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

1 Answer
Sep 27, 2014

The natural domain of #f# is #(-infty,3/2)#.

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 < 3/2#.

Hence, the natural domain is #(-infty, 3/2)#.