Answer
$(-∞, \frac{1}{2}) ∪ [1, ∞)$
Work Step by Step
Consider the function as follows:
$f (x) =√(x/(2x-1)-1)$
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$(x/(2x-1)-1)≥0$ 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:
$f (x)=\frac{x}{2x-1}-1≥0$
$f (x) =\frac{x-1(2x-1)}{2x-1}≥0$
$f (x) =\frac{-x+1}{2x-1}≥0$
The solution set of the above inequality, which is also the domain, is given in interval form as follows:
$(-∞, \frac{1}{2}) ∪ [1, ∞)$
Conclusion: The domain of the given function is given by the interval
$(-∞, \frac{1}{2}) ∪ [1, ∞)$.
