1. **Problem Statement:** Find the root of the function $f(x) = e^x - 2x^2$ in the interval (1, 2) using the Newton-Raphson method up to 6 iterations and calculate the corresponding error for each iteration to four decimal places. 2. **Formula and Explanation:** The Newton-Raphson method formula for root finding is: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ where $f'(x)$ is the derivative of $f(x)$. 3. **Function and Derivative:** $$f(x) = e^x - 2x^2$$ $$f'(x) = e^x - 4x$$ 4. **Initial Guess:** Choose $x_0 = 1.5$ (a value in the interval (1, 2)). 5. **Iterations:** Calculate $x_{n+1}$ and error $|x_{n+1} - x_n|$ for each iteration. - Iteration 1: $$x_1 = 1.5 - \frac{e^{1.5} - 2(1.5)^2}{e^{1.5} - 4(1.5)} = 1.5 - \frac{4.4817 - 4.5}{4.4817 - 6} = 1.5 - \frac{-0.0183}{-1.5183} = 1.5 - 0.0121 = 1.4879$$ Error = $|1.4879 - 1.5| = 0.0121$ - Iteration 2: $$x_2 = 1.4879 - \frac{e^{1.4879} - 2(1.4879)^2}{e^{1.4879} - 4(1.4879)} = 1.4879 - \frac{4.4263 - 4.4243}{4.4263 - 5.9516} = 1.4879 - \frac{0.0020}{-1.5253} = 1.4879 + 0.0013 = 1.4892$$ Error = $|1.4892 - 1.4879| = 0.0013$ - Iteration 3: $$x_3 = 1.4892 - \frac{e^{1.4892} - 2(1.4892)^2}{e^{1.4892} - 4(1.4892)} = 1.4892 - \frac{4.4330 - 4.4360}{4.4330 - 5.9568} = 1.4892 - \frac{-0.0030}{-1.5238} = 1.4892 - 0.0020 = 1.4872$$ Error = $|1.4872 - 1.4892| = 0.0020$ - Iteration 4: $$x_4 = 1.4872 - \frac{e^{1.4872} - 2(1.4872)^2}{e^{1.4872} - 4(1.4872)} = 1.4872 - \frac{4.4230 - 4.4200}{4.4230 - 5.9488} = 1.4872 - \frac{0.0030}{-1.5258} = 1.4872 + 0.0020 = 1.4892$$ Error = $|1.4892 - 1.4872| = 0.0020$ - Iteration 5: $$x_5 = 1.4892 - \frac{e^{1.4892} - 2(1.4892)^2}{e^{1.4892} - 4(1.4892)} = 1.4892 - \frac{4.4330 - 4.4360}{4.4330 - 5.9568} = 1.4892 - \frac{-0.0030}{-1.5238} = 1.4892 - 0.0020 = 1.4872$$ Error = $|1.4872 - 1.4892| = 0.0020$ - Iteration 6: $$x_6 = 1.4872 - \frac{e^{1.4872} - 2(1.4872)^2}{e^{1.4872} - 4(1.4872)} = 1.4872 - \frac{4.4230 - 4.4200}{4.4230 - 5.9488} = 1.4872 - \frac{0.0030}{-1.5258} = 1.4872 + 0.0020 = 1.4892$$ Error = $|1.4892 - 1.4872| = 0.0020$ 6. **Conclusion:** The root oscillates between approximately 1.4872 and 1.4892 with error about 0.0020 after 6 iterations. **Final approximate root:** $x \approx 1.4882$ (average of last two values)