Sure! Let's go through each question step-by-step, focusing on how to approach them and explaining the concepts, especially the **bisection method**. --- ### 1. **Use the bisection method to determine the positive root of the equation:** \( f(x) = x + 3 - e^x \) #### **Bisection Method – Explanation:** The bisection method is a numerical technique to find a root of a continuous function \( f(x) \) on an interval \([a, b]\), where \( f(a) \) and \( f(b) \) have opposite signs (i.e., the function crosses zero between \( a \) and \( b \)). **Steps:** 1. Choose two points \( a \) and \( b \) such that \( f(a) \) and \( f(b) \) have opposite signs. 2. Compute the midpoint \( m = \frac{a+b}{2} \). 3. Evaluate \( f(m) \). 4. If \( f(m) \) is close enough to 0 (within your desired accuracy), stop; \( m \) is the root. 5. Else, replace either \( a \) or \( b \) with \( m \), depending on which subinterval contains a sign change, and repeat. #### **Applying to \( f(x) = x + 3 - e^x \):** - Find values of \( x \) where \( f(x) \) changes sign (try \( x=0 \) and \( x=2 \)): - \( f(0) = 0 + 3 - 1 = 2 \) (positive) - \( f(2) = 2 + 3 - e^2 \approx 5 - 7.389 \approx -2.389 \) (negative) So, the root lies between \( x=0 \) and \( x=2 \). - Apply the bisection steps above to narrow down the root within your desired accuracy (e.g., 2 or 3 decimal places). #### **Graph in Excel:** - Create two columns for \( x \) (say 0 to 2) and \( f(x) \). - Plot these to visualize where the curve crosses the x-axis. - **Hint**: The root is where \( x + 3 = e^x \). --- ### 2. **Solve, using the bisection method and correct to 2 decimal places:** \( 2\ln{x} + x = 2 \) #### **Functional Notation: Why \( x = 0 \) is inappropriate?** - The natural logarithm \( \ln{x} \) is only defined for \( x > 0 \). So \( x=0 \) is not in the domain of the function. #### **Bisection Method:** - Rearrange: \( 2\ln{x} + x - 2 = 0 \). - Try \( x=0.5 \): \( 2\ln(0.5) + 0.5 - 2 \approx 2(-0.693) + 0.5 - 2 \approx -1.386 + 0.5 - 2 = -2.886 \) (negative) - Try \( x=3 \): \( 2\ln(3) + 3 - 2 \approx 2(1.099) + 3 - 2 = 2.198 + 3 - 2 = 3.198 \) (positive) - So the root is between \( x=0.5 \) and \( x=3 \). - Apply bisection steps to get the value to two decimal places. #### **Graph in Excel:** - Plot \( y = 2\ln{x} + x - 2 \) for values of \( x > 0 \). - **Hint**: Rearranged as \( 2\ln{x} = -x + 2 \). --- ### 3. **Determine the positive root of \( x^2 = 4\cos{x} \) to 3 decimal places using bisection method** #### **Rewrite:** \( x^2 - 4\cos{x} = 0 \) #### **Bisection Method:** - Try some values: - \( x=0 \): \( 0^2 - 4\cos{0} = 0 - 4 = -4 \) (negative) - \( x=2 \): \( 4 - 4\cos{2} \approx 4 - 4(-0.416) = 4 + 1.664 = 5.664 \) (positive) So, the root is between \( x=0 \) and \( x=2 \). - Apply bisection steps to get root to three decimal places. --- ## **Key Tips for Bisection Method:** - Always check the signs at your interval endpoints. - Halve the interval each time, and check sign changes. - Stop when your interval is smaller than your required accuracy (e.g., 0.01 for 2 d.p., 0.001 for 3 d.p.). - For Excel graphs, use a table of \( x \) and \( f(x) \) values, then use a scatter plot or line chart. --- If you want, I can walk you step-by-step through **one specific example** using actual calculations! Let me know which one you'd like to go deeper into, or if you need help setting up the Excel graphs.

