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

The domain of 4/((x+1)^2-1) is x 0

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

domain

\frac{4}{( x + 1 ) ^{2} - 1}

x < - 2 or - 2 < x < 0 or x > 0

Interval Notation

(- \infty, - 2) \cup (- 2, 0) \cup (0, \infty)

Solution steps

Find undefined (singularity) points:

x = 0, x = - 2

The function domain

x < - 2 or - 2 < x < 0 or x > 0

f (x) = \frac{( x ^{2} + 25 )}{x}

y = 4 x + 3

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

f (x) = lo g_{\frac{1}{2}} (x)

arctan (e^{x})

What is the domain of 4/((x+1)^2-1) ?
The domain of 4/((x+1)^2-1) is x<-2\lor-2<x<0\lor x>0