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$\left|x_{n+1}-x_n\right|<0.001$

78.$f\left(x\right)=\sqrt{x+1}-\frac{1}{x},x_0=1$

We have the given functions, $f\left(x\right)=\sqrt{x+1}-\frac{1}{x}$and $x_0=1$.

Here we have to find the root of the functions.

Let us consider the function$f\left(x\right)=\sqrt{x+1}-\frac{1}{x}$

We have the equation $x_{k+1}=x_k-\frac{f\left(x_k\right)}{f\text{'}\left(x_k\right)}$.......Equation (1)

Therefore,

$\begin{array}{r}f\left(x_k\right)=\sqrt{x_k+1}-\frac{1}{x_k}\\ f^{\mathrm{\prime }}\left(x_k\right)=\frac{1}{2\sqrt{x_k+1}}+\frac{1}{x_k^2}\end{array}$

Substituting this in equation(1)

$x_{k+1}=x_k-\frac{\sqrt{x_k+1}-\frac{1}{x_k}}{\frac{1}{2\sqrt{x_k+1}}+\frac{1}{x_k^2}}$......Equation (2)

Now to find the value of $x_1$, substitute $k=0$in equation (2)

$\begin{array}{r}{x}_{0+1}=x_0-\frac{\sqrt{x_0+1}-\frac{1}{x_0}}{\frac{1}{2\sqrt{x_0+1}}+\frac{1}{x_0^2}}\\ x_1=x_0-\frac{\sqrt{x_0+1}-\frac{1}{x_0}}{\frac{1}{2\sqrt{x_0+1}}+\frac{1}{x_0^2}}\end{array}$

Substitute$x_0=1$

localid="1649311239425" $$\begin{gathered} x_1=\left(1\right)-\frac{\sqrt{\left(1\right)+1}-\frac{1}{\left(1\right)}}{\frac{1}{2\sqrt{\left(1\right)+1}}+\frac{1}{(1{)}^2}} \\ =0.693981 \end{gathered}$$

Therefore,localid="1649311250474" $x_1=0.693981$

Now to find the value of $x_2$, substitute localid="1649311685428">k=1 in equation (2)

$x_2=x_1-\frac{\sqrt{x_1+1}-\frac{1}{x_1}}{\frac{1}{2\sqrt{x_1+1}}+\frac{1}{x_1^2}}$

Substitute$x_1=0.693981$

$$\begin{gathered} x_2=0.693981-\frac{\sqrt{0.693981+1}-\frac{1}{0.693981}}{\frac{1}{2\sqrt{0.693981+1}}+\frac{1}{(0.693981{)}^2}} \\ =0.750648 \end{gathered}$$

Therefore,$x_2=0.750648$

Now to find the value of $x_3$, substitute $k=2$in equation (2)

$x_3=x_2-\frac{\sqrt{x_2+1}-\frac{1}{x_2}}{\frac{1}{2\sqrt{x_2+1}}+\frac{1}{x_2^2}}$

Substitute $x_2=0.750648$

$$\begin{gathered} x_3=0.750648-\frac{\sqrt{0.750648+1}-\frac{1}{0.750648}}{\frac{1}{2\sqrt{0.750648+1}}+\frac{1}{(0.750648{)}^2}} \\ =0.754858 \end{gathered}$$

Therefore,$x_3=0.754858$

Now to find the value of $x_4$, substitute $k=3$ in equation (2)

localid="1649314112171" $x_4=x_3-\frac{\sqrt{x_3+1}-\frac{1}{x_3}}{\frac{1}{2\sqrt{x_3+1}}+\frac{1}{x_3^2}}$

Substitute$x_3=0.754858$

$x_4=0.754858-\frac{\sqrt{0.754858+1}-\frac{1}{0.754858}}{\frac{1}{2\sqrt{0.754858+1}}+\frac{1}{(0.754858{)}^2}}$

Therefore,$x_4=0.754878$

Now to find the value of $x_5$, substitute $k=4$in equation (2)

$x_5=x_4-\frac{\sqrt{x_4+1}-\frac{1}{x_4}}{\frac{1}{2\sqrt{x_4+1}}+\frac{1}{x_4^2}}$

Substitute,$x_4=0.754878$

localid="1649314429257" $$\begin{gathered} x_5=0.754878-\frac{\sqrt{0.754878+1}-\frac{1}{0.754878}}{\frac{1}{2\sqrt{0.754878+1}}+\frac{1}{(0.754878{)}^2}} \\ =0.754878 \end{gathered}$$

Therefore,$x_5=0.754878$

Here,

$$\begin{gathered} \left|x_5-x_4\right|=|0.754878-0.754878| \\ =\left|0\right| \end{gathered}$$

Since,$\left|x_5-x_4\right|<0.001$let us stop the iteration.

Therefore, the approximate value of the root of$f\left(x\right)=\sqrt{x+1}-\frac{1}{x}$is$0.754878$.