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

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**(c)**  
The function is given by  
$$
f(x)=e^{x-3}+1.
$$  
Since $$ e^{x-3} > 0 $$ for all $$ x $$, the minimum value of $$ f(x) $$ occurs when $$ e^{x-3} $$ is closest to 0. As $$ x \to -\infty $$, $$ e^{x-3} \to 0 $$. Thus,  
$$
f(x) \to 0+1=1.
$$  
So the range of $$ f $$ is  
$$
(1,\infty).
$$  
The horizontal asymptote is determined by the limit:  
$$
\lim_{x\to -\infty} f(x)=1.
$$  
Thus, the horizontal asymptote of $$ f $$ is  
$$
y=1.
$$

**(d)**  
To find the inverse $$ f^{-1} $$, start with  
$$
y = e^{x-3} + 1.
$$  
1. Subtract 1 from both sides:  
$$
y - 1 = e^{x-3}.
$$  
2. Take the natural logarithm of both sides:  
$$
\ln(y - 1) = x - 3.
$$  
3. Solve for $$ x $$:  
$$
x = \ln(y - 1) + 3.
$$  
Thus, the inverse function is:  
$$
f^{-1}(x)=\ln(x-1)+3.
$$

**(e)**  
The domain of the inverse function $$ f^{-1} $$ is the range of the original function $$ f $$, and the range of $$ f^{-1} $$ is the domain of $$ f $$.

- The range of $$ f $$ is $$(1,\infty)$$, so the domain of $$ f^{-1} $$ is  
$$
(1, \infty).
$$

- The domain of $$ f $$ is all real numbers, i.e., $$(-\infty, \infty)$$, so the range of $$ f^{-1} $$ is  
$$
(-\infty, \infty).
$$
**(c)** The range of $$ f $$ is $$ (1, \infty) $$. The horizontal asymptote of $$ f $$ is $$ y=1 $$. **(d)** The inverse function is $$ f^{-1}(x)=\ln(x-1)+3 $$. **(e)** The domain of $$ f^{-1} $$ is $$ (1, \infty) $$ and the range is $$ (-\infty, \infty) $$.

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