Find the domain of g(x) = \; {4x² - 1}{{9x² + 1}}

Correct Answer is : x:x∈Rx: x {R}x:x∈R

Find the domain of g(x) = \; {4x² - 1}{{9x² + 1}}

Correct Answer is : x:x∈Rx: x {R}x:x∈R

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WAEC Further Mathematics 2018 Paper

Question : 2

Total: 40

g(x) = \; \frac{4x^{2} - 1}{\sqrt{9x^{2} + 1}}

$\{x : x \in \mathbb{R}, x = \frac{1}{2}\}$
$x: x \in \mathbb{R}, x \neq \frac{1}{3}$
$x : x \in \mathbb{R}, x = \frac{1}{3}$
$x: x \in \mathbb{R}$

Validate

Solution:

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 $\sqrt{9x^{2} + 1} \ge 0$ .

$\sqrt{9x^{2} + 1} \ge 0 \implies 9x^{2} + 1 \ge 0$ which because of the square sign has a value for all values of x, be it negative or positive.

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