The correct answer is **D) $(-\infty, 2) \cup (3, 4)$**. To determine the domain of the function $f(x)$, we must satisfy all conditions for which each part of the function is defined. The function $f(x)$ consists of two terms: a logarithmic term and a term involving a square root in the denominator. **Condition 1: Domain of the logarithmic term** For the term $\log_2(x^2 - 5x + 6)$ to be defined, its argument must be strictly positive. $x^2 - 5x + 6 > 0$ First, we find the roots of the quadratic equation $x^2 - 5x + 6 = 0$. Factoring the quadratic, we get: $(x-2)(x-3) = 0$ So, the roots are $x=2$ and $x=3$. These are the critical points. To satisfy $(x-2)(x-3) > 0$, we test intervals: * For $x 0$. This interval is part of the domain. * For $2 < x < 3$ (e.g., $x=2.5$): $(2.5-2)(2.5-3) = (0.5)(-0.5) = -0.25 < 0$. This interval is NOT part of the domain. * For $x > 3$ (e.g., $x=4$): $(4-2)(4-3) = (2)(1) = 2 > 0$. This interval is part of the domain. Thus, the domain for the first term, $D_1$, is $(-\infty, 2) \cup (3, \infty)$. **Condition 2: Domain of the term with the square root in the denominator** For the term $\frac{1}{\sqrt{4-x}}$ to be defined, two conditions must be met: 1. The expression under the square root must be non-negative: $4-x \ge 0$. 2. The denominator cannot be zero, which means $\sqrt{4-x} \ne 0$, implying $4-x \ne 0$. Combining these two, the expression under the square root must be strictly positive: $4-x > 0$ Solving for $x$: $4 > x$, or $x < 4$. Thus, the domain for the second term, $D_2$, is $(-\infty, 4)$. **Combined Domain** For the entire function $f(x)$ to be defined, $x$ must satisfy both conditions. Therefore, the domain of $f(x)$ is the intersection of $D_1$ and $D_2$. Domain $f(x) = D_1 \cap D_2 = \left( (-\infty, 2) \cup (3, \infty) \right) \cap (-\infty, 4)$ Let's find the intersection: * The interval $(-\infty, 2)$ intersects with $(-\infty, 4)$ to give $(-\infty, 2)$. * The interval $(3, \infty)$ intersects with $(-\infty, 4)$ to give $(3, 4)$. Combining these intersecting parts, the domain of $f(x)$ is $(-\infty, 2) \cup (3, 4)$. Final Answer is **D**.