Let's break down the problem step by step as requested: --- ### **Step 1: Intermediate Value Theorem (Part a)** #### **Explanation** The Intermediate Value Theorem (IVT) states: If \(f\) is continuous on \([a, b]\), and \(f(a)\) and \(f(b)\) have opposite signs, then there exists at least one root \(c \in (a, b)\) such that \(f(c) = 0\). Given: - \(f(x) = \frac{\cos x}{3} - x - \frac{1}{2}\) - Interval: \([-1, 1]\) Let's check \(f(-1)\) and \(f(1)\): 1. \(f(-1) = \frac{\cos(-1)}{3} - (-1) - \frac{1}{2}\) - \(\cos(-1) \approx 0.5403\) - \(f(-1) = \frac{0.5403}{3} + 1 - 0.5 \approx 0.1801 + 1 - 0.5 = 0.6801\) 2. \(f(1) = \frac{\cos 1}{3} - 1 - \frac{1}{2}\) - \(\cos 1 \approx 0.5403\) - \(f(1) = \frac{0.5403}{3} - 1 - 0.5 \approx 0.1801 - 1 - 0.5 = -1.3199\) - \(f(-1) \approx 0.68 > 0\) - \(f(1) \approx -1.32 < 0\) Since \(f(-1)\) and \(f(1)\) have opposite signs and \(f(x)\) is continuous (sum of continuous functions), by the IVT, there is at least one root in \([-1, 1]\). #### **End of Step 1 Explanation** - **The choice of initial interval is valid.** --- ### **Step 2: Initial Guesses with Bisection (Part b)** #### **Explanation** The bisection method uses: - Initial interval: \([-1, 1]\) - Midpoint: \(x_0 = \frac{a_0 + b_0}{2}\) **Iteration 0:** - \(x_0 = \frac{-1 + 1}{2} = 0\) - \(f(0) = \frac{\cos 0}{3} - 0 - \frac{1}{2} = \frac{1}{3} - 0 - 0.5 = 0.3333 - 0.5 = -0.1667\) Now, check the sign of \(f(-1)\) and \(f(0)\): - \(f(-1) \approx 0.68 > 0\) - \(f(0) \approx -0.1667 < 0\) The root is between \(-1\) and \(0\). **Iteration 1:** - New interval: \([-1, 0]\) - \(x_1 = \frac{-1 + 0}{2} = -0.5\) - \(f(-0.5) = \frac{\cos(-0.5)}{3} - (-0.5) - 0.5\) - \(\cos(-0.5) \approx 0.8776\) - \(f(-0.5) = \frac{0.8776}{3} + 0.5 - 0.5 = 0.2925\) So \(f(-0.5) \approx 0.2925 > 0\) Now, root is between \(-0.5\) and \(0\). **Iteration 2:** - New interval: \([-0.5, 0]\) - \(x_2 = \frac{-0.5 + 0}{2} = -0.25\) - \(f(-0.25) = \frac{\cos(-0.25)}{3} - (-0.25) - 0.5\) - \(\cos(-0.25) \approx 0.9689\) - \(f(-0.25) = \frac{0.9689}{3} + 0.25 - 0.5 \approx 0.3230 + 0.25 - 0.5 = 0.0730\) So \(f(-0.25) \approx 0.0730 > 0\) Now, root is between \(-0.25\) and \(0\). #### **End of Step 2 Explanation** - **Initial guess \(x_0 = 0\), \(x_1 = -0.5\), \(x_2 = -0.25\) with their corresponding \(f(x)\) values calculated above.** --- ### **Final Answer (Summary):** 1. **Step 1:** The initial interval \([-1, 1]\) is valid by the Intermediate Value Theorem since \(f(-1) > 0\) and \(f(1) < 0\). 2. **Step 2:** The first three bisection points and their values are: - \(x_0 = 0\), \(f(0) = -0.1667\) - \(x_1 = -0.5\), \(f(-0.5) \approx 0.2925\) - \(x_2 = -0.25\), \(f(-0.25) \approx 0.0730\) **Final Answer:** The process confirms the root lies between \(-0.25\) and \(0\) after two iterations.

