Use the Newton-Raphson method to find an approximate solution of the equation9x3 – 5x - 3 = 0in the interval [0, 2]. Use the method until successive approximations obtained by a calculator are identical.If c1 = 2, then c2 =

The Newton Raphson iteration formula: xn+1=xn-fxnf'xn Here, fxn=9xn3-5xn-3. Differentiate to obtain…

Answered: Use the Newton-Raphson method to find…

College Algebra

10th Edition

ISBN:9781337282291

Author:Ron Larson

Publisher:Ron Larson

Chapter5: Exponential And Logarithmic Functions

Section: Chapter Questions

Problem 39CT: The population P (in millions) of Texas from 2001 through 2014 can be approximated by the model...

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Use the Newton-Raphson method to find an approximate solution of the equation
9x3 – 5x - 3 = 0
in the interval [0, 2]. Use the method until successive approximations obtained by a calculator are identical.
If c1 = 2, then c2 =

Expert Solution

Step 1

The Newton Raphson iteration formula:

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

Here, $f (x_{n}) = 9 {(x_{n})}^{3} - 5 x_{n} - 3$ .

Differentiate to obtain $f' (x_{n}) = 27 {(x_{n})}^{2} - 5$ .

Starting with $x_{0} = 2$ .

For n=0, the iteration formula becomes,

$\begin{array}{rcl} x_{1} & = & x_{0} - \frac{f (x_{0})}{f' (x_{0})} \\ = & 2 - \frac{9 {(2)}^{3} - 5 (2) - 3}{27 {(2)}^{2} - 5} \\ = & 2 - \frac{59}{103} \\ \approx & 1.4272 \end{array}$

For n=1, the iteration formula becomes,

$\begin{array}{rcl} x_{2} & = & x_{1} - \frac{f (x_{1})}{f' (x_{1})} \\ = & 1.4272 - \frac{9 {(1.4272)}^{3} - 5 (1.4272) - 3}{27 {(1.4272)}^{2} - 5} \\ \approx & 0.9688 \end{array}$

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