`f(x) = ln(1 - ln(x))` (a) Find the vertical and horizontal asymptotes. (b) Find the intervals of increase or decrease.

Textbook Question

Chapter 4, 4.3 - Problem 51 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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The vertical asymptotes of the given function are x = 0 and x = e, such that:

`lim_(x->0) (ln(1 - ln x)) = ln lim_(x->0) (1 - ln x) = ln (1 - lim_(x->0) ln x)`

`lim_(x->0) (ln(1 - ln x)) = ln (1 - (-oo)) `

`lim_(x->0) (ln(1 - ln x)) = ln oo = oo`

Hence, the function has vertical asymptote x = 0.

`lim_(x->e) (ln(1 - ln x)) = ln (1 - lim_(x->e) ln x)`

`lim_(x->e) (ln(1 - ln x)) = ln (1 - ln e)`

`lim_(x->e) (ln(1 - ln x)) = ln (1 - 1)`

`lim_(x->e) (ln(1 - ln x)) = -oo`

Hence, the function has vertical asymptote x = e.

You need to evaluate the horizontal asymptotes of the function:

`lim_(x->oo) (ln(1 - ln x))` impossible to be evaluated since 1 - ln x < 0 as x approaches to oo.

b) You need to evaluate the monotony of the function, hence, you need to determine the intervals for f'(x)>0 or f'(x)<0.

You need to determine the derivative of the function:

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

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

You need to notice that f'(x) > 0, hence the function increases, for `x in (e,oo)` and f'(x) < 0, hence the function decreases, for `x in (0,e).`

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