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f(x)=(x-1)/(x+ 1)=(x+1-2)/(x+1)=1-2/(x+1),
由于:2<=x+1<=4,故1/4<=1/(x+1)<=1/2
-1<=-2/(x+1)<=-1/2
0<=1-2/(x+1)<=1/2
所以,0<=f(x)<=1/2,即值域是[0,1/2]
由于:2<=x+1<=4,故1/4<=1/(x+1)<=1/2
-1<=-2/(x+1)<=-1/2
0<=1-2/(x+1)<=1/2
所以,0<=f(x)<=1/2,即值域是[0,1/2]
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f(x)=(x-1)/(x+1)
=1-2/(x+1)
1<=x<=3
2<=x+1<=4
当x=1时f(1)min=0
当x=3时f(3)max=1/2
所以函数值域[0,1/2]
=1-2/(x+1)
1<=x<=3
2<=x+1<=4
当x=1时f(1)min=0
当x=3时f(3)max=1/2
所以函数值域[0,1/2]
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