Draw the graph of `f(x)="ln" (1-"ln "x)`. Find the point of inflection.

Explanation

`f(x) = "ln"(1-"ln "x)`
`f(x)` is defined if `1-"ln "x gt0`
or `"ln " x lt 1`
or `0 lt x lt e`
So the domain is (0,e).
Also, `f^(')(x)=-1/((1-"ln "x). 1/x lt 0`
So `f(x)` is continous and decreasing `AA x in (0,e)`.
Further `underset(xto0^(+))"lim""ln "(1-" ln "x)="ln "(1+infty)=infty`
`underset(x to e^(-))"lim" "ln "(1-"ln"x) = "ln "(1-1^(-))="ln "0^(+)=-infty`
For the point of inflection, `f^('')(x) = (-"ln "x)/(x^(2)(1-"ln "x)^(2))`
Clearly, `f^('')(1)=0`, so `x=1` is a point of inflection.
From the above discussion, the graph of `y=f(x)` is as shown in the following figure.

From the graph, `y=0` and y=e are asymptotoes.