# Local Minima and Maxima

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

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).
@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.