show that function has a fixed point at P precisely when f(P)=0,where f(x) = x^4+2x^2-x-3

Question

a.g1(x) = (3+x-2x^2)^1/4

Answer 1

When x=-0.8761 P=x, and x^4=3+x-2x^2 (rearranging when f(P)=0), so g1(x)=(3+x-2x^2)^(1/4).

Comments

Note that the negative fourth root is the actual zero of f(x) when x=P.