已知函数f(x)=x^3+bx^2+cx在点M(2,2)处的切线方程为5x-y-8=0 (1)

题目

已知函数f(x)=x^3+bx^2+cx在点M(2,2)处的切线方程为5x-y-8=0 (1)
已知函数f(x)=x^3+bx^2+cx在点M(2,2)处的切线方程为5x-y-8=0 (1)求函数y=f(x)的解析式(2)求函数y=f(x)的单调区间

答案

1)f'(x)=3x²+2bx+cM(2,2)处的切线为y=5x-8则f(2)=8+4b+2c=2,即2b+c=-3f'(2)=12+4b+c=5,即4b+c=-7两式相减：-2b=4,得b=-2故c=-3-2b=-3+4=1所以f(x)=x³-2x²+x2)f'(x)=3x²-4x+1=(3x-1)(x-1)得极值...