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 \(x: x \in R\)
waec 2018furthermathematics

The correct answer is D. \(x: x \in R\)

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.

Previous question Next question