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Suppose the determinant of Hessian matrix is 0. Then it is not self evident whether there exists local minima or maxima or saddle point. Now, how do I figure that out?

Thank You.

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You can first look at simple example like: $x-> \pm x^3$ –  Ilies Zidane Nov 1 '12 at 17:50
    
@IliesZidane You can use \mapsto for $\mapsto$. –  martini Nov 1 '12 at 18:04

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up vote 2 down vote accepted

Try to remember why this works. The idea of the Hessian test was that if we have a critical point $x \in U$ of a sufficiently smooth $f \colon U \to \mathbb R$, then by Taylor \[ f(x+h) = f(x) + \frac 12 f''(x)h^2 + o(h^2), h \to 0 \] that is the behaviour of the Hessian $f''(x)$ determines locally $f$'s behaviour, that is if the quadratic form $h \mapsto f''(x)[h,h]$ is (positive, negative, in-)definite, than $f$ will have a minimum, maximum, saddle locally. If $f''(x)$ is positive semidefinite, say, you cannot conclude from $f''(x)$'s behaviour on $f$'s, what you can do is to look at the next term of the taylor expansion, writing \[ f(x+h) = f(x) + \frac 12 f''(x)h^2 + \frac 16 f'''(x)h^3 + o(h^3), h \to 0 \] If now for example, $f''(x)$ is positive semidefinite, the second term is non-negative allways, if now for example the cubic form $h \mapsto f'''(x)[h,h,h]$ is positive (for $h \ne 0$), $f$ will have a local minimum, if $f'''(x)$ is negative in a direction $h$ where $f''(x)$ vanishes (and $f''(x) \ne 0$), then you will have a saddle (if $f''(x) \le 0$ you can argue analogously).

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perfect! but why do you have f'' as second term and not f'? –  007resu Nov 1 '12 at 19:02
    
@user1710036 As in a critical point $x$ you have $f'(x) =0$. –  martini Nov 1 '12 at 21:25

You may write immediately \begin{gather} d^2{f}=\frac{\partial^2{f}}{\partial{x}^2}dx^2+\frac{\partial^2{f}}{\partial{x}\partial{y}}dxdy+\frac{\partial^2{f}}{\partial{y}^2}dy^2 \tag{*} \end{gather} and check positive or negative definiteness of quadratic form $(*)$ at critical point.

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