作业帮已知函数f(x)=sin(wx+π/3)(w>0),f(π/6)=f(π/2),且f(x)在区间(π/6,π/2)有最大值无最小值,则w
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已知函数f(x)=sin(wx+π/3)(w>0),f(π/6)=f(π/2),且f(x)在区间(π/6,π/2)有最大值无最小值,则w
已知函数f(x)=sin(wx+π/3)(w>0),f(π/6)=f(π/2),且f(x)在区间(π/6,π/2)有最大值无最小值,则w
已知函数f(x)=sin(wx+π/3)(w>0),f(π/6)=f(π/2),且f(x)在区间(π/6,π/2)有最大值无最小值,则w
已知函数f(x)=sin(wx+π/3)(w>0),f(π/6)=f(π/2)且f(x)在区间(π/6, π/2)内有最大值,无最小值,求w
解析：∵函数f(x)=sin(wx+π/3)(w>0),f(π/6)=f(π/2)且f(x)在区间(π/6, π/2)内有最大值,无最小值
则函数f(x)初相为π/3,离Y轴最近的极值点为最大值点
最大值点：wx+π/3=2kπ+π/2==>x=2kπ/w+π/(6w)
最小值点：wx+π/3=2kπ+3π/2==>x=2kπ/w+7π/(6w)
要使f(π/6)=f(π/2)且f(x)在区间(π/6,π/2)内有最大值,无最小值
须使x=π/6,x=π/2关于x=π/(6w),即x=π/3对称
令π/(6w)=π/3==>w=1/2
∴f(x)=sin(1/2x+π/3)
已知函数f(x)=sin(ωx+π/3）（ω>0),，f(π/6）=f(π/3），且f(x)在区间（π/6，π/3）有最小值，无最大值，求ω=?
(1/2)(π/6+π/3)=π/4，∵f(π/6）=f(π/3），∴x=π/4是其对称轴，又因为f(x)有最小值-1，
故有f(π/4)=sin(ωπ/4+π/3)=-1，考虑到ω>0，故有ωπ/4+π/3=3π/2，ωπ/4=3π/2...
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已知函数f(x)=sin(ωx+π/3）（ω>0),，f(π/6）=f(π/3），且f(x)在区间（π/6，π/3）有最小值，无最大值，求ω=?
(1/2)(π/6+π/3)=π/4，∵f(π/6）=f(π/3），∴x=π/4是其对称轴，又因为f(x)有最小值-1，
故有f(π/4)=sin(ωπ/4+π/3)=-1，考虑到ω>0，故有ωπ/4+π/3=3π/2，ωπ/4=3π/2-π/3=7π/6，
∴ω=14/3。
事实上，f(π/6)=sin[(14/3)×(π/6)+π/3)]=sin(10π/9)=sin(π+π/9)=-sin(π/9)
f(π/3)=sin[(14/3)×(π/3)+π/3)]=sin(17π/9)=sin(2π-π/9)=-sin(π/9)=f(π/6)
f(π/4)=sin[(14/3)×(π/4)+π/3]=sin(9π/6)=sin(π+π/2)=-sin(π/2)=-1.
π/6<π/4<π/3.即在区间(π/6，π/3)上确有最小值-1，但无最大值(因为是
).
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