I'm reviewing for a final exam tomorrow. This is an exercise that I am having trouble with:
The function: $f(x,y)=x^2-y^4$
I determined that there is one critical point, at $(0,0)$.
I determined that the Hessian at $(0,0)$ was neither positive definite nor negative definite, and that $(0,0)$ was not a saddle point.
So now I am looking for another way to solve this problem. My attempt:
Proving that it is not a local min: Consider $x=y^2/2$. $f(y^2/2,y)=(3y^2/2)(-y^2/2)=-3y^4/4$. Close to (0,0), this is $<0=f(0,0)$, so $f(0,0)$ cannot be a local minimum.
Proving that it is not a local max: Consider $x=3y^2/2$. $f(3y^2/2,y)=(5y^2/2)(y^2/2)=5y^4/4$. Close to (0,0), this is $>0=f(0,0)$, so $f(0,0)$ cannot be a local maximum.
Is this correct reasoning? What do I call this point (what is its "nature") if it is neither a max, a min, nor a saddle point?
First of all, saddle points are by definition stationary points that are not local extrema. So there are, by definition, only three kinds of stationary points: local maxima, local minima, and saddle points. You have shown that $(0,0)$ is a stationary point and have shown it isn't a local extremum. It is therefore a saddle point. Your argument seems a bit complicated though. Just look along the line $y=0$, and see that $f(x,0) = x^2$ there. So $(0,0)$ can't be a local minimum. On the other hand, looking along $x=0$ we see that $f(0,y) = -y^4$, so $(0,0)$ can't be a local maximum.