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

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

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

domain

f (x) = \frac{9 - x ^{2}}{x + 1}

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

Interval Notation

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) = (- 2.2)

f (x) = \frac{2}{x ^{3} + 27}

x - \frac{2 x - \frac{3}{3 x} - 2}{3 x} 4

y = x - \frac{4}{x}

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

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