(1)
f(-x)=-f(x)
故log a (1+mx)/(-1-x)=log a (x-1)/(1-mx)
所以 (1+mx)/(-1-x)=(x-1)/(1-mx)
解得m=-1
(2)x>1
f(x)=log a (1+x)-log a (x-1)
f'(x)=(((x+1)lna)^-1)-(((x-1)lna)^-1)
f'(x)=-2/(lna(x^2-1))
若0
(1)
f(-x)=-f(x)
故log a (1+mx)/(-1-x)=log a (x-1)/(1-mx)
所以 (1+mx)/(-1-x)=(x-1)/(1-mx)
解得m=-1
(2)x>1
f(x)=log a (1+x)-log a (x-1)
f'(x)=(((x+1)lna)^-1)-(((x-1)lna)^-1)
f'(x)=-2/(lna(x^2-1))
若0