Root Newton Raphson 0D9181

1. **State the problem:** Find the root of the equation $$e^{2x} + 4x - 5 = 0$$ which corresponds to the intersection of the curves $$y = e^{2x}$$ and $$y = 5 - 4x$$. 2. **Show the root lies between 0 and 1:** Evaluate the function $$f(x) = e^{2x} + 4x - 5$$ at $$x=0$$ and $$x=1$$: $$f(0) = e^{0} + 0 - 5 = 1 - 5 = -4$$ $$f(1) = e^{2} + 4 - 5 = e^{2} - 1 \approx 7.389 - 1 = 6.389$$ Since $$f(0) < 0$$ and $$f(1) > 0$$, by the Intermediate Value Theorem, there is at least one root between $$0$$ and $$1$$. 3. **Newton-Raphson Method (NRM):** The formula for NRM is: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ Here, $$f(x) = e^{2x} + 4x - 5$$ $$f'(x) = 2e^{2x} + 4$$ 4. **Start with $$x_0 = 0.5$$:** - Calculate $$f(0.5)$$: $$f(0.5) = e^{1} + 2 - 5 = e^{1} - 3 \approx 2.718 - 3 = -0.282$$ - Calculate $$f'(0.5)$$: $$f'(0.5) = 2e^{1} + 4 = 2 \times 2.718 + 4 = 5.436 + 4 = 9.436$$ - Apply NRM formula: $$x_1 = 0.5 - \frac{-0.282}{9.436} = 0.5 + 0.0299 = 0.5299$$ 5. **Second iteration:** - Calculate $$f(0.5299)$$: $$f(0.5299) = e^{2 \times 0.5299} + 4 \times 0.5299 - 5 = e^{1.0598} + 2.1196 - 5$$ $$\approx 2.886 + 2.1196 - 5 = 0.0056$$ - Calculate $$f'(0.5299)$$: $$f'(0.5299) = 2e^{1.0598} + 4 = 2 \times 2.886 + 4 = 5.772 + 4 = 9.772$$ - Apply NRM formula: $$x_2 = 0.5299 - \frac{0.0056}{9.772} = 0.5299 - 0.00057 = 0.5293$$ 6. **Third iteration:** - Calculate $$f(0.5293)$$: $$f(0.5293) = e^{1.0586} + 4 \times 0.5293 - 5 \approx 2.883 + 2.117 - 5 = 0.0000$$ - Since $$f(0.5293) \approx 0$$, the root is approximately $$x = 0.53$$ to 2 decimal places. **Final answer:** The root of $$e^{2x} + 4x - 5 = 0$$ is approximately $$x = 0.53$$.