# Thread: [SOLVED] Domain of square root

1. ## [SOLVED] Domain of square root

So I have to find f of g which turns out to be

square root of (1 - (square root 1+ x)^2)

I can simplify this to square root of -x I believe.

Original f(x) = square root 1-x^2
Original g(x) = square root 1+x

Domain of f(x) = x less than or equal to 1
Domain of g(x) = x greater or equal to -1
Domain of f of g = x less than or equal to 0

The domain for square root of -x should then be [0,1] right?

2. Originally Posted by thekrown
So I have to find f of g which turns out to be
square root of (1 - (square root 1+ x)^2)

Original f(x) = square root 1-x^2
Original g(x) = square root 1+x
The domain for square root of -x should then be [0,1] right?
Surely this is a counter-example: $\displaystyle f \circ g( - 0.36) = \sqrt {1 - \left( {\sqrt {1 - 0.36} } \right)^2 }$ which is defined.

Here is a post script hint: $\displaystyle \text{dom}_{f \circ g} = \text{dom}_f \cap \text{Rng}_g$

3. I didn't understand that post at all.

4. Originally Posted by thekrown
I didn't understand that post at all.
Then do you understand this probem as as given at all?
You need a live-sit-down with a tutor.
You are simply lacking the very basics of this material.
I truly hope you will try to get the basics.

5. If you're saying the domain is simply the range of it's inverse, I get that but I wanted to try and solve it using the teacher's methods.

Anyways, wish me luck, my test is in a few hours.