What is the domain of v(x)=(x+1)/(x^2-1) ?

The domain of v(x)=(x+1)/(x^2-1) is x 1

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domain of v(x)=(x+1)/(x^2-1)

domain

v (x) = \frac{x + 1}{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

f (x) = \frac{( x - 6 )}{x + 3}

f (x) = - - 15 x^{2} + 14 x - 3

f (x) = \frac{x - 15}{x + 8}

\frac{12 - x - x ^{2}}{∣ 2 x - 5 ∣}

f (x) = e^{\frac{x ^{2} - 1}{x}}

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