I've been working on some problems for critical points on graphs 3D functions (Maxima/minima/saddle points), using the matrix of second partials:

[ ∂<sup>2</sup>f/∂x<sup>2</sup> ∂<sup>2</sup>f/∂x∂y ]

[ ∂<sup>2</sup>f/∂y∂x ∂<sup>2</sup>f/∂y<sup>2</sup> ]

I've been using the fact that the gradient vector is equal to zero to find critical points, and then using the determinant of this matrix to see whether they are maxima, minima or saddles at those points.

Using the notation D(x,y) to be the determinant of the matrix, I've been using these facts to solve the problems:

D(x,y) = (∂<sup>2</sup>f/∂x<sup>2</sup>)(∂<sup>2</sup>f/∂y<sup>2</sup>) - ( ∂<sup>2</sup>f/∂x∂y)<sup>2</sup>

1.) D(x,y)>0 --> Minimum or Maximum at (x,y)

--Then using: ∂<sup>2</sup>f/∂x<sup>2</sup>>0 -->Min, and ∂<sup>2</sup>f/∂x<sup>2</sup><0 -->Max

2.) D(x,y)<0 --> Saddle at (x,y)

3.) D(x,y)=0 --> Could be minimum, maximum or saddle ("or something much more bizzare", the book says) at (x,y), not enough information to know.

I've been answering the problems using 1.) and 2.) with no problems at all, but for the case of 3.), I'm not sure how to do them. I thought about using the second test (as in 1.), but this only takes account of whether it is a maximum or a minimum and not whether it is a saddle point or "something much more bizzare"....unless ∂<sup>2</sup>f/∂x<sup>2</sup>=0 gives a saddle point?

Also, I'm still not clear on exactly how points 1,2 and 3 are derived from the definition of taking the determinant of the matrix. I can see how the ∂<sup>2</sup>f/∂x<sup>2</sup> terms are positive for a minimum, and negative for a maximum, as it's basically the same thing as for a normal second derivative, but the ∂<sup>2</sup>f/∂x∂y terms, with relation to the saddle points and the determinat are confusing me a bit. Because of that, I can't grasp why 3.) is defined as being any of the possibilities.

Thanks.