#### Answer

Domain: $[0, \frac{1}{2})\cup (\frac{1}{2},\ \infty)$

#### Work Step by Step

We are given:
$g(x)= \frac{\sqrt{x}}{2x^{2}+x-1}$
The domain of a function consists of all values that $x$ is allowed to have. In this case, we can not take the square root of a negative number, so:
$x\geq0$
We also can not have division by $0$:
$2x^{2}+x-1\neq 0$
$(2x-1)(x+1)\neq 0$
$ 2x-1\neq 0$ or $x+1 \neq 0$
$ x\neq\frac{1}{2}$ or $x\neq-1$
We can ignore the $x\neq -1$ solution, since we already have $x\geq 0$.
Therefore, the domain is: $[0, \frac{1}{2})\cup (\frac{1}{2},\ \infty)$