B
分析:求出函數(shù)的導(dǎo)數(shù),再令導(dǎo)數(shù)等于0,求出c 值,再檢驗函數(shù)的導(dǎo)數(shù)是否滿足在x=2處左側(cè)為正數(shù),右側(cè)為負(fù)數(shù),
把不滿足條件的 c值舍去.
解答:∵函數(shù)f(x)=x(x-c)
2=x
3-2cx
2+c
2x,它的導(dǎo)數(shù)為f
′(x)=3x
2-4cx+c
2,
由題意知,在x=2處的導(dǎo)數(shù)值為 12-8c+c
2=0,∴c=6,或 c=2,
又函數(shù)f(x)=x(x-c)
2在x=2處有極大值,故導(dǎo)數(shù)值在x=2處左側(cè)為正數(shù),右側(cè)為負(fù)數(shù).
當(dāng)c=2時,f
′(x)=3x
2-8x+4=3(x-
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)(x-2),不滿足導(dǎo)數(shù)值在x=2處左側(cè)為正數(shù),右側(cè)為負(fù)數(shù).
當(dāng)c=6時,f
′(x)=3x
2-24x+36=3(x
2-8x+12)=3(x-2)(x-6),
滿足導(dǎo)數(shù)值在x=2處左側(cè)為正數(shù),右側(cè)為負(fù)數(shù).故 c=6.
故選 B.
點(diǎn)評:本題考查函數(shù)在某點(diǎn)取得極大值的條件:導(dǎo)數(shù)值等于0,且導(dǎo)數(shù)在該點(diǎn)左側(cè)為正數(shù),右側(cè)為負(fù)數(shù).