Given the equation:
4𝑥cubed + 2𝑥squared − 3𝑥 − 7 = 0 
Use the iterative techniques listed below to find the root of the equation, working to 4 decimal places.
Perform 3 iterations for each technique.
a) Newton-Raphson Method.
show detailed workings and explain in the most simple of ways

Let's use the Newton-Raphson method to find the root of the given equation:

\[
4x^3 + 2x^2 - 3x - 7 = 0
\]

### Newton-Raphson Method Overview
The Newton-Raphson method is an iterative method used to find the roots of a function. It starts with an initial guess \( x_0 \) and uses the following formula to improve the guess:

\[
x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
\]

Where:
- \( f(x_n) \) is the value of the function at \( x_n \),
- \( f'(x_n) \) is the value of the derivative of the function at \( x_n \).

### Step 1: Identify \( f(x) \) and \( f'(x) \)

The function is:

\[
f(x) = 4x^3 + 2x^2 - 3x - 7
\]

Now, let's calculate its derivative \( f'(x) \):

\[
f'(x) = \frac{d}{dx}(4x^3 + 2x^2 - 3x - 7)
\]
Using standard differentiation rules:
\[
f'(x) = 12x^2 + 4x - 3
\]

### Step 2: Choose an Initial Guess \( x_0 \)

We will choose \( x_0 = 1 \) as a reasonable initial guess based on intuition or rough graph inspection.

### Step 3: Perform Iterations

#### Iteration 1:
\[
x_0 = 1
\]
First, compute \( f(x_0) \) and \( f'(x_0) \):

\[
f(1) = 4(1)^3 + 2(1)^2 - 3(1) - 7 = 4 + 2 - 3 - 7 = -4
\]

\[
f'(1) = 12(1)^2 + 4(1) - 3 = 12 + 4 - 3 = 13
\]

Now, apply the Newton-Raphson formula:

\[
x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1 - \frac{-4}{13} = 1 + \frac{4}{13} \approx 1 + 0.3077 = 1.3077
\]

#### Iteration 2:
\[
x_1 \approx 1.3077
\]

Compute \( f(x_1) \) and \( f'(x_1) \):

\[
f(1.3077) = 4(1.3077)^3 + 2(1.3077)^2 - 3(1.3077) - 7 \approx 8.9212 + 3.4191 - 3.9231 - 7 \approx 1.4172
\]

\[
f'(1.3077) = 12(1.3077)^2 + 4(1.3077) - 3 \approx 12(1.7101) + 5.2308 - 3 \approx 20.5212 + 5.2308 - 3 = 22.752
\]

Apply the Newton-Raphson formula:

\[
x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 1.3077 - \frac{1.4172}{22.752} \approx 1.3077 - 0.0623 = 1.2454
\]

#### Iteration 3:
\[
x_2 \approx 1.2454
\]

Compute \( f(x_2) \) and \( f'(x_2) \):

\[
f(1.2454) = 4(1.2454)^3 + 2(1.2454)^2 - 3(1.2454) - 7 \approx 7.7194 + 3.1020 - 3.7362 - 7 \approx 0.0852
\]

\[
f'(1.2454) = 12(1.2454)^2 + 4(1.2454) - 3 \approx 18.579 + 4.9816 - 3 = 20.5606
\]

Apply the Newton-Raphson formula:

\[
x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 1.2454 - \frac{0.0852}{20.5606} \approx 1.2454 - 0.0041 = 1.2413
\]

### Final Result after 3 Iterations:
The approximate root after 3 iterations is \( x_3 \approx 1.2413 \).

### Summary of Iterations:
- Iteration 1: \( x_1 \approx 1.3077 \)
- Iteration 2: \( x_2 \approx 1.2454 \)
- Iteration 3: \( x_3 \approx 1.2413 \)

### Explanation in Simple Terms:
1. We started with an initial guess (\( x_0 = 1 \)).
2. We used the Newton-Raphson formula to improve our guess by subtracting the ratio of the function value and its derivative.
3. After each step, we got closer to the root.
4. After 3 iterations, the result is approximately \( x = 1.2413 \), which is the solution to the equation \( 4x^3 + 2x^2 - 3x - 7 = 0 \) to 4 decimal places.

---

Do you have any questions or need further details?

Here are 5 related questions to explore further:
1. What are the conditions for Newton-Raphson method to converge?
2. How does the initial guess impact the convergence of the method?
3. Can the Newton-Raphson method fail? If so, how?
4. What is the geometric interpretation of the Newton-Raphson method?
5. How can we apply Newton-Raphson to systems of equations?

**Tip:** When using the Newton-Raphson method, choosing a good initial guess can help the method converge faster to the solution.

Given the equation: 4𝑥^3 + 2𝑥^2 − 3𝑥 − 7 = 0 Use the iterative techniques listed below to find the root of the equation, working to 4 decimal places. Perform 3 iterations for each technique. a) Newton-Raphson Method. Show detailed workings and explain in the most simple of ways.

Explore how to solve the cubic equation 4x^3 + 2x^2 − 3x − 7 = 0 using the Newton-Raphson method. This detailed step-by-step guide demonstrates 3 iterations and explains the process in simple terms, making it ideal for high school or early college students learning numerical methods.

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