A)$f: \mathbb{R^3} \rightarrow \mathbb{R}$ with $f(x,y,z)=(x+y+z-1)^2$. Find all the critical values and all $c \in \mathbb{R}$ for which $f^{-1}(c)$ is a smooth surface.

B)Now let $f(x,y,z)=z^2$ prove that $0 \in \mathbb{R}$ is a critical value but $f^{-1}(0)$ is still a surface.


A) the jacobian $Df=2((x+y+z-1),(x+y+z-1),(x+y+z-1))$ so its zero at $(x,y,z)$ s.t $x+y+z=1$ Critical points. So the critical values are all $y$ s.t the set

$f^{-1}(y)=\{(x,y,z) \in \mathbb{R^3}:f((x,y,z))=y\}$ contains critical points. The only $y$ that its pre image contains critical points is the zero since whatever critical point you input into the function its image is zero. So only critical value is $y=0$.

And the $c \in \mathbb{R}$ s.t $f^{-1}(c)$ is a smooth surface are all the regular values of $f$ so its for all $c>0$

B) $Df=(0,0,2z)$ so $f^{-1}(0)$ pre image of it contains $(x,y,0)$ x,y anything so it contains the critical point $(0,0,0)$ so its a critical value. But the set $f^{-1}(0)=\{(x,y,0)\}$ $x$,$y$ $\in \mathbb{R}$ is a surface cause it is easyl parametrized by the function $φ(x,y)=(x,y,0)$ which has its jacobian matrx degree 2. So its asurface parametrization.(is this reasoning incomplete?).

Are the answers correct and complete?

  • 1
    $\begingroup$ Part A looks right to me but I have to admit I really do not know. I have no idea about Part B. One minor note, jacobian is normally spelled with a capital J. $\endgroup$ – Bob Aug 31 '18 at 21:12

Yes, $Df$ is as you say and so the set $\lbrace (x,y,z):x+y+z=1\rbrace$ is the set of critical points. When we plug these into $f$, the value of $f$ is always zero. So $0$ is the only critical value of $f$.

For the second part, I think you can get away with saying that $f^{-1}(0)$ is the $xy-$plane, which is a surface that doesn't need a chart.


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