Let's proceed step by step for **each sub-question** for the function: \( f(x) = 3x + \sin(x) - e^x \) on the interval **(0, 2)**. --- ## **1.1: Bisection Method by Hand (Up to 10 Iterations)** ### **Step 1: Check f(x) at Endpoints** Let's compute \( f(0) \) and \( f(2) \): - \( f(0) = 3*0 + \sin(0) - e^0 = 0 + 0 - 1 = -1 \) - \( f(2) = 3*2 + \sin(2) - e^2 \approx 6 + 0.9093 - 7.3891 \approx 6.9093 - 7.3891 = -0.4798 \) Thus, both values are negative. This suggests **no sign change** in (0,2). But let's check at a midpoint, say \( x = 1 \): - \( f(1) = 3*1 + \sin(1) - e^1 \approx 3 + 0.8415 - 2.7183 \approx 3.8415 - 2.7183 = 1.1232 \) Now, \( f(0) = -1 \) and \( f(1) = 1.1232 \): a sign change between \( x=0 \) and \( x=1 \). So, **new interval is (0,1)**. --- ### **Step 2: Bisection Iterations** We'll use: - \( a_0 = 0 \), \( b_0 = 1 \) - \( f(a_0) = -1 \), \( f(b_0) = 1.1232 \) Let’s define: - \( m_n = \frac{a_n + b_n}{2} \) - Decide the next interval based on the sign of \( f(m_n) \). #### **Iteration Table** | Iteration | \( a_n \) | \( b_n \) | \( m_n \) | \( f(m_n) \) | Interval chosen | |-----------|-----------|-----------|------------|--------------|------------------| | 0 | 0 | 1 | 0.5 | ? | | | 1 | | | | | | | 2 | | | | | | Let’s fill it step by step: --- #### **Iteration 1:** - \( m_0 = (0 + 1)/2 = 0.5 \) - \( f(0.5) = 3*0.5 + \sin(0.5) - e^{0.5} \) - \( = 1.5 + 0.4794 - 1.6487 = 1.9794 - 1.6487 = 0.3307 \) Now: \( f(0) = -1 \), \( f(0.5) = 0.3307 \) → sign change between \( x=0 \) and \( x=0.5 \). **New interval**: \( (0, 0.5) \) --- #### **Iteration 2:** - \( m_1 = (0 + 0.5)/2 = 0.25 \) - \( f(0.25) = 3*0.25 + \sin(0.25) - e^{0.25} \) - \( = 0.75 + 0.2474 - 1.2840 = 0.9974 - 1.2840 = -0.2866 \) Now: \( f(0.25) = -0.2866 \), \( f(0.5) = 0.3307 \) → sign change between \( x=0.25 \) and \( x=0.5 \). **New interval**: \( (0.25, 0.5) \) --- #### **Iteration 3:** - \( m_2 = (0.25 + 0.5)/2 = 0.375 \) - \( f(0.375) = 3*0.375 + \sin(0.375) - e^{0.375} \) - \( = 1.125 + 0.3663 - 1.45499 = 1.4913 - 1.45499 = 0.0363 \) Now: \( f(0.25) = -0.2866 \), \( f(0.375) = 0.0363 \) → sign change between \( x=0.25 \) and \( x=0.375 \). **New interval**: \( (0.25, 0.375) \) --- #### **Iteration 4:** - \( m_3 = (0.25 + 0.375)/2 = 0.3125 \) - \( f(0.3125) = 3*0.3125 + \sin(0.3125) - e^{0.3125} \) - \( = 0.9375 + 0.3076 - 1.3667 = 1.2451 - 1.3667 = -0.1216 \) Now: \( f(0.3125) = -0.1216 \), \( f(0.375) = 0.0363 \). **New interval**: \( (0.3125, 0.375) \) --- #### **Iteration 5:** - \( m_4 = (0.3125 + 0.375)/2 = 0.34375 \) - \( f(0.34375) = 3*0.34375 + \sin(0.34375) - e^{0.34375} \) - \( = 1.03125 + 0.3369 - 1.4105 = 1.3682 - 1.4105 = -0.0423 \) Now: \( f(0.34375) = -0.0423 \), \( f(0.375) = 0.0363 \). **New interval**: \( (0.34375, 0.375) \) --- #### **Iteration 6:** - \( m_5 = (0.34375 + 0.375)/2 = 0.359375 \) - \( f(0.359375) = 3*0.359375 + \sin(0.359375) - e^{0.359375} \) - \( = 1.078125 + 0.3518 - 1.4329 = 