WEBVTT
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PROFESSOR: Welcome
back to recitation.
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In this video, what
I'd like us to do
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is answer the
following question.
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Suppose that f is a continuous,
differentiable function.
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And if it's derivative,
if f prime is never 0,
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and a is not equal to
b, then show that f of a
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is not equal to f of b.
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I'm going to let you think
about it for a while,
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see if you can come up with
a good reason for that,
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and then I'll be back
to explain my reasons.
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OK.
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Our object, again,
is to show, if f
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is continuous and differentiable
and its derivative is never 0
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and you're looking at two
x-values that are different,
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show that their y-values
have to be different.
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Show that if the
inputs are different,
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the outputs have
to be different.
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Now, this might remind
you of something
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you saw in lecture about if
the derivative has a sign,
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show the function-- if the
derivative is positive,
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show the function is
always increasing,
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or if the derivative
is negative,
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show the function is
always decreasing.
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So this is a similar
type of problem to that.
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So what we're going to use
is actually the mean value
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theorem.
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If you'll notice, I have f.
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It does satisfy the
mean value theorem
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on an interval from a to b.
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I haven't even specified
which is bigger, a or b.
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But it doesn't
matter in this case.
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So what do we know?
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The mean value theorem tells
us that if we look at-- well,
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let's just write it out-- f of
b minus f of a over b minus a
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is equal to f prime
of c-- and what
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do we know-- for
c between a and b.
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So we want to know whether
or not f of b minus f of a
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can ever be 0.
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We're trying to show
that it cannot be 0.
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So we're going to isolate this
expression and show that this
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subtraction cannot be 0.
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Well, how do we do that?
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Let me come over here to
give us a little more room.
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I'm going to rewrite
the mean value theorem.
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I'm going to multiply
through by b minus a.
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So we get f prime
of c time b minus a.
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Now, we just want
to show, again,
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that f of b minus
f of a cannot be 0.
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What's the only thing--
well, not only thing,
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we know two things-- what
two things do we know?
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We know f prime of c is not 0.
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That was given to you.
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f prime is never 0, so
certainly at any fixed value,
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f prime of c is not 0.
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So we know this term is not 0.
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We also know that b
is not equal to a,
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so we know b minus a is not 0.
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The only way to get a product
of two numbers to be 0
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is if one of them is 0.
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So this in fact, this
product is not equal to 0.
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The fact that this
product is not equal to 0
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tells us f of b minus f
of a is not equal to 0.
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And that alone is enough
to conclude that f of b
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is not equal to f of a.
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So, again, let me
just point out of this
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is probably reminds you very
much of the type of thing
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you've seen where you were
showing if f prime had a sign,
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then you could determine whether
f was increasing or decreasing.
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It's the same type
of problem as that.
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It's exploiting what the
mean value theorem tells you.
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So I think we'll stop there.