Question

If $f\left(x\right)=x^3-9x+1$ and $x_0=3$, then using Newton-Raphson method, first iteration is

• A. $\frac{53}{18}$
• B. $\frac{53}{9}$
• C. $\frac{35}{18}$
• D. $\frac{35}{9}$

Answer 1

The correct option is A

$\frac{53}{18}$

Explanation for the correct option:

Step 1: Find the value of $f\left(3\right)$ and $f\text{'}\left(3\right)$:

The value of the function $f\left(x\right)=x^3-9x+1$ at $x=3$ is given as:

$\begin{array}{rcl}f\left(3\right)& =& 3^3-9\times 3+1\\ & =& 27-27+1\\ & =& 1\end{array}$

Now, the derivative of the function $f\left(x\right)=x^3-9x+1$ is given as:$f\text{'}\left(x\right)=3x^2-9$.

So at $x=3$ the value of $f\text{'}\left(3\right)$ is given as

$\begin{array}{rcl}f\text{'}\left(3\right)& =& 3\times 3^2-9\\ & =& 27-9\\ & =& 18\end{array}$

Step 2: Find the value of the first iteration:

By Newton Raphson method the first iteration is given as $x_1=x_0-\frac{f\left(x_0\right)}{f\text{'}\left(x_0\right)}$.

So for $x_0=3$:

$\begin{array}{rcl}{x}_1& =& 3-\frac{f\left(3\right)}{f\text{'}\left(3\right)}\\ & =& 3-\frac{1}{18}\\ & =& \frac{54-1}{18}\\ & =& \frac{53}{18}\end{array}$

Thus, the first iteration is $\frac{53}{18}$.

Hence, the correct option is A.