Question

已知函数y＝㏑（1+x）/x证明若x≥1,y≤㏑2

Answer 1

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

g'(x)=1/(x+1)^2-1/(x+1)=-x/(x+1)^2