College Algebra, Chapter 3, 3.1, Section 3.1, Problem 58

Determine the domain of the function $\displaystyle g(x) = \frac{\sqrt{x}}{2x^2 x -1}$
The function is not defined when the radicand is a negative value and when the denominator is 0. Since,
$\displaystyle g(x) = \frac{\sqrt{x}}{2x^2 + x -1} = \frac{\sqrt{x}}{(2x-1)(x+1)}$
Then,

$
\begin{equation}
\begin{aligned}
2x^2 + x - 1 &> 0\\
\\
(2x-1)(x+1) &> 0
\end{aligned}
\end{equation}
$

We have,

$
\begin{equation}
\begin{aligned}
2x -1 &> 0 &&\text{and}& x +1 &> 0 \\
\\
x &> \frac{1}{2} &&\text{and}& x &> -1 && \text{(However, negative values are not defined in square root)}
\end{aligned}
\end{equation}
$


So, the domain is...
$\displaystyle \left[ 0, \frac{1}{2} \right)\bigcup \left( \frac{1}{2}, \infty \right)$