# regular surface- solution check

• Feb 17th 2013, 11:24 PM
rayman
regular surface- solution check
For which values of $\displaystyle c\in\mathbb{R}$, is $\displaystyle f^{-1}(c)$ a regular surface , where
$\displaystyle f:\mathbb{R}^3\to \mathbb{R}$ is defined as
$\displaystyle f(x,y,z)=x^2y^2+x^2z^2+y^2z^2-4xyz$

I found 2 theorems I think I should use when solving this problem

thm 1
$\displaystyle c\in f(U)$ is a regular value of the function $\displaystyle f:U\subset\mathbb{R}^3\to \mathbb{R}$ iff the partial derivatives $\displaystyle f_{x},f_{y},f_{z}$ do not disappear instantly for any point of the pre-image $\displaystyle f^{-1}(c)=\{(x,y,z)\in U|f(x,y,z)=c\}$

thm 2

If $\displaystyle f:U\subset\mathbb{R}^3\to \mathbb{R}$is differentiable and $\displaystyle c\in f(U)$ is a regular value , then $\displaystyle f^{-1}(c)$ is a regular surface

So I start off with finding critical points (point where all partial derivatives disappear at the same time)

$\displaystyle \nabla f=(f_{x},f_{y},f_{z})=(2xy^2+2xz^2-4yz,2x^2y+2yz^2-4xz,2x^2z+2y^2z-4xy)$

$\displaystyle \nabla f=0$ I get the following points:

$\displaystyle c_{1}=(-2,0,0)$
$\displaystyle c_{2}=(-1,0,0)$
$\displaystyle c_{3}=(-1,1,-1)$
$\displaystyle c_{4}=(0,0,0)$
$\displaystyle c_{5}=(1,0,0)$

so that means that all point except the above ones are regular points?
the second theorem gives that $\displaystyle f^{-1}(c)$ is a regular surface $\displaystyle \for all c=\{(x,y,z)\in \mathbb{R}^3\setminus \{c_{1},c_{2},c_{3},c_{4},c_{5}\}\}$

Is this solution correct??