Q:

To check if (0,0) is local minima for$F ( x, y) = x (x - 2y^{2}) $

Hello Thanks for your time 
$F ( x, y) = x (x - 2y^{2}) $ . I have applied second derivative test which does not give any result .
By looking at function i see that  when x is greater than  $2y^{2} $ , f is positive otherwise negative  , So according to me should be saddle point  . But textbook states it is minima .I need help.So much Thanks .So long !

A:

The function $F$ is zero on the $y$-axis $x=0$ and on the parabola $x=2y^2$. When you cross one of these curves $F$ changes sign. To be explicit: $F(x,0)=x^2>0$ when $\>x>0$, and  $$F(y^2,y)=y^2(y^2-2y^2)=-y^4<0$$ 
when $y\ne0$. It follows that $F$ assumes as well positive as negative values in any neighborhood of $(0,0)$. Since $F(0,0)=0$ this implies that $F$ has neither a local maximum nor a local minimum at $(0,0)$.
The second derivative test is inconclusive in this example, since $(0,0)$ is a degenerate critical point of $F$. In such cases only a an actual discussion of ${\rm sgn}(f)$ in the neighborhood of the critical point admits  a definitive answer.

