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∵f′(x)=6x2-6x-12,令∵f′(x)=6x2-6x-12=0,求得x=-1或x=2,列表如下:x0(0,2)2(2,3)3f′(x)-0+f(x)5遞減極小-15遞增-4故函數y在[0,3]上的減區間為[0,2),增區間為[2,3),故函數y在[0,3]上的...
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