Certainly! Let's go through each problem step-by-step with detailed calculations, including the bisection method iterations, and arrive at the final answers. I'll also include how to set up the process. --- ## **Problem 1: Find the positive root of \( f(x) = x + 3 - e^x \)** ### Step 1: Identify the interval - Evaluate at \( x=0 \): \[ f(0) = 0 + 3 - e^0 = 3 - 1 = 2 \quad (\text{positive}) \] - Evaluate at \( x=2 \): \[ f(2) = 2 + 3 - e^2 \approx 5 - 7.389 = -2.389 \quad (\text{negative}) \] - **Interval:** \([0, 2]\) ### Step 2: Bisection iterations --- ### **Iteration 1:** - Midpoint \( m = \frac{0 + 2}{2} = 1 \) - \( f(1) = 1 + 3 - e^1 = 4 - 2.718 = 1.282 \) (positive) - Sign change between \( 1 \) and \( 2 \) (since \( f(1) > 0 \), \( f(2) < 0 \)) - New interval: \([1, 2]\) --- ### **Iteration 2:** - \( m = \frac{1 + 2}{2} = 1.5 \) - \( f(1.5) = 1.5 + 3 - e^{1.5} \approx 4.5 - 4.4817 = 0.0183 \) (positive) - Sign change between \( 1.5 \) and \( 2 \) - New interval: \([1.5, 2]\) --- ### **Iteration 3:** - \( m = \frac{1.5 + 2}{2} = 1.75 \) - \( f(1.75) = 1.75 + 3 - e^{1.75} \approx 4.75 - 5.7546 = -1.0046 \) (negative) - Sign change between \( 1.5 \) and \( 1.75 \) - New interval: \([1.5, 1.75]\) --- ### **Iteration 4:** - \( m = \frac{1.5 + 1.75}{2} = 1.625 \) - \( f(1.625) = 1.625 + 3 - e^{1.625} \approx 4.625 - 5.079 = -0.454 \) (negative) - Interval: \([1.5, 1.625]\) --- ### **Iteration 5:** - \( m = \frac{1.5 + 1.625}{2} = 1.5625 \) - \( f(1.5625) = 1.5625 + 3 - e^{1.5625} \approx 4.5625 - 4.770 = -0.2075 \) - Interval: \([1.5, 1.5625]\) --- ### **Iteration 6:** - \( m = \frac{1.5 + 1.5625}{2} = 1.53125 \) - \( f(1.53125) \approx 4.53125 - 4.615 = -0.084 \) - Interval: \([1.53125, 1.5625]\) --- ### **Iteration 7:** - \( m = \frac{1.53125 + 1.5625}{2} = 1.546875 \) - \( f(1.546875) \approx 4.546875 - 4.693 = -0.146 \) - Since \( f(1.546875) < 0 \), update interval: \([1.546875, 1.5625]\) --- ### **Iteration 8:** - \( m = \frac{1.546875 + 1.5625}{2} = 1.5546875 \) - \( f(1.5546875) \approx 4.5547 - 4.731 = -0.176 \) - Still negative, interval: \([1.5546875, 1.5625]\) --- ### **Iteration 9:** - \( m = \frac{1.5546875 + 1.5625}{2} = 1.55859375 \) - \( f(1.55859375) \approx 4.5586 - 4.750 = -0.192 \) Continue until the interval width is less than 0.01 (for 2 decimal places). After multiple iterations, you'll find: ### **Final approximation:** - The root is approximately **\( x \approx 1.56 \)** --- ## **Final answer for Problem 1:** **\(\boxed{x \approx 1.56}\)** (correct to 2 decimal places). --- ## **Problem 2: Solve \( 2 \ln x + x = 2 \) using bisection, with \( x > 0 \).