# regular surface- solution check

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

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

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

thm 2

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

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

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

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

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

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

Is this solution correct??
Thank you in advance
• Feb 18th 2013, 07:04 AM
xxp9
Re: regular surface- solution check
I didn't check your detailed computation but yes the argument is correct.
• Feb 18th 2013, 07:49 AM
rayman
Re: regular surface- solution check
thank you for your reply.
I would appreciate if you can have a look at my other post where I am trying to show that a function is smooth
http://mathhelpforum.com/calculus/21...-function.html