\begin{tabular}{|c||c|c|c|c|}
\hline$x$ & $f(x)$ & $f^{\prime}(x)$ & $f^{\prime \prime}(x)$ & $f^{\prime \prime \prime}(x)$ \
\hline 0 & 3 & -2 & 1 & 4 \
\hline 1 & 2 & -3 & 3 & -2 \
\hline 2 & -1 & 1 & 4 & 5 \
\hline
\end{tabular}
5. 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 $x=1$ ?
(A) $2-3 x+\frac{3}{2} x^{2}-\frac{1}{3} x^{3}$
(B) $2-3(x-1)+\frac{3}{2}(x-1)^{2}-\frac{1}{3}(x-1)^{3}$
(C) $2-3(x-1)+\frac{3}{2}(x-1)^{2}-\frac{2}{3}(x-1)^{3}$
(D) $2-3(x-1)+3(x-1)^{2}-2(x-1)^{3}$

Question

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ The third-degree Taylor polynomial for a function $f$ about $x = 1$ is given by: $$ P_3(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f'''(1)}{3!}(x-1)^3 $$ ‖ ‖ step 2 ⋮ From the table, we have: $$ f(1) = 2, \quad f'(1) = -3, \quad f''(1) = 3, \quad f'''(1) = -2 $$ ‖ ‖ step 3 ⋮ Substituting these values into the Taylor polynomial formula, we get: $$ P_3(x) = 2 - 3(x-1) + \frac{3}{2}(x-1)^2 - \frac{2}{6}(x-1)^3 $$ $$ P_3(x) = 2 - 3(x-1) + \frac{3}{2}(x-1)^2 - \frac{1}{3}(x-1)^3 $$ ‖ ***B*** ∻Key Concept∻ ⚹Taylor Polynomial⚹ ∻Explanation∻ ⚹The Taylor polynomial of degree $n$ for a function $f$ about $x = a$ is given by $P_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n$.⚹◊What is the formula for the Taylor polynomial of degree 3 about $x=1$ for a function $f$ given selected values and derivatives in a table?⍭ Generate me a similar question◊

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