Use the given function $$ f $$ to answer parts (a) through $$ (f) $$ below. $$ f(x)=\ln (x-3) $$ (Type your answer in interval notation.) Determine the vertical asymptote of $$ f $$. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. (Simplify your answer.) B. There is no vertical asymptote. (d) Find $$ f^{-1} $$, the inverse of $$ f $$. $$ f^{-1}(x)=e^{x}+3 $$ (Simplify your answer.) (e) Find the domain and range of $$ f^{-1} $$. The domain of $$ f^{-1} $$ is ( $$ \left.-\infty, \infty\right) $$. (Type your answer in interval notation.) The range of $$ f^{-1} $$ is (Type your answer in interval notation.)

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1. **Vertical Asymptote**

The function is $$ f(x) = \ln (x-3) $$. The domain of $$ f $$ is determined by the condition
   $$
   x-3>0 \quad \Longrightarrow \quad x>3.
   $$
   The vertical asymptote occurs where the argument of the logarithm is zero, that is when
   $$
   x-3=0 \quad \Longrightarrow \quad x=3.
   $$
   **Answer:** $$ (3, \infty) $$ (with vertical asymptote $$ x=3 $$).

2. **Finding the Inverse**

To find $$ f^{-1}(x) $$:
   
   Start with
   $$
   y=\ln (x-3)
   $$
   Swap $$ x $$ and $$ y $$:
   $$
   x=\ln (y-3)
   $$
   Exponentiate both sides to solve for $$ y $$:
   $$
   e^x=y-3 \quad \Longrightarrow \quad y=e^x+3.
   $$
   **Answer:** $$ f^{-1}(x)=e^{x}+3 $$.

3. **Domain and Range of $$ f^{-1}(x) $$**

The domain of $$ f^{-1} $$ is the range of $$ f $$.

For $$ f(x)=\ln (x-3) $$ with domain $$ (3,\infty) $$, the range is all real numbers:
   $$
   (-\infty,\infty).
   $$
   Thus, the domain of $$ f^{-1} $$ is:
   $$
   (-\infty,\infty).
   $$

Next, the range of $$ f^{-1} $$ is the domain of $$ f $$. Since the domain of $$ f $$ is:
   $$
   (3,\infty),
   $$
   the range of $$ f^{-1}(x)=e^{x}+3 $$ is:
   $$
   (3,\infty).
   $$

**Answer:**  
   Domain of $$ f^{-1} $$: $$ (-\infty,\infty) $$  
   Range of $$ f^{-1} $$: $$ (3,\infty) $$
**Vertical Asymptote:** $$ x = 3 $$ **Inverse Function:** $$ f^{-1}(x) = e^{x} + 3 $$ **Domain of $$ f^{-1} $$:** $$ (-\infty, \infty) $$ **Range of $$ f^{-1} $$:** $$ (3, \infty) $$

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