2023 FRQ

Part B

Question 6

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 \(f'''(x) = -2x \cdot f'(x^2)\)

1. Find \(f^{(4)}(x)\), the forth 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.

   Solution

   Forth derivative of \(f\) with respect to \(x\):

   \[ f^{(4)}(x) = -4x^2f''(x^2) - 2f'(x^2) \]

   Derivatives of \(f^n(x)\) at \(x=0\):

   \[ f(0) = 2, f'(0) = 3, f''(0) = -2, f'''(0) = 0, f^{(4)}(0) = -6 \]

   Fourth degree Taylor polynomial for \(f\) about \(x=0\):

   \[ T_4(x) = 2 + 3x -\frac{2}{2!}x^2 - \frac{6}{4!}x^4\\ \]

2. The forth-degree Taylor polynomial for \(f\) about \(x=0\) is used to approximate \(f(0.1)\). Given that \(|f^{(5)}(x)| \leq 15\) for \(0 \leq x \leq 0.5\), use the Lagrange error bound to show that this approximation is within \(\frac{1}{10^5}\) of the exact value of \(f(0.1)\).

   Solution

   By the Lagrange error bound, the error \(|R_n| \leq \frac{f^{(n+1)}(z)|x-a|^{n+1}}{(n+1)!}\) must be less than \(\frac{1}{10^5}\):

   \[ \begin{aligned} \left|\frac{1}{10^5}\right| &\leq \frac{15|0.1-0|^5}{5!}\\ &\leq \frac{1}{8} \cdot \frac{1}{10^5}\\ &\leq \frac{1}{10^5} \end{aligned} \]

3. Let \(g\) be the function such that \(g(0)=4\) and \(g'(x) = e^xf(x)\). Write the second-degree Taylor polynomial for \(g\) about \(x=0\).

   Solution

   The second derivative of \(g\) with respect to \(x\):

   \[ g''(x) = e^xf'(x) + f(x)e^x \]

   Derivatives of \(g^n(x)\) at \(x=0\):

   \[ g(0) = 4, g'(0) = 2, g''(0) = 5 \]

   Second degree Taylor polynomial for \(g\) about \(x=0\):

   \[ T_2(x) = 4 + 2x + \frac{5}{2!}x^2 \]