We are given the equation:

$x^3 - 3x - 5 = 0$

We will use the Newton-Raphson method to find the real root of this equation. The general formula for Newton's method is:

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

Where:

• $f(x) = x^3 - 3x - 5$
• $f'(x) = 3x^2 - 3$

Step 1: Define the function and its derivative

• $f(x) = x^3 - 3x - 5$
• The derivative, $f'(x) = 3x^2 - 3$

Step 2: Initial guess

Let’s start with an initial guess $x_0$. A good initial guess can be obtained by evaluating the function graphically or choosing a reasonable value. In this case, let’s choose $x_0 = 2$ based on the behavior of the function.

Step 3: Apply the Newton-Raphson iteration

We apply the formula iteratively. Let's compute the values step-by-step.

Iteration 1:

1. $x_0 = 2$
2. $f(x_0) = 2^3 - 3(2) - 5 = 8 - 6 - 5 = -3$
3. $f'(x_0) = 3(2)^2 - 3 = 12 - 3 = 9$
4. Calculate $x_1$: $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2 - \frac{-3}{9} = 2 + \frac{1}{3} = 2.3333$

Iteration 2:

1. $x_1 = 2.3333$
2. $f(x_1) = (2.3333)^3 - 3(2.3333) - 5 = 12.7037 - 7.0 - 5 = 0.7037$
3. $f'(x_1) = 3(2.3333)^2 - 3 = 3(5.4444) - 3 = 16.3333 - 3 = 13.3333$
4. Calculate $x_2$: $x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 2.3333 - \frac{0.7037}{13.3333} = 2.3333 - 0.0528 = 2.2805$

Iteration 3:

1. $x_2 = 2.2805$
2. $f(x_2) = (2.2805)^3 - 3(2.2805) - 5 = 11.849 - 6.8415 - 5 = 0.0075$
3. $f'(x_2) = 3(2.2805)^2 - 3 = 3(5.1987) - 3 = 15.5962 - 3 = 12.5962$
4. Calculate $x_3$: $x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 2.2805 - \frac{0.0075}{12.5962} = 2.2805 - 0.0006 = 2.2799$

Iteration 4:

1. $x_3 = 2.2799$
2. $f(x_3) = (2.2799)^3 - 3(2.2799) - 5 = 11.8482 - 6.8397 - 5 = 0.0005$
3. $f'(x_3) = 3(2.2799)^2 - 3 = 3(5.1983) - 3 = 15.5949 - 3 = 12.5949$
4. Calculate $x_4$: $x_4 = x_3 - \frac{f(x_3)}{f'(x_3)} = 2.2799 - \frac{0.0005}{12.5949} = 2.2799 - 0.00004 = 2.27986$

Step 4: Convergence

The values of $x$ have stabilized and the change between iterations is very small. Thus, we can round the result to three decimal places.

The real root of the equation is approximately:

$x = 2.280$

Summary:

The real root of the equation $x^3 - 3x - 5 = 0$ using the Newton-Raphson method is $x \approx 2.280$.

Would you like further details or have any questions? Here's a list of 5 related questions:

1. How do we choose the initial guess for the Newton-Raphson method?
2. What are the conditions for the Newton-Raphson method to converge?
3. How can we verify that a given solution is indeed a root of the equation?
4. How does the Newton-Raphson method compare to other root-finding methods like bisection or secant?
5. How do you handle cases when the derivative $f'(x)$ is zero or very small in Newton's method?

Tip: If you're ever unsure about your initial guess, graphing the function can often help you choose a good starting point.