#### Answer

$(-∞, -1) ∪ [1, ∞)$

#### Work Step by Step

Consider the function as follows:
$f (x) =√(2x/(x+1)-1)$
For a function $f$ defined by an expression with variable $x$, the implied domain of $f$ is the set of all real numbers variable $x$ can take such that the expression defining the function is real. The domain can also be given explicitly.
For $f (x) =\frac{2x}{x+1}-1≥0$ to be real, the radicand (expression under the radical) of the square root function must be positive or equal to 0. Thus, the inequality becomes
$f (x) =\frac{2x}{x+1}-1≥0$
$f (x) =\frac{2x-1(x+1)}{x+1}≥0$
$f (x) =\frac{x-1}{x+1}≥0$
Hence, the solution set of the above inequality, which is also the domain, is given in interval form as follows:
$(-∞, -1) ∪ [1, ∞)$ .
Conclusion: The domain of the given function is given by the interval as follows:
$(-∞, -1) ∪ [1, ∞)$ .