Question

You may use a graphing calculator to solve the following problems. True or False Newton's method will not find the zero of \(f(x)=x /\left(x^{2}+1\right)\) if the first guess is greater than \(1 .\) Justify your answer.

Short answer

The statement can only verified to be true or false by using Newton's method to iterate starting from a first guess greater than 1. The conclusion depends on the results of these iterations.

Step-by-step solution

Define the Function and Its Derivative

Start by defining the function \(f(x) = \frac{x}{x^2+1}\) and its derivative. The derivative, \(f'(x)\), can be computed using quotient rule for derivatives, which for any 2 differentiable functions u(x) and v(x) is \(\frac{d(u/v)}{dx}= \frac{u'v - uv'}{v^2}\). For f(x), u=x and v=\(x^2+1\), thus \(f'(x)=\frac{(1)(x^2+1) - x(2x)}{(x^2+1)^2}=\frac{1-x^2}{(x^2+1)^2}\).

Apply Newton's Method

Newton's method can be stated as follows: for a guess \(x_n\), the next guess, \(x_{n+1}\), is given by \(x_n - \frac{f(x_n)}{f'(x_n)}\). Start with an arbitrary guess greater than 1, e.g., \(x_n = 2\). Using the above equations: \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = 2 - \frac{\frac{2}{2^2+1}}{\frac{1-2^2}{(2^2+1)^2}} = 2 - \frac{2/5}{-3/25}=2 + \frac{10}{3} = \frac{16}{3}\). This value is still greater than 1, hence, it is required to iterate the method until \(x_n\) is lesser than or equal 1 or when the value of \(x_n\) starts repeating or not changing significantly.

Draw Conclusion

If, after repeated iterations, the sequence of \(x_n\) never reaches a value less than or equal to 1 or starts repeating/not changing, then it can be concluded that the statement is true. Newton's method does not find the zero of the function when the first guess is greater than 1. Alternatively, if the iterations converge to a root, the statement is false.