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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 !

tomb_raider
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  • Hint: Consider intersections of the surface $z=x(x-2y^2)$ with vertical planes $ax+by=0$. – Janko Bracic Jan 23 '15 at 17:27
  • @JankoBracic i don't understand . Can you please explain – tomb_raider Jan 23 '15 at 18:03
  • If you typed the question correctly I'd agree with you and say that this point fails to be a local minima. And you almost clearly stated the reason of that. – Evgeny Jan 23 '15 at 18:35
  • @Evgeny i have typed question correctly ,but ì am not sure whether my reasoning is correct\adequate – tomb_raider Jan 23 '15 at 19:01
  • If you know that the value of function at point $(0,0)$ is zero and you can show that in any neighbourhood there are points with function values of different signs -- then your reasoning is of course correct and adequate. – Evgeny Jan 23 '15 at 19:06

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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.

Christian Blatter
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