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dougliedigg

  • 2 years ago

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

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  1. Cutler
    • 2 years ago
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    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

  2. dougliedigg
    • 2 years ago
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    thanks so muchhh :)

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