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

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

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

domain

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

- \infty < x < \infty

Interval Notation

(- \infty, \infty)

Solution steps

Find non-negative values for radicals:

- \infty < x < \infty

The function has no undefined points nor domain constraints. Therefore, the domain is

- \infty < x < \infty

f (x) = (\frac{3}{4})^{x}

f (x) = (\frac{1}{5 x + 2})^{3}

f (x) = ln (5 - x) + lo g_{10} (x + 2)

ln (12 - 5 x^{2})

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