Find the critical points of the function $f(x,y)=(y-x^2)(y-2x^2)$ and examine if they are local extremums.

Question

I have given a function $f(x,y)=(y-x^2)(y-2x^2)$. I have to inspect its critical points and to see if they are actually local extremums. I found the critical points by setting $ \nabla f(x,y)=0$ and I got two critical points at $(0,0)$ and at $(\frac{2}{3},\frac{2}{3})$.

Then I calculated the Hesse-Matrix and I got $ H_{f}(x,y)= \begin{pmatrix} 12x-6y & -6x\\ -6x & 2 \end{pmatrix} $

Therefore at $(0,0)$ : \begin{pmatrix} 0 & 0\\ 0 & 2 \end{pmatrix}

I, then calculated the determinant and the trace since we are in two dimensions and I got $det H_{f}=0$ which means that this matrix is positive semidefinite which ,if I am not wrong, means that we cannot make any statement whether the point $(0,0)$ is a local extremum or a Minimum or Maximum.

So what can I do in this case to further examine if this point is a local extremum, is there any other way to find out.

I would really appreciate some (hopefully not so complicated) method that could help me in such cases.

Thank you in advance,

Annalisa

Answer 1

If $x\ne0$ and $y$ lies between $x^2$ and $2x^2$, then $f(x,y)<0$. But, if $y\ne0$, then $f(0,y)=y^2>0$. Therefore, $(0,0)$ is a saddle point of your function.