What is the domain of f(x)= 1/(sqrt(x^2+4)) ?

The domain of f(x)= 1/(sqrt(x^2+4)) is -infinity <x<infinity

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domain of f(x)= 1/(sqrt(x^2+4))

domain

f (x) = \frac{1}{x ^{2} + 4}

- \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

y = \frac{x ^{2} - x + 6}{x - 2}

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

f (x) = x + \frac{7}{x}

f (x) = \frac{x ^{2} - 8 x + 16}{x - 4}

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

What is the domain of f(x)= 1/(sqrt(x^2+4)) ?
The domain of f(x)= 1/(sqrt(x^2+4)) is -infinity <x<infinity