Correct Answer: (C) To determine the domain of the composite function $h(x) = f(g(x))$, we must consider two fundamental conditions: 1. The inner function, $g(x)$, must be defined for the given $x$ values. 2. The output of the inner function, $g(x)$, must be within the domain of the outer function, $f(x)$. **Step 1: Determine the domain of $g(x)$.** The function $g(x) = \frac{1}{x-5}$ is a rational function. A rational function is undefined when its denominator is zero. Therefore, we must have $x-5 \neq 0$, which implies $x \neq 5$. The domain of $g(x)$ is all real numbers except $5$, i.e., $x \in (-\infty, 5) \cup (5, \infty)$. **Step 2: Determine the domain of $f(x)$.** The function $f(x) = \sqrt{x-2}$ is a square root function. For the square root of a real number to be defined as a real number, the expression under the square root must be non-negative. Therefore, we must have $x-2 \ge 0$, which implies $x \ge 2$. The domain of $f(x)$ is $x \in [2, \infty)$. **Step 3: Apply the domain restriction of $f(x)$ to $g(x)$.** For $h(x) = f(g(x))$ to be defined, the value of $g(x)$ must fall within the domain of $f(x)$. From Step 2, this means $g(x) \ge 2$. Substitute the expression for $g(x)$ into this inequality: $\frac{1}{x-5} \ge 2$ To solve this inequality, we must be careful with the denominator $(x-5)$, as its sign affects the direction of the inequality when multiplying. We consider two cases: **Case A: $x-5 > 0$ (i.e., $x > 5$)** If $x-5$ is positive, we can multiply both sides of the inequality by $(x-5)$ without reversing the inequality sign: $1 \ge 2(x-5)$ $1 \ge 2x - 10$ $11 \ge 2x$ $x \le \frac{11}{2}$ $x \le 5.5$ Combining this with the condition for Case A ($x > 5$), the valid range for $x$ in this case is $5 < x \le 5.5$. **Case B: $x-5 < 0$ (i.e., $x < 5$)** If $x-5$ is negative, we must reverse the inequality sign when multiplying both sides by $(x-5)$: $1 \le 2(x-5)$ $1 \le 2x - 10$ $11 \le 2x$ $x \ge \frac{11}{2}$ $x \ge 5.5$ Combining this with the condition for Case B ($x < 5$), we are looking for values of $x$ such that $x < 5$ AND $x \ge 5.5$. There are no such real numbers, so there are no solutions in this case. **Step 4: Combine all valid conditions.** From Step 1, we know that $x \neq 5$. From Step 3, the only valid range for $x$ that satisfies $g(x) \ge 2$ is $5 < x \le 5.5$. Both conditions are simultaneously satisfied when $x$ is in the interval $(5, 5.5]$. Therefore, the domain of $h(x) = f(g(x))$ is $(5, 5.5]$. **Why other options are incorrect:** * **(A) $(5, \infty)$:** This interval satisfies $x \neq 5$ and $x > 5$, but it does not impose the necessary upper bound $x \le 5.5$ that arises from the condition $g(x) \ge 2$. * **(B) $(-\infty, 5.5]$:** This interval incorrectly includes values where $g(x)$ is undefined (i.e., $x=5$) and values where $g(x)$ is defined but $g(x) < 2$ (e.g., $x=4$, where $g(4) = -1$, which is not in the domain of $f(x)$). * **(D) $[2, 5.5]$:** This interval also incorrectly includes $x=5$, making $g(x)$ undefined. Furthermore, it includes values like $x=2$ or $x=4$ for which $g(x)$ is defined but $g(x) < 2$ (e.g., $g(2) = \frac{1}{2-5} = -\frac{1}{3}$, which is not $\ge 2$).