What is the domain of f(x)=(sqrt(3x-2))/(x^2-9) ?

The domain of f(x)=(sqrt(3x-2))/(x^2-9) is 2/3 3

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domain of f(x)=(sqrt(3x-2))/(x^2-9)

domain

f (x) = \frac{3 x - 2}{x ^{2} - 9}

\frac{2}{3} \leq x < 3 or x > 3

Interval Notation

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

Solution steps

Find non-negative values for radicals:

x \geq \frac{2}{3}

Find undefined (singularity) points:

x = 3, x = - 3

Combine real regions and undefined points for final function domain

\frac{2}{3} \leq x < 3 or x > 3

8 - x

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

\frac{x ^{2}}{1 + x}

y = \frac{5}{x + 2}

0.5 (x - 6)^{2} - 3

What is the domain of f(x)=(sqrt(3x-2))/(x^2-9) ?
The domain of f(x)=(sqrt(3x-2))/(x^2-9) is 2/3 <= x<3\lor x>3