Question

已知f(x)=x+2/x-3,x属于[1,2],求值域

Answer 1

f(x)=x+2/x-3>=2√2-3(当x=√2时等号成立)
f(1)=f(2)=0，所以f(x)的值域是[2√2-3,0]

Answer 2

解：
f(x)=(x+2)/(x-3)
=(x-3+5)/(x-3)
=1+5/(x-3)
∵1≤x≤2
∴-2≤x-3≤-1
-1≤1/x-3≤-1/2
∴-4≤f(x)≤-3/2
f(x)∈[-4,-3/2]
希望对你有帮助，望采纳，谢谢~

Answer 3

解:f(x)=(x+2)/(x-3)
=(x-3+5)/(x-3)
=1+5/(x-3)
设g(x)=5/(x-3)
当1≤x≤2 时,-2≤x-3≤-1
得出 -5≤5/(x-3)≤-5/2
即 -5≤g(x)≤-5/2
f(x)=1+g(x)
其最小值为 1-5=-4
其最大值为 1-5/2=-3/2
f(x)的值域为[-4,-3/2]

Answer 4

[2倍根号2-3，0]最小值在x=1取得，最大值在端点处取得