数学简单三角函数!f(x)=2sin(wx+q)图象过(0,2)(6,0),求函数f(x)的解析式(2)令M=f(x)+
来源:学生作业帮 编辑:拍题作业网作业帮 分类:数学作业 时间:2024/05/01 17:28:42
数学简单三角函数!
f(x)=2sin(wx+q)图象过(0,2)(6,0),求函数f(x)的解析式
(2)令M=f(x)+(1/2)f(-x),求M的最大值
f(x)=2sin(wx+q)图象过(0,2)(6,0),求函数f(x)的解析式
(2)令M=f(x)+(1/2)f(-x),求M的最大值
f(x)=2sin(wx+q)图象过(0,2)(6,0)
2=2sin(0+q)
0=2sin(6w+q)
w=±π/12,q=π/2
f(x) = 2sin(-xπ/12+π/2)或f(x) = 2sin(xπ/12+π/2)
因为sinα=sin(π-α)
∴sin(-xπ/12+π/2)=sin[π-(-xπ/12+π/2)]=sin(xπ/12+π/2)
∴f(x) = 2sin(-xπ/12+π/2) 和 f(x) = 2sin(xπ/12+π/2) 是等效函数,可用f(x) = 2sin(xπ/12+π/2)一个式子表达.
M=f(x)+(1/2)f(-x)=2sin(xπ/12+π/2)+1/2*2sin(-xπ/12+π/2)
=2sin(xπ/12+π/2)+sin(-xπ/12+π/2)
=2sin(xπ/12+π/2)+sin(π+xπ/12-π/2)
=3sin(xπ/12+π/2)
-1 ≤ sin(xπ/12+π/2) ≤ 1
-3 ≤ 3sin(xπ/12+π/2) ≤ 3
所以M最大值3
2=2sin(0+q)
0=2sin(6w+q)
w=±π/12,q=π/2
f(x) = 2sin(-xπ/12+π/2)或f(x) = 2sin(xπ/12+π/2)
因为sinα=sin(π-α)
∴sin(-xπ/12+π/2)=sin[π-(-xπ/12+π/2)]=sin(xπ/12+π/2)
∴f(x) = 2sin(-xπ/12+π/2) 和 f(x) = 2sin(xπ/12+π/2) 是等效函数,可用f(x) = 2sin(xπ/12+π/2)一个式子表达.
M=f(x)+(1/2)f(-x)=2sin(xπ/12+π/2)+1/2*2sin(-xπ/12+π/2)
=2sin(xπ/12+π/2)+sin(-xπ/12+π/2)
=2sin(xπ/12+π/2)+sin(π+xπ/12-π/2)
=3sin(xπ/12+π/2)
-1 ≤ sin(xπ/12+π/2) ≤ 1
-3 ≤ 3sin(xπ/12+π/2) ≤ 3
所以M最大值3
数学简单三角函数!f(x)=2sin(wx+q)图象过(0,2)(6,0),求函数f(x)的解析式(2)令M=f(x)+
设函数f(x)=3sin(wx+圆周率/6),w>0,x属于R,且以圆周率/2为小正周期求f(x)解析式
求三角函数解析式f(x)=sin(wx+q) 已知w大于0小于3,q大于等于0小于等于兀 切函数在R上是偶函数 ,图象关
设函数f(x)=sin(wx+q)+cos(wx+q)(w>0,q的绝对值
高中的三角函数f(x)=sin^2wx-sinwxcoswx(w>0) 要怎么转化
求该图三角函数解析式已知函数f(x)=Asin(wx+∮)【x∈R,A>0,w>0,︱∮︱<π /2】部分图象如图所示,
设函数f(x)=sinwx+sin²wx/2(w>0)的最小正周期为2π/3,求函数解析式
设二次函数f(x)满足f(x+2)=f(2-x)且f(x)=0的两实根平方和为10,图象过点(0,3),求f(x)的解析
已知函数f(x)=sin(wx=π/6)(w>0)的图象相邻两对称轴间的距离为2 ,则f(2009)=
三角函数,求详解已知函数f(x)=sin(wx+φ)(w>0,0
已知函数f(x)=sin(wx+π/4),其中w>0,若函数f(x)图象的相邻两条对称轴距离等于π/3,求函数解析式
已知2f(1/x)+f(x)=x(x不等于0) 求函数f(x)的解析式?