Question

设f(x)在点x=1处取得极值，且点(2，4)是曲线y=f(x)的拐点，又若f(x)=3x2+2ax+b，求f(x)．

A.

Answer 1

【答案】：由f^'(x)=3x²+2ax+b得f^''(x)=6x+2a由f(x)在x=1处取得极值得f^'(1)=0即3+2a+b=0又点(24)是曲线的拐点故f^''(2)=0即12+2a=0由以上两式可得a=-6b=9所以f^'(x)=3x²—12x+9两边积分得f(x)=x³—6x²+9x+C再将点(24)代入得C=2故f(x)=x³—6x²+9x+2．
由f^'(x)=3x²+2ax+b得,f^''(x)=6x+2a,由f(x)在x=1处取得极值,得f^'(1)=0,即3+2a+b=0,又点(2,4)是曲线的拐点,故f^''(2)=0,即12+2a=0,由以上两式可得a=-6,b=9,所以f^'(x)=3x²—12x+9,两边积分得,f(x)=x³—6x²+9x+C,再将点(2,4)代入得C=2,故f(x)=x³—6x²+9x+2．