Question

begin by graphing f(x)= ln x. use transformations of this graph to graph the given function. graph and give the equation of the asymptote. use the graphs to determine the functions domain and range.
g(x) = ln (x + 1)

graph g(x)= ln (x + 1). graph the asymptote of g(x) as a dashed line. use the graphing tool to graph the function.

what is the vertical asymptote of g(x)?
( type an equation. )
what is the domain of g(x)= ln (x + 1)?
( simplify your answer. type your answer in interval notation. )
what is the range of g(x)= ln (x + 1)?
( simplify your answer. type your answer in interval notation. )

Explanation:

Step1: Recall Vertical Asymptote of Logarithm

For \( y = \ln(u) \), vertical asymptote is where \( u = 0 \). For \( g(x)=\ln(x + 1) \), set \( x + 1 = 0 \).
\( x+1 = 0 \implies x=-1 \).

Step2: Determine Domain of \( g(x) \)

Logarithm is defined when argument \(> 0 \). So \( x + 1>0 \implies x>-1 \). In interval notation, \( (-1, \infty) \).

Step3: Determine Range of \( g(x) \)

The range of \( \ln(u) \) for any \( u>0 \) is all real numbers, \( (-\infty, \infty) \), since \( \ln(u) \) can take any real value as \( u \) varies over \( (0, \infty) \).

Answer:

s:
Vertical Asymptote: \( \boldsymbol{x = -1} \)
Domain: \( \boldsymbol{(-1, \infty)} \)
Range: \( \boldsymbol{(-\infty, \infty)} \)