Matrices to solve a system of equations

WEBVTT
Kind: captions
Language: en

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We've done a lot of work
on multiplying, adding,

00:00:03.470 --> 00:00:05.950
subtracting and inverting
matrices.

00:00:05.950 --> 00:00:09.280
So now let's delve a little
into what a matrix is

00:00:09.280 --> 00:00:10.250
actually good for.

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And remember, all a matrix
is is, a way of

00:00:13.800 --> 00:00:14.870
representing data.

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And all of those rules we
learned, you can kind of view

00:00:17.120 --> 00:00:18.890
them as human-created rules.

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There's no fundamental thing in
nature that says matrices

00:00:23.660 --> 00:00:25.490
have to be multiplied
the way we learned.

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But I think you'll see as we
progress into applications,

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that the way that matrix
operations have been defined

00:00:32.220 --> 00:00:34.020
are actually quite useful.

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So let's go back to our Algebra
1 or Algebra 2.

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I forget when you tend
to learn it.

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But let's go back to
linear equations.

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So what were linear equations?

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Systems of linear equations.

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Well you had two lines, and you
essentially had to figure

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out where the two lines
intersected.

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So you might have had something
like-- let me think

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of something-- 3x plus 2y.

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Is equal to 7.

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And then you might have, minus
6x plus 6y is equal to-- I

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need to do this in my head just
to make sure that I get

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numbers that work out
well-- equal to 6.

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I think this will
work out well.

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And what was this problem
essentially?

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Well this is a line,
and this is a line.

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So you had to figure out
where they intersected.

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And if you were to draw
those two lines--

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Actually let's draw them.

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Just because this is all about
getting intuition, and seeing

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how it maps over into
the matrix world.

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And the word 'matrix world'
has a whole new

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meaning after 1999.

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Let's see, so if that's my
coordinate axes, what is this?

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I always have to put everything
into y equals mx

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plus b form for me to-- So
this equation is what?

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It's y is equal to
3/2 x plus 7/2.

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So 7/2 is what?

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It's like 3 1/2 or something?

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So if that's 7/2, that's going
to have a slope of 3 1/2.

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So it's a little bit steeper
than a slope of 1.

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So it's going to look
something like that.

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That's that line.

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And then this line is going
to look like what?

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I'll do a different color.

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It's going to look like-- It's
the same thing is as--

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Oh, you know what?

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I did that wrong.

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Because that line, I just
realized, is equal to

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minus 3x plus 7/2.

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Because when you take this to
the other side, it becomes

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minus 3x divided by 2, so
it's actually going

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to be downward sloping.

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So it will look something
like this.

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It's going to be a little bit
steeper than something that

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has a slope of negative 1, so
I'm just approximating.

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So that line will look
something like that.

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And this line, this will be y--
I'm just rewriting this--

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y is equal to x plus
1, if I'm right.

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Yeah.

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Because this go to
the other side.

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Divide everything by 6.

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y is equal to x plus 1, so its
y intercept will be-- We said

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this was 3 and 1/2, so
maybe if this is 1.

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And it has a slope of 1.

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Then it'll just look like
something like this.

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And so when you solve a system
of equations, you're

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essentially looking for the x
and the y values that satisfy

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both of these equations.

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This magenta line shows us all
the x and y values that

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satisfy this first
linear equation.

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And this green line shows all
of the x's and y's that

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satisfy the second equation.

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And of course where they
intersect shows us the

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particular x and y that
satisfies both equations.

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So that's what we did
in Algebra 1.

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We'd solve both of these
equations for that.

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And we'd either do it by
substitution, or we'd scale

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them and add them together,
et cetera, et cetera.

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As you'll see, that's
essentially just what we

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learned in the Gauss-Jordan
elimination.

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It's the exact same thing.

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It's just when we did the
Gauss-Jordan elimination, we

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just represented it a
little different.

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But I think you know
this much.

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But let's now do it in
the matrix world.

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So how can we represent this
problem as a matrix?

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We could write it like this,
and we'll take out a little

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time to prove to you that
it really is the same

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representation.

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If you define matrices the way
we have defined them in their

00:05:09.740 --> 00:05:10.850
multiplication.

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You can define this problem
as 3, minus 6, 2, 6.

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I just took the coefficients,
3, minus 6, 2, 6.

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And if I were to multiply
that by soon. column

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vector matrix xy.

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And if I were to set that equal
to another column vector

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matrix, 7, 6.

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Now you might want to pause it
and actually just try to

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multiply this out, the way
that we have learned to

00:05:54.650 --> 00:05:55.720
multiply matrices.

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And you will see that you
get the same thing.

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But I will do it now,
in case you don't

00:05:59.660 --> 00:06:00.740
want to do it yourself.

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So let's just multiply
these two matrices.

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Let's multiply this matrix
out and see what happens.

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So what do you do?

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You get your row information
from the first matrix, column

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information from the
second matrix.

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And this is of course
the product matrix.

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So this is saying, 3 times x
plus 2 times y is equal to 7.

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Well that's exactly what
we wrote up here.

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3 times x plus 2 times
y is equal to 7.

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And similarly when you multiply
the bottom row, you

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get minus 6 times x plus 6
times y is equal to 6.

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So if that was a little
confusing to you, go review

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how we multiply matrices.

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But if you just multiply this
out, you'll get these exact

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same equations.

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So hopefully you understand that
this is just another way

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of representing this problem.

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Although we've gotten rid
of the plus signs

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and the equals signs.

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But of course you have to
know the representation.

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But why is this useful?

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Why is this representation
useful?

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Well let's call this matrix a.

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Let's call this vector x.

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It's not a variable.

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It's a vector.

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So maybe we'll bold it, or we'll
put a little vector sign

00:07:17.420 --> 00:07:17.970
there or something.

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Whatever.

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But you'll see it in
your textbook.

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It's bolded real heavy.

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And then we call
this vector b.

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And the general notation-- if
I remember it correctly-- is

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that anything that's a matrix
or a vector is bolded.

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And matrices that are not
vectors, that have more than

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one dimension in either
of the dimensions,

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they're capital letters.

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While lower-case letters
represent vectors.

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So these are matrices, but
they're also vectors.

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So that's why they got the
lowercase letters.

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And that's why this one got
the uppercase letters.

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That's just convention.

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So this equation has the form
ax equals b, where a is this

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matrix, x is this vector-- or
this matrix, same thing-- and

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b is this column vector.

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So what does that do for us?

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Well, what happens if
we knew a inverse?

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Well actually, let me
take a step back.

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If these were numbers,
what would we do?

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If I just gave you an algebra
equation, ax is equal to b.

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How do you solve that?

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Well you would divide both sides
of this equation by a.

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Or another way of saying it, you
would multiply both sides

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of this equation by
the inverse of a.

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So you would essentially say,
1/a times ax is equal

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to 1/a times b.

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And then these would cancel out,
and you would get x is

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equal to b/a.

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That's how we would do
it in a traditional

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simple, linear equation.

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So how would you do it here?

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Well what's the matrix
analogy to division?

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And I'm going to give
you the answer now.

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What's the analogy to
multiplying by your inverse?

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Well, it's multiplying
by your inverse.

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So what if we knew the
matrix a inverse?

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We could just multiply
both sides of this

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equation by a inverse.

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And remember, order matters.

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So it's not like when you're
doing a linear equation you

00:09:13.310 --> 00:09:14.910
could multiply 1/a
on this side.

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But then you can do it on
the right side here.

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But no.

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Notice, I put it in front of
the numbers in both cases.

00:09:19.700 --> 00:09:21.840
So you have to do it in front of
the numbers in both cases.

00:09:21.840 --> 00:09:25.750
But if we know a inverse, and
if a inverse exists, then we

00:09:25.750 --> 00:09:29.240
can multiply both sides-- you
can say the left side of both

00:09:29.240 --> 00:09:31.705
sides of this equation
by a inverse.

00:09:35.160 --> 00:09:44.580
a inverse times a, times the
vector x is equal to a

00:09:44.580 --> 00:09:47.480
inverse times b.

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All I did is I took this
expression, and I multiplied

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both sides by a inverse.

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And what's a inverse times a?

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Well that's just the
identity matrix.

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That's the identity matrix,
times x is equal

00:09:59.070 --> 00:10:03.570
to a inverse b.

00:10:03.570 --> 00:10:05.900
And of course that's just x.

00:10:05.900 --> 00:10:07.720
The identity matrix times
any other matrix

00:10:07.720 --> 00:10:08.720
is just that matrix.

00:10:08.720 --> 00:10:11.540
So that's just the matrix
x, or the vector x

00:10:11.540 --> 00:10:15.310
times a inverse b.

00:10:15.310 --> 00:10:19.010
So, if you're given a linear
equation, and if you know the

00:10:19.010 --> 00:10:21.400
inverse of this matrix, to solve
for x and y, we just

00:10:21.400 --> 00:10:23.740
have to multiply this number
times the inverse.

00:10:23.740 --> 00:10:25.690
And you might say, Sal,
that's such a pain.

00:10:25.690 --> 00:10:28.720
Because this is such a simple
linear equation to solve.

00:10:28.720 --> 00:10:31.210
Why would I go through all the
trouble of getting an inverse,

00:10:31.210 --> 00:10:33.370
and then multiplying the inverse
times this number.

00:10:33.370 --> 00:10:35.570
And I would agree with
you to some degree.

00:10:35.570 --> 00:10:39.960
That for a 2 by 2 system of
equations, it is easier to

00:10:39.960 --> 00:10:44.150
solve it the way that you did it
in Algebra 1 or Algebra 2.

00:10:44.150 --> 00:10:49.170
But if you're doing it for a 3
by 3, well, finding a matrix

00:10:49.170 --> 00:10:50.780
is still pretty difficult
for a 3 by 3,

00:10:50.780 --> 00:10:51.460
so it's still difficult.

00:10:51.460 --> 00:10:53.990
But as you get to larger and
larger numbers, it's

00:10:53.990 --> 00:10:56.900
sometimes-- well, finding a
matrix can be difficult too--

00:10:56.900 --> 00:11:01.370
But actually the real place
where it you really, really

00:11:01.370 --> 00:11:05.980
pays off is, let's say that
you have a bunch of linear

00:11:05.980 --> 00:11:08.220
equations to solve.

00:11:08.220 --> 00:11:10.160
And the left hand side
stays the same.

00:11:10.160 --> 00:11:12.150
But you keep changing
the right hand side.

00:11:12.150 --> 00:11:13.860
So let's say you have
ax equals b.

00:11:13.860 --> 00:11:15.380
And then you have another one
that says, ax equals c,

00:11:15.380 --> 00:11:16.680
and ax equals d.

00:11:16.680 --> 00:11:18.380
And these numbers
keep changing.

00:11:18.380 --> 00:11:19.730
And these numbers
are the same.

00:11:19.730 --> 00:11:23.080
Then it really pays off to
solve for the inverse.

00:11:23.080 --> 00:11:26.200
And then every time you need
to find a new solution, you

00:11:26.200 --> 00:11:29.090
just multiply your new
right-hand side times your

00:11:29.090 --> 00:11:30.880
inverse, and you just
get the answer.

00:11:30.880 --> 00:11:32.810
And that'll really pay
off when we view

00:11:32.810 --> 00:11:33.950
this in another way.

00:11:33.950 --> 00:11:34.870
But anyway, I wanted
to show you that

00:11:34.870 --> 00:11:36.020
this is the same thing.

00:11:36.020 --> 00:11:39.540
And so let's solve
it using what our

00:11:39.540 --> 00:11:41.020
knowledge is of matrices.

00:11:41.020 --> 00:11:43.070
Let me erase this here, and I
know I'm running over time,

00:11:43.070 --> 00:11:46.440
but hopefully I'm not completely
boring you.

