College Algebra 7th Edition

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