Question

use the given function f to answer parts (a) through (f).
f(x)=2+ ln x
...
(c) from the graph, determine the range and any asymptotes of f.
the range of f is (- ∞,∞)
(type your answer in interval notation.)
determine the vertical asymptote of f, if it exists. select the correct choice and, if necessary, fill in the answer box to complete your choice.
a. the vertical asymptote of f is x = 0. (simplify your answer.)
b. there is no vertical asymptote.
(d) find ( f^{-1} ), the inverse of f.
( f^{-1}(x) = e^{x - 2} ) (simplify your answer.)
(e) find the domain and range of ( f^{-1} )
the domain of ( f^{-1} ) is □
(type your answer in interval notation.)

Explanation:

Step1: Recall domain/range of inverse function

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

Step2: Determine domain of \( f^{-1} \)

From part (c), the range of \( f(x) = 2 + \ln x \) is \( (-\infty, \infty) \). So the domain of \( f^{-1}(x) \) is the range of \( f \), which is \( (-\infty, \infty) \). Also, we can analyze \( f^{-1}(x)=e^{x - 2} \). The exponential function \( e^{u} \) has domain \( (-\infty, \infty) \) for any real \( u \), here \( u=x - 2 \), so the domain of \( f^{-1}(x) \) is all real numbers.

Answer:

\((-\infty, \infty)\)