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

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 $x=1$ ?

Summary Submit Skip Question

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 $f'''(1)$

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