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

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

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

domain

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

- \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

\frac{1}{x ^{2} + 4}

f (x) = \frac{x}{5}

f (x) = arccos (2 x)

5 x + 8 y = 16

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

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