Chapter 3, Problem 53RE

To determine

Find the domain, $x$ -intercept and vertical asymptote for the given logarithmic function and draw a graph.

Answer to Problem 53RE

Domain $=x>0$ .

  $x$ -intercept $=\left(\frac{1}{e^3},0\right)$ .

Vertical asymptote is $x=0$ .

Explanation of Solution

Given information:

The given functionis $f\left(x\right)=\ln x+3$ .

Calculation:

Domain $\to$ it is the set of input values for which the function is real and defined.

The function gives the positive values, when $x>0$ .

Domain $=x>0$ .

  $x$ -intercept $\to$ it is the point, where $y=0$ .

  $\begin{array}{l}f\left(x\right)=\ln x+3\\ \Rightarrow 0=\ln x+3\\ \Rightarrow 0-3=\ln x+3-3\\ \Rightarrow -3=\ln x\\ \Rightarrow e^{-3}=x\\ \Rightarrow x=\frac{1}{e^3}\end{array}$

  $x$ -intercept $=\left(\frac{1}{e^3},0\right)$

Vertical asymptote $\to$ the logarithmic function $f\left(x\right)=c{\log }_a\left(x+h\right)+k$ has a vertical asymptote $x=-h$ .

In the function $f\left(x\right)=\ln x+3$ ,

  $h=0$

Vertical asymptote is $x=0$ .

Create a table list for some values for the function.

$x$	$f\left(x\right)=\ln x+3$
$\begin{array}{l}0\\ 0.05\\ 1\\ 6\end{array}$	$\begin{array}{l}-4\\ 0\\ 3\\ 4.792\end{array}$

The graph of the function is given below.