$x=g(x)$ fixed point iteration When you setup to do Fixed point iterations you often end up with several different $g_n(x)$ functions
The $g_n(x)$ functions do not always converge and some converge faster than others
what are the conditions that can be used to select the $g_n(x)$ that is guaranteed to converge?
How will you choose the function that converges the fastest?
 A: There is no general recipe, but there are some general considerations. I'll assume that $g$ is twice continuously differentiable, though there are analogues of most of what I will say when $g$ has less regularity. I'll denote the desired solution by $x^*$.
First of all, if $|g'(x^*)|>1$ then fixed point iteration will not converge (at least not to $x^*$). Typically it won't converge if $|g'(x^*)|=1$ either.
Among $g$s with $g(x^*)=x^*$ and $|g'(x^*)|<1$, there are basically two relevant considerations. The first one is: what is $|g'(x^*)|$? The Banach fixed point theorem tells us that the error after $n$ steps behaves like $|x_0-x^*||g'(x^*)|^n$ if $0<|g'(x^*)|<1$. If $|g'(x^*)|=0$ then a similar analysis gives quadratic convergence. So you generally want $|g'(x^*)|$ to be as small as possible.
The second one is: how large is the region of convergence? Even if $|g'(x^*)|$ is zero, the convergence is still only local. So unless you pretty much know what you are looking for, you should be concerned about how far away your initial guess can be with a method.
Generally speaking, in 1D problems in which explicit differentiation is possible, you find that Newton's method is the way to go. The main reason is that Newton's method gives $g'(x^*)=0$ for free. This gives quadratic local convergence, which is typically quite fast, if you are in the region of convergence. Surprisingly the region of convergence for Newton's method is also often quite large. For example, Newton's method for square roots (i.e. for solving $x^2-r=0$) converges for any nonzero initial guess.
In certain 1D problems, the function in question is given only as a black box. For example, you could have $f(x)=y(1)-a$ where 
$$y''=h(t,y,y'),y(0)=b,y'(0)=x.$$ 
Here $h$ is a given function and $a,b$ are given constants. You might do this because solving $f(x)=0$ gives you a way to solve the BVP 
$$y''=h(t,y,y'),y(0)=b,y(1)=a.$$ 
Anyway, when $f$ is given as a black box, explicit differentiation is impossible. In these cases the secant method, which is technically a 2D fixed point iteration, can be very useful.
