f(x)=Asin(ωx+φ)1f(x)的单调递增区间,2使f(x)取最小值的x的集合.

Question

f(x)=Asin(ωx+φ)
1f(x)的单调递增区间,
2使f(x)取最小值的x的集合.

Answer 1

f(x)=Asin(ωx+φ)
一般假设ω>0,A>0
1、
x=π/4是最大,x=7π/4是最小
则7π/4-π/4=3π/2是半个周期
所以T/2=3π/2
T=3π
T=2π/ω
所以ω=2/3
x=-π/2,y=0
所以0=Asin(-2/3*π/2+φ)
-π/3+φ=0
φ=π/3
所以f(x)=Asin(2x/3+π/3)
sinx增区间是(2kπ-π/2,2kπ+π/2)
所以2kπ-π/2<2x/3+π/3<2kπ+π/2
2kπ-5π/6<2x/3<2kπ+π/6
3kπ-5π/4sinx减区间是(2kπ+π/2,2kπ+3π/2)
同理求得减区间
所以
增区间(3kπ-5π/4,3kπ+π/4)
减区间(3kπ+π/4,3kπ+7π/4)
2、
f(x)=Asin(2x/3+π/3)
A>0则sin(2x/3+π/3)=-1时最小
所以2x/3+π/3=2kπ-π/2
2x/3=2kπ-5π/6
x=3kπ-5π/4
所以集合是{x|x=3kπ-5π/4,k∈Z}