如果僅求單調(diào)區(qū)間
T/2=π/2+π/3=5π/6
∴T=5π/3
圖上左面一段交點(diǎn)是(-π/3-5π/6,0)=(-7π/6,0)
所以增區(qū)間是[-7π/6+k*5π/3,-π/3+k*5π/3],k∈Z
減區(qū)間是[-π/3+k*5π/3,π/2+k*5π/3],k∈Z
如果求解析式
下面
最大值=3
最小值=0
∴a+b=3
-a+b=0
b=3/2
a=3/2
f(x)=3/2sin(ωx+φ)+3/2
T/2=π/2+π/3=5π/6
∴T=5π/3
∴ω=2π/T=6/5
f(x)=3/2sin(6/5x+φ)+3/2
將x=-π/3代入得
-2π/5+φ=π/2+2kπ,k∈Z
∴φ=9π/10+2kπ,k∈Z
∵-π