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I've found the critical points:

$$f(x,y)={ x }^{ 3 }-3x-{ y }^{ 3 }+12y+2\\$$ $$\\ { f }_{ x }=0\\ { x }^{ 2 }-1=0\\ x=\pm 1\\ $$ $$\\ { f }_{ y }=0\\ { y }^{ 2 }-4=0\\ y=\pm 2$$

But I don't know what the y values are for x=±1 and what the x values are for y=±2. Any help? I've tried subbing in x=±1 in the original f(x,y) but I don't know what the y values are to sub in.

I keep getting: This post does not meet our quality standards. What am I supposed to fix? The character count?

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  • $\begingroup$ As you computed, the critial points are $(-1,-2), (-1,2), (1,-2), (1,2)$. $\endgroup$ – Git Gud Apr 15 '13 at 20:31
  • $\begingroup$ Oh. Thank you. I didn't realize it was simply all the combinations. $\endgroup$ – user72708 Apr 15 '13 at 20:32
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    $\begingroup$ Remember that the critical points are the points $(x,y)$ such that $f_x(x,y)=0 \land f_y(x,y)=0$. And you got $$\begin{align} f_x(x,y)=0 \land f_y(x,y)=0 &\iff x=\pm 1 \land y=\pm 2\\ &\iff (x=-1 \lor x=1) \land (y=-2 \lor y=2)\\ &\iff (x=-1 \land y=-2) \lor (x=-1 \land y=2) \lor (x=1 \land y=-2) \lor (x=1 \land y=2) \end{align}.$$ That's why you get all the combinations. $\endgroup$ – Git Gud Apr 15 '13 at 20:38
  • $\begingroup$ Funny! The equivalences above start with $=0$, but I just checked and I really did type f_x(x,y)=0 $\endgroup$ – Git Gud Apr 15 '13 at 20:43
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You found that $x$ is $-1$ or $1$ and $y$ is $2$ or $-2$. Therefore there are four critical points: $(-1,2), (-1,-2), (1, 2), (1, -2)$. That's it. When you set $f_x = 0$ and $f_y = 0$ you got equations with only $x$ or only $y$.

I don't see anything wrong with your question. People post much worse questions here all the time.

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