A) You should solve the equation f'(x)=0 to find if the function has critical values such that:

`f'(x) = ((1 - ln x)')/(1 - ln x)`

`f'(x) = (-1/x)/(1 - ln x)`

You need to solve the equation f'(x)=0 such that:

`(-1/x)/(1 - ln x) = 0 `

Since `-1/x!=0 and 1 - ln x!=0 =gt f'(x)!=0`

Hence, the function has no critical values.

B) The negative values of derivative indicate that the function may only decrease over `(0,oo).`

C) Notice that the domain of the function is `(0,oo)`  and since f'(x)<0, then the function decreases over `(0,oo).`

D) Since the function has no critical values, the function does not have local maximum.

E) Since the function has no critical values, the function does not have local minimum.

F) You need to find f''(x) t`o` check where the function is concave up or concave down.

`f''(x) = ((1/(x^2))(1-lnx) - 1/(x^2))/((1-ln x)^2)`

Reducing like terms yields:

`f''(x) = (-ln x)/((1-ln x)^2)`

Notice that the second derivative is negative for any x, hence the graph of function may only be concave down over `(0,oo).`