We are tasked with finding the natural domain of the function:

$f(x) = \sqrt{3 - x} + \frac{\ln(x - 1)}{x - 2}$

Step 1: Analyze the square root function

The square root function $\sqrt{3 - x}$ requires the argument inside the root to be non-negative. So:

$3 - x \geq 0 \implies x \leq 3$

Thus, the values for $x$ must be $x \in (-\infty, 3]$.

Step 2: Analyze the logarithmic function

The logarithmic function $\ln(x - 1)$ is only defined when the argument inside the logarithm is positive, so:

$x - 1 > 0 \implies x > 1$

Thus, for the logarithm, $x \in (1, \infty)$.

Step 3: Analyze the denominator

The term $\frac{\ln(x - 1)}{x - 2}$ includes a denominator $x - 2$. To avoid division by zero, we need to exclude the value where $x = 2$.

Step 4: Combine the conditions

We now combine the restrictions from all three parts:

• From the square root: $x \leq 3$
• From the logarithm: $x > 1$
• Exclude $x = 2$ to avoid division by zero.

Thus, the domain is:

$x \in (1, 2) \cup (2, 3]$

This is the valid range of values for $x$ that satisfy all the conditions.

Step 5: Conclusion

The correct answer from the given options is:

$[1, 2) \cup (2, 3]$

Do you have any further questions or would like more details?

Related Questions:

1. What would the domain be if the logarithmic term were removed?
2. How does the domain change if $\sqrt{3 - x}$ is replaced with $\sqrt{x + 3}$?
3. What happens to the domain if the denominator becomes $x - 3$ instead of $x - 2$?
4. How would the range of the function change based on this domain?
5. Could the function still be defined at $x = 2$ if $\ln(x - 1)$ were replaced by another term?

Tip:

When solving for domains, break the function into parts, analyze restrictions separately, and combine all conditions for a complete solution.