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
问题描述:
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
答
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/4f(x)=4sin(3x+π/4)单调增区间 :2kπ-π/2...