91影视

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,

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

Substituting this in equation($1$)

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

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

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

Substitute$x_0=-2$

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

Therefore,$x_1=-4.75616$

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

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

Substitute $x_1=-4.75616$

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

Therefore$x_2=鈭�24.0033$

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

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

Substitute$x_2=鈭�24.0033$

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

Therefore,$x_3=鈭�26.1199$

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

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

Substitutelocalid="1649263166094" $x_3=鈭�26.1199$

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

Therefore$x_4==鈭�24.6056$

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

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

Substitute localid="1649262418782" $x_4==鈭�24.6056$

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

Therefore,$x_5=鈭�25.1877$

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

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

Substitute$x_5=鈭�25.1877$

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

Therefore,$x_6=鈭�25.1327$

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

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

Substitute$x_6=鈭�25.1327$

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

Therefore,$x_7=鈭�25.1327$

Here,

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

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

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