What is the domain of 9/((x+1)^2-1) ?

The domain of 9/((x+1)^2-1) is x 0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

domain of 9/((x+1)^2-1)

domain

\frac{9}{( 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

y = \frac{7 x + 3}{x ^{2} + 1}

1 - x

f (x) = x^{4} - 16

f (x) = \frac{2 - x}{x + 5}

f (x) = sin (x) + cos (x)

What is the domain of 9/((x+1)^2-1) ?
The domain of 9/((x+1)^2-1) is x<-2\lor-2<x<0\lor x>0