For $c<1$, let $f_c(x)=x^2+c$, determine the period-2 points? Ok so I know I need to set $$f_c^2(x)=x$$ so:
$$(x_0^2+c)^2+c = x \iff x_0^4+2cx_0^2-x_0+c^2+c=0$$
But how do I then solve this? Ok I have solved for the four set of roots, now you can see there are two roots for $c<\frac{1}{4}$ and four for $c<-\frac{3}{4}$. How do I find the hyperbolicity and therefore the nature of these points?

Notice there is no $x^3$ term.
Try something of the form $(x^2+Ax+B)(x^2-Ax+D)$
$$(x^2+Ax+B)(x^2-Ax+D)=x^4+(D+B-A^2)x^2+(AD-AB)x+BD$$
So we know
$$D+B-A^2=2c$$
$$AD-AB=-1$$
$$BD=c^2+c$$
Which has solution $A=1$, $B=c+1$, and $D=c$.
So $x^4+2cx^2-x+c^2+c=(x^2+x+c+1)(x^2-x+c)$
I assume you can take it from there.
