Question

已知函数fx=Asin(wx+φ)(A>0,w>0,0<φ<π)图像的一部分如图所示:

求函数解析式。求单调递减区间

Answer 1

解析：因为f(x)=Asin(wx+φ)(A>0,w>0,0<φ<π)
由图知：A=2，T/2=π/3+π/6=π/2==>T=π==>w=2
所以，f(x)=2sin(2x+φ)==>f(π/12)=2sin(π/6+φ)=2==>φ=π/3
所以，f(x)=2sin(2x+π/3)
单调减区间：2kπ+π/2<=2x+π/3<=2kπ+3π/2==>kπ+π/12<=x<=kπ+7π/12