I don't even get to !
The critical points of a multivariable function are where it is not differentiable or where .
Here, the function is so the function is always differentiable and . That will be 0 when both components are 0: and .
Those should be easy to solve. Where in the world did you get " "?
Ok, sorry for the delayed response, I moved towards C++ for a bit.
I was just wondering. When you get to:
3x^2 - 3
and
-3y^2 + 12
Do you then solve for x as 1,-1 and then y as 2,-2
so the critical points are:
1,2
1,-2
-1,2
-1,-2?
I'm sorry I was following a youtube video, didn't know what the heck I was doing. Thanks.
Notice that I did NOT just get "3x^2- 3" and "3y^2- 12"! I said that, in order to have a critical point, we must have and from that I got the equations "3x^2- 3= 0" and "3y^2- 12= 0". Yes, and satisfy those equations. Since any pairing makes the gradient 0, the critical points are (1, 2), (1, -2), (-1, 2), and (-1, -2).