Linearization of a quadratic function in $\mathbb{R}^2$

Question

In my research, I came across with a undesirable quadratic function that must be linearized. I am wondering whether it is possible. The problem is described in the following.

Let the function

\begin{array} {lcl} f: (-\varepsilon, \varepsilon) \times [-1,1] \longrightarrow (-\varepsilon, \varepsilon)\\ \hspace{1cm}(x,y)\longmapsto xy. \end{array}

Assuming $\varepsilon << 1$, my question is:

Is there some real values $a,b$ and $c$ such that $$f(x,y) \approx ax +by +c?$$

Thanks in advance!

Answer 1

Look at this picture of your function ($\epsilon = 0.05$ here, but the picture will look indistinguishable for any other choice, as long as you rescale the height). The best you can do is probably the plane $f(x, y) \approx 0$, just from looking at all the symmetry.

A plane that does well for $y$ around $-1$ will do terribly for $y$ close to $1$. Same for $x$. Just picking the middle ground seems as good an option as any.