AP Calculus BC 10.12 Lagrange Error Bound - FRQs - Exam Style Questions
Calc-Ok Question
The function f has derivatives of all orders for all real numbers. It is known that \(f(0)=2\), \(f'(0)=3\), \(f”(x)=-f(x^{2})\), and \(f^{(3)}(x)=-2x\,f'(x^{2})\).
(a) Find \(f^{(4)}(x)\), the fourth derivative of \(f\) with respect to \(x\). Write the fourth–degree Taylor polynomial for \(f\) about \(x=0\). Show the work that leads to your answer.
(b) The fourth–degree Taylor polynomial for \(f\) about \(x=0\) is used to approximate \(f(0.1)\). Given that \(\bigl|f^{(5)}(x)\bigr|\le 15\) for \(0\le x\le 0.5\), use the Lagrange error bound to show that this approximation is within \(\displaystyle \frac{1}{10^{5}}\) of the exact value of \(f(0.1)\).
(c) Let \(g\) be the function such that \(g(0)=4\) and \(g'(x)=e^{x}f(x)\). Write the second–degree Taylor polynomial for \(g\) about \(x=0\).
Most-appropriate topic codes (from the AP Calculus AB/BC CED):
• TOPIC 10.11: Finding Taylor Polynomial Approximations of Functions — parts (a, c)
• TOPIC 10.12: Lagrange Error Bound — part (b)
• TOPIC 10.12: Lagrange Error Bound — part (b)
▶️ Answer/Explanation
(a) Find \(f^{(4)}(x)\) and \(T_{4}(x)\) about \(x=0\)
Product + chain rule on \(f^{(3)}(x)=-2x\,f'(x^{2})\):\[ f^{(4)}(x)=\frac{d}{dx}\!\big[-2x\,f'(x^{2})\big] =-2\,f'(x^{2})+(-2x)\cdot\big(f”(x^{2})\cdot 2x\big) =-2\,f'(x^{2})-4x^{2}f”(x^{2}). \] Values at \(x=0\): \(f(0)=2\), \(f'(0)=3\), \(f”(0)=-f(0)=-2\), \(f^{(3)}(0)=0\), \(f^{(4)}(0)=-2\,f'(0)=-6\).
Taylor polynomial: \[ T_{4}(x)=f(0)+f'(0)x+\frac{f”(0)}{2!}x^{2}+\frac{f^{(3)}(0)}{3!}x^{3}+\frac{f^{(4)}(0)}{4!}x^{4} =2+3x-\!x^{2}-\frac{1}{4}x^{4}. \]
(b) Lagrange error bound at \(x=0.1\)
With \(\displaystyle M=\max_{0\le x\le 0.5}\!\bigl|f^{(5)}(x)\bigr|\le 15\), the remainder after degree 4 satisfies \[ \bigl|f(0.1)-T_{4}(0.1)\bigr| \le \frac{M}{5!}\,(0.1)^{5} \le \frac{15}{120}\cdot 10^{-5} =\frac{1}{8}\cdot 10^{-5} =1.25\times 10^{-6} <\frac{1}{10^{5}}. \] Hence the approximation by \(T_{4}\) is within \(1/10^{5}\) of the exact value.(c) Second-degree Taylor polynomial for \(g\) about \(x=0\)
\(g(0)=4\). \(g'(x)=e^{x}f(x)\Rightarrow g'(0)=e^{0}f(0)=2\). \(g”(x)=e^{x}f(x)+e^{x}f'(x)=e^{x}\!\big(f(x)+f'(x)\big)\Rightarrow g”(0)=2+3=5\).Therefore \[ T_{2,g}(x)=g(0)+g'(0)x+\frac{g”(0)}{2}x^{2} =4+2x+\frac{5}{2}x^{2}. \] (Construction of Taylor polynomials)