展开全部
已知函数f(x)=sin(2x+φ)(|φ|<=π/2)图像的一条对称轴是直线x=-π/8，（1）求φ的值，（2）求函数y=f(x)的单调增区间
(1)解析：∵函数f(x)=sin(2x+φ)，(|φ|<=π/2)，图像的一条对称轴是直线x=-π/8
∴2x+φ=2kπ-π/2==>x=kπ-(π+2φ)/4=-π/8==>φ=-π/4
2x+φ=2kπ+π/2==>x=kπ+(π-2φ)/4=-π/8==>φ=3π/4
∴φ的值为φ=-π/4或φ=3π/4
(2)解析：∵函数f(x)=sin(2x+φ)，(|φ|<=π/2)
2kπ-π/2<=(2x+φ)<=2kπ+π/2
==>2kπ-π/2-φ<=2x<=2kπ+π/2-φ==>2kπ-(π+2φ)/2<=2x<=2kπ+(π-2φ)/2
==>kπ-(π+2φ)/4<=x<=kπ+(π-2φ)/4
∵|φ|<=π/2
当φ=-π/4时，函数f(x)的增区间为kπ-π/8<=x<=kπ+3π/8
当φ=3π/4时，函数f(x)的增区间为kπ-5π/8<=x<=kπ-π/8
(1)解析：∵函数f(x)=sin(2x+φ)，(|φ|<=π/2)，图像的一条对称轴是直线x=-π/8
∴2x+φ=2kπ-π/2==>x=kπ-(π+2φ)/4=-π/8==>φ=-π/4
2x+φ=2kπ+π/2==>x=kπ+(π-2φ)/4=-π/8==>φ=3π/4
∴φ的值为φ=-π/4或φ=3π/4
(2)解析：∵函数f(x)=sin(2x+φ)，(|φ|<=π/2)
2kπ-π/2<=(2x+φ)<=2kπ+π/2
==>2kπ-π/2-φ<=2x<=2kπ+π/2-φ==>2kπ-(π+2φ)/2<=2x<=2kπ+(π-2φ)/2
==>kπ-(π+2φ)/4<=x<=kπ+(π-2φ)/4
∵|φ|<=π/2
当φ=-π/4时，函数f(x)的增区间为kπ-π/8<=x<=kπ+3π/8
当φ=3π/4时，函数f(x)的增区间为kπ-5π/8<=x<=kπ-π/8
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
