## College Algebra (6th Edition)

$(-∞, -1) ∪ [1, ∞)$
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: $(-∞, -1) ∪ [1, ∞)$ .