Let's solve part (a) as an example, step by step as you requested. ### Problem (a): Given \( g(x) = (2x-1)/x^2 \), fixed point \( r = 1 \). We are to use Theorem 1.6 (Fixed Point Iteration Convergence) to determine if iteration is locally convergent to **r**. --- ### **Step 1: Check if \( g(x) \) is continuously differentiable near \( r \)** - \( g(x) = \frac{2x-1}{x^2} \) - \( g'(x) = \frac{(2)(x^2) - (2x-1)(2x)}{(x^2)^2} = \frac{2x^2 - 4x^2 + 2x}{x^4} = \frac{-2x^2 + 2x}{x^4} = \frac{-2x + 2}{x^3} = \frac{2(1-x)}{x^3} \) - \( g'(x) \) is defined and continuous for all \( x \neq 0 \), in particular near \( x = 1 \). **Explanation:** The function and its derivative are defined and continuous near \( x = 1 \), so the first hypothesis of the theorem is satisfied. --- ### **Step 2: Check \( |g'(r)| < 1 \) for local convergence** - \( g'(x) = \frac{2(1-x)}{x^3} \) - At \( x = 1 \), \( g'(1) = \frac{2(1-1)}{1^3} = 0 \) - \( |g'(1)| = 0 < 1 \) **Explanation:** Since the absolute value of the derivative at the fixed point is less than 1, the fixed-point iteration is locally convergent. --- ## **Final Answer:** **Yes, the Fixed-Point Iteration of \( g(x) \) is locally convergent to the fixed point \( r = 1 \).**

Let's analyze the problem step-by-step for each part, based on the given functions and theorems. --- ### **Part (a):** Given \( g(x) = \frac{2x - 1}{x^2} \) with fixed point \( r = 1 \). --- ### **Step 1: Verify differentiability of \( g(x) \) near \( r = 1 \)** **Explanation:** The function \( g(x) \) is a rational function. It is defined and differentiable everywhere except where the denominator is zero, i.e., at \( x=0 \). Since \( r=1 \neq 0 \), \( g(x) \) is differentiable in a neighborhood around \( x=1 \). The derivative \( g'(x) \) can be computed to confirm this. --- ### **Step 2: Calculate \( g'(x) \) and evaluate at \( r = 1 \)** - \( g(x) = \frac{2x - 1}{x^2} \) Using the quotient rule: \[ g'(x) = \frac{(2)(x^2) - (2x - 1)(2x)}{x^4} \] Simplify numerator: \[ 2x^2 - [ (2x - 1) \times 2x ] = 2x^2 - (4x^2 - 2x) = 2x^2 - 4x^2 + 2x = -2x^2 + 2x \] Thus: \[ g'(x) = \frac{-2x^2 + 2x}{x^4} = \frac{2x(1 - x)}{x^4} = \frac{2(1 - x)}{x^3} \] Evaluate at \( x=1 \): \[ g'(1) = \frac{2(1 - 1)}{1^3} = 0 \] **Explanation:** Since \( g'(1) = 0 \), and \( g(x) \) is differentiable near 1, the first condition for local convergence (from the Fixed Point Theorem) is satisfied. --- ### **Conclusion:** Because \( |g'(1)| = 0 < 1 \), the fixed point iteration \( x_{k+1} = g(x_k) \) converges locally to the fixed point \( r=1 \). --- ## **Final answer for (a):** **Yes, the iteration converges locally to \( r=1 \).**