求函数y=2arcsin√x-1/2的定义域和值域
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arcsin定义域是[-1,1]
-1<=√(x-1/2)<=1
则0<=√(x-1/2)<=1
0<=x-1/2<=1
1/2<=x<=3/2
0<=√(x-1/2)<=1
所以arcsin0<=arcsin√(x-1/2)<=arcsin1
0<=arcsin√(x-1/2)<=π/2
所以定义域[1/2,3/2]
值域[0,π/2]
-1<=√(x-1/2)<=1
则0<=√(x-1/2)<=1
0<=x-1/2<=1
1/2<=x<=3/2
0<=√(x-1/2)<=1
所以arcsin0<=arcsin√(x-1/2)<=arcsin1
0<=arcsin√(x-1/2)<=π/2
所以定义域[1/2,3/2]
值域[0,π/2]
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