已知函数f(x)=Asin(3x+φ)(A>0,x∈-∞,+∞),0<φ<π)在x=π/12时取得最大值4.(1)求f(x)单调增区间 (2)求函数f(x)在[0,π/3]上的值域 QAQ
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![已知函数f(x)=Asin(3x+φ)(A>0,x∈-∞,+∞),0<φ<π)在x=π/12时取得最大值4.(1)求f(x)单调增区间 (2)求函数f(x)在[0,π/3]上的值域 QAQ](/uploads/image/z/1980943-7-3.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3DAsin%283x%2B%CF%86%29%28A%EF%BC%9E0%2Cx%E2%88%88-%E2%88%9E%2C%2B%E2%88%9E%29%2C0%EF%BC%9C%CF%86%EF%BC%9C%CF%80%29%E5%9C%A8x%3D%CF%80%2F12%E6%97%B6%E5%8F%96%E5%BE%97%E6%9C%80%E5%A4%A7%E5%80%BC4.%281%29%E6%B1%82f%EF%BC%88x%EF%BC%89%E5%8D%95%E8%B0%83%E5%A2%9E%E5%8C%BA%E9%97%B4+%EF%BC%882%EF%BC%89%E6%B1%82%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%E5%9C%A8%5B0%2C%CF%80%2F3%5D%E4%B8%8A%E7%9A%84%E5%80%BC%E5%9F%9F+QAQ)
已知函数f(x)=Asin(3x+φ)(A>0,x∈-∞,+∞),0<φ<π)在x=π/12时取得最大值4.(1)求f(x)单调增区间 (2)求函数f(x)在[0,π/3]上的值域 QAQ
已知函数f(x)=Asin(3x+φ)(A>0,x∈-∞,+∞),0<φ<π)在x=π/12时取得最大值4.(1)求f(x)单调增区间 (2)求函数f(x)在[0,π/3]上的值域 QAQ
已知函数f(x)=Asin(3x+φ)(A>0,x∈-∞,+∞),0<φ<π)在x=π/12时取得最大值4.(1)求f(x)单调增区间 (2)求函数f(x)在[0,π/3]上的值域 QAQ
解析:因为函数f(x)=Asin(3x+φ)(A>0,x∈-∞,+∞),0<φ<π)在x=π/12时取得最大值4
所以,A=4==>f(x)=4sin(3x+φ)
f(π/12)=4sin(π/4+φ)=4==>π/4+φ=π/2==>φ=π/4
f(x)=4sin(3x+π/4)
单调增区间 :2kπ-π/2
f(x)=Asin(3x+φ)(A>0,x∈-∞,+∞),0<φ<π)在x=π/12时取得最大值4,
∴A=4,π/4+φ=(2k+1/2)π,k∈Z,取k=0,φ=π/4,
∴f(x)=4sin(3x+π/4),
(1)f(x)的增区间由(2k-1/2)π<3x+π/4<(2k+1/2)π确定,
各减π/4,得(2k-3/4)π<3x<(2k+1/4)π,
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f(x)=Asin(3x+φ)(A>0,x∈-∞,+∞),0<φ<π)在x=π/12时取得最大值4,
∴A=4,π/4+φ=(2k+1/2)π,k∈Z,取k=0,φ=π/4,
∴f(x)=4sin(3x+π/4),
(1)f(x)的增区间由(2k-1/2)π<3x+π/4<(2k+1/2)π确定,
各减π/4,得(2k-3/4)π<3x<(2k+1/4)π,
各除以3,得(2k/3-1/4)π
sinu的值域是[-√2/2,1],
∴f(x)=4sinu的值域是[-2√2,4].
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