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

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Answer 1 · 2015-09-24T18:18:40

Refer to 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 $(0,+oo)$

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