AP Calculus BC 10.11 Finding Taylor Polynomial Approximations of Functions - MCQs - Exam Style Questions -
Question
Let \( f(1)=2 \) and for \( n\ge 1 \) let \( f^{(n)}(1)= -\dfrac{n}{(2n-1)^{2}} \). Estimate \( f(2) \) using the third-degree Taylor polynomial for \( f \) about \( x=1 \).
(B) \( \dfrac{4}{9}-\dfrac{4}{25} \)
(C) \( 1-\dfrac{2}{9}-\dfrac{3}{25} \)
(D) \( 1-\dfrac{1}{9}-\dfrac{1}{50} \)
▶️ Answer/Explanation
Correct answer: (D)
Third-degree Taylor at \(x=1\): \[ T_3(2)= f(1) + f'(1)(1) + \frac{f”(1)}{2!}(1)^2 + \frac{f^{(3)}(1)}{3!}(1)^3. \] From the formula, \[ f'(1)=-1,\quad f”(1)=-\frac{2}{9},\quad f^{(3)}(1)=-\frac{3}{25}. \] Hence \[ T_3(2)=2-1+\frac{-2/9}{2}+\frac{-3/25}{6} \;=\; 1-\frac{1}{9}-\frac{1}{50}. \]
No-Calc Question
| \(x\) | \(f(x)\) | \(f'(x)\) | \(f”(x)\) | \(f”'(x)\) |
|---|---|---|---|---|
| 0 | 2 | \(-1\) | 3 | 5 |
| 2 | 3 | \(-2\) | 7 | \(-1\) |
(B) \(2 – x + \dfrac{3}{2}x^{2} + \dfrac{5}{3}x^{3}\)
(C) \(3 – 2(x-2) + \dfrac{7}{2}(x-2)^{2} – \dfrac{1}{6}(x-2)^{3}\)
(D) \(3 – 2(x-2) + \dfrac{7}{2}(x-2)^{2} – \dfrac{1}{3}(x-2)^{3}\)
▶️ Answer/Explanation
Use the Taylor polynomial centered at \(x=2\):
\(T_{3}(x)=f(2)+f'(2)(x-2)+\dfrac{f”(2)}{2!}(x-2)^{2}+\dfrac{f”'(2)}{3!}(x-2)^{3}.\)
From the table: \(f(2)=3,\ f'(2)=-2,\ f”(2)=7,\ f”'(2)=-1\).
Thus \(T_{3}(x)=3-2(x-2)+\dfrac{7}{2}(x-2)^{2}-\dfrac{1}{6}(x-2)^{3}\).
✅ Answer: (C)