AP Calculus BC 10.14 Finding Taylor or Maclaurin Series for a Function - FRQs - Exam Style Questions
No-Calc Question
A function \(f\) has derivatives of all orders for all real numbers \(x\). A portion of the graph of \(f\) is shown above, along with the line tangent to the graph of \(f\) at \(x=0\). Selected derivatives of \(f\) at \(x=0\) are given in the table below.
| \(n\) | \(f^{(n)}(0)\) |
|---|---|
| 2 | \(3\) |
| 3 | \(-\dfrac{23}{2}\) |
| 4 | \(54\) |
Most-appropriate topic codes (CED):
• TOPIC 10.14: Finding Taylor or Maclaurin Series for a Function; using known series like \(e^{x}\) — part (b).
• TOPIC 10.15: Representing Functions as Power Series (operations with series/products) — part (b).
• TOPIC 10.10: Alternating Series Error Bound — part (d).
▶️ Answer/Explanation
Information from the figure & tangent at \(x=0\)
From the graph: \(f(0)=3\). From the tangent line at \(x=0\): slope \(f'(0)=-2\).
(a) Third-degree Taylor polynomial for \(f\) at \(x=0\)
General form: \(T_{3}(x)=f(0)+f'(0)x+\dfrac{f”(0)}{2!}x^{2}+\dfrac{f^{(3)}(0)}{3!}x^{3}\). (Taylor approximation: Topic 10.11.)
Substitute \(f(0)=3\), \(f'(0)=-2\), \(f”(0)=3\), \(f^{(3)}(0)=-\dfrac{23}{2}\):
\[ \boxed{\,T_{3}(x)=3-2x+\frac{3}{2}x^{2}-\frac{23}{12}x^{3}\, }. \]
(b) Series for \(e^{x}\) and the 2nd-degree Taylor polynomial for \(e^{x}f(x)\)
Maclaurin for \(e^{x}\): \(1+x+\dfrac{x^{2}}{2!}+\cdots\). (Known series foundation: Topic 10.14.)
Multiply \(e^{x}\) (through \(x^{2}\)) by \(T_{3}\) (through \(x^{2}\)) and keep terms \(\le x^{2}\):
\[ (1+x+\tfrac{x^{2}}{2})\big(3-2x+\tfrac{3}{2}x^{2}\big) =3\,(1+x+\tfrac{x^{2}}{2})-2x(1+x)+\tfrac{3}{2}x^{2}(1). \] Collect coefficients:
constant \(=3\); \(x\)-term \(=3x-2x=x\); \(x^{2}\)-term \(=\tfrac{3}{2}-2+\tfrac{3}{2}=1\).
Therefore the second-degree Taylor polynomial is \[ \boxed{\,3+x+x^{2}\, }. \] (Operations with series/products: Topic 10.15.)
(c) Approximate \(h(1)\) with \(T_{3}\)
\(h(x)=\displaystyle\int_{0}^{x}f(t)\,dt \approx \int_{0}^{x}T_{3}(t)\,dt.\)
\[ h(1)\approx \int_{0}^{1}\!\Big(3-2t+\tfrac{3}{2}t^{2}-\tfrac{23}{12}t^{3}\Big)\,dt =\Big[3t-t^{2}+\tfrac{1}{2}t^{3}-\tfrac{23}{48}t^{4}\Big]_{0}^{1} =\boxed{\frac{97}{48}}\ \ (\approx 2.020833). \]
(d) Alternating series error bound for \(h(1)\)
The Maclaurin series for \(f\) has next term \(\dfrac{f^{(4)}(0)}{4!}x^{4}=\dfrac{54}{24}x^{4}=\dfrac{9}{4}x^{4}\) (positive). Integrating term-by-term to get the series for \(h(1)\), the first omitted term’s magnitude is \[ \left|\int_{0}^{1}\frac{9}{4}t^{4}\,dt\right|=\frac{9}{4}\cdot\frac{1}{5}=\boxed{\frac{9}{20}}=0.45. \] Because the terms in the series for \(h(1)\) alternate and decrease to \(0\), the alternating series error bound (Topic 10.10) guarantees \[ \big|\,h(1)-\tfrac{97}{48}\,\big|\le \frac{9}{20}\le 0.45. \]