1、求值域要具体步骤:y=cos ①x∈[π/6,π/3]②x∈[-π/6,2π/3]③x∈[-2π/3,4π/3]2、y=-2cos(2x-π/4)+1,x∈[0,π/2]的值域
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1、求值域要具体步骤:y=cos①x∈[π/6,π/3]②x∈[-π/6,2π/3]③x∈[-2π/3,4π/3]2、y=-2cos(2x-π/4)+1,x∈[0,π/2]的值域1、求值域要具体步骤:
1、求值域要具体步骤:y=cos ①x∈[π/6,π/3]②x∈[-π/6,2π/3]③x∈[-2π/3,4π/3]2、y=-2cos(2x-π/4)+1,x∈[0,π/2]的值域
1、求值域要具体步骤:y=cos ①x∈[π/6,π/3]②x∈[-π/6,2π/3]③x∈[-2π/3,4π/3]
2、y=-2cos(2x-π/4)+1,x∈[0,π/2]的值域
1、求值域要具体步骤:y=cos ①x∈[π/6,π/3]②x∈[-π/6,2π/3]③x∈[-2π/3,4π/3]2、y=-2cos(2x-π/4)+1,x∈[0,π/2]的值域
由y=cosx,
①当x∈[π/6,π/3]时,单调减函数,
f(π/6)=√3/2,
f(π/3)=1/2
∴1/2≤y≤√3/2.
②x∈[-π/6,2π/3]时,
x=2π/3时最小值y=-1/2,
x=0时,最大值y=1
∴-1/2≤y≤1
③x∈[-2π/3,4π/3]时,
x=0,最大值y=1
x=π,最小值y=-1.
∴-1≤y≤1.
(2)由y=-2cos(2x-π/4)+1
x∈[0,π/2],
其中:周期T=2π/2=π,
初相:Θ=π/4,
振幅:A=2(取正)
2x-π/4=0,即x=π/8时,
y取最小值y=-1
2x-π/4=π/2,即x=π/8时,
最大值y1=1.