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).` **

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