Discussion
Let \(f: \mathbb{R} \to \mathbb{R}\) be a function such that \(f(x) = x^{3} + x^{2}f'(1) + xf''(2) + f'''(3)\) , then \(f'''(3)\)=
Answer: B
The value is \(f'''(3) = 6\).
The numbers \(f'(1)\), \(f''(2)\) and \(f'''(3)\) are constants, so f is a cubic \(x^{3} + ax^{2} + bx + c\) with leading coefficient 1.
\[ \begin{aligned} f'(x) &= 3x^{2} + 2ax + b \\ f''(x) &= 6x + 2a \\ f'''(x) &= 6 \end{aligned} \]
The third derivative of a cubic with leading coefficient 1 is the constant 3! = 6, whatever a, b and c are, so \(f'''(3) = 6\). 3 and 9 come from stopping at a lower derivative or multiplying by the point 3, and \(f''(2)\) is one of the coefficients, not the answer.
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