Let $b > 0$, $0 < a < 1$, and set $f(x) = ax + b$. Moreover, put $I_0 = [0, b]$, $I_n = f^{◦n}(I_0)$, $n = 1, 2, \dots$ and $$I = \bigcup_{n=0}^\infty I_n.$$
- Find an explicit formula for $I_n$ in terms of $a$ and $b$. The endpoints of In should be polynomial in powers of $a$. Use this to prove that $I$ is an interval. You may assume without proof that the length of this interval is finite. Let $\ell$ denote the length of $I$.
A picture of the question in case my formatting does not make sense picutre.
Any help to even just get started with this question would be very much appreciated.
The function $\,f\,$ is continuous and strictly increasing, and therefore, as suggested by Gary in his comment, we have
$$ I_1=f(I_0)=f([0,b])=[f(0),f(b)]=[b,ab+b]=[b,(a+1)b] $$
and similarly
$$ I_2=f(I_1)=f([b,(a+1)b])=[f(b),f((a+1)b)]=[(a+1)b,(a^2+a+1)b]. $$
Note that $I_0$ and $I_1$, as well as $I_1$ and $I_2$, are consecutive closed intervals, i.e. the right end of the first equals the left end of the second, so $I_0\cup I_1$ and $I_0\cup I_1\cup I_2$ are intervals.
The formulas above seem to suggest that in general one has
$$ I_n=[(a^{n-1}+\ldots+a+1)b,(a^n+a^{n-1}+\ldots+a+1)b], $$
which is very simple to prove by induction:
$$ I_{n+1}=f(I_n)=[(a(a^{n-1}+\ldots+a+1)b+b,a(a^n+a^{n-1}+\ldots+a+1)b+b]=\, =[(a^n+\ldots+a+1)b,(a^{n+1}+\ldots+a+1)b]\qquad \text{q.e.d.}$$
For each $n\in\mathbb N$, the intervals $I_0,I_1,\ldots,I_n$ being consecutive, we have
$$ \bigcup_{k=0}^n I_k=[0,(a^n+\ldots+a+1)b]=\Big[0,\frac{1-a^{n+1}}{1-a}\,b\Big]. $$
Now it is easy to prove that
$$ I=\bigcup_{n=0}^{\infty} I_n=\Big[0,\lim_{n\to\infty}\frac{1-a^{n+1}}{1-a}\,b\Big[=\Big[0,\frac{b}{1-a}\Big[. $$
Indeed, as the sequence $n\mapsto \frac{\displaystyle 1-a^n}{\displaystyle1-a}\,b\;$ is strictly increasing (being $0<a<1$) towards $\frac{\displaystyle b}{\displaystyle 1-a}$, it follows that
$$ x\in I \iff x\in I_\nu=\Big[0,\frac{1-a^{\nu+1}}{1-a}b\Big]\; \text{ for some } \nu\in\mathbb N \iff 0\leq x<\frac{\displaystyle b}{\displaystyle 1-a}. $$