The correct answer is **C)**. To determine the domain of the function $f(x) = \frac{\sqrt{x^2 - 5x + 6}}{|x - 4| - 1}$, we must satisfy two main conditions: 1. The expression under the square root must be non-negative. 2. The denominator of the fraction must not be equal to zero. **Condition 1: Expression under the square root must be non-negative** We require $x^2 - 5x + 6 \ge 0$. Factorizing the quadratic expression: $(x - 2)(x - 3) \ge 0$. To satisfy this inequality, both factors must have the same sign (both non-negative or both non-positive). This occurs in the following intervals: * If $x - 2 \ge 0$ AND $x - 3 \ge 0$, then $x \ge 2$ AND $x \ge 3$. This implies $x \ge 3$. * If $x - 2 \le 0$ AND $x - 3 \le 0$, then $x \le 2$ AND $x \le 3$. This implies $x \le 2$. So, the solution to $x^2 - 5x + 6 \ge 0$ is $x \in (-\infty, 2] \cup [3, \infty)$. Let's call this Domain 1 ($D_1$). **Condition 2: Denominator must not be zero** We require $|x - 4| - 1 \neq 0$. This implies $|x - 4| \neq 1$. From the definition of absolute value, this means: * $x - 4 \neq 1 \Rightarrow x \neq 5$ * $x - 4 \neq -1 \Rightarrow x \neq 3$ So, from the denominator, $x$ cannot be $3$ or $5$. Let's call this Domain 2 ($D_2$), which is $x \in \mathbb{R} \setminus \{3, 5\}$. **Combining the conditions** The domain of $f(x)$ is the intersection of $D_1$ and $D_2$. Domain of $f(x) = ( (-\infty, 2] \cup [3, \infty) ) \cap ( \mathbb{R} \setminus \{3, 5\} )$. We need to exclude $x=3$ and $x=5$ from $D_1$. 1. Excluding $x=3$: The interval $[3, \infty)$ includes $x=3$. Removing $x=3$ from this interval gives $(3, \infty)$. 2. Excluding $x=5$: The value $x=5$ is part of the interval $(3, \infty)$. Removing $x=5$ from $(3, \infty)$ s it into two parts: $(3, 5)$ and $(5, \infty)$. Therefore, the overall domain becomes: $(-\infty, 2] \cup (3, 5) \cup (5, \infty)$. This matches option C. **Why other options are incorrect:** A) $(-\infty, 2] \cup [3, \infty)$: This is $D_1$ but does not account for the denominator being zero at $x=3$ and $x=5$. B) $(-\infty, 2] \cup (3, \infty)$: This correctly excludes $x=3$ from the second interval but fails to exclude $x=5$, which makes the denominator zero. D) $(-\infty, 2] \cup [3, 5) \cup (5, \infty)$: This incorrectly includes $x=3$ in the domain, which would make the denominator zero.