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