Question

Find a polynomial equation of degree 3 such that f(0)=31, f'(1)=4, f"(2)=2, and f'"(3)=6.

Answer 1

You know that the function has to be f(x) = ax^3 + bx^2 + cx + 31.
Next you know that the f'(x) = 3ax^2 + 2bx + c, and that f'(1) = 4, so we know that 3a + 2b + c = 4
Then f''(x) = 6ax + 2b, and f''(2) = 2, Therefore 12a + 2b = 2
Last but not least f'''(x) = 6a, and f'''(3) = 6, and with that you know that a = 1
Go back to 12a + 2b = 2, plug in a, and you get 12 + 2b = 2, therefore b = -5
Now you can go back to 3a + 2b + c = 4, plug in, and you will get 3 + (-10) + c = 4, and c = 11.
Finally, you get the equation that f(x) = x^3 - 5x^2 + 11x + 31

Answer 2

thanks so muchhh :)