已知函数f(x)=2cos(2x-π/3)

①f(-17π/6)=2cos[2(-17π/6)-π/3)]=2cos(-6π) =2cos(0) =2
②x属于[π/3,π],则2x-π/3属于[π/3,5π/3]
而f(x)=2cos(2x-π/3)在 2x-π/3属于[π/3,π]上递减,在[π,5π/3]上递增
故在此区间,f(x)最小值=f(π)=-1,而f(x)最大值=f(π/3)=f(5π/3)=1/2
则函数f（x）在区间[π/3,π]上的值域为[-1,1/2]
③x属于[0,π],则2x-π/3属于[-π/3,5π/3]
故f（x）在 2x-π/3属于[-π/3,0]或[π,5π/3]上递增
即在x属于[0,π/6]或[2π/3,π]上递增
函数f（x）在区间[0,π]上的单调递增区间为[-π/3,0]或[π,5π/3]

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