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

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

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

domain

f (x) = 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) = - x^{3} + 75 x

f (x) = 2 x + 4

f (x) = \frac{6}{x + 6}

f (x) = \frac{1}{x ^{4} - 65 x ^{2} + 64}

f (x) = - x - 14

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