Question

函数f(x)=cos2x-cosx+1在[π/3,5π/6]上的值域

Answer 1

f(x)=cos2x-cosx+1
=2cos²x-1-cosx+1
=2cos²x-cosx
看成二次函数，其对称轴=1/4
cosπ/3=1/2
cos5π/6=-√3/2
所以cosx能取到对称轴
f(x)min=2*1/16-1/4=-1/8
f(x)max=2*(-√3/2)^2+√3/2=(3+√3)/2
值域[-1/8,(3+√3)/2]
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Answer 2

f(x)=2cos^2(x)-1-cosx+1=2cos^2(x)-cosx
令t=cosx,
f(t)=t(2t-1)
x在(π/3,5π/6)上，则t=cosx的取值范围在（-√3/2，1/2)
所以t的最小值在f(t)的对称轴上，为-1/8
最大值为f(-√3/2)=（√3+3)/2
所以函数f(x）的值域在(-1/8,(√3+3)/2)上

Answer 3

f(x)=cos2x-cosx+1=2(cox)^2-1-cosx+1=2(cosx)^2-cosx=2(cosx-1/4)^2-1/8
x在[π/3,5π/6] -√3/2 <=cosx<=1/2 -√3/2 -1/4<=cosx-1/4<=1/4
0 <=(cosx-1/4)^2<=(13-4√3)/16 -1/8 <=(cosx-1/4)^2-1/8<=(11-4√3)/16
函数f(x)=cos2x-cosx+1在[π/3,5π/6]上的值域 [-1/8 ,(11-4√3)/16 ]

Answer 4

∵cos2x=2cos²x-1
∴f(x)=cos2x-cosx+1
=2cos²x-cosx
=2(cos²x-1/2cosx+1/16)-1/8
=2(cosx-1/4)²-1/8
∵x∈[π/3,5π/6]
∴cosx∈[-√3/2,1/2]
∴当cosx=1/4时,f(x)min=-1/8
当cosx=-√3/2时，f(x)max=(3+√3)/2
∴f(x)的值域为[-1/8,(3+√3)/2]
参考http://58.130.5.100//