So who's the domain?

The argument of both square roots must be non-negative, meaning that $x\geq{0}$ and $1-\sqrt{x}\geq{0}$ (or $1\geq{\sqrt{x}}$ ).Thus it is required that $1\geq{x}\geq0$ .

I am quite confused of the possible solutions of this specific problem that include the union of positive real numbers and zero the apparent positive real numbers excluding {0,1} the infinite real numbers contained as decimal and two natural numbers and other possible answers, the specific denotation of the third answer is incorrect to the denotation utilized in mathematics; * is it not? *

as domain means the value for which f(x) is not a negative o now you may proceed as others have said

this problem is based on fact that inside the square root we can not have negative argument ,so first thing x>=0 (for inner squareroot)........(1) now for outer square root : 1-x^1/2>=0 1>=x^1/2 since on both side there is positive number so we can square on both sides 1>=x...(2) now taking intersection of both solution set 1&2 0=<x<=1

Solve for 1-x^1/2>=0 & x>=0 as while finding domain of a composite function we go outside to inside.=>x lies in [0,1] &x>=0 which upon intersection gives x in [0,1]

Since domain means the set of possible values of x with which f(x) will be a real no, therefore, $1-\sqrt{x}$ must be a non-negative value Thus, x has to be less than or equal one.. So we only have the interval [0,1] that can fit.