521lizhi
幼苗
共回答了17个问题采纳率:88.2% 举报
增函数.
依题意可设2≤x1<x2
f(x1)-f(x2)=x1+4/x1-x2-4/x2
=x1-x2 + 4/x1-4/x2
=x1-x2 + (4x2-4x1)/x1x2
=(x1-x2)-4(x1-x2)/x1x2
=(x1-x2)(1 - 4/x1x2)
=(x1-x2)(x1x2/x1x2 - 4/x1x2)
=(x1-x2)[(x1x2-4)/x1x2]
因为2≤x1<x2,x1x2-4>0,x1x2>0.
所以f(x1)-f(x2)=(x1-x2)[(x1x2-4)/x1x2]<0,即f(x1)<f(x2)
所以f(x)=x+4/x在[2,正无穷)上为单调递增函数.
1年前
5