已知函数f(x)=2sin(2x-1/3π)+1,求f(x)在[1/4π,1/2π]上的最大值最小值

X∈[π/4,π/2],2x∈[π/2,π],
2x-π/3∈[π/6,2π/3]
sin(2x-π/3) ∈[1/2,1].
2sin(2x-π/3)+1∈[2,3].
∴x=π/6时,函数取到最小值2.
x=5π/12时,函数取到最小值3.