作业帮已知函数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.
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![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.](/uploads/image/z/3914106-42-6.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%EF%BC%89%3D%E2%88%9A3%2F2sin2x-cos%26%23178%3Bx-1%2F2+%28x%E2%88%88R%EF%BC%89%E5%BD%93x%E2%88%88%5B-%CF%80%2F2%2C5%CF%80%2F12%5D%E6%97%B6%2C%E6%B1%82%E5%87%BD%E6%95%B0%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BC.)
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.
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.
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.
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
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]<...
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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
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