What is the domain of f(x)= 1/(sqrt(3x-2)) ?

The domain of f(x)= 1/(sqrt(3x-2)) is x> 2/3

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domain of f(x)= 1/(sqrt(3x-2))

domain

f (x) = \frac{1}{3 x - 2}

x > \frac{2}{3}

Interval Notation

(\frac{2}{3}, \infty)

Solution steps

Find non-negative values for radicals:

x \geq \frac{2}{3}

Find undefined (singularity) points:

x = \frac{2}{3}

Combine real regions and undefined points for final function domain

x > \frac{2}{3}

f (x) = \frac{x}{3 x - 9}

f (x) = \frac{8}{x - 3}

\frac{x ^{2} - x - 6}{x ^{2} - 2 x - 3}

f (x) = \frac{3 - x}{x ^{2} - 2 x}

\frac{1}{x ln ( x ^{2} - 1 )}

What is the domain of f(x)= 1/(sqrt(3x-2)) ?
The domain of f(x)= 1/(sqrt(3x-2)) is x> 2/3