Question

已知函数f(x)=根号(x^2+1)-ax,a>0

Answer 1

证明:设x1>x2≥0,则

f(x1)-f(x2)=√(x1^2+1)-ax1-√(x^2+1)+ax2

=(x1^2-x2^2)/[√(x1^12+1)+√(x2^2+1)]-a(x1-x2)

=(x1-x2){x1+x2-a[√(x1^2+1)+√(x2^2+1)]}/[√(x1^2+1)+√(x2^2+1)]

又x1>x2≥0,a≥1,即

x1-x2>0,x1