What is the domain of f(x)= 1/(x^2-2)+sqrt(x^2-3) ?

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

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

domain

f (x) = \frac{1}{x ^{2} - 2} + x^{2} - 3

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

f (x) = 15.45 (1.4 6^{x})

(3.1) (4.7)

f (x) = 2^{x} + 6

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

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

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