# Nature of critical point

This is the given function

$$f(x, y) = x^4- 2(x^2)(y^2)+ y^4 +y^3$$

I have calculated the critical points to which I get $(0,0)$ and $(0,-3/4)$.

$$f_{xx} = 12x^2- 8y \\ f_{yy} = -4x^2 +12y^2 +6y \\ f_{xy} = -8$$ but when I calculate hessian for $(0,0)$, I am getting $8$ and $f_{xx},\ f_{yy}=0$ for which I am confused how do I determine the nature of critical point $(0,0)$ Can somebody help?

• i think there are no extrema Commented May 4, 2017 at 14:06

The most straightforward way to answer these kinds of questions is to use the Hessian. $$H(x,y) = \begin{bmatrix} 12x^2 - 4y^2 & -8xy \\ -8xy & -4 x^2 + 12 y^2 + 6 y \end{bmatrix}$$ You will notice that at $(0,0)$, the Hessian is zero and the Hessian test is inconclusive about whether the critical point is a min/max or saddle point.
We can try to evaluate $f(x,y)$ for a specific slice and determine the behavior of that. So let's calculate it for $y=x$. $$f(x,x) = 48x^3$$ Along this slice, we see that $(0,0)$ is an inflection point. Note that any point which is an extrema would stay an extrema when looking along any direction. So if a point is not an extrema along a specific direction, then it cannot be an extrema of the original function.
Thus, $(0,0)$ is a saddle point.