Question

已知过f(x)=x^3+ax^2+1的图像上一点B(1,b)的切线斜率为-3,g(x)=-f(x)-3x^2+tx+1,

Answer 1

f'(x)=3x^2+2ax

k=y'|(x=1)=3+2a=-3 a=-3

g(x)=-x^3+3x^2-1-3x^2+tx+1=-x^3+tx

g'(x)=-3x^2+t

(1) t