To find a real root of the equation $f(x) = x^3 - 2x - 5 = 0$ using the Newton-Raphson method, we follow these steps:

Newton-Raphson Formula:

The formula for Newton-Raphson is: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ 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: Define the function and its derivative

Given the function: $f(x) = x^3 - 2x - 5$ The derivative is: $f'(x) = 3x^2 - 2$

Step 2: Initial guess $x_0$

We need a reasonable initial guess $x_0$. By trial, $x = 2$ is a good starting point because $f(2) = 2^3 - 2(2) - 5 = -1$, which is close to zero.

Step 3: Iteration process

We now apply the Newton-Raphson iteration formula to get closer to the root.

Iteration 1:

$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}$ Substitute $x_0 = 2$: $f(2) = 2^3 - 2(2) - 5 = -1$ $f'(2) = 3(2^2) - 2 = 12 - 2 = 10$ $x_1 = 2 - \frac{-1}{10} = 2 + 0.1 = 2.1$

Iteration 2:

Substitute $x_1 = 2.1$: $f(2.1) = (2.1)^3 - 2(2.1) - 5 = 9.261 - 4.2 - 5 = 0.061$ $f'(2.1) = 3(2.1)^2 - 2 = 3(4.41) - 2 = 13.23 - 2 = 11.23$ $x_2 = 2.1 - \frac{0.061}{11.23} = 2.1 - 0.0054 = 2.0946$

Iteration 3:

Substitute $x_2 = 2.0946$: $f(2.0946) = (2.0946)^3 - 2(2.0946) - 5 \approx 0.00044$ $f'(2.0946) = 3(2.0946)^2 - 2 \approx 11.18$ $x_3 = 2.0946 - \frac{0.00044}{11.18} \approx 2.0946 - 0.00004 = 2.09456$

Conclusion:

After three iterations, the approximate real root of the equation $x^3 - 2x - 5 = 0$ is $x \approx 2.0946$.

Would you like a more detailed explanation of any step?

Here are some related questions:

1. How is the Newton-Raphson method derived?
2. What are the limitations of the Newton-Raphson method?
3. Can the Newton-Raphson method fail to converge?
4. How does the choice of initial guess affect the convergence of the method?
5. What other numerical methods can be used to find roots of equations?

Tip: The accuracy of Newton-Raphson depends on the initial guess. A good initial guess speeds up convergence!