What is the domain of f(x)=\sqrt[3]{1/(x^2-1)} ?

The domain of f(x)=\sqrt[3]{1/(x^2-1)} is x 1

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domain of f(x)=\sqrt[3]{1/(x^2-1)}

domain

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

x + 11 + 5

f (x) = ln (x^{2} - 4) + π, x < - 2

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

f (x) = 1 - x^{2} - 4 x

f (x) = 2 x + 9 x^{- 2}

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