已知函数f（x）等于x加1分之2减x，证明：函数f（x）在（负1，正无穷大）上的减函数

显示全部楼层 · 2013-8-5 14:43:56

证明:设函数上的两个点x1,x2,且x2>x1>-1,则f(x2)-f(x1)=(2-x2)/(x2＋1)-(2-x1)/(1＋x1)=[2＋2x1-x2-x1x2-(2x2-x1x2＋2-x1)]/(x1＋1)(x2＋1)=[2(x1-x2)＋(x1-x2)＋]/(x2＋1)(x1＋1)=3(x1-x2)/(x2＋1)(x1＋1)<0,所以函数f(x2)<f(x1),所以函数f(x)在(-1,正无穷大)上为减函数。...

千问 · 2013-8-5 14:43:56

f(x)=(2-x)/(x+1)=[3-x+1)]/(x+1)=-1+3/(x+1)任取-1<x1<x2f(x1)-f(x2)=-1+3/(x1+1)-[-1+3/(x2+1)]=3/(x1+1)-3/(x2+1)=3[(x2+1)-(x1+1)]/[(x1+1)(x2+1)]=3(x2-x1)/[(x1+1)(x2+1)]∵-1...

千问 · 2013-8-5 14:43:56

f(x)=(2-x)/(x+1)=[3-x+1)]/(x+1)=-1+3/(x+1)任取-1<x1<x2f(x1)-f(x2)=-1+3/(x1+1)-[-1+3/(x2+1)]=3/(x1+1)-3/(x2+1)=3[(x2+1)-(x1+1)]/[(x1+1)(x2+1)]=3(x2-x1)/[(x1+1)(x2+1)]...