设函数f(x)=Asin(ωx+φ)(Aωφ是常数,A>0ω>0)若f(x)在区间{π/6,π/2}上具有单调性且f(π/2)=f(2π/3)f(π/2)=f(2π/3)=-f(π/6),则f(x)的最小正周期为π怎么求的求画图

Question

设函数f(x)=Asin(ωx+φ)(Aωφ是常数,A>0ω>0)若f(x)在区间{π/6,π/2}上具有单调性且f(π/2)=f(2π/3)
f(π/2)=f(2π/3)=-f(π/6),则f(x)的最小正周期为 π怎么求的 求画图

Answer 1

由f(π/2)=f(2π/3)可知函数f(x）一条对称轴为x=(π/2+2π/3)/2=7π/12
       则x=π/2离最近的对称轴距离为7π/12-π/2=π/12
       又f(π/2)=-f(π/6)且f(x)在区间[π/6,π/2]上具有单调性
       ∴x=π/6离最近的对称轴距离也为π/12
       函数图象大致形状如图：





∴T/2=7π/12-π/6+π/12=π/2
    则T=π