WEBVTT

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- [Instructor] The table
gives selected values

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of the differentiable function f.

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All right, can we use
the mean value theorem

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to say that there is a value
c such that f prime of c

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is equal to five and c
is between four and six?

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If so, write a justification.

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Well it meets, to meet the,
to use the mean value theorem,

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you have to be differentiable
over the open interval

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and continuous over the closed interval,

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so it seems like we've met that.

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Because if you're
differentiable over an interval,

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you're definitely continuous
over that interval.

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It's saying that it's just
a generally differentiable

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function f I guess over any interval.

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But the next part is to say,

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all right, if that condition is met,

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then the slope of the secant line

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between four comma f of
four and six comma f of six,

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that some at least one point
in between four and six

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will have a derivative that is equal

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to the slope of the secant line.

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And so let's figure out what the slope

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of the secant line is
between four comma f of four

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and six comma f of six.

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And if it's equal to five,

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then we could use the mean value theorem.

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If it's not equal to five,

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then the mean value
theorem would not apply.

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And so let's do that.

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f of six minus f of four,

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all of that over six
minus four is equal to

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seven minus three

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over two, which is equal to two.

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So two not equal to five.

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So mean value theorem doesn't apply.

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All right let's, I'll
put an exclamation mark

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there for emphasis.

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All right let's do the next part.

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Can we use the mean value theorem to say

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that the equation f prime of
x is equal to negative one

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has a solution?

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And now the interval's from zero to two.

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If so, write a justification.

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All right so let's see this.

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So if we were to take the
slope of the secant line,

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so f of two minus f of zero,

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all that over two minus zero.

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This is equal to negative two minus zero,

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all of that over two,

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which is equal to negative two over two,

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which is equal to negative one.

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And so and we also know that we meet

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the continuity and
differentiability conditions,

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and so we could say and since

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f is generally differentiable,

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generally differentiable,

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differentiable,

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it will be

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differentiable, differentiable,

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and, and continuous

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over the interval from zero to two,

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and I'll say the closed interval.

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You just have to be differentiable
over the open interval,

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but it's even better I guess
if you're differentiable

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over the closed interval because
you have to be continuous

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over the closed interval.

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And so, and since f is
generally differentiable,

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it will be differentiable
and continuous over zero two.

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So the mean value theorem tells us,

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tells us, that there

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is an x

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in that interval from zero to two

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such that

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f prime of x

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is equal to that secant slope,

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or you could say that
average rate of change,

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is equal to negative one.

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And so I could write,

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yes, yes, and then this
would be my justification.

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This is the slope of the secant line

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or the average rate of change,

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and since f is generally differentiable,

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it will be differentiable and continuous

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over the closed interval.

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So the mean value theorem tells us

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that there is an x in this interval

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such that f prime of x
is equal to negative one.

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And we're done.