Question

determine the domain of the function.
$f(x) = \frac{\sqrt{x}}{x^2 - 81}$
\\(\circ\\) $x \geq 0$ and $x \
eq 9$.
\\(\circ\\) all real numbers.
\\(\circ\\) $x \
eq -9$
\\(\circ\\) $x \
eq 9$
\\(\circ\\) $x \geq 0$

Explanation:

Step1: Analyze the square root

For the square root $\sqrt{x}$, the radicand must be non - negative, so $x\geq0$.

Step2: Analyze the denominator

The denominator is $x^{2}-81=(x - 9)(x + 9)$. The denominator cannot be zero, so we set $(x - 9)(x + 9)
eq0$. This gives $x
eq9$ and $x
eq - 9$. But from the square root, we already have $x\geq0$, so we only need to consider $x
eq9$ (since $x\geq0$ implies $x
eq - 9$ is automatically satisfied).
Combining the two conditions: $x\geq0$ and $x
eq9$.

Answer:

$x\geq0$ and $x
eq9$ (corresponding to the option: $x \geq 0$ and $x
eq 9$)