Question

question 1
let
$f(x) = \frac{x - 4}{\sqrt{x + 3}}$
what is the domain of $f(x)$?
$\big -3, \infty \big)$
$\big( -\infty, -3 \big) \cup \big( -3, \infty \big)$
$\big( -\infty, 3 \big) \cup \big( 3, \infty \big)$
$\big( -3, \infty \big)$

Explanation:

Step1: Analyze the square root

For the square root $\sqrt{x + 3}$, the expression inside (radicand) must be non - negative, so $x+3\geq0$. But also, since it's in the denominator, $\sqrt{x + 3}
eq0$. So we have two conditions:

1. The radicand $x + 3>0$ (because if $x + 3 = 0$, the denominator is zero which is undefined) and
2. There are no restrictions from the numerator $x - 4$ (a linear function is defined for all real numbers).

Step2: Solve the inequality

Solve $x+3>0$. Subtract 3 from both sides of the inequality: $x>-3$.

Answer:

$\boldsymbol{(-3,\infty)}$ (corresponding to the option: $(-3,\infty)$)