已知函数f(x)=2sin(2x+π/3)

1个回答

  • (1)当-π/6≤x≤π/6时,由函数f(x)=2sin(2x+π/3)可知,0≤(2x+π/3)≤2π/3,在这个区间内sinx的取值范围为[0,√3/2],所以函数f(x)的值域区间为[0,√3].

    (2).最小正周期T=2π/IωI=π,正弦函数sinx的对称轴方程为x=(2k+1)π/2,所以2x+π/3=(2k+1)π/2,解得x=(k/2+1/12)π,k为整数.

    (3)因为sinx的单调增区间[-π/2 +2kπ ,π/2+2kπ] 减区间[π/2+2kπ ,3π/2+2kπ],故当-π/2 +2kπ≤(2x+π/3)≤π/2+2kπ时f(x)递增,当π/2+2kπ≤(2x+π/3)≤3π/2+2kπ时f(x)递减,解得[-5π/6+kπ ,π/6 +kπ]为函数f(x)增区间,[π/6+kπ ,7π/6 +kπ]为函数f(x)减区间.k∈Z

    (4).g(x)与f(x)同是正弦函数且定义域相同,故单调区间也相同.

    (5).由(3)可知当x∈[-5π/6+kπ ,π/6 +kπ]时f(x)递增,则G(x)递减;当x∈[π/6+kπ ,7π/6 +kπ]时f(x)递减,则G(x)递增.

    (6).f(x)可由①y=sinx先将x缩小2倍,再向左移动π/6个单位,再将值域扩大2倍

    ②先将x向左移动π/3个单位,再缩小2倍,再将值域扩大2倍