00:11:49.750 --> 00:11:51.780
So I'm going to keep that there,
just because I think

00:11:51.780 --> 00:11:53.080
it's nice to have that
visual representation

00:11:53.080 --> 00:11:54.650
of what we're doing.

00:11:54.650 --> 00:11:56.240
Always to remember
what's going on.

00:11:56.240 --> 00:11:57.930
So, what's a inverse?

00:11:57.930 --> 00:12:05.860
So first of all, the a inverse
is equal to 1 over the

00:12:05.860 --> 00:12:12.260
determinant of a times the
adjoint of this matrix.

00:12:12.260 --> 00:12:14.520
I don't want to get fancy with
terminology and all that, but

00:12:14.520 --> 00:12:15.350
what was that?

00:12:15.350 --> 00:12:16.950
2 by 2 is fairly easy.

00:12:16.950 --> 00:12:21.980
You swap these two terms.
You get a 6 and a 3.

00:12:21.980 --> 00:12:23.690
And then you make these
two terms negative.

00:12:23.690 --> 00:12:26.100
So a minus 6 becomes a 6.

00:12:26.100 --> 00:12:27.780
And a 2 becomes a minus 2.

00:12:30.500 --> 00:12:32.870
And what's the determinant
of a?

00:12:32.870 --> 00:12:36.610
The determinant of a is equal to
this times this minus this

00:12:36.610 --> 00:12:37.070
times this.

00:12:37.070 --> 00:12:38.830
So 3 times 6.

00:12:38.830 --> 00:12:45.840
3 times 6 is 18 minus
this times this.

00:12:45.840 --> 00:12:47.540
So 6 times 2 is 12.

00:12:47.540 --> 00:12:48.430
That's a minus 6.

00:12:48.430 --> 00:12:49.320
That's minus 12.

00:12:49.320 --> 00:12:52.460
So minus minus 12.

00:12:52.460 --> 00:12:56.280
It's equal to plus.

00:12:56.280 --> 00:12:58.690
So 18 plus 12 is equal to 30.

00:12:58.690 --> 00:13:00.350
So what does a inverse equal?

00:13:00.350 --> 00:13:02.090
1 over 30 times this thing.

00:13:02.090 --> 00:13:10.870
So a inverse is equal to-- we
could even keep the 1/30 on

00:13:10.870 --> 00:13:12.120
the outside.

00:13:13.830 --> 00:13:15.490
That might simplify things.

00:13:15.490 --> 00:13:16.520
Well actually I'll put it--

00:13:16.520 --> 00:13:19.890
So a inverse is equal to what?

00:13:19.890 --> 00:13:21.120
This divided by 30.

00:13:21.120 --> 00:13:33.300
So that's 1/5, minus-- Actually
I do want to keep it

00:13:33.300 --> 00:13:36.180
on the outside, because that's
going to make the later

00:13:36.180 --> 00:13:38.080
multiplications easier.

00:13:38.080 --> 00:13:46.570
So anyway, a is equal to 1/30
times 6, minus 2, 6, 3.

00:13:46.570 --> 00:13:47.680
That's a inverse.

00:13:47.680 --> 00:13:49.370
So now let's solve
for x and y.

00:13:49.370 --> 00:13:53.340
So we said x and y is equal
to a inverse times b.

00:13:53.340 --> 00:13:58.990
So we could say x-- another way
to write x is like this.

00:13:58.990 --> 00:14:00.820
x is just this vector.

00:14:00.820 --> 00:14:02.090
x and y.

00:14:02.090 --> 00:14:04.785
Not to get confused, this x is
different than that x, even

00:14:04.785 --> 00:14:05.570
though I've written
them the same.

00:14:05.570 --> 00:14:08.430
If I was a typographer, I would
make this really bold,

00:14:08.430 --> 00:14:10.040
so that you know that
this is a vector.

00:14:10.040 --> 00:14:11.890
Maybe I should put a
vector notation.

00:14:11.890 --> 00:14:12.150
I don't know.

00:14:12.150 --> 00:14:14.480
You could do a bunch
of things with it.

00:14:14.480 --> 00:14:17.890
It's equal to a inverse
times this.

00:14:17.890 --> 00:14:20.660
So that's 1/30.

00:14:20.660 --> 00:14:22.490
I did that just for the
matrix addition.

00:14:22.490 --> 00:14:26.920
I didn't divide everything
by 30, just so the matrix

00:14:26.920 --> 00:14:28.580
multiplication's a
little easier.

00:14:28.580 --> 00:14:32.510
Minus 2, 3, times 7/6.

00:14:37.930 --> 00:14:40.616
And so what is this equal to?

00:14:40.616 --> 00:14:45.700
It's equal to 1/30 times-- I
know I'm crowding this down

00:14:45.700 --> 00:14:46.810
here-- let's see.

00:14:46.810 --> 00:14:52.440
6 times 7 minus 2 times 6.

00:14:52.440 --> 00:14:59.240
So 6 times 7 is 42.

00:14:59.240 --> 00:15:01.690
Minus 2 times 6, so minus 12.

00:15:01.690 --> 00:15:04.830
So that's equal to the 30.

00:15:04.830 --> 00:15:07.220
And then 6 times 7
plus 2 times 6.

00:15:07.220 --> 00:15:11.090
So 6 times 7, once
again is 42.

00:15:11.090 --> 00:15:12.820
Plus 2 times 6.

00:15:12.820 --> 00:15:17.420
So 42 plus 12 is 50.

00:15:17.420 --> 00:15:19.620
Is that right?

00:15:19.620 --> 00:15:22.990
6 times 7-- oh I'm sorry.

00:15:22.990 --> 00:15:23.700
This is a 3.

00:15:23.700 --> 00:15:25.320
That's why I was getting
confused.

00:15:25.320 --> 00:15:27.700
See, it's important to
have good penmanship.

00:15:27.700 --> 00:15:33.565
So it's 6 times 7 is
42, plus 3 times 6.

00:15:33.565 --> 00:15:38.120
So it's 42 plus 18,
which is 60.

00:15:38.120 --> 00:15:40.480
And of course you divide
both of them by 30.

00:15:40.480 --> 00:15:43.770
So you get the final xy.

00:15:43.770 --> 00:15:45.450
I'll write it here.

00:15:45.450 --> 00:15:46.480
I don't want to erase
anything.

00:15:46.480 --> 00:15:52.850
So we get xy is equal to--
divide both of those by 30--

00:15:52.850 --> 00:15:58.410
is equal to 1 and 2.

00:15:58.410 --> 00:16:02.420
And so that tells us that these
two linear equations

00:16:02.420 --> 00:16:10.160
intersect at the point x is
equal to 1, y is equal to 2.

00:16:10.160 --> 00:16:12.610
That might seem a little bit
like a lot of work, but that's

00:16:12.610 --> 00:16:14.480
just because I took the time
to explain it and all that.

00:16:14.480 --> 00:16:17.000
But if you just immediately took
that, represented it this

00:16:17.000 --> 00:16:19.240
way, found the inverse, and
multiplied, it wouldn't have

00:16:19.240 --> 00:16:20.050
taken you that much time.

00:16:20.050 --> 00:16:22.100
And I encourage you to do
that as an exercise.

00:16:22.100 --> 00:16:23.530
Anyway, I'll see you
in the next video.

00:16:23.530 --> 00:16:25.610
And in the next video, we're
going to do this exact same

00:16:25.610 --> 00:16:29.010
problem, but we're going to see
that this data represents

00:16:29.010 --> 00:16:30.370
a different problem.

00:16:30.370 --> 00:16:31.620
See


Graphing with Calculus

WEBVTT
Kind: captions
Language: en

00:00:00.530 --> 00:00:03.340
Let's do another problem where
we graph a function based

00:00:03.340 --> 00:00:05.140
on the properties of
its derivatives and

00:00:05.140 --> 00:00:06.010
second derivative.

00:00:06.010 --> 00:00:08.700
And what's especially
interesting about this problem

00:00:08.700 --> 00:00:10.870
I think is the fact they
don't give the function.

00:00:10.870 --> 00:00:12.635
They just give a bunch of
clues about the function.

00:00:12.635 --> 00:00:14.070
So what do they tell us?

00:00:14.070 --> 00:00:24.220
Well they tell us that
f of 0 is equal to 3.

00:00:24.220 --> 00:00:28.725
They also tell us that at
f of 3 is equal to 0.

00:00:34.150 --> 00:00:36.200
They tell us that f
of 6 is equal to 4.

00:00:41.880 --> 00:00:43.590
What I like about this problem
is that you can't use a

00:00:43.590 --> 00:00:45.950
calculator to solve it because
you don't know the function.

00:00:45.950 --> 00:00:51.010
So even if I give you a graphic
calculator you'd be helpless.

00:00:51.010 --> 00:00:52.750
Let's write out all the
information then we

00:00:52.750 --> 00:00:56.020
can plot it out.

00:00:56.020 --> 00:00:59.010
Let me write all the
derivative information here.

00:00:59.010 --> 00:01:06.100
They tell us that f prime--
so the derivative of x--

00:01:06.100 --> 00:01:14.870
is less than 0 on the
open interval 0 comma 3.

00:01:14.870 --> 00:01:18.150
So that means not
including 0 and 3.

00:01:18.150 --> 00:01:28.520
And they also tell us that the
derivative of is greater than 0

00:01:28.520 --> 00:01:32.410
on the open interval
from 3 comma 6.

00:01:32.410 --> 00:01:35.120
So we can already guess that
if we're less than 0 as we

00:01:35.120 --> 00:01:38.900
approach 3 and we're greater
than 0 as we go away from 3,

00:01:38.900 --> 00:01:41.060
that the derivative
must be 0 at 3.

00:01:41.060 --> 00:01:42.620
That's the intuition.

00:01:42.620 --> 00:01:44.390
Although they don't tell us
that for sure, but I'm assuming

00:01:44.390 --> 00:01:47.290
that this function is
continuous and differentiable

00:01:47.290 --> 00:01:49.620
over all the intervals, because
the don't tell us otherwise.

00:01:49.620 --> 00:01:49.880
OK.

00:01:49.880 --> 00:01:51.180
So that's the information
to tell us about the

00:01:51.180 --> 00:01:52.740
first derivative.

00:01:52.740 --> 00:01:55.630
And what do they tell us
about the second derivative?

00:01:55.630 --> 00:02:03.750
They tell us that the second
derivative is greater than

00:02:03.750 --> 00:02:09.720
0 on the interval 0 to 5.

00:02:09.720 --> 00:02:12.030
I did the open
interval from 0 to 5.

00:02:12.030 --> 00:02:15.740
So that means it's
concave upwards there.

00:02:15.740 --> 00:02:21.680
And they also tell us that the
second derivative is less than

00:02:21.680 --> 00:02:28.010
0, so it's concave downwards on
the open interval

00:02:28.010 --> 00:02:29.430
from five to 6.

00:02:33.420 --> 00:02:34.620
You know they don't give
us information what

00:02:34.620 --> 00:02:35.990
happens after 6.

00:02:35.990 --> 00:02:38.450
Maybe it goes to an
inflection point.

00:02:38.450 --> 00:02:39.050
We don't know.

00:02:39.050 --> 00:02:42.470
So maybe you know, we'll just
assume that is doesn't change.

00:02:42.470 --> 00:02:45.030
So what does all this
information tell us?

00:02:45.030 --> 00:02:46.890
Well first let's just graph it.

00:02:46.890 --> 00:02:49.095
So f of 0 is 3, so we
have to graph the

00:02:49.095 --> 00:02:52.700
point 0,3 3,0 and 6,4.

