Question

Use Newton’s method with initial approximation \({x_1} = - 1\) to find \({x_2}\), the second approximation to the root of the equation \({x^3} + x + 3 = 0\). Explain how the method works by first graphing the function and its tangent line (-1, 1).

Short answer

\({x_2} = - 1.25\)

Step-by-step solution

Introduction

The Newton’s method about a function is,

\({x_{n + 1}} = {x_n} - \frac{{f({x_n})}}{{f'({x_n})}}\)

f(x) is the function,\(f'({x_n}),{x_n}{\rm{ and }}{{\rm{x}}_{n + 1}}\) are derivation of the function and last two are roots of the function.

Given

\({x^3} + x + 3 = 0\)

(-1, 1)

Explanation

\(\begin{aligned}{c}f\left( x \right) &= {x^3} + x + 3\\f'\left( x \right) &= 3{x^2} + 1\\{x_1} &= - 1\end{aligned}\)

From the newton’s law,

\(\begin{aligned}{l}{x_{n + 1}} &= {x_n} - \frac{{f({x_n})}}{{f'({x_n})}}\\{x_2} &= - 1 - \frac{{\left( { - 1} \right) - \left( { - 1} \right) + 3}}{{3{{\left( { - 1} \right)}^2} + 1}}\\{x_2} &= - 1.25\end{aligned}\)

This will follow the tangent line (-1, 1) with the x axis coordinate (-1.25,0) by giving the second approximation \({x_2} = - 1.25\).