Let's proceed step-by-step as per the question: --- ### **Step 1: Verify the validity of the initial interval \([a_0, b_0] = [-1, 1]\) using the Intermediate Value Theorem (Part a)** **Explanation:** The Intermediate Value Theorem states that if a function \(f(x)\) is continuous on \([a, b]\) and \(f(a)\) and \(f(b)\) have opposite signs, then there exists at least one root \(c \in (a, b)\) such that \(f(c) = 0\). Given the function: \[ f(x) = \frac{\cos x}{3} - x - \frac{1}{2} \] - Check \(f(-1)\): \[ f(-1) = \frac{\cos(-1)}{3} - (-1) - 0.5 \] \(\cos(-1) = \cos 1 \approx 0.5403\), so: \[ f(-1) = \frac{0.5403}{3} + 1 - 0.5 \approx 0.1801 + 1 - 0.5 = 0.6801 \] - Check \(f(1)\): \[ f(1) = \frac{\cos 1}{3} - 1 - 0.5 \approx 0.1801 - 1 - 0.5 = -1.3199 \] Since \(f(-1) > 0\) and \(f(1) < 0\), the signs are opposite, and \(f(x)\) is continuous on \([-1, 1]\). Therefore, by the IVT, **there exists at least one root in \([-1, 1]\).** --- ### **Step 2: Compute initial guesses \(z_0\), \(z_1\), and \(z_2\) using the bisection method (Part b)** **Explanation:** Initialize with \(a_0 = -1\), \(b_0 = 1\). - **First midpoint:** \[ z_0 = \frac{a_0 + b_0}{2} = \frac{-1 + 1}{2} = 0 \] - Compute \(f(z_0) = f(0)\): \[ f(0) = \frac{\cos 0}{3} - 0 - 0.5 = \frac{1}{3} - 0.5 = 0.3333 - 0.5 = -0.1667 \] Since \(f(-1) > 0\) (from above) and \(f(0) < 0\), the root lies in \([-1, 0]\). - **Next midpoint:** \[ z_1 = \frac{-1 + 0}{2} = -0.5 \] - Compute \(f(z_1) = f(-0.5)\): \[ f(-0.5) = \frac{\cos(-0.5)}{3} - (-0.5) - 0.5 \] \(\cos(-0.5) = \cos 0.5 \approx 0.8776\), so: \[ f(-0.5) = \frac{0.8776}{3} + 0.5 - 0.5 \approx 0.2925 + 0 - 0 = 0.2925 \] Since \(f(-0.5) > 0\), root is in \([-0.5, 0]\). - **Next midpoint:** \[ z_2 = \frac{-0.5 + 0}{2} = -0.25 \] - Compute \(f(z_2) = f(-0.25)\): \[ f(-0.25) = \frac{\cos(-0.25)}{3} - (-0.25) - 0.5 \] \(\cos(-0.25) = \cos 0.25 \approx 0.9689\), so: \[ f(-0.25) = \frac{0.9689}{3} + 0.25 - 0.5 \approx 0.3230 + 0.25 - 0.5 = 0.0730 \] Since \(f(-0.25) > 0\), root is in \([-0.25, 0]\). --- ### **Summary:** - **Initial guesses:** - \(z_0 = 0\), \(f(z_0) \approx -0.1667\) - \(z_1 = -0.5\), \(f(z_1) \approx 0.2925\) - \(z_2 = -0.25\), \(f(z_2) \approx 0.0730\) --- ### **Final notes:** - The initial interval \([-1, 1]\) is valid because the function changes sign within it. - The first three bisection points are as calculated, and the root is narrowing down between \(-0.25\) and \(0\). --- **Final answer:** \[ \boxed{ \textbf{(a)} \quad \text{Interval } [-1,1] \text{ is valid (signs opposite).} \\ \textbf{(b)} \quad \begin{cases} z_0 = 0, \quad f(z_0) \approx -0.1667 \\ z_1 = -0.5, \quad f(z_1) \approx 0.2925 \\ z_2 = -0.25, \quad f(z_2) \approx 0.0730 \end{cases} } \]

Certainly! Here is a MATLAB code implementing the bisection method with the specified criteria. The code prints out the iteration number \(k\) and the current approximation \(z_k\) at each step, stopping when \(|z_k - z_{k-1}| < 10^{-7}\) or when the maximum number of iterations is reached. ```matlab % MATLAB implementation of Bisection Method for f(x) = (cos x)/3 - x - 0.5 % Define the function f = @(x) (cos(x))/3 - x - 0.5; % Initial interval a = -1; b = 1; % Tolerance tol = 1e-7; % Maximum iterations to prevent infinite loop max_iter = 100; % Initial guesses z_prev = a; z_curr = (a + b)/2; % Evaluate function at initial guesses f_a = f(a); f_b = f(b); f_z_curr = f(z_curr); % Check initial signs if f_a * f_b > 0 error('Function does not change sign over the interval. Bisection method cannot proceed.'); end fprintf('Iter\tz_k\n'); fprintf('%d\t%.8f\n', 0, z_prev); fprintf('%d\t%.8f\n', 1, z_curr); for k = 2: max_iter % Midpoint z_next = (a + b)/2; f_z_next = f(z_next); % Print current iteration fprintf('%d\t%.8f\n', k, z_next); % Check stopping criterion if abs(z_next - z_curr) < tol break; end % Update interval if f(z_next) * f(a) < 0 b = z_next; else a = z_next; end % Update guesses z_prev = z_curr; z_curr = z_next; end fprintf('Root approximation after %d iterations: %.8f\n', k, z_next); ``` ### Explanation: - The function `f(x)` is defined as per the problem. - The interval is initialized to \([-1, 1]\). - The loop continues until the change between successive approximations is less than \(10^{-7}\) or the maximum iteration count is reached. - At each iteration, the code prints the iteration number `k` and the current approximation `z_k`. - The interval is updated based on the sign of the function at the midpoint. --- **Note:** You can copy and run this code in MATLAB. It will display each iteration's approximation until the desired accuracy is achieved.