# Is the point at $(0,0)$ of the graph $z=x^3 + y^3$ considered a saddle point?

Is the point at $(0,0)$ of the graph $z=x^3 + y^3$ considered a saddle point?

I was given that function, and I used the second-order derivative test only to find that the $Hf(0,0) = 0$, with the critical point being at 0,0. Is anything that is not a local minimum or a local maximum considered a saddle point? What the graph looks like

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## 1 Answer

1. In your case, the Hessian is a null matrix. This does not guarantee that the point is a saddle point. Although I cannot come up with an example but I think you can have a null matrix as hessian at the minima. But don't quote me for that.

2. Such a point, AFAIK, is given the name "Degenerate Critical Point". Whether it is considered saddle or not, I am not sure.

3. If I am standing at (0,0,0) and you give me an $\epsilon$, I can step $(\epsilon,0)$ and make the function positive or $(-\epsilon,0)$ and make the function negative. So, no matter how small epsilon you give me, I can always move to points that increase the function value and reduce it. This makes the (0,0,0) a saddle point.
4. Does this question help : Classifying singular points as local min, max or saddle points?
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Standard example of a local min with vanishing Hessian is $x^4 + y^4$.  The common definition of saddle point is that it is neither a local min nor local max, and this is defined without any reference to derivatives or Hessian. – Erick Wong Jan 16 at 5:31
@ErickWong, Fair enough. But given that the gradient and hessian is both 0 (in their own dimensions) at that point, how do you know that it is not maxima or minima? (One can say so by observation or by computing the function at other points but that's not robust). Is there any other way I am missing? – Inquest Jan 16 at 5:34
@Inquest You're not missing anything.  Things get harder when the second derivative vanishes, except in simple cases such as these.  This homework problem is meant to point out a limitation of the method. – Michael E2 Jan 17 at 12:43
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