for the given logarithmic function, complete parts a through c.\na. determine 3 points that lie on the graph…

Explanation:

Step1: Recall the property of $y = \ln x$

For $y=\ln x$, when $y = - 1$, we solve $\ln x=-1$. By the definition of the natural - logarithm ($y = \ln x$ means $x = e^{y}$), we have $x = e^{-1}=\frac{1}{e}$. For the function $f(x)=1+\ln x$, when $x=\frac{1}{e}$, $f(\frac{1}{e})=1+\ln\frac{1}{e}=1 - 1=0$. So the point on $y = \ln x$ is $(\frac{1}{e},-1)$ and the corresponding point on $f(x)=1+\ln x$ is $(\frac{1}{e},0)$.

Step2: When $y = 0$ in $y=\ln x$

If $\ln x = 0$, then $x = e^{0}=1$ (since $y=\ln x$ is equivalent to $x = e^{y}$). For the function $f(x)=1+\ln x$, when $x = 1$, $f(1)=1+\ln1=1+0 = 1$. So the point on $y=\ln x$ is $(1,0)$ and the corresponding point on $f(x)=1+\ln x$ is $(1,1)$.

Step3: When $y = 1$ in $y=\ln x$

If $\ln x=1$, then $x = e^{1}=e$. For the function $f(x)=1+\ln x$, when $x = e$, $f(e)=1+\ln e=1 + 1=2$. So the point on $y=\ln x$ is $(e,1)$ and the corresponding point on $f(x)=1+\ln x$ is $(e,2)$.

Step4: Determine the domain of $f(x)=1+\ln x$

The domain of the natural - logarithm function $y = \ln x$ is $(0,\infty)$. Since $f(x)=1+\ln x$ is a transformation of $y=\ln x$, the domain of $f(x)$ is also $(0,\infty)$.

Step5: Find the vertical asymptote

The vertical asymptote of the natural - logarithm function $y=\ln x$ is $x = 0$. The function $f(x)=1+\ln x$ is a vertical shift of $y=\ln x$ by 1 unit up, so the vertical asymptote of $f(x)$ is also $x = 0$.

Answer:

Points that lie on the graph of $y=\ln x$	Corresponding points that lie on the graph of $f(x)=1+\ln x$
$(\frac{1}{e},-1)$	$(\frac{1}{e},0)$
$(1,0)$	$(1,1)$
$(e,1)$	$(e,2)$
Domain: $(0,\infty)$
Vertical asymptote: $x = 0$