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

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

domain

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

- \infty < x < \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{( x - 1 )}{x ( x ^{2} - 9 )}

f (x) = - 4 x^{2} + 12

5 t + 6

- \frac{1}{2} x^{2} + 2 x + 3

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

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