Second derivative test is inconclusive here , given f( x, y) is
$x^{4} + y^{4} - 2x^{2} - 2y^{2} +4xy $
At (0,0) how do i check nature ? Also i would like to know general tactics when things like these happen .Thank You
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Do you know what is Hessian? – Stefan Gyürki Jan 25 '15 at 10:39
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@StefanGyürki its zero – tomb_raider Jan 25 '15 at 10:41
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I think it is -16. – Stefan Gyürki Jan 25 '15 at 10:42
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You are right.. sorry, I was computing by heart. – Stefan Gyürki Jan 25 '15 at 10:43
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Clearly $f(0,0)=0$. What is the sign of the function value if you are going from $(0,0)$ along the line $x=0$? If you are making a small $\varepsilon>0$ step from $(0,0)$ in that direction, then you have $f(0,\varepsilon)=\varepsilon^4-2\varepsilon^2=\varepsilon^2(\varepsilon^2-1)<0$. At the second step, you can go from $(0,0)$ along the line $y=x$ and follow the values on it: $f(\varepsilon, \varepsilon)=2\varepsilon^4>0$. Thus, if you are going in the first direction, you are going downwards on the graph of the function, while if you are going along the other line you are going upwards from zero. Therefore $(0,0)$ is a saddle point of $f(x,y)$.
Stefan Gyürki
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please vote my question so that i may have enough reputations to vote your answer .Thanks – tomb_raider Jan 25 '15 at 12:54