Let's go through each step in detail. --- ### **Step 1: Tangent Line** Given: - \( f(x) = x^2 - d \) - Approximation \( x_n \) **Find the tangent line at \( (x_n, f(x_n)) \):** 1. **Compute the derivative:** \[ f'(x) = 2x \] 2. **Evaluate at \( x_n \):** \[ f'(x_n) = 2x_n \] 3. **Find \( f(x_n) \):** \[ f(x_n) = x_n^2 - d \] 4. **Equation of tangent line:** \[ y - f(x_n) = f'(x_n)(x - x_n) \] Substitute values: \[ y - (x_n^2 - d) = 2x_n(x - x_n) \] \[ y = 2x_n(x - x_n) + x_n^2 - d \] --- ### **Step 2: X-intercept of the Tangent Line** Set \( y = 0 \) and solve for \( x \): \[ 0 = 2x_n(x - x_n) + x_n^2 - d \] \[ 0 = 2x_nx - 2x_n^2 + x_n^2 - d \] \[ 0 = 2x_nx - x_n^2 - d \] \[ 2x_nx = x_n^2 + d \] \[ x = \frac{x_n^2 + d}{2x_n} \] This is the Newton-Raphson formula for approximating \( \sqrt{d} \): \[ x_{n+1} = \frac{x_n^2 + d}{2x_n} \] Or, more commonly, \[ x_{n+1} = \frac{1}{2} \left(x_n + \frac{d}{x_n}\right) \] But **our formula matches the general Newton-Raphson step for \( f(x) = x^2 - d \).** --- ### **Step 3: Use the formula** Given \( x_1 = 3 \), approximate \( \sqrt{d} \) for \( d = 12, 13, 14 \), up to \( x_3 \). #### **For \( d = 12 \):** - \( x_1 = 3 \) \[ x_2 = \frac{3^2 + 12}{2 \times 3} = \frac{9 + 12}{6} = \frac{21}{6} = 3.5 \] \[ x_3 = \frac{3.5^2 + 12}{2 \times 3.5} = \frac{12.25 + 12}{7} = \frac{24.25}{7} \approx 3.46428571 \] #### **For \( d = 13 \):** - \( x_1 = 3 \) \[ x_2 = \frac{9 + 13}{6} = \frac{22}{6} \approx 3.66666667 \] \[ x_3 = \frac{(3.66666667)^2 + 13}{2 \times 3.66666667} \] \[ (3.66666667)^2 \approx 13.44444444 \] \[ x_3 = \frac{13.44444444 + 13}{7.33333334} = \frac{26.44444444}{7.33333334} \approx 3.60544218 \] #### **For \( d = 14 \):** - \( x_1 = 3 \) \[ x_2 = \frac{9 + 14}{6} = \frac{23}{6} \approx 3.83333333 \] \[ x_3 = \frac{(3.83333333)^2 + 14}{2 \times 3.83333333} \] \[ (3.83333333)^2 \approx 14.69444444 \] \[ x_3 = \frac{14.69444444 + 14}{7.66666666} = \frac{28.69444444}{7.66666666} \approx 3.74291052 \] --- #### **Summary Table** | d | \( x_1 \) | \( x_2 \) | \( x_3 \) | |-----|-----------|----------------|------------------| | 12 | 3 | 3.5 | 3.46428571 | | 13 | 3 | 3.66666667 | 3.60544218 | | 14 | 3 | 3.83333333 | 3.74291052 | --- ### **Step 4: Examine the error** Compute actual square roots (to 8 decimal places): - \( \sqrt{12} \approx 3.46410162 \) - \( \sqrt{13} \approx 3.60555128 \) - \( \sqrt{14} \approx 3.74165739 \) Compute errors (actual - approx): #### **For \( d = 12 \):** \[ \text{Error} = 3.46410162 - 3.46428571 = -0.00018409 \] #### **For \( d = 13 \):** \[ \text{Error} = 3.60555128 - 3.60544218 = 0.00010910 \] #### **For \( d = 14 \):** \[ \text{Error} = 3.74165739 - 3.74291052 = -0.00125313 \] --- ### **Why does the error increase as d increases?** The Newton-Raphson method's convergence is fastest when the initial guess is close to the true value. Here, the initial guess \( x_1 = 3 \) is closer to \( \sqrt{12} \) (which is about 3.464), than to \( \sqrt{13} \) (about 3.606), and much closer than to \( \sqrt{14} \) (about 3.742). As the true value increases, but the initial guess remains at 3, the relative distance between the guess and the actual root increases, resulting in slower convergence (i.e., larger errors after the same number of steps). **Summary Table of Errors:** | d | Approx (\( x_3 \)) | Actual sqrt | Error | |-----|---------------------|-------------|-------------| | 12 | 3.46428571 | 3.46410162 | -0.00018409 | | 13 | 3.60544218 | 3.60555128 | 0.00010910 | | 14 | 3.74291052 | 3.74165739 | -0.00125313 | **Conclusion:** The further \( d \) is from 9 (where \( x_1 = 3 \) is the exact root), the less accurate the estimate after a fixed number of steps. This is due to the quadratic convergence nature of Newton's method, which is highly sensitive to the quality of the initial guess.

