# Preimage of $f:(x,y,z)\mapsto x^2+y^2+z^2+2x+5$

I have the function $f:\Bbb R^3 \to \Bbb R$, $f:(x,y,z)\mapsto x^2+y^2+z^2+2x+5$

What does some $A\in\Bbb R$ look like in the preimage of this function? How do I work that out?

It's strange, apparently this is bijective, but then what about $(x,a,b)$ and $(x,b,a)$ these should map to the same place, so not injective? If it's not bijective, that doesn't matter I guess for finding the preimage, but I can't work it out.

• @user3491648 well i was going to lead on to using the preimage to show my standard basis for $\Bbb R$ has open sets in the preimage hence topological continuity – Calculus Jun 16 '15 at 9:48
• @user3491648 yes dats fin this was the crux of the matter – Calculus Jun 16 '15 at 9:49

Complete the square $$f(x,y,z) \equiv (x+1)^2 + y^2 + z^2 + 4$$ Now find the preimage of some $A \in \Bbb R$ $$(x+1)^2 + y^2 + z^2 + 4 = A \\ (x+1)^2 + y^2 + z^2 = (\sqrt{A - 4})^2$$ So it's a sphere centered on $(-1,0,0)$ with radius $\sqrt{A-4}$.
• What do you call a 3D annulus? A hollow sphere? I imagine the preimage of an open interval in $\Bbb R$ looks like this. I.e. a sphere with thickness of it's surface being $\sqrt{b-4}-\sqrt{a-4}$ (for $(a,b)$) – Calculus Jun 16 '15 at 9:57