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 $x_0$) $x_{n+1}=x_n-\cfrac{f(x_n)}{f'(x_n)}$
- 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 $f(c)=0$.