Question

已知函数f(x)=3tan(π/3-x/2)求它的最小正周期和单调区间

Answer 1

问：已知函数f(x)=3tan(π/3-x/2)求它的最小正周期和单调区间

答：最小正周期T=2π计算过程：由函数f(x)=3tan(π/3-x/2)可知，f(x)=3tan(π/3-x/2)=3tan(π/3)-3tan(x/2)=3tan(π/3)-3cot(x/2)=3tan(π/3)-3cot(π/3-x/2)=3tan(π/3)-3cot(π/3-x/2+2π/3)=3tan(π/3)-3cot(π/3-x/2+2π/3+2π/3)=3tan(π/3)-3cot(π/3-x/2+4π/3)=3tan(π/3)-3cot(π/3-x/2+2π)=3tan(π/3)-3cot(π/3-x/2)=3tan(π/3)-3tan(x/2)=3tan(π/3-x/2)可见，f(x)的最小正周期为T=2π。单调区间：当x∈[0,2π]时，f(x)单调递增；当x∈[2π,4π]时，f(x)单调递减。