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

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

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

domain

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

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 = 2, x = - 2

Combine real regions and undefined points for final function domain

x \leq - 3 or x \geq 3

- \frac{5 + a}{3}

f (x) = x^{2} + 2 - x

f (x) = \frac{1}{∣ x ^{2} - 4 ∣}

x^{3} + 4 x

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

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