# Saddle point or not?

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: • a saddle point should have positive curvature along one direction and negative curvature along another – phdmba7of12 Mar 1 at 13:34
• You have to take into account the value $f_{xx}(0,0)=0.$ – user376343 Mar 1 at 13:35
• @phdmba7of12 So, how do we show in this case that it is (or isn't)? Are there any necessary and sufficient conditions? – PGupta Mar 1 at 13:35
• @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 $f_{xx}$. – PGupta Mar 1 at 13:36
• @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? – PGupta Mar 1 at 13:43

## 4 Answers

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.

• 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)? – PGupta Mar 1 at 13:48
• @PGupta: It does look like a saddle, near the origin! You need to zoom in much more in your picture. – Hans Lundmark Mar 1 at 14:23

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.

• 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. – PGupta Mar 2 at 14:25
• I would call it one. Not sure what the book authors are thinking. – Oscar Lanzi Mar 2 at 19:47

here's what the function looks like (would have posted as a comment but can't) • Can we conclude it is not a saddle point without using the 3D graph? – PGupta Mar 1 at 13:41
• of course ... i'm always a fan of graphing functions ... by hand if necessary – phdmba7of12 Mar 1 at 13:42
• 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. – PGupta Mar 1 at 13:46
• agree with your point ... absolutely – phdmba7of12 Mar 1 at 13:48
• Then how should they answer the question!? – PGupta Mar 1 at 13:49

From graphical plot, it appears to be a saddle point having positive and negative curvature along mutually perpendicular directions  