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

77. $f\left(x\right)=e^x+\sin x,x_0=-2$

We have the given functions, $f\left(x\right)=e^x+\sin x$and $x_0=-2$.

Here we have to find the root of the functions.

Let us consider the function$f\left(x\right)=e^x+\sin x$

We have the equationrole="math" localid="1649260873855" $x_{k+1}=x_k-\frac{f\left(x_k\right)}{f\text{'}\left(x_k\right)}$.....Equation($1$)

Therefore,

$\begin{array}{r}f\left(x_k\right)=e^{x_k}+\sin x_k\\ f^{\mathrm{\prime }}\left(x_k\right)=e^{x_k}+\cos x_k\end{array}$

Substituting this in equation($1$)

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

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

$\begin{array}{r}{x}_{0+1}=x_0-\frac{e^{x_0}+\sin x_0}{e^{x_0}+\cos x_0}\\ x_1=x_0-\frac{e^{x_0}+\sin x_0}{e^{x_0}+\cos x_0}\end{array}$

Substitute$x_0=-2$

role="math" localid="1649261327561" $$\begin{gathered} x_1=(-2)-\frac{e^{(-2)}+\sin (-2)}{e^{(-2)}+\cos (-2)} \\ =-4.75616 \end{gathered}$$

Therefore,$x_1=-4.75616$

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

$x_2=x_1-\frac{e^{x_1}+\sin x_1}{e^{x_1}+\cos x_1}$

Substitute $x_1=-4.75616$

localid="1649261941775" $$\begin{gathered} x_2=(-4.75616)-\frac{e^{(-4.75616)}+\sin (-4.75616)}{e^{(-4.75616)}+\cos (-4.75616)} \\ =-24.0033 \end{gathered}$$

Therefore$x_2=-24.0033$

Now to find the value of $x_3$, substitute localid="1649341362938" $k=2$in equation ($2$)

localid="1649261914102" $x_3=x_2-\frac{e^{x_2}+\sin x_2}{e^{x_2}+\cos x_2}$

Substitute$x_2=-24.0033$

$$\begin{gathered} x_3=(-24.0033)-\frac{e^{(-24.0033)}+\sin (-24.0033)}{e^{(-24.0033)}+\cos (-24.0033)} \\ =-26.1199 \end{gathered}$$

Therefore,$x_3=-26.1199$

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

localid="1649341466863" $x_4=x_3-\frac{e^{x_3}+\sin x_3}{e^{x_3}+\cos x_3}$

Substitutelocalid="1649263166094" $x_3=-26.1199$

localid="1649262239593" $$\begin{gathered} x_4=(-26.1199)-\frac{e^{(-26.1199)}+\sin (-26.1199)}{e^{(-26.1199)}+\cos (-26.1199)} \\ =-24.6056 \end{gathered}$$

Therefore$x_4==-24.6056$

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

localid="1649341527802" $x_5=x_4-\frac{e^{x_4}+\sin x_4}{e^{x_4}+\cos x_4}$

Substitute localid="1649262418782" $x_4==-24.6056$

$$\begin{gathered} x_5=(-24.6056)-\frac{e^{(-24.6056)}+\sin (-24.6056)}{e^{(-24.6056)}+\cos (-24.6056)} \\ =-25.1877 \end{gathered}$$

Therefore,$x_5=-25.1877$

Now to find the value of $x_6$, substitute $k=5$in equation ($2$)

localid="1649341576454" $x_6=x_1-\frac{e^{x_5}+\sin x_5}{e^{x_5}+\cos x_5}$

Substitute$x_5=-25.1877$

$$\begin{gathered} x_6=(-25.1877)-\frac{e^{(-25.1877)}+\sin (-25.1877)}{e^{(-25.1877)}+\cos (-25.1877)} \\ =-25.1327 \end{gathered}$$

Therefore,$x_6=-25.1327$

Now to find the value of $x_7$, substitute localid="1649341588715" $k=6$in equation $\left(2\right)$

localid="1649341616429" $x_7=x_6-\frac{e^{x_6}+\sin x_6}{e^{x_6}+\cos x_6}$

Substitute$x_6=-25.1327$

$$\begin{gathered} x_7=(-25.1327)-\frac{e^{(-25.1327)}+\sin (-25.1327)}{e^{(-25.1327)}+\cos (-25.1327)} \\ =-25.1327 \end{gathered}$$

Therefore,$x_7=-25.1327$

Here,

$$\begin{gathered} \left|x_7-x_6\right|=|-25.1327-(-25.1327\left)\right| \\ =\left|0\right| \end{gathered}$$

Since$\left|x_7-x_6\right|<0.001$let us stop the iteration

Therefore, the approximate value of the root of$f\left(x\right)=e^x+\sin x$is$-25.1327$