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6. The function $f$ has derivatives of all orders for all real numbers. It is known that $f(0)=2, f^{\prime}(0)=3$, $f^{\prime \prime}(x)=-f\left(x^{2}\right)$, and $f^{\prime \prime \prime}(x)=-2 x \cdot f^{\prime}\left(x^{2}\right)$.
(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 $\left|f^{(5)}(x)\right| \leq 15$ 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 $f(0.1)$.
(c) Let $g$ be the function such that $g(0)=4$ and $g^{\prime}(x)=e^{x} f(x)$. Write the second-degree Taylor polynomial for $g$ about $x=0$.

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∻Solution by Steps∻ ### Part (a) ‖ step 1 ⋮ Given $f''(x) = -f(x^2)$, we need to find $f^{(4)}(x)$. ‖ ‖ step 2 ⋮ Differentiate $f''(x)$ to find $f'''(x)$: $f'''(x) = -2x \cdot f'(x^2)$. ‖ ‖ step 3 ⋮ Differentiate $f'''(x)$ to find $f^{(4)}(x)$: $f^{(4)}(x) = -2 \cdot f'(x^2) - 4x^2 \cdot f''(x^2)$. ‖ ‖ step 4 ⋮ Evaluate at $x=0$: $f^{(4)}(0) = -2 \cdot f'(0) = -2 \cdot 3 = -6$. ‖ ‖ step 5 ⋮ The fourth-degree Taylor polynomial for $f$ about $x=0$ is $P_4(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4$. ‖ ‖ step 6 ⋮ Substitute the known values: $P_4(x) = 2 + 3x - \frac{1}{2}x^2 - \frac{1}{6}x^3 - \frac{1}{4}x^4$. ‖ ### Part (b) ‖ step 1 ⋮ The Lagrange error bound for the Taylor polynomial is given by $R_n(x) = \frac{M}{(n+1)!} |x|^{n+1}$, where $M$ is the maximum value of $|f^{(n+1)}(x)|$ on the interval. ‖ ‖ step 2 ⋮ Given $|f^{(5)}(x)| \leq 15$ for $0 \leq x \leq 0.5$, we use $M = 15$. ‖ ‖ step 3 ⋮ For $x = 0.1$ and $n = 4$, the error bound is $R_4(0.1) = \frac{15}{5!} (0.1)^5 = \frac{15}{120} (0.1)^5 = \frac{15}{120} \cdot 10^{-5} = \frac{1}{8} \cdot 10^{-5} = \frac{1}{10^5}$. ‖ ‖ step 4 ⋮ Therefore, the approximation $P_4(0.1)$ is within $\frac{1}{10^5}$ of the exact value of $f(0.1)$. ‖ ### Part (c) ‖ step 1 ⋮ Given $g(0) = 4$ and $g'(x) = e^x f(x)$, we need to find the second-degree Taylor polynomial for $g$ about $x=0$. ‖ ‖ step 2 ⋮ The Taylor polynomial is $P_2(x) = g(0) + g'(0)x + \frac{g''(0)}{2!}x^2$. ‖ ‖ step 3 ⋮ Evaluate $g'(0) = e^0 f(0) = 1 \cdot 2 = 2$. ‖ ‖ step 4 ⋮ Differentiate $g'(x)$ to find $g''(x)$: $g''(x) = e^x f(x) + e^x f'(x)$. ‖ ‖ step 5 ⋮ Evaluate $g''(0) = e^0 f(0) + e^0 f'(0) = 1 \cdot 2 + 1 \cdot 3 = 5$. ‖ ‖ step 6 ⋮ Substitute the known values: $P_2(x) = 4 + 2x + \frac{5}{2}x^2$. ‖ ∻Answer∻ ⚹ Part (a): $f^{(4)}(0) = -6$, Taylor polynomial: $P_4(x) = 2 + 3x - \frac{1}{2}x^2 - \frac{1}{6}x^3 - \frac{1}{4}x^4$ ⚹ ⚹ Part (b): The approximation $P_4(0.1)$ is within $\frac{1}{10^5}$ of the exact value of $f(0.1)$. ⚹ ⚹ Part (c): Second-degree Taylor polynomial for $g$: $P_2(x) = 4 + 2x + \frac{5}{2}x^2$ ⚹ ∻Key Concept∻ ⚹Taylor Polynomial and Lagrange Error Bound⚹ ∻Explanation∻ ⚹The Taylor polynomial provides an approximation of a function around a point, and the Lagrange error bound gives an estimate of the error in this approximation◊What is the formula for the Lagrange error bound when approximating a function using a Taylor polynomial?⍭ Generate me a similar question◊⚹

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