00:01
In this problem, we need to find the critical points of the given function where f of x, y is equal to minus x cube plus y cube minus 3 x, y.
00:13
So, we find f y which is derivative of the function with respect to y.
00:19
Hence, we can write this as 0 plus 3 y square minus 3 x.
00:27
Simplifying this, we can write this as 3 into y square minus x.
00:32
Similarly, we are going to differentiate the function with respect to x and hence we obtain the value as equal to minus 3 into x square plus y.
00:44
Now, f of x where x, y is equal to 0 is equal to the f of y into x, y.
00:55
So, now let us consider x square plus y to be equal to 0 and let this be equation 1.
01:05
Similarly, we will consider 3 into y square minus x equal to 0 where y square is equal to x and let this be equation 2.
01:18
At x equal to y square, we could use this in the equation 1...