f`(x)=3ax2+2bx+c
f``(x)=6ax+2b
f```(x)=6a
f(–1)=–a+b–c+d
–a+b–c+d=1
f`(x)=3ax2+2bx+c
f`(–1)=3a–2b+c
3a–2b+c=0
f``(x)=6ax+2b
f``(–1)=–6a+2b
–6+2b=0
f```(x)=6a
f```(–1)=6
6a=6 ⇒a=1
–6a+2b=0 ⇒ –6·1+2b=0 ⇒ b=3
3a–2b+c=0⇒ 3·1 –2·3+c=0⇒c=3
–a+b–c+d=1 ⇒–1+3–3+d=0⇒d=1
О т в е т. f(x)=x3+3x2+3x+1