I am attempting the find $\mu$ for which the map $$x_{n+1} = \mu + x_n^2$$ undergoes a period doubling bifurcation.
I understand that finding the fixed points of the map is the first step towards finding the period doubling, and I find these to be $$x_{*} = \frac{1}{2} (1 \pm \sqrt{1-4\mu})$$ in terms of the parameter. What I don't understand is where to go from here. Do I need to find the fixed point of the second iterate of the map? Or do I need to ensure that the condition that $f_{\mu}'(x_{*}) = -1$ is satisfied?
Your iterated map is given by:
$$f_{\mu}(x)=x^2+\mu$$
Your fixed points are correctly given as:
$$x_{1,2}=\frac{1}{2}(1\pm\sqrt{1-4\mu})$$
If a period change occurs at the fixed point $x_1$, then necessarily
$$|f'(x_1)|\ge 1$$
so it suffices to solve the equation:
$$|f'(x_1)|= 1\Rightarrow$$ $$|2x|_{x=x_1}=1\Rightarrow$$ $$|1+\sqrt{1-4\mu}|=1\Rightarrow$$ $$\sqrt{1-4\mu}=0\Rightarrow$$ $$\mu=\frac{1}{4}$$
And now you can be sure that the iterated map $f_{1/4}(x)$, will suffer some sort of period change at $x_1=\frac{1}{2}$.
Work similarly for the other fixed point.
For later verification, note that this map is the Mandelbrot map for the Julia Set with $\mu=c=1/4$, so the bifurcation point is the right cusp of the set for this $c$.