Question

函数f(x)=x^3+ax^2+x在点(1,f(1))处的切线斜率为6,则实数a=?

Answer 1

f'(x)=3x^2+2ax+1
f'(1)=3+2a+1=6
a=1

Answer 2

f(x)=x^3+ax^2+x
f'(x)=3x^2+2ax+1 k=f'(x)|(x=1)=4+2a=6 a=1

Answer 3

如楼上所说 求导比较好办