Hi, I have this problem.

I need to sketch a bifurcation diagram for the fixed points of the family $\displaystyle \mu \rightarrow f_\mu$ where $\displaystyle f_\mu= \mu x^2(1-x)$.

I have that the fixed points are $\displaystyle x=0$ and $\displaystyle x=\frac{1\pm\sqrt{1-\frac{4}{\mu}}}{2} $

I know that:

When $\displaystyle \mu < 4$ $\displaystyle f_\mu$ drops below the identity mapping and so there is only one fixed point at $\displaystyle x=0$

When $\displaystyle \mu =4$ the identity mapping is tangent to $\displaystyle f_\mu$ and so there are 2 fixed points at $\displaystyle x=0 $and $\displaystyle x= 0.5$

When $\displaystyle \mu > 4$ $\displaystyle f_\mu$ rises above the identity mapping and so there are 3 one fixed points at x=0 and $\displaystyle x=\frac{1\pm\sqrt{1 - \frac{4}{\mu}}}{2}$

So i can sort of picture how the diagram will look, I just can't figure out for what values of $\displaystyle \mu$ the fixed points $\displaystyle x=\frac{1\pm\sqrt{1 - \frac{4}{\mu}}}{2}$ will be repelling/attracting.

I know that $\displaystyle f'_\mu(x) = 3-\frac{\mu}{2}\pm\frac{\sqrt{\mu^2 - 4\mu} }{2}$ at the last two fixed points. And that if $\displaystyle |f'_\mu(x)|<1$ then fixed point is an attractor and if $\displaystyle |f'_\mu(x)|>1$ then fixed point is a repellor. But i'm stuck here.

Please help.

Katy