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

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

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

domain

f (x) = f (x) = 1 - \frac{9}{x ^{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 = 0

Combine real regions and undefined points for final function domain

x \leq - 3 or x \geq 3

3 + \frac{9}{x}

f (x) = - \frac{1}{6} x^{2} + 100 x

\frac{x + 5}{x ^{2} + 3 x - 18}

2^{\frac{1}{x - 1}}

∣ x + 1 ∣ + ∣ x - 1 ∣

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