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

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

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

domain

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

x < - 3 or x > 3

Interval Notation

(- \infty, - 3) \cup (3, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - 3 or x \geq 3

Find undefined (singularity) points:

x = 3, x = - 3

Combine real regions and undefined points for final function domain

x < - 3 or x > 3

- sin (x) - 9

\frac{4}{x - 3}

5 x + 2 y = 6

3 x^{2} - 4

3 - 2 x^{3}

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