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JOEL LEWIS: Hi.

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Welcome back to recitation.

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In lecture, you've been
learning about critical points

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of functions, how to find them
using the first derivatives

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and how to classify them using
the second derivative test.

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So I have a question
here for you about that.

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So we have a function w.

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It's a function of two
variables, x and y,

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and it's given by this
polynomial function of them.

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So w equals x cubed
minus 3xy plus y cubed.

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So what I'd like you
to do is to first find

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the critical values
of this function

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and then classify
them-- are they

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minima or maxima
or saddle points--

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using the second
derivative test.

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So why don't you
pause the video,

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take some time to work that out.

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Come back and we can
work it out together.

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Hopefully, you had some luck
working out the solution

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to this question.

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Let's have a go at it.

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So in order to find
the critical points,

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we need to look at
the first derivative.

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So the critical
points are the points

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where both partial derivatives--
or all partial derivatives,

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if we had a function of more
variables-- are equal to zero.

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So we need to look at
the first partials.

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So the first partials here, w
sub x, the partial with respect

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to x-- well, it's
just a polynomial

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so it's easy to compute
those partial derivatives.

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It's going to be 3
x squared minus 3y,

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and then the last term gets
killed because we treat y

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as a constant, and so we want
that to be equal to zero.

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And similarly, we want the
first partial with respect to y,

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w sub y, to be equal to zero.

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And so that's minus 3x
plus 3 y squared equals 0.

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Now, luckily, these are
fairly simple equations,

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so to solve them, we
could, for example,

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take the first equation and we
could solve the first equation

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for y in terms of x.

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So that'll give us
y equals x squared.

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And now if we plug
y equals x squared

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into this second equation,
well, we get minus 3x plus 3 x

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squared squared-- so that's x
to the fourth-- is equal to 0,

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and we can divide out by
that 3, so that means minus x

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plus x to the fourth equals 0.

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Well, OK, so we could have x
equal to 0, or you can divide,

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and then you get
x cubed equals 1,

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and that has
solution x equals 1.

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So x equals 0 or 1.

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Those are the only
solutions to this equation.

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And then the corresponding
y-values, well, we

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know y is equal to x squared,
so this gives us critical points

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when x is 0, y is 0,
and when x is 1, y is 1.

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So this function has two
critical points: (0, 0)

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and (1, 1).

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Now we need to figure out
whether those critical points

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are minima, maxima,
saddle points, sum

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of several of those.

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So in order to do
that, we're going

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to use this nice
tool that we have:

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the second derivative test.

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So in order to apply the
second derivative test,

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the first thing I need is
the second derivatives.

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So let's compute them.

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So the second-- let's
do the xx first.

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So we take our first partial,
3 x squared minus 3y,

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and we take another partial
of it with respect to x.

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So that's, in this case,
that's just going to be 6x.

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And then we've got the other
pure second partial, yy,

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so we go back over
here, and we look

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at what our first
partial w_y was,

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and then we take another partial
of this with respect to y,

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so that's just going to be 6y.

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And then we have the mixed
partials w_xy and w_yx,

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which, of course, are equal
to each other whenever

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our function is nicely
behaved, like a polynomial.

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So w_xy, we just take
the two mixed partials

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and-- OK, so we take
the partial of w_x

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with respect to y, for example,
and that gives us minus 3.

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So these are our three partials,
and then often, we, you know,

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call this one A and this
one C and this one B. I

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guess I kind of mixed up the
order a little bit there.

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So we look at these
three expressions,

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and now we want to look
at what sometimes people

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call the discriminant, although
I don't know if Professor

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Auroux used that term.

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So we want to study what the
expression A*C minus B squared

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is, so we want to know is this
positive, is this negative?

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At the critical points.

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So at the critical
points, right?

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This is important.

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At the critical points.

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So let's do the
point (1, 1) first.

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So at (1, 1), we have A is equal
to-- well, we put in x is 1,

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y is 1 into the expression for
A here, and that just gives a 6.

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We put x 1, y 1 into
the expression for C,

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and that also gives
a 6, we put x 1, y 1

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into the expression for B,
and that gives us minus 3.

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So A is 6, B is minus 3, C is 6,
so A*C minus B squared is equal

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to, well it's equal to
36 minus 9, so that's 27.

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And, in particular,
it's positive.

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So when this is
positive, that means we

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either have a
maximum or a minimum.

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So in order to
figure out whether we

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have a maximum or a minimum,
we check the sign of A.

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So in this case, the
sign of A is positive.

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A is a positive number.

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So when you have that A*C minus
B squared is positive and A is

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positive, that means
you have a minimum.

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So the critical point
(1, 1) is a local minimum

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for this function.

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All right.

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Now we can do the same thing
for the critical point (0, 0).

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So recall A was
6x, so at (0, 0).

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A is equal to 0, B was equal
to negative 3 everywhere,

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and C was equal to
6y, so at (0, 0),

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that's 6 times 0
so that's also 0.

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So then our quantity that we
want to look at, A*C minus B

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squared, well, that's
0 times 0 minus 9,

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so that's equal to negative 9.

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And negative 9 is less than 0,
so when A*C minus B squared is

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less than 0, that means
we have a saddle point.

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So in this case, the
second derivative test

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was able successfully
to distinguish

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what kinds of critical
points we had,

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and it found that the first
critical point 1, 1 was

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a minimum and that the
second critical point

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0, 0 was a saddle point.

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So just to quickly rehash what
we did, we had a function.

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Back over here, we started
with this function w.

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We had a nice formula for it.

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We computed its
first derivatives.

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We set them both equal
to zero and we solved

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that system of equations.

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So we found two solutions
to that system of equations,

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and those two solutions
are the critical points,

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the points where both partial
derivatives are equal to zero.

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So when you have the
two critical points,

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then you want to apply
the second derivative test

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to figure out for
each critical point

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whether it's a saddle point,
a minimum or a maximum.

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So we took our two
critical points, (1, 1)

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and (0, 0), and at those
points, we evaluated

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the second derivative.

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So A is the xx
second derivative.

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B is the mixed partial w_xy, and
C is the yy second derivative.

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So we evaluate those expressions
at the points in question,

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and then we look at
A*C minus B squared.

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And then the sign of A*C minus
B squared, if it's negative,

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that gives us a saddle point.

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If it's positive, that gives us
either a maximum or a minimum,

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and we check which
one by looking

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at the sign of A. So
here A was positive,

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so we got a minimum at (1, 1).

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So I'll stop there.