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

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

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

domain

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

x < - 1 or - 1 < x < 1 or x > 1

Interval Notation

(- \infty, - 1) \cup (- 1, 1) \cup (1, \infty)

Solution steps

Find undefined (singularity) points:

x = 1, x = - 1

The function domain

x < - 1 or - 1 < x < 1 or x > 1

\frac{x ^{2} + 2 x - 24}{x ^{2} - 2 x - 8}

f (x) = \frac{x ^{2}}{7 - x}

f (x) = 8 + (ln (x^{2} - 25))^{2}

f (x) = x^{- 2}

f (x) = \frac{3}{2} x^{4} - 2 x^{3} - 6 x^{2} + 8

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