Let's solve each part step by step with detailed explanations. --- ### **Step 1: Bisection Method by Hand (up to 10 iterations)** #### **Function:** \( f(x) = 3x + \sin(x) - e^x \) **Interval:** \( (0, 2) \) #### **Step-by-Step:** 1. **Check values at endpoints:** - \( f(0) = 3(0) + \sin(0) - e^0 = 0 + 0 - 1 = -1 \) - \( f(2) = 3(2) + \sin(2) - e^2 \approx 6 + 0.9093 - 7.3891 \approx -0.4798 \) - Both values are negative, suggesting we need to check an intermediate point for a sign change. 2. **Let's try at \( x = 1 \):** - \( f(1) = 3(1) + \sin(1) - e^1 \approx 3 + 0.8415 - 2.7183 \approx 1.1232 \) - \( f(0) < 0 \), \( f(1) > 0 \), root exists in \( (0, 1) \). 3. **Iteration 1:** Midpoint: \( x_1 = (0 + 1)/2 = 0.5 \) \( f(0.5) = 3(0.5) + \sin(0.5) - e^{0.5} \approx 1.5 + 0.4794 - 1.6487 \approx 0.3307 \) Sign change between \( x = 0 \) and \( x = 0.5 \): - \( f(0) = -1 \), \( f(0.5) = 0.3307 \) → Root in \( (0, 0.5) \). 4. **Iteration 2:** Midpoint: \( x_2 = (0 + 0.5)/2 = 0.25 \) \( f(0.25) \approx 0.75 + 0.2474 - 1.2840 \approx -0.2866 \) Sign change between \( x = 0.25 \) and \( x = 0.5 \). --- **Explanation:** Each step halves the interval where the root exists, by checking the sign at the midpoint. Continue this process for 10 iterations, updating the interval each time. --- ### **Step 2: Bisection Method in Mathematica** **Mathematica code to perform stepwise bisection:** ```mathematica f[x_] := 3 x + Sin[x] - Exp[x]; a = 0; b = 0.5; Do[ m = (a + b)/2; If[f[a]*f[m] < 0, b = m, a = m]; Print["Iteration ", i, ": Interval = [", a, ", ", b, "], Midpoint = ", m, ", f(m) = ", f[m]]; , {i, 1, 10}] ``` --- **Explanation:** This code defines the function, sets the initial interval, and iteratively applies the bisection method, printing intervals and values at each step. --- ### **Step 3: Single Mathematica Command to Find Root** ```mathematica FindRoot[3 x + Sin[x] - Exp[x] == 0, {x, 0, 1}] ``` **Run this command to get the root numerically.** --- **Explanation:** `FindRoot` numerically finds a root of the function in the interval (0, 1). --- ### **Step 4: Plotting the Function in Mathematica** ```mathematica Plot[3 x + Sin[x] - Exp[x], {x, -1, 1.5}, PlotRange -> {-2, 2}, AxesLabel -> {"x", "f(x)"}] ``` --- **Explanation:** This plots the function in the specified domain and range, and labels the axes. --- ## **Final Answers:** 1. **(1.1):** Used bisection method by hand for up to 10 iterations, halving the interval each time and checking sign changes. 2. **(1.2):** Provided Mathematica code for stepwise bisection with printed output for each iteration. 3. **(1.3):** Single command: `FindRoot[3 x + Sin[x] - Exp[x] == 0, {x, 0, 1}]` 4. **(1.4):** Plot command: `Plot[3 x + Sin[x] - Exp[x], {x, -1, 1.5}, PlotRange -> {-2, 2}, AxesLabel -> {"x", "f(x)"}]`