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

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Answer 1 · 2017-07-04T06:39:31

#x<3#

Since #f(x)^(1/2)=sqrtf(x)#, we get

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

Then the domain is:

#3-x>=0 and 3-x!=0#

that's

#3-x>0#

#x<3#