Let's solve the problem step by step.

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Part (a): Identify $f(x)$ and $x_0$

The function $f(x)$ for linear approximation needs to satisfy the relationship:

$\sqrt[3]{3124} \approx f(x_0) + f'(x_0)(x - x_0)$

The choice of $f(x)$ is $f(x) = \sqrt[3]{x} = x^{1/3}$.
The given $x_0$ is 3125 because it's close to 3124 and its cube root is easy to compute ($\sqrt[3]{3125} = 15$).

$f(x) = x^{1/3}, \quad x_0 = 3125$

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Part (b): Linear approximation of $\sqrt[3]{3124}$

Using the formula for linear approximation:

$f(x) \approx f(x_0) + f'(x_0)(x - x_0)$

1. Compute $f(x_0) = \sqrt[3]{3125} = 15$.
2. Derivative: $f'(x) = \frac{1}{3}x^{-2/3}$, so $f'(x_0) = \frac{1}{3}(3125)^{-2/3} = \frac{1}{3 \cdot 15^2} = \frac{1}{675}$.
3. Substitute $x = 3124$:

$f(3124) \approx 15 + \frac{1}{675}(3124 - 3125) = 15 - \frac{1}{675} = 15 - 0.00148 = 14.99852$

Thus: $\sqrt[3]{3124} \approx 14.9985$

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Part (c): Choose $g(x)$ for Newton's Method

Newton's Method approximates roots of functions. To approximate $\sqrt[3]{3124}$, we define $g(x)$ such that $g(\sqrt[3]{3124}) = 0$.
The obvious choice is:

$g(x) = x^3 - 3124$

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Part (d): Choose $x_0$ (initial guess for Newton's Method)

To find an integer $x_0$ close to $\sqrt[3]{3124}$, consider $\sqrt[3]{3125} = 15$. Thus:

$x_0 = 15$

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Part (e): Perform two iterations of Newton's Method

Newton's Method formula is:

$x_{n+1} = x_n - \frac{g(x_n)}{g'(x_n)}$

For $g(x) = x^3 - 3124$, we have $g'(x) = 3x^2$. Starting with $x_0 = 15$:

First Iteration:

$x_1 = x_0 - \frac{g(x_0)}{g'(x_0)} = 15 - \frac{15^3 - 3124}{3 \cdot 15^2}$

1. Compute $g(15) = 15^3 - 3124 = 3375 - 3124 = 251$.
2. Compute $g'(15) = 3 \cdot 15^2 = 3 \cdot 225 = 675$.
3. Substitute:

$x_1 = 15 - \frac{251}{675} = 15 - 0.3719 = 14.6281$

Second Iteration:

$x_2 = x_1 - \frac{g(x_1)}{g'(x_1)}$

1. Compute $g(14.6281) = (14.6281)^3 - 3124$: $14.6281^3 \approx 3125.854 \quad \text{so} \quad g(14.6281) = 3125.854 - 3124 = 1.854$
2. Compute $g'(14.6281) = 3 \cdot (14.6281)^2$: $14.6281^2 \approx 214.082, \quad \text{so} \quad g'(14.6281) = 3 \cdot 214.082 = 642.246$
3. Substitute:

$x_2 = 14.6281 - \frac{1.854}{642.246} = 14.6281 - 0.00289 = 14.6252$

Thus, after two iterations: $x_1 \approx 14.6281, \quad x_2 \approx 14.6252$

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Final Answers:

(a) $f(x) = x^{1/3}, \, x_0 = 3125$
(b) $\sqrt[3]{3124} \approx 14.9985$
(c) $g(x) = x^3 - 3124$
(d) $x_0 = 15$
(e) $x_1 \approx 14.6281, \, x_2 \approx 14.6252$

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Would you like detailed steps for any part or further clarifications?

Related Questions:

1. How does linear approximation differ from Newton's Method?
2. Why is $f(x) = x^{1/3}$ chosen in this problem?
3. How is the derivative $f'(x)$ calculated for $x^{1/3}$?
4. What happens if $x_0$ is not close to the true root in Newton's Method?
5. How can the accuracy of Newton's Method be increased?

Tip: Newton's Method converges faster when the initial guess is close to the actual root.