What is the domain of y=sqrt(3x-2) ?

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

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domain of y=sqrt(3x-2)

domain

y = 3 x - 2

x \geq \frac{2}{3}

Interval Notation

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

Solution steps

Find non-negative values for radicals:

x \geq \frac{2}{3}

The function domain

x \geq \frac{2}{3}

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

f (x) = 6^{- x}

f (x) = - x^{2} + x + 2

f (x) = lo g_{3} ((1 + y)^{2})

f (x) = \frac{5}{x ^{2} - 16}