What is the domain of (x^2-4x+3)/(x^2-1) ?

The domain of (x^2-4x+3)/(x^2-1) is x 1

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domain of (x^2-4x+3)/(x^2-1)

domain

\frac{x ^{2} - 4 x + 3}{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 ^{2} - 16}{x ^{2} - x - 6}

arccos (3 x^{2} - 2)

f (x) = e^{0.5 x} + 4 e^{- 0.5 x}

\frac{x ^{2} - 8 x + 15}{x ^{2} - 7 x + 12}

f (x) = - 4 - x

What is the domain of (x^2-4x+3)/(x^2-1) ?
The domain of (x^2-4x+3)/(x^2-1) is x<-1\lor-1<x<1\lor x>1