# Thread: Critical points of a function of 2 variables

1. ## Critical points of a function of 2 variables

$f(x,y)=x^4+y^4-4xy+1$.
Taking the partial derivatives of $f$ and setting them to $0$ gives $y^9=y$ and $x^9=x$.
Then my professor states that the critical points of $f$ are $(0,0)$, $(1,1)$ and $(-1,-1)$. Fine, I understand why, but I don't understand why the points $(0,1)$, $(1,0)$, $(-1,0)$, $(0,-1)$, $(-1,1)$ and $(1,-1)$ are not critical points.
How do you know that when you take $y=0$, then you have to chose $x=0$ and not $x=1$ and $x=-1$?

Because I know of an example ( $f(x,y)= x^2+y^2+\frac{1}{x^2y^2}$) which gives $x=\pm \frac{1}{\sqrt{y\sqrt 2}}$ and $y=\pm \frac{1}{\sqrt{x\sqrt 2}}$. It still remains to solve for $x$ and $y$ (I see that $x=\frac{1}{\sqrt {\sqrt 2}}$ and $y=1$ 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?

2. Originally Posted by arbolis
$f(x,y)=x^4+y^4-4xy+1$.
Taking the partial derivatives of $f$ and setting them to $0$ gives $y^9=y$ and $x^9=x$.
Then my professor states that the critical points of $f$ are $(0,0)$, $(1,1)$ and $(-1,-1)$. Fine, I understand why, but I don't understand why the points $(0,1)$, $(1,0)$, $(-1,0)$, $(0,-1)$, $(-1,1)$ and $(1,-1)$ are not critical points.
How do you know that when you take $y=0$, then you have to chose $x=0$ and not $x=1$ and $x=-1$?

Because I know of an example ( $f(x,y)= x^2+y^2+\frac{1}{x^2y^2}$) which gives $x=\pm \frac{1}{\sqrt{y\sqrt 2}}$ and $y=\pm \frac{1}{\sqrt{x\sqrt 2}}$. It still remains to solve for $x$ and $y$ (I see that $x=\frac{1}{\sqrt {\sqrt 2}}$ and $y=1$ 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?
When you calculate the first partials you obtain

$
f_x = 4(x^3-y),\;\;\; f_y = 4(y^3-x).
$

Critical points are points where both vanish. As you can see many of the points you mention don't satisfy these (for example $(0,1)$ ).

3. Originally Posted by Danny
When you calculate the first partials you obtain

$
f_x = 4(x^3-y),\;\;\; f_y = 4(y^3-x).
$

Critical points are points where both vanish. As you can see many of the points you mention don't satisfy these (for example $(0,1)$ ).
Thank you very much for the quick response!
Ok, so I have to test each point? It does not bothers me by the way.