# Thread: Classifying ciritical point using partial derivatives.

1. ## Classifying ciritical point using partial derivatives.

I'm having some real difficulty classifying the critical points for the following function (sorry LaTex doesn't seem to want to work)
g(x,y)=x^(3)-3xy^(2)+y^(4)

Could someone help me, i've found the critical points to be (0,0) (3/2,3/2) and (3/2, -3/2)

But for (0,0) I find the second derivative test is inconclusive (because g11g22-g12^2=0) equal 0)

3. I get:
g1=3x^2-3y^2
g2=-6xy+4y^3

g11=6x
g12=6y
g22=-6x+12y^2

4. Originally Posted by flashylightsmeow
I get:
g1=3x^2-3y^2
g2=-6xy+4y^3
These are correct, given

$\displaystyle 3x^2-3y^2= 0 \implies 3x^2=3y^2 \implies x=y$

Now use this in the second equation to get $\displaystyle -6xy+4y^3= 0 \implies -6x^2+4x^3= 0\implies x = 0 , \frac{3}{2}$

5. Ah great, thank you. Just for clarification, are you getting finding that the critical point (0,0) gives an inconclusive result when performing tests using second derivatives? Because my book is saying (0,0) is a saddle point :S
Just out of interest, what is happening at a point where the second derivative test is inconclusive?

6. "inconclusive" means just that- we cannot conclude one way or another from this test. It does NOT mean the critical point is not one of max, min, or saddle point- those are the only things a critical point can be. For example, f(x,y)= x^4+ y^4, g(x,y)= -x^4- y^4, h(x,y)= x^4- y^4 all are "inconclusive" at (0, 0) but the first is obviously a minimum, the second a maximum, and the third a saddle point.

With both first and second derivatives being 0 at (0,0), look at the third derivatives: f_{xxx}= 6, f_{xxy}= 0, f_{xyy}= -6, and g_{yyy}= 4y which is 0 at (0, 0). That is, the third degree Taylor's polynomial for g is 6x^3- 6xy^2= 6x(x^2- y^2). Along the line y= 0, that is just 6x^3, both increasing and decreasing and so (0, 0) is a saddle point.