Question

Let f : R^2 --> R be defined by (x, y) -> x^2 + y^2. Find a point (x, y) for which f(x,y) = 4.
Sketch the preimage f^-1(4) in the (x,y)-plane.
The set U = {f^-1(x) | x e R, x=> 0} consisting of all preimage sets is a partition of R2. Find
a subset of R^2 that intersects each element of this partition in exactly one point.

Answer 1

First question:
It is equivalent to : x^2+y^2-4 = 0
wich is circle. One solution would be f(2,0)
Second question :
f^-1(4) is a circle of center 0,0, radius 2.
Third question:
one obvious would be
|x|^2+y^2 = 4, but I'm not sure.

Answer 2

thanks. can you explain where you got f(2,0) and also why f^-1(4) is just plotting the circle?

Answer 3

Of course :

f(2,0) = 2² + 0² = 4.

f^-1(4) is all point (x,y) for wich f(x,y)=4, therefore, we have f^-1(4) -> x^2+y^2-4 = 0, wich is the cartesian equation for a circle of radius 4 (and no offset, so the center is the origin of the grid)

Answer 4

thanks