Let's solve each part step by step: --- ## (a) **Algorithm for Halving the Interval (Bisection Method)** **Bisection Method Algorithm:** 1. **Given:** A continuous function \( f(x) \) on interval \([a, b]\) where \( f(a) \) and \( f(b) \) have opposite signs (i.e., \( f(a)f(b) < 0 \)). 2. **Step 1:** Compute the midpoint \( c = \frac{a + b}{2} \). 3. **Step 2:** Evaluate \( f(c) \). 4. **Step 3:** If \( f(c) = 0 \) or the interval \([a, b]\) is sufficiently small, then \( c \) is the root. Stop. 5. **Step 4:** If \( f(a)f(c) < 0 \), set \( b = c \); else, set \( a = c \). 6. **Step 5:** Repeat steps 1–4 until the desired accuracy is achieved. --- ## (b) **Approximate \( \sqrt[3]{13} \) using Bisection Method** We are to solve \( x^3 - 13 = 0 \) for \( x \) (i.e., \( x = \sqrt[3]{13} \)). Let's find the interval: - \( f(2) = 2^3 - 13 = 8 - 13 = -5 \) - \( f(3) = 3^3 - 13 = 27 - 13 = 14 \) So, \( f(2) < 0 \), \( f(3) > 0 \), root lies between [2, 3]. Let’s apply the bisection method: ### **Iteration 1:** - \( a = 2 \), \( b = 3 \) - \( c_1 = (2+3)/2 = 2.5 \) - \( f(2.5) = (2.5)^3 - 13 = 15.625 - 13 = 2.625 \) \( f(2) < 0 \), \( f(2.5) > 0 \); root is in [2, 2.5] ### **Iteration 2:** - \( a = 2 \), \( b = 2.5 \) - \( c_2 = (2+2.5)/2 = 2.25 \) - \( f(2.25) = (2.25)^3 - 13 \approx 11.3906 - 13 = -1.6094 \) \( f(2.25) < 0 \), so root is in [2.25, 2.5] ### **Iteration 3:** - \( a = 2.25 \), \( b = 2.5 \) - \( c_3 = (2.25+2.5)/2 = 2.375 \) - \( f(2.375) = (2.375)^3 - 13 \approx 13.396 - 13 = 0.396 \) \( f(2.375) > 0 \), so root is in [2.25, 2.375] ### **Iteration 4:** - \( a = 2.25 \), \( b = 2.375 \) - \( c_4 = (2.25+2.375)/2 = 2.3125 \) - \( f(2.3125) = (2.3125)^3 - 13 \approx 12.351 - 13 = -0.649 \) \( f(2.3125) < 0 \), so root is in [2.3125, 2.375] ### **Iteration 5:** - \( a = 2.3125 \), \( b = 2.375 \) - \( c_5 = (2.3125+2.375)/2 = 2.34375 \) - \( f(2.34375) = (2.34375)^3 - 13 \approx 12.869 - 13 = -0.131 \) \( f(2.34375) < 0 \), so root is in [2.34375, 2.375] ### **Iteration 6:** - \( a = 2.34375 \), \( b = 2.375 \) - \( c_6 = (2.34375+2.375)/2 = 2.359375 \) - \( f(2.359375) = (2.359375)^3 - 13 \approx 13.131 - 13 = 0.131 \) \( f(2.359375) > 0 \), so root is in [2.34375, 2.359375] ### **Iteration 7:** - \( a = 2.34375 \), \( b = 2.359375 \) - \( c_7 = (2.34375+2.359375)/2 = 2.3515625 \) - \( f(2.3515625) = (2.3515625)^3 - 13 \approx 13.000 - 13 = 0.000 \) So, the root is approximately **2.351**, correct to three decimal places. **Final Answer:** \[ \boxed{\sqrt[3]{13} \approx 2.351} \] --- ## (c) **Find a root of \( f(x) = x^3 + 2x^2 - 3x - 1 \) starting at \( x_0 = 2 \) using:** ### (i) **Newton’s Formula** \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] First, compute derivative: \[ f'(x) = 3x^2 + 4x - 3 \] Start with \( x_0 = 2 \): - \( f(2) = 8 + 8 - 6 - 1 = 9 \) - \( f'(2) = 12 + 8 - 3 = 17 \) - \( x_1 = 2 - \frac{9}{17} \approx 1.471 \) Second iteration: - \( f(1.471) \approx (1.471)^3 + 2(1.471)^2 - 3(1.471) - 1 \approx 3.18 + 4.32 - 4.41 - 1 = 2.09 \) - \( f'(1.471) \approx 3(2.163) + 4(1.471) - 3 = 6.49 + 5.88 - 3 = 9.37 \) - \( x_2 = 1.471 - \frac{2.09}{9.37} \approx 1.248 \) Continue as needed. ### (ii) **The X = G method (\( x_{n+1} = g(x_n) \))** Choose a form, e.g., from original: \[ f(x) = x^3 + 2x^2 - 3x - 1 = 0 \implies x^3 + 2x^2 = 3x + 1 \implies x = \sqrt[3]{3x + 1 - 2x^2} \] Set \( g(x) = \sqrt[3]{3x + 1 - 2x^2} \) Start at \( x_0 = 2 \): - \( x_1 = \sqrt[3]{3*2 + 1 - 2*4} = \sqrt[3]{6 + 1 - 8} = \sqrt[3]{-1} = -1 \) - \( x_2 = \sqrt[3]{3*(-1) + 1 - 2*1} = \sqrt[3]{-3 + 1 - 2} = \sqrt[3]{-4} \approx -1.587 \) - \( x_3 = \sqrt[3]{3*(-1.587) + 1 - 2*(-1.587)^2} \approx ... \) (continue iterations) --- **Summary of Answers:** **(a)** Step-by-step bisection algorithm provided. **(b)** \( \sqrt[3]{13} \approx \boxed{2.351} \) (to three decimal places). **(c)** Newton's method and X=G method steps shown for finding roots, starting at \( x_0 = 2 \).