We aim to find the 4th-degree Taylor polynomial $T_4(x)$ for $f(x) = x^{1/12}$ centered at $a = 1$, and then calculate the error bound at $x = 1.4$.

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

The function is $f(x) = x^{1/12}$.

1. $f'(x) = \frac{1}{12}x^{-11/12}$
2. $f''(x) = \frac{-11}{144}x^{-23/12}$
3. $f'''(x) = \frac{11 \cdot 23}{1728}x^{-35/12}$
4. $f^{(4)}(x) = \frac{-11 \cdot 23 \cdot 35}{20736}x^{-47/12}$

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Step 2: Evaluate the derivatives at $a = 1$

At $x = 1$, $f^{(n)}(1)$ simplifies since $1^n = 1$:

1. $f(1) = 1^{1/12} = 1$
2. $f'(1) = \frac{1}{12}$
3. $f''(1) = \frac{-11}{144}$
4. $f'''(1) = \frac{11 \cdot 23}{1728} = \frac{253}{1728}$
5. $f^{(4)}(1) = \frac{-11 \cdot 23 \cdot 35}{20736} = \frac{-8855}{20736}$

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Step 3: Write the 4th-degree Taylor polynomial $T_4(x)$

The general form of the Taylor polynomial is:

$T_4(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f'''(1)}{3!}(x-1)^3 + \frac{f^{(4)}(1)}{4!}(x-1)^4$

Substituting the values:

$T_4(x) = 1 + \frac{1}{12}(x-1) + \frac{\frac{-11}{144}}{2}(x-1)^2 + \frac{\frac{253}{1728}}{6}(x-1)^3 + \frac{\frac{-8855}{20736}}{24}(x-1)^4$

Simplify each term:

1. $f'(1)(x-1) = \frac{1}{12}(x-1)$
2. $\frac{f''(1)}{2}(x-1)^2 = \frac{-11}{288}(x-1)^2$
3. $\frac{f'''(1)}{6}(x-1)^3 = \frac{253}{10368}(x-1)^3$
4. $\frac{f^{(4)}(1)}{24}(x-1)^4 = \frac{-8855}{497664}(x-1)^4$

Thus:

$T_4(x) = 1 + \frac{1}{12}(x-1) - \frac{11}{288}(x-1)^2 + \frac{253}{10368}(x-1)^3 - \frac{8855}{497664}(x-1)^4$

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Step 4: Error Bound

The error term $R_4(x)$ for a Taylor polynomial is:

$R_4(x) = \frac{f^{(5)}(c)}{5!}(x-1)^5$

Where $c$ is some number in $[1, 1.4]$, and $f^{(5)}(x)$ is the 5th derivative of $f(x)$. Compute $f^{(5)}(x)$:

$f^{(5)}(x) = \frac{11 \cdot 23 \cdot 35 \cdot 47}{248832}x^{-59/12}$

The maximum of $|f^{(5)}(x)|$ occurs at $x = 1.4$ since $x^{-59/12}$ decreases as $x$ increases. Evaluate at $x = 1.4$:

$f^{(5)}(1.4) = \frac{11 \cdot 23 \cdot 35 \cdot 47}{248832} \cdot 1.4^{-59/12}$

Numerically approximating $1.4^{-59/12}$ gives:

$1.4^{-59/12} \approx 0.676$

Thus:

$f^{(5)}(1.4) \approx \frac{11 \cdot 23 \cdot 35 \cdot 47}{248832} \cdot 0.676 = \frac{408845}{248832} \cdot 0.676 \approx 1.109$

Now calculate the error:

$R_4(1.4) \leq \frac{|f^{(5)}(1.4)|}{120}(1.4-1)^5 = \frac{1.109}{120}(0.4)^5$

$R_4(1.4) \leq \frac{1.109}{120} \cdot 0.01024 \approx 0.0000948$

In scientific notation:

$R_4(1.4) \leq 9.48 \times 10^{-5}$

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Final Answer

1. The 4th-degree Taylor polynomial:

$T_4(x) = 1 + \frac{1}{12}(x-1) - \frac{11}{288}(x-1)^2 + \frac{253}{10368}(x-1)^3 - \frac{8855}{497664}(x-1)^4$

2. The error bound:

$|f(1.4) - T_4(1.4)| \leq 9.48 \times 10^{-5}$