Obtain a solution for x³-0.9x²-0.12x+0.02 = 0 using the newton Raphson method.

To find a solution to the equation \(x^3 - 0.9x^2 - 0.12x + 0.02 = 0\) using the Newton-Raphson method, we follow these steps:

1. **Define the function and its derivative:**
   \[
   f(x) = x^3 - 0.9x^2 - 0.12x + 0.02
   \]
   \[
   f'(x) = 3x^2 - 1.8x - 0.12
   \]

2. **Newton-Raphson formula:**
   The Newton-Raphson iteration formula is:
   \[
   x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
   \]

3. **Choose an initial guess \(x_0\):**
   Let's start with an initial guess. A good initial guess can often be obtained by plotting the function or analyzing its behavior. Let's choose \(x_0 = 0.5\).

4. **Iterate using the Newton-Raphson formula:**

### Iterations:

- **Iteration 1:**
  \[
  x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}
  \]
  \[
  f(0.5) = (0.5)^3 - 0.9(0.5)^2 - 0.12(0.5) + 0.02 = 0.125 - 0.225 - 0.06 + 0.02 = -0.14
  \]
  \[
  f'(0.5) = 3(0.5)^2 - 1.8(0.5) - 0.12 = 0.75 - 0.9 - 0.12 = -0.27
  \]
  \[
  x_1 = 0.5 - \frac{-0.14}{-0.27} = 0.5 + 0.5185 \approx 1.0185
  \]

- **Iteration 2:**
  \[
  x_2 = x_1 - \frac{f(x_1)}{f'(x_1)}
  \]
  \[
  f(1.0185) = (1.0185)^3 - 0.9(1.0185)^2 - 0.12(1.0185) + 0.02 \approx 1.056 - 0.933 - 0.122 + 0.02 \approx 0.021
  \]
  \[
  f'(1.0185) = 3(1.0185)^2 - 1.8(1.0185) - 0.12 \approx 3.112 - 1.834 - 0.12 \approx 1.158
  \]
  \[
  x_2 = 1.0185 - \frac{0.021}{1.158} \approx 1.0185 - 0.0181 \approx 1.0004
  \]

- **Iteration 3:**
  \[
  x_3 = x_2 - \frac{f(x_2)}{f'(x_2)}
  \]
  \[
  f(1.0004) = (1.0004)^3 - 0.9(1.0004)^2 - 0.12(1.0004) + 0.02 \approx 1.0012 - 0.9008 - 0.1200 + 0.02 \approx 0.0004
  \]
  \[
  f'(1.0004) = 3(1.0004)^2 - 1.8(1.0004) - 0.12 \approx 3.0012 - 1.8007 - 0.12 \approx 1.0805
  \]
  \[
  x_3 = 1.0004 - \frac{0.0004}{1.0805} \approx 1.0004 - 0.00037 \approx 1.0000
  \]

- **Iteration 4:**
  \[
  x_4 = x_3 - \frac{f(x_3)}{f'(x_3)}
  \]
  \[
  f(1.0000) = (1.0000)^3 - 0.9(1.0000)^2 - 0.12(1.0000) + 0.02 = 1.000 - 0.900 - 0.120 + 0.02 = 0.000
  \]
  \[
  f'(1.0000) = 3(1.0000)^2 - 1.8(1.0000) - 0.12 = 3.000 - 1.800 - 0.12 = 1.080
  \]
  \[
  x_4 = 1.0000 - \frac{0.0000}{1.080} = 1.0000
  \]

### Conclusion:
After four iterations, the Newton-Raphson method converges to \(x \approx 1.0000\), which is a root of the equation \(x^3 - 0.9x^2 - 0.12x + 0.02 = 0\).

Would you like more details or have any questions about this process?

### Additional Questions:
1. What is the general approach to the Newton-Raphson method?
2. How do you determine a good initial guess for the Newton-Raphson method?
3. Why is it important to find the derivative of the function when using the Newton-Raphson method?
4. How does the convergence of the Newton-Raphson method depend on the initial guess?
5. What are the potential pitfalls of the Newton-Raphson method?
6. Can the Newton-Raphson method be used for functions other than polynomials?
7. How can you check the accuracy of the root found using the Newton-Raphson method?
8. Are there any alternatives to the Newton-Raphson method for finding roots of equations?

### Tip:
Always verify the initial guess and perform a few iterations manually to ensure that the Newton-Raphson method is converging towards the correct root.

Learn how to solve the equation x³ - 0.9x² - 0.12x + 0.02 = 0 using the Newton-Raphson method. Follow a step-by-step process with detailed iterations and calculations. Suitable for undergraduate students studying calculus and numerical methods.

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