WEBVTT
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CHRISTINE BREINER:
Welcome to recitation.
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Today what I'd like us to do is
look at the inequality tangent
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x bigger than x for the x-values
between 0 and pi over 2.
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We want to show that
that is definitely true
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using the mean value theorem.
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And I want to point out that
this was actually something
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that Joel used when he
was a graphing tangent x
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and arctangent x on
the same xy-plane.
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He used this fact in order to
get the right-looking graph.
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So what I'd like
you to do, again,
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is I want you to show that for
any x between 0 and pi over 2,
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tangent x is bigger than x.
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And use the mean value
theorem to do it.
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I'll give you a little time to
think about it, to work on it,
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and then I'll be back
and we'll do it together.
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OK.
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Welcome back.
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So I'm going to take us
through how to do this problem.
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And what I want to do
is I want to point out
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a few things initially.
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So what I'm going
to do is I'm going
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to remind us of the form
of the mean value theorem
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that we need.
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And the form that
we need will be
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f of x is equal to f of a plus
f prime of c times x minus a.
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And here, remember, c has
to be between a and x.
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So in this case, what we want
to consider is the region from 0
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up to some x-value.
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So always, this is, if you
think about it, the a and the x,
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well, a will be 0 and x will be
the right-hand region, always
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less than pi over 2.
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So what we need to
check is, I'm going
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to consider f of
x equal tangent x.
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And what I need to
consider is, does tangent x
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satisfy the hypotheses
of the mean value theorem
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on the region of interest?
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And so our region of interest
will always be a equals 0
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and b is equal to x, which
is less than some pi over 2.
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And it is true, tangent
x is continuous between 0
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and any value less
than pi over 2.
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And it's also
differentiable between 0
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and any value less
than pi over 2.
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So I can apply the mean
value theorem to tangent x.
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So in order to do this
now, what I'd like to see
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is what kind of things I
need for the right-hand side
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of this equation.
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So I obviously need to
know what the output is
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at a, which is equal to 0.
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So let's recall f of 0.
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Well, tangent 0 is 0-- it's sine
of 0 divided by cosine of 0.
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And then I need to evaluate the
derivative somewhere between a,
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which is 0, and pi over 2.
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x can be any value
less than pi over 2.
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So let's evaluate what the
derivative is in terms of x.
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We did this, actually,
in another recitation.
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The derivative of tangent
x is secant squared x.
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And so now let's
plug in what we know
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and then see what else
we need to do in order
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to finish solving this problem.
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OK.
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So what we have.
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f of x, I'm actually
going to write tangent x,
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so we can see what's
happening here.
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f of x is tangent x, and
then I have that that equals,
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well, f of a is 0, plus
f prime evaluated at c--
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so that's going to be secant
squared c-- times x minus a.
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Well, a here is 0, so
this is just times x.
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So we're very close.
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We're very close
to showing tangent
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x is always bigger than x.
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In fact, you can see very
easily what ultimately we
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need to show.
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We just need to show that secant
squared c is bigger than 1
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in our region of interest.
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And that would do
it, because then this
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would be bigger than 1 times x.
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The right-hand side
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would be bigger than 1 times
x, so that would be sufficient.
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So let's make sure
we understand what
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secant squared c can look like.
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Or what the values can be.
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So we have c is between 0 and
x, which is less than pi over 2.
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Right?
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That's where c is.
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So we always need
to remember what
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values c could possibly have.
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And then let's think about
what we know about secant of c,
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there.
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Well, secant of c is equal
to 1 over cosine of c.
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So if you can't remember
what secant's values are,
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think about the values
of cosine in that region.
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So I'm going to draw a rough
sketch of the value of cosine
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between 0 and pi over 2.
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The value of cosine does
something like this.
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This is 0, this is pi over 2.
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This is a very rough sketch.
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But this output is 1.
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So between 0 and pi
over 2, cosine of c
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is always less than 1.
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So 1 over cosine of c
is always bigger than 1.
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OK?
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Because we're taking
that reciprocal value.
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So again, cosine of c from 0
to pi over 2, not including 0,
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is always strictly less than 1.
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And so 1 over cosine c is
always strictly greater than 1.
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And so now I have
the information
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I need to come back
and finish the problem.
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So again, we were looking
at the expression,
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we have tangent x is equal
to secant squared c times x.
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Well, now I know secant
c is bigger than 1,
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so now I know this whole
thing is bigger than 1 times
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x, which is just equal to x.
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So if you noticed,
on the left-hand side
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we have a tangent
x equals something
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which is bigger than x.
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So we've just shown that,
for any value of x between 0
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and pi over 2, tangent
x is bigger than x.
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And I think we'll stop there.