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

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

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

domain

y = \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) = 60 x^{3} - 120 x

y - 5 = 0

(- 9, 10), (- 7, - 1)

f (x) = \frac{x ^{2} + 5 x + 6}{x + 4}

f (x) = lo g_{5} (x) + 2

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