判断函数f(x)=x+4/x在[2,正无穷)上的单调性,并用定义证明尤其是在证明f(x1)-f(x2)>0时要详细~另加金额~
2019-06-02
判断函数f(x)=x+4/x在[2,正无穷)上的单调性,并用定义证明
尤其是在证明f(x1)-f(x2)>0时要详细~另加金额~
优质解答
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.
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.