Question

不等式问题，求高手解答

设函数f(x)=ln(1+x)-mx(x＞0)
(1)若函数f(x)在x=1处有极大值，求函数f(x)的单调区间

Answer 1

f(x) = ln(1+x) -mx
f'(x) = 1/(1+x) - m
f'(1) =0
m = 1/2
f''(x) = - 1/(1+x)^2 < 0 ( for x≠ -1 )
f(x) is defined for x > -1
f(x)的单调区间
增加 (-1, 1/2)
减小(1/2, ∞)