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## Using the mean value theorem

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# Justification with the mean value theorem: table

AP Calc: FUN‑1 (EU), FUN‑1.B (LO), FUN‑1.B.1 (EK)

## Video transcript

- [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.

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