Answer
$x\approx 0.25$ and $x\approx 2.542$
Work Step by Step
Since
$$ f(x) = e^{x}-5 x$$ and $f'(x) = e^x-5$
By using Newton’s Method, we have
\begin{align*}
x_{n+1}&=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}\\
&=x_n- \frac{e^{x_n}-5 x_n}{ e^x_n-5}\\
&= \frac{(x_n-1)e^{x_n}}{ e^x_n-5}
\end{align*}
From the given figure, for the first root, choose $x_0=0$
\begin{align*}
x_1&= \frac{(x_0-1)e^{x_0}}{ e^x_0-5}\approx 0.2\\
x_2&= \frac{(x_1-1)e^{x_1}}{ e^x_1-5}\approx 0.259\\
x_2&= \frac{(x_2-1)e^{x_2}}{ e^x_2-5}\approx 0.259
\end{align*}
Since $x_2=x_3 $, then the root is $x\approx 0.25$.
From the given figure, for the second root, choose $x_0=2.5$
\begin{align*}
x_1&= \frac{(x_0-1)e^{x_0}}{ e^x_0-5}\approx 2.544\\
x_2&= \frac{(x_1-1)e^{x_1}}{ e^x_1-5}\approx 2.542\\
x_2&= \frac{(x_2-1)e^{x_2}}{ e^x_2-5}\approx 2.542
\end{align*}
since $x_2=x_3 $, then the root is $x\approx 2.542$
