Question

use the given function f to answer parts (a) through (f).

( f(x) = 2 + ln x )

...

the range of f is ( (-infty, infty) ).

(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 ( (-infty, infty) ).

(type your answer in interval notation.)

the range of ( f^{-1} ) is ( square ).

(type your answer in interval notation.)

Explanation:

Step1: Recall inverse function domain/range relation

For a function \( f \) and its inverse \( f^{-1} \), the domain of \( f^{-1} \) is the range of \( f \), and the range of \( f^{-1} \) is the domain of \( f \).

Step2: Find domain of original function \( f(x) = 2 + \ln x \)

The function \( \ln x \) is defined for \( x > 0 \), so the domain of \( f \) is \( (0, \infty) \).

Step3: Determine range of \( f^{-1} \)

Since the range of \( f^{-1} \) is the domain of \( f \), and the domain of \( f \) is \( (0, \infty) \), the range of \( f^{-1} \) is \( (0, \infty) \).

Answer:

\((0, \infty)\)