# How do you find the domain for f(x) = 1/(sqrt(3-2x))?

Jun 4, 2016

Whenever you are dealing with domain/range of a function that involves a radical, you need to set the radical to the following: $x - b \ge 0$, since a radical is undefined in the real number system if $x < 0$. Once you have set it in the way explained above you solve the resulting inequality.
$\sqrt{3 - 2 x} \ge 0$

${\left(\sqrt{3 - 2 x}\right)}^{2} \ge {0}^{2}$

$3 - 2 x \ge 0$

$3 \ge 2 x$

$\frac{3}{2} \ge x$

Interestingly, this is also a rational function. Therefore, we must determine any vertical asymptotes. Vertical asymptotes in a rational function can be found by setting the denominator to 0 and solving for x. Doing this:

$\sqrt{3 - 2 x} = 0$

${\left(\sqrt{3 - 2 x}\right)}^{2} = {0}^{2}$

$3 - 2 x = 0$

$x = \frac{3}{2}$

Therefore, $x \ne \frac{3}{2}$ in this function.

In the domain statement, we can say that $x < \frac{3}{2}$, because we had no choice but to remove the $x = \frac{3}{2}$, since this was the equation of the vertical asymptote.

Hopefully this makes sense!