00:01
In this question, we are asked to find the relative xterma and the saddle points of the function f.
00:07
To do that, we first need to find the critical points.
00:11
And to find the critical points, we need to solve the equation f x equals 0 and fy equals 0.
00:20
The derivative of f with respect to x equals to 2x plus 4.
00:27
And the derivative of f with respect to y equals to 2y minus 16.
00:34
And we want both of them to be equal to 0.
00:39
This gives us a system of equations for x and y.
00:42
From the first equation to x equals to negative 4.
00:45
And from the second equation to y equals to 16.
00:51
Therefore, x equals to negative 2 and y equals to 8 is the only critical point of the function f.
01:10
Negative 2 8.
01:14
Now we are going to use the second derivative test.
01:21
By the second derivative test, we first need to calculate the quantity called d, which equals to f double x times f double y minus f xy squared.
01:36
Where f double x is the second derivative of f with respect to x, and to get it we need to differentiate f x with respect to x, and we are going to get 2.
01:50
To get f double y we need to differentiate fy with respect to y and we are also going to get two to get f xy we need to differentiate f x with respect to y so let's get back to f x and if we differentiate with respect to y we are going to get zero because f x doesn't depend on y therefore d equals to 2 2 times 2 minus 0 equals to 4.
02:31
And in particular, in general, d would depend on x and y, and you would have to plug in the coordinates of the critical point.
02:42
But in our case, d of negative to 8 is still going to be 4 because d is a constant.
02:48
It doesn't depend on x and y...