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$f ( x , y ) = \ln ( 2 + \sin ( x y ) )$. Consider only the critical point $(0,0)$.

I have solved for the first and second partial derivatives and I see that they are both equal to $0$ at $(0,0)$. One of the exercises in my textbook mentions the Hessian matrix and I think I should be using that, but I am not sure how it works.

$f ( x , y ) = \left( x ^ { 2 } + 3 y ^ { 2 } \right) e ^ { 1 - x ^ { 2 } - y ^ { 2 } }$

For this problem, I took the first and second partial derivatives and I observed that the critical points are at $(0,1)$, $(0,-1)$, and $(0,0)$. Do I have to use the Hessian here as well? How would that work?

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    $\begingroup$ If $xy > 0$ then $f(x,y) > f(0,0)$ if $xy < 0$ then $f(x,y) < f(0,0)$ This suggests to me that if you have a critical point, it must be a saddle point. $\endgroup$ – Doug M Nov 12 '18 at 18:12
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For the first one we have

$$f ( x , y ) -f(0,0)= \ln ( 2 + \sin ( x y ) )-\ln 2=\frac{xy}2+o(x^2+y^2)$$

and then it is a saddle point.

For the second one we have

$$f ( x , y ) = \left( x ^ { 2 } + 3 y ^ { 2 } \right) e ^ { 1 - x ^ { 2 } - y ^ { 2 } }\ge \left( x ^ { 2 } + y ^ { 2 } \right) e ^ { 1 - x ^ { 2 } - y ^ { 2 } }=r^2e^{1-r^2}=er^2-r^2+o(r^2)$$

that is positive as $r\to 0$ and therefore it is a minimum.

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  • $\begingroup$ Can you please explain why you did $f(x,y) - f(0,0)$? Does this have to do with the Hessian? Also, where did the $r$ come from in the second part? $\endgroup$ – Mohammed Shahid Nov 12 '18 at 18:21
  • $\begingroup$ @MohammedShahid For the first function $f(0,0)=\ln 2$ and we are interested to the change $f(x,y)-f(0,0)$. For the second one $f(0,0)=0$. $\endgroup$ – gimusi Nov 12 '18 at 18:23
  • $\begingroup$ Sorry for my ignorance, but what do you mean by, "we are interested in the change". Also, why does $\frac{xy}{2}+o(\sqrt{x^2+y^2})$ imply that there is a saddle point? Also, what does the $o(\sqrt{x^2+y^2})$ mean? $\endgroup$ – Mohammed Shahid Nov 12 '18 at 18:35
  • $\begingroup$ Are you aware about Taylor's expansion or at least first order expansion? $\endgroup$ – gimusi Nov 12 '18 at 18:36
  • $\begingroup$ I know about Taylor series that give polynomial approximation of a function. Is this similar? $\endgroup$ – Mohammed Shahid Nov 12 '18 at 18:41
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Sometimes a graph is worth a thousand words:

enter image description here

As @gimusi points out, this is clearly a saddle point.

Here's your second function:

enter image description here

Clearly a minimum because for any $r>0$, a point a distance $r$ from $(0,0)$ is above the point at the origin. The second derivative with respect to $x$ evaluated at $(0,0)$ is $e$ ($>0$) and hence the origin is a minimum.

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    $\begingroup$ That's confirm also the second one is a minimum! Thanks $\endgroup$ – gimusi Nov 12 '18 at 18:24

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