Chapter 4.7, Problem 6E

To determine

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

$\frac{1}{3}{x}^3+\frac{1}{2}{x}^2+3=0$ where $x_1=-3$ .

Answer to Problem 6E

The third approximation is $x_3=-2.7186$ .

Explanation of Solution

Given information:

The equation is given as:

  $\frac{1}{3}{x}^3+\frac{1}{2}{x}^2+3=0$

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,

$\frac{1}{3}{x}^3+\frac{1}{2}{x}^2+3=0$ and $x_1=-3$

This implies −

  $\begin{array}{l}f(x)=\frac{1}{3}{x}^3+\frac{1}{2}{x}^2+3\\ f\text{'}(x)=x^2+x\end{array}$

Now, let $n=1$ ,

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

  $x_2=x_1-\frac{\frac{1}{3}{x}_1^3+\frac{1}{2}{x}_1^2+3}{x_1^2+x_1}$

  $\begin{array}{l}{x}_2=(-3)-\left(\frac{\frac{1}{3}{(-3)}^3+\frac{1}{2}{(-3)}^2+3}{{(-3)}^2-3}\right)\\ x_2=(-3)-\left(\frac{-9+\frac{9}{2}+3}{9-3}\right)\\ x_2=(-3)-\left(\frac{\frac{-18+9+6}{2}}{6}\right)\\ x_2=(-3)-\left(\frac{\frac{-3}{2}}{6}\right)\\ x_2=(-3)-\left(\frac{-3}{2}\times \frac{1}{6}\right)\end{array}$

  $\begin{array}{l}{x}_2=(-3)-\left(\frac{-1}{4}\right)\\ x_2=-3+\frac{1}{4}\\ x_2=\frac{-12+1}{4}\\ x_2=\frac{-11}{4}\\ x_2=-2.75\end{array}$

The second approximation is $x_2=-2.75$ .

Let $n=2$ ,

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

  $x_3=x_2-\frac{\frac{1}{3}{x}_2^3+\frac{1}{2}{x}_2^2+3}{x_2^2+x_2}$

  $\begin{array}{l}{x}_3=(-2.75)-\left(\frac{\frac{1}{3}{(-2.75)}^3+\frac{1}{2}{(-2.75)}^2+3}{{(-2.75)}^2+(-2.75)}\right)\\ x_3=(-2.75)-\left(\frac{-6.932291+3.78125+3}{7.5625-2.75}\right)\\ x_3=(-2.75)-\left(\frac{-6.932291+6.78125}{4.8125}\right)\end{array}$

  $\begin{array}{l}{x}_3=-2.75-\left(\frac{-0.151041}{4.8125}\right)\\ x_3=-2.75+0.031385\\ x_3=-2.718615\end{array}$

The third approximation with four decimal places is $x_3=-2.7186$ .