1. The problem involves applying Newton's Method (NM) to find the root of the function defined by its derivative $$f'(x) = x - e^{-x}$$ and second derivative $$f''(x) = 1 + e^{-x}$$ starting with initial guess $$x_1 = 5$$ and tolerance $$\varepsilon$$. 2. Newton's Method iteration formula is: $$ x_{n+1} = x_n - \frac{f'(x_n)}{f''(x_n)} $$ We continue iterating until the change between successive approximations is less than or equal to $$\varepsilon$$, i.e., $$|x_n - x_{n-1}| \leq \varepsilon$$. 3. Using the provided table: | n | x_n | f'(x_{n-1}) | f''(x_{n-1}) | |x_n - x_{n-1}| | |---|------|-------------|--------------|----------------| | 2 | 0.040| 4.993 | 1.006 | 4.959 | | 3 | 0.509| -0.920 | 1.960 | 0.469 | | 4 | 0.566| -0.091 | 1.600 | 0.057 | | 5 | 0.567| -0.001 | 1.567 | 0.001 \lt \varepsilon| 4. The iteration stopped at $$n=5$$ because $$|x_5 - x_4| = 0.001 < \varepsilon$$, which meets the stopping criterion. 5. Therefore, the root approximation is $$x \approx 0.567$$. Final answer: The root found using Newton's Method with initial guess 5 and given tolerance is approximately $$\boxed{0.567}$$.