A function has derivatives of all orders with supplied derivative relationships. Students find a higher-order derivative and build a fourth-degree Taylor polynomial about zero, use the Lagrange error bound to bound the approximation error, and construct a Taylor polynomial for a related function.
A function has derivatives of all orders with supplied derivative relationships. Students find a higher-order derivative and build a fourth-degree Taylor polynomial about zero, use the Lagrange error bound to bound the approximation error, and construct a Taylor polynomial for a related function.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
**Fourth derivative.** Differentiate the given $f'''(x)=-2x\,f'(x^2)$ using the product rule (and the chain rule on $f'(x^2)$): $$f^{(4)}(x)=\frac{d}{dx}\Bigl[-2x\,f'(x^2)\Bigr]=-2\,f'(x^2)+(-2x)\,f''(x^2)\cdot 2x.$$ **Derivatives at $0$.** Use $f(0)=2$, $f'(0)=3$, and the given relations: - $f''(0)=-f(0^2)=-f(0)=-2.$ - $f'''(0)=-2(0)\,f'(0)=0.$ - $f^{(4)}(0)=-2\,f'(0)+0=-2(3)=-6.$ **Fourth-degree Taylor polynomial about $x=0$.** $$T_4(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\frac{f^{(4)}(0)}{4!}x^4.$$ Substituting, $$T_4(x)=2+3x+\frac{-2}{2}x^2+\frac{0}{6}x^3+\frac{-6}{24}x^4=2+3x-x^2-\frac{1}{4}x^4.$$
4 points: P1 for correct product-rule form of f4(x), i.e. -2 f'(x^2) + (-2x) f''(x^2)(2x). P2 only for a completely correct f4(x) expression. P3 for two correct terms of T4 (the 2 + 3x - x^2 portion, using f''(0) = -2). P4 for the remaining terms, i.e. the zero x^3 term and the -(1/4)x^4 term; a nonzero x^3 term, any degree >4 term, or a trailing '+...' forfeits P4.
By the Lagrange error bound, the error in using $T_4$ to approximate $f(0.1)$ satisfies $$\bigl|T_4(0.1)-f(0.1)\bigr|\le\frac{\displaystyle\max_{0\le x\le 0.1}\bigl|f^{(5)}(x)\bigr|}{5!}\,(0.1)^5.$$ We are given $\bigl|f^{(5)}(x)\bigr|\le 15$ on $0\le x\le 0.5$ (hence on $0\le x\le 0.1$), so $$\bigl|T_4(0.1)-f(0.1)\bigr|\le\frac{15}{120}\cdot(0.1)^5=\frac{15}{120}\cdot\frac{1}{10^5}=\frac{1}{8}\cdot\frac{1}{10^5}=\frac{1}{800000}.$$ Since $\dfrac{1}{800000}\le\dfrac{1}{10^5}=\dfrac{1}{100000}$, the approximation is within $\dfrac{1}{10^5}$ of the exact value of $f(0.1)$.
2 points: P1 for the correct error-bound form, max|f5|/5! * (0.1)^5, or equivalently (15/5!)(0.1)^5. P2 for communicating the full inequality Error <= (15/5!)(0.1)^5 <= 1/10^5. Writing Error = (15/5!)(0.1)^5 (an equality) or Error = 1/10^5 does not earn P2 — the chain must be an inequality terminating at <= 1/10^5.
We need the second-degree Taylor polynomial for $g$ about $x=0$, i.e. $g(0)$, $g'(0)$, $g''(0)$. - $g(0)=4$ (given). - $g'(x)=e^{x}f(x)\Rightarrow g'(0)=e^{0}f(0)=1\cdot 2=2.$ - Differentiate with the product rule: $g''(x)=e^{x}f(x)+e^{x}f'(x)$, so $g''(0)=e^{0}f(0)+e^{0}f'(0)=f(0)+f'(0)=2+3=5.$ Then $$T_2(x)=g(0)+g'(0)x+\frac{g''(0)}{2!}x^2=4+2x+\frac{5}{2}x^2.$$
3 points: P1 for g''(x) = e^x f(x) + e^x f'(x) (or g''(0) = f(0)+f'(0)). P2 for the first two terms of the polynomial, the form 4 + 2x + ax^2. P3 for the complete correct polynomial 4 + 2x + (5/2)x^2. An alternate path multiplying the Maclaurin series of e^x by the part-(a) polynomial and integrating g'(x) is fully acceptable. Any degree >2 term or a trailing '+...' forfeits P3.
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