Question

已知函数f(x）=√3/2sin2x-cos²x-1/2 (x∈R）当x∈[-π/2，5π/12]时，求函数的最大值和最小值。

求详细过程！！拜托了

Answer 1

f（x）=(√3/2)sin2x-(1/2)(1+cos2x)-(1/2)

=(√3/2)sin2x-(1/2)cos2x
=sin2xcox(π/6) - cos2xsin(π/6)
=sin(2x - π/6)
∵ x∈[-π/12,5π/12]

∴ 2x - π/6 ∈ [-π/3,2π/3]
令 2x - π/6 =t, 则f(x) = sint ,t∈ [-π/3,2π/3];
画y= sint ,t∈ [-π/3,2π/3]的图像，由图像知道t= -π/3时，y值最小为：sin( -π/3)= -√3/2;
t= π/2时，y值最大为：sin( π/2)= 1；
即f（x）的最大值和最小值分别为：1 ， -√3/2

Answer 2

f(x)=√3/2*sin2x-1/2*(1+cos2x)-1/2
=√3/2*sin2x-1/2*cos2x-1
=sin(2x-π/6)-1
∵-π/2≤x≤5π/12，∴-7π/6≤2x-π/6≤2π/3
∴-1≤sin(2x-π/6)≤1
∴f(x)max=1-1=0，f(x)min=-1-1=-2

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