To use Newton's method to approximate the positive fourth root of $18$, we solve the equation $f(x) = x^4 - 18$. Here's the step-by-step process:

Step 1: Formula for Newton's Method

Newton's iteration formula is: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ where $f'(x)$ is the derivative of $f(x)$.

Step 2: Define $f(x)$ and $f'(x)$

• $f(x) = x^4 - 18$
• $f'(x) = 4x^3$

Step 3: Start with $x_0 = 2$ and compute $x_1$

Using the Newton's method formula: $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}$ Substitute $x_0 = 2$: $f(2) = 2^4 - 18 = 16 - 18 = -2$ $f'(2) = 4 \cdot 2^3 = 4 \cdot 8 = 32$ $x_1 = 2 - \frac{-2}{32} = 2 + \frac{1}{16} = 2.0625$

Step 4: Compute $x_2$

Now use $x_1 = 2.0625$ to find $x_2$: $x_2 = x_1 - \frac{f(x_1)}{f'(x_1)}$ First calculate $f(x_1)$ and $f'(x_1)$: $f(2.0625) = (2.0625)^4 - 18$ $(2.0625)^4 = 2.0625 \cdot 2.0625 \cdot 2.0625 \cdot 2.0625 = 18.0293 \quad (\text{approx.})$ $f(2.0625) = 18.0293 - 18 = 0.0293$ $f'(2.0625) = 4 \cdot (2.0625)^3 = 4 \cdot 8.7809 = 35.1236 \quad (\text{approx.})$ Now substitute these into the Newton's method formula: $x_2 = 2.0625 - \frac{0.0293}{35.1236}$ $x_2 = 2.0625 - 0.000834 \approx 2.0617$

Final Answer

The value of $x_2$, the approximation of the positive fourth root of $18$, is approximately: $\boxed{2.0617}$

---

Do you want further clarifications or more detailed steps?

Here are five related questions to deepen understanding:

1. What is the significance of choosing an initial guess $x_0$ in Newton's method?
2. How does Newton's method behave if the derivative of $f(x)$ is very small or zero?
3. Can Newton's method be applied to equations with no real roots? Why or why not?
4. How does the convergence of Newton's method depend on the choice of $x_0$?
5. What other numerical methods can be used to approximate roots of equations?

Tip: Always check the derivative at each step to ensure it does not approach zero, as this could slow convergence or cause the