# Nature of the critical point $(0,0)$ of the function $f(x,y)=x^6-2x^2y-x^4y+2y^2$

Consider the function $$f(x,y)=x^6-2x^2y-x^4y+2y^2.$$ The point $$(0,0)$$ is a critical point. Observe, \begin{align*} f_x & = 6x^5-4xy-4x^3y, f_x(0,0)=0\\ f_y & = 2x^2-x^4+4y. f_y(0,0)=0\\ f_{xx} & = 30x^4-4y-12x^2y, f_{xx}(0,0)=0\\ f_{xy} & = 4x-4x^3, f_{xy}(0,0)=0\\ f_{yy} & = 4, f_{yy}=4 \end{align*}

So, in order to determine the nature of the above critical point, we need to check the Hessian at $$(0,0)$$ which is $$0$$ and hence the test is inconclusive. $$H(x,y)= \det \begin{pmatrix} f_{xx} & f_{xy}\\ f_{yx} & f_{yy} \end{pmatrix}=\det \begin{pmatrix} 0 & 0 \\ 0 & 4 \end{pmatrix}=0$$So, I tried to see the function on slices like $$y=0$$ and $$y=x$$ but nothing worked. So please suggest me how do I find the nature of the critical point in this case?

• @GeorgeDewhirst Still the Hessia at $(0,0)$ is $0$ and which is again not useful. Do you mean $f(x,x^2)$, because $f(x^2,y)$ for sure is not working? Even the first one is also not working. – Sachchidanand Prasad Jan 4 at 17:50
• you have a typo: last equality should be $f_{yy} = 4$. – peek-a-boo Jan 4 at 17:53
• @peek-a-boo Yeah thanks for that. – Sachchidanand Prasad Jan 4 at 17:56
• @GeorgeDewhirst The Hessian is coming $0$ and therefore it can not be positive definite. Please correct me if I am wrong. – Sachchidanand Prasad Jan 4 at 17:57
• @GeorgeDewhirst that is definitely not positve definite, since it isn't invertible. (more directly, if you call the matrix $H$ then $(e_1)^t H e_1 = 0$ which is not positive) (I think OP meant to say it has determinant zero... i'm not sure though) – peek-a-boo Jan 4 at 18:00

You have that $$g_a(x)=f(x,ax^2 ) = 2\left( {a^2 - a} \right)x^4 + \left( {1 - a} \right)x^6$$ With $$0 the function $$g_a(x)$$ has a local maximum. With $$a>1$$ the function has a local minimum. This means that $$(0,0)$$ is a saddle point.

You might note that your function factors as

$$(x^2-y)(x^4-2y).$$

So there are easy to find regions in the $$xy$$-plane where the function is positive an negative. Close to the origin and between the curves $$y=x^2$$ and $$y=x^4/2$$, the function is negative. This suggests trying the limit along the curve $$y=x^3$$. It's not too horrible to analyse

$$f(x,x^3) = 3x^6-2x^5 -x^7$$

around $$x=0$$ to see that it's negative there.

Comparing with the curve given by $$y=0$$, we get a saddle point at $$(0,0).$$

With $$g(x) = x^2$$ we have

$$f(g(x),y) = g(x)^3-2g(x)y-g(x)^2y +2y^2$$

or

$$f(g,y)=g^3-2g y-g^2y+2y^2$$

and the hessian of $$f(g,y)$$ is

$$H(g,y) = \left( \begin{array}{cc} 6 g-2 y & -2 g-2 \\ -2 g-2 & 4 \\ \end{array} \right)$$

and also

$$H(0,0) = \left( \begin{array}{cc} 0 & -2 \\ -2 & 4 \\ \end{array} \right)$$

with eigenvalues

$$\left\{2 \left(1+\sqrt{2}\right),2 \left(1-\sqrt{2}\right)\right\}$$

You have $$f(0,0)=0$$. You can find both positive and negative values of $$f(x,y)$$ in any region around $$(0,0)$$, which means that you don't have a local extremum at this point: $$f(x,x^3)= -x^5(1-x)(2-x)<0\quad\text{for}\,\, |x|<1, x\ne 0$$ and $$f(x,0)=x^6>0\quad\text{for}\,\, x\ne 0$$