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

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

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

domain

g (x) = \frac{1}{x ^{2} - 9}

x < - 3 or - 3 < x < 3 or x > 3

Interval Notation

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

Solution steps

Find undefined (singularity) points:

x = 3, x = - 3

The function domain

x < - 3 or - 3 < x < 3 or x > 3

f (x) = (x - 16)^{2}, [16, \infty]

x^{2} + 8 x + 3

\frac{x}{x - 6}

f (x) = \frac{x - 1}{x + 6}

lo g_{10} (8 - t)

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