7/26/2019 Calculus 1 answer

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f(x) = ln(9 ln x) is defined when x > 0 (so that ln x is defined) and

9ln x >0 [so that ln(9ln x) is defined]. The second condition is equivalentto9> ln x x < e9 so fhas domain (0, e9).

(a) As x 0+,ln x , so 9 ln x and f(x) . As x e9

,ln x 9, so9 ln x 0+ and f(x) . Thus, x = 0 andx= e9 are vertical asymptotes. There is no horizontal asymptote.

(b) f(x) = 1

9 ln x

1

x

=

1

x(9 ln x) < 0 on (0, e9). Thus, f is

decreasing on its domain, (0, e9).

(c) f

(x) = 0 on (0, e9

), so fhas no local maximum or minimum value.

(d) f(x) = [x(9 ln x)]

[x(9 ln x)]2 =

x(1/x) + (9 ln x)

x2(9 ln x)2 =

ln x 8

x2(9 ln x)2

sof(x)> 0 ln x