Chapter 4.5, Problem 61E

To determine

To calculate: The third estimate provided that the first estimate is 1.

Answer to Problem 61E

The correct option is (D).

Explanation of Solution

Given information: $f\left(x\right)=x-x^3+2$ provided that the first estimate is 1.

Concept used: Use the first approximation to estimate the second approximation and second approximation to get a third approximation. These approximations can be determined by using the formula, $x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}$ .

Calculation:

It is given that the first approximation of the solution is 1.

So, $x_1=1$ .

Differentiate the given function with respect to $x$ .

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left(x-x^3+2\right)\\ =-3x^2+1\end{array}$

  $x_{n+1}=x_n-\frac{x_n-{\left(x_n\right)}^3+2}{-3{\left(x_n\right)}^2+1}$

Substitute 1 for $n$ into the obtained equation and simplify it to determine the value of the second approximation, $x_2$ .

  $\begin{array}{c}{x}_{1+1}=x_1-\frac{x_1-{\left(x_1\right)}^3+2}{-3{\left(x_1\right)}^2+1}\\ x_2=1-\frac{1-{\left(1\right)}^3+2}{-3{\left(1\right)}^2+1}\\ =1-\frac{2}{\left(-2\right)}\\ =2\end{array}$

Now, substitute 2 for $n$ into the obtained equation and simplify it to determine the value of the third approximation, $x_3$ .

  $\begin{array}{c}{x}_{2+1}=x_2-\frac{x_2-{\left(x_2\right)}^3+2}{-3{\left(x_2\right)}^2+1}\\ x_2=2-\frac{2-{\left(2\right)}^3+2}{-3{\left(2\right)}^2+1}\\ =2-\frac{-4}{\left(-11\right)}\\ =\frac{18}{11}\end{array}$

This implies that option (D) is correct.