Answer
$(\frac{1}{2}, +\infty)$
Work Step by Step
(i) The denominator cannot be equal to zero thus,
$\sqrt{2x-1}\ne 0
\\2x-1 \ne 0^2
\\2x-1 \ne 0
\\2x \ne 1
\\x \ne \frac{1}{2}$
(ii) The radicand (number inside the square root symbol) cannot be negative. Thus,
$2x-1 \ge 0
\\2x \ge 1
\\x \ge \frac{1}{2}$
This means that the value of $x$ has to be greater than $\frac{1}{2}$.
Therefore, the domain of the function is:
$(\frac{1}{2}, +\infty)$
