- [Instructor] The table gives selected values
of the differentiable function f.
All right, can we use the mean value theorem
to say that there is a value c such that f prime of c
is equal to five and c is between four and six?
If so, write a justification.
Well it meets, to meet the, to use the mean value theorem,
you have to be differentiable over the open interval
and continuous over the closed interval,
so it seems like we've met that.
Because if you're differentiable over an interval,
you're definitely continuous over that interval.
It's saying that it's just a generally differentiable
function f I guess over any interval.
But the next part is to say,
all right, if that condition is met,
then the slope of the secant line
between four comma f of four and six comma f of six,
that some at least one point in between four and six
will have a derivative that is equal
to the slope of the secant line.
And so let's figure out what the slope
of the secant line is between four comma f of four
and six comma f of six.
And if it's equal to five,
then we could use the mean value theorem.
If it's not equal to five,
then the mean value theorem would not apply.
And so let's do that.
f of six minus f of four,
all of that over six minus four is equal to
seven minus three
over two, which is equal to two.
So two not equal to five.
So mean value theorem doesn't apply.
All right let's, I'll put an exclamation mark
there for emphasis.
All right let's do the next part.
Can we use the mean value theorem to say
that the equation f prime of x is equal to negative one
has a solution?
And now the interval's from zero to two.
If so, write a justification.
All right so let's see this.
So if we were to take the slope of the secant line,
so f of two minus f of zero,
all that over two minus zero.
This is equal to negative two minus zero,
all of that over two,
which is equal to negative two over two,
which is equal to negative one.
And so and we also know that we meet
the continuity and differentiability conditions,
and so we could say and since
f is generally differentiable,
generally differentiable,
differentiable,
it will be
differentiable, differentiable,
and, and continuous
over the interval from zero to two,
and I'll say the closed interval.
You just have to be differentiable over the open interval,
but it's even better I guess if you're differentiable
over the closed interval because you have to be continuous
over the closed interval.
And so, and since f is generally differentiable,
it will be differentiable and continuous over zero two.
So the mean value theorem tells us,
tells us, that there
is an x
in that interval from zero to two
such that
f prime of x
is equal to that secant slope,
or you could say that average rate of change,
is equal to negative one.
And so I could write,
yes, yes, and then this would be my justification.
This is the slope of the secant line
or the average rate of change,
and since f is generally differentiable,
it will be differentiable and continuous
over the closed interval.
So the mean value theorem tells us
that there is an x in this interval
such that f prime of x is equal to negative one.
And we're done.