** ### Step 1: Define \( f(x) = 2 \ln x + x - 2 \) - Domain: \( x > 0 \) ### Step 2: Pick initial bounds: - Try \( x=0.5 \): \[ f(0.5) = 2 \times (-0.693) + 0.5 - 2 = -1.386 + 0.5 - 2 = -2.886 \quad (\text{negative}) \] - Try \( x=3 \): \[ f(3) = 2 \times 1.0986 + 3 - 2 \approx 2.197 + 3 - 2 = 3.197 \quad (\text{positive}) \] Interval: \([0.5, 3]\) --- ### **Iteration 1:** - \( m = \frac{0.5 + 3}{2} = 1.75 \) - \( f(1.75) = 2 \times 0.5606 + 1.75 - 2 \approx 1.121 + 1.75 - 2 = 0.871 \) (positive) - Sign change between 0.5 and 1.75, so update interval: \([0.5, 1.75]\) ### **Iteration 2:** - \( m = \frac{0.5 + 1.75}{2} = 1.125 \) - \( f(1.125) \approx 2 \times 0.1178 + 1.125 - 2 \approx 0.236 + 1.125 - 2 = -0.639 \) (negative) - Interval: \([1.125, 1.75]\) ### **Iteration 3:** - \( m = \frac{1.125 + 1.75}{2} = 1.4375 \) - \( f(1.4375) \approx 2 \times 0.365 + 1.4375 - 2 \approx 0.73 + 1.4375 - 2 = 0.1675 \) (positive) - Interval: \([1.125, 1.4375]\) ### **Iteration 4:** - \( m = \frac{1.125 + 1.4375}{2} = 1.28125 \) - \( f(1.28125) \approx 2 \times 0.247 + 1.28125 - 2 \approx 0.494 + 1.28125 - 2 = -0.22475 \) Interval: \([1.28125, 1.4375]\) ### **Iteration 5:** - \( m = \frac{1.28125 + 1.4375}{2} = 1.359375 \) - \( f(1.359375) \approx 2 \times 0.308 + 1.359375 - 2 \approx 0.616 + 1.359375 - 2 = -0.0246 \) Interval: \([1.359375, 1.4375]\) ### **Iteration 6:** - \( m = \frac{1.359375 + 1.4375}{2} = 1.3984375 \) - \( f(1.3984) \approx 2 \times 0.336 + 1.3984 - 2 \approx 0.672 + 1.3984 - 2 = 0.0704 \) Interval: \([1.359375, 1.3984375]\) ### **Iteration 7:** - \( m = \frac{1.359375 + 1.3984375}{2} = 1.37890625 \) - \( f(1.3789) \approx 2 \times 0.319 + 1.3789 - 2 \approx 0.638 + 1.3789 - 2 = 0.0169 \) Interval: \([1.359375, 1.3789]\) ### **Iteration 8:** - \( m = \frac{1.359375 + 1.3789}{2} = 1.3691 \) - \( f(1.3691) \approx 2 \times 0.319 + 1.3691 - 2 \approx 0.638 + 1.3691 - 2 = 0.0071 \) Interval: \([1.359375, 1.3691]\) ### **Iteration 9:** - \( m = \frac{1.359375 + 1.3691}{2} \approx 1.36424 \) - \( f(1.3642) \approx 2 \times 0.319 + 1.3642 - 2 \approx 0.638 + 1.3642 - 2 = 0.0022 \) ### **Iteration 10:** - \( m \approx 1.3618 \): - \( f(1.3618) \approx 0.001 \) (close enough) **Approximate root:** \( x \approx \boxed{1.36} \) **Answer (to 2 decimal places):** **\( \boxed{1.36} \)** --- ## **Problem 3: Find the positive root of \( x^2 = 4 \cos x \) to 3 decimal places** Rewrite as: \[ f(x) = x^2 - 4 \cos x \] ### Step 1: Pick initial bounds - \( x=0 \): \[ f(0) = 0 - 4 \times 1 = -4 \quad (\text{negative}) \] - \( x=2 \): \[ f(2) = 4 - 4 \times \cos 2 \approx 4 - 4 \times (-0.416) = 4 + 1.664 = 5.664 \quad (\text{positive}) \] - Interval: \([0, 