Originally Posted by

**arbolis** $\displaystyle f(x,y)=x^4+y^4-4xy+1$.

Taking the partial derivatives of $\displaystyle f$ and setting them to $\displaystyle 0$ gives $\displaystyle y^9=y$ and $\displaystyle x^9=x$.

Then my professor states that the critical points of $\displaystyle f$ are $\displaystyle (0,0)$, $\displaystyle (1,1)$ and $\displaystyle (-1,-1)$. Fine, I understand why, but I don't understand why the points $\displaystyle (0,1)$, $\displaystyle (1,0)$, $\displaystyle (-1,0)$, $\displaystyle (0,-1)$, $\displaystyle (-1,1)$ and $\displaystyle (1,-1)$ are not critical points.

How do you know that when you take $\displaystyle y=0$, then you have to chose $\displaystyle x=0$ and not $\displaystyle x=1$ and $\displaystyle x=-1$?

Because I know of an example ( $\displaystyle f(x,y)= x^2+y^2+\frac{1}{x^2y^2}$) which gives $\displaystyle x=\pm \frac{1}{\sqrt{y\sqrt 2}}$ and $\displaystyle y=\pm \frac{1}{\sqrt{x\sqrt 2}}$. It still remains to solve for $\displaystyle x$ and $\displaystyle y$ (I see that $\displaystyle x=\frac{1}{\sqrt {\sqrt 2}}$ and $\displaystyle y=1$ works for example. I've plotted the function and I saw 4 critical points (minimums), instead of 2 as my professor "would" find here.

So what am I overlooking?