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

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

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domain of g(x)=sqrt(x^2+1)

domain

g (x) = 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

x^{2} + 8 x + 15

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

f (x) = - 16 x

f (x) = 23 sin (0.5 x - 2) + 72

x^{3} + 2

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