Question

In Exercises 75–78 use Newton’s method (see Example 8) to approximate a root for the given function with the specified value of $x_0.$Terminate your sequence when $\left|x_{n+1}-x_n\right|<0.001.$

$f\left(x\right)=x^3-2,x_0=1$

Short answer

The approximate value of a root for the given function is$1.25992.$

Step-by-step solution

Step 1. Given Information.

The given function is$f\left(x\right)=x^3-2,x_0=1.$

Step 2. Find the root for the given function.

To find the root of the function, let the equation is $x_{k+1}=x_k-\frac{f\left(x_k\right)}{f\text{'}\left(x_k\right)}.$

Now, $f\left(x_k\right)={\left(x_k\right)}^3-2,$so $f\text{'}\left(x_k\right)=3{\left(x_k\right)}^2.$

Put the values in the equation,

$x_{k+1}=x_k-\frac{{\left(x_k\right)}^3-2}{3{\left(x_k\right)}^2}.$

Step 3. Find the values of x1, x2.

To find the value of $x_1,\mathrm{put}k=0:$

$$\begin{gathered} x_{0+1}=x_0-\frac{{x_0}^3-2}{3{x_0}^2} \\ {x}_1=1-\frac{1^3-2}{3\left(1\right)} \\ {x}_1=1-\frac{-1}{3} \\ {x}_1=1+\frac{1}{3} \\ {x}_1=\frac{4}{3} \\ {x}_1=1.33333 \end{gathered}$$

Now, to find the value of $x_2,\mathrm{put}k=1:$

$$\begin{gathered} x_{1+1}=x_1-\frac{{x_1}^3-2}{3{x_1}^2} \\ {x}_2=1.33333-\frac{{\left(1.33333\right)}^3-2}{3\left(1.33333\right)} \\ {x}_2=1.26388 \end{gathered}$$

Step 4. Find the values of x3 and x4.

Now, let's find the values of $x_3,\mathrm{put}k=2:$

$$\begin{gathered} x_{2+1}=x_2-\frac{{x_2}^3-2}{3{x_2}^2} \\ {x}_3=\left(1.26388\right)-\frac{{\left(1.26388\right)}^3-2}{3\left(1.26388\right)} \\ {x}_3=1.25993 \end{gathered}$$

Let's find the values of $x_4,\mathrm{put}k=3:$

$$\begin{gathered} x_{3+1}=x_2-\frac{{x_4}^3-2}{3{x_4}^2} \\ {x}_4=\left(1.25993\right)-\frac{{\left(1.25993\right)}^3-2}{3\left(1.25993\right)} \\ {x}_4=1.25992 \end{gathered}$$

Step 5. Solve.

To approximate the root of the function, let's calculate $$\begin{gathered} \left|x_4-x_3\right| \\ =\left|1.25993-1.25992\right| \\ =\left|0.00001\right| \end{gathered}$$

Since $\left|x_4-x_3\right|<0.01.$

Therefore, the approximate value of the root is$1.25992.$