How do you determine the behavior of critical points when you have the Hessian?

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SUMMARY

The discussion focuses on determining the behavior of critical points using the Hessian matrix in multivariable calculus. The user correctly identifies that after calculating the Hessian, which is a 2x2 matrix, they can apply rules from linear algebra to assess the definiteness of the matrix. Specifically, they use the determinant and trace to classify critical points as local minima, local maxima, or saddle points based on the conditions: local minimum if both fxx and fyy are positive, local maximum if both are negative, and saddle point if fxxfyy - fxy² is less than zero.

PREREQUISITES
  • Understanding of multivariable calculus concepts, particularly critical points.
  • Knowledge of the Hessian matrix and its properties.
  • Familiarity with linear algebra, specifically determinants and traces.
  • Ability to compute second derivatives of functions.
NEXT STEPS
  • Study the properties of the Hessian matrix in greater detail.
  • Learn about the implications of positive and negative definiteness in optimization problems.
  • Explore examples of critical point classification in multivariable functions.
  • Investigate applications of the Hessian in real-world optimization scenarios.
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus and optimization, as well as anyone involved in fields requiring critical point analysis in multivariable functions.

kelp
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Hello,
I have solved for the critical points using the gradient, and I have solved for the Hession, which yields a 2x2 matrix. I have plugged in my critical points into the gradient.

Now, do I apply the same rules as in linear algebra where I find the determinant and trace to calculate positive definite, negative definite, and indefiinite? I currently add up the diagonal and check the determinant. If they are both positive, then it's a local min, if they are both negative, it's a local max, and a saddle point if none of the above, correct?
 
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if fxxfyy-fxy^2 < 0, saddle point

if fxxfyy-fxy^2 > 0 then

local maximum if fxx,fyy < 0 and local minimum if fxx,fyy > 0
 

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