Question

已知函数f(x)=4cosxsin(x+π/6)-1,求f(x)在区间[-π/6,π/4]上的最大值和最小值

Answer 1

f(x)=4cosxsin(x+π/6)-1
=4cosx(√3/2sinx+1/2cosx)-1
=2√3sinx+2cos^2x-1
=√3sin2x+cos2x
= 2sin(2x+π/6)
x∈[-π/6,π/4]，则（2x+π/6）∈[-π/6,2π/3]
画个单位圆，一比划就出来了
所以f（x）最大值为2，最小值为-1

Answer 2

f(x)=4cosxsin(x+π/6)-1
=4cosx(√3/2sinx+1/2cosx)-1
=2√3sinx+2cos^2x-1
=√3sin2x+cos2x
= 2sin(2x+π/6)
∵-π/6≤x≤π/4
∴-π/3≤2x≤π/2
∴-π/6≤2x+π/6≤2π/3
∴-1/2≤sin(2x+π/6)≤1
∴-1≤2sin(2x+π/6)≤2
∴当x∈[-π/6，π/4]时，f(x)的最大值为2和最小值为-1

Answer 3

最大值应为2+2=4，最小值应为-1+2=1