True of false about partial derivative

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SUMMARY

The discussion centers on the conditions for identifying saddle points in multivariable calculus, specifically at the critical point (2,1) for the function f. It is established that if the second derivative test condition fxx(2,1)fyy(2,1) < (fxy(2,1))^2 holds, then (2,1) is indeed a saddle point. The participants clarify that the necessary condition for a critical point, namely fx(2,1) = fy(2,1) = 0, is already satisfied, confirming the conclusion about the saddle point.

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Homework Statement


if (2,1) is a critical point of f and fxx(2,1)fyy(2,1) < (fxy(2,1))^2
then f has a saddle point at (2,1)


Homework Equations





The Attempt at a Solution


i think its right
but it turns out to be wrong
can someone tell me why?
 
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wouldn't it need fx(2,1) = fy(2,1) = 0 as well?
 
lanedance said:
wouldn't it need fx(2,1) = fy(2,1) = 0 as well?

could you please explain it a little bit more?
 
ok now i see it is already a critical point, i would agree with you then
 

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