2]\) --- ### **Iteration 1:** - \( m = 1.0 \) - \( f(1) = 1 - 4 \times 0.5403 \approx 1 - 2.161 = -1.161 \) (negative) Interval: \([1, 2]\) ### **Iteration 2:** - \( m = 1.5 \) - \( f(1.5) = 2.25 - 4 \times 0.0707 \approx 2.25 - 0.283 = 1.967 \) (positive) Interval: \([1, 1.5]\) ### **Iteration 3:** - \( m = 1.25 \) - \( f(1.25) = 1.5625 - 4 \times 0.315 \approx 1.5625 - 1.26 = 0.3025 \) (positive) Interval: \([1, 1.25]\) ### **Iteration 4:** - \( m = 1.125 \) - \( f(1.125) \approx 1.266 - 4 \times 0.428 \approx 1.266 - 1.712 = -0.446 \) Interval: \([1.125, 1.25]\) ### **Iteration 5:** - \( m = 1.1875 \) - \( f(1.1875) \approx 1.410 - 4 \times 0.376 \approx 1.410 - 1.504 = -0.094 \) Interval: \([1.1875, 1.25]\) ### **Iteration 6:** - \( m = 1.21875 \) - \( f(1.21875) \approx 1.486 - 4 \times 0.344 \approx 1.486 - 1.376 = 0.11 \) Interval: \([1.1875, 1.21875]\) ### **Iteration 7:** - \( m = 1.2031 \) - \( f(1.2031) \approx 1.447 - 4 \times 0.360 \approx 1.447 - 1.440 = 0.007 \)

Certainly! Let's continue with **Problem 3** and provide the detailed calculations, including all bisection iterations, leading to the final answer to 3 decimal places. --- ## **Problem 3: Find the positive root of \( x^2 = 4 \cos x \) to 3 decimal places** **Rewritten as:** \[ f(x) = x^2 - 4 \cos x \] **Domain:** \( x > 0 \) --- ### Step 1: Choose initial bounds - At \( x=0 \): \[ f(0) = 0^2 - 4 \times \cos 0 = 0 - 4 \times 1 = -4 \quad (\text{negative}) \] - At \( x=2 \): \[ f(2) = 4 - 4 \times \cos 2 \approx 4 - 4 \times (-0.416) = 4 + 1.664 = 5.664 \quad (\text{positive}) \] **Interval:** \([0, 2]\) --- ### Step 2: Bisection iterations **Goal:** Narrow down the root between 0 and 2 until the difference is less than 0.001 (for 3 decimal places). --- ### **Iteration 1:** - \( a=0 \), \( b=2 \) - \( m = 1.0 \) - \( f(1.0) = 1 - 4 \times 0.5403 \approx 1 - 2.161 = -1.161 \) (negative) Since \( f(1.0) < 0 \), root is between 1.0 and 2. --- ### **Iteration 2:** - \( a=1.0 \), \( b=2 \) - \( m = 1.5 \) - \( f(1.5) = 2.25 - 4 \times 0.0707 \approx 2.25 - 0.283 = 1.967 \) (positive) Root between 1.0 and 1.5. --- ### **Iteration 3:** - \( a=1.0 \), \( b=1.5 \) - \( m = 1.25 \) - \( f(1.25) \approx 1.5625 - 4 \times 0.315 \approx 1.5625 - 1.26 = 0.3025 \) (positive) Root between 1.0 and 1.25. --- ### **Iteration 4:** - \( a=1.0 \), \( b=1.25 \) - \( m = 1.125 \) - \( f(1.125) \approx 1.266 - 4 \times 0.428 \approx 1.266 - 1.712 = -0.446 \) (negative) Root between 1.125 and 1.25. --- ### **Iteration 5:** - \( a=1.125 \), \( b=1.25 \) - \( m = 1.1875 \) - \( f(1.1875) \approx 1.410 - 4 \times 0.376 \approx 1.410 - 1.504 = -0.094 \) Root between 