
已知函数f(x)=Acos(wx+p)的图像如图所示,f(兀/2)=-2/3,则f(0)=
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解析:∵f(x)=Acos(wx+φ)
由图示可知:T/2=11π/12-7π/12=π/3==>T=2π/3==>w=3
∴f(x)=Acos(3x+φ)
∵f(7π/12)=Acos(21π/12+φ)=0
21π/12+φ=3π/2==>φ=18π/12-21π/12=-π/4
∴f(x)=Acos(3x-π/4)
∵f(π/2)=Acos(3π/2-π/4)=-2/3==>A=2√2/3
∴f(x)=2√2/3cos(3x-π/4)
∴f(0)=2√2/3cos(3x-π/4)=2/3
由图示可知:T/2=11π/12-7π/12=π/3==>T=2π/3==>w=3
∴f(x)=Acos(3x+φ)
∵f(7π/12)=Acos(21π/12+φ)=0
21π/12+φ=3π/2==>φ=18π/12-21π/12=-π/4
∴f(x)=Acos(3x-π/4)
∵f(π/2)=Acos(3π/2-π/4)=-2/3==>A=2√2/3
∴f(x)=2√2/3cos(3x-π/4)
∴f(0)=2√2/3cos(3x-π/4)=2/3
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