Chapter 4.3, Problem 37E

(a)

To determine

To find : The vertical and horizontal asymptotes.

Explanation of Solution

Given: The function is $f\left(x\right)=\ln \left(1-\ln x\right)$

Consider the function.

  $f\left(x\right)=\ln \left(1-\ln x\right)$

The horizontal asymptote is not defined.

The vertical asymptote is calculated as,

  $\begin{array}{c}\underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\ln \left(1-\ln x\right)\\ =\infty \end{array}$

And,

  $\begin{array}{c}\underset{x\to e}{\lim }f\left(x\right)=\underset{x\to e}{\lim }\ln \left(1-\ln x\right)\\ =\infty \end{array}$

So, the vertical asymptotes are $x=0,e$ .

(b)

To determine

To find : The interval of increase or decrease.

Explanation of Solution

Given: The function is $f\left(x\right)=\ln \left(1-\ln x\right)$

Consider the function.

  $f\left(x\right)=\ln \left(1-\ln x\right)$

Differentiate the above expression with respect to $x$ .

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{1}{\left(1-\ln x\right)}\left(\frac{-1}{x}\right)\\ =\frac{-1}{x\left(1-\ln x\right)}\end{array}$

The function is decreasing in the interval $\left(0,e\right)$

(c)

To determine

To find : The local maximum and minimum values..

Explanation of Solution

Given: The function is $f\left(x\right)=\ln \left(1-\ln x\right)$

The function is decreasing in the interval $\left(0,e\right)$

Since, the function is only decreasing.

There is no local maxima and local minima.

(d)

To determine

To find : The intervals of concavity and the inflection points.

Explanation of Solution

Given: The function is $f\left(x\right)=\ln \left(1-\ln x\right)$

Differentiate $f^{\prime }\left(x\right)$ with respect to $x$ .

  $\begin{array}{c}{f}^{″}\left(x\right)=-\left[\frac{1-1+\ln x}{x^2{\left(1-\ln x\right)}^2}\right]\\ =\frac{-\ln x}{x^2{\left(1-\ln x\right)}^2}\end{array}$

The value of $f^{″}\left(c\right)$ is positive for $0<x<1$ . So the concavity is upward in the interval $\left(0,1\right)$ .

The value of $f^{″}\left(c\right)$ is negative for $1<x<e$ . So the concavity is downward in the interval $\left(1,e\right)$ .

The function inflection point is $\left(1,0\right)$ .

(e)

To determine

To sketch : The graph of the function.

Explanation of Solution

Given: The function is $f\left(x\right)=\ln \left(1-\ln x\right)$

Consider the function.

  $f\left(x\right)=x\tan x$

The graph of the above function is shown in figure below.

  

Figure (1)

Therefore, the graph of the function is shown in Figure (1).