Certainly! Here's each part broken down with all calculations shown separately. --- ### **Part 1: Find the equation of the tangent line to \(f(x) = x^2 - d\) at \(x = x_n\)** 1. **Derivative of \(f(x)\):** \[ f'(x) = 2x \] 2. **Evaluate at \(x = x_n\):** \[ f'(x_n) = 2x_n \] 3. **Function value at \(x_n\):** \[ f(x_n) = x_n^2 - d \] 4. **Equation of tangent line at \((x_n, f(x_n))\):** \[ y - f(x_n) = f'(x_n)(x - x_n) \] \[ y - (x_n^2 - d) = 2x_n (x - x_n) \] --- ### **Part 2: Find the x-intercept of that tangent line** Set \( y = 0 \): \[ 0 = 2x_n (x - x_n) + x_n^2 - d \] Expand: \[ 0 = 2x_n x - 2x_n^2 + x_n^2 - d \] \[ 0 = 2x_n x - x_n^2 - d \] Solve for \(x\): \[ 2x_n x = x_n^2 + d \] \[ x = \frac{x_n^2 + d}{2 x_n} \] This gives the iterative formula: \[ x_{n+1} = \frac{x_n^2 + d}{2 x_n} \] --- ### **Part 3: Use the formula with \(x_1 = 3\) to approximate \(\sqrt{d}\) for \(d=12,13,14\), up to \(x_3\)** --- #### **For \(d=12\):** - **Iteration 1:** \[ x_2 = \frac{(x_1)^2 + d}{2 x_1} = \frac{3^2 + 12}{2 \times 3} = \frac{9 + 12}{6} = \frac{21}{6} = 3.5 \] - **Iteration 2:** \[ x_3 = \frac{(x_2)^2 + d}{2 x_2} = \frac{(3.5)^2 + 12}{2 \times 3.5} = \frac{12.25 + 12}{7} = \frac{24.25}{7} \approx 3.46428571 \] --- #### **For \(d=13\):** - **Iteration 1:** \[ x_2 = \frac{9 + 13}{6} = \frac{22}{6} \approx 3.66666667 \] - **Iteration 2:** \[ x_3 = \frac{(3.66666667)^2 + 13}{2 \times 3.66666667} \] Calculate numerator: \[ (3.66666667)^2 \approx 13.44444444 \] \[ 13.44444444 + 13 = 26.44444444 \] Calculate denominator: \[ 2 \times 3.66666667 = 7.33333333 \] Compute: \[ x_3 = \frac{26.44444444}{7.33333333} \approx 3.60544218 \] --- #### **For \(d=14\):** - **Iteration 1:** \[ x_2 = \frac{9 + 14}{6} = \frac{23}{6} \approx 3.83333333 \] - **Iteration 2:** \[ x_3 = \frac{(3.83333333)^2 + 14}{2 \times 3.83333333} \] Calculate numerator: \[ (3.83333333)^2 \approx 14.69444444 \] \[ 14.69444444 + 14 = 28.69444444 \] Calculate denominator: \[ 2 \times 3.83333333 = 7.66666666 \] Compute: \[ x_3 = \frac{28.69444444}{7.66666666} \approx 3.74291052 \] --- ### **Part 4: Actual square roots and error calculation** - Actual values: \[ \sqrt{12} \approx 3.46410162 \] \[ \sqrt{13} \approx 3.60555128 \] \[ \sqrt{14} \approx 3.74165739 \] - Errors: **For \(d=12\):** \[ \text{Error} = \text{Actual} - \text{Approx} = 3.46410162 - 3.46428571 = -0.00018409 \] **For \(d=13\):** \[ 3.60555128 - 3.60544218 = 0.00010910 \] **For \(d=14\):** \[ 3.74165739 - 3.74291052 = -0.00125313 \] --- ### **Summary** | \(d\) | \(x_1\) | \(x_2\) | \(x_3\) | Actual \(\sqrt{d}\) | Error (\( \text{Actual} - \text{Approx} \)) | |--------|---------|---------|---------|---------------------|-----------------------------------------| | 12 | 3.0 | 3.5 | 3.46428571 | 3.46410162 | -0.00018409 | | 13 | 3.0 | 3.66666667 | 3.60544218 | 3.60555128 | 0.00010910 | | 14 | 3.0 | 3.83333333 | 3.74291052 | 3.74165739 | -0.00125313 | --- ### **Final note:** The larger the difference between the initial guess and the actual root, the larger the error after the same number of iterations. This explains why the error is smallest for \(d=12\) and largest for \(d=14\).

