Does point's neighborhood have no local extremum? I have polynomial of some limited degree:
$f(x,y) = a_1 + a_2x + a_3y + a_4xy + a_5x^2 + a_6y^2 + a_7x^2y + \ldots$
There is a point $p_0=(x_0,y_0)$, which is NOT a local extreme NOR inflection point. I have also small neighborhood of this point $|p-p_0|<R$ with radius R. I need some criteria to test if R is small enough, so there are no any other extremes NOR inflections inside this neighborhood. For 1D situation I can take Taylor series near $p_0$ and estimate R by means of comparing of odd and even monomials. But how can I do this for 2D? 
Thanx.
 A: It seems the following. 
We have  $f_x(p_0)\ne 0$ or  $f_y(p_0)\ne 0$. Let, for instance, $|f_x(p_0)|=\varepsilon>0$. Since the function $f_x$ is continous at the point $p_0$, there should be a number $\delta>0$ such that $|f_x(p_0)-f_x(p)|<\varepsilon$ provided $\|(x_0,y_0)-(x,y)\|<\delta$, where $p=(x,y).$ We have to find an explicit value of $\delta$. 
Let $$r>0,$$ $$f_x(x,y)=\sum_{i,j} a_{ij}x^iy^j,$$ $$\mathcal I=\{(i,j):a_{i,j}\ne 0\},$$ $$A=\max\{|a_{ij}|:(i,j)\in A\},$$ $$M_1=\max\{(|x_0|+|r|)^i, |y_0^j|:(i,j)\in A\},$$ $$M_2=\max\{ 1, \max\{i(|x_0|+|r|)^{i-1}, j(|y_0|+|r|)^{j-1}:(i,j)\in A, i,j>1\},$$ and $\delta<r$. 
Then 
$$|f_x(x,y)-f_x(x_0,y_0)|=\left|\sum_{i,j} a_{ij}x^iy^j-\sum_{i,j} a_{ij}x_0^iy_0^j\right|\le 
\sum_{i,j} |a_{ij}||x^iy^j-x_0^iy_0^j|\le \sum_{i,j} |a_{ij}|(|x^iy^j-x^iy_0^j|+|x^iy_0^j-x_0^iy_0^j|)\le
\sum_{i,j} |a_{ij}|(|x^i||y^j-y_0^j|+|y_0^j||x^i-x_0^i|)\le \sum_{i,j} A(M_1M_2\delta+ M_1M_2\delta)\le 
2|\mathcal I|AM_1M_2\delta.$$
So it suffices to put $$\delta=\min\left\{r,\frac{\varepsilon}{2|\mathcal I|AM_1 M_2}\right\}.$$  
