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/4f(x)=4sin(3x+π/4)单调增区间 ：2kπ-π/2...