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∵x>0時,f(x)=xx2+1=1x+1x≤12x•1x=12,當且僅當x=1時“=”成立;∴在x∈(0,1)時,f(x)是增函數,x∈(1,+∞)時,f(x)是減函數;當x<0時,f(x)=xx2+1=1x+1x≥1−2(−x)•1−x=-12,當且僅當x=-1時...
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