(1)
圖象關于原點對稱,則g(x) + f(-x) = 0
g(x) = -f(-x) = -loga(1 - x)
(2)
F(x) = f(x) + g(x) = loga(x + 1) - loga(1 - x) + m = loga[(1+ x)/(1 - x)] + m
F(x)為奇函數(shù),則F(-x) = -F(-x)
loga[(1- x)/(1 + x)] + m = -loga[(1+ x)/(1 - x)] - m
2m = -loga[(1+ x)/(1 - x)] - loga[(1- x)/(1 + x)]
= loga[(1 - x)/(1 + x)] - loga[(1- x)/(1 + x)]
= 0
m = 0
(3)
令G(x) = f(x) + g(x) = loga[(1+ x)/(1 - x)]
G'(x) =[(1- x)/(1 + x)]lna
a > 1,lna > 0
x∈[0,1)時:1-x > 0,1 + x > 0,G'(x) > 0
[0,1)上的最小值為G(0) = 0
n = 0