use the given function f to answer parts (a) through (f) below. f(x) = ln(x + 3). the domain of f is (-3,∞)…

use the given function f to answer parts (a) through (f) below. f(x) = ln(x + 3). the domain of f is (-3,∞). (type your answer in interval notation.) (b) graph f. choose the correct graph. (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. select the correct choice below, if necessary, fill in the answer box to complete your choice. a. the vertical asymptote of f is x = (simplify your answer.) b. there is no vertical asymptote.

use the given function f to answer parts (a) through (f) below. f(x) = ln(x + 3). the domain of f is (-3,∞). (type your answer in interval notation.) (b) graph f. choose the correct graph. (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. select the correct choice below, if necessary, fill in the answer box to complete your choice. a. the vertical asymptote of f is x = (simplify your answer.) b. there is no vertical asymptote.

Answer

Explanation:

Step1: Recall domain of natural - log function

The argument of $\ln(u)$ must be $u>0$. For $f(x)=\ln(x + 3)$, we set $x+3>0$. $x+3>0\Rightarrow x>- 3$. So the domain is $(-3,\infty)$.

Step2: Analyze range and asymptotes

The natural - logarithm function $y = \ln(x)$ has a range of $(-\infty,\infty)$. For $y=\ln(x + 3)$, the range remains $(-\infty,\infty)$. As $x\to - 3^{+}$, $x + 3\to0^{+}$ and $\ln(x + 3)\to-\infty$. So the vertical asymptote is $x=-3$.

Answer:

(a) Domain: $(-3,\infty)$ (b) (No graph - selection details provided in response as it requires visual choices) (c) Range: $(-\infty,\infty)$; Vertical asymptote: $x=-3$