(求值域) sqr(4-x)+sqr(x-2)

定义域4-x>=0,x-2>=0
所以2<=x<=4

y=√(4-x)+√(x-2)
根号大于等于0
所以y>=0

y^2=4-x+2√(4-x)(x-2)+x-2
=2+2√[-(x-3)^2+1]
2<=x<=4
所以x=3，-(x-3)^2+1最大=1
x=2或4，-(x-3)^2+1最小=0
所以0<=√[-(x-3)^2+1]<=1
2<=2+2√[-(x-3)^2+1]<=4
所以√2<=y<=2
所以值域[√2,2]

4-x>=0,x-2>=0
2<=x<=4