已知函数f(x)sinxcosx-√3sin²x+√3/2(1)求函数f(x)的最小正周期
在f(x)在[-π/2,π/2]上的单调减区间;(3)求f(x)在[0,π/4]上的最值及相应的x取值范围...
在f(x)在[-π/2,π/2]上的单调减区间;(3)求f(x)在[0,π/4]上的最值及相应的x取值范围
展开
2个回答
展开全部
答:
f(x)=sinxcosx-√3sinx*sinx+√3/2
=(1/2)sin2x+(√3/2)cos2x
=sin(2x+π/3)
(1)f(x)的周期为kπ,所以最小正周期为π
(2)-π/2<=x<=π/2
-2π/3<=2x+π/3<=4π/3
所以f(x)在[-π/2,π/2]上的单调减区间为:-2π/3<=2x+π/3<=-π/2及 π/2<=2x+π/3<=4π/3
即单调减区间为[-π/2,-5π/12]及[π/12,π/2]
(3)0<=x<=π/4,π/3<=2x+π/3<=5π/6
f(x)min=sin(5π/6)=1/2,此时x=π/4;
f(x)max=sin(π/2)=1,此时x=π/12.
.
f(x)=sinxcosx-√3sinx*sinx+√3/2
=(1/2)sin2x+(√3/2)cos2x
=sin(2x+π/3)
(1)f(x)的周期为kπ,所以最小正周期为π
(2)-π/2<=x<=π/2
-2π/3<=2x+π/3<=4π/3
所以f(x)在[-π/2,π/2]上的单调减区间为:-2π/3<=2x+π/3<=-π/2及 π/2<=2x+π/3<=4π/3
即单调减区间为[-π/2,-5π/12]及[π/12,π/2]
(3)0<=x<=π/4,π/3<=2x+π/3<=5π/6
f(x)min=sin(5π/6)=1/2,此时x=π/4;
f(x)max=sin(π/2)=1,此时x=π/12.
.
展开全部
f(x)=sinxcosx-√3sin²x+√3/2=(1/2)sin2x - (1/2)(1-cos2x)+√3/2
=(1/2)(sin2x+cos2x-1+√3)
=(1/2)(√2sin(2x+π/4)-1+√3)
这样的话最小正周期是π
单调减区间和最值求导易求
难度只在化简上 要灵活运用二倍角公式
=(1/2)(sin2x+cos2x-1+√3)
=(1/2)(√2sin(2x+π/4)-1+√3)
这样的话最小正周期是π
单调减区间和最值求导易求
难度只在化简上 要灵活运用二倍角公式
追问
在f(x)在[-π/2,π/2]上的单调减区间;(3)求f(x)在[0,π/4]上的最值及相应的x取值范围
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
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