1. Statement of the problem: Use the Newton-Raphson method to find a root of the function $f(x)=x^3-5x^2-4x$ near the initial guess $x_0=5$.\n2. Formula and derivative: The Newton-Raphson update is\n$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$\nThe derivative is $f'(x)=3x^2-10x-4$.\nImportant rules: Ensure $f'(x_n)\neq 0$ at iterates, choose a reasonable initial guess, and iterate until the change is below the desired tolerance.\n3. Iteration 1: Evaluate at $x_0=5$.\nCompute $f(5)=5^3-5\cdot5^2-4\cdot5=125-125-20=-20$.\nCompute $f'(5)=3\cdot5^2-10\cdot5-4=75-50-4=21$.\nApply the update: $$x_1=5-\frac{f(5)}{f'(5)}=5-\frac{-20}{21}=5+\frac{20}{21}\approx 5.95238095$$\n4. Iteration 2: Evaluate at $x_1\approx 5.95238095$.\nCompute $f(5.95238095)\approx -0.388576$.\nCompute $f'(5.95238095)\approx 15.51846$.\nApply the update: $$x_2=5.95238095-\frac{-0.388576}{15.51846}\approx 5.977385$$\n5. Iteration 3: Evaluate at $x_2\approx 5.977385$.\nCompute $f(5.977385)\approx -0.001014$.\nCompute $f'(5.977385)\approx 15.66072$.\nApply the update: $$x_3=5.977385-\frac{-0.001014}{15.66072}\approx 5.9774497$$\n6. Conclusion: The iterates converge rapidly and after three iterations the root is approximately $x\approx 5.97745$.\nTo reach a specified accuracy continue iterating until $|x_{n+1}-x_n|$ is below your tolerance.\n