The 3rd and 4th Critical Point?

I must find and classify all the critical points in the following function: $$f(x,y)= x^2 + y^2 +x^2y +4$$ I have said that $$f_x=2x+2xy=0$$ $$2x = -2xy$$ $$\frac{ 2x}{\ -2x}=y$$ $$y=-1$$ $$f_x = 2y +x^2 =0$$ $$-2 +x^2 = 0$$ $$x=±\sqrt{2}$$

Therefore my critical points are $$(±\sqrt{2}, -1)$$

Is this right? I only have two critical points and both can be classified as saddle points so I am missing a max and a min.

• You are missing the solution x=0 when you divide by $2x$ – max Jul 31 '14 at 17:19
• @RagibZaman Thats not right, $f_y = 2y + x^2$ so $x=y=0$ is indeed a critical point. – Winther Jul 31 '14 at 17:33
• @RagibZaman Hehe. Anyway, you didn't have to delete it. The comments about no global max and min points are useful. – Winther Jul 31 '14 at 17:40
• Yes and no, it depends on the domain you choose. A continious function on a compact (closed and bounded) domain always has min and max points. However $\mathbb{R}$ is not compact so it does not have to have global min/max points. For example, the function $f(x) = x^2$ has a global minimum at $x=0$, but no maximum on $\mathbb{R}$. If you ask for the min/max on a given compact domain, say, $x\in [1,5]$ then $x=1$ is a minimum and $x=5$ is a the maximum. Another example which has no global max or min points on $\mathbb{R}$ is $f(x) = x$ . – Winther Jul 31 '14 at 18:02
• As far as I know there is no other way other than making sure you find all the solutions to $f_x = f_y = 0$. Just be careful by dividing by something that might be zero and you should be OK:) – Winther Jul 31 '14 at 18:08