题目
已知函数f(x)=(x-1)/(x+1),x∈[1,3],求函数的最大值和最小值
提问时间：2020-11-03
答案
f(x)=(x+1-2)/(x+1)
=(x+1)/(x+1)-2/(x+1)
=1-2/(x+1)
1<=x<=3
2<=x+1<=4
所以1/4<=1/(x+1)<=1/2
-1<=-2/(x+1)<=-1/2
1-1<=1-2/(x+1)<=1-1/2
0<=f(x)<=1/2
所以最大值=1/2,最小值=0
=(x+1)/(x+1)-2/(x+1)
=1-2/(x+1)
1<=x<=3
2<=x+1<=4
所以1/4<=1/(x+1)<=1/2
-1<=-2/(x+1)<=-1/2
1-1<=1-2/(x+1)<=1-1/2
0<=f(x)<=1/2
所以最大值=1/2,最小值=0
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