00:02:52.700 --> 00:02:55.410
It seems like all of the action
is in the first quadrant, so

00:02:55.410 --> 00:02:58.210
I'll graph it accordingly.

00:02:58.210 --> 00:03:01.190
So let me graph it.

00:03:01.190 --> 00:03:06.600
I want to focus on the first
quadrant, so that's my x-axis.

00:03:06.600 --> 00:03:07.460
That's my y-axis.

00:03:10.190 --> 00:03:11.020
I have to go to 6,4.

00:03:16.960 --> 00:03:26.790
I want to draw this point
6,4, so 1, 2, 3, 4, 5, 6.

00:03:26.790 --> 00:03:31.320
1, 2, 3, 4.

00:03:31.320 --> 00:03:32.520
So that's the point 6,4.

00:03:39.190 --> 00:03:40.300
And let's see.

00:03:40.300 --> 00:03:42.560
3 comma 0.

00:03:42.560 --> 00:03:45.180
So that's 1, 2, 3, so that's
this point right here.

00:03:47.810 --> 00:03:51.050
That point right
there is 3 comma 0.

00:03:54.960 --> 00:03:56.630
I got that from
this information.

00:03:56.630 --> 00:03:59.900
And then we have the
point 0 comma 3.

00:03:59.900 --> 00:04:01.180
0, 1, 2, 3.

00:04:01.180 --> 00:04:01.600
So this is 0,3.

00:04:07.800 --> 00:04:09.580
Well, those are definitely
points on the function.

00:04:09.580 --> 00:04:11.760
That's what this tells us.

00:04:11.760 --> 00:04:12.840
And what do they tell us?

00:04:12.840 --> 00:04:17.030
They tell us that the
derivative is less than 0 on

00:04:17.030 --> 00:04:18.250
the interval from 0 to 3.

00:04:21.400 --> 00:04:23.670
From here to here the
derivative is less than 0.

00:04:23.670 --> 00:04:29.090
And that makes sense, because
we have to obviously decrease

00:04:29.090 --> 00:04:30.810
to go from this value
to this value.

00:04:30.810 --> 00:04:33.020
But what it does tell us is
that the function does not

00:04:33.020 --> 00:04:34.465
do something like this.

00:04:34.465 --> 00:04:37.260
The function does not do this.

00:04:37.260 --> 00:04:38.880
The function does not do that.

00:04:38.880 --> 00:04:40.040
And how do we know that?

00:04:40.040 --> 00:04:44.690
Well because if the function
moved up and then down, then

00:04:44.690 --> 00:04:47.560
you have a positive slope
here, which this says

00:04:47.560 --> 00:04:49.020
is not possible.

00:04:49.020 --> 00:04:52.030
The derivative is less than 0
from this interval to this

00:04:52.030 --> 00:04:55.290
interval, so the slope has to
be negative the entire time.

00:04:58.050 --> 00:05:02.550
And then it tells us that the
slope is positive from this

00:05:02.550 --> 00:05:05.150
interval-- from x is equal
to 3 to x is equal to 6,

00:05:05.150 --> 00:05:06.310
that's what this tells us.

00:05:06.310 --> 00:05:09.580
And so you know the line has to
look something like that, or

00:05:09.580 --> 00:05:12.680
like that, but can't curve down
then go up, because if it

00:05:12.680 --> 00:05:15.350
curved down you'd have a
negative slope at some point.

00:05:15.350 --> 00:05:16.900
So OK, fair enough.

00:05:16.900 --> 00:05:19.740
Essentially the graph is
monotonic from this point

00:05:19.740 --> 00:05:21.720
to this point and this
point to this point.

00:05:21.720 --> 00:05:25.140
And if you want to get
technical, the monotonicity

00:05:25.140 --> 00:05:28.970
theorem tells us that, well
then this must be a point

00:05:28.970 --> 00:05:34.060
either that is a point where
the derivative is 0, or it must

00:05:34.060 --> 00:05:36.150
be a point where there is no
derivative, but I'm assuming

00:05:36.150 --> 00:05:38.670
that this graph is
differentiable over everything.

00:05:38.670 --> 00:05:40.370
And you know you didn't need
a theorem to tell you that.

00:05:40.370 --> 00:05:42.800
We know that the slope is
less than 0 as we approach

00:05:42.800 --> 00:05:44.470
here in this interval.

00:05:44.470 --> 00:05:46.690
We know the slope is greater
than 0 in this interval.

00:05:46.690 --> 00:05:49.110
And if we assume that the
derivative is continuous, which

00:05:49.110 --> 00:05:51.270
I will make that assumption
although they haven't told us,

00:05:51.270 --> 00:05:55.200
but in the absence of that
information, why not.

00:05:55.200 --> 00:05:57.650
If the derivative is always
kind of incrementing or

00:05:57.650 --> 00:06:00.800
decrementing at a continues
rate, the derivative must be 0

00:06:00.800 --> 00:06:03.700
at this point, because it's
less than 0 here and

00:06:03.700 --> 00:06:05.260
greater than 0 here.

00:06:05.260 --> 00:06:09.025
I'll draw a little horizontal
line to indicate that the

00:06:09.025 --> 00:06:10.070
derivative is 0 there.

00:06:10.070 --> 00:06:14.540
So a it's going to be a bit
of a minimum point I think.

00:06:14.540 --> 00:06:16.260
And then let's move on to the
information on the second

00:06:16.260 --> 00:06:18.440
derivatives and see what
they're telling us.

00:06:18.440 --> 00:06:21.740
Well they're saying that the
second derivative is greater

00:06:21.740 --> 00:06:23.810
than 0 from 0 to 5.

00:06:23.810 --> 00:06:26.200
So from 0 to 5.

00:06:26.200 --> 00:06:27.710
So this is 5 right here.

00:06:32.120 --> 00:06:32.770
What does this tells?

00:06:32.770 --> 00:06:37.970
This means that we are
concave upwards from 0 to 5.

00:06:37.970 --> 00:06:39.180
That gives us a lot
of information.

00:06:39.180 --> 00:06:43.700
So that tells us that not only
are we monotonically decreasing

00:06:43.700 --> 00:06:47.890
as we approach 3, but
we're concave upwards.

00:06:47.890 --> 00:06:51.260
And and why is this
concave upwards?

00:06:51.260 --> 00:06:53.120
Well as you see the
slope here is negative.

00:06:53.120 --> 00:06:54.830
It's a little less
negative here.

00:06:54.830 --> 00:06:56.580
It's even less negative here
and then as it slowly

00:06:56.580 --> 00:06:58.410
approaches 0 it gets less and
less negative, and then

00:06:58.410 --> 00:07:03.580
it keeps increasing
until you get to 5.

00:07:03.580 --> 00:07:07.070
And then from 5 to
6 what happens?

00:07:07.070 --> 00:07:10.130
The second derivative becomes
negative, so we enter a point

00:07:10.130 --> 00:07:11.450
that is concave downwards.

00:07:11.450 --> 00:07:14.180
So from 5 to 6-- I don't know
what the value of the function

00:07:14.180 --> 00:07:21.070
is it 5-- but from 5 to 6 the
function becomes concave

00:07:21.070 --> 00:07:23.490
downwards from x equals
5 to x equals 6.

00:07:28.810 --> 00:07:31.583
We don't know what f of 5 is,
but f of 5-- they didn't tell

00:07:31.583 --> 00:07:33.640
us-- is an inflection point.

00:07:33.640 --> 00:07:35.490
That's where we went from
going concave upwards

00:07:35.490 --> 00:07:37.600
to concave downwards.

00:07:37.600 --> 00:07:40.330
And I think that's all we
can do with this graph.

00:07:40.330 --> 00:07:41.770
We don't know what happens.

00:07:41.770 --> 00:07:43.820
I mean actually we made a
couple of assumptions anyway,

00:07:43.820 --> 00:07:45.730
that the graph is continuous
and differentiable

00:07:45.730 --> 00:07:46.535
over the whole thing.

00:07:46.535 --> 00:07:49.640
But I think this will give you
an intuition on what all of

00:07:49.640 --> 00:07:51.650
this first and second
derivative and all this

00:07:51.650 --> 00:07:52.810
stuff does for us.

00:07:52.810 --> 00:07:55.370
And just so you know, this
is an inflection point.

00:07:55.370 --> 00:08:00.580
they could have told us that f
prime prime of 5 is equal to 0.

00:08:00.580 --> 00:08:02.960
We don't know that for sure,
but I'm just assuming.

00:08:02.960 --> 00:08:08.220
Anyway hope you found
that vaguely useful.

00:08:08.220 --> 00:08:10.120
Actually since I have time,
let me do one more bonus

00:08:10.120 --> 00:08:14.300
problem that I was sent.

00:08:14.300 --> 00:08:18.270
And it was written right after
that problem so why not do it.

00:08:18.270 --> 00:08:20.710
And it says prove that
a quadratic function

00:08:20.710 --> 00:08:21.220
has no inflections.

00:08:21.220 --> 00:08:23.980
So what does a quadratic
function look like?

00:08:23.980 --> 00:08:30.180
f of x is equal to ax
squared plus bx plus c.

00:08:30.180 --> 00:08:32.930
And it says prove that this
has no inflection point.

00:08:32.930 --> 00:08:33.990
Well, what's an
inflection point?

00:08:33.990 --> 00:08:38.400
It's a point where the second
derivative is equal to 0.

00:08:38.400 --> 00:08:40.760
So let's figure out what
the second derivative is.

00:08:40.760 --> 00:08:45.300
The first derivative of this
quadratic-- but any quadratic

00:08:45.300 --> 00:08:51.270
can be written like
this-- is 2ax plus b.

00:08:51.270 --> 00:08:52.405
And what's the
second derivative?

00:08:55.280 --> 00:08:58.240
Well that's just equal to 2a.

00:08:58.240 --> 00:09:00.860
Now an inflection point
is when the second

00:09:00.860 --> 00:09:03.040
derivative is equal to 0.

00:09:03.040 --> 00:09:10.010
Now my question to you is, can
this function ever equal 0?

00:09:10.010 --> 00:09:12.780
Well, the only way that this
function would equal 0 is for

00:09:12.780 --> 00:09:15.610
some reason a was 0, but if a
was 0 then it wouldn't be a

00:09:15.610 --> 00:09:17.650
quadratic, because then this
term wouldn't exist, it

00:09:17.650 --> 00:09:19.960
would be bx plus c and
it would be just a line.

00:09:19.960 --> 00:09:22.080
So if we're talking about a
quadratic where this a

00:09:22.080 --> 00:09:24.430
coefficient is non 0, which it
has to be because that's the

00:09:24.430 --> 00:09:27.420
coefficient of the term x,
there's no way that

00:09:27.420 --> 00:09:32.190
this function, this
expression, is non 0.

00:09:32.190 --> 00:09:35.280
So there's no way that the
second derivative can ever be

00:09:35.280 --> 00:09:39.980
0, and thus we have proved
that a quadratic has no

00:09:39.980 --> 00:09:41.530
point of inflection.

00:09:41.530 --> 00:09:42.360
see you in the next video.


Calculus: Maximum and minimum values on an interval

WEBVTT
Kind: captions
Language: en

00:00:00.720 --> 00:00:05.700
I've been requested to do some
problems on critical points and

00:00:05.700 --> 00:00:10.550
maxima, minima, and concavity,
so I will do some problems.

00:00:10.550 --> 00:00:12.430
So these were sent to
me by [? Akosh ?].

00:00:12.430 --> 00:00:16.110
I don't know where he is but
these are interesting problems

00:00:16.110 --> 00:00:18.780
and I haven't done anything
like these before, so I thought

00:00:18.780 --> 00:00:20.400
I would do the exact
problem he's given.

