Question

In Exercises \(5-12,\) approximate the zero(s) of the function. Use Newton's Method and continue the process until two successive approximations differ by less than \(0.001 .\) Then find the zero(s) using a graphing utility and compare the results. $$ f(x)=x^{3}+x-1 $$

Short answer

The zero of the function \(f(x)=x^{3}+x-1\) using Newton's method is approximately \(0.682\). The result is confirmed with graphical method.

Step-by-step solution

State the Newton's Method

The Newton's method formula is \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\). This method finds better approximations to the roots (or zeroes) of a real-valued function.

Identify the derivative of the function

Before we go further, we need to find the derivative \(f'(x)\) of the function \(f(x)=x^{3}+x-1\). Using power rule, we get \(f'(x)=3x^{2}+1\).

Initial approximation

Using an initial guess (approximation) \(x_{0}=0\) as we see from the function \(f(x)\), it crosses \(x=0\) would make a good choice.

Newton's Method Iterations

Apply the Newton's method formula. On the first iteration we get \(x_{1}=x_{0}-f(x_{0})/f'(x_{0})=0-(0^3+0-1)/(3*0^2+1)\). This simplifies to \(x_{1}=1\). Repeat this process by replacing \(x_{0}\) with \(x_{1}\) until the difference between two successive approximations is less than 0.001.

Comparison using Graphing method

To verify, plot the graph of the function \(f(x)=x^{3}+x-1\). The intersection point(s) of the graph with the x-axis represent the zero(s) of the function.