$$ \begin{array} {|c||c|c|c|c|} \hline \hspace{1em}x\hspace{1em} & f(x) & f'(x) & f''(x) & f'''(x) \\ \hline 0 & 3 & -2 & 1 & 4 \\ \hline 1 & 2 & -3 & 3 & -2 \\ \hline 2 & -1 & 1 & 4 & 5 \\ \hline \end{array} $$

Selected values of a function \(f\) and its first three derivatives are indicated in the table above. What is the third-degree Taylor polynomial for \(f\) about \(x=1\) ?

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A Taylor polynomial is given by:

$$ P_n(x) = f(c) + f'(c)(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \frac{f'''(x)}{3!}(x-c)^3 +...+ \frac{f^{(n)}c}{n!}(x-c)^n $$

Substituting \(c=1\) and \(n=3\) gives:

$$ P_3(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f'''(1)}{3!}(x-1)^3 $$ $$ = \boxed{2-3(x-1)+\frac{3}{2}(x-1)^2-\frac{1}{3}(x-1)^3 } $$

One can attempt to take the first, second, and third derivatives of each of the answer choices and evaluate \(f(1), f'(1), f''(1), \text{ and } f'''(1)\). For example, for the correct option:

$$ f(x)=2-3(x-1)+\frac{3}{2}(x-1)^2-\frac{1}{3}(x-1)^3 $$

We first check \(f(1)\).

Note that when plugging in \(x=1\), all of the terms become zero except the constant.

$$ f(1)= 2   \checkmark $$

Checking \(f'(1)\)

$$ f(x)=2-3(x-1)+\frac{3}{2}(x-1)^2-\frac{1}{3}(x-1)^3 $$ $$ f'(x) = -3+3(x-1)-(x-1)^2 $$ $$ f'(1)= -3   \checkmark $$

Checking \(f''(1)\)

$$ f''(x)=3-2(x-1) $$ $$ f''(1)=3   \checkmark $$

Checking \(f'''(1)\)

$$ f'''(x)=-2 $$ $$ f'''(1)=-2   \checkmark $$