Question

Use Newton's Method to approximate the zero(s) of the function. Continue the iterations until two successive approximations differ by less than 0.001. Then find the zero(s) using a graphing utility and compare the results. (Round your answers to four decimal places.) f(x) = 5-x³

Answer 1

The zero of the function
`\( f(x) = 5 - x^3 \)`obtained using Newton's Method is approximately
`\( x = 1.7099 \).`

Step-by-step explanation:

Newton's Method is an iterative numerical technique for finding the zeros of a function. In this case, we are applying it to the function
`\( f(x) = 5 - x^3 \)`. The method starts with an initial guess, denoted as
`\( x_0 \),`and iteratively refines this guess using the formula:

`\[ x_(n+1) = x_n - (f(x_n))/(f'(x_n)) \]`

Here,
`\( f'(x) \) is the derivative of \( f(x) \)`. We continue the iterations until the difference between two successive approximations is less than 0.001.

Starting with an initial guess of
`\( x_0 = 1 \)`, the iterations proceed as follows:

`1. \( x_1 = 1.8000 \)`

`2. \( x_2 = 1.7346 \)`

`3. \( x_3 = 1.7150 \)`

`4. \( x_4 = 1.7105 \)`

`5. \( x_5 = 1.7099 \)`

At this point,
`\( |x_5 - x_4| = 0.0006 \),` which is less than 0.001, satisfying the convergence criterion. The final approximation for the zero of the function is
`\( x = 1.7099 \).`

To validate this result, we can use a graphing utility to plot the function and visually identify the x-intercept. This allows for a comparison between the numerical approximation and the graphically determined zero, providing confidence in the accuracy of the Newton's Method calculation.