1.4299 - 1.4329 = -0.0030 \) Now: \( f(0.359375) = -0.0030 \), \( f(0.375) = 0.0363 \). **New interval**: \( (0.359375, 0.375) \) --- #### **Iteration 7:** - \( m_6 = (0.359375 + 0.375)/2 = 0.3671875 \) - \( f(0.3671875) = 3*0.3671875 + \sin(0.3671875) - e^{0.3671875} \) - \( = 1.1015625 + 0.3590 - 1.4442 = 1.4606 - 1.4442 = 0.0164 \) Now: \( f(0.359375) = -0.0030 \), \( f(0.3671875) = 0.0164 \). **New interval**: \( (0.359375, 0.3671875) \) --- #### **Iteration 8:** - \( m_7 = (0.359375 + 0.3671875)/2 = 0.36328125 \) - \( f(0.36328125) = 3*0.36328125 + \sin(0.36328125) - e^{0.36328125} \) - \( = 1.08984375 + 0.3555 - 1.4386 = 1.4453 - 1.4386 = 0.0067 \) Now: \( f(0.359375) = -0.0030 \), \( f(0.36328125) = 0.0067 \). **New interval**: \( (0.359375, 0.36328125) \) --- #### **Iteration 9:** - \( m_8 = (0.359375 + 0.36328125)/2 = 0.361328125 \) - \( f(0.361328125) = 3*0.361328125 + \sin(0.361328125) - e^{0.361328125} \) - \( = 1.083984375 + 0.3537 - 1.4358 = 1.4377 - 1.4358 = 0.0019 \) Now: \( f(0.359375) = -0.0030 \), \( f(0.361328125) = 0.0019 \). **New interval**: \( (0.359375, 0.361328125) \) --- #### **Iteration 10:** - \( m_9 = (0.359375 + 0.361328125)/2 = 0.3603515625 \) - \( f(0.3603515625) = 3*0.3603515625 + \sin(0.3603515625) - e^{0.3603515625} \) - \( = 1.0810546875 + 0.3528 - 1.4344 = 1.4339 - 1.4344 = -0.0005 \) --- ### **Summary Table** | Iter | a | b | m | f(m) | Next Interval | |------|----------|----------|-----------|----------|------------------------| | 1 | 0 | 1 | 0.5 | 0.3307 | (0, 0.5) | | 2 | 0 | 0.5 | 0.25 | -0.2866 | (0.25, 0.5) | | 3 | 0.25 | 0.5 | 0.375 | 0.0363 | (0.25, 0.375) | | 4 | 0.25 | 0.375 | 0.3125 | -0.1216 | (0.3125, 0.375) | | 5 | 0.3125 | 0.375 | 0.34375 | -0.0423 | (0.34375, 0.375) | | 6 | 0.34375 | 0.375 | 0.359375 | -0.0030 | (0.359375, 0.375) | | 7 | 0.359375 | 0.375 | 0.3671875 | 0.0164 | (0.359375, 0.3671875) | | 8 | 0.359375 | 0.3671875| 0.36328125| 0.0067 | (0.359375, 0.36328125) | | 9 | 0.359375 | 0.36328125| 0.361328125| 0.0019 | (0.359375, 0.361328125)| | 10 | 0.359375 | 0.361328125| 0.3603515625| -0.0005| (0.3603515625,0.361328125)| --- ### **Conclusion for 1.1** - After **10 iterations**, the root is narrowed down to the interval \( (0.36035, 0.36133) \). - The **approximate root** is \( x \approx 0.3604 \), with \( f(x) \approx -0.0005 \). --- ## **1.2: Bisection Method in Mathematica (Step-by-Step Iterations)** You can use the following Mathematica commands for **each iteration**: ```mathematica f[x_] := 3 x + Sin[x] - Exp[x]; (* Iteration 1 *) a = 0; b = 1; m1 = (a + b)/2 f[m1] (* Iteration 2 *) If[f[a] f[m1] < 0, b1 = m1; a1 = a, a1 = m1; b1 = b]; m2 = (a1 + b1)/2 f[m2] (* Iteration 3 *) If[f[a1] f[m2] < 0, b2 = m2; a2 = a1, a2 = m2; b2 = b1]; m3 = (a2 + b2)/2 f[m3] (* ... Continue up to iteration 10, updating a and b each time ... *) ``` Alternatively, you can automate this with a simple loop: ```mathematica f[x_] := 3 x + Sin[x] - Exp[x]; a = 0; b