00:00:20.400 --> 00:00:23.440
So the first problem says,
identify the critical point,

00:00:23.440 --> 00:00:26.140
and find the maximum and
minimum value on the

00:00:26.140 --> 00:00:26.860
given interval.

00:00:26.860 --> 00:00:38.670
And they say f of x is equal
to x squared plus 4x plus 4.

00:00:38.670 --> 00:00:44.210
And the interval that they care
about is from minus 4, and

00:00:44.210 --> 00:00:46.040
including minus 4-- that's why
we have the brackets

00:00:46.040 --> 00:00:49.100
there-- to 0.

00:00:49.100 --> 00:00:51.740
And just so you know, this
interval notation, if it was

00:00:51.740 --> 00:00:54.340
written like this with
parentheses, it would mean all

00:00:54.340 --> 00:00:58.800
the numbers between minus 4 and
0, but not including them.

00:00:58.800 --> 00:01:00.830
This is a closed interval,
so you're including minus

00:01:00.830 --> 00:01:02.290
4 and you're including 0.

00:01:02.290 --> 00:01:03.520
This is called an
open interval.

00:01:03.520 --> 00:01:05.660
But anyway you can ignore
this, because that's not

00:01:05.660 --> 00:01:07.450
what was in the problem.

00:01:07.450 --> 00:01:10.310
So the first thing they ask is
what are the critical points.

00:01:10.310 --> 00:01:13.020
And different people use
different terminology, but my

00:01:13.020 --> 00:01:14.910
understanding is critical
points are all the points

00:01:14.910 --> 00:01:15.720
that are interesting.

00:01:15.720 --> 00:01:19.200
So it's the points where the
derivatives equal to 0, or the

00:01:19.200 --> 00:01:22.300
derivative just doesn't exist.

00:01:22.300 --> 00:01:25.310
But as we see in this interval,
we'll stick to the derivative.

00:01:25.310 --> 00:01:30.330
f prime of x is
equal to 2x plus 4.

00:01:30.330 --> 00:01:32.050
So why does this equal to 0?

00:01:32.050 --> 00:01:32.690
Let's see.

00:01:32.690 --> 00:01:35.550
2x plus 4 is equal to 0.

00:01:35.550 --> 00:01:39.430
2x is equal to minus 4.

00:01:39.430 --> 00:01:41.660
I just almost made
a mistake there.

00:01:41.660 --> 00:01:44.440
x is equal to minus 2.

00:01:44.440 --> 00:01:50.960
When x is equal to minus
2, what is f of minus 2?

00:01:50.960 --> 00:01:53.570
So we know that f prime
of minus 2 is equal 0.

00:01:53.570 --> 00:01:54.560
We just solved for that.

00:01:54.560 --> 00:01:56.170
So we know the slope
is 0 at that point.

00:01:56.170 --> 00:01:58.170
But I just want to know what
the coordinate point is.

00:01:58.170 --> 00:02:02.400
So f of minus 2 is
4 minus 8 plus 4.

00:02:02.400 --> 00:02:05.240
So f minus 2 is 0.

00:02:05.240 --> 00:02:06.940
Let's evaluate the function
at the end points

00:02:06.940 --> 00:02:08.960
of this interval.

00:02:08.960 --> 00:02:16.010
f of minus 4 is equal to
minus 4 squared is 16.

00:02:16.010 --> 00:02:20.170
Minus 16, 4 times, so that's 0.

00:02:20.170 --> 00:02:23.070
And then plus 4, so
f of minus 4 is 4.

00:02:23.070 --> 00:02:25.230
And then what's f of 0?

00:02:25.230 --> 00:02:29.670
f of 0 is equal to-- well
that's 0 0, that's equal to 4.

00:02:29.670 --> 00:02:33.270
So now we could graph this.

00:02:33.270 --> 00:02:35.510
Let me graph the difference.

00:02:35.510 --> 00:02:50.190
So if this is the x-axis and
that is the y-axis, and the

00:02:50.190 --> 00:02:57.010
interval that we care about is
from x is equal to minus 4.

00:02:57.010 --> 00:03:01.420
From minus 4 to 0.

00:03:01.420 --> 00:03:02.930
So the first thing they
said, we'll identify

00:03:02.930 --> 00:03:03.650
the critical points.

00:03:03.650 --> 00:03:05.870
Well the critical points were
when the slope is equal to 0.

00:03:05.870 --> 00:03:09.330
And that's when f is equal to
minus 2, and that happens at

00:03:09.330 --> 00:03:12.620
the point minus 2,0, so it's
right here, which is directly

00:03:12.620 --> 00:03:16.200
in between those two
points incidentally.

00:03:16.200 --> 00:03:20.140
Which is a good intuition
behind why these two values are

00:03:20.140 --> 00:03:23.350
equal, because this is a
parabola and it's symmetric, so

00:03:23.350 --> 00:03:26.140
if you go 2 away on either
side, the function should

00:03:26.140 --> 00:03:26.780
have the same value.

00:03:26.780 --> 00:03:30.180
But anyway, ignore that
if it confused you.

00:03:30.180 --> 00:03:32.440
So at minus 2 we have
a negative slope.

00:03:32.440 --> 00:03:35.110
f of 0 is equal to 4.

00:03:35.110 --> 00:03:38.870
And f of minus 4 is equal to 4.

00:03:38.870 --> 00:03:41.190
So the graph is going to
look something like this.

00:03:41.190 --> 00:03:44.280
And I'll do it in
another color.

00:03:44.280 --> 00:03:46.310
And since we only care about
that interval, let's only

00:03:46.310 --> 00:03:47.790
graph it over that interval.

00:03:47.790 --> 00:03:49.890
It's going to look
something like that.

00:03:52.640 --> 00:03:54.730
And if you said this was a
critical point and they

00:03:54.730 --> 00:03:56.870
want to know the maximum
and the minimum point.

00:03:56.870 --> 00:03:59.010
Well the minimum point
is pretty clear.

00:03:59.010 --> 00:04:01.930
It's this critical point right
here where the derivative

00:04:01.930 --> 00:04:02.980
was equal to 0.

00:04:02.980 --> 00:04:04.930
And if you wanted to use you
know whatever they call

00:04:04.930 --> 00:04:06.190
it, the concavity theorem.

00:04:06.190 --> 00:04:07.770
But hopefully you have
the intuition of why.

00:04:07.770 --> 00:04:09.430
You would see that the
second derivative at

00:04:09.430 --> 00:04:11.360
this point is positive.

00:04:11.360 --> 00:04:12.635
And what's the
second derivative?

00:04:12.635 --> 00:04:19.110
If that's f prime of x, f prime
of x for all x is equal to 2.

00:04:19.110 --> 00:04:22.410
So really at any point you test
you're going to get the second

00:04:22.410 --> 00:04:23.510
derivative to be positive.

00:04:23.510 --> 00:04:26.120
So especially at this critical
point where the first

00:04:26.120 --> 00:04:27.250
derivative is 0.

00:04:27.250 --> 00:04:29.360
The second derivative being
positive tells you that the

00:04:29.360 --> 00:04:31.860
slope is constantly increasing.

00:04:31.860 --> 00:04:33.910
So the rate of change of
the slope is positive.

00:04:33.910 --> 00:04:36.830
The slope is really negative
there, little less negative,

00:04:36.830 --> 00:04:39.300
goes to 0, and then keeps
increasing, becomes more

00:04:39.300 --> 00:04:42.010
positive, even more
positive slope there.

00:04:42.010 --> 00:04:44.620
So I think that was out
of scope of this exact

00:04:44.620 --> 00:04:46.960
problem, but it's good
to have that intuition.

00:04:46.960 --> 00:04:48.670
And they want to know
the maximum points.

00:04:48.670 --> 00:04:53.551
Well the value here, x
is equal to minus 4 and

00:04:53.551 --> 00:04:55.170
the value x equals 0.

00:04:55.170 --> 00:04:57.680
They're both maximum points.

00:04:57.680 --> 00:05:00.000
They tie for first place.

00:05:00.000 --> 00:05:03.710
So those are the maximum
points, x equals 0

00:05:03.710 --> 00:05:04.900
and x equals minus 4.

00:05:04.900 --> 00:05:08.480
You could say 0,4 or minus 4,4.

00:05:08.480 --> 00:05:11.080
What's the next
problem they give?

00:05:11.080 --> 00:05:11.730
Let's see.

00:05:15.280 --> 00:05:19.050
Understanding the problem and
the terminology is often more

00:05:19.050 --> 00:05:21.470
confusing then the
problem itself.

00:05:21.470 --> 00:05:29.720
So they give h of r is equal to
1/r, and they care about the

00:05:29.720 --> 00:05:38.150
interval-- I'm assuming from
this interval on r-- from

00:05:38.150 --> 00:05:40.980
negative 1 to 3, and we're
going to include negative 1 and

00:05:40.980 --> 00:05:43.590
3 because we have brackets.

00:05:43.590 --> 00:05:45.870
Well what are the
critical points?

00:05:45.870 --> 00:05:54.290
They're the points where either
the derivative is 0, or the

00:05:54.290 --> 00:05:55.680
derivative doesn't exist.

00:05:55.680 --> 00:05:59.680
Now this brings up an
interesting point.

00:05:59.680 --> 00:06:01.730
We'll see in a second there's a
point here where the derivative

00:06:01.730 --> 00:06:04.180
doesn't exist, but the function
also isn't defined

00:06:04.180 --> 00:06:05.040
at that point.

00:06:05.040 --> 00:06:06.430
So that's not a critical point.

00:06:06.430 --> 00:06:08.580
So in order to get a critical
point-- and this is a

00:06:08.580 --> 00:06:09.310
bit of a technicality.

00:06:12.910 --> 00:06:15.790
The easy one is
derivative is 0.

00:06:15.790 --> 00:06:19.990
Derivative equals 0.

00:06:19.990 --> 00:06:25.360
The other critical point-- and
I was little bit non-rigorous

00:06:25.360 --> 00:06:27.610
with it when I said before-- is
when there's no derivative.

00:06:34.300 --> 00:06:37.590
But f of x is defined.

00:06:41.370 --> 00:06:44.260
So as you're going to see in
this problem, there is a point

00:06:44.260 --> 00:06:47.470
where the derivative is
undefined but f of x-- or in

00:06:47.470 --> 00:06:50.760
this case, h of r-- is also not
defined at that point, so it

00:06:50.760 --> 00:06:53.210
won't be a critical point.

00:06:53.210 --> 00:06:53.940
Or maybe it is.

00:06:53.940 --> 00:06:54.300
I don't know.

00:06:54.300 --> 00:06:55.530
Depending on how your
teacher defines it.

00:06:55.530 --> 00:06:58.870
But the way I learned it is
a critical point-- if the

00:06:58.870 --> 00:07:02.080
derivative isn't defined,
but the function is.

00:07:02.080 --> 00:07:02.960
But anyway.

00:07:02.960 --> 00:07:04.750
What's the derivative here?

00:07:04.750 --> 00:07:07.076
h prime of r.

00:07:07.076 --> 00:07:10.150
Well that's r to the negative
1, right, so it becomes

00:07:10.150 --> 00:07:13.050
negative r of the negative 2,
or that's minus 1

00:07:13.050 --> 00:07:14.790
over r squared.

00:07:14.790 --> 00:07:16.080
That's the derivative.

00:07:16.080 --> 00:07:18.060
Can this ever equal 0?

00:07:18.060 --> 00:07:20.980
Well no, it's never
going to be equal 0.

00:07:20.980 --> 00:07:22.440
But where is it undefined?

00:07:22.440 --> 00:07:24.930
Well it's undefined
when r is equal to 0.

00:07:24.930 --> 00:07:28.980
And r equal to 0 is
in that interval.

