Question

已知函数f（x）＝㏑（1＋x）－mx.（1）当m＝1时，求函数f（x）的单调递减区间；（2）求函数f（x）的极值.

Answer 1

(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

Answer 2

1.把m=1带进去，f(x)=㏑（1＋x）－x,f'(x)=(1/x-1)-1,令f'(x)＜0，有x＞2.
2.f'(x)=(1/x-1)-m,令f'(x)=0,有x=(1/m)+1.

Answer 3

(1)f'(x)=1/(1+x)-1≤0，故单调递减区域为（-∞，-1）∪（0，+∞）
（2）即f'(x)=1/(1+x)-m=0，即x=1/m-1,带入f(x),即f(x)的极值为ln(1/m)-1+m,其中m>0