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

Refer to explanation

#### Explanation:

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

$3 x - 2 > 0 \implies 3 x > 2 \implies x > \frac{2}{3}$

hence, the domain is $\left(\frac{2}{3} , + \infty\right)$

for the range we set $y = f \left(x\right)$ and we have

$y = \frac{2}{\sqrt{3 x - 2}} \implies \left(\sqrt{3 x - 2}\right) = \frac{2}{y} \implies \frac{2}{y} > 0 \implies y > 0$

So the range is $\left(0 , + \infty\right)$