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

1 Answer
Noah G
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 >= 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 - 2x) >= 0#

#(sqrt(3 - 2x))^2 >= 0^2#

#3 - 2x >= 0#

#3 >= 2x#

#3/2 >= 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 - 2x) = 0#

#(sqrt(3 - 2x))^2 = 0^2#

#3 - 2x = 0#

#x = 3/2#

Therefore, #x != 3/2# in this function.

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

Hopefully this makes sense!

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