Find a third-degree polynomial Q such that Q(1)=0, Q'(1)=0, Q''(1)=2, and Q'''(1)=12.

Answer 1

Q(x)=ax^3+bx^2+cx+d
Q(1)=a+b+c+d=0
Q'(x)=3ax^2+2bx+c
Q'(1)=3a+2b+c=0
Q''(x)=6ax+2b
Q''(1)=6a+2b=2
Q'''(x)=6a
Q'''(1)=6a=12
so u can solve and find the polynom

Answer 2

Another approach might be to work work the problem backwards. For a third degree polynomial, Q''' is just a constant. Set that to 12. Inegrate to get Q''
=12x + C Adjust the constant for the desired result.
Repeat for Q' and Q.

Answer 3

the second answer is better...