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Consider the function $f(x,y)=2xy-x^3-y^2$. One of the stationary points is $(0,0)$. At this point, $f_{xx}f_{yy}-f_{xy}f_{yx}<0$. According to me, this indicates that (0,0) is a saddle point. However, the text I am referring to calls this "neither an extremum nor a saddle point". Am I missing something?

Edit The plot (from GeoGebra) looks like this: enter image description here

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  • $\begingroup$ a saddle point should have positive curvature along one direction and negative curvature along another $\endgroup$ – phdmba7of12 Mar 1 at 13:34
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    $\begingroup$ You have to take into account the value $f_{xx}(0,0)=0.$ $\endgroup$ – user376343 Mar 1 at 13:35
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    $\begingroup$ @phdmba7of12 So, how do we show in this case that it is (or isn't)? Are there any necessary and sufficient conditions? $\endgroup$ – PGupta Mar 1 at 13:35
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    $\begingroup$ @user376343: Please elaborate. One of the texts I am following says that if $f_{xx}f_{yy}-f_{xy}{yx}<0$ then it is a saddle point. No restrictions are given on $f_{xx}$. $\endgroup$ – PGupta Mar 1 at 13:36
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    $\begingroup$ @phdmba7of12: So other than the determinant of the Hessian, we need to check if $f_{xx}$ or $f_{yy}$ is non-zero and only then conclude if it a saddle? $\endgroup$ – PGupta Mar 1 at 13:43
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You're right, and there's a mistake in the example. I'm pretty sure something like $x^3+y^2$ was intended; that's genuinely not a saddle point, despite increasing in some directions and decreasing in others.

This is also dependent on the definition; some sources define a saddle point to be a critical point that's not a maximum or minimum, in which case this situation would be impossible.

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    $\begingroup$ I see. So if we consider the purely geometric meaning of it "looking" like a saddle (max from one side and min from another), then this is not a saddle (as shown in the graph above)? $\endgroup$ – PGupta Mar 1 at 13:48
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    $\begingroup$ @PGupta: It does look like a saddle, near the origin! You need to zoom in much more in your picture. $\endgroup$ – Hans Lundmark Mar 1 at 14:23
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If you follow the path $y=x$ then $f(x,x)=x^2-x^3$ meaning a local minimum. If you follow $y=-x$ then $f(x,-x)=-3x^2-x^3$ meaning a local maximum. These behaviors match $g(x,y)=xy$, an archetypal saddle point at the origin.

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  • $\begingroup$ So it is a saddle point, and not what the book says? I must point out here that there are other examples in the book as well where it is claimed that the stationary point is not a saddle or an extremum, but it doesn't look like there would be so many typos in the text. $\endgroup$ – PGupta Mar 2 at 14:25
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    $\begingroup$ I would call it one. Not sure what the book authors are thinking. $\endgroup$ – Oscar Lanzi Mar 2 at 19:47
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here's what the function looks like (would have posted as a comment but can't) enter image description here

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    $\begingroup$ Can we conclude it is not a saddle point without using the 3D graph? $\endgroup$ – PGupta Mar 1 at 13:41
  • $\begingroup$ of course ... i'm always a fan of graphing functions ... by hand if necessary $\endgroup$ – phdmba7of12 Mar 1 at 13:42
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    $\begingroup$ Definitely, graphing gives visual clarity. But 3D graphs are difficult to visualise without a computer. I plan to show the graphs to my students in the class, but they won't be able to draw them/visualise during an exam. $\endgroup$ – PGupta Mar 1 at 13:46
  • $\begingroup$ agree with your point ... absolutely $\endgroup$ – phdmba7of12 Mar 1 at 13:48
  • $\begingroup$ Then how should they answer the question!? $\endgroup$ – PGupta Mar 1 at 13:49
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From graphical plot, it appears to be a saddle point having positive and negative curvature along mutually perpendicular directions

enter image description here

enter image description here

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