00:07:28.980 --> 00:07:35.040
So h prime of 0 is undefined.

00:07:37.870 --> 00:07:39.870
But h of 0 is also
undefined, right?

00:07:39.870 --> 00:07:41.490
1/0 is undefined.

00:07:41.490 --> 00:07:45.950
h of 0 also undefined.

00:07:45.950 --> 00:07:50.370
So I'm not going to really
consider that to be a

00:07:50.370 --> 00:07:51.560
critical point frankly.

00:07:51.560 --> 00:07:55.240
It's just a point at which
the derivative and the

00:07:55.240 --> 00:07:56.980
function is undefined.

00:07:56.980 --> 00:08:00.110
And so what will the
graph look like?

00:08:00.110 --> 00:08:03.880
Well let's see if we
can draw the interval.

00:08:03.880 --> 00:08:06.430
And we might just want
to draw some points.

00:08:06.430 --> 00:08:08.550
That's what I always used
to do if I never got

00:08:08.550 --> 00:08:10.040
stuck on a math problem.

00:08:10.040 --> 00:08:12.350
I would just graph some points.

00:08:12.350 --> 00:08:14.406
You can ever go
wrong with that.

00:08:14.406 --> 00:08:15.980
Let's see, 1/r.

00:08:15.980 --> 00:08:18.783
So we want to go from negative
1 to 3, so we're going to

00:08:18.783 --> 00:08:19.630
have some negative values.

00:08:22.130 --> 00:08:27.880
So we want to go from r-- this
is the r-axis, right, because

00:08:27.880 --> 00:08:31.510
this is the function of r,
this is the h of r-axis.

00:08:31.510 --> 00:08:33.850
So we're going to go from
r is equal to negative

00:08:33.850 --> 00:08:38.000
1 to r is equal to 3.

00:08:42.960 --> 00:08:44.270
And we could plot some points.

00:08:44.270 --> 00:08:46.360
What is h of negative 1?

00:08:46.360 --> 00:08:49.560
It's 1 over negative 1, so
it's this point right here.

00:08:53.530 --> 00:08:56.430
And then what happens is
we get to smaller and

00:08:56.430 --> 00:08:58.220
smaller negative numbers.

00:08:58.220 --> 00:08:58.750
Well think about it.

00:08:58.750 --> 00:09:01.350
If you have negative 1/2 here.

00:09:01.350 --> 00:09:04.830
When r is negative 1/2, it
goes to negative 2, and it's

00:09:04.830 --> 00:09:06.985
actually going to asymptote.

00:09:06.985 --> 00:09:08.450
We only care about this
interval, so it's going to

00:09:08.450 --> 00:09:09.430
look something like this.

00:09:12.142 --> 00:09:17.400
It's going to asymptote to
negative infinity, right, if I

00:09:17.400 --> 00:09:20.130
kept drawing this all the way
down that would just asymptote.

00:09:20.130 --> 00:09:22.100
And then what is h of 3.

00:09:22.100 --> 00:09:23.830
Well h of 3 is just
going to be 1/3.

00:09:23.830 --> 00:09:25.670
It's going to be
relatively low number.

00:09:25.670 --> 00:09:27.275
And then what's going to
happen as we get closer

00:09:27.275 --> 00:09:28.470
and closer to 0?

00:09:28.470 --> 00:09:32.310
Well if you put a really small
positive value here, you get

00:09:32.310 --> 00:09:34.430
larger and larger numbers.

00:09:34.430 --> 00:09:40.450
So this actually asymptotes to
positive infinity, so it'll

00:09:40.450 --> 00:09:42.780
look something like this.

00:09:42.780 --> 00:09:46.440
I always have trouble
drawing these hyperbolas.

00:09:46.440 --> 00:09:47.680
It's going to look something
like that, it's going to

00:09:47.680 --> 00:09:48.670
go to positive infinity.

00:09:48.670 --> 00:09:51.620
So that goes to positive
infinity, and this will on this

00:09:51.620 --> 00:09:54.090
side go to minus infinity.

00:09:54.090 --> 00:09:56.120
And frankly just looking at
this graph you see why the

00:09:56.120 --> 00:09:57.490
graph is not defined at 0.

00:10:00.450 --> 00:10:02.810
From the right hand side
the limit approaches

00:10:02.810 --> 00:10:03.530
positive infinity.

00:10:03.530 --> 00:10:05.490
From the left hand side it
approaches negative infinity,

00:10:05.490 --> 00:10:07.920
and 1/0 is just undefined.

00:10:10.570 --> 00:10:12.870
Similarly that's why you
couldn't get a derivative there

00:10:12.870 --> 00:10:16.090
either because the definition
of the derivative means you're

00:10:16.090 --> 00:10:18.180
taking a limit and the limit
has to be valid from

00:10:18.180 --> 00:10:18.960
both directions.

00:10:18.960 --> 00:10:21.060
And as we see the limit
is not valid from

00:10:21.060 --> 00:10:23.330
both directions here.

00:10:23.330 --> 00:10:27.990
So what are the maximum
and minimum points here?

00:10:27.990 --> 00:10:34.540
Well if we look at this
value, we might say oh,

00:10:34.540 --> 00:10:36.900
is that a minimum point?

00:10:36.900 --> 00:10:39.440
Well no, because you
have values much lower.

00:10:39.440 --> 00:10:42.290
Actually you have values
that go to minus infinity

00:10:42.290 --> 00:10:43.710
in this interval.

00:10:43.710 --> 00:10:45.190
And then you say, well is
this a maximum point?

00:10:45.190 --> 00:10:47.150
Well no, you have values that
go to positive infinity

00:10:47.150 --> 00:10:48.340
in this interval.

00:10:48.340 --> 00:10:55.200
So it actually turns out that
there is no-- I mean this isn't

00:10:55.200 --> 00:10:58.300
very proper-- but to some
degree at 0 you're at both

00:10:58.300 --> 00:11:00.970
positive and negative infinity
depending on what direction

00:11:00.970 --> 00:11:01.710
you're coming from.

00:11:01.710 --> 00:11:05.870
But this really has no maximum
or minimum points, because

00:11:05.870 --> 00:11:07.990
the graph is undefined at 0.

00:11:07.990 --> 00:11:10.880
And then you know if I said, oh
well what if I get really close

00:11:10.880 --> 00:11:13.370
to 0 and I called that
the maximum point.

00:11:13.370 --> 00:11:15.960
Then you say, no, no, no, no,
but there's a point even closer

00:11:15.960 --> 00:11:18.690
that has even a higher value,
that's even closer to infinity.

00:11:18.690 --> 00:11:22.230
So there's actually no
exact point that is

00:11:22.230 --> 00:11:23.160
the maximum value.

00:11:23.160 --> 00:11:25.740
So that's kind of interesting.

00:11:25.740 --> 00:11:30.540
Anyway, I'm already at 11
minutes and people at YouTube

00:11:30.540 --> 00:11:33.290
apparently let me go well past
the limit now, but people told

00:11:33.290 --> 00:11:37.060
me that they like the 10 minute
nuggets of knowledge, so I will

00:11:37.060 --> 00:11:38.980
stop now and I'll continue
these problems in

00:11:38.980 --> 00:11:39.930
a future video.

00:11:39.930 --> 00:11:40.070
See you soon.


Monotonicity Theorem

WEBVTT
Kind: captions
Language: en

00:00:00.860 --> 00:00:01.530
Welcome back.

00:00:01.530 --> 00:00:04.720
Well, I've been requested
to do several problems by

00:00:04.720 --> 00:00:05.530
our friend [? Akosh. ?]

00:00:05.530 --> 00:00:07.280
So I thought I would
keep doing them.

00:00:07.280 --> 00:00:08.360
I'm skipping around
a little bit.

00:00:08.360 --> 00:00:09.910
Because I think if I did
all of them, it would just

00:00:09.910 --> 00:00:11.520
generate too many videos.

00:00:11.520 --> 00:00:15.030
But I encourage all of you all
to let me know if you feel that

00:00:15.030 --> 00:00:16.450
there's something that
I might have missed.

00:00:16.450 --> 00:00:18.790
But anyway, I'm just going
through some of the

00:00:18.790 --> 00:00:19.550
problems he gave.

00:00:19.550 --> 00:00:21.620
So this one-- and I actually
almost find it funny, because

00:00:21.620 --> 00:00:24.340
they use such formal language
for something that's actually

00:00:24.340 --> 00:00:25.890
a pretty intuitive concept.

00:00:25.890 --> 00:00:29.130
Well they say, use the
monotonicity theorem to find

00:00:29.130 --> 00:00:30.995
where the given function is
increasing, and when

00:00:30.995 --> 00:00:32.260
it is decreasing.

00:00:32.260 --> 00:00:45.990
Monotonicity theorem.

00:00:45.990 --> 00:00:49.840
It make sound very serious.

00:00:49.840 --> 00:00:53.120
Well all the monotonicity
theorem-- at least if I'm

00:00:53.120 --> 00:00:56.750
remembering things properly--
all it says is when the

00:00:56.750 --> 00:01:03.890
derivative is positive, when f
prime of x is greater than 0,

00:01:03.890 --> 00:01:06.540
your function is increasing.

00:01:06.540 --> 00:01:10.540
And when the derivative is
negative, your function

00:01:10.540 --> 00:01:13.100
is decreasing.

00:01:13.100 --> 00:01:16.480
And why am I kind of
disparaging of this?

00:01:16.480 --> 00:01:18.240
Well what's the derivative?

00:01:18.240 --> 00:01:19.730
The derivative is the slope.

00:01:19.730 --> 00:01:22.400
And you learned in Algebra
1, that if the slope is

00:01:22.400 --> 00:01:27.610
positive, the graph of
the line is increasing.

00:01:27.610 --> 00:01:29.500
And if the slope is
negative, the graph of

00:01:29.500 --> 00:01:30.980
the line is decreasing.

00:01:30.980 --> 00:01:33.360
And the only thing that's
different now is that this

00:01:33.360 --> 00:01:35.610
function isn't necessarily
a line, it's a function.

00:01:35.610 --> 00:01:36.400
So it could be a curve.

00:01:36.400 --> 00:01:40.230
But having a positive slope
means that the value of the

00:01:40.230 --> 00:01:41.210
function is increasing.

00:01:41.210 --> 00:01:44.630
For every change it x, there's
a positive change in y.

00:01:44.630 --> 00:01:47.980
And similarly, a negative slope
says that for every change in

00:01:47.980 --> 00:01:50.350
x, no matter how small, there's
a negative change in y.

00:01:50.350 --> 00:01:51.660
So there's nothing fancy here.

00:01:51.660 --> 00:01:53.850
But let's see, they
gave some examples.

00:01:53.850 --> 00:01:57.090
So they want us to use this
theorem to find where the

00:01:57.090 --> 00:01:59.400
given function is increasing,
and when it is decreasing.

00:01:59.400 --> 00:02:03.060
So the first function
they give is f of x is

00:02:03.060 --> 00:02:05.800
equal to 3x plus 3.

00:02:05.800 --> 00:02:11.950
What's the derivative of this?
f prime of x is equal to 3.

00:02:11.950 --> 00:02:14.950
For any value of x
this is a positive.

00:02:14.950 --> 00:02:15.420
Right?

00:02:15.420 --> 00:02:18.470
F prime of x is positive
for all values of x.

00:02:18.470 --> 00:02:21.460
So this thing, using the
monotonicity theorem

00:02:21.460 --> 00:02:23.250
is positive.

00:02:23.250 --> 00:02:26.630
Sorry, it's increasing
for all values of x.

00:02:26.630 --> 00:02:28.910
And you could have done
that in the 9th grade.

00:02:28.910 --> 00:02:29.610
How?

