What is the domain of (1/(sqrt(x)))/(x^2-9) ?

The domain of (1/(sqrt(x)))/(x^2-9) is 0 3

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domain of (1/(sqrt(x)))/(x^2-9)

domain

\frac{\frac{1}{x}}{x ^{2} - 9}

0 < x < 3 or x > 3

Interval Notation

(0, 3) \cup (3, \infty)

Solution steps

Find non-negative values for radicals:

x \geq 0

Find undefined (singularity) points:

x = 3, x = - 3, x = 0

Combine real regions and undefined points for final function domain

0 < x < 3 or x > 3

f (x) = 3 x + 5 y = 15

f (x) = \frac{x ^{3}}{2 x ^{2} - 8}

f (x) = \frac{x - 9}{x ^{2} + 18 x + 81}

f (x) = \frac{x - 3}{x ^{2} + 2 x - 15}

What is the domain of (1/(sqrt(x)))/(x^2-9) ?
The domain of (1/(sqrt(x)))/(x^2-9) is 0<x<3\lor x>3