Question

use the given function f to answer parts (a) through (f).
f(x)=2+\ln x

(b) graph f. choose the correct graph.
\bigcirc a. graph \bigcirc b. graph \bigcirc c. graph \bigcirc d. graph

(c) from the graph, determine the range and any asymptotes of f.
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.
\bigcirc a. the vertical asymptote of f is x = \square (simplify your answer.)
\bigcirc b. there is no vertical asymptote.

Explanation:

Part (c) - Vertical Asymptote

Step 1: Recall the parent function

The parent function of \( f(x) = 2 + \ln x \) is \( y=\ln x \). The domain of \( \ln x \) is \( x>0 \), and it has a vertical asymptote at \( x = 0 \) (since as \( x \to 0^+ \), \( \ln x \to -\infty \)).

Step 2: Analyze the transformation

The function \( f(x)=2+\ln x \) is a vertical shift (up 2 units) of \( \ln x \). Vertical shifts do not affect the vertical asymptote (which is determined by the domain restriction and the behavior as \( x \) approaches the boundary of the domain). Since the domain of \( f(x) \) is still \( x>0 \) (because the \( \ln x \) term requires \( x>0 \)), the vertical asymptote remains at \( x = 0 \).

Answer:

A. The vertical asymptote of \( f \) is \( x = \boldsymbol{0} \)