00:02:29.610 --> 00:02:33.540
Well you would have said, this
is the slope y-intercept.

00:02:33.540 --> 00:02:35.070
The slope is 3.

00:02:35.070 --> 00:02:37.055
So this is an
increasing function.

00:02:37.055 --> 00:02:39.125
The function will look
something like this.

00:02:39.125 --> 00:02:40.990
The y-intercept is 3.

00:02:40.990 --> 00:02:41.870
And the slope is 3.

00:02:41.870 --> 00:02:44.030
So this function looks
something like that.

00:02:44.030 --> 00:02:49.350
And so it's increasing over
from minus infinite to positive

00:02:49.350 --> 00:02:51.580
infinite over that interval,
if you wanted to be fancy.

00:02:51.580 --> 00:02:55.120
But this is really just a lot
of fancy terminology to talk

00:02:55.120 --> 00:02:58.000
about, I think, something
fairly straightforward.

00:02:58.000 --> 00:03:00.755
Anyway, the next one looks a
little bit more interesting.

00:03:03.400 --> 00:03:16.590
H of z is equal to z to the
fourth over 4, minus-- and I

00:03:16.590 --> 00:03:23.420
hope I'm reading this right--
4 z to the third over 6.

00:03:23.420 --> 00:03:26.900
So let's see if we can figure
out when this is increasing,

00:03:26.900 --> 00:03:30.130
and when this is decreasing,
when the function is

00:03:30.130 --> 00:03:31.910
increasing or decreasing.

00:03:31.910 --> 00:03:34.090
So what's the
derivative of this?

00:03:34.090 --> 00:03:38.820
H prime of z is equal to 4.

00:03:38.820 --> 00:03:39.770
This will cancel out.

00:03:39.770 --> 00:03:43.580
So it's z to the third,
4z to the third over 4.

00:03:43.580 --> 00:03:54.510
So it's z to the third minus 3
times 4 is 12z squared over 6.

00:03:54.510 --> 00:03:58.720
So it's 2 2z squared.

00:03:58.720 --> 00:04:02.140
So we just have to figure out
when is this function greater

00:04:02.140 --> 00:04:05.380
than 0, and when is this
function less than 0.

00:04:05.380 --> 00:04:07.960
And to figure this out, we
really just have to break out

00:04:07.960 --> 00:04:10.100
our algebra toolkit to figure
out well one, when does

00:04:10.100 --> 00:04:12.390
this function equal 0?

00:04:12.390 --> 00:04:20.760
So let's set z-cubed minus
2z squared is equal to 0.

00:04:20.760 --> 00:04:23.720
We could factor
out a z-squared.

00:04:23.720 --> 00:04:27.660
And then we have z
minus 2 is equal to 0.

00:04:27.660 --> 00:04:32.250
So we know that either
z-squared is equal to 0, or

00:04:32.250 --> 00:04:34.580
z minus 2 is equal to 0.

00:04:34.580 --> 00:04:39.830
So the points at which the
derivative is 0, so H prime

00:04:39.830 --> 00:04:43.030
of 0 is equal to 0,
and H prime of 2.

00:04:43.030 --> 00:04:47.750
And that's just our Algebra 2.

00:04:47.750 --> 00:04:51.180
So H prime of 2 is equal to 0.

00:04:51.180 --> 00:04:54.010
And so we just have to figure
out what happens in the

00:04:54.010 --> 00:04:58.662
interval maybe when we're
below 0, between 0 and

00:04:58.662 --> 00:05:01.070
2, and then above 2.

00:05:01.070 --> 00:05:02.210
So what is the derivative?

00:05:02.210 --> 00:05:06.520
So when we're below 0,
let's just take a point.

00:05:06.520 --> 00:05:14.340
Let's say H prime of negative
1 is equal to what?

00:05:14.340 --> 00:05:19.890
That is equal to negative 1 to
the third is negative 1, minus

00:05:19.890 --> 00:05:22.660
2 times negative 1 squared.

00:05:22.660 --> 00:05:25.820
Well negative 1 squared
is just positive 1.

00:05:25.820 --> 00:05:29.120
So it's minus 2.

00:05:29.120 --> 00:05:32.400
So it equals minus 3.

00:05:32.400 --> 00:05:34.320
OK, let's take a point
in between these two.

00:05:34.320 --> 00:05:35.380
Well 1 is in between them.

00:05:35.380 --> 00:05:42.950
So let's just take H prime
of 1 is equal to 1 minus 2.

00:05:46.060 --> 00:05:48.870
Lets make sure I'm
doing that right.

00:05:48.870 --> 00:05:50.180
Because it's 1 squared.

00:05:50.180 --> 00:05:55.090
So it equals negative
1, H prime of 1.

00:05:55.090 --> 00:05:59.625
And then what happens when we
go of z values greater than 2.

00:05:59.625 --> 00:06:01.480
So let's try 3.

00:06:01.480 --> 00:06:06.010
H prime of 3 is equal to what?

00:06:06.010 --> 00:06:12.520
H prime of 3 is equal to 27--
what's minus 2 times 9.

00:06:12.520 --> 00:06:15.610
It's equal to 9,
right, 18, right.

00:06:15.610 --> 00:06:17.300
So it's positive.

00:06:17.300 --> 00:06:18.970
So, what do we know?

00:06:18.970 --> 00:06:24.880
We know that in all fairness,
this is kind of an

00:06:24.880 --> 00:06:26.210
interesting problem.

00:06:26.210 --> 00:06:29.370
Because the derivative
is negative.

00:06:29.370 --> 00:06:37.080
So if we were to draw the
derivative of this, what are

00:06:37.080 --> 00:06:38.500
the interesting points?

00:06:38.500 --> 00:06:40.660
It's 0.

00:06:40.660 --> 00:06:41.600
So that's the x-axis.

00:06:46.120 --> 00:06:46.765
That's 0.

00:06:46.765 --> 00:06:50.000
And then we had 2.

00:06:50.000 --> 00:06:51.640
And I'm going to draw
the derivative.

00:06:51.640 --> 00:06:54.900
I'm not drawing the
actual function.

00:06:54.900 --> 00:06:56.690
Because the monotonicity
theorem, we care about

00:06:56.690 --> 00:06:59.480
whether the derivative
is positive or negative.

00:06:59.480 --> 00:07:03.640
So what this tells us is
that the derivative, it

00:07:03.640 --> 00:07:06.560
was a negative when we're
to left of 0, right?

00:07:06.560 --> 00:07:07.650
Because we just
took a test point.

00:07:07.650 --> 00:07:10.160
Because we know it's 0 here.

00:07:10.160 --> 00:07:11.010
But then what happened?

00:07:11.010 --> 00:07:12.560
It's not like it
got positive then.

00:07:12.560 --> 00:07:15.100
It went back negative
again when we tried to

00:07:15.100 --> 00:07:15.940
derivative it once.

00:07:15.940 --> 00:07:17.320
It probably does
something like this.

00:07:30.940 --> 00:07:34.092
Now what would it be?

00:07:34.092 --> 00:07:39.910
Then it goes up at 2 where
it's greater than 0.

00:07:39.910 --> 00:07:41.560
Right.

00:07:41.560 --> 00:07:43.240
I was confusing
myself for a second.

00:07:43.240 --> 00:07:43.470
Right.

00:07:43.470 --> 00:07:46.020
So the derivative, it starts.

00:07:46.020 --> 00:07:47.790
But it never goes
above right there.

00:07:47.790 --> 00:07:52.580
So the derivative probably
has a bit of a maximum

00:07:52.580 --> 00:07:53.680
point there or something.

00:07:53.680 --> 00:07:55.000
This is the
derivative, remember.

00:07:55.000 --> 00:07:56.040
That's why I was
confusing myself.

00:07:56.040 --> 00:07:57.870
And then it probably
flips around, and then

00:07:57.870 --> 00:07:59.590
goes above there.

00:07:59.590 --> 00:08:01.740
But from the monotonicity
theorem, what do we care about?

00:08:01.740 --> 00:08:04.020
We care about the intervals
where the derivative

00:08:04.020 --> 00:08:06.890
is positive, and the
derivative is negative.

00:08:06.890 --> 00:08:10.200
So if the derivative is
positive for all of

00:08:10.200 --> 00:08:14.600
these values, for z
is greater than 2.

00:08:14.600 --> 00:08:18.790
So we could say, using the
monotonicity theorem, this

00:08:18.790 --> 00:08:23.015
function is increasing
when z is greater than 2.

00:08:23.015 --> 00:08:24.230
So let me write that.

00:08:31.250 --> 00:08:36.990
And then we can say that the
function is decreasing.

00:08:36.990 --> 00:08:38.870
We could say flat or
decreasing, right?

00:08:38.870 --> 00:08:41.410
Because the slope of the
function is 0 right here.

00:08:41.410 --> 00:08:43.210
And remember this is the
graph of the derivative.

00:08:43.210 --> 00:08:46.060
So we could say flat
or decreasing when

00:08:46.060 --> 00:08:47.850
z is less than 2.

00:08:50.470 --> 00:08:54.910
You normally do consider flat,
monotonic in one direction.

00:08:54.910 --> 00:08:56.980
So you could say monotonic
increasing would still

00:08:56.980 --> 00:09:00.000
include something that
kind of flattens out.

00:09:00.000 --> 00:09:03.720
Well anyway, I'm
pushing ten minutes.

00:09:03.720 --> 00:09:06.170
And I don't think I have time
for the next problem, which I

00:09:06.170 --> 00:09:06.970
might do in the next video.

00:09:06.970 --> 00:09:09.010
But hopefully you found
that a little bit helpful,

00:09:09.010 --> 00:09:10.740
and not too confusing.

00:09:10.740 --> 00:09:13.220
I will see you in
the next video.


Limit examples w/ brain malfunction on first prob (part 4)

WEBVTT
Kind: captions
Language: en

00:00:00.900 --> 00:00:03.720
My cousin Nadia is taking a
summer calculus course and

00:00:03.720 --> 00:00:04.670
she called me last night.

00:00:04.670 --> 00:00:07.150
And she had some limit
problems and they were

00:00:07.150 --> 00:00:08.020
excellent problem.

00:00:08.020 --> 00:00:11.050
So I thought it was worthwhile
to make videos on the problems

00:00:11.050 --> 00:00:13.040
that she had to figure out.

00:00:13.040 --> 00:00:14.830
Anyway, so let's do them.

00:00:14.830 --> 00:00:15.930
If I remember them correctly.

00:00:15.930 --> 00:00:17.070
I'm doing this on memory.

00:00:17.070 --> 00:00:19.785
So I hope the answers work out
like they did last night when

00:00:19.785 --> 00:00:21.500
we were going over
them on the phone.

00:00:21.500 --> 00:00:26.750
So if I remember correctly,
the first one was the limit

00:00:26.750 --> 00:00:33.530
as z approaches 2 of x.

00:00:33.530 --> 00:00:33.870
Oh no.

00:00:33.870 --> 00:00:35.440
Not that-- There's no x.

00:00:35.440 --> 00:00:37.230
It's a z.

00:00:37.230 --> 00:00:50.080
Of z squared plus 2z minus 8.

00:00:50.080 --> 00:00:57.020
All of that over z to
the fourth minus 16.

00:00:57.020 --> 00:00:58.810
So the first thing when you
want to do when you try a limit

00:00:58.810 --> 00:01:03.270
problem is, well you just try
to substitute the value into--

00:01:03.270 --> 00:01:05.970
Maybe there's no problem when z
equals 2 and you just evaluate

00:01:05.970 --> 00:01:07.820
the function at z equals 2.

00:01:07.820 --> 00:01:09.980
If it's a continuous function,
then the limit as z

00:01:09.980 --> 00:01:13.480
approaches 2 is going to
be the function at 2.

