Question

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

Answer 1

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)<=-π/4==>-1/(2w)<=-1/4==>w<2
函数f(x)最大值点：wx=2kπ+π/2==>x=2kπ/w+π/(2w)
π/(2w)>=2π/3==>1/(2w)>=2/3==>w<=3/4
取二者交w<=3/4
∵w>0
∴w的取值范围0<w<=3/4