## Calculus (3rd Edition)

$x\approx 0.25$ and $x\approx 2.542$
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 $x\approx 2.542$