00:01:13.480 --> 00:01:16.410
But you immediately see a
problem if you put 2 right

00:01:16.410 --> 00:01:18.850
here-- 2 to the fourth minus
16 --and you get a 0 in the

00:01:18.850 --> 00:01:21.970
denominator, which is
undefined, so we have to

00:01:21.970 --> 00:01:24.590
figure out some way
to get around that.

00:01:24.590 --> 00:01:27.600
And nine times out of 10, when
you see a problem like this,

00:01:27.600 --> 00:01:30.650
the solution, because both the
numerator and the denominator

00:01:30.650 --> 00:01:33.330
look factorable, is to factor
the numerator and

00:01:33.330 --> 00:01:34.130
the denominator.

00:01:34.130 --> 00:01:42.130
So this is equal to the
limit as z approaches 2.

00:01:42.130 --> 00:01:43.940
What's the numerator factored?

00:01:43.940 --> 00:01:44.400
Let's see.

00:01:44.400 --> 00:01:48.550
Something when you add two
numbers, you get positive

00:01:48.550 --> 00:01:50.490
2, and you multiply
them, you get minus 8.

00:01:50.490 --> 00:01:54.370
So it's probably plus
4 and minus 2, right?

00:01:54.370 --> 00:01:57.380
So this is going to
be x plus-- No.

00:01:57.380 --> 00:01:58.510
Not an x.

00:01:58.510 --> 00:02:00.140
I'm so conditioned.

00:02:00.140 --> 00:02:01.300
I'm like a dog.

00:02:04.220 --> 00:02:04.380
Oh.

00:02:04.380 --> 00:02:05.730
I can't even undo it anymore.

00:02:05.730 --> 00:02:10.440
Anyway, well z-- Actually
let me erase that.

00:02:10.440 --> 00:02:12.536
I don't want to be messy.

00:02:12.536 --> 00:02:13.810
Let me erase.

00:02:13.810 --> 00:02:18.130
I tried to undo it and it
doesn't remember everything.

00:02:18.130 --> 00:02:24.220
So it's z plus 4
times z minus 2.

00:02:24.220 --> 00:02:25.300
That's just factoring
a quadratic.

00:02:28.460 --> 00:02:29.420
And what is this?

00:02:29.420 --> 00:02:32.130
This has the form a squared
minus b squared, right?

00:02:32.130 --> 00:02:37.380
So that's going to be-- But a
squared-- If this is a squared

00:02:37.380 --> 00:02:41.450
minus b squared, a would
be z squared, right?

00:02:41.450 --> 00:02:48.330
So it's z squared plus 4
times z squared minus 4.

00:02:48.330 --> 00:02:50.830
And then, of course, this
also has the form a squared

00:02:50.830 --> 00:02:52.790
plus b squared, right?

00:02:52.790 --> 00:03:02.110
So this will further factor
into z plus 2 times z minus 2.

00:03:02.110 --> 00:03:03.790
Well, our factoring paid off.

00:03:03.790 --> 00:03:06.280
We see a term in the numerator
and the denominator

00:03:06.280 --> 00:03:07.230
that are the same.

00:03:07.230 --> 00:03:09.630
And not only are they the
same, but they-- This is

00:03:09.630 --> 00:03:11.570
the term that was giving
us the problems, right?

00:03:11.570 --> 00:03:13.510
Because you put a 2 in
here, you got a 0.

00:03:13.510 --> 00:03:18.590
So let's assume that we're not
evaluating it at z equals 2.

00:03:18.590 --> 00:03:23.420
And so, for all other
values, we can divide those

00:03:23.420 --> 00:03:25.430
because those are going
to be the same values.

00:03:25.430 --> 00:03:27.260
And then what are we left with?

00:03:27.260 --> 00:03:30.150
This is equal to the limit--
and I've change colors

00:03:30.150 --> 00:03:40.060
arbitrarily --as z approaches
2 of z plus 4 over z squared

00:03:40.060 --> 00:03:44.870
plus 4 times z plus 2.

00:03:44.870 --> 00:03:46.240
Which is equal what?

00:03:46.240 --> 00:03:50.000
That's equal to 6 over--
what's 2 squared plus 4?

00:03:50.000 --> 00:03:52.880
Is 8.

00:03:52.880 --> 00:03:56.570
And then 2 plus 2 is 4.

00:03:56.570 --> 00:03:59.820
6 over 12 is equal to 1/2.

00:03:59.820 --> 00:04:01.250
There you go.

00:04:01.250 --> 00:04:03.820
I thought that was a pretty
interesting problem.

00:04:03.820 --> 00:04:04.980
Let's do another one.

00:04:04.980 --> 00:04:07.290
And this one I found
even more interesting.

00:04:07.290 --> 00:04:10.160
She gave me-- It's really
testing my memory to see

00:04:10.160 --> 00:04:11.860
if I can-- But I remember
the gist of the problem.

00:04:11.860 --> 00:04:14.640
So I might not get the exact
numbers she'd given me, but

00:04:14.640 --> 00:04:16.930
hopefully I get the
exact properties.

00:04:16.930 --> 00:04:27.250
So it's the limit as x
approaches infinity of the

00:04:27.250 --> 00:04:40.080
square root of x squared
plus 4x plus 1 minus x.

00:04:40.080 --> 00:04:41.860
So when you look at this
you're like well let's see.

00:04:41.860 --> 00:04:43.070
What's happening here?

00:04:43.070 --> 00:04:44.160
Let's see this.

00:04:44.160 --> 00:04:47.460
As you go to infinity, this
term will get really big.

00:04:47.460 --> 00:04:49.750
But then we're taking
the square root of it.

00:04:49.750 --> 00:04:53.810
And it seems like this term
would overpower this term.

00:04:53.810 --> 00:04:57.460
And then, so maybe this kind
of converges to x, but then

00:04:57.460 --> 00:04:58.670
we would subtract an x.

00:04:58.670 --> 00:05:01.700
So maybe it approaches 0.

00:05:01.700 --> 00:05:04.860
So that's, at least, that
was my first intuition

00:05:04.860 --> 00:05:05.860
when I spoke to her.

00:05:05.860 --> 00:05:08.930
But as we will see, the
intuition here is wrong.

00:05:08.930 --> 00:05:11.050
And really to do this you
have to know a little

00:05:11.050 --> 00:05:12.340
bit of a trick.

00:05:12.340 --> 00:05:16.470
And this trick pays off a lot
whenever you see something with

00:05:16.470 --> 00:05:18.950
the square root sign, then
subtracting something else, if

00:05:18.950 --> 00:05:20.730
you want to get rid of
that square root sign.

00:05:20.730 --> 00:05:25.120
So what we are going to do is
going to multiply-- Essentially

00:05:25.120 --> 00:05:27.380
the conjugate we normally
apply to complex numbers.

00:05:27.380 --> 00:05:31.300
But if we have something like
a plus b, the conjugate

00:05:31.300 --> 00:05:32.920
is a minus b, right?

00:05:32.920 --> 00:05:34.630
Or if we have something
like a minus b, the

00:05:34.630 --> 00:05:36.300
conjugate is a plus b.

00:05:36.300 --> 00:05:38.150
And the reason why-- And what
we're going to do is we're

00:05:38.150 --> 00:05:39.650
going to multiply by
the conjugate of this.

00:05:39.650 --> 00:05:41.060
And why does-- Why
is that normal?

00:05:41.060 --> 00:05:41.880
Why is that useful?

00:05:41.880 --> 00:05:43.065
Because we have a minus b.

00:05:43.065 --> 00:05:46.110
If we multiply it times a
plus b, we get a squared

00:05:46.110 --> 00:05:47.440
minus b squared.

00:05:47.440 --> 00:05:50.530
Which will make this radical
sign disappear without

00:05:50.530 --> 00:05:51.240
too much work.

00:05:51.240 --> 00:05:52.190
So let's do that.

00:05:52.190 --> 00:05:55.560
Let's multiply by the
conjugate of this thing.

00:05:55.560 --> 00:05:57.060
But we can't just
multiply it, right?

00:05:57.060 --> 00:05:59.800
We have to multiply it
by it over itself.

00:05:59.800 --> 00:06:01.790
Because you can only-- To not
change the value of something,

00:06:01.790 --> 00:06:03.140
you can only multiply it by 1.

00:06:03.140 --> 00:06:09.020
So let's multiply it by the
conjugate: x squared plus

00:06:09.020 --> 00:06:15.930
4x plus 1 plus x, right?

00:06:15.930 --> 00:06:16.630
That's the conjugate, right?

00:06:16.630 --> 00:06:18.660
Instead of minus x,
we have a plus x.

00:06:18.660 --> 00:06:19.780
And we can't just
multiply that.

00:06:19.780 --> 00:06:21.360
We have to multiply it by
1, otherwise we would

00:06:21.360 --> 00:06:22.240
be changing the value.

00:06:22.240 --> 00:06:25.440
So it's going to be divided
by the same thing.

00:06:25.440 --> 00:06:31.630
x squared plus 4x
plus 1 plus x.

00:06:31.630 --> 00:06:33.610
Let me erase this stuff
down here so we don't

00:06:33.610 --> 00:06:35.970
get distracted.

00:06:35.970 --> 00:06:37.770
Don't want to get distracted.

00:06:37.770 --> 00:06:39.930
And so what do we have?

00:06:39.930 --> 00:06:48.530
This will become the limit
as x approaches infinity.

00:06:48.530 --> 00:06:51.240
Well, this is a minus
b times a plus b.

00:06:51.240 --> 00:06:52.560
So we end up with a squared.

00:06:52.560 --> 00:06:54.340
Well, what's this squared?

00:06:54.340 --> 00:07:01.250
That's x squared plus 4x
plus 1 minus b squared.

00:07:01.250 --> 00:07:02.600
Well what's b squared?

00:07:02.600 --> 00:07:03.350
Well, b is x.

00:07:03.350 --> 00:07:05.400
So it's going to be
minus x squared.

00:07:05.400 --> 00:07:07.740
This is just algebra.

00:07:07.740 --> 00:07:15.150
Divided by this thing: the
square root of x squared

00:07:15.150 --> 00:07:21.995
plus 4x plus 1 plus x.

00:07:21.995 --> 00:07:22.720
So let's see.

00:07:22.720 --> 00:07:24.400
There's a little
simplification we can do.

00:07:24.400 --> 00:07:28.900
We can subtract-- We can--
These top two terms cancel out.

00:07:28.900 --> 00:07:32.820
So x squared minus x squared.

00:07:32.820 --> 00:07:34.890
And now let's see
what we can do.

00:07:34.890 --> 00:07:38.700
Well since we're taking x
to infinity, and this is

00:07:38.700 --> 00:07:40.490
what you normally do what
you take x to infinity.

00:07:40.490 --> 00:07:43.270
We can divide the numerator
and the denominator by

00:07:43.270 --> 00:07:45.730
our highest degree term.

00:07:45.730 --> 00:07:48.800
And in this case, our highest
degree terms is x, right?

00:07:48.800 --> 00:07:50.520
We have an x here
and an x here.

00:07:50.520 --> 00:07:53.080
And then when you take-- And
then, of course, when you

00:07:53.080 --> 00:07:54.725
divide something like this
by x, and you take it to,

00:07:54.725 --> 00:07:56.070
infinity, this will approach 0.

00:07:56.070 --> 00:07:57.180
So let's do that.

00:07:57.180 --> 00:07:59.440
Let's divide the numerator
and the denominator by x.

00:07:59.440 --> 00:08:01.000
And remember: anything you do
in the numerator, you have

00:08:01.000 --> 00:08:01.700
to do in the denominator.

00:08:01.700 --> 00:08:03.420
Otherwise you're
changing the value.

