Inverses, Domains and Ranges!?

Question

My last question i asked got a very nice response that helped me alot with the other questions like it, for this one Domains and ranges have always eluded me, just something I haven't been able to figure out, this here is a very basic one but i still can't 100% figure it out! On a graphing calc I can see what sqrt(x) is, but how do I figure out the domain and range of that and the inverse of it? Any help would be mucho appreciated!

The full question is:

let f(x) = sqrt(x)
Sketch F and F-1(inverse) on the same coordinate plane
Stage the Df(Domain) and Rf(range)

Answer 1

sqrt is defined only for positive numers and 0
You can't have √(-1), well at least in real numbers
So the domain is [0,∞)
the Range in again all positive numbers and 0, since √(x^2) = |x| .

to find inverse, use again that
√(x^2) = x and (√x)^2 = x for any positive x.
to conlude that the inverse of √x is x^2.
Note that the inverse is defined on the interval [0,∞) too and not on real line.
Even if x^2 is defined on all real line, it is injective, therefore it has an inverse, only on [0, ∞)

Answer 2

Sketching these is impossible with this text editor.
f(x) = √(x) is a two-valued relationship. It is a parabola with vertex at (0,0) and opening along the +x axis.
The domain is x ≥ 0
The range is all y

To obtain the inverse, swap variables:
x = √f^-1(x)
f^-1(x) = x^2
The domain is all x
The range is y ≥ 0