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

The domain of (sqrt(9-x^2))/(x+1) is -3<= x<-1\lor-1<x<= 3

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

domain

\frac{9 - x ^{2}}{x + 1}

- 3 \leq x < - 1 or - 1 < x \leq 3

Interval Notation

[- 3, - 1) \cup (- 1, 3]

Solution steps

Find non-negative values for radicals:

- 3 \leq x \leq 3

Find undefined (singularity) points:

x = - 1

Combine real regions and undefined points for final function domain

- 3 \leq x < - 1 or - 1 < x \leq 3

f (x) = - \frac{( 1 )}{2 7 - x}

x + 8 - x

x^{2} - 1

f (x) = \frac{x}{x ^{2} - 16}

\frac{1}{x - 3}

What is the domain of (sqrt(9-x^2))/(x+1) ?
The domain of (sqrt(9-x^2))/(x+1) is -3<= x<-1\lor-1<x<= 3