(1)
m = 1, f(x) = ln(x + 1) - x
f'(x) = 1/(x + 1) - 1 = -x/(x + 1) = 0
x = 0
-1 < x < 0: f'(x) > 0
x > 1: f'(x) < 0, 单调递减
(2)
(i) m = 0
f(x) = ln(x + 1), f'(x) = 1/(x + 1); 在定义域x > -1内, f'(x) > 0, 无极值
(ii) m = 1
见(i), 极大值: f(0) = 0
(ii) m ≠ 0
f'(x) = 1/(x + 1) - m = (1 - m - mx)/(x + 1) = 0
x = (1 - m)/m = 1/m - 1
m < 0时, x = 1/m - 1 < -1, 在定义域外, 无极值
m > 0时, x = 1/m - 1 > -1, 在定义域内; 显然f'(x)左正右负
极大值f(1/m - 1) = ln(1/m - 1 + 1) -m(1/m -1) = ln(1/m) - (1 - m) = m - 1 - lnm
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