Certainly! Here are the final answers for each part, including all calculations and results: --- ### **Part 1: Equation of the tangent line to \(f(x) = x^2 - d\) at \(x = x_n\):** \[ \boxed{ \text{Tangent line:} \quad y = 2x_n (x - x_n) + x_n^2 - d } \] --- ### **Part 2: X-intercept of the tangent line (formula for next approximation):** Set \( y=0 \): \[ 0 = 2x_n (x - x_n) + x_n^2 - d \] Solve for \( x \): \[ 2x_n x - 2x_n^2 + x_n^2 - d = 0 \] \[ 2x_n x = x_n^2 + d \] \[ \boxed{ x_{n+1} = \frac{x_n^2 + d}{2 x_n} } \] which simplifies to the Newton-Raphson iteration for \(\sqrt{d}\). --- ### **Part 3: Approximations of \(\sqrt{d}\) using \(x_1=3\)** **For \( d=12 \):** - \( x_2 = \frac{3^2 + 12}{2 \times 3} = \frac{9 + 12}{6} = \frac{21}{6} = \boxed{3.5} \) - \( x_3 = \frac{(3.5)^2 + 12}{2 \times 3.5} = \frac{12.25 + 12}{7} = \frac{24.25}{7} \approx \boxed{3.46428571} \) **For \( d=13 \):** - \( x_2 = \frac{9 + 13}{6} = \frac{22}{6} \approx \boxed{3.66666667} \) - \( x_3 = \frac{(3.66666667)^2 + 13}{2 \times 3.66666667} \) Calculate numerator: \[ (3.66666667)^2 \approx 13.44444444 \] \[ 13.44444444 + 13 = 26.44444444 \] Calculate denominator: \[ 2 \times 3.66666667 = 7.33333333 \] Approximate: \[ x_3 = \frac{26.44444444}{7.33333333} \approx \boxed{3.60544218} \] **For \( d=14 \):** - \( x_2 = \frac{9 + 14}{6} = \frac{23}{6} \approx \boxed{3.83333333} \) - \( x_3 = \frac{(3.83333333)^2 + 14}{2 \times 3.83333333} \) Calculate numerator: \[ (3.83333333)^2 \approx 14.69444444 \] \[ 14.69444444 + 14 = 28.69444444 \] Calculate denominator: \[ 2 \times 3.83333333 = 7.66666666 \] Approximate: \[ x_3 = \frac{28.69444444}{7.66666666} \approx \boxed{3.74291052} \] --- ### **Part 4: Actual square roots and errors** - \(\sqrt{12} \approx 3.46410162\) - \(\sqrt{13} \approx 3.60555128\) - \(\sqrt{14} \approx 3.74165739\) **Errors (Actual - Approximate):** | \(d\) | Approximate \(x_3\) | Actual \(\sqrt{d}\) | Error (Actual - Approximate) | |--------|---------------------|---------------------|------------------------------| | 12 | 3.46428571 | 3.46410162 | \(-0.00018409\) | | 13 | 3.60544218 | 3.60555128 | \(+0.00010910\) | | 14 | 3.74291052 | 3.74165739 | \(-0.00125313\) | --- ### **Summary of Final Answers:** - **Tangent line at \(x_n\):** \[ \boxed{ y = 2x_n (x - x_n) + x_n^2 - d } \] - **Next approximation (x-intercept):** \[ \boxed{ x_{n+1} = \frac{x_n^2 + d}{2 x_n} } \] - **Approximations after two iterations starting from \(x_1=3\):** | \(d\) | \(x_3\) approximation | Actual \(\sqrt{d}\) | Final Error | |--------|-------------------------|---------------------|--------------| | 12 | **3.46428571** | 3.46410162 | \(-0.00018409\) | | 13 | **3.60544218** | 3.60555128 | \(+0.00010910\) | | 14 | **3.74291052** | 3.74165739 | \(-0.00125313\) | --- **Note:** The smaller the initial guess's distance from the true root, the smaller the error after a fixed number of iterations.

The tangent line at \( x = x_n \) is \( y = 2x_n (x - x_n) + x_n^2 - d \); the x-intercept formula for the tangent line is \( x_{n+1} = \frac{x_n^2 + d}{2 x_n} \); using \( x_1=3 \), the approximations are \( \sqrt{12} \approx 3.46428571 \), \( \sqrt{13} \approx 3.60544218 \), \( \sqrt{14} \approx 3.74291052 \); the errors compared to actual roots are approximately \(-0.00018409\), \(+0.00010910\), \(-0.00125313\) respectively, illustrating increased error for larger \( d \) due to less accurate initial guesses.