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

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

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

domain

f (x) = \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

f (x) = 3 x^{2} ln (\frac{x}{4})

4 x^{2} - 5 x + 7

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

\frac{3}{x + 2}

f (x) = x - 2 + 5

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