Fixed point obtaining g's I currently am working on learning more about fixed point method.
Finding equations that satisfy the constraints of a g function can sometimes
require a bit of engineering.
I have come across one that many would consider simple. Yet, I have been stuck on
it for some time now.
Here it is
$ f(x) = x^2 - x - 2 = 0 $ on $[1.5,3]$.
I have tried many things; however, I have yet to successfully discover
one that maps domain to range for both $g$ and $g^\prime$.
Would anyone be able to give me a guiding hand? Perhaps there are cool
tricks.
 A: When trial and error does not work, try a trick for $g(x)$ using Newton's method
$$x= g(x) = x - \dfrac{f(x)}{f'(x)} = x - \dfrac{x^2-x-2}{2 x-1}$$
This may not always work.
A: As per Renaissance reasoning put terms on both sides so that all coefficients are positive
$$
x^2=2+x
$$
and then solve for the higher-degree side to get the iteration
$$
x_{n+1}=\sqrt{2+x_n}.
$$
This is a standard exercise for a monotonously converging recursive sequence.

As you have the interval, you can also shift the variable as $x=1.5+h$ or $x=3-h$. The resulting equation might be more intuitive to transform into a fixed-point iteration.
$$
0=\frac94+3h+h^2-\frac32-h-2=h^2+2h-\frac54.
$$
As $h$ is now assumed to be small, the focus has to be on the linear and constant term. One could make the iteration $$h_{n+1}=\frac5{4(2+h_n)}$$ out of this.

For the Newton approach it is sometimes interesting to balance the degrees of the terms, for instance by considering $\tilde f(x)=\frac{f(x)}x=x-1-\frac2x$ so that
$$
x_{n+1}=x_n-\frac{x_n-1-\frac2{x_n}}{1+\frac2{x_n^2}}
=x_n\frac{x_n^2+2-(x_n^2-x_n-2)}{x_n^2+2}
=x_n\frac{x_n+4}{x_n^2+2}
$$
The visible advantage here to the original Newton iteration is that the denominator has no singularities on the real line.
