Attractive and repulsive fixed points Consider the function $f(x) = ax(1-x)$. I have to show that if $ 1 < a < 3$ that the fixed point $p_2 = \dfrac{a-1}{a}$ is attractive, and if $ 3 < a < 4$ it is repulsive.
I actually have no idea how to do this. I found that the top of the function is at $ ( \dfrac{1}{2}, \dfrac{a}{4} )$ and I have found that $f'(0) = a$ and $f'(\dfrac{a-1}{a})  = 2-a$, but I don't know what to do with this.
 A: Your function is a 'logistic map' and the trick is to compare $|f'(p_2)|$ to $1$. From Wikipedia he have indeed :
"If the function $f$ is continuously differentiable in an open neighbourhood of a fixed point $x_0$, and $|f'(x_0)|<1$, attraction is guaranteed."
This link to Wikipedia's picture may give you an intuitive explication (warning the vertical scale is a little too elongated). The iterations start at $x=-1$ and clearly converge at the right : now let's suppose that the vertical line at the right would go down a little more (supposing that the slope is steeper with $\ f'(x_0)<-1\ $) then we would go farther than we were at the top and go away from the limit instead of approaching as we do here (the spiral would be divergent). 

You found $\ f'\left(\dfrac{a-1}{a}\right)  = 2-a$ 
But for $\ 1<a<3\ $ we have $\ 2-3<2-a<2-1\ $ and obtain that 
$\ \left|f'\left(\dfrac{a-1}{a}\right)\right|<1$
The repulsive part will be handled the same way.
Btw Wikipedia has a nice animation when the value of $a$ changes.

