Chapter 4.7, Problem 5E

To determine

To calculate: The third approximation to the root of equation,

$x^3+2x-4=0$ where $x_1=1$ .

Answer to Problem 5E

The third approximation is $x_3=1.1797$ .

Explanation of Solution

Given information:

The equation is given as:

  $x^3+2x-4=0$

Formula used:

Newton’s Method:

We seek a solution of $f(x)=0$ , starting from an initial estimate $x=x_1$ .

For $x=x_n$ , compute the next approximation $x_{n+1}$ by

  $x_{n+1}=x_n-\frac{f(x_n)}{f\text{'}(x_n)}$ and so on.

Calculation:

Consider the equation,

$x^3+2x-4=0$ and $x_1=1$

This implies −

  $\begin{array}{l}f(x)=x^3+2x-4\\ f\text{'}(x)=3x^2+2\end{array}$

Now, let $n=1$ ,

  $x_2=x_1-\frac{f(x_1)}{f\text{'}(x_1)}$

  $x_2=x_1-\frac{x_1^3+2x_1-4}{3x_1^2+2}$

  $\begin{array}{l}{x}_2=(1)-\frac{{(1)}^3+2(1)-4}{3{(1)}^2+2}\\ x_2=1-\frac{1+2-4}{3+2}\\ x_2=1-\frac{3-4}{5}\end{array}$

  $\begin{array}{l}{x}_2=1-\left(\frac{-1}{5}\right)\\ x_2=1+\frac{1}{5}\\ x_2=\frac{6}{5}\\ x_2=1.2\end{array}$

The second approximation is $x_2=1.2$ .

Let $n=2$ ,

  $x_3=x_2-\frac{f(x_2)}{f\text{'}(x_2)}$

  $x_3=x_2-\frac{x_2^3+2x_2-4}{3x_2^2+2}$

  $\begin{array}{l}{x}_3=(1.2)-\frac{{(1.2)}^3+2(1.2)-4}{3{(1.2)}^2+2}\\ x_3=1.2-\frac{1.728+2.4-4}{4.32+2}\\ x_3=1.2-\frac{4.128-4}{6.32}\end{array}$

  $\begin{array}{l}{x}_3=1.2-\left(\frac{0.128}{6.32}\right)\\ x_3=1.2-(0.020253164)\\ x_3=1.179746836\end{array}$

The third approximation with four decimal places is $x_3=1.1797$ .