Question

How do you find domain and range for `f(x) = 2 / sqrt(3x-2)`?

Answer 1

Refer to explanation

Explanation:

In order for the function to have meaning in the set of real numbers
we must have

`3x-2>0=>3x>2=>x>2/3`

hence, the domain is `(2/3,+oo)`

for the range we set `y=f(x)` and we have

`y=2/sqrt(3x-2)=>(sqrt(3x-2))=2/y=>2/y>0=>y>0`

So the range is `(0,+oo)`