What is the domain of that inverse function of $$ f(x)=3+\sqrt{x-2} $$. a) $$ [2, \infty) $$ b) $$ [3, \infty) $$ d) $$ (-\infty, 3] $$ d) $$ (-\infty, 2] $$

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To find the domain of the inverse function of $$ f(x) = 3 + \sqrt{x - 2} $$, we first need to determine the range of the original function $$ f(x) $$.

1. **Determine the domain of $$ f(x) $$**:
   The expression $$ \sqrt{x - 2} $$ is defined when $$ x - 2 \geq 0 $$, which means $$ x \geq 2 $$. Therefore, the domain of $$ f(x) $$ is:
   $$
   [2, \infty)
   $$

2. **Calculate the range of $$ f(x) $$**:
   - When $$ x = 2 $$:
     $$
     f(2) = 3 + \sqrt{2 - 2} = 3 + 0 = 3
     $$
   - As $$ x $$ approaches infinity, $$ \sqrt{x - 2} $$ also approaches infinity, so:
     $$
     f(x) \to \infty \text{ as } x \to \infty
     $$
   Therefore, the range of $$ f(x) $$ is:
   $$
   [3, \infty)
   $$

3. **Determine the domain of the inverse function**:
   The domain of the inverse function $$ f^{-1}(x) $$ is the same as the range of the original function $$ f(x) $$. Thus, the domain of the inverse function is:
   $$
   [3, \infty)
   $$

So, the correct answer is:
$$
\text{b) } [3, \infty)
$$
The domain of the inverse function is $$ [3, \infty) $$.

What is the domain of that inverse function of \( f(x)=3+\sqrt{x-2} \). a) \( [2, \infty) \) b) \( [3, \infty) \) d) \( (-\infty, 3] \) d) \( (-\infty, 2] \)

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