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

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

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

domain

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

x < - 3 or x > 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 = 3, x = - 3

Combine real regions and undefined points for final function domain

x < - 3 or x > 3

y = \frac{1}{2} x^{2} + 2

f (x) = 7 - ∣ x - 5 ∣

g (x) = ln (3 - x)

f (x) = \frac{4 x ^{2} - 4 x - 3}{x ^{2} + 2 x - 15}

f (x) = x + 4 + 1

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