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I want to determine if a singular point is a local min, max or saddle point.
We are dealing with singular points so we cannot use the hessian matrix.
What I have written, and I think I must of missed something is :

Say we have a function $f(x,y,z)$. To show that $(2,2,2)$ is a saddle point, we want to show :
$\forall \epsilon >0$, we want to find $k,w,h$ such that if $k^2+w^2+h^2<\epsilon^2$ then
And here is where what I have writen is incomplete:
$f(2+k, 2+w, 2+h) > f(2,2,2)$
...................................$< f(2,2,2)$

Is this correct as $f(2-k, 2-w, 2-h)<f(2,2,2)$?

Edit: I'm not sure how standard these terms are, so to be clear a singular point here is a point at which the partial of the function does not exist.

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1 Answer 1

up vote 1 down vote accepted

A saddle point is neither a local max nor a min. For $(2,2,2)$ to be a saddle point we need the following. For every $\epsilon$$>0$ there exists $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ such that $d=\sqrt{(2-x_k)^2+(2-y_k)^2+(2-z_k)^2}$$<\epsilon$ for $k=1,2$ and $f(x_1,y_1,z_1)<f(2,2,2)<f(x_2,y_2,z_2)$.

ie, no mater how small of an interval we put around the domain point $(2,2,2)$ there are points in that interval producing a larger and smaller value of $f$.

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