We are tasked with finding the 4th-degree Taylor polynomial $T_4(x)$ for $f(x) = x^{11/2}$, centered at $a = 1$, and using the Error Bound to find the maximum possible size of the error at $x = 1.4$.

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Step 1: Compute derivatives of $f(x)$

The function is $f(x) = x^{11/2}$. Its derivatives are:

1. $f'(x) = \frac{11}{2} x^{9/2}$
2. $f''(x) = \frac{11 \cdot 9}{4} x^{7/2} = \frac{99}{4} x^{7/2}$
3. $f'''(x) = \frac{11 \cdot 9 \cdot 7}{8} x^{5/2} = \frac{693}{8} x^{5/2}$
4. $f^{(4)}(x) = \frac{11 \cdot 9 \cdot 7 \cdot 5}{16} x^{3/2} = \frac{3465}{16} x^{3/2}$

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Step 2: Evaluate derivatives at $a = 1$

At $x = 1$:

1. $f(1) = 1^{11/2} = 1$
2. $f'(1) = \frac{11}{2} \cdot 1^{9/2} = \frac{11}{2}$
3. $f''(1) = \frac{99}{4} \cdot 1^{7/2} = \frac{99}{4}$
4. $f'''(1) = \frac{693}{8} \cdot 1^{5/2} = \frac{693}{8}$
5. $f^{(4)}(1) = \frac{3465}{16} \cdot 1^{3/2} = \frac{3465}{16}$

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Step 3: Write the Taylor polynomial $T_4(x)$

The Taylor polynomial centered at $a = 1$ is: $T_4(x) = f(1) + f'(1)(x - 1) + \frac{f''(1)}{2!}(x - 1)^2 + \frac{f'''(1)}{3!}(x - 1)^3 + \frac{f^{(4)}(1)}{4!}(x - 1)^4$

Substitute the evaluated derivatives: $T_4(x) = 1 + \frac{11}{2}(x - 1) + \frac{99}{4} \cdot \frac{1}{2}(x - 1)^2 + \frac{693}{8} \cdot \frac{1}{6}(x - 1)^3 + \frac{3465}{16} \cdot \frac{1}{24}(x - 1)^4$

Simplify coefficients: $T_4(x) = 1 + \frac{11}{2}(x - 1) + \frac{99}{8}(x - 1)^2 + \frac{231}{16}(x - 1)^3 + \frac{1155}{128}(x - 1)^4$

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Step 4: Compute the error bound

The error bound for the Taylor polynomial is given by: $|f(x) - T_4(x)| \leq \frac{|f^{(5)}(c)|}{5!} |x - a|^5$ for some $c$ in $(1, 1.4)$.

1. Compute $f^{(5)}(x)$: $f^{(5)}(x) = \frac{11 \cdot 9 \cdot 7 \cdot 5 \cdot 3}{32} x^{1/2} = \frac{10395}{32} x^{1/2}$

2. Find the maximum $|f^{(5)}(c)|$ for $c \in (1, 1.4)$: Since $f^{(5)}(x)$ is increasing for $x > 0$, the maximum occurs at $c = 1.4$: $f^{(5)}(1.4) = \frac{10395}{32} \sqrt{1.4} = \frac{10395}{32} \cdot 1.183$

3. Compute the error bound: $|f(x) - T_4(x)| \leq \frac{f^{(5)}(1.4)}{120} |1.4 - 1|^5$ $= \frac{\frac{10395}{32} \cdot 1.183}{120} (0.4)^5$ $= \frac{12296.685}{3840} \cdot 0.01024$ $= 0.0327$

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Final Answer

1. $T_4(1.4) = 1 + \frac{11}{2}(0.4) + \frac{99}{8}(0.4)^2 + \frac{231}{16}(0.4)^3 + \frac{1155}{128}(0.4)^4 = 2.066$
2. Error Bound: $|f(1.4) - T_4(1.4)| \leq 0.033 \quad \text{or } 3.3 \times 10^{-2}$

Would you like further clarification or details on any step?

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Related Questions:

1. What is the significance of the error bound in Taylor polynomial approximations?
2. How does the choice of $a$ affect the accuracy of a Taylor polynomial?
3. What happens if we compute $T_5(x)$ instead of $T_4(x)$ in this problem?
4. How would you calculate the exact value of $f(1.4)$ to compare it with $T_4(1.4)$?
5. Why does the error increase as $|x - a|$ increases?

Tip: For better accuracy, always use software or tools to compute factorials and derivatives when working with higher-degree Taylor polynomials.