Question

三次函數(shù)當(dāng)x=1時(shí)有極大值4,當(dāng)x=3時(shí),有極小值0,且函數(shù)過原點(diǎn),則此函數(shù)是(    )

A.y=x³+6x²+9x                                  B.y=x³-6x²+9x

C.y=x³-6x²-9x                                   D.y=x³+6x²-9x

Answer 1

答案：B  三次函數(shù)過原點(diǎn),可設(shè)f(x)=x³+bx²+cx,f′(x)=3x²+2bx+c,由題設(shè)知,f′(1)=3+2b+c=0,f′(3)=27+6b+c=0,

∴b=-6,c=9.∴f(x)=x³-6x²+9x;f′(x)=3x²-12x+9=3(x-1)(x-3).

當(dāng)x=1時(shí),f(x)極大值為4;

當(dāng)x=3時(shí),f(x)極小值為0,滿足條件.