f(x)=2x/(x^2+1)
if x1>x2>=1
so f(x1)-f(x2)
=2X1/(X1^2+1)-2X2/(X2^2+1)
={2X1(X2^2)+2X1-2X2(X1^2)-2X2}/(X1^2+1)(X2^2+1)
因?yàn)?(X1^2+1)(X2^2+1)>0,暫不考慮
2X1(X2^2)+2X1-2X2(X1^2)-2X2=2X1X2(X2-X1)+2(X1-X2)
=2(X1X2-1)(X2-X1)
因?yàn)?X1>X2>=1
X1X2>1,X1>X2
SO 2(X1X2-1)(X2-X1)<0
SO {2X1(X2^2)+2X1-2X2(X1^2)-2X2}/(X1^2+1)(X2^2+1)<0
SO F(X1)由上可知,在〔1,+無(wú)窮）,應(yīng)為減函數(shù).
如果我對(duì)題意沒有理解錯(cuò),F(1)=2/2=1,F(2)=4/5=0.8,該函數(shù)在指定區(qū)間應(yīng)為減函數(shù)