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Consider the function:

$f(x,y,z)=xyz^2$

Its gradient is $\nabla f=(yz^2, xz^2, 2xyz)$ then the critical points are all in the sets $\{(x,y,0): x,y\in \mathbb{R}\}, \{(0,0,z): z\in \mathbb{R}\}$.

My question is: The set defined by $xyz^2=0$ is a regular surface? I thinks it is not, but for $f(x,y,z)=0$ there are points such that its differential is surjective.

Thanks!

EQJ
  • 4,369

1 Answers1

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The set defined by $xyz^2 =0$ is the union of three planes:

$$\{ x=0\} \cup \{y=0\} \cup \{z=0\},$$

thus is not regular.