Why does the Hessian determinant Δ_p = -1 imply that P(0, 0) is a saddle point?

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SUMMARY

The Hessian determinant Δ_p = -1 at the critical point P(0, 0) for the function f(x, y) = xye^[-(x^2 + y^2)] indicates that P(0, 0) is a saddle point. This conclusion arises from the properties of the Hessian matrix, specifically when it is represented as H = [[0, 1], [1, 0]]. The quadratic approximation of the function near this point reveals that the surface behaves like z = xy, confirming the saddle point classification through visual inspection of the graph.

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s3a
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Homework Statement


For the function f(x, y) = xye^[-(x^2 + y^2)] find all the critical points and classify them each as a relative maximum, a relative minimum, or a saddle point.

Homework Equations


Partial differentiation and Hessian determinants.


The Attempt at a Solution


I get how to compute the derivatives. I also get how to compute the Hessian determinants. Basically, I get all the algebraic details but what I would like to ask about (at least for now) is why does the Hessian determinant Δ_p = -1 imply that P(0, 0) is a saddle point?

Any input would be greatly appreciated!
Thanks in advance!
 

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s3a said:

Homework Statement


For the function f(x, y) = xye^[-(x^2 + y^2)] find all the critical points and classify them each as a relative maximum, a relative minimum, or a saddle point.

Homework Equations


Partial differentiation and Hessian determinants.


The Attempt at a Solution


I get how to compute the derivatives. I also get how to compute the Hessian determinants. Basically, I get all the algebraic details but what I would like to ask about (at least for now) is why does the Hessian determinant Δ_p = -1 imply that P(0, 0) is a saddle point?

Any input would be greatly appreciated!
Thanks in advance!

The Hessian at (x0,y0) gives the quadratic terms in the Taylor expansion at that point. In other words, if we write
f(x,y) \approx f(x_0,y_0) + a_0 (x-x_0) + b_0 (y-y_0)<br /> + \frac{1}{2} r_0 (x-x_0)^2 + s_0 (x-x_0)(y-y_0) + \frac{1}{2} t_0 (y-y_0)^2 + \text{ higher order terms }, the constants r_0, s_0, t_0 are the elements of the Hessian at (x0,y0); that is, the Hessian is
H(x_0,y_0) = \left[ \matrix{r_0&amp;s_0\\s_0&amp;t_0} \right]. So, if the Hessian is H = \left[ \matrix{0&amp;1\\1&amp;0}\right], that means that for small x and y we have f(x,y) \approx xy, and if you plot the surface z = xy around (0,0), you will see that (0,0) is, indeed, a saddle point.

Doesn't your textbook or your course notes have this material?

RGV
 

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