# Suppose that f(x)= ln(1-ln(x)) (A) Find all critical values of f. If there are no critical values, enter -1000.B) Use interval notation to indicate where f(x) is increasing.(C) Use interval...

Suppose that f(x)= ln(1-ln(x))

(A) Find all critical values of f. If there are no critical values, enter -1000.

B) Use interval notation to indicate where f(x) is increasing.

(C) Use interval notation to indicate where f(x) is decreasing.

(D) Find the x-coordinates of all local maxima of f. If there are no local maxima, enter -1000.

(E) Find the x-coordinates of all local minima of f. If there are no local minima, enter -1000.

(F) Use interval notation to indicate where f(x) is concave up.

(G) Use interval notation to indicate where f(x) is concave down.

(H) Find all inflection points of f. If there are no inflection points, enter -1000.

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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).` **