1.1875 and 1.25. --- ### **Iteration 6:** - \( a=1.1875 \), \( b=1.25 \) - \( m = 1.21875 \) - \( f(1.21875) \approx 1.485 - 4 \times 0.344 \approx 1.485 - 1.376 = 0.109 \) Root between 1.1875 and 1.21875. --- ### **Iteration 7:** - \( a=1.1875 \), \( b=1.21875 \) - \( m = 1.203125 \) - \( f(1.203125) \approx 1.448 - 4 \times 0.360 \approx 1.448 - 1.440 = 0.008 \) Close to zero! Since \( f(1.203125) \) is positive and very close to zero, and the difference between bounds is \( 1.21875 - 1.1875 = 0.03125 \), we continue to refine. --- ### **Iteration 8:** - \( a=1.1875 \), \( b=1.203125 \) - \( m = 1.1953125 \) - \( f(1.1953125) \approx 1.429 - 4 \times 0.368 \approx 1.429 - 1.472 = -0.043 \) Now, the function at \( m \) is slightly negative; root between 1.1953 and 1.2031. --- ### **Iteration 9:** - \( a=1.1953 \), \( b=1.2031 \) - \( m = 1.1992 \) - \( f(1.1992) \approx 1.440 - 4 \times 0.364 \approx 1.440 - 1.456 = -0.016 \) Close! Now, the interval is \([1.1992, 1.2031]\), width \( \approx 0.0039 \). --- ### **Iteration 10:** - \( a=1.1992 \), \( b=1.2031 \) - \( m = 1.2012 \) - \( f(1.2012) \approx 1.443 - 4 \times 0.363 \approx 1.443 - 1.452 = -0.009 \) Interval is getting smaller. ### **Iteration 11:** - \( a=1.2012 \), \( b=1.2031 \) - \( m = 1.2022 \) - \( f(1.2022) \approx 1.445 - 4 \times 0.3635 \approx 1.445 - 1.454 = -0.009 \) --- ### **Iteration 12:** - \( a=1.2022 \), \( b=1.2031 \) - \( m = 1.20265 \) - \( f(1.20265) \approx 1.446 - 4 \times 0.3637 \approx 1.446 - 1.454 = -0.008 \) --- ### **Iteration 13:** - \( a=1.20265 \), \( b=1.2031 \) - \( m = 1.20288 \) - \( f(1.20288) \approx 1.447 - 4 \times 0.3638 \approx 1.447 - 1.454 = -0.007 \) --- ### **Iteration 14:** - \( a=1.20288 \), \( b=1.2031 \) - \( m = 1.20299 \) - \( f(1.20299) \approx 1.4475 - 4 \times 0.3639 \approx 1.4475 - 1.454 = -0.006 \) --- ### **Iteration 15:** - \( a=1.20299 \), \( b=1.2031 \) - \( m = 1.20305 \) - \( f(1.20305) \approx 1.4476 - 4 \times 0.3640 \approx 1.4476 - 1.454 = -0.006 \) --- Continuing this process (or using a calculator/spreadsheet), the root converges around: ### **Final estimate:** \[ x \approx \boxed{1.203} \] **To 3 decimal places, the root is approximately:** \[ \boxed{1.203} \] --- ## **Summary of Final Answers:** | Problem | Final Root (to specified decimal places) | | -------- | ---------------------------------------- | | **1** | **x ≈ 1.56** (correct to 2 decimal places) | | **2** | **x ≈ 1.36** (correct to 2 decimal places) | | **3** | **x ≈ 1.203** (correct to 3 decimal places) | --- Let me know if you'd like help with setting up Excel sheets for these calculations or further explanations!