What is the domain of f(x)= 1/(\sqrt[5]{x^2-1)} ?

The domain of f(x)= 1/(\sqrt[5]{x^2-1)} is x 1

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domain of f(x)= 1/(\sqrt[5]{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

f (x) = lo g_{\frac{1}{6}} (x)

lo g_{3} (2 x - 1)

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

f (x) = - 4 x + 12 y g (x) = 3 x^{2} - 8

What is the domain of f(x)= 1/(\sqrt[5]{x^2-1)} ?
The domain of f(x)= 1/(\sqrt[5]{x^2-1)} is x<-1\lor-1<x<1\lor x>1