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∵f(x)=2x4-3x2+1,x∈[12,2]∴f′(x)=8x3-6x=0,解得x=0或x=32或x=-32(舍去),∴x∈[12,32)時,f′(x)<0,函數f(x)為减函數;x∈(32,2]時,f′(x)>0,函數f(x)為增函數;∴f(x)= 2x4-3x2+1在x=…
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