已知函数f(X)=根号2sin(wX一兀/4)(w>0)的最小正周期是兀.求w的值,若X
已知函数f(X)=根号2sin(wX一兀/4)(w>0)的最小正周期是兀.求w的值、若X€[0,兀/2].且f(X)=0求X的值.求f(x)的单调区间....
已知函数f(X)=根号2sin(wX一兀/4)(w>0)的最小正周期是兀.求w的值、若X€[0,兀/2].且f(X)=0求X的值. 求f(x)的单调区间.
展开
2个回答
展开全部
T = 2π/w = π
所以w = 2
sin(2x - π/4) =0
2x - π/4 = 0
2x = π/4
x = π/8
(2) f(x) = 2sin(2x - π/4)
由f ‘ (x) = 2cos(2x - π/4) * 2 >0
得:cos(2x - π/4) >0
2kπ - π/2 <2x - π/4 < 2kπ +π/2
2kπ - π/4 < 2x < 2kπ + 3π/4
kπ - π/8 < x < kπ + 3π/8
所以kπ - π/8 < x < kπ + 3π/8 时 f(x) 单调递增.
所以w = 2
sin(2x - π/4) =0
2x - π/4 = 0
2x = π/4
x = π/8
(2) f(x) = 2sin(2x - π/4)
由f ‘ (x) = 2cos(2x - π/4) * 2 >0
得:cos(2x - π/4) >0
2kπ - π/2 <2x - π/4 < 2kπ +π/2
2kπ - π/4 < 2x < 2kπ + 3π/4
kπ - π/8 < x < kπ + 3π/8
所以kπ - π/8 < x < kπ + 3π/8 时 f(x) 单调递增.
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询