I am given the following exercise:
The function $f(x) = x^2$ has two obvious fixed points: $p_0 = 0$ and $p_1 = 1$. Show that there is a $0<\delta <1$ such that $|\,f(x) - p_0| < |x - p_0|$, whenever $|x - p_o| < \delta, x \neq p_0$. Conclude that $f^n(x) \to p_0$ whenever $|x - p_0| < \delta, x \neq p_0$.
In contrast, find a $\delta>0$ such that if $|x - p_1|<\delta, x \neq p_1$, then $|\,f(x) - p_1| > |x - p_1|$. That means that $p_1$ is a repelling fixed point for $f$; orbits that start out near 1 are pushed away from 1. In fact, given any $x\neq1$, we have $f^n(x) \nrightarrow 1$.
Question:
I'm a bit confused by this exercise because it seems to me that 'finding' the $\delta's$ is in fact quite trivial. In the first case every $0<\delta<1$ will do and in the second case any $\delta > 0$ will do, right? Is this exercise just about introducing different kind of fixed points, or am I missing something?
You're missing nothing (except for the constraint that @snulty notes). The author is showing that two fixed points of a particularly simple function have certain properties, and is hoping to generalize these later (I expect).
It's worth noting that if you add some perturbation like $-10x^4$ to your function --- a perturbation that's "small" near $p_0 = x$, then the same claim (about $p_0$) holds, but finding the exact delta becomes considerably harder.