Extrema when Hessian Matrix is $0$ at $(0,2)$

Question

Let $f: \mathbb R^{2}\to \mathbb R$, $f(x,y)=xy^2-4xy+x^4$

Find all critical points of $f$ and determine whether or not they are extrema and their type:

Setting $0=\nabla f(x,y)$

I get the following equations: $y^2-4y+x^4=0$ and $2xy-4x=0\Rightarrow x(y-2)=0$

So either $x=0$ or $y=2$. If $x=0$ , we have $y = 2$ or $y = 0$. Similarly, if $y=2$, we have $x=1$

Therefore critical points $(0,0),(0,2),(1,2)$.

Looking at the Hessian Matrix we have:

$(D^2f)(x,y)=\begin{bmatrix} 12x^2 & 2y-4\\ 2y-4 & 2x \end{bmatrix}$

I immediately get the answers at $(1,2)$ and $(0,0)$. However at $(D^2f)(0,2)=0$

So I took the function under greater consideration and evaluated the behavior of the function in each "direction", namely,

$f(0,b)=0$ for all $b \in \mathbb R$, while for all $a \in \mathbb R$

$f(a,2)=a^4-4a$

$f'(a,2)=4a^3-4$

and $f'(0,2)=-4\neq0$, so therefore $(0,2)$ is not an extremum on $f$, and not even a saddle point. Is this correct?

Any smoother way to prove this?

Answer 1

You are wrong about the critical points. The point $(0,2)$ is not one of them, since $\frac{\partial f}{\partial x}(0,2)=-4$. The critical points of $f$ are $(0,0)$, $(1,2)$, and $(0,4)$. The Hessean matrices at these points are $\left(\begin{smallmatrix}0&-4\\-4&0\end{smallmatrix}\right)$, $\left(\begin{smallmatrix}12&0\\0&2\end{smallmatrix}\right)$, and $\left(\begin{smallmatrix}0&4\\4&0\end{smallmatrix}\right)$ respectively.

Answer 2

I agree with Jose. As a side note, if you did have a Hessian with zero gradient and zero determinant at a point $(x,y)$, a general strategy you could use to determine whether or not $(x,y)$ was an extrema would be to resort back to the definition of an extrema by looking at the difference $f(x + t_1, y + t_2) - f(x, y)$ for $t_1,t_2 \in \mathbb{R}$. If you could show that this difference had a definite sign for $t_1,t_2$ sufficiently small (i.e., $f(x + t_1, y + t_2) \geq f(x,y)$ or $f(x + t_1, y + t_2) \leq f(x,y)$ for $t_1,t_2$ sufficiently small), then the point would be an extrema. On the other hand, if this condition did not hold (say, your critical point is $(0,0)$ and you find a direction $y = g(x)$ such that $f(x, g(x)) = x^3$ [has neither a min nor a max at $x = 0$]), then the point would not be an extrema.