consider the equation (3x^{3}+3x + 1 = 0). if newtons method is applied to the equation with initial guess…

Explanation:

Step1: Define the function and its derivative

Let $f(x)=3x^{3}+3x + 1$, then $f^\prime(x)=9x^{2}+3$.

Step2: Calculate $x_2$ using Newton - Raphson formula

The Newton - Raphson formula is $x_{n + 1}=x_{n}-\frac{f(x_{n})}{f^\prime(x_{n})}$. For $n = 1$, $x_1=-2$. First, find $f(x_1)$ and $f^\prime(x_1)$: $f(-2)=3\times(-2)^{3}+3\times(-2)+1=3\times(-8)-6 + 1=-24-6 + 1=-29$. $f^\prime(-2)=9\times(-2)^{2}+3=9\times4+3=36 + 3=39$. Then $x_2=x_1-\frac{f(x_1)}{f^\prime(x_1)}=-2-\frac{-29}{39}=-2+\frac{29}{39}=\frac{-78 + 29}{39}=-\frac{49}{39}\approx - 1.256410$.

Step3: Calculate $x_3$ using Newton - Raphson formula

Now, for $n = 2$, $x_2=-\frac{49}{39}$. $f(x_2)=3\times(-\frac{49}{39})^{3}+3\times(-\frac{49}{39})+1$. $(-\frac{49}{39})^{3}=-\frac{49^{3}}{39^{3}}=-\frac{117649}{59319}$. $3\times(-\frac{117649}{59319})=-\frac{352947}{59319}$. $3\times(-\frac{49}{39})=-\frac{147}{39}=-\frac{2223}{59319}\times39$. $f(x_2)=-\frac{352947}{59319}-\frac{2223}{59319}+1=\frac{-352947-2223 + 59319}{59319}=\frac{-355170+59319}{59319}=-\frac{295851}{59319}$. $f^\prime(x_2)=9\times(-\frac{49}{39})^{2}+3=9\times\frac{2401}{1521}+3=\frac{21609}{1521}+3=\frac{21609+4563}{1521}=\frac{26172}{1521}$. $x_3=x_2-\frac{f(x_2)}{f^\prime(x_2)}=-\frac{49}{39}-\frac{-\frac{295851}{59319}}{\frac{26172}{1521}}$. $\frac{-\frac{295851}{59319}}{\frac{26172}{1521}}=-\frac{295851\times1521}{59319\times26172}$. $x_3\approx - 0.770248$.

Answer:

$x_2\approx - 1.256410$, $x_3\approx - 0.770248$