求下列函数的单调区间（1）y=sin(3x-π/3)（2）y=-cos(2x-π/3)

显示全部楼层 · 2012-3-2 01:29:50

(1)由图像可得，当-π/2+2kπ≤3x-π/3≤π/2+2kπ,即-π/18+2kπ/3≤x≤5π/18+2kπ/3时函数单调递增;同理,当π/2+2kπ≤3x-π/3≤3π/2+2kπ,即5π/18+2kπ/3≤x≤11π/18+2kπ/3时函数单调递减。其中k∈Z。（2）由图像可得，-π/2+2kπ≤2x-π/3≤2kπ，即-π/12+kπ≤x≤π/6+kπ时函数单调递减；同理，当2kπ≤2x-π/3≤π/2+2kπ，即π/6+kπ≤x≤5π/12+kπ时函数单调递增。其中K∈Z。这里把函数看成y=sinA和y=-cosB，这样问题就简单了。...

千问 · 2012-3-2 01:29:50

（1）y=sin(3x-π/3)的周期T=2π/3,水平向右位移为π/3/3=π/9,则依据正弦函数的图像可知，{xI2kπ/3-π/6+π/9<x<2kπ/3+π/6+π/9为单调增函数，，{xI2kπ/3+π/6+π/9<x<2kπ/3+π/3+π/9为单调减函数。（其中z为自然数）（2）y=-cos(2x-π/3)的周期T=π,水平向右位移为π/...