Attractive fixed-point? I'm sorry in advance for how specific this problem is. I've been trying to make it as generic as possible but this is as far as I could get. I want to prove that a fixed point is attractive.
I didn't find much on Google about proving the attractiveness of a fixed point, but if there is a general methodology, that would be very helpful already.
The function is:
$$
f(x) = (ax + b)^{1/\alpha}
$$
with $a>0$, $\alpha\in(0,1)$ and $b\in(0,b_0)$ where
$\displaystyle{b_0 = (1-\alpha)\left(\frac{\alpha}{a}\right)^{\alpha/(1-\alpha)}}$. 
The domain of this function is $[0,x_0)$, with $\displaystyle{x_0 = \left(\frac{\alpha}{a}\right)^{1/(1-\alpha)}}$.
I proved that this function has one unique fixed point in its domain, but I don't know how to prove that it is attractive. Specifically, the derivative is not absolutely lesser than 1. What I know already about the derivative is that it is strictly positive and increasing, and takes values in $[0,x_0)$. 
EDIT:
I made a mistake in my calculations; the derivative takes values in $[0,1)$ which is sufficient to prove the fixed-point is attractive. Thanks to Julian for his answer.
 A: We will work on $[0,+\infty[$.
First observe that for $x\geq 0$: 
$$
f(x)=x\quad\Leftrightarrow\quad ax+b=x^\alpha.
$$
Now draw the graphs of $ax+b$ and $x^\alpha$ to convince yourself that there could be $0$, $1$ or $2$ positive fixed points in general.
It turns out that the condition $b<b_0$ guarantees that there are $2$ positive fixed points.
Indeed, consider the point
$$
x_2:=\frac{\alpha b}{(1-\alpha)a}.
$$
At this point, we have, after simplifications:
$$
ax_2+b<x_2^\alpha\quad\Leftrightarrow\quad b< (1-\alpha)\left( \frac{a}{\alpha}\right)^\alpha=b_0.
$$
The latter is one hypothesis given by the OP, so now we know why it is here.
Since the graph of $x^\alpha$ is above $ax+b$ at $x_2$ whereas it is below at $0$ and at $+\infty$, we know that there are two fixed points $x_1,x_3$ such that $0<x_1<x_2<x_3$.
Next compute the derivative at $x_1$ and use the fixed point condition:
$$
f'(x_1)=\frac{a}{\alpha}(ax_1+b)^{1/\alpha-1}=\frac{ax_1}{\alpha(ax_1+b)}.
$$
Finally, we see after simplification that:
$$
f'(x_1)<1\quad\Leftrightarrow\quad x_1<x_2.
$$
So the smallest fixed point is attractive.
Note that the last paragraph is actually the consequence of the more general observation made by Will Jagy.
Finally, the condition $x_1<x_0$ is also fulfilled.
Indeed, we have $ax_0+b<x_0^\alpha$ so $x_1<x_0$, since the former turns out to be equivalent to $b<b_0$ again.
A: The second derivative $f''$ is positive as is $f'$ and $f$ itself. In particular, $f(0) > 0.$ So, one possibility is no fixed points. Another is exactly one fixed point, as in $g(x) = e^{x-1}$ at $x=1,$ in which case the derivative is exactly one. 
If, as you say, there are two fixed points, call them $0 < u < v.$ So, $f(u) = u$ and
$ 0 < f'(u) < 1,  $ which is what you wanted to know.  Then $f(x) < x$ for $u < x < v.$ Finally $f(v)=v,$ with $f'(v) > 1,$ and for $x > v$ we get $f(x) > x.$  
EEDDIITT: One way to see all this is simply to notice that $\frac{1}{\alpha} > 1,$ so $f(x) > x$ for large $x.$ So, define
$$  h(x) = f(x) - x.  $$
We know $h(0) > 0,$ also $h(x) > 0$ for large $x.$ And we know $h'' > 0.$ So, if $h(u) = h(v) =0,$ we know from the Mean Value Theorem that there is some $u < w < v$ with $h'(w) = 0.$ Furthermore, $h'(u) < 0, \; h'(v) > 0,$ and $h(w) < 0.$ The part you want is $h'(u) < 0.$
