# Thread: Find the critical points of a multivariable equation

1. ## Find the critical points of a multivariable equation

Picture is attached, may be small
I can't get past x^2 + y^2 = 5

2. I don't even get to $x^2+ y^2= 5$!

The critical points of a multivariable function are where it is not differentiable or where $\nabla f= 0$.

Here, the function is $f(x,y)= x^3- 3x- y^3+ 12y$ so the function is always differentiable and $\nabla f= (3x^2- 3)\vec{i}- (3y^2- 12)\vec{j}$. That will be 0 when both components are 0: $3x^2- 3= 0$ and $3y^2- 12= 0$.

Those should be easy to solve. Where in the world did you get " $x^2+ y^2= 5$"?

3. 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.

4. 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 $\nabla f(x,y)= 0$ and from that I got the equations "3x^2- 3= 0" and "3y^2- 12= 0". Yes, $x= \pm 1$ and $y= \pm 2$ satisfy those equations. Since any pairing makes the gradient 0, the critical points are (1, 2), (1, -2), (-1, 2), and (-1, -2).