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f(x)=2sin(2x+π/6) x属于[π/12,π/2] 求值域
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f(x)=2sin(2x+π/6) x属于[π/12,π/2] 求值域
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答案和解析
π/12<=x<=π/2
π/6<=2x<=π
π/3<=2x+π/6<=7π/6
sinx在[π/3,π/2]是增函数,[π/2,7π/6]是减函数
所以2x+π/6=π/2时最大=sinπ/2=1
2x+π/6=7π/6时最小=sin7π/6=-1/2
所以-1<=2sin(2x+π/6) <=2
值域[-1,2]
π/6<=2x<=π
π/3<=2x+π/6<=7π/6
sinx在[π/3,π/2]是增函数,[π/2,7π/6]是减函数
所以2x+π/6=π/2时最大=sinπ/2=1
2x+π/6=7π/6时最小=sin7π/6=-1/2
所以-1<=2sin(2x+π/6) <=2
值域[-1,2]
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