导数题阿阿阿求解已知导数f(x)=lnx+a(x+1) (1)当a=-1时,求导数f(x)的单调区间 (2)求函数f(x
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(1) a = -1, f(x) = lnx - x - 1

f'(x) = 1/x - 1 = 0, x = 1

0 < x < 1: f(x)递减

x > 1: f(x)递增

(2)

f(x) = lnx + ax + a

f'(x) = 1/x + a = 0

x = -1/a

(i) a = 0

f(x) = lnx,...

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