00:01
Okay, so in this question, we're asked to study the critical points of the function x squared minus 2xy squared plus 4y cubed minus 4x.
00:19
And there's two steps to this, right? so the first, we do the first derivative test to find the critical points, and then we do the second derivative test to classify the critical points.
00:27
So let's go step one, find the critical points.
00:33
And you know that these are the solutions of dxf equals zero and dyf equals zero.
00:40
So let's make the calculation.
00:42
First, the derivative with respect to x, we get 2x.
00:45
From here, we get minus 2y squared.
00:48
From here, we get zero.
00:50
And from here, we get the minus four.
00:52
And with respect to y, we get y squared.
00:55
So 2y times the minus 2x.
00:57
So minus 4xy.
01:00
From here, we get plus, the three comes down, 12y squared, and nothing from there.
01:07
Okay, so let's start by operating on the second equation.
01:10
So we can isolate the y.
01:13
This is y minus 4x plus 12y equals zero.
01:19
So from here, we conclude that we have already two solutions when y equals zero, or when minus 4x plus 12y equals zero, which is the same as x being the same as 3y.
01:31
Okay? so when y equals zero in the first equation, we get 2x equals four.
01:36
So x is equal to two.
01:38
And when x is equal to 3y, we get 6y minus 2y squared minus four equal to zero.
01:46
So let's just focus for now on this equation.
01:49
As a side calculation, i'm not going to carry everything over.
01:52
I'm just going to focus on this equation for now.
01:57
So 6y minus 2y squared minus four is equal to zero.
02:01
Divide everything by minus two, just to make it a more friendly formula.
02:06
Y squared minus 3y plus two is equal to zero.
02:11
So i just divide it by minus two.
02:14
And this one we solve.
02:15
Let's see.
02:16
So y is equal to three plus or minus square root three squared minus four times one times two minus eight over two.
02:27
So this becomes three plus minus one over two, and we get two solutions, right? so y1 is three plus one over two, which is two.
02:36
And then the solution y2, which is three minus one, two over two, which is one.
02:42
Okay? so critical points are, we already had the two zero, right? so then we have another one.
02:54
When y is equal to two, x is three times y.
02:58
So we get six, six two.
03:00
And when y is equal to one, x is three times one, which is equal to three.
03:04
Okay? so these are the critical points of the system.
03:10
And this concludes the first step, which is identifying the critical points.
03:14
Now the second step is to classify the critical points.
03:21
And you know that this goes, we use a second derivative test for this.
03:24
So it's about the sign of our hessian matrix.
03:27
So we start by calculating the hessian matrix.
03:30
This is the matrix of second derivatives.
03:32
So d2xf, dxyf, dxyf, these are the same.
03:38
And then d2fy.
03:41
So let's see what we get.
03:43
Let's look at the first derivative in x and differentiate again with respect to x, which is get two.
03:51
Now let's go derivative with respect to x and differentiate with respect to y and we get minus four y is symmetric...