题目

已知函数f(x)=Asin(ωx+ψ09A>0,ω>0,|ψ|

提问时间：2021-03-31

答案

(1)
1/2T=2π==>T=4π=2π/w
==>w=1/2
f(x)=Asin(1/2x+ψ)
因为(x0,2)在图像上,所以A=2
f(x)=2sin(1/2x+ψ)
因为（0,1）在图像上,所以sinψ=1/2
==>ψ=π/6
f(x)=2sin(1/2x+π/6)
因为(x0,2)在图像上,所以
sin(1/2x0+π/6)=1
1/2x0+π/6=π/2
==>x0=π/6
(2)
将代入到标准正弦y=sint的单调增区间中去求解得：
-π/2+2kπ≤1/2x+π/6≤π/2+2kπ
==>-4π/3+4kπ≤ x ≤2π/3+4kπ
单调增区间为：
【-4π/3+4kπ ,2π/3+4kπ】
(3)
f(x)=2sin(1/2x+π/6)

-π≤x≤π ==> -π/3≤1/2x+π/6≤2π/3
-√3/2≤sin(1/2x+π/6)≤1
-√3≤f(x)≤2
f(x)的值域为：【-√3,2】