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

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

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

domain

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

x \leq - 3 or x \geq 3

Interval Notation

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

Solution steps

Find non-negative values for radicals:

x \leq - 3 or x \geq 3

Find undefined (singularity) points:

x = - 1

Combine real regions and undefined points for final function domain

x \leq - 3 or x \geq 3

x^{2} + 6 x - 5

g (x) = \frac{x}{x ^{2} - 4}

f (x) = \frac{2 x + 6}{- x ^{2} - 2 x + 48}

f (x) = \frac{3 x - 5}{4 - 3 x}

f (x) = \frac{x ( x ^{2} - 9 )}{x ^{2} - 4}

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