Use Newton's Method to find all solutions of the equation correct to six decimal places. $\ln x=\frac1{x−3}$

Question

When I use desmos for graphing, there are two intersections and the smaller one is about $0.653$. According to Newton's method, when I plug $1$ into $x_1$, $x_2$ is $0.6$ which is less than $0.653$, which implies that $x_n$ will be less than $0.653$. Why is this happening?

Answer 1

Nothing about the definition of the sequence $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}, \quad n \ge 0,$$ suggests that it must be monotone, so I am puzzled as to why you believe that if $x_0 = 1$ is your initial guess and $x_1 < x_0$, then $x_2$ would be less than $x_1$. In fact, this is not true: such a recursively defined sequence can be alternating, or it can be partially alternating and partially monotone. Simply continue to calculate the successive iterates of $1$ and you will see the result numerically converge.

Answer 2

Consider $$f(x)=\ln x-\frac1{x−3} \implies f'(x)=\frac{1}{x}+\frac{1}{(x-3)^2}>0$$ The function has two vertical asymptotes at $x=0$ and $x=3$ so there are two roots $0 < x_1 <3$ and $3<x_ 2$.

By inspection $f(1)=\frac12$ and $f(4)=\log (4)-1>0$ which makes $0 < x_1 <1$ and $3<x_ 2<4$.

Start your calculations at the mid point of the interval, namely $x_0=\frac12$ for the first root and $x_0=\frac72$ for the second root.

Newton method works as a charm.