What is the domain of f(x)=sqrt(4x^2+1) ?

The domain of f(x)=sqrt(4x^2+1) is -infinity <x<infinity

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domain of f(x)=sqrt(4x^2+1)

domain

f (x) = 4 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

f (x) = 16 - x^{2} + x - 3

f (x) = lo g_{10} (3) (x) + 2

\frac{x ^{2} - 7 x + 10}{x ^{2} - 5 x + 6}

f (x) = - 4 cos (2 x)

f (x) = \frac{5 - x ^{2}}{x ^{2} - 1}

What is the domain of f(x)=sqrt(4x^2+1) ?
The domain of f(x)=sqrt(4x^2+1) is -infinity <x<infinity