已知函数F(x)=2sin(wx+π/6),w∈R 且w≠0(1)若F(x)的图像经过点(π/6,2),且0(1)中以求得w=2,(2)怎么解呢
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已知函数F(x)=2sin(wx+π/6),w∈R 且w≠0(1)若F(x)的图像经过点(π/6,2),且0(1)中以求得w=2,(2)怎么解呢
已知函数F(x)=2sin(wx+π/6),w∈R 且w≠0
(1)若F(x)的图像经过点(π/6,2),且0
(1)中以求得w=2,(2)怎么解呢
已知函数F(x)=2sin(wx+π/6),w∈R 且w≠0(1)若F(x)的图像经过点(π/6,2),且0(1)中以求得w=2,(2)怎么解呢
(1)∵F(x)的图像经过点(π/6,2)
代入:2=2sin(π/6w+π/6) 由于sin(kπ+π/2)=0,且0
(1)∵F(x)的图像经过点(π/6,2)
代入:2=2sin(π/6w+π/6) 由于sin(kπ+π/2)=0,且0
(2)g(x)= mF(x) + n =2msin(2x+π/6)+n
∵m>0,且x∈[0,π/2],2x+π/6∈[π/6,7π/6],即sin(2x+π/6)∈[-1/2,1],
∴g(x)= 2msin(2x...
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(1)∵F(x)的图像经过点(π/6,2)
代入:2=2sin(π/6w+π/6) 由于sin(kπ+π/2)=0,且0
(2)g(x)= mF(x) + n =2msin(2x+π/6)+n
∵m>0,且x∈[0,π/2],2x+π/6∈[π/6,7π/6],即sin(2x+π/6)∈[-1/2,1],
∴g(x)= 2msin(2x+π/6)+n∈[-m+n,2m+n]
即 -m+n=-5,2m+n=1
解得:m=2,n=-3
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