step makes sense.
The reason I'm going through they say is Max's work correct?
If not, what is his mistake?
So let's look at step two.
The mean value theorem so Max is now saying alright I found
the average rate of change between x equals three and x
equals six and then he says the mean value theorem
guarantees a solution where g prime of x equals two
on this interval.
So we have to be careful here, because he just immediately
applies the mean value theorem without establishing
the conditions for the mean value theorem.
So I would say even though it does say that we're
differentiable over it, I would make it, I think his mistake
is that he did not make this clear.
He should have said step two should have been something
like because g is differentiable over...
It sounds like it's a differentiable function so we could
say for all x is differentiable for all x.
We can say the mean value theorem applies over this interval
if you wanted to be really careful you could've said
because g is differentiable for all x is definitely
differentiable over this closed interval and if it's
differentiable over the closed interval it's also
continuous over the closed interval and so the mean value
theorem definitely applies and then say the mean value
theorem guarantees a solution.
So the reason why it's important to stress is if you're
taking a test especially things like the AP exam,
if you're going to apply one of these, really any theorems,
they're going to want you to put down all the conditions
for the theorem to ensure that you can apply it as part
of your argument.