What to do when second partial derivative test is inconclusive

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SUMMARY

The discussion centers on the challenge of determining the nature of a critical point at (0,0) for the function f(x,y) = (x^2)y + x(y^2) when the second partial derivative test is inconclusive due to all second derivatives being zero. Participants suggest exploring the behavior of the function along specific curves, such as the axes or the line y=x, to identify whether the critical point is a saddle point. They emphasize the importance of analyzing higher-order derivatives to gain insights into the function's behavior near the critical point.

PREREQUISITES
  • Understanding of critical points in multivariable calculus
  • Familiarity with second and higher-order partial derivatives
  • Knowledge of saddle points and their characteristics
  • Ability to analyze functions along specific curves
NEXT STEPS
  • Study the behavior of higher-order derivatives in multivariable functions
  • Learn about the method of curves for analyzing critical points
  • Research the implications of the third derivative in determining extremum
  • Explore the concept of Taylor series expansions for multivariable functions
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Students and educators in multivariable calculus, mathematicians analyzing critical points, and anyone seeking to deepen their understanding of function behavior near critical points.

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Homework Statement


I have two problems where there is a critical point of f(x,y) at (0,0), but the second derivatives and mixed second derivative are all zero. The second partial derivative test is therefore inconclusive- all the information I can find online/in my notes just says it is inconclusive and doesn't offer an alternative method. If anyone could give me a link or even just tell me what I should be googling that would be really helpful! Thanks.


Homework Equations





The Attempt at a Solution


I vaguely think I could show the critical point was a saddle point if I could show along one curve f(x,y) is positive as (x,y) -> (0,0) and along another curve f(x,y) is negative as (x,y) -> (0,0) but I have no theorem that states this.
 
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so in the single variable case you would consider the 3rd derivative

if you consider a passing along a line eg x or y axis, or y=x, as you say you could either consider the 3rd derivative, however whilst you may be able to show a saddle point, it won't be able to confirm an extremum
 
so what was the function?
 
The first one is f(x,y) = (x^2)y + x(y^2)

but I was really after ideas for a method not the solution.
 
well i don't know any cook book methods, but you could have a think about what happens to the higher order derivatives, or the previous idea

however for this case I would consider the line y=x (or the axes)
 
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