How to solve an equation using Newton's method with and without backtracking? Lets assume I have this equation:  $$\log(e^x+e^{-x})=2x+5,\quad x \in (-50,50).$$
As always we have to pick a starting point to solve this by Newton's method, but how can i know for what initial values this equation will or will not converge?
I think i need to use Matlab here, but I am not good at it...
Also, if I have $\log(e^x+e^{-x})=0.1x+5$, how can i specify the region of convergence for each possible solution?
At the end, I know that Newton's method has a quadratic convergence, can we show that it is true for our problem cases? 
Thanks
 A: We have the function $f(x)=\log(e^x+e^{-x})-(2x+5),\quad x \in (-50,50).$ and it's first derivative $f'(x)$ and to solve using the Newton's method, we use the sequence (for an $x_0$)
$$x_{n+1}=x_n-\cfrac{f(x_n)}{f'(x_n)}$$

  
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*It's sufficient to look for a point $x_0$ close to the root and the tangent at that point exists, non zero  i.e. $f'(x_0\ne0)$ and finite.

Look at the function $$g(x)=x-\cfrac{f(x)}{f'(x)}$$ where 
$$g'(x)=\cfrac{f(x)f''(x)}{(f'(x))^2}$$ and $$g''(x)=\cfrac{[f'(x)f''(x)+f(x)f'''(x)]\cdot(f'(x))^2-2f(x)f'(x)(f''(x))^2}{(f'(x))^4}$$
To have convergence for a point $x_0$ such that $f(x_0) \approx  f(r)=0$, $r$ is the root, observe that $g'(x_0) \approx 0$ so we can't set a condition there but with at the second derivative of $g(x)$ we have 
$$g''(x_0)=\cfrac{f''(x_0)}{f'(x_0)}$$
Observe that $g''(x_0)$ has no prioritized reason to be zero, so the condition stated above is necessary but not always sufficient to have a quadratic convergence.
EDIT The most optimal way to guess a good $x_0$ is to use the Intermediate Value Theorem. 

If $f$ is a real valued continuous function  in the interval $[a,b]$ such that $f(a)<0$, $f(b)>0\quad$ OR $\quad f(a)>0$, $f(b)<0$ then $\exists c\in [a,b]$ such that $f(c)=0$.

