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

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

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

domain

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

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

f (x) = \frac{12 x + 4 x ^{2}}{x + 3}

f (x) = 13 - 13 x

\frac{x ^{2} + 4}{x ^{2} - 9}

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