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Question:

Let $f : \mathbb{R}^2 \to \mathbb{R}$ be a function defined by $f(x,y) = (y - x^2)(y - ax^3)$ at each $ (x,y) \in \mathbb{R}^2$, where $a \in \mathbb{R}$. Find all values of $ a $ for which $ f$ does not have a relative extremum at $(0,0)$.

I performed the second derivative test and calculated the determinant of the Hessian matrix at $(0,0)$, and it turned out to be zero irrespective of $a$, (I think if it was negative I would say that $(0,0)$ is saddle point so there is no relative extremum). According to Wolfram MathWorld on Second Derivative Test, I understand that I need to perform further tests since the second derivative test is inconclusive when the determinant of the Hessian is zero. Could someone guide me on what these further tests are and how to apply them to this function?

bruno
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If $y=0$ then $f(x,y)=f(x,0)=ax^5$. Consequently, in case that $a\neq 0$, we will be able to find points on the line $y=0$ with negative and positive values, causing the origin to not be an extremum.

The only case remaining is $a=0$, where we have $f(x,y)=(y-x^2)\cdot y$. If $x=0$ and $y\neq 0$ then $f(x,y)=y^2> 0$, but if $y>0$ and $y<x^2$ the function is negative, so again the function cannot have an extremum at $(0,0)$. For this last subcase you can consider the sequence $\left(\dfrac{2}{\sqrt{n}},\dfrac{1}{n}\right)_{n\geq 1}$ as a way of approaching $(0,0)$.

Julio Puerta
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  • I could not get the sequence part – bruno Apr 29 '24 at 18:53
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    @bruno In the points belonging to that sequence we have $y>0$ and $y<x^2$, so $f$ is negative. It is just an example of the second part of the $a=0$ case, as it is not as trivial as the $x=0, y\neq 0$ 'path'. – Julio Puerta Apr 29 '24 at 18:56
  • Thank you for the response! Your answer has been really helpful in understanding the problem. I'm curious to learn about different approaches to this question, particularly those that might involve the use of Hessians or other methods from multivariable calculus. If you have the time, could you possibly elaborate on any alternative methods? I greatly appreciate your insight and assistance! – bruno Apr 29 '24 at 19:37
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    @bruno When the second derivative test fails for multivariate functions one must usually rely on more 'rudimentary' methods, as the one showed above. See, for instance, this and this. – Julio Puerta Apr 29 '24 at 19:41
  • Now we say that for all real values of a, the desired condition holds right? Did I understand correctly? Also what if we have $ax^2$ instead of $ax^3$, then the first argument does not hold. – bruno Apr 29 '24 at 20:55
  • Yes, but that should be a different question. – Julio Puerta Apr 29 '24 at 21:36
  • I posted it with my solution attempt: https://math.stackexchange.com/q/4908480/1287431. – bruno Apr 30 '24 at 20:09