已知函数f(x)=2sin(2x-π/3)-1 (1)求函数最大值,最小值 (2)求当x∈[-π/2,π/2]是函数的增区间

(1)当2x－π/3＝2kπ＋π/2即x＝kπ＋5π/12时,f(x)取得最大值1
当2x－π/3＝2kπ－π/2即x＝kπ－π/12时,f(x)取得最小值﹣3
(2)∵x∈[﹣π/2,π/2] ∴2x－π/3∈[﹣4π/3,2π/3]
∴当2x－π/3∈[﹣π/2,π/2]即x∈[﹣π/12,5π/12]时,函数单调递增
∴当x∈[﹣π/2,π/2]时函数的增区间为：[﹣π/12,5π/12]
(3)∵f(x)＝2sin(2x－π/3)－1＝2sin[2(x－π/6)]－1
∴函数的图象可以由函数y=sin2x的图像先沿x轴方向向右偏移π/6个单位,再沿y轴方向向下平移1个单位