I have been asked to 'find a fixed point formulation so that the fixed point iteration converges' for
$$x = \frac{2-e^x+x^2}{3}.$$
I'm confused how to do it because past examples have had something like 'in [2,3]' but this question does not have that! How do i complete this question?
Your equation is
$$x^2-3x+2-e^x=0$$ by MVT, There is a solution in $[0,1] .$
it can be written as $$x=g (x)=\frac {x^2-e^x+2}{3} .$$ it is easy to check that $$g ([0,1]\subset [0,1] .$$
$$g'(x)=\frac {2x-e^x}{3} $$
but $0 <x <1 \implies 0 <2x <2 $ and $-e <-e^x<-1$ thus $$\frac {-e}{3}<g'(x)<\frac {1}{3} $$ and $$|g'(x)|<\frac {e}{3}<1$$ from here we are sure that the recurrent sequence
$$x_0\approx \frac 12$$ $$x_{n+1}=g (x_n) $$ will converge to the root.