Find the domain of \(g(x) = \frac{4x^{2} - 1}{\sqrt{9x^{2} + 1}}\)
A. \({x : x \in R, x = \frac{1}{2}}\) B. \(x: x \in R, x\neq \frac{1}{3}\) C. \(x : x \in R, x = \frac{1}{3}\) D. \(x: x \in R\)
Correct Answer: D
Explanation
The domain of a function refers to the regions where the function is defined or has a value on a particular region.
\(\frac{4x^{2} - 1}{\sqrt{9x^{2} + 1}}\) has a domain defined on all set of real numbers because the function is defined on the set of real numbers when the denominator \(\sqrt{9x^{2} + 1} \geq 0\).
\(\sqrt{9x^{2} + 1} \geq 0 \implies 9x^{2} + 1 \geq 0\) which because of the square sign has a value for all values of x, be it negative or positive.
