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

$3-x>0$

$x<3$

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