Question

已知函数f(x)=2sin(1/2x+∅),∅∈(0,π/2),f(0)=根号3.求f(x)的解析式

求f（x）在区间（0，π）上的最大值和最小值

Answer 1

f(0)=√3=2sin（0.5*0+φ）
sinφ=√3/2，又因为φ∈(0,π/2)，所以φ=π/3
f(x)=2sin(1/2x+π/3)
f'(x)=0.5*2*cos(1/2x+π/3)=cos(1/2x+π/3)
令f'(x)=0，解得x=π/3 (就区间（0，π）而言)
所以f（x）在区间（0，π/3）单调递增，在区间(π/3,π)单调递减
f(x)max=f(π/3)=2sin(0.5*π/3+π/3)=2
f(x)min=f(π)=2sin(0.5*π+π/3)=2*1/2=1

Answer 2

f(0)=根号3，代入求出∅=π/3 所以f(x)=2sin(1/2x+π/3)
0<x<π, π/3<1/2x+π/3<5π/6
1<f(x)<=2