已知函数fx=2sin(wx),w>0 若fx在[-π/4,2π/3]上单调递增,求w的取值范围

解析：∵函数f(x)=2sinwx(w>0)在区间[-π/4,2π/3]上单调递增
∵函数f(x)初相为0
∴最小值点在Y轴左,最大值点在Y轴右,二者与Y轴之距相等
函数f(x)最小值点：wx=2kπ-π/2==>x=2kπ/w-π/(2w)
∴-π/(2w)-1/(2w)wx=2kπ/w+π/(2w)
π/(2w)>=2π/3==>1/(2w)>=2/3==>w