Question

急求高一数学函数题

求函数f(x)=x/x-1在区间【2,5】上的最大值与最小值

Answer 1

f(x)=x/x-1
=(x-1+1)/(x-1)
=1+1/(x-1)
因为 2<=x<=5
所以 1<=x-1<=4

所以 1/4<=1/(x-1)<=1

所以 f(x)的最大值为 1+1=2
最小值为 1+1/4=5/4

Answer 2

式子写错了吧？

Answer 3

f(x)=(x-1+1)/(x-1)
=(x-1)/(x-1)+1/(x-1)
=1+1/(x-1)

2<=x<=5
1<=x-1<=4
所以1/4<=1/(x-1)<=1
加上1
5/4<=1+1/(x-1)<=2
所以最大值=2，最小值=5/4

Answer 4

是f(x)=x/(x-1)吧
上式可化为f(x)=(x-1+1)/(x-1),即f(x)=1+1/(x-1).
由于x取值为【2,5】，
2<=x<=5
1<=x-1<=4
所以1/4<= 1/(x-1)<= 1
所以f(x)在区间上最大值为2，最小值为5/4