求函数f(x)=2x+1⼀x+1在区间[1,4]上的最大值最小值

过程给下
2024-12-23 11:15:48
推荐回答(2个)
回答1:

f(x)=2x+1/x+1=(2(x+1)-1)/(x+1)=2-1/(x+1)
1<=x<=4
2<=x+1)<=5
所以,1/5<=1/(x+1)<=1/2
-1/2<=-1/(x+1)<=-1/5
3/2<=2-1/(x+1)<=9/5
即最大值是9/5,最小值是3/2

回答2:

最大为37/4最小为3