$f(x,y)=x^6+2x^4-8x^2y^2+2y^4$ Has f a (strict) local minima in (0,0)?

Question

I need some help with the following task:

Find and classificate all critical points of $f(x,y)=x^6+2x^4-8x^2y^2+2y^4$.

Well I found all critical points and could classificate all of them except one. The critical point I have a problem with is (0,0). Since $f_{xx}(0,0)=0$ and $f_{xx}(0,0)f_{yy}(0,0)-f_{xy}(0,0)^2=0$ I can't say whether if f has a (strict) local minima or maxima in this point.

But what we can do. We take $f(0,y)$ and realize that $f(0,y)=2y^4>0$ $\forall y\in\mathbb{R}$, which means no matter how close we choose y to 0, we get f(0,y)>f(0,0)=0. which means that this can't be a local maxima right?

Since we can't really show that f has a local minima in (0,0), I thought I could do this. I show that (0,0) is not a saddle point, thus f would have a local minima in (0,0)...is that correct?

If (0,0) is a saddle point, $\forall \epsilon >0$ there exist $(x_1,y_1);(x_2,y_2)$ such that $|(x_i),(y_i)|<\epsilon$ for i=1,2 and $f(x_1,y_1)<f(0,0)=0<f(x_2,y_2)$. But if we chose $\epsilon$=2, we can take $(x_1,y_1)=(1,0)$ sucht that $|(1,0)|<2$ but $f(1,0)=3$ is not smaller than $f(0,0)$. Thus (0,0) is not a saddle point of f.

Am I allowed to conclude that since (0,0) is a critical point but not a saddle point nor a local maximum, that (0,0) is a local minimum? If yes, how am I supposed to check whether its a strict local minima or not... strict local minima means (if that is the correct english terminology) that there is a $\delta >0$ such that $f(x_0,y_0)<f(x,y) \forall (x,y)$ with $0<|(x,y)-(x_0,y_0)|<\delta$

Is there anyone who could help me out? I would be very grateful.

Answer 1

You have$$f(x,x)=x^6-4x^4=x^4(x^2-4),$$which is smaller than $0$ if $x\in(-2,2)\setminus\{0\}$. But you also have $f(x,2x)=x^6+2x^4$, which is greater than $0$, unless $x=0$. Therefore, $(0,0)$ is a saddle point.

Answer 2

Of course it's not a local minimum. Just take $x = y = \epsilon>0$, where $\epsilon$ is a very small number. Then $f = -4 \epsilon^4 + \epsilon^6 < f(0,0)$.

Answer 3

$f(x,x)=x^6-4x^4$ which is negative in a neighborhood of $0$, but $f(\sqrt y, y)=2y^4-7y^3+2y^2$ is positive for a small $\epsilon>0$, so it's a saddle point

Answer 4

We know now from various answers that the origin is indeed a saddle point.

The claim that the origin is not a saddle point fails because we do not need the function to alternate above and below the proposed critical value at all distances $\epsilon$ from the point in question. We need only to have this alternation at all distances smaller than some nonzero value. To wit, a critical point C is a saddle point if:

$\exists R>0(\forall\epsilon\le R, d(P,C)=\epsilon\not\implies sign(f(P)-f(C))=constant.$