问题
解答题
设x为正整数,则函数y=x2-x+
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答案
∵y=x2-x+
=x(x-1)+1-1 x
=1+x-1 x
=1+x2(x-1)-(x-1) x
=1+(x-1)(x2-1) x
,(x-1)2(x+1) x
∵x为正整数,
∴
≥0,(x-1)2(x+1) x
当x=1时,
=0,(x-1)2(x+1) x
∴y=1+
≥1.(x-1)2(x+1) x
∴函数y=x2-x+
的最小值是1.1 x