## Precalculus: Mathematics for Calculus, 7th Edition

$(\frac{1}{2}, +\infty)$
(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: $(\frac{1}{2}, +\infty)$