What is the domain of f(x)=sqrt(16+x^2) ?

The domain of f(x)=sqrt(16+x^2) is -infinity <x<infinity

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

domain of f(x)=sqrt(16+x^2)

domain

f (x) = 16 + x^{2}

- \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

(1.9) (1.3)

f (x) = \frac{x}{x ^{2} + 49}

f (x) = \frac{8}{x ^{2} + 64}

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

x^{4} - 2 x^{2}

What is the domain of f(x)=sqrt(16+x^2) ?
The domain of f(x)=sqrt(16+x^2) is -infinity <x<infinity