Undergrad Finding Max and Min Extremes of a Function with Second Derivatives Equal to Zero

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When the second derivatives of a function f(x, y) equal zero, further differentiation is necessary to identify maximum and minimum extremes. The process involves calculating the third partial derivatives and analyzing the behavior of the function in the direction of eigenvectors associated with zero eigenvalues in the Hessian matrix. If both eigenvalues are zero, examining cubic terms becomes essential to understand the function's behavior. Additionally, switching to a different coordinate system, such as polar coordinates, can simplify the analysis and reveal global maxima and local minima more clearly. Ultimately, continued differentiation and alternative coordinate systems are key strategies for resolving the issue of vanishing second derivatives.
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What should I do when the f(x, y) function's second derivatives or Δ=AC-B² is zero? When the function is f(x) then we can differentiate it until it won't be a zero, but if z = some x and y then can I just continue this process to find what max and min (extremes) it has?

What I've done is:
Differentiated z=f(x, y) by partial derivatives with respect to x and y;
Made them equal to zero in the system of numbers;
Where I got x=0, y=0;
Differentiated function by the second partial derivatives;
Used Δ = AC-B² where if Δ=0 then it needs more calculations.
I got Δ=0.

What should I do now? Differentiate again (third partial) as we do to function f(x) with only one variable x until it won't be equal to zero?
 
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What the determinant of the hessian matrix of f vanishing tells you is that is has a zero eigenvalue with a corresponding eigenvector \mathbf{v}. To determine the behaviour of f in this direction, you should look at <br /> \frac{d^3}{dt^3}f(\mathbf{x}_0 + t\mathbf{v}) and higher derivatives until you find one which does not vanish. If both eigenvalues are zero then the hessian is zero, and you need to look at the cubic terms to see what is going on.

Switching to a different coordinate system may give better insight into what is going on, for example https://www.wolframalpha.com/input?i=(x^2+y^2)*(1+-+(x^2+y^2)) where in polar coordinates it is easy to find that r^2(1-r^2) has a ring of global maxima at r = 1/\sqrt{2} and a local minimum at r = 0.
 

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