Let's solve the problem step by step: ### **Step 1: Rewrite the equation as fixed-point forms** The equation is \( e^x - 4x - 1 = 0 \). It is rewritten as \( x = g(x) \) in three ways: 1. \( x_{n+1} = e^{x_n} - 3x_n - 1 \) 2. \( x_{n+1} = \ln(4x_n + 1) \) 3. \( x_{n+1} = \frac{e^{x_n} - 1}{4} \) --- ### **Step 2: Determine which sequence converges** We need to find which function \( g(x) \) will converge to the fixed point in the interval \([1, 2]\). #### **Fixed Point Iteration Convergence** A sufficient condition for convergence is that \(|g'(x)| < 1\) near the fixed point. Let’s analyze each: #### (a) \( g_1(x) = e^x - 3x - 1 \) - \( g_1'(x) = e^x - 3 \) - At \( x=1 \): \( e^1 - 3 \approx 2.718 - 3 = -0.282 \) - At \( x=2 \): \( e^2 - 3 \approx 7.389 - 3 = 4.389 \) - For \( x \) in \([1,2]\), \( g_1'(x) \) goes from negative to greater than 1, so **not always less than 1 in magnitude**. #### (b) \( g_2(x) = \ln(4x + 1) \) - \( g_2'(x) = \frac{4}{4x+1} \) - At \( x=1 \): \( \frac{4}{5} = 0.8 \) - At \( x=2 \): \( \frac{4}{9} \approx 0.444 \) - For \( x \) in \([1,2]\), \( |g_2'(x)| < 1 \), so **this sequence will converge**. #### (c) \( g_3(x) = \frac{e^x - 1}{4} \) - \( g_3'(x) = \frac{e^x}{4} \) - At \( x=1 \): \( \frac{2.718}{4} \approx 0.68 \) - At \( x=2 \): \( \frac{7.389}{4} \approx 1.847 \) - For \( x \) in \([1,2]\), \( g_3'(x) \) exceeds 1, so **may not always converge**. **Conclusion:** The convergent sequence is: \[ x_{n+1} = \ln(4x_n + 1) \] --- ### **Step 3: Find \( x_2 \) using \( x_0 = 1.1 \)** Given \( x_0 = 1.1 \), compute \( x_1 \) and \( x_2 \): #### Compute \( x_1 \): \[ x_1 = \ln(4x_0 + 1) = \ln(4 \times 1.1 + 1) = \ln(4.4 + 1) = \ln(5.4) \] \[ x_1 \approx 1.686 \] #### Compute \( x_2 \): \[ x_2 = \ln(4x_1 + 1) = \ln(4 \times 1.686 + 1) = \ln(6.744 + 1) = \ln(7.744) \] \[ x_2 \approx 2.048 \] --- ## **Final Answers** ### **a) The sequence that converges is:** \[ x_{n+1} = \ln(4x_n + 1) \] ### **b) Using \( x_0 = 1.1 \), the value of \( x_2 \) is:** \[ \boxed{2.048} \]