Answer
(a) $x=0$ and $x=e$ are vertical asymptotes.
(b) $f$ is decreasing on the interval $(0, e)$
(c) There is no local maximum or local minimum.
(d) The graph is concave down on this interval: $(1, e)$
The graph is concave up on this interval: $(0, 1)$
The point of inflection is $(1, 0)$
(e) We can see a sketch of the graph below.
Work Step by Step
(a) $f(x) = ln~(1-ln~x)$
$f(x)$ is defined for values of $x$ such that $0 \lt x$ and $(1-ln~x) \gt 0$
$f(x)$ is defined for values of $x$ such that $0 \lt x$ and $x \lt e$
$x=0$ and $x=e$ are vertical asymptotes.
$\lim\limits_{x \to \infty}f(x) = -\infty$
There is no horizontal asymptote.
(b) We can find the points where $f'(x) = 0$:
$f'(x) = \frac{1}{1-ln~x}\cdot (-\frac{1}{x})$
$f'(x) = -\frac{1}{x~(1-ln~x)} = 0$
There are no values of $x$ such that $f'(x) = 0$
When $0 \lt x$ then $f'(x) \lt 0$
$f$ is decreasing on the interval $(0, e)$ (the domain)
$f$ is not increasing on any interval.
(c) There are no values of $x$ such that $f'(x) = 0$
There is no local maximum or local minimum.
(d) We can find the points where $f''(x) = 0$:
$f''(x) = -\frac{-(1-ln~x)-(x)(-\frac{1}{x})}{x^2~(1-ln~x)^2}$
$f''(x) = \frac{1-ln~x-1}{x^2~(1-ln~x)^2}$
$f''(x) = -\frac{ln~x}{x^2~(1-ln~x)^2} = 0$
$ln~x = 0$
$x = 1$
The graph is concave down when $f''(x) \lt 0$
The graph is concave down on this interval: $(1, e)$
The graph is concave up when $f''(x) \gt 0$
The graph is concave up on this interval: $(0, 1)$
$f(1) = ln~(1-ln~1) = ln(1) = 0$
The point of inflection is $(1, 0)$
(e) We can see a sketch of the graph below.