00:08:03.420 --> 00:08:07.630
So times 1 over x
over 1 over x.

00:08:07.630 --> 00:08:10.350
I'm just dividing the numerator
and the denominator by x.

00:08:10.350 --> 00:08:16.360
So this is equal to the limit
as x approaches infinity of--

00:08:16.360 --> 00:08:24.620
What's-- That's going to
be 4 plus 1 over x over--

00:08:24.620 --> 00:08:25.420
Let me ask you a question.

00:08:25.420 --> 00:08:28.320
What is 1 over x
times this thing?

00:08:28.320 --> 00:08:32.920
What is-- This is a bit of an
algebra review, but 1 over x.

00:08:32.920 --> 00:08:37.230
What's 1 over x times x
squared plus 4x plus 1?

00:08:37.230 --> 00:08:40.150
I'm just doing a little
on the side here.

00:08:40.150 --> 00:08:42.360
Well if we take the 1 over
x and we put it into the

00:08:42.360 --> 00:08:45.310
radical sign, it becomes
1 over x squared, right?

00:08:45.310 --> 00:08:50.630
This is the same thing as--
You can say 1 squared

00:08:50.630 --> 00:08:54.660
over x squared, but--
1 over x squared.

00:08:54.660 --> 00:08:55.890
You could say 1 squared.

00:08:55.890 --> 00:08:59.640
You could put a square there,
but-- Times x squared

00:08:59.640 --> 00:09:02.410
plus 4x plus 1.

00:09:02.410 --> 00:09:04.250
And that should make
sense to you, right?

00:09:04.250 --> 00:09:06.380
Because if we started with
this, we could easily just

00:09:06.380 --> 00:09:07.980
take the square root of
this and take it outside.

00:09:07.980 --> 00:09:09.325
And the square root
of this is 1 over x.

00:09:09.325 --> 00:09:11.640
I'm just going in the
other direction, right?

00:09:11.640 --> 00:09:16.050
So, assuming you're comfortable
with that, everything

00:09:16.050 --> 00:09:17.020
under the radical sign.

00:09:17.020 --> 00:09:19.070
Even though we're actually
dividing by 1 over x, since

00:09:19.070 --> 00:09:22.060
we're going into the radical
sign, we're actually

00:09:22.060 --> 00:09:24.340
dividing by x squared.

00:09:24.340 --> 00:09:29.840
So it becomes-- this is the
radical sign --x squared

00:09:29.840 --> 00:09:33.770
divided by x squared
is 1, right?

00:09:33.770 --> 00:09:35.820
I hope you understand why we're
dividing by x squared here.

00:09:35.820 --> 00:09:36.860
We're actually
dividing by 1 over x.

00:09:36.860 --> 00:09:38.690
But when you put it under the
radical sign, it becomes

00:09:38.690 --> 00:09:39.440
1 over x squared.

00:09:39.440 --> 00:09:41.540
Let me put it this way.

00:09:41.540 --> 00:09:44.800
1 over x times the
square root of a.

00:09:44.800 --> 00:09:46.140
That's the same thing.

00:09:46.140 --> 00:09:51.600
That equals the square
root of 1 squared over x

00:09:51.600 --> 00:09:53.630
squared times a, right?

00:09:53.630 --> 00:09:54.970
And this is just 1
over x squared.

00:09:54.970 --> 00:09:58.280
So that's the property, or the
algebra that we're using.

00:09:58.280 --> 00:10:01.020
So anyway, we divide all
of this by x squared.

00:10:01.020 --> 00:10:11.160
So that becomes a 1 plus 4
over x plus 1 over x squared.

00:10:11.160 --> 00:10:13.770
And then, of course, we divide
this one by-- This term right

00:10:13.770 --> 00:10:16.450
here, we divide by x, right?

00:10:16.450 --> 00:10:18.970
Because it's not in
the radical sign.

00:10:18.970 --> 00:10:21.140
So that just becomes a 1.

00:10:21.140 --> 00:10:24.150
So now what's the limit as
x approaches infinity?

00:10:24.150 --> 00:10:26.900
Well as x approaches infinity,
this term right here

00:10:26.900 --> 00:10:28.720
goes to 0, right?

00:10:28.720 --> 00:10:30.480
1 over infinity is 0.

00:10:30.480 --> 00:10:32.560
This term right here goes to 0.

00:10:32.560 --> 00:10:34.790
This term right here goes to 0.

00:10:34.790 --> 00:10:36.520
And so what are we left with?

00:10:36.520 --> 00:10:44.270
This is equal to 4 over the
square root of 1 plus 1.

00:10:44.270 --> 00:10:45.070
Well that's just 1.

00:10:45.070 --> 00:10:53.040
So that equals 4 over 2,
which is equal to 2.

00:10:53.040 --> 00:10:54.840
There you go.

00:10:54.840 --> 00:10:57.860
Now let's do one more problem.

00:10:57.860 --> 00:10:59.120
This is the third
one she'd given me.

00:10:59.120 --> 00:11:01.720
In it, we had to kind of brush
off our trig identities.

00:11:01.720 --> 00:11:04.560
And really these harder limit
problems, they're all about

00:11:04.560 --> 00:11:06.620
kind of knowing your algebra
and your trigonometry

00:11:06.620 --> 00:11:07.150
really well.

00:11:07.150 --> 00:11:08.750
Just so you know how to
manipulate these functions.

00:11:08.750 --> 00:11:10.790
Because the limit part-- You
just have to get it into a form

00:11:10.790 --> 00:11:12.405
where taking the limit is
fairly straightforward.

00:11:12.405 --> 00:11:13.880
So let's do that trig problem.

00:11:16.980 --> 00:11:19.102
Clear image.

00:11:19.102 --> 00:11:20.950
Invert colors.

00:11:20.950 --> 00:11:27.110
So it was the limit
as x approaches 0

00:11:27.110 --> 00:11:33.170
of cotangent of 2x.

00:11:33.170 --> 00:11:34.370
Was that it?

00:11:34.370 --> 00:11:34.610
Yeah.

00:11:34.610 --> 00:11:43.080
It was cotangent of 2x
over the cosecant of x.

00:11:43.080 --> 00:11:45.870
And this one, just like
previous problems, more than

00:11:45.870 --> 00:11:47.940
knowing your pre-calculus or
your calculus, you need to

00:11:47.940 --> 00:11:50.020
know your trig identities.

00:11:50.020 --> 00:11:53.500
So cosecant of x.

00:11:53.500 --> 00:11:55.070
That's just 1 over sine.

00:11:55.070 --> 00:11:57.650
I remember that by saying
it's not intuitive.

00:11:57.650 --> 00:11:59.800
You would have thought that the
cosecant is 1 over cosine.

00:11:59.800 --> 00:12:00.230
But it's not.

00:12:00.230 --> 00:12:01.060
It's 1 over sine.

00:12:01.060 --> 00:12:02.740
So I remember that
it's not intuitive.

00:12:02.740 --> 00:12:11.480
And cotangent of 2x is equal
to 1 over tangent of 2x.

00:12:11.480 --> 00:12:14.330
And tangent is sine over
cosine, so cotangent's

00:12:14.330 --> 00:12:14.610
the opposite.

00:12:14.610 --> 00:12:22.290
And so that equals cosine of
2x over sine of 2x, right?

00:12:22.290 --> 00:12:23.430
OK.

00:12:23.430 --> 00:12:25.180
So what is this equal to?

00:12:25.180 --> 00:12:30.100
So this is going to be the
limit as x approaches 0.

00:12:30.100 --> 00:12:37.740
Cotangent of 2x, we said, was
cosine of 2x over sine of 2x.

00:12:40.570 --> 00:12:43.490
And then it's going to be all
of that over the cosecant of x.

00:12:43.490 --> 00:12:47.000
Well that's just 1
over sine of x.

00:12:47.000 --> 00:12:49.820
Well if you divide by 1 over
sine of x, that's the same

00:12:49.820 --> 00:12:54.120
thing as multiplying
by sine of x.

00:12:54.120 --> 00:12:55.780
So we have-- What do we have?

00:12:55.780 --> 00:13:08.110
We have cosine-- Well we have
sine of x times cosine of 2x.

00:13:08.110 --> 00:13:10.060
All of that divided
by sine of 2x.

00:13:10.060 --> 00:13:11.610
Just doing a little arithmetic.

00:13:14.640 --> 00:13:15.910
And we have a
problem here still.

00:13:15.910 --> 00:13:18.200
Because when you take x
approaches 0, this term right

00:13:18.200 --> 00:13:21.080
here goes to 0 and we have a 0
in the denominator, which

00:13:21.080 --> 00:13:22.140
is just not acceptable.

00:13:22.140 --> 00:13:23.130
Because it's undefined.

00:13:23.130 --> 00:13:24.520
And that's the whole reason
why we're doing this

00:13:24.520 --> 00:13:25.780
limit to begin with.

00:13:25.780 --> 00:13:27.230
And actually that's the first
thing you should have done.

00:13:27.230 --> 00:13:28.730
You should have tried to put it
and you would have seen that

00:13:28.730 --> 00:13:32.070
you would've gotten a 0 value
in the denominator and it would

00:13:32.070 --> 00:13:32.600
have been unevaluatable.

00:13:35.720 --> 00:13:36.170
Right.

00:13:36.170 --> 00:13:37.390
Because really, this is
just-- We haven't even

00:13:37.390 --> 00:13:38.340
done any manipulation yet.

00:13:38.340 --> 00:13:40.190
This is the same thing as this.

00:13:40.190 --> 00:13:43.870
And if you put the 0 right
here, you get undefined.

00:13:43.870 --> 00:13:45.040
So what can we do?

00:13:45.040 --> 00:13:47.210
Well this is where your
break out the trig or you

00:13:47.210 --> 00:13:48.710
brush off your memory.

00:13:48.710 --> 00:13:50.270
What is the sine of 2x equal?

00:13:50.270 --> 00:13:53.490
And this is your double-- One
of the double angle formulas.

00:13:53.490 --> 00:14:01.660
Sine of 2x is equal to 2
sine of x cosine of x.

00:14:01.660 --> 00:14:05.570
So if you know that, then
you've gone a long way because

00:14:05.570 --> 00:14:07.290
then it becomes pretty
simple to simplify.

00:14:07.290 --> 00:14:13.710
So it becomes 2 sine
of x cosine of x.

00:14:13.710 --> 00:14:16.710
And if we assume that x isn't
0, it's just approaching 0, we

00:14:16.710 --> 00:14:21.140
can divide the numerator and
the denominator by sine of x.

00:14:21.140 --> 00:14:23.340
And what are we left with?

00:14:23.340 --> 00:14:29.870
We're left with the limit as
x approaches 0 of cosine

00:14:29.870 --> 00:14:35.420
of 2x over 2 cosine of x.

00:14:35.420 --> 00:14:38.270
Well what's cosine of 0?

00:14:38.270 --> 00:14:40.400
Cosine of 0 is 1, right?

00:14:40.400 --> 00:14:43.010
So cosine of 2 times 0,
which is 0, that's also 1.

00:14:43.010 --> 00:14:46.560
So that is equal to 1 over--
right, cosine of 0 is

00:14:46.560 --> 00:14:48.490
1 --over 2 times 1.

00:14:48.490 --> 00:14:50.690
So it equals 1/2.

00:14:50.690 --> 00:14:51.160
There you go.

00:14:51.160 --> 00:14:54.840
I think those are three
fairly meaty limit problems.

00:14:54.840 --> 00:14:57.020
And if you know that, you're
probably prepared for something

00:14:57.020 --> 00:14:59.670
that your calculus teacher
might throw at you.

00:14:59.670 --> 00:15:01.960
See you in a future video.