To verify if the statement is true or false, let’s analyze the function $f(x) = \sqrt{\ln(x-1)} + 2$ and determine the domain of $f(x)$.

Step 1: Conditions for the domain

1. Logarithmic function condition:

   • The argument of the logarithmic function, $x - 1$, must be positive: $x - 1 > 0 \implies x > 1.$

2. Square root condition:

   • The expression inside the square root, $\ln(x-1)$, must be non-negative: $\ln(x-1) \geq 0 \implies x - 1 \geq 1 \implies x \geq 2.$

Step 2: Combine the conditions

• From the above, the combined condition is $x \geq 2$.

Step 3: Express the domain

• The domain of $f(x)$ is $[2, \infty)$, which matches the statement in the problem.

Conclusion:

The statement is true.

Would you like additional clarification on this, or any detailed explanation about logarithmic or square root functions?

Here are 5 related questions:

1. How do you find the domain of logarithmic functions in general?
2. Why does the square root function require a non-negative input?
3. What is the range of $f(x) = \sqrt{\ln(x-1)} + 2$?
4. Can the domain of $f(x)$ ever include negative numbers in this case?
5. How would the domain change if the function were $f(x) = \ln(\sqrt{x-1})$ instead?

Tip: Always start by analyzing each individual component of a composite function when determining its domain.