
#1
Dec2812, 09:21 AM

P: 25

For [tex]f(x,y)=x^2+y^3[/tex]
Is there a saddle point at (0,0) or does the function have to have 2 or more sides going down like [tex]g(x,y)=x^2y^2[/tex] 



#2
Dec2812, 09:25 AM

P: 1,339

Okay so have you taken the required derivative and found your critical points?




#3
Dec2812, 09:32 AM

P: 25

The critical point is (0,0). I'm just wondering if f(0,0) is a saddle point when the graph is shaped more like a chair than a saddle.




#4
Dec2812, 09:39 AM

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P: 25,178

Is this a saddle? 



#5
Dec2812, 09:39 AM

P: 1,339

[itex] D = f_{xx}f_{yy}  f_{xy}^{2}[/itex] This will allow you to see if your function has a saddle, min or max at a critical point ( Doesn't alwayyyyys work though ). 



#6
Dec2812, 09:40 AM

P: 25





#7
Dec2812, 01:46 PM

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P: 4,677

Note that there is a bit of mismatch between the necessary and sufficient secondorder conditions: there is no sufficient condition for a nonstrict local min, so functions like that of the OP slip through the cracks. This happens as well in one dimension, where testing functions like f(x) = x^3 for an optimum at x = 0 cannot just rely on the second derivative. For the OP's function f(x,y), the Hessian at (0,0) is positive semidefinite, so (0,0) cannot be a local *maximum *; it must either be a local minimum or a saddle point. In fact, f(x,0) = x^2, so x = 0 is a minimum along the line (x,0), and f(0,y) = y^3, so y = 0 is an inflection point along the line (0,y). Thus, there are points near (0,0) giving f(x,y) > f(0,0) and other points near (0,0) giving f(x,y) < f(0,0); that is moreorless what we mean by a 'saddle point'. 


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