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希望有人愿意帮忙)函数f(x)=sin2x+2根号(2)cos()

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希望有人愿意帮忙)函数f(x)=sin2x+2根号(2)cos()
函数f(x)=sin2x+2根号(2)cos(π/4+x) + 3 的最小值是 _________.
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解:
f(x)=sin2x+2根号(2)cos(π/4+x) + 3 = sin2x + 2(cosx - sinx) + 3.
令cosx - sinx = t, t∈[-√2,√2]
则, f(x) = g(t) = - t^2 + 2t + 4 = - (t - 2)^2 + 5 .
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请问...
这步 t∈[-√2,√2] 是怎么计算来的.
又怎么到 f(x) = g(t) = - t^2 + 2t + 4 = - (t - 2)^2 + 5 . 这步的?.
希望有人愿意帮忙,想要详细演算过程步骤..非常感谢.,短时间里一定采纳,谢谢了.
希望有人愿意帮忙)函数f(x)=sin2x+2根号(2)cos()
f(X)=sin2x+2√2cos(π/4+x)+3
=cos(2x-π/2)+2√2cos(π/4+x)+3
=1-2sin2(x-π/4)-2√2sin(x-π/4)+3
=1-[2sin2(x-π/4)+2√2sin(x-π/4)]+3
=4-[2sin2(x-π/4)+2√2sin(x-π/4)+1]+1
=5-[2sin2(x-π/4)+2√2sin(x-π/4)+1]
=5-[√2sin(x-π/4)+1]^2
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