The correct answer is **B**. To determine the domain of the function $f(x) = \sqrt{\frac{x-2}{x+3}}$, we must ensure two conditions are met: 1. The expression inside the square root must be non-negative: $\frac{x-2}{x+3} \ge 0$. 2. The denominator of the fraction cannot be zero: $x+3 \ne 0$, which implies $x \ne -3$. Let's analyze the inequality $\frac{x-2}{x+3} \ge 0$. We can use a sign analysis by identifying the critical points, which are the values of $x$ that make the numerator or denominator zero. The critical points are $x=2$ (from $x-2=0$) and $x=-3$ (from $x+3=0$). These points divide the number line into three intervals: Interval 1: $x < -3$ Let's pick a test value, for example, $x=-4$. Numerator: $(-4)-2 = -6$ (negative) Denominator: $(-4)+3 = -1$ (negative) Fraction: $\frac{\text{negative}}{\text{negative}} = \text{positive}$. So, $\frac{x-2}{x+3} > 0$ for $x < -3$. This interval $(-\infty, -3)$ is part of the domain. Interval 2: $-3 < x < 2$ Let's pick a test value, for example, $x=0$. Numerator: $(0)-2 = -2$ (negative) Denominator: $(0)+3 = 3$ (positive) Fraction: $\frac{\text{negative}}{\text{positive}} = \text{negative}$. So, $\frac{x-2}{x+3} < 0$ for $-3 < x < 2$. This interval is not part of the domain. Interval 3: $x > 2$ Let's pick a test value, for example, $x=3$. Numerator: $(3)-2 = 1$ (positive) Denominator: $(3)+3 = 6$ (positive) Fraction: $\frac{\text{positive}}{\text{positive}} = \text{positive}$. So, $\frac{x-2}{x+3} > 0$ for $x > 2$. This interval $(2, \infty)$ is part of the domain. Now, we also need to consider the case where $\frac{x-2}{x+3} = 0$. This occurs when the numerator is zero, i.e., $x-2=0$, which means $x=2$. Since $x=2$ does not make the denominator zero, $x=2$ is included in the domain. Combining the intervals where $\frac{x-2}{x+3} \ge 0$, we get $(-\infty, -3) \cup [2, \infty)$. Finally, we must also ensure that $x \ne -3$. Our current interval $(-\infty, -3)$ already excludes $-3$ (it uses a parenthesis), and $[2, \infty)$ does not include $-3$. Thus, the combined domain satisfies all conditions. The domain of $f(x)$ is $(-\infty, -3) \cup [2, \infty)$. **Why other options are incorrect:** A. $(-\infty, -3) \cup (2, \infty)$: This option incorrectly excludes $x=2$. At $x=2$, the function is defined as $f(2) = \sqrt{\frac{2-2}{2+3}} = \sqrt{0} = 0$, so $x=2$ should be included. C. $(-3, 2]$: This option incorrectly includes values where the expression inside the square root is negative and excludes valid regions. D. $[-3, 2]$: This option incorrectly includes $x=-3$ (where the denominator is zero) and values where the expression inside the square root is negative.