已知函数f(x)=Asin(ωx+ψ09A>0,ω>0,|ψ|
1个回答

(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】

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