# How do you find the domain of g(x) = 1/(3-x)^(1/2)?

Jul 4, 2017

$x < 3$

#### Explanation:

Since $f {\left(x\right)}^{\frac{1}{2}} = \sqrt{f} \left(x\right)$, we get

1/(3-x)^(1/2)=1/(sqrt(3-x)

Then the domain is:

$3 - x \ge 0 \mathmr{and} 3 - x \ne 0$

that's

$3 - x > 0$

$x < 3$