Let's solve the given problem step by step.

(a) Calculate $x_2$ using the iterative scheme:

We are given the iterative scheme:

$x_{k+1} = g(x_k) = \sqrt{4x_k + 2}, \quad k = 0,1,\dots$

With the initial guess $x_0 = 1$, we want to calculate $x_2$.

1. Calculate $x_1$: $x_1 = \sqrt{4x_0 + 2} = \sqrt{4(1) + 2} = \sqrt{4 + 2} = \sqrt{6}$ $x_1 \approx 2.45$

2. Calculate $x_2$: Using $x_1 \approx 2.45$, $x_2 = \sqrt{4x_1 + 2} = \sqrt{4(2.45) + 2} = \sqrt{9.8 + 2} = \sqrt{11.8}$ $x_2 \approx 3.43$

Thus, $x_2 \approx 3.43$.

(b) Minimum initial guess $x_0$:

We need to find the minimum value of $x_0$ such that the fixed-point iteration converges.

The function $g(x) = \sqrt{4x + 2}$ will converge if the initial guess is greater than or equal to a certain value. We set the inequality to ensure that $x_0 \geq -2$, because when $x = -2$:

$g(x) = \sqrt{4(-2) + 2} = \sqrt{-8 + 2} = \sqrt{-6}$

At $x = -2$, the function becomes imaginary, so the iteration will break down for values less than $-2$. Therefore, for real solutions, the iteration will work for $x_0 \geq -2$.

Thus, the minimum value of $x_0$ is: $x_0 \geq -2$

Summary:

• $x_2 \approx 3.43$
• $x_0 \geq -2$

Would you like more details, or do you have any other questions?

Here are five related questions:

1. What happens to the iteration process for values of $x_0$ less than -2?
2. How does the convergence behavior change for initial guesses much larger than -2?
3. Can we use other iterative methods for the same equation?
4. What is the general criterion for the convergence of fixed-point iterations?
5. How would the solution change if the function $f(x)$ were different?

Tip: Fixed-point iteration converges faster when the derivative of $g(x)$ is small near the root.