Question

知函数f(x)=ln(1+x)/x证明若x大于等于1则f(x)小于等于ln2

Answer 1

f'(x)=[x/(x+1)-ln(x+1)]/x^2=[x-(x+1)ln(x+1)]/(x+1)x^2

因为x≥1,所以分母(x+1)x^2>0,只需判断分子的符号即可；

令g(x)=x-(x+1)ln(x+1),则g'(x)=1-ln(x+1)-1=-ln(x+1),

因为x≥1,则ln(x+1)>0,所以g'(x)