Question

已知函数f(x)=1/3X3+1/2X2-(2+2a)X+b ①若曲线y=f(x)在点P（1,f(1))处得切线方程为y

Answer 1

.f(x)在P(1,f(1))处切线方程是y=1/2

所以有f(1)=1/2,且f'(1)=0,即

f(1)=1/3a+1/2-(2+2a)+b=1/2------------------①

f'(1)=a+1-(2+2a)=0------------------------------②

联立方程①②,可求出a=3,b=7