In one dimension, a saddle point is a point which is both a stationary point and a point of inflection. Since it is a point of inflection, it is not a local extremum.

Actually, I didn't get it when it said that it could also be a point of inflection. Can anyone give me an example of such a function and its saddle point which is both a stationary point and point of inflection?

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Images at google.com/… –  whuber Jun 15 '12 at 20:32
Thats what I didn't get. I mean the point is a saddle point by definition. That is fine. It is also a stationary point because the derivative at that point is 0. But how come it is a point of inflection? As mentioned in the wiki they are referring to one dimesion. So any example in one dimension –  rajan sthapit Jun 15 '12 at 20:59

Take the function $y=x^3$, $x=0$ is a stationary point and a point of inflection.
For functions of one real variable $t$ they call a critical point $t_0$ a saddle point if $t\mapsto f(t)-f(t_0)$ changes sign a $t_0$, and in most cases this means that the first nonvanishing derivative of $f$ at $t_0$ is of odd order.
For functions $f$ of $n\geq2$ real variables a critical point ${\bf p}=(p_1,\ldots, p_n)$ is a saddle point if the Hessian of $f$ at ${\bf p}$ is indefinite. This means that $$f({\bf p}+{\bf X})-f({\bf p})={\bf X}'\ H\ {\bf X}+o(|{\bf X}|^2)\qquad({\bf X}\to{\bf 0})$$ where the symmetric matrix $$H=\left[{\partial^2 f(x_1,\ldots, x_n)\over \partial x_i\partial x_k}\right]_{i,k}$$ has at least one strictly positive and at least one strictly negative eigenvalue. It follows that $f({\bf p}+{\bf X})-f({\bf p})$ takes as well positive as negative values in the immediate neighborhood of ${\bf p}$.
In the case of $n\geq2$ variables one usually assumes that the critical point in question is nondegenerate, meaning that the rank of $H$ is $=n$. The saddle points in the one-variable case are degenerate critical points in this sense, as the rank of the Hessian is $=0$ there.