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

Sep 27, 2014

The natural domain of $f$ is $\left(- \infty , \frac{3}{2}\right)$.

Let us look at some details.

There are two rules you need to keep in mind when it comes to finding the natural domain of a function. Here are the rules:

Rule 1: Nonzero in the denominator
Rule 2: Nonnegative in the square-root (eventh-root)

By Rule 1,
sqrt{3-2x} ne 0 Rightarrow 3-2x ne 0 Rightarrow -2x ne -3 Rightarrow x ne 3/2
By Rule 2,
3-2x ge0 Rightarrow -2x ge -3 Rightarrow x le 3/2

By combining the two conditions above, $x < \frac{3}{2}$.

Hence, the natural domain is $\left(- \infty , \frac{3}{2}\right)$.