求函数f(x)=2x+1/x+1在区间[1，4]上的最大值最小值

显示全部楼层 · 2009-10-3 18:21:26

由基本不等式可知，f(x)当且仅当x=（根号2）/2时取到最小值2（根号2）+1，所以在区间[1，4]上，f(x)的最大值在两个端点处取得，带入求值，得到，最大值为f(4)=41/4。

千问 · 2009-10-3 18:21:26

f(x)=2x+1/x+1=(2(x+1)-1)/(x+1)=2-1/(x+1)1<=x<=42<=x+1)<=5所以,1/5<=1/(x+1)<=1/2-1/2<=-1/(x+1)<=-1/53/2<=2-1/(x+1)<=9/5即最大值是9/5,最小值是3/2

千问 · 2009-10-3 18:21:26

f(x)=2x+1/x+1=(2(x+1)-1)/(x+1)=2-1/(x+1)1<=x<=42<=x+1)<=5所以,1/5<=1/(x+1)<=1/2-1/2<=-1/(x+1)<=-1/53/2<=2-1/(x+1)<=9/5即最大值是9/5,最小值是3/2